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= m
(2.6)
dt
Thus the relation between expectation values of momentum and velocity is the same as in classical physics.
Now we find how
varies in dt = -
time.
d
i
<[~_m 2 + V(r),
pj]>
= ~
<~V > =-
~x~
J
or d
(2.7)
These identities are extremely general, tation values in any state at all.
for they refer to expec-
Connections with classical physics
can be made only if the state resembles
those of classical physics,
in which we assume that it is possible to specify dynamical variables without uncertainty.
This is an idealization even in classical
physics; ~ in quantum mechanics
it is not even that, since the
indeterminacy relations preclude simultaneous pairs of variables. parameters,
specification of certain
But indeterminacy rarely affects macroscopic
and in all but a few very special situations we may assume
wave packets which are, and remain,
so narrow in all dynamical
variables that the quantum indeterminacy would be swamped by the errors in any physically possible program of measurement. Problem 2.1.
Derive
C2.7) from the basic commutation relations.
Problem 2.2.
A l-gm point mass is dropped onto a target as accurately
as possible from a height of 1 km. rotation,
Neglect air pressure,
the earth's
etc., but assume that a strategy has been derived for ^
is the same number as . difference in meaning:
The notations
suggest a slight
is the expectation value of the dynami^
cal variable A, while is the expectation value of the operator ^
A that represents A.
3
See f o r
e x a m p l e Born 1 9 5 0 .
41
minimizing
the quantum error
attainable
accuracy
contribute
to the error.
with variances
(Aa) z and
the individual
variances.
(2.6) and
of macroscopic
to be made.
if a and b vary at random in a + b is the sum of
that
in appropriate
is the better
Oscillators
of Newtonian mechanics. and more general
theory,
effects which can be similarly
of Newtonian
to apply,
states,
does not require any approximation
(neglecting relativistic the equations
which it was intended
dynamics,
for the
systems
to
agree exactly with it.
and Pendulums
It is not our intention course
we are speaking of macroscopic
that
systems prepared
the derivation
that
Ax and Ap each
Use the fact t h a t
Quantum mechanics
and we now see
2.2
The initial
(2.7) reduce to the basic equations
It is important
treated)
Show that the theoretically
Note:
(~b) 2, the variance
With the understanding parameters
in the aim.
is about 1 fm.
to repeat
in elementary physics
quantum mechanics,
the mechanics
section of a
or to linger over the analogous parts of
but a few remarks here on simple oscillating
systems will be of use later. A simple harmonic
oscillator
by a spring of stiffness
k.
is a mass m bound to a fixed point
For motion in one dimension,
the equation
is m~ + kx = 0 with
(2.8)
the solution
x
:
Acos(mt
For the calculation
-
of sinusoidally
use of a complex exponential generalize
readily
discussed below.
systems
oscillations
like this the
but it does not
like the pendulum
to be
For this reason we shall not use it in this section.
Although nothing integrated,
oscillating
is the method of choice,
to nonlinear
(2.9)
~2= k/m
O0)
could be simpler
it is a prototype
which are appreciably way of integrating to the first order.
harder,
than the equation we have just
of other equations
of the second order
and we should note here that there is a
such equations
that drops
it from the second order
Multiply both sides of (2.8) by x: m~
+ kx~
=
0
42
This can be integrated
at once to give ½m& 2 + ½kx 2 =
(const.)
E
in which the constant E is, of course, energy,
(2.10)
interpreted
as the particle's
and the equation now becomes dx dt
which is integrated
dx (A 2
_
x2) ~
_
~o(2m_i
_2E
(2.11)
_ x2)½
as
-
oJ
I
dt
.2~ .½ A = [m-~J
oJ ( t =
_
t o)
(2.12a)
or
sin -I x / A
=
m(t
x = Asinm(t Several points
in this elementary
-
(2.12b)
to)
(2.12m)
- to)
example
should be noted.
we know when energy is conserved without needing we could have started with second-order E and t0.
(2.10).
equation contains
The general
two arbitrary
to prove
it, and so
solution of the
constants, A and ~0 or
In the second solution we lowered the order of the equation
from 2 to 1 and picked up a constant again and picked up t O .
of integration;
For a system not subject
influences
from outside,
integration
and it is the least interesting,
the arbitrary
then integrated
to time-dependent
t O is always one of the constants
setting of clocks.
from 2 to 1 before solving
it, but more complex systems have more complex equations, their order with the inclusion
means,
of constants
is independent
feature
of these oscillations
of the amplitude;
in the
is that the
that is what "harmonic"
and it comes from the linearity of the equation
In Chapter
and lowering
of integration
equation of lower order will be a useful device.
The characteristic frequency
of
since it refers only to
In this case it is no great thing
to have dropped the order of the equation
remaining
Generally,
of motion.
3 we shall discuss more complicated vibrating
that are still harmonic,
finding that the linearity
systems
is a great help
43 in their solution.
Our next example will be one with a nonlinear
equation. Note
on A r c - S i n e s
In (2.12c),
t may be as large as one likes.
Therefore
this must
also be true in (2.12a),
and how this can be so may be puzzling
one looks at the obvious
boundedness
what
of the integral.
if
To understand
is happening we must go back to (2.11) and note that it is not
really correct, positive
since 4 2 has two square roots,
or negative.
it must go through ±A; we knew that, (2.12a).
For the quantity
zero;
therefore
so the double
The points where
The integral
in (2.12a)
and ~ can be either
(A 2 - xZ) ½ to become negative
it changes
sign as x goes through
sign is not written
the sign changes
is to be understood
explicitly
are called turning points. as going back and forth
along the line from -A to +A, the square root changing be positive when dxis positive negative
(back).
(forth)
in
signs so as to
and negative when
In this way the value of the integral
it is continually
increases. Problem
2.3.
Show that the h a r m o n i c i t y
the linearity of the equation, d4x/dt ~ ~ Problem
-Kx
2.4.
of the oscillation
and that a magic oscillator
comes from obeying
is also harmonic. Find the motion of the oscillator
force proportional
to the velocity.
if it has a resistive
(Note that there is no energy
integral here.) Problem
2.5.
Assuming
A hole is drilled through the earth along a diameter.
no friction,
no rotation,
motion of an object allowed soon does it arrive time between
and uniform density,
to fall from the earth's
at the antipodes?
the same two points
find the
surface.
How
How does this compare with the
for a satellite
in an orbit at
treetop height? The S i m p l e
Pendulum
This p e n d u l u m of length Z. pendulum.
is a point mass m on the end of a weightless
If the mass
is distributed
We shall do so in Chapter
The mass moves
may be taken as x
are not needed to say where
angle @ with the vertical,
suffices.
but w i t h one degree of freedom. in terms of torques
8.
in a plane whose coordinates
and y, but two variables
string
one speaks of a physical
it is.
One,
the
It is a system in two dimensions
To set up Newton's
second law in
is not hard but it is unnecessary,
for we can
44
more easily write down the energy: (2.13)
E = %m£2@ 2 - mgZcos@ where we have taken the zero level of potential suspension.
(This makes E negative
The differential
equation
energy at the point of
for amplitudes
less than 7/2.)
is solved by the integral
(2.14)
(m--~ + c°sO)~ The integration functions
cannot be carried out in terms of elementary
unless we are willing
to assume
that the energy is so small
that @ never gets very large and use the small-angle
approximation
cos@ : 1 - #8 2
T h e n we h a v e
I
de
(~z_.)~.~_
Cm_~-~-z+ ~ _ .to2) "~= which is formally
[2.1s)
I dt
the same as (2.12a).
Problem 2.6.
Complete
the solution
in the small-angle
Problem 2.7.
Starting
from
find the second-order
motion
equation of
in @.
To reduce
the exact equation
angular amplitude so that
(2.13),
approximation.
of swing,
(2.14) requires
@0"
to a standard From
form we introduce
(2.13). with
8= O, E =
the
-mg£cos@ O,
the integral
(cose - cOS@o)½ We have already arc-sine)
seen that t(@) reduces
when @0 is small.
when @0 is small, we write
to something
familiar
To introduce a parameter
(an
that is small
45
cose - cose 0 : 2(sin~½eO
sin2@e)
and finally, to measure sine 0 against the amplitude, sin½e = sin@8osin¢. With this, ½cos%ede
=
k =sin½O 0
kcosCd¢
or
2kcos~d$ de =
(1 - k2sin2¢) ~
SO that finally,
t=
I
2kcos~dO [2k 2(1
sin2¢)(1 - k2sin2¢)]½
-
o
or
I ¢
t
=
(~)½
de
(1
-k2sin~¢) ½
(2.163
0
where we have chosen the limit so that the clock starts as the pendulum goes through its lowest point. The amplitude determines k, and for ordinary pendulums k 2 is pretty small. To find the period T, we note that in a quarter-period ¢ goes from 0 to ~/2: ~/~
T = 4(~) ½ I0
d*
(1 - k 2 s i n 2 ¢ ) ½
(2.17)
= 4(~)½ ]0 (I + ~k2sin2¢ + ~k~sin~¢ + ...)d,
or
T = ~(
~)½ (I
+~
k2
~
+ ~ +
...)
(2.18)
46
As Galileo noted many years ago, the pendulum's period does not depend much on the amplitude.
Even for 00 = ~/2,
with k 2 = ½, the
period is increased by only 17 percent.
Problem
2.8.
This p e n d u l u m may also be regarded as a little ball
that rolls in a trough of circular curvature. curvature is that of a cycloid. amplitude?
How does
It will be convenient,
Suppose now that the
the period depend on the
using the equations
for a cycloid
on page 25, to introduce a new variable u defined by u 2 = I - cos@. The result you are about to obtain was embodied by Huyghens
in the
m a n u f a c t u r e of clocks in 1657, when Isaac Newton was 15 years old. "
2.3
Interlude on Elliptic Functions The integral
that occurs in (2.17)
is called a complete
elliptic integral of the first kind, 5 and the indefinite integral in (2.16) defines an elliptic function.
Since we shall encounter them
again and they are important in many branches
of physics, we shall
b r i e f l y survey them here. If in (2.16) we make the substitution x = sin~,
the integral
takes the form
i
s
t =
dx [(:
- x 2)(:
-
(2.19)
k2x 2)]½
o
where we have omitted the irrelevant constant. standard form. everything
This is another
Let us look at it first in the limit k = 0, w h e n
is familiar.
E v a l u a t i o n of the integral gives
t(s) = s i n - i s The notation suggests
(2.20a)
a more convenient form for the answer,
s(t) = sint since whereas
(2.20a)
(2.20b)
defines a m u l t i - v a l u e d
defines a s i n g l e - v a l u e d
function of s,
(2.20b)
function of t.
Huyghens 1673. The proof involves infinitesimals (as do the proofs in Newton's Principia) but antedates any systematic development of the calculus. It requires a dozen propositions to prove it. s
Formulas and tables in Dwight 1934; W a t s o n 1927.
theory in W h i t t a k e r and
47 If we knew nothing about
it as follows.
differential
about the sine function,
Go back to the integral
equation: dt
1
-~
=
(I
Since we are interested
82) ½
k
in s as a function d8
(1
~ry =
and find the differential
equation
from
(2.19)
that s(O)
The sine is defined by
of t, we invert
(2.21b)
again:
ds
= 0 and
=
(2.21b)
-8
from
(2.21a)
and these boundary
that s(O) = 1.
conditions.
Let us go through the same steps with the elliptic (2.19).
it, (2.21cc)
(1 - ~2)½ ~
-
=
o
for ~ by differentiating s
dr2
=
- 82) ½
d28
It follows
we could find out
and write down its
integral
We find I
dt
[(1 and by the same maneuver
-
(1 + k 2 ) s
.... dt 2
+ 2kZs a
equation and the same boundary
~(0) = I define a periodic
te~
cf. (2.17), its period
(2.22b)
conditions
s(@)
= O,
function of t that was named s(t)
(pronouned e s s - e n
(2.22a)
k282)] ½
as before,
d28
This nonlinear
82)(2
= snt
), by Jacobi
is 4 K ( k ) ,
in 1827.
From our previous work,
where
1
K(k) =
(2.23)
dx [(1
- x2)(I
(2.24)
k ax2)]½
0 (It has another period with respect we shall not need that.) If sn is analogous cosine?
We define
Fig.
to the imaginary part of t, but
2.1 shows snt for
to the sine, what
two values
is analogous
of k.
to the
48
cnt
with
the
same s i g n
cosine.
=
(1 - sn2t) ½ in
conventions
Then by d i f f e r e n t i a t i n g
taking
root
as
for
the
snt dsnt cn--~ --J?--
n
,
/
( X
iO
k = 0.998? KCk) = 4.~6~
k = O. 8660 K(k) = 2.157
Figure 2.1
I f we i n v e r t
square
(2.25)
dcnt d---~- = -
/
the
(2.25)
Jacobian elliptic function snt for three values of k.
(2.22a) dsnt
and w r i t e
it
as
= cntdnt
d n t = (1 - k 2 s n 2 t ) ½
(2.26a)
we h a v e dent --dT-=
These are
easy
analogy
with
(2.26b)
shows
to remember:
sines that
cnt
does not
dt =
dent dntsnt
the
=
-(1
satisfy
relations
and m u l t i p l y
is a s o l u t i o n
dac dt2
Note that
write
and c o s i n e s cnt
(2.26b)
-sntdnt
-
2k2)a
the
you would e x p e c t
by d n t .
Differentiating
of -
(2.27)
2k2c a
same e q u a t i o n
as
snv.
dcnt
--
by
[(i _ c n 2 t ) ( l
_ k 2 + kScnat)]½
From ( 2 . 2 6 b ) ,
49 and since cn(O) = I, this is
dx
t = I '1 [(1
_ =2)(1
(2.283
/<2 + /<2x2 ) ] ½
cnt
Problem
2.9.
for dnt.
Find the differential e q u a t i o n and definite integral
Remember
that
(I - k2) ½ ~ dnt < 1.
By e x p a n d i n g the integrands
in (2.19) and
(2.28),
Problem
2.10.
express
snt and cnt in terms of the sine and cosine to the first
order in k 2 . Solution
for
Expanding
snt
(2.19) and integrating the first two terms gives t = sin-ls + 7I
k2 [sin-ls
(:
-
82)½]
+ 7I k 2 ) s i n _ Z s = t + 7I k 2 8 ( I
-
8 2)½
-
s
or
(I
We reduce this, always omitting higher powers of k2: e = sin[(l - 7: k 2 ) t
= sin(1 - 7I
k2 ) t
+ 7: k 2 e ( : + 7I
k2s(1 _
-
e2) %] s 2 ) ½ c o s ( l - 7I
k2)t
Let ~ = I - 7I k s , and in the small second term let s = sin~t, 1 s = sinmt + ~ k2sin mtcos2mt
so that finally
e = snt = (I Thus,
+ /~ k2)sin~t
+ /~
k2sinSmt + OCk~)
(2.29)
in lowest order, k 2 changes the amplitude and frequency and
introduces an anharmonicity.
