m
u
--Eca
m'
>-u
am >~ ..c ~) E ... c. Q)
..ccaCl)
... u-c
c-- c::::: m __ -W--I ca E "'l--= cae..: .!iUL...
C
-= - -
=-..c
-=...a -=C-c
0
.1::
."
•
.-= Q
C.
c.caQ)
/'
•
,
P
Contents
Pion Limited, 207 Brondesbury Park, London NW2 SJN ,.",
Authors Preface
© 1971 Pion Limited All rights reserved. No part of this book may be reproduced in any form by photostat microfilm or any other means without written permission from the publishers. SBN 85086 023 7
·,1,·, [
( >1
+
[J
J 3
The laws of thermodynamics (P. T.Landsberg)
v Mathematical preliminaries " Quasistatic changes • The First Law The Second Law
"Simple ideal fluids
Joule - Thomson effect
v Thermodynamic cycles v Chemical thermodynamics 'V The Third Law
"' Phase changes
" Thermal and mechanical stability
"oj
v
0 K
(
\
( ... -..~
Set on IBM 72 Composers by Pion Limited, London.
Printed in Great Britain by J.W.Arrowsmith Limited, Bristol.
v
vii
I
I
4
6
9
II
21
23
28
33
34
37
2 Statistical theory of information and of ensembles (P. T.Landsberg) VEntropy maximisation: ensembles
v Partition functions in general
vEnt ropy maximisation: probability distributions "Most probable distribution method
.Some general principles
44
44
48
51
53
60
3 Statistical mechanics of ideal systems (P. T.Landsberg) v Maxwell distribution
~ Classical statistical mechanics
"' Virial theorem v Oscillators and phonons
or The ideal quantum gas
",Constant pressure ensembles v Radiative emission and absorption
63
63
67
72
74
80
92
93
4 Ideal classical gases of polyatomic molecules (C.l. Wormald) -The translational partition function "Thermodynamic properties and the theory of fluctuations v The classical rotational partition function "The quantum mechanical rotational partition function A convenient formula for the high temperature rotational partition function Thermodynamic properties arising from simple harmonic mode" of vibration Corrections to the rigid rotator-harmonic oscillator model Contributions to the therm9dynamic properties arising from low lying
electronic energy levels Calculation of the thermodynamic properties of HCI from spectroscopic data Thermodynamic properties of ethane
98
98
100
103
105
108
110
,((3>
118
121
127
Contents
Contents
ii
5 Ideal relativistic classical and quantum gases (P. T.Landsberg)
132
6 Non-electrolyte liquids and solutions (A.J.E.Cruickshank)
Cell theories of the liquid state
Equation of state treatment of liquids
Binary solutions
140 142 154 168
7 Phase stability, co-existence, and criticality (A .i.E. Cruickshank) Singulary systems Binary systems
193 193
~
..
iii
23 Time dependence of fluctuations: correlation functions, power spectra, Wiener-Khintchine relations (C W.McCombie)
472
24 Nyquist's theorem and its generalisations (C W.McCombie)
489
25 Onsager relations (C W.McCombie)
502
.../ 26 Stochastic methods: master equation and Fokker-Planck equation (I. Oppenheim, K.E.Shuler, G." Weiss)
511
214
27 Ergodic theory, H-theorems, recurrence problems (D. ter Haar)
531
8 Surfaces (/.M.Haynes)
The Gibbs model of a surface
230 230
28 Variational principles and minimum entropy production (S.Simons) Macroscopic principles Electron flow problems
9 The imperfect classical gas (P.CHemmer)
The equation of state
The virial expansion
Pair distribution function. Virial theorem
246 246 254 261
549 549 557
10 The imperfect quantum gas (D. fer Haar)
The equation of state
Second quantisation formalism
v 11 Phase transitions (D. fer Haar) Einstein condensation of a perfect boson gas Vapour condensation Hard sphere gas
270 270 278 282 289
293 303
13 Green function methods (D. fer Haar)
Mathematical preliminaries
General formalism
The Kubo formula
The Heisenberg ferromagnet
319 319 320 324
14 The plasma (D.ter Haar)
333
15 Negative temperatures and population inversion (U.M. Titulaer)
Dynamic polarization
A model of laser action
341 346 348
16 Recombination rate theory in semiconductors (/.S.Blakemore)
350
17 Transport in gases (D.J.Gri{{iths)
378
323
18 Transport in metals (/.M.Honig)
401
19 Transport in semiconductors (/.M.Honig)
432
\(20 Fluctuations of energy and number of particles (C W.McCombie)
448
v21 Fluctuations of general classical mechanical variables (C W.McCombie)
457
Fluctuations of thennodynamic variables: constant pressure systems, isolated systems (C W.McCombie)
~
282
12 Cooperative phenomena (D. fer Haar)
oJ 22
Author index Subject index
465
\
r
563 565
I Authors
.)
•
J .S.Blakemore
Department of Physics, Florida Atlantic University, Boca Raton, Florida
A.J .B.Cruickshank
School of Chemistry, University of Bristol, Bristol
D.J .Griffiths
Department of Physics, University of Exeter, Exeter
J.M.Haynes
School of Chemistry, University of Bristol, Bristol
P.C.Hemmer
Institutt for Teoretisk fysikk, NTH, Trondheim
J.M.Honig
Department of Chemistry, Purdue University, Lafayette, Indiana
P.T.Landsberg
Department of Applied Mathematics and Mathematical Physics, University College, Cardiff
C.W.McCombie
J.J. Thompson Physical Laboratory,
University of Reading, Reading
I.Oppenheim
Department of Chemistry, Massachusetts Institute of Technology, Cambridge, Massachusetts
K.E.shuler
Department of Chemistry, University of California, San Diego, California
S.Simons
Queen Mary College, London
D.ter Haar
Department of Theoretical Physics, University of Oxford, Oxford
U.M.Titulaer
Institut voor Theoretische FYsica der Rijksuniversiteit, Utrecht
G.H.Weiss
National Institute of Health, Bethesda. Maryland
C.J.Wormald
School of Chemistry, University of Bristol• Bristol v
I Preface
1
Problems and solutions! Their production is an annual ritual-feared and unpopular-for the university teacher. Nor are examination questions particularly liked by students. Yet, take away the examination aura, and the technique appears in a new light. The problem-and-solution style of writing presents to the author new difficulties and constraints, and thus offers him a novel challenge. Perhaps it is like teasing a sculptor with a new and promising type of stone, or a poet with a new rhythm. Viewed in this light, the potential author has the opportunities which go with a new medium: for example, a novel way of arranging important results and of setting them up to be seen more clearly than is possible in a uniformly flowing and elegant exposition. As to the reader, he finds before him a series of hurdles. They are easy enough at first to bewitch him, beguile him and persuade him to join into the fun-until he gradually participates with the author in a neo-Socratic dialogue. The existence of these opportunities must surely be one of the considerations which explain how it was possible to find so select and experienced a group of persons to help in the construction of this book. Each author contributed in his own special area of research with the result that the book presents a penetrating view of statistical physics and its uses which, as regards the width of its sweep is, I suspect, beyond any one of today's experts. Thus, in spite of the age of our subject and the love and care which has been bestowed on it by successive generations, this book presents in some sense a new unit. I have planned its outline and have discussed various points with the authors, but have imposed only a limited uniformity of style. I have also attempted to ensure that overlap of material is not extensive, that cross references are adequate, and that the early problems in each chapter are reasonably easy. In this way I wanted the reader of each chapter to feel drawn into a fascinating world which opens up for him once he realises he can actually derive results for himself. Here then is a book for teachers, undergraduates or graduates who want to know what can be done with reasonably simple models in thermodynamics and statistical physics. It is also suitable for self-study. It should appeal to a wide range of numerate readers: mathematicians, physicists, chemists, engi neers and perhaps even economists and biologists. The foundation of the general theory of statistical physics is involved and difficult, and its discussion takes a great deal of time in a lecture course. One can turn to this book for ideas and for inspiration vii
viii
Preface
if one wishes to pass on to areas of application while condensing the time spent on the foundations. This procedure will appeal to those who feel, as I do, that the foundations of the subject are best discussed briefly at first, and then again later from time to time as the effectiveness and the power of the methods are being appreciated. It will come as no surprise to the experienced teachers that the need for care in the setting of a problem can stimulate original work. There are new presentations in various places (e.g. in Problems 1.24 to 1.26, 1.29-1.31, and 3.19 to 3.22), and in several sections of this book previously unpublished ideas will be found. For example, in Section 6 some looseness of logic inherent in previous work has been eliminated, through discussions between the author and the editor, and there are new approaches to equations of state, to the law of corresponding states and to the relationship between reduced equations of state (a) for chain molecules and (b) for their components. Problems 17.7 to 17.9 have also some novelty in the use made there of the random walk approach. There are several other places where recent research work has here been incorporated in book form for the first time (for example, in Problem 5.5). In conclusion, I wish to thank all contributors and the publishers for the cooperation which made this venture possible.
,
P. T.Landsberg
.,
r
1 The laws of thermodynamics P.T.LANDSBERG (University College, Cardiff)
MAJHEMATICAL PRELIMINARIES
VI.I If each of the three variables A, B, C is a differentiable function of the other two, regarded as independent, prove that (a)
(~;) (~~t G~t C
= -1,
(~~)B = I/G~t· Solution
Let the functional dependence of A on Band C be expressed by f(A, B, C) = 0, then
(a~) B,CdA + (:;) A,CdB + (:~) A.BdC
= If A is constant this becomes
(:;) A'CG~) A= - (:~) A,B' i.e.
G~) A= -(:~t.B/(:;) A,C' Similarly
(:~t = (a~)B,cl(:~t'B' (~~) C= -(:;) A,C I(a~) B,C' Multiplication of these three equations yields
(~~)C(~~t (:~t
=-1.
O.
Chapter 1
2
1.1
Interchange of A and C in the second of these equations yields
(~~)
=
B
1/ G~t .
1.2 (a) Integrations over the following two paths in a plane are to be performed: (i) the straight lines (XI, Yl) ~ (x 2 , yd ~ (X2, Y2); (ii) the straight lines (Xl,Yl) ~ (X 1 ,Y2) ~ (X2,Y2)' P(xl>yIl, Q(X2,Y2) are two points, andx 1 =1= X2,Yl =1= Y2' The differen tial forms to be integrated are
1.2
The laws of thermodynamics
Solution
(a) We have
i i
(dx+dy)
(I)
(dx+dy) = (Y2 -Yl)+(X2 -xd·
These two line integrals have the same value, namely u(Q)-u(P) = (X2+Y2)-(Xl +Yl)'
On the other hand
r dv
du
0)
where u
=X
and
=
f
f
dv
(I)
du
= u(Q)-u(P),
(il)
=1=
f
(ii)
~(x~ -Xi)+X2(Y2 -Yl),
J(i)
f
dv == x(dx+dy).
f
= (X2 -X 1 )+(Y2 -YI),
(ii)
du == dx+dy, Show that
3
= X 1 (Y2 -yd+t(x~ -xi).
dv
J(ii)
Since the two results are unequal, there does not exist any function vex, y) of which the differential form dv can be considered as an exact differential, for otherwise both integrals would yield v(Q)-v(P). In the present case av, although inexact, has an integrating factor, i.e. a function g(x, y) exists which converts dv to an exact differential
dv,
dF = gdv .
and discuss the result. [We shall denote differential forms with this property by av instead of dv, and call them inexact, while du is an exact differentiaL In relations of the type du(x,Y, ... ) g(x,Y, ...)crv(x,y, ... ), g(x,y, ... ) will be called integrating factors.] (b)If dF = X(x,y)dx+ Vex,
If one integrating factor exists, then there exists an infinity of them, and a simple example is furnished in the present case by
is an exact differential, show that
so that
(~~t = (~~t· (c) Pfaffian forms have the general form n
dv or av
=
i
L A/(x = I
1> x 2 , ..· xn)dxi'
(dv may be exact or inexact). Show that for n = 2, if the Xj are single-valued, continuous and dif ferentiable functions, av has always an integrating factor provided X 2 is non-zero in the domain of variation considered. 3 need not have an integrat (d) Verify that a Pfaffian form with n ing factor by considering dv = X dy + k dz, where k is a non-zero constant.
g
= 1/x,
dv
du .
(b) For an exact differential dF
ax y dx+ (aF) ay x dy == Xdx+ Ydy, ( aF)
=
aF (ax) axay = ay 2
x
(ay) ax y'
(c) Let
crf
= Xdx+ Ydy
,
where X, Yare continuous differentiable and single-valued functions of the independent variables x, y. This restriction on X and Y together with Y =1= 0 means that the equation dy X -=- dx Y has a solution of the form F(x,
=
C,
where C is a constant.
i.e. dF
=
(~~)y dx+
C;t
dy
0,
Chapter 1
4 Comparing coefficients in df
= 0 and
dF
±(~~t = ~(~~)x Hence gdf
1.2
(a) From the first two expressions for dQ an equation for dt is found:
== g(x,y).
(C p -Cv)dt
(
ax
y,z
= 0,
x
k
(aF)
az
g'
= Iv dv -I p dp
.
Substitution for dt in the second expression for oQ
= gXdx+gYdy = dF
(aF) ay x,Z
5
Solution
= 0,
and dfhas the integrating factor g(x, y). (d) Suppose dv(x, y, z) == xdy+kdz = g(x, y, z)dF. Then aF)
The laws of thermodynamics
1.4
g
x,y
It follows from part (b) that
aF 1 1 (a g ) aXay=0=g-g2 ax y,; aF = ° = -g2k (aaxg ) axaz 2
_
IvCp
(
dQ - C p -C dv + lp v
IpCp ) C -C dp. p v
Comparison with the last form for oQ yields the required result. (b) The first equation under (a) yields this result at once. (c) C v is the heat required per unit rise of empirical temperature at constant volume, also called the heat capacity at constant volume; Ip is the heat required per unit rise of pressure at constant temperature, some times called the latent heat of pressure increase. The other coefficients can be described analogously.
2
1.4 Let
y,z'
a~
x(~)
ayaz = -g2
az
k(~)
-g2 ay
x,y =
cxp
==
K t
QUASISTATIC CHANGES(I)
1.3 For a fluid and other simple materials any three of pressure p, volume v, and empirical temperature t are possible variables. They are connected by an equation of state so that only two of the three variables are independent. An increment of heat added quasistatically may then be expressed in the alternative ways = Cvdt+lvdv = Cpdt+lpdp
= mvdv+mpdp,
where the coefficients are themselves functions and are characteristic of the fluid. The term empirical temperature refers to an arbitrary scale and is used to distinguish it from the absolute temperature, denoted by T. Prove that the following relations hold: (a)
_ IvCp __ IpC v mv '!!.E.. _ . mv - Cp -C v ' mp Cp -C v ' 1v + Ip - 1,
(b)
= Cp -C v (aatp ) v = _ Cp l-Cpv' (av) at p Iv·
(c) Express in words the physical meanings of the coefficients in the expressions for dQ.
\
_.l(av) ap V
t
be the isothermal compressibility of a fluid. The notation of Problem 1.3 will be used. (a) Show that the Griineisen ratio r == cxpv/KTCv of the material satis fies
r = v(~~)jcv.
(b) Show that the ratio of heat capacities 'Y == Cp/Cv satisfies
(~~))(~~)t' - 0 = G~)j G~)p' 'Y~l = (~~))(~~)v' 'Y
=
where ( )a denotes a quantity evaluated under quasisfatic adiabatic con ditions, i.e. for dQ = 0. (c) If Ka is the adiabatic compressibility, prove that
Kt
K (I) Changes consisting of a continuum of equilibrium states.
~(~~ \
be the coefficient of volume expansion at constant pressure, and let
These equations cannot be satisfied by a finite function g(x, y, z).
oQ
==
x,;
a
= 'Y.
Chapter 1
6
(b) Prove that the Griineisen ratio of Problem 1.4 is v
(d) Show from a consideration of d(lnv) that
( ~) ap
= t
_(aKat
t)
7
The laws of
1.6
1.4
r
. p
v
I(~~)o
mp
(c) Find the most general equation of state of a fluid whose Griineisen ratio is independent of pressure.
Solution
(a) We have
r
-vG;)p
=
Solution
v(~)v
where Problem 1.1 (a) has been used. (b) From the first equation in Problem 1.3:
v
ap )
( av
a
_ Cpl mp - Cvlp
[from Problem 1.3(a) 1
G;)a
Iv '
(~~)a
lp ,
(~~)t
Iv Ip '
p
where x
=0
_
y
= pdv+xdy
Tor p. If oW were exact we would have p
) (a ay v
=
(ax) av
= 0 y
•
This yields 1 0 if y is chosen to be p, and hence oW is inexact. (b) We start with oQ = dU+pdv = Cvdt+l"dv whence
We now take the ratios of terms in the first column to corresponding terms in the second column. The first pair yields "I at once, The second pair yields, in conjunction with Problem l.3(a),
.'
Co = Ca~)v (~~)v(~)v' =
"I-I'
The third pair yields "1/("1-1). (c) This follows from (b). (d) We observe that t and p are the independent variables needed. The form of the equation to be proved suggests that the procedure of Problem 1.2(b) may be involved. Fortunately we also have the hint to consider d(lnv). Hence dOnv)
oW
'G~)p = mv a) ( at v
_Cvmv __ Cv~ Cp Iv CpCp-Cv
I
(a) This is obvious since dU = oQ -ow. If dW were an exact differential, one could integrate to find Q = U+ W+constant, in contra diction with the inexact nature of oQ. Alternatively, note that we have two independent variables, so that one could write
= ~dV = ~ [(~~)p dt+(:;)tdP].
This is OI.p dt-Ktdp, and we can indeed apply Problem 1.2(b). THE FIRST LAW
1.5 The first law states that an internal energy function U exists such that for a fluid or similar material we have, in addition to the first equa tion of Problem 1.3, oQ = dU+pdv, where p is the pressure, and p dv is the mechanical work done by the system. fa) Show that the increment of work dW pdv is inexact
Substitute in Problem 1.4(a) to find
r = v(~~)) C
v vI(~~)v' =
(c) Integrate the result of (b) to find
1
U = r(v)[pv+f(v)],
where rand f can be functions of volume, and fjr is a constant of the integration with respect to p. The equation of state of a solid is some times taken in this form: pv = r(v)U(v, T)
1.6 Show that (a)
(~~)v'
mp = mv
(~~)p +p,
=
(~) av
p
=
(amap
v
1.
) v
8
Chapter 1
1.6
(b)
(~~)p - (~~)v (~~\ (~~t·
(c)
Cp -Cv =
1.7
1.7 The second law asserts that the reciprocal absolute temperature
[(~~)t +p] G~\·
liT is an integrating factor of ctQ, the resulting function of state being called the entropy S, so that for quasistatic changes dS = ctQ/T. This
Solution
(a) From
dU+pdv
=
[(~~)p +p] dv+ (~~)v dp
one finds
mp =
(~~)o'
mv =
(~~)p +p.
(omop
(~~)s = -(:~)v' (~~)s (:~t, (~~)p = -(:~)T' (~~)v = (:~)T' [F is the Helmholtz free energy, H is the enthalpy, and G is the Gibbs free energy. J
~_(3) avop - ov p =
I
adds a further relation to those given at the beginning of Problems 1.3 and 1.5. Returning to these, adopt the absolute temperature as the most convenient empirical temperature scale t to establish the following results. [Note that in a quasistatic adiabatic change the entropy is constant.] (a) By considering dU = TdS-pdv, dF == d(U-TS), dH == d(U+pv) and dG == d(U+pv TS) establish Maxwell's relations =
Also, differentiating, we obtain
v)
-1.
(~~)t dv + (~~)v dt
(~~)t[(:;)tdP+(:~)p dt]+(~~)vdt = (~~)t dp+ [(~~)t (:~)p +(~~)Jdt.
(~~)p - (~~)v = (~~)J~~)p'
(c) The first equations of Problems 1.3 and 1.5 yield whence
d:Q = Cpdt+lpdp = dU+pdv, Cp =
=
T(::)v'
G~
Iv
=
T(:~)v'
Ip
-T(:rt,
T(~;)v'
m"
T(~~)p'
0
=
Hence
Cv
,:i
(b)
(b) The independent variables (v, t) and (p, t) are involved, suggesting that a substitution for one in terms of the other two is required SOme where. We have
dU =
9
THE SECOND LAW
=
oQ
The laws of thermodynamics
(~~)p +p(:~)p'
This, combined with the relation proved in (b) and with C v = (oU/ot)v, yields the required result.
mp (c)
Ipmv
=
-TCp,
lump
=
T(::)p'
=
TC o ,
Ivlp = -T(Cp-Cv)·
(d) In view of these relations, and earlier ones, which of the six functions C, I, m give a set of independent quantities in terms of which the others can be expressed? Solution
The stated equation for dU clearly yields the first required result by the process indicated in Problem 1.2(b). This can be applied next to
d(U-TS)
dU-TdS-SdT
=
-pdv-SdT.
This yields the last equation. The other two results are established analogously. (b) These results follow immediately from the equation at the begin ning of Problem 1.3, except that in the case of Iv and lp a Maxwell relation is also needed. For example, TdS = CIl dT+l vdv implies
Iv T(~~)T = T(:~)v' =
1.7
Chapter 1
10
lpmo = (:~)p -TCp, lomp = T2(:~)v = TCv' It was shown in Problem 1.3(a) that
mv mn Iv lp . relations for mo and mp _TCp+TCt} = 1 Ip Iv Ivlp , -+----<:...= 1
which arise from the
Iv
=
from a result of Problem 1.3,
Ip _Cp I-Co T
1.8 With the notation of Problem 1.4 establish the following results:
(a p) _ ~ Cp_Cv T(av) aT p aT v- Tv KT . KT-K Tv:::::.C·!! ~2
a
p
(c) Using Problem 1.5 show that in a quasistatic adiabatic change of a simple fluid
Texp
[1:
rd(lnV)]
remains constant; here r is the Griineisen ratio introduced in Problem lA, and Vo is a standard volume. (a) From
with the result of Problem I. 7(b) the second law enables us to put
mp = T(;;)o = -T(:;t· Hence for an adiabatic process, rd(lnv)+d(lnT) = O. It follows that
Texp
dOnV)]
is constant. SIMPLE IDEAL FLUIDS
from (c) above.
o
Solution
Cp-Co = -T(:;)p (iv)T(:;)P = -T(V~p)2 (- v~J. (b) We have from Problem 1.4(c) that KT/Ka 'Y = Cp/Co. Hence, from part (a), (KT-Ka)Cp= KT (I -~;)cp = Tv~~. (c).Jt was shown in Problem 1.5 that r v/mp. But in conjunction
from (c) above,
Iv
mo=~ Cp -Cv
(b)
11
;=;
which is the required result.
Retain and the heat capacities. The other quantities are
mp
The laws of thermodynamics
equation. Now use the identity of Problem 1.1 (b) to write
(c) Frpm part (b) we have
Substituting the new second law, we obtain
1.9
Co dT+l odv = CpdT+l pdp it follows that Cp = lo(:~)p = T(:~)o(:;t,
where a result of Problem 1.7(b) has been used. This is the first
1.9 (a) Establish the result
p+ (au) av
= T
T(as) av
= T
T(aaTp) ; 0
P(;;)T +(~~)T = -T(;;)p' (b) Joule's law for a fluid states that (aU/aV)T "'\ O. Show from (a) that it implies the existence of a function fCv) of the volume and that pf(v) = T. (c) A fluid which satisfies the equation pv = AT, where A is a con stant, is called an ideal classical gas. SlJ.ow from (a) that it satisfies Joule's law. Is the relation pv = At, in which an empirical temperature t
is used, adequate to infer Joule's law?
(d) Verify that for an ideal classical gas and with the notation of Problem 104 ~p
T'
Cp-Co
KT =
I p'
r
;=;
A, 'Y-1.
12
1.9
Chapter 1
1.10
(e) A fluid which satisfies the equation pv = gU, where g is a constant, is called an ideal quantum gas (2). Show from Problem 1.5(b) that for such a system the Griineisen ratio is r = g, and so is a constant. [This is called Griineisen's law.] (0 Prove that an ideal classical gas for which Cn is a constant is an ideal quantum gas with g = "1- I and Cv = A/g, provided its internal energy vanishes at the absolute zero of temperature.
(e) Problem 1.5(b) yields, if pv
= gU,
(0 If Cp is constant, then it follows from part (d) that C v = Cp-A and "I = Cp/Cv are also constants. From part (c) we note that Joule's law holds, so that U(v, T) depends on T only and dU = C v dT; there fore U = CvT+constant. The constant becomes zero given that U vanishes with T. Now, from part (d) we have
(a) The independent variables are v and T. Hence the first and second laws combined can be used in the form
=
13
r = v Ie~)v = g.
Solution
dU
The laws of thermodynamics
CvAT AT pv CT= =-= v Cp-C v "1-1 "1-1·
TdS-pdv ,
It follows that
whence
pv=("(-I)U,
(~~)T dv+ (~~)v dT = TG~t dV+T(::)v dT-pdv.
which is characteristic of an ideal quantum gas. Also,
Equating coefficients of dv we find the required result with the aid of a Maxwell relation. The second result is found similarly, using p and T as independent variables and equating coefficients of dp. (b) Joule's law reduces the result of (a) to p = T(ap/aT)v. For constant volume this integrates to yield Inp + In f = In T, where In f is the constant of integration. This can of course depend on the volume. (c) The result (a) depends on the use of the absolute temperature T and yields for a system satisfying pv = At au) ( Tv T
Thus if t
=
T
(ap) at
dt vdT - P
=
T
p dt dT - P
t
=P
=
(dint) din T - 1 .
p(:;t Cv -
dU _ A dT - g.
Cv
A/s(
(~-i),
1.4,
= "I-I.
(2) This macroscopic definition was given in P.T.Landsberg, Am.J.Phys.• 29, 695 (1961). See also G.E.Uhlenbeck and E.A.Uehling. Phys.Rev.• 39, 1014 (1932) and H.Einbinder. ibid. 74, 805 (1948); G.Siissmann and E.Hilf. Proc.lntern.Conf.on Thermodynamics. Cardiff 1970. PureAppl.Chem. 22,243 (1970).
\
=
(b) Assume that the fluid is an ideal classical gas of constant "I, i.e. that the absolute temperature scale is chosen in (a). Show that each Bi can be expressed in terms of the entropy S of the fluid through
B2 = A exp
r, given in Problem
r = ~ = .!!.~ = .:i. _ Cp -Cv TCv
=
Bi = AB1- 1
= A.
Substitution of these results in the definition for gives KTCv
Cv
where the Bi are constants, by assuming that "I is a constant for the fluid. Show that
(m-l)p
and Joule's law results only if t = T. (d) cxp and KT follow at once from pv = AT. For Cp -Cv , Prob lem 1.6(c) may be used with (au/aT)v = 0 from part (b) above. This yields Cp -Cv =
= gU = AT
1.10 (a) A fluid whose equation of state is pv = At, where A is a con stant, undergoes quasistatic adiabatic changes. Use the result of Prob lem 1.4(b) to establish the laws t pv'Y = Bl, tV'Y- 1 = Bi-l, pb- l)h = B 3 ,
= Tm, ( au) av T
pv
yields
(:
I
where i is a thermodynamically unidentifiable constant (not V-I !). (c) If the fluid envisaged in part (b) expands from an initial state (T1,V 1) into a vacuum, so that its volume increases to V2, obtain an expression for the increase f:j.S in its entropy, and show that the work done by it is T1f:j.S. [It is desirable to choose the constants in part (a) as Bi and Bl- 1 , and not simply as Bland B 2 ; see Problem 1.23.]
14
1.10
Chapter 1
15
The laws of thermodynamics
1.11
Since Joule's law holds [see Problem 1.9(b)], U1 = U1 , i.e.
Solution
r
T
(a) Problem 1.4(b) together with pv = At yields
(aav) p
= "I a
(aav )
= _ "I~ == _ "IP
p
t
V
so that 11
V
= 12.
= Cv ("1 -1)1n V1V1 V1
= (Cp -C., )InVI
where B7 is a constant of integration. Under the same conditions we have also
~. pv1'
tV1'-l
=
= AlnVIV1
~7 == Bl"1 .
Lastly T pFr-OI1'
TV1'-1 BlIA _ (pv1')h-1)ly = Bl-l = A = B3 .
Alternatively one can start from one of the other relations established in Problem 1.4(b). (b) The first and second laws yield dS =
~(dU+Pdv) = ~ [C.,dT+ (~~)T dV+PdV]
Since (aU/aV)T = 0 [see Problem 1.9(c)] and piT = A/v, we have dS
CvdlnT+Adlnv = C"dln(Tv1'-I)
= C.,("I-l)dln(T
1
/(1'-I)v)
= Adln(TI/(1'-I)v).
Now, if "I is a constant, integration yields
S = Aln{Tl/(1'-I)v)-ln(A-A/e i )
Bl
Aexp(~-i) A
1
(CS "1-1)
=AlI1'B~1'-l)h=Aexp - - i - p
B3
Cp
~ C., l)
"I
"1-1) = BdA = exp (Csp -i--;:;.
(c) From parts (a) and (b), S = Co In(pv1')+ constant. Hence f1S = Cvln [ -P2 ~V2)1'] -
PI
VI
C"ln [.. -12 ~V1)1' 11 VI
1 V
VI
1.11 (a) The van der Waals equation of state is
(p+~ )(V-b)
At,
where a, b, A are constants. On a (p, v) diagram the extrema of this equa tion, which are obtained by choosing various values of t, lie on a curve. Find the equation of this curve. Show that the maximum of the curve found in part (a) is given by a Sa Ve = 3b, Pe = 27b2, tc 27Ab' so that Ate/Peve = 2·667.
[This point is called the critical point.]
(c) Show that the van der Waals equation can be written
,
= Aexp (
[Problem 1.9(d)].
The work done by the gas in the expansion is
2 f2 1 V2
f pdv = ATI -dv:= A111n- == T1f1S .
(1T+ :2)<31/J-l)
where the last term is a constant of integration. It follows that B2 = vT 1/h- 1)
CvdT,
It follows that LlS
Hence, for quasistatic adiabatic processes with "I constant, we have dv -dp i •. e pv1' ~ B1'1 , P = -'V-,/: I }!I.,:o'
f
T z·
,
JoC"dT
ST,
where
I/J == V/V e ,
1T
== piPe,
T
== t/t e •
(d) One sometimes writes a general equation of state in the form pv = E 1+Ezp+E3p Z + ... , where E 1 , E2 , E3, ... are functions of t and are called first, second, third, ... virial coefficients. If a, b are small enough, show that a van der Waals gas has an approxi mate second virial coefficient a
E2 = b- At'
16
Chapter 1
1.11
(e) The Boyle temperature tB of a fluid is defined by [a(pv )/ap ]p=o = 0, so that Ei = 0 and the fluid approximates a Boyle's law gas (for small n ;;,-" 3) in the neighbourhood of this temperature.
Show that for a van der Waals gas
tB
tc
3·375.
(f) Show that CII for a van der Waals gas is independent of volume. (g) Show that at v = Vc
4b
2
Kt
= 3(( - t e )
Cip
diverge at t = te' of state is regarded here as given. A closely related equation is derived from statistical mechanics in Problem 9.7. Different definitions of virial coefficients are in use; the most common is perhaps that given in Problem 9.S. See also Problem 10.1.] (a) From the equation of state At
(:~)t
To check that this is a maximum, observe that 2 pi _ 6a _ 24ab _ 6a 4b)
d
dvr - vf
The van der Waals equation at the critical point furnishes now a value of tc: Sa Ate = 27b2+ 9b2 (3b -b) = 27b
( a a)
(c) An easy algebraic result. (d) We write a At ( 1+-2 Pv pv
)-1 (b)-I 1--
a ,
At 2a -(v-_-b-)-2 + v 3
(e) Ez = 0 if t
tB
,
te -b)2 (Pi +a/vf)(vi -b)
=
Hence the required curve on the (p, v)-diagram is given by a Pi
2ab
;;- v? 1
I
To check that these extrema are maxima, note that ~_6a aZp ) ( ov 2 t (V-b)3 v4 22a6a I
aAt bAt At-+ pvz V
= a/Abo Hence = a . 27Ab :::: 27 Ab
3.375
Sa
.
(~~)T +p = T(;~) v Differentiating with respect to temperature at constant volume yields Zp p oZU (Op) (a ) (a ) ovaT+ aT II = T OT2 ,,+ aT v i.e. (
2a
- v~ R:l-v~ . 1
1
(b) This last curve has a maximum given by 2a 6ab 2a --+- -(3b -vJ = 0 .
vr v1
R:l
acv)
av
(a2p)
II
= T aT2
v
The right-hand side vanishes for a van der Waals gas. (g) From the solution of (a), and by differentiating we obtain
which at an extremum becomes v·-
V
(f) Recall from Problem 1.9(a) that
(Vi
2a ::::
Vi
At the point at which dpddvj is zero, this quantity is negative as required. Pc at Vi = 3b, so that Hence, at this maximum, Ve = 3b. Also Pi a 2ab a Pc = 9b2-27b3::::
Thus for extrema, labelled by 'i',
vr
(1 _
vr -
For small a, b the approximate equation is pv = At. Using this in the two correction terms, we have a ap i.e. E z = b - At pv ~At- At+ bp ,
Solution
p:::: v-b -v 2
17
The laws of thermodynamics
1.11,
v1
Kt
I Avt 2a' (v-b)Z--;;:
Cip
On using the value of tc and putting v = Ve follows.
3b the required result
1.12
18
Chapter 1
pv
1.12 (a) The ideal quantum gas, introduced in Problem 1.9(c) through gU, has an internal energy which obeys the equation
The laws of thermodynamics
1.13
Note that the 'Y, in the analogous adiabatic laws for an ideal classical gas of constant heat capacity ratio 'Y, is replaced here by I +g. This might have been expected from Problem 1.9(f).
aU) v(oU) U = T ( aT v -g Tv T'
1:13 A polytropic fluid of index n is one for which pv n is a con stant (3). (a) An ideal classical gas of constant heat capacities Cp , Co undergoes a quasistatic change for which dQ CdT, where C is some constant. Show that it is then a polytropic fluid of index
Establish this result [the equation of Problem 1.9(a) may be used], and verify that the condition is satisfied by U = v-gf(Tv g), where f(x) is some function ofits argument. (b) Show that if this gas undergoes a quasistatic adiabatic change, then the following quantities are constants:
Cp-C
n
T
Tv g ,
pv
pg{(t+ g )
C o -C'
Further, establish that during this change the following quantities remain constant T Bl pv n = B~, Tv n - 1 A' A' p(n- I)/n
•
Br
Solution
The result in Problem 1.9(a) is p =
TG~)v -(~~t
(b) In what sense does part (a) generalise the result of Problem 1.IO(a)? (c) Explain why the heat capacity C for the polytropic change can be negative. (d) An ideal quantum gas undergoes a change such that dQ = bCodT, where b is a constant, but Co need not be a constant. Show that TvK/ O - b ) is a constant for this change.
Multiplying it by v/g and using pv/g = U, one finds
U- - T - 0 ( aU) aT Suppose U
19
v (aU) g ov T
•
v-gf(Tv g ); then, iff'(z) == df/dz, one has
Solution
(a) From the basic equations of Problems 1.3 and 1.5 we have
(~~)o = r, aU) ( ov T
dQ It is easily seen that
_gU+gT r v V '
so that the equation for U is satisfied. (b) Multiply dS = T- 1 (dU+pdv) by Tv g to find
TvgdS
10 = p+ (~~)T
For the change envisaged, therefore,
vg(dU+g~dV) = d(Uv g).
From part (a) the right-hand side is d[f(Tv g )], so that the entropy is seen to be a function of Tv g only. Hence in a change in which the entropy is constant, Tv g is also constant. From part (a), the constancy of Tv g implies the constancy of I Uv g = - pv 1 +g g Dividing the constant Tv g by the constant pv1+ g yields as a further constant during the change the quantity Tl/g/pl/(I+g), so that TpK/(1+ g)is also constant.
dU+pdv = CvdT+lodv .
CdT
CvdT+
IP+(~~)J dv
Using the result of Problem 1.9(c) to neglect the differential coefficient, dividing by T, and noting that piT A/v = (Cp -Col/v, one finds dT dv (Co -C)-;y+(Cp -Co)/) = O. (3) A detailed discussion of poly tropics is due to G.Zeuner [Grundziige der mechanischen
l
Wiirmetheorie, 2nd edn., p.143 (A.Felix, Leipzig), 1866). The theory was developed notably by R.Emden in the first quarter of this century. Its use in stellar problems was reviewed by E.A.Milne [Handbuch der Astrophysik, VoU (1930)]. This review is reprinted in Selected Papers on the Transfer ofRadiation (Ed. D.H.Menzel) (Dover Publications, New York), 1966.
Chapter 1
20
1.13
On introducing the stated value of n this becomes
The last form of writing arises from the observation that
dT dv -+(n-l)- = 0 Tv'
so that, if B2 is a constant,
= Bf- 1
Tv n - I
as) v = Tv g (as) C v = T ( aT az v = z (as) az v
=
r;
=
Tv n- I = pvn ,
Integration shows that zT- b
(b) For C = 0 the change is adiabatic, whilst for C = 00 the change is isothermal, and n = "I and 1 respectively. Other possibilities exist since -00:S;;; C:s;;; 00. (c) From the solution for part (a)
I
-
n
= AT it follows that
B? .
Hence an increment of work done by the ideal classical gas in the poly tropic change is
Solution
(a) Observe that under the conditions stated
vn A A C -C dW == pdv = - p - - . -dT = ---dT = - P v dT. n-l
n-l
pvn
n-l
(b) Take A dW
= -n-l - C dT v
A}'"
2
>
V2 -VI
"1-1
= --dUo
n-l
Vaporization Melting (normal) Melting of ice
Thus if 1 < n < "I the work done in an increment exceeds the drop in the internal energy of the fluid. Hence dQ = dU+dW is positive while dT is negative. Now C
Ap AS AT Av 0; then we have
T-=-=
Hence "1-1
VI
where A}'" 2 is the heat required to take a mass m of material isother mally from the state of aggregation 1 to the stage of aggregation 2, and V 1 and v 2 are the volumes occupied by this material in the two states of aggregation. [If m is taken as the unit mass, then A is the latent heat, and the v's are specific volumes.] (b) Show that increase of pressure raises the boiling point and the freezing point of a normal liquid, but that, in the case of water, pressure lowers the freezing point.
AT =
TI-bvg is a constant.
V2
dQ = dU+pdv = Cv dT+pdv ,
v
=
1.14 (a) If a fluid is in equilibrium with its saturated vapour pressure, then the pressure in the system is independent of volume and depends only on temperature. Show from the Maxwell equation given in Prob lem 1.9(a) that under these conditions AI'" 2 dp T dT = ~ (Clausius-Clapeyron equation),
Tv n- I (pvn)(n-I)/n
whence dU = Cv dT. Also, from pvn = B?, pv
= f ,(z) .
b-=-= T Cv z
and this makes the fluid isotropic of index n. The last result required follows from T prn:TfTii
= z· Z-I f'(z)
The change taking place is defined by TdS = bCv dT, whence dT dS dz
.
It follows that another constant is
B? == Atv n- I
21
The laws of thermodynamics
1.15
n -"I
= --ICv n
state 1
state 2
liquid solid solid
vapour liqUid liquid
dp/dT V2 V2 V2
> VI > VI
< VI
positive positive negative
JOULE-THOMSON EFFECT
1.15 If gas is allowed to flow slowly through a porous plug between two containers, which are otherwise isolated from each other and from their surroundings, the enthalpy HI == U I + P IV 1 before the process is equal to H 2 , the value after the process. The temperature change is measured by the Joule-Thomson coefficientj == (aTjap)H.
and one observes that, for 1 < n < "I, C is in fact negative. (d) In the solution of Problem 1.12(b) it was noted that if z == Tv g then Uv g = fez), and also dS = z-ldf(z) = z-If'(z)dz = z-lCv dz .
\
(IJ
22
1.16
1.15
Chapter 1
The laws of thermodynamics
(b) From part (a) and a Maxwell relation we obtain
(a) Show that dB
= TdS+vdp, j=
and hence that
.
J
v
= -(Tcx C p p
-1) •
j
=-
[T(:i)
v
J
+v(~~) jC(~~) p
=
(2av _3ab -b P) v 2
j
v
v
p
PiO
=
2a 3a a 3b 2- 9b 2 = 3b2
.
=
0, i.e. at Vi
a
PiO a 27b 2 Pc - 3b 2 ' -a =
= ViO = 3b.
The
p
It follows that
j < 0
3b 2
9.
Also
31'.:.
A1io
a) 8a a = ( 3b2+ 9b2 (3b -b) = 9b
and
lio 8a 27Ab 1'.: = 9Ab . sa- = 3 .
Solution
(a) Using dU = TdS-pdv, we obtain
= dU+pdv+vdp = TdS+vdp,
3b
v
Figure l.lS.1
THERMODYNAMIC CYCLES
1.16 (a) In an incremental process a fluid gains heat energy aQ at temperature T. Establish that in a closed cycle
as required. It follows that dB
T
[T(~) -v] aT p
(f) Its maximum occurs at dpddvi corresponding value of Pi is
[The maximum of the inversion curve lies at a pressure and tempera ture which are above their critical values.]
dB
p
2a 3a Pi = bv.1 - v?-1 .
Show also that =
J
(~~) jC(~~)
(e) From the result in part (d) the curve is
PiO = 9pc'
1io
+v
= _1 Cp
2ab)c . j (p_~+ 2 3 ~
v
T'
(e) Obtain the curve on a (p, v)-diagram separating the region j > 0 from the region j < 0 for a van der Waals gas. This is the inversion curve. [Expansion leads to cooling only for states lying in the j > 0 region.] (f) If Pw is the pressure at the maximum of the inversion curve, show that for a van der Waals fluid.
[T(:i)
This is more convenient than the result in (a) if p = p(v, T) is known.
(c) From Problem 1.9(d) we have Tcx p = 1. (d) The expression is obtained by a direct calculation of
(c) Establish that j = 0 for an ideal classical gas. (d) Show that for a van der Waals gas introduced in Problem 1.11 j
[T(:~)p(~~t -v(~~)J/Cp(~~)T
=-
(b) Show that, alternatively,
fdi .:;
= [V+TG!)JdP+T(::)p dT,
so that j
23
(Clausius inequality),
where the equality holds for a quasistatic cycle. (b) In a general quasistatic cycle a working fluid receives heat at various temperatures and gives up heat at various temperatures. The efficiency 1/ of the cycle is defined as the mechanical work done divided by the sum of all the (positive) increments of heat gained. Prove, using
== (aT) ap H = _ [T(as) ap T +v] jT(as) aT p = ...!...IT(~) Cp aT p -v],
where one of the Maxwell relations of Problem 1.7(a) has been used. The required result follows.
\
0
,
(-J
Chapter 1
24
1.17
1.16
1't-12 ----y;-
11c, the Carnot efficiency),
the heat gained is equal to the work done. Hence
where Tl is the highest temperature at which heat is gained, and 12 is the lowest temperature at which it is given up. (c) Show that the maximum efficiency 11c can be attained by using as a working fluid an ideal classical gas of constant heat capacities, which is working between isothermals at temperatures Tl and 12 separated by an adiabatic expansion and an adiabatic compression. rThis is the Carnot cycle.] (d) If you have used the properties of the ideal classical gas in (c), generalise the argument so that it applies to any fluid describing a Carnot cycle quasistatically.
QI =
==
= f
jT
Hence Q2/Ql ~ p
dQ-f dQ +T
_T
12/Tl , so that 11 ==
~
1 dQ_,
-12
1't
12'
l-Q2/Ql ~ 1-12/1't.
(c) With the notation shown in Figure 1.16.1, TV'Y- 1 is constant on the adiabatics be and da [Problem l.lO(a)]. Hence
a
1 = 'F-V'Y-I T,V'Y-l T,V'Yla 2 d , lb
c
.... v
Figure 1.16.1
=
Wbe
--
W~
:::: -
pdv
La
Ve
Vd
= -I
dU = C v
~
dU:::: Cv(TI -12),
(12 -1't).
As in the solution of Problem 1.16, the amount of heat absorbed from a reservoir during the isothermal change ab is
1 'F-v'Y2e,
Qab
The heat rejected on cd is Va
~
Solution
i.e. Vb
Vb
-A 12 In- :::: A121n
1.17 The working fluid of a thermodynamic engine is an ideal classical gas of constant heat capacity Ct). It works quasistatically in a cyclic process as follows: isothermal expansion at temperature TI from volume VI to V2; cooling at constant volume from temperature 1't to 12; (iii) isothermal compression at temperature 12 from volume V2 to v I; (iv) heating at constant volume from temperature 12 to TI . Obtain expressions for the amount of heat (Qt) supplied to the gas in steps (i) and (iv) and the amount (Q2) rejected in steps (ii) and (iii). Show that for the above cycle the efficiency 11 < (1't -12)/TI . How is the efficiency affected if (ii) and (iv) are replaced by adiabatic expansion to volume V2 and by adiabatic compression to volume v 1> respectively.
f dQ-f dQ = QI_Q2
+1't
Va
Vd
(d) From the first law work done by the engine is equal to the heat gained by the fluid in a cycle: W = QI Q2. Also from part (a) we have Q'l = 1't 12 Hence W Q I -Q'l _ l_Q2 12 1-11 = Ql QI Q1 Tl
the integrals extending over the appropriate increments. QI is the total (positive) heat gained; Q2 is the total heat rejected, counted positively. Energy conservation gives the mechanical work done as W = QI - Q2 ; the efficiency is 11 (QI - Q2)/Ql' Now
o idQ
pdv =
a V
e
For a quasistatic cycle the equality holds for each increment and hence for the cycle as a whole. (b) Divide the increments of the cycle into those in which heat is gained, dQ+, and those in which heat is lost, dQ_. Then define Q2
d
Vb A1't fbdV -:::: ATlln,
e
fd~ ':;;;0.
1 dQ+,
a
pdv =
Hence 11 = 11c. Note that the work done on the adiabatics cancels out:
We have dQ/T':;;; dS, where dS is the incremental change of the entropy of the fluid. The entropy is a function of the state of the fluid and in a cyclic process its final value equals its initial value, whence
==
f f b
Q2 = -
Solution
QI
25
It follows from Joule's law [Problem 1.9(b)] that the internal energy remains constant on each isothermal, so that dQ = dU + (tW implies that
the Clausius inequality, that 11 .:;;;
The laws of thermodynamics
Qed
-=
('j
Vb
AT1ln-. Va
Vd
= A12lnVe
Vb
A12ln
Va
1.17
Chapter 1
26 since Va
Vd
and Vb =
V c'
1.19
During the cooling bd the heat rejected is
The internal energy gained by F2 is
C,,(TI -1;) .
Qbc
Q2 = C2 (To 1;).
During the heating da the heat supplied is
The overall loss of internal energy is
Qda = C v (11-1;) .
QI -Q2 = CITI -~12 -(CI +C2)To . It follows from energy conservation that this must be equal to the total amount of work done by the Carnot engine. (b) In the absence of the performance of work, energy conservation yields CI (I1-To) = C2 (To-1;) so that
To aTI +b1; (a+b I) .
The entropy lost by FI is
Hence
= Qda+Qab AI1 1n(vb/va)+C,,(TI -1;) QI -Q2 = A(I1-1;)ln(vb/va)' QI Q2 TI -72 11-1; 11 = < -TI QI T, + Cv(TI -1;) QI
A In (V2/V I) If steps (b) and (d) are changed as suggested, a Carnot cycle results, and the conclusions of Problem 1.16(c), (d) apply. I
1.18 (a) Two fluids FJ, F2 of fixed volumes and constant heat capaci ties CI , C2 are initially at temperatures TI, 12 (11 > 1;), respectively. They are adiabatically insulated from each other. A quasistatica11y acting Carnot engine E uses F 1 as heat source and F2 as heat sink, and acts between the systems until they reach a common temperature, To say. Obtain an expression for To and for the work done by the Carnot engine. (b) If a common temperature is established by allowing direct heat flow between F I and F 2, what is the final temperature and what is the change in entropy? (c) Show that for all positive C I , C2 this change is an increase. Solution
(a) Since CI = T(aS/aT)", the entropy lost by FI is _ fTo dT TI SI = -CI T = ClInT, . T,
0
The entropy gained by F2 is S2 == C21n(To/1;). If the working substance of the Carnot engine is in the same state finally as it was initially, the entropy gain of the whole system is S2 -SI = In
l(~r2 (~rlJ =
0.
Hence To where
l1a1;b,
CI
a == CI+C.l '
b
==
C2 CI +C2
The internal energy lost by FI is in the form of heat QI = -CI
f
To
T,
27
The laws of thermodynamics
dT = CIOi-To) .
f
~dT
f
To
~==-~
l
The entropy gained by
~
11
~=~~T,' 0
is
,,)
S2
C2
dT
T,
To
= C2 ln T, . 2
The gain of entropy for the whole system is S2- S 1
To)b (To)aJ = (CI +C2)ln [( 1; TI
To (CI+C2)ln11'1i"
Hence S2 -Sl
aTI +b1;
= (CI +C2)ln 11'1i'
(c) For aU positive a, b such that a+b = I, consider y = all +b12 -l1a
This quantity has a minimum when 11 = 1; and is therefore never nega tive. Hence S2 -SI ;;;.. O. 1.19 (a) Show that the Jacobian
a(p, v) _ (a p)
aCT, S)
=
(av)
aT s 'as
(a p) T -
as
T
(av) aT
s
I.
(b) A thermodynamic engine uses a cycle which is represented by closed curves on a (p, v)· and on a (T, S)-diagram. Use the fact that the work W done by the working fluid is equal to the heat Q adsorbed, expressed in the same units of energy, to obtain the result of part (a).
28
1.19
Chapter 1
Solution
J
(a) We have dp
p
)
Solution
l
)
T -
T
(b) We have for corresponding domains of integration in the (p, and (T, S)-planes a(p, _, W aCT,S) dTdS
ffdpdV
where the Jacobian arises because of the change of variables. Also
Q=
Hence, since W
njdlli = O.
among the intensive variables.
(aTa s (av) as (aas (av) aT s p
=
(d) Establish the Gibbs-Duhem equation SdT-vdp+
Git (~~)v +(~~)TG!)V
G:)v
29
The laws of thermodynamics
Gi)sdT+(~~tdS.
Hence, using two Maxwell relations of Problem 1.7(a) =
1.20
~
The equation states that, if the independent variables (which can be chosen to be all extensive variables) are multiplied by a factor a, then the entropy is multiplied by the same factor, and is therefore also extensive. A similar equation' for U can be deduced from this equation, as is necessary since U is also extensive. For suppose S(U, v, nl, ... ) = So can be solved for U; then if g is some function, the solution is, say, U
It follows that the solution of S(aU, av, anI' ...) aU
Q for any domain, the result follows. But aU
CHEMICAL THERMODYNAMICS
1.20 If two identical systems are joined together and considered as one system, the quantities whose values double are called extensive, and those whose values remain the same are called intensive. A phase is a thermodynamically homogeneous region of space, thermodynamically uniquely specified by its internal energy U, its volume v, and the num bers of molecules n I, n2, ... of the chemical species contained in it. (a) Verify that the equation for all a > 0
= g(So, v, nl' ...) . = aSo is
= g(aSo,av,an1, ... ).
= ageS, v, nt, ... ), so that we have g(aSo, av, an 1> ... ) = ag(So, v, n 1> ... )
•
This is the analogue of the given equation for S. Similar equations hold for v and the nj. (b)
TdS = T
(auM) .
v,ni dU+T
(M) a V
U,ni dv+T~ I
M
an,.
dnj
- L Iljdnj .
S(aU, av, anI, an2, ... ) = as(U, v, n l , n2, ... )
If Sa
is consistent with the extensive nature of S, U, v, and the nj. (b) Assuming the entropy to be a function of the state of a phase, and that it satisfies
G~)v.ni
T'
== S(aU, av, anI' ... ), then dSa = aSa d(aU) + aSa d(av) + L aSa d(anj) da a(aU) da a(av) da I a(anj) da
(~) U,n = f,
Hence
i
S
even if the nj are variable, show that
I
= r(U+pv -
L Illnl) . j
TdS = dU+pdv- LIl;dnj, j
Ili
== _T(;S)
nj u, v, other n's
.
(d) From part (c) we have
Ilj defined here is called the chemical potential of species i.] (c) Show by differentiating with respect to a, that G
== U-TS+pv
dU-TdS-SdT+pdv+vdp- L(lljdnj+ntdlli) i
From part (b),
Lllinj. j
dU-TdS+pdv- Llljdnj
[G is called the Gibbs free energy.]
i
lu
O.
= O.
O.
oUfpov - ToS
I
'"\ 1.21 Each of n phases I, 2, ... n consists of one and the same chemical species. A new system without inhibiting constraints is formed of these n phases by combining them so that the internal energy U, volume v, and particle number n of the resulting system are each the sum of the original quantities for the phases:
L~,
V= LVI,
1= I
i
=I
n =
L nl' 1= I
Show from the result in Problem 1.20(b) that for equilibrium in the new system
Ii
T2
1;"
PI =
pz
= ... = Pn,
III = lIz
=
=
LlltOn/; LPjOVj ~
0,
f
where the two sums involve pressures and volumes which can change owing to internal processes in the system(4). (a) A system consists of two phases, labelled I and 2, which are . .. u... u~.uJ separated mechanically at pressures PI and P2- If they are then coupled mechanically, but their compositions are fixed, show that the phase whose pressure is greater will expand. If the volumes of the phases are fixed, but the molecules of a certain chemical species can suddenly be transported from one phase to the other, show that the molecules will migrate from a region of higher to one of lower chemical potential. [The tendency to equalisation of intensive variables in equilibrium is illustrated by both Problems 1.21 and 1.22.J
n
n
n
U=
31
whose mass is fixed). Suppose that, with a generalised interpretation of 0, one has
Hence
SdT-vdp+ Ln/dlll =
The laws of thermodynamics
1.23
1.20
30
= lIn·
Solution
Solution
(a) If one phase expands, it does so at the expense of the other, -ovz. Hence so that ov I
The entropy change for phase i is
OSI
I
= r;(oUj+pIOVj-lliOnj).
-P Z OV2 = -(PI-P2)OVI ~ 0. then PI > P2; if ov I < 0, then PI < P2' -PIOVI
I
Thus if OVI > 0, In either case the result stated in the problem follows. (b) Let the phases be denoted by I and 2, so that on I -on2, since the particles lost by one phase are gained by the other. Hence
We seek to maximise LoSt subject to i
j = LOVt = L ont = LOU i t i
°.
1l1 0n l +ll zonJ, = (Ill -1l2)onl ~ 0.
Using undetermined multipliers, consider
If on l > 0, then 112 > Ill; if onl < 0, then 112 result stated in the oroblem follows.
~ [(*-a)Ui+(~-~)Vi-(~-~)njJ
f
Here Ti, Ill, PI, a, ~,~ are constants. The quantity fhas to be maximised for arbitrary and independent variations of £1;, VI, and nj. One finds
t [(~-a)OUt+(~-~)OVj-(~-~)ontJ PI = PI
III
= II:!.
= ... =
= O.
PV
gU=AT,
Cv
~,
1;, = l/a,
= ... = Pn
T"t
= ~Ti =
= ... = lIn = ~Ti
In either case the
1.23 An ideal classical gas of constant heat capacities is considered in this problem and it is recalled that for such a system
This implies
TI = T2
< Ill'
=
Cp =
[S
exp C
(l+~)A,
J
-(~-I)i ,
g =
~-l,
}
(Problem 1.9)
(Problem 1.10)
v
where i is a constant (and not V-I !).
1.22 A form of the second law suggested by Problems 1.5 and 1.16 is 0, where 0 denotes an incremental change the end points of which are equilibrium states of a closed system (i.e. a system
8U - pov - ToS ~
(4) For a discussion of this generalisation see P.T.Landsberg, Thermodynamics with Quantum Statistical Illustrations (lnterscience, New York), 1961, p.156.
I
Chapter 1
32
1.23
(a) Two such systems having identical heat capacities satisfy pv = AT, pv A'T. Given only that A is an extensive variable, prove from Problem 1.10 that Bland B2 are extensive while B3 and i are intensive. [Because A is extensive, it is written as kN, where k is Boltzmann's constant and N is the number of molecules.] (b) Show that for one such gas
S = A [(1+
~)lnT-lnp+iJ = A(~lnT+lnv+i-lnA).
[The above is a generalised Sackur-Tetrode vapour pressure equation, and i is here called the chemical constant.] (c) Show that the chemical potential of the gas satisfies the condition
:T
= lnp+
(I +~)O-lnT)-i.
[It is not possible to identify the chemical constant thermodynamically. But it is possible to do so from statistical mechanics. See Problem 3.13.] Solution
1.24
The laws of thermodynamics
(c) Since pv = gU
G
U+pv-TS
Since G
B1
AB?-l
!:iN (Problem 1.20) and A
(B~)"r BI
=
X(B~ - )"rB2
X"r (B~ B3 )
l
=
(1)
= p&"r-I)I"Y
.
THIRD LAW
1.24 Suppose the entropy remains finite and continuous as T -+ O. [This is from reference 4, p.112 equivalent to Nernst's heat theorem.] Show that then X = Cv , Cp , III, Ip, m v , mp each tends to zero in this limit. Solution (i) Recall from Problem 1.7(a) that
dF = dU-TdS-SdT = -SdT-pdv ,
rr = Bl = p"r-l
(BA2 )"r-
(:~)p = -S =
= exp
J
[S CII -('Y-1);
dH
.
'YlnT-('Y-l)lnp = C -('Y-1)I.
dG
= A/g (Problem 1.9), it follows further, on multi
S = A ( g+ I InT-lnp+i) .
g
T->O
[(aF) - -(au) ] aT aT v ' II
= TdS+vdp
= CpdT+(lp+v)dp,
-SdT+vdp
[Problem 1.7(a)] .
Hence the following quantity is indeterminate as T -+ 0:
_(~;)p = S =
H;G .
Hence
;i~ [- (~;)J = ~~ [C (~;)J
v CII
=lim II
whence Cp = (aH/aT)p' Also
It follows that
Since'Y = g+ I and plying by A/g, that
F-U .
i.e. ST= 0 = ST= o+Cv, T= o. Hence Cv -+ O. (ii) From H = U + pv, we have
= B3
S
(aF)
limaT
T->O
for both gases so that B3 B~. Hence B~/BI = B~/B2 X, and they are therefore extensive. The substitution of i for B2 through B2 AeSIA-i was therefore reasonable, and i is intensive. (b) In Problem 1.10 it was shown that l
I
Now, as T -+ 0, F -+ U for finite S, so that the right-hand side becomes indeterminate. Differentiating numerator and denominator and sub stituting the values appropriate to T = 0, we obtain
The adiabatic through a given point (Po, To) satisfies t to
ph-Oh
Nk [part (a)], this yields
Ii kT = Inp - 1+g InT-i+ I + g .
A"r Bl .
Hence, if A '/ A = X, we have
= AT, the Gibbs free energy of the system is
(1 +~)pv -TS = AT [1 + ~ - (1 + ~ )tnT+lnp-i].
=
so that
(a) We have
33
p -
so that Cp -+ O. (iii) From Problem 1.3 we have
TdS
= CvdT+lvdv = CpdT+lpdp
'
= mvdv+mpdp.
Chapter 1
34
1.24
If S remains finite and continuous
T(~!t = CO(~~)y +Iv(~:) y . , = ... 40. Consider various cases as T (x,Y)
Result
4
O. One finds
(T,v) (T,p) (v, T) (p, T) (v,p) (p, Cv 4 0 Cp 40 Iv 4 0 Ip 4 0 mv 4 0 mp 4 0
1.27
The laws of thermodynamics
(a) Show that the slope PL of the line L satisfies op) (op) , (aatp ) T = -PL' ( oT t = aT L PL; (b) Show also that for any function of state y
(:~)p
(it)p;
X/To
Oy) (oT
t
, (Oy)
(it)
= PL ap t;
G;)
=
T
(c) If Wand X are two functions of state for the
This argument incorporates independent proofs of (i) and (ii), above. 1.25 Take a strong heat theorem in the form (as/ax)y 40 as T 40, where x, yare any two independent variables (other than S). Show that the result of Problem 1.24 can then be strengthened to yield the vanishing of
35
(OW) (aw) oT x = aT
establish that
(OW) (OX) oX ap t'
I
t -
T'
PL
T
Write down the equations resulting from (W, X) = (S, p), (S, v), (v, p) and from the first and last of these deduce that
Solution
S. _ (OS) T - oT t +PLVO:p ,
We have now (as/ox)y 4 0 in the solution (iii) of Problem 1.24, so that it is clear that the quantities (Cv/T), (Cp/T), etc. will vanish.
O:p
I
I (ov)
= -;; aT t
I +KTPL,
. 1.26 Show that an ideal classical gas of constant heat capacities (defined in Problem 1.9) cannot exist indefinitely close to the absolute zero, even if the third law is taken in the weak form given to it in Problem 1.24. Show that an ideal quantum gas (defined in Problem 1.9) does not violate the third law even if it is taken in the strong form given to it in provided one takes (x. y) (v, T). Problem I
where O:p and KT have the meanings given to them in Problem 104. Interpret these results in terms of curves which might be plotted. Cd) Recover the result
Solution
Solution
Consider an ideal classical gas. The result Cp -C v = A of Problem 1.9 will be violated near T 0, since from Problem 1.24 we have Cp 40, Cv 4 O. If pv gU (ideal quantum gas), then
g oP) ( aT v = -;;Cv 40 from Problem 1.24. Hence a Maxwell relation yields (as/OV)T 4 0 as T 4 O. Note that for (x, y) == (T, IJ) the strong theorem fails already for an ideal electron gas for which (as/aT)v approaches a non-zero value as T 4 O. To show this, use Problem 3.15(d).
C-Cv p
=
T(:~},(:;)
{Problem 1.8(a)]
p
by choosing the line tOto be a line of constant volume. (a) Taking the two independent variables to be T and p, 1 dt = dT- - I dp. PL The stated results follow. (b) We have
(Oy) at
and from part (a)
(OyaT)
p
oT at
(Oy) oT
p
p
Oy)T = (Oy oP) (Oy) (at op at T = - PL op T (Oy oP) (Oy) ay ) (oT op aT PL op I
'
I
t
PHASE CHANGES
1.27 Let Tdp) be a line L in the phase space of a fluid with two independent variables which assigns a temperature TL to any pressure p. Let temperatures T be measured from this line: t T- TL(p).
t
t
.
(c) For two independent variables the following result is general:
(~~)x = (~~t -(~~)y (~~t
.
Putting Y
('OaTW)x
~
=
T, Z
~
Solution
t
(a) Integrate
('OaTW)t - (apax) ('OW) ('OW) , ('OW) (ax) aT ap t ax T aT t -PL ax T ap t
The three special results are, using Maxwell's relations, as)
( aT
~
_
_ (as)
p - T - aT
t +PL, (av) aT p p\
=
t
=
T
=-
(1.27.2) (1.27.3)
I
Ca~ t = (aa~}~,
.
Choosing W = T and X = S or V, we obtain
(~~tp
(~~)s (~~)II on L .
This ratio is l/p~ on L. Also
(aa~) p~
p
=-
(~~)
II ( : )
= -
T
pl~ ( : )
T
is finite and non-zero and the left-hand side vanishes on L, = 0 on L. Also
(ap/aV)T
1.28 Suppose the line L in the preceding problem marks a transition from a phase I to a phase 2. (a) Assuming (as/aT)t to be continuous across L, recover the Clausius Clapeyron equation of Problem 1.14 by integrating the Cp/Tequation of Problem 1.27(c). (b) Show that, if p~ is finite and non-zero, on the line L
(:) T
aT t+PL (av) aT v
Since P is the only variable on L, the last term vanishes, and
As
II
(as)
(~~t (~~)Cp (~~)p (~;)C~
II
p
aT p
The first term in (as/aT)t does not contribute. (b) As in Problem 1.27(c), we have
p (ax) aT t('OaxW)T ~ (aaT) (av) aT p .
p ap)s = (ap) (aT aT c = (aaT) (~~)p (~~)p
(as)
(d P)
The last equation yields, on multiplying by V-I, the connection between CY.p and K T . Equation (1.27.1) shows that when Cp/T is plotted against (av/aT)p one obtains a curve which close to the line L approaches a straight line of slope pL and intercept (as/aTk· The notation (k means: evaluated at t :=: 0, Le. at the line L. Equation (1.27.2) shows that the same line is an asymptote of the curve obtained by plotting C./T against -(ap/aT)II(av/aph· The last relation under (c) shows that the curve of CY.p versus KT has an asymptote of slope p~ and intercept V-I (av/ aT)L' (d) Choose W = S, x = p in the first equation of part (c), noting that
, ('oaxW) (ax) ap t
~ T
at constant pressure from a point just inside phase I to a point just inside phase 2. Then 2 A}'" (V2- V I ) ' T dt L
(1.27.1)
,
I TCv (as) aT -PL, (aaT)1I (av) ap t ' (;;)p (;;)t -p~ (~;)T . as) ( aT
-PL
37
The laws of thermodynamics
1.29
1.27
Chapter 1
36
(~~)p = Ca~)p
0 on L.
THERMAL AND MECHANICAL STABILITY
,
1.29 Two macroscopic thermodynamic fluids I and 2, free from external forces, are in thermal and mechanical contact and the total system is isolated. The four independent variables, say the volumes and entropies VI> V2, SI, S2, are constrained by the given total volumelJ the given internal energy U: 0.29.1) () IJt+V2,
= PL ' O.
spite of the divergence of Cp , CY.p , and KT on L, which follows from
part (b), the equations between them given in Problem l.27(c) remain
valid. These are generalisations of relations due to Pip pard (5) • J
U
(5) M.J.Buckingham and W.M.Fairbanks in Progress in Low Temperature Physics (Ed. c.J.Gorter).
since for each fluid U variables.
3, 89 (North·Holland, Amsterdam), 1961. and A.B.Pippard, Phil.Mag., 1,473 (1956). See also
J.Wilks, The Properties of Liquid and Solid Helium (Oxford University Press, Oxford), 1967,
p.302.
\
=
U,+U2
,
(1.29.2)
U(S, (), n) there are no other independent
38
Chapter 1
1.29
VI
(a) Taking the remammg two independent variables as establish the useful results
aV2) 81 = -1, (~ as2) f(aU2) (au J/(au2) (~ 81 = L aV2 82 av I 81 aS2 aV2) ( aS =0, as2) (aUI) /(au2 ) ( asI = - asI aS2 l
I
(1.29.3)
vI
0.29.4)
)
V2'
(1.29.5) 0·29.6)
V2·
l )
i.e.
PI
=
(au2 ) aV2 82'
0.29.7)
P2 (= Psay) (mechanical equilibrium) and (au 2 ) ( au asI = aS 2 ~ (= T say) (thermal equilibrium).
(1.29.8)
V2'
i.e. TI = (c) Show that the conditions for mechanical and thermal stability, i.e. for S to be a maximum rather than only an extremum, include
p
p
- [( -a ) + ( -a ) av I 81 aV2 8
J = [(-vK81) I+ (1) J>0 VK8 2 0
39
1+ (-av2)
aV I 81 = 0, (aU2) (av2) (au2) (as 2)
( au aV I 81 + aV2 82 aVI 81 + aS 2 ~ 81 = O. l)
0.29.13)
The first of these is already Equation 0.29.3). Using it in the second one yields Equation 0.29.4). Next, differentiate Equations 0.29.11) and 0.29.12) with respect to SI at constant v I:
avI) + (av2) asI ( as I
(av2) asI = 0 , 2 -as2) = O. ( au as I + (aU2) a;-2 82 (av2) asI + (au as 2) (as I l
0
[(:~)v+(:£)Jo= [(~)I+(~)JO>O'
0 29 9)
'
•
.
(1.29.10)
where the suffix 0 indicates evaluation at equilibrium conditions. Solution
VI +V2[V I, Sd = v, UdvI,Sd+ U2[V2(VI, SI), S2(VI, SI)] = U.
0.29.11) (1.29.12)
Since V and U are constant, and S I and v I are independent, the partial derivatives with respect to S I and v I are independently zero. Differen tiating Equations 0.29.11) and (1.29.12) with respect to VI at constant
= 0 ,
VI
. I.e.
VI
vI
v2
0.29.14)
vI
The first of these is already Equation 0.12.5). Using it in the second one yields Equation 0.29.6). (b) Rewrite Equations 0.29.4) and 0.29.6) in the forms
(:'~)81 [(~~: t2-(~~:)8J /(~~:) (:%JVI
1-
2 f(au L asI /(au aS 2) J = )
v\
V2
PI-P2 ~
v2
1-
TI~
0.29.15)
,
;
both quantities must vanish for equilibrium. (c) Necessary conditions to ensure oS < 0 include that the second derivatives of S with respect to S I and v I be negative. Differentiating Equation 0.29.13) with respect to v I, at constant S I, we obtain 2 2I f(a2U2) (aV2) (a U2 ) (as 2) J(av2) a U ) ( avi 81 + L av~ 82 aV I 81 + aV2aS2 ~ 81 aVI 81 2 2) (a2v2) f(a U2) (as2) (a 2U2 )(av2) J (as 2) au +( aV2 82 avi 81 + L as~ ~ 8 1+ as 2aV 2 av I 8\ ~ 81 2) (a2S2) + -2 =0 ( au aS-2 -aVI 81 • V2
v2
On using
(a) Write Equations 0.29.1) and (l.29.2) as
VI
)
l
l )
VI
SI yields
VI
(b) The appropriate form of the second law of thermodynamics is, from Problem 1.22, 0 U+ pov - ToS ; ; . 0, i.e. oS at constant Uand Vis to be a minimum. Show that for equilibrium between fluids 1 and 2, i.e. for S = S1+ S2 to be an extremum, the necessary conditions are
( au av I 81
The laws of thermodynamics
v2
vI
VI
SI
and
1.29
av2) (-aVI 81
=-1
(a2v2) --2 'avi 81
=0 '
and Equation 0.29.15), the results of parts (a) and (b) yield the simplification, valid at equilibrium,
22 2) (a--2 a2UI) + (a2U2) (au --2 --2 + -~( aV I 81 aV2 82 aS2 aVIS ) 81J 0 v2
=0 '
40
1.29
Chapter 1
whence 2 {(-aZs) aVI 81
(aZsz) --z aVI
81
1 == - T
[(aZUl) (aZu z)]} --2 aVI + --z avz 81
82
.
(1.29.16)
0
Next, on differentiating Equation (1.29.14) with respect to omitting now terms which will not contribute, we obtain
aZul (aZuzas z azuz avz)asz au a s aSi + as~ as! + aszavz aS I aS + aS 2 aSi 2
Sl, and
2
2
I
1.31
The laws of thermodynamics
constraints TJ = constant, Tz = constant, on intensive variables. This reduces the number of independent thermodynamic variables by one. Whereas in Problem 1.29 two independent variables lead to two stability conditions, one independent variable in this problem may be expected to lead to only one stability condition. Of the four variables, /)1' /)z, TI , Tz only one is independent, and we chose Ii J. Condition (1.29.3) becomes
O.
dll z -= -1
dV I
Using the results of parts (a) and (b) we find
aZUl azuz auzaZs ) ( aSi + as~ + asz aSiz
0
whence
{( aZs) aSi
VI
(aZS2) aSi
'VI
-
1 T
[(aZUl) aSi
VI
+
For equilibrium at constant temperature the free energy may be minimised. So, in analogy to S = SI + S2, we have
= 0 ,
(a 2U2)]} as~
V2
o'
0.29.1
Now the condition for S to be a maximum, rather than just an extremum,
is that the quantities (1.29.16) and (1.29.17) be negative, whence
identities (1.29.9) and (1.29.10) follow.
[Notes.
1. We have shown that the sum of two terms must be positive from both Equation 0.29.16) and Equation 0.29.17). Each of the four terms depends on the variables of one system only and it therefore is reasonable to infer that the following inequalities hold for each system individually:
(v~Jo = (-~~t,o (~:~)8.0>0, (~)o (~~)v.o (~~~)v,o > O.
(1.29.18)
2. For completeness the sign of the cross term a2 Slav Ias I must also be considered (see Problem 1.31). 3. The important case of stability in a two-fluid system in the presence of mass transfer is treated in Chapter 7. Note particularly Problem 7.3.] 1.30 Reconsider the derivation of the stability conditions in Problem 1.29 on the assumption that, instead of the whole system being isolated
energetically, each part system is ins contact with a heat reservoir at
temperature T.
[Hint: The appropriate form of the second law is 6F+ S6T+ p61) ;;;.. 0.]
P = FI(vl, 1',. F~(1J2' T2 ), and Equation ( I .29.1 5) is replaced by
f(aFI LaVI
a/'~) d1!2] + ( aV2 T2 dv 0 J
This problem differs from Problem 1.29 in that one condition + Uz = U on an extensive variable has been replaced by two
=
(aF'z) av.
f(aFl)
L 3vJ
T]-
T2
l
0
O.
Thus the variability of II I and I) 2 yields o
P2,O
as an equilibrium condition (free energy an extremum). For a minimum rather than an extremum, the yields
p)
a2 ( avt
Thus
(iJ
2
2
f
f2)
( aav~
. 'l
ad
TI
condition
>0. T2
ap 2
+ ( aU2
o
dv
> o.
0.30.1 )
1.31 The relation between the conditions (1.29.9), (1.29.10), and 0.30.1) is elucidated by considering one of the fluids in the preceding two problems. (a) Let a, b. h be real constants, let x, y be variables, and let I(x. y)
+ 2hxy + by2
0=
.
Show that I(x. y) has one and the same sign for all real x, y if ab > h 2 • (b) If a function f(x. y) is subjected to a Taylor expansion about its value at (x o, Yo), say, show that the condition (a) applied to the second order terms is eauivalent to a condition on the Jacobian
a f) 2 ( ax 0 a2f ( axay 2
Solution (6)
(6) J.T.Lopuszanski, Acta Phy~. Polonica, 33,953 (1968).
41
J
a(fx , fy) I I a(x. y)
f)
a2 ( axay 0 a2f) ( ay2 0
> o.
1.31
Chapter 1
42
Here I(x, y) is to be reinterpreted as f(x, y) f(x 0, Yo) and the suffix 0 denotes an evaluation of derivatives at the point (x 0' Show that the conditions of Problems 1.29 and 1.30 for thermo dynamic stability applied to a single phase are met by the appropriate form of the condition (b) above, together with the condition that one of the diagonal terms of the determinant is positive.
1.31
The laws of thermodynamics
We conclude with some remarks concerning the relationship between the results (1.31.1) to (1.31.3) of this problem and the result (1.31 A) of Problem 1.29. An interpretation of 1 > U is obtained as follows: a(-p, T) a(v, S)
a(Uv , Us) a(lI, S)
= - -T
Solution
(a) Write 1 a [(ax
I(x,
+ }zy)2 + (ab -
nc vK
v
an interpretation of a, b, and and the result follows. above is fxxfvv > (fxy}2, i.e. The condition of parts (a) and
1 == I
fxy fyy
fyx
s)
one must have written as
~
1
[1> 0,
fxx
>
01
> 0]
>
=>
C
v
yy
T v
>0
(1.31.1)
-
(1.31.2)
> 0,
ap ) ( an s
In Equation (1.29.18), on the other for a single phase, fxx>O,
( 1.31.5)
IJ
T
,
T
>0
=>
T -C
>0
=>
> 0,
v
> 0,
vKs
vK T
> 0,
>0
vKT
>0
(1.31.6) ( 1.31
where the last inequalities in (1.31.6) and (1.31.7) are clearly fulfilled. Indeed, Problem IA(c) enables one to infer also that in both cases C 2C > 0 , i.e. TICp > 0 . K v s ~v
TIC v
>0
=>
>0
but because of Ks
a2
+ T/1:::::E C
[Problem 1.8(b) 1
this requires a condition for the right-hand side of this equation (e.g. TIlICp > 0). The case of two fluids IS considered in Problems 7.3 and 7.4. GENERAL REFERENCES
= C
=>
>0.
v
p
i.e. f
TIC v -K
y~s
T
f yy
CT v KT
v 1<.s
KT
Uo; the stability conditions can therefore be
from which it follows that fxx [1> 0,
X~IJ,
>0,
T
> 0,
oKs> 0,
Then f, fxx, and fyy have like signs. For thermodynamic stability the sign itself is prescribed. For example, with the interpretation F~U,
l(al1) ap T
The result (1.3104) does not go quite as far. In this case one would like to argue
> O.
I
(ap ) /( as) all T aT v
a(-p, T)/a(IJ, s) a(v, T) a(v, T)
The implications (1.31.2), ( 1.31.3) become
}z2)y2] .
of a if ab > h 2 • This implies that a and b have like Thus I has the signs. (b) We have, if x and yare measured from x 0 and Yo respectively, and first order terms are zero, 2 2 2 ( a f ) a2f) (a f) f(x, 0) = ( ax2 y, 0 x + 2 axay /y+ ayL x, This
43
I
vKs
>0.
(1.31
Problem 1.29, interpreted
fyy>O.
(1.31
It is now seen that these conditions are to be supplemented by 1
which implies a condition on the cross term a 2 u;avas.
> 0,
M.W.Zemanski, Heat and Thennodynamics, 5th Edn. (McGraw-Hill, New York), 1968. R.Becker, Theory o[Heat, 2nd Edn. (Springer-Verlag, Berlin), 1967. M.E.Fisher, Rep.Progr.Phys., 30,615 (1967). J.s.Rowlinson, Liquids and Liquid Mixtures, 2nd Edn. (Butterworths, 1969.
2.2
Statistical theory of information and of ensembles
45
~Suppose that an extensive variable x takes on the value XI when ~ system considered in the preceding problem is in its ith state. Recon sider the search for the maximum &ntropy distribution as carried out in
2
the preceding problem if the average value of X( =
Statistical theory of information and of ensembles
~:
;11
1 \1
P,T,LANDSBERG
(University College, Cardiff)
ENTROPV MAXIMISATION: ENSEMBLES
In these problems a statistical &ntropy is defined, and is distinguished from the thermodynamic entropy by the symbol for its initial letter, Though k can here be any constant with the dimension of entropy, it is usually taken to be Boltzmann's constant.
~.1
L i=
Pi
where Z(x) == L exp(-t3xj)' Also show that
Solution
The preceding solution is generalised by considering
I. Use the
1
f = -kL(PilnPi
where 13 is the second multiplier. We have
I
af
I
.., = PN
=N
= -k(lnp;+ I -a+{3xj)
apj
and
I Z(X)ex p (-t3x j)
Pj
Solution
We need to maximise S = -kLP;lnp; subject to LPt l. Let a be an undetermined multiplier; then consider the maximisation of
Z(X) == L exp(-t3xj) , j
i
Also
with respect to each Pi' One finds the condition
, -k(lnpj+l
a)
0,
Xo
\;
= LPjXj =
I
Z(x)~xjexp(-t3xj) J
= [alnZ(X)] a{3
Lastly
Hence lnpj = a-I for all j. Therefore all Pi are equal. Normalisation yields the required result for Pj' The maximum &ntropy is
S2 == Smax = +kLPj[{3xj+lnZ(x)] i
SI == Smax
= -k f (-kln~) = klnN. 44
(j = I, 2,
where
f = -kL(p/lnp/-apt)
apj
,
It follows that
S = SI == klnN.
af
apj+t3PiXj)
i
&ntropy S == -k LPi InPi of the probability distribution,
= P2
[alna~(.~211,x2''''
S = S2 == kt3xo+klnZ.
method of undetermined multipliers to show that for the maximum
PI
=
and
N
I, 2, .." N), where
= I, 2, .." N and an undeter
I
Pj = Z(x)ex p (-t3x j)
Xo
A system can be in anyone of N states. The probability of it
being in its ith state is PI (i
have the given value xo' Show that for j mined multiplier 13, one finds
it/IXi) is known to
= kt3xo +klnZ ,
l
~
x I,X2,
...
46
Chapter 2
2.3
v 2.3 The extensive variables x and Y take on values Xi and Yi, respec tively, when the system considered in the preceding two problems is in state i. If the average values of x and yare fixed at Xo and Yo for the system, show that for the maximum s,ntropy of the distribution function I Pj = :::exp(-tJxj -')'Yj) (j = I, 2, ... N) , where:::
2.5
47
The term grand ensembles is sometimes used if only the mean total number of particles is fixed. In petit ensembles the total number of particles itself is fixed. With this interpretation, verify that the following identifications are consistent with thermodynamics: tJ
L exp (-(3xt - ')'Yi), and tJ and')' are undetermined multipliers. i
Statistical theory of information and of ensembles
-!>
I kT '
')'-!>
Jl
-kTlnZ
-!>
F,
kTln:::
-!>
pv ,
where F is the Helmholtz free energy, provided only that the statistical s,ntropy and the thermodynamic entropy can be identified.
Show also that 0 In:::) x - -( ootJXl>
Yo
",YN,'Y
=- (
In:::) ar 0
Solution X"""YN,{J
In the case of S2,
and S
S3
==
S2 ktJxo+k')'Yo+kln:::.
is now the internal energy U, and we write
XO
k~U+klnZ
U-F T
[cfProblem 1.7(a)].
In the case of S3, Yo is the mean number of particles n, and we write
Solution
Proceeding as in the preceding problems, we find
S3 = ktJU+k')'n+klnZ
U-I.m+pv T
(cf Problem 1.20).
f = -k L(PilnPi-apj+tJPiXi+')'PiYi), j
The identifications proposed in the problem follow.
of
:;- = -k(lnpj+ l-a+(3xj+')'Yj)' UPj
VI' n = n l' 2.5 (a) A system is specified by the values of U = UI> V Suppose its entropy is then SI' Under different conditions it is specified by v = VI' 11 11 1, and a temperature 7;. Suppose its average internal energy is then U = U2(7;) and its entropy S2(7;). Choosing the appro priate ensembles, prove that if U2(7;) ~ UI then S2(7;) > SI' (b) Discuss this result qualitatively in terms of probability distribu tions(1).
Hence 1
Pj
= Zexp(-tJxj -')'Yj)
where
Z
Also Xo =
~PiXi 1
== Smax
L exp(-tJxi-')'Yi)' j
0 In:::) = - ( ~ x"Y,,'Y
Lastly S3
==
'
Yo = - (
Solution
OlnZ)
ar
x"Yj,{J'
(a) We have, using a canonical ensemble, for the second condition of the system
kLPi(tJXi+')'Yi+lnZ) = k(3xo+k')'Yo+klnZ,
S2 =
i
2.4 A collection of copies of the system which are at any given time distributed over their states in proportion to the probabilities Pi obtained in the preceding three problems is called an ensemble. Problem 2.1 describes a microcanonical ensemble, Problem 2.2 a canonical ensemble if x is the internal energy of the system, and Problem 2.3 a grand canonical ensemble if x is the internal energy and Y the number of (identical) particles in the system. Z and Z are called partition functions.
=
U T;+klnZ2 .
Now, instead of summing over states in Zz (as in Problem 2.2), one can sum over energies E j by inserting the degeneracies gj of these energy levels. Then, if the suffix '0' refers to a particular energy ,
Z2
tgjexp
(-{f;) > goexp (- :~).
Choose now Eo as the energy U1 of the first description of the system. (1) See also E.A.Guggenheim,
,Research. 2,450 (1949).
2.5
Chapter 2
48
Then go = g 1 becomes the N value of Problem 2. I for the microcanonical ensemble appropriate to the first condition, and Sl = king!. Hence S2
U2
> 12 +klngl 12
# Sl .
Statistical theory of information and of ensembles
2.6
49
(b) From Problem 1.7(a) we have
dF = d(U-TS)
= -SdT-pdv+p.dn
and from Problem 2.4
In the second condition the energy of the system can fluctuate and the probability distribution is spread over many more states than is possible when the energy is fixed. Hence S2( 12) > St.
dF
-d(kTlnZ)
Hence
-klnZ dT kTd(1nZ).
U
P
p.
kT2dT+ kT dv - kT dn ,
dlnZ PARTITION FUNCTIONS IN GENERAL
2.6 (a) If the number n of identical particles in a system, its internal energy U, and its volume v are given, the microcanonical ensemble of Problem 2. I is appropriate. If these quantities are varied one has to consider a set of ensembles with neighbouring values of n, U, 0, which become the independent variables. Show from Problem 1.20(b) that the partition function k InN satisfies alnN) ( ~ V,n
p
= kT'
(alnN)
~
V,v
=
kT'
(
P,
n
alnZ) ( av T,IoI
L kT'
an
v, T
=
(alnZ)
ap'
v,T
p.
fl (dlnZ) = U-p.n
kT' aT V,1oI kT 2 '
.
(
PV)
d kT
=
n i p kTdP.+ U-p.n)dT+ kTd/J .
(d) This follows since the last result is an exact differential. See also Problem 20.4. d,7 A system is kept at fixed chemical potential and temperature. Show that the -logarithm of its grand partition function is proportional to the volume. [Hint: The first result of Problem 2.6(c) is useful.]
'I.
Solution
From Problem 2.6(c) x
InZ satisfies
(
ax)
av
x
p fJ., T
v
= kT
Hence for systems at constant p. and T x canst. v Alternatively, if p. and T are constants, the Gibbs-Duhem relation of Problem 1.20(d) shows that p is constant. Hence
(d) Establish
e~)T' v-p.
S)
(P.fl U == d k1'- kT+I
Now write down the derivative on the right and use Problem l.20Cb) to simplify the result to
= kT'
U alnZ) ( aT P, n (c) For a grand canonical ensemble the independent variables are v, T, and p., only average values of U and n being now given. Systems specified in this way are supposed to be in equilibrium with a large heat reservoir at temperature T and a large particle reservoir at chemical potential p.. Show that the partition function satisfies
(alnZ)
(pv)
d kT
dlnN)
au
Systems specified in this way can be regarded as isolated systems. (b) For a canonical ensemble the independent variables are 0, n, and T, only an average value of U being now given. Systems specified in this way can be regarded as isolated apart from a large heat reservoir at temperature T with which they are supposed to be in eauilibrium. Show that the partition function satisfies dlnZ) ( ---a;- 1;n = kT'
and the results follow. (c) From Problems 2.4 and l.20(c) we need
T,v
T(:;)IoI,v'
In.:.
kT
0:
v.
Solution
(a) The results follow from Problems l. 20(b) and 2.4 which yield TdS
kTlnN = dU+pdv -p.dn .
2.8 The replacement of the partition functions Z and Z in Problem 2.6 by their thermodynamic equivalents, given in Problem 2.4. leads to thermodynamic results of some interest.
50
Chapter 2
2.8
(a) Infer from
Statistical theory of information and of ensembles
2.9
EN}R0PY MAXIMISATION. PROBABILITY DISTRIBUTIONS
( that
n
alnZ)
a;-
n
/2.9 A one-dimensional normal distribution of zero mean and standard deviation 0 is given by
V.r = kT
v(~~)
OlnZ) =
( oT V,!l
show that
-----ar
S = k [ InZ+ T ( OlnZ)V,!l
(a) Show that its G;ntropy is ~kln(21Te02), where e is the base of the natural logarithms. (b) Show that for given L:x 2p (X)dx == 0 2 the normalised probability distribution having the largest G;ntropy is the one-dimensional normal distri bu tion.
l.
Hence establish the relation
Solution
S = v
(a) The G;ntropy is
(:i) !l . v,
(c) Establish the thermodynamic relations as found in parts by purely thermodynamic methods.
S
and (b)
(a) Replacing InZ by
we obtain
0fJ.
v, T
as required. (b) From the equation stated OlnZ)
( aT
!l
which is the first result stated under (b). we obtain O(PvlkT)] S = T kT [ aT pv
= r+ vT
[Ir
(oP) oT
_
-rln .::.
S =
-1 v. J J .
v(~~)T
V
•
We must maximise with respect to arbitrary variations of p(x) in the integrand of f[p(x)] == -kL:p(X)lnp(X)dX-aJ:p(X)dX-t3L:x2p(X)dX
=
(a
p ) v aT
(:i)!l .
_~-t3x2 =
-k-k
l'. JJ. .
O.
It follows that
p(x) = aexp(-t3x 2 )
for the maximum. The normalisation
SdT-vdp+ndfJ. = 0
n
x 2p(x)dx
where ~, t3 are Lagrangian multipliers. Hence
!l
(c) From the Gibbs-Duhem relation of Problem I
and
1
Replacing InZ as in part (a),
v,
f
~ k In (21Te0 2 )
v, T
S pv S kT - kT2 = kT
TS v,
kT 0fJ.
In 21T02 x 2 ) 2 -202 dx
= '2k ln (21T0 2 )+ 20k 2
= ~(op)
[O(PvlkT)]
kT
-kfp(x)lnp(x)dx
= -k fp(x)
Solution
~=
-ex> <x <ex>.
(21ra2fVzexP(-2:22)
p(x)
v, T'
(b) From
one finds
51
a (2)
f '" exp(-bx
00
-00
x 2r exp(-/lx 2 )dx
For r = 0 the result is (1T/(3)Y'.
1X
2
)dx
= a (%)Y' =
= ~~-"":"':--':':'':'::'-'''--'''':''':-'-I
I. for r
=
1,2,
52
Chapter 2
2.9
Also 2 0
f
=a
Hence
= 2
2
_oox exp(-bx )dx
p(x) =
="2a(7r)Yl [33 =
-r~-_-_-,12'-'-'-'-Y2exp (-
I
Statistical tl1eory of information and of ensembles
2.11
(a) Maximise as in Problem 2.5 the integrand in
2[3
;;2) .
under the conditions stated.
f
p(x)[ -klnp(x) -a'-[3lx Ir] dx ,
f whence
-k-klnp(x)-a'-[3lxl' = O.
The distribution of largest &ntropy compatible with the specifications is Po(x)
2.10 Let By p(x)
2rllT
==
(I + +), where r
"" j~ M, == [f x I'p(x)dx S[p(x)]
is the Gamma function,
I = f-:P(X)dx M r = 2a r
11,
(r> 0), the rth moment, and
1
= aexp(-[3lx
The constants a, [3 can be identified by
pC-x), a probability distribution,
_00
53
f" 0
= 2a
too exp(-[3xr)dx = 2ar (I ++)[3-1Ir ,
2a (r+l) x r exp(-[3x')dx = -[3-('+l)/T -- = I r r r[3
It follows that _
I -lIr
a- 2r. (r. + I) M r
= -kIp Inpdx, the &ntropy of probability distribution.
r
Br
My '
r
Show that
M, 9~' Byexp {S[P(x)J k
_l.}
and that the equality sign holds when Br
Ir)
(IX uexp rM;
=
p(x)
and
r
.
p()(x)
k
The following integral will be needed: irs
==
L""x ex P (-[3x r )dX. S
Put y
x r , dx
I I = -X-(r-I)dy = _y-(r-1)/rdy. r
Then i r. s =
r
If""Po(x) (B, Ix-Ir) dx--I = In Mr In--
0
r
_bO
Mr
rMyr
r
Br
It follows that for given Mr all other distributions have a smaller &ntropy, i.e. Mr exp(So_!);;, exp{§[P(XH_!}
Br k r k r'
which is the required result. MOST PROBABLE DISTRIBUTION METHOD
If""
-;:
(Ixlr)
as required. The &ntropy of the distribution Po satisfies the relation So -.-- = -
Solution
By
M;.ex p - rM; ,
yP exp (-[3y)dy
If'{p+l) [3P+ 1
=r
(S+ I)
I = -;:[3-(st l)/T -r- .
S+ I ) (P -r--I
2.11 The states of a quantum-mechanical system are labelled by a complete set of quantum numbers. Suppose one of these determines its energy Ej and thatthe corresponding energy degeneracy is gj (j = I, 2, ... ). Consider an ensemble of N copies of this system in the sense of Problem 2.4. Let one State of such an ensemble be specified by the numbers (n 1> n2, ... ), where nj is the number of systems with energy E j • The capital S is a reminder that a State of an ensemble is considered.
54
2.11
Chapter 2 (a) Prove that the number of ways of realising this State is G
n!
=
(c) From part (b) and the definition of "'(n), we have
Nfgrtlgrt2 I 2'" nl!nry! ...
d "'(N) """ dn[nlnn-n+!ln(21Tn)]
(b) If n is large enough, check from a book on special functions that Stirling's approximation holds:
r(l +n)
nrt e-rt (21Tn)Y'
where r is the Gamma function and n is a positive integer. (c) Make the continuity assumption (3) with regard to n, and define Gauss' "'-function by
dnr(l+n).
"'(n)
From tables we find "'(3)= 1·2561,ln(3·5) 1'2528, so that the error in "'(3) """ In(3·5)isO·0033/l·2561 = 0·26%. It is less for n > 3. (d) For the most probable State of the ensemble, we have to consider InG(n 1 ,n2, ... }-exLn;-{3LE;n;,
f
j
J
n· ::1
N
J
J
= -"'(nj)+lng;-cx Hence, if (n!,
{3Ej = O.
ni, ...) is the most probable State, nt = gjexp(-cx-{3Ej )-!
J
where (3 is a Lagrangian multiplier. (e) Discuss the relation between this result and that of Problem 2.2. (a) In any State of the ensemble, let us give systems with the same energy the same letter, and systems with different energies different letters. Hence G(n I, n2, ... ) gives the number of distinguishable arrange ments of N letters, n 1 of one type, n2 of another type, etc., and is equal to N!/n I !n2 L.. if the degeneracies are neglected. As a result of the degeneracy gi' each of the G arrangements gives rise to a number of further arrangements equal to the number of ways of assigning nj systems to gj states, i.e. to gfi arrangements. The result of part (a) is thus obtained. This matter is discussed in many books. 'l!(n)
was pointed out in
.
If these are M energy levels, cx may be identified by summing over the energy levels to find
N+!M
Solution
(3) The importance of this assumption and the usefulness of hoc. Natl. Acad. Sci., U.S., 40, 149 (l954).
j
Instead of maximising G with respect to each ni' it is more convenient to maximise In G. We have af a an. = an J-Inn;! + nj Ingj - an; - {3E.jn;]
i.e.
Lg.e-iJbj+O( j
LEjn; = Eo·
w(nt) = In[gjexp(-ex-/3Ej ] ,
n'!' """ _ _.:::.1-_ _ ::L
N
;
where cx and {3 are undetermined multipliers to take account of the conditions that N and Eo are given, i.e.
2: n; = N,
Verify, by the use of tables of "'(n), that for n ~ 3 this approximation holds with an error of less than 0·26%. (d) Make an assumption of equal a priori probabilities of different ways of realising a State, and assume that the energy of the ensemble is given. Hence show, using part (c), that for the most probable State of the ensemble the probabilities nj!N are given by
./
= Inn+ 1-1 +fn
""" Inn+ln(l +(n) = In(n+!).
Prove from (b) that "'(n) """ In(n+-!).
55
Statistical theory of information and of ensembles
2.11
= exp(-cx)4.gjexp (-/3E;)
Zexp(-cx) ,
J
where the canonical partition function has been introduced. It follows that '!l _ ( ~) g; exp (-/3}'j) N- 1+2N Z When the terms in 1/N are collected together, the desired result is obtained. (e) The most probable State of the ensem ble leads to the most probable values nt. The canonical ensemble of Problem 2.2 yields the mean values of the n; as averaged over all States of the ensemble. In this average the most probable State makes a dominant contribution, but less probable States will also contribute, and hence the d·ifferent results which are, strictly speaking, obtained by the two methods.
56
Chapter 2
2.11
The present method requires the nj to be large enough for the assump tion of continuity to be justifiable. This condition is difficult to fulfil unless N ~ 00. Although M is often also infinitely large the usual result follows only if MIN ~ O. These difficulties are often overlooked. It is particularly easy to overlook them if the coarser approximation \}f(n) - Inn is used, which yields at once
Solution
(a) Let u refer to the upper level, and I to the lower level. Then nl +nu
G( )
N! n = [!(N+n)]![!(N-n)]!
ni _ exp(-{3Ej)
Z
Hence using Problem 2.11 (b) InG(n)
2.12 (4) Develop the ideas of Problem 2.11 for an ensemble of N identical systems each of which has two non-degenerate states. Take N to be even, and specify a State of the ensemble by the number n (which is even) giving the difference between the num ber of systems in the lower energy state and the number in the upper energy state. Assume that N, but not the total energy, of the ensemble is given. (a) For n <{ N show that the number of distinct ways of realising a State n of the ensemble is, within the Stirling approximation, liz
pen)
= Nln
n)] +!(N - n) -1 In [rr(N - n}]
2+! In(rr~ )-!(N +n + I ) In (I +~) -!(N -n+ I )In(1 -~)
Using In(l +x) = x
lIor x
<{
I, we find
(2)
2
whence the result follows. (b) The number of States of the ensemble is 2N since each of N systems can be in one of two states. Hence P( n)
G(n) ---;:;-IV "
)lIz exp (n2) - 2N
To test the normalisation, put y = !CN+n), when HN-n) = N-y. Then y = 0 at n = -N and y = N at n N, and the corresponding values are given below:
I'
y
o
11
-N
-N+2
~N
1N + 1
N
o
2
N
Observe also that (a+b)
N
~
L
Y =
N! ,.N '(N- ),aYu-Y ,
oY·
Y .
whence (4) Problems 2.12 and 2.13 are of interest in various contexts. See, for example, J.E.Mayer and M.Goeppert Mayer, Statistical Mechanics (John Wiley, New York), 1940, p.75, and C.Kittel, Elementary Statistical Physics (John Wiley, New York), 1958, p.22.
2
,_ I n n (N+ I) InG(n) - Nln2+!ln rrN - N+ 2N2
exp (n2) - 2N .
where A is a normalisation constant. Determine its value, and hence verify that P(n) is a one-dimensional normal distribution of zero mean and standard deviationvN (defined in Problem 2.9). (c) Compare the mean value of n and the most probable value of n. (d) Compare the total number of States of the ensemble, GT , with the number of ways G(O) of realising the most probable State, as N becomes very large. Repeat this process for the corresponding entropies, and use the result to discuss the key properties of the method of the most probable distribution for this example.
NlnN-N+! In (2rrN) !(N+n)ln[!CN+n)]+!(N+n)
-! In [rr(N +n) j-!(N -n) In [!(N -
(b) Assuming that distinct ways of realising a State of the ensemble are eq uiprobable, show that the probability of a state n (<{ N) is 2 A ( rrN
!(N-n).
Since N is even, it follows that n is even. The number of ways of realising a state n is, from Problem 2.11 (a) withgl = g2 I,
Z
2) G(n) = 2N ( rrN
n,
nu
nl = !(N+n) ,
This is the standard result of Problems 2.2 and 2.4 if we rewrite it for quantum states instead of for energy levels. It then becomes N -
nl-n u
N,
whence
'.!l- gjexp(-(JEj) N -
57
Statistical theory of information and of ensembles
2.12
= 2N
Using all these
we can
N
2.13
2.12
Chapter 2
58
N
(even n)
2
2N ( 1JV N'h
I .
(even n)
(rrN2 )112 (2Nrr)'h = 2 .
N
InG(n) = O. C(n)
(even n)
Direct algebraic study of the original expression for G(n) in terms of factorials gives the most probable value of n as n O. This is also the value obtained from the expression given in part (b). Hence the mean and the most probable values coincide in this case. (d)
rr
N
2 )112
2N ( rrN
(n2)
exp - 2N .
(b) If the system is in equilibrium at temperature T show that the probability of finding a state n in an external magnetic field H is Gaussian with the mean value (n) and the standard deviation a given by (n)
N! - [(tN)!
(~) \12 2N
I) weakly 2.13 Each of a system of N atoms align themselves either parallel or antiparallel to the applied magnetic field H. Let n be the difference between the number of atoms in the lower level and the number of atoms in the upper level, and let Jl be the magnetic moment of an atom, then a state of the system can be specified by the integer n. The energy of the system in a typical state n, referred to an energy zero at n 0, is En = -nJlH and the magnetic moment is Jln. Show by using Problem 2. 12 that in the limit of zero magnetic field the number of arrangements which can give rise to a state n <{ N is
This suggests that the normalisation is incorrect. However, an approxi mate formula for n <{ N has here been used for -N ~ n ~ N as nand N -j. 00, and the fact that n takes only every other integral value has been ignored. This supplies a correction factor of!. (c) The symmetry between the two states of the system gives G(n) G(-n), whence the mean value of n is
I nP(n) = 2-N n=-N
)'12
and it indicates infinite 'steepness' as N -j. 00. The replacement of all States by the most probable States is therefore justifiable in this case.
Hence the value of A introduced in the problem is unity. One may be tempted to check the normalisation by evaluating the integral 2 )~ roo J.r=., f(n)dn = (rrN J~ exp (- ;N2) dn
59
take P(O) divided by the standard deviation. This ratio is
the correct normalisation of Pen):
I Pen) = 2-N n I=-N G(n) = n =-N
JlHN
kT
J2
and one has
'
a
yiN.
Solution
fl ==
C T -C(O) C T
=
2
1- (rrN)'h
-j.
(a) The value of C(n) is derived in Problem 2.12(a). However, there is a change in the point of view since only one system is contemplated. The number N is now the number of particles in this system. But the formula
I .
Also ST
Hence
Statistical theory of information and of ensembles
= kNln 2,
f2 == ST -S(O) ST
S(O)
== klnG(O) .
In(trrN) 2Nln 2
N! n 1!
G(n) -j.
O.
Thus the value of f, which gives the fractional error in replacing all States by the most probable States in a calculation, is large in a calcula tion of the number of States, but small for the &ntropy. This circum stance is typical and shows that the &ntropy is a very insensitive function. The probability pen) has a maximum at n = O. This is not, however, a sufficient condition for an average over the ensemble to be approxi mated well by the most probable States. One must show that this maximum is also very steep. As a measure of the 'steepness' one could
.
still applies. In Problem 2.12(a) one divides by the factorial factors since it clearly does not matter to the specification of the State of the ensemble which of the systems are in state 1 and which are in state 2, so long as the number in each state is definite. In the present problem the division arises again because the atoms are assumed indistinguishable. We have for a canonical ensemble that the probability of state n is
k:
(-E)
2N( 2 )'h Z rrN exp[f(n)],
60
Chapter 2
2.13
where Z is the partition function and
np.H kT
n
fen) =
2.14
Statistical theory of information and of ensembles
(e) Verify that the implications proved under (c) and (d) belong to a wider scheme of implications given
2
M+D~
D/ xip
Writing nl == p.HNlkT, we now have
fen)
nr
(n -nd 2
2N
2N
=
Aex p [
Here A is a normalisation constant which depends on N. SOME GENERAL PRINCIPLES
2.14 The states of a system form W groups labelled by the suffixes i i , 2, ... , W, the ith group having Gi equiprobable states. The proba
bility that the system is in any of these states is Pi' Let the probability per unit time that the system makes a transition from a particular state of group i to a particular state of group j be a constant which we shall denote by Aij (with Au == 0). (a) Prove that the transition rate from group i to group j is
Rij
=
"S'i
C)
Solution
(a) Multiply the probability of finding any state in group i by This gives only the transition probability per unit time into a particular state of group j, and has to be multiplied by the factor Gj • (b) We have
A=
PiA jjGj .
Gj
L(Rji-R jj ) j
4: GjAjj (R J.- R) G. J
J
I
where Fj
V
equation for Pi is sometimes called the master equation and was first discussed by W.Pauli in 1928. Its generalisations are currently topics of research. The principle X generalises the principle M, the principle P generalises the principle of equiprobability of states. Note that S is weaker than D, and these therefore are not equivalent principles. The subjects of master equations and detailed balance are pursued further in Chapter 26.]
L{Ajj-Aji)Gj. j
(c) A system is in a steady state if the following principle holds: Prove that S is satisfied if the following two principles 0; and P: the suffixes i fall into classes a., (J, ... such that hold: X: F; within each class there holds Aij 0 if and only if ~/Gi = ljlG j K", say (i, j within a.th group). Interpret this result. Show that the 8ntropy is R S -kf~lnGi
s: Pi = 0 for all i.
*
and prove that the 8ntropy production rate satisfies H: S ~ 0 if the principle of microscopic reversibility M (A ij Aii for all i, j) holds.
Ll}AjPi - L ~Ai;Gj j
j
Ll}AjiGi -~F; - ~ LAjjGj
(The principle of detailed balance asserts D: Rij Rji for all i, (b) Prove that the rate of change of a typical ~ can be written as
A -=
M
~H~)
so that
Pen)
61
j
j
P.) = L GjAji G;( i- G. -~F;. p,
J
J
I
(c) The principles X and P clearly imply Pi 0 if we take into account the results of part (b). If the principle P holds, the W groups of states decompose into a smaller number of classes of states between which transitions are not possible. The principle S is fulfilled if the probability per state has the same value for each class of states. If fi = 0, the equiprobability of all states leads to detailed balance only if one assumes also that all states are interconnected. The 8ntropy is
-k "L. Pi -;-InPi
all states
G;
Gi
Sum over all states in group i first. Their probabilities are all equal to so that the expression given in the problem is found. Next we have that all F; vanish, and
~/G;,
s
G +Pi Pi . PiJ G [ . (P.)
-k~ Piln
i
j
.
62
Chapter 2
Since
LI Pi
we find
2.14
0 by normalisation, the last sum vanishes. Using part (b)
. _'\' ' ,(!l_~) I"
"
'\' [GIGj (Pt,- P) P (P.G.-t,P)Ajiln:LP G. AjilnG~+GiGj .J G
= -!k!-,. 1,1 _
3 Statistical mechanics of ideal systems
Pi G. Aj/ln G. I
k!-,.GiGj G
S
TI
(I}
1,1
I
I
,
I
I,
T{
P.,) ~G· (;.- G, AjilnpG~
'\'
- -!k !-,.GIGi
I
P.T.LANDSBERG (University Col/ege, Cardiff)
,
I
=
If Pt/G i ~/Gj the (i, j) contribution to the ~/Gi =1= ~/Gj then the contribution is positive,
The statistical &ntropy of Chapter 2 and the thermodynamic entropy of Chapter I are regarded as identical in the rest of this book.
double ~um vanishes. If Hence S ~ o.
(e) The proofs are simple (5). MAXWELL DISTRIBUTION
2.15 (a) Show that, for all positive x, lnx ~ I - I/x, where the equality holds if, and only if, x I. (b) Show from (a) that if {Pi}, (pP} are two probability distributions for the same set of states and such that PP > 0 for all i, then K(p,pO)
3.1 (a) A system is specified by variables Xl' Xl' .." XN which have in dependent normalised probability distributions PI (x 1), Pl(X2), "', PN(XN)' Show that the entropy may be written as
== k~PilnpP~ I
S
I
PP for all 0, [Suppose the PP are the equilibrium probabilities of a possibly non isolated system, for example it could be the canonical distribution if the system is in contact with a heat reservoir. Then the knowledge that the actual distribution is (say) Pi under these conditions, represents an 'information gain' K(p, pO). Conversely, the passage from Pi to PP corresponds to an internally produced entropy K(p, pO). The quantity K(p, pO) was introduced by A. Renyi (6).] is positive unless the distributions are identical (Pi
(b) The Xi are interpreted as the three Cartesian velocity components V; of a particle in a gas with point interactions (Le. the particles are points and interact only if they are at the same point) in equilibrium at temper ature T. Assuming that the mean kinetic energy associated with each component is'; kT, show that for a state of maximum entropy
V;) =
(a) Let y == lnx - 1+ l/x. Then
dy I-x
dx
This is negative for 0 < x < I and positive for I y occurs therefore for x = I, so that y ~ O. (b) With Xi pdp?,
k~ p;oxilnXi ~ k ~PiOXi (1- xl) I
I
m)Y2 exp(mv,2) (21rkT - 2kT '
where m is the mass of a molecule. (c) Derive from (b) the probability that a molecule in this gas has a speed in the range (V, V + d V) is p( V)d V, where V can have any value from 0 to 00 and ,I
Solution
K(p,pO)
= -kitJJpi(Xi)lnPi(Xi)dXl
I
< x. = k
p(V) = 41rV
The least value of
2;. (Pi -
2
Vl) ( m)" exp (m 21rkT
2kT .
This is the Maxwell velocity distribution. (d) Discuss the range of validity of these results by inspecting the assumptions needed to obtain them.
pP) = 0,
I
Solution
Hence K is non-negative. If all x/s are unity, K = O. If one Xi =1= I this term will contribute a positive quantity to K and K > O.
(a) The independence of the probability distributions Pi(Xt) implies that the probability of finding a state XI' X2, ... , Xn of the system is
(5) P.T.Landsberg, Phys.Rev.• 96, 1420 (1954) and Section 34 of P.T.Landsberg, Thermo dynamics with Quantum Statistical Illustrations (lnterscience, New York), 1961.
N
P(Xl> X2, ... , XN)
(6) For recent discussions, see F.Schlogi, J.Phys.Soc.Jap., 26 Supplement, 215 (1969).
=
n
Pi(Xi)'
i = 1
63
I
64
Chapter 3
3.1
Hence the entropy is
r
I
Statistical mechanics of ideal systems
65
(d) The main assumption is that PI (Vd, P2 (V2), P3( V3 ) are independent probability distributions and that the point particles have only point interactions. In a dense gas the interactions cannot be approximated in this way. It has also been assumed that one particle can be treated separately from the rest. This is invalid for a system of indistinguishable particles when exchange effects must be expected.
S = -kf..JPlnPdT = -kf..JPI ...PNln(Pl",PN)dx1 ... dXN
-k
3.2
~ (fPi InPidX;).
(b) The following constraints apply to the p;'s:
L~Pi(V;)dV;
J~Pi(!mW)dV;=!kT
I,
~
(i
3.2 Obtain the following quantities for a Maxwell distribution of velocities: (a) The average of the nth power of the velocity is
1,2,3).
-~
Maximising the entropy subject to these constraints as in Problem 2.2,
consider f= Hence
-kitl
(fp;lnp;dV;+o:;jPidV;+13Jpi WdV;).
:;;
-k (Inp; +0:;
f
W+ 1) dlj
where n is real, n > I, and (b) The average speed is
exp(-131 W-0:;),
O:j
f This yields
~
2
lj exp(-13llj
2
1,
i.e.
( ~)
8) .
«V-(V»)2) = kT( m 3--; (d) The 'fluctuation' in the kinetic energy is
\6
exp(-O:j)
1,
(;m)2«V2_
O:;)dlj - m'
(e) The most probable speed is
-~
1
;1,
(
Vo -_ (2kT)Y'
m
m (1r)\6 2kT13j 13 exp(-O:/)'
so that exp(-O:/)
(~~r
kT
_
1r)% ( 13; exp(-o:;) = 1
is the gamma function.
(c) The 'fluctuation' in speed is
== 0:;+ 1.
To identify the Lagrangian multipliers observe, using the integral given in the solution of Problem 2.9, that
i~exp(-13j W-O:j)dlj
r
0,
whence P;
(n 3)
- 2 (2kT)/'I/2 r -+2 ' V1r m
2~T' )
131 =
Solution (a)
m 2kT'
m)V, exp (mV2) p(V)dV= ( 21rkT -2kT 41rV2dV, since
J J J d V; d V; d V3
=
= 41r(~)'I'f~ Vn+2 exp (- mV2)dV
and the stated expression for Pi fonows. (c) On integrating the probability PIP2P3 of a velocity with compon ents in the ranges (VI' VI + d VI), (V;, V; + d V;), (V3 , V3 + d V3) over a shell of essentially positive radius V = (J1 + Vi + J1 )'h, we find the required probability
21rkT 0 2kT = 41r ( -m-)'hkT(2kT)(n+I)/2f~ -- - x(n+ 1)/2 e-x dx. 21rkT m m 0
(n+3) and the result follows.
The integral is
r -2-
(b) Put n
=
I in (a).
(c)
41r V2 d V.
«V -( V»)2)
shell
,
\
3kT m
= (V2 -
2 V< VH« V»)2) = (V2)-« V»2
= 3kT m
8kT 1rm
66
3.2
Chapter 3
3.4
Statistical mechanics of ideal systems
67
Normalisation of the distribution according to
(d) The 'fluctuation' is
= (!m)2
+«J72»2)
mJ72-<-tmJ72»2)
= (!m)2[
47r J72g( J72)d V
::=
I
4
CLASSICAL STATISTICAL MECHANICS
The probability has the form A J72exp (- ;:;).
P
Hence dp/d V
leads back to the Maxwell distribution.
0 yields A ( 2 V - J72 mv) kT exp (mJ72) - 2kT
=
O.
The required result follows. 3.3 Maxwell's original argument. Let 47r V2g( V2)d V be the probability of finding a molecule in the gas with velocity magnitude in the range (V, V + d V). Here g( J72) is an unidentified differentiable function. Obtain the Maxwell velocity distribution on the assumption that the probability distribution for the three Cartesian components of a velocity vector are (a) independent, and (b) identical.
3.4 The canonical partition function in classical statistical mechanics is given by an integral over a phase space instead of a sum over states. In fact, each state is specified by a point in the space in which generalised coordinates q 1, ... , qf and generalised momenta PI' "., Pfare the axes, A factor gh-f , where II is a constant with the dimension of action, reduces Z again to a dimensionless number and allows each state to be g-fold degenerate. If fl is the Hamiltonian of the system, the prob ability of finding the system in an element dr of phase space is propor tional to exp(-H/kT)dr and
- -rf ... fexp (- kT H) dPl,,·dqr·
Z - gil
(a) Show that for one particle of mass m moving classically and nonin a field-free container of volume v (with g I) ZI = h- 3 v(27rmkT),h,
Solution
Let f( ~) be the probability distribution function for the x-compon ent of the velocity. Then one can put g( Vi
g( J72 )
+ ~ + T1) = f( Vi) f( Vi )f( T1).
where T is the temperature of this system. For n distinguishable particles moving independently but with point interactions as in (a), show that one would expect classically Z"
It follows from these two expressions for g( J72) that
(oOg) Vi v
Y
'
=
(0
dg J72) d J72 0 Vi Vyo ~~
=
dg d J72
and that
COn)
I~::=
f(~)f(T1).
Explain the qualitative effect on Z" if the particles are indistinguishable. (c) Writing Zn = (ZI1~}n for generality, where in depends only on n, consider the restrictions on 1~ if the Helmholtz free energy of the system is to be an extensive quantity (cf Problem 1.20). In particular, show that for large n the classical partition function (Zn)cl satisfies Zn
Hence I dg g
df(Vi) -dVi
(:= -(3),
The quantity (3 can depend only on Vx, but similar results hold for the y and z-components. Hence (3 is independent of the velocity components. It follows that f(Vi) = exp(a:-(3Vi), where a: is a constant of integration. Finally
g(J72)
exp(3a:-(3J72).
= Z?
=
(Z" }cdn!
(d) Obtain Boyle's law for the system specified above and show that it does not depend on the values of g, h, and in. {The correction factor 1;, is needed because particles are indistinguish able in a simple gas. It has an interesting history. The quantum-statistical approach has no need for such corrections. See Problem 3.12.] Solution
(a) Take the potential energy of the particle as zero. Then p2 H=
2m'
68
Chapter 3
3.5
3.4
where K is the kinetic energy and M is the potential energy. Let
The integrations over the three coordinates, conveniently taken to be Cartesian coordinates, yields v. Hence h
-3 V
"'f"'f"" exp (- PI22mkT + P22+ P32) dpi dp
f
_'" -co
-00
2 dP3
3
f
(KI
+ K2kT + ... + Kn) dpi ... dP3n
eXP(-:;)dPldP2dP3T
= Zf·
For indistinguishable particles the value of Zn must be smaller. For example, if n 2, the above integral treats KI = I eV and K2 = 2 eV as contributing equally with KI 2 eV and K2 I eV. For indistinguish able particles there can be only one such contribution. (c) From Problem 2.4 Fn = -kTInZn
=
-nkTIn(ZJn)
= -kTn[ln(vj~)+~
In T+ 1In(2rrmk/h2)J.
Fn , n, and v are extensive; T and also constant terms can be regarded as intensive. It follows that In cannot be unity; instead it must make vln intensive. So, if D is a constant, D
In
= n .
=
(~r,
and this is for large n approximately lin! if D is interpreted as the base of the natural logarithms. (d) It follows from the solution of Problem 1.7(a), which dealt with a closed system n = constant, that the pressure is P =
-(!:)T.n
kTn v
3.5 We are retaining the notation of Problem 3.4. A classical statistical mechanical system is at temperature T and has Hamiltonian H
= K(PI, ''',
Pp = Bexp(-K/kT),
Cexp
(b) Check the Pp-formula by applying it to a simple gas of n distin guishable particles of mass m whose momenta can range from -00 to +00 and show that B = (2rrmkTr%.
Also check that the Maxwell velocity distribution of/Problems 3.1, 3.2, and 3.3 is obtainable by this method. (c) Using Problem 3.4, show that the canonical partition function is for g l,f = 3n Zn
(2rrmkT)3nI2Qn /z2 n! '
n
provided that K =
L pl!2m i= I
and that the so-called collfigurational -~
integral or partition function is
~.
f... Jexp(-~)dql
- Qn
+M(ql' ... , qt),
...dq3n,
where the integration is over the volume of the fluid. Show also that the grand partition function (Problem 2.4) satisfies pv ::: = exp kT
It follows that the correction factor for Zn is
I~
coordina tes. (a) If Band C are normalisation constants, prove that Pq
(b) In this case if Ki is the kinetic energy of the ith particle h- nvn exp
\
(Pi' Pi+dpi) and let Pq dql ... dqt be corresponding probabilities for the
1
= h- 3v(2rrmkT)'iz.
Zn
Pp dp I'" dpt be the probability of finding the momenta in the ranges
I
h-3v(2mkT)'1{J~ exp(-x 2 )dxJ3
69
Statistical mechanics of ideal systems
'" /;:'0
n
where the so-called fugacity is
z
2rrmkT)% (
112
f.1
exp kT .
(d) A right circular cylinder of base area A, of great height, and at uniform temperature T contains n particles each of mass m. They are acted upon by a gravitational acceleration g, independent of height. Use the Pq -formula to show that the particle concentration at height z is given by the barometer-formula mgn (mgz) p(z) = AkT exp - kT .
If the law pv = nkT holds at all levels show that the pressure variation is p(z) =
exp ( - mgz) kT .
70
3.6
3.5
Chapter 3
Statistical mechanics of ideal systems
Hence the probability of being in the range (z, z + dz) for anyone particle is
[The configurational integrals for real fluids are studied in Problems 9.2 to 9.7.] Solution
K+M) dqj ... dpj (----u
To find the number of particles in range dz, we must multiply by n, and to obtain the concentration we must divide the result by A dz. The stated result is then found. To obtain the pressure at level z, put
is a normalisation constant) over all q's to find the stated result with a new normalisation constant B. A similar argumen t leads to Pq . 3n 2 (b) Putf = 3n and K = L ~i • It follows that i
j
p(z)
2 j3n (2mkT)3nl{f~exPC-X2)dX]3n
3.6 Let ri be any of the generalised momenta PI' ... ,IPj or coordinates q 1, ... , qj. Suppose that it can range between the values a and b in a certainsystemandthata OorH(a) = ooorboth,andb = OorH(b) 00 or both. Let () denote an average over the classical canonical distribution discussed in Problems 3.4 and 3.5. This yields a number of equipartition theorems. (a) Prove that (1) = kT.
where a = 1/2mkT. The integral is .J1T and this yields the required result. Next take n I to find
m3
= (2mnkT)IJ,ex p ==
Since dVjdV2 dV3
Pv,v,v, dVl dV2 dV3
In(V? + Vl + V:?)] dV dV;dV3 j
==
/ aH)
'{i ari
PvdV.
(b) If
41TVZdV, one finds Pv
"
m)'h exp (InVZ) = 41T V2 ( 21TkT - 2kT .
~! lz;nQn
~
_
p.n
nLoZnexPkT =
f...f
exp (-
_
=j
(cx + (3)kT
k~) dPl ... dPj'
For a classical relativistic gas of particles with point interactions, H = c(pi+p~+p~+m5c2)'Iz applies to one particle with rest mass Ino· Prove that
00
(~)
I (21TmkT)3nI2 p.n h2 QnexPkT
n~on!
, (-kTi~lqi mgn) = Cexp
kT
(m;)
1kT ,
Solution
(a) The normalisation integral for the canonical distribution is
f1n [mg exp (-'!!.gq;) 1
=
where m = (3mo and (3 == (I - VZ/c 2 [In quantum statistics these theorems hold only in the classical limit; see Problem 3.9.]
dqj ... dqn
'.n- i"" exp(-mgq;/kT)dql ,-'I
j
r
n [ exp(-mgq;/kT)dqi]
'0
{3
L bjqj ,
(H) = - -
This is the required result. (d) The probability that particle I is in the range (qj, qj +dqd, part icle 2 in dq2, etc., is Pq ,'I, ... dq 1 dq2··· dqn
alP, +
; .., I
show that
This is integrated as in part (b). From Problem 2.4 ,!!,
L
H
(c) From Problem 3.4(c)
Zn =
n(z)
= -vk T = p(z)kT.
~fn
B- 1 = [I:exP C- ap )d P
Pp dp 1 dp2 d p3
(mgz) = mg kT exp - kT dz.
Pz dz
To obtain Pp integrate the probability distribution Aexp
71
kT
I = A
.J
f.. .f
exp (-
:n
dq, '
0
(I)
J
In this general form the result is due to R.C.Tolman.
dqj ... dpj.
12
Chapter 3
3.6
A 1... 1
=
Statistical mechanics of ideal systems
3.8
Iql exp (- :r) I: dq2·.· dPt
+ :Tf..Jql
:~ exp (- :r)dqj ... dPt .
The first integral does not contribute by hypothesis and the result fol lows. (b) It follows from part (a) that each quadratic term in the energy contributes an amount kT/r. (c) In this case aH c c c 2 p? 2P kT Piapi = P("2 H i =
Solution
(a) From Problem 3.6
(qij:~) 1
(
2'~. I,
The right-hand side is, from the principles of relativistic mechanics, 132m~Jfc2
_ -
I3m~Jfc (m~VZ+m~c2/{32)n
I3moV?
=m
_ -
{3m~Jfc
n
- " ( pt) ,
nkT,
3
= 2'nkT.
(c) Use the fact that dW
r~
== uVV.
Also
E = K+ W
(1 + ~ )7<.
1
(a) Assuming (i) the Hamiltonian equations of motion (dq Iii dt aHlapii' dpI)dt = -aHlaqij; i = 1,2, ..., n; j = x, y, z), and Oi) the ergodic hypo thesis that ensemble and time averages yield identical results, prove that C = ~nkT. (b) If the forces are derivable from a potential W, -aWlaq,j, and the momenta are involved only in a kinetic energy of the form n
=
3
1" ( a11) 2'1-! Piiy;-:. I,J P'I
L Fi • rl, where the bar denotes a time average.
== I
K
aw.. qlj ) aq 'I
K = 4- 2m
VIRIAL THEOREM
n particles is C
I
VW =
(q ix' q Iy' q iz) when the force acting 3.7 A particle i has coordinate ri on it is Fi = dp,./dt, where Pi = (Pix, Ply, PIZ)' The virial of a system of
kT.
Also
(m~VZ+m~c2-m~VZ)Y'
m
(-qijFii)
This becomes C == 1nkT. The above result is
f·
c({32m~VZ+m~c2)Yl
73
[The virial theorem K C is due to Clausius, and can be derived independently of the equipartition theore:rn. This is demonstrated in Problem 9.16, which deals with closely related questions.]
Integrate partially with respect to q 1 : 1
.~
L pt/2m, prove that
3.8 The particles of a gas interact with fo:rces[(lrj -rk I) [(rjk) which depend only on the distance between the particles. For the results to be established, use the assumptions and inferences given in Problem 3.7. Show that the interaction forces contribute -!Lrjkfrrjk) to the where the sum extends over all pairs of particles. (b) Show that the force exerted by a container of volume v on a gas at pressure P contributes ~pv to the virial. (c) Show that for a classical imperfect gas <:)f n particles at temperature T
1=1
K
!LVW' rl = 'nkT.
pv = nkT+, L rjk[(rjk)' (pairs)
(c) For a single particle (n = 1) under a central force of potential energy W = arU prove that u_ =2'W= where E denotes the total energy.
Cd) Establish the virial theorem (u+2)K = uE+ 3pv
for a gas whose interaction forces are derived from a potential energy which is a homogeneous function of order u in the coordinates.
"
3.8
Chapter 3
74 Solution (a)
F+-V----
-+G
F=
(Zl)q = ( I
G
~
I divrdv =
~pv,
where Gauss's theorem and divr = 3 has been used. (c) We have from Problem 3.7 and the above results
L
~pv -~
K = !nkT
[rjkf(rjk)j.