The change
in frequency corresponds
the change in the p e r i o d of a p e n d u l u m found earlier. same p r o c e d u r e + cn2t = 1,
to
Now follow the
for cnt and check your results by v e r i f y i n g that sn2t
50
Problem
The general
2.11.
contains
solution of a second-order
two arbitrary constants.
equation
Show that the general
(2.8) can be written either as the sum of two elementary (x = A c o s ~ t + Bsinmt) (x = Csin~(t nonlinear Problem
- to)).
equations,
restoring
Remembering construct
An oscillator
2.12.
x = A.
or as a solution with displaced
force, F = - a x
solutions
origin
solution
of
(2.22b).
is made with a spring having a nonlinear It is started t = 0 with & = 0,
- b x 3.
How will
of
that snt and cnt satisfy different,
the general
Show that its subsequent
suitable ~ and k.
solution
motion
is given by x = A c n ~ t
the oscillator's
frequency
with a
depend on A
for small A? Problem
Show that the perimeter
2.13.
of an ellipse with semi-axes
a and b is
i~/2 P =
4a
where k is the eccentricity, complete
elliptic
integral
kind of function gets Problem
2.14
aided by
(I
- k2sin2~)½d~
k2 =
1 - b 2 / a 2.
t
The integral
of the second kind,
defines
its name.
Starting from the equation derived in Prob.
.
(2.13),
a
and shows how this
derive
(2.22)
[Remember that the constant
2.7 and
directly by change of variable.
factor
(g/h) ½
was omitted
in writing
(2.19) .] Problem
The amplitude,
2.15.
function
am, of the argument of an elliptic
is defined by snt = sin amt,
cnt = cos amt
Show that amt is the ¢ defined by amt =
t + ~1 k 2 s i n 2 ~ t
+
...
d -d~ a m t
2.4
(2.16) and that
~ =
1
- ~1 k 2 +
...
(2.30) = dnt
Driven Oscillators When an oscillator
resonance occur.
is driven by an outside
The case familiar
which the driving force f(t)
force, phenomena
from elementary physics
is sinusoidal
and applied
of
is that in
for a long time.
51 We shall discuss the harmonic oscillator arbitrary f(t)
in the more general case of
and then give an elementary treatment of resonance to
an applied sinusoidal
force in an anharmonic oscillator,
in which
some entirely new phenomena occur.
Driven Harmonic
Oscillator
If a damped harmonic oscillator
is driven by a sinusoidal
it will resonate if the frequency is right.
force
Here we shall discuss
the more general situation in which the force per unit mass is an arbitrary function f(t).
The equation to be solved is
+ 2r~ + ~ x where r represents
(2.31)
= f(t)
the resistance due to friction.
The general
solution for the motion without an applied force is
x(t) = (acosmlt + bsinmit)e -rt
(2.32)
where a and b are constants, ~i = ~
- r21½
(2.33)
and the cosine and sine are replaced by hyperbolic large enough to make ~i imaginary.
functions
We are going to solve
if r is
(2.311 by
assuming that the oscillating particle's present position at time t is produced by the entire past history of the force f(t), and there is an influence
function ~ which tells how the value of f at a particular
past moment t' influences
the present value of x(t).
The obvious
formula that embodies these ideals is
(t
x(t)=
- t')f(t')dt'
(2.541
~
where
the
lower
limit
is
taken
as
-~ for
convenience
only,
any fixed value. Differentiating
~(t)
(2.341 gives
= ¢(O)f(t)
~(t I = ~(O)f(t)
+ f$(t
+ $(O)f(tl
- t'lf(t')dt'
+ I~(t - t')f(t')dt' J
and may be
52
Substitution +
into the original
2 r .~ + ~
equation
= ~(o)~(t) 2r$(t
+
(2.31) gives
[~(o3 + 2r~CO)]fCt) + I [ $ ( t
- t') + o~#(t
and this is satisfied provided
- t')]f(t')dt'
that ~ satisfies
t') +
= fCt)
(2.31) with f = O,
together with the boundary conditions ~(0)
Given the general
= O,
solution
$(0)
= 1
(2.35)
(2.32), this is easy to arrange:
O(z) -- ~'le-rrsinmlr
T = t - t'
(2.36)
I
and the general
solution
for all the oscillator's
x(t)
of (2.31),
assuming
present motion,
Ite_r(t_t,)sinm1(t
=
that the force f accounts
is
-
t')
Note the role of the resistance
r here:
long,
and the corresponding
the exponential
have little forgets
feature
2.16.
that
if the interval
on the present x(t).
its ancient history.
peculiar
Problem
influence
is small,
t - t' is
values
of f(t')
That is, the oscillator
The idealized case with r = 0 has the
such an oscillator
Show that
(2.373
f(t')dt'
(2.32)
forgets nothing.
is the general
solution
of (2.31)
when f = 0.
Problem
2.17.
A hammer blow at time tostarted
the oscillator moving.
The force of the blow was such that the hammer, with velocity v, was brought
Problem
2.18.
additional
A sinusoidally
cube-law restoring
to rest.
Where
is the oscillator
driven anharmonic force
-ex 3.
of mass m travelling
oscillator
now?
has an
Write the equation
of
motion as
+ ~ X = -ex 3 + focos~t and find the first-order solution x 0 without
perturbed
f0 = const. solution x = x 0 + xlby finding a
s and then letting
53
x2(t)
We shall discuss Problem
=
(t - t ' ) [ - E x a ( t ' ) ] d t
a nonperturbative
Let the applied
2.19.
f(t)
(2.38)
'
solution below.
force be a slow push of the form
= ct~le-(t-to)~/t~
centered at to, and let t - t o >> t I so that the push was completed long ago and the upper Discuss
in the light of common experience
according Problem
Discuss
2.20.
driving
Use the example
2.21.
discuss what oscillators Problem
the response
force f0cosmt
with negative Discuss
2.22.
If an anharmonic more interesting we illustrate
the response
to a to the
of a resistanceless
(i.e., nonlinear)
oscillator
is driven by a For convenience
case.
anharmonicity the equation
+ ~x the p a r a m e t e r
to
but they are considerably
than those of the linear oscillator.
so t h a t
oscillator
that was turned on at time t o .
with a simple
force,
to
in which mechanical
Oscillator
Let the oscillator's
be produced by a cubic
of motion is of the form
+ ex 3 = f0cos~t
e will not be considered
damping w o u l d only complicate (2.29)
of a driven harmonic oscillator
force there are resonance phenomena,
restoring
oscillator
r could be arranged.
force f0cos~t
A Driven Anharmonic
where
of the harmonic
(f0 const.), paying attention
life would be like in a laboratory
a sinusoidal
periodic
in behavior
to unity.
that r ~ may exceed m02 in (2.33)
possibility Problem
the difference
as ~Qt I is small or large compared
sinusoidal
2.5
limit of the integral may be replaced by ~.
the picture
(2.39) small.
To include
(Stoker 1950).
Equation
suggests we assume a solution of the form x = acos~t + b c o s 3 m t
Substitute
this
into
(2.39)
and reduce
+
...
the nonlinear
term by trigono-
54 metric -(~
-
+¥3 +
where
keeping only terms of frequency m and 3~:
identities, o~)~cos~t
-
(9o~ ~ -
a 3
g[(a 3 + a2b + 2ab2)cos~t
terms are of frequency
there is coupling between frequencies.
3)
+ b
cos3~t]
5m, 7~, and
the various modes, resonant
9~.
and driving
responses
Evidently the oscillator
at all the higher
Another way to put this is to say that the system has
resonant
frequencies
and can be made to oscillate
driving it at lower frequencies
called subharmonics.
in them by
The multiple
can be clearly heard in the ear, which for protective
purposes
is made highly nonlinear.
(monitor
it with an oscilloscope
still in the linear range!) heard corresponding resonating
+ 2a2b
(2.40)
with frequency ~ will produce
resonances
+ (~
foCOS~t
. . . .
the omitted
several
~)bcos3~t
to a sine-wave
and turns up the volume,
to the excitation
structures
The analysis
If one listens
to make sure that the amplifier of higher
tone is
a shrillness
frequencies
is
in the
of the inner ear.
of an equation like
study only the fundamental
(2.40)
resonance.
is wearisome;
let us
To do this we set b = 0 and
ignore the higher frequency:
(-~+¥s
~a2)a
=
A
fo
=
m2
2
-
¢00
or
3 ~a2 A = 2" This function
is sketched
easily-identified to a horizontal
points
in Fig.
-
f_o a
(2.41)
2.2 for E > 0 and ~
are labelled.
line at A = ~2 _ ~ .
A given value of ~ corresponds If one varies ~2, starting
negative values of A, there will be a steady-state remaining
motion,
at
but with a
finite as long as E is not zero, along the right-hand
branch of the curve.
If one starts
at large positive
there are three values of a possible, marked unstable
does not represent
shall see below that the slightest onto one of the stable branches correspond
> 0, and a few
to different modes,
values of A,
but the part of the curve
an observable
motion,
disturbance will
of the curve.
since we
throw the system
The two stable branches
in one of which the motion
is in phase
55
with the driving force while
in the other it is out of phase.
system is started at positive
\~
A in the out-of-phase
If the
mode and ~ is
#
ks~J
Figure 2.2
Sketch of A(a) with salient points noted.
decreased below the minimum value allowed for that mode, oscillator
will make an abrupt transition
Problem 2.23. Problem 2.24.
neighborhood difficult
Derive Eq.
C2.40).
Use (2.40)
to discuss
of ~1 m0"
CQuestions
the
to in-phase motion.
resonance
of stability
at frequencies
in the
of these resonances
are
and can be ignored.)
Problem 2.25.
Analyze
the nonlinear
taking both the anharmonicity
system by quantum mechanics,
and the driving force as perturbations.
(The way to do this is to let both perturbations
be turned on
adiabatically
from t = -~ and use time-dependent
perturbation
Problem 2.26.
What happens when the anharmonic
force is -ex 2
theory.)
instead of -ex3? The Question of Stability
Imagine that some part of a moving dynamical gently with a hammer. displacement
Three kinds of behavior
from the unperturbed
system is tapped
are possible:
path may become smaller
the
and smaller,
56 so that the effect of the perturbation ultimately disappears.
Or the
system may adopt a new motion that fluctuates around the unperturbed path but never goes far from it. away in a new path which, the old one.
Or finally, the system may wander
in time, takes it as far as one likes from
In this case the motion is said to be unstable;
two represent different kinds of stability.
the other
The criteria for different
kinds of stability have been variously stated and studied by some of the best mathematicians Lyapounov), mathematical
of the last hundred years
(Poincar6, Kolmogorov,
and it would be folly to go into generalities here, but facts known to most readers of this book are enough to
treat the stability of the anharmonic oscillator discussed above, 6 and in Sec. 8.4 we shall briefly mention a simpler case involving the free precession of a solid body. Let us suppose that the motion of an oscillator satisfying has been slightly perturbed, initially small,
(2.39)
so that x(t) becomes x(t)
(2.39)
+ ~(t), with
After the perturbation the system still satisfies
:
+ ~ + codex + ~) + sCx + ~)3 = focoscot If we keep only the first two terms in (x + ~)3 we find that ~, as long as it is small,
+
With x = a c o s c o t ,
(co~
+
(2.42)
and subtract
(2.39),
satisfies the linear equation
zex2)~
=
o
this is "~ +
(co:
3
3
+ "~ Ea 2 + ~
¢a2cOS2cot)(
(2.43)
= 0
The prototype of this equation is known as Mathieu's (Whittaker and Watson 1927, Mathews and Walker 1970).
equation
It is often
written in the standard form
d2~ dx 2 +
(a
+
6cos2x)~
=
0
x
= cot
(2.44)
It is a somewhat advanced topic in the theory of special functions, but the solutions have a qualitative feature that the reader may have encountered in another context.
d2~ + ~ m dx~ 6
Write it as
[E _ V(x)]~ = 0
(2.4S)
For a complete discussion see Bogoliubov and Mitropolsky 1961.
57
where
E
=
~2~ 2m
"
V
It is now the Schr6dinger periodic
lattice,
equation
has solutions
~2~ cos2x 2m
for a particle
8 in (2.44),
in a sinusoidally
x represents
is stable
is not especially
the time,
for large values of t.
series of which a few terms
!
t
-1
E for which the and we see
that,
there are only certain ranges of a for which
the solution
-,Z
(2 " 46)
remain finite in x lie in bands.
that
context,
of stability
Figure 2.3
-
and we know that the eigenvalues
In the present for a given
equation
=
simple,
To derive
these limits
but they can be expressed
are shown in Fig.
as
2.3.
|
0
1
Stable and unstable regions
2
3
~x
of the ~8 plane together with approximate
expressions for their boundaries.
The question before us is to see under what circumstances solutions
of (2.43)
comparing with
are stable.
(2.44) £02C~ =
3 6002 + ~e(~ 2 ,
L028
=
3 ~- c a 2
(2.47)
that ~ lie in the lowest stable band can be
in Fig.
~2~ or by (2.47)
the
and
gives
Since a > 0, the condition read off the curves
Setting x = ~t in (2.43)
2.3, <
~2
_
~
B
-
.
.
.
58
~
By (2.41),
3 ~ 2 < ~2 - ~3
+ ~
and neglecting
+
the higher terms, this is Ea
2
<
-
2
a
We can express f0 in terms of the amplitude the left-hand branch of the resonance
fo = - ~
...
s ~a~
a I in Fig.
2.2 at which
curve has its minimum,
given by
(aI < o)
Thus
a ~ < ~-=
whence
lal 3 < la, l 3
or
I~I for stability. placed
<
I~I
(2.48)
If the oscillator were to start with parameters
it exactly on the unstable branch of Fig.
that
2.2, it would in
principle
remain there, but only in the sense that a pencil can in
principle
(if one ignores quantum effects)
The slightest perturbation stable branches--how
the transition
choice would be an interesting Problem
2.27.
be balanced
on its point.
would cause it to jump to one of the is made and what determines
exercise
in machine
the
computation.