(pairs)
In this case rjki'(rjk) = -rjk(oW/orjk) = -uW. It follows that
K= Multiply
2 and add
~pv+!uW. =
3pv+
3.9 The state of/ a I w j-oscillator of angular frequency w = 21rv is specified by w quantum rrumbers n = (n 1, .•. , n w ), its energy in this state being 1 +!)+ (n2+t)+ ... +(nw +t »)hw. E(n) = The nj are positive integers or zero and h is Planck's constant divided by 21r. (a) Show that the energies may be expressed in the form
= (tw+j)hw,
j
=
(c) Obtain expressions for the classical partition function and mean energy of a one-dimensional oscillator of mass m, angular frequency w, and temperature T, according to the prescription of Problem 3.4, given the Hamiltonian H(p, q) = ap2+bq2
j!(w-I)!'
(-00
< p, q < 00),
where a == 1/2m, b tmw 2 • Check that (Zl)q, (Uda go over into these expressions in the limit T -+ 00 (classicallimit). (d) Prove that N identical and distinguishable [w )-oscillators are equiv alent to Nw [ I)-oscillators. (e) Show that the heat capacity of N identical, distinguishable quantum [I)-oscillators is (Cv)q = NkE(2x), where y 2 expy (Einstein function). (expy-I)2 [Part (c) is an independent check on the equipartition theorem in a special case.] w
L nj = j.
i = 1
Clearly j can assume all positive integral values or zero. Given one of these, the degeneracy gj of this level is given by the following number theory problem: gj is the number of ways of expressing an integer j as a sum of w integers, zero and repetitions being aIlowed and order being important. This in turn is clearly the same as the number of ways of placing j indistinguishable balls into w distinguishable boxes. This num ber is (w +j - l)!/j!(w - I )! Y) Note that the ground state j = 0 is always non-degenerate, go 1. Also all states of the [I ]-oscillator (Le. the one-dimensional oscillator) are non-degenerate. The quantum-statistical partition function is
0, 1,2, .. , j
(w+j-l)! gj
}q = hww[!+(exp2x-l)-l].
(Zl)q = exp(-wx)
the degeneracy of this level being
)W
I 2sinhx'
By rewriting E(n) as E j we make the substitution
OSCILLATORS AND PHONONS
Ej
)w (
Solution
uK to both sides to find +2)K
exp(-x) exp(-2x)
where x == hw/2kT. Hence show that its mean energy is
-F.
The contribution to the virial due to the pair (j, k) is -1(rj . F-rk • F) = -!(f; fk)· F = -!rj!l(rjk)' Hence the result. (b) The force exerted by the container on an element of area da is -pnda, where n is the unit outwards drawn normal. The contribution to the virial is In· rda
75
Discuss some simple special cases. Prove that the canonical partition function of one such quantum oscillator at temperature Tis
as positive for repulsive forces,
C=
Statistical mechanics of ideal systems
fk
fj
We have, taking
3.9
L0gjexp(-2jx)
(w+j-I)! L exp(-2jx) ; =0 00
exp(-wx)
;!f,,,_l\!
(2) P.T.Landsberg, Thermodynamics with Quantum Statistical Illustrations (Interscience, New
York), 1961, p.447.
76
3.9
Chapter 3
The binomial theorem
l+qa+q(q~,I)a2+ ... +q(q
(l+a)q
l)",~,q-j+l)ai+ ...
.
(I -a)
_ W
=
I +wa+
J.
-l)ai
l)a 2
w(w
+ ... +
+ ...
~(w -I)!_ jLoj!(w-l)! a'.
Hence (Zd q = exp(-wx)[l-exp(-2x}rw. -',
It follows from Problem 2.6(b) that the internal energy is
(a-----aT InZ1)
"'2
k1
)q
v
hw 2
~ exp (_~p2 _~q2) dpdq
kT
kT
kT
21fkT)v, h- 1 ( 21fmkT mw 2
The limit T -+
00
hI"
Also classically
(U1
kT
2
E z +E th == L thw(q, s)+ q, s
= (Zl )cl'
kT.
From part (b) we have I
(Ul-)q -+
= kT
(Udel'
(d) The correction for indistinguishability (Problem 3.4) does not apply and I ZN = zf = ( 2' I Sin lX
)NW
is the partition function of N [w ]-oscillators. It is equal to the partition function of Nw [1 )-oscillators. (e) This result follows from au C v = aT =
NhwaU
ax .
fff=~~~,
_ I
q 2 dqdQ
(Y == ~i),
~ f q 2dqdQ = 3rN.
v
aT
hw k
L hw(q, s)(n(q, s»,
q, s
where q is the length of q, and the surface q = qo(O, ¢) up to which the integrations are carried out satisfies
alnZ _
= kT2 kT hw
77
where E z occurs at all temperatures (zero-point energy) and E' h is the thermal energy and vanishes at T = O. (c) Let 0, ¢ specify the direction of q, let dQ = d( cosO) d¢ be an element of solid allgle in q-space, and note that v/81f 3 is the number of wave vectors per unit volume of q-space in a crystal of volume v for each polarisation. Replacing the q-sum by an integration obtain the formal expression
= vkT ~
or x -+ 0 of (ZI)q is for w 1 (Zl)q -+
Statistical mechanics of ideal systems
3.10 Consider a basic set of r points in space called a unit cell. If it is repeated N times it yields a periodic set of points called a lattice. If atoms are placed on the rN points, one has an idealised representation of a crystal. If each atom is a [3)-oscillator and if its internal structure is disregarded, one has an ideal crystal with 3rN degrees of freedom. Each degree of freedom gives rise to a so-called mode of oscillation (q, s). The wave vector q assumes N values; the polarisation index s takes on 3r values. (a) Prove that the (canonical) average for temperature T of the quan tum number n(q, s) of an oscillator of mode (q, s) is [exp(hw/kT) I w being its angular frequency. (b) Show that the energy of the ideal crystal can be expressed as
aax'
and this gives the stated results. (c) We have
(Zl )e1 = h- 1
3.10
Iinstead of an oscillator in state n(q, s) one can think of n(q, s) quanta of excitation with wave number q and polarisation s. They are related to sound waves as photons are to light waves, and are called phonons. More generally, excitations or particles whose mean occupation numbers per quantum state satisfy (a) are called bosons of zero chemical potential. But see Problem 3.12 for a generalisation to arbitrary chemical potential.] Solution
(a) Put Z =
L exp[-(n +t )tlhw)
h=O
= (I +a
+"')exp(-ttlhw)
exp(-!tlhw) I-a
a
== exp(-tlhw)
3.10
Chapter 3
78
-~{3hw-In[l-exp(-{3hw)].
Now (n)
L
=
+lWhw]= 2
nexp
n=O
I olnZ I -2 hw
Eth
[exp ({3hw) - 1] .
Lf,(q, s)
one finds E
e, r/>, s).
by
2
dqdn,
hw y -, - kT' L. expy 1 L.expy-l q,s q,s
Iff
(a ~ 1) is the Riemann
zeta function, and you may assume that D(m,x) ....., mx-mr(m+ 1)t(m+ 1) for large x. For t = I and temperatures putting t( 4) 11'4/90, verify that the above model yields 4 lI' S vk 4 T 4 3r 1 E th :;;:; L 3·
las [In the simple Debye theory of specific heats (phonons) one has r = t = 1, a1 = az is the transverse sound velocity, and a3 is the long itudinal ~ound velocity. For black body radiation (photons) the low temperature theory applies, with t = I, 3r = 2, a 1 = az being the velocity of light. For spin waves in a ferromagnetic solid (magnons) one has t 2. The laws C" 0: T3 (De bye solid at low temperatures, and black body radiation) and C v 0: T% (magnons) are implied.] s=
th -
vkT 811'3 ~
L ra j = I
q
8~3Jf(q, s)q
79
kTv 3r (kT)3/t r (I +T3) t (1 +T3) 211'Zts~1 has
=
in agreement with part (a). Here na)
This follows from Problem 3.9. (c) The angular frequency depends on q and s, i.e. on (q, With this understanding and replacing
Statistical mechanics of ideal systems
(c) Establish the high- and low-temperature approximations for the result of part (b), verifying that the former is in agreement with the equi partition theorem, while the latter yields
Then InZ
3.11
y - 1 q 2 dqdn. ~ .. ~"
The limiting surface qo(e, r/» in q-space must he such as to give the cor rect number of modes of oscillation.
Solution
(a) Write, with{3
3.11 A system of non-interacting bosons is confined to a volume v. The state of a boson is specified by labels q and s, V/811'3 being again the number of wave vectors per unit volume in q-space. The results and notation of Problem 3.10 must be used. (a) If the dispersion relation between wand q is w = a(s, r/> )qt , where t > 0 is a constant, establish the law C v 0: T3/t
e,
for the low-tempeljature heat capacity of the system. (b) If a is a constant as, and the surface q qo(e, r/» is a sphere of radius qs for polarisation s, show that kTv 3r (kTX) s 3/t
S~l
Eth
where
ha s
D
(3 ) t ' XS
'
Xs
== asq!,
m IX ymdy Dm 0 expy I ( , x) = - xm
is a generalised Debye function. The values of Xs are subject to IS1I'2 rN _ V
-
3 _
Lqs - Ls s
(
kTX s)3/t ha . s
=
l/kT,x =
/3hw = {3haqt, so that
kT)3/t 2 q dq = ( ha
Hence =
::~t
(1) T
X
3 t I / -
dx.
e:y't ~f:~tJ-~3/~dx, .
For fixed e, r/>, i.e. for fixed solid angle, the upper limit of the x-integra tion is given by the limiting surface in q-space. However, at low enough temperatures the integration extends to x = 00 so that the last integral is replaced by a constant numerical value. The temperature dependence is therefore given by the factor in front of the summation: Eth 0: T3/t+l. This leads to the stated form of the heat capacity. (b) We now have, with Xs = hWs/kT = haqi/kT, kTv (kT)3/t x 3 / t dx Eth
If
= 811' 3t h
f 411'(a f 3/ Jorxs --s
,.
t
mIx expymdy y - 1'
Dm ( , x) = - xm
0
then, in the classical limit (see Problem 3.9), T -+
00,
x
-+ 0 and so D -+ 1.
80
3.11
Chapter 3
This is a useful way of normalising the new function D. The stated equa tion for Elh now follows. Also the values of q" are subject to v 3rN = -62Lq;'
3.12
Statistical mechanics of ideal systems
are given by (n)
s
1f
3) ~ ( 3) S~l 1+t
3r
(kT)3/t
has
== \
f
nj) = A(kT)s+1 r(s+ l)1().L/kT, s, f),
F = ).L(n)-A(kT)s+2r(s+ I)I().L/kT, s+ 1, f),
(c) At high temperature the two results of part (b) are easily com bined to yield Eth = 3rNkT. This is the equipartition result for 3rN one dimensional oscillators (Problem 3.9). At low temperature Xs 00; and also D(m, x) --+ mx- m r(m + 1)~(m + 1) is required, where ~ is the Riemann zeta function (ref. 2, p.263). One finds kTv ( Eth = 21f2tr 1+t
1().L/kT, s+ 1, ±)
U = (s+ 1)(n}kT I().L/kT, s, ±)
S (s+ l)pv
.
= k(n)
r.L(s+2) 1().L/kT, s+ I, ±) ).L J 1().L/kT, s, ±) - kT '
= U. x
THE IDEAL QUANTUM GAS(3)
s + 1, ±)
== 1().L/kT, --------,-'---'-:--' 1().L/kT, s, ±)
check that
3.12 The grand canonical mean occupation number
alnZ
(nj) a(E) kT)'
U == (s+ l)pv == (s+ I)(n)kTx,
F
).L(n)-kT(n)x,
S = k(n)[(s+2)x -).L/kT].
(e) Check that the expressions for (n), F, S, and U in part (c) are exten sive, that the system is an ideal quantum gas in the sense of Problem 1.9 with g = I/(s+ 1), and that U - TS+ pv is equal to ).L(n). (1) If s ~ and A == 41f1Jgmh- 3 v(2m) (as in Problem 3.10), where g is the spin degeneracy of each state and m the mass of a particle, show that
and hence prove from the given information that kT
InZ
= ± ~ In(l
p
± tj),
I
where tj == exp[().L-Ej)/kT]. (b) Assume that the number of single-particle quantum states in the energy range (E, E+dE) is in a continuous spectrum approximation AEs dE, where s > 1 is a constant and A is independent of energy. Show that then InZ = A(kT)s+ lr(S+ 1)1().L/kT, s+ 1, f), where 1
lea, s, ±)
,
(d) With
(d) The result is immediate.
pv
81
r(s+ 1)
f"" 0
== kTg (
[Non-interacting particles means particles with point interactions. This term implies that the particles are pictured as geometrical spheres of radius r, and that they do not interact except when they collide. The limit is then taken as r --+ 0.] Solution
(a) From Problem 2.4
pv kT'
Also the probability of a set of occupation numbers n 1, n2, ... is
Hence establish from Problem 2.7 that A is proportional to volume. (c) Show that the mean total number of particles, the Helmholtz free energy, the entropy, the internal energy, and the pressure of the system (3) Problems 3.12 to 3.17 follow the exposition introduced in Section 28 of reference 2.
_
In..:.
xSdx
exp(x -a) ± I'
21fmkT)'h 1().L/kT,~, f). h2
--1
..:.
("niE; ,,).Ln;) kT + kT '
exp -
7'
7'
since LniEi is the energy of the system and Inj the number of particles i
j
/"'h~_+_.
82
3
3.12
~
in a general state. Hence
I
=
(nj)
f!
Substituting for (nj), the equation for Z is
a
II
,..
alnZ
I
=
f
dt· ~
InO ± tj)+constant.
The general expression is therefore that stated in the problem. (b) We have
InZ
=
In[ 1 ±
S+
dx ± exp(JL/kT- x)]
I'"
A(kTY+ I r(s+ I )1(JL/kT, s+ I, ±).
f
V can be obtained from
I
_
(nj)E j - A(kT)
s+2
'" f 0
pv = kTlnZ.
These results are immediate.
A
= 41Tvmgh- 3Y(2m),
-lo- - 0 0
, ,-
I res + I)
f'" 0
X'~
e a - x d •X
where the definition of the gamma function (Problem 1.20)
res)
XS
e- x dx
has been used. (b) We have I
I.
k(n)[(s+2)lnT-Inp+ij.
J(a s +) :::::>
dX /,~,
pv
U
= s+ I
A(kTy+2r(S+ I
Hence JL kT
The pressure is obtainable from
S can be obtained from
-lo-
Solution
xSdx
F is obtained from
JL(n)-pv = JL(n)-kTlnZ.
and hence x
[This problem supplements the thermodynamic results of Problem 1.23. These are seen to be valid only in the classical aDOfoximation. Note that the chemical constant is sometimes defined
exp(x -JL/kT) .
= U-TS =
ea
t,
S
As JL
(c) As in part (b) 1
-lo-
where g is the spin degeneracy of each level. Hence find an expression for i. (d) Show that in the classical approximation
0
I ('" X'~+ldx } ±s+IJoexp(x-JL/kT)± I
\t
!-Y1T.
for the chemical constant. (c) Show that in the simple case of a gas of structureless particles in a box of volume v with V/81T 3 wave vectors k = p/h per unit volume of k-space ras in Problem 3.1
s
~+ Ilin [1
F
lea, s,
±A fpIno ± li)dE
XS
) =
(b) By using the classical approximation and Problem 1.23(c) obtain the expression s+2+ln[r(s+ l)k~+2A/vl
if all t's are kept fixed, except for t j , InZ
(n)(JL-xkT+xkT)
3.13 Develop the classical implications of the model of Problem 3.12 as follows. The distributions are then referred to as non-degenerate. Show that in the classical approximation (JL -lo- -(0)
=tj~.
~I
F+kTlnZ
as required. (f) This result follows from a
alnZ
I
(e) We have U-TS+pv
exp[;~(JL-Ej)l'J
... Inj
n).rlz,
83
Statistical mechanics of ideal systems
3.13
provided i
= Inp
(s+2)lnT+(s+2)-j
= s+2+1nr(s+I)~+2A v
ea ,
84
Chapter 3 (c) We have
~41Tk2dk
AP dE Since p2
g 81T 3
3.13 2 = 41Tvgp dp 3 h
v(2Em)mdE and the quoted result follows. ~ +In[(21Tm)%k7l gh- 3 j.
(d) Substitute the expression for p from part (b) into that for the entropy: S
= k(n >(s + 2 k(n>[s+2
:T)
(n >
Solution
(n> TS
ns+l}.
D :;: res + 1)/(0, s + 1, ± )/(S+ 2 A/v, show that if p
(s+ l)pv
0
(s+ l)vDp+2
A(kT)S
A = 81TVh- 3 c- 3 ,
where c is the velocity oflight. Given that n4) = D
U
U
A(kT)s+2I'(s+2)/(0, s+ 1, ±).
The stated results follow. (c) As before in Problem 3.13(c) AEs dE = 41Tvgh- 3p 2 dp. For a photon h E E E = hv, P = I = VA = C·
so that t
ck
for photons and phonons, and show that for a sound field or phonon gas at low temperature
= n1TSVh-3(kT)4( 23+~) Ct
hc
hv =
= hck,
= 1 in Problem 3.11.
3.15 Thermodynamic properties of the model of Problem 3.12. (a) Writing for simplicity I(s) for l(p/kT, s, ±) whenever this causes no confusion, and 1 :;: p/ kT, prove
For the model of Problem 3.11 establish the dispersion relation
Eth
hw
verify that
-,&1T s k 4 h- 3 c- 3 , -h1T svh- 3 c- 3 (kT)4. w
1)/(0, s, ±),
= (s+2)A(kT)s+2res+l)/(0,s+I,±),
(c) Show by an argument analogous to that used in Problem 3.13(c) that for a gas of photons (i.e. black body radiation)
= 2,
+ 1 res +
The required result follows noting that g = 2 for the two directions of polarisation. As in part (c), since the wave vector k = 21T/A, we have
= ---'--'--'---'-
s
85
a 1T 2 k 4/60h 3 c 2 ", S'7x 10-Sergcm-2 sec- 1 deg- 4 is Stefan's constant. These laws apply also to low-temperature phonons in a Debye-type theory. The number of phonons or photons (n> is not generally con served, but it does remain constant in a quasistatic adiabatic change, since S is then constant. The generalised black body relations of part (b) do not presuppose a boson gas and the question arises if they can be illustrated also by a simple model of a fermion gas; see Problem
Inp+(s+2)lnT-(s+2)+i].
s,-)
Statistical mechanics of ideal systems
(a), (b) Use Problem 3.12 to find
3.14 Develop the implications of a zero chemical potential for the model of Problem 3.12. Ca) Show from Problem 3.11 (c) that
U
3.15
c.
where Ct and Cl are the velocities of sound for transverse and longitudinal directions. [For black body radiation based on bosons the characteristic laws U 3pv = iTS = 3vDT 4 hold. If one puts U = (4a/c)vT 4 , then
a) ( aT P. a1 ) ( av T, ('I) 1
(aTal)
1
V
I(s) I(s - I) ,
s+
I I(s)
v
v,(n)
I(s-I)'
(b) The coefficient of volume expansion
1 v
(av) Jl(s-1)' I(s)
v aT
T
(av) aT
p, <'I>
p, ('I)
"I
3.15
Chapter 3
86
is
Statistical mechanics of ideal systems _ (n>kTI(s+ 1) p v I(s)
(c) The isothermal
ap) ( av
1 (av) v ap T is
(d) The heat capacity is Cv
=
U
r
[2(s)
(e) Check that the Griineisen ratio f' = cxpvlKtCv 1.9(e), 3.12(e)] is I/(s+ I).
(a1')
+A(kTy+lf'(S+ 1)/(s-l) aT
p, (n)
0
I(s)
av
U T, (n)'
I
CX p
P I(s)
L
[/(S)
I(S+I)/(S-l)JS+I~--:-: Pes)
T I(s
1)'
P
Pes)
T
(c) Obtain an expression for K T • Using (n>/v = 2·56 x 1022 cm- 3 (sodium), show that
KT 11·6 x 1 cm 2 dyn- l •
v aT
J~ p,(n)
I(s
1)'
I(s+ I)l [s+ I I (av) T + ; aT p,
I [
2m
2 (n> (3(n» ';' ~ V 41Tvg 2m'
Pes) ] 1-(s+I)/(s-I)/(s+1)+s+l.
Solution
We have
roo
I(c,s, +)
xSdx
= ---:-:-Jo exp(x-c)+l
J:
~ cS+ 1 ~
Hence the required result.
1-
3 (3(n»'13 h = -(n> - - - 5 41Tvg 2m
=
= T-/(s+l) l(s-l)-l(S)J
Pes) ] cxp [ 1+/(s+I)/(s-1)-1
[
( 41Tvg
P
I(s+ 1)
- I(s) ,
=~_(n>kTrl I(S+I)/(S-I)][s+l+l(av) T
U
= T-(s+l)(n>kT
2
/(s - I) (a1')
using the first result of part (a), we have p.(n)
I(s)
3(n»'I3~
This gives the second result. The third result is obtained similarly. From
( aT
+1)
= (s+ I
J1
v
av)
1)
Show also that for the values of A and s given in Problem 3. 13
(n
v
v
I(c, s,
o(n» T,(n)
Pes)
(a) is now needed. We argue that
.
This gives the first result. Also ( a;-
L
3.16 The model of Problem 3.12 will be applied to non-interacting electrons of mass m in a volume v at low temperature T. (a) Show that for this case
is zero and this yields
p, (n)
rl _ I(s + 1)/(s - 1)J l-::---,-
The stated equation follows. (e) The proof is immediate.
A(kT)S+ 1 r(s+ 1)/(s).
s+ (I n H(n>(av) 0= T - aT v
CV
Problems 1.4(a),
Use
p, (n)
V
]
Solution
a(n» Hence ( aT
(n >kT V
(d) The third result of
(s+l)k(n --~+2-(s+l)/(s+l)/(s-l) .
(n>
= T,(n)
p -----'::-- V I(s + I )/(s - 1) .
I(s+ 1)/(s-l) p/2(S)
KT
87
The second equation of part (a) is now needed. The argument is
I(s+ I)/(s- 1) ] +2) Pes) -s-l.
T
CX p
3.16
xSdx
3.16
Chapter 3
88
(b) From Problem 3.13 we have (n) ~ vg (
d'Y) ( dT
r(~)
h2
3y'1f
h2
(n)kTI(s+ I) ----v ~s)
F s+2jJ.(n), jJ.
(c) It follows from Problem 3.15, puttingg
=
3
Kr
=
4m (n -
=
pv = (n) kT
2 -(n)jJ.. 5v
2, that
»)-y, (31f2)
Inserting the values m = 9· 11 X 10- g, h (n)/v = 2·56 x 10 22 cm- 3 , we obtain
h.
= 6·62 X
10- 27 erg sec, and
Kr '" 11·6 x 10- 12 cm 2 dyn- 1 •
£.) ' exp ( -ff l'
fo'''''[exP(x 11')_1
kT' a+ 1
Solution (a)
r"'[___ex...!.p.....:.(x_-...!-'Y)~_
1 r(s+2)Jo
u
d
- [/(,)" s+ 1, d'Y
xS
+1
I
,
v =
exp(x -1') . [exp(x-'Y)-Ij2' I
I
J
yields
[ I
I +(s+ I) Jofooexp(x -1')
xs+ I
Xs
OQ
0
I
= 1(1', s, a).
(a + Ifexp[(a+ l)(x -'Y)]]xs+ Idx {exp[(a+l)(x 'Y)]-IF .
u = (s + l)x s , v = - ---.-----,-----.,..
=- -
> -I
[exp(x-'Y)-Ij2
A partial integration on the first term with
-=-c-___
(a) Show that for s
J
r
1('Y,s+l,a) k(n)L(s+2) 1('Y,s,a) 1'.
S
jJ.
l'
1(1', s + 1, a) 1 I( 'Y,s,a ) = s + 1 U
d d'Y/('Y, s+ 1, a)
3.17 The expressions for the Fermi and Bose mean occupation num bers according to the grand canonical ensemble have been used in Prob lem 3.12 without being proved. They can be obtained as part of a more general procedure. Each quantum state of a particle in a system of weakly interacting indistinguishable particles can accommodate 0, 1, 2, ... , a particles, where a is a constant. Let
1('Y,s,a) =
(0 Verify that for a = 1 and 00 the results for fermions and bosons respectively are obtained. These are the expressions for bosons and fermions given in Problem 3.12(a).(4)
I
V 28
tj
s + I I( 1', s, a)
= -T 1(1', s 1, a)'
jJ.(n)-A(kT)s+2r(s+ 1)/(1', s+ 1,
s+ I
(n)kT v
~
v, (n)
(e) Also show that
jJ.,
r(s+2) U~(s+I)kT(n)kTr(s+3) ~
89
(d) Prove that
jJ.
p
Statistical mechanics of ideal systems
21fmkT )'I2(jJ./ kT)'f,
4gv (21fm)% liz
=
3.17
= -=r:7(s-+:-:-:-I)
f'" 0
x'
dx .
In the second term we use (b) Show that the mean occupation number per quantum state is _ HI ' _ (a+1)2 exp [(a+I)(x-'Y)]. 1 a+ I u - x , v - {exp[(a+ l)(x-'Y)]-IF' (n·)=---- 1 exp exp[(a+ l)(x -1')] -1 . a+l u' = (s+ l)x s , v = exp [(a + l)(x -1')] -1 . (c) Show that the mean total number of particles in the system is for a continuous spectrum approximation [peE) AP as investigated in Problem 3.1 (n) = A(kTy+ 1 r(s+ 1)/(1', s, a).
(4) P.T.Landsberg, Moiec.Phys., 6, 341 (1963).
3.17
Chapter 3
90
3.17
Statistical mechanics of ideal systems
Using next the solution of part (b), we obtain
One finds a vanishing uv-term, and the surviving term is S
(a+ l)x r(s+ I)Jo exp[(a+ l)(x-,),)]-l
dx,
kT~ln(l-t;+l)+kT~ In(l-tj)
F-Jl(n)
J
'J
I j) Iexp [ kTt-(llnj-Ejn
Iexp pN-E kT
-::.,
L" ti,
I II
S
InZ
I
aInZ
0
s+ I
-(kT)S+2Ar(s+ I
F-Jl(n)
a) = n I I- _ ~.
0
pv
J
kTlnZ
s+l,a).
(kT)s+2Ar(s+1)/(,)"s+l,
1(,)" s, a)
Also
alnZ
TS
= ti----at;
=
TC~)v,
(n)
+
n.,H
a+l
2)A(kT)S+2f(s + 1)/(,)"
s+ I, a)
(n)
+
+l)/(,)"s,a)T(:;)
(c) This follows from
rJ,
.
a')' T (a')' Jl) = -Jl(n) kT(n)T-kT-+aT aT T .
j
with the continuous spectrum approximation. Cd) From (n) A(kTy+ 1 r(s+ 1)/(,)" s, a) it follows that
Hence TS
I(,)"s,a)
aT
U = F+ TS
v, (n)'
The stated result follows. (e) From Problem 2.4, we find an expression for the Helmholtz free energy F = Jl(n)-pv = Jl(n)-kTlnZ.
=
r
1(,)" s+ I, a) kT
]
')'.
Lastly observe that
= (S+l)(n\/(,)"S-I,a)(a')') T
(n)
The first and last term add to yield
(n) = I(nj)
(a(n») aT o,(n}
dx.
= kT(n/(,)"s+l,a)
1
(a+l) I ] tj [ l-t;+lf! + l-ti =
o
I
Next, on the basis of the argument just given, we can write
t~+ 1
i
a(E/kT)
1= +I- i~
This gives the required expression
Hence from Problem 3.12(a) (nj)
1
S
" - · ·tlllt"2 1 '12"·
I - t?+ 1)
dx}.
1
X + - '-In[l-exp(,),-x)] s+ I
0 II, = 0
J
t)]
X + 1 i~xS+l--7~:-'--cl --In{-exp[(a+ l)(')'-x)]} I~ - s+ 1 0 s+ I 0
a
I otpi
ta +1) -In(1
The integral is
Ifill
where the sum is over all single-particle quantum states. Here and nj are respectively the energies and occupation numbers of quantum states j. Also tj = exp[(Jl-Ej)/kT]. Specifying the system quantum states by the nj, we have a
J
- (kT)S+2 A {t~x S [ln(1 -
whence the result follows. (b) For indistinguishable and nnn_intl>,." .::.
91
Jl(n>
(s+ I
,
(s+ l)pv.
kT(n)
I(,)"s+l,a) I(,)"s+l,a) T( ') -Jl(n)+kT(n)Cs+2)-----r(- - ) l',)"S,a l'')',s,a
/C')', s+ I, a) 1(,)" s, a)
92
Chapter 3
For a
Fora =
I the solution of part
3.17
yields
z
no+
:z =
II(I-tjf l .
These are the partition functions for fermions and bosons. CONSTANT PRESSURE ENSEMBLES
3.18 The (petit) pressure ensemble is one for which p, T, and the num ber of particles n are given, together with the mean energy and the mean volume Vo. (a) Show from Problem 2.3 that the probability of a state (E, v) is exp l-(E + pv)/kT] pee, v) = Zp ( p, T , ,n) where Zp is a normalising constant related to the Gibbs free energy by exp(-G/kT).
(b) Discuss the difficulties associated with the partition function ,
L.
[Em(VI)+PVI] exp kT .
m, I
(c) Show that these difficulties do not arise if one takes Zp
T, n) = VoI
J' r ;;-
exp .. Em(V)+PV] kT dv
I roo ( Pv) voJo Z(v, T, n)exp, - kT dv,
where Z is the canonical partition function, and v 0 is a suitably chosen constant volume, the definition of which presents difficulty. (d) Find an expression analogous to that under (b) for the classical partition function Zp , and verify that there are no difficulties in this case. Show also that the ensemble average of the volume is -kT [
1)
alnZp(p, T, ap
n)]
alnZ(V, T, av
In Problem 2.3 identify x with the energy E and y with the volume v. A state i of the system in the ensemble is specified by the volume VI and one of the energies Em(v/) available for that volume. Thus i imolies two specifications. The result of Problem 2.3
S kt3Eo + k')Vo + kIn Z and the thermodynamic result TS = Eo+pvo-G
must be equivalent for all Eo and Vo and this leads to -(E+pv)/kT P(E,v) = exp Z ( ) p P, n (b) In general the 'eigenvolumes' VI form a continuous set so that the {-summation diverges. (c) By integration over v one avoids the divergence, but shifts the difficulty to the question of how Vo should be defined. However, the precise value of Vo will not affect those thermodynamic quantities which depend on a differentiation of InZp. The reference cited gives an intro duction to the current situation. (d) In the classical case E may be replaced by the classical Hamiltonian H expressed in terms of the f generalised coordinates and momenta. Fol lowing the procedure in Problem 3.4,
.
- -If... fexp [H(PI .kT .. qr)+pv] dpl ... dqf·
Zp(p, T, n) - h
There is no divergence and no dimensional difficulty here. The differen tiation indicated in the problem now yields the average volume. RADIATIVE EMISSION AND ABSORPTION
3.19 In a certain fermion system transitions occur from a group i of single-particle states I(ei) to a group j of states J(ej) at a rate Uij
T,n
in analogy with the result P = kT [
93
Solution
j
Zp(p, T, n)
Statistical mechanics of ideal systems
(b) and (c) are discussed by Lloyd and O'Dwyer (5) who give references to earlier work.]
00,
Zp(p, T, n)
3.19
n)] T,n
of Problem 2.6(b). [The classical ensemble is used in Problem 9.4. The difficulties under
=
L [PrSuqJ-pJSJIqrl
f(ei) J(ej)
,
where PI is the mean occupation number of a quantum state I, (I -qJ) is· the mean occupation number of a quantum state J, and Su is a transition probability per unit time. (5) P.Lloyd and J.J.O'Dwyer, Mo/ec.Phys., 6, 573 (1963). See also D.R.Cruise, J.Phys.Chem., 74,405 (1970).
94
Chapter 3
3.19
(a) Show that for free fermions in equilibrium at temperature denoted by C.. )0,
to be
I)J
[N (qrPJ II PlqJ
has the value unity. (c) Assuming the result (a) and that Su probabilities, show that
Pi/
= (Su)o for all transition
Su
SJI Su
. Vl.l.UVa
number of any
Xl
qJ PI
kT
- exp Il/k-:i)
Hence Uj!
[ ( E I -llr+IlJ-EJ ) kT 0 -1
exp
Pij - Pji,
J.
In thermal equilibrium all chemical potentials are equal, as follows from Problem 1.21. This establishes the result.
11; - Ili kT
These results are of use in the recombination theory of semiconductors where the Il/s are driving forces for transitions. They are then referred to as quasi-Fermi levels, see Chapter 16.
3.20 Stimulated and spontaneous emission rates and absorption rates of photons due to single particle transitions between states I and J are, respectively, in the notation of Problem 3.19:
u sp AUPIqJ' u abs == BJIN"PJqI ' Here J corresponds to a lower energy, and it will be assumed that Bu
(Bu)o.
= (Au)o . Bu = BJ[ may
Au
The quantum-mechanical result that be used in these equations. Note that the spontaneous rate does not depend on (a) Show from Problem 3.19 that Au Bu. (b) Show that u st -u abs u'P
= I + XI
The result follows. (b) Radiation in thermal equilibrium corresponds to black body radiation (see Problem 3.14), Hence, using also part (a), 1 exp
kT
E - Er + 111 - E J + Er - Ili exp J
= -PJ -qI
Pi/
(a) Writing Xl == exp [(EI -111)/ kT] one has essentially from Problem 3.12
= 1 (PI)O
= exp EJ -EI
(PIqJ) PJqr 0
0
where
Solution
1 Y"u = exp(hv/kT)
= (SJI)
= I,J LPISUqJ
that the ratio of the reverse rate
Since one fermion is the maximum mean state I,
Hence, since
(d) Reverse to forward rates are in the ratio
0
[They are equal in thermal equilibrium in accordance with the principle of detailed balance.]
1-'
J is zero.
It follows that
- Ili ) exp (111kT .
(PI)O
~
the net rate I
95
= 11/,
(EJ -EI) kT '
exp
where EI is the energy of the state /. Show with the assumptions of to forward rate 0" is
!!Ji.
In thermal in equilibrium Ili
SJ[ Su
where III is the chemical potential appropriate to state I. Let N" be the photon occupation number of a mode of frequency v where hv = EI-EJ . Assuming (a) valid, show that the quantity Y"u defined by
SJI Su
Statistical mechanics of ideal systems
EI -Ill) -u
exp (
Y "u
3.20
exp
IlJ -1l1+ hV )
kT
'
(c) Show that the ratio of reverse to forward rate is N" N + I exp
"
+hv kT
(d) Show that the result (c) satisfies the equation of Problem 3.19(d) only if the radiation field is that of a black body at temperature T. [All these results are important and are used in semiconductor theory, Problem 16.3, and in the theory of the laser, Problems 15.6 to 15,8.]
3.19
Chapter 3
94
(a) Show that for free fermions in equilibrium at temperature T, to be denoted by (.")0, EI-J.J.I) =exp ( ~ where J.J.I is the chemical potential appropriate to state I. Let N v be the photon occupation number of a mode of frequency v where hv = Assuming (a) valid, show that the quantity Y vIJ defined Y vIJ
I)]
PlqJ
exp (
J.J.'-J.J.') Tr .
Solution
(a) Writing Xl == exp [(EI - J.J.I)/ kT] one has essentially from Problem 3.12 l'
Since one fermion is the maximum mean occupation number of any state I, 1
SJI = (SJI) SIJ SIJ 0
--r
J is zero. Hence, since
= (PlqJ) = expEJ-EI PJql
.
kT
0
(d) Reverse to forward rates are in the ratio SJI = PJ ql = expEJ-EI+J.J.j-EJ+EI-J.J.i SIJ q, PI kT . J . j -J.J.i) Uij -_ LPISIJqJ (1J-exp----;(T I,J
Hence Uij
where
=
The result follows. Radiation in thermal equilibrium corresponds to black body radiation (see Problem 3.14). Hence, using also part (a),
I (EI - J.J.I + J.J.J - EJ) ] 1 kT 0 -I . YvIJ = exp(hv/kT)-llex p In thermal equilibrium all chemical potentials are equal, as follows from Problem 1.21. This establishes the result.
= Pij-Pji,
P'i J.J.·-J.J.i ~=exp~ Pij kT
,
[They are equal in thermal equilibrium in accordance with the principle of detailed balance.]
(PI)O
(c) In thermal equilibrium the net rate I in equilibrium J.J.i = J.J.j,
0
where EI is the energy of the state I. (d) Show with the assumptions of (c) that the ratio of the reverse rate Pji to forward rate Pij is Pij
95
Statistical mechanics of ideal systems
It follows that
has the value unity. (c) Assuming the result (a) and that SIJ = (SIJ)o for all transition probabilities, show that RJ -EI) exp ( --;:y-
3.20
. of semiconductors These results are of use in the recombination where the J.J./s are driving forces for transitions. They are then referred to as quasi-Fermi levels, see Chaoter 16. 3.20 Stimulated and spontaneous emission rates and absorption rates of photons due to single particle transitions between states I and J are, in the notation of Problem 3.19:
== BIJNvPIqJ,
u
sp
== AIJPIqJ,
u
abs
== BJIN"PJqI .
Here J corresponds to a lower energy, and it will be assumed that BIJ = (BIJ)o,
AIJ = (AIJ)o·
The quantum-mechanical result that B IJ = B JI may be used in these equations. Note that the spontaneous rate does not depend on N". (a) Show from Problem 3.19 that AIJ = B IJ. (b) Show that ust-uabs
(
N" I -exp
J.J.J-J.J.I+hV) kT .
(c) Show that the ratio of reverse to forward rate is N" N
"
+ 1exp
J.J.J -J.J.I+hv kT
(d) Show that the result (c) satisfies the equation of Problem 3.1 only if the radiation field is that of a black body at temperature T. [All these results are important and are used in semiconductor theory, Problem 16.3, and in the theory of the laser, Problems 15.6 to IS
96
Chapter 3
3.20
Solution
3.21
Statistical mechanics of ideal systems
Solution
(a) In this case the rate is zero in thermal equilibrium, i.e.
(a) Let XIJ == PJqJip[qJ. Then Au/Bu (xIJBJJiBIJ)-1
(BIJNvo+AIJ)(pJqJ)o = BIJNvo(pJqJ)o·
It follows that (
NvqJPJ) = N + AIJ . vo PJqJ 0 BIJ
It follows from Problem 3.19(a) that in equilibrium
Applying the result of Problem 3.19(b), we find BIJ = AIJ. (b) The ratio of the net stimulated emission rate to the spontaneous emission rate is BIJNvP[qJ(l-PJqJip[qJ) BIJp[qJ
( I -exp
= N v ( I exp
EJ - EJ -Ih +IlJ ) kT IlJ-IlJ+hV) kT .
(c) The forward rate is B IJ(Nv + l)PJqJ' The reverse rate is BJ./VvPJqJ. Hence the required ratio is reverse rate Nv IlJ - Il[ + hv --exp forward rate N v + 1 kT (d) For black body radiation N
Nv v
+I
( hV) = exp - kT .
The results of Problem 3.19 apply in this special case with S IJ = B IJ(Nvo + I) , SJ[ = B IJNvo .
3.21 In Problem 3.20 (Nv)o was assumed known because the result of Problem 3.19(b) was used. Assume now the existence of the three rates, with unknown (Nv)o and unknown. temperature-independent transition probabilities BIJ , A IJ, BJJ ; assume also that fermions make the transitions with EJ hv and that the system is in equilibrium at temperature T. (a) Show that BIJ BJI by supposing that (Nv)o -+ 00 as T -+ 00. (b) Assuming Wien's law (Nv)o <X v 3 exp(-hv/kT) for v -+ 00 show that <X
v3
[This is Einstein's original argument (6) which established the existence of spontaneous emission. The coefficients introduced here are called Einstein's A and B coefficients.] (6) A.Einstein, Ann.Physik, 18,121 (1917).
hv (XIJ)o = exp kT
Hence as T -+ 00, (x/J)o -+ I; so that the result follows. (b) One finds (Nv)o
AIJ/BIJ exp(hv/kT)-1
so that the result follows.
GENERAL REFERENCES
G.H.Wannier, Statistical Physics (John Wiley, New York), 1966.
97
4.1
Ideal classical gases of polyatomic molecules
99
Solution
(a) The total energy E of any system can be written as the sum of the E(t) + E(2) + €(3)' The total partition different types of energy, E function is therefore
4 Ideal classical gases of polyatomic molecules
L exp[(-E(I)f- €(2)j- €(3)k)lkT] ilk
Ztot =
,
where i, j, and k are the numbers of energy levels associated with the energies of types (I), (2), and (3). We can write Ztot in the form
C.J.WORMALD
(University ofBristol, Bristol)
Ztot
=
Li exp(-€(I)dkT) L exp(-E(2)dkT)Lk exp(-€(3)klkT) , j
so that
THE TRANSLATIONAL PARTITION FUNCTION
4.1 From Problem 2.2 we have that the total energy E of an assembly of N systems is given E
OlnZ) = -N ( of)
V,N
= NkT2 (OlnZ) - oT
(b) We calculate first the Helmholtz free energy F InZN
V,N'
L exp(-EjlkT) , j
I+
in which Ej is the energy of the jth quantum state. (a) For an assembly of systems each of which possesses energy levels of the form E(I)i+ E(2)j+ E(3)l" where i, j, k label the levels of essentially three independent spectra, show that the total partition function Ztot can be written as the product Ztot = Z(1 )Z(2)Z(3)' (b) For one mole of gaseous molecules confined within a container of fixed volume V we have from Problem 3.5 the translational partition function for one mole of an ideal classical gas in the form ZN(T, V) = (
2'1fmkT)3NI2 VN h2 i(T~ ,
where i is the internal partition function for a single molecule and No. Show that the molar Gibbs free energy and the entropy of the gas are given by the equations
N
G= -
S
-NokT[tlnT+~lnM
= Nok[iInT+~lnM
InP-3'6605]-NokTlni(T) ,
InP-I·1605]+Nok Ilni(T)+ T
Olni(T)] :>''1'
'
where P is in atmospheres and M is the molecular weight of the gas. [Note: I atmosphere = 1·01325 x 106 dyne cm- 2.] (c) Would you expect these formulae to be valid at all temperatures and pressures?
98
I
InN! + ~ Nin
2'1fmkT
1.2
-kTlnZN'
+ Nln V + Nlni(T) .
Using Stirling's approximation that InN! NlnN - N [Problem 2.11 (b)], putting M mN and V NkTIP we obtain
where the partition function Z of a single system is given by
Z =
= Z(I)Z(2)Z(3) .
~ In(~:2) + Hnk-ln(1'01325 x 106 ) +HnT+~lnM-lnP+lni(T)
Using N = No = 6·0225 X 10 23 mole-I, h k = 1·38054 X 10- 16 erg deg- I we obtain F
Finally
6·6256
X
.
10-27 erg s, and
= -NokT[~InT+iInM-lnP-2·6605]-NokTlni(T).
G=
F+ NokT, so that
G=
-NokT[~lnT+~InM
InP-3·6605]
NokTlni(T).
The entropy S is obtained using S = -(dG/dT)p,N' Noting that the term T x ~ In T yields ~ In T+~, we have
S
Nok[~ In T+ ~ InM -lnP- I ,1605] + Nok
T
Olni(T)] oT
.
(c) We note that as T approaches zero the entropy approaches -00, whereas it should approach the value zero, which corresponds to the system being in a single quantum state. However, if we have only one quantum state we cannot write a partition function for the canonical ensemble of the form ZN = ZNIN! as this formula is conditional upon the number of available quantum states being much greater than the number of molecules. For gases at low temperatures and high densities it is necessary to replace Boltzmann statistics by Bose-Einstein or Fermi-Dirac statistics, and when this is done the paradox is resolved.
100
Chapter 4
4.2
THERMODYNAMIC PROPERTIES AND THE THEORY OF FLUCTUATIONS
4.2 (a) For a single molecule the distribution function for the energy is broad, whereas for an assembly of many (N) molecules the distribution function for the total energy is sharp. This contrast arises because in the assembly the relative fluctuations in the energies of individual molecules almost cancel each other out, and only a small relative fluctuation in the energy remains. For an assembly, in which there is a Boltzmann distribution, show that the mean square deviation of the energy £ of the individual molecules, which is fluctuating about the average energy (£) of all the molecules, is given by (£- (£»2
= N«e 2 ) - «(;-)2),
where e is the energy of an individual molecule and (e) is its mean value. (b) Show that the heat capacity C v of an assembly can be written in the form
~
=mrTZ
N 2 kT 2 «e)
Show that the fluctuation in the total energy £ of an assembly from its equilibrium value V is the heat capacity Cv given in part (c). Solution
£
= Ie;,
(£)
= I
(el)
=1 (note that £ is the internal energy V), where ej is the energy of the ith molecule and (ej) is the energy of the ith molecule averaged over the whole assembly. 1= 1
so that
N
j
'
N
I I «ej-(ej»(ej-(ej»). 1=lj=1
(£-(£»)2
In a Boltzmann distribution the molecules i and j are statistically independent, and we can readily show that terms for which i =1= j average to zero. Using the Boltzmann equation to express the fraction of the molecules i and j in the assembly which are in the states PI and Pj we have «ej - (ej »( ej - (ej»)
= [dI(e i -
(€j »exp(-ej. pJ kT)
".J Pi
x
-f-
L(ej - (ej»exp(-€;.
= (€i -
(€j»(€j - (€ = «€i)- (€i»«€;)- (ej» = 0 . Retaining only terms for which i = j we have N
(£_(£»2
= L
(€ j -(ej)}2
= N<e-(e»)2
1
j
N«e 2 ) - 2(e)(e)+ (e)2)
= N«e 2)
(e)2).
Alternative expressions for the energy £ can readily be obtained from £ F + TS, which we can write in terms of Z as £
ZN -kTln N!
£
NkT
hence
(olnZ) J
r
ZN + T LklnN! + NkT --aT v ;
2(0InZ) __ Nkl~\J __ Z oT v a(T7TJ v - Nk Z
From this we have
C" = ( oT v
I ( = -
o£) Nko(Z'/Z) oOIT) v = T2 oOIT)
Differentiation using the quotient rule immediately yields Co
=
Nk [Z"Z' (Z'Z )2J . T2
The average energy of a molecule is given by (1) This holds for a system in contact with a heat reservoir at temperature T. It can be obtained from the modification of the result p 0: cxp[S(X)/kj (Problem 22.4) to this situation, when it reads p 0: exp [-F(X)/kTl, where F is the free energy.
pJkT)]
Pj PI
0£)
N
N
N
I (ej-(el» 1= 1
E-(£)
2
1(
101
We can write
(e».
(d) The fluctuation at eqUilibrium in any thermodynamic quantity x from its equilibrium value xo, at constant y, is given by the general formula (1) 02£ 02S) (X-x o)2 = kT ox2 - T ox 2 Y •
(a) We have
Ideal classical gases of polvatomic molecules
(Z)2] Z '
where Z is the partition function of a single system, Z' = oZ/o( liT), and Z" = 0 2 Z/o( I/T)2 , both taken at constant volume. (c) Use the formula obtained in (b) above to show that the heat capacity C v of an assembly is proportional to the rate of change of the average energy (£) with temperature so that (see also Problem 22.3) Co
4.2
(e)
Lniet j
4.2
Chapter 4
102
Ideal classical gases of polyatomic molecules
103
THE CLASSICAL ROTATIONAL PARTITION FUNCTION
where exp(-edkT)
ni
4.3 (a) For most polyatomic molecules at temperatures above the normal boiling point of the fluid, the rotational energy levels are so closely spaced that they approximate to a continuum, and rotational energy may be calculated accurately on the basis that the molecule is a rigid body obeying the laws of classical mechanics. Show that the rotational kinetic energy of a heteronuclear diatomic molecule about its centre of mass is given by
L exp(-edkT) i
so that
Leiexp(-edkT)
=
(e)
---"i_ _ _ __
Lexp(-edkT) i
From
4.3
Z=
L exp(-t;/kT) we have
_~( 2 ~) e - 21 Pe + sin 2 e
i
dZ dOlT)
, / k1L..tiexp(-ei kT)
=-
so that
1
k(e) =
and
[ ~J Z
aO/T) v
=Z ,
= Z = z" _ (Z')2 Z
Z
Zro!
,,_ az' _
1, 2 - aO/T) - k2 'lei exp(-edkT) ,
L elexp(-edkT)
(e 2 )
i
k2
k 2L exp(-edkT)
,
i
and the ratio
Z')2 (
=
Z
1
= h2
Joe
1T
f J-~J-~ 1T
0
r~ r~
(e)2
k2
Now
Zro!
Cv
=
Nk (Z" T2 Z
-
Z') Z
2
=
Nk(e ) (e)2) T2 122 -122
=
N kT2(e2)- (e)2).
(d) Putting x = E and y = V and noting that the average value of the total energy (E) is the internal energy U, we have (E-U)2 =kT
Using
(:~)v = ~
( a2S)
S)
E a2 aE2- T aij2 v'
a2 (
I
aij2 v
=-
P2)] dPe dPq, de dljJ .
=
~~(81T2kT)'/,(!a/b/etl
,
where a is the symmetry number of the molecule and is defined as the number of values of the rotational coordinates which all correspond to one orientation of the molecule. The quantity la/b/e is the product of the three principal moments of inertia of the molecule. Obtain expressions for the rotational energy, heat capacity, free energy, and entropy from the above partition function. Solution
1 (aT) T2 au v
we obtain (E - U)2
[
I ( exp - 21kT PJ+ sin~ e
(c) The general quantum mechanical solution for the rotational coordinates of a rigid body with three degrees of rotational freedom is complicated but, since the moments of inertia of all but the very lightest of molecules are large, and as the quantum levels are closely spaced compared with kT at temperatures above the normal boiling point of the substance, the classical partition function may be safely used. Evaluation of the classical phase integral yields
the ratio Z" Z
'
where I is the moment of inertia, and Pe and Pq, are the moments conjugate to e and 1jJ. (b) Obtain the rotational partition function for the rotation of a heteronuclear diatomic molecule by evaluating the phase integral introduced in Problem 3.4.
Z'
Z dOlT)
= [a(Z'/Z)J
k aO/T) v
Here
dZ
'
= kT 2 C v .
Using the result obtained in part (a) we find N Cv = kT 2(e 2 )- (e)2) .
=-
1 T 2C v
(a) The molecule rotates about the centre of gravity c. The total energy is the sum of the precessional energy, in which the atoms of mass mA and mB follow the circular orbits shown in Figure 4.3.1 with velocity dljJ/dt, and the energy of end-over-end rotation with velocity de/dt. The moment of inertia is 2
2
l=mArA+mBrB
mAmB + (rA+rB) 2 mA mB
4.3
Chapter 4
104
= mA(rAsinO)2+ mB(rB sinO)2 = IsinzO
IQ = mA(Aa)2+ mB(Bb)2
.
The precessional kinetic energy E", is
HmA(rAsinO)2+mB(rBSinO)21~~ VSin20(~~y
2
dO)Z +tmB (dO)2 rBdt V (dO)2 dt
The total energy €
is therefore
+
=
V(IP+ tihin 2 0) . Differentiating with respect to () and if; we obtain the angular momenta €
Po and Prp. Po
O€
M
10
P
O€
• I¢sin 2 0
f"'foo exp (= 211' h2 (211'IkT)'h 0
3.
Zrot
4.
Zrot :;
-QO
211'
€
(211'IkT)'h .
Zrot
{'II'
Jo
P~ ) 2IkTsin2 0 dPrp dO . sinO dO
=
811'2IkT h2
811'2
21 [, I )V'(kT)% = -(811' ah3 abc .
We at once obtain the molar rotational thermodynamic functions -
E rot
-
NokT
J
_ L~Jl 811'2 'h 3 aT - NokT dTL1lnah3(lalbIJ +:2"lnkT ,
2 0lnZ
whence E = !NokT and F rot
~j+
= I
h2 (211'IkT),-I'(211'IkT)Y'
(C )
_
so that
L~ exp ( - 2~!T) dPo
Using the same standard integral again we have
For the end-over-end rotation energy Ell we have simply EIl=tmA ( rAdt
105
molecules
Ideal classical gases of
so that
The precessional inertia is
E .. =
4.4
tv
~Nok.
2
811' k) = -NokTlnZ = -NokT ( ~ In T+ -! Inlalblc -In a + t In 11' + ~ Infi2
and
_ Srot
+! In T+ :Pnlalblc
E
+ NokInZ
2 811' k) Ina + tIn 11'+! Infi2
Collecting together the numerical constants we obtain ~
.srot
InT+:2"1 In Ia Ib Ie
Ina+ 134·684).
THE QUANTUM MECHANICAL ROTATIONAL PARTITION FUNCTION
Figure 4.3.1.
(b) The fourfold integration is straightforward.
[211'
Jo
1.
2.
roo
d¢
211'
(pj)
J~ exp - 2IkT dPe .
This has the form of the Gaussian error integral
f:
eXP (-dx 2 )dx
t(~)~
4.4 Polyatomic molecules in the gaseous phase are free to rotate about the centre of mass of the molecule. This rotational motion is strictly independent of the translational motion, and at moderate temperatures in which the vibrational modes of the molecule are excited only to a negligible extent, it can be assumed to be independent of vibrational motion within the molecule. A heteronuc1ear diatomic molecule has two degrees of rotational freedom, and the rotational energy is quantized such that the energy €J in the Jth rotational state is given by h2 €J J(J+l)811'2I=J(J+I)kOr,
where Or = h 2/811'2Ik and has the dimensions of temperature. Each rotational state except the first (J 0) is doubly degenerate, as the
4.4
Chapter 4
106
rotation can be either left or right handed, and the degeneracy WJ of the Jth level is in this case (2J + 1). (a) Assuming that the rotational energy levels are close enough together to be approximated to a continuum (i.e. when 8 r lT < 1) show that the quantum mechanical rotational partition function is the same as the classical rotational partition function obtained in Problem 4.3(b). From this partition function obtain expressions for the rotational energy, heat capacity, and entropy. (b) At low temperatures when 8 IT ~ 1 the rotational energy cannot be approximated to a continuum and a summation must be used. Investigate the form of the rotational energy and heat capacity in the low temperature limit. Sketch the temperature dependence of the rotational energy, and deduce the form of the rotational heat capacity curve.
Ideal classical gases of polyatomic molecules
4.4
107
The rotational entropy is given by
2 - = T E rat Srat + NoklnZ rat = Nok+ Nokln (81T h2IkT) 81T2k) = Nok ( 1 + lnIT+ lnll"2 ' which can be written in the alternative form
Srot
= Nok ( 1 + In;) .
(b) The rotational energy is given by
E
rat
= NkT2dlnZrat dT
Nk8
dZ
-Z d(81T)
Using the rotational partition function in the form
Solution
(a) The rotational energy €J and the degeneracy WJ are given by
h
€J = J(J+ 1)81T 21 ' WJ = (21+ 1). Hence
r
Zrat = ~(21+ 1)exp L00
When 81T
<
J(J+ l)8rJ T
r
roo
r
J(J+ l)8rJ roo (J2+J)8rJ T (2J+l)dJ= JoexPLT d(J2+J)
J
L (21+ 1)exp rL-
r
Nk8 J(J+ l)8rJ E rat = ZL(21+ 1)(J2+J)exp LT and
r
In the limit of low temperature the partition function approaches unity, and only the first term of the summations is important. When J = 1 we have
_
oo
_
T
0
.12Nok(~) exp (-2~) .
r
81T 2IkT Zrot
-
I
r_ (J2+J)8rJ}J=oo =:£ T J= 8
l
But 8 r = h 2/81T 2Ik so that
Erot - NokT
erot =
0
From the rotational partition
(8) exp (- 2r8) '
Erot = 6Nok r
_ -:£i [_ (J2+J)8rJ [_ (J2+J)8r] Zrat 8 exp T d T = -:£{ex p 8r
h2
f~ction
we obtain
2 2(alnZrat )\ ~ 2 d ( 81T IkT)_ aT - NokT dT In h2 - NokT
""I""~
1'0
(")1"" ~ OLI--~--------------------------~~ high o low
temperature
so that C\ot
= Nok .
J
_Nk(8)2 2 J(J+1)8 r - Z2 r L (21+ 1)(J2+J) exp LT .
We have this integral in the form e-"d(-a) = Id(e-") = e- a so that
r
J(J+ l)8rJ T '
we obtain
erot
1 we can replace the sum by an integral
Zrat= JoexPL-
Zrot =
2
Figure 4.4.1.
108
4.4
Chapter 4
As we know that the high temperature limits are E rot = NokT and Crot = Nok, we can sketch the rotational energy and heat capacity curves (see Figure 4.4.1). The point of inflection in the energy curve results in a maximum in the corresponding heat capacity curve. A CONVENIENT FORMULA FOR THE HIGH TEMPERATURE ROTATIONAL PARTITION FUNCTION
4.5 The partition function for a molecule in which two degrees of rotational freedom are possible is given by the equation
4.5
Ideal classical gases of polyatomic molecules
109
Solution
(a) Putting
fOCI) = (21+ 1)exp [ -
we have that
I(I+ lWrJ
T
~ f(I) dJ~s of the form
i°
eT i~° C· da = -8T
/'
r
and when
r
I = 0, fO(O) = 1. The higher derivatives are readily obtained 1
[_(21+l)2 8 r] [_I(I+lWrJ 2 T exp T '
_
f (0) -
~ [ I(I+ l)8rJ Zrot = ~(21+ l)exp T .
so that when I = 0 If 8 r lT is large, Zrot can be evaluated only by direct summation. If 8 r lT < 1, it is possible to construct a simple and convenient expression for Zrot which does not contain a summation of exponential terms. Mulholland (2) used the Euler-Maclaurin expansion, which expresses the difference between the (unknown) sum and the corresponding (known) integral in polynomial form, to obtain a simple expression for Zrot. (a) Obtain an expression for the rotational partition function of a heteronuclear diatomic molecule as a polynomial in 8 r lT using the Euler-Maclaurin summation formula in the form ~ r~ 1 1 1 n ~/(n) = fO(I)dJ+ HO(O) - 12 fl (O)+ 720 f3 (0) - 30240 f5 (0) ,
Jo
where fk (0) is the kth derivative of the (I) function with I = O. (b) Show that the polynomial obtained in part (a) is given by the general formula T
(8)[ 1 +
Zrot = 8 r exp 4 ~ where
~
n~ an
°
(8i )n+IJ '
B4 =
-lo,
B6 = i2, Bs =
-lo,
i6, ....
(c) Show that the high temperature heat capacity of an assembly of N = No molecules each of which has two degrees of rotational freedom is given by 1
Cot = Nk [ 1 +~
(8)2 i + 945 (8i )3 + ... ] 16
and obtain a similar expression for the rotational entropy.
8
+
8
(8
(8
8 7(8
(8
T r [ 1 r r) 2 3 1 r) 3 ] Zrot = 8 exp 4T 1 + 12T + 480 T + 8064 T +... ; r expanding the exponential we have
8
(8
(8
On multiplying out and collecting terms we obtain the formula derived in part (a). (c) To obtain the heat capacity we make use of the formulae
F = -NkTlnZrot Now
E = -T2 d(FIT) dT
,
8T
(8T
C 2
v= (~~)v . 3
x x X4 234
lnZ(T)=-ln~-ln(1-x)=- ln~+x+-+-+-+'"
where x =
(2) H.P.Mulholland, Proc.Cambridge Phil. Soc., 24,280 (1928).
l80(i
Substituting into the Euler-Maclaurin formula and collecting terms we obtain T [ 1 r 1 r) 2 4 r) 3 ] + 315 T + ... . Zrot = 8 1 + 3T + 15 T r (b) The general formula gives
8
B 10 =
r- r 30(ir-(ir
r)3 r 1 1 r 1 r) 2 eXP4T=1+4T+32T +384T + ....
and where B2n are the Bernoulli numbers
i,
8r + 12 (8T)r\ 2- (8Tr) 3
T
f5(0) = l20(i
B 2n + 2
(n+ I)!
8
r T ,
f3(0) = -12
(_l)n(1_T2n-l)
an
B2 =
f l (0)=2-
_[l3 8T ~ (8T r+
15
r )2 +
----±(8 r )3 + ... ] 315 T
.
)
4.5
Chapter 4
110
MUltlplymg out and collecting terms, we find
I -,
r
InZ(T)
I
- In-;r-
I
from which we obtain
E=NkT[I-~
- I
8 2835
4.6
vibrations and these generally have frequencies > 1000 cm- I . The remaining 2n 5 are bending modes, and generally have frequencies considerably lower. Vibrational modes are excited at high temperatures, and an error in the assignment of these modes does not usually introduce an appreciable error into the calculated thermodynamic quantities. It was shown in Problem 3. 9( e) that the heat capacity of N identical quantum_Ul-oscillators is ("'vib = NkE(2x) where and
(eT )3 + r
(iY- (ir+ . -]. 8
Note that as T -+ 00 the energy approaches the limiting value of Nk(T-je r ) rather than the classical value NkT. Terms in ie r do not occur in either the entropy or the heat capacity:
r
I
Cv = Nk [1 + 45
Using S S
1)2'
By graphical interpolation of the Einstein function E(y) tabulated below calculate the molar vibrational heat capacity and entropy at 298·15°K of 02, C1 2, and Br2 for which the fundamental vibrational frequencies are 1580, 565,and 323 cm- 1 respectively.
(e)2 16 (e )3 i + 945 i +
E/T+ NklnZ we obtain
2 8 (e )3] I e I e e I (e Nk [ 145 T T -Nk ln T - 3T 3T I
r
r)
r
r
8 - 2835 16
(i) 3+ ..]
y
r
L
Nkll-lni-9~(ir-
(eT
r)
.
The limiting value Nk of the heat capacity as T -+ 00 is approached from above. This is in accordance with the ideas of Problem 4.4. THERMODYNAMIC PROPERTIES ARISING FROM SIMPLE HARMONIC MODES OF VIBRATION
4.6 (a) In Problem 3.9 it was shown that the partition function for a [I)-oscillator of angular frequency w = 27rv is given by exp(-!hw/kT) Zvib = I exp(-hw/kT) Show that the heat capacity and entropy of a [I)-oscillator are independ ent of the amount of energy which may be possessed bv the oscillator in its lowest vibrational state. (b) The vibrational partition function for a polyatomic molecule with i vibrational modes may be written as the product of the functions of i distinguishable [I)-oscillators [see Problem Zvib =
111
Ideal classical gases of polyatomic molecules
r;) -eXp(~hwJkT) .
A non-linear polyatomic molecule has 3n - 6 vibrational modes, and a linear molecule has 3n - 5. Of the 3n - 6 modes n - I are stretching
y
Nok
1·000 00 0 0·50·979 ],704 1-0 0·921 1 '041 1·50·8320·683 2·0 0·724 0-458 2·5 0-609 0-309
Cvib
SVib
Nok
Nok
3'0 0'496 3·50·393 4·0 0-304 4'50'230 5·0 0·171
0'208 0-)39 0·093 0-066 0-041
y
Cvib
Svib
Nok
Nok
5'5 0·125 0·028 6'00-090 0'017 6- 5 0-064 0'011 7,00-0450,007 7·5 0'031 0-005
y
8·0 8' 5 9'0 9·5 10·0
Cvib Nok
0·021 0'015 0-010 0·007 O-OOS
0'003 0- 002 0'001 0·001 0·000
(c) By expanding the exponential term in the vibrational partition function and retaining only terms up to (e v /T)2, obtain simple approxi mate formulae for the heat capacity and entropy of a [I)-oscillator which are valid in the region of temperature for which (e v /T) < 1. Use these formulae to calculate the vibrational heat capacity of iodine vapour at 100°C and compare with the values obtained using the Einstein functions. (The fundamental vibrational frequency of 12 is 214·6 cm- I .) Solution
(a) Rather than rewrite the numerator of the function in the form exp (E min/ kT), we will carry the calculation E min thv. The energy of the [I)-oscillator is given by _
EVib -
NkT
(a InaTZVib)
2
v
so that = NkT2 =
d~{ln I
.. Nkl-;-I
eXP/-hv/kT) + In[eXp(-thV/kT)J} I
_] +!Nhv.
4.6
Chapter 4
112
The right hand term, which is the minimum (zero point) energy of the oscillator, disappears on subsequent differentiation. For the heat capacity we have Cvib
aE) = (aT
v
(hV) d I = Nk k dTexp(hv/kT)
For the entropy of the [1 ]-oscillator we have alnZVib ) aT + NklnZvib,
so that
_ hv 1 ~_1 Svib - NkT kT2exp(hv/kT)- 1 + NkT dT1n[exp( !hv/kT)]
+
Nhv
-Nkln[l
Nhv exp(-hv/kT)l-!r
Terms arising from the zero point energy cancel out. (b) We will denote the spectroscopic units, which are wave numbers, by w (cm- l ). Multiplication by the velocity of light e converts wave numbers into frequency v. Using the numerical values of the funda mental constants we have that hw hv hew Ov 2x = kT = kT = = I· T ' where Ov is the vibrational characteristic temperature of the [I)-oscillator and Ov/T is y in the table given. For oxygen at 298·15°K, Ov/T 1·4387 x 1580/298·15 7·62. Graphical interpolation of the Einstein functions gives Cvib/Nok O' 027 and SVib/Nok = 0'0044, so that Cvib 0·054 cal mole- 1 deg- 1 and Svib = 0·0088 cal mole- 1 deg- 1 • Similarly, for chlorine and bromine we obtain Cvib = 1·095 and 1·64 cal mole- 1 deg- l , and Svib = 0·52 and 1· 32 cal mole- 1 deg- 1 , respectively. These values agree well with experi mental values. However, at temperatures for which Ov/T ~ 1, thermo dynamic functions calculated from the harmonic oscillator model are not in good agreement with experiment. (c) Expanding Zvib = [1 - exp(-8 v/T]-1 we obtain Z-l
= 8;
[1
t(8;)+ie;y
...
J.
x
r,
In ~ 2_
2 •• _
The free energy F, energy E, and vibrational heat capacity Cvib, and the entropy Svib are now readily obtained.
F E
= -NkTlnZ
NkT[ln (8;)
NkT2 d~lnz
CVibG~) Svib =
hV) I S· =Nk ( -kT exp(hv/kT)-l Vlb
2 x3
{(8) t ; - Ii (0)2 ; + ... -t (0;) + i (8;) k(0;)
In(1- x) = -x
-Nkln[l- exp(-hv/kT)]+Nkln[exp(-thv/kT)] , which yields
113
We use the logarithmic series expansion (which is valid only when Ov/T < I) to obtain InZ- 1 -lnZ;
1
hV)2 exp(hv/kT) -Nk ( kT [exp(hv/kT) 1]2
Svib = NkT (
Ideal classical gases of polyatomic molecules
4.7
-"'1,
(8;) + -b c;r-.. J, Nk [1 - -b (0; + ... J,
NkT[l-! v
-t (8;) +f4 (8;)2
NkTe~/)+ NklnZ
r
Nk [I-In (8; )+ f4 (8;) 2- ...
For iodine vapour (12) we have 214·6 x 1-4387/373,15 From the approximate equation we have
J.
= 0·827_
Cvib = Nok[ I -b(0-827)2] = 0·943Nok = 1·873 cal mole- 1 deg- 1 and by graphical interpolation of the Einstein function we have
,
CVib 0·945Nok = 1·877 cal mole- 1 deg- 1 • At 8v /T 1 the approximate equation yields Cvib/Nok = 0'917, which differs from the value given by the Einstein function (0-921) by about a quarter of one percent. _ Similarly, for the entropy the approximate equation yields SVib/Nok 1·218 whereas the value obtained by graphical interpolation of the Einstein function is 1 . 215. CORRECTIONS TO THE RIGID ROTATOR-HARMONIC OSCILLATOR MODEL
4.7 (a) To treat the vibration of a diatomic molecule as simple harmonic motion is unrealistic because on this model dissociation of the molecule can never occur. A more realistic potential function was given by Morse, who proposed the equation u(r)
= D e {1-exp[-(3(r-re )j2},
where De is the dissociation energy of the molecule, (3 is an empirical constant, r is the separation of the nuclei of the two atoms, and re is the equilibrium separation.
~r
t 4.7
Chapter 4
114
Show that at low amplitudes of vibration this model allows simple harmonic oscillation, and show that the constant {3 is then given by 21T2 P.c 2 )% {3 = We ( -D- e where We is the frequency of vibration of the molecule, expressed in wave numbers, about its equilibrium position, and where p. is the reduced mass of the molecule. (b) When the amplitude of vibration of a diatomic molecule becomes large the simple harmonic oscillator model is inadequate for precise work and a contribution to the thermodynamic properties due to anharmonic vibration must be considered. The vibrational spectra of diatomic molecules are often represented empirically by energy levels which are described by the equation Ev
= hVe(V+!)-xehve(V+!)2+Yehve(V+!)3,
where V is the vibrational quantum number and where Xe , Ye, etc. are called anharmonicity constants. When the Morse potential is put into the Schrodinger equation, the following approximate expression for the allowed energy levels is obtained Ev
= h ve (V+ !I) _
( hv )2 ( V + 1)2 e
4D e
- exp(ixe u - !u)
Zanh-
l-eu
r
o
where 0 == 6!:
InZo =eo/T-I '
r(W;~e f - I J
81' _ Be InZ)' = e/T' where l' = We Be is the rotational constant (h/81T 2 /) calculated with the atoms at their equilibrium separation re. For the above terms in X e , 0, and l' obtain expressions for the corresponding terms which must be added to the heat capacity of the molecule. (e) The equilibrium separation of the nuclei in a chlorine molecule is 1·988x IO- B cm and the corresponding value of Be is 0·2438cm- l . Calculate the contribution to the heat capacity of chlorine gas at efT = 2 (i.e. 133· 2°C) arising from the above terms, and express these contributions as a percentage of the heat capacity obtained from the simple harmonic oscillator model. Solution
(a) At low amplitudes of vibration we may expand the exponential of the Morse potential and examine the leading term
= D e (l-exp[-{3(r-re )j2}. = l-x+!x2-i x 3 and putting X = (3(r-r e ) we obtain u(r)
= hv(V+!)-x ehv(V+!)2,
the corresponding partition function is
115
of the partition function:
2"
Given that the fundamental vibration frequency of 35Cl2 is 564·9 cm- I and the coefficient (3 of the Morse potential is 2·05 X lOB cm- I , calculate the molar dissociation energy Jj e in kilo calories and the anharmonicity constant Xe of chlorine gas. (c) Show that for an anharmonic oscillator, whose energy levels are described by the equation Ev
Ideal classical gases of polyatomic molecules
4.7
Using
e- x
u(r)
= -De[{32(r-re)2-{33(r-re)3+-&{34(r-re)4- ... ].
From the leading term for the potential energy we obtain the force F: F
= -2De{32(r- re)
.
The equations of motion for the two atoms of mass m 1 and m2 at distances r 1 and r 2 from the centre of mass are given by d2rl _
2
m l dt2 - -2D e{3 (rl+r2- re) ,
2xeu
]
2 d r2 _ _ 2 _ m2 dt 2 2De{3 (r l + r 2 r e ),
1+(eU-I)2,
where u == hv/kT == efT. [Hint: Expand the exponential expression containing Xe and retain only the leading term. Simplify the summation over this term by utilizing the first and second derivatives with respect to u of the simple harmonic oscillator function.]
which can be written
(d) It can be shown that terms which allow for the fact that the moment of inertia of a vibrating rotator is greater than that of a nonvibrating rotator (0 term), and which allow for the centrifugal stretching of the molecule (1' term), must also be added to the logarithm
wherer l +r 2 = r,and l/ml+l/m2 = l/p.. When a mass p. is constrained by a force constant of -2De{32 the
so that
2 d (rl +2 r 2 ) dt
=
-2D {32(r + r _ r )(_1_ + _1 ) e 1 2 e m 1 m2
d2 r p. dt 2 = -W e{32(r- re) ,
I 4.7
Chapter 4
116 frequency of vibration v is
=
CWe
1 (2{32De)~ 21f
v
Using v =
Ideal classical gases of polyatomic molecules
4.7
The first term is the simple harmonic oscillator (SHO) partition function equal to I/O - e- U). We write the anharmonicity term in the form
J.L
we obtain
xe u
{3 =
We
L Ve- uv + V 2 e- uV .
Now
(b) The reduced mass J.L of 3sC12 is given by
"
and
)2 (2
X
3.1416 2
10 12 erg mole-I
X
2~~
(1
e_U)3
which can be written X
17·5
X
2.9979
2
X
10
20
)
2·9979
X
1010
(1
so that
56·34 kcal mole-I.
Comparison of the empirical equation for the vibrational energy levels of a diatomic molecule with the form obtained from the Morse potential gives hv" hcw e xe = 4De 4De/N
10-
e- U )2+
)3 ,
564.9 ( 2.05 X lOS
X
~
Using these expressions the anharmonicity term becomes 2e- 2u
w
6·6256
LVe-uv = - (1 _ e-U )2
2 e- uV = e- U (1 "e-uv = "V du 2 L.. L..
= {32 (21f 2J.LC 2 )
27
U
e- U
~ Le-uV = du
2
X
-e
and
where m l = m2 = 35/No, so that J.L 17· . We note that if in the numerator of our equation for {3 we use 17· SINo, then in the denominator we use the dissociation energy per molecule De De/No, and the No cancels out. Hence we obtain for the molar dissociation energy
= 2· 3574
=
L..
ml m 2 J.L = ml+ m2 '
De
117
X
564·9
X
6·0225
X
10
2xeu e-u )(e U
-
1)2 ,
we obtain - exp
(d) The correction term which must be added to the SHO partition function is obtained by quite straightforward differentiation: ~
23
lnZcorr = !xe u
tu+
u(e U
-
1)2+
b U
e -
1+
u
Using = 7·166
X
10- 3
NkT
•
(c) The partition function for the anharmonic oscillator is
we obtain for the four terms
L exp[-u(V+-!)+UXe(V+-! V=o
Zanh
= NkO(-!x e +!)+Nk
..
=
L
v
exp{(!xeu-tU)[-uV+XeuV(V+ 1)J} 0
..
= exp(!xeu-tU)
L e-UV{l+exp[xe(V+ I)]}. v 0
Expanding the exponential term in xe we obtain Zanh = const[
£ ev=o
uv
+
£ xeuV(V+ l)e- UV] v=o
2dlnZcorr dT
.
e 8IT(20/T- 1 + 1 x 2xeO .. ---:-....,..,.,.,-~
+NkbOe8IT(eOIT-l)2+NkT2 x
and finally Ccorr
(dE (ff"'" corr
) V
Nk x 4xe(O/T)2e O/T+ 1 +Nkb(O/T)2 e O/T e (e'l/T - 1)3
2
+ Nk
x 16')' O/T
J
4.7
Chapter 4
118
(e) The constants 'Y and 0 for chlorine are
Be 0·2438
'Y = - = = 4·316 X 10-4 We 564·9
o=
J
r(WeXe)n 6 Be L T -1 We
7·979
10- 3
X
Using Nok = 1 -987 cal mole- 1 deg- 1 , (J /T :::: 2 -0, e OIT 7 -389, we obtain the following values for the contributions to the heat capacity arising from the terms in X e , 0, and 'Y. CXe
::::
0·0190 cal mole- t deg- 1 1
Co = 0·0151 cal mole- deg-
= O· 0071
C"(
1
cal mole- 1 deg- 1
L '/
I
4.8
Ideal classical gases of polyatomic molecules
(d) The nitric oxide molecule has two doubly-degenerate energy levels which are separated by the unusually small gap of e/k 174°K. The oxygen molecule has two levels which are separated by an energy gap of e/k = 11 300o K, and at high temperatures the lower level is triply degenerate and the upper level is doubly degenerate. Estimate the temperature at which the electronic heat capacity has its maximum and calculate the corresponding molar electronic heat capacity for these molecules. [Slide rule accuracy is adequate for parts (c) and (d) of this problem.] Solution
(a) The partition function for a system with two energy levels is
,
Taking the zero of energy eo = 0 we have
_
Z = woe 1+ we- elkT ) ,
t
where w = w d Wo and e = e 1 - eo. The energy and heat capacity follow directly £
CONTRIBUTIONS TO THE THERMODYNAMIC PROPERTIES ARISING FROM LOW LYING ELECTRONIC ENERGY LEVELS
= NkT21
,~.Jnwo+ d~ln(l + we- e/kT )] = wNe------"r-::;
and 4.8 (a) For certain molecules, notably oxygen and nitric oxide, there is an additional contribution to the thermodynamic properties, which arises from the presence of two electronic energy levels separated by an energy which is small compared with kT, so that thermal excitation is sufficient to populate appreciably the upper level. Construct the partition function for this two-level system and show that the electronic contribution to the heat capacity is given by
Cel
wNke
IkT(
C=
C At low temperatures e/ kT C
(~kT+2)
v
wNe +w
= wNkee/kT(---,e,.-;;/k"T_)2
ee/kT+ w
~
1, and
~ WNk( le~k~r ~ 0 . ~
1, and
~ wNk(~kT)2e€/kT --+
O.
We locate the maximum in the heat capacity by putting the temperature derivative of In C equal to zero:
where w is the ratio wdwo of the degeneracies, e is the energy of the upper level, and the energy of the lower level is taken as zero. (b) Investigate the high- and the low-temperature limits of the electronic heat capacity, and show that lnCe ) has a maximum value when e
(_a£) aT
(b) At high temperatures e/ kT
e/kT )2 eelkT + w '
kT = lnw+ In e/kT- 2
= woe-eolkT+ wle-el/kT .
Z
,
The heat capacity of a simple harmonic oscillator with OfT 2·0 (from Problem 4.6) is 1·438 cal mole- 1 deg- 1 , so the above terms contribute 1 ·32%, 1 -05%, and 0·49% respectively, a total of 2· 86%.
119
~ _ ee/kT (e/kT)2 wNk -
(ee/kT + W)2
In(~) = ~+ 2ln (~)- 2In(ee/kT + w) wNk kT kT
.
d (C) e 2 dT ln wNk = - kT2 - T
(c) Obtain an expression for the electronic heat capacity of a molecule with three equally-spaced singly-degenerate energy levels. By taking numerical values of e/kT between 1·0 and 4·0 compare the magnitude of the maximum in the electronic heat capacity of this molecule with that of a molecule in which the lower energy level is singly degenerate and the upper level has threefold degeneracy.
+
2(e/kT2)ee/kT ~€/kT ..L • •
multiplying by T we have
+ 2) (e€/kT+ w) = 2~e€/kT (~ kT kT I
I(
'
0;
,
I
4.8
Chapter 4
120 and
2(e/kT)e €jkT e €jkT + w = --"--'----'-- e/kT+ 2 2{e/kT) w 1+ = (e/kT+ 2) ,
so that we
~
4.9
4.9 The fundamental vibration-rotation spectrum of isotopically pure H 35 Cl is shown in the lower part of Figure 4.9.1. The relative intensity I rel and frequency w of each line, expressed in wave numbers, is indicated below the spectrum. This type of spectrum arises in the following way. When the vibrational quantum number V increases by unity (~V = + I) as a result of the absorption of radiation, the rotational energy levels may be affected in two ways. In the transition to a higher vibrational level, the increase in bond length of the molecule is accompanied by an increase in the moment of inertia. If the molecule absorbs only a small amount of energy, then it may fall into a lower rotational energy level (AI = -1) and give rise to the P branch of the spectrum. If a large amount of energy is absorbed, there may be an increase in the rotational energy, sufficient to raise the molecule into a higher rotational level (AI + I). This gives rise to the R branch of the spectrum. The upper part of Figure 4.9.1 shows the first eight rotational energy levels of the molecule and the double headed arrows indicate the transitions which give rise to the line in the spectrum below the arrow. For ~ V = + I the rotational transitions which we have just discussed are indicated by upward pointing arrows, whereas for ~ V I the downward pointing arrows will pertain. The selection rule for vibration-rotation spectra is AI 0, ± 1. Now the vibration-rotation energy of a diatomic molecule on the rigid rotator-harmonic oscillator model is given by
taking logs we obtain as the condition for the maximum
(e/kT+ 2) Inw+ In e/kT- f
.
(c) The energies of the three level system are 0, e, and partition function is Z = I + e-ejkT + e-2ejkT .
and the
The energy and heat capacity follow directly _ e-ejkT + E = Noe I + e-ejkT + e 2ejkT' , from which we obtain _ ( e )2 e-€jkT(1 + e- 2€jkT + 4e-€jkT) C = Nok kT (I + e-€jkT + e-2ejkT)2 Working to three significant figures we calculate for the two-level system with w = 3 and for the three-level system with w = I the following values: €/kT
1·0
1'5
2'0
2'5
3'0
3'5
4'0
0'249 0·424
0·540 0·602
0·821
0·991
1·02
0'933
0·790
0'634
0·578
0'486
0·390
0·303
C/Nok (tWO-level) (three-level)
121
CALCULATION OF THE THERMODYNAMIC PROPERTIES OF HCI FROM SPECTROSCOPIC DATA
-€jkT _ 2(e/kT) _ _ (e/kT- 2) - e/kT+ 2 I - (e/kT+ 2)
e kT
Ideal classical gases of polyatomic molecules
Ev, R
(V+t)hew+J(J+ l)heB,
where w is the fundamental vibrational frequency expressed in wave numbers and B is the rotational constant h/8rr 2 Ie for a particular vibrational state. The constant e is the velocity of light so that B is also in wave numbers. The above equation represents just one energy level, so that the total vibrational-rotational energy of the molecule is given by the sum over all V of such terms. We now use the equation for EV,R together with the Bohr frequency condition
The maximum for the two-level system is 1·023 at e/kT = 2·8 and the maximum for the three-level system is 0·637 at e/kT = 1·9. (d) We can locate the maximum in the electronic heat capacity using the equation e kT = Inw+ln(e/kT+2) e/kT- 2 .
.
E'y
R
E'~, R
= hew ,
where the single prime and double primes indicate upper and lower vibrational states, and w is the frequency expressed in wave numbers, in conjunction with the selection rule (~V = + I and AI = -I) for the P branch, to obtain a general expression for the difference of energy between any two levels. The set of frequencies P(J) in the P branch is then given by p(J) = (B' - Bn)J + (B' B").f2,
For nitric oxide w 1, and for oxygen w 2/'l...j By trial-and-error substitution we find that the electronic heat capacity of nitric oxide has its maximum value at 72° K and the corresponding maximum heat capacity is O' 875 cal mole- 1 deg- 1 • The maximum for oxygen is at 4900° K and it has the value 0·616 cal mole- 1 deg- 1 •
w
where J may have any integral values I, 2, 3 etc. other than zero.
'tl
I
122
4.9
Chapter 4
,I
(
4.9 by
8
R(J)
7 6
4
3 2 01
I
I
,
_
I
I
I
I I I
I
I
I
I
I
I
1
,
,
i
'I
8
'
,
~
- 61
5 -4
'I
_0 1
J
'"..... .::,
'"
M 0
00 N
M
N
N
N
'"t-
N
N
N
v v 00 N
00
t-
'"
on
V
M
N
M 0
'"
'"
r-
V'l
6
v
0"1
6
.::,
on tt-
00
00 0\
.::,
N", 0"1
'--' "---' '-"' \...J ' - - ' \...J ' - ' ' - -
r-
t"'VI r- 0\
on
~
M '"
6'::'
.::, .::,
OMM
0
'"
00
v v
N
0 00
;...::, 6
1: ...,
t-
t-
..... on t-
P branch
.....
oo
00 '" N l:sl
-
'" '" N
N
:;1l
0
'" N '" N'"
N
M
Solution
(a) Using the equations for R(J) and P(J) we readily obtain the combination terms
v_N \0 v N
66'::'
00 v 0 "'_ M "'00 NM M
V
R branch
on '" t
B").! + (B' + B").f2 ,
where J may now have any integral value including zero. Analyse the spectrum of H35CI in the following way. Calculate the combination terms R(J) - P(J) and R(J - I) - P(J + I), and by considering them as a function ofU+ I determine B' and B" respectively, hence obtain the fundamental vibrational frequency w. (b) The rotational constant in a vibrational state may be related to the equilibrium value Be by the equation B v Be - (V + t)a where a is a vibration-rotation constant. Calculate Be and hence obtain the equilib rium moment of inertia for H35CL [BI h!8rr 2 = 5· 0553 X 1011 atomic mass units A2 Hz.] (c) Given that the degeneracy of the Jth rotational energy level is 2J + I (cf Problem 4.4), derive an expression, based on the Boltzmann distribution law, for the population density of the rotational energy levels. Show that the most densely populated level is that for which kT )V2 J max = ( 2Bh -t·
R(J) - P(J)
'E
= w+ 2B' + (3B'
(d) The area beneath an observed peak is proportional to the intensity of the line, and this is in turn proportional to the population density of the energy level. Absolute intensities are difficult to measure, but relative intensities are readily available from the observed spectrum. Show that a plot of In(IreJ/U+ 1) againstJ(J+ I) has a slope of-hB/kT. Analyse the R branch of the spectrum and determine the temperature of the gas. (e) Calculate the entropy, Helmholtz free energy, and heat capacity of H 35 CI at 3000 K and 1 atm pressure. [The molecular weight of H 35 CI is 35 ·9877.J
-3
-2
' - - - - ' \...-.....J " - - ' " - - ' '--'''--------I
123
In the same way the set of frequencies R(J) in the R branch is given
J
5
Ideal classical gases of polyatomic molecules
= 2B' + (3B' + B").! + (B' + B").! + (B' -
B").f2 - (B'
= 2B'(J+ 1), and
..
R(J-I) = w+J(B'+B")+.f2(B'-B"), p(J+ 1) = w+J(B'
3B")+J2(B'-B")-2B";
on subtraction we obtain
Figure 4.9.1. The vibration-rotation spectrum of H35 0 showing the energy levels and transitions which give rise to the P and R branches.
R(J-I)-p(J+ 1)
L
=
2B"(U+ 1).
B").f2
4.9
Chapter 4
124
We can obtain the combination terms from the spectrum in the following manner: 0
1 3 2865 p(J) 2865 2844 p(J+ l) R(J) 2906 2926 R(J-I) 2906 R(J)-p(J) 61 R(J-I)-p(J+ I) 62 J
21+ 1
2 5 2844 2821 2944 2926 100 105
3 7 2821 2798 2962 2944 141 146
4 9 2798 2775 2980 2962 182 187
5
6
11
13
2775 2751 2998 2980 223 229
2751 2728 3014 2998 263 270
4.9
"
J(J+ I)B .
7 8 17 15 2728 2703 2703 3030 3014 3030 302 311
The total number of molecules occupying all the rotational levels is simply the rotational partition function, which can be approximated by the integral kT (21+ l)exp -J(J+ I) hB] kT dJ hB'
Jor'"
h B = 8rr 2Ic
n = (21+
substitution
,("dO
n
d (nJ) dJ
so that ]2+J+
a
(i- 2hB kT)
gives
X
J
=
0
kT)'h - ~ . ( 2hB
(d) For convenience we select the line J(lmax) of greatest intensity the R branch, and we scale all other intensities relative to it IJ
hB/kT(21+ 1) exp[-J(J+ l)hB/kT] rel
I = [max = hB/kT(21m + I) exp[-Jm(Jm + l)hB/kT] so that hB InJ;.el In(21+ I)-ln(21m + 1)+ [Jm(Jm + l)-J(J+ 1)) ,
Imax in
I
where the integer J m is a constant.
Analysing the lines of the R branch we have:
10- 40 g cm- 2 .
(c) The relative population of the rotational energy levels is given by the Boltzmann distribution law n
-(21 + 1)(21+ 1) hB kT exp [ -J(J+ 1) hB] kT + 2exp [ -J(J+ 1) hB kT
o
16·8628 Ie = 10.4888 = 1·60770 atomic mass units A2
nJ
and of
= 16·8628 atomic mass units A2 cm-
1-60770 x 10- 16 6-0225 X 1023 = 2-6695
hB] .
[
The value of J for which nJ/n is a maximum is given by
we calculate the equilibrium moment of inertia to be
-
hB
1) kT exp -J(J + 1) kT
Jmax
5·0553 x 10 11
=,.,. "(""'"' '
back
[
The fraction of molecules in the Jth rotational state is therefore
-10·0558 == -Be+(l+!)a; and
125
so that
We now plot R(J) - P(J) to 21+ I and obtain from the slope B' = 10·0558 cm- I , and from a graph of R(J - 1)- P(J+ 1) to 21+1 we obtain B" = 10·3448 cm- I . Using these constants and the value of P(J) 2865 cm- I at J = 1, we obtain w = 2885·7 cm- I . (b) For V = 0 we have Bvo 10·3448 cm- I , and for V 1 we have I BVI = 10·0558 cm- . We obtain the vibration-rotation constant a by eliminating Be between the equations 10·3448 = Be - (O+!)a, addition gives a = O· 289, Be = 10·4888 cm- I . Using
Ideal classical gases of polyatomic molecules
J(J+ 1)
0 1 0·76 In 100Irel /2J + I 4"33
21+ I I rel
= gJ exp(Er/kT) .
For Hel the statistical weight gJ is simply the degeneracy of the level is given by 21 + I, and the energy J(J+ l)h 2 Er rr2I
2 3 0'95 3-94
6 5 1'00 2'99
12 7 0·93 2'58
20 9 0·83 2·22
30
42
11
13
0'64 1'76
0'41 1 ·15
56 15 0·22 0'40
Plotting In 100fret/21 + I against J(J + 1) for values of J > J(l max) we obtain a straight line of slope 0·05 ± 0-001, so that Bhc 10·488 x 6·6256 x 10-27 x 2·9979 x = 302 ± 15°K. k x 0·05 1·38054 x 10- 16 x 0-05
L
126
4.9
Chapter 4
The intensity of the spectral lines is not a simple exponential function as the Boltzmann distribution law alone suggests, as there is a term which involves the dependence of the transition probability upon J which we have not considered. However when J > J(I max) this term is small compared with the Boltzmann term (3). (e) Formulae for the contributions to thermodynamic properties of a gas which arise from translational, rotational, and vibrational motion of the molecule have been obtained in Problems 4.1, 4.4 and 4.6. We shall consider first the vibrational contributions. From the fundamental 2885·7 cm- 1 we obtain the ratio vibrational frequency w Ov/T 1·4387 x 2885· 7/300 13· 839. The value of exp(Ov/T) which occurs in the formulae for the vibrational thermodynamic functions is then approximately 1 x 10 6 , so that the magnitude of these thermo dynamic functions is exceedingly small, and they can be completely neglected. For the translational and rotational contributions we have for the entropy: Strans = 1'98717[~ln300
= 36·71
cal
mole- 1
Inl +!ln35'9877-1'1605] deg- 1
,
Srot = 1·98717[1+1n2·66948 x 1O-4o+1n 300+1n(8'lT 2 k/h 2 )] =
1·98717[1-91·1215+5'70378+88
:;;: 7·9288 cal mole- 1 deg- 1
,
_ -1 Strans + Srot - 44 • 6391 cal mole- 1deg .
= 10118·8 cal mole-I, F rot =
-1,98717 x 300[ln300+ln2'66948 x 1O-4°+ln(8'lT 2 k/h 2 )] -1782'5 cal mole-I,
= -11901 cal mole-I.
Finally for the heat capacity Cp we have:
Cv+R
= iNoK:.rans+!Nokrot+Nok
6·956 cal
mole- 1
127
Ideal classical gases of polyatomic molecules
THERMODYNAMIC PROPERTIES OF ETHANE
4.10 At 500 0 K and a pressure of I atm the calorimetric heat capacity Cp of ethane is 18·66 cal mole- 1 deg- 1 and the calorimetric entropy is 62·79 cal mole- 1 deg- 1 • The molecular weight of ethane is 30'047, and the principal moments of inertia of the molecule are 42'23, 42·23 and 10·81 x 10-40 g cm 2 • The following vibrational frequencies (cm- 1 ) have been assigned from the spectrum: C-H stretching C-C stretching C-C bending CH 3 group deformation
2955 993
2954
2996 d
2963 d
1375
S2I d 1472d
1190 d
1375
1460 d
Frequencies with the subscript d are doubly degenerate. (a) The only assignment which has not been made is that of the twisting mode about the C -C bond. Calculate the contribution to the heat capacity and to the entropy which arises from this mode. Investigate the following three models (denoted b, c, and d below) which purport to describe the nature of the motion in this mode by choosing (where possible) the parameters of the model to fit the heat capacity, and using these parameters to calculate the entropy: (b) The free rotational model in which the methyl groups rotate freely about the C C bond. The reduced moment of inertia lr of the rotating group is given by 1m Ir = Igy+Im , where Ig is the moment of inertia of the rotating group about the axis of rotation and 1m is the moment of inertia of the rest of the molecule about the same axis. The kinetic energy € of the rotating group is given by 1
-1'98717 x 300[!ln300+!ln35'9877-1nl-2'6605]
Ftrans
Cp
4.10
g
This value is in excellent agreement with the calorimetric value of 44·5 cal mole- 1 deg- 1 obtained at 298· 10 K for a mixture of H 3s CI and H 37CI in their natural abundancies of 75·4% and 24·6% respectively. Similarly for the Helmholtz free energy we have:
Frot
(
= 3·5 x 1·98717
deg- 1 •
The calorimetric value of the heat capacity is 6· 96 cal mole- 1 deg- 1 • (3) G.Herzberg, Molecular Spectra and Molecular Structure, 2nd Edn. VoU, Spectra of Diatomic Molecules (Van Nostrand, Princeton), 1950, p.l25.
€
= 21r P~
where Po is the momentum conjugate to rotation about this axis at an angle O. [Note: Include a symmetry number in the derived thermo dynamic formulae.] (c) The harmonic oscillator model in which the two methyl groups undergo torsional oscillation. (d) The restricted rotator model in which at low temperatures only torsional oscillations take place, but as the temperature is increased the amplitude of these oscillations increases until a certain energy barrier is overcome, and thereafter rotation of the hindered group becomes possible. The potential barrier V is given by V
! Vo(l -
cosnO) ,
where n = 3 for ethane and Vo is the maximum energy of the barrier.
4.10
Chapter 4
128
Pitzer and Gwinn (4) tabulated thennodynamic functions for the hindered rotator model, and a section of their tables is reproduced below. In these tables Z is the numerical value of the classical partition function for the group with reduced moment of inertia I r , and the tabulated functions have the dimensions of cal mole- 1 deg- 1 • RT 0·25 3-355 3'180 3·008 2·838 2·678
2-0 2·5 3·0 3·5 4'0
0'30 3'004 2·836 2-667 2'500 2·343
0'35 2'709 2·548 2-380 2'218 2'069
0-40 2'458 2-303 2'138 1'978 1-834
0-25 1'632 1-840 1·996 2-106 2'168
0-30 1·606 1-801 1-952 2'054 2·110
4.10
c.nblNok = 1-181, w(cm8:yIT
0-40
Total for doubly degenerate vibrations:
1'541 1·717 1·846 1·934 1'980
so that
CviblNok
= 3· 246, SINok
Cvib = 1,9872(1·181+3-246) Svib
=
0-430_
1190 1460 1472 2963 2996 3-42 4-20 4'24 8'51 8-63 0-410 0-275 0-013 0'270 0-010 0-148 0-079 0-074 0-002 0·002
0-35
= 38·75 cal mole- 1 deg- 1 , Ctrans
C~b
1-9872 (0·430+ 1·302)
= 1· 302 ,
8-796 cal mole- 1 deg- 1 3-45 cal mole-I deg-
1
,
_
= ~Nok = 2-980 cal mole- 1 deg- 1 _
For rotational motion we have [cf Problem 4.3(c)]: 3 I = Nokb-lnT+!lnI"IbIc
Ina+ 134-68] = 1·9872 x 2·3026[i Ig500+! Ig(42·23 x 42-23 x 10-81 x 10- 120 ) -lg6] + 1-9872 x 134-68 17-83 cal
= tNok
mole- 1
deg- 1 ,
= 2·980 cal mole- 1 deg- 1 _
For vibrational motion [cf. Problem 4.6(a), (b)] we obtain the entropy and heat capacity of each vibrational mode by graphical interpolation of the hannonic oscillator functions tabulated in Problem 4.6_ Now = I -4387w/500 so that we have w(cm- 1 ) 8:yIT C~tllVok
SVib/lVok
993 2-86 0-525 0-232
2955 1375 1375 2954 8·50 8'50 3-95 3'95 0·013 0-013 0-315 0-315 0-002 0-002 0-097 0-097
(4) K.S.Pitzer and W.O.Gwinn,J. Chern. Phys., 10,428 (1942).
= (Cp -Nok)expt-Cvcalc = 16-67-(2'980+2-980+8-796)
= 1·91 cal mole- 1 deg- I Svib
,
= 62-79-(38'75+ 17-83+3'45)
= 2·76 cal mole- 1 deg- 1 _
= Nok[~ In T -lnP+ ~ InM - 1-164] = 1·9872 x 2-3026[~ Ig500 -Ig I + ~ Ig30-047 - 1-164]
Crot
821-5 2·36 0-645 0-346
)
SINok
For motion about the C-C bond we therefore have,
Strans
=
1
1'574 1-756 1'900 1-995 2·048
(a) We calculate first the contributions to the heat capacity and the entropy which arise from motion within the ethane molecule about the c-c bond. This is done by subtracting the statistical thermodynamic quantity from the calorimetric quantity_ For translational motion we have [cf.Problem 4_I(b)]:
-
129
Total for singly degenerate vibrations:
S~b/lVok
Solution
Srot
Ideal classical gases of polyatomic molecules
C~b/lVok
Heat capacity Z-l
Entropy Z-l
Vo
c
We can now investigate the proposed models. (b) The free rotational model. We must first obtain expressions for the thennodynamic functions of a rotor with one degree of freedom_ We follow the methods of Problem 4.3. The kinetic energy of a part of a molecule which is rotating about a single axis with respect to the rest of the molecule is given by 1 e = 2l/~ . The classical partition function is therefore Z
= -I 1211' ah
0
J-
-_
(p2)
21T exp _ _8_ dO dP8 == -(21TI kT)'h. r 2Ir kT ah
3
(81T !r kT)'Iz = __
The relevant thennodynamic fonnulae follow at once: -_ 2 alnZ _ E - NokT aT - !NokT , Cv = !Nok,
F = -NokTlnZ _ E
= -NokT ( !lnT+!InI;.
(
S= T+NoklnZ=Nok !lnT+!Inlr
81T2 k)
Ina+!lnJiZ ' 2
81T k ) Ina+!lnJiZ+! .
130
Chapter 4
4.10
The smallest moment of inertia listed above for the ethane molecule is 10·81 x 10-40 g cm 2 , and this clearly corresponds to joint rotation of the two methyl groups along the axis formed by the C-C bond. The reduced moment for one methyl group is consequently (5 ·405 x 10-40 )2 Ir = 2 x 5.405 x 10 40 2·702 X 10-40 g cm 2 In the course of the complete rotation of one methyl group with respect to the other we see that there are three identical configurations; hence (J 3. From the above equations we obtain Crot 0·99 cal mole-I deg- I , Srot = 2-46 cal mole-I deg- 1
0·7 ------:----lnO·49659 = 1·377 . Syib = 2·736 cal mole- 1 deg- I .
(d) The restricted rotator model. We calculate first the numerical value of the classical rotational partition function obtained in part (b). 3, we obtain Putting Ir = 2·702 X 10-40 g cm 2 , T = 500 o K, and (J I/Z = 0·292. Plotting and cross plotting the tabulated thermodynamic functions we find that a heat capacity of I ·91 corresponds to a value of V/RT = 2-80, and that 2-66 cal mole- 1 deg- I is the corresponding entropy. On comparison of the calorimetric entropy, 2·76 cal mole-I deg- I, with that obtained for the three models: (b) 2'46, (c) 2'74, Cd) 2·66 cal mole-I deg- I , it would appear that the harmonic oscillator model is marginally better than the restricted rotator model. However when the above comparison is made over a wide temperature range it becomes evident that the restricted rotator model is superior. In Problems 4.9 and 4.10 we have calculated some thermodynamic properties of ideal classical gases of polyatomic molecules from statistical
4.10
Ideal classical gases of polyatomic molecules
131
mechanical formulae based on simple molecular models. We have used in these formulae constants derived from experimental spectroscopic data, and have made a comparison of the calculated values with those obtained from direct calorimetric measurements. Agreement between the values obtained via these two routes validates the molecular models upon which the statistical mechanical formulae are based.
~I
•
For this model there is no parameter which we can adjust to improve agreement with the experimental values. (c) The harmonic oscillator model. Using the Einstein functions (Problem 4.6) we find that CyiJNok = 1·91/1·987 = 0·96 corresponds to () /T = O' 70, and this in turn is equivalent to assigning a frequency of 243·3 cm- 1 to the torsional oscillation mode. We obtain an accurate value of the vibrational entropy using the formula obtained in Problem 4.6(a) rather than by interpolation of the tabulated Einstein functions, so that for ()y/T = 0·70 we have, Svib ()y/T
Nok = ~~_{IJ 17'\_ 1 -In[1 exp(-()y/T)]
Hence
t.
~
(
5.1
Ideal relativistic classical and quantum gases
133
Solution
(a) The grand partition function is, from Problems 2.4 and 3.12,
5 P.T.LANDSBERG
(University College, Cardiff)
5.1 A system of non-interacting identical particles of rest mass mo is in a cubic box of side 1. In a single-particle quantum state (j I, h, h, a) the energy is e(jl, h, h) and the momentum components are Pr = (h/L)ir (r = I, 2, 3), where the j's are integers which cannot all be zero. The chemical potential J.I. is written as ')'kT, T being the temperature of the system. The spin label a assumes g values if each single-particle state has spin degeneracy g. (a) Prove that
pV = 'l7kTiLL L In[l +'I7tCiI,i2,ja)]
hhh where '17 = + 1 for fermions and '17 = -} for bosons, and
(. . .)
t h,/z,/3
[e U kT ,i2,h)] .
exp ')'
1
(b) Using a continuous spectrum approximation, assuming the energy to depend only on the magnitude of the momentum, and treating the magnitude of the momentum, pee), as a function of the energy e, show that the pressure is 41TgfP "" [p(e)pde p = 3h 3 P= 0 exp(e/kT-,),)+'17
j
where the sum extends over all many-particle states of the system. If the system is in a typical state i, it contains a total number of particles and a total energy given respectively by Cil = i2 i3 a excluded) g
= E;
= kTg (
(1
=
L L = 1 i1
::::::-00
L
L
it ,h,i,
(1
""
""
'L
L
h
::::::-00
h
:::::-QCI
n;CiI,i2,i3, a)
e(jl,h,i3)njCiI,i2,j3, a).
Hence
L nnnn [t(jl,i2,h)r M1 ,!2,h,a)
Z
;
i l h i3
(1
A many-particle state i can for indistinguishable particles be specified by the set of occupation numbers n;(l,a,a,a l
n;(a,
a,
I,a l ), nj(a,l,a,a l ),
(S.1.I) These njCil>h,j3, a) can have values a or I for fermions and a, 1,2, ... ,00 for bosons. A summation over i is then equivalent to summing over all admissible values of the numbers (S, 1.1). Hence ),
lor""
Z
I or"
= n(l, 0, 0,LI) = 0
L
n(O, 1,0, I)
....
... nnnn [t(jI.i2,ia)]n(j"h,i"a) i h h
O.
(1
.
l
The product goes over all quantum numbers. Carrying out the summa tions first
(c) Show that in the non-relativistic case P
= Lexp{(pNj-Ej)/kT}
Z
Ideal relativistic classical and quantum gases
Z
21TmgkT)';' h I(,)" " ±)
= nnnn[I +'I7tUI,h,h)]'1 (1
h i2 ;,
n
[I +'17fCil>/z,ia)j'1l.
j"j2J,
Hence
in agreement with Problem 3.12(f). (d) Given that in the relativistic case e2 = p 2C 2 +e5, where eo = moc2 show that 41Tg r~ (E 2 + 2Eeo) 'hdE 3 3 p = 3h c exp[(E+eo)/kT-,),] +'17 whereE == e-e o. (e) Show from n = v(op/oJ.l.h (Problem 2.8) that the mean number of particles in the above case is (n) _ 41Tvg r""(E+eo)(E2 +2Eeo)'hde - h 3c 3 exp[(E+eo)/kT-,),] +'17
Jo
Jo
132
kTlnZ
pv
= 'l7kTg L
LLln[ I +'17fCi I ,iz,ia)] .
i l h ia
(b) We can put
L~~ ... II
h
IJ
-+
f
oo
41TL3f""
~ ... dildi2dj3 =Jl3
... p 2 dp.
0
Applying this to Equation (S.1.2), with V = L 3 , we obtain
p
=
41T'I7kTg roo 1.3
Jo In[1 +'I7t(p)]p
1
dp
(S .1.2)
Chapter 5
134
5.1
t
5.2
Ideal relativistic classical and quantum gases
135
form suitable for substituting the L/s. In this way one finds (n >and pv. where t is regarded to be a function of p through its dependence on the The entropy is found, by using S v(op/oT)v,/.L (see Problem 2.8) from Also the expression for pv, t(p) = exp['Y €(p)/kTj.
jr'
TS = ~{r"(E2+2E€0),J,EeXP(-a+E/kT)dE Integrating partially one finds the stated relation. (c) We have pee) = (2mo€)¥. so that the pressure is kT [exp(E/kT-a)+llF 41rg roo (2mo)'/'€%d€
(E2 + 2EEo)'hexp (-a+ E/kT) dE} -akT '
p = 3h 3 exp(€/kT-'Y)+ll o [exp(E/kT-a)+llJ2 whence the result follows. = (€2 -€6)'A" so that (d) We have cp(€) B[ 4L4 + 13EoL3 + 10€ -akT(3L 3 +9€ 2 +6E6L 2 c 3 [p(€)]3 [(€ - moc 2 )( € + moc )]'h It is convenient to make a table of coefficients as follows: = E'h(E + 2€0)'/' = (E 2 + 2E€o)'h . This gives the result. L4 L3 L2 L1 (e) The stated result is found after a partial integration. TS 4 13€ IO€6 9akTEo -6akTEa S.2 Let B E'dE Lr== o'~o,~~ ,1;,r (r=0,1,2,3,4) l1(n> 0 311 9€ 6€ B where a (11- m oc 2)/kT, and B 41rvg/3h 3C 3. _pv (a) Establish -1 - 4E o -4E~ 0 B 3B[L3+3€oL2+2€6Ld, (n) _ (n)€o 0 - 3Eo - 9E6 -6€ 6 pv = B[L4+4€oL3+4€aL2] ' B
Jo
i
Jo
f
.
/r/L'7"
OO
_.\1._
Ld , B[4L4 +(I3€ 0 -3akT)L3 +(1 O€o -9akT)€oL2 -6akT€6 (11 ~€o)(n)+ TS-pv 3 9€o B L 2] ' 2 €a V = 3B[L4+3€oL3+
TS
6E6
0
where V is the internal energy excluding the energy due to the rest mass The last line gives V/B. This shows that the relativistic theory leads to a of the particles. renormaIisation of the chemical potential from 11 to 11' :::: 11- moc 2 • (b) Verify that Alternatively one can keep the 11 of non-relativistic theory and renormal V-TS+pv = (11-€o)(n) ,
ise the internal energy from V to 0' = V +€ by the energy due to and discuss this result.
the rest mass of the particles. (c) Verify that this system is not an ideal quantum gas according to the (c) Neither the relation pv gV nor the relation pv gO' is satisfied definition in Problem 1.9(e). with a constant g. Cd) A system is said to be ultrabaric if its energy (including its rest (d) A calculation of V+Eo(n)-pv yields the following sum of positive mass) per unit volume is exceeded by its pressure. Show that this does terms: not apply here. B(2L 4 +8E OL 3+ 1 IEaL2+6E6Lt). [If interactions are taken into account then a system may become ultrabaric. (1) 1 5.3 Using equations of the preceding problem, establish the following' results: Solution (a) In the non-relativistic limit, U == moc 2/kT;p I, (a) and (b) If one multiplies the integrand in the expression for (n) in 2 Problem S.l(e) by (E +2Eo)'l'!(E2 + 2Eo)¥. one has the denominator in a kT )1':1 Lr "'" (kT)r ( 2moc2 f'(r+,)I(a,r-,,±). (1) S.A.Bludman and M.Ruderman,Phys.Rev., 170,1176 (1968).
5.3
Chapter 5
136
In the extreme relativistic limit, u
« I,
Lr "'" (kTyr(r)l(a, r-l, ±) .
Discuss the meaning of these approximations. (b) Use these approximations to establish the results given in the fol lowing table. u~1
(n)
u
2rrm okT)% vg ( h 2 l(a,!,±)
kT)3 8rrvg ( hc l(a,2,±)
2rrm okT )'Iz ,vg ( 1.2 kTI(a,,, ±)
kT)3 24rrvgkT ( hc l(a, 3, ±)
~u
pv
u
u«l
1U
(c) Discuss the approximation e'" 1, and show that it leads to
« 1 when made in conjunction with
(
5.4
Ideal relativistic classical and quantum gases
(c) The approximation implies 'non-degeneracy' as used in connection with gases in statistical mechanics. It means that moc 2 -p. ~ kT or u - p./ kT ~ I. Since u ~ I is given, the restriction on p. is that it can be negative, but if it is positive it must satisfy p./kT « u. 5.4 A four-vector (cb, a) in an inertial frame I and the corresponding one (cb', a') in an inertial frame I' are related by
(5.4.1)
a' ;:;:: f3(a+v' b),
V'bl=f3(v'b+~:a)
, (5.4.2)
v x b'
V
x b,
where f3 = (l -v 2/c 2 rv>, v is the velocity of the origin of I in I', and c is the velocity of light. (a) Prove the reciprocal relations
~
a = f3(a ' -v • b') ,
kT )V> Lr ;:;:: (kT)r ( 2moc2 r(r+!) e '" . [The results of Problem 3.12 for an ideal quantum gas have here been recovered with stand s = 2 respectively. The extreme relativistic thermodynamic properties of a Fermi gas are seen to be rather similar to the thermodynamic properties of black-body radiation. An important difference is, however, that a = 0 in the latter case. The theory can be taken further by obtaining expressions for other quantities. (2)]
137
(5.4.3)
2
v'b
I v ') . f3v'b-c2a (
(5.4.4)
(b) An inertial frame 10 can be defined by specifying that the vector b shall be zero, Le. b o O. If the origin of 10 has velocity w in I' and the corresponding value of f3 is denoted by f3w, prove that
a' ;:;:: f3w ao , b'
Solution
a'
w,
(5.4.5) (5.4.6)
The integral is L = kT r -
(
)
r
f
~
0 (x 2
xr dx +2xu)V,[exp(x-a)+17]
The square root in the denominator is (2xurV> or X-I in the two approxi mations. The approximation u == moc2/kT ~ I implies relatively low tempera tures, or relatively heavy particles, or both. The thermal velocities of these particles will therefore be small enough for a non-relativistic treat ment to be valid. In the limit mo -+ 0, however, any temperature above absolute zero is capable of imparting high thermal velocities to the par ticles, and such a situation may be called 'extreme relativistic'. (b) The results stated are simple algebraic consequences of part (a) and the results of Problem 5.2. (2) P.T.Landsberg and J.Dunning-Davies, in Proceedings of the International Symposium on Statistical Mechanics and Thermodynamics (Ed. J.Meixner) (North-Holland, Amsterdam), 1965.
a' = ; : +w • b ' .
(5.4.7)
If I~ be a frame with velocity w+ dw in I so that (a, b) in I corresponds to + b o+ db o) in I~ then, assuming an equation of the type (5.4.7) to apply to I and I~, establish
(a o dao,
-aod(L)
b ' • dw
(5.4.8)
and hence that I da o I da = -+w· db f3w
(5.4.9)
[A physical application of these results is given in the following problem.] Solution
(a) From Equation (5.4.1) substitute for V' b in Equation (5.4.2). Similarly from Equation (5.4.2) substitute for a in Equation (5.4.1).
138
5.4
Chapter 5
(b) Equations (5.4.5) and (5.4.6) follow immediately from Equations (5.4.1) and (5.4.2). Equation (5.4.7) is then a useful consequence of these results. (c) Use the relation
-d(JJ~) = ~~
W'
(5.4.10)
dw .
f
[These considerations, basic to relativistic thermodynamics, have been recently a subject of controversy. (3) Note that the transformation of dQ is independent of the value of A.] Solution
(a) From Equation (5.4.7) it follows that
Now note that
,
E'+'Ap'v'
a' b' • dw = c2 W • dw
JJwao
=7
( 1) w • dw = -ao d JJw '
which is Equation (5.4.8). Now differentiate Equation (5.4.7) to find da' =
~o+aod(L) +b' • dw+w' db' .
The second and third terms cancel as a consequence of Equation (5.4.8), yielding Equation (5.4.9). 5.5 The quantity a = E+ APV and linear momentum b = P of a system transform like a four-vector (cb, a) of Problem 5.4. The suffix '0' refers in this problem to the inertial frame in which the centre of mass of the system is initially at rest, JJ := (1 -w 2/c 2 and w is the initial velocity of the centre of mass in an inertial frame I. For a fluid in a container A = I, and A = 0 for a solid, a box, or a fluid with the energy and momentum (but not the rest mass) due to the stresses in the container included with the system. Assume v = vo/JJ, p = Po for the transforma tion of volume and pressure. (a) Establish the result E-w' P = Eo JJ (b) Establish the relation
r'h,
,
dE-w'dP+Apodv+
(1 - A)Po
JJ
I
~
dvo=l3(dEo+Podvo).
(c) Justify physically the following expression for the compressional work done on the system:
dWc = -Po
-(l-A)vOd(~ )]
( I-A) .
= -Po Adv+-JJ-dvo
(d) Justify the expression for the translational work dWtr W' dP if the non-relativistic definitions of force and work are adopted. (e) From the first law in the form dQ = dE-dW = dE-dWc -dWtr show that the heat transforms as dQ = dQo/JJ.
139
Ideal relativistic classical and quantum gases
5.5
EO+JJ~ovo+w'
P'.
Hence the transformation of P and v makes the terms involving A cancel out. On changing the notation to the present problem, this yields E Eo/JJw +w • P. (b) We use Equation (5.4.9) and then evaluate da' -da o/(3 = W' db': 1 AVo W' dP = dE+APodv+Avdpo-j3dEo dVo-Tdpo
= dE ,
1 , APO JJ dEo+APodv-Tdvo I
P
= dE-j3dEo -{fdv o +A Po dv+
(1 -A)pO
t3
dv o
as required. (c) An increment of volume transforms as dv
= idvo+vod(i)
so that there are two causes of compressional work in frame I for A = 1. One is compression of the system; the other is due to Lorentz contrac tion of the volume when it suffers acceleration. As the system is 'bare' and the walls are not included, the only way of producing an accelera tion is by regarding the pressures on the system as doing the work. If A 0 the system behaves like an accelerated box and compressional work arises only if dv o =1= O. This work is -Podvo in 10 , but -Podvo/JJ in I. So we must combine -Podv for A = I with -Podvo/JJ for A = O. dWtr = f· ds
dP dt . ds = w . dP .
(e) The right-hand side of part (b) is I
I,
I
j3(dEo+ Podv o) = j3(dEo -dWo) = j3d Qo . The left-hand side is dE-dWtr-dWc = dQ. (3) P.T.Landsberg and K.A.lohns, J.Phys.Soc.Jap., 26 Supplement, 310-312 (1969) andAnnals of Physics. 56, 299 (1970). P.T.Landsberg, Essays in PhYSics, 2, 93 (1970) and papers in A Critical Review of Thermodynamics, Eds. E.B.Stuart, B.Gal-Or, and A.I.Brainard (Mono Press, Baltimore), 1970.
t 6 Non-electrolyte liquids and solutions A.J.B.CRUICKSHANK (University ofBristol. Bristol)
6.0 The condensed fluid is generally discussed in terms of so-called configurational thermodynamic properties, this designation implying the parts of the corresponding total properties which may be attributed to the inter-relations between the molecules. Unfortunately, the conven tional definitions (I) of the configurational properties give rise to difficulties which are inimical to the purposes of this book. For indistinguishable molecules possessing neither rotational nor vibrational degrees of freedom, the canonical partition function takes the form stated for proof in Problem 3.S(c), namely, _ (2rrmkT)3N/2QN ZN h2 N! '
where QN is the configurational integral defined by QN=
f..Jexp(-~;)drl ... drN; v
here the potential energy M of Problem 3.S(c) is identified as the configurational intrinsic energy U*, and the r i are three-dimensional position vectors. ZN is dimensionless, and it may be used to study solutions either as it stands, or by factorizing it in either of two ways: 2rrmkT )3N/2U N QN _ (a) ZN = ( h2 N! uN = ZtrsZ * ;
=
ZN
(b)
(
2rrmkT)3N/2Q , ; ; == h2
ZmolZconf'
The first factors represent the translational or molecular part, and the second factors the configurational part. The Helmholtz free energy F (= -kTlnZN, cf Problem 2.4) can accordingly be exhibited as a sum in either of two ways:
+ F* ,
(a) F
=
(b) F
= Fmol + Fconf
~rs
Ftrs == ,
-kTlnZtrs ,
Fmol == -kTlnZmol ,
F* == -kTlnZ* ; Fconf == -kTlnZconf'
6.0
Non-electrolyte liquids and solutions
141
If necessary we may define, in addition, rotational and vibrational Helmholtz free energies. The corresponding partition functions may be assumed independent of molecular environment in the fluid state, at least in the absence of hydrogen bonding etc., so they include no volume dependent terms, and do not contribute to the change in the free energy on mixing. The procedure usual in solution theory is based on (b), the justification being that (2rrmkT/h 2 )3N/2 is independent of volume, and so cannot change in an isothermal mixing process; its dl~merit is that the factors of ZN are not dimensionless, but have the dimensions of [urN and [vjN respectively. The corresponding 'free energies' are, however, individually extensive, cf solution to Problem 3.4(c). The consequent logical difficulties can be resolved by using molar quantities and referring volumes to unit molar volume, but since the procedure based on (a) avoids these difficulties altogether, it will be adopted here. The two procedures necessarily give equivalent results for the thermodynamic functions of mixing, as does using ZN itself, see solution to Problem 6.11. To describe liquid-vapour equilibria, i.e. the possibility of two phases of different densities at the same p, T, an equation of state t(p, u, T) = 0 must have three real roots in u over some region of P and T (see Chapter 7). It is consequently convenient to use T, u rather than T, p as independent variables, and we therefore separate p (rather than u) into configurational and translational parts. Thus we define
_ (aF* ----a;- )
p* = -
T'
_ (aF au
Ptrs = -
trs )
T'
respectively, as the configurational or internal (2) pressure and the translational (3) or kinetic pressure. The latter is the quantity described by the kinetic theory of gases, i.e. Ptrs == NkT/u. F* may be evaluated for the ideal gas by putting U* = 0 (the necessary and sufficient condition to define the ideal gas) in the configurational integral, giving QN = v N , whence Z* = I. Thus the configurational Helmholtz free energy for the ideal gas, F.d = 0, whence ptd == -( aF.dl aU)T = 0, and the total pressure for the ideal gas is Pid ==Ptrs+pTd = NkT/u, see Problem 6.I(a). It follows that all the configurational thermodynamic potentials for the ideal gas are zero; since 'configurational' is always taken to mean 'pertaining to or resulting from the forces between molecules', this conclusion has at least the virtue of semantic consistency. For the general case of the non-ideal fluid, F* may be evaluated (i) by proceeding in the same general way as above, but using a molecular model to specify U* in the configuratio'nal integral; or (ii) for a specific
(I) R.H.Fowler and E.A.Guggenheim, Statistical Thermodynamics (Cambridge University Press,
(2) J.O.Hirschfelder, C.F.Curtiss, and R.B.Bird, Molecular Theory of Gases and Liquids (Wiley, New York), 1954, Chapter 4, Section 2, p.255.
Cambridge), 1952, pp.700-701. J .S.Rowlinson, Liquids and Liquid Mixtures, 2nd Edn. (Butterworths, London), 1969, pp.250-252.
(3) J.O.Hirschfelder, C.F.Curtiss, and R.B.Bird, Molecular Theory of Gases and Liquids (Wiley, New York), 1954, Chapter 2, Section 6, p.116.
140
II
6.0
Chapter 6
142
(
f:
P = NkT,
+
pr v ;
Ut
pt dv .
F; + TSt .
F*
The configurational properties X; as defined above are identical to the 'residual' properties (4) obtained by comparing the volume derivative of the total property to that for the ideal gas over the range 00 to v, id ) ] Vr(ax~) X*==I -av T -(aX dv, ~ av T as may be seen on subtracting (aXtrs/aV)T from both terms of the integrand, since (aXid/aV)T is zero. For example, in the case of G F+pv,
p
( aG)T = (aF) & T + P + v (aav ) T and
p
v (aav
f
(a
pd v (apt) Gt = '"vrIV (apav~ ) T V a;J. ) ] dv "" v Tv dv, implying Gt = + pt v. In this argument, however, it is assumed that (aG trs! av h v(aptrs /al)T, i.e. that Gtrs == Firs + PtrsV, which is equivalent to defining G* == F* + p*v at the outset.
f
CELL THEORIES OF THE LIQUID STATE
"
= NkT-3NkTln[ 1-11(v:
f'J,
(6.1.1)
where 11 == (-./211'/6) '/', and Va is the volume of the system at maximum Na 3/..J2. density (close packing) Va (c) Derive the equation of state, P = p(v, T,N) corresponding to the configurational Helmholtz free energy expression in part (b) and prove that the isothermal bulk modulus B T -v(ap/aV)T is I (Va) B T = P [ I +}11 -;;-
pv ] NkT .
'Il
Solution
(a) Assume each cell to be of average size v/N. Consider first a particular arrangement, X, of the molecules among the cells, and evaluate the contribution Q~) of this arrangement to QN' The N independently ranging vectors r l , ... , rN are weighted by exp(-U*/kT), and this is zero whenever rj is outside the cell to which molecule i has been assigned. Thus Q~) breaks up into separate integrals each over a cell of volume v/N. Then
Q~)
= (
dr l )
(LINdr2)..·(LINdrN)
=(Nf
There are N! such arrangements avoiding multiple occupancy, so
6.1 Simple cell theory. The cell theories all assume that each molecule is, during most of the time, confined to a cell whose boundaries are determined by the potential due to the neighbouring molecules. For a 00 whenever a molecule overlaps hard-sphere fluid, this implies that U* its cell boundary, and U* = 0 otherwise. Suppose the cell boundary to be defined by planes which perpendicularly bisect the lines joining the centre of the molecule to the centres of its neighbours when all are at rest at the centres of their cells, so that the cells completely fill the volume without overlap. (a) If the volume of each molecule is negligibly small, and the mole cules are identical, prove that the configurational Helmholtz free energy, P
N!
and compare the model to the ideal gas. Remember that there are ways of arranging N particles among N cells. (b) Assuming that the polyhedral cell may be approximated by the sphere of the same volume, Le. of radius b, ~1I'b3 == v/N, and that the molecule, diameter a, cannot approach within ~a of the boundary, prove that
Since all equations of state necessarily approach ideality as v -+ 00, it follows that F{(oo) = Fj~(oo) O. The other configurational thermo dynamic. potentials are then defined as, for example,
Gt
143
is, for this model,
equation of state, Hp, v, T) = 0 which yields a total pressure p~(v, T), by integrating the corresponding internal pressure, pt == p~ - NkT/v, with respect to v, as
F{(v) - F;(oo) =
Non·electrolyte liquids and solutions
6.1
QN
= -kTln vN ,
(4) l.S.Rowlinson, Liquids and Liquid Mixtures, 2nd Edn. (Butterworth., London), 1969, pp.57-58.
QN =
~ Q~) = N!( ~
f'
whence QN N! vN =
using Stirling's approximation, QN v
-N
-W = e
,
which leads directly to the required result. In the case of the ideal gas, the total volume is accessible to every molecule, i.e. exp(-U* /kT) is unity for all values of each rj' Then QN VN, and Z* QN/V N = 1, so P = U* = O. The difference in P between this cell model and the ideal gas corresponds to S* = -Nk in
11
6.1
Chapter 6
144
the former case. This is the so-called communal entropy (5), which would be gained by the assembly in the cell model if the molecules were allowed freely to interchange between cells, i.e. if multiple occupancy were allowed. (b) Since U* = 00 whenever a molecule overlaps its cell boundary, its centre cannot approach within! 0 of the boundary. The average volume accessible to the centre of the molecule-the so-called free volume per molecule-vf , is Vf
3
Now j1Tb == vIN. and Vf
0
3
=
so
= ~[V IIl_ (j1T)
The
1 N -1'IVa IV
1'1 (
= -(v'
gives
of
QN =N! whence
=
r
N{ ~ [I
1'1(
v: )
V
v: yilT
3N
QN = NW-N [1- (Va) IhJ3N v 1'1 N
a
V
(Va) J-\ . NkT{ (I) )1/,[ I 1'1 (I)/; )lhJ-l + I }, P = P* + Ptrs = -v- 1'1 /; P*
and
V
= NkT -1'1 (Va) V V
13
III [
1-1'1 -
V
1\ f ~I
6.2 Hirschfelder's cell theory for hard spheres assumes that the cell is bounded by the polyhedron whose apices are the centres of the neighbouring molecules when the latter are at rest at the centres of their own cells, mutually distant a, in close-packed array; it is further assumed that this cell may be approximated by the sphere of radius a, and that the centre of the molecule may not approach within 0 of the cell boundary (6). (a) Taking a3 ..j2(v/N), 0 3 ..j2(v a /N), as for close packing, prove that the configurational entropy is
-Nk+ 3Nkln [1
Compare this to the configurational entropy according to Problem 6.I(b). [Hint: it is necessary to take account of the fact that this way of defining the cells leads to a multiple overlap, so that each element of volume is counted more than once. J (b) Express the isobaric thermal expansivity and the isothermal (Xp
(5) J.O.Hirschfelder, C.F.Curtiss, and R.B.Bird, Molecular Theory of Gases and Liquids (Wiley, New York), 1954, pp.213-216.
I
== -;;-
(av) aT
Ky == -
p ,
~(:;)T
(cf Problem 1.4) as functions of (T, v), (p, v), respectively, and compare these to the corresponding properties derived from the equation of state obtained in Problem 6.l(c). Solution (a) Dr
= (j1T)(a - 0)3, Le. the volume over which the centre of the molecule can move. With
(v )Ih '
V)lh
a~ll_1'I(V: )
II -1'IC:) lhft} , IVa)
' 0 == V2 N ..j32 x ! 1T(V I/, - V:')3 .
a == V2 (N As in Problem 6.1,
= -P
145
and the given result follows. It is clear that for this model By approaches (from above) the ideal gas value in the limit of large v. As (v decreases towards unity, BT values lie in the range found experimentally for solids rather than in that for liquids.
'
To obtain the isothermal bulk modulus, differentiate both sides with respect to V at constant T:
and solutions
(V: ) I/,J
P{I+
I
PV = NkTll-1'I(V:) V'TI P+V(~)y = -NkT[I-1'I(v:)
-V(~)T
S*
(V )IIlJ-l 1-1'1 (-al'av*) -_-3NkT-3v1'1 (Va)lhl T
Non-electrolyte
'
and the given result follows. (c) From the result of part (b) we have
hence
Hence
j1T(b-!0)3.
==
6.2
Ntl r
QN N!vf,
=
NW-N(N;r)N
(6) R.J.Buehler, R.H.Wentorf, LO.Hirschfelder, and C.F.Curtiss,J.Chem.Phys.. 19,61 (1951).
146
6.2
Chapter 6
so
P
= NkT[ I-In(v
V:,)3+
\I,
Inv -In(v32 x J1T)]
.
To obtain the given result, we have to remove the term -NkTln(v32 x J1T) .
In close-packed array, the polyhedral cells defined by the twelve neighbouring centres clearly have a fourfold overlap. The actual figure here is 5 ,92; the difference arises from approximating the cells by spheres, increasing the overlap. Thus the normalized configurational free energy is
3In [I _ (
N kT{ I -
F*
v; ) },
and the required result follows directly from S* From Equation (6.1.1)
3In
S* = -Nk{ 1-
-(aF*/aT)v'
r 11 (v; )\/']} . I
The only difference between these equations is in the value of the coefficient -q, which increases from 0·905 to 1·0 as the amount of overlap between neighbouring cells is increased from zero. (b) It is simplest to consider the general equation of state for both models as p -_ NkT[ - - 1-11 v V or NkT
(Va )II>J-I
V
'I1V "1)';'
a
'I
= -
p
6.3
V
whence I
(Xp
(av )
== -;;- aT
Nk pv
p
QM = MM(e- a)M .
Use this expression to obtain QN for a K-tunnel model, where N == MK. For the derivation of QM see Problem 9.3. (b) For given 1)IN, only one of band e can vary independently, but the ratio ble can take any positive value consistent with b, e > a; thus QN must be maximized with respect to ble. Show that this maximization gives
v%
In Q
Similarly,
KT
NkT
-Vp2
i[I-11(v~)
-i11(v; f'J
=
Solution
increases.
(a) There are clearly N!/(M!)K ways of arranging N molecules into K sets of M; thus, by an argument exactly parallel to that of Problem 6.2(a) we have
-j11 (va)' hJ-I I)
![I_11(v;)'h][I_i11(v;) (Xp
rLI -v~ () Vl~ '('J
Va
rjl-i11(V; ) %]-1
It is clear that at given T and v, p increases and
3Nln
where, as in preceding problems, == Na3/v2. (c) Comment on the general features of the equations of state of Problems 6.1 to 6.3.
p '
and substituting for pv from the equation of state, (Xp
147
6.3 The tunnel theories for the hard sphere fluid replace the regularly packed polyhedral cells of the preceding problems by hexagonal prisms or tunnels, stacked parallel in two-dimensional array. The longitudinal motions of the spheres in neighbouring tunnels are taken to be mutually independent, the transverse motion of each sphere being limited only by the geometry of the tunnel. Barker's tunnel theory (7) is analogous to the cell theory of Problem 6.2: the tunnel cross-section is defined by the hexagon whose apices are the centres of neighbouring tunnels; it is assumed that the cross-section may be approximated by the circle of radius b, b being the centre-to-centre distance in the two-dimensional array, and that U* = 00 whenever the centre of a molecule approaches within a of its tunnel boundary. (a) Formulate the configurational integral for transverse motion in terms of a free area af per molecule. For motion parallel to the axis of the tunnel (longitudinal motion), the problem is that of M hard spheres, diameter a, arranged with their centres on a straight line (the tunnel axis) so that they lie always within a distance eM, where e, the mean distance per molecule, is greater than a. If the positions of two spheres, i, j, are defined by Xi' Xi' then U* = 00 whenever IXi - Xj I < u. The configura tional integral, QM, for one tunnel is
Differentiating with respect to V we have
(Va) 1/3 Nk(aT) 1- 2 -p av 3 11 -
Non-electrolyte liquids and solutions
'IlJI
QN(transverse) whence QN
and KT decrease as 11 (7)
= [afN(e-
I.A.Barker, Australian J.Chem.. 13, 187 (1960).
N!
af (M!)K (6.3.1 )
r
i
6.3
Chapter 6
148
j
(b) The area of the hexagon of side b is h/3 3 b 2 , so the volume per tunnel is h/3 3b 2 cM, and the volume of N/M tunnels is h/3 3 b 2cN; there is a threefold overlap, so v (6.3.2) N = h/3b 2 c == exb 2c .
6.4
and so, for the total pressure, PV
= [7rN(b -
The three equations of state of Problems 6.1-6.3 differ only in the value of 11, which is determined by the degree of overlap of neighbouring cells or tunnels. The index of (va Iv) is! because the restriction on molecular movement is expressed by a radial quantity, i.e. in dimensions oflength.
Substituting for c according to Equation (6.3.2) into Equation (6.3.3) we obtain N QN
a)2(N~b2
= [7rN(b
It is convenient to maximize lnQN rather than QN:
~_,,'_v b - a IV\Nexb 2
dlnQN) ( db v This is zero when
_a)-l
6.4 Smoothed-potential theory. Suppose that the molecules of Prob lem 6.2 interact according to a spherically symmetrical potential tjJ(r),
2v
tjJ(r) -+ 00 as r -+ 0 , '; ~,
j!
(7rN)N(b*-a)3N
Now
= L7rNb*3(1-:* YJN
so
a ( b* =
V2 {/3
(
Va )'h
V
{/~
and b*3
w = Wo,
v 1 Na.
whence
Q = Nln7r+Nlnv'3 2 + 3Nln Inv~
w
rI-V!: (v ) 'Il] '
and the required result follows on elimination of the term Nln(27r/v'3), which is due to incomplete normalization, cf.solution to Problem 6.2(a). Note that this model includes the communal entropy; this is because the summation of QM allows a molecule to be anywhere within its tunnel. (c) Differentiating In (QN/V N ) with respect to vat constant T, we find P*
NkT (v )'1> [ 1-11: (v) =-v-11:
t LtjJ(r;j) . i, i
The configurational intrinsic energy w(s) of a molecule in its cell obviously is negative when the molecule is at the cell centre, increasing toward +00 as the radial displacement from the cell centre approaches the nearest-neighbour distance a. The smoothed potential, or square well, theory approximates w(s) by
a- -N
Va) 'I, '1.j2
V
u*
_(VO)'/'
(~r'(±
b*
< 0 for a < r < 00
and that the configurational intrinsic energy is the sum of the interaction potentials of all pairs of molecules without extra contributions from groups of more than two, i.e.
or Ncxb alv = a; cxb v/N. But v/N exb c, so the extremum of InQN is at b = c. The second derivative confirms that this is a maximum. Then QN
tjJ(r)
'~
2
3
3
tjJ(r) -+ 0 as r -+ 00 ,
~
3 Nexb a v
b-a=b
(v: f'Tl
13 I ()' NkTll-11 1): J-I
pv
(6.3.3)
a)2(c-
-{It
=
149
This is clearly a member of the class
Given af = 7r(b - a)2, we have QN
Non-electrolyte liquids and solutions
1\
=
00,
< s < (a- a) (a - a) < s , 0
;
where Wo = Wo (a) is the configurational intrinsic energy of a molecule at rest at its cell centre, and a is defined by tjJ(a) == O. (a) Let z be the coordination number of the cell lattice, i.e. the number of neighbours of a particular molecule with which it is in contact when a = a. Then z has the maximum value 12 for face-centred cubic and hexagonal close-packed lattices. If tjJ(r)
4e*
l( 7y2 (7 rl
where -e* is the minimum value of the pair potential corresponding to
\
150
6.4
Chapter 6
6.4
Non-electrolyte liquids and solutions
151
r = r* = !<j2a, prove that if only nearest neighbour interactions are counted, either
when the numerical term h/3211", due to overlap, is dropped. Differen tiating with respect to v at constant T, and adding Ptrs gives, cf.Problem 6.3(c), U* = 2zN€* (v ~ (V4 V2) P NkT[ V 1+ 2zN€ 4 v~ - 2 vU3 , or U* +00, whence the required result follows. according to whether or not any molecule has Sj > (a - a); Vu has the (c) Putting P = 0 in this equation of state gives same meaning as in the preceding problems. (b) Derive the explicit form of QN = QN(v, T) for this theory, and [1_(V~ = 4z€ _2(V~) thence prove that the equation of state is
r
[(V; _(V;)
)';'J-l
l
pv
V NkTll -( ;
Y13Tl
(vu/v) )'/,J-I + 4zN€*2 [- (V)4 ; -(I); )2J .
Settingy =
p
Sj«a
a),
r(~r2 -(~rJ
whence
[(V;
!zrp(a), with
Wo
= 4€* [( v;
r (V;) 2J '
t:,
r-( rJ .
(b) Since the cell is bounded by an infinite potential barrier at a - a, the free volume is, as in Problem 6.2(a), Vf = ~1I"(a - a)3, with a V2(v/N) '13, a == V2(v u /N) 'h. Then 4211"( 'il '13)3 3N V -Vu
JI
.
(NWo) '
QN = N!(Vf) N exp - kT
'Ii
v
a
IJ
Sj
In Q v~ = -N+ 3Nln [
+
I I I
+
= 2z€* v;
and
+
i=I,2, ...,N,
pair-interaction distances are rij = a. Then
Wo
kT
4z€
2y2)(l- Y
c
In the first case, corresponding
U* has the same value as when every molecule is at its cell centre, Le. all
rp(a) = 4€*
4J .
+
Solution (a) It follows from the postulates of the model that U* has only two
= Nwo or U* = 00.
r
transforms this into y2(l
(c) Prove that equation of state of part (b) predicts 'condensation', i.e. the possibility of co-existence of stable high and low density phases at the same pressure and temperature. Note that it is a sufficient, though not a necessary, condition for two-phase equilibrium that the equation of state has two real positive roots in IJ for p = 0 over some range of T.
values open to it, U* to
[(V;
+
Figure 6.4.1. Comparison of van der Waals and Dieterici equationfl of state for two substances having similar vapour pressure (indicated by the broken lines): a, van der Waals; b, Dieterici; c, ideal gas. The vapour pressure has been located by the equal area rule, see Problem 7.5(b), and Figure 11.14.1.
There is clearly a range of T, 0 < T < 1", for which there are two real roots, 0 < y < 1/v'2, while for T> 1" there are no real roots. Note that the van der Waals equation, see Problem l.ll(a), also shows this behaviour; the Dieterici equation of state,
RT exp (a) P v-b -RTv '
(V) Ill] ----;:r2zN€* [(V)4 has no real roots in v for P 0, but nevertheless predicts condensation. 1-; ; - (v; ) The difference is illustrated by Figure 6.4.1.
I
152
6.5
Chapter 6
6.5 Hole theory. Instead of accounting for variation of total volume by variation of the cell radius, we consider an assembly of fixed-volume cells which may be either occupied or vacant. With N molecules distributed among (N + No) cells in close-packing, the total volume cf Problem 6.1, (N+ NO)a3 N+ No . _ Na 3 v ..)2 = Va ----y;r- , wIth Va = ..)2 where, as in the preceding problems, a is the parameter of the cell lattice. It is assumed that only nearest-neighbour attractions are significant, and
that the configurational energy per cell has the value corresponding to the molecule being at the cell centre [i.e. !(a) per neighbour] every where within a free volume Vf, and the value +00 outside Vf. (a) Show that for this theory the general form of the configurational integral is Nzx(a)] exp QN vdi, 2kT '
~ [tUI
/\ f
6.5
(b) In the expression for QN, put Vr (i, X) = Vr, and take the average coordination number for every configuration to be that for all configura z zN/(N+ No). Then tions, Zx
I\1
l
Vr
QN = -In--Nv Nv (NV Vr zNva(a) Inr;r - - N ) In (NV) - - N +Nln--N, V Va Va Va Va V 2kTv
r
V V (V = NkT .va .-In-Va Va
P
* = NkTlln_V_ Va V-v a
= NkTf_ln Va
_
and the
result follows.
(I
V
- 1] .
z(a) (Va) 2kT V
Va) + z(a) (Va)2 _ NkTJ. V 2kT V V
"'" NkT(1 _ Va 2v
)-1 + Nz(a)V -2-:
Recall that (a) is necessarily negative for a pair potential like that of Problem 6.4. The corresponding form of the van der Waals equation is
Solution
f-
NVf In (V-Va) - - + In-Va V
Adding Ptrs gives the required result. Expand the logarithmic term in the equation of state, retaining the second power in (va/v), since this occurs in the second main term; the result reduces to 1+ Va)+ Nz(a)Va pv 2v 2 V
where z is the coordination number of the cell lattice. (c) Establish the approximate form of this equation of state appro priate to V IVa ~ 1, and compare it to the van der Waals equation, see Problem 1.11 (a), and to the equation of state of Problem 6.4(b). (d) Show that the equation of state of part (b) predicts a two-phase region; the argument of Problem 6.4(c) is not appropriate because of the logarithmic term; use instead the condition (ap/av h = 0, when two real roots ensure phase separation.
m:
.
Differentiating with respect to V at constant T, we find
2kT V
(a) The assumption of uniform potential over each Vr (i, X) for the ith molecule in the configuration X allows the configurational energy = zx (a)N/2kT to be taken outside the integral. As in the solution to Problem 6.1 (a), each integration dr j yields Vf(i), within the configura tion X. Then (X) _ Nz x(a)] flN . QN - exp 2kT i = 1 Vf (I, X) ,
J~ I
r
ZN2 (a) eXPL - 2(N + No)kT
The number of arrangements of N molecules among N+ No cells is clearly (N+ No)!/(No!), so, since N+ N_ = _ (Nv/v a )! N QN - (Nv/v a N)!vr exp Using Stirling's approximation in the logarithmic form and simplifying, we obtain
QN kTlnr;r V
NkT[ z(a)(Va)2] p == - -In ( 1 - Va) - +- V
N
QN
where zx is the average number of occupied cells neighbouring an occupied cell in the configuration X (a member of the set appropriate to N,N+No)· Assuming that z x may be replaced by its average over all configura that Vr (i, X) may be treated likewise, and that the average, Vr, so obtained is independent of overall density, i.e. is a function only of the lattice parameter a, confirm that the equation of state is
Va
153
Non·electrolyte liquids and solutions
pv
'\
b
RT ( 1--;
)-1 -~.
Thus the statistical theory produces a result formally the same as that obtained by empirical modification of the ideal gas law. If we compare this equation of state with that obtained in Problem 6.4(b), we find that the salient points are that the first term in the former suggests a second class of equations of state similar to those of the cell theories, but having the index of Va /v equal to unity rather than
Chapter 6
154
6.5
In fact, one may construct models for which this index is 1, so that defining Vr by a radius gives (va/v)v\ by a cross-section gives (va/v)';', and by a volume gives (Va/v)l. In the low density limit the cohesive energy term from the equation of state of Problem 6.4(b) takes the form -a/v 2 instead of -a/vas in the hole theory. Thus a family of equations of state, having the same sort of 'theoretical' justification as the van der Waals equation, may be set up; cf Problems 9.7,9.17, and 11.l2 to ILlS. (d) Differentiating the equation of state of part (b) with respect to volume, we obtain ap = NkT[ZIj>(a)v~ _ Va ] av Va kT v 3 v 2 -v a v .
II
1.
This is zero when 2
V
+
What are the corresponding extensive quantities? (b) Problem I.S(a) states for proof
cp -Cv where I
t;
KT == -
!(:;)
T •
C;
C:
Note that while is defined by (au* /aT)v, c; identity given in Problem 1.1 (a), a p ) (al)) (aT) ( av T aT p ap v = - I
(aH* /aT)p' The
is useful in avoiding the difficulties due to p being treated as the dependent variable.
EQUATION OF STATE TREATMENT OF LIQUIDS
Solution
(a) It is simplest first to establish F* , and then to use G* == F* + p*v, U* = F* + TS*, etc., for the others; but all four may be established independently. From the equation of state,
6.6 The van der Waals liquid (i). The van der Waals equation of state is usually written for one mole, as (P+;2)CV-b) = RT
p*(vdW)
with R == Nok, No being the Avogadro number, and v the molar volume, v == Nov/N. It is the simplest explicit equation of state capable of describing qualitatively the observed phase behaviour of fluids and their mixtures (8). (a) Use the methods suggested in the Introduction to this chapter to prove that for one mole of fluid obeying this equation of state,
(v Ib -ifI) -v V( _ I -=I)dV-a v-b v v a
RT
and F*(vdW) = -RT
2
f
co
since F*CV
= (0)
=
~ =
O. Hence
F*(vdW)
.
V RTln=--b v-
a
v
v' 2a
_V
Since S* == -( a F* / a T)v, we have V +RT_- U*(vdW) S*(vdW) = -Rln=--' v-b v-b b
H*
=
F*
Similarly = -=V + RTln=- v-b
a
v
II
a
= (F*+TS*)vdW=-=v
v a b a G*(vdW) = RTln=-b -=+ v v RT=--b-= v v b v G* = -V + RT=--b v - + RTln =--b v a b a H*(vdW) = -=+RT=--b-= v vv . R.L.Scott and P.H.van Konynenberg, Disc.Faraday Soc., 49, 87 (1970).
2a
(8)
KT '
Thus (Cp - C v ) is specified completely by the equation of state and derivatives. Use a simple adaptation of the method of part (a) to prove that for one mole of the van der Waals fluid 2aR(v-b)2 (C; - C~) = -v~3R-T---"-2a-CV =- bP
This has two real roots for T < -zlj>(a)/4k, otherwise two imaginary roots; the two real roots are coincident for T = -zlj>(a)/4k. The locus of the real roots is the boundary of the region within which a single phase is unstable, and -zlj>(a)/4k defines the 'critical temperature' above which phase instability does not occur.
U*
Tva;
~(:~
ap
II
vazlj>(a) v~zlj>(a) _ kT V + kT - 0 .
a
155
Non-electrolyte liquids and solutions
6.6
156
6.6
Chapter 6
1\
-p+
T(:i)
(~~t v(~~)T' =
v '
so
we have
I V[-P{+TC:j)JdV.
U*
I
For the van der Waals fluid, p
-p*+T (-a
*)
aT
v
*
Gt For the van der Waals fluid
G*(vdW)
=-R
The expression for identities,
v
a
v
f f
D[ V I RT = (v-b)2dv
(aH/av)T
,~
thence, using the identity
1b V+(V-b)2 I bJ dv
= - v
is obtained from the partial differential
(~~)T = [,J-TGi)J (~)T;
..
2a v b -=- + R Tln=--b v - + R T=--b v-
v
(~~t (~:t(~)T; (~:)T = (~:)s + (~~)p G~t which give
2a
-
_ J.. _ap* v Ov dv .
2a RT -v-
and U*(vdW) =
157
Non-electrolyte liquids and solutions
From the solution to Problem 1.24,
Alternatively, for U*, since Problem 1.9(a) gives
(~~)T
6.6
I
,
The corresponding extensive properties may be discussed in terms of G* , since it includes all three types of term: 2a/v 2an/v; the correspond ing extensive term must be linear in n, i.e. v b nb In=-- = RT--- = RT- v -b v- b v nb Thus 2an 2 v nb G*(vdW) = + nRTln-+ nRT- v v-n b v-nb
(b) The quantity f:(~~) do used in the argument of the introduc (~t(:it(~~)v 1, p) tion gives precisely the volume-dependent part of X. If, as with aH) (ap) (a -v +T ( av (C C there is no other part, the definition av aT
I
T -
and so
H{ =
T
f:[v(a:J)T +
v '
T(W)J dV.
f [(ax) -(aXav au =
id )
T
] dlJ
T
reduces to (Cp
-
C v );
(Cp
-
Cv )
Now
=
(Cp
vTa:~
Cp)t - (Cp
~ = -T
-
Cv)id .
(a/))2( aavp ) aT p
T'
but this involves differentiation at constant pressure, as with Cp itself. This is avoided bv substituting for (al)/aT)p according to
TC:;)v = 2a
(!~)p
(!i))( ~~)T : yielding 2a b -v + RT=--b C -T(!i):; (a:v) v b
= v2 - R T-(v---b-)-::"2 ,
H*(vdW)
v ),
t
ve:;)T = ~-RT[(v~b)2-~ J, ve:;)T+ RT[(V~b)2 -~J+R~V~b -~) and
p -
X*
For the van der Waals fluid,
2a
T
p -
'I
=
T .
158
)\
Chapter 6
For the van der Waals fluid, R
(:~)v
( oP)
v-b
aV
6.6
Cv
RT
T
2aJ-I
R2T [RT rv-b)2-,J3
= rv-b)2
For the ideal gas, yielding
.2a
+ -3 V
Y(r) = I
=R
[1-(#)r]'I2.
,
(b) Apply this expression for G*/RT to two van der Waals liquids, 0, I, at the same reduced temperature r, and thence prove
[2a rv -b)2]-1 1---=v"'"3R=-T=
Gj(T)
= flG~ T
,
where
(:~)v
(oP)
R
ov
v
(Cp -Cv )
=
h == Vel ~ I VcO
'T'
RT
fl
_1£!
=
T
'T'
leO
(c) Show that the analogous relations for the configurational enthalpy and the molar volume (at p = 0) are
R
and
(C;
159
where = _
yielding Cp
Non·electrolyte liquids and solutions
6.7
C:)(vdW)
LI -
R
Hi(T)
2a(v- b)2]-1
v 3 RT -R
[o(a1v)] - ---aT
* C (vdW) v
v
Solution
(a) When p = 0, v/(v - b) a/RTv. This equation is clearly quadratic in v, having two real positive roots for T less than some determinate value: the smaller represents the stable phase, for which (ap/av h > O. Substituting into the expression for G* given in Problem 6.6(a), for v/(v b)-since a/RTv is the simpler function-we obtain
0.
[Note: The methods explored in this problem are applicable to any explicit equation of state Hp, v, T)
G*(v T) /,
6.7 The van der Waals liquid (ii). The 'liquid' phase of the van der Waals fluid at a temperature below the boiling point may conveniently be specified by the condition p = 0, taking the smaller of the two positive roots in v. In this problem, since we shall deal in molar quantities throughout, we shall omit the bar signifying this. (a) Use the equation given in Problem 6.6(a) to express G*, the molar configurational Gibbs free energy for the van der Waals liquid at p = 0 as a function of T, the liquid molar volume VI> and the parameter a. Thence use the reduced variables
r
T
8a 27Rb'
Vc
is the smaller root of the quadratic:
[I (I _4RTb) a
On transforming into reduced variables, this becomes
a
VI
3b
[( I
)l J I - 8a/27Rb h
#T
or 9 16r
Ve
cf> = - [ I Vc
for the
or
a v/RT
3b,
(I
32 r)'I2]
'ft,
2 l-(l-#r)'iz
2
= Y(r)
Considering the other terms in G*(VI> T) we have
to show that for the van der Waals liquid G* RT=
v/
a ) = RT(-RTv/ -2a - + a -I +In- RTIJ/
v = _a / 2RT
cf> = -
where the critical critical molar volume van der Waals fluid. cf.Problem 1.11 (b), are
Te
where
v
Te ' temperature, Te, and
(f).
hIV/o(~).
_ [ 2a rv-b)2][ _ 2a rv-b)2JI - R v 3 RT I v 3 RT It is obvious that
flHti
2
a In-v/RT
InY(r)+ln2-1,
t
In2-1n[I-(l-~r)'I:t] == In2
InY(r).
160
Chapter 6
6.7
)
Then, substitution into the expression for G*(vj, T) gives the required result. (b) At temperature T, for substance I
GT
RT
2 yeT) -In Y(T)+ In2
r
f l G6
Y(T)-lnY(T)+ln2-1
, 'J
2a RTb H* = --+-v v-b
0, this reduces to H* RT
= 4e*
RT'
and the required result follows. From Problem 6.6(a), f0r the van der Waals liquid 2a RTv = --+ -RT v v-b
III
IJIRT
a
2
1)/RT
yeT)
f
\ I
*
2RT yeT) - RT ,
U*Cl\)
I
I~
and
m; ( IIT) = 2RT yeT)
f1
RT = H'(T) .
Similarly, since
= 1\ yeT)
we have VII(T) = v CI
vlO(~) = hIVIO(~)
VeO
9 Y(T), 16T
hi
=!:'..s.! VeO
(aaol)3
vlI(T).
! i,2.(rij) , j
cf Problem 6.4.] (b) Show that, at p = 0, the result of part (a) implies, for one mole, N=No, F,(T,No ) =
fIF6(~' No);
Gj(T, No)
f1G'tJ(
f,
No) ;
just as when the comparison is made in terms of a specific equation of state, cf Problem 6. 7(b), , . (c) Show that the corresponding relation for the Ft, defined as the sum of the translational and configurational parts of the molar Helmholtz free energy, is
I!TYCT)'
VCII!TY(T)
Tel _ eT
Teo - e6
[Hint: consider a particular configuration of the N molecules of species fl> ... , fN, in volume Vi. and take the configurational intrinsic energy in this configuration to be
, 2RT' , R*(T) = - - - R T o yeT)
1
I
i,
whence HdT)
h~Q~)(~, N),
where
.
>n,
m
then the critical temperature Te correlates directly(ll) with e*lk, and the critical volume v c with a 3 . (a) By considering the general form of the configurational integral, QN, show that if the form of (r) for two substances, I and 0, is as 0, above, then at p
Now, from part (a),
and
r(~ r-(;rJ,
Q}P(T,N) =
a ---1
161
6.8 The principle of corresponding states (i). The relations proven in Problem 6.7(b), (c) can be derived also by considering the configura tional integral itself(iO). Problems 6.4(c) and 6.S(d) exemplify the proposition that if (r) takes the form
2
RT
and at p
Non-electrolyte liquids and solutions
irrespective of its explicit form. The second provides a powerful method of testing whether two liquids do in fact follow the same reduced equation of state, at least in the region p - 0(9).
I.
For the reduced temperature to be the same, T = T/Tel T'ITeo, so we = Tlfl' must consider the reference substance, 0, at temperature Then
6.B
(T)
=fIFJ(~)-RT1nhl-!RT1nfl ~RTln~l
(9) A.J.B.Cruickshank and C.P.Hicks,
,
Disc.Faraday Soc., 49,106 (1970).
(10) K.S.Pitzer, J. Chem.Phys., 7, 583 (1939).
These two relations, together with that proven in part (b), are valid for any pair of liquids which follow the same reduced equation of state,
(11) J.S.Rowlinson,
pp.265-266.
Liquids and Liquid Mixtures, 2nd Edn. (Butterworths, London), 1960,
6.8
Chapter 6
162
where MI and p = 0,
( , i~
are the two molecular weights; thence prove that at
6.8
Ft(T,N)
=
r-
(a) Any pair of molecules of species 0 mutually distant r contribute $o(r) = 4€6 [(
to
:0 (:0r]
E'trs =
t'l
Similarly, a pair of molecules of species I mutually distant contribute
to Vr Then $I(radao) = II$o(r). Thus if the volumes occupied by No molecules of species 0 and species 1, respectively, are so related that in geometrically similar configurations the separations of all pairs of species 1 are v I / a 0 times the corresponding separations of all pairs of species 0, then for that configuration (A),
F' = -RTln.[ (
"l,1
this condition clearly implies that every dimension of the container of I is VI/VO times the corresponding dimension of the container of O. This hIVo, or Vo vI/h l . Thus, if the assembly of in turn implies VI species I is defined by T, I), N, and that of species 0 by Till' VIh I, N, then Vti(vlh l , A,N) Vi(v, A,N) kTlit kT and this will be true of every configuration, as long as we may assume a I : 1 correspondence between configurations in the two assemblies. Since QN has the dimensions of v N , it follows that in this case
(I) _ QN (v, T,N) $(r) and
2rrmkT)% v ] h2 N -NkT,
2rrmkT)'/' v ] h2 No
~(27rmkT)'!,V] L h2 ~
{ RT = -RT In
N
hi
rl = T/'Fcl
Q)..P(T,N) =
(O)(~!. ) QN hI' II ' N . = r2
= Till
hrQ~)(f,
so N).
I F1(T) = -RT { 1n$+ 1 + In [(2rrmkT)'hvcI]} h2 No '
and for substance 0 at the same reduced temperature, i.e. at Till' ,(T) Fo it =
RT{ [(2rrmokT)'hvco]} 7: In$+ I+ln 1h No ' 2
whence fiFo,(T) h
= -RT { In$+ I + In
Q}J)
hl>( Q~)
vN
---;;n-
Q~)
= (vlhd N
[(27rmlkT)'l:VCI]} h2 No + !RTIn mIll mo
+RTln~
(';
VcO
and I 3 mIll FI(T) = lIFor(T) h - zRTln mo - RTlnhi .
Now, from part (b), Ft(T)
IIFti(f) ,
and simple addition gives the required result. Differentiating, (aF' I av h, gives p Irs N kT/v, so G,
(b) It follows directly that
+ 1n$+ I} .
Then, for substance I at T, p = 0,
q
VnA) = II uti (A) ;
= 0, $
-NkTln [(
4€i[(:0)m (:or]
$1('::)
(f, N).
when Stirling's approximation is used for N!. Then, for one mole, denoting per mole by F',
v~.
At P
Q}J) IlkT QJS) =
-kTIn-;:v- = --y:ln(vlh)N
Changing N to No, with NokT = R T, the required result follows. (c) From the translational partition function,
Generalize these relations to the case p =1= O. Solution
163
whence
III!J(~).
(T)
Non-electrolyte liquids and solutions
V ] = -RT1n [( 2rrmkT)% 2 h
No
and the same argument gives , GI(T)
IIGo,(T):3. fi -"RTln mIll mo -RTInh l
•
6.8
Chapter 6
164
~ I
f We can now either differentiate the given expression with respect to T, = -st and obtain U t == Ft + TSt , or obtain S' == -caF'/aT)v
6.9
Non-electrolyte liquids and solutions
165
The rigorous derivation for C T is as follows:
(aFt laT)v
cl
as 2rrmkT)'12 v ] S' = Rln [ ( h 2 No +RT(r'iz x ~TY')+R,
Fl +lJIP = Fl +¢VCIP = Ft +h 1
Ct = Ft o
+ voP
Pot + ¢vcoP
kl
0
k
0
'
l
whence
whence TS' = RTln [( 2rr:2kT
v
J+ ~RT
= II f ~t0 + ¢VCOp/I kI
t
II C 0
=
'Pot
JI
0
+ h I ¢v coP
.
Now U'(T) = ~RT;
Ft == IIFJ -RTlnh l -
as H'(T) we have 3
J.RT
.:i RT Uo'( T) 21;7: h '
6.9 The principle of corresponding states (ii). The necessary condition for the relations proven in Problem 6.8 to be useful is that there exists (over the relevant range of T and v) an approximation to QA?) or Fti, which can be expressed by a Taylor series expansion about a datum v. Because QW) determines f7;, but not vice versa, we examine first the expansions for the configurational thermodynamic potentials. For simplicity, we shaH consider only the case P O. (a) Expand Hti(T/f) about T == () to obtain
and similarly for H'(T). Adding Ur(T) and Hr(T), respectively, gives the required result. (d) When p =1= 0, v is no longer a function only of T, but is either itself an independent variable, or a function of T, p, according to ~(p, v, T) = O. Then, to ensure I and 0 are at the same reduced volume, compare F!(T,v) and FJCT/II,v/h 1 ). Similarly compare CiCT,p) and CJ(T/II' p/k l ), where kl == Pcdpco. The procedure is exemplified by extending the van der Waals case, cf Problem 6. 7(b), as F*(T, v) = RT(-
v~T+ Inv ~b)
whence Fr(T, v) T
f6 ( h Thus t
Fl (T, v) Ci(T,p)
8:r + In¢~1) RT( 9 7: - 8¢r + In ¢ ¢ t ) .
' hiv) =
(T v) liFo h' hI t
RTlnh
l
~RTln/l
I\UJ(f,:J.
o
co
ml 2RTInmo
p,
tn TO'
T) rn co (-ey-n(arHti) = L--L I n=orn!r n (r-n)! aTr p,T 00
JI.* ( o
Ht ( -IT) 3
~-
where t == 1- T/el; and thence prove the alternative expansions: i)
=/Ic;r(f ' :J-RTlnhl-~RTln/l-!RTln::
_ t( IIT ' k\12-)
Ht) L (-e)n (an -aT" (IT) = non!
11.* -
8:r + In¢ ¢ t)
= RTC
(T, p) - IIHo (T,lJ)
RT(-
1
o and addition of hi ¢v coP to both sides converts this to the required result for ct.
~RT
U'(T)+ p'v == U'(T)+ RT
Tln/l-!RTln:
0
Tn (anHti) L= 01-n-, --n. n. aT p, T = 0 co
n
(b) Use the Gibbs-Helmholtz relation, C = H+ T(ac/aT)p, to show that for n > 2, anc) ( - aTn p,T=O
(
I )n-I (n ()
--
n-I(-eY-I(arH) 2)' L . . 'r= I (r l)! aT' p,T=O'
and thence prove
f
C* (I) = R*«()+I(ac
t r +2
Xr=~_ dr+
l)(r+ 2)
166
6.9
Chapter 6 Show that the leading term (n
6.9
Non-electrolyte liquids and solutions
167
Thence
I) in the summation above is
(aaTG) p fI(aH) aT p , I (aZH) (aaTG)3 p + T2I (aH) aT p -f aT 2 , 2 3 2(aH) l±l(a H) !(a H) 4 ( aaTG) 2 T aT 3 p ,
aT p + TZ aT p Z 4
~(aH) _ 2+4(a H) 2+ I (a H) _!(a H) (asG) aTs p + T4 aT p T3 aT2 p + aT3 p T aT 4 p , ~(a2H) _2 x 3! (a H) ~(a4H) (aaTG)6 p ~(aH) TS aT p + T4 aT z p aT 3 p + T2 aT 4 p I (as H) -f aTs p , S! (aH) SI (a H) S! (a H) (a?G) aT? p + 0!T6 aT p -IITs aT p +2!T 4 aT3 p
2
(I -0 ( -aHt) aT T=IJ
T
T
T
2
3
Solution
(a) The first series is obtained by expanding in powers of [(T/f) 0 J and then factorizing as [(T/f) - o]n (-o)n tn. This, rather than on (_t)n , is chosen to conform to the series in part (b) below. The most direct way of converting the series in tn into a series in (T/f)n is to multiply out the t n according to the binomial series:
T- )n - L n n! ( 1 -Of - r= o(n
3
(T)n -Of
the following array, where all (an H* laTn)p are at TO: 3 Z T) (aH*) 02(a H*) H* ( f = H*(O) - f) aT p + 2' aT 2 p - 3' aT3 p + .,. 4 2 3 T l~(aH*) 02(a H*) 03(a + aT p - 0 aT 2 p + 21 aT3 p - 31 aT 4 p + 2 4 S 3 TZl(a H*) (J2(a H*) + 2!J2 aT 2 p - 0 aT 3 p + 2! aT 4 p - 3! aT s p + ... 6 2 ~ l~(a3H*) 0 H*) _ f)3(a H*) ... 3 4 + 3!J2 aT p 0 aT p + aTS p 3! aT6 p +
This
7
(a H*) (a H*) _ (a 4H*)
6
(p(a H*) H*)
03(a H*)
(as
]
J
(aG)
it is clear that (aG/aT)p cannot be functionally related to H alone. Differentiate both sides with respect to T: =
(a
2
4
S! H) -3!T 3 aT4 p+"" etc. On putting T = 0, inspection shows the stated result to be correct. Expanding G'6(TIf) about T = 0 and factorizing as in part (a), we find
G6 ( 7T)
G'6(O)-O
(aG6) T(aG6) 02(a2G6) aT p +7 aT p +2! aT 2 /2 3
_ 03(a G'6) (3+ ... 3! aT3 p , and putting Ht(f) = G'6(O)-0(aG6/aT)p gives
T)
G'6 ( f
T(aG6)
(-o)n(anG6)
= Ht(O)+7 aT p + n~2 ----n! aT n /n.
Substituting for (an G6IaTn)p, n
H+TaTp;
(aH) (aG) (aZG) (aG) aT p aT p + aT p + T aT p .
3
2
which is the required form. By inspection, each of the constituent series gives directly the value of its leading term at T = 0, whence the third alternative form of the expansion may be written down. Measured heat quantities, especially those pertaining to solution processes, chemical reactions, and ionisation processes, have most often been analysed using truncations of series belonging to the same family as the third series established here, namely that for which the reference state is T 0. This form has no special advantage, however, and the choice between the first and third forms should be made in each situation according to their relative rapidity of convergence. G
3
2
+ ... ,
(b)
4
> 2, as above, gives
_ * I( aGaT6)p,T Go*(I) f -Ho(O)+f + L OQ
OQ
IJ
(-OW
n=2
(n-2)!n-I(-OY-I(a rn.*) L _0 n! r=l(r-I)! aTrp,TIJ
double summation transforms directly into the required result, as is
6.9
Chapter 6
168
6.10
Non-electrolyte liquids and solutions
for the solution, and
clear on writing it as the array:
is the mole fraction of species i: - N,
Xi
_~(OHti) 2x I
Xi
oT p
_~(OH6) + ()Zt (OZHti)
z oT
__ ~(OHti) + ()Zt (oZH'6) 5 x 4 oT p 5 x 4 OT2 P S
S
S
()3t (03Hti) + ()4t (04Hti) 5 x 4 aT 3 p 5 x 4 oT 4 P
and collecting columns. (c) For n = I, the summation over t r + 2 becomes t't Z
tZ t3 t4 t5 r~o(r+ 1)(r+2) = TX2+ x + 3 x 4 + 4 x 5 + ...
tZ t 3 t3 223
---+---+ t
=
t+(1
=
t+ (1- t)InO
t)[-t-!t Z _jt 3 _!t4
OZH'6) [T ( aT 2 p,T=O I
where N
Nl + Nz , X == N 1IN, and thence prove that for such a mixture: ., Q (id)
I'M
T
(
T)2]
():rn()I-~ l-()!
'
= -kllnQ(id)Q"(id) xN
NkT[xlnx+(l-x)ln(l
where Qjjd) = v N [see Problem 6.1 (a)] and Q~~ == (xv yN; remember that QN is not an extensive quantity. (b) Show that if the molecules of species I and species 2 are now taken to be identical, the procedure of part (a) gives the necessary result, FM = O. (c) Show that the above results for obtain also for the cell model of Problem 6.1 (a). Cd) Deduce from the above expression for FM that the chemical potentials of the two species in an ideal solution Problem I IJ
(aF)
= (OG
= -
i - a n i T,
P,
lJi =
Theories of solu tions deal with the molar functions of mixing defined by
Xx - LXiXi , i
G, or v; the subscript x denotes the molar function
x)],
(I -x)N
other n -
(where ni is the number of moles of i,
XM
/) ,
•.• ]
BINARY SOLUTIONS
where X is U, H,
h2
t)-!t 4(1-t) ...
...
t).
T
21fmkT)'Iz
use the general argument of Problem 3.4(c) to prove that for a mixture of Nl molecules of ideal gas species I, and Nz molecules of ideal gas species 2, in volume v at temperature T, _ (21fmlkT)3XN/2(21fm2kT)3(I-X)N/2 vN Z 2 ZN h h (xN)![(1 - x)N]!'
t4 t4 t 5 t 5 -+---+ 344 5
t+tZ-~t2(1-t)-it3(l
=
6.10 Ideal solutions. Ideal solutions may be defined as those for which the functions of mixing are the same as those for a mixture of ideal gases. (a) Commencing from the partition function for one ideal gas molecule in volume v at temperature T, ZI = (
Replacing t by (l TI()f), I - t = T() II, gives the required result. Since > 0, the series is always absolutely or conditionally convergent. It is interesting to note that the expansion for G'6(TIf) of part cannot be transformed into terms of (anHti/aTn)p, Toby using the binomial series for tn; this procedure produces an array whose lines alternate in sign, but (except for the top two) are divergent series. The reason is that, in general, G* is non-differentiable at T = 0, as is obvious from the general form for (an Gla Tn)p, see solution to part (b). The difficulty is avoided, however, by proceeding as in part (c) for all values of n in the general series for G'6 (T/f), e.g. writing the second term as ()Z
LN;
X M specifies the change in the value of the function when one mole of solution is formed from Xi mole of each species. The mixing process is taken to be (i) isothermal and (iO either isobaric or isochoric.
x 2 oT p
4 4
_ _ (OHti) + ()Zt (oZHti) _ ()3t (03H'6) 4x3 oT p 4x3 oT z p 4x3 oT 3 p p
tZ I
=
i
3
3x2
169
nj
ani IT, p, other n
== N;INo ) follow
1J?+RTlnxi;
here IJ? refers to pure i at the temperature and pressure of the solution. Thence show that the equivalent expression in terms of the partial pressure, Pi == XiP, IS lJi
IJ! I
+ R Tin pt Pi
'
170
6.10
Chapter 6
pt being the arbitrarily chosen standard pressure at which, for pure i, IlP(T, p)
III (T).
Putting v FM
Solution
(a) Considering at first all the N molecules to be distinguishable, (Z) N Cl
1(1)
[Zl(l)a l
]xN[Z 1(2)a2
Then
-x)N
One might proceed either by putting at = D/xN, a z = D' /(1 - x)N, whence, taking D D' = e gives the required result; or by putting at = a2 e/N, giving _ (21rmlkT )3XNI2(21rm2kT)3(1-XlNI21P' h2 hZ ZN -
•
That this second result is incorrect follows from its giving an absurd result for part (d) below, i.e. that the chemical potential of species i in a binary mixture is independent of x, remaining finite as x -+ O. Thus the extensivity of F is, alone, an inadequate criterion. If we interpret the factors ai as expressing the fact, [cf solution to Problem 3.4(b)] that molecules of the same species are indistinguishable in the sense of having the same set of allowed energies (for quantized translational motion) then-since interchanging energies between two molecules of different species changes, in general, the list of momenta-the result is a distinct arrangement. The correct result for ZN is therefore the one stated in the problem. The required result for FM is most simply obtained by writing, for xN molecules of species I in volume xv xNkT/p,
FM = NkTln[xx(l
N ZN v PM == -kTln ZxNZ(1 x)N = -kTln[xv rN[(1- xlv ]0 x)N
=
since Q}!,d)
vN ,
QjJd) -kTlnQ~igQ~ild~x)N
etc., see Problem 6.1 (a).
x)(i x)]
kTln
[xN]'[(I-x)N]' . 1\11 •
,
and, applying Stirling's approximation to the factorials, we see that the two contributions to FM cancel; but recall that QN vN , so (id)
-kTln~(id~id) xN
{1-x)N
NkTln[xx(l-x)(t-x)],
irrespective of whether or not the two species are indistinguishable. (c) For the simple cell model, QN N!(v/N~, QxN = [xN]!(v/N)",N, Q(l-x)N = [(l-x)N]!(v/N)(I-x)N, since for the isochoric process, (v/N) is the same for both pure species and for the mixture. Then, for two distinguishable species _ (21rmlkT)3XNI2(21rm2kT)3(I-X)NI2 N!(,)/N~ h2 h2 [xN]![(1- x)N]! ' ZN _ (21rm 1 kT)3XNI 2[XN]!(I)/N)XN
ZxN h2 [xN]!'
Z(1-x)N
(
21rmzkT)3(I-X)NI2[(1- x)N]!(v/N)
h2 [(1 x)N]!
and ZN ZxNZ(I-x)N
_ (21rm 1kT )3XN12 (xv )xN ZxN h2 (xN)! ' x)N molecules of species 2 in volume (l -
= (21rmkT)3XNIZ( 21rmkT)3(1-X)N1Z(NkT/P)N h2 h2 .-
ZN
The requirement that the Helmholtz free energy shall be ex tensive is met by writing ZN =
= xNkT/p, etc., we obtain
-kTln(xNrN[(l'-x)N](1 x)N = NkTln[xx(1-x)(l-x)].
1(2)
21rm kT)3XNI2 xN (21rm 2 kT)3(l-X)N I 2 (l-xlN __=-1_ h2 V h2 v. (
and similarly for (1 Then
NkT/p, xv
(b) If the two species are identical, then ZxN and Z(1-x)N are formally unchanged, whereas ZN is clearly
- ZXN Z(I-x)N
-
171
Non-electrolyte liquids and solutions
6.10
QN N! QxNQ(I-X)N = [xN]![(l-x)N]! ' N!
FM = -kTln[xN]![(l
xlv.
x)N]!'
and Stirling's approximation leads to the required result. If the two species are now taken to be indistinguishable, _ (21rmkT)3XNIZ(21rmkT)3(I-X)NI2N!(V/N~ ZN h2 h2 N!
and
ZN _ (v/N)N ZxNZO-X)N - (V/NyN(V/N)O-x)N
I.
6.10
Chapter 6
172 (d) Now, from the definition,
= No (-aF)
/li
aNi
.
T, v, other N
Since NI
N2
= kT ( Niln NI + N2 + N21n NI + N2
FM
we have
)
'
NI
( aFM) aNI T,v,N2 /lIM /ll
=
kT[lnNI+ I-ln(NI+N2)- N +N 2 I
= NokTInx = RTlnx = /l?+ RTlnx ,
N2 ] N I+N2 '
,
where /l? is the value for pure 1 at the same T and p. The same result is obtained from G = GM + X I /l?+ x 2/l~· Since, for a one-component system, cfProblem 1.20(d), d/l
= -SdT+vdp
,
a/l)
ap
T
p'
= /It(p = pt)+RTIn(:t).
Thus, in a mixture at p, t
(
p)
_
Pi
t
/li = /li +RTIn pt +RTInxi - /li (T)+RTInpt
This result is widely used in approximate treatments of gas-phase chemical equilibria. The ideal solution result, FM = NkT[xlnx+(l-x)ln(l-x)] general izes for multi-component systems to FM
G
- -kTI { QCJ)[T/lx, p/kx , N] } n Q~~[T/II' p/k l , xN]Q~~~X)N[T/f2, p/k2 , (1 - x)N]
M -
- R TIn Eh + pv M
= RT~::Xilnxi .
P) -xIIG~ (7;' T
T GM(T,p)==lxG~ ( lx' kx
verify that this is equivalent to the result of part (a). (c) Similarly, show that in terms of Gt [see Problem 6.8(d)] GM
It is clear from G == F + pv that, similarly,
:J-
t(f, ~2 )
(1 - X)12 G
-RTInEh - ~RTInE I;
6.11 Non-ideal solutions. One general way of constructing a theory for non-ideal solutions is to start with the result of Problem 6.1 O(c), Q~) M
G~ + G~d) ,
G~(T, p) == Ix Gb(l ' ~) - xII Gb(f,
i
for mixing at constant pressure.
F
=
where
= RT~::Xilnxi
*( P)
p) -(l-X)12 GO h'k2 T kl
InEv == Invx(T, p)-xlnvl(T,p)-(l-x)lnv2(T, p), GUd) = R T[ x Inx + (1 - x) In (1 - x)] ;
i
GM
,
InEh == Inhx -xlnh l -(l-x)lnh 2 ; and derive an explicit form for VM in terms of the properties of the reference substance (a first-order Taylor expansion in T, p suffices). (b) Show that, in terms of G* and G' (see Problem 6.8(c)], G M may be written GM = GM - RTInEv + G~d) , where
= v = RT
so /lO(p)
whose generality may be inferred from the general form of ZN [cf Prob lem 3.S(c)]. Alternatively, FM may be expressed in terms of the configurational quantities Ft, together with the translational quantities F! [see solution to Problem 6.8(c)]; or FM may be expressed directly in terms of the configurational-translational properties Fit. In all three cases the principle of corresponding states plays an essential role in the derivation of usable explicit forms for F M . Because the direct outcome of experiment is always GM , HM for the isothermal, isobaric process, we concentrate on this, rather than the isothermal, isochoric process. It is convenient to use molar quantities throughout. (a) Starting from the above expression for F M , prove that for the isothermal, isobaric process,
where, as before, supersclipt (0) identifies the reference substance, and
for an ideal gas (
173
Non-electrolyte liquids and solutions
6.11
(I)Q(2) )N = -kTIn Q xN (I-X
'
verify that this is equivalent to the result of part (b). (d) Suggest a procedure for calculating FM for the isochoric and isobaric processes, applicable to the model of Problem 6.4, for the case
6.11
Chapter 6
174
6.11
Non·electrolyte liquids and solutions
175
Gtt: follows. From Problems 6.8(c),
start with the general expression for FM quoted in the preamble to this problem.
etc., and the result quoted for 6.IO(b), for one mole of solution,
Solution
, _ _ {(21f'mlkT)3x/2(21f'm2kT)3(1-X)/2 h2 h2 Gx(T, p) - NokTln
V crl
V cr2
V crx;
The appropriate form of the principle of corresponding states is obviously that of Problem 6.8(a): T ~) (I)(T, v) = hNQ(O)( QN I N r.' h I. . J
and, since G' is extensive, xG'I(T,p,No) = G~(T,p,xNo) so
1
Substituting into the expression for F M , we obtain hx QjJ)(T/fx, vx/h x ) FM = -kTln hixNQxN(Tf"xVI (O)! / (0), hdh2(1 x)NQ(l-x)N(T/12 , (l X) IJ 2I h 2)' (O)! / ] -kTln rl(ii}{ QN (Tk'v x h x ) NkTlnEh. QxN(T/fl, xv l/hl)Q~I-X)N(T/f2' (I - X)V2/h,) Note that the v retain their identifying subscripts because, even if hi h2 hx,
· l
vi(T, p.N} = h;vo(f,
~ , N) #; hjv o(; , ~ , N)
= Vj(T, p,N},
cf.Problems 6.7(c}, 6.8(d). But since v;(T,p,N)/h; vo(T/fl' p/kj , N), QC§)(T/fx, IJx/h x ) may equivalently be written QC§)(Tlfx , p/kx , N). Now G F+ pv, and adding the appropriate terms to FM for the isobaric process gives p[v x - xv I - (I - x)v 2] for No molecules of solution. These arguments lead to the required result, with NokT R T. It is convenient to expand vo(T/f;, p/ki ) about the standard tempera I bar): to the ture and pressure, T = 0, P = I (e.g. 0 = 298°K, p first order:
(avo) (T P) =vo(O,I)+ (avo) aT p,T=() (T) 7- 0 + ap T,p
v07'k
and h;vo
!. (fi.'
I(P
E.) _ (!!2 _OOlp _P(3Tk; +(3T ) k; - hjvo(O, 1+ fi 1)
since for corresponding-states substances hk
1),
. ,
= f,
VM(T,P)=v oO -OOlp +(3T)(fr +IJOTOlp(*r
voP(3T(~r,
where Vo, Olp, (3T all refer to the reference substance at 0, I, N, this implies only that the equation of state ~o(p, v, T) is known in the neighbourhood of 0, 1, and that it may be derived from Q~)(T, v,N). (b) From Problem 6.8(b)
G:(T'P)=fxG6(~' ~),
xGj(T, p)
Vx } x [xNo]X [(1- x)No](l-x) ,
XfIGi;(;,
:J,
xG',(T, p)
-No k Tln
etc.; thus the contributions to may write:
eM == G~
XG'1
(l
G~
l(
';0
21f'm kT)3X/2(V )X] h~
are comparable directly at T, p, so we
= -NokT[lnv x -
xlnv 1(1- X)lnv2] +NokTln[xx(l-x)(l-x)] ,
and the required result follows from GM = GM: + G~ . By definition, Q(x) Q%) Fx* == -kTln ~ , Gi = -kTln---w- + PVx - NokT , IT; Vx since p*v x
pvx
NokT. Now G* is an extensive quantity, so Q(O
xGt(T, p,No) Gf(T,p,xNo) = kTln( x~N+XPVI-XNokT; XVI whence QW vN
GM: = -kTlnQ(l)Q%.) + kTln[ ]XN[(l ~ X}1)2 ](l-x)N + PVM
xN (l-x)N XI) I Q(x) = -kTlnQ(J) Qf2) +NokTlnEv-NokTln[xx(1 x)(l-x)]+PVM xN (l-x)N and QN(X) * RTln'v+ .1" G(id) + PVM . GM M = - kTIn Q (j)/l(2) xN¥(l-x)N This particular relation suggests that a pseudo-ideal solution having GM: = RTlnEv, would not have VM = 0; the only ideal solution is one fl 12, hx hi = h 2. for whichfx (c) From the solution to Problem 6.8(d) we have xGt(T,p) (l
x)Gi (T, p)
P) -RTxlnh1-!RTxlnfl-jRTln (ml)X T kl xflGJ ( fl' mo '
= (1- X)f2GJ(~,
:J-
RT(1- x)lnh2
-iRT(l-x)ln12
m )(l-X)
~RTln ( m:
6.11
Chapter 6
176
t p T ) - RTlnhx Ix Go ( Ix 'kx
~RT ( Inf" -In
1m2 m XI-X)
mo
G~ - R TIn E v
G~.
This requires that G'b be expressed in terms of G6. From the definitions [see solution to Problem
G
t
= -3NkT[1
r(
-
.
-(I - X )E2
0
t(I.. L) -- fxG o*(I..Ix ' ) + ~RTInlx
RTln
[c1r~~kTrVo(T/~o' P/kx )],
vo(T/lx,p/kx )
G~ + ~RTIn E 1- RTInv~(T/ll' p/k d V6 X(T/12, p/k 2 )
- R TIn E h .
R
r.~.
While these equivalence proofs are essentially emphasise the importance in corresponding states arguments of precise definitions, and (ii) the temperature and pressure to which each extensive quantity relates. (d) Consider first FM for the general process. The solution to Problem 6.4(b) gives
= -N + 3Nln
I: ( )'/)J
l
V
"'ZNE ~kT
y] ~-NkTIn
Ev(T, p) +
p~d) .
Cl~) !Il]
In{ 4Z€
r
[C; -C,; f]},
and this simplifies the expression for F M • A possible line of progress is to use vi(T,O) = vo(T/li' 0), Ii EdEo to express the /)i in terms of vo(T, p) and the fi. Some tedious algebra and drastic approximation yield expressions for PM in terms of E1> E2, when a recipe is assumed for Ex(EI' E2, x) (see Molecular Theory 01 Solutions (I 2) for detailed exposi tions). Alternatively, one might use Taylor expansions for QYPCT/I, /J/h) about I h = I, but this requires an explicit form for vo(T, p), as a solution to the equation of state. Probably the simplest procedure is to solve the equation of state for each v i and thence to evaluate QA,7), (2) . • Iar reCIpes . f or Ex, Vax m . terms 0 f EI, E2; /Jal, an d Q (l-x)N usmg partIcu V a 2'
On separating -RTInEh into its components and incorporating these into the Vo term, this equation reduces to =
I ~:
ctJJ
whence
- ~ R TIn E I
II
This relation is formally the same for both the isochoric and the isobaric processes. In the former case, however, lJ x = X IV 1 + (I X)V2, while in the latter, Vx = hxvo(T/lx, p/kx )' Thus FM (isochoric) clearly differs from FM (isobaric) in all three terms. It is obvious from the equation of state [Problem 6.4(b)] that for the isobaric process at p 0, -In [I -
21rmk T)';' v ] h2 N,'
In [I - (va /v x ) '/, 1 (Va /VI )!!']X[I_(Va /V2 )'h)I-X
+2ZN{Ex[(~:r-G:r]-XEl [(~~r (~~rJ
t
IxGo Ix 'kx
Gt
Q}:) -kTIn QxN (1)Q(2) (J-x)N
== G* + G , = -RTIn
177
whence
I
+
whence the required result follows. To establish the equivalence of this result to that of part (b), it suffices to prove
Gt
Non-electrolyte liquids and solutions
6.11
,
G;(T,p) to IxG'b(T/lx,p/kx), as with G~(T,p) and ) in part (b), it becomes clear that G1 (T, p) also includes the extra term G~d). The remainder is said to describe the equivalen t substance. which differs from the solution onlv hv G~!d). Then
t Gx(T, p)
l
I
[(v: )4 -(v: )2] '
A second type of solution theory follows the method of Problem 6.6. Expressions are derived for the X* (or xt) in terms of the parameters of the chosen equation of ~(p,v, T) = 0, e.g. the parameters a and b of the van del' Waals equation. The values of the parameters which best reproduce the observed p, v, T behaviour of the reference been decided, the principle of corresponding states is used to estimate the ii, hi values characterizing the components of the solution (13). Given a recipe for Ix =lx(k/2,h I ,h 2,x) and another for hx, either (i) X* are calculated for I, 2, and x as in Problem 6.6, and thence the X Mare calculated directlv. or Oi) the corresponding-states formulae for Xi in cf Problem 6.9, are used, and the X M calculated and (c) above. A variation on method (ii) is to (12) LPrigogine, Molecular Theory of Solutions (North Holland, Amsterdam), 1957. (13) A.J.B.Cruickshank and C.P.Hicks, Disc.Faraday Soc., 49, 106 (1940); D.Patterson and 1.M.Bardin, Trans.Faraday Soc., 66, 321 (1970).
6.11
Chapter 6
178
Non-electrolyte liquids and solutions
6.12
treat the X6' as adjustable parameters to fit experimental H M , GM , and values for one system; the X6 thus specified are then used to predict X M for other systems in terms of their Ii and hi' This procedure does not assume an explicit equation of state. Note that the Xr. are calculable from measured p, Il, T data, although not very to this section). Detailed exposition of these various procedures is a matter of computation rather than
to prove
= dY Y-T dT
UM
(c) Consider, as a first approximation, the case of completely random mixing, corresponding formally to the limiting case as w/kT -+ O. Show that to this approximation -
6.12 The regular solution (i). This theory(14) is related to a simplified version of the smoothed potential cell model (see Problem 6.4), in which the variation of U* with volume is ignored, i.e. it takes the form
U*
UM
FM
i, j
where, as in preceding problems, z is the co-ordination number of the (implied) cell lattice. Thus, if a molecule of species I from pure liquid 1 is interchanged with a molecule of species 2 from pure liquid 2, the net increase in the energy of the two systems is 2w. (a) Write down the configurational intrinsic energy U* for an assembly of N, molecules of species I and N2 molecules of species 2, N, + N2 == N, such that there are z Y I, 2 pairwise interactions, remembering that there are -! zN pairwise interactions in total, and thence prove that. for the of Y by binary solution, U M is related to the average value, UM
= Yw.
(
L: ~~~.:,--.-:. A
By extracting the mean, exp(- Yw/kT), show that FM
and use the relation
('4) E.A.Guggenheim,
= P!~d)+ Yw,
aFM) UM = FM - T ( aT
x(l-x)N,
x(l-x)Nw,
= FMd) + x(1 -
x)Nw .
(d) Show that these results are obtained also by taking only the first term of each of the corresponding-states expansions for FM and obtained from the expansions in F~(O) and U6(O) analogous to those in Gt(O) and H~(O) in Problem 6.9(a), (b), provided that
0
+2x(1-
which is appropriate to random mixing of molecules of equal size.
f\
L
;1
Solution
(a) A molecule of species 1, with z nearest neighbours is engaged in z pairwise interactions. Let az of these, on average, be I, 2 interactions be I, I interactions. Then there are azN, I, 2 interactions and (I Y. The total number of 1, I interactions is clearly in all. i.e. aN. !(1-ex)zN,
-!z(N,-Y).
if the average number of 2, I interactions per molecule of 2 is
~~
fJz, the total is fJZN2 z Y, and the total number of 2, 2 interactions is ! z(N2 - Y). The total num ber of pairwise interactions is then ~ z(N l + N 2 )
as required. The total configuratio!lal intrinsic energy is
(b) Assuming the free volume per cell to be Uf for all cells, and considering each possible configuration in turn. as in Problems 6.1 6.5(a), show that QN QxNQ(i-x)N
N,N2
Y=~
= L: €;j (nearest neighbours only) .
While obviously invalid as a model for the liquid state, this provides a way of describing solutions which, although itself of limited applicability, is of interest as a basis for introducing the concept of incompletely random mixing (see Problem 6.13). The three pairwise interaction energies are denoted by -€ I, -€ 2 , -€ 12, and we define the interchange energy w by w == -Z€ 1 2+ -!z(€, + € 2 ) ,
179
,CN1 - y)-!z€2(N2 - y)-Z€12 Y '
U~
j
and for the average of all configurations
•
U*
"1
~i
Since Ut
-iZ€ I N
1,
UM v
iZ€2(N2 -Y)-Z€12 Y '
-!z€,(N, U;
!Z€2N 2,
!Z€2Y-U12Y= Yw.
(b) Since U~ is constant for each arrangement of the molecules in their cells, as in Problem 6.I(b), QN breaks up into separate integrals each over
Mixtures (Clarendon Press, Oxford), 1952, pp.30-32.
Ir
:ll
Df,
6.13
6.12
Chapter 6
180
The assumption that the free volume is the same for species I and 2, and for the solution, when the reduced temperatures all differ, implies a curious mixing process in which the components are initially at different pressures, while the final solution is at a third pressure; it is obviously closer to the isochoric process than to the isobaric process. the leading term in the US(T/f) expansion to that of Problem 6.9(a),
so QN
= v7L: ex p (- Ut ) ;>. kT
v N '\' ex !zE,N,+! f
1:'
P
ZE N 2 2- YAW kT '
_ N' xN !ZE,N, Q xN ,.vf eXP-rrQ(1-x)N
N 2 !v},-X)N
.
and so
QN
exp(-Y;>.w/kT)
QxNQ(I-x)N
Nl!N2!
I!
Ix and, since UM UM
the mean, and remembering that there are N! arrangements,
F~d)
whence
= RT[xlnx+(l
orU ») v
Y
Yw =
since fi. but U6'(O) this approximation
v'
since R Tin E 11 obtain
x), and the mean number of z(i - x), so the total number is
)Nz+ !x 2Nz+ !Nz - xNz
Y = x(l of temperature, the differential equation of part (b) satisfied by
and the required results follow immediately.
+ TC:o(O)lnEj+ F~d)
0 for this process.
,
Neglecting the second term we
1
of I, I interactions is ! zxN I , and x)N2 • The total number is then
x)N
xf,- (I - x)fi 1
= !zNoEo in the regular solution model, so to
FM = US(O)f E
U
Y = Y = x(l
X)2 fi ,
UM = x(1 - x)Now for one mole of solution. For the Helmholtz free energy, isochoric process, it is clear from Problem 6.9(c) that
+!x 2 Nz
is
x 2 fl + 2x(l - x )f12 + (l
1 - x)w/!ZEo ,
- (dY)
(x
x)ji1·
I - x) - f2(1- x)x + 2fl2X(l - x)j
Y T dT .
x 2 )Nz+ !x 2Nz+ !(l- x)2Nz
t(f),
X)f2 U
+ 2x(l- X)fI2+ (I
x)ln(i-x)1,
OYW) -T (aT dYW) T ( dT
(c) In the case of random mixing ex I, 2 interactions per molecule of 1 is Nl z( 1 x) = x( I - x)Nz. The number the number of 2,2 interactions is !z(1
and
u~(l) -xfl ut(~) - (1
fx
-FrS/d) ,
Yw, so that Yw
(x
= fx
then
d
-T ( aT
f~ U6(0),
U~,
Iffx is taken as
Yw/kT)N! NI!N2!
and the required result for FM follows at once.
But
U6{X)
U6'(OHfx -xfl-(l
we obtain QxNQ(1-x)N
181
Non-electrolyte liquids and solutions
,
f \
6.13 The regular solution (ii). In the model of Problem 6.12, instead of random mixing, the quasi-chemical equilibrium: (1, I) + (2, 2)
~
2(1, 2) ,
for which the energy increase per molecular unit of reaction is clearly = 2w/z. The equilibrium 'concentrations' of the three types of interaction are assumed to accord with a quasi-chemical equilibrium constant K, defined by
flu
flF
==
-NokTinK ,
where fll" is the Helmholtz free energy change per molar unit of
6.13
Chapter 6
182
Non-electrolyte liquids and solutions
6.13
reaction, or, equivalently, by
holds at T = 00. Is this last assertion necessarily valid? Suggest an alternative way of evaluating FM which avoids the logical difficulties associated with the reference temperature T = 00.
-t::.U) (-t::.U) K == nexp ( NokT = nexp kT ' where t::.U, the intrinsic energy change per molar unit of reaction, is given by the usual relation
Solution
(a) From Problem 6.12(a), the numbers of interactions are
t::.U = t::.F-T ( dt::.F) dT ' and n is the change in the degeneracy factor such that
2,2:
!z(N2 - Y); zY,
4Y2 (Nl - Y)(N2 - y)
(
= nexp -
2W) zkT .
For a I, I interaction, there is clearly only one distinguishable arrange ment, and likewise for a 2, 2 interaction; but two I, 2 interactions may be achieved in four ways. This is confirmed by the fact that for w/z kT = 0 we must have y2 = (N l - Y)(N2 - Y) to conform to random mixing, Y = N l N 2/(N l + N 2 ) = x(1- x)N. On putting exp(-2w/zkT) = 1/7'/2, the equilibrium relation becomes I y2 (N l - y)(N2 - y) = 7'/2 giving (N l - Y)(N2 - Y) = 7'/2 y2 .
[C 12 F [C l l ][C22 ]
(a) Express the concentrations of the different types of pair inter actions in terms of z Y [see Problem 6.12(a)], the number of I, 2 interactions, and show that the equilibrium value of Yobeys (Nl - Y)(N2-Y)-7'/2y2
1Z(Nl- Y);
and the concentrations are given by dividing each number by v. Then
K is equal to the ratio of the product of the concentrations of the reaction-product species (each raised to the power equal to its numerical coefficient in the reaction equation) to the corresponding product of the concentrations of the reactants. Thus, denoting the equilibrium concen tration Cji , we have for the above reaction
=
I, I:
1,2:
kTlnn == Tt::.S.
K
183
=0
where 7'/ == exp(w/zkT). If UM be written 2 UM == ((3+ l)x(1- x)Nw ,
Putting Y
= 2x(1 -
x)N/((3+ 1), so that,
- = N rLX - 2x(1X)] (3+ I ' - r 2X(1-X)] N2 - Y = N LI - X (3+ I '
Nl - Y
so that UM
-+
x(1-x)Nw as (3
(3
-+
I, prove that
= [I +4(7'/2_I)x(1-x)]Y'
.
(b) Since (3 js a fU!lction of T, Y is also a function of T, and the relation Y = Y - T(dY/dTlcannot be solved for Y directly. Verify the alternative (1 5) form Y = d(Y/T)/d(1/T) and show that substituting for Y and for d(1 / T) in terms of (3, and integrating between (3 = I and (3 = (3 gives for FM _
(id)
1
_
FM -FM +"J.zNokT(1
4x 2(1 - X)2
_
y2
=
N2
((3+ 1)2
,
the equilibrium relation becomes N2
r (3 + I - 2x (3 - I + 2x] x) ln L(1-x)((3+1)+xln x((3+I) ,
((3+ 1)2[((3+ 1)x-2x(1-x)][((3+ 1)(1-x)-2x(1-x)] 47'/ 2x 2(1 - x)2N2
,I
when Y((3 = 1) i.§.. put equal to zero. (c) Show that Y((3 = I) = 0 is the necessary condition that
-
-
((3+ 1)2 which reduces to
dY
(32-4(7'/2_1)x(1-x)-1
Y = Y-T
dT
and the required result follows.
(15) E.A.Guggenheim, Mixtures (Clarendon Press, Oxford), 1952, pp.38, 39.
v
\
=
0
184
Chapter 6
6.13
6.14
(b) The required form follows exactly the standard alternative form of the Gibbs-Helmholtz relation: H = [3( G/T)I 30IT)]p' Thus -
dY T dT =
y=
dY
jd(1/T)
Td(1/T),~
1 dY Y+Td(1/T)
T(dY/dT) at T
(32
4(17 2 -1)x(l-x)- I
d{3 dT
= 0
dY
132_ (1- 2X)2 17 = 4x(l x)
T dT
2
2w zkT = In[j32
When (3 (1 - 2x)2 d(3 .
df
zNkx(1
2w
X){
(l-X Xl)
Integrating between 13 I and (3 13 we obtain = zNkT Y = ~[(1- x)ln(j3+ 1- 2x)+ xln(j3- 1+ 2x)-ln«(3+ 1)]1 + Y«(3 = 1) .
r
=
=
Y - Y( 13
l)x(l
x)+ I .
=
I)
X)N(
13
I, then T ==
00,
1
I
x
{3+ I
I)
x 2w 2x + {3- 1 + 2x - 13+ I exp zkT '
exp(2w/zkT)
==
Y({3 = 1)-
I, and
(3
=
_
1)- Y«(3
1).
Thus.llt T =_00, Y({3 = 1) = 0 is the only assumption which satisfies Y Y - T(dYldT). In fact, to assume that the general relation [3(AG/T)/3(l/T)]p = AH holds at T 00 implies categorically that AG(T = 00) = O. This is to assert that AG/T remains differentiable with respect to I/T at liT == O. There is no simple way to establish the validity of this assertion. It can be avoided, in principle, by_choosing a reference temperature T = e, other than T 00. Then Y must be evaluated at T = e through the ratio QN/QXNQ(1-X)N' Various approxi mations have been used on this problem, and it has been proven(16) that the result of part (b) is obtained by assuming that all pairwise interactions are mutually independent. Higher-order approximations give slightly different results; the algebra is too tedious for inclusion here; the methods are described in the reference cited. The result of part (b) is important because it suggests a temperature dependence of FM - F~d) which is different from that of UM , as is usually observed when experimental results are converted to give FM (isochoric) and UM (iso choric). Note the qualitative similarity to the inclusion of the term TC~o(O) In E f in the corresponding-states formula [cf Problem 6. 12(a) ].
4j3dj3 } [13+ l][(3L (1- 2X)2 J .
Separating the integrand on the right-hand side into partial fractions, we find 4(3 [(3+ l][j3+(l-2x)][j3-(l-2x)] I x I ) x(1 x) 13+ 1 2x + - 13+ 1 whence Y ZNk( 1 x d T = 2w (3+ I - 2x + 13- 1 + 2x - 13+ I d(3.
=~L(l-x)ln
4(17 L 2
dY) ( T dT /3=
Substituting for Yand for d( liT),
zNkT
=
2x)2]-ln4x(1-x) ,
2(3
(32
) d{3
2x(l - x) (d17 ) = 4x(1 - x) (_~) 2w 13 dT (3 zkT2 exp zkT '
-
whence 2w zk d(l/T)
I
x
1)+~ {3+1-2x+(3-1+2x {3+1 dT'
2x(1 (1
I- x
ZNkT2(
=
Y-Y({3
Hence
and, from part (a),
Y
From the result of part (b)
(32
'
giving
00.
185
From the result of part (a)
y == 2x(1 x)N
13+ I
=
dY TdT
d(Y/T)
=
Non·electrolyte liquids and solutions
6.14 Solutions of chain mo1ecules (i). Problem 6.10 shows that the ideal solution formulae may be derived from the simple cell model. For the case Val = tJ a2 Vax the same result is obtained from the models of Problems 6.2 and 6.3. There is thus some justification for using the simple cell model to derive the thermodynamic functions of mixing for
(3 + I - 2x (3 - I + 2x ] = I-x +xln x -In((3+I) +Y(j3= 1).
The given result follows at once when Y«(3 I) is put equal to zero. 1 corresponds to 17 2 I, 2w/zkT = 0, i.e. T 00, in the (c) (3 general case w =1= O. The simplest way to proceed is to evaluate
(16) E.A.Guggenheirn, Mixtures (Clarendon Press, Oxford), 1952, pp.42-47. I
1
186
Chapter 6
6.14
I
for the first segment, there are
solutions of non-interacting chain molecules. The essential assumption is that the molecules of both species are chains whose links, or segments, can each, interchangeably, occupy a single cell. The simplest case is a solution of 'monomer' (species 1) and 'r-mer' (species 2), the molecules of the latter each occupying a chain of r cells. The total number of cells is then N NI + rN2 • As in Problem 6.1 (b) and Problem 6.2, QN
z z - I (N - rn)
=
=2:A
(_Z_)(I
z -
InrP+ (I
(I
-
z
r
rnJ ,-
1N)-- N
2
=
Z
z
N
ways of placing the complete r-mer. Since there are N
=
n (N-r )
N2 -
r
t
11
g(N2' N)NI !vr ;
while for the pure r-mer [rN2 cells,
= (1 -
rP)N]:
g(N2' rN2)vjl q,)N; cells, Nl
=
Qq,N = NI!vtN .
Then -kTin
t
(r-l)
n=0
QN
rP)].
- 1N-rnJ' )--
2
Then, for the solution (N cells, N = NI + rN2 ):
Solution
rn) (z N -'" "' ( N
f
Since [(N Ir) - N 2 ] (NIr) - (1 - rP)NIr = rPNIr, the given result follows.
For N2 r-mers distributed among rN2 cells, the result (a) evidently becomes z )N2( Z - I )N2(r I) N g(N2' rN2 ) = ( z _ rN2 (ry 2 (N2
,
(a) Once the first segment of the (n + 1)th r-mer has been placed, there are zeN - rn)/N possibilities for the second segment, i.e. z times the average fraction of vacant cells. For the third segment there are (z - 1) cells accessible, since one of the z adjacent cells is occupied by the first segment, so the number of possibilities is (z- 1)(N-rn)/N. Thus, for every siting of the first segment, there are
I
(Z~I
-. . N rnJr- J (z-l)-xr r
2
I )(1-
[rPNlrj!
N
z - I )(I-
__ . "" (_Z )(1 N r [(Nlr) N ]!
z-I
and rP (the where Z is the co-ordination number of the cell segment fraction, or volume fraction of monomer) is defined by rP == NI/N. A generally valid way of proceeding is to consider the situation after n r-mers have been placed; count the number of choices open for the placing of each segment after the first, and so formulate the
number of ways of placing the (n + I )th r-mer. (b) Repeat the procedure of part (a) for N2 r-mers among rN2 cells,
and thence prove that for the binary mixture, of N2 r-mers and NI monomers, Nt + N2 == FM "" R
IllC z I)(N-
g(N2 ,N)
I
N-rnJr-l
1)-
ways of placing the (n + I )th r-mer. Thus the total number of ways of arranging = (I - rP )Nlr r-mers among N "" rN2 + NI cells is
The present problem is then to formulate the number of ways of arranging N2 r-mers among N cells; there are clearly Nl ! ways of arranging the Nl monomers among the remaining NI cells. The problem is greatly simplified if we make the general assumption that the average proportion
of occupied cells neighbouring any empty cell is the same as the proportion for the whole assembly of N cells. Only a trivial error is introduced by taking this average proportion as constant during the
placing of one complete r-mer. (a) Using these two assumptions, prove that the number of ways of arranging N2 r-mers among N Nl + rN2 cells is g(N2)
187
Non-electrolyte liquids and solutions
6.14
QN = -kTln g(N2' N) Q(1-q,)NQq,N g(N2' rN2) _ _ { [N/r]! }'(rN2)N2(' - kTln [(Nlr)-N2]!N2! N
I)
Now (rN2IN) N 2(r-l) = (l
r
I'M
=
I rP, (Nlr) N2 rPNlr [see part (a) above], and rP)N-N2. Thus [N/rll -1<.1 rln [rPNlr]![(l ' rP)Nlr]! -(\-rP)NkTln(\ +N2 kTln(l-rP) .
rn possibilities I
I
Chapter 6
188
6.14
Using Stirling's approximation for the factorials, we obtain
N rjJN -kT [Nln,-rjJNln
r
FM
(l-rjJ)Nln
rjJ)Nln(1-rjJ)+rjJNlnrjJ-(l
O-rjJ)NJ ,
rjJ)Nln(1
W12 rjJ) } ,
rjJ)+Nz ln(1
UM
= NdNo,
I
x
= Nz/No,
It is clear that an exactly parallel argument for N J 'I-mers and N 2 '2-mers will give the same formal result if rjJ is now defined by 'INI 'INt +'2 N 2 The effect of the second assumption in the derivation vanishes as z -+ 00, and this is held to justify using the general result for solutions. In fact, just as with the ideal solution formula, alternative derivations, independ ent of the cell model, give the same general result(l rjJ ==
Solution
€
-Z€12+~Z(€I+€2)'
(b) Defining W12 as the interchange energy per molecule of species I, so that when '2 molecules of species I from pure I are interchanged with '1 molecules of species 2 from pure 2, the net gain in interaction energy (I 7)
aZ)Zwab'
(d) Verify that the result.for U M obtained by substituting the result of part (c) into the result of part (b) is obtained also by direct counting of all types of segment interactions in a solution of I and 2 characterized by the volume fractions rjJ, 1- rjJ, respectively, and of all types of segment interactions in pure I and in pure 2. (a) The given definition of qi leads to
zqi = Z'i qi
=,. I
2'i+ 2, 2(,; - I)
z
and, in the limit z -+ 00, q; = 'i' Consider first a molecule of I in the solution: the number of I, I interactions per segment is simply zrjJ, and the number per molecule of I is then zrjJq I, so the total number is t zq I rjJN 1 • Similarly the number of 1,2 interactions is zql(l rjJ)N I • The corresponding numbers derived by considering a molecule of 2 are !zqz(l-rjJ)Nz and zq2rjJNz . It is clear that adding all types thus enumerated will give twice the number of I, 2 interactions, so take half of each count, Le. ~zql(1 rjJ)NI+JzZq2rjJN2 The total is now !Z(qlNl+q2NZ) as required. The corresponding
is the interchange energy per segment, defined (cf Problem 6.12)
==
rjJ)WI2 N o·
wab == -Z€ab + tZ(€aa+ Ebb) .
where zqi is the number of external contacts per molecule of species i, comprising (z - I) for each end segment and (z - 2) for each segment other than the end ones. Thence show that taking the formal limit z -+ 00 gives UM = ('IN I + 'zNz)rjJ(l- rjJ)w ,
W
X)JrjJ(l
where Wab is the segment interchange energy defined by
in which the enthalpy of mixing is significantly different from zero, formulae may be derived which bear the same relation to that of Problem 6.14(b), as do the regular solution formulae to those for ideal solutions. The essentials of the method are apparent in the first approximation (random mixing) based on the cell model of Problem 6.14. (a) For a solution of species I ('I identical segments) and species 2 ('2 identical segments), prove that
W
[x+~:(1
Wl2 = 'I(at
6.15 Solutions of chain molecules (ii). For chain-molecule solutions
where by
,
(c) The resul t of part (b) holds, in fact, whether or not all the segments in each species are identical; it therefore suffices to derive WIZ for each type of solution. A case of particular interest is the solution of species I and species 2, each of which has two types of segment, type A and type B (e.g. middles and ends, respectively). Suppose that the molecule of species I has , I segments, comprising ai' I type A and b l '1 type B; and that species 2 has '2 segments, comprising aZ'2 type A and b 2'2 type B. Evaluate for a molecule of species 1 in pure I, for a molecule of species 2 in pure 2, and EI2 for a molecule of species 1 in pure 2, and thence show that
= NokT[xlnrjJ+ (1 - x)ln(l
UM = tzq tNt (1 - rjJ )(€ 1 €lZ)+ ~ zq2 N2rjJ(€2
==-ZEI2+!Z(EI+~E2)
where -tzEINo is the configurational intrinsic energy per mole of species I; show that
= kT[NllnrjJ+Nzln(1-rjJ)] ,
and for one mole of solution, No = NI + N z , x
189
is'2 times W12 , Le.
- kT{ I - rjJ )Nln (I - rjJ) - Nzln ( 1 = kT[(1
Non-electrolyte liquids and solutions
6.15
(
j
I
P.I.Flory, Spiers Memorial Lecture 1970, Disc. Faraday Soc., 49,7 (1970).
J
I
190
Chapter 6
Then
numbers for pure 1 and pure 2 are clearly tzq1N1 and tzqzNz , and so UM = -!z[q 1N 1
[
whence the required result follows on simple rearrangement. Replacing qj by rj, corresponding to the limit z ~ 00, we obtain =
whence W12 = -!zrl[(al
tz[rINI(l
E12+E2
=
EIl- C2l) ;
I
r2 Lx+r;(l-
El
== Z ri
0:,13
exj
L exj (3j E a+l '
",.il
W11 rj
•
For a molecule of I in pure 2 (rial Ea + rl b I Eab)za2 + (rial Cab + r l b l cb )zb t .
(31
=
aj,
bj,
Cj, .. , .
a2)(b l
b 2)w ab+ (b l - b2)(cj-C2)Wbc+
(CI
c2)(al-a2)wca '
(d) For an assembly of NI molecules of species 1 and N2 molecules of species 2 in random mixing, list the energies associated with each type of interaction, dividing each by 2 to normalize the total number of inter actions to !z(rINI+ r2N2):
1-
(rlalEa + rlblEab)zal + (rlaIEab+ rlblEb )zb l
= a;, bi' Ci' ... ;
Thus for three types of segment we obtain
,
= (r Za 2Ea + rZb2Eab)zal+ (r 2a2Eab+ r2b2Cb )zb 2 .
EI2
•
where the summation extends over all ex, (3;
The importance of this apparently trivial result is that, being in terms of W12 (defined in terms of Hll , E I , and E2 ), it is independent of the segmental constitution of species 1 and species 2. (c) EI may be constructed as the sum of the two types of interaction open to type A segments, and the two types of interaction open to type B segments, thus Similarly
ZrjL exd3 j c",;
Eii
No, for one mole of solution,
W12
The argument generalizes at once to any number of segment types on writing
this is stated to be equal to r z W12 , so again, W12 = rlw. The two definitions are thus consistent. Substituting W12/r I for w in the result of part (a) yields
NI rlN2) ( No +~No
r2 rl
2
Alternatively, interchanging r2 molecules of 1 from pure 1 with rl molecules of 2 from pure 2, the energy increment is clearly
and, taking NI + Nl
rj(al-al)2 wab ·
r2(al-a2) Wab
Wl2 = -ZrIE12+tzrICI+tzrIE2 = rlw,
= (NI+ ~~N2 )
Cb),
The symmetry of these equations ensures that if 2 is taken as reference species instead of I,
(b) Since -! zE t No is the configurational intrinsic energy per mole of 1, clearly EI = rIEl' Similarly E2 = r2E2, and En = rtEIZ, since it refers to a molecule of species 1. It follows then from the definition of WIZ ,
UM
-!zrl(al a2)2(2E ab- Ea W12
[r t N I+r2 N 2 1
-t zr lr2( 2E 12
-(b l -b 2 ),sothat
we have
and then introducing the definition of w, we find UM
b 2 ) 2c bl,
a2)2Ea+2(al-a2)(bl-bl)Cab+(bl
W12
E
b~ Eb
-aiEa - 2a2b2Eab
1,a2+b2 == 1,(al-a2)
sinceal+b l
taking out the total number of segments, rINI + r2NZ, we obtain UM = tz[r I N I +r2 N2H
2ala2Ea + 2b1a2Cab+ 2atb2Eab+ 2blb2Ebj -a1Ea-2albjEab-b1cb
= !zrl
+tZ[q IN I EI+q2N2E21,
UM
191
Non-electrolyte liquids and solutions
6.15
6.15
abO ):
-! rl NI ad<pal + (l -!r INl atl
ba(1):
-!rjNlb tl<pa l + (l -
bb(1 ):
-trtNlbtl
aa(2):
-tr2N2aZ[(1-
ab(2):
-tr2NZal[(1 -
ba(2):
-tr1N1b1[(l-
bb(2):
-tr2Nlbl[(1-
aa(l ):
,I
192
6.15
Chapter 6
I
Then the total energy of aa interactions is U:a
= {-trjNj[I/Jar+ (1- l/J)aja21- t r 2N 2[(l
=
I/J)a~+ l/Ja t a 2]}u a
7
[rt N t+ r 2N 2][I/Ja l +(l-I/J)a2
Similarly for the bb interactions U~b = -trr,N,+
t+(1
Phase stability, co-existence, and criticality
F
l/J)b 2 u b, A.J.B.CRUICKSHANK (University ofBristol, Bristol)
and for the ab interactions U:b
-[rtNl+r2N2][l/Jaj+{l-I/J)a2J[l/Jbj+(l-I/J)b2]ZEab' SINGULARY SYSTEMS
For N j molecules of pure I and N2 molecules of pure 2, the total energy is
+r
U* - _lz[r N 1,2 -
2
1
j
7.1 Thermodynamic stability of the single phase. Consider a uniform system (i.e. in which each intensive variable, T. p, has the same value at all points) of n moles of a singie component, constrained to constant entropy and volume. Phases a and /3 will be identified by suffixes, and the initial (unperturbed) state by index (0). Initially the system is entirely in phase a. The perturbation is o~ mole going over into phase /3, only slightly different from a. The condition for equilibrium in such a system at constant S, v, diffusional processes being allowed, is that (1) (au/ans, v > 0, i.e. that U is a minimum. That this is equivalent to T(as/a~)u v < 0, i.e. that S is a maximum(i) is obvious also from the second law as stated in the preamble to Problem 1.22. The derivation of the particular conditions for phase stability is simplified by: (i) using molar (intensive) rather than total quantities for each phase, since == [f(S, v) wher~as U = U(S, v, n), cf Problem 1.20(a), _ expressing both Ua. and U(3 by Taylor expansions about U~O), valid because phase /3 is similar to phase a. Then, initially,
[I/Jar+(l-I/J)a~]Ea+ [I/Jbi+(l-I/J)b~lEb}
N ] { +2[I/Ja j b j +(l
22
1/J)a2 b 2]E b a
.
Straightforward algebra then gives for UM , as before, UM
= -tz(rINI+r2N2)I/J(l-I/J)[(al-a2)2Ea+(bl-b2 + 2(a l + )I/J(l-I/J)(a l -a2)2 wab ·
a2)(b l - b 2 )Eab]
GENERAL REFERENCES
In addition to the works cited in the Introduction to this chapter, the following books provide comprehensive surveys, including detailed expositions of the theories explored here in a preliminary way. They also introduce the group of theories based upon the radial distribution function (cf Chapter 9) which has been excluded here, not because they are mistakenly thought to be conceptually difficult, but because (i) an essential prerequisite is to become familiar with an extensive and specialized vocabulary, and (ii) they involve a 'surprisingly formidable' amount of algebra which is not
n (O) ct
=n
n(O)
= O'
,(3'
S(O)
nS(O) ex.
,
v(O)
==
nv(O) ., Cl
and as a result of the perturbation
5:"
THE LIQUID STATE
n or.
n
Barker, J. A., Lattice Theories of the Liquid State (Pergamon Press, New York), 1963. Egelstaff, P. A., An Introduction to the Liquid State (Academic Press, New York), 1967. Fisher, I. Z., Statistical Theory of Liquids (trans. T.M.Switz, with supplement by S.A.Rice and P.Gray) (Chicago University Press, Chicago), 1964. Frisch, H. L., and Lebowitz, J. L., The Equilibrium Theory of Classical Fluids (Benjamin, New York), 1964. Tempedey, H. N. Y., Rowlinson, J. S., and Rushbrooke, G. S., The Physics ofSimple Liquids (North-Holland, Amsterdam), 1968.
n(3
= 5:"
Sa. == S(O) + oS S(3 = S(O) a. + .<:lS ,
().; ,
Cl
U", ,
,
= v(O)+ 5v v- ( 3 = ,v(O) A •• . + t..W
v
Ct
Ct
,
•
The distinction between .<:lS and oS emphasizes that as o~ -+ 0, oS -+ 0, but .<:lS does not, being the difference between the molar entropy of phase a and that of the hypothetical neighbouring phase /3. (a) Prove that for the complete system,
au) S,v (1}f
I
{(alU) aS v,n(.<:lS?+2 (a asauU) n.<:lS.<:llJ+ (aav2U) s,}.<:lV)2 } , 2
2
l
where all derivatives refer to phase a in the unperturbed state.
SOLUTIONS
Guggenheim, E. A., Mixtures (Clarendon Press, Oxford), 1952. I., Molecular Theory ofSolutions (North-Holland, Amsterdam), 1957.
(I) J.W.Gibbs, Collected Works. Yo1.1 (Yale University Press), 1948, p.56, Eqns.l, 2. l.Prigogine and R.Defay, Chemical Thermodynamics (trs.D.H.Everett), (Longmans, London), 1954, pp.35. 36.
I
I
193
7.1
Chapter 7
194
7.1
Show that the necessary and sufficient conditions to ensure v > 0 with respect to fonnation of any phase (3 differing only a: are
iPU) ( aS 2 together with either of
0,
n
2
(a U) 2 asa'J n > 0
(a2U) av 2 S, n
similarly (n
(n
2
S, n
\, !
> O.
C7.1.3)
Show that the conditions (7.1.1-7.1.3) may be expressed in the equivalent forms T I (7.1.4) Ks Cp ' v > 0, T 1 (7.1.5) ->0. Co ' vKT Solution
(a) The Taylor expansion for U/3 is
=
LlS +
(au) au
sLlv+
(a iJ)
/Lll)2.
and similarly for Ua., but with oS, ov, instead of LlS, Ll,). The net change in U (total) is o~)Ua. + o~l!a -
(~¥)v[(n - O~)OS+ o~LlS] + (~~)sr(n - OnOl) + o~Ll/)]
(a U) iJ.r
+21 aS2
+
oS,
s[(n- on(01J)2+ oHLltJ)2]
aZU) (asav [en -
oOoSov + o~LlSLl/J
OV to LlS, Llv, as
(n-o~)CS!O)+oS)+oHS!O)+LlS) ==
o~)oS+ S~LlS
= 0;
(n
Co0 2 LlSLlv.
2
I
(7.1.8)
2
2
p ( aav ) S (aT) as v - (aT)2 a;;- s' or
p p) 2 (aav ) s (aT) (a as v - as v
•
The first may be transformed into a simple product by using
(~~t = (~:t e~t
;
it then becomes
(7.1.6)
+
(~~)s q;)
J'
when the appropriate Maxwell relation is used, and this in turn reduces to -cap/aIJ)s(aT/aS)p in virtue of the standard change-of-constraint formula. The definition of Ks, with (as/a T)p = Cp /T, gives the required result. Similarly, with cap/aS)v = (ap/aT)vcaT/aS)v, the second form becomes
(~~)v
nSJO)
S02(OS)2 = (on2(LlS)2 ;
2
(~~t
whence
-
002 0S01)
Z
2
S=
(7.1.7)
(~~t l(~~)v + (~~)s (~:tJ
nU!O)
The conditions S, v, constant relate
= (on 2 (Llv)2,
and taking the limit as o~ -+ 0 gives the required result. To ensure equilibrium, the right hand side of the result of part (a) must be positive, irrespective of the higher order terms from the Taylor expansions. (b) The right hand side of this result is a simple quadratic in either Llv/LlS or LlS/Llv; it is everywhere positive provided (i) there are no real roots, and (ii) it is positive as LlS/Llv -+ <X'!. The first is ensured by condition 1.1) expressed in molar quantities, i.e. multiplied through 11 2 , Now, condition (7.1.1) implies that (a 2 u/aS 2 )v and ca 2 u/av 2 )s have the same sign, so the second provision is ensured by either one of conditions (7.1.2) or C7.1.3) being met. (c) The left hand side of condition (7.1.1) reduces at once to
2
+21 afJ2
(n
L
ou
(a U) v(LlS)2+ asaii LlSLlv
1
2
OU
(11-0n 2 (01)2
[··(a U) (a iJ) (a U) .. aS2 vCLlS)2+2 asa'J LlSLlv+ av /Llv
(7.1.2)
as 2 v,n > 0 ,
as v (au)
== 0;
Clearly the first two terms in OU are zero; using Equations C7.1.6) 1.8) as appropriate on the last three tenns gives
2
aV 2
onSv+o~Llv
whence
1.I)
(a U) ( a U)
195
Phase stability, co-existence, and criticality
S
+
(~~) vG~)
J
and the change-of-constraint formula
(~~)v [(~)
S
+
(~~) v(~~)J
gives the required result.
7.1
Chapter 7
196
7.2
Phase stability, co·existence, and "riti""lit"
197
An alternative procedure is apparent on writing condition (7.I .1) as
02U) U2S == ( OS2 v'
>0,
(7.2.3) or
Continue via the Jacobian, as
U2S
I Uvs
Usvl U2v
=O(T,-P)/O(S,IJ)=_(OT) /(Ov)
o(S, p) o(S, p) oS p op s
or equivalently,
oeUv, Us) o(-p, T)/O(I!, S) (Op) /(OS) O ( I I , S ) - O ( v , T) O(V,T) = - a;; T oT v' Note that the phase stability conditions (7.1.4) or (7.1.5) may be derived by considering the case where the total intrinsic er,ergy and the total volume are kept constant(2). The results (7.1.4) or (7.1.5), whose equivalence reflects the identity KslKT = CvlCp , indicate that in a one-component system diffusional stability is secured by the conditions for thermal stability (TICv > 0) and for mechanical stability (III) KT > 0). It will be seen (Problem 7.7) that this is not so for a two-component system, an additional condition of diffusional stability being required.
Solution
(a) Proceeding as in the solution to Problem 7.1 (a), write the Taylor series for and Fi3 about FJO), including terms up to (04FloV 4 )T' Then
ljF
(n
o~)Fa. + lj~p;, - n/?JO) ,
7.2 Higher order stability conditions. For simplicity we consider uniform systems in which one intensive variable is, additionally, held constant. The appropriate forms of the equilibrium condition are, cf.preamble to Problem 1.22,
( ~~) t;
> 0, T, v
F == U- TS, or
(OO~) > 0, S, p
H
U + PI) .
It is necessary also to validate the first-order stability conditions thus
by comparison with conditions (7.1.4) and (7.1.5). Prove that the first-order stability condition for a one-component, one-phase system constrained to constant volume and uniform and constant temperature is
02F) ( -ov 1 T, n
- (OP) (1)
I
T,n
liKT
and that the condition for stability when IIIJKT
02p) =0, ( ov 2 T, n
03 p ) ( ov 3
>0
(7.2.1)
I (02F) -lj~)O/)+O~d/J]+2! av 2 T[(n
I(a F) I(a F) + 4! 'av 3
+
a/J3
4
(7.2.2)
T, n
(~~)T
F)
(aaiJF) a F)
( aiJ
." .
4
0,
1
2
no~
2!(dV) n-lj~
T
I 3 n2lj~ ( 3!(Llv) (n-lj~)2 I
T
I n3lj~ [ 4! (dv) (n _lj~)3 1
3
3
I
T
4
(2) l.Prigogine and R.Defay, Chemical Thermodynamics (trs.D.H.Everett), (Longmans, London), 1954, pp.209-212.
-lj~)(O/J)4+ O~(d/! )4]+
T
Using the relations (7, 1.7), together with the corresponding higher-order relations for increasing powers of dv gives the coefficients of the derivatives of FJO), leaving off the redundant indices and suffixes, as
2
<0.
T[(n-o~)(lj/)3+ljHd/J)3] 4
( aaf.i 2
0 is
o~)(ljv)2+lj~(dli)2]
4
n20~) n30~ + 3 (lj~)2J -;::
198
7.2
Chapter 7
Thus, taking the limit as 8~ -+ 0, 4 2 3 aF) I (a I (a I (a afT,v=2! au 2 /Llv)2+ ! au /Llv)3+ ! av4r(LlV)4+ .... 3 ( 4 3
F)
F)
F)
This makes it obvious that while condition (7.1.5) implies condition (7.2.1), the reverse is not true; rather, condition (7.2.1) implies only that the two conditions (7.1.5) are met, or not, together. Taking conditions (7.2.1) and (7.2.3) together adds only that l/vKs , T/C v have the same sign. The algebraic relation between the F(T, v) and U(S, v) stability conditions gives the same information as condition (7.2.5), but deriving it establishes a general method which is useful in wider contexts. Commence with
C:)T,
2 a ( aoF)
[a~) (~~)J
2 ' T, n
T, n
n
(~~t,
=
=
[a~) (~~)
n =
J
s, /
-p,
[aas(~~tl, (~~)T, n
according to the change of constraint formula. Making use of
(aT) (as) av T, n (an) aT s, n as v, n = - I
,
and putting
(-~~t,
n
2 a U) ( avas n
(aaSU) v, n ' 2
(aT) '
as
v, n
2
we obtain
(aavF) 2
2
T, n
2 ( aauU) 2 S,
(aT) (as) asav n a;- s, n aT v, n 2 2 (aasavU) n2/(aaSU) 2 v, n '
(a2U) n-
2 (aavU) 2 S, n -
Phase stability, co-existence, and criticality
199
i.e.
The condition that (aF/aUr v > 0 is clearly either (7.2.1) or, if (a 2F/aV 2)T = 0, since (LlV)3 c~n take either sign, (7.2.2). In the general case, the condition for stability is that the first non-vanishing derivative of p must be (i) of odd order, (ii) negative. (b) The proof is exactly analogous to that of part (a). (c) The various sets of conditions are related as follows. That either condition (7.1.4) or condition (7.1.5) ensure all four of I/KT' I/Ks, T/C p , T/C v positive is obvious from the relations cited. Note that condition (7.1.1) alone does not ensure all four parameters positive; nor does I/Ks, I/KT > 0, etc. The relation between condition (7.2.1) and conditions (7.1.1), (7.1.2), (7.1.3) is obvious on writing I T/CvvKT a 2F) (7.2.5) ( av 2 T = vKT = T/C v > 0 .
whence
7.3
n
Sv F 2v U2S = U2S U2V - UvS UvS = U2S U I . I USv U2V (d) The conditions (7.1.4), or (7.1.5), equivalently imply that all four of the reciprocals of Cp , Cv , KT , Ks are positive; (ap/aT)v, n being positive and non-singular implies that (Xp has the same sign as KT , and has the same zeros and/or infinite discontinuities. Then, in virtue of the relations Cp = Cv + T(X~v/ KT and KT = Ks + T(X~v/Cp, a stable phase has Cp > Cv > 0, T/C v > T/C p > 0; KT > Ks > 0, l/vKs > l/vKT > O. This suggests that in the absence of an infinite discontinuity in T/C v , stability is likely to fail first by T/Cp , I /v KT going through zero simultaneously. If Cp , KT go through zero, however, then Cv , Ks approach zero faster. (e) In the van der Waals equation, using molar quantities,
ap ) (aT
v =
R v- b
If v = 3b, T
<
>0
for all v
(aavp ) T
>b.
= -
RT 2a (v - b)2 + -;?
8a/27 b, we have for the uniform-density state (3)
ap ) (-ao T >0,
I -K
V
T
<0,
(Xp <0.
Now C~ = 0, from Problem 6.6(b); C~ = ~R, Cv(internal) > 0; thus > Cv > 0 for all T, v > 0, i.e. T/C v > O. Further, Cp - Cv < 0 at the point specified, since I /v KT < 0, so Cp may have either sign, and Ks has the sign opposite to that of Cp . Along this isotherm, (ap/aV)T has two zeros which, with C v > 0 everywhere, necessarily define stability limits. Thus the stability condition which fails, at the so-called spinodal locus, is T/C p > 0, l/vK T > 0, and it fails by passing through zero; if Cp is positive anywhere in the unstable region, the part of that region within which Cp > 0 is bounded by zeros of Cp , the locus coinciding with zeros of Ks. It is the fact that phase stability normally breaks down by T/Cp , 1/ vK T passing through zero simultaneously, which justifies using conditions (7.2.1) or (7.2.3)-usually the former-rather than the complete conditions (7.1.4) or (7.1.5) when searching for critical points or for spinodal loci (see following problems). 00
7.3 Co-existence of two phases (i). A one-component system, con strained to constant total entropy, volume, and mass, is supposed to comprise two phases, (X and (3, in thermal, mechanical, and diffusional contact. The phases are both internally uniform, but they may differ in (3) That this state has a higher free energy than the two-phase state may be deduced from the results of Problem 6.6(a). For an analogous problem in statistical mechanics see Problems 11.14 and 11.15.
7.3
Chapter 7
200
T, p, and IJ.. The perturbation to be considered is the transfer of oS, Oli,
7.3
201
Phase stability, co-existence, and criticality
where
on (extensive quantities) from phase (3 to phase ex.
(aaSU)v,n'
(a2F) av T.n'
(a G)2 2
This is equivalent to a generalization of Problem 1.29. The general condition for equilibrium, as in the preceding problems, is (0 U)s, v > o. (a) Taking S"-, Ii a., and na. as independent variables, express Ua. and by Taylor series expansions about U!O) and 0°), and thence show conditions for equilibrium, Le. for U Ua. + UR to be an extremum, are (4) Ta. = 1/3, Pa. Pp, 1J.a. 1J.{3; (7.3.1)
(d) Use the same general method as in part (b) to prove that the conditions for stability under constant and uniform temperature,
and that those for stability, i.e. for U to be a minimum rather than merely an extremum, are
are secured by
U;s
ifsv
ifsn
U:;s
U;v
U:;n
u:.s
ifnv
U;n
I U;s U:;s
ifsv U;v
I >0 ;
U;s=
eu.) 2 ast 0;
2
v, n
+
Va.
I > 0;
IJ.
(7.3.3)
U)
+ (a~ (3
I)
(
~~) s,
=
pl"TP
, Fvn Uw F2v Fnv F2n
I
v. n
,
etc.
p
(
~~)
T, v
(
~~
G ) T, p
(:~)
I=
>0.
'
p
(~~)T
=
n
0, that
= V ;
and that the molar entropy and volume may be defined also by
s = (~~t. p' (7.3.5)
U2sF2vG2n = 0,
(7.3.6)
av) ( an T, p
v
,
with
U= Solution
(a) Since in OU are:
oU =
That the co-existence condition p. = p~ may be deduced equivalently by maximizing the canonical partition function for non-uniform density at T < 7;, is seen from Problem 11.14. A related analysis using the grand partition function establishes also that,... ,..~.
(4)
(~~)s,
v
(7.3.4)
The results of Problem 7.I(c) are useful. (c) To prove that condition (7.3.2) also is secured by condition (7.3 .5) is both formidable and tedious, but the method is similar to that of part (b). Prove the following general proposition, which is a pre requisite for proving that condition (7.3.2) is secured by condition (7.3.5),
U2S Us"
UvS U2v Uvn UnS Uno U2n
etc.
T,p'
> O.
V{3
also, it follows from the Gibbs equation, SdT- II dp + ndlJ.
> 0;
TIT -C' > 0; v -;;va.
an
2n
lla.KTa. '
(7.3.2)
G2n =
2
Fv~ I > 0, F~+n > 0 , F+
F;v
IFnv
(b) Prove that the conditions (7.3.4) and (7.3.3) are secured by I ' K T,,-
F2v =
2
Note that it follows from Problem 1.20(b) that
U;s where
2
U2S =
oSp
-OS,,-,
ua.) v,n [(-aasa.
oVp
=
(~~
p
-ov"-, onp
=
-on,,-, the first-order terms
(au) ]oS a. + [(aua.) ]00 " ~ - (au.) ~ asp v,n _ S,n allp S.n aV,,-
+
f,(aua.) v - (~) lona. + .... anp s.
Lana. s,
Clearly, conditions (7.3.1) are the necessary and ensure 0 U 0 to the first order.
conditions to
202
7.3
Chapter 7
When these conditions are met, I
ou = "2
[(aZu,,-) as~
+ as~
(32v CX2SCXlv", ..... 20
(oS,,
v,
(oS,.01J,.) I
+2
I·
+"2l(
n
aU,. an~
(a~ anffi 1
s,
II
+
0
)]
s,
II
2
na.)
(
+
I~~( a2 ua.
~
ana.as,.
CX2v
+ (32S(32va1'20
+ «(32S(32V
_
2 CXS v _ (3sva I'Sv
2 CX2v (3sv) a I'2v
2
+ CXSv
((32V _ (3SV) CX
2v
CX
Sv
2
(CX2V _ CX s v )
+ (3sv a
I'2v
a
I'Sv
.
(7.3.7)
aS2"
CX2S,
"
~
asp a,) p
(3sv ,
we can write the determinant (7.3.3)
= (CX2S+
+ (32V)
(cxs v + (3sv)2 •
out and collecting like terms we obtain cx~v + (32S(32V - (3~v + CX 2S(32V
cxs v(3sv
+ CX2v(32S -
cxs v (3sv •
According to the solution to Problem 7.1 (c), the first four terms become
T T --,:-:----- + -=--=- v a.KrfL C v"
vpKrpCvp
CX2V)2 ( CXSv - (3SvR I'2v
(32 v a' 2v
( 1 + x)A + ( I +
~ )B + xC 2 ,
Thus D has the form D
Arraying the coefficients in the expression (7.3.7) so that the three 'even' coefficients, i.e. those of (OS,,-)2, (8V,,-)2, and (on,.)2, are diagonal, gives the determinant on the left hand side of inequality (7.3 .2). Since 0 U is to be positive for all possible combinations of oS,., 81)", on", it must be zero in particular for any two of them being zero, so that the three diagonal elements must be positive; when anyone is zero Equation (7.3.7) reduces to a simple quadratic, and the argument of Problem 7.l(b) thus adds that the 2 x 2 minors of the determinant shall also be positive. Since the determinant comprises the totality of the coefficients in Equation (7.3.7), it follows that the conditions (7.3.2), (7.3.3), and (7.3.4) are both necessary and sufficient. (b) Using the notation
a2 u a2 u,
CX1SCX 2v -
2 (32 v CXS v )::;;""20
v
+ terms in (OS,.)3 , etc.
D =
( CX 2SCX 2v
2 (3sv CXSv", .....So
)
+ (~) anpasp v ](On,.oSfL)
D
203
2
+( a~:~JJ (olJ,.on fL ) )
Phase stability, cO'existence, and criticality
The last two terms rearrange to the quadratic form
I' aZU,.) ,.) + I av ,. an ,. S
+
Z
7.3
The last four are partially separable as
(aZup)
v, n
11
A
K
T
C
,etc.,
_ v(JKsf3 K
x -
r,. v" Va. S" Evidently D > 0 is secured by x > 0, [(1 + x)A + (1 +1/x)B] > O. According as x ;:s; I, however, the latter condition may be met with either A or B < 0, and this is not pre-empted by condition (7.3.4), T/Cv,,- + T/C vp > O. Thus the conditions (7.3.3) and (7.3.4) are secured by the condition (7.3.5), though not by (T/v a. Kr"C v,. + T/v pKrpCvp ), > 0; but conditions (7.3.3) and (7.3.4) do not require (T/C v ,.+ as necessary. The further inequalities from condition cond!tion (7.3.2) also are secured by condition (7.3.5). Note that condition (7.3.2) includes terms in I)"
a2 U) - - I ( 1+---4- Scxn T) (avan S - nKs Cp ' 2 a ( u)
S. v
-;(~+ TS2).
n
Kr
Cv
A two-way perturbation always leads to a condition like (7.3.3), implying through its cross-terms that some property has the same sign for cx and (3, whereas the diagonal terms of condition (7.3.2) being positive implies only that the sum for cx and (3 is positive. This is because a determinant of sums is not equal to the sum of the corresponding determinants for cx and (3 separately. (c) The form in which the problem is stated suggests working from right to left in Equations (7.3.6). The method suggested by the solution to Problem 7.2(c) is to use the change-of-constraint formula. To prove (a2G/an2>r,p 0, start with the definition G == U- TS+pv, whence given the Gibbs equation [Problem 1.20(d)] for one Mtnl">Af'lpn1 SdT-v dp+ ndJ.l. 0, we have dG
-SdT+vdp+J.l.dn
204
7.3
Chapter 7
whence
( OG) on T.p =IA,
(02G) on 2 T,p
=
(01A) on T,p;
but dIA -SdT+ 15 (dlA)T, pO. 2 To relate (02G/on )T, p to second derivatives of F, write
( 02G) on 2 T, p
(01A) on T, p
(01A) (01A) (ov) -on T, v+ & T, n on T, p ,
7.3
Phase stability, co-existence, and criticality
Forming the determinant in F and multiplying through by (0 2 U/oS 2 )v, n gives the required result. The result (7.3.6) has the following implication in regard to expression (7.3.7). If the latter is separated into two quadratic forms, one for each phase, then the corresponding pair of 3 x 3 determinants are each zero by (7.3.6). This means that for each phase there is a particular combination of oS, (1), on (such that oS/on S, ol)/on = Ii) for which the quadratic form has two coincident real roots. Thus, provided that the principal minors of the determinant for that phase are positive, Le.
(02F) (01A) ( 02F) or12 T, v + ovon T op T, n ' (01A) l(oP) 2 T,V+ (02F) ( 02F) =(jn ooon T OV T,n & T,n' 02F) = ( On 2 T,v
(02F)2j(02F)· ol)on T 00 2 T,n'
and the required result follows on multiplying through by (02F/OI)2)T, no Alternatively, use the method of Jacobians,cfsolution to Problem 7.1(c),
F2n Fvn I
f~vl
F2v
o(IA,-ph O(IA,-p)/o(n,l) (01A) /(OV) o(n,t)T = o(n, -p) o(n, -p) = - on T, p op T, n 02G) (02F)
= ( on 2 T,p Ot)2 T,n'
The third step necessitates expressing the three second derivatives of = F(ll, n) at constant temperature in terms of the second derivatives of U = U(S, P, n). The above procedure gives
U2S Usvl Uvs U2v
I
> 0,
U2S
v-
(02U) /(02U) oSon v OS2 v, n '
and from Problem 7.1(c) we obtain
Similarly
( 02F) Ot) 2 T, n
(02U)
S, n -
(02U)2/(02U) oSOV n OS2 v, n .
2 (01A) (01A) (01A) (OS) 0 F) ( Of) on T = OV T = OV S + oS v, n a;; T, n 02U) (02U) (op) = ( ovon s + oSon v oT v,n =
02U) (02U)(Op) (OS) ( ovon s + oSon v oS v,n oT v,n 02U) (02U) (02U) /(02U) ( ovon s - oSon v ovoS n OS2 v,n'
> 0,
the quadratic form for that contribution to 0 U is everywhere positive or zero. These are the conditions (7.1.1)-(7.1.3) secured by conditions 1.4) or (7.1.5). The same applies to the other phase; but unless the two phases are identical, there is no combination of oS, ov, on for which OU overall is zero, since non-identity implies oS/on = Sa. S/3, etc. Thus condition (7.1.4) or (7.1. 5) for each phase secures overall stability. (d) Using the notation
*
2
Fa.) = a , 2n ( 0on! T, v we can write the determinant
02p, ) ( 01J/30~/3 T
(a2n+{32n)(a2v+{32V) (a vn +{3vnf
= a2na2v+{32n{32v
F
(02U) ( 02F) on 2 T, v = on 2 S,
205
{3v. n, etc., a~n-{3~n
+a2v{32n+ a2n{32v- 2a vn {3vn . The first four terms are zero by Equation (7.3.6). That equation enables the last three terms to be put into quadratic form as:
(avn{32n - {3vn a 2n)2 a 2n {3 2» This is positive if, and only if (02Fa./on;,)T, v and (02f~/on3h, v have the 2 {32n a vn ",....2n
+ {32
_ 2
(01A)
(011) T, n on T, p
vn{3 2n
{3
a vn vn
=
same sign. Now
( 02F) on 2 T, v = cfpart (c), so
(1)
1")
02 ( on 2 T,
(01A) (on) (oP) - op T, n on T, p on T, n' v-
v =
vKT
v
n2KT
Thus the determinant is positive if, and only if, KTa. and KT/3 have the same sign, negative volume being excluded, and the other condition
02F) + (02P,) ~ >0 (_a. on~ T, v on! T, v
206
Chapter 7
'.
7.3
requires that sign to be positive. Compare this result with that of Problem 1.30; again, a two-way perturbation requires for stability that the two phases or fluids be individually stable, whereas a one-way perturbation requires only that they be collectively stable. 7.4 Co-existence of two phases (U). Reconsider the system of Problem 7.3, but use molar rather than extensive properties as the variables. The perturbation may then be described in terms similar to those of Problem 7.1, i.e. as o~ mole passing from phase (X to phase {3, with consequent changes in the molar entropies and volumes of the phases imposed by the constraints to constant total entropy and volume. (a) Show that (n«-o~)oS(t+(np+o~)oS(3-(S£O))o~ = 0, (7.4.1) (n(t
O~)ov",+(np+o~)ov(3-(v~O)-vJO»o~
OU
n",(U(t -
U~O»+
np(Up -
~O»+
oHU(3 -
0,
7.4
where the constraints on the partial derivatives refer to the molar quantities for phase (X; and similarly for . Substituting in 0 U we obtain
-o~ [u,(0) + (~) '" as",
r
-
oU = (n ... - o~)U'" + (n{3 + o~)Up- n... mO) - n(3~O) . Note that if n(3 -l" 0, and ~O) -l" U£O), this reduces to the corresponding relation of Problem 7.I(a). The Taylor expansion for
-(0) ) (au... ) u.. _-u. . + (au... as... F OS ... + av,,-sOv..
Then, by virtue of Equations (7.4.1), (7.4.2),
OU
[(n", -
o~)oS", - S~O)o~{e~:) F - (~~)
+[(n", - onov", -v~O)o~] [(~~:)s -
(a 2v.. ) +2 as! F(OS",) + 1
as... av",
~SbO) + pp
(UJO)- T",S~O)+ p",v~O»]
°
(:~) (~~)p (:;) v
T
-I
to obtain [.1 U- TS+ pv. Thus the third condition is [.1", = [.1p. (c) The second-order terms in 0 U are
2-
...) (0- )2+ ... os.. ov ... + ~(a2U 2 av! s 1)... ,
(~)s J
+second-order terms. (7.4.4) Clearly, for OU = (to first order) T", = ~ and p", ~pp. _Now G == U- TS+ pv so, on dividing by n, we obtain [.1 = U- TS+ pv; Alternatively, take (au/an)s v = [.1, change variable twice to give (au/an)r,p = U, and use .
-
2 2 (a u... )
(au.)
(all)
U",) _ (0£0) i +v~O) ( au",) av", s - v&O} (aUp) avp "8] + second-order terms.
a 2u'" (n",-0~)L2(aS!)/oS",)
U.. is
(a~",) Of) J a!) '" s '"
+ o~ rU!O) - uJO) + S~O) (a S&O) as", F as(3
+ o~[ (uJO)
Rearrangement gives the required result, and similarly for 00 = 0. Since U... = U... (S... , v.. ),
'"
+[(np+o~)oS(3+SbO)o~] ~ ii+[(n(3+0nOI)(3+V&0)o~] ~ S
Solution
°
+
oS if
+o~ ~O)+ (~~)OS{3 + (~~)sOVIlJ + second-order terms = [(n", - o~)oS'" -SiO)o~] (~~:)ii + [(n", - onov", -~O)O~](~~:)s
(7.4.3)
_ (a) !nitial state: S = n",S!;.O) + n(3S~O); perturbed state: SIt = S~O)+ oS", Sp = S!O) + oS(3, nIt ot etc., and this specification is complete, since oS"" O!)(t' o~ are independent variables. Then oS = (n ... - o~)(S~O)+ oS(t)+ (np + o~)(SbO) + oSp)- n... S~O)- npSbO) .
[(~) as(3 i OSp + (~) aV(3 O!)(3Jl
J
i
oU -- n... [(au«) as... OS ",+ (au... av", ) sO!) ... +np
(7.4.2)
where the indices and suffixes have the same meanings as in Problem 7.1, with the incidental difference that neither oSp nor OS(t necessarily tends to zero as S~ tends to zero ( and simila~ for the 01) t;.l 01) (31 (b) Use Taylor series expansions for U(t' U(3 about U~o>, UJO) respectively to confirm that the conditions for equilibrium are (7.3,1). (c) Thence show that thermodynamic stability is secured by the condition that each phase be stable by the criteria (7.1.4) or (7.1.5), (d) Give a physical interpretation of condition (7.3.1) in terms of the tangent planes to the surfaces U'" = U(t(S(t' v",), Up = U(3(S(3' vp). What is the maximum number of phases which can, in general, co-exist in a one-component system?
207
Phase stability, co-existence, and criticality
+
2
J
a 2U'" ) 1 ( a 2-) U'" 2 + (as",av« oS",ov"'+2 av", 8(0/) ... ) 2~
( 2-)
~ lau(3 o~) [2lau, (a:sl)F(OS(3) + (aspavJOSpOV(3 + 2 avp 2
/Ol)p)
J
2
(7.4.5)
The simplest condition ensuring that this sum is positive is that both square brackets be positive, i.e. that each phase fulfils (7.104). Since 5S(3' 5v (3, are dependent variables, it follows that while this condition is sufficient, it may not be necessary. Substituting for 5S(3 and 5v (3 according to (704.1) and (704.2) does not reduce the second-order terms to a single bracket, except if the two phases are identical. Thus the conditions which may be formulated in terms of sums
1
[(o;f£) + e;~~)J ' etc.,
are not sufficient. (d) The stability conditions (7.104), (7.1.5) ensure that for a stable
phase the U U(S, v) suiface is concave upwards. For two co-existing
phases, the equili~!ium c0I!.ditions Ta..:::' 1{J, Pa. ;= P(3 ensure that the
tangent planes to Ua., U(3 at Sa., Va. and S(3' v(3 respectively are parallel; it
is evident from the third term of expression (70404) that 11", = 11(3 ensures
additionally that the tangent planes are identical. Thus two phases can
co-exist in a one-component system over a range of temperature, but
fixing the co-existence temperature fixes the co-existence pressure.
There is in general one, and only one, way of putting a common tangent
plane to three uniformly curved surfaces. Thus three phases can co exist at a determinate temperature and pressure. It might seem that four phases could co-exist fortuitously, but since the tangency must be geometrically exact, this occurrence has zero probability. 7.5 Criticality, continuous equation of state. The region of liquid vapour co-existence of real fluids is invariably bounded in temperature, and the same is true of the van der Waals and similar continuous equations of state, ~(p, ii, T) ;= O. Designate this bounding temperature 'Fe. Then for all T > 'Fe the fluid is stable, by the criteria of Problem 7.l(b), ( c), over the accessible range of volume; and for T < Te, the continuity of ~(p,fJ, T) = 0 implies a region of instability bounded by the so-called spinodal locus, U2S _
IUvs
_Us.
I
= 0,
U 2V
cf.Problem 7.2(e). The locus joining the molar entropies and volumes of co-existing phases is called the binodal locus, and defined by the co existence conditions (7.3.1). (a) Consider two co-existing phases, ex and ~. Use the three co existence conditions in terms of the F = F(T,v) surface to show that if the tie-line connecting Fa.(T,va.) to F(3(T,ff(3) becomes vanishingly short as T -:;. Te, then at T = Te
OF) ( oU T
< 0 , (02P) ov 2 T
=
(03P) (04F) 01)3 T = 0 , 01)4 T
>0,
Phase stability, co-existence, and criticality
7.5
7.4
Chapter 7
208
I
I
i
209
and that the binodallocus is tangential to the spinodal locus. (b) Consider a temperature !:J.T below 'Fe. The co-existing phases are defined by !:J.p, 1:J.1),., !:J.v(3; and since the equation of state is continuous, !:J.p is expressible as an explicit function of !:J.v, !:J.T along this isotherm from & a. to !:J.1)(3' Use a Taylor series expansion for !:J.p in terms of !:J.v and !:J.T, together with the co-existence condition 11a. = 11(3 to show that p\ 2 {(02p) 1 (03 2 [(!:J.va.) 2 -(!:J.v(3)] ovoT c!:J.T+4' ov3)c[(!:J.Va.) + (!:J.V(3) 2 ] } = O. Derivatives with respect to T of order higher than first may be neglected. (c) Use Taylor series expansions for !:J.p;= !:J.p(!:J.v a.' !:J.T) and !:J.p !:J.p(!:J.v(3,!:J.T), together with the result of part (b) to show that p 02p) 1(03 ) JYz !:J.v IX = -!:J.v(3 [ -6 ( oToiJ c!:J.T oU 3 T, C i.e. that the locus of the mid-points of the tie-lines connecting the molar volumes of co-existing phases passes through the critical point, the binodallocus near the critical point being quadratic in volume, and that
~~ = (~~) v
.'1
c'
i.e. that the vapour-pressure curve is co-linear with the isochore passing through the critical point. (d) In real fluids, the binodal locus near the critical point is described better by !:J.va. :::::: -1:J.1)(3 :::::: X(!:J.T)'J, than by the result of part (c); what does this imply about Hp, D, T) = 0 in the critical region? Solution
(a) The co-existence conditions are:
Ta.
1{J,
Pa.=P(3,
F)=F(3-p(v(3-va.),
~ that at each T,
and P(3 have a common tangent of slope -po If F F(T, v) is continuous betwe~ v_and v(3 there is necessarily an intervening region of instability (F = F(iJh convex upwards) bounded by points of inflexion (02P/ov 2h = 0, and these are necessarily both between iJ a. and U(3, i.e. the spinodal locus is necessarily inside the binodal = F( T, v) being everywhere locus. If the binodal locus is to close, analYtic implies that the points Va. and v{3 and the two points of inflexion (a 2 F/oV 2 )T = 0 must coincide at T = Te, i.e. the spinodal locus coincides with the binodallocus at T = Te, with
02F) ( -ou 2 T,
-
c
(op) -ov T,
-0
c -
•
7.5
Chapter 7
210
J
If this point is to be stable, then the given conditions on the higher derivatives of F must hold. The remaining condition, p > 0, follows
from the requirement that all continuous equations of state necessarily converge to pu = RT as fj -+ 00. Thus a dilute phase (vapour) can never sustain negative pressure, and so liquid-vapour co-existence can only occur at p > 0; and this applies also at the liquid-vapour critical point. The critical point is the only point on the spinodal locus at which Cd 2 F/ai5 2 h and (a 3 F/afj3)T are simultaneously zero. (b) Since
p (aau ) T,
C
(a2p) au2 T, =
0,
C
7.5 or, since 11/J"
f
udp =
i
I
t
2 } + 41 [ (I1v,,) 2 +(l1v{3)]
;1
1
Ll.va
&
/J
T, C
d(l1v)
r.
0,
3p
(aarau2p ) l1TJ + 811 f(03au c
v..
p)
Vp
3
J
v.
e(l1v)4 V~
,
and the required result follows. (c) The Taylor expansions are as in part (b) except that & is replaced by !1t)", 11/J{3' respectively. Subtracting the second series from the first eliminates I1p as
(a 3p
0,
03 p ) ( OV3 T, c (I1v" + &(3)(111) " _11/J(3)3 = 0, and 111)" =
-&{3 .