Develop the theory of the driven anharmonic
oscillator
for e < 0. Problem
2.28.
tions made, Problem
2.29.
oscillator
Show that in the region of validity
the entire right-hand branch
of the approxima-
is stable.
Study the effect of damping proportional
considered
to & in the
above. ~
See Landau and Lifschitz 1969; Stoker 1960. nonlinear mechanics is given by N. Minorsky Murphy 1964.
A good survey of in Margenau and
59
2.6
,
Lagrange's
Equations
The last two calculations linear and angular, be used at once, polar
eventually
and in the next one, planetary motion,
since the planet's position
coordinates.
ing successively
have used two kinds of coordinates,
We could perfectly well go on for a while,
more complicated
systems by ad-hoc methods,
the returns would diminish,
start using more general methods. here.
in treat-
but
and it would be necessary
to
We are going to start using them
The first will be a simple and general way of introducing
different
kinds of coordinates
into the equations
of motion.
subsequent
chapters we shall start looking at ways
solutions,
not just the equations,
x, y, and z coordinates
to construct
system in which
function
the
of all the particles
kinetic energy can be written
the forces can
We shail denote
V(x).
giving it an index where necessary:
system's
In
with a minimum of labor.
Let us start with an N-particle be derived from a potential x,
both will
is to be specified
i = I ... 3N.
xi,
the
by the general
letter The
in the form
3N
T =
T__.
(2.49)
i=1 and w i t h this we define the lagrangian
L(x, resembling
function
(z .so)
~] = T(~:] - V(x)
the energy except for a change
in sign.
Newton's
equations
of motion can now be written as
~L ~x i
which we abbreviate
d
~L
- avaxi-~'-i~-'~--= 0
=
0
i
Two facts flow from this formula. a variational
I ....
in our earlier notation 6L 6x.
analogous
i =
formulation
=
3N
(l.40b) 1
. . . .
as
3N
(2.51]
The first is that L can be used in
of the laws of motion that will be closely
to Fermat's principle
fundamental
similarity
description
in terms of waves.
in optics.
of mechanics
We have already
and optics starting
This new analogy
seen the
from their
is historically
far
60
older, and it will enable us to approach from the principles
the same similarity
starting
formulated by Fermat and Newton in the 17th
century. The second important be generalized
fact about
(2.51)
is that it can immediately
to other kinds of coordinates.
f degrees of freedom;
Suppose the system has
that is, suppose that it requires f numbers
ql' ... qf to specify where everything is. If the system is a gas, every atom moves separately and f will be about 1023 If it is a rigid pendulum
Ca real one, not a point mass on a weightless
cord)
there may still be 1023 atoms but the position of every one is determined
as soon as the pendulum's
Thus, f ~ 3N, and in expressing ql'
angular orientation
the coordinates
is given.
x i in terms of
... qf we will usually be able to introduce economies.
obviously
true for solid bodies,
but even for a system
system we shall profit from being able to separate the center of mass,
This is
like the solar
out the motion of
since no aspect of the solar system
that interests
us depends on how the whole thing is moving through space. The remarkable
feature of Lagrange's
equations
that is now to be
proved is that if we write
x i = xiCql . . . . or xCq)
for short,
qf)
the functional
i = I .... derivative
3N
of L with respect
to q
becomes ~N 6L
~qn so that from
i=I
9x. n
=
I ....
f
(2.51) we can conclude
exactly the same form in generalized
SZCq, ~) _6q n
that the laws of motion take coordinates,
o
n
=
i ....
f
In writing down the proof we shall use the summation in any expression
a certain index occurs twice,
be summed over the range of that index:
6L 6qn
6L
convention:
equation
<2.52),
~x i
=
n
6xi ~qn
=
I,
(2.53)
the expression
becomes - -
<2.s2)
~qn
...
f
if
is to
for example,
81
In these expressions summation
i runs
from I to 3N; m and n from I to f.
is over i and not over n.
~L
~L
To change variables we have
~x.
~L
-
~xi ~qn
In the last term interchange
xi
~.
+
~qn
Since
The
~xi ~qn
order of differentiation:
~L
9L
~x.
~qn
~xi ~qn
~L
+
d
~x.
~xi dt ~qn
does not depend on @, we have next
To simplify
~L
~L
~.
~qn
~xi ~qn
this we note that
qm ~qm so that differentiating
with respect
to one of the q's, say
qn'
gives
We can now assemble
~B.
~x.
~qn
~qn
the foregoing
~xz. +
~L
d ~L
~L
~qn
dt ~qn
~xi ~qn
dt
which
is (2.52),
1760 he actually
as promised.
into
~L d 9x.z
d f~L ~xil
~xi dt ~qn
dt \~x i
K/
~qn
When Lagrange
gave these equations
gave more than we have given here,
in
since he included
62 the possibility of nonconservative
forces.
its derivation are given in many books,
The more general form and
e.g. Goldstein
(1951).
Before applying Lagrange's equations we note by comparison with (1.47) the single variational principle by which all f equations of motion can be expressed:
I L[q(t), ~(t)]dt
=
0
(2.S4)
where the q's are not to be varied at the endpoints. was given by Hamilton in 1834. 8 the principle of least action;
the former is more exact because the
integral is not necessarily a minimum. segment of the path minimizes
In this form it
It is called Hamilton's principle or As in optics,
any small
the integral, but the whole path may
make it a maximum or a point of inflection.
"Stationary action" would
be better, but it is rarely used. There is obviously a certain difficulty of physical tion if Hamilton's principle mechanics:
is taken as the fundamental
interpretalaw of
since both endpoints are fixed a system has to know where
it is going to end up before it starts, merely chooses
and the dynamical principle
the right way to get there.
P. L. M. de Maupertuis,
The French mathematician
who made the first sketch of a principle of
least action in 1744, was well aware of the implication and was proud to have shown how the science of mechanics explains the working of God's plan for the universe and so beautiful,
(Maupertuis
1752):
"These laws, so simple
are perhaps the only ones that the Creator needed to
produce all the phenomena of the visible world."
We shall see in
Sec. 6.7 that there is another way to explain the physical content of the action principle:
a particle tries everything.
Suppose we add to L the time derivative
of any function of q and
t:
L(q,q~ ÷ LCq, 4) + FCq, t) Since the integral in (2.54) is not varied,
involves F evaluated at the ends where q
the addition does not change the content of the
equations of motion in any way,
though it changes their form and may,
with suitable F, be used to simplify them.
We shall see how this is
done in Sec. 4.3. Finally, we note that since by hypothesis neither T nor V involves
t, we can use the theorem of (1.43) to derive the existence
of a constant of the motion, 8
W. R. Hamilton,
Phil. Trans. Roy. Soc.
(1834),
II, 247.
63 ~L qn ~
~qn This
(2.ss)
- L = E (const.)
is the energy integral.
Problem
2.30.
coordinates
Verify
Then set up Lagrange's and show t h a t Problem
masses
that if the q's are taken to be the Cartesian
x i of an N-particle
equations
E is again
for the simple pendulum
of the m o y a S l e ~J~LWJ2
total energy. of Sec.
2.2
the energy.
The device of Fig.
2.31.
E is the system's
system,
elements,
2.4,
in which
the numbers
is constructed
and let go.
give the Neglecting
2J JFi2 IWF2W/WI Pl
Figure 2,4
the rotational
inertia of the pulleys,
To ~llustrate ProSlem 2.31.
find the acceleration
of the
5-unit mass. There is another way in which constants found.
Suppose that L does not explicitly
coordinates,
say qm" d
9L
dt
-~qm
Then
of the motion can be
involve
one of the
(2.53) becomes ~L
- O,
= Pm
(const.)
~qm
A familiar example is the flight of a projectile, where the potential energy depends
on the height but not the
horizontal
coordinate x.
Thus
L = ½ m(,~ 2 + #2)
VCy),
and Px = m~ = const.
9L/~x
= 0
64 This is the momentum corresponding general Pm a momentum also; ing to (or conjugate
to x, and in its honor we call the
it is the generalized momentum correspond-
to) the generalized
coordinate
qm"
The missing
qm is traditionally called an ignorable coordinate or a coordinate. We shall look carefully for them in what follows,
coordinate
cyclic
for conservation
2.7
laws are of great importance
in physics.
The Double Pendulum To show how Lagrange's
equations
worked out in complete detail. suspended
are used, here is an example
Suppose that one simple pendulum
from another and both are set swinging
in such a way that they swing periodically other conditions Sec. 3.3.)
is
through small angles,
thereafter.
(There are
after which they do not swing periodically--see
The periodic modes of motion are called normal modes.
Figure 2,5 Ca) shows the coordinates.
To simplify
formulas we shall
O
(b) Figure 2,5
Double pendulum.
assume that m I >> m 2 and that £2 > £I" the Cartesian coordinates x I
=
x 2 =
With respect to the origin 0,
of the two masses
are
h/sine/
Yl
hlsin@/ + Z2sine~
y~ = - (£1cos@l + £gcose~)
=
-
~lCOS@l
and from these we find easily that ,2
Xl
.2
+
Yl
2"2
=
£101
• 2 ,2 2~2 2'2 x2 + Y2 = £1 + £282 + £1£2c°s(01
*
- 02)@102
65 so that the L a g r a n g i a n L = T - V is 2.2
2
2
m2(Z1~)l
L = ½ m1,£101 + ½
2.2
+
,£202
+
£1~2cos(01 - 02)5102)
+ mlg£1cosO 1 + mgg(~lCOS01 + £2COS02) Now we make the s m a l l - a n g l e
approximation.
Retaining only quadratic
terms, we have 2,2
2"2
-
L = ½ m129~i01 + ½ m2(~202 + .£i.%20102)
½ m12g£101
- ½ m2g£2O 2
(2.s6) plus a constant,
where m12 = m I + m 2.
Lagrange's
equations
of motion
are
m12£lO 1 + ½ m2~20 2 + ml2gO 1 = 0
(2.57) m2£20 2 + ½ m2£101 Except
for the small-angle
Now we state that
O1 where
=
approximation,
these are exact.
the desired solution
alcos(mt
@ is arbitrary
(2.57)
+ m2gO 2 = 0
+
¢),
is periodic
in ~,
02 = a 2 c o s ( ~ t + @)
and ~ is to be determined.
Substitution
into
gives
m12(g
- Stlos2)aI - ½ m2£2~o2a 2 = 0
(2.s8) -
For these, at all,
½ m2£1~o2a I + m2Cg - ;L2~o2)a2 = 0
considered
in e2 whose solutions
092 =
as equations
their determinant
/
in a I and a2,
must vanish.
to have any solution
This gives a quadratic
equation
are
{m12(~,1 + ~2 ) + [m;2(~ 2 - ~,1)~ + m12m2~1~2 ]½} (2.59a)
66
Our simplifying
assumptions
allow us to write
I m2 + ~ )
(I 2m12LaZ ~
+-m12(~ ~ - ~ : ) [ :
+
m211£ 2
m12(ft2 - £i)2:
<<
{m12(~1 + ~2) m2~l~ 2 ]} + ... 2m:2(~ 2 _ ~ : ) 2
or
I~
m2
(I +
~2
+ ...)
4m12 ~2 - ~I (2.59b)
Li2
(I -
m2 ~I 4m12 ~2 - ~1
+ ...)
There are thus two normal modes, with frequencies of the individual pendulums. the first
(larger)
move, we set
value of ~2 into the first of (2.58):
a2 = _ whereas
quite close to those
To see how the pendumums
_
al
~2 = 9___
(2.60a)
al
~2 = .q
(2.60b)
for the lower frequency,
2m12 ~2 - ~I
a2
m2 The first one looks
like Fig.
the two normal modes pendulums
~2
i2 2.5(b);
the second looks like
are clearly distinguished,
are exactly
(a), and
for in one the
in phase and in the other they are exactly out
of phase.
Note:
In the following problems
instabilities
potential
may arise as in Fig.
2.6a,b.
energy curves with central In the first case we have
an integral of the form
I
dx
1
(2.61a)
wxI
w)1 I
67 where
k = xl/x 2.
To reduce
this to standard
form, make the substitu-
tion u =
(I
-
k'2v2)
k '2
½
=
I
-
k2
(2.61b)
The second case gives rise to 1
du (2.61o)
[(I
- u2)CZ< '2 + k 2 u 2 ) ] ~
0 In this,
let u =
The 24 possible Tables
substitutions
(I
-
v~) ~
C2.61d)
of this type are given in Peirce's
(Peirce and Foster 1957).
Problem
2.32.
What have
the formulas
the situations
illustrated
Problem
Analyze
2.33.
in Fig.
the behavior
(2.61a)
and (2.61c)
of the system shown
,
VC~i
to do with
2.6a,b? in Fig. 2.6c as
}~
k
m
V(x)
~
E
E
Figure 2.6
F
i
Xi
Xz
~¢
(a,b) Potential energy curve with a central maximum, showing discontinuous behavior as E increases. (c) To illustrate Problem 2.33.
completely
as you can.
Assume
enough so that the springs Problem
2.34.
Fig.
length is ~.
@ and find the equilibrium. equilibrium.
distances
are long
horizontal.
2.7 shows an inverted pendulum of length
supported by a stiff spring unstressed
that the horizontal
remain effectively
(to keep the angles small) whose
Sketch
(approximate)
the potential
points
Find the frequency
at which
energy as a function of the pendulum
of small oscillations
is in
around the
68
"II]IIIII
lllllllilllJll
III
II
e -. . . . . . . . . . . . . ~ , ~ ,JJ ,~..I rJISS" Figure 2.7
Problem 2.35.
Discuss
To illustrate Problem 2.34.
the motion of the system in Fig. 2.7 if the
ceiling height is increased
to 3Z but none of the other parameters
are
changed.
2.8
Planets
and Atoms
The determination
of orbits under a central
inverse square of the radius problem in physics. calculated
is historically
In Newton's
Prineipia,
and observed positions
the agreement
system.
In the history of quantum mechanics formula successively
the entire
it was the derivation
by Bohr using his semi-classical
Pauli and Dirac using quantum mechanics,
as the
of the
of the planets validated
Balmer's mechanics
force varying
the most important
and Schr~dinger
that first convinced people of the importance
of
theory,
using wave of these
theories. In quantum mechanics energies
the calculation
is straightforward.
the planet an equation
of wave functions
In classical mechanics
is at a given time.
scattering
Let us first assume
it was hard to solve
We shall derive an approximation.
if the energy of the system is positive, formula for Coulomb
one asks where
The answer hinges on the solution of
first given by Kepler in about 1609;
then and it is hard now.
and
Finally,
we can derive Rutherford's
as a bonus.
that the sun, or the nucleus,
is clamped.