02p ) c(&,,-!1t){3)I1T+ 1 oiJ3 ) e[(l1v,,)3 0= ( aTov 3
T,
C
(!1t) ,,)2 ,
and the required result follows. To obtain I1p/I1T, add the two Taylor series expansions for I1p, and put &" = -l1u(3; since (110,,) 3 -( I1v {3 )3, the second and third terms on the right hand side are each zero, and the required result follows. (d) The results of part (c) demonstrate that the physical consequence of the assumption that is a continuous differentiable function of T and v is that the binodal locus has a rounded top which is quadratic in v. The order of the curve depends on the order of the first non-vanishing derivative of F with respect to u at constant T. If the fourth derivative is zero, then stability requires that the fifth be likewise, with the sixth being positive. If the first non-vanishing derivative is of the order 2n (since it must be even), an analysis similar to the above must show that !1t)" is proportional to (11 T) 1/(2n - 2). In other words, no continuous 0 can give I1v" proportional to (11T)'j, or equation of state ~(p, iJ, T) to any similar power of I1T. Note that if the equation of state is not continuous from liquid-like to gas-like states, the spinodal locus ceases to have any simple meaning; but the critical isotherm is necessarily continuous, even if non-analytic, and, at least
Substituting into the argument of the integral and integrating gives for the first two terms I
(I1V{3) 2 ](110,,-11/J{3) 2
02 p \ I (03 p) (oTou) c I1T = - 3! ov 3
a -) I1T+-1 (a- ) , (11/J)2+ .., a(I1P») = ((a( I1v) T. c aTau T, c 2 aiJ 3 T, c
2L(I1/J)2
3p) (aou) T}(&") 2
Substituting back into the equation for (a2pjaTor5)cI1T, we obtain
and [a(l1p )/a(l1v )JT. c is obtained by differentiating the Taylor series as
2p
0,
whence I
The physical meaning of this condition-known as Maxwell's equal area rule-is obvious from the diagram of Problem 6.4. Expressed in terms of & as independent variable it becomes
f
r) (I1v ,,)2+ I1v ,,11/J{3 + (11/) (3)2] .
p [(11/),,)2 - (111) (3)2] ( 03 ov 3 ) T, e { - 3!I [(l1v ,,)2 + I1v ,,&{3 + (11/J (3)2]
I'
&dp = O.
[a(I1p)J '0(11)
1 (03 p ) 3! av3
Substituting for (o2p/orafJ)cI1T in the result of part (b), we obtain
P~
Ll.V.
211
*" I1v{3'
a2p) (oTou CI1T
the expansion for I1p is
2 3p ( a p\ ap) I (a ) I1p = ( aT ii, CI1T+ arav) C I1v 11T+ 3 au 3 T, c(I1v )3+ .... The condition IJ." = IJ.{3 has to be expressed in terms of I1v" and I1v {3'
given I1p" I1P{3' and I1T. Now (aIJ.japh = iJ, cfProblem 7.3(d), so IJ." = IJ.{3 implies Ll.v.
pa
Phase stability, co-existence, and criticality
p
(aoiJ)
(I1V{3)3],
t
= (02p) C
OU 2
c
= 0
•
Chapter 7
212
7.6
7.6 Stability at a critical point. The preceding problem examines the critical point in terms of derivatives of F = iJ), i.e. in terms of the volume derivatives of pressure. To elucidate the relevance of the condition of thermal stability, T/C v > 0, requires the [1 = [1(8, surface. (a) Express the slope of the projected tie-lines (on the 8, v plane) by alternative equations derived, respectively, from the co-existence 1{J, Pa. = Pp, by using Taylor series expansions for AT conditions Ttl. and Ap in terms of AS a. , A,}tI.; ASp, Avp. Thence show that the binodal locus at T = 'Fe is characterized by -(Op/aU)T = 0, irrespective of whether or not T/C. = 0; i.e. that the binodallocus on the [J = [1(8, surface coincides, at the critical point, with the spinodal locus as specified by the condition of mechanical stability alone. (tJ} Us~t!!.e equation of the tangent at T = 'Fe to the spinodal locus on the U = U(S, v) surface, D
s
I r:: Uvs
~Sv I
U2iJ
0,
c
= 0 .
Solution
(a) The expansions for AT and Ap are [leaving off the bar, since all extensive variables are to be read as molar (intensive)]: 2 (a2U) (a U) AT = A as • = aS2 v, cAS + asa,) c A!J + ... , 2 2 -Ap = A ( -a u) = (a - - AS+ (a -2 • Av + .... av s asov c ou s, c
(OU)
U)
U)
Putting Ttl. = 1{J, Ptl. = PP' and subtracting the series for {3 from that for ex, and neglecting terms of order higher than the first, we obtain 2 a U) (a2U) ( aS2 v, c(AStI. - ASp) + aSa,} c (Av... Aup) = 0, 2 (a2U) a U) ( asav e (AS... ASp)+ ov2 s)Av tI. - Avp) = O. At the critical point, the limiting forms of these equations equivalently define the tangent to the binodallocus: 2 (a2U) (a U) OS2 v,e dS+ asav e dv = 0, oSov c dS+ ov 2 s,c dlJ O.
(02U)
(02U)
02u
asov =
(aT) av s
-
Phase stability, co·existence, and criticality
213
The conditions of stability above the critical point require that (a 2 p/avanc < 0; dividing through we obtain
(~): dS- (~): (~~)
v, c dv
+
I
l(~) a~:TJ(~:~)T' c
c th)
= O. (7.6.3)
Comparing Equation (7.6.3) with (7.6.1) and (7.6.2) evidently justifies the assertion of the problem. It is sometimes argued (5) that the identity of (7.6.1), (7.6.2), (7.6.3) requires that the ratio (ap/av h c/(TICv)c = 0, so that if Co is infinite at the critical point, the infinity is of a lower order than that in KT • Certainly, for (T/Cv)c finite, even if vanishingly small, the first and second isothermal derivatives of p with respect to v must be zero, and criticality is determined by the conditions of mechanical stability, i.e. in the one-phase region near 'Fe,
TTl >-C ~O, >-K p v T
~O
cf.solution to Problem 7.2(d). It is doubtful, however, whether the foregoing analysis, being based on a continuous analytic equation of state [for which (T/Co)c is finite, cfProblem 7.2(e)] is applicable to real fluids in which, apparently, (T/C,)c = 0, because the critical isotherm in such systems is non-anaiytic at the critical point.
to show that at the critical point
~ (~~~) r,
7.6
p (as) (aT) T (a ) ov T as v = - c. aT
v '
BINARY SYSTEMS
7.7 Diffusional stability of the single phase. In a binary system the local composition is completely specified by a single intensive variable, the mole fraction of component 1,
n - n+m
x=-
N N+M'
where n is the number of moles and N the number of molecules of component 1, and m the number of moles and M the number of molecules of component 2, within an arbitrarily small element of volume. We therefore expect stability with respect to diffusion of either or both components in a binary system at uniform temperature and pressure to be ensured by a single condition. Suppose a two-component, one-phase system to be maintained at constant and uniform temperature and pressure. The appropriate form of the general equilibrium condition of Problem 1.22 is 8Gr,
p
>0.
Consider two regions, ex and {3, each internally uniform with respect to the chemical potentials III and 1l2' ex containing n ... and mtl. moles of the (5) J.S.Rowlinson, Liquids and Liquid Mixtures. 2nd Edn. (Butterworths, London), 1969, p.83.
7.7
7.6
Chapter 7
214
111 = 111(T, p, x);
- .G(O)+ G(O) -- n..111.. + m ..112.. + n{3111{3 + m{3I1~{3 . . {3 G(O) The perturbation to be considered is the passing of on mole of component 1 from a to 13 and, simultaneously, the passing of om mole of component
=
/I
,...1{3'
"
(7.7.1)
/I
,...2..
,... 2{3 •
(b) Use the definition G nl11 + m112, together with the Gibbs-Duhem equation, cf Problem 1.20Cd), SdT+lJdp+
Li nidl1i
11 1
( '0ax )
_
T, p
T, p, m
0111 ) ( om n(0111)
an
am
(7.7.2)
0,
(7.7.3)
0111) an
I x Ol1l
m =
0111 ) ( om n 02111) ( on 2
n + max; -
x 0111 n+ m
ax
an
=0.
(7.7.4)
(0 112) om
n
om~
n
02111 --= omon 02112
x oil' = - n+ m
-a:x:
ndl11 + mdl12 = O.
Put
an
dill = ( 0111) m dn + (0111) om
2 ( x ) 2 0 11 2x Oil 2 n+ m OX22 + (n + m)2 ox
x(1-x)02
2x-I oI12
n
dm ,
dl12 = ( -0112) dn+ (OI1Z) an m am n dm;
x(1-X)02 111 2x-lol11 --+ (n + m)2 oX (n + m)2 OX2
112 - - - (n+ m)2 --+ omon ox 2 (n+ m)2 ox
(7.7.8)
The last two terms include only the independent variables, so to ensure that the first-order terms sum to zero, these terms must be independently zero. The condition (7.7.1) follows. Thus, in addition to being uniform with respect to temperature and pressure, the system has to be uniform with respect to each of the chemical potentials. Note that condition (7.7.1) must apply also in the case where a and 13 are co-existing phases. (b) For a binary system at constant temperature and pressure the Gibbs-Duhem equation reduces to
n + m ax ;
(I-X\202111 2(1-X)0111 n+ m) ox 2 (n+ m)2 ox 2 = (1- x) 2 0 112 _ 2(1- x) 0112 02112) ( on 2 m n+m ox 2 (n+m)2 ox 2 02111) . (X)2 0 111 2x 0111 ( om 2 n· \n+m ox 2 + (n+m)2 ox (
(7.7.7)
+(1111l 111Jon - (1121l 112",)om.
m
0211 )
>0
oG = (n", - on)ol1l", + (m", + om)0I12", + (nil + on)ol1lll + (mil - om)oI1Z{3
1 x 0112
m =
(7.7.6)
•
(a) Write down the exptession for G = G", + Gil (subsequently to the perturbation) corresponding to the given expression for G (0). By inspection
T,p,n
(0 112)
>0
Solution
(c) Formulate the first and second derivatives of x with respect to n and m, at constant temperature and pressure, and thence prove
(
T, p
and express it also in extensive variables.
T.p,m
+m(Ol1l) T,p,m
aX
02e
+m(0112) = 0, on 1', p, m
_ (0 112) T,p,n on
(0 112)
(f) Show that condition (7.7.6) is exactly equivalent to
= 0,
to prove the results n(Ol1l) on
(7.7.5)
112 = I1zeT. p, X) .
(d) Expand 0111 .., 01111l' 0112 .. , 0l121l as Taylor series in powers of on, om, and use the results of part (b) to show that the condition (7.1.1) being met ensures that the first-order terms in oG = oG(on, om) sum to zero. (e) Use the results of part (c) to convert the second-order terms in oG = oG(on, om) into derivatives with respect to X; derive a result from Equation (7.7.2) which reduces the second derivatives of 11; with respect to X to first derivatives; and thence show that for stability (terms second-order in on, om, to sum to a result greater than zero)
2 from 13 to a, without restriction on the ratio on/om. (a) Express oG in terms of on, om, and thence show th~ t a necessary condition for equilibrium (terms first order in on, om to surrJ to zero) is "
215
Thence establish that 111 and 112 are completely specified by
two components respectively, and 13 containing n{3 and m{3 rJ'loles, so that the extensive Gibbs free energies are, initially,
""1..
Phase stability, co-existence, and criticality
then
l
111 ( 0an )
m
an J [('0am111) n + m (aamI12)
+ m (0112) m dn + n
I dm
= 0 .
Chapter 7
216
7.7
7.7
2
ax
dS- (T) (a (:£) C C aT p
2n ~]
+
(-CT)
v c
J
an
(-CT) (a-aTp) v
= (1
m
oM! x)a:;,:-,
(~)
aM! -X-, ax
+m) (-aMI) am II
5G
=
(n",
lX
m
1 8n + (aM2<¥.) am n 8m J+'"
II,
c
c
dS-
(a
c
C::)
n 5mJ
(aD) as
+ ... ; (7.7.9)
II,
c dv
] [(~\ m 5n+ nil an}
m
'(~\ In(3 am}
n
+ [ n", ( aMIIX) am) n ] 8m am n +m", (aMZ/i\
(T) (a-aTp) C
2 dv v, c
dv- T, c
+
aT)
an
= - I ,
II
v
c
•
> 0;
[(~) IG~) T, c
= 0
II,
through
J
dv
O. (7.6.2)
(aT)2 alp as vavaT .
v
- ( as
- [ n rL (an)
aT
0, changing the differentiating variable from S to
8G = (aM2/i) m +m rL
(as)
p aD) (aT) [a (a ) ( a;;8 = as II av av
the terms which are actually first order are
aMIIX\
T
p
dS+ - ) alJ
(~) (:~)
and since (ap/a!) h T, we obtain
·anj
[e:~I!) m 8n -
c
(av) as
-(~~)II(~)T' e~) v= - (~DJa~(~~) Jv - (~~)v (~~)T'
M ] + (nil + 8n) I,(a !(3\ m 8n - (aMI~\ am} n 5m + ...
Hm" - 8m)
'
D
J
r(
dv = 0
(ap/a/) h, c must be zero irrespective of (T/Cv)c' (b) From Problem 7.
5n+ (aMI<¥.) - - 8m + ... m am n
aM2/i) +(m +5m)-.-an
8, c
If Equations (7.6.1) and (7.6.2) are to describe the same line, then
etc. Note that, consequently, condition (7.7.1) implies x'" = x(3' (d) Substituting the Taylor series expansions for the 5M into the residue of expression (7.7.8) (less the last two terms) gives aMI-IX} 5n)l- ( an
(7.6.1)
II, C
The result of Problem 7.5(c) establishes (ap/aT)v,c gives
This is a tautology; it reflects the possibility of segregating variables, as aMI) an
dv = 0 ,
)
c
p (ap) dS+ (a - ) aT II, c av
aT) (~ 8
aaxMl dx . (n+ m) (
II
or, changing constraint on the last term, and using
aMI [(ax) ( aM!) m dn+ (OM!) am n dm = a:;,:an m dn+ (ax) am n dm =
C
II
etc.; and similarly for M2' To establish the functions (7.7.5), dMI
217
so the equations to the tangent become
a M! [ a (OM!)] { a [OM! -n ]} anam = an am n m = an a:;,:- (n+ m)l m n ,a (OM!)] OM! I·· I - (n+m)l Lan m +a:;,:-- (n+ x(l- x) a2M! aMI -+ 2x I ' - (n+ m)2 ax 2 (n+ ax'
Phase stability, co-existence, and criticality
MZ + mil (a (3\
an} m
+m(3
and Equation (7.7.2) shows that all four brackets are zero.
(~\ am}
J8n
v
aT)
= - ( as
II
T
J
8 -
p aZT (a ) av
T'
r(aZp) + alJazpa T (aT) ] ~ 2 (as) (aT)2 (aZp) ap av + av a T alJ as L av 2 2
T
8
T
T
'
II •
Thus the equation to the tangent to the spinodal locus is
18m, w
l
2p ) (T) (a Zp ) - dS+ (-CT)2 (aavaT C av -2
v c
c
II
c
T, c
dv
p (T)2 (a - ) C aT v c
v, c
2p (aavaT - -)
dv = 0
c
.
218
Chapter 7
7.7
7.7
Since this is to be true for all dn, dm, the two terms must be independently zero. Differentiating G = nl11 + m112, we obtain
Phase stability co-existence, and criticality
(e) The second-order terms from the expression (7.7.9) are
[(o~~~ m + (~) mJ (on)2
112 ( OG) m =111+ n (0111) m+m (0 ) m =111,
an
an
(0 111 ) om n
0111",
OX) ( om n
n (n+m)2 =
I ( n", + m"
n+m '
{o[-nl(n+ m)2]} am n
=
2(1- x) (n+my 2m 2x-1 (n+m)3=(n+m)2
2n (n+ m)3
n",
ox
n~
\IX
m
n (om)
lJ
02111@ onom 1 (~) onom +"2 om l n (om) 1 2111 111'" 1 (alI12~) 21 +m",l"2I (0 "') m(on) 2 - 0onom onom+"2 om 2 n (8m) J 2 211 ) (on)2 -~onom+0 11 1 (0~2) (8m)2 +m [ -1 (0211 -----'Yi 2 ~ 2 on onam 2 om
~ (n+ m)2 .
p 2 L"2 (~) on l m(8n)
lJ
--anz
_
m
n
1
(7.7.11)
m
[(~1) men:m)i]}m
[0 (011 \1 1
0111 {a
r
m
~ an oxjJ m +ax on!(n+m)2
J2
02 111" rL"2l (02111a.) I (02111~\ onom +"2 --anz m(on) 2- -onom om 2 ) + n~
111\ J rLan0 (0an)
m
0112~ _ 0112@
-ax- -
0112 } . -ax[(1-x)onom+x(om)Z]
The first four results quoted follow at once. The simplest method for the other six is as
= {oon
(7.7.10)
Of the second-order terms in the Taylor series for the Oil, the following contribute second-order terms to oG:
02X) = {0[ml(n+m)2]} = _ 2m ( on 2 m on m + 02X {o[ml(n+m)2]} I oman am n =(n+
211 1 ( 0on 2 )
lonom
{0111 . ~[(l-x)(on)2+xonom]
+ +I
x n+m'
The second derivatives are
02X) ( om 2 n
(o:~@)
The terms of(7.7.10) thus reduce to
1- x
=
0111@ ox'
-ax-
whence Equation (7.7.3) follows, and Equations (7.7.2) and (7.7.3) give Equation (7.7 A). (c) The first derivatives of x == n/(m+ n) are:
n n+m - (n+m)2
+
On changing variables to x, note that if T, P are uniform, and if 111~' 112~= 112~,itfollows,sinceI11 =111(T,p,X),112 = 112(T,p, x), that x'" = x~ and thence that
02G _ (0 112) onom on m'
I
n
111~
•
Similarly,
ox) ( an m
-l C~~~)
[e~~~) m + e~~~) mJ onom+le:~J n + e:~~) J(Om)2.
an
and
02G oman
219
I
On changing the differentiating variables to x and collecting terms in pairs this becomes, when the terms from (7.7.10) are 2 11 [!O x)l(on)l + x(1- x)onom oG { [ x 0211 OXll + (1- x) 0OXll
J
J} m
(112) [(1- x)2(on)2 + 2x(1 ax ax
2
+!x 2(om)2]+ ( 0111
m 0 111 2m [ (n+ m)2 ox 2 - (n+ (1 X)202111 2(l-X)0111 = (n+m)2 ox 2 (n+m)2 oX
+x2(8m)2]}(n~lm", + n~;m~)' 11
x)8n8m (7.7.12)
220
Chapter 7
7.8
7.7
From Equation (7.7.2) we obtain
Thus, in extensive variables, they become
OJ.l.l OJ.l.2 x-+(1-x)- 0 ox ax'
OZG) ( on2 m
and thence by differentiating
aZJ.l.l
OJ.l.l
OJ.l.2
ax + x ax ax 2 -
this gives
1 2 ox
1 + - -I) +ma. n/3 + m/3
ax
1 + 1 ), +m... n/3+m/3
+ (_OJ.l._I __ OJ.l._2)fO- X)Dn+xDmP ox ax and since n.... m .... n/3' the inequality (7.7.6). From
...
=
nlJ.l.I + n2J.1.2
0= XJ.l.l +(l-X)J.l.2 •
T,p
J.I. + J.l.l + (1- x) ox l,p
0)
0 ( ox 2
ax
ax)
T, p -
(l
OJ.l.l
=ax' or
(OJ.l.2) -;;uX
OJ.l.2
ax ' T, p
m
> 0,
or
(OJ.l.I) om n
< 0,
or
(0on J.I. z) m
( 2-) 0 G
T, p
>0.
>0
_ OJ.l.2 ox
>0
for the system.
T, p •
oZG x ox2
= 0,
2
ax '
7.8 Criticality, continuous equation of state. The region of two-phase co-existence in a binary system is frequently bounded in temperature, either by an upper critical solution temperature (UCST), as with liquid vapour co-existence, or by a lower critical solution temperature (LCST), as when a one-phase binary system separates into two liquid phases with increasing temperature. In either case, if G = G(p, T, x) is continuous, there is necessarily a region of instability bounded by a spinodal locus, (ozG/ox2h~ p = 0, analogous to the locus (02F/oij2) 0 of Problem 7.5; the corresponding binodallocus is defined by the co-existence conditions, cf conditions (7.7.
(7.7.13)
<0.
By the results of part (c) these are in turn equivalent to
OJ.l. 1\ ( an)
Glm
OJ.l.l
so that the stability condition may be expressed equivalently as
OJ.l.l) >0 (~ uX 1: p
I Gnm
for each region, Le.
It follows that
azo x)ox 1
Gmn
OJ.l.l) , ( on m ( OJ.l.2) om n
(OJ.l.~
(OJ.l.l) T, p
T,p
< O.
so that, in respect of two regions, [cf solution to Problem 7.3(c)] stability is ensured by either G 2n > 0 or Gzm > 0, for each region. Note that for a homogeneous region, adding material at the same T, p, and x is a reversible process. Note that proceeding by analogy to Problem 7.2(a), Le. using o G(T, p, x), and the perturbation Dna., -Dma., leads directly to
- J.l.l ,
by Equation (7.7.2) ,
J.l.l - J.l.2 2
(0 l}
ax "
or omon
A third procedure is to truncate the expansions for J.l.l a., J.l.l/3' etc. at the first derivative. This gives, for the second order terms, (7.7.10) rather than the full list (7.7.12). Separating the terms (7.7.10) into two inequalities, one for each region, gives simple quadratic forms for both. The corresponding determinants are zero by Equation (7.7.4), and, by an argument similar to that of the solution to Problem 7.3(c), the stability condition is
and hence
x (OJ.l.l)
G2n
2
we have
(00) ox
02G
> 0,
H(n ... + ma.)(Dxa.) + (nB + ma )(DXB)] ox2
cannot be negative, the stability condition is
G
(02G) om2 n
or
0 ;
2 +(I_X)a J.1.z=_(OJ.l.I "OJ.l.2\. ax 2 ax ox)
= - - (OJ.l.l -OJ.l.2) - [(1- x)Dn+ xDm]2
> 0,
conditions correspond to the fact, cf Equation (7.3.6),
These that
02J.1.2 + (1 - x) oX 2
Thus expression (7.7.12) becomes
DG
221
Phase stability, co-existence, and criticality
<0
or
2)
J.I. ( 0om
> O. n
T...
i'.
= 1/3,
P...
P/3' J.l.la.
= J.l.l/3'
J.l.za.
J.l.z/3·
(7.8.1 )
Write Equation (7.8.4) as
(a) Consider two co-existing phases, O! and (3. Use the conditions (7.8.1), together with the relations derived in the solution to Problem 7.7(0 to show that the tie-line connecting Ga.(T, p, xa.) to Gp(T, p, xp) has the equation
Ga.
aG p _ aGa.
= 0,
(03G) ox 3 T,
P
4 (0 G) 3x 4 T,
= 0,
P
and the equation to the tie-line is Ga.- Gf3
and that the binodallocus is tangential to the spinodal locus. (c) Use a Taylor series expansion for G about the critical point, at constant pressure (omitting powers of l1T higher than the first), with the results of part (a) to show that (7.8.2)
ox4 (04~
3
(l1X)2
T,Ac
i.e. that l1x is quadratic in (T-
G)
3 = -6 ( 32xoT
AC
l1T,
(aG/ax)e
(7.8.3)
Solution
(a) The solution to Problem 7.7(0 gives [leaving off the bar, since all extensive variables are to be read as molar (intensive)],
= (PI-P2)e'
4
33G ) I ( aG\
( ax 2aT c l1T+ 2! ax 3 aij e l1T(l1x tt + l1x(3)
3G tt
I (aax4G) c[(l1xa.) + l1xa.l1xf3 + (l1x{3) ) = o.
P2tt+3x IX
4
2
+31
whence (1- xa.)P1a.
= (1- Xtt )P2tt+ (l -
3G tt
xa.) 3x
a.
=
'
2
2
oG .. G.. + (l - x .. ) ox a.
Ga.-Gp
l1x a.- l1x {3)+ (a G) e l1T(l1x tt -l1x{3) ax c( axaT ( aG) 3
Thus, from the co-existence condition PItt
oG tt
Ga.+(l-XaJ-~-=Gf3+(l
uXa.
a G ) c l1T[(l1xa.) 2 -(l1xf3) 2 ] + I ( ax20T
PI{3'
oG@
uXf3
X(3)~
(7.8.4)
+
Similarly _ oGa. P2 .. - G.. -Xa.-~uXa.
= G{3
3G
x ~ f3 aXf3
(7.8.8)
The forms obtained using Equation (7.8.5) are more symmetrical than those obtained from Equation (7.8.7). The relation may be set up as follows
3Ga. x .. )ox ' a.
Pltr.=X"Pla.+(l-Xa.)P2tt+(l PI tt
(7.8.7)
X(3)G;)a.,f3·
(c) The co-existence conditions should be used in terms of G and its derivatives; Equations (7.8.6) and (7.8.5) or (7.8.7) are suitable forms. On using Taylor series expansions to fourth order for (aG/ax), and recalling [cfProblem 7.S(c») that l1xa. =1= l1xp, and that (a 2 G/ax 2 )e, (3 3 G/ax 3 )c = 0, we obtain from Equation (7.8.6), on dividing through by (l1x a. -l1x (3 )
T.J near the critical point.
PIa.
(x ..
Thus the tie-line is the common tangent at the points x tt' x{3 to the curve G = G(x), for constant temperature and pressure. Since, for stability, (a 2G/ox)2 is positive at X tt , xf3' and the continuity of G = G(x) requires a range of x for which (3 2G/ax2) is negative, this range must be between Xtt and xf3' and be smaller than xf3 - x tt · (b) The argument is exactly parallel to that of Problem 7.S(a), except for the last part of the latter; there is no restriction on the sign of
> 0,
l1xa. = -l1x p ,
(7.8.6)
oXf3 - oXa. '
and that the binodal locus spans a larger range of x, (xp - xa.) than does the spinodal locus. (b) Thence show that, if the tie-line becomes vanishingly short as T -,>- 'Fe, then at T 'Fe , p
_ _ 3G@_aGa._ aG p 3G .. Gf3 - ~ ~ X{3 ~ + Xtt ~ vX{3 lIX a. uXf3 vX a.
On comparing this with Equation (7.8.5) we obtain
Gp-Ga. _ (OG) . xp-xa.- ox a.,p'
2 0 2 T, ( oxG)
223
Phase stability, co-existence, and criticality
7.8
7.8
Chapter 7
222
I (
4 a G\ 3 ax3aij c l171(l1x tt ) -
3
(l1x{3) )
(a3xG)4 c[(l1Xa.)4_( l1X f3 t)+ ..·,
+ 4!1
(7.8.5)
t
4
1.8
Chapter 1
224
and
2
aG", (aG) -x'" ax", = - ax eX",
( a G)
1.9
Phase stability, co-existence, and criticality
(a) Prove that
3
( a 2G ) axaT c fj"Txa. ax aT c fj"Txa.fj.xa.
a
(a
4
F215 Pox
IFxo
4
G ) c fj"Tx a.(fj.xa.)2- 311 ax4 G) c X a.(fj.xa.)3+ ... , - 2'1 ( ax 3aT 2
G)
+xll ~ aXil -_ + (aG) ax c XIl + (a axaT c fj"TXIl+ ...
where
4
o.
(fj.x1l)4- 4Xa.(fj.xa.)3+ 4xll(fj.x1l)3]
(7.8.9)
Substituting for (a 3 G/ax 2 oT)c from Equation (7.8.8) gives finally 3 3 a4G ) 2 (ax 3 aT c fj"T[ (fj.x a.) + (fj.xll) ] 4
+
G)
a ( ax4
-
c [(fj"x a.)2 -
(fj.x1l)2 j[fj"x a. - fj.xll j2
0
3
3
c
2
-(fj"xa.)(fj"xll)+ (fj"xll)
(aax4G) c[fj.x '" + fj.xIl j[ fj.x
2]
T. 0'
etc.,
G) T, P. c
(- ) 3
=0
.
Solution
(a) The change of constraint formula gives, in molar quantities,
4
+
,
(- )2
a (-ax3
or
04G) , 2 (ax aT fj"T[fj"xa. + fj.xll][(fj"xa.)
T. x
and thence deduce the stability conditions for a phase at constant temperature and volume. (b) Proceed by analogy to Problem 7.6 to show that for the alternative equations for the (ii, x) projection of the tangent to the binodallocus at the gitical point to be equivalent requires that G2X be zero whether or not F 20 is zero. (c) Differentiate the determinant of part (a) to obtain the equation for the tangent to the spinodal locus at the critical point, and use the relations obtained in the solution to part (b) to reduce it, for (ap/au)c =1= 0, to F xv Fxv F xo - 3F2xo F + 3Fx2v F - F 3v 1 = O. 2v 2v 2v Prove that the result of part (c) is simply
(fj.xIl)L 2xa.fj.xa. + 2xllfj.xlll
4
c [(fj.x",)4_
== ( a-IJ 2 2
G ) c fj"T[(fj.xa.)L (fj.xIl)L 3xa.(fj.x"Y+ 3xll(fj.x1l)2] + 31I ( axa3 aT
1(aax4G)
a2 2
3
+ 4'
,
P2x
p) _F 2x == (aaxp) F 20
and, on addition, Equation (7.8.5) becomes 1( aG ) 21 ax2aT c fj"T[(fj.xa.
225
IX -
fj"x (3 po,
(~~)
and relation (7.8.2) follows unambiguously. Substitution from Equation (7.8.2) in Equation (7.8.8) gives the result (7,8.3). The analogy to Problem 7.5(b) and (c) is obvious. Note that, extending Equation (7.8.8) we have 4 I (a G) I ( aSG ) a3G ) ( ax 2aT c fj"T+ 31 ax4 c (fj"X)2+ 3 ox4aT c fj"T(fj.x)2
I (a G) (fj.x)4 c
T. p
+ (~~) T. v (~~) T. x T. v '
and from F == G - PI},
( aF) ax T, Hence
O.
7.9 Stability at a critical point. To elucidate the relevance of the condition of mechanical stability at a critical point in a binary system ,x) surface. In terms of this surface, the evidently requires the P = phase ccrexistence conditions include, in addition to conditions (7.7.1), the condition p", = Pll'
(~~)
(ap) +v( -aG) ax T, p ax
6
+ 51 ax 6
T. v
2 (aaxG) T. p 2
af}) ( ax T,
"~Ii
I \
p
v
=
(aG) ax T.
v-
p (a )
v ax T.
(aG) oX
v
2 (a2F) F) (av) = ax 2 T. v + OVOX T ax
(a
=
(a2G) axap T
=
la: (~~) J G;)
= T,
ra(aF) Op ax T. T. x
T, P •
T, p ;
J
v T, x
(a: I(~:~) 2
T. x
::) T
T. x
226 and
7.9
Chapter 7
2 2 2 (3 F) (3 F) (3 F) 2 2 2 OX T,v 3v T,x- 3v3x T
2 2 T,p (32F) ( 33xG) 3v 2 T,x
IT
3(Fx, Fv) _ o(x, v) -
Phase stability, co-existence, and criticality
we have
2 3 F) ( -0 2 X
Alternatively, use the method of lacobians:
F2x Fxv Fxv F2v
7.9
v)
1L2), -p ]/ 3(x, v) o(x, -p) o(x, -p) 2 3 G) (02F) = ( 3x 2 T,p ov 2 T,x'
= _ r3( ILl .
1L2 )]
oX
T, p
/(31))
op T, x
-
dx ( -0 2G) 2 ox T, p, c
I)
->0 vKT
T, p
T, p
>0 ,
V
I K
T
(ILIa. _ 1L2a;) -_ (OFa;) ox a.T,v
T >0 . > 0, -C v
(7.9.2)
(OFfl) 3vfl T, x = -Pfl '
(~) 3x
flT,"
-_ (1L1fl _ 1L2fl)'
(~) oVfl T, x
~) T, v (oXfl
1L2fl,
p' The Taylor series expansion for -t::.p,
t::.v, is
02F) dx+ (02F) -2 dv = 0 (3v 3x T, c (1) T, x, C
t::.(1L1 -1L2)
=
o.
(7.9.4)
dx
(~~)
dv = 0,
r(3P)2 I(a- p) ]d x (ap) -ox T, D, av T, x, ax T, v, ..
T, v, c
T. x. c
c
C
(7.9.5) dlJ C
0 -
(7.9.6)
ap
ap
(7.9.7)
and equivalence of Equations (7.9.3), (7.9.5) requires (02G/ax 2 ) = 0; if (op/ovlr x, c = 0, then (op/ox)T, v,c = 0, and Equation (7.9.6) requires (02G/OX 2 )T, p, c = O. The situation differs from that in a one-component system (continuous equation of state) in that
02G) ( ax 2 T,
p, c
(ap) T, x, (1)
c
= 0
(-3D) oX or, explicitly,
(OD)
T, v, c
dx+ -
av
T, x,
dv = 0 C
'
(F3x F2v + Fx2vF2x - 2F2xv Fxv)dx + (F3v F2X+ F 2x vF2V - 2Fx2vFxD)dv
O.
(7.9.8)
-t::.p = t::. ( -3F) = (02F) -t::.x+ (02F) -2 t::.v+ ..·. 3v T, x OV oX T, c ov T, x, C Subtracting the series for (3 from that for <:Y, and proceeding to the limit,
Similarly for
do
can occur, e.g. at a liquid-vapour critical point; the critical point is, in this case, also an azeotrope, and the occurrence is unique in T, p, x, like one-component critical point, whereas normal binary-system critical points define a continuous locus on the p, x, or T, p, projections. (c) Denoting the determinant by D, the tangent to the projection of the spinodal locus on the I), x plane is
OF",) (OFa.) 1L2'" = Fa. -Va. ( ~ -x"':;- vi} a; 1~ x vX a. T,"
since \VI-/VA)T v to first order in: t::.x,
T, c
( ax 2 3v 3x dx-axdx- ov dv = 0,
In terms of F and its derivatives (molar quantities being understood) Pfl and (7.7.1) are: the co-existence conditions p",
0Fa;) -Pa; = ( OVa; T, x
(~~)
d2 G op/oP)
(7.9.1)
'
and, correspondingly, when fluctuations in entropy also are allowed,
oX (~l~
V X
=1= 0, then it follows from Equation (7.9.5) that and Equation (7.9.6) becomes
I
>0,
--a-a
dx+ (a2F)
x, c
Evidently the stability conditions are,
OIL ( -3x-
D, C
Equations (7.9.3) and (7.9.4) may be written, respectively, cf.part (a) above:
-"-'-'---'--''''---'---''
= 3[( ILl -
T,
227
= t::. (~~) T, p
(7.9.3) •
t::.e~) T, D
To obtain the given result, this must be reduced to terms of one variable. This may be done using Equations (7.9.3) and (7.9.4); the two possible ways are equivalent. Thus, changing terms in dv to terms in dx, we obtain
-2[Fx2v Fxv]dv = 2[Fx2vF2x]dx, [F2xvF2D]dv = -[F2xvFxv]dx, and Equation (7.9.8) becomes [F3xF2v-3F2xvFxv+ 3Fx2 vF2X]dx+ [F3v F2x ]dv = 0;
228
7.9
Chapter 7
but, D
7.9 whence
= 0 gives
rO(Fxv)2]
(Fx.Y Fzv
and
Hence
[F3xF2v
Finally,
LO ax
1; P
L
7: p
r
(Fxv)2J 3Fzxv F xv+ 3Fx2vF2X]dx+ LF3V~ dv
(Fxv)2 F2X J - - FwFxvF dx [.F 3x F 2v - 3F2xvFxv+ 3Fx2v F zo 20
[
1 ]
rO(Fzvt
l
T, p
Thus -..[
(7.9.9~
---ax-
from the standard change of constraint formula. But
(~~)
T, v
T, x
0(FxV)2J 0x
_Fxv FlV'
-(.F
xv
)2
T, p
[0(F2v t ox
= p
F2xvFxv Fl.
(7.9.10)
L
oX
T, P
2Fxv (oFxv) ox 1; P
(Of))
]
7; x ox T,
I
T, P
Fzv
,
Fxv
(7.9.11)
+ Fx2v (Fxv) 2 _ F (Fxv) 3 FZv 3V F 2V
(7.9.12)
= T, p
V
Fxv (Fxv) 2 = -2F2XVP ' 2v + 2Fx2v F--:20
1 ]
, p
= -(FZv )2 + F 3v (F2V )3
oSz ./8S)Z8~, (OZV) or
Now
rO(FXV )2J
1
2V
Adding expressions (7.9.10), (7.9.11), and (7.9.12) gives the required result. Thus, as in Problem 7.6, the higher-order stability condition ensures that the binodal locus and the spinodal locus coincide at the critical point. It is clear from Problems 7.1 and 7.2, and the note to Problem that the method of partial molar quantities is well suited to the problem of the single phase, since it suppresses the difficulties associated with the 'reversible' perturbation, adding or subtracting matter at the same values of the intensive variables. Problem 7.4 shows, however, that this method is incompletely rigorous for co-existing phases; it always establishes the correct stability conditions as sufficient, but may fail to establish them as necessary. This is because the list of second-order terms obtained by this method may not correspond exactly to the list obtained using extensive variables, cf.the expressions (7.3.7) and (7.4.5), noting that the latter contains reducible third-order terms such as
whence
OF2X) ( ox T,
T, v
oX
[0(Fov f
Fxlv
the identity of Equation (7.9.8) implies, cf.Equation (7.9.9), that x c is finite; if so, Equation (7.9.9) leads at once to the required res~lt; if not (critical azeotrope) then Equation (7.9.9) is not usable (G is singular) and behaviour in the critical region is determined by (02p/OV Z)y. x = 0, etc., as for a pure substance. (d) To express G 3x =: (03Glox3h; p in terms of F = F(T, v, x), start from the result of part (a) in the form (Fxv)2 GT2x F 2X - - F 2v Now OF2X) (Ov) ( F zxv ox 1,' P ' T, P
-
+
]
i 2v
20
p
1
J
(Fxv)2 2Fx2v F . 2v
whence
= 0,
-
(:~) 1;
= 2F2xvFxv -
F -(F1vf 1F x1v + (F2V f 2F 3v
2v
2
ox
ox
= 0,
r,F3X F 3Fzx vFxv + (FxV)3] F F3v (F )2 dx 0, 2V Fxv (FFlOxv) - 3F3v (F'xV) I dx = 0 . F2v r ~3X - 3F2xv F + F 2V 2V -
0(F2vt
~lXV + Fx2v (:~)
= 2Fxv
r (Fxv)2]
dv = FJv(Fxv)2 l7 dv. zv
229
Phase stability, co-existence, and criticality
n (6) page 221.
z [~(OU) oSz on T, p
J
v, n
(8S)18~ .
8.0
231
the heavy line may represent the real variation of the volume density x of some extensive thermodynamic quantity in the direction z normal to the interface between two bulk phases (distinguished by single and double primes), whilst the fine line represents the model situation. The total quantity of X within the whole system of volume V is then made up by assigning any discrepancy between model and real systems to the surface itself. Thus (using a superscript s to distinguish quantities associated with the surface),
8 Surfaces J.M.HAYNES (University ofBristol, Bristol)
X = V'x' + VI/x" +Axs ,
(8.0.1 )
where A is the area of the interface, Axs is proportional to the shaded area in Figure 8. 0.1, and the volumes V' and V' are defined by the position of the model surface of discontinuity and by the conservation equation (8.0.2) V+V' V.
THE GIBBS MODEL OF A SURFACE (1)
8.0 When two bulk phases are placed in contact, their various properties do not, in general, remain uniform right up to their common boundary orinterface. Rather, the operation oflong-range intermolecular forces will cause the transition at the molecular level from one phase to the other to be more or less gradual, within an interfacial region. This is so even in systems of only one component; in multicomponent systems, especially, the net composition of the interfacial region may differ very markedly from that of either bulk phase-the phenomenon of adsorption.
For example, ni, the number of moles of the ith chemical component in the system, may be subdivided into contributions n; and n;' from the two bulk phases, and an additional number of moles nf which can be attributed to the surface. Thus, applying Equation (8.0.1), we have
'phase
ni = V'c;+ V"C;'+AI'i
(8.0.3)
where c;, c;' are concentrations, and r i n~ fA) is the surface excess or adsorption of i per unit area of interface. Since such a dividing surface is entirely hypothetical, and its location quite arbitrary, the surface excess quantities defined with reference to it are not susceptible to experimental measurement (unless by methods involving direct observation of the interfacial region). It is therefore necessary to define relative surface excess quantities which are indepen dent of the location of the dividing surface, and which may be deduced directly from observations of the two bulk phases alone. Since the location chosen in the model for the surface dividing the two bulk phases determines the relative magnitudes of V' and V", this location also controls the values of the surface excess quantities X S (which may be energy, entropy, or the mass of any chemical component).
"phase
x'
Surfaces
x
x Figure 8.0.1. Real (heavy line) and hypothetical (fine line) v~riation of the quantity x in the z direction normal to the interface.
Development of a quantitative surface thermodynamics based on a realistic molecular-statistical picture of the interfacial region is impeded by our ignorance of its microscopic structure. For many purposes, how ever, it is an acceptable simplification to imagine the two homogeneous bulk phases to be separated by a surface of zero thickness, across which the thermodynamic properties change abruptly. Thus, in Figure 8.0.1
8.1 Show that the relative adsorption of i,
r t, 1
ri -
,
1'1
rio I, defined by "
c·-c· +-----+., CI- c 1
(8.1. 1)
is independent of the posltion of the dividing surface. [Proceed by using Equation (8.0.3) to express r i and 1\ in terms of V' and V", which are then eliminated using Equation (8.0.2).]
J.W.Gibbs, Scientific Papers (Dover reprint, New York), 1961, VoLl. R.Defay, I.Prigogine, A.Bellemans, and D.H.Everett, Surface Tension and Adsorption (Longmans, London), 1966.
(1)
230
f
232
8.1
Chapter 8
Surfaces
is defined [ef.Equation (8.0.3)J by
dQ
r1
AI [ nl- V" c 1 - V""J CI'
(8.1.2)
Using Equation (8.0.2) to eliminate V' and V" between Equations (8.0.3) and (8.1.2) yields
= A I·.ni I
r· I
1
, VCi-(n
l
, c; - c;' ] Vcd~ . CI
cl
All the quantities on the right-hand side are experimentally accessible and are independent of the position of the hypothetical dividing surface. So, therefore, is r i , I, as defined by Equation (8.1.1).
z
= dU+p'dV'+p"dV"-adA.
Solution
Let the liquid drop (" phase) and its surrounding vapour (' phase) constitute an isolated system of constant total volume, in which for any infinitesimal fluctuation about the equilibrium state, dQ
and -dV'
=
V'C'I+
dV"
dU
= trdA .
p
r
the Laplace equation. This result is also obtainable as a mechanical equilibrium condition of curved membranes under uniform tension. The present derivation demonstrates that the model surface of zero thickness in which the surface tension appears to act mechanically is the same as the one that correctly defines the volumes V' and V". The distinction is only of importance when the radius r is comparable with the thickness of the actual interfacial region. More generally, the factor 21r in the Laplace equation may be replaced by the mean curvature e defined as the reciprocal of the harmonic mean of the principal radii: 1 1
ci'
o.
The values of all the r j (including r I) depend critically on the position of the dividing surface with respect to which they are defined. In particular, there exists a dividing surface for which r 1 = 0; that is to say the relative magnitudes of V' and V" may be chosen so that nl
:=
from the given statement of the first law, 2a 0:= -p' dV" + p" dV" --dV" r " , 2a p
Figure 8.1.1. Choice of dividing surface which gives r l =
(8.2.1)
The new intensive variable a is called the surface tension. Show from this that a spherical liquid drop of radius r, at equilibrium with its vapour, is under an excess pressure 2a/r.
or
Cl
233
8.2 The first law of thermodynamics applied to a single-component system of two phases separated by an interface of area A states that
Solution
rl
8.2
V"c;' ,
and the shaded and stippled areas of Figure 8.1.1 then become equal. In this case, the last term of Equation (8.1.1) vanishes, thus giving the physical meaning of r i , 1 as the adsorption of the ith component relative to the dividing surface at which the adsorption of component 1 is zero. Relative surface energy and entropy may be similarly defined.
e
rl
+r2
When the effects of external fields such as gravity are negligible (e.g. small systems, or nearly equal densities in the two phases), mechanical equilibrium requires that e be uniform over the interface. Mathematical studies of such surfaces of constant mean curvature are important in dealing with problems of capillarity. It may be shown variationally (2) that for reversible fluctuations in such a system of constant curvature e, dA dV" = e, (2) C.F.Gauss, Theone der Gestalt von Ji1ii.ssigkeiten (Wilhelm Engelmann, Leipzig), 1903, p.46. See also l.W.Gibbs, loc.cit., pp.219-229.
234
8.4
8.2
Chapter 8
Surfaces
235
where 0 is the contact angle. Since, evidently, cosO = I h/r, it is readily shown that for variations in which the contact angle is unchanged
which leads immediately to the balance of the p V and aA work terms in the first law statement.
d V"
8.3 Consider a system involving equilibrium between liquid (1) and
vapour (v) phases each in contact with an inert solid (s) (such as the case of a liquid droplet resting on a plane solid surface, as in Figure 8.3.1). Assume obedience to Young's equation
and
dA sl
= -d V' = 1rr2 (l - cosO)2(2 + cosO)dr, = -dA sv = 21rrsin 2 0 dr,
dAly = 41rr( I - cosO)dr .
in which 0, the contact angle, is the angle included by the sll and interfaces at their intersection. Show that in such cases an effective area, defined by n = A1v - AslcosO ,
Insertion of these quantities into the first law statement (8.2.1), as in the preceding solution, yields the Laplace equation once again. If the l/v interface in such a system has constant mean curvature, then (3)
dn
dV' = C.
replaces the area A in the First Law statement of Equation (8.2.1).
This equation is the counterpart of dA/d
a sv
asl+ alvcosO ,
~
(8.3.1 )
= C in Problem 8.2.
8.4 Derive the Kelvin equation, RTln(p/pO) 2a/r, relating p, the equilibrium vapour pressure of the liquid drop of Problem 8.2, with pO, that over a plane surface of the same liquid at the same temperature T.
Solution
The hydrostatic pressure inside the drop exceeds that at a plane interface by II " , I 20
Figure 8.3.1. Three-phase system showing contact angle (J.
Solution
In this system the term -dA of Equation (8.2.1) must be replaced by the three terms -(alvdA 1v + asidA sl + asvdA sv ). These are not independent however, since alv, ash and asv are related by Young's equation (8.3.1), and, furthermore, for any variation (for example, of the radius of the drop at constant 0) it must be true that dA,"V
= -dA
S\ •
The three terms above may then be reduced to the form -alv(dA 1v - dAslcosO) , which, according to the definition of the effective area n, is equal to -a\vdn. This quantity therefore replaces the surface work term in the first law. If r is the radius of curvature of the drop (assumed to be unaffected by gravity), and h is its maximum height (Figure 8.3.1), then (using single and double primes as before to denote the vapour and phases)
V'
= j1rh2(3r- h) ,
A\v
= 21rrh
,
and ASI = 1rr2 sin 2 0 ,
Pd - Pp = Pd - Pp
+
r
where subscripts d and p refer to droplet and plane interfaces. The chemical Dotential in the liquid phase is thus enhanced by
(
,
2a)
Pp + r
)=
1l.1l"
(assuming the molar volume v" to be independent of pressure). That in the vapour phase is changed by 1l.Il'
= RTln(p~/p~)
(assuming ideality).
The vapour pressure over the droplet, p~ (written simply as p) is obtained by equating 1l.Il' and 1l.1l". The approximation (p~ - p~) ~ 2a/ r yields the Kelvin equation in the form RTln(p/pO)
2a r
in which pO has been written for p~, the vapour pressure over the plane interface at the temperature T. Thus, for droplets in equilibrium with their vapour, p > pO. The effect is quite small for radii greater than a (3) C.F.Gauss, loe.cit.
236
8,4
Chapter 8
micron or so. For radii less than about 100 A, corrections may be significant for compressibility, vapour non-ideality, and the variation of surface tension with curvature. The same result can be obtained by considering equilibrium at the meniscus in a capillary rise experiment, in which the reduced liquid hydrostatic pressure is given by the Laplace equation, and the vapour pressure in the adjacent vapour column of equal height is given by the hypsometric equation. Thus, in a column of vapour of average density pi, the pressure p at a height h above the level where p pO is given by If the same substance, existing as a liquid of density p", attains a capillary rise height of h when its meniscus curvature is C, then balancing the hydrostatio pressures at the meniscus yields
pO_p;:::,:: O.
(The same approximation was made in the alternative derivation.) Eliminating h we obtain In(plpO) = aC pO p"
i.-
Surfaces
8.5
237
can be subdivided, according to the Gibbs model, into bulk and surface contributions: F = F'+F"+F s • Show the relationship of the specific surface free energy to the surface tension. (b) Deduce the Gibbs adsorption equation da
r
= -sSdT- :[rjdp-i,
(= PIA)
(8.5.1 )
i
where sS
p = pOexp(-p'ghlpO) .
hp"g+aC
I
SO lA, and the chemical potential P-i is defined by
( aF)
P-- -
i - a n i T. V,ni,A .
(c) Rewrite this result in a form which is independent of the position of the Gibbs dividing surface. Solution
(a) Writing
+ :[p-;dn;+ Lp-;'dn;'+ Lp-fdnf
dF = -SdT-p'dV'-p"dV"+adA
i i i
Since p'IJ'
RT for an ideal gas, it follows that
p"v", and pv'
RT In(p Ip) ° -II
v
'
dE" + dE'"
and expressing dE", dE'" in the form
aC ,
< pO. The Kelvin equation has been experimentally verified for moderate positive curvatures(4) (droplets in the micron range), but not for negative curvatures (5). Despite this and other objections, it is frequently used to estimate pore sizes in porous media, from measurements of the vapour pressure of volatile liquids condensed therein (6). Note that droplets, having dpjd V" < 0, are unstable in equilibrium with their vapour. Such equilibria may be stabilized if the droplet contains an involatile solute. Systems of this kind are of meteorological interest (7).
dF'
as before. In this case the interfacial curvature is negative and p
8.S In a system of two bulk phases and their interface, containing i components, the Helmholtz free energy
F
F(T, V, V', A, n;, n;',
+ dP
nD
(4) V.K.La Mer and R.Gruen, Trans. Faraday Soc., 48,410 (1952). (5) J.L.Shereshefsky,J.Amer.ChemSoc.. 55,3149 (1933). (6) S.J.Gregg and K.S.W.Sing, Adsorption, Surface Area and Porosity (Academic Press, London), 1967. (7) L.Dufour and R.Defay, The Thermodynamics of Clouds (Academic Press, New York), 1963.
= -S' dT- p' d V' + Lp-;dn; , etc.,
yields dP
-SOdT+adA+ LP-fdn1. i
Holding the intensive variables T, p-1, and a constant and integrating we obtain
P = aA + Lp-~nf . i
The constant of integration is clearly zero, since F S must be zero in a system with no interface (A n~ = 0). Dividing by the area gives
r
a+ LP-fri
.
i
Thus, the surface tension (an experimental quantity) is not in general equal to the specific surface free energy (whose definition, like that of r t , depends on the location of the Gibbs dividing surface). Only for that choice of dividing surface for which Lp-f r i = 0 is a = i
r.
238
8.5
Chapter 8
(b) Differentiating the last expression for Ps we obtain dPS = adA+Ada+-Illidn~+ In~dlli' i
i
Comparing with the previous expression for
dF s ,
we find
Ada = -8" dT- I nl dill i
or da = -ssdT- Ifldlll, i
which is the surface analogue of the Gibbs-Duhem equation [Problem 1.20(d)]. (c)
da
= -sjdT- Li fj ' J dill
J
8.6
Surfaces
239
In comparing this equation with its three-dimensional form (Problem 1.11), it will be seen that cjJ and Iff are the surface analogues of the pressure and volume, respectively, of a bulk phase, and the temperature dependent constants a and b have the same significance as before. (Hint: Differentiate the equation of state to find an expression for dcjJ at constant temperature.) (b) Examine the special cases of (i) an ideal two-dimensional surface film, and (ii) a film in which the adsorbate molecules are non-interacting hard spheres of finite size. Discuss in each case the behaviour of the adsorbed phase predicted by these equations. Solution
Differentiating the two-dimensional van der Waals equation and introducing the Gibbs equation gives df
(see Problem 8.1),
2a
= f(l- br)2 - RT df .
8.6 The Gibbs adsorption equation (8,5.1) may also be written in terms of the spreading pressure cjJ, which appears thermodynamically as an intensive variable equal, to within an additive constant, to -a, (An operational definition for systems in which surface tensions are experimentally accessible, such as solutions, is
cjJ=ao-a, where ao is the surface tension of the pure solvent.) Consider the adsorption of a single gaseous component, at pressure p, at the surface of an involatile inert solid. For isothermal variations, dll in the last term of Equation (8,5.1) is equal to RTdlnp for the gas (assumed ideal), and is zero for the solid; the Gibbs equation then becomes dcjJ = RTfdlnp , where f is the surface excess of the gas defined relative to a dividing surface essentially coincident with the solid surface. The spreading pressure may also be expressed as a function of temperature and adsorption by means of a surface equation of state: cjJ = cjJ(f, T) . An adsorption isotherm equation can now be formed by elimination of cjJ between the preceding two equations:
Integration yields the Hill-de Boer equation: (8) P
f
(cjJ+af 2 )(l-br)
= RTf.
2af)
'
where K is an integration constant. Because the surface equation of state on which this isotherm is based is of the van der Waals type, there will be a two-dimensional phase transition below a critical point defined by dp d 2p df = df2 = O. Performing the differentiation we find I
fc
3b'
4:
8a 27R '
as for the van der Waals equation itself [compare Problem l.ll(b)]. For ideal behaviour of the surface film, we set a = b 0, yielding the linear isotherm p = Kr, which is the two-dimensional form of Henry's Law. For non-interacting hard spheres we retain only the co-volume term which yields the Volmer equation (9)
f = f(T,p).
(a) Derive by this method the adsorption isotherm equation corresp onding to the surface equation of state of the van der Waals type
(bf
Kl-bfexp I-bf-RT
p
=K
f bf l-bfexPI-bf
(8) J .H.de Boer, The Dynamical Character ofAdsorption (Oxford University Press, Oxford), 1953. (9) M.Volmer, Z.physik.Chem. 115,253 (1925).
~
Chapter 8
240
8.6
The larger the value of b, the greater the pressure needed to reach a given surface coverage, as expected. 8.7 In the treatment of the previous problem the adsorbed phase is regarded as a continuum. By contrast, the adsorbate molecules may be regarded as occupying definite sites in the surface. (An identical formalism describes reaction occurring at fixed sites on a macromolecule.) The simplest such lattice model leads to the Langmuir equation
e
cx = -:":(l:-_-x""'):-:(-::-I-:"+-c-x---x-:") ,
where c is a constant and x = plpo, po being the saturated vapour pressure of the bulk condensed adsorbate. Solution
Extending the Langmuir assumptions slightly, we suppose that the rate of evaporation of species A is unaffected by the presence of species B, and vice versa, and that each species condenses, at a rate proportional to its partial pressure, only on those parts of the surface not covered by either species. Hence (}A aAPA(i-e), where
eA
Surfaces
is the fraction of sites occupied by particles of species A,
241
aA aAlvA, and 8 = 8 A +(}B' Combining this with an analogous equation for species B gives e = _a,-"A-,--p-,-,A~+_a=-B, -P,:::.B I +aAPA +aBPB Note that in this model a particle of species A is supposed to cover the same area (Le. one site) as a particle of species B. (10) (b) The steady-state condition for the ith layer is
~ I +ap ,
in which () is the fraction of the surface sites occupied when the pressure of unadsorbed gas is p, and a is a temperature-dependent constant. Such an equation describes monolayer (i.e. () < I) adsorption of non-inter acting particles at identical localized sites. It may be derived by a 'steady-state' method, in which the rate of condensation on the fraction (l ()) of unoccupied surface, ap(l e), is equated to the rate of evaporation from the occupied sites, v(}. The temperature-dependent proportionality constant v may be factorized to show a term in exp(-qlkT), where q is the heat evolved when one molecule is adsorbed from the gas phase (in some suitably defined standard state). The Langmuir model postulates q independent of e, from which the Langmuir equation follows immediately, with a a/v. Extend this derivation to the case of adsorption from a mixture of two gaseous species. (b) Multilayer (Le. (}-unrestricted) adsorption may be included by permitting filled sites in the first layer to act as adsorption sites for the formation of a second layer, and so on. By assuming that the evapora tion and condensation properties of the second and all higher layers are those of the bulk condensed adsorbate, show that
e
8.7
aiPsi_l
qi
= bjsjexPkT'
where Sj is the number of ith layer sites filled but not covered by occupied sites in the (i + 1)th layer, and a" b i are constants. This set of simultaneous equations defines the Si' which give the total number n of adsorbed particles in the expression = isi n = no L i= 0
1=L j
Si .
0
Here, no is the number of particles which would completely fill anyone layer (assumed independent of 0. The assumption suggested in the problem may be formulated as follows: (i) If a· qj gj = b;exp kT
then gz 1 goo
... = goo;
g3
boo
= Po = aoo exp
(qL) - kT '
where pO is the saturated vapour pressure of the bulk adsorbate at the is the corresponding latent heat of evaporation temperature T, and (per molecule). Thus, it is only the adsorbed layer nearest the solid surface that is assumed to behave differently from the bulk adsorbate. Hence, Sl = glPsO (for layer 1) and Sj XS i _ 1 (i>l), where x = plpo. The second of these equations becomes
qL
Sj
1S1
(10) For an experimental test of this equation see F.e.Tompkins and D.M.Young, Trans.Faraday Soc., 47,88 (1951).
r
242 (since Sj _ I =
8.7
Chapter 8 XSj _ 2,
f
etc.) and hence
alb""
g
-
a.. b I exp
(ql kTqL) . + cx
:E: x)'
x)
noc
+
(1 + Aq)n 0
;
= (l+AAqA +ABqB)n O ;
:E:
(c) multilayer adsorption: :E:
c noc
(l+Aql+A2qtq2+A3qtq2q3+ ... )no;
where qi is the partition function for a particle adsorbed in the ith layer. Solution (a)
Experimental data giving n as a function of x at constant T can thus be plotted as x/n(1- x) against x and, from the slope and intercept of such a plot, the quantities no and c can be calculated. If the area per site is known, no may in turn be used to evaluate the surface area of the solid adsorbent. (12)
n
A~[noln(l+Aq)l = nO OA
oln:E:
I
Aq + , Aq
as before. For either species considered separately, In:E:) AA- ( aOAA AB, 1,
nA
8.8 The results of the preceding problem may also be obtained by statistical methods. For example, if the surface is assumed to contain no independent sites, then, for the case of monolayer adsorption, each site may contain 0 or 1 particles. Let <Xo and <XI be the respective probabili ties of zero and unit occupancy of a site. Then, if A is the absolute activity of the adsorbed species, and q the ordinary partition function for a single particle, we have <Xo l. <Xl Aq
leading to
V, A
AAqA(l-(J),
(JA
where
e ()
+ (JB .
A
Hence,
e (c) With the assumption
e, the fraction of filled sites, and using <Xo+ <Xt = ()
.
(b) mixed monolayer adsorption (species A and B):
For x > I/O +yc), () > 1, corresponding to multilayer adsorption, and (J -+ 00 as x -+ 1 (i.e. p -+ pO). This isotherm equation, known as the Brunauer, Emmett, and Teller (BET) equation (II>, may be rearranged in the linear form x
T, V,A
Re-derive the results of Problem 8.7 by this method, given the following expressions for the grand partition function: (a) monolayer adsorption (single species):
cx
n - = () no
On identifying <Xt with it follows that
Oln:E:)
A~ (
n
The summations for n can now be effected, yielding eventually
nO -
243
number of filled sites, from the identity
_gl _
C -
Surfaces
cxis o ,
Si
where
8.8
ql
'Aq 1+'Aq
solution to Problem 8.7) that
= cq, q3 = ... = q,
q2
the grand partition function becomes
Since A is proportional to p, and q is independent of it, this is the Langmuir result. A more general method consists of expressing :E:, the grand partition function, in terms of A and q for a given model, and determining n, the
Z
(l+Acq+A2cq2+A3cq3+ ...)no = [
Ifwe use
e=
(11) S.Brunauer, P.H.Emmett, and E.TeUer,J.A mer. ChemSoc.• 60, 309 (1938).
n no
= ~(Oln:E:)
no
OA
p/po
= x
and assume further that
(12) S.J.Gregg and K.S.W.Sing, loc.cit.
Aq ~
l+(C-I)AqJn o I-Aq
Chapter 8
244
we obtain
o=
8.8
ex +ex-x) ,
as before. (l3)
8.9
Surfaces
Hence 11 kT
-=
and
(alnQ)
-an
0
Solution
Let the fractional surface occupation be 0 = nino; then, of the z sites surrounding any adsorbed particle, an average number znlno will be occupied by other particles, yielding altogether zn2/2no interacting pairs. (The 2 in the denominator accounts for each pair having been counted twice.) Thus, the average total interaction energy will be wzn2/2no, where w is the interaction energy for one pair. The canonical ensemble partition function is then no! ( wzn2) Q = n!(no- n)!ql'l exp - 2nokT . It has been assumed that the distribution of filled sites is random, in evaluating both the pre-exponential factor and the number of inter acting pairs (Bragg-Williams approximation). Use of Stirling's approximation yields
InQ
wzn 2 - nlnn - (no- n)ln(no- n)+ nlnq - 2nokT
(13) For a more detailed statistical treatment see T.L.HUl, J.ChemPhys., 14,263 (1946). See also E.A.Guggenheim, Applications of Statistical Mechanics (Oxford University Press, Oxford), 1966.
A.
If qA.
0
-= exp kT
wzO
In(l_ O)q + kT
T
1'10,
11
8.9 A more realistic model of monolayer adsorption would permit a particle in a surface site to interact with particles in neighbouring sites. Introduction of an appropriate interaction potential might be expected to reproduce in the model such co-operative phenomena as surface phase transitions (cf Problem 8.5). An exact treatment is impeded by the difficulty of calculating the configurational part of the partition function, and averaging the inter action energy, whilst allowing for the re-arrangement of the adsorbed particles under their mutual influence. The Bragg-Williams approximate treatment simply assumes that the arrangement of particles is random, as it would be in the absence of interaction. This somewhat severe approximation, which is the same as that underlying the van der Waals treatment of gas imperfection Problem 1.11) is not quantitatively satisfactory; nevertheless it succeeds in predicting the occurrence of surface phase transitions below a certain critical temperature, as does van der Waals' equation. Write down the canonical partition function for such a system, calculate the chemical potential of the adsorbed species, and hence derive the adsorption isotherm equation and find the two-dimensional critical temperature.
245
wzO
(1- O)qex p kT
plpo (see solution 8.8) then
p pO
0
nC1Tnllir
wzO
form for w
p(O)p( I
0) =
O.
rp(t >F ,
where p(O) denotes the pressure at which the fractional coverage is 0, we deduce that the critical point lies at Oe = t. From the condition it follows that
dP) ( dO 0 T.
== 0 Oc
zw
c
Thus a phase transition, giving a vertical discontinuity in the isotherm, is to be expected for an attractive interaction (negative w).
The imperfect classical gas
9.3
247
9.2 Assume that for a class of substances the pair potential is of the form IP<:r) = eljJ(r/o),
9 The imperfect classical gas
where ljJ is a universal function, and where 0 (of dimension length) and e may differ from one substance to another. (a) Prove from the configuration integral that in terms of appropriately scaled thermodynamic variables all these substances have the same equa tion of state. This is one form of the law of corresponding states. (b) Prove also that if these substances have a critical point, then they have the same critical ratio I<e = Pevc/k1'c.
P.C.HEMMER (lnstitutt for Teoretisk Fysikk, NTH, Trondheim)
THE EQUATION OF STATE
Solution
9.1 The intermolecular potential energy U{rl, r:h ... rN) of a real gas of N particles is a homogeneous function of degree "Y in the position coordi nates of the particles. Show that the equation of state is of the form pT-l+ 3/'Y == f(v i 3/'Y)
r
J
= ..·tXP
[u(r1, ... rN)] kT
J
... r1 , . . . , exp [_A'Y U(rj, kT vA
3
)J drl* ... drN, *
using the homogeneity property U(Arj, ... M1) sider now the following function of v and T:
f [VA'3
exp
= Al' U(rj, ... r1).
Con
U(rj, ... r1)1 '" '" kTA--Y Jdrl'" drN .
The substitution T -+ n-Y, v -+ VAl leaves this function invariant, hence it must depend upon v and T through the combination v i 3h : i3NhQN(V, T) = g(vT-3 /'Y)
with an unknown function g. The pressure
a
p = kT-InQN
= kTl- 3 /'Yg'(vT- 3 /'Y)g-1
av is of the required form, with kg'(x)/g(x) == f(x). 246
QN
03N
'
and v*
v 3
0
'
f
v·
drj." drN- exp[-.L ljJ(rrk)/T*] 1< k
is therefore the same for all substances, cf solution to Problem 6.8(a). For the pressure, p = kTa InQN/av, introduce the dimensionless quan tity, p* = ko 3 e- 1 p . The resulting expression for the reduced pressure p* , alnQ1 p*
_
- (TX-'Y)-3N/'YJ T -3NI'Y QN (T) v, oA'3'"
kT
= -e
Q1 = QNO- 3N
Ifwe introduce new integration variables by rn = Ar~ in the configura tion integral then we obtain J
T*
The dimensionless configuration integral
Solution
vA
r
= -o
drl·.. drN,
contributes to the pressure. QN is called the configuration integral.]
- 3NJ QN(V,T) - A
r*
into the configuration integral for N particles
where f is an undetermined function of one variable. [Proceed broadly as in Problem 3.5. In the partition function only the integral over the position variables, QN
(a) Introduce the reduced (dimensionless) variables
= T*---a;;* ::;; p*(v*, T*) ,
contains no reference to the specific properties of the substances, and thus the law of corresponding states holds. Two substances are in cor responding states if the reduced variables are the same for the two sub stances. (b) Since (with ve = ve/N) PeV~
I< c = k T.e =
p~v~
-----r:* ' c
the critical ratio is a dimensionless number, the same for all these sub stances. 9.3 A gas of hard rods. N particles interacting via a hard core of length d are moving on a tine segment of length v. Calculate the con figuration integral and find the equation of state in the thermodynamic limit N -+ 00, v -+ 00, N/v fixed.
9.3
Chapter 9
248 Solution
Since the integrand of the configuration integral is symmetric in the positions XI, ... XN of the N particles we may choose one particular ordering of the particles and multiply by N!:
QN
f··· f
N!
9.4
The imperfect classical gas
(a) Use the constant pressure ensemble of Problem 3.18 to show that the equation of state (in the limit N ~ 00) is given by
L"" exp [-Ilpu - J)p(u)] du
v
dx1···dxN·
SoOQ exp [-Ilpu - J)p(u)] du '
O<Xl <x",,<XN
The existence of the hard core puts the restrictions Xi+ 1 -Xi> d upon the domain of integration. Introducing the new variables Yn :;;: xn
we obtain
QN
-en -l)d ,
f ···f
N!
249
where 13 = 1/ kT. (b) Show that this system does not exhibit a phase transition. Calculate the equation of state for the pure hard core potential, defined by 00 for X .;;;;; d ip(x)=
dYl ... dYN
{
o
for X
> d.
O
Solution
jv-(N-l)d
... dYN
[v -(N _l)d]N .
(a) The partition function Zp of the constant pressure canonical ensemble, related to the canonical partition function Z by
The equation of state:
a
p
kT av In QN
In the thermodynamic limit N ~ equation of state, (l) p
00,
=
NkT
= v - (N v ~
00,
I r"" exp (pv) voJo -kT Z(v,T)dv,
Zp(p,T) l)d
N/v
= p,
we obtain Tonks'
kTp I-pd'
may be evaluated for this one-dimensional gas. A suitably chosen volume is inserted to make Zp dimensionless. In Z(v, T) symmetry allows us to choose one ordering of the particles, 0 < XI < X 2 •.• < XN v, and by N!. The energy is then
Vo
N
904 A one-dimensional gas consists of N particles that interact pairwise via the nearest neighbour potential (see Figure 904.1 ). 00 for x.;;;;; d ip(x) arbitrary for d < X < 2d o for 2d';;;;; X
1
The external pressure is p, the temperature is T, and the volume (length) v is defined as the distance between the end particle on the right hand side and a wall on the left hand side.
g •• L 1
N
1
.,
N
p
p~
N-l
I -2 + I n=1 m n
I
ip(xn+l-x n ),
Integrating over the momenta we obtain Zp
=
with
A-Nvel
13
r:
Jo
xP (-llpXN)dx N rXNdxN_l ... (X'dx1exp [-13
.10
Jo
= 1/ kT and A = h(21rmkTr'h.
we may write
Zp = A-Nvel Figure 9.4.1.
(I) L. Tonks, Phys. Rev., 50,955 (1936).
E =
UI
=
Un+ 1
= Xn+l-X n
f"
fOQ
o
...
0
I
ip(Xn+1 -Xn)]
Introducing relative coordinates
Xl
(
exp -IlP
(n = I, 2, ... N - I) ,
IN
n=l
= A-N (vopllf 1{ fo" exp [-Ilpu -
)
Un dUN
n
N-I n=l
exp[-J)p(un)]du n
J)p(u) ldU} N - 1 .
250
9.4
Chapter 9
The average volume v is given by v volume per particle D = v/N:
- kTa InZp/ ap. We obtain forthe
9.5
The imperfect classical gas
using the neutrality condition Lqj = O. In terms of the two-dimensional i
'volume'v
L2 of the container,
uexp[-J3pu -/3Ip(u)]du D::;;;
Q
~---------
exp (-J3pu) du
p - = -lnQ = kT av
or p :::::: kT/(v - d). This is Tonks' equation of state obtained by a dif ferent method in Problem 9.3. 9.S Determine the equation of state of a two-dimensional gas of N positive and N negative charges interacting via the pair potential
-qjqjlnlrj-rjl'
Here qj = ±q is the charge (in suitable units) of particle i. Assume the container to be the square 0 < x < L, 0 < y < L. Show that the canonical partition function does not exist if T < q2/2k.
v
p = (kT-!q2)p
in tenns of the number density p = 2N/v. We observe that this imper fect gas follows the equation of state of an ideal gas with the temperature scale shifted by an amount q2
10
:::::: d+ pJ3 '
-~~-
or
I
uexp(-J3pu)du
2
a
neglecting terms that vanish in the thennodynamic limit N ~ 00. (b) At a given temperature v is detennined uniquely by the pressure through the last formula in part (a). Consequently the isotherms in a (p, v)-diagram cannot exhibit the horizontal part characteristic of a first order phase transition. One can also show directly by differentiation that (aD/ ap h is always finite and negative when p > O. (c) In the case I{) = 00 for r < d and zero otherwise, we obtain
D=
= /v2N - t{3q N
This yields the pressure
exp[-J3pu - /3Ip(u)] du
1{)(lrj-rjl)
251
= 4k
The configuration integral may diverge for configurations in which oppositely charged particles coalesce. If qi -qj then the integrand contains the divergent factor rt!q2. The (two-dimensional) integral over r/> say, exists only if J3q 2 < 2, or q2 T> 2k = 2To •
For T < 210 the gas collapses into neutral pairs. Three isochores at densities P3 > P2 > P 1 are shown in Figure 9.5.1 where also the pressure of N neutral noninteracting pairs is indicated. p
P3
Solution
In the configuration integral
Q
=
f..·fdr1 ...
~
drwexp(J3 L qiqjlnrjj) i
"8
introduce the dimensionless two-dimensional vector variable Rj = rdL . The integration limits now become independent of L and we may write
Q
///:YVI --- -
= L4N exp(J3
L qjqjlnL)/, i<j where/is the factor independent of L. We have 2 L q/qj i
<
j
= (Lqj)2 i
Lqr
= -2Nq 2
P2
0.
S
/
/-:::-// // ;:~-I
o
,
j
To
I . ~~ ________
2To
Figure 9.5.1.
"
PI
/'
T
252
9.6
Chapter 9
9.6 As a simple model of a gas with hard core interaction consider the following lattice gas: N particles move in a volume v divided into cells, each of volume b. The potential energy is assumed to be +00 if two or more particles are in the same cell, zero otherwise. Determine the equation of state in the thermodynamic limit.
Calculate the nth virial coefficient.
Solution
In this case the configuration integral QN is simply determined by the number of different ways of distributing the N particles over the v v/b cells without multiple occupancy, each allowed configuration contribut ing a factor bN/N! to QN' The number of allowed configurations is v(v -I) ... (v - N + I), since the first particle has v cells available, the second particle (v - I), etc. Hence, QN
= v(v-l) ... (v-N+ l)b N =
/..
1l T\
(1- ~brV/b(~_b)N
This yields the following equation of state:
(1-
J!... = alnQN = -~In
av
kT
b
The imperfect classical gas
Solution
For this gas model the attractive part of the total potential energy is independent of the configuration of the particles and is equal to (2a/v) times tN(N - I), the number of pairs of particles. This yields a factor 2aN(N-l)IJ 2 kT
exp [ v
in the configuration integral QN' Hence, when both the repulsion and the attraction are taken into account (see the preceding problem),
-l)J
QN _ N( v/b ) [aN(N N! - b N exp vkT
kT
f bn-
1
n=1
av
ap ) ( aO T
(
in the thermodynamic limit.
b)
v - i}2a
(I-~)-~2 0 v kT
2a
ii(ii-b)+~
0
together with
9.7 As a model of a real gas with an intermolecular interaction con sisting of a hard-core repulsio)l pl~s a long-range attraction consider the lattice gas of the preceding problem, and assume an additional attractive interaction - 2a/v between each pair of the N particles. Here a is a constant. (a) Show that this gas model obeys the equation of state kT
kT ln b
is of a similar structure as van der Waals' equation. (b) The critical point is determined by
n'
shows that the nth virial coefficient equals bn- 1 En n
P = -I) In 1 -
kT ( 1 Nb) = --In - -aN2 b v v2
or (with v
P=
pn
V ==
v. = fimte
.
In the last expression both N and v are assumed to be very large. The equation of state
Expansion of the logarithm (with N/v = p),
J!... ==
)N exp (vkT aN2 )
Nb )-V/b( v ~ ( 1 --;; N- b
a P = kT-lnQN Nb). /)
253
(b) Determine also the critical point and the critical ratio Ke = Peve/ k'Fe for this gas model. [Note the analogies with the van der Waals gas, Problem 1.11, and with the hole theory of liquids, Problem 6.5.]
I .
Using Stirling's formula N! "" NN e-N for the factorials we obtain for large N and v
~~ =
9.7
(~~)T = k{z~~V~~~I - ~~ = O. Elimination of T yields Ve = 2b . The critical temperature a 'Fe = 2a(ve -b)
k-Ve2
2kb
The equation of state furnishes the critical pressure
_ k'Fe In 2
Pc -
b
a _ !!..-(In 2 4b 2
-
b2
2
_l) 4
'
9.7
Chapter 9
254
9.8
The imperfect classical gas
255
Show that the first vitial coefficients Bn are given by
The critical ratio is a dimensionless num ber, in this case
BI = 1 ; Kc
=
= In4-1
= 0·386 .... B2 =
This is close to the critical ratio for a van der Waals gas, which is Kc i = 0·375 (see Problem 1.11).
9.8 (a) The expression for the pressure in the grand canonical en semble given in the comment on Problem 3.5 is (Qo
Solution
•
(9.8.1)
Q2
-Qr
2v
2
b2 =
b - Q3- 3 QIQ2+ 2 Q1
6v
3 -
b3
(b) For an intermolecular pair potential of finite range the b/s stay finite in the thermodynamic limit v -+ 00. Verify this for the three b/s above. Show that the grand canonical average number density p = (N)/v is given by p =
I
Ib1z l
•
(9.8.2)
I = I
(d) Inverting this series expansion, to obtain an expansion z z(p), and inserting into Equation (9.8.1) one obtains the virial expansion (or density expansion) of the pressure P
kT =
is assumed.)
I
n = I
v
exp
'-
I
'!!!ilJ drl ... drN' kT
In particular QI = v, Le. b I = 1. Inserting for the configuration integrals we may write the second and third cluster function
v'
b
I In'
-f [- "
QN -
QI. I
-
and keeps terms up to order Z3 • (b) Recall the definition of the configuration integrals,
Establish that the three simplest cluster integrals b1 are given by b
(-x)n
~
l+x)
I~I biz I
kT
fer) = exp[-\;?(r)/kTj-1 .
== 1).
It may be rearranged to yield a power series expansion in the fugacity z,
00
f(r12)f(r23)f(r31) dr 2 dr3
(a) An easy algebraic result if one makes use of the expansion
= In N~oQN N!
-+
f
B3 =
where we have introduced the Mayer function
THE VIRIAL EXPANSION
(Here the limit v
ff(r)dr ;
-t
Bnpn .
2~f(eI2-l)drldr2
= 6~f(e12e23e31 -e12
e23 -e31 +2)dr l dr2drJ
with \;?(rik) elk = exp [ -----u-l
J.
Note that the first integrand vanishes if r12 > R, and that the second integrand vanishes if two relative distances exceed R. Integration over the variables r 2 and r3 yields a result independent of fl and the subse quent integration of rl cancels the factor V-I. The argument fails when particle 1 is within a distance R (2R in the last case) from the wall, but this possibility can be neglected when v -+ 00. Hence, b2
-+
b 3 -+
t fee 12 -
I)dr2
i f(eI2e2Je;:u
in the thermodynamic limit.
3e12+2)dr2dfa,
9.8
Chapter 9
256
(c) Use the probability of finding precisely N particles in the grand canonical ensemble ZNQN/N! PN =
2- (ZNQN/N!)
The
9.10
classical gas
t
Hence,
d
B2
257
1)4'ITr2 dr = j'ITd 3
In the expression for B3 ,
N
a
=2-
integrate over r3 first. Because of the step function character of the Mayer function, this integral is precisely the intersection volume I(r) of two spheres each with radius d centred a distance r = Irl -r 2 1 apart. It is an easy exercise to show that for r ~ d
= zaz
N
a
= zaz
I(r)
Hence (N)
p
=-
V
=
=
I
2-= lb,z' .
P
kT
= P -b 2p 2 +(4b~ 2b 3 )p3.
Using the results for b 2 and b 3 obtained in part (b), we obtain (in the thermodynamic limit) B2
= -b 2 = -!
J
f(r)dr ,
-2b3+4b~ = -! J(el2e23e31+3el2
3el2e31)dr2dr3' This
is easily shown to be identical to B3
=
f(r23) f(r3ddrz dr3 ,
9.9 Using the results of Problem 9.8(d), calculate the second and third virial coefficient for a gas of hard spheres of diameter d. Solution
The expressions for B2 and B3 are given in the preceding problem. For hard spheres the Mayer function is simply
=
a
B3 =
1)3
<.p(r)] exp [- - -1 kT
= {I
ifr
I(r)4'ITr 2 dr
J
0
5'IT2d6
5Bi
= -----rg- = 8
The equation of state for hard spheres thus has the virial expansion P
kTp(1 +Bzp+jBip2+ ... ) ,
where B2 is four times the volume of one sphere. 9.10 (a) Prove that the second virial coefficient B 2 (T) as a function of temperature can have at most one extremum. (b) Prove that B2 (T) is monotonically decreasing with temperature if the pair potential <.p(r) is non-negative. (c) Find the necessary and sufficient condition for B 2 (T) to exhibit a maximum. Solution
(a) It is convenient to use {3 perature:
using f(rik) = eik-1.
fer)
,'IT(4d 3 -3d 2 r+ir 3 ).
The final integration over r2 yields
I
(d) Write z = p + C2p 2 + C3p 3, and determine the unknown coefficients C2 and C3 by inserting into p = z + 2b 2z 2 + 3b 3z 3. This gives C2 = , C3 8b~ - 3b 3 • Ifnow z = p -2b 2p 2 +(8b~ -3b 3)p3 is inserted into the expression for the pressure, Equation (9.8.1), we get to order p3
andB3 =
-i ff(r l z ) f(r23)f(r3d dr 2dr3 ,
B3
to obtain the average number of particles
B 2 ({3)
=
1/ kT as variable rather than the tem
t J{ 1 -exp[-{3cp(r)]}dr .
If the interaction potential contains a hard core of diameter d we write 3 2'ITd +! B2 ({3) {l-exp[-{3cp(r)]}dr.
J
r> d
The second derivative with respect to {3 equals
B~({3) = -t J<.p2exp(-{3cp)dr < 0, and the derivative can therefore change sign at most once.
258
9.10
Chapter 9
(b) The assertion follows from the fact that the derivative,
f
B~({3) t
3
B2
B~(oo)
-00
In part (a) we showed that B~({3) is a monotonically decreasing function of {3, and consequently it has, by continuity, a zero if and only if
f
!
.pdr
>
0.
f
.pdr
for cO,
gas.
9.12 Calculate the second vidal coefficient for the following inter molecular potentials:
c
(a)
ip(r) = r6
(b)
'P(r) =
> 0 . (2)
1f: .0
The two necessary and sufficient conditions for a maximum in B2 (T) are thus
< 0 for some distances, but
-00,
as it must be since both these limiting cases correspond to a hard sphere
r>d
.p
for c
={ j1TR 3 i1Td
is always positive if .p(r) is non-negative. (c) Since it follows from part (b) that a maximum is only possible when the potential is negative somewhere we have for T --J>. 0 that
259
Note that
ip(r)exp(-{3ip)dr,
r> d
B;(O)
The imperfect classical gas
9.13
\O\r)
(c)
for r";;; d
< r";;; Xd for Xd < r.
ford
~ {-U(2:-r1d)
for
r";;;
d
< r";;; 2d
for 2d < r.
ford
r>d
Solution
9.11 Calculate the second virial coefficient for two gases, one with the intermolecular potential
'Pl(r) =
{
for r
+00
c(r3~R3)
< r,
the other with .p2(r)
i c(d -r 3
for r 3
)
0
Introducing r3
= s as new integration variable we find in both cases
B2 (T)
21TR3
21TkTr
3c
21Td 3
(c)
I -exp
C(R3_d3)]
(2) H. L. Frisch and E. Helfand, J. Chem. Phys., 32,269 (1959).
kT
[ia 3 -2-4a-4a 2 -(2+2a+a2 )e"'] , with a =
9.13 (a) Show that the difference F- Fi between the free energy for a real gas and for an ideal classical gas of N particles can be written
F-Fj
< r.
[The identity of the results demonstrates the important fact that the function B2 (T) does not determine the intermolecular potential uniquely.] Solution
21Td 3
ford";;; r";;;R
for R
)Yl
-3-[X 3 +(l-X 3 )exp(€ / kT)]
(b)
for d";;; r";;; R for R
+00
C1T 41T( kT
L
Here U i
<
= -kTln{l+v-Nf(exP(-U/kT)-I]dr! ... drN}'
ip(ru,) is the interaction energy. 11:
Assume now that the real gas is so dilute that only one pair of particles are close together (Le. within the range of the potential.p) at the same time. More precisely assume that in the expression for F- fj the integrand (which vanishes if no pair of particles are close together) may be replaced by the sum over all pairs,
L
exp(-U/kT)-1 ~ i
<
{exp[-ip(ri/c)]-l}. 11:
260
9.13
Chapter 9
Assume further that there are so few particles in the volume v that the first term in the expansion of the logarithm above suffices. Show that these approximations lead to
The imperfect classical gas
9.14
yields a factor v, leading to }-ri ~ ,
F-Fi
------y;;-
NkTf
~ ~
{I - exp [-I;?(r)/kT)}dr .
(c) Show that this expression for the free energy yields a first-order correction to the pressure of an ideal gas in agreement with the virial expansion of Problem 9.8. Cd) Calculate the corresponding lowest order correction term to the ideal gas results for each of the following quantities for a real gas: Gibbs free energy G, the entropy S, the internal energy U, the enthalpy and the heat capacity at constant volume C v .
1
l· I-v-B2(T) ~ J~ -v-B2CT). ~kT
-kTln
We have used the fact that ln(l +x) ~ x for x ~ 1. The free energy per particle, F-F NkT ~~ v B2 (T) , is an intensive quantity, as it should be. - ('oF/aV)T' In this case the excess The pressure is given by p pressure (compared with the ideal gas) is ~kT
Solution
p - Pi
(a) The free energy F is given by the partition function exp(- F/kT) = ZN =
N!~3N f exp (- E/kT)dPI ... dpNdr l ... drN
N!~3N fex p
K/kT)dPl'" dpNfex p
U/kT)drl'" drN'
This is the equation of state of the gas in the approximation we are con sidering. It is identical with the first two terms of the virial expansion· of Problem 9.8. (d) Using the results obtained above, = j
L {exp[-I;?Crik)/kT]
I
and
== _ I}
i < k
yields !N(N - I) equal contributions. In each term N - 2 of the inte grations are trivial, leading to +N(N-I)f ~ .. 2 texp [-lPCrlz)/kT]-
v2
v
G = F+pv = G.+ 2NkTB2
S by
P -Pi = kTB 2(T)
NkTB 2 (T) ,
we find easily
+v-Nf [exp(- U/kT) -I] dr 1 ... drN } •
(b) Replacement of exp(-U/kT)-1
F- Fi = - kTln III
B 2 (T)
F-F
Putting exp(-U/kT) = 1+[exp(-U/kT)-I], and taking the logarithm, we have
{I
---;;r- B2 (T) , =~
_ F- Fi) _ fexp(- U/kT)drl ... drN _ -Nf _ kT f - v exp( U/kT)drl'" drl ... drN
F- rj == - kTln
=:::
or
We have stated that the energy is the sum of the kinetic energy K(p) and the interaction energy U(r). For an ideal gas U = O. Hence
exp (
261
l}dr l dr2] .
Here we may replace N-l by N. For a large volume the integration over r 1 may be taken to be independent of r2 and equal to
U
(aF) aT
F+TS
H = U +pv C = v
(au) aT
S. v
=
=:::
V
Nk[Bz(T)+TB~(T)]
V
I
'
~ Nkrz~;(T) ~
+
v NkT[B z
TB;(T)]
v
'
C. Nk112B~(T)+ TB~(T)] I)
V,l
()
PAIR DISTRIBUTION FUNCTION. VIRIAL THEOREM
f{exP[-I;?(r)/kTl-I}dr = -2Bz(T), in the notation of Problem 9.8. The subsequent integration over r2
9.14 (a) Show that, if the configuration of a gas of N molecules is observed, the probability of finding one particle within the volume
262
Chapter 9
9.14
element drl at rl and another particle in the volume element dr2 at r 2 is given by n2(rf,rz)drl drl
=
The
9.14
classical gas
263
Solution
(a) In the canonical ensemble the probability distribution of all posi tions and momenta is given by
1 drl drzI ... I dr) .,. drNexp(- UN/kT).
=
p( PI'" r N)
-,----'=--''---'-'---'-'-'------''
"
Here UN is the interaction energy of the N particles and QN is the con figuration integral. No external fields are present. In the thermodynamic limit £ this distribution function will depend upon the relative distance between r I and rz: £nz(rl> rz)
Integration over all momenta and over r), r4, ...rN yields the
Per I, r2) .£l.4~!.L2f finding particle I in a volume element dr 1 at r I and paftTCIe2 in drJ at rA
dr) ... drNexp(- U/kT)
nZ(r12) .
P(rl, rz)
nz(r) is called the pair distribution function and the function g(r)
Wi! obtain
vZnz(r) - I
is often called the correlation function. Show that g(r) vanishes for an ideal gas. Explain why g(r) should vanish when r -+ 00 in a homogeneous system (one phase). (b) Consider a system consisting of a gas and a liquid phase, both of macroscopic extent. Give a probabilistic argument showing that one expects the following form for the pair distribution function in the two phase region: n2(r; v) V-I[XlvlnZ(r; VI)+ XgVgnl(r; Vg)] . Here VI and i)g are the specific volumes of the coexisting liquid and gas phases (see Figure 9.14.1). Xl and Xg are the corresponding mole frac tions so that v = xlvl+xgV g • The pair correlation function nz(r) de pends clearly upon the state variables (ii, T), and the notation nz(r; iI) is used above to Indicate explicitly the value of ii. p
drz ... drNexp(- U/kT)
(v ==
QN dr3 ... drN exp(-U/kT)
The probability of finding one particle (not necessarily particle I) in dr l and another in dr2 is simply !!.(N-I)P(tlt2)dr,dr21 since there areN choices for the particle in dr I, and then (N I) possibilities for the second particle. Hence n2(rlr2)
... drNexp(- U/kT) .
For the ideal gas this reduces to n2(r l ,rZ) since U
=
1)
= 0 in this case.
Taking the thermodynamic limit we obtain 1 n:z(r) ; g(r) =v 2nz -1 = 0 .
Qne expects two particles in a gas or liquid to become completely inde endent as the distance between them Increases. Hence the JOInt proBa bilIty 0 In Ing one par IC e In drl and another particle in dr2 approac es :tEe product of the probabilIties of each event, namely (
Figure 9.14.1.
~ dr I) ( ~ dr2) = vI2 dr
I
dr 2
•
(b) The probability of finding a particle in drl is still O/v)dr l , since the positions of the two phases are random when there is no outside field of force. Now Xg and XI are the a priori probabilities that the first particle belongs to the gas and liquid phase. If the first particle belongs to the gas phase then the other particle at a microscopic distance r will also be in the gas phase (of macroscopic size). Hence the conditional
9.14
Chapter 9
264
probability of finding this second particle in drz, given that the first particle is at rb is Vg112(r; vg)dr2' The same argument applies to the liquid phase. We obtain thus for the joint probability distribution func tion in the two-phase system the following expression: _ Xg l)gn2(r; Vg )+ Xl Vl nZ(r; VI) nz(r;v) = _ v 9.15 Prove that
If
kT P = ij-6
n{)
I
9.16
The imoerfect classical gas
265
In the thermodynamic limit the pair distribution function is a func r. The virial theorem of sta tis tical mechanics, tion of r 12 p
kTp
-i
f
drrr.p/(r)n 2 (r) ,
also called the thermal equation of state, follows. 9.16 (a) Prove Clausius' virial theorem in classical mechanics (for bounded motion) N
Ekin = -! E r n F n ,
(r)nz(r)dr
n" I
by differentiating the configuration integral with respect to the volume. Here n2(r) is the pair distribution function of Problem 9.14 and r.p(r) is the intermolecular pair potential. To facilitate the differentiation procedure introduce new vari ables ii = riv-1I3 in the configuration integral. The above result is often called the virial theorem of statistical mechanics.] Solution
Introduce the new variables ii = rJL, with L 3 tion integral: QN
f
= L3N
di l ... diNexp [- i
v, in the configura
~kr.p(Lfik)/kT]
where Fn is the force acting upon particle n. The bar denotes an aver age over time. (b) Apply Clausius' theorem to an imperfect gas in equilibrium, assum ing that the total force can be derived from the wall potential and the intermolecular potential r.p(rik) acting between each pair i, k of particles. Assuming that the time averages involved may be evaluated as phase averages in the canonical ensemble, show that the result of Problem 9.15 follows. [For related considerations see Problems 3.7 and 3.8.] Solution (a) Use
Note that the integration limits are now independent of v. Hence
1 aQN
aQN =---=--L kT QN av 3VQN aL
__ I
f
~
rnFn
I
A
[N
3N _
- QN dr l ... drN v L
L 3N+ I ar.p(Lfik )] [_ '\' ..p(Lfik)] 3kTv aL exp /(~k kT
In the second term use a..p(Lfik ) ~ ar.p(Lfik ) aL = rik a(Lrjk) ,
r
when T
E i<
r.p(r ik )] k kT .
The first term is simply N/v = p, and by symmetry the second term consists of ;N(N -1) equal integrals. Hence, N(N - I p - kTp 6 Q _
v
N
)fdr
1 ...
drNr l2
ar.p(r 12)
ar12
I fdrl dr r a..p(r 12) nz ( r1> r 2 ) -- k Tp--6 Z 12 v r l2
a
exp
E i
r.p(r ik )]
kT
d
drn
d
dt(rnPn)-
-+
00.
m
Summation over all particles yields
= -2Ekin .
Ern n
(b) The kinetic energy consists of 3N quadratic terms and the equipar tition theorem yields the ensemble average
Ekin
"NkT. The force exerted by a wall element dS equals -p dSn. Here n is a unit vector normal to dS pointing outward. The contribution of the wall force to ErnFn equals
,
using the definition of the pair distribution function (see Problem 9.14).
P~
The last term is twice the kinetic energy of particle n. The time average of the first term on the right hand side vanishes for a motion where rn and Pn are bounded since I (T d I TJo dt(rnPn)dt T[rn(T)Pn(T)-rn(O)Pn(O)] -+ 0
and transform back to the original coordinates: p 1 [N I ar.p(r(k)] kT= QNJdrl ... drN ;-3kTvi~/ik arik exp
dPn
= rnTt = d/rnPn)-PnTt =
n
-pfrndS
-pLvrdr=-3PV.
266
9.16
Chapter 9
Here we have used Gauss' theorem and Vr = 3. The contribution to L rn F n from the intermolecular forces is the sum of !N(N - I) equal terms, one from each pair of particles, -!N(N-l) (r i V I +rz V 2) ip(r ll )
Solution
Introduce in the virial theorem of Problem 9.IS the low density form of the pair distribution function, exp[-ip(r)/kT] nz
.
Here V k operates on the coordinates rk. Using aip r l -rl VIip(r I2) = (V I r I2)-ar = -r-ip (rI2) , 12
12
ri2, we may write for this contribution
1:.. -! kT - ij + 6kTV2
pv = NkT-iN(N -1) rI2ip/(rn) .
The ensemble average of rnip/(rll) is given by rI2ip'(rn)
'
f
1
drl dr2rnip (r12)N(N_l)n2(rl, r2),
This agrees with the usual expression for the second virial coefficient given in Problem 9.8(d).
J
kT
which is the same form of the virial theorem as was obtained in the solution of the preceding problem.
1
2
3
~-2 exp [ip(r) ] - kT
.
Here ip(r) is the interaction energy of the two particles. The constant in front of the exponential is chosen so as to yield the correct limiting behaviour n2 -+ (N/V)2 = ij-2 asr -+ 00 (see Problem 9.14). Use this in conjunction with the virial theorem of Problem 9.IS to obtain the following first order correction to the ideal gas equation of state: p I 21T -exp [ _ ip(r)] kT }r2 dr . kT = j}+fj2 that this agrees with the virial expansion of Problem 9.8(d) to order p2.]
roo
3'
1:.. -! ~ J"" 3 [ip(r)J~ [_ ip(r)] kT - ij + 3 1T 0 drr nl(r)exp kT dr exp kT
'
and show that the first bracket is finite everywhere, while the second bracket vanishes for r < d and has a 5-function contribution at r = d.] Solution
9.17 Assume that for a very dilute gas the pair distribution function n2(r) is well approximated by the Boltzmann factor
21T
j}+3'1Td nz(d+) - 3kTJd+ drr ip (r)n2(r)
when the interaction potential ip contains a hard core of diameter d. Here nz(d+) means the limiting value when r -+ d from above. [Hint: Rewrite the virial theorem of Problem 9.1S as follows:
= NkT-i dr l dr2rnip'(rI2)n2(rlr2) ,
v
2
_
9.18 Show that the vitial theorem of Problem 9.IS may be written
from the definition of the pair distribution function (Problem 9.14). Thus, pv
I
rip (r)exp( ip/kT)41Tr dr.
I 21Tf"" j}+ ij2 0 [I -exp(-ip/kT)]r2 dr.
p kT
= dNfdrl ... drNrl2ip'(rn)exp (-UN/kT)
=
0
v
An integration by parts yields
-!N(N-l) r I2ip/(rn)'
Collecting the three contributions we obtain
-2
_I_J""
We obtain
I
and rl(rl -r2 )-r2(r2 -rl) =
267
The imperfect classical gas
9.18
The function exp(-ip/kT) vanishes for r < d and jumps at r zero to the value exp [-t,p(d+)/kT], so that its derivative equals
= d from
{ex p [-ip(d+)/ kT] 5(r -d) for r ~ d exp [-ip(r)/ kT] for r > d ~ where 5(r-d) is Dirac's delta function. Recall also the definition of the pair distribution function (Prob lem 9.14): d drexp(-ip/kT)
nl(rl>r2) == N(N-l)f .n dra·.. drNexp [ -i~kip(rik)/kT] . Here a factor exp [-ip(rI2)/kT] can be taken outside the integral, which shows that exp [ip(r)/ kTlnl(r) is for all r finite (and even continuous).
268
9.18
Chapter 9
Splitting the range of integration in the virial theorem at r = d+ we obtain P I fd+ kT = "D+ ~7T 0 drr3n2(r)exp [1P(r)/kTj exp [-IP(d+)/kTj cS(r-d)
-j7Tf~ drr 3IP'(r)n2(r) d+
I f~ drr3IP'(r)n2(r). = -;:-+~7Td3n2(d+)-~7T
v
d+
In particular, the equation of state of a gas of hard spheres is deter mined completely by the pair distribution function at contact, P _ I 2 3 kT - "D+"j7Td n 2(d+)
9.19 Consider an interaction potential with a hard-core repulsion plus a weak exponential attraction of long range (see Figure 9.19.1)
J
IP
00
=t --'Y a e--yr 3
47T
for r
for r
>d
where a and 'Yare constants. Assume only one phase to be present so that the pair distribution function n2 ~ v- 2 when r ~ 00 (Problem 9.14).
269
The imperfect classical gas
9.19 Solution
The virial theorem yields P
kT f~r 3e--yr (r)dr. = -_-+~7Td3n2(d+)-!'Y4 n2
v
d
In the integral introduce the new variable s 'Y ~ 0: 'Y4 L~r3e--Yrn2(r)dr
= 'Yr,
and take the limit
= f-y~s3e-Sn2(S'Y-l)ds
~ t~s3e-Sn2(00)ds = 6v- 2 , using n2(00) = l/v 2. With the assumption that when 'Y ~ 0, n2(d+) ~ n2,s(d+), the pair distribution function of a pure hard sphere gas, we obtain in the limit kT 2 3 a P = T+ "j7Td n2,s(d+)-ij2 Note that the first two terms are the pressure Ps of a hard sphere gas. Hence, a
P=Ps- V2 Note that the last term which embodies the effect of the attraction is precisely the same term that occurs in van der Waals' equation(3).
'P
d
Figure 9.19.1.
Use the virial theorem of the preceding problem to obtain the follow ing equation of state in the limit 'Y ~ 0 (very weak attraction of very long range): a
P
= Ps -ij2
Here Ps is the pressure of a gas of hard spheres at the same temperature and density. Assume that because of the weakness of the attraction n2(d+) is determined by the hard core alone.
(3) For a rigorous proof of this result from the partition function see J. L. Lebowitz and O. Penrose, J. Math. Phys., 7, 98 (1966). For a complete discussion of the one-dimensional version of this model see M. Kac, G. E. Uhlenbeck, and P. C. Hemmer, J. Math. Phys., 4,216 (1963).
The imperfect quantum gas
10.2
271
To obtain the equation of state, we must evaluate the grand partition function :E: Tr exp(ooVop -(3H) , (10.2.4)
10
The imperfec t quantum gas
where (3 is again 1/ kT, 01./(3 the partial thermal potential p, and the number operator. In fact, it is more convenient to work with socalled q potential q = In:E: , (10.2.5)
(I)
D.ter HAAR (University of Oxford. Oxford)
which satisfies the
(see Problems 2.4 and 2.6c) q = (3pv ,
THE EQUATION OF STATE
and
10.1 In Problem 1.11 the virial expansion pv EI +E2P +E3P2 + ... has been considered. An alternative virial expansion is B C T)V = NkT{A +-+-+ t" V v2 where the A, B, ... are volume-independent virial coefficients. Find expressions for B, C in terms of the first three E's.
EIE2
B
C
= NkT'
= -=--"----'=--=
1O. 2 Consider a system of identical particles governed by the following Hamiltonian: (l0.2.1) H = Ho+HI' where Ho is given by the
Let 'Pn be a complete orthonormal set of functions for a system of N identical fermions-which means that they are completely antisymmetric in the coordinates of the various particles. We now introduce a set of functions WN(r I, ... , rN) defined by the equations ... ,
rN)
,,(-~V'~) T 2m I
,
(10.2.2)
and HI by the equation HI
=
! .).U(rjj ) /1/
,
(10.2.3)
with rij = Irj-ril. Here m is the mass of the particles, rj their position (we assume them to be point particles which may have spin), and we have assumed that the interactions between the particles are binary in nature so that HI gives the complete interaction Hamiltonian: we thus neglect any three body forces. For a further discussion, see D.ter Haar, Blements of Statistical Mechanics (Holt, Rinehaxt, and Winston, New York),
(I)
270
N1L 'Pn(rj)exp(-(1HN)'Pn(ri),
(10.2.8)
n
L 1/I:(ri)exp(-(1HN)1/In(r;) n A(r;; r;) = L 1/1: (rj) 1/In (r;) ,
(10.2.10)
N\ L€p L 1/1: (rpi) 1/In (ri) ,
(10.2.11)
(10.2.9)
W(rj;r;)
n
and Aqu(ri; r;>
Ro
(10.2.7)
where we have written HN to emphasise that there are N particles in the system. [The sum on the right-hand side of Equation (10.2.8) is called a Slater sum.] We shall introduce the quantities
Straightforward series expansions give
A=
G~)V.T
N =
WN(r l ,
Solution
(10.2.6)
. P
n
where the summation is over all Nl permutations Pi of the N values of i and where €p = + I or I for even or odd permutations, and where in Equations (10.2.9) to (10.2.11) the 1/In form some complete orthonormal set which does not satisfy any symmetry conditions. Prove that (10.2.12) N1W(rj;rj) ; W(rj;r;)
exp(-(3H')A(rj;r;),
(10.2.13)
where the prime on the H' indicates that it operates on the r; but not on the rj, (iii) if HI = 0, that is, for the case of a perfect fermion gas, then
WA?)(ri)=N!~1Y~€pexp [
L(ri -rPi)2] i
}..?
'
(10.2.14)
10.2
Chapter 10
272
, ~
'j
where
e~2r
A=
/'
(10.2.15)
I
(10.2.16)
,
I
rj ~
is essentially the thermal de Broglie wavelength, and Vo = 1T'h A3 •
.a(ri; r;)
.aqu(ri; r;)
,
1,
N'L... . P
, If'
we have ' = \7 '2 {)(r-r)
exp[i(k' r-r )]d 3 k,
f
-
i
exp(-{3Ho) {)(r-r')
I
[(r-r A2
Vo exp-
l
L... €Plexp
Vo PI
[_ ,(rl 1 -rP lil )2] x _1_, [_, (rl z -rp 212)2] L... A2 N2 L... P € 2 exp L... A2 Vo P2
II
IZ
N2'
where the sums over PI (and i I) and P2 (and i 2 ) extend only over the first and second cluster, respectively, and where we have used the fact , that €p = €PI€ 2 The above argument suffices for the case of a perfect gas. If there are interactions, we note that, if the N particles form two clusters, we have HN = H NI +HN2 or eXJ,?(-{llIN) = exp(-{3HNI)exp(-{llIN2) ' and because of this and the product property of the WN for the case of a perfect gas, Equation (10.4.1) follows. 10.5 Show that the W satisfy the Bloch equation aW(ri;r;) , , a{3 = -H W(ri;ri) ,
2
k (21T)3exp[i(k' r-r , )]d 3 k.
and hence
NI
NI
'
i
_1_,
W(O) • W{O)
€p 11 {)(rpi -ri) .
In the case of a perfect gasH is given by Equation (10.2.2), and, if we write the three-dimensional Dirac delta function in the form {)(r-r) =
The proof consists of two parts. First of all, we note that, if the N particles form two clusters, all terms where the kth particle and the Pkth particle are in different clusters will have a factor (rk -rpk)2/ A2 in the exponent and this will mean that those terms can be neglected, as Ir-rpki > D;i!> A. We have thus
,
n
273
Solution
W(O) N -
when the proper symmetry requirements have been taken into considera tion. In Equation (10.2.11) the summation over n is over a complete set irrespective of symmetry requirements. In that case, we can use the completeness relation for the lJJn and write instead of Equation (10.2.11) .aqu(ri;ri) =
The imperfect quantum gas
~
~
Solution (iii) We note that
10.6
(10.5.1)
where H' operates on the r; and not on the ri'
)2J ,
Solution
From Equation (10.2.13) we see that
where Vo and A are given by Equations (10.2.16) and (10.2.15).
Combining the various results, we now get the result (10.2.14).
W(rl; rD lp
0
= .a(ri; r;) ,
and hence that
10.3 Show that the above results remain valid for a system of N W(ri;r;;{3) = exp(-{3H')W(rl;r;;0) . bosons, when the IPn are symmetric in the particles, and €p = 1. Equation (l 0.5.1) follows immediately. Solution
The proof proceeds as in Problem 10.2.
10.4 N identical particles are said to form separate clusters of NI and + N2 = N) if the particles of the first are separated from each of the particles of the second by at least a distance D, where D is such that VCr) 0 when r > D, and also D ;i!> A. Prove from Problem 10.2 that (10.4.1) = WNI WN2 ·
I
1
10.6 Prove that q and WN are related through the equation e q =
N2 particles (N!
en"'JWN(r/)d3rl'" I-, n n.
d 3 rN .
(10.6.1 )
Solution
Equation (10.6.1) follows immediately by using Equation (10.2.8).
274
10.7
Chapter 10
~
10.7 Use the result of Problems 10.5 and 10.6 to prove that in the classical limit (h ~ 0 or A ~ 0) the WN go over into the WN of classical theory. From Equation (10.6.1) and the expressions in Chapters 2 and 9 we see that W~ = 1JON exp(-{3Hd. (l0.7.1) In the proof it is convenient to write the WN(ri) in the form WN(ri)
(10.7.2)
LEpexpg(rj;rPi), p
where g(rj;r;)
= In[exP (-Illi')I!6(ri -r;)]
(l0.7.3)
,
We note that quantum effects and interaction effects are completely mixed up. 10.9 Expand the expression for B found in the preceding problem, for high temperatures, in a power series in h, up to the term in h 2. Solution
The second term in the square brackets can be found easily using the result of Problem 10.2 and we can thus rewrite the expression for B as follows: B =
Solution
The function g satisfies the equation 2
275
The imperfect quantum gas
10.9
~f[I-2VMexp(-{3m).6.(rl>r2;r'l>r~)}]f;
fjd3rld3r2'
Introducing centre of mass and relative coordinates, we can integrate over the former which gives a factor v, and we are left with
2
og '" h V j' 2 g+72m( '" h V',goVig)-Hlg. o{3= 72m I
B
This equation can be solved as follows: g= -
2;2~ (ri -r;)2 - Nlnvo -Illil + power series in ~
The first part is the solution when HI = constant and for details of obtaining the power series, we refer to the solution to Problem 10.9. In the limit as A ~ 0 (or Vo ~ 0) we need retain in the sum over P in the expression for WN(ri) only the term with the identical permutation. Moreover, in the same limit, we can drop the power series in {3h 2/2m. Hence we find for WN WN
~
exp g(rj, ri)
~
~
tl
I
10.8 As the equation of state follows from Equations (l 0.2.6) and (10.2.7) and as Equation (10.6.1) is formally the same as in the classical case with the WN of Equation (10.2.8) replacing the WN of Equation (10.7.1), we get formally the same expressions for the virial coefficients. Find the expression for the second virial coefficient B for a quantum gas.
where 2 f(r;{3) = 2'1· vo fexp ({3h V'2 -{3U){6(r-r') ± 6(r+r')}] , L m r=r, where the upper (lower) sign refers to the case of bosons (fermions). We stress that it will pay the reader to derive the above results in detail. We note that the above equation for fer; {3) can be written in the form
!'
fer; {3)
I
11
Nlnvo -Illil .
\
•
and that f(r;{3) thus satisfies a Bloch-type equation Of h 2 o{3 = m V2f-Uf.
l!l:
i
If we introduce a function g(r; r') by the equation g(r;r') = In we have f(r;{3)
From the cluster expansion for the equation of state we have
= NvH 2v J[W2(rl, r2)
3
3
WI(r l ) WI(r2)]d rid r2,
which can be written in the form I I B = - Nv~f 2v [2exp(-{3H2I ).6.(rl>r2;rl,r2) -exp(-{3H~O)I).6.(rl;r'd.6.(r2;r;)Jf; rjd3rld3r2'
= exp ({3~V2 - {3U) fer; 0)
'I
Solution
B
= !NJ[l-f(rI2;{3)]d3rI2 ,
) ,I
~
\
~
lexp({3~V'2-{3U)6(r-r')J
= i"vo{exp[g(r;r)] ± exp[g(r;-r)]},
while g satisfies the equation h2 h2 -og = -V 2g+-(Vg o{3 m m
0
Vg)-U.
We look for a solution in the form , "
m(r-r')2 4'/l'{3h2 g(r;r) = AI?1o.2 - , In----;:n- - n'Y;lanff' .
276
10.9
Chapter 10
Substituting this expression into the equation for g we can solve it by successive approximations, and we find for g(r; r) 2 2h2 41T13h 13 [ 13 g(rr)+~ I n = -I3U+'VU)] + .... , m m -f~;2U+-('VU' 12
_Nvo
Nv 2
or
2
10.10 Split the expression for B found in Problem 10.8 into two parts: Bperf corresponding to a perfect quantum gas, and B imp containing both quantum and interaction effects. Prove that the second part can be written in the form 2 2 ood 'l1 exp (13h k ) dk, (10.10.1) Bimp = -1TNvo~ (21+ 1)0 , ° dk ---;n-
= exp [ h (P'
€k
= -+E 4m n,l,
2
d R
dr 2
(10.10.3)
Birnp
,
with
- -Nvijf 'uto), . ' r, 2v [exp( - l3u2 ){2.6.(r 1 ,r2,rl> 2)
Bperf -
Nvijf[{ex d3 r2 P(-l3m) -exp(-t3.HiO)')}.6.(rl, r2 ;r'1, r~)]J'!I = J'.d3r1 v I
2
l-
l(l+l)} U(r}]--rR = 0, 2 -
-ihNvoL (21+ I
,
}5,f ""exp(-I3E)p(E, l)dE , °
h 2k 2 E=
m
- .6.(1' 1; r'd.6.(r2 ; r~ )})ri = rid3r 1 d 3 r 2 , -
{mh [En.
where we have assumed 0) that there are no discrete energy levels, (ii) that we may replace the summation over n by an integration over E, and (iii) that the numbers of energy levels Enl and E~~) between E and E+dE is given by Pl(E;l)dE and Po(E;l)dE with peE;/) = Pl(E;l)- Po(E;I). Introducing the wave num ber k by
We can rewrite the second expression as follows:
Bimp
+
and the same equation with U = 0 for R(O). The contribution from the centre of mass gives a factor 2% Vivo, and we have
(10.10.2)
Solution
B = B perf +Bimp
lRnl(r12) r12 0/ Y,m(w) ,
R)J
where w = rI2Ir12, y,m is a spherical harmonic, and 0/ = 1(0), I = even; 0/ = O( I), I = odd for bosons (fermions). The R n " satisfy the equations
and 00 •
i
lPk(rl, r 2 )
where 0, = 1(0) when I is even and 0(1) when I is odd for the case of bosons (fermions), and where '11 is the phase of the asymptotic solution of the radial Schr6dinger equation,
as r -..
,
p2
f
R "'" sin[kr+'I1(k,
]
where the €k(€~O» are the eigenvalues of H 2 (H!°». Introducing centre of mass [R = Hrl +r2)] and relative (r12) coordi nates, we get the eigenfunctions of H2 and H1°) in the form
where Bel is the classical result [see Problem 9.8(d»).
OJ R = 0
r
Nv -_o.L exp(-I3€k)- L exp(-/3€~O» v k k
B imp
B = BcI +h 2 B1 + ... ,
".k 2 - m U(r)--r1(1+2 h2
,
Bimp = -~[Trexp(-t3.H2)-Trexp(-I3HJO»] ,
Substituting this expression into the equation for B we find
2
20/,
with the upper (lower) sign again referring to bosons (fermions). In fact, it is more convenient to write Bimp in the form
h2 [ --V 13 2 U+-(VU' 13 f(r'l3) 'VU) + ... . , = exp[-I3U(r») { 1+m 6 12
d R
+
Bperf
J}
3
277
It is a straightforward exercise to evaluate Bperf with the result
We do not need the expression for g(r; -r) as we are looking for a high-temperature expression in which case exp [g(r; -r») will contain a factor exp mr2/13h1 ) which will lead to an extra factor Vo which is negligible. We finally find for fer; 13) 2
The imperfect quantum gas
10.10
'
we have Bimp = -2';'Nvo~(21+ 1)0,
•
Ii
Jo exp 00
[
13h2 k 2] --;;g(k,l)dk,
278
10.10
Chapter 10
10.12
(a) Prove that the sets (10.11.1) and (10.11.2) satisfy the ortho normality relation
and the equations for the R n, i are now 2
d R [ k 2 --lJI(r) m 1)J R -+ -l(l+ dr 2 1'l2 r2
= 0
1
•
(i'1> i~, ... , i'tvl i I> i 2, ..., iN) = N'
and a completeness or closure relation. We assumed here that the index If it is a discrete one, the Dirac 0 functions must be replaced by Kronecker ones. (b) If 1'11) is a properly symmetrised wave function of the N-particle system, express 1'11) in terms of the set (10.11.1 ). Also give an expression for an operator n operating on functions in the Hilbert space spanned by the set (10.11.1).
i is a continuous parameter.
with the boundary condition (corresponding to the vanishing of the wave function at the boundary of the volume) R(ro) = R(O)(ro) = 0, where ro is large. This boundary condition leads to
= mr ,
kr 0 + 71(O)(k, I) = mr .
L €po(i I -i~l) ... o(iN-i~N) , (10.11.3)
.p
For large values of r, U(r) will vanish, and the asymptotic solution for R is R = sin [kr+71(k, I)] ,
kro +71(k, I)
(10.10.4)
Solution
f...
If 11k and I1k(O) are the changes in k when n changes by unity, we have clearly 1 1 p(k, I) = 11k - I1k(O) ,
where
and from Equation (10.10.4) we have
(and if necessary summing over its spin variables).
SECOND QUANTISATION FORMALlSM(2)
)
= N!~€Pl{Jil(P1}.pi2(P2) ... l{JiN(PN).
fdil ... diNlil, ... ,iNHiI, ...,iNI'I1),
di indicates integrating over the coordinates of the ith particle
... fd''I'"
d'IN d"'I'" d'" . )(.'I"",'N~" . Inl"'I"",'N ")(.,'1"",IN ., I . 'Nill,··,IN
10), Ii>, li l ,i 2 ),···, li l ,i2 ,···,iN ),···,
a+(i)lil, ... ,iN ) =(N+1)Y>li,i l , ...,iN )·
(10.12.2)
Express the Iii' ... , iN) in terms of the a+(i) and the vacuum state. (10.11.1)
(b) If [A, BL = AB - BA is the commutator, and [A, B]. = AB+BA the anticommutator, of the two operators A and B, prove that
This is a properly symmetrised set. The bra set corresponding to the ket set (10.11.1) is
[a+(i), a+(j)L = 0
for bosons,
[a+(i), a+(j)]. = 0
for fermions.
(10.12.3)
and 1 (il>i 2, ... ,iN i = N!~€pl{J~(p1)I{J~(P2) ... I{J:;/PN).
(10.12.1)
where 10) is the vacuum state. We now introduce creation (or construc tion) operators which will produce an eigenvector corresponding to N + 1 particles from one corresponding to N particles, as follows:
10.11 For many purposes it is convenient to use a formalism which is independent of the number of particles in the system. Let I{Jn be now a complete orthonormal set (c.o.&) of single-particle functions (including the spin dependence where necessary). For a system of N identical systems one can then use as a c.o.s. of basis functions the set ••• ,iN
=
10.12 (a) Consider systems with an arbitrary number of particles, that is, consider the Hilbert space which is the (direct) product space of the Hilbert spaces corresponding to 0, 1, ..., N, ... particles. The c.o.s. span ning this Hilbert space will be the set
and hence Equation (10.10.1) follows.
lil>i 2
J
n f ~,,=
=~'
1
1'11)
(b)
(ro+ ~~)l1k = ~, d71(O») - - I1k(O) ( r o + dk
279
The imperfect quantum gas
(10.11.2)
(c) We now introduce an operator a(i) by the equation (2) For a general reference, see l.de Boer, Progress in Low Temperature Physics, Vol III, (North-HOiland, Amsterdam), 1965, p.2l5.
(il, ...,iNla(i) = (N+1)Y>(i,il> ... ,iN I.
,i,
(10.12.4)
280
Chapter 10
10.12
10.14
Prove that [a(i), a(j)L
(10.12.5)
[aU), a(j)l.
0
n/ = -
forfermions.
(d) Find an expression for a(Oli l , ... , in>. (e) Prove that
a+(j)L = aU for bosons,
and [a(t), a+(j) 1. a(i - j) for fermions.
Solution
(10.12.6)
,n =
i
i,;
n ij =
V(r/;),
(10.13.2)
2 2 "fi k + L. 2m
k
where
"nP)+l " n(~) L. I '! L. I, '
fi2 2m vj2+U(r) ,
find an expression for n in terms of the at and ak'
(f) If n is an operator of the form
n
281
In that case, one usually writes a; and ak rather than a +(k) and a(k). If n/ and nil are of the form
= 0 for bosons,
and
The imperfect quantum gas
U(q) =
(l 0.12.7)
where the n}l) and n~2) are, respectively, single-particle and two·particle operators which differ only in the particles on which they operate, express n in terms of the a+(i) and a(i).
+ 1" + 1 '\ + + akak+-; L.U(q)ak a k-q+2v f.t V(q)akak,ak'+qak_q, k,q k,k ,q
f
d 3 rexp[-i(q· r)]U(r),
V(q)
f
d 3 rexp[-i(q· r)]V(r).
10.14 Give an expression for
a +(i)a(i) Ii I> ..., iN> , and discuss the physical meaning of the operator
Solution (a)
IiI, ..., iN>
nO)
1
N! a+Uda+(i2) ... a+(iN) 10)
a+(i)a(i)!il, .. ·,iN )
where the upper (lower) sign again refers to the case of bosons (ferm· ions). From this equation it follows that the aU) are annihilation operators. (e) This follows from Equations (10.12.1), (10.12.4), and the equation given under (f) n = fdidi'a+(i)a(i')
f
00.12.8)
10.13 Consider a system contained in a finite volume v so that we can take for the original c.o.s. a set of plane waves, '-Pi
-T
v-Y>exp [i(k .
(l 0.13.1)
N(i)lil> ... ,iN >,
where N(i) is an integer which tells us how often i occurs among the numbers iI, ... , iN' The operator n(t) is thus an occupation number operator.
N-Y>{a(i - i d li2' ... , iNH ... +(± l)N -I aO - iN)Ii 1> ••• , iN. I >},
+! di dj di' dj' (iii n(2)! i'j')a+ei)a +(j)a(j')aU') .
(10.14.1)
Solution
(b) This follows from Equations (l 0.12.1) and (10.11.1). (c) This follows from Equations (10.12.4) and (10.11.1 ). (d) By considering the matrix element (i'J, ... , i~_1 Ia(t)Ii'1> ... , iN> and bearing in mind that the resulting expression is valid for any
a +(i)a(i) .
i.
t
.I'
11.1
Phase transitions
283
and show that the equation of state follows from the equations
= f(y)
(3pvo
11 Phase transitions
Vo
where
where n
(~~ ) v,
(10.2.7')
2m)'/'V q = -21T ( h2
(11.1.8)
0
,jEln(1- e" -(3E)dE .
(11.1.9)
(11.1.10)
From Equation (10.2.7') it then follows that 21Tm) '/' - e nil (11.1.11) N = ( (3h 2 V L ----;;, n=1 n and Equations (11.1.6) and (11.1.7) follow immediately. To obtain the virial expansion, we must express y as a power series in Vii from Equation (11.1.11) and substitute this into Equation (11.1.10). This can be done either by tedious sorting out, or more elegantly as follows. Ifwe write x = volvl' we have
11.1 For a perfect boson gas with single-particle quantum states j of energy Ej we have (see Problem 3.l2a) (11.1.1 )
/
tj = exp(ex-{3Ej ).
f-
Expanding the logarithm, and integrating, we have 21Tm)'/' - en" q = ( (3h 2 V n~1 n'/,
EINSTEIN CONDENSATION OF A PERFECT BOSON GAS (I)
where
yn
n'/, '
Combining Equations (11.1.1), (11.1.2), (11.1.3), and (10.2.6) and going over to an integral from the sum in Equations( 11.1.1), we find
= Nlv = llvl, with VI = vlN the specific volume.
q = -~ln(1-tj),
L n=I
Solution
(10.2.6)
T '
(11.1.7)
and where Vo is given by Equation (10.2.16). Obtain a virial expansion in the form of a series expansion of (3pv I in a power series of Vii.
and
=;
-
f(y) =
11.0 Let us remind the reader that for a grand canonical ensemble we have (Problem 10.2) q = In:=: , (10.2.5)
n
= yf'(y),
VI
D.ter HAAR (University of Oxford, Oxford)
q = {3pv ,
(11.1.6)
,
(11.1.2)
If we consider the case of spin-zero point particles in a force-free volume, the number of energy levels d.Z lying between E and E+dE is given by the expression (compare Problem 3.l3c)
x = yf'(y) ,
(11.1.12)
and we want to find the expansion dZ = 21T ( 2h":. )'/'vEY'dE.
(11.1.3)
f(y)
From Equation (11.1.2) it is clear that we must always have
< {3Ej
(11.1.4)
so that tj is always less than unity. Put y = e"
(11.1.5)
ex
f
--=-=La xn yf'(y) x n'
(11.1.13)
Let Yo be the value of y satisfying Equation (11.1.12). From the theory of functions of a complex variable it follows that 1 "r f(y)dln[yf,(y) -xl 21Ti
=
[(Yo) ,
(11.1.14)
where the contour lies in the complex y-plane and is going around the origin and around Yo.
(I) For a general discussion see, e.g. D.ter Haar, Elements of Thermostatistics (Holt, Rinehart, and Winston, New York), 1966.
282
j
284 If we write
= yf'(y)
[I Yf~y)J
we can expand the logarithm and we find I
I
= 27Ti jf(Y) d
{Inyf'(y) - J;~ 0 n +I 1
l J x yf' (y)
From Equations (11.1.1), 01.1.2), and 00.2.7') we have
OLLIS)
n+
=_1_ ~
xn+'llf'(y)rndy 27Tin~on+lj .. n+1 ,
1
N = ~ -ex-p-(--a-+ (3Ej )
I}
(3pv I
= N [1 -
11.2 It has been shown(2) that when y can be written in the form
;/2 + .. J. <
1
exp(-a+{3Eo) -1 + j~O exp(-a+{3Ej )-l
01.1.16)
1 N= -"-1+) e jr'o
and using Equation 01.2.3) we find 1
defined for y > 1 (the power series no longer converge). As long as the specific volume VI is larger than a critical value Vc given by the equation 1, 2·61 1 - = -f(1) = , (11.2.3)
=
No v Iyf'(y) N + Vc f'(f)
01.2.7)
If VI > vc, we expect y to be about equal to 1. Let us put y = 1 in the last term on the right-hand side of Equation 01.2.7) and afterwards verify that the errors made are negligible. For No we write
Vo
the parametric form of the equation of state of Problem 11.1, Equations (11.1.6) and (11.1. 7), can be used. We must now discuss what happens when VI < Vc' The difficulty arises because, when V I is decreased from very large values (v I ~ v J, the solution y of Equation (11.1.7) will increase and will approach 1 as v I approaches v c' This means that a approaches zero starting from large negative values for v I ~ Vc' This in turn means that for the lowest energy level Eo, which we had put equal to zero when deriving Equation (11.1.6), tj = to will approach unity, and In(1-t o) will approach-oo. This entails that the corresponding term in the sum on the right-hand side of Equation (11.1.1) dominates. This term represents the number of particles in the level Eo (see Problem 3.12). Assuming N to be large but finite, find the value of y for the case v I < v c , by splitting off from the sum in Equation (11.1.2) the term with j = O. Assume the state j = 0 to be non-degenerate. Hence find an expression for that part of the isotherm for which v I < V c'
(11.2.5 )
Vo
01.2.1)
= -3·54 (-lny)Y>+2'61-1'461ny-0'1O (lny)2+ .... (11.2.2) We see that these functions have a branch point at y = 1 and are not
Vo
(11.2.4 )
Using the fact that the first term is the number No of bosons in the lowest energy state, and replacing the sum in Equation 01.2.5) by an integral, we get v N = No+-Yf'(y) , (11.2.6)
(11.1.17)
f'(y)
Vc
1
If we choose Eo = 0 [corresponding to the lower limit on the integral in Equation 01.1.9)], we can write Equation 01. 2.4) in the form
1 the functions fey) and f'(y)
= 2·36 (-lny)J/,+ 1·34+ 2·611ny -0·73 (lny)2 + ... ,
-
1
=
from which it follows that (n+ l)a n is the coefficient of yn in [f'(y)rn. In this way we find
fey)
285
Solution
yf'(y) x
[(Yo)
Phase transitions
11.2
11.1
Chapter 11
1 y Noo = ~1 e - = -1-y
1
-1-y .
01.2.8)
(VI
01.2.9)
R::
We then have from Equation 01.2.7) 1
1
vI
= NO-Y)+vc
,
or 1 Vc y = 1-Nvc-vl '
I
I'
which, indeed, as N -+ 00 leads to y -+ 1. From the expansions (11.2.1) and (11.2.2) we can easily estimate the magnitude of the terms which we have neglected. Substituting Equation 01.2.10) into Equation 01.1.6) we find that for v I < Vc the isotherm is given by the equation 1
(3p = -fO) , Vo
that is, the isotherm is horizontal.
(2) W.Opechowski, Physica, 4,722 (1937).
I.'
01.2.10)
01.2.11)
286
Chapter 11
11.5
11.2
No VI - = 1- N Ve
(11.2.12)
1'c=3'2°K,
(11.4.3) a value sufficiently close to the A-temperature, where 4He I goes over into 4He II, to suggest to many people that the A-transition of liquid 4He is, indeed, something in the nature of an Einstein condensation. Among other things, experiment diverges from this simple theory by the fact that for a perfect boson gas the specific heat stays finite at 1'c (see Prob lem 11.6), while at the A-point the specific heat of liquid helium becomes infinite.
that is, a finite fraction of the system is i'1 the lowest energy state. This is called the Einstein condensation. 11.3 Using the expansions (11.2.1) and (11.2.2) show that on the iso therm at V = V e , anp/av n is of the order of N(n - 2)/3 for large N. (3) Solution
If v I ~ V e , we must take, instead of only the leading term, 2·61 == f(1), on the right-hand side of Equation (11.2.2), the first two terms, and we get instead of Equation (11.2.9) the equation
so that forvi
287
If we substitute p = 0·15 g cm- 3 ,M = 4, we find
We also note that now
N
Phase transitions
"J ,
I+ NVI =- rvo --3'54(-a)" -a Vo Ve
11.S Find the locus in the (p, I) I) diagram of all condensation points and prove that it is an isentrope.
(11.3.1 )
Solution
= Ve we have
y
-a =
(3.5:~Ver~J
0/., _
pv I
The terms neglected in Equation (11.3.1) can readily be shown to be small compared with those retained. From Equation (11.1.6) we can now find the partial derivatives of p with respect to a and we find that ap/aa is finite and amp/act" (m> I) is of the order N(2n-I)13. Similarly, from Equation (11.1. 7) it follows that anv daan is of the order N(2n - 1)/3 • Combining these results we find by straightforward calculation that anp/av7 is of the order N(n - 2)13. This means that the isotherm has a horizontal tangent at v = Ve'
-
~ f(1) 2rrm If{T)j%
.
(11.5.1)
The q-potential, q = (3pv, is related to a(= (3/1), 13(= l/kT), and the entropy S by the equation (compare Problem 1.20)
q
S
= 'k+ aN -{3U ,
(11.5.2)
so that we have for the entropy S = kq -kaN+k{3U.
(11.5.3)
For a classical perfect gas we have (see Problem 3.12)
11.4 Estimate at what temperature a perfect boson gas of molecular weight 4 and density O· 15 g cm- 3 (the density of liquid helium) will, at constant volume, show the Einstein condensation phenomenon.
U = ~pv ,
(11.5.4)
and using Equation (11.1.1) we finally obtain
Solution
The transition temperature follows from Equations (10.2.16) and (11.2.3) and we find h2 ( P )~3 (11.4.1 ) 1'c = 2rrmk 2·61m where p is the density, or 1'c = 115p~'Ar'/l, (11.4.2) where p is the density in g in oK.
The locus is obtained from Equations (11.1.6) and (I 1.1.7) by putting I and eliminating 13. The result is
=
cm- 3 ,
S kN =
! 13pv I -
(11.5.5)
a.
Using Equations (11.1.6) and (11.1.7) we can write Equation (11.5.5) in the form S kN
M the molecular weight, and where 1'c is
f(y)
=! yf(y)-In y
,
(11.5.6)
so that the isentropes are the curves y = constant. As the locus (11.5.1) is one of those curves, it must be an isentrope.
(3) See D.ter Haar, hoc. Roy. Soc.• A212, 552 (1952).
.'I
~ }
,<
Equation (11.1.7) withy = 1, and Equation (10.2.16) we find
11.6 Calculate the specific heat of a perfect boson gas.
No
Solution
As long as the temperature is above have for the pressure kT P = -;;;f(Y)
4, given by Equation (I 104.1), we
= Crhf(y) , h2
01.7.2)
In a two-dimensional system, the energy levels satisfy the equation h2 Ej = 8mL2(ni +nn , 01.8.1)
(11.6.2)
k,
< 'Fe .
Solution
with C-
T
11.8 Show that a two-dimensional perfect boson gas does not show an Einstein condensation,
01.6.1 )
_ (2rrmk)'h
~ - (~r2l
=N
and for the energy, from Equation (11.504),
U = !CvT'Y"f(y) , while the specific heat per particle is given by I dU dp Cv NdT = dT '
dZ
,vl
which is a monotonically decreasing function of T for T> Below 'Fe we have y = 1 and hence p
where the ni are positive integers. Instead of Equation (11.1.3) we now have (now v = L2, the two dimensional volume)
(11.6.3)
(11.604)
4.
q
,.
'I
N v
As
L
)'h I 2rrm vf'(l)
c
-
--
,
2rrm
~
en"
L- . n =I n
(11.8.5)
given volume, and there will be no condensation. l The occurrence or non-recurrence of condensation in perfect boson systems depends on the density of states function (4).] VAPOUR CONDENSATION
11.9 Consider a model (5) of a vapour consisting of non-interacting drops where each drop takes up a volume WI in which [ atoms move independently in a smoothed-out negative potential -XI (l = I, 2, ... ) and where XI and WI vary with [ as follows (X, WI are constants): [-I
',. WI
01.7.1)
01.804)
n- I diverges, we can accommodate any number of particles in a
Solution
T. ,V2 - (h2 --
en"
);:1 Ji2
n= I
From Equation (11.604) and the behaviour of p (compare Problem 11.3) it follows that Cv is a continuous function of T, while aCv/aT is discontinuous.
From Equations (11.2.2) and (11.4.1) written in the form
~
{3h 2 v
- = {3h2
(11.6.7)
11.7 Calculate the fraction of particles in a perfect boson gas, which is in the lowest energy state, as function of temperature below the transi tion temperature.
=
(11.8.3)
and from Equation (l 0.2.7') we now get
We note that in the perfect boson gas C y behaves as T'i> at low tempera tures, while the specific heat of liquid helium behaves as T3, another reason to hesitate before identifying the 'A-transition with the Einstein condensation. At T = 4 w~ find from Equations (11.6.6) and (1104.1): f(1) C y = ljkf'(l) = 1·9k.
2rrm q
01.6.6)
C v = 4"CT 1IlfO).
2rrm v f~ 0 In(l-e,,-{1E)dE,
leading to
and hence II
01.8.2)
Equation (11,1.9) becomes
01.6.5)
CT%fO) ,
2rrm yvdE.
XI = -Z-X ,
01.9.1
= WI[(l- 2)/(/-1)
01.9.2)
(4) P.T.Landsberg, Thermodynamics (Interscience, New York), 1961, p.313 and Appendix D.
(5) See H.Wergeland, Avhandl. Norske Videnskaps·Akad. Oslo, Mat.·Naturv. Kl., No.ll (1943).
j
.~
Prove that the free energy II of a drop is given by the expression exp(-t3II)
=
v exp(-t3X) rWI WI l! L~exp(t3x)
jI l
l-2
.
Solution
(a) Equation (11.10.1 )follows, if we bear in mind that from the general thermodynamics of reactions it follows that the chemical potential of a drop of I atoms must be I times the chemical potential of an atom, cx, expressed by Equation (11.10.2). (b) Combining Equations (11.9.3), (11.10.1), and (11.10.2) we obtain V !-2 q = -e-{3XL- I (1l.l0.7) vI I l! z .
(11.9.3)
Solution
From the theory of canonical ensembles it follows that II is given by the equation dn , exp(-t3II) = T! exp(-t3e) h31 (11.9.4)
If
The equation of state (11.10.3) and (11.10.4) then follows from Equa tions (10.2.6) and (10.2.7'). (c) To eliminate z we define quantities ~, 'T/, t by the equations ~ 11- 2 ~ = JIl!zl , (1l.l0.8)
where dn is an element of the 61-dimensional phase space and where e is the energy of the drop. The integration over the momenta gives a factor (2rrm/t3)31/2 as shown in Problem 3.5. Together with the factor h 31 this leads to v"(} with Vo given by Equation (10.2.16). To evaluate the integral over the coordinates we note that the drop itself has the whole volume v at its disposal, but the atoms in the drop only the volume WI. Moreover, each atom in the drop contributes a factor exp (t3XI). The final result is then given by Equation (11.9.3), if we use Equations (11.9.1) and (11.9.2).
~
'T/=
~
L exp(cxl-t3ld , 1=1
with CXI
= Icx .
We note that
f
1
l+t=-· 2m or
(11.10.3)
1 11
L-z I!
i
'
I
The pole
Uo
(11.10.4 )
eliminate z. Discuss briefly the shape of the isotherms.
du ~
u
1=0
(11.10.12)
U
- du u
=
t=
(11.10.5)
(11.10.13)
(1l.l0.l4)
zeuo ,
Uo
(1l.l0.l5)
1 -Uo
From Equation (11.10.11) we now get 'T/
1.
1 du Xu x! = 2rrij ux+ I e
z
and we get from Equations (11.10.13) and (11.10.14)
(c) Using the relation XX
o
thus satisfies the equation
where WI a+{3x
z -- -Voe
1
dz
(11.1 0.11) t(z)
f - L(zeU)1 f u-ze
1 1+ = t 2rri
Uo
V N = -e-{3x vI
'T/ =
Using Equation (11.10.6) we find (11.10.1 )
"
.
(1l.l0.l0) z
'T/(z)~ ,
0
(b) Show that the equation of state follows from the equations
,.. I I
II
dz
z
~ =
(11.10.2)
e-{3x 11- 2
p= ' -z I . (.Iv_L.-I-'
(1l.l0.9)
t = IL= II., Z l .
ll.lO (a) If cxdt3 is the chemical potential of a drop of I atoms in Problem 11.9, justify the following relation for the q-potential of the vapour [as the number of drops can change, we must use a grand canonical ensem ble to discuss the system]: q =
11- I
L-, Zl, I = I I.
,
(11.10.6)
and ~ =
z__ u 0 _dz _ = ol-uoz
f
Zu ~dz =
Jo z
fUo 0
fU odu 00
= u 0,
(1 -uo)duo = Uo -!U6 .
(1l.l 0.16) (1l.l0.l7)
Chapter 11
292
11.10
From Equations (1Ll O. 7), (11.10.4), (11.10.8), (11.10.9), (11.10.16), and (11.1 0.17) we now get by eliminating Uo the (exact) equation of state (v 1 = v/N) R WI (11.10.18) ",pv 1 - e t3x
Phase transitions
11.12 Solution
From the general theory of grand ensembles it follows that the average value of m, satisfies the
oq
(m,) = -;- ,
and if we write Equation (11.10.7) in the form
we have
V 11-2~V.2 -PX )1 e al q = -e-PxL-e VI I l! Vo
VI
I! ~ I'
01.]0.20)
we see that the general term in the sum on the right-hand side of Equa tion (1l.10.19) is of the form (z/e),r%, as compared to yn n-'h in f'(y) occurring in Equation (11.1.7). One would expect that the analysis of Problem 11.2 can be repeated with minor alterations. The isotherm will be given by Equation (11.10.18) as long as VI is larger than a critical volume Vc given by the equation Vc
e- t3x WI'
e-t3x Pc = 2t3vI
11.11 number Hon
(11.1 0.22)
the results of the preceding problem show that the mean of drops of 1 atoms at temperature T is given by the equa1 -2
' (m) = N-,.,I-Ie-Ir)
I! '/
(11.11.1)
with 1'/ the solution of the equation 1'/e-r) = z
(11.11.2)
I
and z given by Equation (11.10.5).
(11.11.4) (11.11.5)
We note in passing that combining Equations (11.11.5) and (11.1 0.3) we obtain t3pv = L (m,) , (11.11.6) I
which is, of course, a consequence of our assumption of non-interacting drops. Using Equations (11.10.14), (11.10.16), (11.10.7), and (11.10.9), we can write Equation (11.11.5) in the form (11.11.1). We saw earlier that at the condensation point 1'/ = 1, while 1'/ -+ 0 as VI -+ 00 [compare Equation (11.10.19)] so that 1'/ measures the degree of saturation. We note that, as v I -+ 00, we find (m l )
(11.10.21)
This relation follows from the fact that for z = lie, Uo 1 from Equa 1 so that the tion (11.10.14), and thus from Equation (11.10.16), 1'/ relationship (11.1 0.21) follows from Equations (11.10.4) and 01.10.9). Similarly, we find that for V I < Vc the isotherm will be horizontal with a pressure P Pc given by the equation
'
V p-2 (m ) = -e- t3x - z1 vI l! . I
(1Ll 0.19)
we note its resemblance to Equation (I] .1.7). Indeed, using the Stirling approximation (see Problem 2.1])
( I 1.1 1.3)
vtY.1
VI
The derivation of Equation (11.10.18) is valid as long as the series for P and N converge. This is the case as long as z < lie. If we write Equation (11.10.4) in the form
293
=
= 0, I> 1 ,
01.11.7)
that every atom is Even at the condensation point where it follows from Equation (11.10.20) that (m,) is proportional to I-'!l, only the smallest drops are present in appreciable amounts. HARD SPHERE GAS(6)
11.12 Consider a gas of N small hard spheres interacting through two particle interactions which are long-range and smoothly varying. More over, assume the interaction potential, rJ>, to be everywhere negative or zero. Consider the canonical partition function for such a gas. If we want to derive the equation of state from it, we must find its volume dependence. We have seen on many occasions that the integration over momenta which occurs in the expression yields only a multiplying factor VON (compare Problems 3.5 and 11.9), which we can leave out as it is unimportant for our discussion. To evaluate the configurational partition function, divide the volume V into cells of volume Ll so small as to make rJ> practically constant inside Ll, but large enough for each cell to contain a large number of particles. Let f; be the position of the ith cell and the number of particles in it. If {j measures the volume of the hard (6) See N.G.van Kampen,Phys.Rel'., 135, A362 (1964).
294
11.12
Chapter 11
spheres, and if w(N;) is the amount of phase space of Ni hard spheres in a volume Ll, we have for wi(Ni ) in one dimension (see Problem 9.3) w(Ni ) = (Ll- N;O)Ni . (11.12.1) Assume that this expression can still be used for the three-dimensional case. Van Kampen has shown that this assumption is sufficient for the discussion of the condensation of the system considered. Write down an expression for the canonical partition function, QN, omitting the kinetic energy contribution in the form of a summation over configurations {Ni }, that is sets of numbers N i ,
I
QN =
exp4>{NJ.
(11.12.2)
Give an expression for cp{N;}. Use the standard procedure of statistical mechanics (see Problem 2.11) to evaluate the sum over configurations by finding its largest term. Find the condition to be satisfied by the configuration corresponding to this maximum of CP, say CPs. Assuming that the density is homogeneous, show that CPs satisfies the equation v -No N2 (11.12.3) CPs = Nln~+N-~(3rJ>o--;; , with rJ>o
=
t
rJ>ijLl
=
f
3
rJ>(r)d r ,
(l1.l2.4)
where we have assumed that the interaction forces are central forces,
= rJ>(r) ,
(11.12.5)
= rJ>(ri,rj) = rJ>(lri-r;I).
(11.12.6)
rJ>(r) and where rJ>ij
Hence derive the van der Waals equation
N
N2
(3p = v -No +!(3rJ>o~ ,
(11.12.7)
and discuss the result, comparing Equation (11.12.7) with the usual form of the van der Waals equation. Solution
We first of all note that the N; must satisfy the condition
IN; = N,
(11.12.8)
i
and that the energy of a given {N;} configuration is
~ ~ rJ>(ri, rj)NiN; = ~ "r.rJ>ijN;N; . " "
(11.12.9)
11.12
Phase transitions
295
The partition function will thus be
- ~" ~ n [n. J {-
QN - NIL... .
N;!'
w(Ni ) exp
'!(34 1 " rJ>ijN;N; } ,
01.l2.1O)
q
i
which is of the form (11.12.2) and where the sum is over all configura tions satisfying condition (11.12.8). The function cp{N;} in Equation (l1.l2.2) is then given by cp{N;}
=
Ii
[N;ln(Ll- N;o) - NilnN; + N; ]-!(3 L rJ>ijN;N;, 01.12.11) ij
where we have used Equation (11.12.1) and Stirling's formula for the factorial. The maximum term in Equation (11.12.10) with the Ni satisfying condition (11.12.8) is found in the usual way by varying the Ni and using a Lagrangian multiplier 'Y, to be determined from the condition (11.12.8), to take the subsidiary condition into account. The result is Ll-No No In .,.' Ll-'No (3"i;.rJ>ijN;='Y. 01.l2.12)
,
"
If the density is homogeneous we put
NLl N·=, v'
and if we determine 'Y from the equation v -No No N In~- v-No-(3rJ>o-;; = 'Y,
01.l2.13)
01.l2.14)
with rJ>o given by Equation (11.12.4), all Equations (11.12.12) are satisfied. This means that relation 01.12.13) is a possible solution of Equation (11.12.12), and that the uniform-density states are thermo dynamic states as they make QN stationary. It remains to be deter mined whether these states are stable, metastable, or unstable. Substi tuting Equation (11.12.13) into Equation (11.12.11) we obtain the result (11.12.3). If CPs, given by Equation (11.12.3), is an absolute maximum, we have QN = expcps and the free energy F is given by -(3F = CPs, so that one finds p by taking the derivative with respect to v from which the relation (11.12.7) follows. We can write this equation in the form
(3(p-!rJ>o~:)(V-NO) =N,
01.12.15)
which is the usual van der Waals form. We see clearly the excluded volume term and the term due to attractive forces (rJ>o is negative!): -~rJ>o is the work that a particle has to do against the attraction of the other particles to reach the boundary.
o
296
11.13
Chapter 11
I-n) fen) = nln ( -n- -!tl4>on 2 ,
which is related to 4>s of Equation 01.12.3), when we put write n = N/v. (a) Prove that, if A is a real symmetrical matrix of the form
= ail)ij-b ij ,
bij
b ii
> 0,
01.13.1) l) =
I and
(11.13.3)
Ai> Lbij j
for all i. Prove also that A is not positive definite, if for all i. (b) Use the results of (a) to show that, when f"(n) is negative, the solution leading to condition 01.12.3) corresponds to a (relative) maxi mum of 4>, but, if f"(n) is positive, this is not the case. (c) Show that the condition on tl that f"(n) is always negative is
-tl4>o <
Solution
If we write n = N/v and use units in which 0 = 1 so that 0 < n < I, we have 4>s = vf(n) with f given by Equation 01.13.1). Equation 01.12.14) for'Y then has the form
f'(n)
= 'Y,
01.13.5)
I-n n
= In----64> n I -n . 0 n.
(11.13.6)
In the following we also need f" (n) which is given by the equation ( ' (n)
= nO -1 nF tl4>o·
01.13.7)
To prove this let A be an eigenvalue and {Xi} the corresponding eigenvector of A so that we have Axi
= ~Ajjxl I
= d{X; "-'
< XI Lb j; < xlal
.
(11.13.9)
Hence each eigenvalue is positive, and A is positive definite. On the other hand, if one considers the vector {I, I, ..., I} == {yd we if
< 0 if aj <
Lb ij for all i. J
(b) To find out whether a solution gives a maximum of 4> we construct the matrix of the second functional derivatives 2 a24> .<1 ] (11.13.10) aN;aNj = - LOq N;(.<1-N;)2+tl4>// '
r.
a24>
t
bijxj .
(11.13.8)
I
Let X I be the component of {Xi} with the largest absolute magnitude and choose the arbitrary multiplying factor of all Xi such that X I is real and
-or
n(1-~)2.<1 +tl4>ij .
aN;aNj =
01.13.11)
This matrix will be negative definite so that the solution leads to a (relative) maximum-and hence to a stable or at least metastable thermo dynamic state-if
(11.13.4)
and show that the maximum is an absolute maximum. Discuss condition 01.13.4) on tl.
with
297
or, if we consider the uniform-density case
jj < Lb j
f(n)
~btjXii
(al -A)XI
see that LYiA/jYj 01.13.2)
it is positive definite if
ai
Phase transitions
positive. From Equation (11.13.3) it then follows that
11.13 Consider the function fen)
Aij
11.13
.
1
-.~-.
" > -tl4-4>ij ,
tl4>o
-~,
(11.13.l2)
and we see that this condition is satisfied if f"(n) is negative. On the other hand, if ('(n) is positive, the condition for the second part of the lemma holds, and the solution does not correspond to a maximum: the corresponding thermodynamic state is unstable. (c) From Equation (11.13.7) it follows easily that f" (n) has a maximum which is equal to -Z; -Mo, from which condition01.l3.4) for n = follows. From a similar calculation it also follows that .<12 27 (11.13.13) N;(.<1-N;)2 ~ 4.<1 '
t
so that the matrix 01.13.10) is negative definite for any configuration {N;}. The function 4>{N;} is thus convex and can have only one maximum. Provided condition (11.13.4) holds, the stable state is thus the uniform density state. In the previous problem we noted the relation between Equation (11.12.7) and the van der Waals equation of state. If we express van der Waals' constants a and b in terms of our parameters, we have a =
-! 4>~ ,
b = Nl) ,
(11.13.14)
and condition (11.13.4) becomes
tl < tlc or T > 'L
01.13.15)
298
Chapter 11
with the critical temperature
11.13
Tc satisfying the van der Waals relation 8a 27b'
NkTc
11.14
Phase transitions
while the condition (11.12.8) becomes
f
n(r)d 3r = N .
01.13.16)
This means that the uniform-density state is thus the stable state above the critical temperature. Temperatures below Tc are considered in the next two problems. 11.14Consider temperatures for which condition (11.13.4) is not satisfied. Introduce now a position-dependent density nCr), so that 4> becomes a functional of nCr). Assume that nCr) varies so slowly that n(r)-n(r') can be expanded in powers ofr-r'. Show that 4> becomes a maximum, if nCr) satisfies the equation 2
01.14.7)
Equation 01.14.6) can be written in the form
=
f
f(n)d 3r+!t3 fftl>(r,r')[n(r)-n(r')]2d 3rd 3r' , 01.14.8) where fen) is given by Equation 01.13.1). To find the extremum we take functional derivatives and take condi tion (11.14.7) into account through a Lagrangian mUltiplier, and the result is 4>{n(r)}
f'(n)+t3 ftl>(r,r')[n(r)-n(r')]d 3r' = 'Y.
(l1.14.1)
f'(n) -!t3t1>zV n = 'Y ,
299
(11.14.9)
where 'Y is a Lagrangian multiplier and tl>z =
i
f
t'(n)
r2 t1>(r)d 3 r .
01.14.2)
Assume now that nCr) depends on only one coordinate, x say, so that Equation (11.14.1) becomes d 2n -!t3t1> zdx 2 = -f'(n)+'Y . (11.14.3) Note that this equation can be interpreted as the equation of motion of a classical point mass with coordinate n, mass -!t3t1>z(tI>z < Ol), time x, moving in the potential (11.14.4) "'(n) f(n)-'Yn. Prove that f'(n) has one mInImUm and one maximum in the range o < n < 1. Consider a value of'Y lying between those two extrema and find (i) the two solutions of Equation 01.14.3) with n(x) = constant and (ti) the solution with n = nl as x ~ -00 and n = nu nl as x ~ +00. Prove that this last solution is possible only, when one applies Maxwell's equal-area rule.
\»>P'/",d.
01
n.
lOY
,
I
... n
nil
'*
Solution
If we assume that the density depends on position, so that we can write (11.14.5) Nj = n(r)A, we get instead of Equation ( 11.12.11 ) 4>{n(r)} =
f
nCr) {In I :(:jr) +
I} dlr fftI>(r, r')n(r)n(r')dlrd r' , -!t3
3
01.14.6)
Figure 11.14.1.
If nCr) is varying sufficiently slowly so that we can expand nCr) - nCr'), Equation (11.14.9) leads to the result 01.14.1 ). If we then assume that nCr) depends on x only, we get the solution (11.14.3). In the preceding problem we considered some of the properties of fen) and we saw there that f"(n) had one extremum at n = ! and, if condition (11.13.4) is not satisfied, f" will be positive for some range of n-values. As f"(n) is negative for n = 0 and n = I, we have proved that f'(n) behaves as shown in Figure 1l.l4.1 and hence "'(n) given by Equation (11.14.4) behaves as shown in Figure 11.14.2.
Chapter 11
300
11.14
°
= n( = constant or n
11.15
Phase transitions
301
nIl = constant, d 2n/dx 2 as they correspond to maxima of l/I(n), and these two values of n thus correspond to uniform-density states. To find a solution with n nI as x -+ -<Xl and n = nIl as x -+ + <Xl, we consider the problem posed by Equation (11.14.3) and see that such a solution is possible only if the 'particle', initially at rest on top of the nI), 'moves' to the other 'peak' and comes to first 'potential peak' (n rest there. From the analogy used here it follows that this is possible only if the two peaks have the same 'height', or
11.1 S For given values of the total volume v and total number of particles N find the fractions of space occupied by the densities nI and nu, respectively, of the previous problem, neglecting the transition region. Give an expression for the free energy of this two-density state, neglecting the free energy of the transition region. Prove that the pressure is the same for all two-density states. Also prove that the two-density state has a lower free energy than the uniform-density state for the same Nand (3(7).
01.14.10)
Let VI (VII) be the volume occupied by density nI (nn). Neglecting the transition region we have
It follows that if n
l/I(nI)
l/I(nu) .
Solution
1/1(11)
VI> 0,
VII> 0,
VI+Vn = v.
01.15.1)
For any value of VI we have a solution of condition (11.15.1) correspond ing to the same 'Y 'Yo. The number of particles satisfies the equation N = nIvI + nnVn ,
01.15.2)
so that for all values of N/v in the range ,
o
I
_
nJ
11
01.15.3)
there is a two-density state. The densities nI and nIl are determined, as we have seen, by the conditions
flU
III
< N/v < nIl
f'(nd
= f'(nIl)
(11.15.4)
and )-n1f(nd = f(nn)-nIlf'(nn).
From Equation (11.14.8) it follows that, neglecting the transition region, we have for the free energy
Figure 11.14.2.
This is an equation for 'Y. There is thus only one solution 'Y = 'Yo such that (11.14.11) f(nd-f(nn) = 'YO(nI -nil), or
f
ilII
f'(n)dn
= 'Yo(nll -nd,
(11.14.12)
111
a
-(3F =
-'Yn
l/I(n) ,
) +vu f(nIl) ,
(11.15.6)
or F=Fr+Fn, and for the pressure we find a
indicating that 'Yo is found by the equal-area construction applied to fen): the two shaded areas in Figure 11.14.1 must be equal. This condition implies that P is the same for nI and nIl, as in the uniform density case we have
01.15.5)
Tv
=
TvPI +
avn av Pn = (3PI
01.15.7) 01.15.8)
which proves that the pressure is the same for all the two-density states. To prove that for all values of N/v satisfying condition 01.15.3) the free energy of the two-density state is higher than that for the uniform density state for the same Nand (3, we must prove that
01.14.13)
where we have used the fact that for the uniform-density case f'(n) 'Y. The equality of l/I(nd and l/I(nu) thus implies that PI Pu, cf Equation (7.3.1) of Problem 7.3.
vrf(nd+vnf(nn) > Vf(VJnJ :vllnn) ,
01.15.9)
(7) Other characteristics of the unstable uniform-density state are examined in Problem 7.2(e).
Ch!!pter 11
302
11.15
or, that in an fen )-n diagram (see Figure 11.15.1) the double tangent lies above the curve f(n) which is obviously the case. In conclusion, we note that Problems 11.12 to 11.15 give an (approxi mate) proof of the van der Waals isotherm, including the horizontal part.
12 Cooperative phenomena (1)
fen)
D.ter HAAR (University of Oxford, Oxford)
12.1 For our purpose we define a ferromagnet as a lattice on which spins are situated. In this chapter we shall be extensively concerned with two models of ferromagnets: the Ising model and the Heisenberg model. In general we can write for the Hamiltonian of the interaction between the spins
n.
H
nil
= -2~[Ix(f-g)S;S;+Iy(f-g)sfS~+Iz(f-g)S:S:],
(12.1.1)
where the f and g are the position vectors of the lattice sites, the S;, Sf, S: the components of Sr, the spin vector corresponding to lattice site f (we express all spins in units h) and the lief - g) are functions of f - g only. The Ising model in its simplest form is obtained by putting Figure 11.15.1.
l (f _ ) z g
The treatment is approximate but can be made rigorous (8). However, it is restricted to long-range attractive forces while actual interatomic forces are usually short-range.
{==
Ix(f-g) = Iy(f-g) = O.
(12.1.2)
I, if f and g are nearest neighbours, 0 , otherwise ,
(12.1.3)
and assuming that the particles have spin t so that Sf only takes on the values +t and -t· The Heisenberg model in its simplest, isotropic form is obtained by putting
Ix =
I z = ICf-g)
Iy
(12.1.4)
so that the Hamiltonian (12.1.1) becomes H == -2 ~I(f-g)(Sr . S,) .
(12.1.5)
"
If there is an external magnetic field (induction B), we must write instead
of the expression (12.1.5) H == -gJ.!BL(B· S,)-2> I(f-g)(Sr' S,),
t.i
f
(12.1.6)
(I) See F.C.Nix and W.Shockley, Rev.Mod.Phys., 10,1 (1938); D.ter Haar, Elements ofStatistical Mechanics (Holt, Rinehart, and Winston, New York), 1954, Ch.12; H.J .Goldsmid (Ed.), Problems in Solid State Physics (Pion, London), 1968.
(8)N.G.van Kampen,Phys.Rev., 135, A362 (1964).
l
303
12.1
Chapter 12
304
where P.B is the Bohr magneton and g the Lande g~factor and where we have assumed the field to be uniform. The simplest way of progressing further is through the molecular field approximation. In this approximation one writes the Hamiltonian ( 12.1.6) first of all in the form H
-gP.B
If
[Sf' (B +~ I
gP.B •
l(f -
g)s.)] .
(12.1.7)
One now replaces the S, in the sum over g by their average value {S} which must be independent of g, as the system is invariant under translations. We see from expression (12.1.7) that the external field is replaced by an effective field B + B' ,
Beff
(12.1.8)
where B' is the so-called molecular field. To find the behaviour of the system, we must first study the behaviour of a magnetic dipole of moment p. in an external field B, when its energy is -(p. • B). Show that its average value at a given temperature is
12.3
Cooperative phenomena
12.2 A spin -S particle in a field B has spin energy levels = -gp.BBM , M = -S, .. "
P.(cothZ-! ),
Use Equation (12.2.2) to find an expression for its average magnetic moment. Solution
The partition function (12.2.2) follows directly from the equation
IM exp ({3gp. BBM) .
Z=
02.1.10)
Z = (3p.B .
The average magnetic moment is given by the equation I alnZ
7i
= gp.BSBs(z) ,
where the Brillouin function Bs is given by the equation 2S+ I I Bs(z) = cothl(2S+ 1)z]-2Scothz. At high temperatures, z
Solution
<{
(l2.1.1I)
-(p.' B),
f
(p.) =
e-f1H dcosO
'
(l2.1.12)
where 0 is the angle between p. and B. This yields expression (12.1.9). At high temperatures the Langevin function, cothz -I/z, behaves as !z and we have therefore p.2B (p.) -
3kT
(z <{ 1),
02.1.13)
while at low temperatures we get {p.} -
p.
(z ~ 1) .
(12.1.14)
(12.2.6)
j(S+ 1)z ,
(12.2.7)
I )g2p.fi B 3kT '
(12.2.8)
and thus S(S+
and the average value of p. in the direction of B is given by the expression
fe- i3Hp.cosO d cosO
(12.2.5)
I, we fmd
Bs(z) -
The energy of the dipole is
=
02.2.4)
(12.1.9)
with
H
02.2,1)
+S .
Neglecting all other effects, show that its partition function Z is given by the equation sinh (2S + 1)z (12.2.2) Z . hz sm where now ( 12.2.3) z ~{3gp.BB .
{p.} =
{p.} =
305
{p.} -
which is the equivalent of expression (12.1.13), if we bear in mind that is the quantum-mechanical equivalent of p.2 in the classical case.
S(S + I )g2p.fi
12.3 Use the molecular field approximation to find an expression for the average magnetisation (M) of a spin-t ferromagnet. Show that in this approximation there exists a transition point or Curie point Tc such that in the limit as B -1- 0, (M) = 0 is the only value of (M) for T> while (M) can be non-zero for T < If (M) 0 in the limit asB -+ 0 the material is said to exhibit spontaneous magnetisation. Find an expression for (2)
re,
re.
*'
(2) The terms transition temperature and Curie temperature are used synonymously in this section.
306
Chapter 12
12.3
Cooperative phenomena
12.3
Determine the behaviour of the average spontaneous magnetisation
307
near
has a slope equal to Ll(g)/2kT at the origin in the same diagram, and as
Determine the behaviour of the average spontaneous magnetisation in the low-temperature region. (iv) Determine the susceptibility (M)/B at very high temperatures.
the latter curve is convex, the solution long as
g
Solution
kT>! LI(g).
Instead of the Hamiltonian (12.1.1) we now have H
-(M' B eff )
However, for temperatures below Tc given by the equation
02.3.1)
,
= gIlBLSf,
M
Te (12.3.2)
2 B +- (S) Ll(g) .
Beff
gllB
(12.3.3)
g
From Equations (12.2.5) and 02.2.6) we get a system of N spin-! particles (M)
= NgIlB(S)
Te (M) Td (M)3
02.3.4)
! Ngll B tanh(! BgllBB eff) •
Hence from Equations (I equation for (M) (M)
or, using the fact that T
02.3.5)
~
.4) and (12.3.3) we get the following !NgIl B tanh [!{3gIlB(B +q(M»] ,
T
2LI(g)
0,
= Mo(Tc ~T)
(M)
Mo
(12.3.7)
(M) I 02.3.8) = tanh Mo = -1 Ngll B in the magnetisation at T O. We note first of all = 0 is always a solution. Also, we see that as the curve
~
.i\
lh
[1 -2exp (- i) J' gllBB
(M)
2kT
l
Ll(g)
y = tanh --L- (M) 2kT -Mo
(12.3.13)
(12.3.14)
+
or ) ,
diagram, while the curve
(12.3.12)
while experimentally it is found that Mo - (M) behaves as T'12. (iv) At temperatures above Tc there is no spontaneous magnetisation . When T ~ Tc , we can expand the tanh in Equation (12.3.6) and we have
y
has unit slope in a y versus
(12.3.11 )
As T....;. 0, (M) ....;. Mo. Expanding the tanh for large values of its argument, we find
(12.3.6)
(i) To find the Curie temperature we must solve Equation (12.3.6) for the case where B = 0, that is, we must solve the
Mo - T3 MJ + ...
Tc and (M) (M)
where
q
g
=
or (M)
(12.3.
= LI(g)/2k ,
there will also be a non-vanishing solution to Equation 02.3.8). One can show by evaluating the free energy (compare Problem 12.4) that this solution is the equilibrium solution, so that the system shows spontaneous magnetisation below the Curie temperature Te. Near, but below, Te, (M) will be small so that we can expand the tanh in Equation (l2.3.8):
f
and where
(1
g
where the magnetic moment M of the whole system is
where that
0 will be the only one as
(M)
(12.3.15)
whence we get for the susceptibility per spin X X
]
the so-called Curie-Weiss law.
(12.3.16)
308
12.5
12.4
Chapter 12
12.4 An Ising ferromagnet contains N spin--l particles. Let be the number of spins with z-components (-!) and let the up- and down-spins be distributed randomly over the lattice. Let
R
N.-N_
(12.4.1)
-kN[-l(1 +R)ln10 +R)+!(l E -izINR 2 ,
In!(1-R)],
!zN_p_
(12.4.3)
Q._
F== E
(12.4.7)
E == -VCQ•• +Q-- -Q.-) ,
R
where Q++ (Q__ ) is the number of nearest neighbour pairs where both spins are in the positive (negative) z direction while Q._ is the number of pairs with one spin in the positive and the other spin in the negative
1+R)+!(l-R)ln!( I-R»). 02.4.13)
(12.4.14)
tanh
zI R. 2kT
(12.4.15)
Bearing in mind (i) that for the Ising model I:.I(g) zI, and (ij) that R from expression (12.4.1) is directly proportional to (/11). we see that we have rederived the molecular field equation (12.3.8). 12.5 Use the results of Problem 12.4 to find a parametric expression for the specific heat of an,. Ising ferromagnet in the molecular field approximation. Find the jump in the specific heat at Te , and the behaviour of the specific heat as T -+ O.
(12.4.8)
( 12.4.9)
TS
or
whence introducing R from Equation (12.4.1), and using the fact that N.+N_ == N, we get expression 02.4.2). To find the free energy, we must evaluate the internal energy. We have
where the summation is only over nearest neighbour pairs and where we have introduced new variables f.1r which are + 1 (-I) if the spin on the f site is in the positive (negative) z direction. From Equation (12.4.8) we can write the energy in the form
(12.4.12)
0 we then get
I +R zJ
In I -R
we have
g ,
N.N_
From the condition 'OF/'OR
Using the Stirling approximation (see Problem 2.11) in the form
S == k(NlnN-N.lnN+-N_
!zN+p-+t.zN_p.
== -!zJNR2 +NkT[!(1 +R)
(12.4.5) (12.4.6)
=
Combining Equations (12.4.9), (12.4.12), and (12.4.1), we obtain expression (12.4.3). The free energy is thus given by the equation
(12.4.4)
InN! == NlnN - N ,
(12.4.11 )
and
and the entropy S is given by the expression
-HIr,g f.1ff.1
02.4.10)
N '
(12.4.2)
If the distribution of + and spins is assumed to be completely random, the probability W for N. + spins and N_ - spins is
II
N
2 ----.:':
where p. is the probability that a spin is in the positive z direction. Similarly we have
Solution
S == 'dn W == k(lnN! -InN+! -lnN_!} .
1
2Z
Q++ =
where z is the coordination number, that is, the number of nearest neighbours per spin. Minimise the free energy with respect to R to find the equilibrium value of R.
N! W == N.W_! '
309
z direction. On the assumption that the + and spins are randomly distributed, we find for Q++ (z is the coordination number, that is, the number of nearest neighbours per spin):
be an order parameter. Show that the entropy and internal energy are given by S
Cooperative phenomena
~!
Solution
From Equation (12.4.13) we find for the specific heat 1
Cv
dR
-2 zINR dT '
Cv :
02.5.1)
while R satisfies Equation (12.4.15) which for our present purpose we shall write in the form RTe R tanh (12.5.2) T ,
12.5
Chapter 12
310
whence RTc /,(
dR
- T2f I cosh
2R
TC)
T'
(12.5.3)
so that the specific heat is determined by the parametric equations Nkr 2 (12.5.4) v 2 C cosh r-rcothr' rcothr
=
(12.5.5)
T '
where r is related to R by the equation r R Tc/T. To find the behaviour near we note that as T -+ Tc, r -+ 0, so that we can expand the various hyperbolic functions occurring in Equation (12.5.4); the final result is Cv -+ ~Nk
as T
-+
(12.5.6)
Tc .
As Co 0 for T> Tc in the molecular field approximation [E = 0; see Equation (12.4.3)] the jump in C v is ~Nk at Tc . When T ~ ,we see from Equation (12.5.5) that r "'" Tc/T and we thus obtain from Equation (12.5.4) Co "'"
Tc)2 exp (2Tc) 4Nk ( T -T
.
II
M·1
{Sf1
•
-gIlBL{Sf2
'
f).
[B+2 LI(f) -g) )S"g, + 2 LICf) -g2)SI2]} gllB II .' gllB 12
(12.6.2) (12.6.3)
gllsLSf' fi
I
and
>,}
Bm
B ql<M»
B~ii
B -q2(M»-ql(M2> '
q2<M 2
(12.6.4)
with (M j
)
( 12.6.5)
-!Nglls(Sj>,
the extra factor -! deriving from the fact that the N spins are divided evenly over the two sub lattices, and qI
_ "I(f) -gl) __ "I(fz-gz) } 2 L. N 2 2 L. N( 2, II (gIlB) 12 gllB) "Ie f 2-gd "I(f) -g2) - 2 L. N(g )2 = - 2 L. N(g )2 I, IlB 12 IlB
(12.6.6)
where we have included an extra minus sign to take into account that the dominant interaction leads to antiferromagnetism. As the two sublattices are now decoupled in Equation (12.6.2) the results of Problem 12.2 can be used for each of the sublattices, and we get > = iNgllstanh[!j3gIlB(B-q)<M»-q2(M2»],} _ .1) (12.6.7) - 'lNgIl B tanh[2j3gIlB(B-Q 2 <M»-q)(M2>)]· At high temperatures we can write tanhx "'" x, or "'" !j3N(gIlB)2(B-Q)<M»-Q2<M2~r;r "'" !j3M(gIlB)2(B -Q2(M 1 >-Q) (M2»
(12.6.8)
,f
so that we get for the total magnetisation
Solution
= -gllB L
-(MI' Bm) -(M 2 • B~if) ,
where
q2
Taking into account that there are two sublattices, we write the Hamiltonian in the form
311
where the f), gl (f2' g2) are lattice sites on the first (second) sublattice. In the molecular field approximation we replace S'I and S'2 by their averages (S» and (S2>, which will now be different, and we rewrite the Hamiltonian (12.6.1) in the form
(12.5.7)
12.6 In a ferromagnetic substance I(f - g) is positive when f and g are nearest neighbours. If this quantity is negative, we are dealing with an antiferromagnetic. To consider an antiferromagnetic we assume that the crystal can be divided into two sub lattices such that all the nearest neighbours of 3. spin on one sublattice are on the other sublattice. We shall now have two averages which will be different for the two sublattices, as a negative value of I(f - g) implies a preference for an antiparallel alignment of nearest neighbours. Use the molecular field approximation to find the susceptibility of an antiferromagnetic at high temperatures and to find the Neel temperature, that is, the temperature below which both sublattices show spontaneous magnetisation. Consider again the spin--! case.
II
Cooperative phenomena
12.6
<M>
=
"'"
2B9 T+9) ,
(12.6.9)
with 2LI(g) Q = QI+Q2 = -N{gIlB)2 ,
(12.6.10)
which is the same as expression (12.3.7) apart from the sign, and f2
[B+2LI(fZ-g)Sgl+ 2 gllB II gllB
ICf2-g2)SIJ}, (12.6.1)
9
QN(gIlB )2
(12.6.11)
Chapter 12
312
12.6
To find the Neel temperature, TN, we put B = 0 and expand the tanh, as just below TN the sublattice magnetisations will be small [compare the solution of Problem 12.3 (in]. We then have . (MI> ~ -T(ql(M I >+q2(M2»+term
q
of third order in (Mi>,
e
I
12.7
Cooperative phenomena
313
parallel to (M 2 > (see Figure 12.7.1). The total magnetisation (M I >+(M 2> will be parallel to B, and its magnitude, divided by B, will give us Xl From Figure 12.7.1 we then see that Xl
.
(12.7.3)
q2
(12.6.12)
(M2) ~ - qT(q2(M2 >+ql(M I »+·...
(M 1 )
For an antiferromagnetic we always have a vanishing total magnetisation when there is no external field so that we can put in Equations (12.6.12) (M I >::; - (M2 >and we then see that the sublattice magnetisations vanish at TN given by the equation (12.6.13)
TN =
We note that if we had assumed that there were only nearest neighbour interactions q I would be zero and q2 = q so that TN and e would be the same. We also note that in the unlikely case where q I > q2 we would have no transition; this is not surprising, as q I > q 2 would mean that there would be a strong tendency for spins in the same sublattice to align antiparallel. In that case, the model used would clearly be a very poor one. 12.7 Find for an antiferromagnetic of the type considered in the preceding problem the perpendicular and parallel susceptibility Xl and XII (for the cases where the applied field is perpendicular or parallel to the sublattice magnetisations) at temperatures below the Neel temperature. Solution
Let
(M 2 >
Figure 12.7.1.
Consider now We then expect all vectors to be parallel (or antiparalIel) and, moreover, for small B we expect (M 1 > and -(M 2> to be nearly equal to m so that we can write (12.7.5) (M 1 > = m+oml , (M 2> = m+om2' We use again Equation (12.6.7) and write for the argument of the first tanh !{3gI1B(B -q I (M 1 >-q2(M2»
(M1>o =
(M 2 )o
m,
(12.7.4)
Bllm.
=
(12.7.1)
! (3gl1Bm(q2 -q I) +!(3gI1B(B -ql om. -q2 om 2),
(12.7.6)
where the index 0 indicates the absence of an external field. Consider first the case B1m. (12.7.2)
and similarly for the argument of the second tanh. Expanding in terms of the small quantity which contains B, omh and om2, we obtain from Equation 02.6.7)
The effective fields are again given by Equations (12.6.4). We are interested in the susceptibility, that is, in the value of the magnetisation when there is a small external field present, and as B1 m, we expect that (M 1 > and (M 2 ) will be no longer strictly antiparallel. Let € be the angle between (M t ) and -(M 2 >. We expect € <{ I. The effective fields B~?f will be parallel to the (Mj > so that from Equation (12.6.4) it follows that B -q2(M 2>must be parallel to (M 1 >and B -q2(M 1 >
om l om2
=
i{3N(gI1B)2(B -qlomt -q2om2)cosh- 2[!{3gI1Bm(q2 i{3N(gI1B)2(B -ql om2
-qdJ , }
ql om I) cosh- 2[! {3gl1Bm(q2 -qdl .
(12.7.7)
Adding these equations, we find the total magnetisation (MI>+
om l +om2 ,
12.7
Chapter 12
314
and dividing by B we find Xl!. The result is
2A q(T+ A) ,
XII
where
ecosh- 2 rtiJgJ.LBm(q2 -ql)] ,
A
(12.7.8) (12.7.9)
q2
XII
Xl) ,
(12.7.1
where we have used Equation (12.6.13). As T -+ 0, A vanishes exponentially so that XII -+ O. Finally, we note from Equation (12.6.9) that as T approaches T,.. from above, the susceptibility also approaches 1/q 2' 12.8 Consider the Ising model. We can write the Hamiltonian in the form H = -!gJ.LBLJ.Lr f
L lJ.LfJ.Lg ,
= -!gJ.LBBLJ.Lr-!zLIJ.Lfii, r
f
where J.L is the average value of J.L and z is the coordination number, that is, the number of nearest neighbours per spin. This approximation essentially reduces the problem to that of a single spin problem, or, as we saw in Problem 12.4 to assuming that the + I and -I values are randomly distributed over the lattice. The next approximation takes into account that because of the interaction there will be a preference for ++ pairs or -- pairs, rather than + pairs. Let Q be the total number of pairs in the lattice, so that Q =
HZ+l =
(
!gJ.LBB J.Lo+
tzN.
( 12.8.3)
Let Q++, Q+_, Q__ be, respectively, the number of ++, + ,and -- pairs. Prove the following relations for Q++, Q+_, and Q--:
2Q+++Q+_ : zN+, } 2Q__ + Q+_ - zN_ ,
(12.8.4)
z) -!gJ.L
L; J.Lj j l
where 0 indicates the central spin and (i) Prove that in this approximation
(12.8.1)
(12.8.2)
( 12.8.6)
A given state of the system is now characterised by the long-range parameter R and the short-range parameter a. To proceed further we consider instead of a one-spin system the system of z + 1 spins where we take into account explicitly the interactions between a central spin and its z neighbours, but take the influence of the other spins in the lattice into account through a mean field B'. That is, we consider the Hamiltonian. z
BB'J J.Lj-HJ.L0 1 j
i
z
L; 1J.Lf,
(12.8.7)
= I, ..., z its nearest neighbours.
(f,g)
where now the sum over f and g is such that only terms where f and g are nearest neighbours occur, and where we have introduced new variables J.Lr which can take on the values + I and -I. The molecular field approximation consists in replacing the Hamiltonian (12.8.1) by H
where N+ and N_ were defined in Problem 12.4. We have, of course, also the relation (12.8.5) Q+++Q--+Q+Q, which is a consequence of Equations (12.8.4). We now introduce a short-range parameter a by the equation
Q+++Q__ -Q+_ = aQ.
with e and q given by Equations (12.6.10) and (12.6.11). We note that as T -+ TN, m -+ 0, so that A -+ e, and
2e
315
Cooperative phenomena
12.8
(12.8.8)
where (12.8.9)
x (ii) Find a as a function of x and R.
(iii) Use the fact that for self-consistency (J.Lo) must equal <J.Lj), which determines the mean field B', to find the Curie temperature in this approximation. (iv) Show that in the limit as z -+ 00 the critical values for x of the present approximation and of the molecular field approximation become the same and that also in that limit a = R2 so that the expressions for the energy in the present and in the molecular field approximation become the same. Note that as z -+ 00 one must assume that 1-+0 while zl remains finite, as zl determines the transition temperature which we want to keep finite. Solution
Equation (12.8.4) follows immediately by seeing that each spin has z nearest neighbours and that each +-spin must lead to either a ++ or a +- pair, and that, if we are not careful, we shall count certain pairs twice. (i) To prove the Equation (12.8.8) we write down the partition function Z:
Z=
L
"O~±'l
III
Ilz
L±.1 exp(-iJHz+t>· ~
(12.8.10)
316
Chapter 12
12.8
If we write
Cooperative phenomena
12.8 We note that when T
K
~{3gPsB,
K'
!(3gPs(B+B') , J=~{3I,
>
,R
Z
!J.oL±'1
!J.J±.l
"·!J.z
~±'1 exp (KPO+ K'~Pj+JLPOPj)
02.8.12)
Z =
exp(Kpo+K'Pl +JpopdD exp[(K'+Jpo)Pj] (12.8.13) I
!J.O=±.1 !J.l " " 1.... 1....
/ , z exp(Kpo+Kpl+JpOP1)[2cosh(K+Jpo)]
= exp(K - K' -J)Z~~l
1 l
,
exp(-K + K' -J)~-) 1, Z-- -- exp (-K-K' + J)Z(-) z- 1 , Z_ +
where
j
(12.8.15)
exp(Kpo)[2cosh(K + K' +JpoW
= Z++Z_,
1.02.8.14)
Let us denote the four terms corresponding, respectively, to the Po, Pj values: + I, + I; + I, -1; 1, + I; -I, -1 by Z++, Z_ +, Z__. We have Z++ = exp(K + K' +J)Z~+ll ,
L
02.8.22)
!J.o ±.l
2
!J.o ±.1!J.I =±l
(12.8.21)
To find the transition temperature, one carries out first the Pl summation in Equation (12.8.14) to find
z
=
I-x 1 +x .
o
L
=
= 0 so that then
(12.8.11)
we have
311
( 12.8.23)
with Z±
= e±'KZ~±') .
(12.8.24)
From Equation (12.8.12) it follows, on the one hand, that
( ) _ olnZ _ Z+-Z_
(12.8.25) Po Z On the other hand, we have Z+tanh(K'+J)+Z_tanh(K'-J)
" _ olnZ 02.8.26)
L,.. (Pj) - oK' z Z I
Z~~)l = l2cosh(K' ± J) jZ-
(12.8.16)
1 .
From the consistency condition
From the general theory of partition functions (compare Chapters 4 and 6) it follows that Z++:Z+_:Z_+:Z__
(Q++):~(Q+_):!(Q_+):(Q __ ).
02.8.17)
Equation (12.8.8) follows immediately. In this derivation we have used a symmetry argument to argue that Z+_ must equal Z_+. One could say that this is the argument which determines K' in terms of K and J, and we shall see presently, that indeed, a consistency requirement leads to K' such that Z+_ Z_+ (see part (iii)l. From Equations (12.8.3), (12.8.4), and (12.4.1) we find easily Q++ Q--
= izN(l +0+2R), = izN(1 +0 2R),
1
02.8.18)
Q+- = !zN(l -0) , and hence from Equation (12.8.8) we find (1 +0+2R)(1 +0-2R)
(1-0)2
whence
=
l-x 2
1 L(Pj) ,
z
(12.8.27)
j
we then find Z+ I +tanh(K' -J) Z_ = 1 tanh(K' +J) ,
or, using Equations (12.8.24) and 02.8.16), cosh(K' +J) 2(K' - K) - exp cosh(K' -J) z 1
(12.8.28)
(12.8.29)
from which the equality of and Z_+ also follows. To find the transition temperature we put B = 0 in Equation (12.8.29) so that now (12.8.30) K' = ~ {3gP s B' , while the equation for determining K' becomes
x
-2
,
1 +x 2 -2x[I-R 2 +R 2x 2 j'h o
=
(12819) ..
2K' cosh (K' +J) cosh(K' -J) = exp z -I
(12.8.20)
We note that K' 0 is always a solution. The transition temperature is that temperature for which a non-vanishing solution for K' becomes
(12.8.31 )
Chapter 12
318
12.8
Expanding both sides of Equation (12.8.31) for small values of K' up to K'3 we obtain, after taking the cosh(l + K) 2K' Incosh (J - K') = Z (12.8.32) whence In (I + 2K' tanhl + 2K '2 tanh 2 1 -1 K'3tanhl + ...) ( 12.8.33)
13
Green junction methods(1)
or
D.ter HAAR (University of Oxford, Oxford)
, 2 '3 2 2K' 2K tanhJ-JK' tanhl(l +2tanh l)+··· = z-I . (12.8.34)
As K'
-+
MATHEMATICAL PRELIMINARIES
O. it follows that tanhlc
=z I '
13.1 Show that, if [A,B1_ ==AB-BA :::: K, where A and Bare operators and K is a c-number, and if A is a c-number, e X(A+B) e Me AB e-Y>x2K (l3.1.1)
(I
or I +xc - z
whence
(12.8.36)
I '
Let .p(A) == eXAe AB • We then have
2 Xc::::
At temperatures just below K'2 satisfies the equation
K'2
Solution
a~~A):::: AeMe
(12.8.37)
z
, we find from Equation (12.8.34) that
We further have ~
[eM,BI_=
3(tanhl - tanhlc ) tanhl + 2 tanh 3 1
(12.8.38)
eMBeAB = (A+B).p(A)+[eM,B1_eAB. 03.1.2) An
L n on
becomes and the definition (12.8.9) of molecular field approximation Xc
= exp(-/3cJ)
X
it follows that in the
exp(-2/z) ,
( 12.8.39)
which is the same as expression (12.8.37) in the limit as z 00. To find a in that limit we note that it follows from Equations (12.8.9) and (12.8.11) that X exp(-21) ~ 1-21+ ..· (I in the limit as / -+ 0 (and also 1 -+ 0). We then get from Equation (12.8.20) after some expansions a ~ R2 . {I 2.8.41)
I ..
l\1KAn-l=AKe M . (13.1.3)
Hence or
a.p = [A+B+AK1.p, aA
(13.1.4)
.p = eX(A+B)+'I\X 2 KC ,
(J 3.1
where C is an arbitrary constant operator. By letting A -+ 0, we find that C is the unit operator, which concludes the proof. If [A, B1- D, where D is an operator, expression (13.1.1) becomes more complicated. We note that we can only commute e A and e B , if A and B themselves commute. 13.2 Show that if
is the unit step function, 1,
t>O;
e(t)
0,
t
< 0,
(13.2.1 )
we have
We note that we can write Equation (12.4.9) in the form E = -VQa,
L n 0 =
(iv) From Equation (12.3.10), the fact that for the Ising model L/(g) g
An
~
[An,B1_=
( 12.8.42)
which is now the same as Equation (12.4.3), if (12.8.41) holds. In fact, one can prove generally that all results in the present approximation become those of the molecular field approximation in the limit as z -+ 00.
8(t-t')
== :t[8(t-t')1
-(}(t'-t)
= o(t-t').
(13.2.2)
(1) For a general discussion see: V.L.Bonch-Bruevich and S.V.Tyablikov, Green Function Methods in Statistical Mechanics (North-Holland, Amsterdam), 1962, and D.ter Haar in Fluctuation, Reio.xation and Resonance in Magnetic Systems (Ed.D.ter Haar) (Oliver and Boyd, Edinburgh), 1961. ~HI
320
Chapter 13
13.3 Make it plausible that
+
== lim -+0 X
y
E
13.3
f!II(~) + i1l"D(X), X
- IE
(13.3.1)
Solution
To prove the equality (13.3.1) we consider the integral < .. 0
J
f(x)
- - . dx,
-00
X
03.3.2)
IE
= lim lim €
-+ 0 8 -+ 0
(i
-8
+
f+OO) +8
-00
f(x)
--. dx + lim lim X -IE
€ ..,. 0
8 -+ 0
i
+8
-&
f(x)
- - . dx.
X -IE
ACt)
=
321
The first term gives the principal value of
I
r
iCH - JiNo )tJ i(H - p.IVo h P A exph P
)t] ,
(13.4.3)
a
are functions of t
(13.3.3)
exp
with JJ. = O:/{3. The generalisation from the usual Heisenberg operators is necessary whenever Hand Nop do 110t commute, as should be clear from the solution of the present problem and the result obtained in Problem 13.1. Prove that «AU); B(t'»), as well as the correlation functions FBA
where we assume that has no singularities on the real axis. We write this in the form I
Green function methods
according to our convenience, and the A(t) and B(t') are time-dependent (Heisenberg) operators, defined by the equation
where x and E are real and (!jJ indicates that in an x-integration the principal value of the integral must be taken.
I = lim
13.6
= (B(t')A(t»
and
F~B
= (A(t)B(t'»
(13.4.4)
[' only.
Solution
This follows by writing out explicitly the grand canonical averages, using the facts that (i) inside a trace operators can be cyclically permuted, and (ij) the various factors commute.
f~f(X~dX 13.5 Prove that the «A motion:
and the second term gives
. .f
hm hm
+1
<-+0& -.0 -1
f(oy)dy . _.( I~) = f(O) hm
Y
lEu
B(t'»), satisfy the following equation of a
o(t - t')(AB - ~BA >+ «[A, H - JiNop]-(t);
ih«A(t); B(t'»)
<-+0
03.5.1)
GENERAL FORMALISM
Solution
13.4 In statistical mechanics one is often interested in averages taken over a grand canonical ensemble. Sometimes these are averages of products of operators _ TrABcxp(-{3H + aNop ) (13.4.1) (AB) _ ,
Equation (13.5.1) follows from the result of Problem 13.2 and the relation (13.5.2) iM [A,B -pNopl-.
.=.
where the symbols have the same meaning as in Equation (10.2.4). It is usually difficult to evaluate (AB) exactly and approximation methods must be used. Apart from (AB) one is also often interested in correla tion functions such as (A(t)B(t'» or (B(t')A(t», which arc not the same. Although one can write down an equation of motion for (A(t)B(t'», it turns out that it is more convenient to use the so-called retarded and advanced Green functions defined by the equations
i i~ «AU); B(t'»)~ = +h"e(±t + t')(A(t)B(t'» ± lle(±t + t')(B(t')A(t» , (13.4.2) where e(t) is the step-function of Equation (13.2.1), ~ is at the moment a disposable parameter which we shall choose to be either + 1 or -1
13.6 Introducing the Fourier transform by the equation «A; B»E
lif" '"
-2 11"
so that «A(t); B(t'»)
B(t'»)exp
; B»rE of
B(t'») r
iE(t- t') ... d(t-t')
(13.6.1 )
a
a
_QQ
-00
r iEU
«A; B»E exp ~ -
h
tl)] d (E) h '
(13.6.2)
find the equation for motion for «A; B Solution
Straightforward Fourier transform of Equation (13.5.1) leads to I E«A; B»E = 211"(AB-~BA>+«[A,H-JiNop]_; B»E' (13.6.3)
322
Chapter 13
13.7
13.10
13.7 Introducing the spectral representation (Fourier transform) J(w) of FBA by the equation
=
t')
FBACt,
l~J(w)e-iw(t-t')dw,
Green function methods
where the
n i are the eigenvalues of nUl. Apply this to the operator
H - pNop for the case of a perfect gas, and use the properties of the
operator n(i), to write this operator in the form
(13.7.1)
H-pNop
and similarly introducing the Fourier transform l' (w) of F AB , prove that 1'(w)
= J(w)exp((3hw)
.
=
'I!
h
="7 I(vIBI/l)(/lIA -V,J-L
«ak; a~»E
oVIJ.E v ,
we find J(w)
(nk)
13.8 If G(E) is an analytic function of E which is equal to «A; B»r (2) in the upper half-plane and equal to «A; B»a in the lower half-plane, use the equation e-£t
(e
-+
0+), t
> 0;
au) = 0, t
< 0,
(13.9.2)
21T(E- ek +/l) ,
(13.9.3)
where the quantum number k includes spin dependence, if necessary. Hint: Use 71 = + 1 for bosons and 71 = -1 for fermions. (c) Use the result of (b) to prove
Iv)exp(-(3E v )o(w-E v +EIJ.)'
Repeating the calculation for F AB , we find expression (13.7.2).
aCt) =
(en -/l)a~an ,
where the en are the single-particle energies and where we have assumed that the en form a discrete set. (b) Prove that
Solution (vIH-pNopl/l)
=I
n
(13.7.2)
If the I/l) are the eigenfunctions of H - pNop and if
323
=
1
+
= exp [,..,a ()] ek - /l -71
(ak a k )
(13.9.4)
Solution f,
(13.8.1)
and (13.3.1) to express J(w) in terms of G(E) .
(b) Using the relations (10.12.3), (10.12.4), and (10.12.6) as well as the equation of motion (13.6.3), one finds the expression (13.9.3). (c) From the expression (13.8.3) for J(w) and Equations (13.3.1) and (13.7.1) we get +
Solution
exp[-i(ek-/l)(t'-t)]
,
(akCt )adt)
From Equations (13.6.1), (13.4.2), (13.7.1) and the similar equation for F ab , the relation between J( w) and l' (w), and Equation (13.8.1), we find 1 f+OO dw «A; B»r = -2 [exp((3hw)-71]J(w)E_h W +' (e -+ 0+). a 1T _~ - Ie
and hence, by putting t distributions (13.9.4).
=
= [exp,..,a( ek -
/l ) -71] '
t' for (nk), the usual boson and fermion
THE KUBO FORMULA ~
The function G(E) defined in the problem is thus given by 1 f+~ dw G(E) = 21T _~ [exp((3hw)-71]J(W) E-hw '
(13.8.2)
and using Equation (13.3.1) one finds I
J(w) =
(ah) lim [G(w +ie) -G(w -ie)] . exp ,.., w -71 0-+0+
(13.8.3)
13.9 (a) If we use for the '-Pi in Equation (10.11.1) the eigenfunctions of n(I), and if we are dealing with a system of non-interacting particles so that n(2) = 0, we have instead of Equation (10.12.8)
n =
fdinia+(i)a(i) ,
(2) The suffix E, introduced in Equation (13.6.1), is now dropped.
(13.9.1)
13.10 (a) Apart from their application in studying equilibrium proper ties, Green functions are also useful for deriving kinetic coefficients. To see how this can be done consider a system with a Hamiltonian H(O) perturbed by a periodic term Ueiwt+£t (e -+ 0+). Let at t = - 0 0 the density matrix p(-oo) be the equilibrium density matrix p(O) correspond ing to H(O), p(-oo)
=
p(O)
= Z(O)-lexp(-(3H(O»), Z(O) =
Trexp(-(3H(O»).
(13.10.1)
Using the equation of motion for pct), ihp
=
[H,p]_,
(13.10.2)
p(O) +!!.p ,
(13.10.3)
writing Pct)
=
and neglecting second order terms in U and !!.p, write down the equation of motion for !!.p. Solve that equation, using as an intermediate
Chapter 13
324
ql.antity .6.p' given by the equation .6.p'
=
iH(O)
t)
13.10
(
iJI(O)
t)
(13.10.4)
exp ( --h- .6.p exp - - h - .
(b) Using the solution for .6.p found under (a) prove the so-called Kubo formula (3). (O) 21feiwtHt«G; U»-w , (13.10.5) for the average value of a physical quantity G, where <..'><0) indicates the average taken with the equilibrium density matrix p(O) and where the Green function is the retarded Green function with T/ + I. Solution
(a) From Equations (13.10.2) and (13.10.3) we get, neglecting a term involving [ U, .6.p 1-, ih.6.p = [U,p(O)]_e iwt +e1+[H(O),.6.p]_. Substituting expression (13.10.4) into this equation, integrating the resulting equation and again using expression (13.10.4), we find t [ iH(O)(r - t)1 .6.p - fl _~ exp L h J [U, p(O)l- exp[ - iH(O)(r h e IWT+eT dr.
if
13.11
Green function methods
where g is the Lande g factor, Il-B the Bohr magneton, and B the strength of a uniform magnetic induction. We shall take the z axis along the magnetic induction and we shall restrict our discussion to the spin t case. In that case Sf is the spin-operator vector for the spin on lattice site f which has the components
Si
= th ( 0
1
I) , Sl = th (0 -i) ,Sf
0
i
'J
~~([G'(t), lI(r)l_)(O)eiwT+ETdr, t
iH(O)t)
n,
(iH(O)t)
iH(O)r) (iH(O)r) lI(r) = exp ( - h - Uexp - - h - ,
g
THE HEISENBERG FERROMAGNET
13.11 We shall now apply the Green function formalism to the case of the isotropic Heisenberg ferromagnet which is described by the Hamil tonian (compare Chapter 12): f
(13.11.5)
beb, .
(13.11.6)
2,
L..1(f-g)(S,· Sg), f,1
(3) A simple introduction to this result is given in Problem 24.9.
[be, b g l-,
[b~, b;l-,
[bf, b;l- ,
[br, b g 1+,
[b~, b; 1+,
[be, b; 1+
.
r),
Si = th(be+b Sl th(b; be), Sf = thO 2n,). If we write 1+ I . + I I . br = hSr = h(Si +iSn, b f = = h(Si-ISn (a)
f-~f(r)dr = L~8(t-r)f(r)dr.
gll-BB,
0) , (13.11.2) 1
(c) Express the Hamiltonian (13.4.1) in terms of the be and b;, putting 2 for the sake of simplicity.
Solution
and in. deriving expression (13.10.5) one uses the relation
--h-L..Si
0
(b) Find ex pressions for
G'(t) = exp -h- Gexp - - h - ,
H
(1
(~ ~), b~ = (~ ~),
b,
from which expression (13.10.5) follows. In this equation we have
(
ttl
express the components of Sr in terms of b r and br, and discuss the physical meaning of br, b~ and the operator nr given by
(G(t» = TrpG = Trp(O)G + Tr.6.pG
= «(~)\V!
0
where we have omitted the unit two-by-two matrices, which are factors of the components of S, and which refer to a111attice sites bar f. The l(f-g) are exchange integrals which depend only on f-g; assuming that the system has inversion symmetry-as we shall assume to be the case we have l(f-g) = l(g f), (13.11.3) while we can also put 1(0) = 0 . (13.11.4) (a) Introducing the operators
t)] .
(b) For (G(t» we find
325
(13.11.1)
we see that the b, and bf are the usual raising and lowering operators. The physical meaning of nr follows from its connection with Si . (b)
[b"b g l-=[b;:,b;l_=O
[b"b,l+=2b,b,(1-or,),
[br, b;]_
[be, b;l+
[I -2nrlorg ,
2b;b;(l-org) ,
[be, b; 1+ = 2b r b;(l-.5 rg) + Ofg .
13.11
Chapter 13
326
We note t hat if f and g are different lattice sites, the b c and b; behave as boson ('PCl ,.ltors, but if f and g are the same site, behave as fermion operators. They are sometimes called Pauli operators. (cl
H == -J.l.BBN 2J.l.BBLllr-2Ll(f g)b;b g -!NLI(f)
r,
f
13.13
Green function methods
Using now Equations (13.12.3) and 03.12.4) we get
_ 1 2n
G q {E-2J.l. BB-2(l-2n)[K(O)-K(q)]} = K(q) = LI(f)ei(c, q)
C
g
321
,
C
+2 r.L!(O] rLng]
Lc
2LI(f-g)ncn g
I'g
c, g whert: N is the total number of lattice sites in the system.
Gq
»
»·
n
(independent off),
(13.12.2)
and we can Fourier transform with respect to the lattice sites. Usinl!; the fact that OCg ==
Eq = 2J.l.BB + 2(1
~Lei(q.g-f),
oj.
I
»_ -
V
~Lei(q·g-f)Gq
where v'
vlN is the volume per spin, we get the equation for
I (13.12.4)
q
find the equation for Gq . In Equations (13.12.3) and (13.12.4) the summation over q is over the first Brillouin zone. (c) From the equation for Gq find an implicit equation for nand hence for the quantity 2CS'z )/h a.
n
2n =
(a) E«b g ; bf» =
p
p
(b) From Equations (13.12.1) and (13.12.2) and the last equation we
r
E«b g ; bm
3
v' d q (21f)3 exp({1Eq )-1 .
(13.12.5)
I
=
v'
fd3q,
and we find then the following implicit equation for a:
v' fcoth(!{1Eq)d3q.
2~Ocg[I-2
-2LI(p-g)«b p ; b;»+4LI(p-g)r«n g b p ; b;»-«npb g ;
get
f
n
Using the fact that the number of reciprocal lattice sites in the first Brillouin zone is N, we have
a
Solution
I
-)3' ... -+ (21f
q
«b g; b;»
I -2n" exp[i(q . g-f)-iEq(t-t')/h] 7t exp({1Eq )-1
g, and changing from a summation over q to an
(13.12.3)
where N is the total number of spins in the system, and writing
2n)[K(O) - K(q)] .
From Equation (13.12.4), the equation for Gq , and Equations (13.S.3) and (13.3.1) we find
(13.12.1)
Using the fact that H is translationally invariant, we can put
21f(E-Eq) '
with
13.12 (a) Find the equation of motion for the Green function «b g ; bf», putting T/ = + l. (b) In order to solve the equations found under (a) we must make approximations. The so-called random-phase approximation (RPA) first introduced by Bogolyubov and Tyablikov consists in writing «nCl b r2 ; b (3
1-2n
,
]
= 1 ;~n OCg+ L2J.l. BB+2(l-2n) ~I(f) -2(1
bi»
2n)LI(p-g)«b p ; bi»· p
13.13 Ca) Use the equation for a obtained in the last problem to evaluate the transition temperature Tc, that is, the temperature at which the magnetisation vanishes in zero magnetic field. Find the first term in the expansion of a in terms of Tc - T for temperatures satisfying the inequality (Tc - T)/Tc Tc.
328
Chapter 13
13.13
Solution
(a) Near, but below sponding to {3c, a will that Eq will be small. power series in a. The
the Curie (or transition) temperature Tc , corre be small and in zero magnetic field, which means We can thus expand the hyperbolic tangent in a result is
where
+!{3aK(O)T/(q)+".] '
(13.13.1)
-
,
(211)
or
~(n)
3
(21T)3 T/ (q)d q,
1 _ 2F(-l)
1
'
(3aK(O) +6{3aK(O)F(l)+"',
(13.13.2)
and hence, letting a tend to zero, for (3c 2F(-I)
{3c;':: K(O) (b) From Equation (13.13.2) we find for {3 3 a;:::::: F(l
1)
> {3c, but ({3-{3c)/{3c
~
1
J{3- (3c ~+""
(c) At low temperatures we expect a to be about equal to unity and we expect only q-values close to the origin to contribute to the integral in Equation (13.12.5) so that we can without loss of accuracy extend the integration over the whole of q-space. Using polar angles 0 and .p to characterise q we have 11 2-n
vI
(2 )3 1T
i i 211'
0
d.p
11'
0
sinO dO
i~ 0
2
q dq
L exp[-2r{3aK(O)T/(q)J . r=1 gO
Expanding T/( q) in powers of q, retaining only the first term (which is proportional to q2), and integrating, we get a power series in pl. From the definition of K(q) we get, for instance, for the case of a simple cubic lattice K(q)
=
iK(O)[cos(qxa)+cos(qya)+cos(qz
where a is the nearest-neighbour distance. (One can easily write down the analogous expressions for the cases of face-centred and body-centred lattices. We leave this as an exercise to the reader.) Hence we get in this case for T/(q) 2 T/(q)
I-AT'h- ..·
!a q 2+ ... ,
k [ 41Tl(O)
A
will depend on the crystal structure, and have been computed for several lattices and values of n. From Equation (13.13.1) we find
0-
'"
r~t%
2
where
fn
Vi
rIO
I
The functions F(n), defined by the equations F(n)
329
and hence (note that for a simple cubic lattice Vi a3 ) ~ f~ n v 2 ;:::::: --341T L e-ra.q q2 dq, Q' = {3aK(O)a 2
a
K{q) K(O)
= 1-
T/(q)
Green function methods
;:::::: (;1T)3 41T [{3aK(O)a
(;:)3 Id3q[a{3K(~}I'/{q)
a
13.14
J'h ~G);
is the Riemann zeta function, and kB the Boltzmann constant. (d) In this case we must have B 0 as otherwise a would vanish. We write
'*
1 + totl
coth'2I{3E'q -- to+tl ,
where to
= tanh[{3aK(O)T/{q)].
tanh{3jJBB, tl
Expanding the hyperbolic cotangent in powers of t I and then t I in powers of {3, we find from Equation (13.12.5), again for the simple cubic case, 11 I {3K(O) -[1---+'''] I 211 2to to ' and finally for the susceptibility X per spin X = (3jJ§ [ I +! K(O){3 +
which can be written (approximately) in the Curie-Weiss form jJ§ K(O) X
kB(T 0 ) , 8 = 2kB
13.14 When one studies ferromagnetic resonance, one is dealing with a sample which has a finite size. To take boundary effects into account most simply, even though only approximately, we add to the Hamiltonian a term involving the demagnetisation factors, Nx , Ny, and Nz , which depend on the shape of the sample assumed to be ellipsoidal with the principal axes along the x, y, and z axes. We must thus add to the Hamiltonian (13.11.1) a term
!(NxM;+ NyM} + NzMn , where M is the magnetisation vector, M
h
~Sf'
(13.14.1)
(13.14.2)
Chapter 13
330
13.14
(a) Express the term (13.14.1) in the b r and M. (b) This is a situation envisaged in Problem 13.10. Apart from a steady field B along the z axis there will be an r.f. field B 1 in the xy plane. If that field has components bcoswt, bsinwt, 0, the perturbing Hamiltonian H' will be of the form
. ' H = -L(S . B ) = Ue1wt+U* e- 1wt h r r I
Green function methods
.+_a 2 . + E«b g, br» - 27TOrg+[2JlBB-2aNNzJlB +2aK(0)]«b g , br»
+I
[-2aI(g-p)+aJl~(Nx +Ny)]«b p
(13.14.3)
,
;
M»
+aJl~(Nx -Ny)I ~(b;; M»,
with
p
/lB b 'SU=-h7 r
(S± 'SY) . r -- sxr±lr
(13.14.4)
E«b;; be»
=
-[2JlBB-2aNNzJl~ +2aK(0)]«b;; be»
The r.f. field will produce an additional magnetisation in the xy plane:
oM±
o(Mx ± iMy) = Nx±be±iwt . Express X± in terms of a Green function.
-I[2aI(g-p)+aJl~(Nx+Ny)1«b;; be»
(13.14.5)
=
p
-aJl~ (Nx - Ny) I «b p ; be». p
(c) Writing
X, +. - IX " ,
X+
Using again Equations (13.12.3) and (13.12.4) and the equations
(13.14.6)
express the energy absorbed per unit time, W, in terms of X' and Xl/. (d) Determine X±, using the equation of motion for the relevant Green function and using not only Equation (13.12.1) but also the approximate equation (13.14.7) «nr l b e2 ; b (3 » = (nrl )«b r2 ; b (3 »·
«b;;
OqO
!{NxM} + NyM; +NzM/)
=
r
r, g
{"'-»-47TJlB2 x+ -- x-* -- - 47TJl~ h2 '«s+. L g' ur r
(c)
W = (B 1
•
r, g
ei(r· q)
'
a
= 27T
+aJl~(Nx -Ny)Noqorq ,
where
+ (Nx + Ny )Jl~ I brb g + NzJl~
[! Jl~ (Nx -Ny) I
N r
(E + Eq}rq = -aJl~ (Nx - Ny )No qoG q ,
!N2NzJl~ -2NNzJl~ Inr
+
= -1 I
we find
(E-Eq)G q
(a)
M»~Iei(q.r-q)rq, q
Solution
(b)
331
provided we use Equations (13.12.1) and (13.14.7):
p
2JlB ,
I
13.14
Ir, g nrng
(brb g + Mb;)]
'« L br,. b +» r
g
Eq
*'
We note that for q 0, Eq differs from Eq in that B is replaced by B - N z (NaJlB), that is by the field including the demagnetisation. Solving the equations for r q and Gq' we find
.
q
*' 0:
a
l'q = 0, G q = 27T(E-E ) q
• b2 w 2 1/ oM) = N2i(X+-X-) = Nb wX q
(d) If we use the decoup1ing expressions (13.12.1) and (13.14.7) we find that the equation of motion for «b g ; M» contains the Green functions «b;; M». However, if we write down the equations of motion for both the «b g ; M» and the «b;; be», we get 3 closed set,
= Eq + NaJl~ [(Nx + Ny )Oqo - 2Nz 1.
=
(much as in Problem 13.12)
0:
a E+Eo
Go
= 27T E2 - E; ,
where
E; =
4Jl~[B+NaJlB(Nx -Nz)][B+NaJlB(Ny -Nz )].
Chapter 13
332
13.14 IIi
For X+ we get now X+
=
-47TJ.12BL.. '«b'g'
b+» f
f
=-
47TJ.11, 'G q N L..L.. f
. [47TJ.11' = hm - - - L.. 0 e..o+
N
qcf0
ei(f-g·g)
14
q
.1
27Th(W+IE)-E q
, L.. . el(q·
f-g)
The plasma(I)
f
- 47TJ.11 L~ h(w+iE)-Eo N f 27Th 2(w+iE)2 -E;
[1+. )-E
. {0J.11 . hm --E [h(w+IE)-Eol h( W r
E .. 0+
IE
r
-h(
W
J
1 J} '
+.IE )+Er
and, as both wand Er are positive so that 8(hw + Er) = 0, we get for X" "
X
=
27TJ.11 (hw + Eo) E 8(hw-E r ), r
showing that absorption takes place only at w = Er/h, with zero line width in the present approximation.
D.ter HAAR (University of Oxford, Oxford)
14.1 A plasma is a fluid consisting of positively and negatively charged particles. We shall consider the following model of a plasma: a gas of negatively charged particles, charge -e, is moving in a neutralising background with uniform charge density noe. Let the average number density of the negatively charged particles be no. Consider now an additional infinitesimal point charge q which we shall-for the sake of simplicity-assume to be at the origin. Let this charge give rise to a small spherically symmetric change ¢(r) in the electrostatic potential. Show that ¢(r)
= 3... e-Kr
(14.1.1)
r
where K2
47Tno{3e2 ,
1 (3
= kT
(14.1.2)
Discuss under what conditions the argument used is valid, and show that one of the conditions is that ro
~
d,
(14.1.3)
where the Debye radius ro is given by the equation K- 1
(14.1.4)
~ 11-';'
(14.1.5)
ro = while d
where n is the particle density, that is, d is a length of the order of the interparticle density. A plasma for which the condition (14.1.3) is satisfied is called a hot dilute plasma. In the following we shall assume that we are dealing with hot dilute plasmas. This kind of theory is called the Debye theory or Debye approximation. From expression (14.1.1) we see that ¢ changes appreciably in the so called Debye sphere, which is the sphere of radius ro within which the potential ¢(r) changes appreciably. (I) For general references see, e.g. N.G.van Kampen and B.U.Felderhof, TheoreticalMethods in Plasma Physics (North-Holland, Amsterdam), 1967.
333
Chapter 14
334
14.1
The electrostatic potential rj>(r) must satisfy Poisson's equation
04.1.7)
with A to be determined from the condition
f
n (r)d 3r = no/).
14.2 Prove that the number density of negative charges satisfies the equation q,,2 n(r) = n O + - (r) , (14.2.1) 47Te when the additional infinitesimal point charge is present.
(14.1.8)
In the thermodynamic limit as the volume of the system /) -l> 00, will tend at infinity to a constant value (00), which we can take to be zero, so that we have A = no. Substituting Boltzmann's formula (14.1.7) into Poisson's equation (14.1.6), and using the assumption that (Jerj> ~ I to expand the exponential and to retain only the term linear in 4>, we V 2- ,,2rj> = -47Tqo(r) , (14.1 where" is given by Equation (14.1.2). The solution of Equation (14.1.9) which has the required spherical symmetry is the expression (14.1.1). We note that this is a shielded potential with range rD. The density n(r) is the average number of negatively charged particles in a volume element divided by the volume of the element. This only has a physical meaning, if the element can be chosen sufficiently large so that it contains many particles. This in turn means that, if d is the average interparticle distance, so that the relation ship (14.1.5) holds, we have as a first condition for the applicability of the present considerations: (14.1.10) d I~n or IVnl~n'% 04.1.11 ) As we have to consider n(r) only as an average over a sufficiently large volume, it is a coarse-grained quantity. The same is true for (r), and we must also require for that it varies slowly over a distance d. This means that rD must be large compared to d, so that expression ( 14. 1.3) is the second condition. Using expressions (14.1.4) and ( 14.1 we get from it the condition (14.1.12) nO{Je 2 d 2 ~ I , or from expression 04.1.5)
Solution
Equation 04.2.1) follows from Boltzmann's formula (14.1. 7) under the assumption that {3e ~ I.
14.3 Find in the Debye theory the total number n exc of excess negatively charged particles in the Debye sphere. Solution
n exc
kT~ e
04.1.13)
e-K'rdr = !I e
04.3.1)
Solution
01 d3r = ql" . E int -- fqe[n(r)-n r
(14.4.1)
14.5 Find in the Debye theory the energy ED of the charge density in the Debye sphere. Solution
:!
ED
rd3rJdVe2Ln(r)-no][n(r')-noJ Ir-r'l
J'
04.5.1 )
Using the expansion in Legendre functions Pn(/J.), where /J. is the cosine of the angle between rand r',
1
ED = :!q2,,4
or
e
fln(r)-noJd3r
14.4 Find in the Debye theory the energy E int of the interaction between q and the charge in the Debye sphere.
we get
2
kT~
335
04.1.6)
At equilibrium, n(r) is related to 4>(r) through Boltzmann's formula n(r) == Ae!lerp,
The plasma
which shows that the average kinetic energy of a particle must be large compared to its average potential energy. Systems obeying this condition' are called hot dilute plasmas.
Solution
V2rj> = 47Ten(r) 47Teno - 47Tqo(r) .
14.5
= !q2" .
L~ dre-
~
I r,)n .
Pn(!.J.),
r
,
-;r1 n I~= 0 ( !,r )nPn(/J.),
r'
>r,
I
-rn=o\r I 1-
K '
L'"
dr'
L~rdre-"r
S:
(14.5.2)
dr' e- Kr ' (14.5.3)
14.6
Chapter 14
336
14.6 Estimate for the Debye theory the average number no of negative particles in the Debye sphere.
14.10
The plasma
337
Use Equation 04.9.1) to derive the following plasma equation of state: (14.9.3)
P = nokT [1 - 18!zo ] '
Solution
no ;::::; trrnorb
I
3V41T
(kT) 'h e 2 n 'Il
with no the quantity defined in Problem 14.6.
,
which, as follows from Equation (14. 2.6), is large compared to unity for all cases where the Debye theory is applicable. 14.7 Estimate for the Debye theot:"Y the ratio of the energy found in the preceding problem to the fluctuations in energy in the Debye sphere. Solution
From Problems 14.4 and 14.5 it follows that the energy attributable to each particle is of the order of e 2 #< which is of the same order as Eo· The total electrostatic energy in th~ Debye sphere will thus be of the order noe 2K. and the fluctuations of the order vnne2K., so that the required ratio is of order which, as we saw in Problem 14.6, is a large number.
vnn
14.8 Find for a hot dilute plasma an expression for the Debye length in a mixture of ionised gases.
Solution
For a plasma the evaluation of <W)tav is simplified by the fact that all forces are Coulomb forces. For W we have W
kr j
I
I'
where the summation is over all ki:J1 d S of ions while ej and nj are their respective charges and densities. 14.9 In Problem 9.16 it is shown that one can use the virial theorem to write the equation of state in the form pv = NkT +~<W)tav , (14.9.1) where W is the virial deriving from in termolecular forces, that is,
W=
L (F
j •
rj) ,
(14.9.2)
Li (rj . 'ViU) ,
-
i
(14.9.4)
where the superscript C indicates that where we are dealing with Coulomb forces and where U is the total electrostatic potential energy. As U is a homogeneous function of degree -1, we thus have from Equation (14.9.4) W= U.
(14.9.5)
To get the time average of U, we write it in the form -
t
1
U-
2 .. I
I
ejej
r.. II
I '\'
2
'
i,..ei"'; ,
(14.9.6)
I
where e; is the charge of the ith particle, rij = Irj-rjl, and "'; is the potential at ri due to all other ions, so that we can write,
-e;J
, ["'i(r) ",,(ri) = Ir-r.1
Solution
Using arguments similar to those 'Which led to expression (14.1. 2) we find 2 411"L e·n· (14.8.1 ) ~ K. 2
L(ri· Ff)
I
r
',.
'
04.9.7)
which is the total potential acting at r; less the potential due to the ith ion itself. For "'i(r) we can take the Debye potential (14.1.1) and taking the average [compare expression (14.4.1)] we get <W)tav
! Le;<"';)tav = !N<",')av = -!Ne 2 K..
(14.9.8)
I
From Equations (14.9.1) and (14.1.2) we thus get for the equation of state P
-:=
nokT
(1- 18~0) ,
(14.9.9)
where no is given in the solution to Problem 14.6. We note that the deviations from the perfect gas law are small whenever the Debye theory is applicable. 14.10 Write the pressure p of a plasma in the following form:
where the summation is over all particles in the system, ri is the position of the ith particle, and Fi the force on the ith particle due to inter molecular forces. Finally C'>tav indicates an average both over the time and over all particles in the system.
P
Pc o+pe,
(14.10.1)
where the first term is the perfect gas term (we assume that the only interactions are Coulomb interactions) while the second term derives
338
Chapter 14
14.10
14.12
from the electrostatic interactions. Prove that
The plasma
Solution
If we use expression (14.10.2) and write
(14.10.2)
ue
where ve is the total electrostatic potential energy, by considering the configurational partition function, introducing new variables r; where L3 = the volume occupied by the system, and finally using the fact that ve is a homogeneous function. Solution
We have the following relations:
,•
with the solution ve
~I
d3rl ... d 3 rN exp(-{3ve) ,
and
(iJiJvS
_ ~~
12
I
ve(rJ = ve(r;L)
a
Q 3L2 aL L
3N
f' d3r
l ...
I
,
(14.10.6)
r
J' (14.10.7)
(3ve(r·) I L I
where the integration is now over a unit volume. The only L-dependence behind the alai sign lies now in the factor L3N and in the L in the index of the exponential, and there is nowhere an implicit dependence. Hence we get, after taking the derivative with respect to L and changing back to the old coordinates, {3p
QIf3 Jd r l
...
(14.11
(14.11. 7)
3 1 I ne d rN;-[N+ J {3ve]exp(-(3u ),
I
I
14.12 Considering the charges in a plasma to be quantities which can be changed adiabatically (scale transformation), and considering the total electrostatic potential energy ve as a function of the entropy S, the volume, and the magnitude of the charges e, find a general expression for Uin the form Ne 2 ve f(x) (14.12.1) where
po = NkT+~(ve).
(14.10.9)
prove that for a hot dilute plasma
= adiabatic invariant.
I)
CiJ~t'lJ
(14.1
04.11.1)
= IN, and x is a dimensionless adiabatic invariant.
From Equation (14.10.4) it follows that 2ve e
iJve) __ I ve ( iJv S.e 1 v
(14.12.2)
Applying a general procedure, explained for instance in Problem 1.20, we have
14.11 From Equation (14.10.2) and the thermodynamic equation of state Problem 1.9a) ave) (iJpe) ( Tv T = T iJT IJ
VI
Solution
or
vT 3
(iJpe) iJT IJ'
which proves the expression (14.11.2).
I
d 3rNexp
T
(14.
1 = -Lve(r;)
I
)
and Equations (14.10.2) and (14.1 1.5) we get se = 4ve jaT 3 v,
(14.10.4)
,
j
rtlL, and using the fact that
_ I _ I _
(14.11.5)
Using the Maxwell relation (see Problem I
(14.10.3)
alnQ {3p=-- . av
(3p
aT 4 1).
uev
e
i
we have
=
1_,1
ve =
= U, r;
(14.11.3)
(14.11.4)
where the integration is over the volume v of the system,
Putting v
ve v '
we can write Equation (14.11.1) in the form
;1"
f
Q
339
I
2ve ve -de--dv+TdSe e 3v
dve or
oev
I;'
=
adiabatic invariant.
Relation (14.12.1) now follows immediately.
(14.11.2)
\!
(14.12.3) (14.1
14.13
Chapter 14
340
14.13 Use the result of Problem 14.11 and Equation (14.9.8) to find first an expression for x and then the form of f(x) in the Oebye approximation. [Use dimensional analysis.]
15
Solution
From Equation (14.11. 2), and the fact that the only natural constants which can occur in x are e and k, we find by dimensional analysis {3e2 ---v,
x
Negative temperatures and population inversion
(14.13.1)
VI
It is of some interest to write this adiabatic invariant in a form by using Equations (14.1. 2) and (14.
pi
J
x :::.::
2
')'J
K VI
2rD
(14.13.2)
The adiabatic invariance of v/rb is plausible, as one would expect that a simultaneous slow change in the volume and in the Oebye cloud would preserve the degree of order and hence the entropy. From Equations (14.13.1) and (14.9.8) it now follows at once that in the Debye approximation f(x)
=
U.M.TITULAER (Rijksuniversiteit, Utrecht)
different
04.13.3)
15.0 The concept of negative temperature is introduced by means of a statistical model of a spin system. The thermodynamic peculiarities are discussed and the possible use of negative temperature systems and systems with non-thermal level occupation as amplifiers of radiation is indicated. Ramsey's classical paper on the subject(l) may serve as a general reference, especially for subjects treated in the first few problems in this chapter. 15.1 Consider a system of n identical weakly interacting spins. Each of the spins occupies one out of 2s + I equally spaced non-degenerate energy levels with energy mW (m = -s, -s+ I, ..., +s). The average energy of each spin is uo. Find the maximum entropy distribution and the relation between Uo and the temperature parameter fl Calculate the partition function of the system and the specific heat as functions of {3 for positive as well as negative values of {3. [Use the relation {3 = 1/kT to define temperature for negative {3.] Find the simplified expressions for the case s = !. Solution
As for an ideal gas the partition function is simply the product of n identical factors, one for each spin. A straightforward application of the procedure used in Problem 2.2 gives for the probability Pm of occupying level m of any single spin -I e-/lmW Pm = Z({3)
with Z({3)
-/lmW _
m
e
-
sinh(s+-!){3W . hi RW gn ~~
(3 will be determined from the relation Uo
=-
alnZ({3)
:\(./
=
W[(s+!)coth(s+!){3W
(1) N.F.Ramsey, Phys. Rev., 103,20 (1956).
341
coth-!{3W].
342
Chapter 15
15.1
15.3
As a result of the concave character of the function cothx the sign of U o is negative when ~ is positive and conversely. For the specific heat we obtain by differentiating once more auo c = -nk~23i3 =
For s
Negative temperatures and population inversion
Solution
(a) Let p(E)dE denote the number of states with energy between E and E +dE. The average value of the energy is given (in a canonical ensemble) by
nk~2W2L!cosech2!~W-(s+!)2cosech2(s+!)~W].
Uo
! we obtain the simplified expressions Z(~) = 2cosht~W ,
The entropy for a system with s
S
!nk~2W2sech2!~W.
t is given by
= n(k~uo+klnZ) nk[~Wtanh!~W+ In(2cosh!~W)] .
For the system with s = ! the expressions are simpler and their general features (signs of Uo and ~, zero of C at ~ = 0) are more easily found. On the other hand any system which can be described by means of a set of occupation probabilities P-'h and P+'h can also be described by means of some inverse temperature k~; the additional information that lnpm is proportional to the energy of level m, which follows from entropy !. maximisation, does not imply any additional constraint in the case s This was our reason for not taking s = t from the beginning. 15.2 (a) Show from general formulae for the canonical ensemble that negative temperature distributions are possible if and only if the density of possible states of the system decreases fast enough as a function of energy; specify 'fast enough' quantitatively. (b) Show that this condition is violated if the energy of the system includes the kinetic energy of a particle or the energy of a mode of the radiation field. [The results of Problems 3.13(c) and 3.14(c) may be used.] For a spin system this implies that negative temperatures are meaningful only if the coupling of the spins to the kinetic degrees of freedom in the 'lattice' and to the radiation field can be neglected. In practical terms this means that the time in which the spins reach equilibrium among themselves should be much shorter than the time in which equilibrium with the lattice or the radiation field is established. J (c) Consider a system for which the density of possible states increases that only temperatures exponentially: peE) 0: e o£, a > 0; show T < l/ka are permissible. [An exponentially rising density of states was postulated (2) in a statistical model of strong interactions at high energies. The possible consequences of the resulting maximal temperature were also discussed. J (2) R.Hagedorn, Nuava Omenta Suppl. III, 147 (1965).
= fe-IlEEP(E) dE
I
f
e- IlE p(E) dE .
Unless peE) decreases at least exponentially with E this expression diverges for any negative value of~. Even the total probability that the system occupies any state with energy lower than any given value Eo would be vanishingly small. Such distributions are physically unaccept able. (b) For the kinetic energy of a particle we found in Problem 3.1 that the number of states with a velocity between Vand V +d V is proportional to V2d V. This corresponds to EY'dE, which is not exponentially decreasing. For the quantum mechanical case a similar result was derived in Problem 3 .13( c). For a single mode of the radiation field peE) is a constant; for the photon field as a whole peE) 0: E2 [see Problem 3.14( c) J. Finally, if a system consists of two parts such that
uo(~) = - Wtanht~W ,
C
343
E
E(1)+E(2), then peE) = IE dE'p(1)(E')p(2)(E-E').
In this formula
p(l)(E) and p(2)(E) are the density of states for the subsystems. 1t is obvious that peE) will only decrease exponentially if both p(l) and p (2)
decrease at least exponentially. (c) If peE) 0: e aE a comparison with (a) shows that values of ~ such tha t ~ .;;; a are unacceptable. Temperatures higher than 10 = 11 ka are therefore impossible in this model. 15.3 Suppose that the level spacing constant W in a system of n spins t is proportional to an external magnetic field: W = /lH. Calculate the amount of heat and work which has to be supplied to the system if we want to keep it at a constant temperature Tl while the field is changed from HI to H 2 • Find also the amount of work needed to increase the field adiabatically and the relation between temperature and field during such a process. What is the net effect of a Carnot cycle operating between two negative temperatures? [The heat absorbed is given by the relation dQ = TdS, for both positive and negative temperatures. J Solution
From the formulae derived in Problem 15.1 we find for the increase in U and S during isothermal magnetisation ~
(ll.U)TI
= n/l[H2 tanhx 2 -
(LlSh l
=
k~l(ll.U)TI +nk[lnZ(x2)-lnZ(xdJ ,
=
!/l~IHI,2 .
,
'i
I
\
Xl.2
HI tanhx d
,
15.3
Chapter 15
344
The amount of heat supplied to the system equals TI (LlS)Yl' The amount of work exerted on the system is given by (M)Yl
=
-nkTIllnZ(x2)-lnZ(x l
)]·
The relation between T and H along an adiabatic line follows from the observation that S depends only on the combination x = !{3J,J.H. Thus x must be constant along an adiabatic and Hand T are proportional to one another. The amount of work absorbed during an adiabatic increase of His (LlA)SI
=
nJ,J.(H2 - Hdtanhx
=
T2 -TI
nJ,J.HI-------y;-tanhx .
We now consider the Carnot cycle
(H2, T I )
adiabatic
(
~ H3
=
T2 ) TI H 2, T2
X2
= !{3IJ,J.H2
j iw'''om,,'
adiabatic
(
H4
=
~
TIHI' T2
)
XI = !(3IJ,J.HI
TI-~
(LlA)tot = -r,-(LlQ)Yl = -r--(LlQ)Y2 . I
345
15.4 (a) Two systems of n l and n 2 spins! respectively, are initially at different temperatures and are then allowed to exchange energy. Determine the final equilibrium temperature and the direction of heat flow. [The level spacing may be taken the same in both systems.] (b) Combine this result with that of the preceding problem and see whether Kelvin's and Clausius' formulation of the second law can be maintained if negative absolute temperatures are admitted. If not, propose a modification. It is reasonable to call a temperature Tl hotter than T2 if the energy content of the system at TI is higher than at T2. With this convention, negative temperatures are hotter than positive ones and -0 indicates the hottest conceivable temperature. Like +0 it cannot be reached in a finite number of steps. (a) The total energy content of the system is equal to n lu({3d+ n 2u({32)' At equilibrium each spin is at the same temperature; this final tempera ture corresponds to u({3d
By adding the amounts of work done on the system in each of the four stages one finally obtains ~-TI
Negative temperatures and population inversion
Solution
1i,o,"'=" (HI> T I )...
15.4
2
Both for positive and for negative values of TI and ~ there are two possible results: (1) an amount of heat LlQ is absorbed from a bath with temperature I;.; a fraction (1 - ~/Td is converted into work, the rest is discarded into a bath with I ~ I < I Til; (2) work is exerted on the system while heat is transported from a bath with low I TI to one with higher ITI. The meaning of these results and their relation to various formulations of the second law will be discussed in the next problem. Notice that we do not consider Carnot cycles between positive and negative values of T. In our model such processes would be possible if the adiabatics would cross at H = 0, T = O. However in this region our model is no longer thermodynamically correct. In real spin systems the interaction between the spins would no longer be negligible and some sort of magnetic ordering would occur in the immediate neighbourhood of H = 0, T = O. A similar phenomenon was observed in the case of the classical ideal gas (Problem 1.26).
=
[n IU({3I) + n2 u({32)]
(n l +n2)
__ ~ (_ nltanh!{3lw+n2tanh!{32W) wargtanh +n . n
(3f -
l
2
Since u({3) is a monotonically decreasing function, equilibrium can be reached only if energy is transported from lower to higher values of {3. If we adopt the definition of hotter and colder proposed in the problem, then the natural direction of energy flow is from hotter to colder temperatures. If we recall that dQ = TdS and notice that (dU/dS)w < 0 for T < 0 we see that the same is true for the natural direction of heat flow. The sign conventions used for dQ and the ordering of temperatures according to 'hotness' are reasonable but not unavoidable (3). (b) With our definition of 'hotter' the possible results of Carnot cycle between negative temperatures as calculated in Problem 15.3 can be viewed differently: (1) when heat is converted into work there is a heat flow from a cool to a hot reservoir; (2) when work is converted into heat some heat has to flow from a hot to a cool reservoir. By combining this with the possibility of natural heat flow from a hot to a cool reservoir we see that Kelvin's principle can be violated. It must be changed into the modified statament: It is impossible to devise an engine which, working in a cycle, would produce no effect other than: (1) the extraction of heat from a positive temperature reservoir and the production of an equal amount of mechanical work, (3) The consequences of alternative choices are discussed in Section 13 of P.T.Landsberg, Thermodynamics with Quantum Statistical Illustrations (lnterscience, New York), 1961.
15.4
Chapter 15
346
15.7
Negative temperatures and population inversion
347
(2) the rejection of heat into a negative temperature reservoir while an equal amount of work is done on the engine. Clausius' formulation of the second law remains unaltered.
The model discussed in this problem was introduced with reference to thermodynamical aspects of laser action (4). The relation with laser amplification will be discussed in the next problems.
DYNAMIC POLARIZATION
15.6 A beam of N photons is directed at a sample containing n two level systems at temperature T. The frequency I-' of the photons is resonant with the level splitting (hI-' = W); the probability that a single photon is absorbed by a system in its lower state and excites it to its upper state is equal to A. Using Problem 3.20, calculate the attenuation or amplification of the photon beam. Neglect spontaneous emission by the system and effects of the size of the sample. [The attenuation or ulI;phft...:ation of resonant radiation is a sensitive indicator of the temperature of a spin system. The presence of negative spin temperatures in a sample and the relaxation to positive temperatures J were first demonstrated in this way by Purcell and Pound
The most direct way of bringing a spin system from a positive to a negative temperature is by means of a sudden reversal of the magnetic field. The spins cannot follow the reversal and are left with a polarization opposite to the field. Since this method is essentially not quasistatic and not completely reversible it is not suitable for obtaining very hot negative temperatures. The alternative method of dynamic polarization, sketched in this problem, is analogous to the techniques of optical pumping. 15.5 Consider an impurity in a magnetic crystal with three relevant energy levels. The transitions between the levels may occur via different mechanisms. Those inducing transi (3) • tions between levels 1 and 3 and between levels 2 and 3 are repre sented by heat baths with tempera tures 11 and ~. The transition between levels I and 2 can occur by means of energy exchange with the spin system. Find the temperature --+--(2) of the spin system at which there is no longer any net heat flow between the reservoirs, and between the (I) I f reservoirs and the soin system. Solution
The transition between levels 1 and 3 is in equilibrium with a heat bath at temperature 11 if the occupation probabilities of the two levels satisfy the relation P3 PI exp(-.81 WI)' In the same manner we find P2
=
P3 exp( +.82 W2 )
= PI exp(-.81 WI +.82 W2 )
•
Equilibrium is possible only if the temperature of the spin system satisfies the relation .81 WI -.82 temperature will be negative if .81 is small enough This . compared to .82 ; in terms of the temperatures TI and ~ we must have WI
11 > 12W2 .
Solution
According to Problem 3.20 the probability of absorption by a system in the ground state is equal to the probability of stimulated emission by the same system in the excited state. The net increase (or decrease) in the number of photons is therefore equal to I:J..N == NnA(-p_% +p+%).
In this formula P-lh and P+% are the probabilities that a two-level system is in its upper or lower state. A comparison with the result of Problem 15.1 gives I:J..N == -nA t anh 1.8W . If the temperature of the two-level systems is negative the beam of photons is amplified by it. This is the principle of maser or laser amplification. Neglect of spontaneous emission is justified if the number of photons in the relevant modes of the radiation field is large compared to unity. For radio and microwave frequencies this is usually the case; in the optical region it is often necessary to include effects of spontaneous emission. 15.7 In Problem 15.6 we saw that the energy in a two-level system at negative temperature can be extracted by means of stimulated emission. In this way it may be used for the build-up of a coherent electromagnetic oscillation. The system discussed in Problem 15.5 could be operated in such a way that heat from the reservoir at T, is partially converted into coherent oscillation. Show that the efficiency of this energy conversion cannot exceed that of a Carnot process operating between TI and ~. ( 4) P.Aigrain in Quantum Optics and Electronics, Les Houches 1964, Eds. C.DeWitt, A.Blandin, and C.Cohen-Tannoudji (Gordon and Breach, New York), 1965, p.S27. (5) E.M.Purcell and R.V.Pound, Phys. Rev., 8t, 279 (1961).
Chapter 15
348
15.7
15.8
Negative temperatures and population inversion
349
For a stationary state we must have
Solution
From the total energy furnished by the heat bath at TI a fraction WdW I has to go into the reservoir at~. Stimulated emission can only exceed absorption if at least the equilibrium temperature of the spin system is negative. According to Problem 15.5 this implies that W2 /W I > ~/Tt· Consequently the conversion efficiency 17 has to obey the relation (6) W3 72 17';;;;-<1- WI ~
1 n(P2 - Pder = B 21 Te
(b) The equations for the change in the occupation probabilities are dpi dt A 2l P2 +NB 21 (P2 -pd-P(PI-P3), dp2
dt = -A 21 P2 +NB21 (PI -P2)+A 32 P3
,
= P(PI -P3)-A 32 P3 .
A MODEL OF LASER ACTION
15.8 The model of Problem 15.5 can be extended to provide a simple model of a laser. For this purpose we suppose that the sample of n three-level systems is enclosed in a resonant cavity. Losses of this cavity are taken into account by giving a finite lifetime Te to the photons in the resonant mode of the cavity. (a) Find the rate of inversion (P2 -pd at which the increase of the number of photons from stimulated emission compensates the cavity losses. Notice that the expression contains the stimulated emission rate B2l> but not the number of photons N. Next we consider the equations for the three-level systems. Suppose the temperature Tl is so large that the upward and downward transition rates between levels I and 3 caused by this pumping system are equal: A 13 A 31 = P. Suppose further that the reservoir at 72 and the systems causing transitions between levels I and 2 (other than the radiation field) are so cold that the upward transition rates can be neglected. Give equations for the change in occupation probability of the three level., for given values of P, A 32, A Zb and B21 and a given number N of photons in the resonant cavity mode. By putting the rates of change equal to zero the degree of inversion in the stationary state can be derived. Simplify this expression by assuming that A 32 is much larger than the other transition rates. (c) By comparing the results of parts (a) and (b) the number of photons in the cavity N can be found. For which values of the pumping rate P is laser action possible? Express the laser output in terms of P-Per • Solution
(a) By adding a term for the losses to the result derived in Problem 15.6 we find for the change in the number of photons dN dt
Llner
N
- Tc + nB21 (P2 -PI)N .
(6) H.E.D.Scovil and E.O.Schulz-DuBois, Phys. Rev. Letters, 2, 262 (1959).
Putting the left hand sides equal to zero and using PI + P2 + P3 = 1 we obtain for the stationary value of the inversion P(A 32 - A zl ) - A21 A 32 (P2 - PI},t = (P+ A 32)(2NB21 + A 21 )+ P(2NB21 + A32 + A2d . If we assume that A 32 is much larger than all other transition rates, this expression becomes P-A 21 (pz-pdst 2NB21 +A 21 +P' (c) A stationary regime is only possible if this inversion rate is equal to the critical inversion rate calculated in part (a): I P-A 21 nB21 Te 2NB21 +A z1 +P or nTc 1 N T(P-A 2d 2B21 (P+A 2 tl· The critical pumping rate is equal to (put N
0):
1 + l/nB21Tc -1/nB21 Te The power output of the laser equals N /Tc • It can be rewritten in the form N/Tc Hn-Llncr)(P-Pcr )' p. - A cr -
211
Notice that t (n - Lln er ) is exactly the number of systems in level I when the critical inversion is reached. The quantity nplP is the number of transitions to level 3 induced by the pumping heat bath. A fixed number of these excited systems is evidently needed to overcome losses due to spontaneous transitions; the remaining ones give rise to photons leaking out of the cavity. Modifications and improvements of this laser model can be found in any textbook on the subject (7). (7) e.g. A. Yariv, Quantum Electronics (John Wiley, New York), 1967, Chapter 15.
16.1
16
Recombination rate theory in semiconductors J.S.BLAKEMORE (Florida Atlantic University, Boca Raton, Florida)
Recombination rate theory in semiconductors
enjoy the same translational wavefunction; we regard these as distin guishable quantum states.) Fermi-Dirac statistics are appropriate for describing the occupancy of 'one-electron' states in accordance with the Pauli principle. An electron which occupies a 'one-electron' state acts dynamically as an independent particle, not interacting with the other electrons in 'one-electron' states, although the properties (e.g. energies) of the 'one-electron' states are affected by the average behaviour of all of the electrons in the system. As discussed in Chapter 3, the time-averaged probability that a state of energy E be occupied by an electron is of the form 1
ftE) 16.1 (a) For non-interacting or 'one-electron' states at energy E, write down an expression for the probability that any such state is occupied. You may employ the knowledge of Fermi-Dirac statistics gained from Problem 3.12. In the present chapter, we shall want to express occupancy probabilities in terms of an electrochemical potential or Fermi energy EF for equilibrium situations. (b) Given that a band of permitted electron states extends upwards )'h, express in from energy Ec with a density of states geE) = A(E integral form a condition for EF at temperature T when no electrons occupy the band at equilibrium. How can this be simplified for the case of no quite small? (c) The kind of band we have considered is usually referred to as the conduction band of a semiconductor, containing a few electrons in the lowest states, and many empty states. When thermal equilibrium is perturbed, the conduction electron density n may well be different from no. Use a quasi-Fermi level Fn as a normalizing parameter for the electron conduction density under such circumstances. At an energy well below Ec in a semiconductor we expect to find a band of allowed electron states which is almost completely full, extending downwards from energy Ev' Describe the total number of 'holes' in this valence band in terms of the Fermi energy EF for equilibrium (Po for holes) and of a hole quasi-Fermi level Fp for a non-equilibrium situation (p holes). Show the sense of departure of Fn and Fp from EF when the free electron and hole populations are enlarged. (d) Show that at eqUilibrium the product nopo depends on tempera ture but not on no or Po provided that the Fermi level lies within the intrinsic energy gap, i.e. between the top of the valence band and the bottom of the conduction band. Why is the quantity nj = (nopo)'h called the intrinsic pair density? Relate the energy difference Fn - Fp for a non-equilibrium situation to the product np (mass action law). Solution
(a) According to the Pauli principle, no two electrons in a system can be in the same quantum state. (Two electrons of opposing spin can 350
351
(16.1.1)
which has the necessary maximum occupancy of one electron per state for states of low energy. Conformity with Boltzmann statistics for the high-energy limit of small occupancy enables us to identify the parameter ~ with kT. The parameter a in Equation (16.1.1) performs a normaliza tion function, for if geE) dE distinguishable one-electron states are found in an energy range dE at energy E then in equilibrium at temperature T we can always find one (and only one) value for a such that
-
g(E)f(E)dE
N
(16.1.2)
is equal to the total number of electrons in the system. The normalization concept can in practice be expressed much more conveniently in terms of a parameter with the dimensions of energy. This is the electrochemical potential, or Fermi energy EF
= -kTln(a)
.
(16.1.3)
This can replace a in Equation (16.1.1) in writing the equilibrium proba bility of electron occupancy as feE)
=
I +exp[(E-EF)/kT]
(16.1.4)
We may note from this that the occupation probability is 50% for a state at energy EF itself. States well above EF are sparsely occupied with electrons. States well below EF are almost all filled; they contain few 'holes' among the ranks of filled states. (b) Given that a band of states starts from minimum energy Ec and that geE) = A(E-Ec)Y' for higher energies, we know that at thermal equilibrium the distribution of the total electron supply no over states of various energies must conform with _ no - I
A(E-Ec)Y' , l+exp[(E-EF)/kT]dE.
(16.1.5)
352
Chapter 16
16.1
This inserts the form of Equation (l6.1.4) into Equation (16.1.2). Equa tion (16.1.5) serves as a definition of the resulting Fermi energy for any combination of no and temperature. Given further that no is very small, we can see that Equation (16.1.5) must be satisfied with much lower in energy than Ec. In this asymp totic case, no
A(kT)'hexp[(E F -Ec)/kT]
r"
16.1
The same figure shows a valence band of states lying below the equi librium Fermi energy. Such a band will be almost filled with electrons at equilibrium, i.e. will have a relatively small density Po of 'free holes'.
exp(y)dY ... Ec :u~ fn.
= A (kT)'12 (:br)Yz exp [(E F -Ec)/kT]
Nc exp [(E F
-
geE)
1+exp[(E-Fn)/kT]dE
(16.1.7)
as an adaptation of Equation (16.1.5). Equation (16.1.7) places no apparent restriction on the value of n. The quantity Fn serves as a quasi Fermi level for non-equilibrium situations, and becomes coincident with EF for equilibrium itself. Provided that the total non-equilibrium population is rather small (so that Fn is lower in energy than Ec), the procedure of Equation (16.1.6) can be applied to Equation (16.1.7) in writing n
= Nc exp [(Fn -
Ec)/ kT]
5 t'F
= §g FP
(16.1.6)
Ec)/ kT] .
It may be noted that no of Equation (16.1.6) is the same as though the band were replaced by Nc states, all at energy Ec. The conditions of Equation (16.1.6) are met provided that no 4;. Nc ; this is a temperature dependent inequality which requires that EF be several kT lower than Ec· (c) We now consider the possibility of a departure from equilibrium, to provide for n =1= no electrons in this conduction band. It is apparent that a normalizing parameter Fn with the dimensions of energy can be used to define the relationship of n and the crystal temperature by
n
353
Recombination rate theory in semiconductors
(n 4;. Nc ) •
(16.1.8)
It is worth observing that the electrons distributed over band states for a non-equilibrium situation may have a velocity distribution which is incom patible with a thermal one for any real temperature. However, Equa tions (16.1.7) and (16.1.8) are concerned only with characterizing the total conduction electron density for a given crystal temperature T and a given density of states. In practice, we hope that excess electrons thrown into a conduction band will thermalize their speeds within a time of 10- 11 seconds or less, while in many semiconductors the time taken for excess conduction electrons to return to states in lower bands is very much longer. When equilibrium is disturbed in a semiconductor, it is usually in the sense of making n > no. From Equations (16.1.7) or (16.1.8), this makes Fn higher in energy than the equilibrium E F , as shown in Figure 16.1.1.
ELl Ev
Density of states g(E), and density of occupied states (shaded) Figure 16.1.1. An idealized yalence band (almost full) and conduction band (almost empty) for a semiconductor with an intrinsic gap extending from energy Ev to energy Eo. 'The equilibrium Fermi energy EF is compatible with the densities no, Po for free electrons and holes at temperature T. For any non-equilibrium densities n, p, which represent an enlargement of the free carrier densities, the quasi-Fermi leyels separate from EF in the sense Fn > EF > Fp.
From the preceding arguments, it will be clear that Po, E F , and Tare related by a condition
f
geE)
Ev
Po =
--«>
1+exp[(E F
-
(16.1.9)
E)/kT]dE ,
since the factor of g(E)dE is clearly the probability of a state not being occupied by an electron. When the Fermi energy is considerably higher than Ey (so that even the uppermost states in the band have rather few free holes), then we can expect to apply the procedure used in connection with Equation (16.1.6) for electrons, and write Po = Nyexp[(Ey -EF )/kT]
(Po
4;.
Ny).
(16.1.10)
In this limiting situation, the free hole density is the same as though the valence band were replaced by Ny states all at energy Ev' Evidently, if the valence band contains P =1= Po free holes for a non equilibrium situation, the density p can be described in terms of a hole quasi-Fermi level}t~ by
f
Ev
p
-00
g{E)
I +exp[(Fp -E)/kT]dE
p = Ny exp [(Ey -}t~)/ kT]
(arbitrary p) (small p)
1
(16.1.11)
Chapter 16
354
16.1
16.2
as adaptations of Equations (16.1. 9) and (16.1.1 0). The quantity Fp will coincide with EF at equilibrium, but will be lower than EF (as shown in Figure 16.1.1) when P > Po. (d) For equilibrium conditions, we can use Equations (16.1.6) and (16.1.10) to write nopo = NcNyexp[(Ey -Ec)/kTj (no ~Nc, Po ~Ny) (16.1.12) which depends only on the width of the valence band/conduction band intrinsic gap, on the density of states configurations near the extrema of these bands, and on the temperature. The product nopo does not depend explicitly on the value of no or of Po if the cited inequalities are met, and these equalities require that the Fermi energy should be appreciably higher than Ey yet appreciably lower than An intrinsic semiconductor is one for which the densities of free electrons and free holes are equal, which of course is the case if excita tion of electrons from the valence band into the conduction band is the dominant reason for the existence of free holes and free electrons. An intrinsic semiconductor is for practical purposes independent of the existence of any extrinsic features (such as the presence of foreign impurities or lattice defects which may in less perfect materials provide localized states for the provision of either free holes or free electrons). Thus in an intrinsic semiconductor, the densities of free holes and free electrons are each the quantity nj
(nopo),/,
(NcNv )% exp
-Ec )/2kTj.
NcNvexp[(Fn -Ec)/kTjexp[(Ev -}~)lkTj
= nrexp[(}~ -}~)lkTj
at
..
f\
I
( II
(16.1.1
by employing Equations (16.1.8), (16.1.11), and (16.1.13). Instead of nr on the right side of Equation (16.1.14) we could equally well write nopo, remembering that nand p must be written in the integral forms of Equations (16.1.7) and (16.1.11) if the free carrier densities are large enough to bring a quasi-Fermi level close to the edge of a band. At any rate, when p~ and Fp sti11lie within the intrinsic gap, -}~
kTln (np/nopo) .
(16.1.15)
16.2 Consider a semiconducting material for which the natural pro cesses of energy transformation at a certain temperature are sufficient to maintain a population of no free electrons per unit volume in a conduc tion band which can accommodate many more electrons. The density no results from a balance at thermal equilibrium of two opposing
355
processes: (i) the generation (at rate = g) of conduction electrons by excitation from filled electron states at lower energies, and (ii) the recombination (at rate r) of free electrons by their de-excitation to any empty states at lower energies. Now suppose that thermal equilibrium is perturbed, and the conduction electron density changed to n no + ne' We refer to ne as the excess electron density. Whenever ne is non-zero, then g =1= r, and usually both g and r are then modified from their equilibrium values. Such a pertur bation may result from externally induced excess generation at a rate ge, or by a flow of excess electrons from elsewhere. The net effect of the various influences can be expressed in an electron continuity equation an -=
(16.1.13)
For a semiconductor (which mayor may not be intrinsic) containing excess electrons and holes as a result of some departure from equilibrium, the product np can be written as np
II I
Recombination rate theory in semiconductors
I
=ge +g-r+-V-I e n
where In denotes the current density due to electron flow. In discussions of recombination in a semiconductor, it is convenient to define the electron lifetime (I) T as the excess electron density per unit rate of (r In terms of T, the electron continuity equation is ane ne I at ge --+-V-I Ten' (a) Comment on the conditions under which this becomes an ordinary and linear differential equation. What is then the general form of solu tion for any period of time-invariant excess generation? (b) Describe the build-up of ne when ge starts abruptly at time t = TI and continues at a constant rate for a long time thereafter. What happens when ge ceases equally abruptly at time t = 12? (c) From these results, show how ne varies when ge changes abruptly T3 . Sketch the time dependence for ge2 larger from gel to ge2 at time t than or smaller than gel' (d) When in practice the excess generation rate ge is an arbitrary func tion of time, express ne as an integral with respect to prior generation. Use this to show how ne responds to a pulse of generation which is an isolated half sine wave as a function of time. (e) In the same way, show how ne declines when ge is an exponentially decreasing function of time. Solution
..,i
(a) The continuity equation for electrons becomes an ordinary differ ential equation if the excess electron density ne is a function of time but not of location. This can be true if we consider a region well within a large, homogeneous single crystal, a region far from any surfaces, p-·n junctions, or contacts. It must also be presumed that the mechanism (I) The various meanings of 'excess carrier lifetime' are explored in detail with respect to the various processes of generation and recombination in the books by Ryvkin and by Blakemore cited in the bibliography at the end of the section.
Chapter 16
356
16.2
responsible for the excess generation rate ge is spatially uniform; thus if ge results from the absorption of photons of suitable energy directed from without, the photons must be ones which are absorbed rather weakly by the solid. Since we assume spatial uniformity and remoteness from surfaces and contacts, the electron current will be solenoidal (non-divergent) and the continuity equation reduces to dne ne (16.2.1) dt = ge--:;: This equation is also linear if the lifetime T is a constant, i.e. if the difference between the natural rates of recombination and generation is directly proportional to the departure from an equilibrium free electron density. (In practice it is found that T has some dependence on ne for any dominant recombination mechanism, though for certain recombina tion regimes T behaves as a constant over a fairly wide range of ne') When conditions of spatial uniformity and constant T are imposed, Equation (16.2.1) can be integrated immediately for any period of time in which ge is maintained at a constant value. The general solution is of the form (16.2.2) ne = geT+Cexp(-t/T) where the quantity C is determined by the initial conditions. (b) If. there is no excess generation prior to time t = T" and the constant rate ge thereafter, then C in Equation (16.2.2) must have such a value that ne = 0 at t = T,. This condition is satisfied if
357
Recombination rate theory in semiconductors
16.2
acts of generation prior to T3 , and the latter to the consequences of the more recent generation. Thus ne
= ge,T[exp(13/T)-exp(TdT)]exp(-t/T) +ge2T{ l-exp[(T3 -t)/T] (16.2.5)
(t> T3 ) .
This can be separated into the sum of a constant component and a transient component in the form ne
= ge2 T -
[(g e2 - ge,)Texp(13/T) +ge,Texp(TdT)] exp(-t/T) (t> T3)' (16.2.6)
The consequences of this equation are illustrated by the curves of Figure 16.2.1 for situations in which ge increases, decreases, or remains unchanged at the time 13. 2'0 ~!....
ge,T
1'5
1'0
0'5
C = -geTexp(TdT)
so that ne
= geT{l-exp[(T, -t)/T]}
(t> T,) .
(16.2.3)
This has reached the value n e (T2)
= geT{l-exp[(T, -12)/T]}
at the time 12 when excess generation abruptly ends. From Equation (16.2.2) it is clear that ne will subsequently decay in accordance with ne
= Cexp(-t/T) = n e(T2)exp[(12 -t)r] = T[exp (12/T) -exp(TdT)] exp(-t/T)
ge
(t
>
T2)' (16.2.4)
(c) In this problem, we assume that the lifetime is independent of n e , so that the continuity equation (16.2.1) is linear. Thus if ge operates at a value gel during the period T, .;;;; t';;;; T3 , and at a different value ge2 for t> T3 , then ne at times later than T3 can be described as a simple sum of contributions in the forms of Equations (16.2.4) and (16.2.3). The former of these will describe that component of ne which derives from
0 0
4 3 2 Time scale (t - T 1 ) in units of T
5
6
Figure 16.2.1. The transient behaviour of the excess pair density when excess generation starts at a time Tl and changes to a different rate at time T3 · The curves here follow Equations (16.2.5) and (16.2.6) after time T3 for situations in which ge2/gel = 2, I, and 0·3.
(d) We must now consider the solution of Equation (16.2.1) wheng e is an arbitrary function of time. This solution can be envisaged by considering ge to be a sequence of instantaneous generation events, each of which can be described by a delta function. Thus suppose that
l:ge dt = NfJ(t - to) . For this generation alone, Equation (16.2.1) has the solution ne(t)
= Nexp[(to-t)/T].
Now since the lifetime t is assumed to be a constant in this problem, Equation (16.2.1) is linear, and the solution for ne(t) for many delta function acts of generation at different times is simply the sum of the
358
16.2
Chapter 16
values for
lle (t)
16.2
the integral representation for ne(t) is
=
o·s
I~ge(to)exp[(to-t)/r]dto
(16.2.7)
~
Gr
which is a very simple example of a Green's function form of solution. We can use Equation (16.2.7) to determine how ne responds to sinusoidally varied creation. Suppose that we have a single half sine wave of generation: ge(to) = Gsin(wt o )
(0
0·6
0'4
< to < 1r/w) ,
\ g,
\
and that there is no excess generation at other times. Then, from Equation (16.2.7), we have ne(t) = Gexp(-t/r)
I
sin(wto)exp(to/r)dto
(0
\
\
< t < 1r/w)
0
4
5
6
As a corollary of the preceding discussion, it can readily be established that generation in the form of a continuous sine wave results in an excess carrier density comprising the sum of a constant term and a sinusoidal term. The sinusoidal component lags behind the phase of the generating sine wave by the angle 8 = tan-1(wr). (e) We should now like to know how ne decays when ge is itself an exponentially decreasing function of time. Suppose that ge has been maintained at the constant value G for all negative time (so that ne = Gr at time t = 0), and that ge = Gexp(-t/T) for all positive t. Then from Equation (16.2.7) we have
ne(t) = G exp(-t/r) fwtexp(x/wr)SinXdx 0
< t < 1r/w) , (16.2.9)
nell)
while the generation is in progress, and
(Gr)exp(-t/r) +
f:
Gexp(-to/T)exp[(to -t)/r]dt o (t
> 0) .