We shall see in Sec. 3.2 that nothing changes much if it is allowed to move. We choose the origin at the center of force polar coordinates, plane
(why?).
(Fig. 2.8), use
and note that the orbit all lies in the same
The lagrangian
½ m(~,2
is + r202)
-
V(r)
(2.62)
69
where
V(r) is the gravitational
may be some other central
or electrostatic
force altogether.
potential,
Lagrange's
or it
equations
are
W%
9
Figure 2.8
second-order
Particle of mass m orbiting an attractive center.
equations
for r and 8; there are four time derivatives
and there will be four constants as a fourth-order complicated,
system.
of motion.
We speak of the equations
Such a thing is liable to be quite
and we search for ways to solve
it stepwise.
The Angular Motion Let us look first at the equation on 8, @ is cyclic,
in 8.
Since V does not depend
and Pe = mr2e = L ( c o n s t . )
Pe = O,
(2.62;)
The quantity P8 is the m o m e n t u m
conjugate
briefly,
The use of L for the numerical
the angular momentum.
of the angular momentum confusion.
as well
Its conservation
integration.
as the lagrangian
should cause no of
is of the third order.
Problem 2.36.
It
with direction
as well as a magnitude
usual
value
in this case has given us a constant
The remaining p r o b l e m
is
to the angle 8; more
in physics
to equip angular momentum
and make it a vector perpendicu-
lar to the orbital plane by writing Pe = m r x ~ Show that the magnitude
Problem 2.39.
of P8 thus defined
(2.64) is mr2e.
Show that if the force is directed
towards
or away
from the point r = 0,
@8 dt
-
0
(2.65)
70 The angular m o m e n t u m geometrical meaning (Fig. 2.9)
is a dynamical
also.
quantity but it has a
Suppose the planet moves from a to b
in a time dt, so that its radius
Figure 2.9
sweeps out an area
Relation between area and angular momentum.
½ r2dO. The rate at which this area is swept out is dd A t
This is Kepler's
second law
general derivation force;
-
r 2 0• =
½
(const.)
L
(Kepler 1609, p. 165) 9
than the other two,
laws 1 and 3 require
It is of more
for it requires
only a central
that the force be inverse-square.
To find r, we use another constant are usually called integrals
of the motion
of the motion).
= >~ m ( ~ 2 + r ~
and incorporating
(2.66)
the integral
(such constants
This one is the energy
(2.67)
2) + v(r)
just found gives t 2
(2.68)
E = ½ m~ 2 + ~-NF~+ v ( r )
Comparing
this with
taking place
(2.10), we see that it is as if the motion were
in one dimension
instead of two,
in an effective
potential t 2
Vef f = ~
If V(r) is an attractive Vef f looks like Fig. s
potential
2.10.
+ V(r)
(2.69)
of the form -y/r, a graph of
Suppose the energy E of the orbiting
Readers who have momentarily forgotten in almost any elementary physics text.
Kepler's
laws can find them
71
planet
is negative.
oscillator:
The situation
outside
resembles
that of a harmonic
the region between r I and r 2 the kinetic energy
P
E
Figure 2.10
Effective potential for an attractive inverse-square force.
w o u l d be negative. quantum mechanics between r I and r2, there
In classical mechanics the wave function
the planet oscillates
linear oscillator? at a varying appear
rate,
strongly.
back and forth.
is still a minimum r I but the planet How would one have to observe
this is impossible while
decreases
in
Trapped If E ~ 0
is unbound.
the planet
for it to look like a
One would have to stand at the sun and turn around always
to be subject
facing the planet.
The planet would then
to the force d~ff Feff . . . .
L2
dV
mr 3 - d-~
(2.69a)
dr
In terms of angular velocity,
the first part of Fef f is
L 2
mrS2
mr3
This
is the centrifugal
in elementary effects,
an object of confusion
We see that it is a real force, with observable
which one encounters
coordinate
Problem
classes.
force that is sometimes
if one places oneself
in a rotating
system.
2.38.
The planet
in the potential well will oscillate back
and forth at a certain frequency. same as the frequency
If this happens
of the angular motion,
to be exactly the
the planet will return to
its original position with its original velocity
at the end of a cycle.
72 The orbit repeats only vary slightly
and is called re-entrant. from a circle,
Considering
orbits
fit a harmonic potential
that
to the
bottom of the effective potential well and see if this is true for inverse-square
motion.
The R a d i a l M o t i o n
The next step in solving the Kepler problem is to solve for ~ and separate
(2.68)
the variables:
dr
v(r) ) - ~
[~(E-
There are now two reasonable Program
A.
Integrate
]%
programs
t
-
t o
( 2 . 703
that might be followed:
(2.70) to get t(r).
to find @(t) integrate
=
(2.713
o - e o = ~- r ~ [ t )
Program
B.
Then
Invert to find r(t).
~ = L/mr2:
Find the orbit first.
Since dr/dt
= (L/mr2)dr/dO,
(2.68)
-
(2.72)
gives
dr [2mr~ L2 Integrate writing
(E - V ( r ) )
this to get O(r),
r~] ½
=
13
0 0
invert to get r(O) and introduce
t by
L = mr20:
t-to= This gives t(@).
/m
r 2 (O)dO
(2.753
Solve for @(t) and put into r(@) to get r(t).
Program A fails at the second step
(Problem 2.39).
t(r) but cannot solve it (at least, not in closed form) Program B can be carried a little further.
= ~±~ 2 V(r)
=
-x/r
y
L ~Mm
Putting
We can find to find
r(t).
73
we find an integral
that can be e v a l u a t e d and inverted to give
r(O)
= L,2/m,,Y 1 - ecosO
(2.74a)
w h e r e e r e p r e s e n t s not electric charge but eccentricity:
2EL 2 ]½ e =
The orbit is a conic section: 2.41),
[1
+
for E < 0 ,
an ellipse
e < I
(see Prob.
a p a r a b o l a for E = 0, e = 1, and a h y p e r b o l a for E > 0, e > 1.
For b o u n d orbits we have derived Kepler's p.
(2.74b)
mY 2
first law
(Kepler 1609,
28S).
For e l l i p t i c a l orbits
(E < 0), the semi-axes
are
L a =
Y -2E
b =
"
(2.75a,b)
(_2mE)½
From the first one, we see that E = -y/2a;
the energy depends only on
the major axis and not at all on the eccentricity. Bohr theory can derive all the energy
(This is why the
levels of h y d r o g e n by consider-
ing only the circular orbits.)
Problem
2.39.
Verify the failure of P r o g r a m A.
Problem
2.40.
Derive
Problem
2.41.
S t a r t i n g from x2/a 2 + y2/b2
(2.74a,b).
e c c e n t r i c i t y as e = (I
= I and the d e f i n i t i o n of
b2/a2) ½, show that (2.74a) r e p r e s e n t s an
ellipse w i t h the origin at one focus.
Find the expressions
(2.75a,b)
for the semi-axes.
Problem For what
2.42.
Suppose that V(r) varies as a power of r: V(r)= kr n.
(positive and negative)
(i.e., closed)?
integral in (2.72) giving @(r) trigonometric in
values of n are the orbits re-entrant
This is the same as:
functions?
for what values of n can the
[or r(@)]
be expressed in terms of
It will be convenient
to change the v a r i a b l e
(2.72) to u = r -I. Having found the orbits, we must now find how they are traversed.
The total time of r e v o l u t i o n is easily d e t e r m i n e d from the area law (2.66): 2~ym ½
T = 2m x ~ab L
(2.76) (_2E)3~
74
and s u b s t i t u t i n g
-2E
from
(2.75a)
gives m ½ a 3~ (F)
r = 2~
(2.773
The period depends only on the energy or only on the major axis. p l a n e t a r y motion,
y = GMm
and 2~
r
This
is
almost
Kepler's
For
third
= -(GM) - ½ law.
instead of the semi-major axis
a3/2
Kepler
(2.783 has
the
mean o r b i t a l
radius
(Kepler 1619, p. 189), but the p l a n e t a r y
orbits are so nearly circular that it makes
little difference.
To continue with P r o g r a m B, we return to
(2.73), w h i c h w i t h
t o = 0 is
This
c a n be
t =
(:
integrated,
but
-
[
e 2 ) 3/2
the
de
(2.79)
(1 _ e c o s O ) 2
results
cannot
be solved
for
oct),
so
Program B comes to a halt unless we are w i l l i n g to get our hands dirty.
Problem
3.43.
Carry out the integration in (2.79) and expand the
result to show that for small e,
3
~t = @ + 2esin@ +-: e2sin2@ +
•
•
•
4
where ~ is the mean orbital
(2.8o)
frequency (2.81)
Problem
3.44.
Solve
@ =
Problem
2.45.
(2.80)
~t
-
to get
2esin~t
8
+ -~ e 2 s i n 2 ~ t
+
...
(2.82)
Complete Program B by e v a l u a t i n g
r =
(I
- e2)a(1
~2
+
ecose + e2cos2@ + ...) e 2
r -- a(1 + ~ - + ecos~t - ~ - cos2~t + ...)
(2.83)
75
and verify that 0 and r are correctly elliptical
given at the two ends of the
orbit, where they can be evaluated by elementary
At this point we run out of results pencil
and paper.
developments
Astronomers
much further
of great m a t h e m a t i c a l has largely
2.9
and introduced
interest
Oscillations
We have m e n t i o n e d radial
(More c o m p l i c a t e d The situation
methods
in Problem
2.38 that a chosed orbit can be
while
carries
out a whole number of
the angular variable
closed orbits
result
runs from 0 to 2~.
if the angle is 4~, 6~, etc.)
questions
seen in (2.69a)
force is subject
of stability
that a particle
to a radial
orbits, and we are
as well. in orbital motion under
force
L2 ~-T - V' (r)
(m = I)
and the orbit will be a circle of radius a if Fef f is zero, to increase nor decrease
tending
the radius,
n 2
a--~- - V' (a) = 0 To study small variations
(2.84)
from circular motion,
(a << a) and expand Fef f around r = a. equal
that are
and Stability
Fef f =
neither
to
1892), but today the computer
is easily studied for nearly circular
led to interesting We have
computational
(Poincar~
as one in which a particle
oscillations
a central
that are easily accessible
of the last century carried the series
taken over the work of calculation.
Orbital
regarded
geometry.
to 1 for simplicity
Setting
let r = a + a the particle's
mass
gives L 2
3L 2
o r by ( 2 . 8 4 ] , =
-
[v"(a)
+ ka
V'(a)]c~
(2.8s]
If
~osc2 a oscillates
~ V"(a)
+ 73 V' (a) > 0
with a frequency ~osc"
ture from circular motion will
(stable)
If m o s ~ < 0 the slightest
grow exponentially
(2.86) depar-
and the particle
76
will wander
away from its
The orbital
circular
frequency
~orb and one finds frequencies
orbit.
is readily obtained
from
[2.84),
2= L v' (a) a
at once that if V(r)
the orbital
= -y/r,
and oscillatory
are the same.
Suppose now that a slight inverse-cube
the force is not exactly
inverse-square
but has
part,
v(r)
=
-
J
morb
~ r- -
r2
e > 0
(2.87)
We find 2
2
LOOSC = ~
The period of a radial oscillation
T°sc and in this time the planet's
morbTosc
=
_y_ +
2_~
a3
a4
is
~osc angular position
advances
by
= 2~(a__=_~)9(a_~ 3 + 2e]ga,, = 2~(1 + ~_~)2~½ 2~ + Ya
Because morbTosc
> 2~, the orbit has
precessed
forward through an
angle ~8 when r resumes general
its initial value.
relativity
introduces
2~e =
ya In Sec.
5.3 it will be seen that
just such a p e r t u r b a t i o n into V(r),
with =
if, a s here,
the particle
3¥2/o
2
has unit mass.
60 = s n a g ao
2
This yields
a precession
(2.88)
77
an estimate still not exact enough to compare with observation since it applies only to infinitesimal and the orbit of Mercury, appreciably flattened. Sec.
deviations
from a circular orbit,
in which the effect is largest,
is
An exact calculation will be performed in
5.3.
Problem 2.46.
For what power-law central forces are all circular
orbits stable? Problem
2.10
For
2.47.
potential
what values of a a r e
V(r) = -ge-Mr/r
Orbital Motion:
circular orbits
in a Yukawa
stable?
Vectorial
Integrals and Hyperbolic Orbits
We close the discussion of orbital motion by showing that
there
is another integral independent of E and L which can be used to simplify the integration of the orbital equations
even further.
We
start with the equation of motion in vectorial form, r
and the formula
(2.64) for the constant vectorial angular momentum, mrX~ :
L
Form ~XL :
By the vector
m~X[rx~]
-
rx[rx~]
~
identity ax[bxc]
we
:
-- ( a . c ) b
-
(a-b)(
(2.89)
have
I r-7
rxCrx~]
:
- F IT
:
-
[ra~-
(r.~)r]
r.~ [7-
-yr
r]
:
-
d
r
dt
r
(2.90)
Therefore 'rXL =
d r
(rXL)
= Y dt r
Lx~
y r
and so the vector A
=
+
(2.91)
r
is
a constant
vector
in
the
plane
of
the
orbit.
To f i n d
the
78 planetary
orbit,
take the dot product
of this equation with mr.
Since mr-~xL
= mrx~.L = L 2
we have = mA.r = m A r c o s O
-L 2 + m y r
where
@ is the angle between A and r.
Solving
for r gives
LZ/my
r
(2.92)
=
1 - y-IAcos@ This
is the same as
(2.74a)
if A = ye
and we see also from ellipse
in the direction
aphelion, vol.
(2.92)
Fig.
i, pp.
2.11.
160-168.
that A points
of the point
The foregoing LaplaceV~
Figure 2.11 here,
(2.93)
furthest
analysis
analysis
from the "sun," the
is due to Laplace
is essentially
in 1710.
Because
lifetime,
letter commenting lation into modern Hamilton,
terms.
Laplace's Pauli,
What has been accomplished
values.
by so well
of integration were used
we give in Box I Bernoulli's
Lenz,
integral of the equations integrals
analysis
discussion
in a
on an earlier attempt by Jakob Hermann and a trans-
Runge,
vector
is known by every name but
....
here by the discovery
of motion
A, L, and E are specified
mentioned below),
that used
the method used illustrates
the virtuosity with which simple methods during Newton's
(1798),
The vectorial integral A.
but the result is already latent in a brilliant
Johann Bernoulli
his:
from the origin of the
is that,
of a new
if the values of the
(subject to restrictions
there is only one orbit compatible with these
We shall not often be so lucky, but in Sec.