Provided that T and r are not exactly the same, the integration yields
in
G ne(t) = -exp(-t/r) exp (x/wr) sinxdx w 0 Gwr 2 . ~ ~ [1 +exp(1r/wr)]exp(-t/r)
ne(t) = GrTeXp(-t/Ti=;exp(-t/r) (t
> 1r/w)
(16.2.10)
while for the special case of T
for the subsequent monotonic decay. The form of the response while the generation is still in progress can be seen more clearly by defining a phase angle 8 = tan-l (wr), and rewriting Equation (16.2.9) as ne(t) = Grcos8[sin(wt-8)+exp(-t/r)sin8]
3
Figure 16.2.2. The response of the excess carrier pair density to a half sine wave of generation, in accordance with Equations (16.2.9) through (16.2.11). The dashed curve illustrates g. for a situation of WT = I, and the solid curve is the result for n,.
and Equation (16.2.8) is of this form when we change to the dimension less variable x = wt o. Thus
rSi~~t) -COs(wt)+exp(-t/r)]
\ 2
0
Time t in units of r
f
-----:;:--..-
\
0·2
(16.2.8) as a description of ne while the generation process is still going on. In order to describe ne after the half sine wave of generation has ceased, we must write Equation (16.2.8) with 1r/ w as the upper limit of integration. Now through integration by parts we can establish that exp(ax) . exp(ax)sinxdx = 1 +a2 (asmx cosx) ,
w
359
Figure 16.2.2 illustrates the rise and fall of ne in accordance with Equations (16.2.9) through (16.2.11) when wr = 1, to make the phase angle 8 = tan-I (wr) = !1r.
derived from each generative act separately. Thus when
ge is any arbitrary (not necessarily continuous) function of time, then
ne(t)
Recombination rate theory in semiconductors
(0
ne(t)
(16.2.11)
\
r the result is
G(t+r)exp(-t/r)
(t> 0) .
(16.2.13)
Figure 16.2.3 displays the decay of ne according to Equation (16.2.13) for the particular case of T = r, compared with the conventional expo nential decay for T = 0 and the greatly slowed decay when T = 3r. Only if T is smaller than, or equal to, r is it possible to determine r from the decay rate at large values of t.
< t < 1r/w).
:~
(t> 0), (16.2.12)
,
360
Chapter 16
16.2
16.3
Recombination rate theory in semiconductors
361
stimulated recombination? (See also Problems 3.20, 15.6, and 15.7 for related considerations.)
0'6
r.:.
0'4
~
'"
<:
'":::: E
0-2
ill
~
o
o
4 2 3 Time t in units of T Figure 16.2.3. The decay of the excess carrier density upon the cessation of or attenuation of the excess generation rate. The lowest curve is the exponential decay for generation which stops abruptly. The middle curve follows Equation (16.2.13) for generation which declines exponentially with time constant T = r. The upper curve fits Equation (16.2.12) for T = 3r.
16.3 In this problem we consider the direct radiative transitions in a semiconductor which result in the creation of, or annihilation of, a hole-electron pair. The term direct radiative means that a photon can create a hole and an electron whose difference in momenta is just the (negligibly small) momentum of the photon itself. An electron state and hole state which are connected by a direct transition probability have the same wavevector, as for example the states of energy Eu and El in Figure 16.3.1. The semiconductor drawn in this figure is a direct gap semiconductor in that the allowed states forming the upper and lower borders of the intrinsic gap E j are of the same wavevector. (a) Use considerations of mass action and detailed balance in discussing the rates at which transitions between El and Eu will occur by stimulated and spontaneous radiative processes. Show that the ratio of stimulated to spontaneous rates of radiative recombination between these energies is just the number N of photons per mode for photon energy hv =
Eu -E1 ,
From this result, show that stimulated transitions have an effect on the radiative recombination rate between Eu and El which can be described by multiplying the spontaneous rate by a factor of
-l"
I-N exp
(hV+FpkT -Fn) - IJ.
What does this tell us about a threshold condition for supremacy of
Wavevector Figure 16.3.1. The model of a direct gap semiconductor assumed in Problem 16.3. The electrochemical potential or Fermi level is EF for thermodynamic equilibrium, and the gross of the two bands for relatively steady-state non-equilibrium are characterized and Fp.
course, operation of a semiconductor laser is also contingent on the elimination of competing (non-radiative) recombination mechanisms, and on an optical configuration which takes proper advantage of recom bination radiation.] SOlution
(a) The radiative transitions we must consider between E, and lie in three categories: drsp = rate of spontaneous events of electron-hole recombination; drst = rate of recombination events stimulated by the recombination radiation field; dg st = rate of generative events stimulated by the recombination radiation field. For the spontaneous recombination process, we may use the principle of mass action in writing the rate in a form dr sp
= AfuO-li)·
(16.3.1)
The parameter A contains information about the densities of states at energies Eu and E., and about the possibilities for electron-hole annihila tion when an electron of kinetic energy Eu - Ec encounters a hole of kinetic energy Ev . In Equation (l6.3.l),!u denotes the that a state at Eu contains an electron, and I - fl is the probability that a state at El contains a hole. If it can be assumed that a disturbance of thermodynamic equilibrium changes the magnitudes of the electron and hole populations but not the character of the velocity distributions within
Chapter 16
362
16.3
the co-existing electron and hole gases, then fu
=
I +exp[(E u -Fn)/kT]
I-fl = 1 +exp{(Fp -E()/kT] for any values of Eu and E( within the bands. We shall denote by fuo and 1 - flO the values of these quantities for equilibrium. The rate of stimulated recombination from to E( has the same functional dependence on fu and fi :
= Bfu(l-fi)N,
(16.3.3)
since (as with spontaneous recombination) this rate depends on the co existence of an occupied conduction state and an empty valence state. N is the number of photons per mode for the photon energy
hv
Recombination rate theory in semiconductors
363
Thus at equilibrium, or away from it, the ratio of stimulated to sponta neous recombination rates between Eu and Eu - hv is
and
drs!
16.3
Eu -E1 •
and we know that in thermodynamic equilibrium N must reduce to the Planck result 1 (16.3.4) No exp(hv/kT) 1 . We are interested in the relationship of the parameter B in Equation (16.3.3) to A in Equation (16.3.1). The quantity B must be used again in describing the rate of radiative transitions stimulated upwards by the presence of a recombination radiation field. Since this latter rate depends upon a full valence state and an empty conduction state, we have (16.3.5) dgst = Bfi( I - fu)N from E( to Eu' The principle of detailed balance-which is tantamount to a statement of the second law of thermodynamics-requires that rsp + rst-gst vanish in thermodynamic equilibrium, not only in toto, but also between any selected groups of initial and final states. From Equations (16.3.1), (16.3.3), and (16.3.5) we can write that (Afuo{l-flO)+Bfuo(1-flO)No -BflO(1-fuo)No ] = 0
at equilibrium. Thus the ratio of stimulated to spontaneous coefficients is Ii _ fuo(l - fio) A - No(f1O - fuo) An insertion offuo, fio, and No from Equations (16.3.2) and (16.3.4) and a brief manipulation of these factors yields the very simple result that B A I . (16.3.6)
drs! = N (16.3.7) hv dr sp from the ratio of the right sides of Equations (16.3.3) and (16.3 .1). (b) When the conditions are of a steady-state non-equilibrium, then the net recombination rate from Eu to E( is dr net = drsp + dr s! - dg st = drsp
= drs p
(I
dgst -drst ) drsp
[1 -
--'-'-'---'-=-:"
since we have demonstrated from detailed balance arguments at equilib rium that A and B must be the same. Substitutions of fu and fi from Equation (16.3.2) then immediately yields that
_ {
r
exp drnet - drsp I -N L
(hV+Fp-Fn ) kT -
}
(16.3.8)
A modest departure from thermal equilibrium will make Fn - Fp a positive quantity, but a quantity smaller than hv = Eu E(. Under these circumstances, the net recombination rate dr net is smaller than drsp alone. A more severe violation of equilibrium which makes Fn - Fp just equal to hv (that is, a violation which makes fu the same as fi) causes stimulated transitions to occur in the upwards and downwards directions at exactly cancelling rates. For this threshold value of Fn - Fp , drnet is identical to drsp. When the excess densities of free electrons and holes are made still Fp > hv. larger, fu is actually larger than fl' This is the case when Obviously this condition can be met most easily for the states of energy Ev and Ec + Eb and with progressively more difficulty for states separated by larger photon energies. The creation of lasing conditions in a semiconductor requires among other things(2) that there be a population inversion (upper energy states more heavily occupied with electrons than lower states, or fu > fi) between the highest valence band states and the lowest conduction band states. When a population inversion is in exist ence, then the net radiative transition rate dr net is appreciably larger than drsp and is dominated by stimulated downward transitions. (2) In order that the recombination radiation produced by downward transitions should be amplified by stimulating further downward transitions, a population inversion is a necessary but not sufficient condition. The semiconductor must also be one for which non-radiative recombi nation processes are rather feeble and present little competition to the radiative reactions. Mirror surfaces must be arranged to give an 'optical gain' which outweighs the inevitable optical losses for some direction of propagation of the photon stream.
364
Chapter 16
16.4
16.41n this problem, as in Problem 16.3, we concern ourselves with direct band-to-band generation of hole-electron pairs and their direct recombination. For convenience, we shall again assume that radiative transitions provide the dominant mechanism for energy transformation. Suppose a large homogeneous semiconducting crystal with no non uniform currents, so that the continuity equation for excess electrons and Po holes is not a function of position. This medium contains no electrons and Po holes per cm 3 at equilibrium, and no + ne electrons and Po + ne holes in a non-equilibrium situation. The quantities no, Po, and ne are all small enough to ensure that , Fn, and Fp all lie several kT within the intrinsic gap (as was true for the situation illus trated in Figure 16.3.1). Show that the radiative lifetime TR of excess carrier pairs is proportional to
nopo no+po+n e
16.4
Recombination rate theory in semiconductors
If the electron and hole populations have quasi-Maxwellian speed distributions, even when excess free carriers are present, then I fu = l+exp[(E -Fn)/kT] ~exp[(Fn -Eu)/kT] ~ I (16.4.3) u
for any conduction band states, and 1 I-fi = l+exp[(Fp-E,)/kT] ~exp[(EI-Fp)/kT] ~ I
for any photon energy relevant to band-to-band transitions. Moreover, since Fn - Fp is required to be appreciably smaller than Ej, it seems probable from the conclusions of Problem 16.3 that N in a non equilibrium situation is unlikely to be substantially different from No of equilibrium itself. Under these circumstances, stimulated recombination dr st from an upper state Eu to a lower state E, can be ignored in com parison with the spontaneous recombination rate drsp between the same states. Thus the net radiative recombination rate is drnet
drsp -dgst ~ A f..
Afu(l-fi)-Afi(1-fu)N
(1 - fi) -Afi(l- fu)No ,
)/kT] exp [(E F -}<~ )/kT] - I}
no+ne) (po+n = A exp(-hv/kT) [( -;;-;;--;;;;-) -I ] . e
(16.4.2)
where we have demonstrated in Problem 16.3 that the same coefficient A should be used for the spontaneous and stimulated terms.
(16.4.5)
Note that the factor inside the square brackets depends only on the total carrier densities, not on the energy separation of the participating states. Thus this factor is preserved in common when a summation is made over all transition energies. Accordingly,
Solution
Since we are given that the equilibrium free carrier densities and the excess pair density ne are small enough to keep Ec - Fn and Fp - Ev both larger than kT, it is apparent that the intrinsic gap E j must be considerably larger than kT. Accordingly, the photon occupancy number per mode under equilibrium conditions must be small compared with unity, (16.4.1) No ~ exp(-hvjkT) ~ I
(16.4.4)
for any valence band states. When Equations (16.4.3) and (16.4.4) are inserted into Equation (16.4.2), the net recombination rate can be written as dr net Aexp(-hv/kT){exp[(Fn -Fp)/kT] -I} = A exp(-hv/kT){ exp [(Fn
Solve the equation for transient decay when externally provoked genera tion is terminated (using the symbol TRO for the small-modulation life time when ne is sufficiently small). Show that an arbitrarily large disturbance of equilibrium must decay to a small-modulation situation within the time TRO'
365
r net
'd
l.... r net
-
TR
r(no+ne)(po+n e ) - C
nopo
I] .
(16.4.6)
The quantity C in the above equation is determined by the interband matrix element, the densities of states, etc. Thus for given densities of free carriers, the value of C will determine whether the radiative lifetime TR is large or small. We can arrange Equation (16.4.6) to write the radiative lifetime in the form TR:::::
_-:--_n-,o,-,-p-,o,---..,.. C( no+po+n e) '
(16.4.7)
which is functionally controlled by the factor cited in the question. Since we are permitted to assume that the continuity equations for free electrons and free holes are spatially homogeneous in this problem, the continuity equation for each free carrier species is
dne
dt = ge
ne TR
(16.4.8)
if non-radiative processes can be ignored. In Equation (16.4.8), ge is the rate of pair generation caused by some external provocation, as discussed in Problem 16.2. Solutions of Equation (16.4.8) are aided by writing the lifetime in the form TRO TR
= I +ne!(no+ Po)'
(16.4.9)
16.4
Chapter 16
366 where T
-
RO -
__ n---"o,--p-,,-o_ C(no+Po)
(16.4.10)
is the small-modulation lifetime which is operative whenever ne is small compared with either one of no and Po. Insertion of Equation (16.4.9) into Equation (16.4.8) gives us the continuity equation
= ge - TRO ~
+
tdt TO
r
N
= TO
In
e
dne ne[l+ne!(no+po)]
=
no
Recombination rate theory in semiconductors
367
an expression for the density of neutral donor impurities in terms of the Fermi energy (at equilibrium) or a quasi-Fermi level (away from equilib rium). Note from the figure that any donor can accommodate one electron at energy E. = -Ed for neutrality, and that there are (31 choices for the wavefunction of this localized electron. We ignore the possibility that any donor may be neutral but with its electron in an excited state.
+ )
(16.4.11) Po and this can be solved simply by separation of variables following the end of any period of excess generation. [Solution during a period of time varying generation is often possible, but can become involved!] If ge is set as zero for all positive time, and it is assumed that ne = N at time t = 0, then separation of variables gives dt
(1
16.5
N(ne+no+po) Inne(N+no+po)
~
~
<= <=
(16.4.12)
Q)
E,
g
lVd donors
<.)
&1
Accordingly, ne is given explicitly by (no+po)N n --e - (N+no+po)exp(t/To)-N
lVa compensators
(16.4.13)
No matter how large N is made, ne is forced to become smaller than no + Po in a time of 0· 7To or less, and the subsequent decay is essentially exponential. The initial decay of ne is hyperbolic if N ~ 16.5 Consider the semiconductor situation of Figure 16.5.1. We say that this semiconductor is extrinsic in that the free electron density is derived from impurities (flaws) rather than by excitation from the con duction band, and that it is n-type extrinsic in that negatively charged free electrons dominate. Assume that localized flaw states other than those shown in the figure can be neglected, and that the valence band has a negligible number of free holes. Assume further that the free electron population is non-degenerate(3) both at equilibrium and in the presence of an excess electron density ne (cf also Problem 3.13). (a) Obtain a relation between the density of free electrons and the quasi-Fermi level for the conduction band, using the symbol Ne to denote the quantity 2(27rmckT/h2)'h as a density of states which could be imagined at Ee as a replacement for the distribution of allowed states in the band. (This is a repetition of part of Problem 16.1.) Also obtain (3) A degenerate electron gas has properties (such as specific heat) which are degenerated from classical predictions because no is very large, and the Fermi energy lies above the bottom of the band. Thus a non·degenerate situation is one for which the electron density is small enough to keep the Fermi energy below the bottom of the band; the Fermi occupancy factor is then essentially a Boltzmann factor for any energy corresponding with band states.
Wavevector Figure 16.5.1. The model of a simple extrinsic semiconductor assumed for Problem 16.5. The solid contains lVd monovalent donor impurities per unit volume, and compensation equivalent to lV. monovalent compensating acceptors (lV. < lVd ). Electrons can be excited from donor bound states into the conduction band, and we arbitrarily set zero energy to be O. The band itself is isotropic, characterized by a the base of the conduction band, Ee scalar effective mass me' The donor ground state energy is EI = -Ed' and a donor can accommodate an electron at this energy in anyone of PI ways; excited states of donors are to be ignored in this problem. The quasi-Fermi levels Fn , Fd , for conduction and donor states respectively, converge on the Fermi level EF for thermodynamic equilibrium.
(b) Show accordingly that the free electron density no for satisfies ) Nc -N. = "Rexp(-Ed/kT) = nl d
a
no
1-'1
as a conservation law for electrons. Solution
(a) The conduction band assumed in this problem is a simple one, characterized by a scalar effective mass me' This mass parameter sets the density of states per unit energy interval to beg(E) = 47r(2m c /h 2 )'hE'h from the known density of quantum states in reciprocal space (4). The (4) Quantum mechanics tells us that a maximum of two electrons (of oppOSing spin) can be associated with a volume h 3 of momentum space in a crystal of unit volume. The quoted result for g(E) follows by transforming to the energy variable F =
368
Chapter 16
16.5
16.5
total number of conduction band electrons for equilibrium at tempera ture Tis no = So""f(E)g(E) dE ,
(16.5.1)
Ndn:(jINdi = exp[(EF -Ed/kT]: I.
Rj
exp[(EF -E)/kT] .
This ratio requires that the total density of neutral donors be
t
I +exp[(E-E F )/kT]
Ndn
-Ndi = - - - - - - - -
EF )/kT]
(16.5.2) Nd 1 1+ (jl exp[(-Ed - EF )/kT]
Equation (16.5.1) then reduces to
no = 411' (
2m kT)% hC2
211' m c kT = 2( h2
foo
exp(EF/kT) Jo yv,exp(-Y)dY
Fd as the energy which must be substituted for EF in Equation (16.5.5)
in order to reproduce correctly the density of neutral donors. For the extrinsic semiconductor of this problem, an electron removed from a donor impurity must be either on one of the Na compensator centers or be one of the no electrons in the conduction band. Thus we have
(16.5.3)
since the integral on the first line of Equation (16.5.3) has a value of (tll')Yl.
In a non-equilibrium situation, Equation (16.5.3) must be modified to take into account a total of n = no+ne free electrons, and then defines the electron quasi-Fermi energy Fn through n
/
no+ne = Nc exp(Fn/kT)
= noexp[(Fn - EF )/kT].
(16.5.5)
For a non-equilibrium situation we can define a 'donor quasi-Fermi level'
)% exp(EF/kT) ,
Nc exp(EF/kT) ,
369
For thermal equilibrium at temperature T, the ratio of occupied to available states is set by the energy separation of EI and the electro chemical potential:
where feE) denotes the Fermi-Dirac probability of occupancy for non interacting states of energy E. Thus, when the free electron gas is non degenerate, feE)
Recombination rate theory in semiconductors
I
(16.5.4)
I
However, we remember that Fn can properly be substituted back into Equation (16.5.2) to give the occupation probability for a particular energy only if the recombination time (lifetime) is very long compared with the time taken for a disturbed free electron distribution to thermalize itself. We now consider the occupancy of the bound donor states in terms of a Fermi level or quasi-Fermi level. If Ndi donors per unit volume are ionized, then N dn Nd - Ndi are neutral; since excited states of the donors are to be ignored, then each of the neutral donors must be in one of its ground states. We note that a previously ionized donor can be made neutral by placement of an electron in anyone of the (jl states at -Ed' Thus we may say that at energy EI we have N dn the energy EI occupied states, compared with (jINdi unoccupied and available states(5). (5) Only one of the PI states of a given donor can be occupied at anyone time. However, for an ionized donor all PI states are available for the initiation of recombinative transitions.
N dn
Nd -Na -no}
Ndi
Na +no
(16.5.6)
in equilibrium. Away from equilibrium, the expressions of Equation (16.5.6) must be written with n = no+ne replacing no (b) We can develop a simple expression for no which does not explicitly
involve EF for our non-degenerate n-type semiconductor at equilibrium,
by using Equations (16.5.3), (16.5.5), and (16.5.6). The first of these
permits us to write no/Nc for exp(EF/kT) in the expression of Equa
tion (16.5.5) for the density of neutral donors. Since this density is also
Nd-Na no,then
Nd -Na -no
=
Nd N I +--%-exp(-Ed/kT)
(16.5.7)
nOl-'I
which can be simply rearranged to the form no(no+Na ) N. -N. d a no
Nc -aexp(-Ed/kT) 1-'1
(16.5.8)
as required. The symbol n I is used in the question to denote the quantity
t
(Nc/(jI)exp(-Ed/kT). nl is often referred to as the mass action constant for the interaction between the donors and the band, and we shall find n I
a useful symbol in Problem 16.6.
370
Chapter 16
16.6
16.6 This problem continues to study free and localized electron den sities associated with the extrinsic semiconductor situation of Figure 16.5.1. (a) Show that the electron conservation equation in a non-degenerate situation (16.5.7) is a consequence of the law of mass action, from the balance of natural generation and recombination in equilibrium. Show further that the total thermal generation rate has the form g = Cnl(Nd -Na -n) whether n is the equilibrium value or not, provided that stimulated radia tive recombination can be ignored, and that recombination events involv ing more than one conduction band state can similarly be ignored. Identify the quantity C in microscopic terms. (b) Show that, for these conditions, the lifetime characterizing an excess electron density ne is inversely proportional to Na + 2no + n 1 + ne' Imagine a steady-state spatially homogeneous situation for which light of intensity 10 results in a quasi-unifonn excess generation rate ge; show that ne is proportional to 10 for weak illumination and to I/f for stronger illumination, and that ne must eventually saturate no matter how intense the light. Solution
The conservation law for electrons in a non-degenerate extrinsic semi conductor controlled by a single species of impurity has been shown to be noCno + Na) Nc r -N =-{3 exp(-Ed/kT)=n l d a no I
(16.6.1)
in the terminology of Problem 16.5. The above result was derived in that problem from the requirement that the same electrochemical poten tial should dictate the occupancies of free and bound states at equilibrium. However, Equation (16.6.1) is of the fonn to be expected from mass action considerations. For we know that the rates of thermal generation and of recombination must balance in detail and in toto at equilibrium. Since we can assume for this problem that the free electron gas is non-degenerate (i.e. states near the bottom of the band are more likely to be empty than full), then the recombination rate should be proportional to the total free electron density(6) n and to the number (31(n+Na ) of available states on ionized donors. We can write this rate as (16.6.2) r = iio{31(n+Na)n where ii denotes the thermal speed of a free electron (averaged with respect to a Boltzmann distribution) and 0 is the (similarly averaged) (6) If an electron could suffer capture at an impurity site by donating the recombination energy to one or more other electrons (in a manner which conserves energy and momentum) then the recombination would have terms dependent on n 2 , n 3 , etc. The question specifically excludes such multi·electron processes from our consideration.
16.6
Recombination rate theory in semiconductors
371
capture cross-section for a free electron presented by an unoccupied donor state. Expression of r in the form of Equation (16.6.2) would not be possible in such a simple way if stimulated radiative recombination were important, but this complication is explicitly excluded. The opposing rate of generation is determined by Ndn> the number of electrons capable of excitation, and the thermal environment for such excitation. However, it will not depend on the occupancy of states within the band, provided that this occupancy is small for all band states. Thus g = A(Nd -Na -n) (16.6.3) where the quantity A should depend on the temperature, on the minimum excitation energy Ed, and on the density of conduction states, but not on Nd , N a , or n. Since g r at equilibrium, Equations (16.6.2) and (16.6.3) can be equated for this condition to yield no(no+Na ) A (16.6.4) Nd - Na -no = iio{3. ' which is of the form of Equation (16.6.1). Evidently we should identify A/iio{31 with the mass action constant n 1, so that the rate of generation is g = iiO{3l n lCNd -Na -n)
(16.6.5)
whether or not the conditions are of eqUilibrium. It may be noted that Equation (16.6.1) would be recreated from mass action considerations even if generation and recombination processes involving two or more electrons were considered; but then Equations (16.6.2) and (16.6.3) would involve higher powers of n. (b) The difference between the rates of natural recombination and generation can be used in the definition of an excess electron lifetime T in the continuity equation ne dne dt = ge +(g-r) = ge -T"
66
(1 . .6)
for excess electrons in a situation of spatial uniformity. Here ge is once again an excess generation rate caused by an external influence. Provided that rand g are given by Equations (16.6.2) and (16.6.5), then ne ne T r-g = iio{31[(nO+n e )(no+n e +Na)-n1(Nd -Na -no-nell There is a major cancellation among the terms in the denominator of the right side, since noCno + Na ) is equal to n 1(Nd - Na - no). Thus I T = vo{31(Na + 2n O+nl +ne) (16.6.7) in the continuity equation (16.6.6).
16.6
Chapter 16
372
ne(Na +2no+nl +ne) = _~ep' va"'l as a response to spatially uniform generation. Now suppose that excess electrons are created at the rate ge (per unit volume and time) as a result of an incident stream of 10 photons (per unit area and time), each with an energy of at least Ed' Excitation can take place only from neutral donors, thus we may expect a quasi uniform (7) excess generation rate aphNdn/o,
where aph is the photo-ionization cross section of a neutral donor. If not all incident photons have the same energy, then aph is a suitable a\erage with respect to the excitation probabilities to the various upper statesEu = hv-E d • SinceNdn = Nd -Na -n, then
ne(Na +2n o +nl +ne) _ !. ~ 0 Nd-Na-no-ne va""p.
Recombination rate theory in semiconductors
373
the condition
This continuity equation requires in steady state that
ge
16.7
(16.6.8)
controls the relationship between 10 and ne' This relationship is shown graphically in the log-log plot of Figure 16.6.1. Part (a) of the curve shows the linear dependence of ne on light intensity for weak illumina tion. The curve changes character at point (b), which corresponds with
(d) (e)
,,: bJl
.£
10gIo
Figure 16.6.1. Variation ofne with incident photon flux 10 as required by Equation (16.6.8).
(7) In a more rigorous version of this problem, we should have to concern ourselves with the attenuation of the photon flux with increasing distance through the crystal, to provide a depth dependent ge'
ne "'" Na +2n o +n l !. "'" o
}
2va(3,(Na + 2n O+nl)2
----:-::-':-'--'---"--:--::---=-----'-''------: aph(Nd -2Na - 3n o-nd
(16.6.9)
For a semiconductor crystal with relatively weak compensation, in which Na ~ N d , and for a temperature low enough to make the mass action constant n I small compared with Nd , it is possible for ne to be substantially larger than Na + 2no + n I yet substantially smaller than N d . Such a combination of requirements is necessary in order that the region (c) in the curve of Figure 16.6.1 be prominent. Within that range (when it exists), the left side of Equation (16.6.8) is approximately n: /Nd ; thus under these conditions ne varies essentially as 16 The upper limit of the range occurs at the point (d) on the curve, when 2
•
10
"'"
ne
~
NdV(31~ aph
}
(16.6.10)
Nd -Na -no
and as section (e) of the curve indicates, a further increase of photon flux cannot make no + ne any larger than the available electron supply of Nd - Na . The behaviour of the system when no+ ne becomes comparable with Nd - Na would be more complicated if stimulated recombination played a significant role. 16.7 Consider steady state conditions in a semiconductor crystal which has free carrier densities no and Po at equilibrium. We shall assume again that the semiconducting medium is spatially homogeneous, and shall assume further that the free carrier populations are non-degenerate (so that nopo = n:). Electron-hole recombination in this material is assumed to be dominated by the activity of N r monovalent recombina tion centres. Each centre has just two states of charge: 'empty', or 'filled' with a single electron. We assume that the centre can be filled in only one way, so that the filling probability is a Fermi-Dirac occupancy factor. The energy of the localized state is such that at equilibrium the ratio of filled centres to empty centres is PI/PO' Along with the definition of the quantity p" we can define the companion quantity n I = nt/PI' so that the ratio of filled to empty centres is also no/n I' It will be obvious that the free carrier densities are just n I and P I when the Fermi energy coincides with the energy of the localized state. Suppose that an empty center has a cross-section an for the capture of a free electron of speed v n . Then we say that the centre has a capture coefficient vna n for a free electron of this speed. Averaged over the
374
Chapter 16
16.7
thermal distribution of electron speeds, the capture coefficient of an empty centre may usefully be written Cn == as the capture coefficient of a full centre for a free hole, averaged over the speeds of all free holes. (a) When an externally provoked process of excess generation causes the free carrier densities to change to n = no + ne and to P = Po + Pe, then show that the fraction of filled recombination centres is ~+Pe-ne
nl +no
Nr
(b) Demonstrate that a steady excess generation rate values for nand P such that Nrcncp(np -nr) ge = cn(n+nl)+c ( ) p P+Pl
ge
produces
(c) Provided that Nr is small enough so that ne and Pe will never be appreciably different, show that the common lifetime for excess electrons and holes is of the form 1'o{no + Po) +1'<>one 1'= no+Po+n e where 1'0 is the lifetime for very small ne and 1'00 is the limiting lifetime for very large ne' Solution
(a) The energy of the localized state associated with a recombination centre in this problem is defined in terms of the numbers n 1 and PI' From the manner of the definition, it is evident that niP 1 = n oPo ;:. Since the ratio of filled centres to empty centres at equilibrium is no/n l ' the fraction of all centres filled at equilibrium is
nr.
nO/nl
Iro = l+nO/nl '
16.7
Recombination rate theory in semiconductors
(b) Assuming spatial homogeneity for this problem, then continuity equations for excess electrons and excess holes can be expressed as
=
no no+nl .
ge
d Pe dt
(g") ge + -r
r
no +nl
N.
ne Tn
}
Pe = ge--:;:
(16.7.2)
p
r = nNr(1-fr)c n into the Nr(l -/r) available centers. The opposing rate of thermal generation must be proportional to the density of occupied centres: g
= ANrfr .
The thermal activation parameter A can be expressed in terms of n 1 and since it is necessary that g = r at equilibrium when fr reduces to fro = no/(n o + n I), and n reduces to no. Thus Cn ,
ANrno no+nl
noN.nlcn = no+nt
A = n1c n •
*'
= ~+Pe-ne
+(g-r)
In these equations, ge is an externally provoked generation rate, and Tn and 1'p are the lifetimes associated with excess populations of electrons and holes. Since electron-hole recombination is dominated by the given set of recombination centres, then r capture rate of free electrons by empty recombination centres, g = thermal generation rate of free electrons from occupied centres, r' = capture rate of free holes by electron-occupied centres, g' = thermal generation rate of free holes from 'empty' centres. It is clear from Equation (16.7.2) that g = rand g' r' at thermal eqUilibrium. Moreover, g- r must equal g' - r' for a steady-state non equilibrium situation. We concentrate first on the generation-recombination traffic between the centres and the conduction band. In terms of the capture coefficient Cn , we can write an electron capture rate
or
This fraction will remain unchanged when equilibrium is disturbed if conditions are such that ne = Pe. However, if ne Pe, the recombina tion centres must assume Pe - ne additional electronic charges (assuming that the occupancy changes of other species of localized states can be ignored). Thus the non-equilibrium situation involves a/ractional occu pancy change for recombination centers of (Pe -ne)/Nr • The resulting fraction of 'full' recombination centres is
f.
= ge -
dne = dt
or
Iro
375
(16.7.1)
In view of this, we can summarize the electron traffic as
= nlNr/rc n r = nNr (1-/r)c n
g
r-g =
}
(16.7.3)
N.cn[n-/r(n+nl)]
in terms of an occupancy factor Ir which remains to be detennined. A similar appeal to the balance of r' and g' at equilibrium permits us to describe the rate of hole generation in terms of PI- The generation
Chapter 16
376
I
16.7
recombination traffic of holes can then be summarized by g' = P I N r (1 - Ir)c p
r' r'
pNrfrcp
.
(16.7.4)
= Nrcp[fr(P+pd-pd
16.7
Recombination rate theory in semiconductors
lifetime. Making ne = Pe in Equation (16.7.7), we have that cn(no+n l +ne)+cp(po+ Pl+n e ) (16.7.8) Tn Tp Nrc n (nO+po+n e ) For a very small departure from equilibrium, the low-level lifetime is
The occupancy factor Ir for a steady-state non-equilibrium situation is then constrained by the requirement that r-g = r' -g', or
Nrcn[n-/r(n+nl)] = NrcD[(p+pd/r -pd which yields Ir
nC n +P1C p cn(n+nl)+Cp(P+PI)
(16.7.5)
Equation (16.7.5) can then be reinserted into Equation (16.7.3) to obtain (r - g) in terms of nand p: (n+nd(ncn +PIC p ) r-g = N C n r n. Cn(n+nl)+Cp(P+PI)
~
Nrcncp(np -nIPI) cn (n +n 1)+ c p (p + PI)
J
(16.7.6)
Since the situation under discussion is a steady state one, it is apparent from Equation (16.7.2) that the result of Equation (16.7.6) must be identified with the excess generation rate ge, and with ne/Tn and Pe/Tp. If we choose to write nlPI as the square of the intrinsic carrier density, we obtain g =ne=Pe Nrcncp(np-nf) (16.7.7) e Tn Tp Cn(n+nl)+Cp(P+PI) The result contained in Equation (16.7.7) was described first by HaU(8) and by Shockley and Read (9). (c) As it stands, Equation (16.7.7) specifies Tn or Tp in terms of both ne and Pe. By a further use of Equations (16.7.1) and (16.7.5) it is possible to secure a relationship between Tn and ne which does not involve Pe, or a relationship between Tp and ffe which does not involve n e , but such expressions are far from simple (10 • However, if Nr is small, then ne and Pe are substantially the same, whether the excess density is large or small compared with the thermal densities of free carriers. The fraction Ir of occupied centers is still a function of the departure from equilibrium, but if Nr(/r - Iro) is small compared with 11. then electrons and holes enjoy a common steady-state (8) R.N.Hall, Phys. Rev., 87,387 (1952). (9) W.shockley and
W.T.Read,Phys.Rev., 87,835 (1952).
(10) See, for example, J.S.Blakemore, Semiconductor Statistics (Pergamon Press, Oxford), 1962, pp.277-28 1.
377
To
cn (no +n d+cp(Po +Pl) N \ rCncp (.J...~ no
(16.7.9)
while for a departure very large compared with the thermal free carrier densities the lifetime asymptoticaUy approaches 1 1 T~="
(16.7.10)
Then, for any magnitude of modulation, the quantities defined as To and T~ can be used in a description of the carrier pair lifetime as
To(no +po)+T""n e no+po+n e
T=
(16.7.11)
as suggested in the question. GENERAL REFERENCES
S. M. Ryvkin, Photoelectric i'.jJects in Semiconductors
Bureau, New
1964.
J. S. Blakemore, Semiconductor Statistics (Pergamon Press, Oxford), 1962. P. T. Landsberg, "Problems in recombination statistics", in Festkorperprobleme, Vol.6 (Ed. O. Madelung) (Academic Press, New York), 1967. A. Rose, Concepts in Photoconductivity and Allied Problems (Wiley-Interscience, NewYork),1963. A. Many and R. "Lifetime of excess carriers in semiconductors", in Progress in (Ed. A. Gibson) (John Wiley, New York), 1958. Semiconductors,
17.2
Transport in gases
Now
379
N-I
L1 tuj.
XN-I
j=
17
Transport in gases
Therefore
N-I
L1(!:J.Xj!:J.XN) = 0
(XN -ltu N ) =
j=
since successive displacements are uncorrelated. Therefore D.l.GRIFFITHS (University of Exeter, Exeter)
(xJ.,)
By induction, since Xo 17.1 (a) A foreign molecule is initially at the point r = 0 in an iso tropic, stationary gas. After making N collisions with other molecules it is at the point rN = (XN,YN, ZN)' If !:J.XN == (XN-XN-d, assume for the averages over many possible paths that (i) (!:J.xJ.,) = (!:J.ylr) = (!:J.zJ.,) (i.e. the gas is isotropic), (ii) (tuJ.,) (== -i A?, say) is independent of N (i.e. the gas is homogeneous and stationary), (iii) (tuj!:J.Xj) = 0 for i =1= j (i.e. successive free paths are uncorrelated). Show that the mean square displacement (xJ.,) in the x direction is -iNA? Hence show that A?t (X2(t)
= 3i
if l/r is the mean collision frequency. Express in words the meaning of
A. (b) Let f(x, t) be the probability distribution of the foreign molecule at time t [i.e. f(x, t)dx is the probability that the molecule lies between the planes x and x + dx at time t, irrespective of its y- and z-coordinates]. Calculate the mean and the variance of f(x, I). (c) According to the central limit theorem of statistics, f(x, t) is, asymptotically for large t, a normal (Gaussian) distribution with zero mean, as introduced in Problem 2.9. Show that in this case it satisfies the equation 2
af
af
at = Dax 2
where D = A2/6r. [The foreign molecule is said to perform a random walk{l).] Solution
(a) Since XN = XN-I+tu N ,
then (xJ.,)
= (x}., - 1)+ 2(XN - 1tuN )+ (tuJ.,) .
(I) F.Reif, Fundamentals of Statistical and Thermal Physics, Ch.12 (McGraw-Hill, New York), 1965.
378
= (XJ.,-I)+!A?
= 0, (XJ.,) =
-iNA?
In time t, the foreign molecule makes N Therefore A?t (X2(t)
A? =
= t/r
collisions on average.
=~ ;
(LlxJ.,)+(!:J.ylr)+(!:J.zJ.,)
=
(LlxJ., +!:J.ylr
+ LlzJ.,) .
Thus A? is the mean square displacement between collisions, i.e. Ais the root mean square free path. (b) Since the gas is isotropic and stationary, the mean displacement is zero, by symmetry. The variance is (X2(t) = A2 t/3r, by definition. (c) The normal distribution for a variable x with mean zero and variance a 2 is f(x) = (27Ta
2r exp (- ;;2) Yz
(cf Problem 2.9) .
Taking logarithms, differentiating, and using the result [from part (b)] that a 2 = A2 t/3r, the desired result follows at once, with a2
A2
D=-= 2t 6r' 17.2 (a) A uniform, isotropic, stationary gas contains a small propor tion of foreign molecules, whose number density n(x, t) is independent of y and z. Assuming that the foreign molecules perform independent random walks of the type discussed in the preceding problem, construct an integral equation for n(x, t) (for large t) in terms of n(x, 0). [Hint: f(x,O) = o(x).] (b) Using the result of Problem l7.I(c) transform this integral equa tion to a partial differential equation for an/at. (c) Hence show that the mean flux of foreign molecules in the x direc tion is proportional to the concentration gradient an/ax, and identify the diffusion coefficient. [This is Fick's law of diffusion.] (d) What is the equilibrium distribution of foreign molecules?
Chapter 17
380
17.2
Solution
(a) For t 0 the distribution function f(x, t) becomes a b function, describing a molecule located with certainty at x == 0, i.e. f(x,O) Now n(x, 0)
b(x) ,
= l:n(~, O)o(x-~)d~
by a standard property of the b-function. This expresses n(x, 0) as a weighted sum of b-functions, each representing one molecule on the plane ~ x. Thus L:n(tO)f(x-t
n(x,O)
O)d~.
Since the proportion of foreign molecules is small, each behaves inde pendently, and each f(x to) evolves independently in time as discussed in Problem 17. I, i.e. n(x, t)
= i:n(~, O)f(x
t)d~ ,
where for large t ~ T, f(x, t) is the normal distribution with zero mean and variance "A.2 t/3T. This is the required integral equation. (b) From part (a), we have
roo
an(x,t)_
at
I-
- 1oon{.;; , 0)
af{x-~,t)
at
dL
and from Problem 17.1(c) af(x-t t) at Therefore
a2
an(x, t) = D
at
Da2f(x-t t) _ Da2f(X-t t) a(x - ~)2 --'-=a-x"'2'-'--'
f'"
n(~,
-2
ax
f{x-t t)d~
_00
= DaZn(x, t) ax z
This is the required differential equation. (c) Let the flux of foreign molecules in the x direction be x per unit area. By symmetry arguments y = z = o.
a x
an(x, t)
at
= -
17.3
381
Integrating, and remembering that, by symmetry, x vanishes when an/ax is zero, an = -D x ax The diffusion coefficient is D, where, from Problem 17.1, "A.2
D (d) In equilibrium, x
= 6T
0 and so an ax = O.
Similarly an
an
-=-=0
ay
az
.
Therefore the equilibrium density distribution is uniform. 17.3 (a) Assuming that all molecules behave as rigid elastic spheres, show that the mean free path of a foreign molecule of radius r 1 in a gas of molecules of number density n2 and radius r2 is approximately 1 1T(r 1 + r 2 )2 n 2
.
(b) Hence show that the coefficient of mutual diffusion D for a small proportion of foreign molecules of mass ml and radius r l in a gas of molecules of number density n2 and radius r2 at temperature T can be written as a D 61Tn2(rl+r2)2 m 1 '
(3kT)'h
where a is a dimensionless constant of order unity and k is the Boltzmann constant. [Use the result of Problem 17.2(c) and the equipartition theorem of Problem 3.6(c).1 {c) Carbon monoxide (CO), ethylene (C2H4), and nitrogen (N 2 ) each have molecular weight 28. The coefficient of mutual diffusion for a small proportion of CO in C2H4 gas at NTP is D(CO; C H ) = 0·129 cm z S-I • Z
4
Similarly, under the same conditions
ax '
since the number of molecules is conserved. Hence, from part (b), ax aZn - - = D z ax ax
Transport in gases
and
D(CO; N z ) = 0·176 cm 2 D(C 2H 4 ; N z ) = 0·129 cm 2
S-1 S-1 .
Show that the coefficient of self-diffusion for nitrogen gas at NTP is D(N z ; N z ) = 0·176 cm z S-1 .
Chapter 17
382
17.3
(d) Can one speak of a coefficient of self-diffusion for a gas of molecules which are quantum-mechanically indistinguishable? Solution
(a) The foreign molecule collides with all molecules whose centres would otherwise come within (rl +r2) of its centre, that is, with all molecules whose centres lie within a circular cylinder of which the cross section is 11'(r 1 +r2)2, and of which the axis coincides with the path of the centre of the foreign molecule. The mean free path is, to a good approxi mation, equal to the mean separation between molecules whose centres lie within this cylinder, which is 11'(r 1 +r2)2 n2
Because of differences in averaging, the root mean square free path A may differ from this value by a factor of order unity. (b) From Problem 17.2(c),
'A?
A A 6 T
D=-=-o
6T
But A/T is of the order of the mean molecular speed, i.e.
A
T = a(v2t" where a is a dimensionless constant of order unity. From the equi partition theorem [cf Problem 3.6(c)],
-!m 1 v2 Therefore
= ~kT.
~ = a(3kT)% T
m1
From this result and that of part (a) the expression for D follows. (c) All three gases have the same molecular mass and temperature. The major constituent has the same pressure in each case and therefore the same number density-to the approximation that the ideal gas equa tion is obeyed. From part (c) the diffusion coefficients differ only through the effective molecular radii. Since D(CO; C2H4 ) = D(C 2H 4 ; N2 ) ,
Transport in gases
17.4 whence
N2 ) = 0·176 cm 2 S-1 . (d) Experimentally, all diffusion is mutual diffusion. Self-diffusion is a convenient term for the mutual diffusion of two distinguishable groups of molecules, where there is no significant difference between the groups in the properties which determine the diffusion coefficient (i.e. the molecular mass and the interaction with other molecules of either group). Thus at an experimental level the question does not arise. At a deeper level, diffusion is an irreversible process associated with an increase in entropy. The mixing (self-diffusion) of identical molecules does not lead to an increase in entropy. Hence one cannot, strictly, speak of a coeffi cient of self-diffusion. [This is the origin of the Gibbs paradox (2).] D(N 2 , N 2 )
= D(CO;
17.4 (a) To what extent can thermal conduction be treated as a random-walk process? (b) A pure gas has coefficient of thermal conductivity K, mass density p, and specific heat at constant volume C v per unit mass. By exploiting the analogy between Fick's law for diffusion [Problem 17.2(c)] and the heat conduction equation, extend the arguments of Problems 17.1 and 17.2 to show that K A2 -=~ pC v 6T' where Au is the effective root mean square free path for the transport of thermal energy. (c) To what extent can the phenomenon of viscosity be treated as a random-walk process? (d) If a pure gas has coefficient of viscosity 1/, show by an argument similar to that in part (b), that 1/ A2 _=:.J!.. P 6T' where Ap is the effective root mean square free path for momentum transport. (e) Hence calculate the ratio K/1/ for a gas and compare it critically with the data shown in Table 17.4.1, for three gases at NTP. C v is the molar specific heat at constant volume and R is the gas constant.
we have r(CO)+r(C 2H 4 ) = r(C 2H 4 )+r(N 2) ,
i.e. r(CO)
= r(N 1 )
.
In the corresponding expression to that in part (c) for the self-diffusion coefficient of nitrogen D(N 2 ; N 2 ), r 1 +r1 = 2r(N 2) = r(N 2)+r(CO) ,
383
Table 17.4.1. Gas
Molecular Weight
Ne N2 Kr
20'2 28·0 82'9
K (J cm- 1 S-l degK-l)
4'54 x 10- 4 2'43 x 10-4 0'89 x 10-4
1)
(g cm- 1 S-l) Cv/R
2 '98 1·67 2 ·33
X X X
10-4 10- 4 10- 4
(2) See, for example, K.Huang, Statistical Mechanics (Wiley, New York), 1963, p.153.
l' 5 2'5 1'5
384
17.4
Chapter 17
Solution
Thermal conduction is the result of the transfer of thermal energy from molecule to molecule in collisions. Like the number of foreign molecules in the case of mutual diffusion, thermal energy is conserved in a collision. Unlike foreign molecules, it is necessarily present in a gas in equilibrium at temperature T. Thus it is any local excess of thermal energy which spreads by a random-walk process. Further, a given excess of thermal energy can (classically) be indefinitely subdivided amongst the molecules. Hence the units which perform the random walk must be taken to be of infinitesimal size. These differences from the diffusion case should leave unaltered the asymptotic behaviour of the probability distributions, and therefore of the differential equations governing the transport, provided one substitutes 'excess thermal energy' for 'number of foreign molecules', and Au for A, where Au is the effective root mean square free path for the transfer of thermal energy. One expects that Au IA is of order unity, the exact value depending on the efficiency of thermal energy transfer in collisions. (b) The analogue of Fick's law of diffusion is Qx
-Du
17.4
Transport in gases
(d) Consider the transport of y momentum in the x direction, where x and y axes are normal Cartesian axes. The analogue of Fick's law is
aT
Qx = -DuPCI! ax
Qx
aT ax
-K-
by definition of K. Therefore K A~ -=D PCI! U = 67 Like thermal energy, a given momentum component is conserved in a collision. Again, classically, it is indefinitely subdivisible amongst the molecules. The transport of a given momentum component normal to its own direction only can be treated as a straightforward random-walk process. Parallel transport can occur only when divv =1= 0: thus density variations appear and the behaviour is dominated by the propagation of sound waves.
~ ax
'
where Pyx is the flux of y momentum in the x direction per unit area normal to the x axis, and Py is the density of y momentum. The diffu sion coefficient for transverse momentum, D p ' is given by A2 D p -- 67p ' where Ap is the effective root mean square free path for the process. Ap IA should be of order unity. Since Py = pU y , where v y is the mean molecular velocity in the y direction, we have
--Dpp ~ ax
PYx -
But
P
(1)
-'fl:::..J::.
ax
yx
by definition of 'fl, whence '!l=D P
p
~ 67
(e) From parts (a) and K CIIA~ CIIA~ -:;:;- = -A-:= M )..2 , 'j
But
-Dp
PYX
au
where Qx is the excess thermal energy flux per unit area in the x direc tion, and u is the excess thermal energy density. (It is assumed for simplicity that aulay = au/az = 0.) The diffusion coefficient for ther mal energy, D u , is given by A2u Du = 67 Now du = pcvdT, whence
385
P
p
where M is the molecular weight of the gas. A~/)..~ should be a number of order unity, the same for all gases with similar collision transfer processes for energy and momentum. From the figures given,
KM -C 'fl II
= 2·47 for Ne, = 1·96 for N 2 ,
= 2·54 for Kr,
i.e, for both the monatomic gases A;;)"~ is approximately 2· 5. For the diatomic gas the lower value reflects the difference between the transfer of the thermal energy associated with the internal (rotational) degrees of freedom, and that associated with the translational degrees of freedom.
386
Chapter 17
17.5
17.6
Transport in gases
Hence the effective rigid-sphere radius r decreases, and 71 varies more rapidly with temperature than Tv,., as observed. (The effect of the longer-range attractive interaction between 4He atoms is negligible at temperatures much above the critical temperature of about SOK.) (d) The mean atomic kinetic energy at temperature T is approximately kT. The force between two atoms with separation x is -C/x n , where Cis constant. Therefore their potential energy at separation x is given by C/(n 1)x n - I (n > I). The effective rigid-sphere radius r is determined by these two energy terms. The only dimensionless combination of r, C, and kT is a function of C/(kTrn I). Hence
17.5 (a) Use the results of Problems 17.3 and 17.4(b) to show that the coefficient of viscosity, at temperature T, of a gas of atoms which behave as rigid elastic spheres of radius r and mass m is approximately
(3mkT)Yl 241Tr2 (b) How should the coefficient of viscosity depend on the pressure of the gas? (c) The variation of the coefficient of viscosity of 4He gas with tem perature is given in Table 17.5.1. Explain qualitatively this temperature variation. Assuming that the force between two helium atoms is predomi nantly repulsive, and varies inversely as the nth power of their separation, use a dimensional argument to determine n from the data.
rex:
and r} ex:
Table 17.5.1. Coefficient of viscosity of 4He gas as a function of temperature.
15'0 29'5
75·5 81·8
171 139
291 197
457 268
665 339
1090 470
2
t,
~
p
6r
Tv,.+2/(n -I)
0·66 ± 0·02
whence n
r}
ex:
-+-= 2 n-I
Solution
From Problem 17.4(b),
Tv,.
I )I/(n-I) (T
From the data, I
TCK) 1/(gcm- 1 S-I) x 10 6
387
13·7±1·6.
[The value n 13 is used for the repulsive part of the familiar Lennard Jones potential. The dimensional argument was first given by Lord Rayleigh (3) .]
')0.,2 ~-=D
6r
'
where D is the coefficient of self-diffusion of the gas. From Problem 17.3(c) and (e), 0:
= 61Tn(2r)2
D
Hence
P
r}
~ 241Tnr2
(3kT)v,.
m
(3kT)Yl
m
'I
(3mkT)'h
(b) r} is independent of density, and hence of pressure, at a given temperature. (c) For a gas of rigid spheres r} ex: • From a plot of 19r} versus 19T for 4He gas, we have 19r}
~
slgT+ constant,
where s ;: O· 66 ± 0·02
> O· 5 .
4He atoms interact through a repulsive force which varies rapidly with their separation. As the temperature is raised the atoms have greater mean kinetic energy and therefore approach more closely at collisions.
'v
17.6 (a) Deuterium and helium, both of molecular weight 4·0, have coefficients of viscosity at O°C of 1·2 X 10-4 and 1·9 x 10-4 g cm- I S-I respectively. Estimate the ratio of their second virial coefficients Ba at this temperature, assuming that their molecular interactions may be treated as those of rigid spheres. [Note from Problem 9.9 that B2 is proportional to the molecular volume.] (b) An ideal gas of given molecular weight at a given temperature and pressure may be considered as the limit of a series of hypothetical imperfect gases with the same molecular weight, temperature and pressure. What is the theoretical value of the coefficient of viscosity of the ideal gas, considered as this limit? (c) The coefficient of viscosity of a gas may be determined from its rate of flow through a capillary tube of diameter d under a pressure gradient. Outline qualitatively the result of a series of such determina tions on the hypothetical series of gases. Cd) In what range of (mean) pressure would 4He behave like an ideal gas in this experiment, if d = 10-4 cm? (3)
Lord Rayleigh. Proc.Roy.Soc., 66, 68 (1900).
388
Chapter 17
(a) From Problem 17.5(a)
r
r(D 2 )
J
1· 2
2
X
10- 4
Ibe second virial coefficient B2 is proportional to the molecular volume for a rigid-sphere interaction (cf Problem 9.9). Hence the estimated ratio is
B(He) _ (1.2)' h ...__ 0 50. 1.9
I[
ff(V)dV = n ,
2
B 2 (D 2 )
-
(b) In the ideal-gas limit, B2 (and higher fore r ~ O. Hence the theoretical coefficient of [cfProblem 17.5(a)J 1/
~ fVf(V)dV = v= 0, is given IfV 21'( v) dv
;;
y-+O
i.e. 1/ ~ 00 as r ~ O. (c) 1/ ~ 00 because A. ~ 00 as r ~ 0 [cfProblem 17.3(a)]. In a tube of diameter d, the coefficient of viscosity is a valid concept only when A. ~ d. In this case, 1/ a:: l/r2. For A. ~ d the flow rate is determined by the collisions of molecules with the tube walls, and is independent of r. Thus the apparent viscosity (as judged from the flow rate) would reach a finite limiting value as r ~ O. (d) 4He would behave like an ideal gas in the pressure range for which intermolecular collisions were unimportant, i.e. for which A. ~ d. From Problem 17.3(a),
of state pRT
p::::: --::::: nkT M
where the symbols have been defined in previous required pressure range is kT
p ~ 41l'r 2d ~
61/ (kT)Yz
d
3m
= V- 2 = V2 .
In a collision between two atoms, the distribution of e is independent of their relative velocity, and the mean value of () is ¢. Calculate the average value (il) of Iill for the collisions of an atom with initial velocity u = 0, to first order in ¢. [<> denotes an average over collisions.] (d) Hence show that a group of atoms initially having zero velocity will begin to 'diffuse' outwards in velocity space, with a 'diffusion coeffi cient' Dv equal to A 2 V2¢2/ T , where liT is the mean collision frequency for an atom with zero velocity, and A is a dimensionless constant. [Use the results of Problem 17.1.] Solution
(a) In the centre-of-mass reference frame the atoms have initial veloci ties ±WI = ±t(V'1 -vd. From conservation of total momentum, the final velocities also are equal and opposite. From conservation of total kinetic energy the magnitudes of the initial and final velocities are equal. Hence the initial velocities are equal and opposite, and so are the final velocities, and all four lie on a sphere in velocity space of radius WI' centred at the centre-of-mass velocity.
1 A. ~ 41l'r 2n
From the ideal gas
'
since
Hence the
~
where
2W2
W2
W2
=
! Iv'l -VII
WI
and =
WI W 2 cos
Resolving ~ parallel and perpendicular to -Wl(l
With the given figures for 4He this gives the pressure range ~
5 x lOs dyn cm- 2
WI,
is the relative velocity after the collision, i.e.
WI' W 2
(3mkT)'h, 1/ ~ 241l'r2
p
~
389
17.7 (a) Two atoms of equal mass, with initial velocities v 1 and respectively, collide elastically. What velocities are possible after the collision? (b) If the angle of scattering in the centre-of-mass reference frame is e, what is the change .6. in the velocity of one of the atoms? (c) A spatially uniform monatomic gas of number density n has an isotropic velocity distribution f(v) with the following properties:
Solution
r(He).
Tr:mc:nnrt in gases
17.7
17.6
tatm .
ill
WI
e.
we 0 btain
cose) ,
390
Chapter 17
17.7
(c) To first order in 0, A1 = w10,
All
17.8
Transport in gases
17.8 (a) For the situation of Problem 17.7(c), calculate the average value of ~ for the collisions of an atom with low velocity u, to second order in q'>. (b) Hence show that a group of atoms with low velocity u will acquire through collisions an average acceleration
O.
Therefore
AI=w,O,
and after averaging over 0
IAI = w1q'> •
2
(Ii> = _ 8 q'>2 -7
The probability per unit time that an atom of velocity zero is struck by one of velocity v is proportional to v f(v). Thus the mean value of IA I is
f
fAVf(V)dV (A)
(averaged over 8) .
vf(v)dv
Since
WI
391
U
where B is a dimensionless constant. (c) Assuming that the expressions obtained for Dv and (iJ> are correct for arbitrary u, show that in general there is a flux of atoms (per unit 'area' of velocity space) given by
(A 2V2!! + B2Vf)
=v 1 f
(A) = -q'>
2
v2f (V)dV = I
J72
fVf(V)dV
Now Ii = (v 2
l'" = .!:::. A'
,
per unit volume of real space per unit time. Hence show that the velocity distribution function f(v) in the equilibrium state of the gas is the Maxwellian distribution of Problem 3.1. (e) From your results identify fo and estimate t* in the following common approximation for the effect of collisions on f(v):
of
where A' is a dimensionless constant. Hence
f-f o
ot=-~
= !A'q'> V . (d) Because the gas is isotropic, .1 is random in direction. Hence an atom with zero velocity begins to perform a random walk in velocity space with steps of average length (.1>, occurring with mean frequency liT. In consequence its velocity departs from zero, and the step length may change as the random walk progresses. The initial diffusion coeffi cient is, from Problem 17.1,
D :::::;
whence 2
[This is the relaxation-time approximation (5). (f) Estimate the order of magnitude of t* in a gas of classical rigid spheres, for which all scattering angles are equally probable in the centre of-mass reference frame. Solution
(a) Consider a collision of two atoms with initial velocities u and v. Let w = Hu -v). Resolve the change A in velocity of the first atom into vectors parallel and perpendicular to;: Since..6.1 may lie in any direc tion in the plane lw with equal probability,
Dv = A q'>2J72 7
where A is a new dimensionless constant. [Some processes in metals where such a random-walk treatment of collisions in velocity space is particularly appropriate have been discussed recently (4).] (4) A.B.Pippard,
hoc. Roy. Soc. , A30S, 291 (1968).
(~1> =
.
0;
from Problem 17.7(b), All = -w(l- cosO) = 1W02 to second order in O. Since the gas is isotropic, (..6. > = <..6.11 > must be parallel to u: let this ~ be the x-direction. Then (A>
(All'
ulu> = -i<02>=
(5) It is discussed further in F.Reif,loco cit., Ch.13.
-i(8 2 )(u -(v x » .
392
Chapter 17
17.8
From here the result follows by a dimensional argument, or from the following more detailed analysis:
17.8
Transport in gases
The net drift calculated in part (b) causes a flux through velocity space given by
B21jJ2
f(v) (v) = --vf(v) ,
fV xIv - u If(v)dv
r
(v)=-"----=----
For small u,
in the same units. Hence the result for the total net flux follows. (d) In equilibrium there is zero net flux for all v. From the result of part (c), af/avllv. This is consistent with the assumed isotropy of f(v), i.e. f(v) is a function of Iv I only. Thus
f,v -u If(v)dv
x
av 2 Vx = v -u-+O(u ) = v -u-+O(u 2). ax v
Iv-ul
A2J72
Hence fIV-Ulf(V)dV = n[V+O(u 2 )]
(17.8.1)
vx/v = O.
fVx
f[Vxv -u v! + O(U 2)] f(v)dv = n [-u( v~) +O(u 2]
=
Also
(~)
C~;)
!(v:) = 1v.
Therefore (v ) x
Hence (~) _
2). = -1(8 4
B 2v 2 In f = - 2A 2J72 + constant
or
Since f(v) is isotropic,
(V~2)
f(v)
f(v)
(u)
=
I
r(u) = -;:-+O(u
where r = r(O). Hence (Ii) =
( mv2)
constant x exp - 2kT .
ff(V)dV
(~)/r(u)
,
where I/r(u) is the mean collision frequency for an atom with velocity u. I/r(u) is proportional to the integrated molecular flux given by Equation (17.8.1). Hence _
=
The constant can be determined, from the normalisation condition
where B2 is a dimensionless constant. (b)
B2V2)
constant x exp ( - 2A 2J72
But v 2 = J72 by definition. Therefore A 2/B2 = j. Using the theorem of the equipartition of energy to relate J72 to the temperature T,
•
=-B 21jJ2 U+O(U 2 ) ,
=
It is easily shown that, with this f(v), 3A 2J72 v 2 -- B2
nuv
= ------::. + O(u 2 ) 3nv
~U+O(U2) 3
af +B 2vf = O. av
i.e. ,
since, by symmetry, Iv -u If(v) dv
393
2
) ,
B21jJ2 ---u+O(u 2 ). r
(c) The diffusive flux of atoms through velocity space is given by the analogue of Fick's law (in three dimensions) as -Dvaf/ov atoms per unit 'area' of velocity space per unit volume of real space per unit time, where Dv has been calculated in part (a).
as
=
n ,
( )
'/'
n 2::T
.
Thus the equilibrium distribution is Maxwellian. [The result is correct despite our approximations.] (e) fo(v) is the equilibrium form towards which f(v) relaxes through collisions, i.e. the Maxwellian distribution from part (d). (For a moving gas, modification of (d) is required.) The relaxation time t* is the time necessary for local departures from fo(v) to be communicated, by a diffusive process, to the whole velocity distribution, which has linear dimensions of order V in velocity space. From Problem 17.1 (b) and (c), an initially localised irregularity is distri buted after time t, over a region of mean square width of order 2Dvt.
Chapter 17
394
17.8
Hence
2Dut* : : : ; V2
i.e.
t*::::::;
r
r ¢2
~-
(f) For rigid spheres, ¢ ::::::; I radian. Hence the condition ¢ satisfied. To order of magnitude, however,
-<
I is not
Transport in gases
17.9
(c) In an electric field the ionic velocity distribution is distorted from the equilibrium Maxwellian form given in Problem 17.8(d), because of the flux calculated in part (a). Collisions of ions with atoms of the stationary gas tend to restore this equilibrium distribution through the flux calculated in Problem l7.8(c). Hence the net flux is (remembering that B2 3A 2)
-A2¢2 - - (Of V2_+ 3vf) r OV
t* : : : ; r.
is the value of t* usually assumed in the relaxation-time approxi mation.] 17.9 (a) Atoms of a gas with velocity distribution function f(v) are acted upon by a uniform force F. What is the resulting flux (i.e. that due to F alone) of atoms per unit 'area' of velocity space, per unit volume of real space per unit time? (b) Using the result of Problem 17.8(d), write down the equilibrium distribution function for a gas at temperature T with number density n, moving with uniform mean velocity v(-
395
eEf +
m
In the steady state, tne net flux is zero. Hence
of
V2 av =
(eEr
)
mA2¢2-3v f.
For the ionic velocity components normal to E this equation is identical with that in Problem 17.8(d) and these components vanish. For the ionic velocity component u parallel to E we have
V2af 3
(eEr
au
)
3mA 2¢2-U f(u).
If the electric field is small, ! V2 is approximately equal to its equilibrium value of kT/m (theorem of equipartition of energy, Problem 3.6). Hence mu( eEr u Inf kT 3mA 2¢2 . i.e. eEru) mu2 f constant x exp ( - 2kT+ 3kTA 2¢2 .
Solution
(a) Under a force F each atom acquires an acceleration v = F/m. Hence the required flux is H(v) = f(v)F m in the given units. (b) The required distribution function is obtained by shifting the origin of velocity coordinates from zero to Y, i.e. f(v) = constant x exp
l-
2;T(V-V)2] ,
(d) This expression may be rewritten as f
This transformation is correct for v are important.
-< c.
= n .
For large v relativistic effects
---''-:-:--=:-''
where
_ u
eEr
= 3mA 2¢2
Hence the mean ionic drift velocity is
u=
where the value of the constant is unchanged, since jr(V)dV = ff(V)d(V
constant x exp [ -
eEr
(e)
er JJ. = 3mA2¢2
396
17.10
Chapter 17
17.10 (a) A gas of atoms of mass m, subject to no externally applied has a velocity distribution function which is a function of position r, i.e. its mean temperature, density and velocity vary with position. If the velocity distribution function at time to is f(v, r, to), what is the distribution function at time t I, if the effect of collisions is neglected? (b) What differential eGuation governs the rate of change of the distri bution function, if it is unaffected by collisions? In the steady state, what is the distribution function f(v, r), ifit is assumed that collisions establish local equilibrium at each point? Show that this distribution function is not, in general, an exact steady solution of the differential equation set up in part (b). When is it an approximate steady solution? (e) Use the relaxation-time approximation of Problem 17.8(e) and the differential equation of part (b) to set up the following better approxi mation for the rate of change of f(v, r, t): of af f-fo
at
-v'
(d) In the steady state of/at = O. Therefore v . of/or = 0, for all v, whence af/or = O. The function written down in part (c) does not satisfy this condition. Therefore it is not a steady solution of the differential equation, but it becomes a better approximation as of/or -l> 0, i.e. as the spatial non uniformity decreases. (e) When the effect of collisions is considered, the equation set up in part (b) becomes -af = -v' of +
(Of)
at
at
Solution
(a) At time tl an atom which was at r at to with velocity v will be at r' = r+ v(t I - to), still with velocity v, if collisions are neglected. Since the number of atoms remains constant, r', tl )dr' dv = f(v, r, to)drdv, where dr' is the volume element at time t 1 corresponding to dr at time to. Since v is constant, dr' = dr. Therefore f(v, r+v(tl -to), td-f(v, r, t ... )
=
O.
(b) Let t 1 -l> to. Then the equation becomes of of v'-+-=O
or ot
'
i.e. of/at is equal to -v' of/or, a term due entirely to the unimpeded motion of the atoms with the given distribution (c) If there is local equilibrium at each point, the distribution is every where Maxwellian with the appropriate local values of temperature, density and mean velocity. We have therefore
af
of f-f o
-v' or -T
.
Since collisions tend to establish local equilibrium quickly over dis tances of order A (the mean free path) such a local approximation is good when [(v. r, t) varies slowly over such distances, i.e. as af/or -l> O. 17.11 (a) A stationary gas of atoms of mass m at uniform temperature x = 0, from which foreign atoms of the same mass are released at a steady rate. Hence a steady concentra tion gradient an/ax exists, where n(x), the number density of foreign atoms at distance x from the wall, is everywhere small in comparison· with total number density of the gas. Write down the steady velocity distribution function fo(v, x) for the foreign atoms, assuming that local equilibrium exists. Calculate the first-order correction to this distribution, using the method of Problem 17.1 O(e). (c) Hence calculate the mean velocity of the foreign atoms, to first order. (d) Hence calculate the coefficient of mutual diffusion D for the foreign atoms in the gas. Compare the result with that of Problem 17.I(c). Why is the present result more valuable?
T is bounded by a wall in the
Solution
(a)
- 2kT . ( m)';' exp (mll2)
r m J'I exp {m[2kT(r) v -V(rW} . 2
fo(v,x) = n(x) 21rkT
f(v, r) = fo(v, r) = n(r\ 211'kT(r)
(6) Further discussion of the equation and its approximations is given in K.Huang, Statistical Mechanics (John Wiley, New York), 1963, Ch.5 and 6, and in F.Reif, loco cit., Ch.13.
collisions'
The next approximation after (b) is to assume that the second term on the right hand side depends only on local departures from the Maxwellian distribution fo(v, r). Thus, in the relaxation-time approximation,
or-T
In what circumstances is this approximation a good one? [The equation under (e) is one approximate form of the Boltzmann transport equation. The equation under (b) is a form of the 'collision less' Boltzmann equation (6).
397
Transport in gases
17.11
[Note that v
0 in this approximation.]
398
Chapter 17
17.11
(b) Let the first-order corrected velocity distribution function be fl' Then, from Problem 17.1 v'
since of/at = 0 in the steady state. To first order, afdar can be replaced by afo/ar on the left hand side. Now afo V' Or =
Vx
m )'hexp (mv2) - 2kT .
fl -fo
t*
n
i.e.
t*vx an) = fo(v,x) ( l-~n~ax
(c) IJ x
Now
=
.
*f
eD
fl.
= kT
Solution
(a) The mean ionic flux due to the electric field is nfl.E. That due to diffusion is -Dan/or. Hence, in the stationary state,
vxf1 (v.X)dV.
an
JVxfodV = O.
vx
1 = -t* - an --
n ax n
f
o2f x 0
dv where VCr) is the electrostatic potential at point r. Therefore
and
2
l'l:::
n
('AIr)?, this expression is essentially the same as the previous one.
been taken into account in its derivation, and because t* is generally a well-defined quantity, whereas 'A can be defined precisely only for atoms which interact like rigid bodies.
O.
= noexp (-~),
where no is a constant. However, the stationary state is one of equilib rium in the given electric field. Therefore the ionic number density is given by the classical Boltzmann (canonical) distribution n
r, for atoms scattering through large angles at collisions, and
It is more valuable, however, because the atomic velocity distribution has
-1f+constant
av -nWar
i.e.
ox
0
fl.V
Inn
using the theorem of the equipartition of energy (Problem 3.6). It is easily verified that Vy = Ii; = O. (d) The flux of molecules in the x direction per unit area per unit time is t*kTan an
= O.
av
E
t*kTan mn ax '
Since t*
'
where k is the Boltzmann constant. [This is the Nernst-Einstein relation (c) By applying the Nernst-Einstein relation to the results of Prob lems 17.9( e) and 17.11 (d) find a more exact relation between t* and r than the estimate obtained in Problem 17.8(e).
nfl.E-D ar
Therefore
399
17.12 (a) A gas of atoms of mass m at temperature T contains a small proportion of ions with the same mass, carrying charge e. The mobility of the ions in the gas is fl., their coefficient of mu tual diffusion is D, and their number density at the point r is nCr). If an electric field E(r) exists in the gas, what is the condition that the mean ionic velocity is everywhere zero? (b) Assuming that the ions obey classical statistics, show that
an ( ax 2rrkT
Therefore, to first order,
f 1 (v,x)
Transport in gases
17.12
= noexp (- ~~),
where e V(r) is the potential energy of an ion at r. By comparison, the Nernst-Einstein relation follows. (7) This is discussed further in F.Reif,
AlI3, 226 (1952).
loco cit., Ch.15, and in P.T.Landsberg, Proc.Roy.Soc.,
,
Chapter 17
400
17.12 )!
(c) From Problem l7.9(e), er Jl = 3mA2rJ>2
From Problem l7.11(d) D
t*kT
18
Transport in metals (J)
m J.M.HONIG
whence Jl _ e r jj - kT3A 2rJ>2 t *
This gives t*
r
IThe previous estimate was t* :::::; r/A 2rJ>2.J GENERAL REFERENCES
A comprehensive account of elementary (mean free path) kinetic theory is given in: J. Jeans, Kinetic Theory of Gases (Cambridge University Press, Cambridge), 1940. The following textbooks deal with the Boltzmann equation: E. A. Desloge, Statistical Physics, Part III (Holt, Rinehart, and Winston, New York),1966. F. Reif, Fundamentals of Statistical and Thermal Physics (McGraw-Hill, New York),1965,Ch.12-14. K. Huang, Statistical Mechanics (John Wiley, New York), 1963, Ch.3-6. The classical treatise on the subject is perhaps S. Chapman and T. G. Cowling, Mathematical Theory of Non-uniform Gases (Cambridge University Press, Cambridge), 1939. For recent advances (c.g. the treatment of dense gases) consult: L Prigogine (Ed.), Pmc. International Symposium on Transport Processes in Statistical Mechanics (lnterscience, New York), 1958. W. E. Brittin (Ed.), Lectures in Theoretical PhYSics, VoLlX C (Kinetic Theory) (Gordon and Breach, New York), 1967.
(Purdue University, Lafayette, Indiana)
INTRODUCTORY COMMENTS
18.0 In Chapters 18 and 19 we consider problems which deal with the response of mobile electrons in solids to externally applied electric fields, magnetic fields, and temperature gradients. In the standard approximation utilized here the electrons are considered to be independent particles subject to Fermi-Dirac statistics. In zero order approximation the solid is viewed as a 'box' or container, within which the electrons move as a 'gas'; this is the so-called Sommerfeld modeL The effect of the solid is introduced more realistically in first order approximation by regarding the periodic potential of the lattice as a perturbation on the nearly free electrons. Alternatively, one may proceed from the opposite assumption: the electrons are considered rather tightly bound to the atomic cores in the solid but able to move through the lattice by virtue of some overlap among orbitals associated with adjacent atoms. In either case the following conclusions are reached: there is an alternation between closely spaced energy levels (energy bands) and forbidden energy ranges (energy gaps), corresponding to ranges where the Schr6dinger wave equation does or does not admit of solutions. The demarcation line between allowed and forbidden levels is termed a band edge. The if; functions can always be represented as free electron wave functions modulated by a function which has the lattice periodici ty. Of cardinal importance is the specification of the energy 8, of the electrons in the solid and of the dependence of 8, on independent variables and on parameters. As in the case of free electrons the energy depends on the wave number vector k. In what follows we shall always treat a very special case, namely bands of standard form for which h2 k 2 8, =: 8,c + 2m = 8,c + EO (18.0.1) where 8,c is the lower band edge and the second term is a kinetic energy formally identical to the expression obtained for free particles, in which m is the mass of the particle, h Planck's constant, and h == h/27r. (\) In Chapters 18 and 19 Boltzmann's constant is denoted by modulus of the wave vector k = Ik I.
401
Ii;
to avoid confusion with the
r
402
Chapter 18
18.0
However, in the present context, m is not the free electron mass but, rather, an effective mass which depends on the band structure of the solid. By this method it is possible to dispense with the explicit consideration of interactions of charge carriers with the lattice. The dynamics of the particle is introduced as follows: it may be shown from very general considerations that the velocity of an electron in the crystal is specified by Vk &(k) (18.0.2) v h where Vk is the gradient operator with respect to the independent variable k. The time derivative Ii of the wave number vector is related to the externally applied force F through the equation Ii = F/h. The interaction of an electron to an externally applied field is handled in the present context by setting up and solving a Boltzmann transport equation, in which the electric, magnetic, and temperature fields appear explicitly as parameters. In a perfect periodic lattice the electron encounters no resistance to its motion; however, impurities, lattice vibrations, and other types of imperfections introduce scattering mechanisms which must also be handled through the Boltzmann equation. A standard procedure here consists in introducing a relaxation time T related to the mean free path I by T 1/\ v I. This approach can be shown to apply under certain rather restrictive conditions, and the results are equivalent to the linear theory of irreversible thermodynamics. Adjustment of T along the lines shown in later problems simulates different scattering mechanisms. As will be shown in the last problem of this section, a set of band states that is very nearly completely occupied may equally well be described in terms of the remaining unoccupied band states which may be associated with fictitious particles, termed holes. These may be regarded as charge carriers that have a positive charge and a kineti," energy, and Fermi level which is referred to the upper band edge. Thus, for holes, h2 k 2 (18.0.3) & 2m :;: &o-e
18.1
Transport in metals
403
(a) The electrochemical potential t for electrons is defined by
t :;: f.l n
(
-e!{:ls
(18.1.1)
where f.l n is the chemical potential, -e the charge on the electron, and !{:Is the electrostatic potential. Express the chemical potential in terms of the activity of the electrons in the system, which is related to the concentration of electrons, C n , by an = 'Yn C n , where 'Yn is the activity coefficient. Show that for a uniform material at constant temperature the gradient of the electrochemical potential per unit electronic charge coincides with the electrostatic field. (b) Let I" be the length of the sample specimen in the direction X = x, y, z; let v:;: Vet/e), U:;: V!{:Is; let fA and CA be the current density and heat flux along the direction X; and let T be the temperature. Referring to Figure 18.3.1 (page 414), note the orientation of the rectangular parallelepiped relative to the Cartesian axes; let L be the distance between the voltage probes at C and D, and let Hz denote the magnetic field, aligned along the z direction. The following definitions will now be introduced for uniform isotropic materials under the open- . circuit conditions fy = fz 0 and for the isothermal conditions VyT = VzT = 0: 1. The electrical resistivity Vx(t/e) . p= J wlthVxT=O. (18.1.2) x
Show that this reduces to the common formulation of p in the terms of the sample resistance Rs. 2. The Hall coefficient = Vy(t/ e ) With . _ R - J 11 VxT - o. (18.1.3) x
z
Show how this quantity is related to the potential difference across y, and to the current along x. Demonstrate why it is desirable to make the samples as thin as possible along the magnetic field direction. Interpret the results. 3. The Seebeck coefficient _ VxG/e). _ IX= VxT wlthfx=O. (18.1.4)
where &v is the upper band edge. The reader should carefully distinguish in subsequent sections between the total energy & of a charge carrier and its 'kinetic energy' h2k2/2m which is designated as e. Finally, a specimen is said to be n type or p type according as electrons or holes are responsible for conduction .
State whether IX may be re-expressed in terms of electrostatic potential differences; discuss the significance of Equation (18.104). 4. The Nernst coefficient Vy(t/e ) . _ N H V T Wlthfx = o. (18.1.5)
18.1 This problem is designed to acquaint the reader (a) with the concept of electrochemical potential which is extensively used in Chapters 18 and 19, and (b) with methods commonly used to describe on a macroscopic thermodynamic scale the response of charge carriers in conductors to externally applied forces.
%
x
State whether N may be re-expressed in terms of electrostatic potential ;I
18.1
Chapter 18
404
differences; discuss the significance of Equation (18.1.5). 5: The thermal conductivity is defined by
-c
" == V'
x
Twi th Ix
0.
(18.1.6)
Interpret this relation. Solution
(a) From standard thermodynamics we find that I1n = 11;; + RTlna n
(18.1
where an is the activity of the electrons in the system under study, and 11~ is the standard chemical potential of the electrons when an = 1. Taking the gradient of the potential (18.1.1) at constant temperature we find RT (18.1.8) V'(~/e) = -eV'(lna n ) V''Ps where V' 'Ps is the gradient of the electrosta tic potential. Since T is constant and the material is uniform, V'lna n == 0; then U V V'(~/e) -V''Ps -1=E (18.1.9) where U is the electrostatic potential difference, L the distance over which this quantity is measured, and E is the electrostatic field. (b) 1. Let Ix be the current along the x direction and utilize the Ixllvlz and defining symbols introduced earlier. Then, with Ix Iylz == Ax we may rewrite the definition (18.1.2) as _ Vxlylz _ Uxlyl z _ UxAx I PI I L I L (18.1. x
On introducing Ohm's law,
x
x
Ux /Ix = R s ' we find
P=
z
Transport in metals
405
to the definition (18.1.3) based on a phenomenological approach to irreversible thermodynamics, the application of a current along the positive x direction, and of a magnetic field along the positive z direction, leads to the establishment of a gradient in electrochemical potential (or of a difference in electrostatic potential in the case of uniform materials) along the y axis. The magnitude of this gradient (or difference in potential) is given by the Hall coefficient under the assumed boundary conditions. 3. Even for uniform materials we may no longer set V'x(~/e) = -Uxllx because in the presence of a temperature gradient it is no longer appropriate to drop the first term on the right in Equation (18.1.8). According to the definition (18.104) based on the phenomenological approach to irreversible thermodynamics, the establishment, within a sample, of a temperature gradient, subject to the conditions detailed earlier, leads to the existence of a gradient of electrochemical potential Fermi level) within the sample, whose magnitude is specified by (1'. In view of Equation (18.1.8) this results in the concomitant establishment of a gradient in activity (or concentration) and of an electric field for electrons within the sample. 4. For reasons discussed above, it is not permissible to replace V'y(~/e) with -Ully here. According to Equation (18.1.5) based on the phenomenological approach to irreversible thermodynamics, the establishment of a temper ature gradient along x and of a magnetic field along z leads to the presence of a gradient in electrochemical potential along the y axis. For reasons given in Part 3, this produces both an electric field and a non uniform distribution of activity (or charge density) along y. The magnitude of this effect in isotropic materials is specified by N. 5. According to irreversible thermodynamics the thermal conductivity measures the heat flux in a material in response to an imposed tempera ture gradient. Note that the heat flow is in a direction opposite to the temperature gradient.
(18.1.11)
L
which is the required formulation connecting p and Rs· 2. In terms of the symbols introduced earlier, we have from definition (18.1.3) and Figure 18.1.1 Vylylz Uyl z R I R I R (18.1.12) x
18.2
x
18.2 In Problem 18.1 we considered the phenomenological description of the response of charge carriers to applied forces. Here we attack the problem from the microscopic viewpoint. It is assumed that the model described in the introduction holds and that the Boltzmann transport equation also alluded to there has been solved (2) for the distribution function
z
where lz is the thickness of the sample. Note that according to Equation (18.1.12), for a given material with fixed R, the potential difference is proportional to l/lz; hence, the thinner the sample, the larger 1Uy I, and the easier it is to measure Uy • Difficulties arise, however, when lz becomes comparable to the mean free path of the electrons. According
A
to
a ) to - V' '" (ae
(18.2.1 )
(2) T.e.Harman and J.M.Honig, Thermoelectric and Thermomagnetic Effects and Applications (McGraw-Hili, New York), 1967, pp.178-183.
18.2
Chapter 18
406 with
1 +exp [(€ - 1l)lkT I
(18.2.2)
and q, = T[F+(ZeTlmc)F x H+(eTlmc)2H(F· H)] 1+ (eTHlmc)2 -
(18.2.3)
and F
ZeVr(sle)- (€ - Il)Vr T
T
(18.2.4)
The quantity A represents the probability that an electron with wave number vector k is actually encountered in the crystaL As is seen from Equation (18.2.1), this quantity is given in terms of the equilibrium Fermi-Dirac distribution function (18.2.2) and further involves a term which represents in first order the departure from equilibrium. Here v is the carrier velocity, € the energy, and Il the Fermi energy relative to the lower band edge if one deals with electrons (Z = I) or relative to the upper band edge if one deals with holes (Z = + 1); K is Boltzmann's constant, and T is the temperature. The correction term in Equation (18.2.1) also involves a function q, given by Equation (18.2.3), in which the electric charge 1e I, carrier mass m, velocity of light c, applied magnetic field H, and relaxation time T appear explicitly. q, also depends on a quantity F which in turn involves the spatial gradient of the electro chemical potential per unit charge V r (S Ie) and of the temperature, Vr T. It is through the quantity F that the externally applied forces are introduced into the problem. As stated in the introduction, the resistive effect of the medium enters through the relaxation time T. A detailed theoretical analysis shows that where this concept is applicable the relaxation time is specified by T = To€r- v,, where To is a collection of constants and r is a scattering parameter which has the values 0, 1, or 2 according as scattering through acoustic vibrational modes, optical vibrational modes, or ionized impurities predominates. In proceeding with the problems cited in this section the reader should note how the function (18.2.1) is used to construct an expression for the current density and heat flux in (a), and how the definitions of Problem 18.1 are used to formulate the transport coefficients in terms of a certain set of integrals, in (b). These integrals are then evaluated under the special set of conditions referred to in (c) and (d). On the basis of the above: (a) Derive phenomenological equations which specify current and heat flux in a crystal subjected to magnetic fields and to gradients in electro chemical potential and in temperature. Utilize the distribution function specified above.
18.2
Transport in metals
407
From (a) identify in terms of appropriate transport integrals the following transport coefficients: 1. the resistivity for H = 0; 2. the resistivity for H =1= 0; 3. the Hall coefficient; 4. the Seebeck coefficient for H = 0; S. the Seebeck coefficient for H =1= 0; 6. the Nernst coefficient; 7. the thermal conductivity for H 8. the charge carrier density. (c) Specialize part (b) as follows: introduce the mobility defined by u eTlm. Further, specify the relaxation time through the assumed relation T To€r- y, referred to in the introduction. Finally, take the limit Hz ~ 0 and apply the limiting case of either classical (-11 ;p 1) or highly degenerate (11 ;p I) statistics in Equation (18.2.2). Tabulate the resulting transport coefficients in terms of atomic parameters. Note that for highly degenerate statistics and for r = 0 the results turn out to be identical with those based on the Sommerfeld model. (d) Repeat (c) for the limit Hz ~ 00. Solution
(a) The rate of transport of charge past unit cross-section is given by
"r Zevdk 41TzejvkA d k .
J
=
3
3
(18.2.5)
On the right hand side the summation over discrete k is replaced by the integration
4~3 fd k; 3
the numerical factor arises from the counting
of states with discrete pseudomomentum hk (3). Now write out the dot product in Equation (18.2.2) and switch from Cartesian to spherical coordinates in k space; the latter step is permissible if the material under study is isotropic. Thus, we replace v x , v y , V z with v,: == V, Ve, Vq,; we then introduce expression 08.2.2) in Equation (18.2.1) and use kr == k, Ok,
-
4zef"" (de)2 (a/o) 3h2 0 q,A dk a€ k 2 dk.
(18.2.6)
We now specialize to the 'transverse case' where the 'forces' F, subject to experimental control, are restricted to lie in a plane normal to the magnetic field which is aligned with the z axis. Introduction to Solid State Physics (Wiley, New York), 1966, 3rd Edition, p.20?; J.S.Blakemore, Solid State Physics (Saunders, Philadelphia), 1969, p.1S?; T.C.Harman and J.M.Honig, loco cit., p.1S?
(3) C.Kittel,
408
18.2
Chapter 18
Specializing Equations (18.2.3) and (18.2.4) in this manner, substi tutin,g (18.2.4) in (18.2.3) and the resultant for \(I~ in Equation (18.2.6), we then obtain by straightforward algebraic manipulations: _ 2 I 3 . Ze Jx - e KIV'A~ e)+Ze HzGIV'y(~/e)+T(KIJ.LB -K2)'Vx T
e 2H
+--;j-(G1J.LB
(18.2.7a)
G2 )V'y T
e2 H Jy == -Ze2HzGIV'A~/e)+e2KIV'y(Ve)--T(GIJ.LB -G2rV'XT Ze
Transport in metals
where ~ may be identified with the electrochemical potential (18.1.1), and where we have introduced the general transport integrals 4 r~ e i - 1rk 2 af, de (18.2.8a) Ki == - 3h 2 I +w 2r2 dk de
a:
Jo
4 f""e i 1r 2 k 2 afode 2 - 3h me 0 1+ w 2r2 dk de
ae
p
= 0 = e 2K 1 •
1e2; 1 1
(18.2.8b) and V'y(~/e)
w==
~ ek Vt!k 4~3 fekvt!k d
3
k.
(18.2.9)
This leads to the relations _
2
.
Cx - ZeK2V'A~/e)+e HzGzV'y(~/e)+ +
ZeHz
---r- (G2J.LB - G
3 ) V'y
T
K2J.LB -K3 T V'x T (18.2.lOa)
(18.2.1 Ob)
Observe that Equations (18.2.7) and (18.2.10) specify the transport of
_ Ze 3 G1 2 HzJx - (e KJ )2+(e 3 G1Hz )2
(l8.2.12b)
4. The Seebeck coefficient is given by a V'x(Ve)/V'x T = V'y(Ve)/V'y T with Jx = Jy = O. Since we also set Hz = 0, we see that either Equation (I8.2.7a) or (l8.2.7b) yields
a
J.
Z [K2 eT Kl - J.LB
(18.2.13)
5,6. The extension of the above procedure to non-vanishing fields is Jy = V'yT = 0 in Equation (18.2.7) and straightforward. Set Jx eliminate either V'y(~/e) or V'xC~/e) from the resultant pair of equations. One can readily determine
a(Hz )
, 2 ZeHz Cy = -e H'ZG2V'X(~/e)+ZeK2V'y(~/e)-r-(G2J.LB -G3)V'xT
2 + K J.LB - K3V'y T .
R(Hz )
(I8.2.8c)
It should be noted that this particular formulation can be used only for materials characterized by parabolic band shapes in which the effective carrier mass m is a constant. A parallel analysis may be carried out for the heat flux C due to the motion of charge carriers: C
(I8.2.1l)
2,3. The extension of this procedure to non-vanishing fields is straight forward. We set V'xT V'yT = 0 in Equations (l8.2.7a,b) and Jy = 0 in (l8.2.7b). This permits us to eliminate V'y(~/e) in favour of V'X<~/e) or vice versa in (l8.2.7a,b). We then find (e 2 K ·)2+( 3G H )2 o(Hz ) Z (18.2.12a)
and
ZeHz me
409
charge and of heat in response to externally applied gradients of electro chemical potential and of temperature, and thus represent phenomeno logical relations. (b) Using the boundary conditions of Problem 18.1 and the definitions listed there, we can now determine the various transport coefficients in terms of the integrals K j and G( 1. The conductivity is given by lip == 0 Jx/V'A~/e) Jy/V'y(~/e) with V'x T = V'y T = O. With Hz 0, Equations (18.2.7) show that J~ = e2KIV'~(~/e) (" = x or y), whence
(l8.2.7b)
+y(K1J.LB -K2 )V'y T
Gi
18.2
=
V'A~/e)
Ze 3 (KJJ.LB K2)K1 +Ze SHi(G 1J.LB -Gz)G J V'xT = T[(e2KJ)2+(e3GJHz)2] (18.2.14) 4 _ V'y(~/e) _ e (Kz G1- K1G Z ) (18.2.15) N(Hz ) - V'x T - n(e 2KJ)2+(e 3G1Hz )Z]
7. To determine the thermal conductivity in zero field, we shall set Jx = Jy = V' y T = Hz = 0 and define Ke = /V'x T. On applying the boundary conditions to Equations (18.2.7a) and (18.2.lOa) one can eliminate V'x(~/e) between them and thus express Cx solely in terms of V'x T.
18.2
Chapter 18
410 This yields
(I<.I
= 1<.) + I<. e
1<.)
Transport in metals
+ K 3 K 1 - K 22
(18.2.16)
TKI
L(UnXI)
= N~r(l+n(r--!)+~)
8. The charge carrier density is given by
n
2 f~
= 81T3
I f~ 2 0 lod k = 1T2 o/o k dk
a/o) k dE; (18.2.17) \-a;
1 f~(
3
31T2
0
L~g(U'H,e)k3(e)(- ~)dE.
==
where f( q) is the gamma fun ction
3
the term on the right is found by integrating j~k2 by parts; the term lok 3 /31T 2 vanishes at both limits of the integral. (c) To evaluate the quantities listed in (b), we introduce the 1llUUlIllY u eT/m and the relation (dE/dk) h 2 k/m. We further utilize the Y, assumed relation for the relaxation time T = TOEr-'Iz = To(kTtwhere x e/ kT. Next, we define a new transport integral L(g)
==
Kj
=
rU+u JP/e 'u ] 2
2
G· = -2I-2L 31T
J
ee
I+
Table 18.2.1.
(
Hz --* 0,
Transport coefficient
x == itT'
1/
Il kT'
[,(U»)
_
L(I)
U
Transport coefficient
I 3rr 2L (l )
0(0)
C:i)L(U)
= ne~~:;
(l8.2.20b)
Expression 2rrm1<:T)%
2( · · h2
e1J
(18.2.19)
1 e1Jr(r+2)
neu ==
3h 3
R(O)
I 3vrr -Ze-nc x 4 -;::---,
a(O)
z/r+2-1/)
1<:
3vrr ec (! - r) x 4 --;:---,
1<:u
N(O) .
K(O), electronic contribution
Expressions in terms of L
n
'12)
-
l
Then in the limit of vanishing magnetic field we obtain the entries in Table 18.2.1, which are found by applying relations (18.2.19) in this limit to Equations (18.2.11)-(18.2.17). The designation L(xu) is to imply that we have set g = Eu/k T in Equation (18.2.18) before carrying out the integration over e. €
e1J h 3 (2mkT)'iz
n
[e/-
eX x q - I dx and
L(u) ] Table 18.2.2. Hz --* 0, classical statistics, U == L(I) .
(l8.2.18)
1 2 U ] 2 u JP /e 2 •
fo~
08.2.20a)
With this result and the identities I'(n + I) = nr(n) the entries in Table 18.2.1 specialize to those listed in Table 18.2.2.
This permits us to write I
411
For classical statistics where -1/» I, we have -aj~/aE ~ (kT)-l e1J e- X; with u = (ero/m)(kTxY - Yz, the transport integrals L(u n Xl) become
is the lattice I<.
18.2
1<:2 2 To(0)(r+2)
e
For highly degenerate materials, 1/» 1, and it then is appropriate to employ the so-called Bethe-Sommerfeld approximation to represent L as neu
L(UnXI)
N,~1/q[l+i1T2q(q-l)1/-2+"'1
(18.2.2Ia)
2
R(O)
1 L(u )L(l) Zenc L2(U) ~rL(xu)
ZeLL(u) N(O) K(O), electronic contribution
feu L(xu)L(u 2)-L(u)L(xu 2 ) - I (1 )------c___-'--- ec ' L\u) 1t 2 T
7
rL(x2u) I}(XU)] 0 (0) L(u) - L2(U)
l
with d
Nn
(2mkT
h3
)'/,()n eTo (kT)n(r- v, m
q == l+n(r-! )+!
I)
.
(l8.2.2Ib) (l8.2.2Ic)
The scattering index r appears because we have again used the relation u (ero/m)(kTxY- Yz. Applying Equation (I8.2.2l) to the entries in
18.2
Chapter 18
412
Table 18.2.1, we obtain the entries in Table 18.2.3. Table 18.2.3. [ Hz Transport coefficient
-+
-
0, degenerate statistics, U
L(U)]
L(l) .
Expression
(2)
~(2m)1Iz(KTY+
0(0)
If, R(O)
Zenc
[1 + 1T21{r-; 1)J
2
II + 31T112(r-~)2J
611
ZeL
Zenc K Ze (~-11)
ExpreSSion in terms of L
R(oo)
e L2(1) -31T-2-L(-u--1)
a(oo)
L2(1) a(O)C:-L(C--:u)L-(-'---u 1)
31T2 ZecL(1 )
Zenc
J
K [L(x) ZelL(l) -11
N(oo)
0
0 (4)
(4) T.C.Harman and J.M.Honig, loco cit., p.22S.
-+
Transport coefficient
(0).
-+
a(oo)
K(oo), electronic contribution
Table 18.2.6. (Hz
a(oo)
Table 18.2.4. (Hz
R(oo)
R(oo)
o o
K(OO), electronic contribution
(d) In the limit of very large magnetic fields we obtain the entries shown in Table 18.2.4 by applying Equation (18.2.19) in this limit to Equations (18.2.11)-(18.2.17).
a(oo)
t&1T 0(0) r(3 - r)I'(2 + r)
0(00)
N(oo)
X3
e2
Transport coefficient
classical statistics). ExpreSSion
a(oo)
L
K(O), electronic contribution
00,
311
KU [1T2(! - r)]
ec 311 k;2To(0) 1T2
N(O)
413
Using the procedure described in (c), the above results specialize in the limit of classical statistics as shown in Table 18.2.5, and for highly degenerate statistics as shown in Table 18.2.6.
Transport coefficient
~r1T2(l+r)]
0:(0)
Transport in metals
Table 18.2.5. (Hz
81T (2mkT)'i2 '/z I + ~ 3h 11 3 811 2 161Te 2 1 Tol1 r + 1
n
18.3
00,
degenerate statistics). Expression 0(0)
[1
1
2
1T (r- W] 2 311
K 1T2 Ze211
N(oo)
0
K(oo),
0
electronic contribution
18.3 Problems 18.3 and 18.4 are designed to provide readers with practice in evaluating certain of the transport coefficients, introduced in Problems 18.1 and 18.2, in terms of experimental measurement as carried out under typical conditions. (a) A homogeneous isotropic sample in the form of a rectangular parallelepiped of dimensions Ix = 3 cm, ly = 1 cm, lz 0·5 cm is maintained at a uniform temperature and connected to a power supply. A current of 10 rnA is passed through the sample along the x direction.
Chapter 18
414
18.3
Voltage leads, 1· 5 cm apart, are connected to a potentiometer, which registers a potential difference of 73 m V. Calculate the sample resistivity and conductivity. (b) The sample is now placed in a magnetic field of 25 kG, oriented along the positive z axis. With a lOrnA current passing along the positive x direction the potential difference along the positive y direction is -85 JJ.V. Calculate the magnitude of the Hall coefficient, the Hall mobility, and the corresponding density of charge carriers. (c) Is the above sample n type or p type? Explain your answer. (d) Describe sources of experimental errors and how they may be minimized.
Transport in metals
18.3
415
(b) The Hall coefficient R is defined by [see Vy(~/e)
(18.1
Vy
HzRJx (18.3.3) where Vy(~/e) == Vy is the gradient of the electrochemical potential (n per unit electronic charge which develops under steady state and open circuit conditions along the y direction when a current density Jx is maintained along the x axis and a magnetic field is impressed along the z axis. Now write Vy = -Uy/ly , where Uy is the potential drop across the y axis (see Figure 18.3.1) and set Jx Ix fly lz. On substitution in Equation (18.3.3) and rearrangement we obtain R
=
lz (-Uy )
(18.3.4)
Ix Hz
Solution
(a) The resistivity p is given by the relation [see Equation (18.1. RsA
(18.3.1)
P=T=
where A = Iy I z is the cross-sectional area perpendicular to the current flow direction, - Ux is the potential drop across the voltage leads along the x direction, L is the separation distance between the voltage Ix is the current, and Rs the total measured resistance of that portion of the sample whose dimensions are L, Iy, lz. Note that the quantity Ix does not enter this problem. The various dimensions are depicted in 1·5 em, Iy 1 cm, Figure 18.3.1. On inserting the values of L Iz = 0·5 cm, Ix = 0·010 A and Ux = 0·073 V one obtains p
== 2·4 ohm cm;
a
=
I/o
= 0·42
}r
Figure 18.3.1.
cm- I
•
(18.3.2)
For consistency we use the c.g.s. e.m.u. system of units. Here, 8 1 X 10 abV, Ix = 0·010 X 10- abA, and 14 = 25000 G. Insertion into the above formula yields for the magni tude of the Hall coefficient Iz = 0·5 cm, Uy = -0'000085
IR 1= 1·7
X
102 cm 3 abC- 1
=
17 cm 3 C- 1
(18.3.5)
•
Using this result in the relation IR I = line, where n is the density of carriers, and writing e = 1·60 X 10- 19 coulombs we find n ==
l/iR Ie
= 3'7 x 10 17 cm- 3
(18.3.6)
•
u
The mobility is computed according to the relations a IR I = line, whence
u = Ra
17 cm 3 C- 1 x 0·42 ohm- I cm- I
= 7·1
cm 2 V-I
S-I •
neu,
(18.3.7)
(c) The fact that, with Ix and Hz positive, Uy is found to be negative implies that the potential at A in Figure 18.3.1 is higher than the potential at B, and that the electric field within the sample points along the positive y axis. Therefore under steady state conditions excess positive charge accumulates at A and excess negative charge accumulates at B. But according to the Lorentz relation, the force on charge carriers, regardless of their sign, is given by F == 1 x H/e = IxHz/c = -F~. If the sample were n type, electrons would be pushed in the direction of face A of Figure 18.3.1; if the sample were p type, holes would be pushed in this direction. Since the latter alternative is consistent with the observation that excess positive charge accumulates on face A, it is concluded that the sample is p type. We show that this is consistent with the definition of R via (18.3.3). For a homogeneous sample at constant temperature we have Vy(~/e) -Vy'Ps Uy/ly, where 'Ps is the electrostatic potential. Since Uy < 0 under the experimental conditions, Vy(~/e) > 0; more over, both Hz and are positive by assumption. Equation (18.3.3) now shows that R > 0, as is consistent with our earlier conclusion.
416
18.3
Chapter 18
(d) Aside from the implication that the material is homogeneous and isotr9Pic, as stated in the problem, the following tacit assumptions were introduced: 1. The open circuit voltage Uy is assumed to be due solely to the Hall voltage and no errors arise from probe misalignments (failure of the leads to lie on an equipotential line) or from thermal e.m.f. To minimize these sources of error one conventionally takes Uy readings with current flow and magnetic fields in the positive and negative directions. Suitable averaging of the results will largely eliminate these errors. 2. The formula n l/IR Ie was assumed to be applicable; this is the if the solid can be characterized by carriers in a single band to conduction. For the more general case of carriers in several bands participating in such processes the reader is referred to Problem 19.6. Moreover, the above formula applies to one-band models only in the limit of high magnetic fields; a more rigorous analysis shows that n = AIIR Ie, where the coefficient A may deviate from unity by as much as 20% in extreme cases. For calculations requiring high precision this factor must be taken into consideration. 3. Steady stat~ conditions are assumed. 18.4 (a) A homogeneous, isotropic sample is cut in the form of a rectangular parallelepiped of length I 0 . 8 cm, width w = 0·2 cm, and thickness t = O' I cm. Copper-Constantan thermocouples attached to the sample indicate a temperature of 273·1 oK at one end and a temperature of 278·4°K at the other end. When a potentiometer is connected across the copper leads of the thermocouples, a potential difference of 0·104 mV is registered on the instrument. To achieve the negative end of the potentiometer had to be connected to the cold end of the sample. What is the Seebeck coefficient for the sample? Is the material n or p type? (b) Discuss three experimental errors incurred in this type of measure ment and how each can be circumvented. What other assumptions are made as regards these measurements? (c) The above sample is placed in a transverse magnetic field of 30000 G which is oriented parallel to the t dimension. The thermo couples register a temperature difference of 4 deg along the positive I dimension. A potentiometer hooked to leads attached across the positive w dimension registers a potential difference of -0' 036 mV. What is the magnitude of the Nernst coefficient for the sample? (d) Describe briefly several sources of experimental error in the above measurement. Solution
(a) For homogeneous, isotropic materials the Seebeck coefficient is defined through the relation [see Equation (18.1.4)] V'x(S-le)
== IS:
cxV'xT
08.4.1)
18.4
Transport in metals
411
where in the notation of Problem 18.1, is the gradient of electro chemical potential per unit charge, and T the gradient of temperature the length of the sample. We now approximate V'x T with Ax Til and Vx with Ux/I where Ax T is the temperature difference between the ends of the sample and Ux is the corresponding potential difference. We can then rewrite Equation (18.4.1) as Ux cx = - AT' (18.4.2) x
Inserting the values Ax T we find CX
= 278· 4 -
273· I
5· 3 deg and Ux = 104 pV
= -104/5,3 = -20pV
(18.4.3)
Since the cold end is connected to the negative end of the potentio meter, cx < 0 and the material is n type. One should imbed the thermocouples away from the ends of the sample to avoid spurious temperature readings due to end effects. Also, instead of relying on a single Ux measurement at one Ax T, one should values corresponding to different Ax T, and compute cx take a set of as the slope from a plot of these data. This procedure circumvents problems which arise if, due to experimental difficulties, Ux does not vanish when Ax T O. Finally, the value of cx obtained through the above procedure includes a contribution due to the lead wires; this matter is discussed in great detail in several sources in the literature (5). If CXx and CXL are the Seebeck coefficients of the test specimen and of the lead wire at the average temperature of 275'7°K, then cx = CXx -CXL, whence CXx = CX+CXL' For copper lead wires CXL "'" 2 pV deg- 1 , showing that the correction is by no means negligible. Other assumptions are implied in the statement of the problem. The material is assumed to be isotropic and homogeneous; this permits us to use Equation (18.4.1) and the approximation implied in (18.4.2). Further, 'isothermal' conditions are assumed to hold along the two directions perpendicular to the temperature gradient. (c) The Nernst coefficient may be defined by [see Equation (18.1.5) 1 N
'iLl
V'y(s-/e) __ Ax T w Hz I 'VxT -
(18.4.4)
2 On inserting the values Uy - 36 p V = - 36 x 10 abV, Ax T 4 deg, w = 0·2 cm, 1= 0·8 em, Hz 30000 G, we obtain for the magnitude of the Nernst coefficient N = 0·12 abV G- 1 deg- I = 0·12 x 10-8 V G- 1 deg- 1 • (18.4.5)
(5) C.A.Domenicali, Rev. Mod. Phys., 26, 237 (1954); T.C.Harman and J.M.Honig, loco cit., Section 1.18, pp.41-47.
Chapter 18
418
18.4
(d) The effects discussed in (b) must be modified to the extent that 'isotl:termal' conditions are assumed only parallel to the applied magnetic field. A further source of experimental error arises from spurious voltages due to misalignment of the voltage probes across the middle of the sample. On reversing both V'x T and/or Hz and taking suitable averages this particular error may be cancelled out. IS.5 This problem is designed to acquaint the reader with the relation between the electronic contribution to the thermal conductivity and the electrical conductivity of a metal. According to the last entry of Table 18.2.1, 1«0) )£0(0); the proportionality bctween I< and a is referred to as the Wiedemann-Franz law, and the proportionality constant is called the Lorenz number. [The symbol £ introduced here is not to be confused with the symbol L introduced in Problem IS.2 for the transport integral (IS.2.18).] In tht' last line of Table 18.2.1 £ is specified by the quantity in brackets. (a) Determine the Lorenz number £ in zero order approximation for a highly degenerate metal (Sommerfcld model). 0 (b) Calculate the electronic contribution at 300 K to the total thermal conductivity of a metal (Sommerfeld model) whose resistivity at that temperature is 5·01 x 10- 6 ohm cm. (c) Show that £ as determined in part (b) has as appropriate units the l dimensions of W i cm- . Solution
(a) For a highly degenerate material the Lorenz number is a universal constant given by Table IS.2.3, last entry) /{2rr2
£ 2 - -3e
(18.5.1 )
the various natural constants, one obtains £ = 2·45 x 10 6 erg2 deg- 2 C- 2 •
(18.5.2)
(b) Since
18.6
Transport in metals
419
Since I erg
10-7 (J
5- 1 )
= 10-7 W
and since ohm-I = I A2 ohm = I W . we see that I erg2 C- 2 ohm- 1
= 10- 14 W ,
which leads us directly to expression (lS.5.4b).
I S.6 This problem is designed to familiarize the reader further with information available from the determination of the Seebeck and Nernst transport coefficients (see Problem IS.I). For simplicity it is to be assumed that all of the conditions leading up to the entries of Table 8.2.3 apply. The Seebeck coefficient of a highly degenerate metal is found to be -7'1 p.V at 77'SoK under conditions where acoustic mode Determine the Fermi level for this material. the nature of the Fermi level calculated according to describe the basic assumptions made in the numerical evaluation of the Fermi level. The metal is now placed in a very strong magnetic field. Neglecting quantization effects and any variation of Fermi energy with magnetic field, redetermine the Seebeck coefficient; observe the magnitude of the change. (d) Calculate the isothermal transverse voltage developed across the width of the sample (y direction) placed in a magnetic field of 10000 G which is oriented along the z axis. The two ends of the sample along the x axis are maintained at 302 and 29SoK respectively, and the drift mobility is 12000 cm 2 V-I S-I; the sample dimensions are Ix = S mm, ly I mm, lz = 1·5 mm. Solution
I<e
= £Ta = £T
(1S.5.3)
p
insertion of Equation (lS.5.2), T = 300 oK, and p = 5·0 I yields I<e 1·46 X 10 14 erg2 deg- I ohm-I cm- l I· 46 W deg- I em-I.
X
10-6 ohm em (18.5 (1
straight (c) In expression (l8.5.4a) we adopted the units obtained forward substitution of the usual dimensions for K, e, T, and a. We now divide both numerator and denominator by S2 to obtain (erg S-I)2 (C S-I r 2 ohm-I deg- I em-I.
(a) According to the Sommerfeld model introduced in Problem IS.2 the Seebeck coefficient is related to the Fermi level, p., by(6) [see Table 18.2.3, fourth entry, with r = 0] rr2/{2 T 0' ± (lS.6.1) Solving for p., inserting k/e S6·4p.V deg- I , K T= 77'SoK,O' = -7'1 p.Vdeg- 1 we obtain p.=O·27eV.
S'62x lO-SeVdeg, (1S.6.2)
N.Cusack, The Electrical and Magnetic Properties of Solids (Longmans, Green & Co., London), 1958, Section 5.9, pp.ll1-117; T.C.Hannan and J.M.Honig, loco cit., Section 3.14,
(6)
18.6
Chapter 18
420
(b) The quantity p. specified in Equation (lS.6.1) is the Fermi level relative to the appropriate band edge, 8 8 , For n type materials, is the conduction band edge; for p type materials, it is the valence band edge. Equation (IS.6.1) applies only to a highly degenerate electron hole) gas. Thus the above relation holds only if the Fermi level lies well inside a single band of standard form whose associated charge carriers are responsible for conduction phenomena. Furthermore, acoustic phonon scattering is presumed to be the dominant scattering mechanism, which means that r == 0 in Problem IS.2(c). (c) In the limit of very high magnetic fields and for the model discussed in part (b), Equation (1S.6.1) must be altered to read [see Table 18.2.6, third entry] 1[2/(2 T
a = ±-2- . ep.
(IS.6.3)
Hence, on substituting the values cited in part (a) and p. :::: 0·27 eV we obtain a 11 p.V deg- 1 . (1S.6.4) This represents an increase by a factor ~ over the value observed in zero magnetic field. (d) The transverse voltage arises from the Nernst effect according to the relation Equation (IS.1 NHz V'xT. (1S.6.5) On the assumption that the material is homogeneous and isotropic, the above may be reduced to V'y(rle)
-- NHzll x T I~
(1S.6.6)
x
Assuming that one is operating in the regime of low magnetic fields, that the material is highly degenerate and characterized by a single band of standard form, and that acoustic phonon scattering predominates, the Nernst coefficient may be written as [see Table 18.2.3, fifth entry with
18.7 This yields N
= 0·16
cm 2
== 8·1
X
(;2) (~) e;)~
(18.6.7)
where u is the drift mobility; and p. the Fermi level relative to the appropriate band edge. For consistency, we employ the c.g.s. e.m.u. system of units. Values we shall insert are: kle = S6·4 p.V deg- 1 = 8·64 x 10 3 abV deg- 1 , ulc 12 x 10 3 cm 2 V-I S-1 1·2 X 10-4 cm 2 abV- 1 kT = 0·0259 eV for T 300o K, p. = 0·27 eV (assuming that there is a negligible shift in Fermi level from 78 to 300o K) .
(18.6.8)
•
102 abV = 8·1 x 10- 6 V
=
4 deg, ly Ilx ==
8·1 p.V.
k,
(18.6.9)
One should observe that Equation (18.6.7) is only marginally applicable since the 'low field regime' holds only as long as uHlc < 1. In our case uHIc I· 2, but it may be presumed that the above calculation is a reasonable first order approximation to the correct answer. 18.7 This problem is designed to introduce the reader to the general linear theory of irreversible thermodynamics as applied to the now of current and of heat through a solid. One may show from rather general considerations (7) that when a solid is subjected to a temperature gradient V'T and a gradient in electro chemical potential per unit charge V'(r/e) == V, there occurs a nux of entropy Js and of electric charge (current density) J. In the above TJ s = C+ aJ; as earlier, C denotes the heat flux and a represents the Seebeck coefficient of the specimen. It should be noted that in the absence of current, TJ s and C coincide. It is generally assumed that TJ s and J depend linearly on V and V'T; thus, relations of the form TJ s = AV'T+BV; J = CV'T+DV may be expected to apply, where A, 8, C, D are constants. However, these interrelations may be simplified by the so-called Onsager reciprocity theorem: as applied to the present case, the theorem states that if one replaces V'T with V' (1 IT) and V with V IT then the equations specifying the response to these particular 'forces' read TJ s
L12
LuV'(lIT)+TV
J L 12V'(1/T)+
!
I S-1
Using this value in Equation (18.6.6), along with llx T we obtain
r = 0]
N =
421
Transport in metals
V
(18.7.1)
(1S.7.2)
where the off diagonal coefficients are identical. The above equations form the basis of our subsequent development. On the basis of the above: Show how the various Lij may be identified in terms of transport coefficients that are experimentally measurable. (b) Rewrite the phenomenological equations with the various Lij replaced by transport coefficients. Sec also D.D.Fitts, Non-equilibrium Thermodynamics (McGraw Hil!, New York), 1962, Ch.3; T_C-Harman and J.M.Honig, loco cit.• Section 1.10, pp.20-22.
(7) See Chapters 25 and 28.
422
18.7
Chapter 18
Solution
(a) At constant temperature and for homogeneous materials we have
J = (L 22 IT)V = (L 22 1T)E as was demonstrated in Equation (18.1.9). This relation between J and E represents a formulation of Ohm's law, whence
L22
T
=
0
or
L22
=
To
(18.7.3)
where 0 is the electrical conductivity in isotropic media. In the absence of Qet current flow Equation (18.7.2) reads (L 21
v =-
TL 21 '11(1/T) L22
= L21 'I1T. L22T
= L 12 ) (18.7.4)
The Seebeck coefficient a is defined by V = a'l1T, for J = 0 [see Equation (18.1.4)]; comparison with Equations (18.7.4) and (18.7.3) shows that (18.7.5) L21 = LJ2 T 2ao.
18.8 In this problem we consider the transport phenomena in solids where electrons in a conduction band and holes in a valence band simultaneously participate in conduction processes. In what follows let the subscript or superscript i = 1 or i = 2 refer to holes or electrons respectively, and assume steady state conditions so that V = '11 (tie) and 'I1T is the same for charge carriers in each band. All of the simplifications leading up to Table 18.2.3 (the Sommerfeld model) are assumed to apply. (a) Express the total 0, a, K in terms of densities, effective masses, and mobilities of carriers in each of the two bands, assuming sufficient carriers in each band to render the statistics highly degenerate, and assuming acoustic mode scattering to predominate for both sets of carriers. Utilize the results shown in Table 18.2.3. (b) Specialize part (a) to a metal with 'mirror image' bands. Solution
(a) Equations (18.7.10) and (18.7.11) apply to each band separately. Let these be denoted by subscripts or superscripts 1 and 2 respectively. Then in the notation of Problem 18.7,
The thermal conductivity K is defined by the relation [see Equation (18.1.6)] TJ s = -K'I1T = KT 2'11(1/T), with the subsidiary condition J = O. The latter condition allows us to substitute for V in Equation (18.7.1) from (18.7.4), and thereby to obtain TJ s -_ ( LlI - L~2) L22 'I1(1/T) -_ KT 2 'I1(1/T).
(18.7.6)
On introducing the results (18.7.3) and (18.7.5) and solving for Ll1 we find L 11 = T2(K + T( 20) . (18.7.7) (b) On substituting the expressions (18.7.3), (18.7.5), and (18.7.7) in Equations (18.7.1) and (18.7.2) one obtains TJ s
= T2(K + T( 20)'I1(1 /T) + T 2ao(VIT)
(18.7.8)
T 2ao'l1(1/T) + To(VIT)
(18.7.9)
J
=
Js
= - (~ + a J
2
0) '11 T + ao V
= -ao'l1T+oV .
J s(i) --
_
2)
(~ T + ai Oi/
J(i) =
(18.7.11)
v
(18.8.1)
T +aioi V ,
= 1, 2) (18.8.2)
-ai o i'l1T+oi V .
Steady-state conditions are assumed which allow us to assign to the gradients of T and of tie the same value in each band. Since the fluxes are additive we find that _ (I) (2) _ K1+K2 2 2 J s - J s +J s - --T-'I1T-(alol+a202)'I1T+(alol+a202)V(18.8.3) J
For T
= J(1)+J(2) = -(a1ol+a202)'I1T+(01+02)V,
(18.8.4)
= 0 and for uniform materials we have (see Problem 18.7) (18.8.5) J = (01+02)V = (0 1 + 02)E.
Since this is a formulation of Ohm's Law, For J
=
0 1 +02
.
(18.8.6)
= 0, Equation (18.8.4) becomes V =
(18.7.10)
t7
(i
o
or
423
Transport in metals
18.8
a 101 + a2 02 'I1T 0 1+02
(18.8.7)
With the definition V = a'l1T and in view of Equation (18.8.6) it is evident that a 1o l +a2 02 (18.8.8) a= o
424
18.8
Chapter 18
The thermal conductivity is found from the relation Tls = -KVT with· J O. In view of this latter requirement we may use Equation (18.8.7) to eliminate V from Equation (18.8.3). This yields
Tls
z z (a 10 1+a20Z)Z] I(Kl+K2)+T(a 1o l +a2 oZ)-T 0 VT.
(18.8.9)
On introducing relations (18.8.12,15,16) into Equations (18.8.6,8,11) and simplifying one obtains the following results: o
a
In conjunction with Equation (18.8.8) this leads to K
=
KI+Kz+T(aiol+a~o2-a2o).
(18.8.10)
On the basis of Equation (18.8.6) this may be rewritten as °IOZ 2 K = KI+K2+T--(al-a2) .
o
(18.8.11 )
Actually, the above relation is strictly correct only for the electronic contributions to the thermal conductivity, since it is not possible to partition the contribution of the lattice K in the same manner as for the electrons. Thus, for the total thermal conductivity we must adjoin the lattice contribution to the value given by expression (18.8.11). For the two bands in question we now introduce the following specific relations njeUj
OJ
(i
= 1,2)
(18.8.12)
where Uj is the drift mobility of the carriers and nj their density in the lth band. The Seebeck coefficient for a degenerate electron gas in a band of standard form is given by (see Table 18.2.3) ZirrlklT
a· I
= 3elli
(18.8.13)
where Ilj is the position of the Fermi level relative to the valence band edge (i = 1) or conduction band edge (i = 2), Z I 1 and = -1, and where we have set r = O. For Ili we substitute the relation appropriate for a highly degenerate electron gas in a band of standard form (see Table 18.2.3, neglecting the term in 1]-2): 2 h (3ni)2h III == 2mi
8rr
(18.8.14)
(..!!...-)'h
Z
2Z k Tmj 3en 2 3n i
a. I
(18.8.15)
where n == hj2rr. Finally, for the electronic contribution to the thermal conductivity we use the Wiedemann-Franz law (see Problem 18.5): k2 K·I
rr2
To·I
=
k Zrr2 TnjUj
3e
K
e(nlul +nzuz) 2 2k T [nYlm1uI -n il m 2U 2] -3 ~ -3 + 11 e n1uI nZu2 4 k 4 T3 k 2Trr2 lUI + n2 U2 ) + -9 ~ ell -3
(18.8.17)
(rr)'h
(18.8.18)
(rr)% (u tU2)(m1niJ+ mlni')2
(n n ) I/J(\nlUI + n1u ) .
Z 1 2 (18.8.19)
We now specialize to the case of 'mirror image bands', where nl == n z n', ml = ml == m', UI U2 u'. Equations (18.8.17,18,19) then simplify to (18.8.20)
0= 2n'eu'
a
0
K
k2 T e
08.8.2l)
2rr2 3
I
I
nU
8 k4 T 3(rr)'YJ(ml)2u
+ 9"
en 4
3'
(n') 'll
l
(18.8.22)
18.9 So far we have utilized special approximations in transport theory that permitted us to obtain final results in closed, analytical form. The remaining problems, 18.9 to 18.11, are designed to provide the reader with a less specialized approach to the subject matter. The basic model discussed in Problem 18.2 is utilized below. The principal change relative to Problem 18.2 is that we do not evaluate the transport integrals for various limiting cases. (a) Obtain an expression for the electrical conductivity of a material characterized by a single band of standard form, in terms of integrals pertaining to Fermi-Dirac statistics. Assume that the relaxation time formalism is applicable, and that no magnetic field is present. (b) Obtain an expression for the Seebeck coefficient of a material in the same terms as under part (a). Show that this measurement specifies the location of the Fermi level relative to the appropriate band edge. (c) Discuss briefly in further detail how the present treatment differs from that of Problem 18.2. Solution
where mj is the effective mass of carriers in the lth band. Then j
425
Transport in metals
18.9
(18.8.16)
(a) Standard analysis (8) as well as the exposition leading to Equation 08.2.11) shows that the electrical conductivity is given by (18.9,1) o e2 K 1o (8) A.H.Wilson, The Theory of Metals (Cambridge University Press, Cambridge), 1954, 2nd Edition, Section 3.3, pp.70-73; E.H.Putley, The Hall Effect and Related Phenomena (Butter worths, London), 1960, Section 8.2, pp.196-198; T.C.Harman and J.M.Honig, loco cit., Sections 4,11, 4.14, pp.183-186, 193-195.
426
Chapter 18
18.9
where K lo is the transport integral specified according to Equation (l8.2.8a) with Hz W = O. afod€ ITk 2 --dE. a€ dk
KO ,
(18.9.2)
As in Problem 18.2, we introduce the relation To€' - v,
T
(18.9.3)
as well as expression (18.2.2) for fo. Then
ai'o
aE =
standard form the functional relation is
E
on k; for bands of
h 2k 2
(18.9.5)
E
where m is the effective mass of carriers in band b; in the case under study this quantity is a constant. Inserting expressions (18.9.3,4,5) into (18.9.1,2) 2
161Te .J2j1i
a
3h 3
(kTY
+1
To
r" A
----~ dx
(18.9.6)
where we have also introduced the changes in variable x == €/kT, 11 The integral on the right may be determined by parts: x' 1)
f
(r+I)F,(11)
0
161Te 2 .J2j1i 3h 3
(18.9.7)
+1
To(kTY
(r+1)Fr (11).
(18.9.8)
(b) The Seebeck coefficient may be expressed in terms of transport integrals as follows (9) [see Equation (18.2.13) 1: IX
k(kTKl K2 ) = ±e -11 .
Solution
(a) In an extension(IO) of the theory leading to Equation (18.9.6) [see also Equation (18.2.8a»), the conductivity of an isotropic material in a magnetic field is given by 2
_ 161Te .J2j1i a -
3h3
r
(kT)
fO<>
r+ I To
0
+
I
(_
af o)
ax
dx
(18.10.1)
where the nomenclature of Problem 18.9 is retained. Here w == ±eH/mc, where H is the externally applied field and c is the velocity of light. Furthermore, fo = I I[ I +exp(x -11B)] is the Fermi-Dirac function, and x €/kT is the reduced energy. It is assumed that the material is characterized by a band of standard form. With T To€r - % and for (WT)2 ~ I, we can approximate expression (l8.1 0.1) by setting I -------:;--;;:::::: I 2
1 +W
J
k [(r+ 2)Fr+ 1(11) ± e (r+ I)Fr (11) -11
.
T2
W 2 T 2 (kT)2r-l 0
x 2r-1
and by writing for later convenience
(18.9.9)
On substituting expressions (18.9.3-7) one obtains IX
18.10 (a) On the basis of transport integrals which are valid for isotropic materials characterized by a band of standard form, show that the conductivity in the presence of a small magnetic field H varies as H2, and establish the condition under which this approximation holds. Specialize the above to the case of a highly degenerate electron gas. Obtain an expression for the conductivity in the absence of a magnetic field and relate it to the more general treatment of problem. (c) Determine the magnetoresistance in the approximation of part (b).
O<>
where the integral defined by Fr is termed the Fermi-Dirac integral. numerical evaluations of Fr are available in the literature. In view of expression (18.9.7) we may write
a
Equation (18.9.10) shows explicitly that IX depends functionally on 11 alone; once r is specified, a measurement of IX fixes 11 experimentally. (c) Aside from the fact that H 0 here, the transport integral KP, Equation (18.9.2), has not been reformulated in terms of the mobility, as was done in Problem 18.2(c), where the related integrals L(x/u n ) were introduced. As a result, it is possible to specify a and IX (as well as K) solely in terms of the integrals Fr defined in Equation (18.9.7), which do not involve u(€). Tabulations of Fr (11) are available in the literature, so that numerical calculations based on the use of Fr (11) are more readily done than those based on the integrals L.
(18.9.4)
}2RT
I t is also necessary to specify the dependence of
427
Transport in metals
18.10
a = C L>Oh(X) (-
~)cL'\:
(18.10.2)
where (18.9.10)
(9) A.H.Wilson, loco cit., Section 3.4, pp.73-77; T.e.Harman and J.M.Honig, loco cit., Sections 4.13,4.14, pp.190-197.
_ I 61Te 2 .J2j1i(kTY+ ITo
c=
3h 3
(18.1
(10) E.H.PutIey, loco cit., Section 3.5, pp.71-77; T.C.Harman and J.M.Honig, loco cit., Section 4.14, pp.193-198.
428
Chapter 18
18.10
and
==
hex)
1[1-W276(/(T)2r-IX2r-I].
(18.10.4)
Insertion of expression (18.10.4) into (18.10.2) shows that we may write
o
C[t~ 1(_~)dx-(e~:o)\RT)2r-It~ (-~)dxJ
(18.10.5)
where the first term represents the conductivity in zero magnetic field and the second represents the first order correction which is valid for 2 2 W 7 < 1. Note that 0 varies as H2. For a highly degenerate electron gas (-(JloI(Jx) approaches the Dirac delta function. Accordingly, we substitute for Equation (18.10.2) the Bethe-Sommerfeld approximation o
Ch(1'/B)+
Cn,2 (J2h(1'/B) 6 (Jx2
(18.10.6)
where 1'/B is the Fermi level relative to the appropriate band edge, divided by /(T. On substituting into expression (18.10.6) from (18.10.4) we obtain 0(H)
=
+!C1r 2 [r(r+
1)17B- 1 - w 2 7J(/(T)2r- 13r(3r - 1)17~r-2]
C17B+ 1 U1 + (i1r )r(r+ 1)1'/132 ] 2
w27J(/(T)2r-I17~r-I[1
+ (i1r 2 )3r(3r- l)17B2]).
(18.10.7)
We note that the only factor depending on His w. Equation (18.10.7) is the specialization of Equation (18.10.5) to a highly degenerate electron gas. When w = H == 0, Equation (18.1 0.7) reduces to 0(0) = C1'/B+ 1 [ 1 + (-l;1r2 )r(r+ 1)17132] (18.10.8) which represents the specialization of Equation (18.9.8) to a degenerate electron gas. (c) The magnetoresistance is defined by D..p
Po
p(H) - p(O) _ 0(0) -o(H) p(O)
-
0(H)
On introducing expressions (18.1 0.7,8) into this definition we obtain D..p
Po
Under the original assumption the second term in the denominator is small compared to the first. Since we intend to keep only terms of order w 2 , we may drop the second term in the denominator and simplify Equation (18.10.9):
C17t W 27J(/(T)2r - 1[I + (-l;1r2 )3r(3r- 1)17132 ]
C17B+ 1[[1 + (~1r2 )r(r + l)17ill- W 2 7J(/(T)2r- 117 f{- I[ I + (!1r 2 )3r(3r - 1)17B2])
(18.10.9)
2
2
D..p = W 276(RT)2r
Po
I
1+ (i1r )3r(3r- 1)1713 2 1 + Ci 11'2 )r(r + 1)17B
.
(18.10.10)
Equation (18.10.10) shows that the magnetoresistance increases para bolically with w or H, so long as the fundamental assumption W 2 7 2 < 1 holds. 18.11 By examination of the energy flux establish that if &'(k, r) is the energy associated with an electron of wave number vector k at position r, then the energy which must be associated with the corresponding hole is -&'(-k, r). Solution
This question may be settled by determining the rate of transport of energy through a group of electrons associated with wave number vectors in the range k to k+d 3k Je (k)
C1'/B+l[1-w276(/(T)2r-I17~r-l]
429
Transport in metals
18.11
In (k, r)
3
&'(k, r)v n (k) 41f3 d k .
.
(18.lLl)
Here In(k, r)d 3k/(41r 3 ) is the density of electrons at r whose wave number vectors lie in the range d 3 k about k. The total energy &'(k, r) of the electron may be rewritten as (18.11.2) &'(k,r) = ~'c(r)+En(k) where &.C is the lower (conduction) band edge energy and En is the energy of the electron relative to &.C. The velocity of these electrons is given by the standard expression (11) (18.11.3) = h-IV k En(k). Vn
It should be noted that &'(k, r) = ~(-k, r), En(k) = En (-k) are even functions in the wave number vector, while Vn (k) = -Vn (-k) is an odd function in k. Consider now the total energy flux obtained from Equation (18.11.1) as (18.11.4) (k,r)d 3 k JE 4: 3 hL&'(k, r)VdEn and subtract from this the quantity
J
0= 41rI3 h b&'(k,r)Vk[en(k)]d 3 k
(18.11.5)
(II) A.H.Wilson, loco cit.• Section 2.8, pp.43-45; T.e.Harman and J.M.Honig, loco cit.• Section 4.4, pp.164-165.
Chapter 18
430
18.11
which vanishes since & is even and its derivative is odd in k. have
4:
JE =
3 hL-&(k,
r)V'd€n(k)][I- fn(k, r)]d 3 k.
We now (18.11.6)
At this stage it is important to relate the properties of various functions at k to their properties at -k. Towards this end we rewrite expression (18.11.2) as &(k, r)
= &(-k, r)
&v(r) - €p(-k)
(18.11
where is the upper (valence) band edge of the same band and €p is the energy of the electrons relative to &v. Comparison with expression (18.11.2) establishes that €p(k) = €p(-k)
&v(r)- &c(r)- €n(k).
(18.11.8)
It now follows that
V'k€p(k)
-V'k€n (k) = -V'_k€p(-k)
(18.11.9)
where the term on the right is obtained from that on the left by the substitution of €p(-k) for €p(k) and of -V'-k for V'k' Finally, we study a new distribution function introduced by fp(-k, r)
1 - fn (k, r) .
(18.11.10)
On inserting expressions (18.11.7,9,10) into (18.11.6) we find that the energy flux is given by JE
= 4:3ht -&(-k, r)[-V'_k€p(-k)]fp(-k, r)d 3 (-k).
In carrying out the final step we have replaced
(18.11.11)
--
f1,s dk x dky dkz
by
- Sf f d(-k x )d(-ky)d(-kz ), and then reversed the integration limits to
--
obtain
fif d(-kx)d(-ky)d(-k z )'
With the understanding that the
integration is always carried out in the direction of increasing values for the variable of integration, this procedure allowed us to replace
fb
d 3 (-k).
f
d 3k by b
Examination shows that the original formulation for JE , Equation (18.1104), has been rewritten in Equation (18.11.11) in such a way that the integration over positive k is now replaced by one over negative k. Further, the distribution function fn(k, r) over occupied states of different k has been replaced with one, fp(-k, r), over unoccupied states [see Equation (l8.1l.l 0)] of different -k. The velocity function for electrons has been replaced with one involving entities associated with -k.
18.11
Transport in metals
431
Finally, as is seen from Equation (18.11.8), the 'kinetic' energy €n(k) is replaced with a negative 'kinetic' energy -€p (-k). The above leads to the important conclusion that a summation of contributions of all electrons to the transport of energy & in a given band can be replaced by a summation over hole states in the same band. However, comparison of Equation (18.11.11) with (18.11.4) shows that in so doing &(k, r) must be replaced with -&(-k, r). In other words the total energy associated with a hole must be the negative of that associated with the electron. Moreover, since k in expression (18.1104) is converted to -k in expression (18.11.11), the momentum of free holes, -Uk, is the negative of the momentum of free electrons, +hk.
19.1
19
Transport in semiconductors LM.HONIG (Purdue University, Lafayette, Indiana)
INTRODUCTORY COMMENTS
Transport in semiconductors
bottom of the conduction band. As already discussed at length in Problem 18.11 (see also Problem 19.4), the voids left in the valence band may be considered as holes. The intrinsic semiconductor must thus be treated as a material in which two sets of charge carriers are simulta neously present. A number of the above concepts have already been explained in connection with various problems in Chapter 16. In what follows we shall always consider bands of standard form; the energy €, taken relative to the band edge energy varies quadratically with wave number vector k. Specifically, we set €
19.0 The reader is referred to Problems 16.1,16.5, Section 18.0, and Problems 18.1 and 18.2 for a review of basic concepts required in later problems. The following summary and additional commentary is offered for subsequent use. Both metals and semiconductors are characterized by partially filled energy band§. However, for metals the degree of filling is so extensive that quantum (degenerate, or Fermi-Dirac) statistics must be invoked in handling problems in statistical thermodynamics or in transport theory. The Fermi level falls within one or more bands so that even at the absolute zero of temperature the metal is a conductor (in many instances superconductivity also sets in at low temperatures). By contrast, the degree of filling of energy bands of semiconductors can be so small that classical statistics may be an excellent first approximation in problems pertaining to statistical thermodynamics (see Problem 16.5) or transport theory. The Fermi level now lies within a gap, so that at the absolute zero of temperature all bands are either completely filled or completely empty; the material is thus an insulator at OOK. We distinguish between an extrinsic (Problem 16.5) and an intrinsic (Problem 16.1) semiconductor at temperature T according to whether the band gap separating the nearly filled from the nearly empty bands is considerably greater than, or nearly comparable to, P:T. In the former case the material would normally still be an insulator, were it not for the fact that impurities are always present which act as a source of electrons or holes. The mechanism by which this occurs is explored in Problem 19.1, together with an elementary sample calculation showing that charge carriers are freed from such impurity centres at very low temperatures. The extrinsic semiconductor is said to be n or p type according to whether the impurities create electrons in the lowest lying empty band, termed the conduction band, or create holes in the highest lying filled band, termed the valence band. A semiconductor is said to be compensated if both types of impurities are present in nearly equal concen tra tion. When P:T becomes comparable to the band gap it is possible to promote electrons thermally from the top of the valence band to the 432
433
h2 k2 --
2m
(19.0.1)
where m is the effective mass of the charge carrier in the lattice (see Section 18.0). Further, we assume that the relaxation time formalism described in Section 18.0 and Problem 18.2 may be invoked. For a definition of the various transport coefficients the reader is referred to Problem 18.1. Problem 19.7 deals with a different type of semiconductor. Here orbital overlap is sufficiently small, relative to the internuclear distances in the lattice, that charge carriers can no longer move freely through the lattice in states characterized by energy bands. Rather, in first approximation, the carriers remain in residence at certain lattice sites and require an activation energy €a to overcome the energy barrier that separates this site from equivalent ones in the immediate neighbourhood. Materials in which conduction occurs through a diffusion-type transfer are said to be characterized by a hopping mechanism. Clearly, both excess carriers and equivalent empty lattice sites must be present for this mechanism to be operative. 19.1 This problem deals with the processes whereby charge carriers are promoted from impurity levels to appropriate bands. (a) Give qualitative arguments to show that the presence of arsenic impurities in germanium converts this material into an n type extrinsic semiconductor. Show how the hydrogen atom model may be invoked to estimate the energy required to generate freely mobile charge carriers. (b) Estimate the temperature range in which complete ionization can be expected for donor impurities in germanium. Within the framework of the hydrogenic model what factors alter this range in other host lattices such as Si, InSb, or ZnTe? (c) Give qualitative arguments to show that the presence of gallium impurities converts germanium into a p type extrinsic semiconductor. Show that the hydrogenic model may be invoked to estimate the energy required to free holes in the valence band.
Chapter 19
434
19.1
l
if
Solution
Arsenic possesses one more valence electron than germanium; hence, for each arsenic atom incorporated substitutionally in the germanium lattice there remains one extra electron that does not fit the bonding pattern of the host lattice. At low temperatures the extra electrons remain localized at their respective impurity centres. At sufficiently high temperatures the electrons become detached and move freely through the lattice in energy states falling within the conduction band. The detachment can be represented symbolically by the relation As ~ As+ + e-, which is akin to the ionization of a hydrogen atom embedded in the germanium host lattice. Impurities that generate free electrons in a conduction band are termed donor impurities. In this very simple model, the energy required to free the electron from the impurity site is given by the Bohr formula m n e4 (19.1.1) ej = 2h 2 K2 where e is the electronic charge, h == h/2rr, h is Planck's constant, mn is the effective free electron mass in germanium (m n :::::; O· 25 mol, mo is the rest mass of the electron in free space), and K :::::; 16 is the dielectric constant of the lattice. Thus, we may write mn I
ej
eH ----2' mo K
(19.1.2)
I'
,I -,
f
where eH (:::::; 13·6 e V) is the ionization energy of the hydrogen atom. It is evident that e;/eH :::::; ! x , whence ej ~ 0·013 eV
435
Transport in semiconductors
19.2
but at elevated temperatures the void may be filled by electrons from neighbouring bonds. In this process the vacant site is shifted to the point of origin of the electron that took its place. Other electrons now are free to move to the location of the displaced void, which thereby is further displaced in the opposite direction. As has been argued in Problem IS. II , the absence of an electron in a 'sea' of surrounding electrons may be treated as a positively charged particle, termed a hole. Hence, the process just described may be represented symbolically by the equation Ga ~ Ga-+ e+, where e+ represents the hole and Ga-, the impurity site with one more electron than its normal complement of three. It is intuitively evident that the process is akin to the ionization of an atom of 'antihydrogen', embedded in germanium, into an anti proton and a hole. Hence, Equations (19.1.1) and (19.1.3) should again be applicable, with mn replaced by mp' Impurities that create holes in valence bands are termed acceptor impurities. 19.2 (a) Specialize the standard transport integral K; as defined by Equation (1S.2.Sa) to the case of zero magnetic field, bands of standard form, and classical statistics. Obtain an expression in terms of the mean free path parameters and the scattering index r introduced in Problem 18.2(c). (b) With the transport integral as found in part (a), obtain an expression for the conductivity a of the solid in terms of 0) the Fermi level and (ii) the charge carrier density. (c) Employing the expressions for the transport integrals K j obtained in part (a), obtain an expression for the Seebeck coefficient a.
(19.1.3) Solution
for the ionization energy of As in Ge. (b) In the crude model used here, nO distinction is made between various impurity ions in the germanium lattice. Accordingly, it is predicted that they would all be characterized by ionization energies in the range of 0·01-0·02 eV, as is indeed verified experimentally for at least a large class of impurities. These energies correspond to tempera tures in the range 100-200oK, in which range one can expect most of the impurity centres to have been ionized. As one passes to other types of host lattices the quantities mn and K are altered, and the temperature range where complete ionization is encountered is altered accordingly. The temperature region where ionization of all impurities is essentially complete before the onset of the intrinsic regime is termed the exhaustion range. (c) Gallium has one less valence electron than germanium; hence, when it is incorporated substitutionally in germanium, there occurs one lacuna per Ga in the bonding pattern of the host lattice. At low temperatures each void remains in the vicinity of its respective impurity,
Under the provisions of the problem, w
0,
I
k2 2me I de 1'=-==/h v dk'
V
Here I is the mean free path, v == (1/h) (de/dk) is the velocity of the carrier whose wave number vector is k, m is the effective mass, e is the energy, and h == h/2rr, where h is Planck's constant. Equation (18.2.Sa) then reduces to l6rrm . a/o (19.2.1) K; =-3ji3 ell ae de. Now write I = loe' [this corresponds to the formulation l' = 1'oe r - y, of Problem IS.2(c)] and /0 ~ e (ll-e)/kT; e and Jl. both refer to the appropriate band edge, k is Boltzmann's constant, and T is the temperature. Then K{
\!
16rrml 0 ell/kT r~ ei+ r e-e/kTde . 3h 3 kT
Jo
(19.2.2)
19
436
19.2
converges to the value (kT)i+r+ 1r(i + r+ I) gamma function. Accordingly, ep./~T
Ki
161Tmlo(kTrlnr+l+i)3h3'
(19.2.3)
i. In zero magnetic field one finds, according to Equation e2 K 1 • Utilizing Equation (19.2.3) one obtains (18.2. 12a), that a ep./~T
161Tme2Io(kTY+ I r(r+ 2) 3h 3
a
.
( 19.2.4)
ii. One may now eliminate the exponential term from Equation (19.2.4) by use of Entry I of Table 18.2.2 or Equation (16.5.3); this yields nr+ 2) a - 4n1oe (kT) 3(21TmkT)'h _
Z
2
437
holes). On the right, A == Nd - Na is the net number of donor centres per unit volume occupied by electrons when l' = 0° K, and Aoee is the same quantity for T > 0, Nd and Na being the densities of donor and acceptor impurities. Let A+ = A - Aoee represent the net density of ionized donor centres. On this basis, consider the ratio A+nn / Aoee when nn is the density of charge carriers in the conduction band. Rewrite this quantity as (A)A)nn/(A"cc/A) and substitute for nn from Equation (16.5.3) and utilize Equation (19.3.1) in the limit where -Pelk1' 'if;> I Since nn A+, we find A+nn oec
~
When k1'
(19.2.5)
(c) According to Equation (18.2.13), a = (ZleT)(K2 IK I -p), where On inserting Equation (19.2.3) one obtains
r(r+ 3)
Transport in semiconductors
~
n~ K=nn
= 2gn
Ed, it follows that nn
(21Tmn kT)'h h2 exp(-Edl kT ) . ~
(19.3.2)
A; we then obtain
r
= ±l.
Z I,
19.4
1
a=eTL'tTr(r+2)-PJ
k(
P) .
Ze r+2- k1'
Solution
(a) According to the standard statistical theory (I) pertaining to impurity levels in semiconductors, the probability of finding a given level occupied by a charge carrier is given by (19.3.1)
In the above Pc == ~ - Ec is the Fermi level and Ed is the ionization energy taken relative to the conduction band edge (see Problem 19.1), and gn is a statistical weight factor with a value of! for electrons (gp 2 for (1) P.T.Landsberg in Semiconductors and Phosphors, M.Schon and H.Welker, Eds. (Vieweg, Braunschweig and Interscience, New York), 1958, pAS. E.Spenke, Electronic Semiconductors (McGraw-Hill, New York), 1958, pp.393-394.
exp(-Ed
[2gn
(19.3.3)
(b) Starting with entries I and 2 of Table 18.2.2 we may eliminate e1? in favour of n to find the following expression for the conductivity
(19.2.6)
19.3 (a) On the basis of the discussion in Problem 16.5 obtain an expression for the density of electrons nn promoted from donor impurities into the conduction band of an extrinsic semiconductor at temperature T. Assume that classical statistics is applicable, that bands are of standard form, and write the final results in terms of densities of donors N d , densities of acceptors N a , the effective mass mn of the promoted electrons, and the activation energy Ed, required for the promotion. For simplicity, set r = O. (b) Using the results of part (a) find an expression for the conductivity a of an extrinsic n type semiconductor in terms of the above mentioned parameters and in terms of To introduced in Table 18.2.2.
1 f= l+gnexp[(Ed-Pc)lkTI== A
nn
4e2Tolln 3mn (kT)'h _ 1T
a On find that
a=j
now for
lin
(r
= 0).
(19.3.4)
from Equation (19.3.3) and simplifying, we
(Nd
Na)]'h(21TmkT) 14 exp(-Ed/ 2k T)
(19.3.5)
which is the desired relation. If the variation of the product To 1'14 with l' may be neglected relative to that of the exponential term, then a plot of Ina versus III' should yield a straight line with slope -Ed /2k; Ed may then be deter mined from the optimal fit of the data to Equation (19.3.5). 19.4 This problem is designed to acquaint the reader with the electrical conductivity of intrinsic semiconductors. (a) Derive an expression for the conductivity a of an intrinsic semiconductor in terms of the effective masses (m n , m p ), mobilities (un, up), and energy gap (Eg), under conditions where extrinsic contribu tions from impurity centres may be neglected. Assume classical statistics to hold and that the bands are of standard form. (b) Show from the results derived in part (a) how the energy gap of the intrinsic semiconductor may be determined from the observed variation of a with temperature. (c) Assuming that the material is electrically neutral, derive a general expression for the conductivity of a semiconductor in which electrons and holes in comparable numbers participate in the conduction process, writing the result in terms of donor and acceptor densities. Assume a
Chapter 19
438
19.4
temperature range such that most of the impurities are ionized. Under what' conditions does this expression reduce to that obtained in part (d) Specialize part (c) to the temperature range before the intrinsic characteristics become significant, i.e. to the exhaustion range. Comment briefly on the temperature dependence of conductivity for this class of materials. Solution
19.5
Transport in semiconductors
Here, n~ and n~ are densities of holes and electrons bound to their respective impurities, and Nd and Na are the densities of dORor and acceptor levels. Under the conditions specified in the problem state ment, n~ and n~ may be neglected relative to Na and Nd respectively. On replacing np with nr Inn in the simplified electroneutrality equation and solving for nn one obtains
nn
According to the discussion of Problem 16.5, the density of electrons in the conduction band multiplied by the density of holes in the valence band is a quantity given by Equation (16.1.12). In the slightly different notation employed here (2), this expression reads
ni == (nnnp)'f2 = Nc Nv exp(-€ g /2kT)
(19.4.1)
where g€ c€ - v€ is the energy gap. For the problem under discussion, Equation (16.5.3) or Entry I of Table 18.2.2 is applicable, according to which Nc ==
2c1T~~kTr
(19.4.2a)
N
2(21TmpkT)'I.
(19.4.2b)
v
The symbols in the above expression have been introduced in Problems 19.2 and 19.3. Then _ nj -_ (nnnp) y, 2 (21TkT)'h }z2 (mnmp) % exp( €g/2kT). (19.4.3) For an intrinsic material nn = np == nj; hence the conductivity of the sample due to the holes in the valence band and due to the electrons in the conduction band is given by a
= nneUn +npeu p = nje(u n +u p ) 21TkT)!(' 2e ( ---;:;'2 (mnmp)%(u n + u p )exp(-€ g /2kT)
(19.4.4)
where U is the mobility of the carrier. If the variation of TJ/'(u n + up) with temperature may be neglected relative to that of the exponential term, then a plot of Ina versus I should yield a straight line with the slope -€ g /2k. (c) We base the derivation on the law of electroneutrality np -nn +Nd Na + n~ - n~ = 0 . Comparison of symbols
Problem 16.5 no Po E Chapter 19 nn rip e
(19.4.5)
(2)
E0
E, €v
€
0
me m. No mn mp No
Nv Nv
EF
~
Na )+ [ieNd - Na )2+ nflV, .
teNd
(19.4.6)
By similar techniques it may be shown that np = !-(Na -Nd)+[!
(19.4.7)
where nj is specified by Equation (19.4.3). The conductivity due to electrons in the conduction band and to holes in the valence band is then given by
a = nneUn +npeu p = eUp(bn n +np) (19.4.8) where Un and up are the mobilities, and b == Un /u p . On substituting from Equations (19.4.6) and (19.4.7) one finds a = eUpH(b
}z2
439
1)(Nd
-
Na )+ (b+ l)[i(Nd
-
Na )2+ nfl'lZ} .(19.4.9)
When Nd = Na the above reduces to Equation (19.4.4). (d) In the 'exhaustion range', nj < INc - Na I and Equation (19.4.9) reduces to a ~ (Nd - Na)beu p = (Nd - Na)eu n . (19.4.10) In this approximation, a varies with temperature to the same degree as Un does; the charge carrier density is effectively given by the constant quantity (Nd - N a ). This is to be contrasted with the cases dealt with in the earlier parts, where the charge carrier density varies sensitively with temperature. 19.5 This problem deals with thermoelectric phenomena in extrinsic and intrinsic semiconductors. Derive an expression for the Seebeck coefficient an of an n type extrinsic semiconductor in terms of the charge carrier density n. Derive a similar expression in terms of the density of donor and acceptor impurity centres, N d , Na • Do the same for a p type extrinsic semi conductor. Assume classical statistics to hold and that the bands are of standard form. (b) Show how the ionization energy d€ or a€ for creation of free holes or electrons may be determined from a knowledge of the variation of an or a p with temperature. (c) Obtain an expression for the Seebeck coefficient an of an extrinsic n type semiconductor in the exhaustion range. How does this coefficient vary with temperature in this range?
19.5
Chapter 19
440
(d) Derive expressions for the Seebeck coefficient 0:' of a nearly intri.nsic semiconductor in terms of appropriate band parameters, effective masses, densities, and mobilities for carriers in two bands. Specialize to the case of intrinsic semiconductors. (e) How may the band-gap be determined from the variation of 0:' with temperature?
19.6
Transport in semiconductors
contributions. On introducing Equation (19.2.6) and its counterpart for holes into the above, one obtains (o == On + Op) 0:'=
e~[-On~+2 ~~)+Op~+2-~~)J.
But with ~ = Pc + Ec = ev
Solution
(a) Under the stated conditions, Entry 4 of Table 18.2.2 or Equation (19.2.6) is relevant. When we substitute from entry 1 of Table 18.2.2 to eliminate PclkT == T/ we obtain h3 nn kT)'iz , (19.5.1) O:'n = --;; r+ 2 In 2(21Tm n
J
k['
where all symbols have been defined in Problems 19.2 and 19.3. On applying Equation (19.3.3) this may be rewritten as k( Ed 3 [h3gn(Nd-Na)/2],h} 2+2kT+41nT-ln (21Tm k)% .(19.5.2) n For holes we set T/ == (ev - nkT yields O:'p and Il'p
= -;; I.r+ 2 k
Pv/kT and proceed similarly. This n h
3
In 2(2m;;pkT)'1,
J
(19.5.4)
(b) In ordinary circumstances the term Ec/2kT or Ea/2kT outweighs the term i In T; hence, in first approximation, a plot of -O:'n versus 11 T, or of O:'p versus liT, should yield a straight line with slope ed/21< or Ea/2k respectively. In the exhaustion range nn = Nd Na, as is intuitively evident; the relative same result is also found from Equation (19.4.6) on neglecting to :HN" N.)2. Inserting this result into Equation (19.5.1) yields
nr
J
r
k (Nd Na)h 3 3 , - -;; Lr + 2 -In 2(21Tm k)'/2 + "2ln T (19.5.5) n which shows that in the exhaustion range -O:'n varies with temperature as ~(K/e) In T. (d) For intrinsic semiconductors one must take into account the participation of both electrons and holes in conduction processes. According to Equation (18.8.8), the overall Seebeck coefficient is given by 0:' = O:'n On + O:'p P (19.5.6) On+
O:'n
°
where the
coefficients with subscripts refer to the one-band
Pv
it follows that Pv = -Pc -
k
(19.5.7) , whence
EgJ
Pc
2)(op
eo
0:' =
(19.5.8)
on)+okT+oPkT .
We may simplify further by dividing expression (16.1.8) by (16.1.11); this yields IVe Pc - Pv = -exp--N I
np Pc _ -Pv - 2pc .=
kT
(19.5.9)
v
which may be solved for
+ Eg
kT
l
3 mn nnJ -In-+ In2 mp np
(19.5.10)
'
where the definitions (19.4.2a) and (19.4.2b) were utilized. Now solve Equation (19.5.10) for Pc and substitute this result in (19.5.8) or (19.5.7) to find
(19.5.3)
1 lnT _ [h3gp(Na-Nd)/2]'h} ek{ r+2+ ~ In (21TmpK)% . 2kT + 4
441
0:'
I< [On = --
0p (
e On + Op
J
3 mn 1 nn +-In---ln-. (19.5.11) 4 mp 2 np
r+2+
We now consider the quantity (b Un Iu p ) On Op _ nnUn - npu p bnnlnp - I nnUn+npu p bn nlnp+l' On+Op and on substitution in Equation (19.5.11) we obtain
lbn
(19.5.12)
J
E")
3 mn --In 1 nn . nIn P - 1 ( r+ 2+-"- +-Ine bn In + 1 21
k
0:'
I< [b - 1 ( ~) 3 mn b+ 1 r+ 2+ 2kT +4 ln m p
--;;
J
•
(19.5.14)
(e) If the dependence of b on T is neglected, Equation (19.5.14) shows that a plot of -0:' versus liT should yield a straight line with slope ~b-l~
21< b+ Ie' 19.6 (a) Show that the Wiedemann-Franz law cited in Problem 18.5 applies to extrinsic semiconductors in the limit of classical statistics and derive an expression for the Lorenz number.
19.6
Chapter 19
442
Contrast the numerical values of the Lorenz number when classical or when highly degenerate statistics is applicable to materials where carriers in a single band participate in conduction. Discuss the relative contribution of charge carriers and of the lattice to the overall thermal conductivity of extrinsic semiconductors and of metals. (c) Derive a general expression for the electronic contribution due to holes and electrons in an intrinsic semiconductor in terms of the one band thermal conductivities, one-band electrical conductivities, and band gap. Assume classical statistics to apply. (d) (s the total electronic contribution to the thermal conductivity of a solid invariably greater than the sum of the one-band contributions? Under what conditions may one expect this total to be very much greater than the one-band contributions? Solution
Under the specified conditions, the last Entry of Table 18.2.2 may be utilized: (19.6.1) Ke T(r+ 2)u ,
(~r
from which it immediately follows that the Wiedemann-Franz law applies and that the Lorenz number is given by £
=
(:r
(r+ 2).
(19.6.2)
(b) On setting r 0, and inserting standard numerical values the Lorenz numbers for the two extreme cases specified by Equations we obtain: (18.5.1) and (1 (l9.6.3a) £ I ·49 X 10- 8 y2 deg-Z, classical statistics,
£ = 2·45
X
10- 8
y2
deg- 2 , degenerate statistics,
(l9.6.3b)
where we have used the value k/e = 86·4 liY deg- 1 • In both cases the one-band thermal conductivities arising from the circulation of charge carriers are proportional to the corresponding electrical conductivities; the latter differ by many orders of magnitude as one passes from semiconductors to metals. By contrast, the lattice thermal conductivities of semiconductors do not differ enormously from those of metals. Hence, as a rough rule of thumb, one may state that the electronic contribution to the total thermal conductivity exceeds that of the lattice contribution in the case of metals, but that the opposite situation obtains in extrinsic semiconductors. Here we apply Equation (18.8.11), into which we substitute from Equation C19.2.6) and its counterpart for holes; this yields Ke
where
tg
Kn+Kp+ (
tc - tv'
e
k)2 TUn up
Un+U
n
[
(tg)~2
~(r+2)+ kT~
,
4 (19.6.)
19.7
Transport in semiconductors
443
Cd) Since the last term on the right-hand side of Equation (19.6.4) is positive, the 'am bipolar' thermal conductivity Ke always exceeds the sum of the one-band contributions Kn + Kp. When (i) up and Un are both large, (ii) the temperature is high, and (iii) tg/kT is large, the ambipolar contribution may make Ke much larger than the sum Kn + Kp . 19.7 This problem relates to the physical characteristics of solids in which electrons are transferred in the 'hopping-type' process discussed in Section 19.0. (a) Starting with the generalized version of Ohm's law, Problem 18.1 (b), and with Equation (18.0.1) for the electrochemical potential, derive the Einstein formula for the diffusion constant of a particle in terms of its mobility, for a nondegenerate material. (b) Obtain an expression for the diffusion constant of particles essentially localized about lattice sites, but being able to jump to equivalent neighbouring sites along a certain specified direction, in terms of the lattice spacing b and attempt frequency" for hopping. (c) Derive an expression for the electrical conductivity of a solid in which excess charge carriers can move between equivalent lattice sites by an activated diffusion-type process. Show explicitly how the tempera ture enters this expression. (d) Discuss briefly the variation of the electrical conductivity derived in the last part: (i) with temperature when the number of excess carriers is fixed, and (ii) with excess carrier density at a fixed temperature. Interpret these results physically. Solution
(a) As discussed in Problem 18.1 (b), Ohm's law in generalized form reads: J = uV(~/e) where J is the current density, u is the conductivity, V the gradient operator, e the electronic charge, and ~, as specified by Equation (18.1.1), is the electrochemical potential. Now, in the relation ~ lin eifJs, substitute for the chemical potential lin the standard expression of thermodynamics(3) lin
=
kTina n
(19.7.1)
,
where k is Boltzmann's constant, T is the temperature, an is the activity of the electron, and lio is the chemical potential of the standard state. Then Ohm's law may be reformulated as follows: J
= uV(~/e) = u("T Van -VifJs) ea n
-eJ n
(19.7.2)
where on the right we have introduced the particle flux I n == J/(-e). In (3) F.A.MacDougall, Thermodynamics and Chemistry (Wiley, New York), 1939, Chapter 13. G.N.Lewis and M.Randall (revised by K.S.Pitzer and L.Brewer), Thermodynamics (McGraw-Hili, New York), 1961, 2nd Ed., Chapter 20. S.Giasstone, Thermodynamics for Chemists (Van Nostrand, New York), 1947, Chapter 15. E.A.Guggenheim, Thermodynamics (North HoUand, Amsterdam), 1967, Chapter 5.
;1
l 444
19.7
Chapter 19
Transport in semiconductors
19.8
445
the absence of an externally applied electric field Equation (19.7.2) reduces to J akT (19.7.3) n e 2 a Van
(c) Assuming that the density of carriers remains constant and that b is sensibly independent of temperature, D varies with temperature as v does. According to the standard theory of reaction rates, we specify this dependence by use of the relation
The above relation may be compared with Fick's law of diffusion in the formulation J = -DVa n , where D is the diffusion coefficient. Compar ison with Equation (19.7.3) shows that akT D = -2(19.7.4) e an Writing the conductivity a as an eu n , where Un is the mobility, we obtain Einstein's formulation for the diffusion coefficient as
v = voexp(-€a/kT) ,
n
D=kTu n e
(19.7.5)
A generalization of the above result to cover the case of degenerate materials has been discussed by Landsberg(4). (b) Consider three parallel adjacent planes passing through rows of atoms in a lattice. Let e be the fraction of atomic sites surrounded by excess charge carriers, let v be the attempt frequency for a jump to adjacent sites, and Ox be the probability that a successful jump occurs in the two directions perpendicular to the above-mentioned planes and along the direction of the applied field. For a transfer to be successful the prospective site in either direction must be empty. Accordingly, the probability of successful jumps in unit time along the specified direction is vox(l- e). Let n(x) be the density of carriers on the plane x at time t, and let n(x ± b) be the corresponding density on the two adjacent planes separated from the central one by the distance b. Then, expanding n(x ± b) in Taylor's series we find n(x ± b)
= n(x) ± bn'(x)+ !b 2 n"(x)+ ....
(19.7.6)
The net change in carrier density on plane x in time dt is proportional to the difference in net inflow and net outflow, which in turn is governed by the probability factor derived earlier. Thus, dn(x)
= dt[n(x+b)+n(x-b)-2n(x)]ox(l-e)v.
(19.7.7)
On substituting from (19.7.6) and simplifying we find
an_ at -
2a2n
Ox (I - e)vb ax2
The above differential equation is of the form an/at encountered in diffusion theory. Hence we may set D
=
ox(l-e)vb 2
•
(4) P.T.Landsberg, Proc. Roy. Soc. (London), A2l3, 226,1952.
(19.7.8)
= D(a 2 n/ax 2 ), (19.7.9)
(19.7.10)
where €a is the activation energy needed to effect the transition of the electron from one site to an adjacent one. On substituting expression (19.7.10) into (19.7.9) and the resultant into expression (19.7.4), one obtains (1- e)v o e 2 ox b2 (19.7.11) a = an ITT exp(-€a/kT) . Let there be C lattice sites which can accommodate the electrons in a crystal of total volume V, and let 'Yn be the activity coefficient of the charge carriers in the lattice; then, an
ce
ce
n - 'Yn
V
Vo
n =-=-=-
(19.7.12)
where c is the number of centres available in a unit cell whose volume is Vo. Finally, define a quantity 0 3 == Vo/b 3 ; then an
=
ce'Yn
b3 0
(19.7.13)
3
On substituting expression (19.7.13) in (19.7.11) we find a
=
cVO'Yn Ox (e(l - e)e
0
bkT
3
2 )
exp(-€a/kT) .
(19.7.14)
°
(d) If one ignores the dependence of 'Y nO x / 3 on T, the above analysis predicts that a should vary with temperature as r1exp(-€a/kT), i.e. essentially exponentially. This is consistent with a diffusion controlled activated mobility process. At constant temperature, a varies with the fraction e of lattice sites that have associated excess carriers, according to the relation e(l - e); thus, a ~ 0 for e ~ 0 and e ~ I, and a is maximal for e = !. This may be understood on the basis that the excess carriers must have empty sites to which they can move. If there are either no excess carriers or no available empty sites there can be no movement of carriers; the case e =! represents the optimal balance between carriers and empty sites into which they can move.
19.8 This problem is designed to acquaint readers with the thermo electric properties of materials in which conduction occurs by a 'hopping' mechanism. (a) Show, from the thermodynamic interpretation concerning the Fermi level, how the Seebeck coefficient a of a charge carrier is related to its partial molal entropy S.
:i --
Chapter 19
446
19.8
Derive an expression for the configurational entropy corresponding to the random distributions of excess charge carriers among lattice sites. From this result and the expression obtained in part (a), derive an expression for the Seebeck coefficient in terms of the thermal contribu tion to the partial molal entropy of the carriers, and in terms of the fractional occupation of available lattice sites by such carriers. (c) Discuss the variation of the Seebeck coefficient in materials characterized by a 'hopping' conduction mechanism: (i) as a function of the fraction of cations in valence states, (n) and (n + 1), at constant temperature; (ii) as a function of temperature at constant ratio of cations in two valence states, (n) and (n + 1). (d) Provide a physical interpretation of the result discussed in part (c, Solution
(a) According to standard thermodynamic theory (5), the expression for the differential of the electrochemical potential is given by d~
-SdT+ Vdp+qd4ls ,
(19.8.1)
where S, V, and q represent the partial molal entropy, partial molal volume, an~ charge respectively. Thus, the derivative (a~ /aT)p, <1>8 is given by -So To the extent to which it is permissible to replace a == V(~/e)/VT by a a(~/e)/aT we may write
a=
S
(19.8.2)
e
In the diffusion-type transport model we may in first approxima tion regard the n carriers as nearly localized about the N fixed lattice sites. Using the standard Boltzmann combinatorial formula we then find for this system (19.8.3) S = Sr+Sc = Sr+klnW where Sr represents the thermal and Sc the configurational contribution to the entropy of the charge carriers on the lattice sites. According to the usual combinatorial statistics, the total configurational entropy reads N! Sc = kIn W = kln(N_ n)'.n., . (19.8.4) On introducing Stirling's approximation, Ina! a, we obtain
- = Sc (5)
See footnote (3).
(asun ":>.
a Ina - a, valid for large
n)ln(N - n) - nlnn .
Sc = NlnN-
Accordingly,
~
=
c) N
n
(19.8.5) (19.8.6)
19.8
Transport in semiconductors
447
On inserting Equation (19.8.6) into (19.8.3) and employing expression (19.8.2) we find
a
= - Sr -+ e
n
k
Sr
k
e
e
0
-+-In-
1-0'
(19.8.7)
where n
o
(c) According to Equation (19.8.7), €Xc == a+
e
k e
e
= -In--
1- 0
(19.8.8)
is a large positive quantity for e -+ 1, passes through zero at 0 !, and becomes a large negative quantity for 0 -+ O. Let us take as a simple model a material, such as an oxide in which the cation M can be in one of two valence states, for example M(n+ 1) and M(n). Here the superscripts (n+ 1) or (n) refer to the formal valence states of the cation. Since the higher-valent cation contains one less electron, 0 = [M(n)]/{[M(n)] + [M (n+ I)]}, where the square brackets denote concentrations. This predicts that a c as specified by Equation (19.8.8) is negative for [M(n)] < {[M(n)] + [M(n+ l)J} and positive other wise; ac -+ -00 as [M(n)] -+ 0 and a c -+ 00 as [M(n+l)] -+ O. Inasmuch as, at fixed [M(n)] and [M (n+ 1)], a c is independent of T, a varies with T only to the extent that Sr does. Generally, this is a weak dependence, so that in zero order approximation a is nearly independent ofT. (d) The variation of a with e may be understood on the basis that e -+ 0 corresponds to the presence of cations primarily in the valence state (n + I), with a sprinkling of cations in the valence state (n), containing an additional electron each. These excess carriers, being able to move to corresponding empty sites on adjacent cations in the (n + I) valence state, are clearly n type carriers. Since the sign of a reflects the sign of the dominant carrier species, one would expect a to be negative for e < !. When e -+ 1, nearly all cations are in the valence state (n), with a sprinkling of cations in the (n+ 1) state. The latter may be characterized by the absence of an electron, in a background of a preponderant majority of cations, all containing an additional electron. The carriers here may be considered to be holes; hence one expects a to be positive for e < !. The precise e value where a shifts from the positive to the negative range depends on the value of -Sr/e relative to that ofac •
20.2
Fluctuations of energy and number of particles
449
i.e.
dE _ = E2 - E2 I::J.E2 dT ' which is the required result. The above considerations clearly apply to a system at fixed volume or at zero pressure, since in both these cases the Hamiltonian operator is such that the eigenvalue will be just the internal energy in the state. In the case of constant non-zero pressure, however, the Hamiltonian of the system will include the potential associated with the forces producing the pressure. The corresponding eigenvalues cannot be interpreted as the internal energy and one must proceed differently. [See Problem 22.1 for further discussion.] kT2
20
Fluc tuations of energy and number of particles C. W.McCOMBIE (University of Reading, Reading)
20.1 A system, which may be macroscopic or microscopic, IS m contact with a heat bath at temperature T, so that the probability Pr of the system being in its rth quantum state, energy En is given by the canonical expression exp(-Er/kT) p = . r L exp(-Es/kT) s
20.2 Apply the preceding result to obtain the mean square fluctuation
in J~!!ergy, and:lthe mean square fractional fluctuation in energy of: (a) a
harmonic oscillator (it is convenient to take the zero of energy to coincide with the ground-state energy so that the zero-point energy does not appear explicitly), (b) a collection of N identical harmonic oscillators, and (c) a non-degenerate perfect gas of N particles in a container of fixed volume, all supposed in contact with a heat bath at temperature T. In cases (a) and (b) consider the form of the results in the high-temperature limit.
Write down expressions for the mean energy E, the mean square energy E2 and the mean square fluctuation in energy I::J.E 2 (where I::J.E E - E) and establish the relation - 2 dE Solution I::J.E = kT2 dT (a) For the oscillator, natural angular frequency wo, one has (r = 0, 1,2 ...) and one finds from the <:;anonical expression for Pr Discuss the applicability of this result, with E interpreted as the internal energy, to a macroscopic system: (a) at fixed volume, (b) at zero -E - L E nwo pre:;sure, and (c) under a fixed non-zero pressure. - r = 0 rPr exp(nwo/kT) - I so that Solution 2 dE _ 2 exp(hwo/kT) Recall first the very simple but important relation between mean kT dT (hwo) [exp( hw o/kT)-Ij2 value, mean square value, and mean square fluctuation from the mean which is obtained as follows: The lli;ll.LlVl fluctuation I::J.E(ll has mean square value (E-E)2 =
= E2_2E2+E2
E2.
1::J.E2
E2
Substituting the expression given for Pr in E = LErPr, we r
E Lexp(-Es/kT)
LEsexp(-EslkT) .
s
s
Differentiating each side with respect to T and dividing the result throughout by Lexp(-Es/kT) gives
Expressed in terms of E this becomes I::J.E 2 nwo =l+~ E At high temperatures (kT ;.>.;> nwo) these results become
E
kT, (kT)2,
s
dE 1 + dT = kT2E2 448
= I .
rhwo
450
Chapter 20
20.2
Note that the fractional fluctuations tend to unity, not to zero, as the mean energy becomes large. (b) Writing EN for the energy of N oscillators, and El for the energy of one of the oscillators, so that the results for El are given by the expressions already obtained, we have, N]!,l,
- tl£2N
and
=
E (1+ hWo)
Note that if N is large the fractional fluctuations are small. This is consistent with the requirement that the fractional fluctuations in energy of a macroscopic system in contact with a heat bath should be predicted to be small, in agreement with experience. (c) The equipartition result (Problem 3.6) gives at once for the non degenerate gas E = !NkT. It follows that !J.E2
and
dE = kT2= dT
~Nk2
!J.E2
1
2
T2
=jXN Note again the decrease in fractional fluctuations with increasing N. 20.3 Use the Debye model to find how the fractional fluctuations in vibrational energy of a crystal of fixed volume depend on temperature at low temperatures. Determine the fraction of the Debye temperature at which the root-mean-square fractional fluctuations become 1% for 10- 4 gram-molecule of a monatomic crystal. Solution
For s gram-molecules the Debye model specific heat at constant volume is given by C = S(1f1T 4R) v
~ = kT2
(!!..) 3
(fY ,
where () is the Debye temperature and R is the gas constant. The corresponding energy is T4 E= Cv dT = s(1f1T4 R)
(!!..)3
= 16 W T ~¥~N)T where N is Avogadro's number. Putting !J.E2/E2 10-4, s = 10- 4 and using N = 6 X 1023 , we find T
- 2 NkT2dT = N!J.E I ,
451
So we have
0' ~
dE\
kT2
-
Fluctuations of energy and number of particles
20.4
2
X
10- 6
•
[The expression used for C v can be linked up with the result rs1TSvk4T4 3 1 Eth ::::: 1.3 l: a3 .1'= I S of Problem 3.11 (d) if 1 3 I 1 -l: "3=3'" 3.1'= las aD Then ~1T5Vk4T4 T4 Eth =' z.3~3 = ~1T4Rs(j3 provided
=
k()
3 RS)'''' ( 41T vk haD'
Rs/k can be interpreted as the number of atoms, N, in the crystal. This formula gives an expression for the Debye temperature e.]
20.4 Derive the canonical-ensemble result,
-
dE
tl£2 = kT2 dT ' starting from the general expression for the mean value of an observable A (corresponding operator also denoted by A) A
with
= Tr(pA) exp(-H/kT)
p
= Tr[exp(-H/kT)] ,
where H is the Hamiltonian operator. Solution
We have E = H
= -::T...::r[-:-ex..:,p...:,(-.,-H-=-/:-;-k:-::T):.,. H-::.]
Tr[exp(-H/kT)] Since all operators which appear are either H or functions of H, they all commute and so the ordinary manipulations can be carried through. Thus _ dE I Tr[exp(-H/kT)H 2 ] I (Tr[exp(-H/kT)H]}2 dT = kT2 Tr [exp (-H/kT)] kT2 (Tr[exp(-H/kT)]}2
452
Chapter 20
20.4
20.5
If we write down the expression for E2,
with respect to J,l at fixed T gives (writing n for
_ Tr[exp(-H/kT)H2]
-
an) (aJ,l
Tr[exp(-H/kT)]
and that for E given above, then comparing t:.E2 expression for dE/dT gives the required result.
E2- £2 and the
1 ~n;exp[-(Er-J,lnr)/kT]
=
kT
T
kT
(b) Differentiating the expression for
LErexp[-(Er r
I
•
aE) _1 ~Ernr exp (aJ,l exp
(an)
aJ,l (aE) aJ,l
(b)
(c)
= kT2
T
kT
L
•
t:.n 2 = kT T
exp[-(Es - J,ln.)/kT]
(aE) aT
JJ.
T
•
1--
kT(En - En)
,
and
(a) Differentiating the expression for
1
kT(E - E)(n - n)
=
Lexp [-(E. - J,lns)/kT]
1-
kTt:.Et:.n .
(c) Differentiating the above expression for E with respect to T at fixed J,l gives, in a very similar way,
aE) (aT -
I
JJ.
= -[E(E kT2
1-
Solution
J,ln.)/kT]
_______________
~.
kT L exp[-(Es - J,lns)/kT]
'
If, corresponding to the fluctuations in energy and number of particles, we define fluctuations t:.T in temperature of the system by using the thermodynamic expression relating the temperature to energy and number of particles, show further that
t:.nt:.T
(E. - J,ln.)/kT]
r
+ kTJ,l(aE) aJ,l T
= kT2
[-(Er - J,lnr )/kT]
1 LErexp[-(Er J,lnr)/kT] Ln.exp[-(E.
'
t:.Et:.n = kT
--
J,ln)-E(E-J,ln)] =
-_
1 - - -
[E 2-E 2-J,l(En-En)]
::::: kT2 [t:.E2 - J,lt:.Et:.n] .
0.
Combining this with the expression for t:.Et:.n derived above gives the required result. (d) We get a reasonably convenient form for the coefficients in the expression for t:.T in terms of t:.E and t:.n as follows:
n,
L nrexp[-(Er - J,lnr)/kT]
•
I
-t:.n 2 kT
with respect to J,l at fixed T gives similarly
-
n
n2)
s
•
(e)
1 -
-(n 2kT
Lexp(-(E.-J,lns)/kT}
exp[-(Er - J,lnr)/kT] ~ . L exp [-(E. - J,ln.)/kT]
(d)
nin the derivative)
__ 1 [~nrexp(-(Er-J,lnr)/kT}]2
Deduce the following results for the fluctuations in the number of particles n, and the energy E [n and E being written nand E in the derivative (1 (a)
453
s
20.5 A system with fixed volume (or at zero pressure) is in contact with a heat bath at temperature T and a particle reservoir with which it can exchange one kind of particle, the chemical potential of such particles in the reservoir being J,l. The probability Pr of the rth quantum state of the system (number of particles nr and energy Er) is given by the grand canonical ensemble expression (cf Problem 2.4) Pr
Fluctuations of energy and number of particles
exp[-(E. - J,ln.)/kT]
t:.E
Tt:.S+ J,lD.n =
n
(I) For a macroscopic system this can be understood as identifying and E with the thermo dynamic equilibrium values nand E. In applying the results to a microscopic system nand E must be replaced by n and E.
= Cnt:.T+
TG~t t:.T+ [TG~)T + J,l] t:.n
LT(~~t +J,lJt:.n,
454
20.5
Chapter 20
where
en
is the heat capacity at constant n. This gives
D.T =
~n D.E ~n [TG~t + IlJ D.n .
Hence
MiD.T =
1
=
=
=
_1 en {kT2(aE) aT Il. + kTIl(aE) all T _ kT [T~S) \an T+ IlJ (aE) all T} kT2 [/aE) (as) (aE) J en \ aT Il. - an T all T .
=
dE-lldn
G~)
Il.
G~t dT+(~!)Tdll-lldn
=
dT+
We take the system of the previous problem to be the single-particle state concerned and suppose this state has energy e. In the Fermi-Dirac case n
exp[(e-Il)/kT] + I
and so, since the system is microscopic, n must be replaced by Ii in the partial derivative (cf footnote to Problem 20.5), -2
D.n
(~!) T(;it dT + (~!) T G~) Tdn - Ildn .
en T(;:)n e~t + (~!)T (;it (:~)Il. - (~!t G~)T ' =
where we have used the relation (as/anh == -(all/ann which is derivable from d(E- TS) = Ildn SdT. (e) Using the above expression for D.T we obtain
[).nD.T
~n { D.nD.E-
(~~) T+ IlJ D.n 2 }
= ~{(~!)T
[T(;~)T +IlJ G:)T}
= kT { (~!) T-
T(~~)T - Il (~:)
J
This is zero, since dE- TdS- Ildn = 0 gives de
so that
[TG~)T +1l(~:)TJdll+ [TG~t +1l(;;)J dT
(~!)T = TG~)T +IlG:)T'
(an) _~I'[(e 1l)/kT] _ (exp[(e 1l)/kT]+ IF - n
-_:::-2
kT all T -
n .
In the Einstein-Bose case I
n = ---::-:---- exp[(e- 1l)/kT] and so
D.n 2 kT(oli) ==
The required expression for Cn is now easily obtained: =
455
20.6 Apply the preceding result to obtain the fluctuations in the occupation number for a single particle state when the particles are Fermi-Dirac and when they are Einstein-Bose.
One can show that the last expression in square brackets is Cn , which is T(aS/ann, as follows:
TdS
Fluctuations of energy and number of particles
Solution
[TG~) T+ IlJ
{D.E2 -
20.7
all
T
exp[(e 1l)/kT] {exp[(e-Il)/kT]-I}2=
n+
20.7 If, in the system considered, the particles which can be exchanged with the surroundings are non-interacting and the temperature is high enough for corrected classical counting of states [cf Problem 3.4(c)] to be used, show that the fluctuations in the number of particles obey a Poisson distribution and verify the expression for N which follows from comparison with the standard form of the Poisson distribution. Solution
Let the single particle partition function be z 1, so that Zl
=
I exp(-es/kT) , Ii
where s numbers the single-particle states and ea is the energy of the sth single-particle state. If r numbers the states of the whole system, the energy and number of particles in the rth state being Er and Nr respectively, the probability P(N) that there will be N particles in the system is
peN) = C (Nr
I
exp[-(Er - IlNr)/kT] , N)
where the notation implies that the summation is over all states (Le. all r) for which NT = N. The normalisation constant C is just the inverse of the sum over all states (all r).
456
Chapter 20
20.7
This gives, on writing ZN for the canonical partition function of the system when it is constrained to have precisely N particles in it, zN P(N) = Cexp(pN/kT)ZN = Cexp(pN/kT) ~! .
This is of the form P(N)
21 Fluctuations of general classical mechanical variables
NN
C N! '
with
N = exp(ll/kT)zl
C.W.McCOMBIE
(University ofReading, Reading)
and normalisation gives C = exp(-N)
21.1 Show that if X is a coordinate of an entirely classical system in contact with a heat bath at temperature T and ~X = X - X then
so that the distribution is the Poisson distribution
_NN
peN) = exp(-N) N! .
We can calculate IV directly by writing
= kTe;;)T' where F is the generalised force on X due to external forces. [Hint: Consider the canonical distribution functions for the undisturbed system, and for the system with the force F applied, so that there is an extra term -FX in the Hamiltonian.j
N = i):zs where ns is the if
occupation number of the sth single particle state. The Fermi-Dirac and Einstein-Bose expressions for the mean occupation number both take the form /1s = exp[-(es - 1l)/kT] in the circumstances under which classical counting is appropriate. Thus
N = L/1s
= exp(ll/kT)Lexp(-es/kT)
8
Solution
In the presence of the force F we have
exp(ll/kT)zl
_
8
in agreement with the expression already obtained.
f exp[-(Ho- FX)/kTjdpdq
20.8 Compare the fluctuation in occupation number of a single particle state for non-interacting Einstein - Bose particles with the fluctuations in quantum number for a harmonic oscillator. Identical probability distributions imply the same mean square deviations from the mean, i.e. the same mean square fluctuations.
where we have introduced canonical coordinates and momenta and have expressed X as a function of them. Differentiation gives at once
ax) (aF
Solution
T
For Einstein-Bose particles the probability distribution for the occupation number n of a single particle state of energy e is given, as follows from the grand canonical result, by PIl = Cexp[-(ne
= C'exp[-(r+ tlhw/kT]
=
1 fX 2 (p, q) exp [-(Ho- FX)/kT] dpdq
f
kT
exp[-(Ho- FX)/kT]dpdq
__ I {fx(p, q) exp [-(Ho- FX)/kT]dpdq
Iln)/kTj = C{exp[-(e-Il)/kT]}1l .
The canonical distribution result gives for the probability Pr that an oscillator of angular frequency w will be in its rth quantum state Pr
fX(p, q)exp[-(Ho FX)/kTjdpdq
X
kT 1 -
= kT(X2 -
C"[exp(-nw/kT))'.
f exp[-CHo- FX)/kTjdpdq 1 ~X2
r 1
.
We shall frequently wish to apply this result to determine the magnitude of the fluctuations at F = 0, in which case the derivative with respect to F will be taken at F = 0, but it is useful to note that the result holds
These distributions agree if we take e = hw and 11 = O. The constants C and C", since they are determined by normalisation, must clearly be the same when e and 11 are chosen as above.
457 I
I
Chapter 21
458
21.1
21.3
more generally. The more general result will be applied later (see Problem 21.3) to determining the fluctuations in volume of a system under a fixed non-zero pressure.
X :;:
Lexp(-E,/kT)(rIXexp(FX/kT)lr)
-
ax (aF )T
r
L
(sIHo- FXlpl)(P1IHo- FXlpz) ... (Pn-IIHo- FXlr).
X and therefore Ho FX (since Ho has only diagonal matrix elements) will have appreciable matrix elements only between states which differ in energy by much less than kT. It follows that all the states which appear in terms contributing appreciably to the above sum will have eigenvalues (I) L.D.Landau and E.M.Lifshitz, Statistical Physics (Pergamon Press, London), 1968, p.345.
rL exp(-E,/kT)(rlexp(FX/kT)lr) r
Lexp(-Er/kT)(rlexp(FX/kT)lr) ,
But it is clear that the argument apQlied above to throw X into an alternative form will, when applied to X2, yield just the first term in the preceding equation. Consequently
L(rlexp[-(Ho
PI ·.. Pn-l
:;: kT
kT
r
(_1)n
1 Lexp(-Er/kT)(rIX 2 exp(FX/kT)lr}
__ 1 [~eXp(-E'/kT)(rIXeXp(FX/kT)lr}]2
L(rIXexp[-(Ho- FX)/kTjlr)
an equivalent form which can be differentiated without difficulty. If we suppose the exponential operator expanded in powers of its exponent, a typical term in (slexp[-(Ho- FX)/kTjlr) will be
Lexp(-Er/kT)(rlexp(FX/kT)lr) ,
Since there is now no difficulty about commuting operators we can differentiate with respect to F in a straightforward way and find
x _ Tr{Xexp[-(Ho- FX)/kT]} - -=T=-'r-;-(e-x-=p[::-"-_-:"(R-::-::-='--:F=X=)-7c/k:-:::T=j7-} o
It is not a straightforward matter to differentiate this with respect to F, because X and Ho need not commute. The classical behaviour which has been postulated for X allows us, however, to throw the expression into
_ _ _ _ _ _ _ _ _ _--:-_
_ -'-:r=-_ _ _ _ _ _ _ __
We again write the Hamiltonian of the system as Ho - FX and we let r number the eigenstates Ir), eigenvalues En of Ho. The mean value of X when the force is F can then be written
(rlexp [-(Ho- FX)/kTjlr)
L exp(-E,/kT)(rIXls><sl exp(FX/kT) Ir) ~"~S-==
L, exp(-E,/kT)(r Iexp(FX/kT) Ir}
Solution
-
459
of Ho which differ negligibly from Er . We can, therefore, replace Ho by Er in all such terms and so in the exponential. Noting that exp(-Er/kT) is then simply a numerical multiplying factor, we have
21.2 We consider now the derivation of the result of Problem 21.1 in the case in which the system has to be described quantum mechanically but the variable X behaves classically. This means that all components of frequency w in the time variation of X will be negligible unless nw < kT(1). Contributions of frequency w in the time dependence of X arise from matrix elements of X between stationary states which differ in energy by nw, so take the classical behaviour of X to imply it has negligible matrix elements between states which differ in energy by more than a very small fraction of kT. Use this assumption to derive the the result of Problem 21.1.
L (rIXls)(slexp [-(Ho- FX)'kTjlr)
Fluctuations of general classical variables
I
I,
(aX) = -
kT aF
T
X2_,r
=
.IlX2.
It may be convenient to include part of the term FX in Ho: this will not affect the above argument. If, for example, X is the volume so that F is the negative of the pressure one may wish to avoid considering a system under zero pressure since such a system (e.g. a perfect gas) may not have finite volume.
21.3 Use the result of Problem 21.2 to find (a) the fluctuations in volume of a system, at a fixed external pressure which need not be zero, (b) the mean square fluctuation in deflection of a galvanometer suspension, (c) the mean square fluctuation in charge on a condenser of capacity C with a resistor connected across its plates, and (d) the mean square fluctuation in the distance r between adjacent atoms in a one dimensional chain of atoms of one kind interacting by nearest neighbour harmonic forces. All the systems are supposed in contact with a heat bath at temperature T.
460
21.3
Chapter 21
Solution
(a) The generalised force associated with volume pressure, -po Thus the general result gives
= -kT(~) op T
1)
is the negative
.
This relates the volume fluctuations to the isothermal compressibility. The generalised force associated with deflection 0 is the couple M. The general result therefore gives =
kT(;~) T
21.4
Fluctuations of general classical variables
21.4 Show that the result established in Problems 2l.l and 21.2 can break down if X does not behave as a classical variable by supposing that X is the x component of the dipole moment of a large number N of negligibly interacting charged particles of mass m and charge e bound to centres such that their frequency of oscillation is woo Solution Xr
If we denote the x component of displacement of the rth particle by and the x component of the total dipole moment by Px , we have
Px = Lex, .
.
If c is the (isothermal) torsion constant this gives
-flO 2 kT c
461
r
'*
Since Xr and Xs (r s) are independent and, in the absence of an external field, have mean value zero x,xs O. Thus
This agrees, of course, with the equipartition result
p2 x
= e2 "x L..-, x
$
=
2 2 e 2 "x L.. r = Ne
7,S
t.kT
tcfl0 2 =
The mean potential energy of an oscillator is half the mean energy, so
but the derivation from the general formula avoids doubts about the validity of treating the macroscopic suspended system as having just one degree of freedom. (c) The generalised force corresponding to charge q on the condenser plates may be taken to be the potential difference V across a battery introduced in series with the condenser, the positive terminal of the battery being connected to the condenser plate on which the charge is q (the work done by the battery is then Voq). The general result therefore gives Oq
=
kT ( oV
2
22 mw OX
This gives
I rI hwo ] "2lz hwo+ exp(hwo/kT)- I
.
[1
lYe hwo ] - --hw +--~--~~~ - mW52 0 exp(hwo/kT)- I .
Since Px is zero in the absence of a field, this is also M;. Turning now to an application of the result derived for a classical variable, we note that the generalised force associated with Px is Ex, the x component of the electric field. In an electric field
= kTC.
This agrees with the equipartition result which is 1 flq2
1
2
jix = N ex- = Ne- E x mW5
so that
0Ji"
--- =
1
2C 2kT.
oEx
Ne 2
---2
mwo
We see that the result (d) The generalised force associated with the distance r between the neighbours is F if a pair of forces of magnitude F act outward on the two atoms. If the increase in energy of the bond between the neighbouring atoms is ! 1<:( then the result
or)2
-tlr
2
(or) kT- of
T
becomes flr 2 = I<:kT, it being assumed that the chain has free ends or is very long with fixed ends.
oPx )
= kT ( aEx
T
holds only for kT » hwo. Notice that this means that the states of the system between which has non-zero matrix elements, namely states in which one oscillator has changed its quantum number by ± I, will then differ in energy by much less than kT. This agrees with the form of the criterion for classical behaviour which we have used in preceding problems.
21.5
Chapter 21
462
21.5 Suppose that X can be expressed as a linear superposition of
normal coordinates qr of the system (kinetic energy ~]q;, potential energy
r
L!w;q;): r
L(Xrqr
X
21.7
Fluctuations of general classical variables
[Note: There is no difficulty in establishing the same result for a quantum mechanical system, Xl and X 2 being assumed to behave classically. It is necessary only to modify slightly the argument used in Problem 21.2.] Solution
r
(X might be, for example, the transverse displacement of the middle
point of a stretched string). The mean square fluctuation in X may be calculated either by~pplying the result of Problem 21.2 or by using the familiar results for Show that if X is classical (and what this for the a's should be considered), then the two methods will lead to the same result.
q;.
We take the Hamiltonian to be
Ho- FIXI F 2 X 2
Consider first the requirement that X is classicaL It will have matrix elements proportional to a r /w'f2 between states differing in energy by hw r . These matrix elements will be negligible for states differing in energy by more than a small fraction of kT, provided ar/w'/.' is negligible, except when hWr ~ kT. This means that all oscillators which play an appreciable part in the problem can be treated classically. To apply the result of Problem 21.2 we note that a generalised force Fx applied to X will result in a generalised force Fxar on the coordinate qr and so a displacement Fx(Xr/w; of the mean value of this coordinate. It follows that the resulting displacement of X (from zero) will be given by
I"'
JxleXP[-(Ho-
feXP[-(Ho- FIX I F2X2)/kT]dqdp This gives
(aXI) aF 2
X2 exp[-(Ho- FIX I F 2 X2 )/kTjdqdp
T
The second method (since qrqs
=
0, r
exp[-(Ho- FIXI - F2X2)/kT]dqdp
,. JXI exp[-(Ho- FIXI - F 2 X 2 )/kTjdqdp
a;
aX
- kTf exp [-(H - FIX - F X )/ kT]dqdp o I 2 2
a2 kTL-1
'*
fX 2
eX P[-(HO-
r Wr
s)
X I - F2X2)/kT]dqdp
Xl
2 Fx Lw r r
= kTaFx
,
where Ho, Xl, and X 2 are all functions of the generalised coordinates and conjugate momenta q, p. Then
Solution
This implies
463
x
gives
f
FIX I - F 2 X2 )/kT]dqdp
exp[-(Ho- FI
Ia;q;
1 ____
r
Now !w;q; HT, since, as already discussed, all oscillators which contribute appreciably can be treated classically. Substituting for gives the same final result as the first method.
=
kT(XI X 2-XI X2)
F2 X 2 )/kT] dqdp
I--;--;--;~;-;;-
kT~XI~X2
q;
21.6 Show that if Xl and X 2 are two coordinates of an entirely classical system and the corresponding generalised forces are FI and F2 then the correlation function for the fluctuations of the two coordinates when the system is in contact with a heat bath at temperature T is given by
~XI ~X2
(aXI) aF2 T
(aX2)
aF
I
T
Clearly the same calculation applied to (aX2 /aF I )T leads to the same expression. 21. 7 A conductor is connected by a thin wire to a very large conduc tor, the two conductors and the connecting wire all being formed from the same metal. The large conductor acts as a heat bath (temper ature T) and electron reservoir (chemical potential J.I.) with which the smaller conductor is in contact via the wire. The smaller conductor, in the presence of the large one, has electrical capacity 8.
21.7
Chapter 21
464
The fluctuations iln in the number of electrons on the smaller conductor may be determined either by applying the result for the fluctuations in charge on a capacitor (Problem 21.3) or by using the expression for the fluctuations in n derived from the grand canonical ensemble (Problem 20.5). Verify that the two approaches yield the same result. Solution
The mean square value of the fluctuation eiln in the charge is given by I e2 iln 2 - - - = lkT 2 -C 2 or iln 2
2
=
C.W.McCOMBIE (University ofReading. Reading)
-CkT
~
On the other hand the result derived from the grand canonical ensemble gives iln
I
22 Fluctuations of thermodynamic variables: constant pressure systems, isolated systems
kT(~:) T
.
The chemical potential which appears here is of course the chemical potential of the electrons in the particle reservoir, which will also be the chemical potential of the electrons in the smaller conductor (fluctuations being irrelevant to evaluating the right-hand side). The chemical potential can be raised, at constant temperature, by adding electrons to the system. These additional electrons will reside on the surface of the metal system and will raise the chemical potential merely by adding electrostatic energy to the electrons of the system. If 8n is the increase in the number of electrons on the smaller conductor, the electrostatic potential will increase from zero to e8n/-C, and the additional energy of the electrons will be e2 8n/-C. This isjust the increase 8J,1 in the chemical potential, so that e2 8n 8J,1 ~. Thus an) -C ( aJ,1 T = e 2
22.1 In Problem 20.1 we considered the fluctuations in energy of a system at zero external pressure in contact with a heat bath. We turn now to considering the fluctuations in thermodynamic quantities for a system at arbitrary fixed external pressure in contact with a heat bath. To avoid the difficulties associated with the introduction of a pressure ensemble (cf Problem 3.18) we shall regard the fixed external pressure p as introducing an extra term pv into the Hamiltonian, where /) is the operator associated with the volume. This enables us to use the canonical ensemble. (a) The observable associated with the Hamiltonian of the system Ho + P'} is E + pv where E is the internal energy. Fluctuations in this quantity, at fixed p, can be investigated in exactly the same way as were energy fluctuations in the zero pressure case (Problem 20.1). Show that this leads to the result for entropy fluctuations ilS2 = kCp • (b) Show also that
------ilSilv =
av )p kT ( aT
[Recall the result (Problem 21.3): ilv 2
.
-kT(av/ap)T']
Solution
(a) The argument which gave for the zero pressure case (Problem 20.1)
and the two expressions derived above are consistent.
-ilE 2 = kT2 (aE) aT now gives, since p is fixed, (ilE+pilv)2 = kT2[a(Ea;PIJ)1 But the first law of thermodynamics gives TilS = ilE+ pill) ,
465
22.1
Chapter 22
466
so that the left-hand side becomes T2 kT2 Cp . We thus have = kCp
while the right-hand side is
~
f [(R - 0 + pv )/kTl1, Trlvexp Tr(exp[-(Ho+ pv)/kTJ} ~~ ~
We have
11£
11£2 = T 211S 2- 2pTI1SI1v + p211v 2
kT 2 C
P
Also
al)) (aT
2pkT2 -
_
p
(av) ap
- p 2 kT
(av)
= kT2 -aT
= TI1Sl1v-
p
T
(av) ap
+pkT -
T
.
IIJ Is)(slexp[-(Ho+pv)/kT] Ir)
:L (rlexp[-(Ho+ pv)/kT]lr)
~ ~~-~~~
Differentiating with respect to T at constant p, and noting that (Ho+ P') and exp[-(Ho+ pv)/kT] have only diagonal elements, we have Iv Ir>(rIHo+ pvlr>
p
= TI1S- pl1v .
It follows that
r
av) (aT
Solution
.
Notice that we have made use of the following way of attaching a meaning to the fluctuation of a thermodynamic quantity such as entropy. We determine from ordinary thermodynamic considerations a linear relation between variations in the thermodynamic quantity and variations in energy and volume. We then use this linear relation to associate fluctuations in the thermodynamic quantity with the unambiguous fluctuations in the purely mechanical quantities energy and volume. (b) We have, denoting the eigenstates of HO+P1J by Ir>, v
467
Fluctuations of thermodynamic variables
22.3
r
i
I I
:L
These results can, of course, be put in a variety of different forms.
Note that if p is zero we reproduce the result kT 2Cp for 11£2, which
we derived in Problem 20.1 .
22.3 Deduce from the general result of the previous question that the fluctuations in internal energy and volume of a perfect classical gas at fixed external pressure and in contact with a heat bath at temperature T satisfy 11£2 kT 2 Cv and
11£11,) = 0 .
I
Lr
Solution
For a perfect gas of N particles
x
Ir> r
I
= kT 2 [v(£+ pv)-
1
I1vl1S =
(~) aT
Nk p
p
av) = ( ap T
= kTzI1IJ(I1£+ pl1v)
NkT p2
Substituting these expressions and
I kT2Tl1v11S.
Thus
NkT
PI)
and so
Cp
av) kT (aT
Cv+Nk
in the general expressions for and leads at once to the required results. These results may be obtained directly from the fact that the classical canonical distribution is such that the total internal energy of a perfect gas (translational kinetic energy of the molecules plus the internal energy of the molecules) has a probability distribution independent of any requirement placed on the positions of the molecules.
p
Deduce from the results of the previous question the values of and 11£l1v for a system at fixed external pressure in contact with a heat bath.
j
f'
22.4
Chapter 22
468
22.5
22.4 Starting from the result that, if X is a classical variable associated with, an isolated system, then the probability p(X)dX that the variable has a value between X and X + dX is given by
p(X)
ex:
Fluctuations of thermodynamic variables
When F is non-zero, and is supposed to have been applied reversibly,
.<:lE
!F.<:lX,
and the entropy is given, to second order in the small quantities F and LlX, by F S(Eo+ !F.<:lX, .<:lX) S(Eo, X o)+ !"T.<:lX - !G.<:lX 2 •
exp
where SeX) is the entropy of the system constrained to have the value X for the variable, show that for an isolated system
Since the process is reversible, the entropy must remain -.<:lX2 = kT (ax)
aF s' initial value S(Eo, Xo), and this implies where F is a generalised force associated with the variable, this force being supposed applied externally (cf Problems 21.1 and 22.6). [Hint: Expand the entropy as far as quadratic terms in .<:lE and LlX about the energy Eo and the equilibrium value Xo of the variable appropriate to the energy Eo. Show that the coefficients of .<:lX, and -.<:lX2 are zero, liT, and !k/.<:lX2 respectively. To determine the relation between F and .<:lX suppose that F is applied reversibly. This means that the entropy stays constant and that (for small F) the increase in energy M will be !F.<:lX. The required relation follows at once.]
LlX=
at F = 0,
B Also
where
Lar$XrX$ is positive definite. Corresponding to each Xr define a force Fr by
Xo
as
LarsXs s
so that each Fr is a linear combination of the XS" Starting from the exp(S/k) form for the probability prove that FuXv k/ju,v,
A,
dU+ pdv (Problem 1.5)
where we have used in analogy with dQ
= dE
-! r,$LarsXrXs
r,s:
= O.
T = aE TdS
LlX2 kT
quadratic terms have been retained and the quadratic form
Fr ::::: -
(as)x
1
(ax) aF s
S = So
S(Eo, Xo)+ A.<:lE+ B.<:lX - HCM2+ 2DM.<:lX + G.<:lX2) . the entropy has a maximum at X
.<:lX2 =F kT
22.5 Suppose that the entropy S is a function of a number of classical variables X., X 2 , ... , X n , which we shall take to be zero at equilibrium (i.e. at maximum entropy). Thus
Since, when the force F is applied, the internal energy E can vary from its initial value Eo as the force does work on the system, we must consider the entropy to be a function of E as well as X, say SeE, X). Expanding this about Eo and Xo (the equilibrium value of X when F is zero) we get, retaining only terms up to the quadratic,
Since, when E = Eo (M = (.<:lX = 0), we have
F
to its
which is the required result.
Solution
SeE, X)
469
where /ju,v is unity if u = v, and zero otherwise. result plays an important part in establishing the Onsager relations (Problem 25.7).]
FdX.
Again, when F is zero, so that M is zero, the stated relation between probability and entropy gives that the probability distribution p(X) for X will be proportional to exp[--t(G/k)LlX2]. Comparing this with the standard form of the normal distribution, exp (- .<:lX2 /2.<:lX 2 ), we get
Solution
We have
FuXv
k G = =2
.<:lX
d
= fFuXvCeXP(S/k)I)dXs
r
470
22.5
Chapter 22
22.6
where C is a normalising constant such that
f
Cexp(S/k)
Since
We have therefore )T = (~V2)s
1) dXs =
provided
I
as
a
Fu exp(S/k) = - axu exp(S/k) = -k axu exp(S/k) ,
s
dXs
If we recall that the generalised force associated with volume I) is -p, we can use the general results of Problems 22.4 and 21.1 to write down the mean square fluctuations in the two cases:
---
-kT (av) ap s ' (~V2)T = -kTG;) (~V2)S =
T '
so that the fluctuations in the two cases depend on the adiabatic and isothermal compressibilities, respectively. The volume fluctuations due to energy fluctuations may be determined as follows ~V2 = ~E2 kT2 .
(av) 2 aT
p
(at}) 2 aT
p
Cp
+ Cp
(av)2 aT
p ,
Ks
= TvZfp
'
where a p is the coefficient of volume expansion. This relation is established in Problem 1.8(b).]
ko u ,.
Solution
(b)
T
T
,
a2
KT
22.6 We can consider the fluctuations in volume of a body under zero pressure: (a) if it is isolated with an energy E such that its temperature is T, and (b) when it is in contact with a heat bath at temperature T. Write down the mean square fluctuations in volume in the two cases in terms of appropriate compressibilities. It seems reasonable to suppose that the fluctuations in the case of contact with a heat bath will be greater than in the case of isolation, because there will be fluctuations in energy of the body in the first case. These can be interpreted as fluctuations in temperature which, by producing thermal expansion, will give rise to fluctuations in volume. Verify by direct calculation that the mean square fluctuation in volume, due to energy fluctuations and resulting thermal expansion, is equal to the excess of isothermal mean square fluctuations (case (b) over adiabatic fluctuations (case
(a)
(apav) s (av) ap
+ ~V2
471
and this is a standard thermodynamic result. [Since the adiabatic compressibility Ka == Ks is (l/v)(ov/oP)a (see Problem 1.4), this thermodynamic relation is
we can integrate the expression for by parts with respect to Xu' The integrated part vanishes because exp(S/k) goes to zero at the limits, and we get
- = kfax FuXv axuvc exp
Fluctuations of tnl>rmnr!\ln<>m variables
,
23.2
Time dependence of fluctuations: correlation functions, power spectra, Wiener- Khintchine relations
(L~ ily(t)dt
y(t)-y
we define the correlation function I/J y (T) by
Vx
I
s
IS y(t)dt-
ily(t)dt
)2 = Jo[S Jo[S
dt, is clearly
kT (T) = -exp(-KITI/m). m ily(tdily(t2)dt 1 dt2
(b) The macroscopic equation of motion is now
Ie + 'YO + cO = 0 .
Jo Jo ~y(tl)ily(t2)dtldt2' On changing variables to t t l and T =
fS-t
dt -t
the correlation function is proportional to
But we have A I/J"x (0) and by a previous result From the equipartition result we have ! mv; Thus
0
[S [8
S
T
vx(O)exp(-Kt/m) .
ifJ"x(T) = Aexp(-KIT
The required mean square value is therefore
( Jo
= O.
This has the solution Thus for positive
0
[S
+ KVx
exp(-KT/m) and, since I/JvJT) = I/J", (-T), this implies
fluctuation from the mean,
ily(t)dt.
m1jx
w,,(T)dT.
Solution
rJosThe
""l/Jy(T)dT = 2S So""l/Jv(T)dT.
l/Jy(T)dT
(a) The macroscopic equation of motion is
from which it follows at once that l/Jy(O) = ily2 and l/Jy(T) = I/Jy(-T). It will be assumed that there is a time Tc such that I/Jv(T) is negligibly small for ITI > Te . Assuming S > Tc , show that the mean square fluctuation from the
J:
loS
Solution
l/Jy(T) = ily(t)ily(t+T) ,
y(t)dt is 2S
=
T
23.1 We consider a fluctuating quantity y(t) (deflection of suspended mirror executing Brownian fluctuations, component of velocity of a diffusing particle, or the like) with statistical properties independent of time. We shall assume that time averages and ensemble averages are equal. Putting
J:
J
23.2 Assuming in each case(l) that the correlation function for positive is proportional to the way in which the quantity decays macroscopically (i.e. with neglect of fluctuations) from an initial non-zero value to zero (the initial rate of change of the quantity, if necessary for specifying the macroscopic decay, being taken to be zero), determine the correlation functions at temperature T for (a) the x component of the velocity v of a particle of mass m moving in a medium such that the force resisting its motion is -KV, (b) the deflection 0 of a suspended system of moment of inertia I subject to a damping couple -'YO and a restoring couple cO. Assume damping less than critical.
C.W.McCOMBIE (University of Reading, Reading)
mean of
473
Provided both t and Stare greater than Tc (and, since S > Tc this holds for all except negligibly small portions of the range of t values between 0 and S) one can extend the range of integration of the integral over T to -00 and +00 without changing the value of the whole expression. Thus
23
ily(t)
Time dependence of fluctuations
t 1, this becomes
l/Jy(T)dT.
472
, 11
(1) It can be shown by general arguments or-as we shall see later (Problem 24.1)-in special cases by detailed calculation, that the macroscopic-decay correlation-function assumption made here is equivalent to assuming that the fluctuations can be regarded as due to the action of a fluctuating force with correlation time 7 c which is negligible on the time scale of the macroscopic decay.
Chapter 23
474
23.5
= 0 is easily found to be
The solution of this with (J(t)
23.2
2DS,
and comparing with the previous expression for x~ gives
where {3 = "Y/21 and w' = (41c-"Y 2 )Y'/21.
It follows that the correlation function is given by
Wo(r)
D
kT. J.1 This of course, the Einstein relation, which can be ways, see for example Problem 17.12.
= k;exP(-f3l r l) LcosW',r,+ ~,sinw'lrIJ.
in other
23.4 Assuming the relation between correlation function and macro scopic decay for the current in a circuit of inductance L and resistance R, determine the mean square fluctuation in the charge passing round the circuit in time S.
23.3 Use the results of Problems 23.1 and 23.2(a) to determine the mean square x displacement in time S of a particle moving at temperature Fx T in a medium such that its mobility is J.1 Combine this with the solution per, t)
475
Comparing this with the standard normal distribution exp(-x 2 /2Llx2) gives 2Dt for the mean square displacement. Thus
f3. ,t ] , + w,smw
exp(-{3t)
Time dependence of fluctuations
Solution
= n(4nDtr'hexp (-r 2 /4Dt)
The macroscopic
of the diffusion equation for n particles concentrated at the origin at t = 0 to relate mobility J.1 and diffusion constant D.
for the current 1, L
d1 dt
+ R1
0,
gives Solution
1
Since the x component of the displacement in time S is related to the x component of velocity by Xs
=
f:
we have (f:1YxU)dt
Combining this with
vx(t)dt,
r
= 2S
t""
x~
00
.
tLf2
tkTwe get kT WI(r) = yexpr-(RIL)lrl].
r
It follows, using the result of Problem 23.1, that
Q2 == (foS 1(t)dt
w(r)dr,
where w(r) is the correlation function for V x ' Since Fx = (l/J.1)vx we see that 1/J.1 plays the part of the damping constant K, so that the result of Problem 23.2(a) gives here kT w(r) = -exp(-irl/J.1m). m The integral of this from 0 to xi becomes
= 1oexp[-(R/L)t]
2S
t""
2kTS R
23.S Our previous discussion of the fluctuations in charge passing round a circuit of resistance R was based on the relation between correlation function and macroscopic decay function. We now consider a kinetic theory calculation which verifies the result of Problem 23.4 for a very particular model of the circuit and the resistive element in it. Suppose that the electrons in the circuit can be treated classically and that the resistance arises entirely from the presence of a symmetric potential barrier of height Vo. It will be supposed that the velocity distribution of particles in planes on either side of the barrier can be treated as equilibrium distributions for the purpose of calculating the rates at which particles move into the barrier region. Collisions in the barrier region will be supposed negligible. Show that the number of particles capable of surmounting a barrier of height V which pass into the barrier from one side per unit time is equal to V/kT) where C is independent of V. Determine, in terms of
is J.1kT and the above expression for
= 2J.1kTS.
Turning now to the given solution of the diffusion equation and y2+ Z2 we note that integrating over the y and z substituting r2 coordinates gives that the probability distribution at time t of the x coordinate of particles at the origin at t 0 is proportional to exp(-x 2 /4Dt) .
t
J
Chapter 23
476
23.5
23.6
Vo and C, the net current across the barrier when an electrostatic pot~ntial
Time dependence of fluctuations
and this will also be the mean square fluctuation in the number. If the numbers are denoted by n 1 and n 2 , and the charge passed by Q, then
20E is applied across it and hence determine the resistance of
the barrier. Determine also the fluctuations in net current across the barrier integrated over the time S and so verify that the previously derived relation is satisfied.
i:l.Q2
=
Q = enl-en2, 2 e i:l.ni+ e2i:l.ni = 2e 2 Cexp(- Vo/kT)S .
Comparison with the expression for 1/R gives
Solution
The number of particles per unit volume with momenta in the range dpx, dpy, dpz at Px, Py, pz is
i:l.Q2
A exp [-(p;+ P;'+ p;)/2mkT]dpx dpy dpz .
The number per unit time with z component of momentum in this range which pass into the barrier (supposed of area a) will be the number having z component of momentum in the required range which lie in a volume (pz/m)a. Define p~ by p? /2m = V. The number of electrons with momentum greater than p~ which pass into the barrier per unit time will be
m
=
y(t)
=
v
i:l.n r2 =
.
i:l.nri:l.n s
=
M
=0
r =F s ,
where A is the mean number of pulses per second. Deduce that y(t)
Ceexp[-( Vo eoE)/ kT] - Ceexp [-( Vo+ eoE)/ kT] .
= Al~~ f(t')dt'
and
For small oE this becomes
l/Jy(r)
oE 2Ce 2exp(- Vo/kT) kT
The resistance R of the barrier is therefore given by I
11r
and
If an electrostatic potential 20E is applied across the symmetric barrier, the heights of the barrier viewed from the two sides will be Vo eoE and Vo+ eo£. The mean net current across will therefore be
=
AL~ f(t)f(t+r)dt.
Solution
The first result is easily obtained:
2
R=
L nrf(t - t r ) .
Since the quantities nr are independent and each has a Poisson distribu tion we have
Baf~ exp(-x/kT)dx
= BakTexp(- V/kT) = Cexp(- V/kT)
S,
23.6 A fluctuating quantity y(t) is made up from a series of identical pulses of form f(t) randomly distributed in time. If we suppose the time scale divided up by a series of closely and equally spaced instants t" such that tr+ 1 - tr = 0, and that nr denotes the number of pulses 'centred on' times between tr and tr+ b we may write [if f(t) is 'centred on' t = 0]:
Bexp(-p;/2mkT)dpz .
Bf~ exp(-p;/2mkT)PZ adpz
2kT
=R
This agrees with the previous result.
The number with pz in dpz at pz (z will be supposed normal to the plane of the barrier) will therefore be
p',
477
Ce kT exp (- Vo/kT) .
y(t)
Because of the independence of the classical particles we can suppose that the number of particles crossing the barrier from either side will have a Poisson distribution. For each direction the mean number crossing in time S will be
=
~nrf(t -
tr)
= M ~ f(t -
tr )
= AL~ f(t')dt'
,
where we have let 0 tend to zero to turn the sum into an integral. The second is derived as follows. We have i:l.y(t)
CSexp(- VolkT)
t I
Ii
= y(t)-y(t) =
Lnrf(t-tr )- Lnrf(t-tr )
=
Li:l.nrf(t-tr )
23.8
23.6
Chapter 23
478
r~Llnrf(t
~ Llnsf(t+ T- t s) ]
t r )] [
LLlnrLlnJ(t
tr)f(t+T-ts)
"s
Solution
We have, from a standard Fourier transform result,
= ALOf(t-tr)f(t+T t r )
i+= _~ I Ys (w)1 dw.
+=
AL~ f(t)f(t+T)dt,
=
2
In the limit of large S-.!he left-hand side, which is the integral of y2 over a YS'(-w) so that time S, becomes Sy2. Since YsU) is real Ys(w) YS (w)1 2 = I YS {-w)1 2. Thus the right-hand side can be expressed as an integral from zero to infinity and we have
where we have again let 0 tend to zero to make the last step.
-
23.7 A current I(t) consists of a random series of pulses occurring at a mean rate A per second, each consisting of a constant current 10 lasting for a time t". Find the correlation function.
y2
= lim -2f"" s .. oo S
From the previous result we have
= {~6Uo-ITI)
ITI < to otherwise.
.
So""Gy(w)dw .
y(t)
=
r
L
f(t
r
the summation being over all r such that tr lies in the interval of length S used in defining y s (t). This gives
f+= Ys(w)exp(iwt)dw . _00
Ys(w) =
=
= lim ~.,----
f+=
I
~.J2rr _~ f(t- tr)exp(-iwt)dt
f
1 _= f(t- tr)exp[-iw{t- tr)]d(t- tr)exp(-iwtr ) ~v'21T
s... ~S
= F(w) Lexp(-iwtr ) ,
where the bar denotes an ensemble average. Show that y2
2 dw
YsU) = Lf(t- t r ) ,
Then Gy (w) is defined by
Gy
1
where the tr occur at random, there being on average A values of tr per unit time. If S is very long compared with the duration of a pulse we may write (neglecting only very small end effects)
23.8 The power spectrum Gy(w) of a real fluctuating quantity y(t), for which y has been made zero by suitable choice of origin, is defined as follows. We first define a truncated quantity y s (t) which is equal to y(t) in a range of t of length S and zero elsewhere. The Fourier transform of ys(t) is denoted by Ys(w), so that 1 = .J2rr
Ys{w)
y2 =
Suppose now that
since the integrand has the value IJ throughout the region for which two pulses, displaced with respect to one another by an amount T, overlap, and is zero elsewhere. The length of the overlap region will be to - IT I if IT I < to and it will be zero otherwise.
YsCt)
I
0
Clearly, taking an ensemble average of the right-hand side will not invalidate this equation, so we get
Solution
\fII(T)
479
Evaluate Gy (w) for the case in which y(t) is formed from a series of identical pulses occurring at random times, a pulse being assumed for will be zero and simplicity to have zero time integral, so that readjustment of the zero is not necessary. Show from this example that it is necessary to take both the limit of large S and the ensemble average in order that the definition of G(w) should be ';'Ilti"f",..tnru
and so . " . , . '=
Time dependence of fluctuations
= tOO Gy(w)dw
r
where F( w) is the Fourier transform of a pulse 'centred on' the origin,
.
:1
480
23.8
Chapter 23
Le. F(w)
23.10
Time dependence of fluctuations
For sufficiently long S the output for the truncated input will differ from the truncated output ys(t) only by a negligible end effect. We can, therefore, write
1
I _~ f(t)exp(-iwt)dt . v2rr
Ys(w) = A(w)Xs(w)
This gives
I Ys(w)1 2
I F(w) 121
~exp(-iwtrf
and the output power spectrum Go(w) is then given by 2 -~~~~~:--
If S is very long compared with l/w, the summation on the right-hand side will be over complex numbers of modulus unity with random phases, there being on average AS terms in the sum. The square of the modulus of this sum will be the square of the distance from the origin to the final point of a two-dimensional random walk consisting of on average AS steps of unit length. No matter how large S, these distances squared will fluctuate wildly. It is only if we average over the directions of the steps by averaging over the possible values of the tr (i.e. taking an ensemble average) that we get a definite value, namely the mean square displace ment in a random walk of A.') unit steps, which is iust AS. Thus for sufficiently large S I Y (w)1 2 = ASIF(w)1 2 S
and so
2-----=
Gy(w) = 51 Ys (w)1 2 = 2AIF(w)1 2
•
It is clear that it has been necessary both to take S large and to take an ensemble average. Thus a definition of Gy (w) which did not include both the limit of large S and an ensemble average would not lead to a sensible result in the particular example considered, and so would not be a satisfactory definition.
23.9 A linear system has a response function A(w), i.e. an input exp(iwt) gives an outputy(t) = A(w)exp(iwt). Determine the power spectrum Go{w) of the output, if a fluctuating input with power spectrum G1(w) is applied. Hence express the mean square value of the output in terms of the -input power spectrum.
xU)
Go(w)
2
lim SIYs (w)1 2 = IA(w)1 2 lim SIXs(w)
s~~
s~~
If the Fourier transform of the truncated input (defined as in Problem 23.8) is denoted by Xs(w), we have 2
lim S IXs{w) 12 .
s-+""
The Fourier transform of the output produced by this truncated input will be A(w)Xs(w) .
= IA(w)1 2 G1(W).
The importance of the power spectrum concept stems in large measure from the simplicity of this relation between output and input power spectra. The relation may of course be regarded as obvious since one may think of G( w)5 w as being the mean square value of the narrow band fluctuations obtained by eliminating all Fourier components of the fluctuations except those with frequency in the range 5w at w. (From the point of view adopted here, this follows from the expression which has been derived for the mean square value in terms of an integral over the power spectrum; one may, alternatively, start the discussion from this less formal definition of the power spectrum.) Since a narrow band fluctuation is sinusoidal with slowly varying amplitude and phase, IA( w) I will give the ratio of the output amplitude to the input amplitude for the frequency range considered and the result follows at once. The mean square value of the output is given by
L~Go(w)dw = I
IA(w)1 2 G1(w)dw.
23.10 Calculate the response function A(w) where the system is a capacitance and resistance in series, the input being the voltage across them and the output the current through them. Calculate also the response function for a damped suspended system, the input being the applied couple and the output the deflection. Solution
In the case of the capacitance C and resistance R in series, the voltage V = VOe iwt and the current 1 = loe iwt are related by
VO Solution
G1(w)
481
Z(w)/o ,
where Z(w), the impedance, is given by elementary a.c. theory, 1
Z(w) =
R+-.
10
A(w)Vo
IWC
But, by the definition of A so that I A(w) = Z(w) = R
I
+ l/iwC
.
482
23.10
Chapter 23
For the damped suspended system the deflection applied couple P by an equation of the form 18+K8+ce
If we put P(t) by
e iwt, then
e is
23.13
related to the
It follows that
P(t).
so that
+ iKW
The inverse of this relation then gives
.J2i -2- Gx (w)
23.11 Calculate A( w) for the case in which the input xU) produces an output y(t) xU + T) x(t). Hence obtain the mean square value of the output if the power spectrum of the input is GA w), and so obtain the correlation function for x, 1/1 x (T), in terms of the power spectrum of x, Gx (w). This gives a somewhat unorthodox _derivation of the__Wi~l1~r Khintchirie--relation between correlation function. andpQwer spectrum. -snow·· that the relation obtained can be inverted to give the power spectrum in terms of the correlation function. Note: No physical system could have the response assumed here being assumed positive), since it implies a response to, say, an input [) function which occurs before the input is applied. This is, however, unimportant for the purely mathematical discussion involved in the question.
or
Gx(w)
G(w) = 2 1r
0
23.13 Evaluate the power spectrum of the fluctuations associated with a random series of identical pulses by applying the Wiener-Khintchine relation (Problem 23.11) to the expression for the correlation function obtained in Problem 23.6. Check that the result agrees with that obtained by direct application of the definition of the power spectrum in Problem 23.8: notice that in this alternative derivation we do not make the assumption (introduced in Problem 23.8 for simplicity) that the time integral of a pulse is zero.
cosWT)GAw)dw.
= 2x L
foo 1/I(T)dT
for values of w up to a value less than (but of the order of) liTe· This expression is independent of w.
But the left-hand side is X(t+T)2+ X(t)2-
0
Because 1/I(T) is negligible for T > Tc , we can restrict the range of integration over T to that between zero and Te' If w ~ I lTc, WT will be much less than unity for all T in the above range, so that cos WT may be replaced by unity in the integrand. Thus to a good approximation
L~leiWr - 112 GAw)dw (1
0
2f'"1/I(T)coswTdT. 1r
e iwr - 1
I
= -2 foo 1/Ix(T)coswTdT.
G(w) =
Expressing the mean square value of the output in terms of its power spectrum, IA(W) 12 GAw), gives
= 2
.
1/Ix(T)e- 'Wr dT
We have the Wiener-Khintchine relation derived in Problem 23.11
so that
=
_00
Solution
eiw(t+r)_eiwt = (eiWr-l)eiwt,
[x(t+ T) - x(t)]2
1-
I .J2i
23.12 Show that if the correlation function has dropped practically to zero for T > Tc , then the power spectrum will be constant up to frequencies of the order 1ITc . Tc is called the correlation time.
An input x(t) = e iwt will produce an output
=
=
1r
Solution
A (w)
Gx(w)coswTdw.
f-
c Iw 2 + iKw '
c
1
To invert this, starting from the more familiar exponential form of the Fourier transform relations, we may proceed as follows. We define Gx(w) for negative W by Gx(-w) Gx(w) and rewrite the above relation as 1 _'" -2-Gx(w)eiWr t::r=. 1/Ix(T) = .J2i ~ v21r dw.
e iwt
A(w) =
483
00
1/Ix(T) =
e will be given, after the decay of transients,
e
Time dependence of fluctuations
21/1x(T)
2 t~Gx(W)dW-21/1x(T).
'I
23.13
Chapter 23
484 We have from Problem 23.6
Af-~f(t)f(t+r)dt.
=
The Wiener-Khintchine relation gives
2 i~
Gy(w) = -
~
1
l/Jy(r)coswrdr = -
~
0
Ar-r
= -; J-~ J-~
f +~ l/Jy(r)eiWTdr -~
A(w) = VFiF(w)
which is the required result. 23.15 Consider a system for which the response to an input oCt) is a 'delta function delayed by a time r, oct - r). Determine the response functionA(w) for the system and show that ifit is written in the form A(w)
f(t)f(t+ r)eiWTdtdr
= A'(w)-iA"(w)
with A' (w) and A" (w) real, then A' (w) and A" (w) satisfy the relations
r-f _l ~rL~f(t)eiwtd~ lf-~f(t)e;wtdtJ i -·---xdx
A = -;J-~
=
485
Time dependence of fluctuations
But the ratio of the Fourier component of the output at frequency w to that of the input at the same frequency must be A( w), and it follows that
Solution
l/Jy(r)
23.15
,
_~ f(tdf(t2)eiw(trtl)dtldt2
= 2AI F(w)
12 ,
where F(w) denotes the Fourier transform of f(t), and we used in the last step the fact that f(t) is real. This agrees with the result obtained previously.
1 i-A"(W')dW'
A (w)
= -P
A"(w)
=
~
~
,
w-w
_~
,
pi-A'(W')dW' _~ w'-w '
if, and only if, the delay time r is positive. The symbol P indicates that the Cauchy principal value of the integral which follows is to be taken. smax [Note: _~ has the value ~, 0, or -~ respectively when a is positive, zero, or negative.]
Solution
23.14 The remaining problems of this section lead up to an important result concerning the response functions of physical systems. This result will be made use of in the following section. The response of a linear system may be specified by giving the output f(t) produced by a delta function input Set). By taking the Fourier transforms of the output and the input, when the input is Set), show that the response functionA(w) for the system is the Fourier transform of f(t) multiplied by .J2ir. The Fourier transform F( w) of f(t) is taken to be given by F( w) = - -1-
VFi
i- . -~
f(t) e- 1wt dt .
In this case f(t) A(w)
= =
o(t- r) and, we have from Problem 23.14 PC y2~F(w)
e- iWT A'(w)
i-.
VFi
-~
1 S(t)e-1wtdt =---
vr:rr
=-1
f
VFi
+00
-~
f(t)e-iwtdt.
,
0(t-r)e-1wtdt
= coswr- isinwr .
i
P
-A"(w')dw' , w-w
_~
f f
=P
-
{
A"(w)
= sinwr .
-sinw'rdw'
_~
--~,----
w-w
-sin[(w'- w)r+ wr]d(w'- w)
,
w-w
~
= coswr
The Fourier transform of the output f(t) will be F(w)
1-
VFi~
= coswr ,
It follows that
The Fourier transform of the input o(t) is 1 ---
-- 1 v2~-
Thus
=P
Solution
=
f-' _~
smrx ----dx+ sinwrP
x
~coswr
r
>0
0
r
=0
-~coswr
r
<0
f ~
cosrx ---dx
x
486
Chapter 23
L~[(COSTX)/Xldx is
zero because (COSTX)/X is odd, and that, since
[(sin TX )/x] dxhas no
A(w)-A'(oo).
singUlarity, the operation of taking the principal value does not affect it. The other relation can be verified similarly.
. 23.17 Verify by direct calculation that the response function relating input voltage to output current for a capacitance and resistance in series (Problem 23.10) satisfies the Kramers-Kronig relations (Problem 23.16).
23.16 If f(t) describes the response of a physical system to an input in the form of a delta- function at zero time, it is clear that f(t) must be zero for negative t as the system cannot respond before the signal is applied. But if f(t) is zero for negative t it can be represented as a superposition of delta functions o(t T) delayed by times r, with all T positive or zero. Assuming that the instantaneous response of the system may be neglected, so that only T > 0 need be considered, use the result of the previous problem to show that the real and imaginary parts of the response function for the physical system will satisfy the Kramers Kronig relations , 1 i-A"(WI)dW' A (w) = -p , , 'IT
_00
Solution
We have (Problem 23.10) I A(w) = R + l/iwC = iwC 1+ iwRC Thus w 2 RC 2 wC A"(w) =
w-w W
1+ w 2R 2C 2
Note that
1 l-A'(W')dW' A (w) = --p / . ~
w 2RC2+ iwC 1+ w 2R 2 C 2
A'(w) = 1+ w 2 R2(;2 ,
"
'IT
481
On the other hand a part Co(O in f(t) will contribute a part C to A(w). This does lead to a contribution at infinite frequency so we can write A(oo) = C. Thus the contribution of CO(t) to A(w) may be written A'(oo). To get the part of A'(w) which satisfies the above relations we must subtract out the instantaneous response part, i.e. replace A(w) by
and since A'(w) is COSWT, we see that the first relation is satisfied if, and only if, T is positive. We have used the fact that P
Time dependence of fluctuations
23.11
23.15
A'(oo) =
w
I
Ii
so that
Show that, if one wishes to get relations which take account of the possibility that the function f(t) contains an instantaneous response part Co(t), one has to replaceA'(w) by A'(w)- A'(oo) in the relations written above.
I
A'(w) - A'(oo) = - R(1
I
We verify first that
Solution
As A(w) is ..J2ii times the Fourier transform of f(t), the A(w) corresponding to a superposition (i.e. linear combination) of delta functions will be the same linear combination of the associated response functions. Since the delta functions all have delay r > 0, the associated response functions will all satisfy the given relations and, as the relations are linear in A /(w) and A" (w), a superposition of response functions which satisfy these will also satisfy them. This establishes the result, on the assumption of no instantaneous response. It will be useful to note here that, if there is no instantaneous response, A (00) will be zero: an individual delta function with delay T will contribute coswr- isinwT to A(w) and this oscillates as w tends to infinity; the total A(w) will, however, be obtained by integrating over r the product of this into some reasonably smooth function of r [in fact the multiplying function is f(T)]: the rapid oscillations with T for large w will cause the integral to tend to zero as w tends to infinity.
A'(w)- A'(oo)
The right-hand side is C
+ w2R2(;2)
r'OOA"(w')dw'
;P J~
w' - w
1-
;P
-00
w'dw ' (w' - w)(l + w'2R 2 C 2)
The integral can be evaluated either by contour integration or by elemen tary methods. We outline both procedures, contour integration first. The integrand has a pole at w' = w with residue w(1 + w 2R 2C 2r 1 and poles at w' ±i/RC with residues (w+ i/RC)(l + R2C 2w 2rl and 2 2 2 -1(w i/RC)(1 + R C W r' respectively. We consider a contour which traverses the real axis from -L to +L except that it passes above the pole on the real axis in a small semicircle centred on the pole. The contour is completed by a large semicircle of radius L centred on the origin and lying in the upper half plane. The contribution of the large semicircle
,t
488
23.17
Chapter 23
part of the contour integral will go to zero as L increases and we there fore have by the residue theorem co w'dw' 1Tiw 1Ti(w+i!RC)
r+
Pj..""
24
l+w2R2C2=-I+w2R2C2,
i.e.
P[
(w'
Nyquist's theorem and its generalisations
w'dw' 1T w)(l + w'2RC 2) = RC 1+ w 2 R 2C2
Consequently the right-hand side in the Kramers-Kronig relation being considered is -(l/R)(l + w 2 R2C2 1 , which is also the value of the left hand side. The other relation can be verified similarly. An elementary evaluation can be based on expressing the integrand in partial fractions by means of the identity
C.W.McCOMBIE (University ofReading. Reading)
r
w'
I
(w'-
Since dw' and
(W
= 1+R 2C 2w 2 w'-w
P
f-
+ 1+
dW - ,' . hm (S-€dW' -,+ __ W 1" ... 0 -co W
-
---7"--.,.-= =
l""dW') -W, -0 I"
0,
the required integral is equal to I dw' 1+ R2C2 W '2 = RCO
in agreement with what we had previously.
2 2wW')
I-R C
1T
+ w 2R2C2)
.
24.1 The electrons in a resistor may be supposed to receive impulses from the atoms of the lattice. These impulses are equivalent to a fluctuating voltage with correlation time (Le. the 7c of Problem 23.12) which will be of the order of the duration of an impulse. Since this duration will be very short we may suppose the power spectrum of the equivalent voltage to be constant up to very high frequencies (constant in fact for all frequencies for which the resistor has a resistance independent of frequency). By considering a circuit consisting of a resistance and capacitance in series, and requiring that the power spectrum of the voltage fluctuations associated with the resistor should lead to the equipartition result for the fluctuations in charge on the capacitance, determine the constant value of the power spectrum (i.e. obtain Nyquist's Show that this implies a correlation function for the fluctuations in charge which agrees with the decay function assumption made earlier (Problem 23.2). Solution
As we saw in Problem 23.10 the response function relating input voltage to output current for a resistance R and capacity C in series is I R+ l/iwC .
The response function relating (input) current to (output) charge is I/iw since charge is the integral of the current. Thus the required response function A (w), relating voltage to charge, is given by
I) 1 Aw () = 1 I: ( iw n-L-
,.1'
,
so that IA(w)i2 =
...
C2
-1').....-.1')
'l
•
If the constant power spectrum of the fluctuating voltage is G y it follows that the power spectrum Gq (w) of the fluctuating charge will be given by Gq(w) = IA(w)1 2G y
489
24.3
24.1
Chapter 24
490 This gives
2
Gq(w)dw
r""(l/R )G v dw _
Jo w 2 + (l/RC)2 -
=
1
7.
s:.
In systems in thermodynamic equilibrium each source of fluctuating force will have a damping associated with it. Thus, so far as the mean square value of the fluctuations of the system is concerned, the tendency of an added fluctuating force to increase the fluctuations is just compensated by the tendency of the associated damping to reduce them. The ratio of the power spectrum of the total fluctuating force to the total damping force remains fixed. It is instructive to examine the constancy of the ratio of power spectrum to damping more explicitly in the case of the couple acting on a galvanometer coil. If the coil has effective area A in a magnetic field H, the field direction being in the plane of the coil when the deflection is zero, then a current I will give rise to a couple AHI. An angular velocity (j will give rise, if the deflection is small, to an induced e.m.f. of magnitude AH8: if the resistance in the circuit of the coil is R this will give rise to a current A1l8/R and so to a couple of magnitude (A 2H2/R)O. The actual sign can be determined from the requirement that the net result is a damping. I f we denote the air damping constant by K, the fluctuating couple due to bombardment by air molecules by P(t), and the fluctuating voltage associated with the resistance in the circuit by V(t), the fluctuating deflection e(t) will be determined by
1f G v . R
But the charge fluctuations on the capacitor satisfy CkT.
Equating the two expressions for Gv
=
This is Nyquist's theorem. The correlation function for q, 1/Iq(T), is obtained from Gq(w) by the Wiener-Khintchine relation
-f-
1/I q(T) -
0
Gq(w)cosWTdw
2
C Gv
f"" 1+COSWT R2C2w2dw . 0
The integral is easily evaluated by contour integration
" f
491
Solution
f:
o
Nyquist's theorem and its generalisations
-
1
1+ R2C2w2dw -
COSWT
7.
r-exp(iWT)dw 1 + R2C 2 W2 -
J--
If
'2
c
exp(izT)
1+ R2C2Z2d.z
.
where, for positive 1', the contour C is the real axis and the semicircle at infinity in the upper half plane: the integral round this semicircle is zero. The only pole inside the contour C is at z i/RC and the residue there is exp(-T/RC)/2iRC. The value of the required integral is. there fore, !1fexp(-T/RC)/RC and so, substituting also for G v , CkTexp(-T/RC)
= q2 exp (-T/RC)
(1'
> 0) .
AH
V(t).
A2H22RkT
The power spectra of the independent sources of fluctuating couple will add, so the power spectrum of the fluctuating couple in the right-hand side will be 2 ( A2H2)
Now q satisfies the equation
+ RC
P(t)+
= [i2-1f
1/I q (T) = q2 exp (-ITI/RC) .
dt
A2H2)
The power spectrum of P(t), Gp , will be (2/1f )KkT (this can be derived by requiring that the mean square deflection of a system with, for simplicity, zero K should have its equipartition value) while that of (AH/R)V(t) will be
The result for negative l' is obtained similarly and we find
dq
(
KO+ K+ ~ 8+ce
0
;kT K+~
which has the solution q = qoexp(-t/RC) .
.
This has a ratio to the damping constant K + A21l 2/R which is independent of the values of K, R, A, and H.
Thus the correlation function and the decay function are related in the way postulated earlier (Problem 23.2).
24.3 A body of heat capacity C loses heat to its surroundings at a rate cxl:.T, where flT is the excess of its temperature over the temperature of its surroundings. Determine the power spectrum (assumed independent of frequency) of the fluctuations in the net exchange of energy with the surroundings in this situation, which corresponds to Newton's law of cooling.
24.2 The fluctuations of a galvanometer mirror may be regarded as arising from both bombardment by air molecules and current fluctuations in the circuit containing its coil. Either of these acting alone would presumably maintain the fluctuations at equipartition level. The two acting together must also result in equipartition fluctuations. Explain.
I.
~
24.3
Chapter 24
492 Solution
If the net rate of receipt of energy from the surroundings is denoted by J(t) one has
d(.~T) + OI.~T
= J(t) .
C dt An input receipt of energy exp(iwt) will give the steady state variation in ~T
exp(iwt)
~T
~T(t)
is
This gives
-
~T2
r~
GI
GI
= 01.2+ w2(f2
I
Jo 01.2+ w 2 C2dw
and, since M
CAT,
= -4AoT 1f
X
kT2 .
IS
From Problem 23.1 the mean square value of
J(t)dt will be
2S I~ 1/II(r)dT,
so that the mean square value of S1
is 0
J(t)dt, the simulated part of the
rate of incidence of signal energy and hence the error in the intensity measurement, will be
~ L~;;(T)dT .
1f
201.C GI '
This therefore gives the mlntnlUm possible mean square error in the intensity measurement. Now the Wiener-Khintchine theorem (Problem 23.11) relating power spectrum to correlation function gives
But the result of Problem 20.1 gave ~E2 = CkT 2
493
The rate of receipt of energy from the surroundings, assumed at temperature T, will be AoT4. The rate of emission of energy will be Ao(T+ AT)4. To first order in AT the net rate ofloss will therefore be 4AoT 3 AT so that 01. of Problem 24.3 is 4AoT 3 • The power spectrum of J(t), the fluctuating net rate of receipt of energy, will therefore be 2 3 GI
If the power spectrum of the fluctuating I(t) is denoted by Gland that of ~T(t) by G.6.r<w), one has
= IA(w)1 2 GI
Nyquist's theorem and its generalisations
Solution
OI.+iwC'
so that the response function A(w) relating input J(t) to output I A(w) = 01.+ iwC .
G.6.T(W)
24.5
,
this implies
GI(O)
_ kT2
AT2=C
= -1f2i~0 1/II(r)dr.
Therefore the minim!:1tJ:ll11e(~p}~uare error ~~y be written. Equating the two expressions for AT2 gives 2 2 GI
1fGI ,'
/
= -OI.kT . 1f
24.4 Consider a radiation detector which operates by measuring the rise in temperature of a black body, of area A, when the radiation is incident upon it. If the black body exchanges energy with its surround ings only by the.emission and absorption of radiation, determine the minimum possible mean square error in the measurement of intensity of a steady stream of radiation incident for time S. " [Hill!:' If the fluctuating nefrate of receipt of energy is denoted by J(t), the quantity
sI is J(t)dt
8AokTs
"
<',
S! ../
L "
24.5 If a resistor R is connected across the terminals of an impedance Z = X + iY, both being at temperature T, the rate of transfer of power from R to Z must equal the rate of transfer of power from Z to R. Assuming that this equality must hold in each frequency range (one may think of connecting the two elements by way of a fIlter) show that, if.th.~ fluctuating forces associated with the impedance are represented by a generafor of fluctuating voltage in series with Z, the power spectrtlm of this voltage will be 2 G Vz (w)=-XkT. 1f
0
will be indistinguishable from a part of the intensity of the incident radiation. This quantity therefore simulates part of the intensity of the radiation being measured. ]
Solution
The current due to a voltage Ve iwt in the circuit will be Ve iwt/(R + Z) and the resulting power diSSipated in R will be !R I VI 2/IR + Z12, while that dissipated in Z will be !XIV1 2 /IR+ZI 2 • This means that the
494
Chapter 24
24.5
power in frequency range dw generated in Z by the fluctuating voltage Gv (w) associated with R will be tXGvR dw/lR + ZI2, while that ge~erated in R by the fluctuating voltage G v associated with Z will be !RG vL dw/IR + Z12. Equating these expressi~ns gives X Gurz = -Gv. R R
;=:
2 -XkT. 11"
24.6 Verify the relation between the power spectrum of the fluctu force associated with an impedance and the real part of the impedance, for an impedance formed from a resistor and capacitor connected in parallel. Start from the fluctuating voltage associated with the resistor.
24.8
Nyquist's theorem and its generalisations
using the analogy with Nyquist's result for an electrical impedance, obtain an expression for the power spectrum GF(w) of the fluctuating force Fwhich must be regarded as acting on the system at temperature T. Deduce the power spectrum GAw) of the fluctuations in x. Finally, writing the rate of absorption of energy by the system from a force express the generalised Nyquist relation in Foe in the form !a(w) the form of a relation between GAw) and a(w): it is this form of the relation which we shall find it convenient to justify theoretically in the following problem.
iwt
1F01z,
Solution
Writing x =
xoe iwt, we have x0
Solution
If the impedance is open circuited, a voltage Ve iwt in series with the resistor will result in a voltage
1 )Jeiwt [ ~/(R+-. lWC lWC appearing across the terminals. If the impedance is short-circuited the same voltage will result in a current (V/R)e iwt through the short circuit. It follows from Thevenin's theorem that the voltage Ve iwt in series with the resistor is equivalent to a voltage I )Je iwt lWC !(R+-. lWC l~
in series with the impedance. Thus a fluctuating voltage with power spectrum (2/1I")RkT in series with R is equivalent to a fluctuating voltage in series with the impedance with power spectrum 2 RkT
1I"1iw-C1 2
1/ R+ iwC I /Z 2 R 2 = 11" 1+ wZRzCzkT = ;XkT,
where X is the real part of the impedance of the resistor and capacitor in parallel.
24.7 In this example we consider the extension, by analogy of Nyquist's result for an electrical impedance, to a general impedance and express the result in a variety of ways. [Some consideration of a more fundamental justification of the generalised Nyquist result will be the subject of Problem 24.8.] If F is the generalised force associated with a generalised coordinate x of a system and F ;=: Foe iwt results in x = x oe iwt we write Xo = [P(w)-iQ(w)]Fo · Taking x to be the analogue of current and F to be the analogue of voltage, express the impedance in terms of P( w) and Q( w). Hence, by
495
Fo
= iwx 0, and so
Xo iw(P- iQ)
Thus the impedance is l/iw(P- iQ). Taking the real part of this and applying the analogue of Nyquist's result we have . 2 Q(w) 1 GF (w) = -; -W p=z"(-w-'-)+---=Q-;;-2(-:-w--:") kT . The power spectrum of the reSUlting fluctuations in x will be
,
Gx(w) = (P 2+ Q2)GF (W) =
2 Q(w) ;-W kT .
Elementary considerations of the work done by a generalised force, when it and the corresponding generalised coordinate vary sinusoidally, allow one to deduce from the initial relation between Xo and Fo that the rate of absorption of energy is twQ(w)IFol z. Thus a(w) = wQ(w) and we have Gx(w)
2 a(w) --2- kT . 11"
w
24.8 We now consider a problem which contains the essence of a quantum statistical justification of the generalised Nyquist relation. The special assumptions we make about the system concerned merely avoid minor complications. A system has a non-degenerate ground state of energy Eo and above it a quasi-continuum of states, there being p(E) dE states of the continuum with energy between E and E + dE. The coordinate x, which isassumed to be a classical variable (cf Problem 21.2), has matrix elements only between the ground state and the states of the continuum, the matrix element of x between the ground state and a state of the continuum with energy E being denoted by x'(E). Find for temperature T (a) the power spectrum Gx(w) of the fluctuations in x, and (b) the rate of absorption of energy by the system when external forces resulting in a generalised force Foe on x are applied.
iwt
496
24.8
Chapter 24
Show that these quantities satisfy the form of Nyquist theorem established at the end of the previous problem. Hint: For any particular quantum state of the system the value of Ii classical variable as a function of time can be identified with its quantum mechanical mean value calculated as a function of time. Fourier components of frequency w in the time dependence of such mean values arise from pairs of stationary states (in the expansion of the given state in terms of stationary states) with energy difference hw between which the variable has non-zero matrix elements. The contribution to the mean square fluctuations of x from components with frequency in the range w to w + dw can therefore be found by putting equal to zero the matrix element of x between all pairs of states for which the energy difference does not lie between h wand hw + h dw, and calculating the mean square fluctuations in the ordinary way. Note: x can be regarded as a classical variable only if components of the fluctuation with frequencies which do not satisfy hw ~ kT are negligible. To be consistent it must therefore be assumed that we restrict w in this way in considering both the power spectrum of the fluctuations and the absorption. It should, perhaps, be emphasised that the definition we have given of a power spectrum breaks down if the variable concerned shows important quantum effects. We cannot then attach a meaning to the value of the quantity as a function of time, since an attempt to observe this would disturb the system. One can, of course, find relations between entities defined in suitable quantum mechanical terms(l) which have essentially the form of the Nyquist relation and which do not assume classical behaviour of the variables concerned. Direct physical significance, however, attaches to expressions, such as scattering cross sections, of which these quantum-mechanically defined entities form part, rather than to the entities themselves. To treat such matters would take us too far afield and we restrict ourselves to the classical form of Nyquist theorem. It is also possible to give a derivation of Nyquist's result for systems described by classical statistical mechanics, but we shall not consider this here. Problems 24.4 and 24.lO, however, amount to a demonstration of the result for a very special classical system. Solution
(a) If H denotes the Hamiltonian and Ir> one of its eigenstates with energy E r , we have x 2 = Tr[x 2 exp(-H/kT)] , Trexp (-H/k T)
Ix Is><slx Ir>exp(-Er/kT)
-
I
exp(-Eq/kT)
q
r
24.8
Nyquist's theorem and its generalisations
We find CAw)dw if in evaluating the sum in the numerator we consider only pairs of states with non-vanishing matrix elements for x and energy difference between hw and hw+ hdw. There are two possibilities: (I) r is the ground state and s corresponds to one of the hp(Eo + hw) dw states of the continuum with energy difference from the ground state in the necessary range, and (2) s is the ground state and r represents the same set of states in the continuum as did s in the previous case. Putting Z for the denominator we find, writing x' for x'(Eo+ hw), Gx(w)dw
~p(Eo+ hw)lx'
2 1
{exp(-Eo/kT)+ exp[-(Eo+ hw)/kT]}hdw
With the assumption hw ~ kT, this becomes CxCw)
=
;eXP(-Eo/kT)P(Eo+ hw)lx'1 2 h.
The applied generalised force which we take to be the real part of Foexp(iwt) introduces into the Hamiltonian a term -tx(Foe iwt + Foe- iwt )
.
Of this, the part in e-iwt gives rise to absorption of energy by transitions to states of higher energy, while the part in e iwt gives rise to emission of energy by transitions to lower energy states. Absorptive transitions can take place only from the ground state to the continuum (since x has no non-zero matrix elements between states of the continuum), and the rate of absorption of energy in this way will be hw
exp(-Eo/kT) 21T
Z
,
2
IFol2
hlx I p(Eo+ hw)~
Here hw is the energy absorbed per transition, the next term in brackets is the probability of the system being in its ground state, and the rest is the usual expression for the rate of transition between a discrete state and a continuum resulting from a sinusoidal perturbation. The calculation of emission of energy due to transitions from continuum states to the ground state will go in exactly the same way, except that the probability of occupation of a continuum state, exp [-(Eo + hw)/ kT liZ, will replace the probability of occupation of the ground state. The net rate of absorption will be the difference of the two expressions and this is just the first expression multiplied by [I - exp(-hw/kT)], i.e. by hw/kT since hw ~ kT. Thus the net absorption rate is
1F0 12 h w exp(-Eo/kT) 21T Ix' 12p(Eo+ hW)~4kT Z tl 2
(1) L.D.Landau and E.M.Lifshitz, Statistical Physics (Pergamon Press, London), 1968, Ch.XII.
497
2
498
24.8
Chapter 24
and comparing this with ta(w)IFoI2 gives , . w2 a(w) = 7ThIx 12p(Eo+ hw)
24.10
'I! •
In the present case the generalised Nyquist result (Problem the form Gpx (w)
We see that
G (w)
2 a(w) kT
x
in agreement with the generalised form of
result.
I\
Since X" (w) is an odd function of w, the fact that the above expression defines X" for only positive w presents no difficulties when we come to determine X'(w) from the Kramers-Kronig relations. We have , , I x(w)-x(oo) =-p
= -7TI P fOOX"(W')dW' + -I P 0 W W 7T 7T
that the kinetic energy is t I,q;. It is assumed that all the normal modes form a continuum, Le. there are no localised modes. Relate the power spectrum of the fluctuations of Px at a temperature T, high enough for all the modes to behave classically, to a function a(w) defined by
I
=
,.
Solution
With the q,. normalised in the way described, the potential energy will be lI,w;q;. It follows that at the high temperature T, -2 qr
_kT _
-
w;
w
< w, < w + dw
'-'-",---'--
w-w
0
w
W
7T
°
w+w
.lpf 7T
=p
0
i
~a(W')dW'
o
W
'2
-w ')
Since we take X'(oo) to be zero, this is the required result. 24.10 Verify the results of the previous problem by a direct dynamical calculation using the following procedure. Assuming that an oscillating electric field E = Eoe iwt is applied in the x direction determine the generalised force on each normal coordinate. Introducing a small damping of constant" (independent of r) on each mode, determine the resulting time dependence after decay of transients of each q,. and hence of Px' Replace the sum over modes in the resulting expression by an integral, making use of the function a(w). Finally determine the real and imaginary parts of the resulting response function in the limit in which" tends to zero. Solution
If the power spectrum of the fluctuations of Px is denoted by Gpx(w), then Gpx(w)dw is the contribution to pi from modes with frequency in the range w to w + dw. Thus " 22 kT "L a(w) Gp (w)dw L a;.q,. a r2 = kT-dw . 2 x
°x"(w')dw'
-OQ
I,
Use the generalisation of Nyquist's result to obtain the imaginary part X"(W) of the polarisability of the system. Apply the Kramers-Kronig relations (Problem 23.16) to obtain X'(w): it may be assumed that X' (00) is zero since the response of each oscillator will tend to zero as the forcing frequency tends to infinitv
,.
I
= -I Pi~x"(w')dW' , + -I P iOOX"(W')dW' -'-'-"--,..--:'--
w,., are so normalised
a;. w < w, < w+dw
dw' w
._~
I
!'
a(w)dw
I
7T
.1\
I,a"q", ,.
where the normal coordinates q,., angular frequency
2
= 7T
X"(w)
II
24.9 The generalised Nyquist relation can be used to determine the impedance from a knowledge of the power spectrum (or, what is equivalent, the correlation function) of the associated fluctuations. General formulae for transport coefficients, the Kubo relations, embody this idea. This problem illustrates it in a simple case. The normal modes of a particular imperfect crystal all contribute to the total x component Px of the electric dipole moment of the crystal so that
Px
takes
and comparison of the two expressions for Gpx (w) gives
w2
7T
499
Nyquist's theorem and its generalisations
w
< w, < w + dw
w
The work done by the electric field E in the x direction when q,. increases by fJq,., qs remaining fixed for s r, is EfJpx = EarDq,.. The generalised force on the rth mode is therefore a,.Eoe iwt. The coordinate qr will satisfy the equation
'*
q,,+ "q,,+ w;q" = arEoeiwt,
500
Chapter 24
24.10
of which the solution, after decay of transients, is
ar
qr
w;-w2+iwKEoe
iwt
'I
.
We have, therefore, for the time dependence of the dipole moment Px
-
Larqr r
(
2Lw r r
r2 W2 +
a
iWK
)
E
0
e iwt .
The sum in brackets is the complex polarisability; replacing the sum by an integral, we have X'(w)-iX"(w) =
J~o
r
I
+.lWK
~a(W')dW' '2 2 oW W
f
2 a( w) kT.
rr
G , (w)
w
p
Consequently "2 _
Px -
i
~
0
2
2 ~ dw Gpx (w)dw = -kT ( a(w)A
Jo
11'
=
= X'(O)kT
Jo
w2
•
Since we have from Problem 24.7 a(w) = wQ(w), which in the present case becomes a(w) = WX"(W), we see that the result obtained is a special case of the form of the Kramers-Kronig relation established in Problem 24.9.
These two portions of the integral can therefore be seen to give the real and imaginary parts respectively. Consequently
and
(Problem 24.7)
We had
11'
2 w
P
Solution
where X/CO) is the static polarisability. Comparison of the two expres sions for p~ gives x/CO) = 2 (""a(w)dw
11' ia(w) ---
X (w)
501
response of the polarization and it is therefore reasonable (Problem 23.16).
p~
while that along the small semicircle is
1
Nyquist's theorem and its generalisations
w But we have from the general result (Problem 21.1)
The denominator of the integrand has a pole of residue a(w)/2w just below the point w on the real axis. We can shift this pole on to the real axis and take the path of integration above it in a small semicircle without altering the value of the integral. The integral along the real axis becomes p
24.11
I i""a(WI)dW '2 ~ o W -w
/I 1I'a(w) X (w) = -
2 w
in agreement with what we obtained before. 24.11 The x component of dipole moment of a particular system will be denoted by Px' The corresponding generalised force is then Ex, the x component of the electric field. If the absorption of energy when = Eoe iwt is denoted by ta(w)IEoI2, show that the integral of a(w) over all positive w can be related to the static polarisability: the derivation should be based on the relation (Problem 24.7) between the power spectrum of Px and a( w) together with the relation between the mean square fluctuation in Px and the electrical polarisability derivable from the general result (Problem 21.1). Show that the result can also be derived from the Kramers-Kronig relations, provided x'(oo) is assumed zero: this assumption is equivalent to assuming no instantaneous
il
25.2
Onsager relations
503
The limit of [ >./Ix (r) >./Ix (0) ]/r as r tends to zero is just the zero r value of d>./lAr)/dr, and this is easily seen to be zero. If on the other hand l/pz ~ Irl ~ l/p!> exp(-pzlrl) becomes negligible while exp(-PIlrl) may be replaced by 1 P II r I. If we take r to be positive we have then
25 Onsager relations
,I,
'/Ix
C.W.McCOMBIE (University oj'Reading, Reading)
25.1 This example is intended to bring out in a simple context a which is important for the derivation. of the Onsager relations. We consider a particle of very small mass m moving in a straight line under the action of a restoring force -ex and a damping force -KX, where x is the coordinate of the particle. It will be assumed that mlK, which is of the order of the time taken by the particle to attain the terminal velocity appropriate to the force acting on it, is very small compared with Kle, the relaxation time for return to the undisplaced position. Use the relation between correlation function and macroscopic decay to write down the correlation function >./IAr). It will be sufficient to do this in terms of the two roots PI, pz of the equation mpz-Kp+C = 0
and to note that in the circumstances considered one of these, pz say, will be approximately Kim, and the other will go to elK in the limit of small m. Consider now x(t)[x(t+ r) - x(t)]/r
which is I >./Ix(r) >./IAO)]/r. Show that if r goes to zero this expression goes to zero, but that if r, taken to be positive, becomes very small compared with Kle, while remaining large compared with mlK, the expression will be given to a very good approximation by -kTIK. We may express this result by saying that xx is zero, as is required by statistical mechanics, but if we work on a time-scale on which the process of achieving terminal velocity takes negligible time and take the 'derivative' of x forward from the time at which x is evaluated, we will obtain the value -kTIK for xx.
(r)
,I,
'/Ix
(0) == kT(_ PIPZ r e PZ-PI
+
PI ) . PZ-PI
With P';Pz negligible, PI replaced by elK and r such that pzr > 1, the right-hand side becomes -kTrIK. Notice that if r were taken negative one would get kTIK for (r)- >./Ix
25.2 Obtain the expression -kTIK for xx in the circumstances of the previous question with negligible m, by starting from the macroscopic equation e x = --x. K
Avoid explicit determination of the correlation function, as this is very easy here but more troublesome in the systems with a number of independent variables which are considered in establishing the Onsager relations. Solution
Consider a large number of records in all of which x has the value Xo at time t. Then the average over the records of the values at time t+ r will be, for small r, e x o --x r K 0 since the change in x, produced by fluctuating forces associated with the damping, will average to zero. Thus averaging the x(t+ r) - xU)
x
(t ) --'---r..:------'--'
over these records will give (since xU) is just the constant x 0 for the records considered)
e
Solution
We obtain from the solution of the equation for x which has finite x but zero x at t 0 >./Ix(r)
-l X
Z
pz -_-exp(-PIlri) _ pz PI pz PI
with pz - Kim, PI - elK, and Xl = kTle.