5.7 it will
79 IIIII1111
IIII
II
BOX
From Bernoulli~s
letter to Hermann,
7 Oct.
I
1710"
Dans v6tre 6quation differentio- differentiel~e
-
-
~eddx - - -
(ydx -- xdy) × (xydx -- xxdy): (xxq-yy) ~ je ne mets pas feulement [ comme vous] - - a d x pour l'integrale de _ _ a d d x , mais ~ a d x ~___une quantit~ conftante, c ' e f i - a dire, --adx ~ e (ydx--xdy); pour le retie je le fais comme vous ; de forte qu'en int6grant vbtre pr6c6dente 6quation differentio-differentielle, je trouve ~ ,*dx -+ e ( y d x y (ydx__xdy): s/ ( x x q - - y y ) ~__ (xydy ¢ (xx+yy), ou--abdx: x x -~- el, ( y d x x x ~ ( k x T d ; -- b y y d x ) : x x ~ / ( x x . + y y ) , d o n t l'intbgrale eft a b : x ~___eby : x ~__ c ~ l, q' ( x x - b y y ) : x c'cfi-~-dire ( en prenant k ~ e b , & en reduifant l'6quation) b ~ by ~ c x ~ _ b" ~/( x x "b y y ) : laquelle bquation, quoiqu'elle renkbrme ,by, que la v6tre ne renfermoit pas, e~ ce--xdy)~__--
~yydx): -xd;):
pendant [comme elle] aUK trois Se&ions Coniques. 1:he e q u a t i o n
in x is
m~ = - ~r Write this, for reasons w h i c h will be clear in a moment, rn~ = - -~--~-xQYF:r-3 x'v "~"2 Z2
or
=
-a~
x(~k
-
(y&
= ft/m = const.)
-x~
a
~.)2
=--
r3 This integrates
directly
as
£2
my
to -c&~ = - Y (Yx - x ~ ) + const. r
The trick is to w r i t e the constant as ; e ~ x factor b x - 2 : _ ab~
± eb.
and introduce
y~ - x~ = _ by C~
X2 Once more, everything
- ~)
X2
can be integrated. ab ¥ h ~ = x x
the integrating
- x~)
x2r Let e b = h.
b ~ ± const, x
c
Fhus br
with h and c arbitrary. h = ebcos~,
+- hy +- c x = a b
Take C = ebsin~,
y = rsinS,
m = rcosO
Then a r = I + esin(@ + ~) By introducing the arbitrary constants h and c we have introduced e and ~, w h i c h specify the eccentricity and the direction of the line of apsides.
J. Bernoulli.
Collected Works
~Lausanne.
1742) v. i, p. 471.
80 become clear that even if the orbit is not uniquely is at least restricted within certain
determined,
it
that
limits by the integrals
can be found.
Problem 2.48.
Show that A2
=
2E
y2
L z +
(2.94)
m
It has been mentioned fourth-order
problem,
earlier
requiring
we can count the constants lies in a plane, second one.
L is p e r p e n d i c u l a r
A lies in the plane,
the clock started.
algebraic
stationary
in space;
one introduces (2.99)
easy because
it is very difficult is similar.
amount of calculation
the quantum operator
the solution
the third integral.
is reduced
And
to, the time in (2.92)
had been
algebra.
and energy as is
remains.
that corresponds
to quantum
or impossible. If one character-
atom by angular momentum
a considerable
E is the
the orbit is a figure
the situation
of the hydrogen
usually done,
one integral.
that r(@) could be obtained
in general
Now
Since the orbit
is that all the integrals
process
In quantum mechanics izes states
of integration.
the last one is trivial:
The reason
It was relatively
to it:
is a
but since its length is given by
is independent:
as we have already mentioned,
found.
four constants
that have been found.
L and E, only one component
by a purely
that the Kepler problem
But if
to A (see This
is, in
fact, how Pauli first solved the problem I° in the days before Schr~dinger's
Hyperbolic
equation had simplified
Orbits
When the energy
is positive
tor of (2.74a) vanishes The orbits Fig.
2.12.
the calculation.
and by
or becomes
are now hyperbolas, The asymptotes
negative
i0
e > I, the denomina-
for certain values
of 8.
still with the sun at one focus,
are at the angles cos@ = e -I
By (2.74b)
(2.74b)
this can also be written
±@, where (2.95)
as
W.Pauli, Zeits. f~r Physik 36 336 (1926). A little later Dirac did it independently: P.A.M. Dirac, Proc. Roy. Soc. (A) 110, 561 (1926).
81
tan@ = (==)½~ m At great distances and
from the origin,
L
the energy is entirely kinetic
(2E/m) ½ is just the asymptotic v e l o c i t y Vo, so that VoL tan@ =
(2.96)
IYi 12
Figure 2.12
Hyperbolic orbits corresponding to positive energies.
If y > 0 (attractive force) we must have cos@ > e -l and @ is bounded below.
This is orbit 1 in Figure
the inequality
is the other way,
Problem 2.48.
Fig.
2.13 shows
Figure 2.13
calculations
of scattering:
the scattering angle.
2.12.
When the force is repulsive
and orbit 2 results.
(for y > 0) the parameters used in
Scattering by a center of force.
s is the collision p a r a m e t e r and
The incident particle's
m o m e n t u m is Pas = (2mE)@"
asymptotic
O s is
linear
Show that the angular m o m e n t u m is SPa s .
82 Problem
2.49.
Show that these orbits
Problem
2.50.
Show from the calculations s = ~
The differential
scattering
are hyperbolas. just done that
cot~@ 8
cross
section
is given
(Park,
1974,
p. 316) by d~ e d~ - sin@ Derive
the Rutherford
Problem
Derive
scattering
The vector
2.51.
(2.95) by forming
d8 d@
8
8
formula.
integral
A gives
the product
a simple derivation
p.A and considering
of @ . 8 the limit of
large r. 2.11
Other Forces
Problem
A particle moves
2.52.
force given by the potential and r(@) Problem
in a central
V(r)
=
for a particle projected Calculate
2.53.
-hr -2.
the orbits
Problem
Show in quantum mechanics
Problem
inverse-cube
to find orbits make
inferences
object orbiting
potential
given the force.
Figure
2.56.
through a hole
table moves the system
its original 2.57.
around
of small oscillations.
from a fixed point by a cord of length £.
is slightly disturbed,
oscillations
(Mg/ma) ½ so that
the orbit of m now oscillates
It is set in motion so that it moves motion
(frictionless)
The mass M is now given a
radius a and find the frequency A mass hangs
Deduce
connected by a cord that
The mass on the
equilibrium.
Show that
that has the heavy
Assume p@ constant.
two masses
in a table top.
to
finds a light
instead of in the middle.
energy.
in a circle with angular velocity ~ = tug.
in an
bound states.
Suppose an astronomer
2.14 shows
is in dynamical
sharp downward
Problem
has no stationary
heavy one in a circle
the force law and the s y s t e m ' s
passes
a particle
Show how to invert the argument
the other way. a motionless
that
in this chapter have all attempted
one located on its circumference
Problem
@(t)
at t = 0.
force.
The calculations
2.55.
direction
of particles with E > 0 scattered
inverse-cube
attractive
inverse-cube
Find and sketch r
in a tangential
by a repulsive 2.54.
attractive
in a horizontal
find the frequency
circle.
If the
of the resulting
around the circular path and sketch the resulting motion.
83
Figure 2.14
Problem
2.58.
To illustrate Problem 2.56.
How must the field of a central force vary with r in
order that it be possible for a body to move in it describing a spiral r = (c@) -I, with c a constant? Problem
2.59.
Several physical theories have proposed that the
Newtonian G varies by a few parts in 1011 per year: = -hG How will the radius, period and eccentricity of a circular planetary orbit vary if this is so? 2.12
(Hint:
for e, find A.)
Bohr Orbits and Quantum Mechanics:
Degeneracy
Planetary orbits and Bohr orbits are, in general,
ellipses.
The
particle moves slowly when it is far from the center and tends to spend most of its time to one side of it. expectation values,
<x>
But
That is, in terms of
if A is in the direction of positive x,
>
O,
=
=
0
(2.97)
(Problem 2.61) all these values are zero in any stationary state
that is an eigenstate of the angular momentum.
It is therefore not
clear at first what quantum mechanics has to do with elliptical If the orbit is not re-entrant there is no difficulty, will swing around and, on the average, directions space,
in its plane.
orbits.
for the orbit
spend equal times in all
Keplerian orbits, being stationary
in
are special, but how?
The solution to the puzzle is the degeneracy of the orbits. Degeneracy
in quantum mechanics refers to the existence of different
states of motion corresponding
to the same energy,
definition will do for classical mechanics
also.
and the same For example,
the
84 plane of an orbit can have different to the plane,
this is reflected L 2.
in the 2~ + 1 values
But a re-entrant
orbit has still
given L there are still d i f f e r e n t directions L).
in which
number
orbits corresponding
degeneracy
to different
(always p e r p e n d i c u l a r
in quantum mechanics?
to
In
not only in L z but also in
all orbits with the same value of the principal
A general
discussion
is laborious;
level, which belongs
¢0
quantum
2.60.
functions
=
_~
(I
-
degenerate
2-~o)e-r/~aO
Show that all four of the functions
of L 2.
states:
(2.98a)
(32~)-½a0-~(x'y'z)e-r/2aO
(2.98b)
(2.98)
are eigen-
Show that ~0 and ~z are also eigenfunctions
and that two further eigenfunctions combinations
we look only at the first
to four different
(8=)-%%
=
@x,y,z
Problem
to any with a
are degenerate.
excited
ProbZem
of L corresponding z further degeneracy:
atom there is a degeneracy
L 2 itself:
L , perpendicular
and in quantum mechanics
the vector A may point
What is the additional
the hydrogen
orientations:
can point in any direction,
of Lz,
can be formed from linear
of ~x and ~y.
2.61.
Show that in eigenstates
of ~ Lz,
and L 2, <x> =
(2.97),
of L 2 and L z do not give expec-
let us look at eigenfunctions
formula for A, A
1 = ~ r - ~ pxL
is not quite ready to take over into quantum mechanics, involves
the product
such a product
of two noncommuting
is itself not hermitian.
try in such cases: Remembering
average
the products
that the vector product
interchanged,
of A.
quantities, There
since it
p and L, and
is a general recipe
to
taken in the two orders.
changes
sign when the factors
are
we write A = Z r - "~" 1 (pxL - t x p )
(2 99)
85 Problem
Verify that this operator commutes with the hydrogen-
2.72.
atom hamiltonian and therefore represents
a conserved quantity.
It is now easy to see that the components of A do not commute, that a state can diagonalize only one of them. tedious calculation using the wave functions
Choose A x.
(2.98)
so
A somewhat
now shows that
^
~x~O where y i s Ax
= -½ Y~x
"
( e l e c t r o n charge) 2
by combining ~0 and
Ax~x
= -½Y~o
Thus we can form two eigenfunctions
of
~x: (2.100)
^
AxCVJo- Cx) By (2.93) we see that these
= ½ Y(¢o
-
Cx)
quantum states correspond to oppositely
oriented orbits with eccentricity e = ½. Thus the spatial degeneracy in the classical theory which allows us to have lop-sided orbits corresponds exactly to a degeneracy in the quantum levels which allows us to combine states with different L to produce a similarly lop-sided wave function.
This happens to be
the wave function for a hydrogen atom polarized by being immersed in an electric field, and there is a picture of it in Park 2.13
The Principle of Maupertuis In its primitive
(1974), p.277.
and its Practical Utility
form, the principle
that a body traces out its orbit between ¥
stated by Maupertuis asserts two given points in such a
way that Iv.ds has a value smaller than it has for any other possible J
path at the same energy. venience,
For most purposes
this proviso
is an incon-
and there is no such restriction on the varied paths in
Hamilton's
principle,
but it sometimes
facilitates practical calcu-
lations as we shall show. In Cartesian coordinates
ofv.
= ofv.
and it is this integral, stationary. path,
E
=
=
+
called the action, which is to be held
is not to vary, nor are the spatial endpoints of the
though the time required for the motion may vary,
schematic manner, principles
In a
showing only one coordinate q, we contrast the
of Hamilton and Maupertuis
in Pig. 2.15.
We have now to
86 prove
that when the path is varied as in Fig.
2.15b the natural motion
~t
1 to Figure 2.15
....
I t,
I
,,
%
t
t,
Natural and varied paths for the variational principles of (a) Hamilton; (b) Maupertuis.
is described by
6
X
(L + E ) d t
=
I<
(q2,~2)dt
-
Jto
Edt
The second pair of integrals
-
Edt
since
unimportant treat
the
L(ql,ql)dt
+
=
0
Jtg
is
E ( t 2 - t o - t I + to)
and we write
r
Jto
Jtg
where
~t
= E6t
the first pair as
the
second
whether first
integral
we w r i t e
integral
as
Ill
has it
in
as
an i n f i n i t e s i m a l a function
Sec.
6L__ 6 q n d t + 6qn
1.5
~L
to
of
q2
get
~q
tl to
value or
of
it
q2"
is We
87
in which ~qn = q2n the figure that
- qln
6qn = 0
and
6qn[tl) Putting
the integrals
61(L +
together
The first term vanishes the second vanishes conclude
if the equations
that if the variation
the
time
of the motion
endpoint
principle.
This
oscillating
system.
Example:
is
of motion
of the as
integral
useful
are obeyed; Thus we
+ E]dt is
a variable,
especially
(2.53)
as specified above.
A = I[L(q,4)
that
the
3L ~n~6 t 8q~ ~ tl
of energy in (2.55).
is p e r f o r m e d
6A = O,
take
it is clear from
gives
from the definition
The f a c t
tl,
-~n(tl)6t
=
6L-~-.6qndt + [E + L 6qn
=
E) dt
At
at t o .
varied
(2.102) means t h a t
as we c a n n o t
in determining
the period
can be w r i t t e n L = ½(~2 _ ~ x 2)
A mathematician approximate
who h a d n e v e r
a half-period
heard
of the x(t)
=
is calculated
from
(2.102) I
a 3
a and b are
chosen
at
function
might
try
bt 2
-
to the half-period x = 0 to
1
2
a s
~0 ~
so as
a
+ ~ E
t o make A s t a t i o n a r y :
~A ~A 0 ~a - ~b and the resulting complete
algebraic
equations yield for the period of a
cycle and the energy
to
by a p a r a b o l a ,
as
A = ~ ~-- - ~
parameters
(2.103)
of a sine
vibration
with which the action corresponding
the
o f an
The Harmonic Oscillator
The lagrangian
x = a/b
we c a n
in Hamilton's
(2.1041
88
T
=
2a b
the m a x i m u m value of
-
x(t)
2 I/-1~ ~0 "
E
=
1
a 2
X2
_
2E ~0 2
is a z
X
-
4b
w h i c h satisfies X 2 =
15
8 The exact solutions T
=
E ~02
are, of course, 2_.[_~ ~0 2 "
(2.105)
The frequency is in error by 0.7 percent.
Problem
2.63.
The h a r m o n i c oscillator will be better solved if we take
x(t)
=
Xsin~t
as a trial function and adjust ~ so as to make A stationary. that
(2.105)
Problem
Show
follows.
2.64.
Show that the lagrangian for a simple p e n d u l u m can be
written in the form
L
=
½[~2
_
~02(02
~
@4
+
...)]
~0 2
=
g/Z
and use the v a r i a t i o n a l m e t h o d to show how the period depends on the amplitude of swing.
I suggest you use more than one trial function.
How can you tell w h i c h one is the best?
CHAPTER 3
Newton's
N-PARTICLE
SYSTEMS
laws really refer only to particles,
reference
to properties
momentum.
The objects
position.
The properties
theory by a process
like size,
elasticity,
described by Newton's of extended
laws have only mass and
of summation or integration.
integrates.
it mentally
into infinitesimal
N means more than one.
there are two classes the bodies
of problems
like pressure, body system.
temperature Chemists,
N and then consider
some general small values.
rigid and elastic bodies, Center-of-Mass
in Figure
in a few-
have to deal with
These calculations them here.
are often
In this chapter
theorems valid for all values of Large values,
as exhibited
in
8 and 9.
at the center of mass.
m 7.
I
The condition
I'}'I~,
I
I
I
X
A loaded weightless beam balanced at its center of mass.
3.1 is
mlg(.~
or
and astronomers
starting
quantities
Theorems
mI
balance
are very different,
will occur in Chapters
A loaded beam balances
Figure 3.1
generally
interest:
and density have no counterparts
engineers,
and we shall not discuss
first prove
In physics
to ask--macroscopic
in w h i c h N is in the dozens.
very difficult,
3.1
of volume and
say up to 4 or 5, and when there
The situations
with the kind of question one wishes
we shall
and sums over them, elements
that are of special
are few in number,
are about 1022 of them.
situations
in the
Either one thinks
In that case one does not speak of N.
In this chapter, where
angular
objects must be included
of the object as being composed of N particles or one divides
for they make no
and internal
-
Xl) +
m2g(x
-
x2)
= mSg(X 3 -
~,)
for
90
mlx I + m2x 2 + m3x 3
(3.1)
=
mI + m2 + m3 We take this a s
the definition of the center of mass and write the
general formula as N
N
Let us suppose that the jth particle has two sorts of forces acting on it:
a force fij exerted by the ith p a r t i c l e and a force fj
exerted from outside the system.
The total force on the jth p a r t i c l e
is then
Fj = f. F +. E. j ~j
(3.3)
i/j Linear Momentum Let us write
the total m o m e n t u m of all the particles of a system
as
N
By
P = E mj~j j=1
(3.4a)
P = Mr
(3.4b)
(3.2), this is
Its rate of change is
j
i/j
The last sum involves pairs like f12 says that this is zero.
Further,
+
f21' and Newton's
the sum of the external
third law forces
exerted on all the particles of the system is the total force from outside,
91
We have now
= M~ = F and we see that if Newton's
(3.6)
laws hold for each particle
(or element of
volume)
of a body or set of bodies,
whole.
This is why one can treat planets as if they were point
particles,
they apply to the motion of the
and incidentally it provides
the experimental v a l i d a t i o n
of the third law.
Angular Momentum If we p u s h an extended
object the object's center of mass moves
off, but also, unless the line of action of the force passes
through
the center of mass, the object will begin to turn. Let ~. be the g angular m o m e n t u m of the jth p a r t i c l e around an origin of coordinate located arbitrarily, mass,
Fig.
and let ~. be that taken around the center of
3.2:
~j = mj~j×3j
£j = m j r j x ~ j ,
...:..
q-r
0
/ Figure 3.2
Coordinates used to locate the center of mass (_circle) and the jth particle.
To see the r e l a t i o n between them, write
(3.7) Then the total angular m o m e n t u m around 0 is L = ZZj = Zmj(pj + ~ ) × ( ~ j +
~)
= Zmj~j×~j + ~xZmj~j - ~'×Zmj~j + r×~Zmj
92 In the third sum,
zmjpj = by ( 3 . 2 )
and t h e
vanishes
also,
second
Zmjrj
-
(3.8)
~ zmj = o
term involves
the
derivative
of this
sum w h i c h
so t h a t L = A + [ A = Zkj,
The angular m o m e n t u m
(3.9a) [ = ~ x P
(3.9b)
is the sum of that around the center of mass and
that of the entire system around the fixed point 0. The rate of change of L is x p$ = ~ r j . x f j . +
[ = _v~ ~ r $
r .J x
Z~
f .~J .
i/$ The double
sum represents
internal
stresses,
particle
forces
angular m o m e n t u m created by the system's
and it vanishes.
This is easy to see if the inter-
lie along the lines of centers,
sum come in pairs
in the
like
r I x f12 + r2 x f21 = ( r l
The more general
for the terms
case requires
can hardly be in doubt:
- r2)
x f12 = 0
a closer examination, I but the result
a skater cannot start to spin by standing
still and flexing his muscles.
By (3.7),
the remaining
term of
gives [ = Zp$ x f.j + ~ x F
If a force f acts on a particle torque,
or moment,
at the point r it produces
about the origin of coordinates
The first term in (3.10)
(3.10)
is the sum of the moments
applied forces about the center of mass.
a
defined as r x f. of all externally
Calling this M , we write O
(3.10)
as I[ = M
0
It is often useful
+ ~ x F
to choose the origin of coordinates
E. Gerjuoy, Am. J. Phys.
I?, 477
(1949).
(3.11)
at the center
93
of mass.
Then r = 0 and only the first term of (3.11)
A final N-particle
remark will be useful
system,
center of mass In studying
system,
specifying
about a fixed point
particles
the point
LQ, = ~imj[(rj P = 0.
Problem
3.1.
or a system
translational
discuss
it in coordinates
is at rest.
In such a system
Q is independent
around which
it is taken.
If LQ and LQ, are the two angular momenta,
because
an
of Q, and
speak of the angular momentum of such a system
and if Q' is at a constant
distance
+ ro)Xrj]
The proof takes about Q and Q',
r 0 from Q, then = LQ + r 0 x p = LQ
Prove that the force of attraction
for a spherical planet exactly
for we naturally
to which its center of mass
one can therefore one line.
of a gas of interacting
stars and planets we do not consider
the angular m o m e n t u m without
in considering
and motion about the center of mass at the same time.
motion of the whole with respect
Usually,
one does not need to bother about motion of the
the behavior
of interacting
later.
survives.
of a spherical
is directed along the line of centers
inverse-square.
(5.12)
sun
and is
(This is a problem from elementary physics
but a very important one.)
Kinetic
Energy
The final theorem of this set concerns write
the kinetic
energy of an N-particle
it breaks up according
kinetic
If we
to
T = ½ Zmj6} + ~.Zmj~ 5 + ½ ~ or, using
energy.
system as
Zm 5
(3.8), '2
r = % zmj~j + % ~ 2 Once more,
there is a term corresponding
as a whole
and another corresponding
(3.13)
to the motion of the object
to its internal motions.
94 The results
just proved
dynamical problems,
lead to simplifications
if we are interested
in the internal motions
system attached
to a system's
already:
that
of a system we need not
bother about its motion as a whole and vice versa, coordinate
in solving
for they tell us what we suspected
and that a
center of mass has special
advantages. The formulas hydrogen atom,
developed
above are very general,
applying
a container of gas, or a spinning baseball.
to look more familiar
if we apply them to special
baseball
spinning
for example,
with angular velocity
~.
cases.
around the z axis through
~, its linear velocity
axis,
then
~y = o~px
total kinetic
T
=
½
p
(3.14]
Px =-~°Py" and the baseball's
the
its center
is
= ~xp
z
They begin Consider
If one of its particles with coordinates
moves with angular velocity
If ~ is along the
to a
energy
Zmj(pj~
is
pjy )
•
+
•
½
I~ 2 +
2
+
½ Mr 2
or
T
=
½
(3.15a)
Mr 2
where
(3.1Sb] is the baseball's of mass.
moment of inertia about an axis through
The internal part of the angular momentum A = T,m j p j X l ~ j
=
~:mj[pj2o~
Working out the components
its center
is
= ~]mjpjx[Loxpj]
-
(~.pj)pj]
by (2.89)
of A with ~ parallel
to the z axis gives
95
A x = -~ozmjpjzPjx
,
Az =~o zmj(Pjx2
Ay = -Zmjpjzpjy
+ pjy2 + Pjz 2 - Pjz2 )
or
A For a spherical
distribution
z~
(3.16)
of mass A
and A vanish by symmetry so x y to the angular velocity m. Note that
that A is in this case parallel this is not always
=
z
the case, however
(see Chapter
inertia of a sphere of mass M, density
p, and radius
M
I = ] p(x2 + Y2)dv If the density
is uniform,
I =
Example from rest.
3.1
dr
=
of the slope
The moment
of
R is
4 ~R3o
~
this is
d@
d~(r2sin2O)rZsin@
= ~ MR a
(3.17)
A sphere rolls down an incline of height h, starting
How fast is it moving when it reaches
Solution.
Energy is Mgh.
is conserved. The kinetic
½I ~2+ ½M~ 2, where V = ~R.
Problem
8).
With
3.3.
A hoop and a solid cylinder
y
by
(3.14)
is
this gives ~2 = _g7h . 1 0
Show that A
and A
energy at the top
energy at the bottom, (3.17)
3.2.
x
the bottom?
The potential
are zero for the baseball
discussed
above.
Problem
rolled down the same slope. reaches
the bottom?
Problem
3.4.
the period
3.5
are
Find 2/3
the ends.
around a horizontal
The mass
acceleration
a) at the end and b) at a point
A rope of negligible
end is wrapped middle.
of the same diameter
is the hoop when the cylinder
A uniform bar of length £ is used as a pendulum.
if it is suspended
the way between
Problem
Where
is allowed
of the mass.
mass supporting
a mass m at the
log of mass M with an axle down the
to fall and the rope unwinds.
Pind the
96
Problem
3.6.
Find the p o s i t i o n
of the mass m in the p r e c e d i n g
as a function
of time
if the mass
Problem
A small
cylinder
What
3.7.
is the p e r i o d
of small
as a useful
generalized
serve
Figure 3.3
3.2
Two Particle We treated
vector
planetary
as usual,
if M and m are the masses
Mr
Solving
these
+ mr
e
for r e
energy
a large one,
about
motion
in Chapter
3.5. 0 will
p
be at rest
2 on the hypothesis
the assumption. at the origin,
that
Let the center
and let r be a
from the sun to the planet.
Then by
(3.2),
of sun and planet,
: O,
r
- r
p
and rp, we can calculate
: r
s
the total
(3.18)
kinetic
as
is w r i t t e n
2 = ½
Mm
½ ~2
I _ I + ! M
~ is called
the reduced mass.
= m(1
and when,
~2
as T =
where
Fig.
equilibrium?
To illustrate Problem 3.7.
2 + ½ m~
This
inside
coordinate.
We can now relax
of the two bodies running,
rolls
oscillations
problem
is not negligible.
Systems
the sun is clamped. of mass
of the rope
as w i t h
atoms
(3.19)
m
We can write
it as
m - M---C--~m)
and p l a n e t a r y
systems,
m << M, we see that
97 is slightly
less than the smaller mass.
The lagrangian
is
(3.20)
L = ½ Vf:2 _ V ( r )
and the
only change
Problem
3.8.
from
(2.62)
is
the substitution
The t w o - b o d y H a m i l t o n i a n
o f ~ f o r m.
in quantum mechanics
is
~2
Change
the variables
factors
to r and ~ and show that the wave function
into a term describing
and a term describing
the motion around the center of mass
the motion of the atom as a whole.
Show that
the effective mass enters here just as it does in classical
3.3 Vibrating
Systems
In several branches molecular motions vibrate. pendulum
theory.
chemistry,
of study,
from structural
it is important
engineering
to know how to analyze
to
the
of a system whose parts are coupled together so that they can We have already seen an example of this in the double of Sec.
2.7, where we found the modes of motion in which it
moves periodically.
In this Section we shall study the periodic
motions
in more general
situations,
motions
that are not periodic.
be only a generalization
and then see how to deal with
The first part of the discussion will
of what we have already done.
Suppose there are f degrees of freedom with coordinates qn' and that the system has some position of stability around which small oscillations
take place.
The potential
energy of the entire system
is V(q), and at the position of stability, where q = q0' V(q) has an absolute minimum or at least a local one, at which it is smaller than at any neighboring
points.
@-K 'V I @qn
Since V is a minimum at q0' we have
= 0
and if we expand V in the neighborhood double pendulum)
n = I ....
f
(3.21)
0
of q0 (as was done with the
the linear terms in the Taylor series vanish and we
have
V(q 0 + y) = V(qo)
+ ½ ~ f E m, n=1
~2V Bqm~qn
I qo
YmYn
+ ...
98
where the y's measure the system's departure from equilibrium. first term is a constant and we can ignore it.
The
The second t e r m
involves a set of constant coefficients w h i c h we shall call Vmn , so that
(3.22)
V(q 0 + y) = ½ VmnYmY n ~... in which the summation c o n v e n t i o n is used. The kinetic energy is T=
½ Z m.&. 2
in which the Cartesian components x. are some functions of the q's and therefore of the y's.
Since
x i = x i ( Y 1 ... Yf) we have 8x.