502
Ir
--xz K
o·
Now doing a further average over x 0 gives the required result. It should be noted that we get the correct result by the formal procedure of multiplying the given macroscopic equation by x and averaging. It will be useful to make use of this formal procedure in the slightly more complicated cases to be considered later.
504
25.3
Chapter 25
25.3 X(t) and YU) are two variables associated with a system both of
which behave classically. The statistical character of the fluctuations will be unchanged by time reversal [i.e. X'(t) XC-f) and Y'(t) y(-t) will have the same statistical properties as X(t) and Y(t)]. Deduce that AX(t)[AY(t+r)-AY(t)] = AY(t)[AX(t+T)-AX(t)],
so that dividing by r and taking the limit of small AX(t)A}Tct)
T
gives
L\Y(t)AX(t).
Here, and in the examples which follow, it is to be understood that T becomes small on one time-scale but remains large on another (see Problem 25.2). Solution --------
Adding AX(t)A YU) to both sides of the first form of the result shows that what we have to prove is AX(t)AY(t+r) = AY(t)AX(t+r).
But the invariance of the statistical properties with respect to time means that such average values are independent of time f, so that we can write in particular AX(t)AY(t+r)
AX(t-r)AY(t).
The invariance under time reversal means that we can replace r by -r on the right-hand side, so that this result becomes AX(t)AY(t+ r) = AY(t)AX(t + r)
as required. 25.4 Suppose that the rates of change of the variables X(t) and YU) are related to their deviations from their mean values by AX(t) = aAX + bAY, A1Tct)
cAX+dAY.
25.5
Onsager relations
This is a special case of the type of relation derived by On sager. It is special because the rates of change of AX and A Yare expressed, in the equations under consideration, in terms of AX and AY themselves. More general equations, to be considered later, relate AX and AY to the departures from the equilibrium values of other quantities. If it happens that AXA Y = a and AX 2 A Yz then the above rela tion reduces to the typical On sager relation form be. It must, however, be emphasised that the physical content of the relation between the coefficients is in no way diminished by its not taking this special form. The same is true in the case of the more general equations mentioned above: the fact that. it is ahyays possible to choose the quantities in terms of which AX and AY are expressed so that the Onsager relation takes its typical form is to some extent incidental.
25.5 A passive electrical network has two pairs of terminals, the d.c. voltage and current at one pair being Ii'; , II and at the other V;, I z . These are related by the equations II ali'; + b V; , Iz =
qi
ClkT,
and qlqz = O.
(b) Express the equations giving II, I z in terms of Ii';, V; as equations relating qI, q2 to q I, qz. Use the time reversal result qlqz
AXAY.
Solution
aAXAY+bAY z = cAX2 +dAXAY.
+ d V; .
q~ = ~kT,
to obtain the required relation.
Using the formal procedure discussed in Problem 25.2 we multiply the first equation by AY(t) and average, the second by AX(t) and average. The left-hand sides are then equal by the time reversal result established in the previous problem, and equating the right-hand sides gives
C VI
The network is assumed to be of such a nature that the determinant of the coefficients is non-zero: this means that I, and 12 cannot both be zero unless Ii'; and V; both vanish. Obtain the On sager relation for the coefficients in these equations by the following steps. (a) Suppose capacitances CI and Cz are connected across the two pairs of terminals. Denote the charges on these capacitances by q I and q2' Show that if the whole system is at temperature T,
Obtain an equation which is satisfied by the coefficients a, b, c, d, the only other quantities involved in the equation being AX z , A y2 and Solution
505
q2ql
(a) The generalised forces associated with q I and qz are voltages, v I and V1 say, introduced in series with the capacitors (the voltages being so connected that the work done by them is VI Dq I in one case and V1Dq1 in the other). The voltages across the terminals will then be 11 I - q dC] and 1I1-q2/~' and when the currents are zero (as they must be in equilib rium) these voltages must be zero. Thus ql
=
CIVI,
qz
=
C2 V z .
506
Chapter 25
25.5
It follows that (ql and q2 are, of course, zero for zero applied voltages, so that t:..ql = ql' t:..q2 == q2)
qlq2 (b)
q1
kT
a (-a I) I
kT
q2 (aav 2) T
= kTC2 ,
kT
aq l ) (-a 1)2 T
= O.
)
= kTCI
T
= -q2
II
-ql,
12
V.
ql CI
V2 =
'
,
,
q2 C 2
The equations relating II and 12 to VI and V; can therefore be rewritten . qI =
a b C q 1 - c,. q 2
c tt2
,
I
d
= - C1ql- C q2' 2
(c) As in Problem 25.4 multiply the first of these equations by q2, the second by q I, and average b
a __ 1= -
Clqlq2-
c2qi,
d_ - C QIQ2'
25.6
Onsager relations
gradient d T! dx in a metal: the metal is assumed isotropic so that it is not necessary to work with vectors and tensors. The coefficients are those in the equations dV dT I = -a dx + a dx ' dV dT J=bdx+c dx '
where a is the electrical conductivity. To obtain the required relation consider, as in Problem 21. 7, a piece of the same metal connected by a wire of the metal to a very large block, also of the same metal, which acts as a heat bath (temperature T) and electron reservoir (chemical potential /l). Suppose that the wire has cross section s and length I and that the capacity of the first piece of metal in the presence of the large block is C. Write down, in terms of the coefficients defined above, equations relating the rate of change of the number of electrons on the piece of metal, and of the energy of the metal to the excess number of electrons t:..n and excess temperature t:..T for the piece of metal. Multiply the first equation by t:..E (the excess energy of the piece of metal), the second by t:..n, and average. Apply the time inversion relationship (Problem 25.3) and some of the mean values determined in Problem 20.5 to obtain the required relationship. Solution
We have Lln =
-~
dV t:..v dx = -[-
t:..E = -Js,
e
=
dT dx
t:..n e IC
2
The left-hand sides are equal, and equating the right-hand sides gives, on substituting from b c,. kT C2
c
= - CI C1 kT ,
that is
b
=
c.
In this case, then, we get the standard Onsager reciprocal relation. 25.6 This problem illustrates the use of the Onsager procedure to obtain a relationship between transport coefficients for a material. The coefficients considered will be those relating current density I and thermal current density J to voltage gradient dV!dx and temperature
501
t:..T
The equations given for I and J can, therefore, be rewritten as
sa
sa
t:..n = +-t:..n--t:..T IC el ' . esb sc t:..E = - TC t:..n - T t:.. T .
Multiplying the first and second equation by t:..E and t:..n respectively, averaging, and using the time inversion relationship t:..ELln = t:..nLlE gives
a--
- Ct:..Et:..n +
-eat:..Et:..T =
-
+ ct:..nt:..T
We saw in Problem 20.5 that t:..Et:..T = and t:..Tt:..n = O. We also saw 2 2 in Problem 21.7 that t:..n = (C/e )kT. This leaves only t:..Et:..n to be put
Chapter 25
508
25.6
25.8
Onsager relations
509
in appropriate form, and we had in Problem 20.5 that
all
all
-MDon = kT (aE) T = kT (aE) an T (an) T· But
= T(as) + 11 = _T(a ll ) + 11 = ( aE) an T an T aT n
-T2~(~) , aT T n
and at constant temperature 2 dll = e d V = e dn C
so that
(~:)T = ~
Thus
-MDon =
.
2
-kT 3 - a (11) C aT T n e 2
3 a (11) a b 2 = -kT +akT -2- +-kT
i.e.
aT T n e e
-(::)n
'
Fn = - (
This relation is not of the standard reciprocal form, but it contains all the physical information derivable from Onsager type considerations. 25.7 Deduce from Problem 22.5 that if we write the rates of change of the coordinates Xr as linear combinations of the associated forces Fs ' defined by Fs = -as/axs , then the array of coefficients is symmetric, i.e. if we write Xr = Lf3rs Fs s then f3rs = f3sr . This is the orthodox form of the Onsager relations. Multiplying the equation for XI by Xm and averaging gives Lf3ls X m Fs s
=
Lf3ls ko ms s
Similarly XIX m
=
kf3ml .
So the time reversal relation gives 131m
=
f3ml .
~~)
E
=
Do (~) .
Re-express the equations relating E and Ii to Don and DoT, with coefficients as given in Problem 25.6, as equations relating E and Ii to FE and Fn. Show that the Onsager relation, when applied to these equations, yields the same relation between transport coefficients as was obtained before. Solution
DoS in the expressions for FE gives a s ) Don = - aEa(as) a(as) aE M- an aE Don (aaE2s)nDoE- (aEan
Substituting the expression for 2
FE = -
2
= -Do (::) = -Do(~) =
~~ .
Similarly
Solution
=
2
Show that the forces associated in the Onsager theory with E and n, FE and F n , are given by DoT FE = T2 ,
aT2 a (11) -e-aT T n +aT = b.
XmX I
2
rI(a s) nDoE 2+2\aEan ( a s) DoEDon+ (aan2s) E Don2]. DoS = !~aE2
The relation between the coefficients is therefore
e
25.8 The aim of this problem is to derive, by an application of the On sager relations in their conventional form, the relation between the electro-thermal transport coefficients of a metal already derived in a less orthodox way in Problem 25.6. As in the earlier problem consider a metal body connected by a metal wire to a much larger metal body, all the metals being the same. If the energy and number of electrons on the smaller metal body increase from their equilibrium values by DoE and Don respectively, the increase in entropy of the whole system will be, to lowest non-vanishing order, a quadratic function of DoE and Don,
=
kf3lm .
F
n
= _~(aS)DoE_~(aS)Don = _Do(as) = aE an
an an
an
Do '\T) (Il~
It is necessary to express both FE and Fn in terms of Don and DoT. FE is
already given in this form and Fn is easily expressed similarly if we recall (Problem 21.7) that, for the system considered, (all/anh = e 2/C, where C is the capacity. Thus
a (11) a (11) T nDoT+ an T
Fn = aT
a (11\ e Don T) nDoT+ Tc . 2
T
Don = aT
510
25.8
Chapter 25
Solving for l:1T and l:1n in terms of FE and Fn gives
l:1T
T2FE'
l:1n
TC ~
(71 -
T
2
a (11.) aT T
71
26 Stochastic methods: master equation and Fokker-Planck equation
]
FE
In the notation of Problem 25.6 we had sa sa .::In ICl:1n - eZl:1T ,
. 1lE
-
esb
1.0PPENHEIM (Massachusetts Institute of Technology. Cambridge, Massachusetts) K.E.SHULER (University of California, San Diego, California) G.H.WEISS (National Institute of Health, Bethesda, Maryland)
sc
Tc l:1n - T l:1T .
In terms of FE and Fn these become
l:1fl =
s T
1 e2aFn
J
(Jl)
s lr:~ a I e 2 aT\T
+
s I
71
lT3 (11.) e aTa \r
T2 a+ea 71
b-
]
Pc FE .
In this case (Problem 25.7) the Onsager relation is just the equality of the off-diagonal coefficients. This gives at once b
a (11.) = Ta + e aT T
26.0 In these problems (1) the following notation is used: Wen, tim, s) is the conditional probability that a system is in state n at time t, given that it was in state m at time s < t. It is normalised so that L W(n, tim, s) = I, where the sum is over all possible states n. 71
Pen, t) is the probability that the system is in state n at time t. For all cases of present interest the condition LP(n, t) = I is valid. 71
P2 (m, s; n, t) is the joint probability that the system is in state m at
T2
71
a,
which is the result obtained before (Problem 25.6).
time s, and that it is in state n at time t. , 1 A(n,m;t) = hm A W(n,t+l:1lm,t)-onml
(26.0.1)
A"'O.u.
is the transition probability per unit time (i.e. the transition rate) for a transition from state m to state n at time t. All of the processes to be considered will be such that A(n, m; t) exists. The probabilities Pen, t) and P2 (m, s; n, t) are related by
pen, t) LP2 (m,s; n, t).
(26.0.2)
m
The conditional probability wen, tim, s) is related to P and P2 by
P2 (m, s; n, t)
Wen, tim, s)P(m, s) ,
(26.0.3)
The reader is referred to Problem 2.14 for an introduction to the idea of master equations.
26.1
Show that the master equation has the form ap(n, t) A(n, m, t)p(m, t) .
L m
(26.1.1)
(I) A general reference to this section is I.Oppenheim, K.E.Shuler, and G.H.Weiss, "Stochastic
theory of multistage relaxation processes", in Advances in Molecular Relaxation Processes, Vol. 1, 1967-8, pp.13-68.
511
512
26.1
Chapter 26
In a radioactivity decay process the probability of emission of a single particle in the time interval (t, t+dt) is Adt, where A is a constant and the probability of emission of two or more particles in the same interval is zero. The state of the system is described by the number of particles n that have been emitted between time 0 and t. (a) Calculate the conditional probability Wen, t+dtlm, t) for this process. (b) Calculate the transition rate A(n, m, t) for this process. (c) Show that pen, t) satisfies the master equation open, t)
at
= A[P(n -1, t) - pen, t)],
n
1,2,3, ...
Stochastic methods
513
(b) By combining expressions (26.0.1) with (26.1.7) one obtains
1
A(n+l;n;t)=~~
A(n,n; t)
A
-A
A(n,m; t) = 0
(26.1.8) form =1= n, n-I.
(c) Substituting the transition rates (26.1.8) into the general master equation (26.1.1) one obtains the specific master equation (26.1.2) (d) Define a generating function
(26.1.2)
ap(o, t)
at
26.2
G(Z, t) =
L pen, t)zn .
(26.1.9)
n=O
= -AP(O, t) .
(d) Calculate the probability pen, t) from these last equations assuming that P(O, 0) = 1.
If the nth term of the master equation (26.1.2) is multiplied by zn and the sum taken, it is found that G(z, t) obeys the equation
aG at =
Solution
By definition
A(Z - l)G
(26.l.1
which has the solution
L m
p(n,t+~) =
(26.1.3)
G(Z,t) = G(z,O)exp[-(1-Z)At].
L onmP(m, t) , m
(26.1.4)
where onm 0 for n =1= m and onm = I for n One then obtains
m, and divide by ~.
For the initial conditions P(O, 0) = I and pen, 0) 0 for n = 1, 2, 3, one finds G(z, 0) = I and G(z, t) = exp[-( I-Z)Atj. Expansion of this function in a power series in Z and comparison with Equation (26.1.9) yields Atn (26.1.12) pen, t) = exp(-At)-,
t+ ~ jm, t)P(m, t) .
Subtract from this the identity pen, t)
pen,
t+~)-p(n, t) ~
Taking the limit one finds
=
L ±[W(n, t+~lm,t)-onmjP(m, t).
(26.1.5)
m
~
= 0 and using the definition of Equation (26.0.1) oP\n, t) =
L A(n, m; t)p(m, t)
(26.1.6)
m
as asserted. (ii) Since the state can only change by I during the time interval (t, t+ dt) one has
(26.1.11)
n.
for the probability that n particles have been emitted at time t. The distribution function of Equation (26.1.12) is known as the Poisson distribution. 26.2 Consider a master equation characterised by a time independent matrix A = (Anm) in which the elements Anm are the rates of transition from states m to states n. Define the vector of state probabilities P(t) by P(O, t)
P(l, t)
P(t) =
I P(2, t)
(26.2.1)
W(n+l,t+dt!n,t) = Adt Wen, t+dt[m, t)
0
Wen, t+dtln, t) = I
(26.1.7)
for m =1= n,n-I W(n+ 1, t+dtin, t)
I-Adt.
Let the matrix A have non-degenerate eigenvalues Aj (j = 0, I, 2, ... ),
~
26.2
Chapter 26
514
right eigenvectors Rj and left eigenvectors Lj , defined by ARj = AiR"
26.2
Stochastic methods
Substitution of this solution into the master equation (26.2.2)
L;A = A,L, .
L [a·(t)-A.a.(t)JR, = I I ,
;=0
The master equation can be written in terms of pet) and A as
P = AP.
(26.2.3)
Show that pet) has the formal solution P(t) =
"" .L (L P(O»Rj exp(A; t) ,=0 i
(26.2.4)
Define a conservative system as one in which A(n, m) n =1= m, and for which A(n, n)
-
L
I
A(m, n),
~
(26.2.5)
00
L pen, t)
,.=0
I
(26.2.6)
= I
(26.2.7)
for all t [the A(n, m) may depend on time]. Most systems of physical interest are conservative, and all systems to be discussed here are conser vative. (iii) Show that there exists a zero eigenvalue. (iv) If Ao = 0 and Al =fo 0, show that P(oo)
= Roj i~O (Ro}j ,
P(O) =
L a,(O)R,
(26.2.13)
in which it is assumed that P(O) is a known vector. To find the a/CO) we show that the inner product L,. R, = 0 when n =fo j. From Equation (26.2.2) it follows that L,. AR/ = \L,. R/ (26.2.14) L,. AR/ A,. L,. R/ or, subtracting, o = (A,. -X;)(L,.R,). (26.2.15) From the hypothesis of non-degenerate eigenvalues it follows that (L,. Rj ) 0 when n =fo j. In all cases of interest the vectors L,. and R,. can be scaled so that L,. R,. = I, and we can assume that the eigen vectors are orthonormal. If we now multiply Equation (26.2.13) by L,. and use the relation of orthonormality we find that a/CO) = L,P(O)
(26.2.8) P(t) =
Solution (i) Assume a solution for P(t) of the form
L a/(t)R, .
(26.2.11)
Thus, the a/CO) can be calculated from the relation
(26.2.16)
,=L (L/p(O»R, exp(A/t) . 0
(ij) Write the master equation out in full as
p(O,t) = AooP(O, t)+A 01 P(l,t)+A 02 P(2,t)+···
PO, t)
= A lOP(O, t) +A llPO, t) +A 12P(2, t) +...
p(2, t)
= A 2oP(0, t)+ A 21 P(l, t)+A 22P(2, t)+···
Show that, when Equation (26.2.9) is satisfied, the eigenvalues Ai are real and negative, and Ao = O. [See also Problem 2.14 for additional explanations of detailed balance.}
; =0
O.
which leads to the desired formal solution (26.2.4)
where (RO)i is the ith component of Ro. Notice that the equilibrium distribution P(oo) is thus independent of the initial condition P(O). (v) The condition of detailed balance at equilibrium can be expressed as A(m, n)P(n, 00) A(n, m)P(m, 00) . (26.2.9)
P(t) =
AP yields
f =0
where the prime denotes omission of the m = n term. Show that if
it follows that
P=
Since the R/ are independent when the Ai are non-degenerate, one may set each term separately to zero and solve the resulting equations a,(t) A/ai(t) to obtain a/(t) a/(O)exp(A/t). (26.2.12)
0 for
m
L P(n,O) = ,.=0
515
(26.2.17)
If these equations are added, taking account of the condition (26.2.5), it
is found that (26.2.10)
..
L Pen, t) = O.
,.=0
(26.2.18)
26.3
26.2
Chapter 26
516
L pen, t) = constant = L P(n,O) =
n~O
One can now verify from Equation (26.2.9) that Snm = Smm Le. S is a symmetric matrix. It is possible to conclude at this point that the AI are negative for j;;;' I if they were not, P(t) as given by expression (26.2.22), could not represent a vector of probabilities for large t. However, the fact that the Aj must be negative can also be proved by showing that S corresponds to a negative semi-definite quadratic form. This can readily be done by expressing the matrix elements A(n, m) in terms of non-negative elements B(n, m) by
n~O
L A(m,n)
= a (I, I, I, ...) .
(26.2.20) (26.2.5),
O,whichisequivalentto
m=O
can be written I)
LoA = 0;
hence the eigenvalue 11.0 corresponding to Lo is O. The existence of the non-degenerate eigenvalue 11.0 = 0 insures that the master equation (26.2.3) with the general solution (26.2.4) has a non-zero equilibrium solution P(oo). (iv) Recall the expansion of the solution (26.2.4) which expresses P(t) in terms of eigenfunctions and eigenvalues. It must be the case that Re Aj ~ 0 for all j, as otherwise P(t) would have unbounded components for sufficiently long times. One can therefore separate out the j = 0 term to find that P(t)
L [LjP(O)]Rjexp(Ajt),
[LoP(O)1Ro+ j
=1
where all terms in the second sum tend to zero as t can be written
Lo = a(l , I, I, ...) , where a is a constant, so that
-jo
00.
S(y) =
(26.2.25)
= I ,
<'i nm
L B(r,n) r
cf
(26.2.27)
n
L
(26.2.28)
SmnYnYm n,m
must be non-positive. But an explicit calculation shows that S(y) can be expressed as
L B(r, n)y~ + L B(n, m)[P(m, oo»)'h[P(n, ooW'Ilynym . (26.2.29)
S(y)
r,n
n,m
rcfn
nFm
Using the principle of detailed balance expressed in Equation (26.2.9) one can rewrite Equation (26.2.29) as
(26.2.23)
Thus P(oo) is nrnnAl'hnt'\
= (l-<'inm)B(n,m)
with B(n, m) = A(n, m) ;;;. 0 for n =t= m. If S corresponds to a negative definite quadratic form then the function
Recall that Lo
(26.2.24)
L [P(oo)]j
A(n,m)
(26.2.22)
= a.
(26.2.26)
Snm = [P(n,ooW':4A(n,m)[P(m,oo)]'h.
(26.2.19)
I .
(iii) Consider the eigenvector
Thecondition
511
easily inverted, and the elements of S are found to be
An integration of this equation yields
Lo
Stochastic methods
\
S(y)
! n~mB(n, m)P(m, OO){[p(n-~:')]Y:.
[P(:~)l'h
r
(26.2.30)
which is manifestly negative semi-definite. The condition of detailed balance, which is met with so frequently in physical systems, therefore suffices to insure the real and negative form of the eigenvalues Aj and so leads to mathematically reasonable forms for the state probabilities.
j
and
a
1/~(Ro)i'
(v) In order to prove reality of the eigenvalues Aj it is sufficient to
show that the transition matrix A is similar to a symmetric matrix S.
Since the eigenvalues of a real symmetric matrix are all real, it follows
that the eigenvalues of A are real. Define a matrix U whose ijth element
is ~j = [P(i,oo)]':4<'i", where <'i jj is a Kronecker delta, and a matrix S
constructed from A by S U-lAU. Since U is a diagonal matrix, it is
26.3 The master equation for the relaxation of an ensemble of non interacting harmonic oscillators in contact with a heat bath at tempera ture T(oo) is
ap~n, r)
{ne-e(oo>P(n -1, r) - [n +(n + I )e-e(oo)jP(n, r) +(n+I)P(n+l,r)},
P(-I,r)
0,
n
0,1,2, ... (26.3.1)
518
26.3
Chapter 26
where Pen, T) is the probability that the oscillator is in vibrational state n with energy nhv at (dimensionless) time T, and where 0(00) hv/lkT(oo)]. Obtain a solution to Equation (26.3.1) for an initial Boltzmann distribu tion of oscillators P(n,O) = (l -e-O(O»e- IIO (O) , (26.3.2) where 0(0) = hv/[kT(O)] and where T(O) is the vibrational temperature of the ensemble of oscillators at T = O. Show that Pen, T) is a Boltzmann distribution at all times T, i.e. that Pen, r) =
( l - e - O(r»e- IIO (r)
(26.3.3)
and find an explicit expression for O(r).
26.4
519
Stochastic methods
Substituting for Kl and K2 from Equations (26.3.8) and (26.3.9) one finds G(z,r) =
O(co)_z } l-z exp[-r(l-e-O(oo»] .
1/1
,
(26.3.10)
The function 1/I(x) can be found by setting r = 0 in this equation. For the initial condition given in Equation (26.3.2), G(z, 0) is obtained from Equation (26.3.4) as G(z,O) = 1 -e-O(O) (26.3.11 ) I-ze-O(O) and after some algebra it is found that G(z, r) takes the form I G(z,r) l-A() = [I-A(T)] LA"(r)zl, 00
Solution
As in Problem 26.1 (d), the solution to the difference equation can be obtained through the use of a generating function G(z, t)
=
"" L z" Pen, t) . 11
If one multiplies the nth line of Equation (26.3.1) by z" and sums over all n, one finds that G satisfies a first order partial differential equation
aG
aG
ar
az
I)G.
(26.3.5)
dr
dz = (l -z)(z
dG =
The second and third of these equations lead to dG dz
G
= eO(oo)-z
A(r)
l)G
11
=0
e1"(l-e IO(OO)-O(O)))-(l-e-O(O)) e-r(l-eIO(oo)-O(o>l)-eO(OO)(l-e-O(O»
(26.3.12)
(26.3.
The coefficient of z" in Equation (26.3.12) is Pen, r) which can be written Pen, r) [1 - A(r)]A"(r) = (l -e-O(r»e- IIO (r) , (26.3.14 ) where O(r)
This equation can be solved by the method of characteristics, which requires the solution of the system of ordinary differential equations
1
where
(26.3.4)
0
-+(l-z)(ze-O("")-l)- = e-O(oo)(z
r z
= -logA(r) .
(26.3.15)
Thus, Pen, r) can be written in the Boltzmann form for all values of the time. The result in Equation (26.3.14) can be interpreted by saying that a vibrational temperature T(r) can be defined at every instant of time by
(26.3.6) T(r)
=
hv kO(r) .
(26.3.16)
It is easily verified from Equations (26.3.13)-(26.3.16) that
lim T(r) = T(oo)
(26.3.7)
r~oo
with the solution
as required. (eO(CO)-z)G(z,r) = KI ,
(26.3.8)
where K 1 is a constant of integration. The first and second equations of (26.3.6) can be solved in a similar way and lead to eO(CO) -z exp[-T(l-e-O(oo»] K2 (26.3.9) where K2 is also a constant of integration. The general solution to Equation (26.3.5) by the method of characteristics can be represented by Kl = 1/I(K2 ) where 1/I(x) is any arbitrary differentiable function.
26.4 Derive the second-order partial differential equation to which the set of differential difference equations (26.3.1) reduce in the limit as 0(00) ~ O. This differential equation is called the Fokker-Planck equa tion for this particular system which corresponds to the classical con tinuum limit of the harmonic oscillator. Solution
The singlet probability Pen, t) is also a function of the parameter 0 so that we rewrite it as Pen, 0, t) which, in turn, we can write as some function of nO, 0, t as P(nO, 0, t). We define a function p(x, 0, t) of the
T
26.4
Chapter 26
520
26.5
(26.4.1 )
Op(x, 0, t) = P(nO, 0, t)
p.(t) =
when x = nO. Equation (26.3.1) can now be rewritten as ===
521
Give necessary and sufficient conditions for the first moment
continuous variable x which has the property that
ap(::., 0, t)
Stochastic methods
f~xp(x, t)dx
(26.5.5)
to have the simple relaxation form given in Equation (26.5.3). Solution
j.{xe-8 p (x - 0,0, t) +(x + O)p(x + e, e, t)
(i) It follows from the master equation (26.5.1) that p.(t) satisfies the differential equation (26.4.2)
-[x+(x+e)e-ejp(x, e,
[J,(t)
=
""
L
P(m, t)
m"O
Expansion of the right hand side of Equation (26.4.2) in a power series in e yields
L nA(n, m) .
(26.5.6)
n=O
From this equation it is clear that if
a2p(x, t) apex, t) 2 e [x "'.. 2 +(x+ 1) "'.. +p(x, t) I +o(e ) ,
nA(n,m)
a-bm
(26.5.7)
n
where p(x, t) p(x, 0, t). We define a new time variable r = et, and take the limit as e -+ 0 to obtain the Fokker-Planck equation apex, r) ---
apex, r)
ax
p(x, t)+(x+ I)
aZ
for all values of m, where a and b are constants, then [J,(t)
azp(x, r) +x
ax2
a
x
lim (et)
r
and
lim (ne)
e->o
(26.4.4)
p(t)lt=o
e->o
t .....
n->""
(26.5.8)
which gives rise to a simple exponential relaxation of the first moment. The identification b = A, alb = p.(oo) then leads to the desired form. In addition to being sufficient, the condition (26.5.7) is necessary. Con sider the initial condition per, 0) = 1, P(j, 0) = 0 for j =1= r. From Equation (26.5.6) it follows that
(26.4.3)
= axz[xp(x,r)]+a)(x-1)p(x,r)] ,
where
a-bp.(t)
n~onA(n,r).
(26.5.9)
But if pet) is to be the solution to Equation (26.5.8) it must satisfy 26.S (i) Let the probability pen, t) satisfy a master equation Pen, t) =
L A(n, m)p(m, t) , m=0
[J,(t)
(26.5. I)
t=0
n
L0nP(n, t)
(26.5.2)
to have an exponential relaxation of the form p(t)
= p(oo)+[p(O)-p(oo)]e- At
(26.5.3)
valid for all values of t. (ii) Let the probability density p(x, t) be the solution to a FokkerPlanck equation apex, t)
at
=
a -ax[b
2 a (x)p(x,t)]+!ax t
t
2[b z(x)p(x,
(26.5.4)
(26.5.10)
at t O. Equating (26.5.9) and (26.5.10) we find that the condition (26.5.7) is necessary as well as sufficient. The preceding calculations can be generalized to derive necessary and sufficient conditions for the vector of the first k moments, p(t) Pt(t), P2(t), ..., Pk (t) to be the solution to [J,(t) A - Bp.(t), where A and Bare constant matrices. Multiplication of the Fokker-Planck equation (26.5.4) by x and integration over all x, followed by a partial integration of the right hand side of the resulting equation, leads to
where the transition ratesA(n, m) are independent of time. Give neces sary and sufficient conditions for the first moment pet) =
=a-br
pet) =
'I
1
"f
[ a b 2(X) apex, t) _""b1(x)p(x, t)dx+; xp(x, t)--ax-+xbz(x) ax -b z(X)P(X,t)-2xb t (X)p(x,t)11".
(26.5.11)
J
\
522
Chapter 26
26.6
26.5
In order that the moment relaxation satisfies ii(t) = a-bIlU)
f~bl(X)P(X,t)dx.
ris
(26.5.12)
T/(T)
a
T/(T) = D(T)dT+T/(T+dT) .
(26.5.13)
(26.6.3)
(26.6.4)
That is, if the molecule is not dissociated at time T, it either dissociates in (T, T+dr) or it is still undissociated at T+dT. Passing to the limit dT = 0 one finds
(26.5.14)
bx
= P(O,T)+P(l,T)+"·+P(N,T).
The probability that dissociation occurs in the time interval (T, T+dT) will be denoted by D(T)dT and can be calculated from the identity
From Problem 26.S(i) it is clear that the form bl(x)
523
The probability that the molecule is undissociated at dimensionless time
we must impose the condition that p(x, t) goes to zero sufficiently rapidly as x --,\> ± 00 so that the bracketed terms go to zero at too. Equation (26.5.11) then reduces to iiU)
Stochastic methods
D(T) = _ dT/(T)
with b = A and alb = Il(oo) is the necessary and sufficient condition for exponential relaxation of the first moment.
(26.6.5)
dT The dimensionless mean time TD to dissociation is(2)
26.6 Assume that an ensemble of diatomic molecules can be modelled by a system of one-dimensional harmonic oscillators whose time depen dent distribution function over the semi-infinite energy level system n = 0, 1, "', N, ... is determined by the master equation of Problem 26.3. Assume that an oscillator dissociates irreversibly whenever its vibrational energy reaches (N + 1)hv. Calculate the mean time to dissociation for an initial delta function distribution of oscillators, pen, 0) = on" where r is an integer satisfying 0 ",.;; r ",.;; N.
TD
JorooTD(T)dT =
-
S" d~dT 0
T
d
=
JorooT/(T)dT
= PO+Pl +P2 +"'+PN,
(26.6.6)
where we have set Pi =
Loop(j, T)dr .
(26.6.7)
Solution
The PI can be calculated by integrating both sides of Equation (26.6.2) over r from 0 to 00. One first notes that
Since for the harmonic oscillator employed here transitions can take place only between nearest neighbour levels, an = ± I, one need calculate only the mean dissociation time from level O. To see this, let (1',. N+ 1 > be the expected time for the energy to reach the value (N + 1)0 st~rting from rO. Then
= (To,,}+<1',.,N+ I>
(TO,N+l>
C""dP(O,r)dT dr
Jo
(26.6.1)
- - = ne-1.1 pen -I, T) n
(To,N+I)-(To,,).
dP(N, T)
=
(2) In the integration by parts which leads from the second integral to the third the term This procedure is valid provided that TV is finite. To see this we note that
!T.! TT/(T) was set equal to zero.
[n+ (n+ l)e- ]P(n, T)+ (n+ l)P(n+ I, T)
Ne-6p(N-I, T)-[N+(N+
I,2, ... ,N,
since P(j, (0) = 0 for all j = 0, 1, ..., N owing to the dissociation, and P(j, 0) = 0 for all j =1= 0, owing to the initial condition P(O, 0) = I.
6
= 0, I,2, ... ,N-I
j
Jo
Hence we need only discuss the case r = O. The master equation with dissociation at level N+ I is (cfProblem 26.3) dP(n, T)
(26.6.8)
r"dP(j, r) dT = P(j, (0) - P(j, 0) = 0 dr
or (1',.,N+l)
P(O,oo)-P(O,O) = -I,
TT/(T)
(26.6.2)
Since
r-dT/(x)
I J7 dXdx T
i-
I .;;; IJ7r-xdXdx dT/(x)
dT/ x-dx o dx
I)e-6 1P(N, T).
is flllite the right side of this equation tends to zero as
j'
T
-+ co,
T
26.6
Chapter 26
524
26.7
When the integrations are performed it is found that the Pi satisfy -I
o o
Stochastic methods
26.7 Calculate the mean chain length at time t of a polymer which can be in one of two states: active, Le. growing; or inactive, i.e. incapable of further growth. Assume that the concentration of monomer is held constant at its initial value M, that the active polymer can change irrever sibly into an inactive state, and that the initial state of the system con sists of active monomers. If E~ represents the concentration of active n-mer, E" that of inactive n-mer, the equations describing the reaction are then
= PI -e-O Po e-O Po-(l +2e-6)PI +2P2 2e-6 PI (2 +3 e-O)P2 +3P2
(26.6.9) 0= (N-l)e-o PN_z-[(N-l)+Ne-o]PN-I+NPN
M+E"*
o = Ne-o PN_l-[N+(N+ l)e-o]PN'
HI +2e-8 )Pl-te-Opo
= e-ZOpo-(e-o+t)
P3 =
i(2+3e-8 )p2-je-OPl
= e-30po-(e-2B+!e-6+!)
e-Nllpo- ( e-(N-l)6+
e-(N- 2)11 e-(N- 3)6 1) 2 + 3 +."+;V.
The general term can be written n
Pn = e-nB(po-sn) ,
where
n
Sn
1,2, ... ,N,
ei6
= j=L-· . 11
E*,,+1
and
'Y E n ' E"* ~
Let P;(t) be the probability that a chain chosen at random at time t is an active n-mer, and let P" (t) be the probability that it is an inactive n-mer. The probability that an active chain will add a monomer in the time interval (t, t+dt) is kMdt and the probability that an active n-mer will become an inactive n-mer in (t, l+ dt) is 'Ydt. A set of equations for the F;(t) and P,,(t) can be derived by writing the various possible events that can occur in (t, t+dt). These are: 1. M +E~ -l> E:; + 1 with probability kM dt; 2. E~ -l> E" with probability 'Ydt; 3. no transition with probability 1-(kM+'Y)dt. Hence
(26.6.10) PN
k -+
SOlution
These equations can be solved recursively for the Pi' This yields PI = e-6po-l
pz =
525
(26.6.11 )
P:;U+dt) = kMF;_l(t)dt+P:;(t)[I-(kM+'Y)dt]
n
~
I
(26.7.1)
P,,(t+dt) = F;(t)-ydt+P,,(t)
(26.6.12)
Passing to the limit dt equations:
To obtain Po, one notes that the last line of Equation (26.6.9) can be rewritten as Ne- B[e-(N-I)II(po -SN-l)] = [N +(N + 1)e- II ][ e-NII(po -SN)] (26.6.13)
-l>
0 one finds the following set of differential
P;(t)
= kMF;_l(t)-(kM+'Y)P:;U)
P" (t)
= 'YP:;(t)
n~1
(26.7.2) (26.7.3)
so that
from which Po is found to be Po = SN+ 1 .
(26.6.14)
P,,(t) = 'Y J:P:;(t)dt.
Using Equations (26.6.11), (26.6.12), and (26.6.14), we can now rewrite Equation (26.6.6) as n N N N+l ejll "L . , . (26.6.15) TD -_ " L., Pn "-nB( L., e SN+l-Sn ) " L., e-nli
" =0 " = 0 " =0 j = ,,+ 1 J
Carrying out the indicated summation yields I N+l e i 6 -1 TD 1- -II e i=1 J
for the mean dimensionless time to dissociation.
L -.
(26.7.4)
Equation (26.7.2) can be solved through the introduction of the generat ing function
L p:;(t)zn .
G(z, t) = "
The function G(z, t) satisfies aG -
at
(26.6.16)
(26.7.5)
I
= [kM(z-l)-'Y]G
(26.7.6)
with the solution G(z, t) .e" '.'.. ' ','
= G(z, O)exp{-[-y+kM(l- z)]t}.
(26.7.7)
~ 526
Chapter 26
26.7
one finds G(z, 0)
I for n = 1 = { 0 for n *- 1
*
(26.7.9)
raG(z, t) (taG(z, t) ]
az +'YJo az dt Z=I' (26.7.10)
= n~ln[Pn(t)+Pn(t)j = L
The second equality on the right hand side of Equation (26.7.10) follows immediately from the definition of the generating function, Equation (26.7.5). Evaluation of Equation (26.7.10) with G(z, t) given by Equa tion (26.7.9) yields III (t)
= e-'Y t (1 + kMt) +'Y
I
t
o
e-'Y t(1 + kMt)dt
=
kM
1 + -(1 - e-'Y t )
'Y
(26.7.11)
for the mean chain length of the polymer at time t. The higher moments Ilr(t)
z)' G(z, t) +'Y J: (z :x)' G(z, t)dtl
= nt Inr[p;(t) + Pn (t)j = [(z aa
= I
(26.7.12) can also readily be obtained so that it is possible to determine the disper sion of the mean polymer length. The probabilities P;(t) and Pn(t) can be obtained by expanding the generating function G(z, t) in Equation (26.7.9). This yields P;(t)
=
(kMt)n-1 (n -I)! exp[-('Y+kM)tj
(26.7.13)
and from Equation (26.7.4) Pn(t)
=
527
and
!(l+~).
The possible positions of the particle are limited by the condition
= z so that
The mean chain length III (t) is given by 00
!(l-~)
(26.7.8)
G(z, t) = zexp{-[-y+kM(l-z)jt}.
Ill(t)
Stochastic methods
probabilities of moving a distance e to the right or left are, respectively,
For the assumed initial condition P;(O)
26.8
(kM)n- I (t 'Y (n -l)! Jo rn-Iexp[-('Y+kM)rjdr.
(26.7.14)
26.8 Consider a one-dimensional random walk where a particle can move the distance e either to the right or to the left. The duration of each step is I::.t. The probability of moving in either direction depends upon the position of the particle; if the particle is at the point k, the
- N ~ k ~ N, where k and N are integers. (i) By appropriate scaling of the variables and in the limit N
~
00,
derive a partial differential equation, the Fokker-Planck equation, in continuous time and space for p(x, r) dx, the probability of finding the particle between x and x + dx at time r. (ii) The one-dimensional Ornstein-Uhlenbeck equation can be written in the form a2p ap(x,r) 1 a (26.8.1) ar = -Yax[F(X)p(X,r)j+D (x,r) 2
ax
where f is the 'friction coefficient', F(x) is an outside force acting along the x-axis, and D is the diffusion coefficient. Compare your Fokker Planck equation with the Ornstein-Uhlenbeck equation and from the explicit expression for F(x) discuss the physics of the random walk of part (i). Solution
(i) Let peke, sl::.t) be the probability that the particle is at point ke at time sl::.t. Since the probabilities for moving right or left add up to unity, the probability that the particle remains at point k during the time interval I::.t is zero, and the difference equation for the random walk therefore is
P(ke,sl::.t)=!
Ne P[(k-l)e, (s-l)l::.tj {rLl - (k-l)e] l)e] } +Lrl + (k+ Ne P[(k+l)e,(s-l)l::.tj.
(26.8.2)
Subtracting P[ke, (s -l)l::.tj from both sides and dividing by I::.t we obtain, after some algebra, P[ke, sl::.tj-P[ke, (s-l)l::.tj I::.t = ~{P[k -l)e, (s -l)l::.tj- 2P[ke, (s -l)l::.tj + P[(k+ l)e, (S-l)l::.t j } 21::.t e2 _l_{(k + 1)eP[(k+ l)e, (s -1 )I::.tj- (k - 1)c:P[(k - l)e, (s - l)l::.t j} + I::.tN 2e .
(26.8.3) In the limit as I::.t
~
0,
e
~
0,
N
~
00 ,
(26.8.4)
T
528
Chapter 26
26.9
26.8
and subject to the validity of the limiting processes IltN -+ 'Y, silt -+ T,
k€
-+
(26.8.5)
X ,
PI(Xt> t l )
~[P(
'Y ox XX, T
).]+IDOZP(X,T) ~ oxz
(26.8.6)
-'Yfx
-ax
- 2al
a Z = <XZ(t»
[X (p/a Z)yj2 }, (26.9.4) 2a Z(l-p2/a 4 )
{
= <XZ(O» = constant
P
= <X(t)X(s»
= p(lt-sl) = p(T).
(26.8.7)
peT) = exp(-~!TI), peT)
== O.
(26.9.8)
Solution
(26.9.1)
P(t3 -ttl
<X(t3)X(tI»
or more.explicitly as
The term wrdx, represents the probability that x, ,X(tr ) < xr+dx, given the information that X(td = XI, X(tl) = Xl, ... , X(tr_ d = X,_I' A Markov process is defined by the requirement that
p(t3- t d = fff-:XIX3P3(Xl,tl; X2,t l ; X3,t 3 )dx l dx l dx 3 tl
h
t,
1)
(26.9.2)
< t z < t3 .
a Zp(t3 -td = P(t3 -tz)p(t z -tl)
or
= PI(Xt> tl)WZ(Xl, tllxI' t 1 )Wl(X3, t3lxz, t l ) x ",w2(xr,trlx,-bt,-I)
for all r
(26.9.9)
If use is made of Equations (26.9.3) and (26.9.4) the integrals can be evaluated explicitly as
for all r> 2. The combination of Equations (26.9.1) and (26.9.2) for a Markov process yield the general relation p,(X 1 , t I ; ... ; X" t,)
(26.9.7)
(J> 0
or
We shall derive and solve a functional equation for the correlation function peT). One can compute p(t 3 -tl) as
x Pr-I(XI,tt; ... ; Xr-t,tr-t)·
= Wz(x"t,lx,
(26.9.6)
The brackets indicate an ensemble average. Correlation functions are of great importance in stochastic theory and statistical mechanics. In particular, transport coefficients can be written in terms of time integrals of correlation functions, see Chapters 13, 23, and 24. Show that for a Gaussian Markov process the correlation function peT) of Equation (26.9.6) has the form (~ is a constant) of Problem 23.3, i.e.
wr(xr,trlx,-l,tr-I; ",XI,tI}
I; ,,,XI,!l)
(26.9.5)
and the correlation function
26.9 For a general stochastic process in continuous state space the state probabilities are defined to be Pr(xt> II; Xz, tl;'" Xr , t r ) where Pr dx I dx 2 .. , dx r is the probability that the random function X(t) satisfies Xl ,XUI) < XI +dx1,xz ,X(tl) < Xz +dx l , ..., Xr ,X(tr) < Xr +dx r . Conditional probabilities wr(x r , tr i x r _ 1 ,tr_ I; ... X I, t 1) can be defined in terms of the Pr by
W,(x"I,!x,_h t ,
,
where the variance
so that the random walk (i.e. the Brownian motion) is that of an elasti cally bound particle.
=
(Xi)
a[27T(l-pl/a4)]% exp
Wl(X, tly,s)
where x and T are the continuous space and time variables. [Note that the passage to the limits as given by Equation (26.8.4) is the crucial step in transforming a discrete difference equation to the corresponding par tial differential equation.] A comparison of Equation (26.8.6) with the general Ornstein Uhlenbeck equation shows that F(x)
1 p = ay'TiTex
I
the above equation reduces to the Fokker-Planck equation OP(X,T) OT
529
A stationary Gaussian Markov process is defined by requiring that
I
€z
Ilt -+ D,
Pr(Xt>t I ; ... , xr,t,}
Stochastic methods
aZp(T+T') = p(T)p(T'),
T, T'
> O.
It follows from Equation (26.9.11) that either peT) peT) must have the form peT) = a 2 exp(-{JT),
(26.9.3)
> 2.
..41.
j
(26.9.
(26.9.11)
== 0 for all T or that (26.9.12)
530
Chapter 26
26.9
where (J is a constant. Since for a stationary process per) = p(-r), Equation (26.9.12) is more properly written per) = a 2 exp(-(Jlrl).
1
(26.9.13)
27
The constant (J is chosen positive in all cases of physical interest so that the correlation does not go to infinity as r -l>- 00. The correlation func tion for the important class of stationary Gaussian Markov processes is thus a simple exponentially decreasing function of time.
Ergodic theory, H-theorems, recurrence problems (I) D.ter HAAR (University of Oxford, Oxford)
27.1 In the absence of external forces, the number of atoms of a monatomic gas whose representative points lie in a volume element dudvdw d 3c of velocity space isf(u,v, w)dudvdw f(c)d 3cl, where u, V, ware the Cartesian components of the velocity c of an atom. Let A(B) be the number of atoms which leave (enter) this volume element per unit time due to collisions. Assuming (i) that the atomic collisions are equivalent to collisions between elastic spheres, and (in that the expression f(c)d 3c is correct for any point in configuration space, obtain expressions for A and B and show that
af
-d 3c
B-A
at
.
(27.1.1 )
The assumption (ii) is the assumption of molecular chaos or the so-called Stosszahlansa tz. Solution
Consider the collision between two identical atoms. Let Cl> C2, C'l> c~ be the velocities of the two atoms before and after the collision, respectively, and let w be the centre-of-mass velocity,
w
HCI +C2) .
(27.1.2)
The velocities c~ and c~ are not completely determined by CI and C2, since we only have four equations, cr +c~ = C'!2 + C~2 ") C! +C2 = C'I +c~
(conservation of energy),
(27.1.3)
(conservation of momentum)
(27.1.4)
for six components. (I) For general references see: R.Jancel, The Foundations of Classical and Quantum Statistical Mechanics (Pergamon Press, Oxford), 1969; D.ter Haar, Elements of Statistical Mechanics (Holt, Rinehart, and Winston, New York), 1954, especially Appendix I; D.ter Haar, Rev.Mod. Phys. , 27, 289 (1955); I.E.Farquhar, Ergodic Theory in Statistical Mechanics (Interscience, New York), 1964.
£,1
531
532
Chapter 27
27.1
l
Let w be the unit vector in the direction of the line of centres, that is,
the vector connecting the centre of atom 1 with the centre of atom 2
:c,ee Figures 27.1.1 and 27.1.2); we have then
c, -C't w ICt -c'd . (27.1.5)
27.1
Ergodic theory, H-theorems, recurrence
533
There is a one-to-one correspondence between the original and the inverse collision, since c" cz, and w completely determine C'I, c;, and w'. ~ e rel I
,..JII' U 2 ",'"
""
,.'"
,."
,."
-",,"w
u;
'" Figure 27.1.3.
u, Figure 27.1. L
Figure 27.1.2.
For B we now get
In the centre-of-mass system the description is much simpler. If u"
Uz, u;, u~ are the velocities in the centre-of-mass system, we have Uj
Ut+UZ ui+u~
= Cj -W,
,
Ui
Cj-W
U'I+U~
= 0, U;2+U;Z, U,
w
,
B (27.1.6) 1.8)
-u,
1.9)
Iul-u'll
ff(C,,)d3C~f(C~)d3c~ fal'2'
-+
tZd2w' ,
1.12)
where the prime on the integration sign indicates that the integration over c'" c~, and w' is such that one of the final velocities falls into the previously fixed volume element d3cI' Let J be the Jacobian of the transformation of C'l' c;, w' to c" cz, w, au; av; aw~ aw; aw;· aUl au,····· au, aUl au,
(27.1
,
Figure 27.1.4.
Simple considerations now show (see Figure 27.1.3) that A
= f(Cdd3CJf(cz)d3C2fal2-+
l'z,d zw,
(27.1.10) J =
where we consider in Equation (27.1.1) a value CI of the velocity, and where a 12'" t' z' is given by alZ-+
1'2'
= DZCrelCOSO { =0
(cosO> 0) , (cosO
< 0) .
ac C'l> c~, w') acc cz,
---'-...::..:..-=..;- w) t,
aU't awz
, (27.1.13) aW 2
(27.1.11) au', aW2
Here D is the diameter of the elastic sphere, Crel C1 C2, and 0 is the angle between wand crel . [Note that in writing down Equation (27.1.10) we have tacitly used the Stosszahlansatz.] To find B we must consider inverse collisions, which are collisions for which the velocities after the collision are c, and C2 and before the collision C'l and c~. The line of centres will now be -w (see Figure 27.1.4 which gives the inverse collision in the centre-of-mass system).
aw; . aW 2
where u, v, ware the Cartesian components of c and where WI and W2 are two quantities which determine the unit vector w (we could, for instance, use the polar angles {} and 'P). In the simple case considered here, where we have two identical atoms in collision, we easily find J = 1,
&,1
T 534
Chapter 27
27.1
a result which is generally true and is a consequence of Liouville's theorem (see Problem As a I' 2' -+ 12 and a 12 -+ l' 2' depend only on C reI and (), which are the same for the original and the inverse collision, quite clearly we have al'2'-+12 == al2-+I'2', and changing in expression (27.1.12) from C'l , c~, Wi to C\, c 2 , W, we obtain B
f
d3clf f(C'I)f(c~)d3C2 ad 2w ,
where.r.
=- fCc;),!;
at =- fCc;).
(27.1.15)
27.2 Show that with the assumptions of Problem 27.1 the Maxwell distribution of Problem 3.1 is left unchanged by collisions. that this result is necessary for the Maxwell distribution to be an equilibrium Solution
If f is the Maxwell distribution, we have (" temperature) from Problem 3.1.
( 2: )'h exp(-~~mc2) ,
f =n ~
where T is the (27.2.1)
(27.2.2)
Hence from Equation (27.1.15)
of
at ==
0,
535
Solution
From Equations (27.3.1) and (27.1.15) we have dB = fOf -(lnf+ 1)d3 c
at
or d 3cd 3c I d 2 w .
dB = f(!'f: dt
(27.3.3)
If we interchange c and C I and bear in mind that this leaves crel and thus a unchanged, we can also write
: = fC!'f; -1i'1)(1nfl + l)ad 3cd 3c 1 d2w.
(27.3.4)
If we now transform from c, CI, W to c', C'I' w' and use the fact that the jacobian of this transformation is equal to 1, we can write instead of Equations (27.3.3) and (27.3.4)
:
= f Ufl -!'fD Onf'+ l)ad 3cd 3c\d 2 w,
(27.3.5)
:
=
fUfl -fU;)(lnf; + l)ad 3cd 3c l d 2 w.
(27.3.6)
or
Taking the average of Equations (27.3.3) to (27.3.6) we have dB d3cd3cl d 2w . )In dt
(27.3.7)
As for p and q non-negative,
and by virtue of Equation (27.1.3)
fti2 == fU; .
Ergodic theory, H·theorems, recurrence problems
C27.1.14)
where we have dropped the index of al'2' -+ 12' Combining Equations (27.1.1), (27.1.10), and (27.1.11), we finally have
f2 -fU;)a d3c :zd 2w ,
27.4
(27.2.3)
p{> 0 (p'* q), q = 0 (p = q),
(p-q)ln
and as a is a positive quantity (compare Problem 27 .1), we see that dB dt < 0 . (27.3.8)
which proves that the Maxwell distribution is left unchanged by collisions. 27.3 If H == fflnfd3c,
prove that for a gas of elastic spheres dB -~O dt ...... and discuss this result. [H is Boltzmann's H function.]
(27.3.1)
One can prove that H is a bounded function and the result (27.3.8) thus means that H will decrease until the distribution function satisfies Equation (27.2.3). Equation (27.2.3) is thus not only a sufficient. but also a necessary condition for f to be an equilibrium distribution. 27.4 The equilibrium distribution function of a gas of non-interacting particles is Maxwellian. Show that the entropy of such a system is
(27.3.2)
S
-kH+K,
where K is some additive constant.
C27.4.1)
Chapter 27
536
27.4
Solution
For a gas of non-interacting particles, the distribution function f is the Maxwell function given by Equation (27.2.1), whence Inf = Inn+~In{3-~{3mc2+const.
(27.4.2)
If v is the volume per unit mass,
27.6
Ergodic theory, H-theorems, recurrence problems
537
27.6 Defining the equilibrium distribution as the one which makes W(Nk ) of Problem 27.5 a maximum for given values of N and of the total energy E, find the equilibrium distribution and find also a connection between the corresponding probability and Boltzmann's H function for the case of a perfect classical gas. [Hint: see Problem 2.11.] Solution
v
(27.4.3)
nm
we find from Equations (27.4.2) and (27.3.1) H - = -Inv In{3 n
From
(27.5.1) we have In W
= In N! + L [N; In Zj -lnN;!] + const.
(27.6.1 )
i
(27.4.4)
If the N; are sufficiently large that we can use the Stirling formula Inx!
where the bar indicates an average,
= xlnx-x,
(27.6.2)
We have G
fGfd3c.
(27.4.5)
N· In W = ~ N j In NZ. + const I
Evaluating mc 2 we find from Equation (27.4.4) H
-n = -lnv+Vn{3+const
or
H
n = -lnv -i In T+ const.
where we have used the fact that the total number of particles is fixed, (27.4.6)
N=L (27.4.7)
(27.4.8)
where the constant D involves the chemical constant. Thus, apart from a possible addition constant, Sv -kH.
E = 2.N;€i ,
oN
=a=
a=
oE
t oN; (In ~i + 1) ,
"oN L /,
(27.6.7)
L€joN; .
(27.6.8)
i
;
toN;(-In:ij +a-{3ei) = 0, = CN!
where C is a normalisation constant.
n i
(27.6.6)
Taking (27.6.6)+(a+ 1)(27.6.7)-{3(27.6.8) we get
Solution
W(Nk )
(27.6.5)
where €i is the energy of an atom in the cell Zj. Looking for a maximum of In W under the conditions (27.6.4) and (27.6.5), we use the method of Lagrangian multipliers. For a variation oN; in the N; we have •
oln W = a = 27.5 Divide velocity space into non-overlapping cells of size Zb each cell being a volume element d3 c. Let the representative points of the N atoms in a gas be distributed over the cells in such a way that there are Nk points in the kth cell. Let W(Nk ) be the probability for a given Nk distribution, defined as the fraction of all possible arrangements for which this Nk distribution is realised. Find an expression for W(Nk ) assuming that the a priori probability for a representative point to into the kth cell will be proportional to its size. rHin t: see Problem 2.11.]
(27.6.4)
The other condition to be satisfied is that of a fixed total energy,
For the entropy Sv per unit volume of a perfect classical gas, we have (see Problem 1.23 with A = nk and g 1) Sv = nk(} In T+ Inv)+ D ,
(27.6.3)
I
(27.6.9)
whence (27.5.1 )
N;
NZ j exp(a-{3ei) ,
which is the Maxwell distribution.
(27.6.10)
538
Chapter 27
27.6
For the case of a perfect gas we have 3
d c,
(27.6.11 )
I 3 N = nf(c)d c ,
(27.6.12)
!mc 2
(27.6.13)
IV; Ej
=
.
From Equations (27.6.3), (27.6.10), (27.6.11), (27.6.12), we then get In W = - fi1nfd3c+const,
I
I
27.8
Ergodic theory, H·theorems, recurrence problems
function will consist of the four numbers fl, fz, f3, f4 which are the numbers of P molecules per unit area in the four possible directions. If we assume that the number IV;jt:.t of P molecules which during a time interval t:.t change direction from ito j per unit area is given by the Stosszahlansatz expression Nilt:.t = fiSiln , (27.7.1) where Sjl is the area of a parallelogram of length ct:.t on that edge ofone of the Q molecules which is in the -i,j quadrant (see Figure 27.7.1), prove that the fi will approach equilibrium values. As there is no preferential direction, we expect that the equilibrium distribution is given by the equation
klnW.
ffq = j!fq
(27.6.14)
27.7 Consider the following simplified model of a gas, due to the Ehrenfests. In the plane of the paper there may be a large number of point particles, N per unit area, which are called P molecules. They do not interact with each other, but they collide elastically with another set of entities which are called Q molecules; these are squares of edge length a, distributed at random over the plane, and fixed in position in such a way that their diagonals are exactly parallel to the x and y axes. Their average surface density is n, and it is assumed that their mean distance apart is large compared to a. Suppose that at a certain moment all the P molecules have velocities which are of the same absolute magnitude, c, and limited in direction to (1) the positive x axis, (2) the positive y axis, (3) the negative x axis, and the negative y axis (see Figure 27.7.1). As a result the distribution y
t
f;q
=
f: q = ~N.
(27.7.2)
From Equation (27.7.1) it follows that we have (Ail = S/in/ t:.t = A) dfl
dt
-N12-N14+Nzl+N41 = A[fz +f4
2fd
dfz = AU1 +f3 -2fz]
dt
df3
dt
A[fz+f4- 2f 3 ]
df4
= AUI + f3 - 2f4]
dt
with the solutions
fi(t) =
fl'q +
-fr]e- 2At •
3 t2 I
27.8 Consider the Lorentz model of a metal in which the electrons, which are supposed to be non-interacting, are scattered in such a way that the number per unit volume Nw.w dtd 2 wd 2 w' which during a time interval dt change their directions from within an element of solid angle d2 w to within an element of solid angle d 2 w' is given by the equation
~4
d 2 d2 ' 2 2 ,_ ...-!:::!.~ Nw,w,dtd wd w - Af({},i()) 41f 41f dt,
......--
-
... x Figure 27.7.1.
539
Solution
and hence, using also Equation (27.4.1),
-kH= Sv
1
(27.8, I)
where f({}, i())sin {}d{}di()/41f f(w)d zw/41f] is the number of electrons per unit volume with velocities in a direction within the solid angle d 2 w. In this model all electrons are moving with the same speed. Find the equilibrium distribution of the electrons and show that the scattering mechanism will produce an exponential approach to this equilibrium distribution.
Chapter 27
540
27.8
Solution
As there is no preferential direction, we expect that the equilibrium distribution is an isotropic one:
27.11
Ergodic theory, H-theorems, recurrence problems
The difference oN arises from the flow of the points in phase space and, by analogy with ordinary flow of a gas or liquid, we find
q
j
j"q(w)
N,
(27.8.2)
where N is the number of electrons per unit volume. From Equation (27.8.1) we find df(w)
A'ff(WI)d2W'-f'(
L
dt
41T
w
N-[N
~=A(N-f)
J
o(w)]e- At
'
aD'
at+~
This equation shows that the reaction in flow and is a form of Liouville's theorem.
r
r
space which
o.
(27.9.1)
aH . aH 7 2) a' qj a ,i 1,2, ... , sN , (2 .9. qi Pi where to simplify matters we have numbered the p's and q's continuously. If dD / dt is the convected rate of change in D, that is, the rate of change when we follow a representative point along its orbit in r space and aD/at the local rate of change, we have
N
"laD.
aD'J
(27.9.3)
Let N be the number of representative points in dn at time t so that = Ddn. At t+ ot we have N+oN = (D+
dt
~~ ot) dn .
(27.9.4)
(27.9.7)
O.
27.10 Let v be the 2sN-dimensional 'velocity' vector in r space with components qb ...,qsN,p" ... ,psN,and let a be the 'projection' ofa cell on in r space on a plane at right angles to v. Find an expression for the mean time T spent by a representative point inside on. Solution T
is clearly given by the equation I V
+ f Lapl i + aqiqj .
(27.9.6)
dD
T=
. Pi
aD
(27.9.5)
whence
space is an incompressible
If a 80int in r space is determined by the values .of the 2s/l! variables q~'. pi (k = 1, ... , s; j = 1, ... , N) where the q~) and p~) are the generalised coordinates and momenta of the jth particle, the motion in r space is governed by the Hamiltonian equations of motion
dD
,,(aD. aD.) aqi qi + aplj ,
at = f
The time
Solution
dt =
aD
(27.8.4)
fIdp;dqi) of
(aD. aD.) apli+aqjqi
dnot.
we find from Equations (27.9.4) and (27.9.3) that
i
dt
-+-=0. aqi api .
(27.8.3)
represent an ensemble of such systems, prove that dD
Pi
a ) (aD. aD.)] f,,[-D (aaqj + api + aqjqi+ apli
aqj api
27.9 A classical system contains N particles, each of s degrees of freedom so that its development can be described by a (representative) point in the 2sN-dimensional phase space (r space) with coordinates PI, ... , PsN, q" ... , qsN' If D(PI, ... , qsN, t)dn is the number of represen
tative points in a volume element dn
oN = -
As it follows from Equation (27.9.2) that
whence j(w)
541
'
(27.10.1)
where v [qi +'''+PiN]'h and [the mean segment of the orbit inside on. From the definition of a it follows that on 1= (27.10.2)
a
and hence on av
T=
(27.10.3)
27.11 Consider a system with a fixed energy E. We shall assume that the energy surface E(p, q) = constant is an invariant indecomposable region n of r space, which means that (n for any point P also the total orbit through P lies in nand (li) it cannot be divided into two parts n' and nil which are separately invariant. We shall not prove, but only state, the plausible fact that indecomposability of n is equivalent to the transitivity of the motion, which means that the orbit from any point P in n will come arbitrarily close to any other poin t P' in n. Not only will any point come arbitrarily close to any other point in n, but if we look at a cell on in r space, the point P will traverse this cell over and over
Chapter 27
542
27.11
again, provided the total volume of n is finite as we shall assume to be the case (Poincare's recurrence fheorem). Give an expression for the mean recurrence time T and estimate this quantity for the case of a gas of 10 18 atoms of mass 1 g in a volume of 1 cm 3 with an average velocity of 104 cm S-I where the cell on has dimensions 10- 7 cm for its position coordinates and 102 g cm S-I for its momentum coordinates.
27.13
Ergodic theory. H·theorems, recurrence problems
Solution
In Equation (27.12.1) we write
f
with
n
t
T=
n
~ (l0-7)3N(l02)3N,
and
n Hence
~
T ~ (3
E )3N/2
l3N ( m X
10 18 )(1'5
~
V
~
104 (3N)"",
(3N)3N/2(l04)3N.
x 10 18) S ~ 10(10
19
)
(27.12.4)
(27.12.5)
.
n/on
L f(tn)Ot n
j
Ni
L= 0 r=L0 f(Pj)otjr.
(27.12.6)
where r numbers the recurrences of passage of P through on; and i numbers the cells, and where N; is the number of recurrences in the time interval (0, t). We have thus n/6n Ot. (27.12.7) f* = lim lim f(~) tiN:.
To estimate T for the case considered, we note that
o
= LOtn
f(tn)Ot n
Choosing for the Ot n the transit times of P through the cells on; and using Poincare's theorem that every cell will be traversed several times, we have
(27.11.1)
av
= lim L
t f(t)dt
o
Solution
The volume swept out per unit time by the orbits passing through on is clearly 01) and by Liouville's theorem this will be the volume swept out by those orbits at any time. Moreover, if n is indecomposable, all orbits will pass through on and they will fill n completely. This means, there fore, that the mean recurrence time will be
543
t .. ""
We clearly have lim Ot;
years.
.L
on .. 0 I = 0
I
mean life time of P in onj, which according
t .. ""
on/ov. We also have lim tiN; = mean t .. "" recurrence time, which according to Equation (27.11.1) is equal to n/ov. to Equation (27.10.3) is equal to
27.12 Using the assumptions of the previous problems, we can now attack a simplified proof of the ergodic theorem, that is, the equality of ensemble and time averages (2). Consider a phase function f(P) which is a function of the representative point P in r space. Its time average f* which is the average which is measured physically-is given by the equation 1 f* = lim fIP(r)] dr, (27.12.1)
t ft
t .. ""
0
while it is usually only possible to evaluate the phase average f given by
f
~L f(p)dn.
(27.12.2)
The ergodic theorem now states that
f* =f.
(27.12.3)
By dividing n into cells on; and the integral in Equation (27.12.1) into time intervals corresponding to the passage of P through the different cells prove the theorem (27.12.3).
Hence we have
f*
on; lim "L. f(Pj)n
6n .. 0
i
See KWergeland,Acta Chem Scand., 12,1117,1958.
f
n
f(p)dn
= f.
27.13 To study the approach to equilibrium and the recurrence problem, let us again consider the Lorentz model of Problem 27.8. We have already seen that the system in that model has an equilibrium distribution (27.8.2). We now divide phase space into 2m + 1 cells of equal volume, which we number from -m to +m. Each cell corresponds to an element of solid angle ow with 411' ow = 2m+ 1 (27.13.1) The distribution function is now a function of a discrete argument, -m, ..., +m), and its equilibrium value is
fv(v
I'eq (2)
= n1
Jv
-
_N __
2m+ 1 .
(27.13.2)
544
Chapter 27
27.13
and assuming Ifv - f:;q 1/fv ~ 1,
The 2m + I fv satisfy the condition +m
2: fv(t) V=-m
== N.
(27.13.3)
To measure the departure from equilibrium we introduce a function A given by the equation
if;(fv)
2m+ I)m [ (f, -rq)2] =(~ (2m+ l)Yl exp -(2m+ 1)2: v 2N . (27.14.4)
2:
(fv_t;q)2 .
Assuming that Ifv - t;q I ~ f:;q, find a relation between If and A, where we define If as If =
A(p) =
(27.13.4)
v=-m
2: fv 1nfv .
v
We then find for the function A(p)
+m
A
545
Ergodic theory, H·theorems, recurrence problems
27.16
f...J
if;(fv)exp(-ipA)
JI
df-vdtv ,
(27.14.5)
where the integral is 2m-fold since the fv satisfy condition (27.13.3). The integration is straightforward, but tedious, and the result is Np
(27.13.5)
A(p)
= (1-2i 2m+ 1
)-m
(27.14.6)
Solution
If
= 2:fv Infv == v
A
Substituting Equation (27.14.6) into Equation (27.14.2) and evaluating the integral by contour integration, we find finally 2m+l)m Am-l [(2m+l)A] w(A) = ( (m_l)!ex p 2N ' (27.14.7)
Ifeq+ 2N '
where Ifeq ==
2:v fv
eq
----m-
Inf:q .
which shoWS that w(A) decreases steeply with increasing A, provided N is large.
27.14 Use the result from probability theory that if
2: ¢k(qi) '
(27.14.1)
k
where the qi are random variables such that if;(qJ ndqj is the probability of finding the qj within ranges qi to qj +dqj. The probability of finding
d
where A(p)
=
+""
exp(-ip
_00
27.15 Find the average value of A and its dispersion (or standard deviation). Solution
From Equation (27.14.7) we find
Aav =
(27.14.2)
2
r
_
2mN
2
P 2:¢k) if;(qj)Ddqj.
(27.14.3)
and hence, using the fact that m ~ I,
Find the normalised probability w(A)dA that A lies within the interval A, A + dA and discuss the result. Solution
In the present case, using Equation (27.13.3), we find easily for the probability distribution function if;(fv)
nN!fv! (2m + 1r
N
or, using the Stirling formula for the factorial in the form (see Problem 2.11 b) lox! == xlnx-x+!lnx+!lo21T,
'
2m(2m + 2) 2
(2m+IP N ,
_
(A )av - JA w(A)dA -
f...J exp 0
if;(fv) :::
r
Jw(A)AdA = 2m+ I
[( A-A av )2] av
~
~-
m
27.16 Using the model of Problem 27.8 calculate the probability w(A, A') that A changes its value from A to A' in a time interval T, assuming that the distribution in space of the scatterers is random. Solution
The change in A is related to the change in the fv and we have
A'-A == 22:(fv-f:q)(f~-fv) v
where we have used Equation (27.13.3).
=
22:fvU;-fv), v
(27.16.1)
546
Chapter 21
21.16
Let Xvv' be the number of particles passing from cell v to cell v' during r. Its average value will be given by the equation rcompare Equations (27.8.1) and (27.8.3)] x av vv ,
=
27_17 Use the results from the preceding problem to find the average value Ll' of Ll a time r after its value was Ll, and also to find the average value Ll" of Ll a time r before its value becomes Ll. (3) Solution
16.2)
We have
As the distribution in space of the scatterers is random, we can, provided r is sufficiently small and aN sufficiently large so that Ar
<{
I and N;'P aN;'P I ,
f
LX.v
weLl, Ll')Ll' dLl'
or
(27.16.3)
Ll~v
take for the distribution of the Xvv' a Gaussian distribution corresponding to
-lh I (x vv2al ' -alv )2J (2rrlv) exp .
1/I(x vv ')
(27.1
v
=
Lv' (xv'v-xvv')'
with A(p,a) =
4~2ffexp[-iaLl-iP(Ll'
(27.16.5)
f...
Ll)]A(p,u)dpdu
m ..•
d/1 .. • dfm dx- m- m .. · dx mm 1/I(fv)
x exp ria L (Iv - Iveq) 2+ 2ip L
(Xv
=
Solution
To find T(Ll) we first of all find the mean time of persistence in a state Ll. Let IjJA (kr) be the probability that Ll is observed at times 0, r, 2r, 3r, ..., (k - I )r, but not at time kr. We clearly have
A(p, a) =
Jm
m+1 4Nap(2Nap+2m+ l)-2Nia+2m+ 1
1
(Ll, Ll)[I-w(Ll, Ll)].
(27.18.1)
The mean time of persistence S(Ll) is clearly given by the equation S(Ll)
L krq,A (kr) k ='1
(27.16.9)
=
L kr[ I
weLl, Ll)]W k - 1(Ll, Ll)
k = 1
r I - weLl, Ll) .
and from Equations (27.16.6) and (27.14.7) finally ,
Wk-
IjJA(kT)
and neglect cubic terms in the exponent in comparison with quadratic terms. We then obtain
"l
= Ll~v
27.18 Evaluate the recurrence time T(Ll) of a state characterised by Ll.
(27.16.8)
Iv'''.J
= (1- 2a)Ll
showing the fact that the Ll curve is symmetric in time.
uv')
Using Equations (27.16.4) and (27.16.5), the integration over the (2m + 1)2 variables Xvv' is straightforward. To evaluate the integral over the 2m variables (,,(v =1= 0) we introduce new variables (Xv,
weLl, Ll)
~r------
which leads to
-lv)J. (27.16.7)
eq (Iv - Iv )
2A
w(Ll")w(Ll", Ll) dLl"
(27.16.6)
n1/1 (x
r
fLl"W(Ll")W(Ll", Ll)dLl" Ll~v
Ll~v
fdf
2a)Ll,
(I
[compare Equation (27.8.3)]. For Ll~v we have
We now are in a position similar to the one in Problem 27.14 with w(Ll)w(Ll, Ll') for ~. We then get w(Ll)w(Ll,Ll')
=
-2a~ =
dLl) ( dt av
We then have rcompare Equation (27.8.3)] I~
541
Ergodic theory, H-theorems, recurrence problems
21.18
{[Ll'-Ll+2(2m+ l)aLlF} (32rrNaLlr v'exp 32NaLl . (27.16.10) 1
(3) Compare S.Chandrasekhar, Rev. Mod. Phys., 15, l, 1943.
,. ii,
(27.18.2)
27.18
Chapter 27
548
Similarly, if 1/In(k1') is the probability that starting from an arbitrary state /I. which is not a we shall observe states /I. at 0,1',21', ... , (k I but a at time kr, we have ""
T(a)= Lkr1/l/1(kr).
28
(27.18.3)
Variational principles and minimum entropy production
"='I
If w(/I., /I.) is the probability that from any /I. there has occurred in the interval l' a transition to some other /I., we have clearly 1/1/1 (k1') = W"-I(/I., /1.)[ I-w(/I., /I.)] ,
(I)
S.sIMONS (Queen Mary College, London)
(27.18.4) i
and hence
l'
T(a)
1- w(/I., /I.) .
(27.18.5)
MACROSCOPIC PRINCIPLES
(27.18.6)
28.1 Let the rate of entropy production a in a system be expressed in the form N
We have clearly I
w(/I., /I.)
w(/I., a) ,
and as at equilibrium the number of transitions /I. ~ a must be the same as the number of transitions a ~ /I., we have (27.18.7) [I-w(a)]w(/I., a) = w(a)[l w(a, a)] . Combining Equations (27.18.5), (27 .18.6), and (27.18.7) we have finally T(a)
1'[ I-w(a)] wi A
\r
1 _,,,(
A
A \1
I weal = 9(a)--w(-a-=-)-'
and we see that T(a) decreases steeply with increasing a.
L IpXp
a
(28.1.1)
P'" I
where Ip and Xp (1 ~ p ~ N) are respectively the fluxes and forces for the system. It is known from Onsager's theory of irreversible thermo dynamics that if in the steady state the relationship between Xp and Ip is of the form (see Problem 25.7) N
Ip =
L
LpqXq
(28.1.2)
q=' 1
(27.18.8)
then, in the absence of a magnetic field and rotation of the system, the phenomenological coefficient matrix L is symmetric; that is Lpq Lqp. Prove that if XI, X 2 , ... , Xs (s < N) are kept constant, then the minimum rate of entropy production for variable X S +l> X s + 2 , ... , X N corresponds to the steady state of the system in which Is~ I, 1$+2, ... , IN are all zero. Solution
We see from Equations (28.1.1) and (28.1.2) that the rate of entropy production may be expressed in the form N
a=
L LpqXpXq. p,q = 1
(28.1.3)
In order to minimise this for variable X S +I> X S + 2 , turn aa/axp to zero for s+ 1 ~ P ~ N. This gives
aa ax p for s+ I
~
P
~
... ,
X N , we equate in
N
=
L (Lpq+ Lqp)Xq
= 0
q=' I
N. Finally, making use of Onsager's symmetry relation
(I) A useful background reference on entropy production is Thermodynamics of Irreversible Processes by LPrigogine (Interscience, New York), 1967.
549
550 Lpq
Chapter 28
28.1
L qp , we obtain N
r
28.2
..
2 L LpqXq = 0 q = I
Variational principles and minimum entropy production
551
Express the left hand side as a quadratic in A and consider the condition for it to be non-negative for real A.J
Solution
for s+ I ,,;;;: p ,,;;;: N, which is equivalent to ls+1 = 0 = ls+2 = ... = IN' To confirm that this indeed corresponds to a minimum value of a (and not to some other stationary value) we need only remark that it follows from the Second Law of Thermodynamics that a is always positive and thus, since it possesses only a single stationary value, this must be a minimum.
(a) Making use of Equation (28.2.4) and of the symmetry of M we find LMpqKpKq -2 LMpqKplq p,q p,q
+ p,q LMpqlplq
:;;;. O.
(28.2.5)
Since
28.2 In the notation of the last problem, the rate of entropy production
Xp = LMpqlq
1\
N
a = L IpXp ,
(28.2.1)
p'" I
and it is clear that in the steady state, when Equation (28.1.2) is applicable, we have
(28.2.6)
q
this gives LM pqKpKq -2L KpXp :;;;. - LMpqlplq p, q p p, q
LMpqlplq p, q
2LlpXp. p
N
Xp =
(28.2.2)
L Mpqlq , q"'l
where M is the inverse of L. We then have N
a
L
Mpqlplq
(28.2.3)
p, q '" I
and, as a is always:;;;' 0, M is a symmetric positive semi-definite matrix (2). If the system is not in a steady state so that Equation (28.2.2) does not hold, the two expressions for a, (28.2.1) and (28.2.3), will in general be different. We shall then refer to these expressions respectively as the extrinsic and intrinsic rates of entropy production and will denote them bya e and aj. (a) If the matrix Mpq and the forces Fp (1 ,,;;;: p ,,;;;: N) are given, prove that the steady state relationship (28.2.2) corresponds to minimising aj 2a e with respect to varying Ip (1 ,,;;;: p ,,;;;: N). (b) Show also that relationship (28.2.2) is given by maximising aj subject to it being kept equal to a e . [Hint: Since M is positive definite, N
L Mpq(Kp -lp)(Kq -lq):;;;' 0, (28.2.4) p,q = I where Kp are arbitrary fluxes and Ip satisfies Equation (28.2.2).J
(c) Prove that a further variational principle for Equation (28.2.2) is given by minimising ada;. [Hint: As M is positive semi-definite, N
L p,q
Mpq(AKp +lp)(AKq +lq):;;;' 0 I
(2) A matrix M is positive semi-definite if
Thus aj -2ae calculated with general fluxes is greater than or equal to its value when calculated with the steady states fluxes satisfying Equation (28.2.6), and hence these latter fluxes minimise aj 2a e . (b) To prove the second variational principle, we note that keeping aj equal to ae corresponds to the constraint
L MpqXpXq
P.q
for all real A.
;;;. 0 for all XI"
LMpqKpKq = LXpKp = LMpqKplq . P.q p p,q
(28.2.7)
If we make use of this, inequality (28.2.1) becomes LMpqlplq :;;;. LMpqKpKq . p,q p,q
This implies that, subject to the constraint aj = a e, aj is maximised by the fluxes satisfying Equation (28.2.6). (c) Proof of the third variational principle proceeds by noting that LMpq(AKp +lp)(AKq +lq):;;;' 0 pq
is equivalent to A2 LMpqKpKq+2ALMpqKplq+ LMpq1plq:;;;' O. p. q
P.q
P.q
For this to be true for all real A, the condition 2
( LMpqKplq ) P.q
,,;;;:
LMpqKpKq LMpqlplq p, q p,q
must hold. As Equation (28.2.6) implies that
I
LM pqKplq = p Xp Kp , p,q
(28.2.8)
28.2
Chapter 28
552
it follows from inequality (28.2.8) that 2. MpqKpKq
P,L.
2.XpKp)2 ( p
~.
Variational principles and minimum entropy production
I
I!.L:;Lq_ _~
f2. X pJp)2 \ p
This means that a;/a: is a minimum when Kp = Jp. Alternative proofs of these results can be formulated using the techniques of calculus. Thus, for the first of the above variational principles, we can find the minimum of aj 2a e (considered as a function of Kp) by equating to zero in turn 3(aj 2a e )/'OKp for I ~ p ~ N. This gives
Let Jp (I ~ P ~ N) be the currents in the system characterised by matrix M + M' corresponding to the given forces Fp . Then I a· ~ in the steady state aM+M' ae N
_
-
2. P,q '"
1
(Mpq + M~q )JpJq N
2. JpXp p
1
2. MpqJpJq
(I ~ p ~ N)
2.MpqKq -Xp = 0
553
Solution
2. M pqfpJq
2.MpqJpJq p,q
P-
28.4
= P,q
q
2. Jp X p
on using the symmetry of M, and this corresponds to Equation (28.2.6). Jp + tJ.Jp , when To show that this is indeed a minimum let Kp tJ.(aj - 2a e ) = 2.Mpq(Jp + tJ.Jp)(Jq + tJ.Jq ) - 22. (Jp p,q p
+ tJ.Jp )Xp
2. MpqfpJq + 22. JpXp p,q p 22. MpqtJ.JpJq + 2. MpqtJ.JptJ.Jq pq P.q
22. Xp tJ.Jp p
p
I
aM+M'
= 2.MpqtJ.JptJ.Jq
p,q
28.3 Consider the steady state of three systems characterised in turn by an inverse phenomenological coefficient matrix M, M', M + M' and suppose the same forces F to be applied to each. If in the steady state the rates of entropy production for the systems are respectively aM, aM', aM+M', use the third variational principle of the last problem to prove that I
I
I
aM+M'
aM
aM'
--~-+
[Hint: Consider adai for the third system, evaluated with the corresponding steady state currents for that system.]
(28.3.1)
p
Now the first term on the right hand side of Equation (28.3.1) is aJai for the system characterised by matrix M, but calculated with currents which in general are different to the steady state currents for that system. Thus, according to the third variational principle of Problem 28.2, this term is ~ aj/a: for the steady state distribution in the system; that is, this term is ~ I laM. Similarly the second term on the right hand side of Equation (28.3.1) is ~ I laM" Hence it follows from Equation (28.3.1) that --~
on making use of Equation (28.2.6). Since M is a positive definite matrix, it follows that tJ.(aj-2a e ) > 0 for all tJ.Jp and therefore that the stationary value is a true minimum. Similar, though rather more lengthy, proofs can be obtained for the other two variational principles.
LM~qJpJq __ 2. JpXp
+~P.-;!;q:;--
aM
1 aM'
+
28.4 In Problem 28.2 we saw that the solution of the steady state equation (28.2.2) is given by certain variational principles (for example, the minimisation of aj - 2ae with respect to the fluxes Jp )' This offers a technique for obtaining approximate solutions of Equation (28.2.2) for the fluxes (in terms of the given forces) in cases where N is large and the matrix M cannot therefore be readily inverted. We assume a solution of Equation (28.2.2) containing a set of undetermined parameters (in number less than N) and proceed to find the best values for these parameters by minimising aj 2a e with respect to variation of the parameters. It is convenient to introduce these parameters as undeter mined constants in a linear combination of given vectors, and we therefore assume a solution for Jp in the form T
Jp
=
2. ~srJs)
s=
(l ~ p ~ N) .
(28.4.1 )
I
Here r~s) (I ~ p ~ N, 1 ~ s ~ T, T < N) are a set of T given vectors and ~s are the arbitrary ·constants. Using this form for Jp , together
,.
with the above variational principle, prove that the values of as are determined by equations of the form
L Rstat T
= Qs
(1
28.5
28.4
Chapter 28
554
"
Variational principles and minimum entropy production
problem for showing that OJ - 20e is minimised by Jp satisfying Equation (28.2.2), it now follows that OJ - 20 e is minimised by as taking values determined by the equation
< s < T)
T
t= I
L Rstat
and obtain explicit forms for R st and Qs in terms of Xp, Mpq, and r~s). Show that, with the present approximation, the rate of entropy production in the steady state is given by
L
fR-1]stQsQt.
.
s, t= 1
Solution
Since T
1.p = ~ L. as r(s) p s =I
OJ
=
T
L Rstasat
s, t '"
(28.4.2)
I
where
N
~ L.
R st =
p,q=
Also N
Oe
(28.4.3)
M pq r(s)r(t) p q . I
N'
~
T
P~lxpC~IO!sr~S»)
== PJ;I JpXp = T
(28.4.4)
L-Qsas
s= I where
N
Qs =
Thus
L xpr~s) p=l
T OJ
-20 e =
i S
o
T (T
sL1Qs Jl[R-1]sQt
)=
r
s,~I[R-I]stQsQt.
(28.4.8)
We may note that this method of obtaining an approximate solution is of particular value when the ultimate purpose is to calculate the rate of entropy production in the steady state. This is because the quantity being minimised, OJ - 20 e , reduces to -0 in the steady state. Hence, if the solution for the currents is in error by a small amount e, the error in the value of 0 obtained will be proportional to e 2 , which will be much smaller. Also, since the correct value of -0 in the steady state is the minimum value of 0i - 20 e , it follows that the value of a as given by Equation (28.4.8) will be less than the true value of a; that is, Equation (28.4.8) provides a lower limit for o. 28.5 In a conducting medium the electric current may be described by a vector Jp (I < P < 3) where J 1 , J 2 , J3 are respectively the components of current density along the x, y, z axes of a Cartesian co-ordinate system. Likewise the electric field may be described by a vector 8,p (l < P < 3) where 8,1> 8,2, 8,3 are respectively the electric field components along the x, y, z directions. The rate of entropy production per unit volume 0 is given by 3·
T
L Rstasat -2
s, t
(28.4.5)
.
(28.4.7)
with Rand Q given by Equations (28.4.2) and (28.4.5). Alternatively, Equation (28.4.7) can be derived by equating to zero o(Oj -20 e )/oas for I < s < T, with OJ -20e given by Equation (28.4.6). In the steady state, the rate of entropy production is given by Equation (28.4.4), where as is obtained from Equation (28.4.7). Thus, in the steady state,
,
p'~IMpqJpJq = pJ=IMpqctlasr~S»)(JlatrJt»)
Qs
t= 1
T
o=
555
L= Qsas '
(28.4.6)
To =
P
I
Now, it readily follows from Equations (28.4.2) and (28.4.3) that R st is a symmetric positive definite matrix, since Mpq has these properties. The expression (28.4.6) is therefore of the same form as the expression for OJ - 20 e in terms of the original fluxes Jp , if we identify a with J, R with M, and Q with X. Thus, by employing the same proof as used in the last
L
Jp 8,p ,
(28.5.1)
I
where T is the absolute temperature, and in the steady state 8, and J are connected by the general relationship 3
&p =
L rpqJq , q = I
(28.5.2)
r ·it
28.6
28.5
Chapter 28
556
Ta = 3·40
&~m r = ( :
(
.
Obtain expressions for aj and a e in terms of J1> J 2 , J3 • Hence, if an approximate solution of Equation (28.5 .2) is given by Jp = O!rp ' where
r
~(~)
and O! is an arbitrary constant, use the method of the last problem to estimate a value for Ta in the steady state. Compare this with the true value and comment. Solution
aj
=
I
T
1 a e = -T
L3
I
p,q= 1 3
L p=l
rpqJpJq = -T(3.!f+3J?+3.!1+ 2J1J 2 + 2JZJ 3 + 2J3 J 1 )
I
tnpJp ::;: -T(il + 2J2 + 3J3 )
(28.5.4)
and, as expected, the estimated value (28.5.3) is less than the true value. The numerical error in the approximate steady state current is quite large since the latter equals
}
3
557
On making use of Equation (28.5.3), we then find
where rpq (l .,;;; p, q .,;;; 3) is the resistance tensor. In a particular case, with a given set of units, the electric field vector tn is
and the resistance matrix takes the form 3 I
Variational principles and minimum entropy production
•
Using the given trial solution, we let J I = O!, J2 20!, J 3 = 30!, and obtainT(aj 2a e )::;: 640!2_280!. To find the minimum value ofaj-2a e , we equate o(aj-2a e )/oO! to zero and obtain 12801-28=0, whence O! = ;,;. With this value of O!,
T(2a e -aj) = Ta = 3·06.
(28.5.3)
To obtain the true value of Ta we must first solve Equation (28.5.2) for Jp , and this gives
J~{:)
0'22) (-0'1) 0·44 while the true solution is 0·4. 0·66
0·9
Nevertheless the error in Ta is much less, being about 10%, and this agrees with the discussion at the end of the solution to Problem 28.4. ELECTRON FLOW PROBLEMS
28.6 We now wish to consider the formulation of variational principles concerning entropy production, based on a description of the system at an atomic level, rather than at the macroscopic level employed hitherto. To be quite definite we shall confine ourselves to particles, such as electrons, which obey Fermi-Dirac statistics, but similar results apply to particles obeying both Maxwell and Bose-Einstein statistics. In a metal each electron is characterised by a three-dimensional wave vector k and the number of electrons of given spin with wave number k is denoted by an occupation number f(k). In thermal equilibrium at temperature T, f(k) is equal to the Fermi-Dirac distribution function [derived in the solutions to Problem 3.12(a)], 1 (28.6.1) Jl) + 1 j'O(k) = exp "T
(E -
where E(k) is the energy of an electron with wave number k, Jl is a constant-the Fermi potential, and" is Boltzmann's constant. In the steady state, however, where a net flow of electrons occurs, f(k) differs from the equilibrium value j'O(k). Now it is known that the entropy S of an assembly of electrons is given by S::;: -"jU{k)ln.t\k)+[I-f(k)]ln[l-f(k)]}dk
(28.6.2)
where the integration is taken over the whole of k space (3). For f(k) close to the Fermi-Dirac distribution we may then let f(k) = fO(k) - ,,-lj'O(l - j'O)r{>(k) ,
(28.6.3)
where r{>(k), which measures the deviation of ffromj'O, is small. (3) P.T.Landsberg, Thermodynamics with Quantum Statistical Applications (lnterscience, New York), 1961, p.233.
r
Chapter 28
558
Making use of Equations (28.6.1), (28.6.2), and (28.6.3) show that the rate of entropy production (J (= as/at) is given by
f
af If(E-p)-dk af
(J
28.7
28.6
"
(28.6.4)
Variational principles and minimum entropy production
Secondly, the value of f will change due to collisions which the electrons undergo. If we restrict our attention to the important case of collisions with impurities in the conductor, it may be shown that the corresponding 'collision' rate of change of f is given by
~~)coli =
on the assumption that terms in
Jdk
af af af -as = -K f[af -+-lnf----ln(1-f) at at at at at
~
(28.6.5)
for all k, k'. Now, in the steady state of the conductor this corresponds to
af ) eony + at af ) coli
Now, it follows from the definition of
I+K lr
.
(28.6.6)
LI-f = exp(P-E) (I-!). KT K This may be substituted into Equation (28.6.5) to give the final result
= -as = at
f
af If(E-p)-dk af
(28.6.7)
on using the first term of the expansion of In (1 -
(28.7.3)
af/at
= 0, and
0
at
On expanding the second factor in Equation (28.6.6) as a power series and retaining only terms linear in
(J
(28.7.2)
L(k, k') ~ 0
L(k, k') = L(k', k),
-Kf~~lnC~f)dk. I-f
fL(k, k')l
where
Solution
L = [~J[I-K-l(1-fO)
559
.
Thus we obtain the steady state condition
f
L(k, k')[
a;:
(28.7.4)
This is the celebrated Boltzmann equation from whose solution the electric current in the conductor may be calculated. It is closely related to the Boltzmann equation for a gas derived in Problem 17.10. If we take each of the expressions on the right hand sides of Equations (28.7.1) and (28.7.2) we can, with the aid of Equation (28..6.4), calculate -v (+k) the corresponding forms for as/at(= (J). Assuming that v(-k) andE(-k) = E(+k), prove that I 3 (J = - L &pl.p , (28.7.5) cony T p= 1 where J is the current density. By making use of the first of Equations (28.7.3), prove that (Jeoll
=
!-
II
L(k, k')[
It may be shown that in calculating (Jcoll the contribution of the second
term on the right hand side of Equation (28.6.3) is zero; for present purposes assume this to be so. Solution
From Equations (28.6.7) and (28.7.1) it follows that (Jcony =
-e&'
f
ar
v
e& -T'
f'
O
(E -p)v af aEdk.
(28.7.6)
r.· "
560
Chapter 28
28.7
Since v(-k) -v( +k) and E(-k) = +E( +k), the integrand in the second term on the right hand side of Equation (28.7.6) is an odd function of k, and thus the value of the integral is zero. As far as the first term is concerned, it follows from the fact that ajO/aE = -jOO - jO)1 ICT (which is readily shown from the definition of jO) that this term may be expressed as
~ 8. •
-
f
=
ff
1
3
-f p L= 1 Jp 8. p
•
(28.7.8)
L(k, k')[rt>(k) -rt>(k')]rt>(k) dk dk'
(28.7.9)
since it may be assumed that the second term in Equation (28.6.7) gives no contribution to as/at) coli, On exchanging k and k' in Equati0J1 (28.7.9) we obtain
~~)COll =
If
L(k, k')[rt>(k')-rt>(k)]rt>(k')dkdk'
561
The analogue of the positive definite character of Mpq and its symmetry are the relations (28.7.3) now satisfied by L(k, k'), so that for all O(k) and rt>(k), fIL(k, k')(O(k)-O(k')-rt>(k)+rt>(k')Fdkdk'
~ 0.]
Consider a general function O(k) and a function rt>(k) satisfying Equation (28.7.4). Then since L(k, k') ~ 0 for all k, k',
ff
tfL(k, k')[rt>(k)-rt>(k')j2dkdk' .
28.8 It was shown in Equation (28.7.5) that -u conv calculated for electron flow is identical with the form for U e as given by Equation (28.5.1) for the case of macroscopic variables. This suggests that just as for macroscopic variables it was possible to formulate variational principles for the relationship between forces and fluxes, based on a stationary value of the entropy production, so similarly it may be possible to obtain the Boltzmann flow equation (28.7.4) by a similar device with rt>(k) as the analogue of Jp • Now for the steady state, it follows from Equation (28.7.4) that -u conv = Ucoll and therefore by analogy with the macro scopic approach we suggest the formulation of variational principles by using -U conv instead of Ue and U coli instead of Uj. Prove that the Boltzmann Equation (28.7.4) is given by minimising U coil + 2u cony with respect to varying rt>(k), or by maximising Ueoll subject to it equalling -u conv , or by minimising ucon/u2onv. [Hint: Follow the same approach as in Problem 28.2 using the continuous variable k instead of the discrete variable p, and replacing summations by integrations.
~0
L(k, k')[O(k) -O(k') -rt>(k) + rt>(k')j2 dk dk'
~
and thus
f
f
L(k, k')[O(k) -O(k')j2dkdk' - 2 L(k, k')[O(k) -O(k')][rt>(k) -rt>(k')] dkdk'
+fL(k, k')[rt>(k)-rt>(k'Wdkdk'
~ O.
(28.8.1)
Now
f
L(k, k')[O(k)-O(k')][rt>(k) -rt>(k')]dkdk' =
(28.7.10)
L(k', k). Finally by adding Equations (28.7.9) and since L(k, k') (28.7.10) we find
~St) coil =
••
Variational principles and minimum entropy production
Solution
From Equations (28.6.7) and (28.7.2) we have Ueoll =
28.8
(28.7.7)
ev(t- to) dk ,
making use of Equation (28.6.3). The integral in expression (28.7.7) is just the total charge transported per unit area per second and is thus equal to the current density J. We obtain the required result
as) at cony
"
f
L(k, k')O(k)[rt>(k) -rt>(k')] dk dk' -
J
L(k, k')O(k')[rt>(k) -rt>(k')]dkdk'
(28.8.2) and exchanging k and k' in the second term, bearing in mind that this leaves L unchanged, we obtain the right hand side of Equation (28.8.2) as 2fO(k){fL(k, k')[rt>(k) -rt>(k')]dk' }dk .
On substituting from Equation (28.7.4) this becomes
ajO f aE O(k)dk .
2e8.. v
We substitute this expression in the second term of the ihequality (28.8.1) and obtain
f
L(k, k')[O(k)-O(k')j2dkdk' -4e&. Iv
~ =
-I I
~:O(k)dk
L(k, k')[rt>(k)-rt>(k')]2dkdk'
L(k, k')[rt>(k)-rt>(k')j2dkdk'-4e&'
f ~rt>(k)dk v
-n . ... . I
562
Chapter 28
28.8
on applying the above transformation to the case e =
.. Author Index Aigrain, P. 347
,
Bardin, J.M. 177
Barker, J.A. 147, 192
Becker, R. 43
Bellemans, A. 230
Bird,R.B.141,144 Blakemore, J.S. 376,377,407 Bludman, S.A. 134 de Boer, J.H. 278 Bonch·Bruevich, V.L. 319 Bray, R. 377 Brewer, L. 443,446 Brittin, W.E. 400 Brunauer, S. 242 Buckingham, M.J. 36 Buehler, R.I. 145 Chandrasekhar, S. 547 Chapman, S. 400 Cowling, T.G. 400 Cruickshank, A.J.B. 161,177 Cruise, D.R. 93 Curtiss,C.F. 141,144,145 Cusack, N. 419 Defay, R. 193,196,230,236 Desloge, E.A. 400
Domenicali, C.A. 417
Dufour, L. 236
Dunning·Davies, J. 136
Egelstaff, P .A. 192
Einbinder, H. 12
Einstein, A. 96
Emden, R. 19
Emmett, P.H. 242 Everett, D.H. 230 Fairbanks, W.M. 36
Farquhar, I.E. 531
Felderhof, B.U. 333 Fisher, I.Z. 192
Fisher, M.E. 43
Fitts, D.D. 421
Flory, P.J. 188
Fowler, R.H. 140
Frisch, H.L. 192, 258
Gauss, C.F. 233,235 Gibbs, J.W. 193,230,233 Glasstone, S. 443,446 Goeppert Mayer, M. see Mayer, M.Goeppert
Goldsmid, H.J. 303
Gregg, S.I. 236,242
Gruen, R. 236
Guggenheim, E.A. 47, 140, 178, 182,
185,192,244,443
GWinn, W.D. 128
Hall, R.N. 376
Harman, T.C. 405,407,412,417,419,
421,425,426,427,429
Helfand, E. 258
Hemmer, P.C. 269
Herzberg, G. 126
Hicks, C.P. 161, 177
Hilf, E. 12
Hill, T.L. 244
Hirschfelder, J.O. 141,144,145 Honig, J.M. 405,407,412,417,419, 421,425,426,427,429 Huang,K. 383,396,400
Jancel,R.531 Jeans, J. 400
Johns, K.A. 139
Kac, M. 269
van Kampen, N.G. 293,302,333
Kittel, C. 56,407 van Konynenberg, P.H. 154
laMer, V.K. 236
Landau, L.D. 458,496
563
564
Author Index
Landsberg,P.T. 12,31,62,75,89,136, 139,289,345,377,399,436,444, 557
Lebowitz, J.L. 192,269
Lewis, G.N. 443,446
Lifshitz, E.M. 458, 496
Lloyd,P. 93
Lopuszanski, J.T. 40
Macdougall, F.A. 443,446
Many, A. 377
Mayer, J.E. 56
Mayer, M.Goeppert 56
Milne, E.A. 19
Mulholland, H.P. 108
Nix, F.C. 303
O'Dwyer, JJ. 93
Opechowski, W. 284
Oppenheim, I. 511
Patterson, D. 177
Pauli, W. 61
Penrose, O. 269
Pippard, A.B. 36, 390
Pitzer,K.S. 128,161,443,446
Pound, R.V. 347
Prigogine, I. 177, 192, 196,230,400,
549
Purcell, E.M. 347
PutIey, E.H. 425,427
Ruderman, M. 134
Rushbrooke, G.S. 192 Ryvkin, S.M. 377
....
Subject Index
Schiegl, F. 62
Schulz-Dubois, E.O. 348
Scott, R.L. 154
Scovil, H.E.D. 348
Shereshefsky, J.L. 236
Shockley, W. 303,376
Shuler, K.E. 511
Sing, K.S.W. 236,242
Spenke, E. 436
SUssmann, G. 12
Teller, B. 242
Temperley, H.N.V. 192
ter Haar, D. 270,282,286,303,319,
531
Tolman, R.C. 71
Tompkins, F.C. 241
Tonks, L. 248
Tyablikov, S.V. 319
Uehling, E.A. 12
Uhlenbeck, G.B. 12, 269
Volmer, M. 239
Wannier, G.H. 97
Weiss, G.H. 511
Wentorf, R.H. 145
Wergeland, H. 289,542
Wllks, J. 36
Wilson, A.H. 425,426,429
Ramsey, N.F. 341
Randall, M. 443,446
Rayleigh, Lord 387
Read, W.T. 376
Yariv, A. 349
Reif, F. 378,391,396,399,400
Young, D.M. 241
Renyi, A. 62
Zemanski, M.W. 43
Rose, A. 377
Rowlinson,J.S. 43,140,142,161,192, Zeuner, G. 19
Ziman, J.M. 558
214
Problem numbers, not pages, are shown. Bold print indicates whole Chapter numbers.
i
.
·•.1;' ',SiI
..
Absorption of radiation 3.19,15.7 rate of 3.20 Absorption of energy, fluctuations in 24.8
Acoustic scattering 18.2,18.6
Activity 18.1
Adiabatic conditions 104
Adiabatic invariant 14.11,14.12
Adiabatic magnetisation 15.3
Adsorption 8.0
isotherm equation 8.6 relative 8.1 Advanced Green function 13 A Ambipolar thermal conductivity 19.6 Amplification of photon beam 15.6 Angular momenta of diatomic molecule 4.3 Anharmonic oscillator 4.7 Annihilation operators 10.12 Antiferromagnetic 12.6, 12.7 Attenuation of photon beam 15.6 Averages ensemble 204,23.1,27.12 time 23.1,27.12 Barometer formula 3.5 Bernoulli numbers 4.5 BET equation 8.7,8.8 Binary solutions 6.10-6.15 Binary systems 7.7 -7.9 Binodallocus 7.5 Black body radiation 3.11,3.14 Black body radiation and Fermi gas 5.3 Bloch equation 10.5 Bohr magneton 12.1, 13.11 Boltzmann's constant 1.23,2.1 Boltzmann's distribution see Maxwell distribution
Boltzmann's equation for distribution function 17.8-17.12, 18.0,18.2 for electric current 28.7 from variational principle 28.8 Boltzmann'sH·function 27.3
Boson condensation 11.1 -11.8
Bosons 3.10-3.12,3.17, 10.3
Boyle's law 304
Boyle's temperature 1.11
Bragg-Williams approximation for
adsorption 8.9
Brillouin function 12.2
Brillouin zone 13.12
Brownian fluctuations 23.1
Bulk modulus, isothermal 6.1
Canonical distribution 3.6 Capture coefficient of an impurity 16.7 Carnot cycle 1.16,1.17 Carnot efficiency 1.16 Carnot engine 1.17 Carrier distribution, configurational entropy 19.8 Cell model 6.10 Cell theories of the liquid state 6.0-6.5 Central limit theorem 17.1 Centrifugal stretching of molecule 4.7 Chain molecules 6.14 Chemical constant 1.23,3.13,2704 Chemical potential 1.20,2.7,3.19,18.1 Classical approximation 3.13 Classical limit 3.6,3.9 Classical statistical mechanics 3.4-3.6 Classical variable 24.8 correlations for 25.3,25.4 fluctuation of, in quantum mechanical system 21.2
565
566
Subject Index 1.16 negative temperature
Oausius - Clapeyron equation 1.14 Clusters 10.4 Ouster expansion 10.8 Ouster integrals 9.8 Coefficient of mutual diffusion 17.3 for foreign atoms in a gas 17.11 Coefficient of thermal conductivity 17.4 Co-existence of phases 7.3, 7.4, 9.14 Coherent oscillation 15.7 Colder, definition of 15.4 Collisions, effect on transport phenomena 17.10 Compensation in semiconductors 16.5, 16.6 Complete orthonormal set of functions 10.2,10.11 Compressibility adiabatic 1.4, 1.8 isothermal 1.4,1.8,3.15,6.2 Compressional work 5.5 Condensation Bose-Einstein 1l.l -11.8 vapour 11.9 -11.1 1 Conduction band 16.1
Conduction, by hopping 19.7
Conductivity in semiconductors 19.3,
19.4
Configurational entropy for carrier
distribution 19.8
Configurational integral 3.5,6,9
for hole theory of liquids 6.5
Conservative system, stochastic
definition 26.2
Constant pressure ensembles 3.18, 9.4
Contact angle 8.3
Continuity assumption 2.11
Continuity equation, for electrons
16.2-16.4,16.6
Continuous spectrum approximation
3.12,3.17,5.1
Cooperative phenomena 12
Co-ordination number 12.8
Correlation function 9.14, 13.4, 23.1,
23.11,23.12,24.1,26.9 for circuit 23.4 for classical variables 21.6,25.3,25.4
Correlation function for particle velocity 23.2 for pulses 23.6,23.7 for simple particle motion 25.1 for suspended system 23.2 Correlation time 23.2,23.12,24.1 Corresponding states, principle of 6.8, 6.9,9.2 Creation operators 10.12 Critical point 1.11,9.7 Critical ratio 9.2,9.7 Criticality, component system 7.5,7.6 two component system 7.8,7.9 Crystal, distinguished from lattice 3.10 Curie temperature 12.3,12.8,13.13 Cycles, Carnot 1.16-1.18 general 1.19 Damping and fluctuations 24.2 de Broglie wavelength, thermal 10.2 Debye length 14.8 Debye model, for fluctuations in energy 20.3
Debye potential 14.9
Debye theory of plasmas 14.1,14.3 14.9,14.13 Debye theory of specific hea ts 3.11 Decay function assumption 23.2,24.1 Delta function input 23.14,23.15 Delta functions, superposition of 23.16 Demagnetisation factors 13.14 Density matrix 13.10,20.4 Density of states and boson condensation 11.8 and negative temperature 15.2 Detailed balance 2.14,3.19,16.3,26.2 Detector, radiation, sensitivity of 24.4 Dieterici equation of state 6.4 exact and inexact 1.2
Diffusion of atoms in velocity space
17.7
Diffusion equation 23.3
Diffusion, by hopping 19.7
Diffusion coefficient
for foreign atoms in a gas 17.11 mutual and self- 17.3 Diffusional stability of the single phase 7.7
1
.,.
Subject Index
567
Dilute gas 9.17
Ensemble Dipole moment of a crystal 24.9, 24.10,
canonical and microcanonical 2.4-2.6 24.11 constant pressure 3.18,9.4,22.1 Direct energy gap 16.3 grand canonical 2.4-2.8,11.0 DisperSion relation 3.11 fluctuations in 20.3 Dissociation of diatomic molecules, most probable state of 2.11, 2.12, stochastic model for 26.6 27.6 Dissociation energy 4.7 petit and grand 2.4 Distribution function in an electric Enthalpy 1.7, US field 17.9, 17.12 Entropy 1.7,2.1,2.2,2.14,3.1,15.1 Drops communal 6.1, 6.3 excess pressure of 8.2 configurational 6.2 statistical mechanics of non fluctuations 22.1 interacting 11.9-11.11 insensi tivi ty of 2.I 2 internally produced 2.14 Dynamic polarisation 15.5 -15.7 maximisation 2.1-2.5,2.9,2.10 of Maxwell gas 27.4 Effective area 8.3 statistical 2.1, 2.4 Effective mass 18.0 thermodynamic 2.4 Ehrenfest's model of a gas 27.7 Entropy production 28.1,28.2 Einstein's A and B coefficients 3.21, approximate, for large number of 16.3 forces 28.4 Einstein condensation 11.1- 11.8 in current flow 28.5 Einstein function 3.9,4.6 in electron flow 28.6-28.8 Einstein relation 17.12, 19.7,23.3 internal 2.14 Elastic scattering 17.7 minimum rate of 28.1 Electric circuits 23.4,23.10.23.17, relations for different systems 28.3 24.1, 24.S, 24.6 Equation of state 1.5, 1.10,9.1,9.6, Electric circuits, Onsager relations for 9.7, 11.2, 14.9 25.5 for cell theory of liquid 6.2, 6.3 Electrical resistivity 18.1 for hole theory of liquids 6.5 Electrochemical potential 16.1, 18.1 for smoothed potential theory of Electron concentration as a function of liquids 6.4
illumination 16.6 for surface 8.6
Electron density, intrinsic 16.1 for tunnel theory of liquids 6.3
Electron flow 28.6-28.8 of imperfect quantum gas 10.1 Electron gas 3.16,16.1,16.5 10.10 Electron-hole pairs, transient decay in a see also Dieterici, Tonks, van der semiconductor 16.2, 16.4 Emission of energy, fluctuations in Waals Equation of state treatment of liquids 24.8 6.6-6.9 Emission, stimulated 3.20,15.7, 16.3 Equilibrium, approach to, for metal Energy absorption, fluctuations in 24.8 27.8,27.13-27.18 Energy emission, fluctuations in 24.8 Equilibrium conditions for spin systems Energy gap 15.5 direct 16.3
Equilibrium distribution from master intrinsic 16.1
equation 26.2 Energy, rotational, of diatomic molecule, Equipartition 3.11,21.3,24.1 quantum mechanical 4.4 Equipartition fluctuations 24.2 classical 4.3,4.4,4.5
568
Subject Index
Subject Index
Equipartition theorem 3.6,3.7 Equiprobability of states 2.14 Ergodic hypothesis 3.7, 27.12 Ethane, thermodynamic properties of 4.10 Euler-McLaurin summation formula 4.5 Excess pressure of a drop 8.2 Exhaustion range in semiconductors 19.1,19,4 Expansion, volume coefficient of lA, 3.15 Exponential relaxation from a master equation 26.5 Exponential relaxation from a FokkerPlanck equation 26.5
Extensive quantity 1.20,3.4
Extreme relativistic limit 5.3
Extrinsic semiconductor 16.5
Fluctuation in the number of particles 20.7 in occupation number 20.8
in volume 21.3
of classical variable in quantum
mechanical system 21.2 Fluctuations damping of 24.2
isothermal and isentropic 22.6
time dependence 23
Fokker-Planck equation 2604,26.5
for the random walk 26.8
Four-vector 5 A
Free energy, rotational, translational
and vibrational 6.0
Fugacity 3.5,9.8
Functions of mixing 6.9
Gamma function 2.1 0, 2.11 Gas
ideal and imperfect 9.5
Fermi - Dirac statistics 16.1 ideal classical 1.9, 1.1 0, 1.26
Fermi energy 16.1 ideal quantum 1.9, 1.12, 1.26,3.12 Fermi gas and black body radiation 5.3 ideal relativistic 5 Fermi level 16.1, 19.0 imperfect classical 9 Fermions 3.12,3.17,3.19 imperfect quantum 10 Ferromagnetic 12.6 of elastic rigid spheres, diffusion and Ferromagnetic resonance 13 .14 mean free path in 17.3 Fick's law of diffusion 17.2 of hard spheres 9.9-9.12, 11.12 First law of thermodynamics 1.5 11.15
Fluctuation thermal conduction in 1704
in charge 23,4,23.5,24.1 viscosity in 1704, 17.5
in charge on a condenser 21.3 Gauss it function 2.11
in classical co-ordinates 21.1, 21 .2 Gauss theorem 3.8,9.16
and damping 24.2 Gaussian distribution 2.9,2.12
in deflection of a galvanometer Gaussian Markov process, stationary
suspension 21.3 26.9 in dipole moment 21.4 Generalised co-ordinates and momenta in the distance between adjacent 3.4,21.6 atoms 21.3 Generalised forces 21.6 in energy 14.2, 20, 22.2, 22.3 Generalised Nyquist relation 24.8, 24.9 in energy, Debye model 20.3 Generation of electrons in semiconductors in a grand canonical ensemble 20.5 16.2 in internal energy and volume of a Generating function 26.1, 26.7, 26.3 perfect classical gas 3.2,22.3 Gibbs adsorption equation 8.5,8.6 in macroscopic variable 4.2, 22,4,22.5 Gibbs-Duhem equation 1.20,2.7,2.8, in non-mechanical variables 22.1 8.5 in speed, Maxwell gas 3.2 Gibbs free energy 1.7, 1.20 in the number of electrons in Gibbs model of a surface 8.0 conductor 21.7
Gibbs paradox 17.3 Grand partition function 2.4 - 2.7, 10.2 Green function 13.1, 16.2 Grtineisen's law 1.9 Grtineisen's ratio 1,4, 1.5, 1.8,3.1 5
~
~
I
H-function of Maxwell gas 27,4 Hall coefficient 18.1 experiment 18.3 8.3 Hall mobility, experiment 18.3 Hamiltonian, spin 12.1 Hamiltonian equations of motion 3.7 Hard core 9.1 8 potential 9,4 repulsion 9.19 repulsion plus long-range attraction 9.7 Hard rods 9.3 Hard sphere gas 9.9-9.12, 11.12-11.15 Hard sphere model for surface 8.6 Harmonic oscillators 20.2 Heat capacity 1.4, 3.15 due to two or three discrete levels 4.8 Heat and work in negative temperature system 15.3 Heat theorem, Nernst's 1.24 strong 1.25 Heisenberg ferromagnet 13.1 1-13.14 Heisenberg model 12.1 Helmholtz free energy 1.7 Henry's law 8.6 Hill-de Boer equation 8.6 Holes, nature of 18.0,18.11 Hopping mechanism 19.0 Hot dilute plasma 14.1,14.8, 14.11 Hotter, definition of 15,4 Hydrogen atom, ionisation energy 19.1 Ideal classical gas 1.9,1.10,1.26,9.5 relativistic 5 Ideal fluids, specified by equations of state 1.9 -1.14 Ideal quantum gas 1.9, Ll2, 1.26, 3.12-3.17 rela tivistic 5 Ideal solutions 6.1 0 ImpUrities in semiconductors 16.5, 19.1 Impurity, capture coefficient for 16.7
569
Impurity ionisation in semiconductors 19.3 Indecomposability 27.11 Indistinguishability and partition functions 3.4 Inertial frame 5,4 Instantaneous response 23.16 Integrating factors 1.2 Intensive quantity 1.20 Interchange energy in solutions 6.12 Interconnected states 2.14 Intermolecular pair potential 9.15 Intermolecular potential energy 9.1 Internal energy function 1.5 Internal processes 1.22 Internally produced entropy 2.14 Intrinsic electron density 16.1 Intrinsic energy gap 16.1 In trinsic semiconductor 16.1 Inversion curves 1. 15 Ionic drift velocity 17.9, 17.1 Ionic velocities 17.9, 17.12 Ionised impurity scattering 18.2 Isentropic fluctuations 22.6 Ising ferromagnet 12.4,12.5 Ising model 12.1 Isotherm equation for adsorption 8.6 Isotherm of a boson gas 11.3 Isothermal bulk modulus 6.1 Isothermal fluctuations 22.6 Isothermal magnetisation 15.3 Jacobian 1.19, 1.31 Joule's law 1.9, 1.10, 1.16 JouIe - Thomson, effect 1.15 Kelvin equation 8.4 Kelvin law, for negative temperature 15,4 Kinetic coefficients 13.10 Kramers-Kronig relations 23.16, 23.17,24.9,24.11 Kubo formula 13.10,24.9 Lagrangian multipliers 2.9,2.11 A.-transition of He 11.4 Lande g-factor 12.1, 13.11 Langevin function 12.1 Langmuir equation 8.7
570
Subject Index
Subject Index
Laplace equation 8.2 Laser 3.20 amplification 15.6, 16.3 cavity, model for 15.8 power output 15.8 Latent heat 1.3 Lattice, distinguished from crystal 3.1 0 Lattice gas 9.6 Lattice model 8.7 Lennard-Jones potential 17.5 Lifetime electrons in semiconductor 16.6 excess carriers 16.7 photons in a cavity 15.8 radiative, in a semiconductor 16.4
Line-width 13.14
theory of 6
Liouville's theorem 27.9,27.11
Longitudinal motions 6.3
Lorentz model of a metal 27.8
Lorenz number 18.5,19.6
MagnetiC moment 2.13
Magnetisation 12.7
Magnetisation vector 13.14
Magnetoresistance 18.10
Magnons 3.11
Many-particle states 5.1,10.11-10.14 Markov process, definition 26.9 Maser amplification 15.6 Mass action constant 16.5 Mass action law 16.1,16.3,16.6 2.14,26 3.1-3.3,14.1, 17.8 from master equation 26.3 left unchanged by collisions 27.2 H-function and entropy of 27.4 Maxwell's equal-area rule 11.13 Maxwell's relations 1.7 Mayerfunction 9.8,9.9 Mean free path of foreign molecule 17.3 Mean occupation number 3.12, 3.17 Mean velocity offoreign atoms 17.11 Metal, Lorentz model of 27.8,27.13 27.18
Metals, transport in 18
Metastable state 11 .13
Microscopic reversibility 2.14
Minimum rate of entropy production 28.1 Mixing, functions of 6.9 Mobility 23.3 Mode of oscillation 3.1 0 Molecular chaos assumption 27.1 Molecular field approximation 12.1, 12.6,12.8 Molecular models, to account for calorimetric d.ata 4.1 0 Molecules moments 4.3
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of, stochastic model for
26.6 Moment of a distribution function 2.10
Moment, magnetic 2.13
Moment of inertia of molecule 4.3,4.7
Morse potential 4.7
Most probable distribution method
2.11-2.13
Most probable state of an ensemble
2.11,2.12,27.6
Multilayer adsorption 8.7,8.8
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Neel temperature 12.6, 12.7 Negative temperature 15
and density of states 15.2
Kelvin and Clausius laws for 15.4
Negative temperature system, heat and workin 15.3 Nernst coefficient 18.1, 18.6 experiment 18.4 Nernst's heat theorem 1.24, 1.25 Non-degeneracy of electron gas 16.5 Non-ideal solutions 6.11 Non-relativistic limit 5.3 Normal co-ordinates 21.5 Normal distribution 2.9,2.12
Number operator 10.14
Nyquist's theorem 21.1,21.5,24
Nyquist's theorem for generalised
impedance 24.6-24.8 Occupation number operator 13.9 Occupation probabilities of electrons in impurity states 16.5 Onsager reciprocity theorem 18.7
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Onsager relations 25, 18.1 Operators, time-dependent 13.4 Optical pumping 15.4 Optical scattering 18.2 Order parameters 12.8 Orstein-Uhlenbeck equation, for the random walk 26.8 Orthonormal set of functions 10.2, 10.11 Oscillator 3.9-3.11,4.1,4.6 Oscillator fluctuation in energy 20.2 Oscillators, master equation for 26.3 Pair distribu tion function 9.14 - 9 .19 Pair potential 9.2,9.5,9.8 Partition function 2.4-2.8, 12.2 canonical 9.5 configurational 3.5,6,9 factorisation of 3.4, 3.5,4.1, 6.0 internal, translational 4.1 oscillator 3.9-3.11,4.1,4.6 rotational, classical 4.3 rotational, high temperature 4.5 rotational, quantum mechanical 4.4 Partition functions and indistinguish . 3.4 Pauli operators 13 .11 Pauli principle 16.1 Perfect gas, fluctuation in energy 20.2 Periodic potential, as perturbation on the nearly free electrons 18.0 Permutations 10.2,10.7 Pfaffian forms 1.2 Phase 1.20, 1.21 Phase transition 1.27,1.28,9.4,11 Phonons 3.10,3.14, 18.2 Photo·ionisation cross section of a neutral donor 16.6 Photon beam 15.6 Photon gas 3. I I , 3.14 Plane waves IO.l3 Plasma 14 hot dilute 14.1 Poincare's recurrence theorem 27.11 Poisson distribution 20.7,23.5,23.6, 26.1 Poisson equation 14.1 Polarisability of a crystal 24.9 - 24.11
571
Polarisation index 3.10 Polyatomic molecules 4 Polymer, mean chain stochastic model for 26.7
vtronir rh~na~ 1.13
1.13 . inversion 15 Population inversion, in semiconductor laser 16.3 Potential barrier 23.5 Potential energy, intermolecular 9.1 Potential, pair intermolecular 9.15 Power spectrum 23.8-23.13,24.7 for pulses 23.8 for Newton's law of cooling 24.3 Pressure, configurational or internal and translational or kinetic 6.0 Pressure ensemble 3.18,9.4,22.1 Probabilities, relation between various types of 26.0 Probability distributions 2.5, 2.9 - 2.14, 9.14 )uantum numbers 2.11 Quasi-chemical equilibrium 6.13 Quasi-Fermi level 16.1 Quasistatic changes 1.3 Radiation detector, sensitivity of 24.4 Radiative emission 3.19-3.21,15.7, 16.3 Radiative lifetime in a semiconductor 16.4 Radioactive decay process, stochastic treatment of 26.1 Random-phase approximation 13.12 Random walk 17.1, 17.2 in velocity space 17.7, 17.8 stochastic model for 26.8 Reciprocallattice 13.12 Recombination rate theory in semiconductors 3.1 9, 16.1 - 16.7 Recurrence time 27.18 Regular solution 6.12 Relative adsorption 8.1 Relative surface excess quantity 8.0 Relativistic gas 5 Relativistic mechanics 3.6 Relativistic thermodynamics 5.5
572
Subject Index
Relaxation-time approximation in transport equation 17.8,17.10 Residual properties of fluids 6.0 Response function 23.9,24.1 for circuit 23.1 0,23.17
for delayed delta function input
23.15
for delta function input 23.14
instantaneous 23.16
for suspended system 23.10
Retarded Green function 13.4 Reversibility, microscopic 2.14 Riemann zeta function 3.11, 13.13 Rigid rotator-harmonic oscillator model 4.7,4.9 Root mean square free path 17.1 for transport of momentum and thermal energy 17.4 Rotational energy of diatomic molecule, quantum mechanical 4.4 Rotational energy of diatomic molecule, classical 4.3 -4.5 Rotational partition function, classical 4.3
high temperature 4.5
quantum mechanical 4.4
Sackur - Tetrode equation 1.23 Scattering parameter IB.2 SchrOdinger equation 10.10 Second law of thermodynamics 1.7 Second quantisation 10.1 1-10.14 Seebeck coefficient 18.1,18.6 due to hopping 19.B
experiment 18.4
in semiconductors 19.5
Self diffusion 17.3 Semiconductors 3.19,16,19 Singulary systems 7.1 - 7.6 Slater sum 10.2 Smoothed potential theory of liquid 6.4 Solutions, theory of 6 Sommerfeld model of a metal 18 Sound field 3.14 Specific heats 3.11,15.1 of a perfect boson gas 11.6 Spectral representation 13.7 Spectroscopic and calorimetric values compared 4.9,4.10
Spectroscopic data, yielding statistical thermodynamic functions 4.9 Spectroscopic units 4.6 Spin systems, approach to equilibrium 15.4 Spinodal locus 7.2, 7.5 Spins, interacting 12, 13.11 -13.14, 15 Spontaneous emission 3.20,15.6,16.3 Spontaneous magnetisation 12.3 Stability diffusional 7.1
mechanical 1.29-1.31,7.1
thermal 1.29,7.1
thermodynamic 7.1
Stability conditions, higher order 7.2 Stability at critical point binary system 7.9 singulary system 7.6 Stable state 11.13 State of an ensemble, most probable 2.11,2.12,27.6 Statistical entropy 2.1, 2.4 Statistics of adsorption isotherms 8.8 Steady state 2.14,28.1-2B.4 Stimulated emission 3.20, 15.7, 16.3 Stirling's approximation 2.11, 2.12,9.6 Stochastic model, for dissociation 26.6 Stochastic process in continuous state space, general definition 26.9 Stosszahlansatz 27.1, 27.7 Stretching, centrifugal of molecule 4.7 Superposition of delta functions 23.16 Surface adsorption, microscopic models for 8.6,8.7 Surface equation of state 8.6 Surface free energy 8.5 Surface tension 8.2 Susceptibility 12.3,12.6,13.13 antiferromagnetic 12.7 Systems, closed 1.22 Temperature absolute 1.3, 1.7 Boyle 1.11 critical 6.5 Curie 12.3, 12.B, 13.13 empirical 1.3 Neel 12.6, 12.7 transition 11.4, 13.13
Subject Index
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Thermal condUction in a gas 17.4
Thermal conductivity IB.I
ambipolar, in a semiconductor 19.6 Thermal conductivity coefficient 17.4 Thermal de Broglie wavelength 10.2 Thermal eql1ation of state 9.15 Thermal stability 7.1 Thermodynamic functions, vibrational 4.6
Thermodynamic limit 9.3,9.4, 9.B,
9.14,14.1
Thermodynamic stability, single phase
7.1 Thermodynamics, laws of 1.5, 1.7, 1.24 Thevenin's theorem 24.6 Third law of thermodynamics 1.24-1.26 Time-dependent operators 13.4 Time reversal 25.6
Tonks' equation of state 9.3,9.4
Transient decay of electron - hole pairs
in a semiconductor 16.2, 16.4 Transition probability 3.19 ,,-transition of He 11.4 TransitiVity, of a motion 27.11 Transport coeffiCients 26.9 in terms of Fermi integrals 18.9
in a magnetic field 18.2
Onsager relations for 25.6,25.8
summary of 18.2
for two-band models 18.B
Transport integral 18.2, 19.2 Transport phenomena, effect of collisions on 17.10 Tunnel theories of liqUids 6.3 Two-density state IU5 U1trabaric system 5.2
Undetermined multipliers 2.1,2.2,2.3,
2.11
END
573
Unit cell 3.10
Unit step function 13.2
Valence band 16.1
Van der Waals gas 1.11,1.15,6.4-6.6,
7.2,9.7,9.19, 11.l2, 11.13
Van der Waals surface analogue 8.6
Vapour condensation 11.9 -11.11
Variational techniques for electron
flow problems 2B.7,28.8 ~J;ydistribution, probabilities of 27.5 Vibrational thermodynamic functions 4.6 corrections to 4.7 Virial, f(um viSCOsity in a gas 17.4 Viriarcoefficient, third 9.9 Virial coefficients 1.11,3.7,9,10.8, 14.9 Virial expansion 9.8 -9.13,9.17, 10.1 Virial theorem 3.7,3.8,9.15-9.19,14.9 Viscosity 17.5,17.6 of ideal gas 17.6 of 4 He 17.5,17.6 of deuterium 17.6 Volmer equation 8.6 Voltage fluctuations 24.1 Waves, plane 10.13 Wiedemann-Franz law 18.5,19.6 Wiener-Khintchine relation 23.11, 23.12,23.13 Work, in special relativity 5.5 Work, mechanical 1.5 Young's equation B.3 Zero chemical potential 3.14 Zero point energy 20.2