~Yn and the kinetic
energy will
be some q u d r a t i c
function
of the ~'s,
(3.23)
T = ½ Tmn~m~n In (3.22) and (3.23),
the arrays V mn
and T
are not n e c e s s a r i l y mn
symmetric in m and n to begin with, but in (3.22) we have
VmnYmY n = VnmYnY m = VnmYmY n so that
VmnYmY n = ½ The coefficient
½(Vmn
+ Vnm),
enters the calculation,
~Vmn + V n m ) Y n Y m
the symmetric part of Vmn , is all that
and so we may assume that Vmn'
and Tmn , are
symmetrized to start with. For small oscillations,
the lagrangian of the system is
L = ½ T m n Y m Y n - ½ VmnYmY n To take the derivatives, we vary the variables one at a time,
dL = ½
(Tmn~md~n
+ Tmn~nd~m
= TmnYndY m - VmnYndYm
- VmnYmdY n - VmnYndYm ~
99 so that
the equations Tmn~n
of motion
~L/6y m = 0 are
+ VmnY n = 0
m = I ....
This is a set of linear homogeneous cients.
Their solutions
form A c o s ( ~ t
equations with constant coeffi-
depend on time through expressions
Since boundary
+ ~).
(3.24)
f
of the
conditions will involve us with
different values of ~, it is s t r a t e g i c to consider Yn as the real part of an a ne i(~t + ~n) with complex an. Now e i~n can be absorbed in the a and all motions of the system have the same n Substituting this form of Yn into (3.24) gives
constant
time dependence.
(3.25)
(Vmn - ~ 2 T m n ) a n = 0
which is a set of f linear homogeneous solutions
only if the determinant
equations
IVmn - ~2Tmnl
This is called the characteristic Its solutions
give us nothing new,
to the system's eigenvalues values of
of the equations
of
normal modes of
are ~the normal mode frequen-
of the system.
(3.25) can be simplified.
m. are zero,
is,
frequencies
frequencies
system.
values or, more
and since negative
are called the system's
and the corresponding
Equation that
to be equal)
there are f solutions
These solutions
cies, or characteristic masses
equation of the vibrating
There are in general f different
~2 (though some may happen
motion,
(3.26)
They are called c h a r a c t e r i s t i c
commonly, eigenvalues.
It has
vanishes,
= 0
are the values of w z that correspond
periodic motions.
motion.
in f unknows.
of the coefficients
If we assume that none of the
it can be shown that the matrix T
it has an inverse.
Multiplying
is nonsingular, mn (3.25) by this inverse gives
[(T-iV)m n - ~26mn]a n = 0 and the characteristic
equation
I(T'lV)mn The simplification Further,
and positive. e
is
- oS28mnl = 0
(3.28)
is that m 2 now occurs only along the main diagonal.
this standard
readily available
(3.27)
form of eigenvalue
computer programs.
This follows
equation is solved by
The eigenvalues
from the conservation
~2 are all real
of energy:
had an imaginary part the energy would mysteriously
disappear.
It can also be proved from the mathematical
if e in
augment or properties
100
and Vm n " 2
of the matrices T m n Example
~I = 5, ~2 = 8. that
The double p e n d u l u m of Fig.
3.1.
2.4 has m I = 3, m 2 = 2,
To find the frequencies of the normal modes, note
from (2.56), T = 12S 2
=
so t h a t
the matrices
T
~2 6 + S4 "~ I + 4°ei 2 e2
--01+
for T and V are
= mn
V ~ 20
64
"
= g mn
0
o1 8
and
T-I mn
=
3SO----~
-20
125
"
(T-iV)ran = g
- -
The c h a r a c t e r i s t i c
~
~5/
e q u a t i o n (3.78) becomes 2 8
~2 g
2 45 =
5
5
o~2
72
36
g
0
of w h i c h the solutions are
g
as can be v e r i f i e d from
4" 9
(2.59a).
Once the eigenvalues have been found, we can put them back into (3.25) and determine the relative sizes of the amplitudes a n as in Sec.
2,2.
(There is an arbitrary scale factor that is d e t e r m i n e d by
the initial conditons.)
Since all the c o e f f i c i e n t s
real,
are real or, if not, have the same complex
phase. 2
all the amplitudes
in (5.25) are
Thus all parts of a system in a normal mode vibrate exactly in
See for example Goldstein
(1951).
~0~ or exactly out of phase with each other.
Problem 3.9. with masses
Figure
3.4 shows a model of a linear triatomic molecule
and spring constants
longitudinal
3
Z
k.
relations
Coordinates in one of its normal modes,
its
move in or out of phase with each other with definite between
the amplitudes.
system so that these conditions conclude
zCz
To i l l u s t r a t e Problem 3.9.
If a system is oscillating components
of
7
6
Figure 3.4
Normal
Find the frequencies
oscillations.
~-
3.4
as shown.
that we would have
What happens
if we start the
are not satisfied?
to start to calculate
a glance at an example we have already The double p e n d u l u m of Sec.
A pessimist might all over again, but
studied dispels
such fears.
2.7 had two characteristic
frequen-
cies
:
~z
and, correspondingly, the amplitudes
: (-2-) %-
£I :
~2)
=
£Z
4m12 £2
two alternative
)
(3.29)
£2 - ~1
m2
(I
~2
~2
) £i
conditions
to be satisfied by
of motion,
a(1)
d
m2 4m12
(~--)½-- (1 + ~1
2(~ 2 - ~z) ~m12 ~2 - h m2
The t i m e b e h a v i o r
(ii __ aI =
. (i) -Ala 1
(3.30)
~c2~z ~ ~ 2 ~ 2~
£2
assumed there
could just as well have been e
was o f t h e
±i~t
form cos(~t
+ ¢);
it
The point is that the equations
102 of motion are linear;
any linear combination
and e ±i~t = cos~t ± icos(~t
solution,
and superpose solutions most general
corresponding
such superposition
Ol(t ) = A e ~ l
of solutions
- ~/2).
is again a
But we can go further
to different values
of ~, the
being of the form
t + Be-i~l t + Ceil2 t + De-i~2 t (3.31)
= _ k l ( A e i ~ l t + B e - i ~ l t ) + t 2 ( C e i ~ 2 t + De-i~2 t )
02(t)
where A, B, C, D are arbitrary (2.56)
constants.
The original
equations
are a set of the fourth order, which we know to require
arbitrary
constants
solution;
nothing more than the normal modes need be found.
Suppose hand,
for its solution.
for example
displace pendulum
release
that we hold pendulum
-kl(A
+ m2(O -
four
is the general
1 vertical with one
This gives four initial
A + B + C + D = O B)
(3.31)
2 through an angle 80 with the other,
them both at t = 0.
ml(A -
Thus
+ B)
and
conditions,
+ k2(C
+ D) = O0
-hlml(A - B) + ~2~J2 (C - D) = 0
D) = 0
with solution
@ 0
A = B = -C = -D
2 ( l I + 12) and so the motion satisfying
the given boundary
conditions
is
@0
@ICt) = - ~i + h2 (C°Smlt - c°sm2t)
(3.32) 82(t)
The pendulums periodically modes
@0 - hl + X2 ( I I C O S ~ I t + 12COS~2t)
swing w i t h varying amplitudes,
and the upper one returns
to rest w i t h an angular frequency ½(ml
of motion are taking place
simultaneously,
- ~2)"
The two
and we are seeing
the phenomena of beats.
Problem
3.10.
Fig.
3.5 shows an arrangement
oscillators
that can very easily be made.
parameters,
calculate
the two basic
of coupled torsion
Assigning
suitable
frequencies when one bar is held
103
still and the other is allowed to oscillate. other needed parameters,
discuss
the motion
In terms of these and that begins when both
bars are initially at rest and one is given a sudden displacement. Then build the thing and see what it does.
A
Figure 3.5
Problem
To illustrate Problem 3.10. A torsion oscillator consisting of two weighted staffs supported by a wire.
3.11.
Figure
a weak spring.
3.6 shows
Analyze
two identical pendulums
coupled with
the motion when they are started a) in phase
with the same amplitude,
b) out of phase with the same amplitude,
c) with both vertical but one in motion. Vll.~v l/llllli#llll
! IIIIII/II
II!I/Jll////l
II I I
t
Figure 3.6 Problem
cords,
3.12.
Fig.
Three equal masses
3.7.
corresponding
To illustrate Problem 3.11. elastic
of normal oscillations
and the
normal modes.
Figure 3.7 Problem 3.13.
are connected by identical
Find the frequencies
To illustrate Problem 3.12.
A mass m is suspended
from a spring of s t i f f n e s s k and
104
swings
like a pendulum
the pendulum Comment
moves
on special
equal or nearly conclusions
through
in a plane, cases
equal
small
angles.
b) when
Find the motion,
it moves
that may arise when
to the pendulum
a) when
in three dimensions.
the spring
frequency,
frequency
is
c) Verify your main
by experiment.
Normal Coordinates The theory coordinates,
in Section
lar set of q's or y's. Suppose
for example
we had chosen designed
3.3 was set up in terms of generalized
and there was no necessity that instead
linear combinations
to produce
of starting with any particu-
But some choices
quantities
are better
than others.
of @i and @ 2 in the double pendulum of the two solutions
in (3.31)
each of which oscillates
at a single
frequency:
u I = 82 - 1281 = 80cOsmlt
(3.33a) u 2 = 82 + ll@ I = @0cosm2t with
l's as in (3.30).
U2 uI l I + 12 "
@i We would have
With these variables, llU I + X2u 2 82 = l I + 12
found u I to have the single
for u2, as if there were no coupling can be seen if we use terms of u I and u 2. m12
(3.33b) Omitting
~
between
to express small
m2~2(2ii-~2).
+ 4m12 (~2-~z)2[(I
like the lagrangian
+
of two uncoupled
(2.55)
in
we find
@ _
2 1 2
.
~]
q (1 + 4mlm(f~m_~l)2)u2 ]J
oscillators.
calculation
we find that u I and u 2 oscillate
m2 given by
(2.59b).
The new variables
How this occurs
@ m2%1~ 2 ~i (I + 4m12(~ 2- il)2U~] - --
~
m2]~i(21%2-9~i)] 4m12(f~2.£1)2Ju22 •
them.
m I and similarly
the lagrangian
terms,
L = 2(~i + x2)~ ~ i ~ [(I + 4mlZ(~2_£1)2J
m2
frequency
(3.33b)
with
u I and u 2 are called normal
By the simplest frequencies coordinates,
~i and and
105
there to
is
find
a mathematical a set
for
Our p r o c e d u r e constructed it
is
reduce
to
do i t
encounter
discussing Problem
here
the normal
easier
We s h a l l
theorem
any s y s t e m
the
Problem
3.15.
Problem
3.16.
a nd d i s c u s s
was b a c k w a r d s - - w e
the
normal
other
Do t h e
but
solved
for
K
system.
coordinates
3
systems
methods.
9.2 while
o f P r o b l e m 3.10, an d v e r i f y
the
use
them t o
frequencies.
Problem 3.8. solution
occurs
m
Figure 3.8
always possible
t h e p r o b l e m and t h e n
in Sec.
general that
is
more c o m p l i c a t e d
again
same f o r
the motion
it
of a continuous
t o a sum o f s q u a r e s ,
Find the
that
way by a d v a n c e d a l g e b r a i c
coordinates
the normal
lagrangian
states
k i n d we h a v e b e e n d i s c u s s i n g .
coordinates,
oscillations Find
3.14.
the
that
of the
if
the
for
the system
system
k'
is
of Fig.
initially
at
3.8 rest
rn
To illustrate Problem 3.16.
and the r i g h t - h a n d cart is struck towards
the right w i t h a hammer.
What are the general and special solutions when the left-hand spring is missing,
i.e., what are the limiting forms of your solutions w h e n
K=O? Problem
3.17.
Fig.
the CO 2 molecule.
3.9 shows a scheme for the linear vibrations
of
The following steps will lead to a d e t e r m i n a t i o n of
ml
k
M
,
k
i
,
m
1o
11
I X Figure 3.9
3
To illustrate Problem 3.17. Mechanical model for the linear vibrations of CO2. The ~'s measure the stretch of the two springs, to left and right respectively, from their equilibrium positions.
Ter Haar 1961, Ch. 3. The use of symmetry to simplify a p r o b l e m is d i s c u s s e d in Corben and Stehle 1960, App. III.
106
the normal modes and thus to the entire solution.
You should work
them out in detail. i.
The lagrangian
is
L = ½(M + 2m)& 2 + m(~ 2
2.
"2
~1)~ + ½ m ( [ l ~ +
g2 ) - ½k(%12 + ~2 )
Since the momentum of the system is conserved,
the coordinate ~ will have special
significance•
(3.34)
one can guess that The transformation
m
(El
x = ~ + M + 2m
- ~2 )
eliminates x and gives m 2
L = ½(M
2
+ 2 m ] ~ 2 - ½ M + 2m •
.
•
(~i - ~2 )
+ ½m(~l + ~ ]
½k(~ 1 +
.
To remove
3. The terms in ~2 x and {i = are gone but [i~2 remains. rotate the ~ axes by ~/4,
u I = 2 - ½ ( { 1 + ~2 ) •
it,
u 2 = 2 - ½ ( { 1 - ~2 )
tO g e t L = ½(M
4.
+ 2m)~ z + ½ m~iZ
The lagrangian
the corresponding
(X,
El,
Problem
is now decoupled. normal modes,
$2 ) tO ( 7 ,
the
foregoing
just discussed
Find the stationary
3.5
problem
describe from
directly
from the
lagrangian
with those just found.
Write down Schr6dinger's
3.19.
vibrations
Find the frequencies,
and find the transformations
(3.34) and compare the frequencies Problem
+ u22 )
u 1, u 2 ) .
Solve
3.18.
+ ½ M Mm + 2m ~2 z _ ½ k ( U l 2
equation
for the CO 2
and transform it to the variables
E, ul, u 2.
energy levels of the molecule.
The Virial Theorem The methods
described
in this chapter become useless when N is
very large unless the particles organization,
have some specially
as they have in a solid body.
simple spatial
They are useless because
the calculations
are too complicated
results anyhow.
If the system is a gas, it would not be interesting
to know the positions
and velocities
and because nobody wants of all the particles,
could, because what we want to learn are the properties
the
even if one
of gases in
107
general, not those of a p a r t i c u l a r start in a p a r t i c u l a r way.
sample in which the molecules
We want a statistical
theory of specific systems.
We cannot in this book make a serious
d i s c u s s i o n of s t a t i s t i c a l mechanics, theorem,
theory, not a
the virial theorem,
but there is a simple statistical
that illustrates
view one adopts w h e n N is large and permits
the changed point of
some useful averages to
be obtained. Consider a finite, b o u n d e d N - p a r t i c l e w i t h respect to the origin,
system at rest as a whole
and define the quantity
N
G =
Since no p a r t i c l e wanders
ever reaches
infinitely far away,
Z i=1
Pi.ri
(3.35)
an infinite value of m o m e n t u m or
G remains finite.
Further,
if we shift
the origin by r 0 the new G is
G' = [ pi'(r - r O] = G - r 0
E Pi = G
and so G does not depend on the choice of origin.
The time derivative
of G is
= Z (Pi.ri
=
2T+
~
+ ~i.r£)
F.-r.
where T is the kinetic energy and the quantity W = ~ F.*r. was called by Clausius
the virial of the system.
w h o l e remains at rest,
W, like G, is independent of the choice of
If the system as a
origin of coordinates. Let us find the time average of 2T + W.
It is
Gdt = Lim~7
Li~ ! T ._~oo T
0
But we have noted above that G(z)
- G(O)
always remains
finite;
the limit is zero, and the r e l a t i o n b e t w e e n time averages =
-
½~'
thus
is (3.36)
108
This is the virial
theorem
(Clausius,
Ig70).
The theorem has a useful application square forces--for evolving clusters in (3.3),
example,
to systems bound by inverse-
an atom, certain plasmas,
of stars.
Introducing
or slowly-
the interparticle
forces as
we have
Y
EE f . . . r .
=
i~j
~J
J
A typical pair of terms is
f12"r2
If the interparticle
+
~1
potential
"rl
= f12 "(r2
"
~1 )
=
V12
is
v12 --
Y -
Ir 2
rll
-
the force is YCr 2 -
f12
rl)
Ir 2 - rl I~
and Y f12-(r2
-
rl)
= _
Thus, Y = V, the total potential
Since by hypothesis
the system's
-
rtl
energy of the system, = -
conserved,
Ir 2
and we have
½~
(3.37)
total energy ~ + ~ = E is
this gives also
(3.38)
E = -~ = % F
Clusters
of Galaxies
The virial
theorem has an important
mass of a cluster of galaxies. of order R, the gravitational
application
in finding the
If the cluster has a mass M and a size potential ~" = -
G-~M2 R
energy is of order (3.39)
109
if v ~ is the mean square velocity of a member of the cluster,
then
and the virial theorem gives M
The n u m b e r s larger
right
does not further
a large know a t
(3.4o)
can be m e a s u r e d ;
t h a n w o u l d be e x p e c t e d
Apparently
Ideal
on t h e
= R Vs2
from the
the masses
luminosity
amount o f n o n l u m i n o u s
present
what
form it
mass
takes.
is
come o u t much
of the present,
See P e e b l e s
clusters. but
one
(1971)
for
details.
Gas The virial theorem also gives insight into the behavior of gases.
In an ideal gas the only force on a molecule is that exerted by the walls of the container.
In magnitude
and direction this is given by
dF = -PdA where P is the pressure and dA is a directed element area.
of
The virial is
W = -PIr.d
A
Evaluate this by Gauss's theorem:
W = -PIV.rdV where V is the volume of the container.
= -3PV The virial theorem now gives
3
= ~ PV
(3.41)
a relation familiar from elementary physics but here proved rather neatly.
Problem
3.20.
How is the result
(5.38) generalized if the force
follows an inverse-nth power law?
Problem
3.21.
Let the system be a hydrogen atom (N = 2) and the
electron and proton in circular orbits around their common center of mass.
Problem
Verify that the virial theorem is satisfied. 3.82.
If the system considered above is immersed in a uniform
magnetic field B, the force on each particle is augmented by evxB. Show that the virial theorem for charged particles now reads
m
110
where L is the system's total angular momentum. Problem
3.23.
It is generally true that expectation values in quantum
mechanics correspond to time averages
in classical mechanics.
and prove a virial theorem for quantum mechanics. (3.41) take in the quantum-mechanical
State
What form does
theory of an ideal gas?
CHAPTER 4
HAMILTONIAN DYNAMICS
The Lagrangian procedure of motion and simplifying and their constant
is very helpful
them through the use of cyclic coordinates
conjugate momenta.
is up to us to solve the equations. more powerful methods which will
and except
not get involved
in fact a large class of calcula-
5 to the evaluation
are of an altogether
for the chapter
on perturbation
to choose
the right variables
problems
in dynamics--our
depended
on the use of polar coordinates A solution
We need a technique cyclic coordinates improvement quantity.
as possible.
hamiltonian connections
There momentum
there is one more small is not an ordinary physical In the Newtonian
to be developed here start with the
that is with the energy, which than the lagrangian,
is a more funda-
and of course
in its usual
they lead to
formulation.
Equations
are two general ways to depict the time behavior
conjugate
a q, t diagram and a q, p diagram, to a given qn"
Pn equations
(2.53)
=
is defined as (4.1)
BLCq,~)
Bqn
Bqn
oscillator
(a) is a graph of x = acos(mt + ~). (2.10),
(4.2)
aLCq,~)
Fig. 4.1 shows how a simple harmonic of the energy equation
of a
where Pn' the
then become
Pn = Diagram
would be far harder.
so far it was simple to find, but this
with quantum mechanics
system:
Lagrange's
for example,
so as to find as many
it is not measured.
The techniques
function,
The Canonical
dynamical
Finally,
The lagrangian
we have performed so.
for solving
to give us the cyclic angular
coordinates
for changing variables
mental part of physics
4.1
in cartesian
It is not conserved;
is not always
order of
theory we shall
solution of the Kepler problem,
to be sought.
calculations
of a few integrals.
different
in them.)
It is important
coordinate.
it
later lead to the solution of problems
tions will be reduced in Chapter (Problems not so reducible
But once this has been done,
In this chapter we shall develop
with a m i n i m u m of equation-solving;
difficulty,
in setting up equations
is described
Diagram
in each.
(b) is a graph
the ellipse
~2
2mE + ~
X2 = /
(4.3)
112 It represents
time behavior because p itself represents
The point P representing on (a); on ~)
it moves
the oscillator's
clockwise
state moves
time behavior.
to the right
(why?) around the ellipse.
Lagrangian
P
/k/ ca) Figurg 4.1
Trajectory of an oscillator, (a) in
dynamics,
and the Hamilton-Jacobi
use the
q,t
representation.
The hamiltonian
qt
space; (b) in qp space.
theory to be developed
Hamiltonian
in Chapter
5,
dynamics uses q and p.
function corresponding
to a given lagrangian
is
defined as
H(p,q) (summed).
Pn4n
=
-
L(q,~)
(4.4)
To see why we call it a function of p and q, imagine
the system is varied, i.e., changed so
that p, q,
that
and ~ all change
slightly:
6H by
(4.1) and
=
(4.2),
pnB~n
+
~nBPn
3L
-
9q---~6qn
~L -
•
Bqn
~'---'~
this is
BH
~n~Pn
=
-
~n6qn
(4.5)
so that we need specify only the p's and q's in order to evaluate H. Changing one p at a time and one q at a time gives Hamilton's equations I ~H qn
=
~H
~Pn
The first of these,
"
Pn
=
anticipated
-
~qn
n
in (1.24),
=
I ....
H(p,q).
again, .the generalized
second law.
these equations i
(4.6)
tells how ~ is related
q and p; we need this in order to write form of Newton's
f
The second is
are no great advance over Lagrange's
Phil. Trans. Roy. Soc. 1835, Pt. I, p. 95. them many years earlier without recognizing
to
(4.2)
Computationally, equations.
Lagrange had published their value.
113
Formally however,
by virtue of their transparent structure,
first-order equations, they open the laws of dynamics.
they
from here,
Example
being 2f
way for a profound study of the
To express the fact that all further studies start
are called the canonical equations.
4.1.
Set up the problem of plane orbits by Hamilton's
equations.
Solution.
Use the coordinates r, 8.
L(q,~)
Then
= ½ m(~ 2 + r202)
- V(r)
The momenta are ~L/a~n , or t
p@ = mr2@
Pr = mr,
The Hamiltonian function is (2.67) expressed in terms of p and q,
(4.7)
H = --~ (pr 2 + po2/r 2) + V(r) and Hamilton's
equations
Pr =
--"
m
~
are
PO =
mr
2 "
PO
=
Since 0 is cyclic, PO is constant.
O,
Pr
po 2
3V
mr 3
~r
Further solution takes us along
essentially the same path we have travelled before. To investigate
the rate
at which energy in a system changes,
let
us consider the general situation in which the external interactions are time-dependent and H is a function of p, q, and, explicity,
t.
We have
d~ - ~Pn aH Pn + as'-'k'-qn + ~_A dt aqn ~t From
(4.6),
this is
dH dt and if the interactions
-
~H ~t
(4.8a)
are not time-dependent, dR
d---{ = o
(4.8b)
The system moves in such a way that its energy is conserved. That L sometimes has physical meaning even if it is not conserved can be illustrated by the example of special relativity. dimension,
In one
the kinetic energy of a free particle can be written as
114
=
ma2[(:
-
v2/o2)
-%
_
1]
(4.9)
m = rest mass
The relation w i t h L is
E = - ~dL v-L
L
=
U
plus a c o n s t a n t m u l t i p l e
I ~Edv
=
mc211
-
(I
-
V2/e2) ½]
o f v which we can omit
(4.10)
{Prob.
4.1).
There i s
an important point that can be made if we drop the irrelevant additive constant in L and write the action as
W =-mc2
I [ 1 - v2/c2]½dt = - mc2 I [dt2 - c-2dxZ]½
This is
(4. l l a )
W = -me 2 I d s and the law of motion becomes 6W = 6 I d s
= 0
(4.11b)
where d8 is the invariant element of proper time.
Let us suppose that
the integrand to be varied in (2.54) were not a relativistic
invariant.
Then we could make the integral larger or smaller by evaluating one coordinate system or another, some p r e f e r r e d coordinate system. are covariant
The only way to have results that
(independent of the coordinate system)
is to have the
integrand an invariant,
as has h a p p e n e d here automatically.
is an interaction term,
it too must be an invariant.
be noted again following Eq.
Problem 4.1.
If there
(This fact will
(4.15).)
Show that no generality
multiple of v that appears
it in
and setting 6W = 0 would pick out
in L.
is lost by omitting the constant
Set up the equations
problem and sketch their solution to the point where
for an actual
it can be seen
that the extra term does not matter. Problem 4.2.
Show that H and L are equal in the low-velocity limit of
the relativistic formulas just given.
Problem 4.3.
(This and the following problems
are d e s i g n e d to allow
the reader to explore one aspect of the relation between classical and quantum mechanics.)
Suppose that the equation of m o t i o n for matter
waves is taken to be of the general form
i~ ~
= H(p n, qn)~
n = 1 ... f
115
where the p's and q's are q-numbers does the ordinary probabilistic
and H is a function of them.
Why
interpretation of ~ require H to be
hermitian?
Problem 4.4.
Assume t h a t
[pm,qn ] = -ig6mn , and prove
[pm,Pn ] = 0
that
[qnk,pn ] = ki~qnk-1 = i~ ~-~--qnk ~qn Problem 4.5.
Show via Taylorts
series that
[f(qn),Pn] S u p p o s e now t h a t differentiation
Problem
f
is
a function
be c a r r i e d
4.6.
=
i~
~f(qn ) ~qn
of the p's
and q's.
How s h o u l d t h e
out?
Prove that
d
-ZY--
i
-- ~
<[H'Pn]>
-- - < ~ H
>
~qn provided that the differentiation
Problem 6 . 7 .
is carried out properly.
Show that
d
4.2
~H ~Pn
Magnetic Forces We have not yet considered the effect of an applied magnetic
field because it does not fit obviously into considerations energy.
based on
This is because the force on a charged particle F =
(e/c in Gaussian units)
is perpendicular
hence cannot affect its energy.
(4.12)
evxB
Further,
to the particle's motion and since we do not introduce
electric forces into the lagrangian formalism by means of the field, we should not expect to do so for the magnetic forces. of force,
it is the potential
Electrostatic by E : -V~.
For both kinds
that we must use.
fields are given in terms of the scalar potential
Magnetic fields are formed from a vector potential A by B =
If the potentials (in MKS units)
VxA
are time-dependent,
(4.13a)
(4.13a) still holds and E becomes
116
E =-V@
-
@A
~)---{
(4.13b)
Since the lagrangian is a scalar, A must enter as a dot product,
and
since the forces are v e l o c i t y - d e p e n d e n t we form the product with the velocity x.,
Let I be a constant to be determined later. L = ½mx 2 -
Suppose that
XA.~<)
e(~-
We h a v e ~A. ~L • Pi = ~x. = mx + eXAi,
3L - e( ~ ~x. = ~x.
- X
xj)
In taking the time derivative of 3 L / ~ i we must remember that a moving p a r t i c l e encounters a changing A even if A is time-independent, so we consider A. as a function of x.(t)
d
9L
and t:
~A.
~A.
J and the Lagrange equations are
mxi
Let i = i.
9A. ~j+ + el( m-~.
~A. ~ ~t
~A. • ~ . mj)
~ + e ~.
= 0
Then in this e x p r e s s i o n
~A i
~Aj •
8A I
~A~)~2
@Ai _ 8A 3 .
-- -B3& 2 + B2& 3
= (B×~)I=-
(~×B)I
and the equation of m o t i o n is
m~ : e ~ x B
@A
- e CW + ~ ~-~)
117
If we take
k = 1 this becomes
the equation
for a charged particle
in
MKS units, ./~ = exxB + eE
whereas
Gaussian units come from taking
The n o n r e l a t i v i s t i c magnetic
lagrangian
(4.14)
~ = c-*
of a particle
in an electric
and
field is therefore
L = ½ mx 2 -
In special
relativity
-
A.X)
the first term becomes
we know gives a relativistically contribution
e(@
of the interaction
invariant
C4.15)
-moZCl
-
X2/O2)½
contribution
to w.
which The
term is
Wint = -e [ C@- A'x)dt = -e [ (¢dt - A'dx}
Since
(¢,A)
forms one four-vector
(dt, x)
Win t is already invariant in form and need not be changed
is another, in writing
(Jackson 197S, Ch. Ii) and
down the relativistic
To find the h a m i l t o n i a n
version of (4.15).
corresponding
to (4.15) we calculate
@L
"~- = p = mx + eA
C4.16)
so that the energy is
E = :x. Cmx + eA)
- ½ m x 2 + eC@ -
A.~:])
or E = ½ mx 2 + e@ The magnetic
field seems
this expression
to have disappeared
for E merely repeats
change the energy.
(4.17) from the theory, but
that a magnetic field does not
To find the h a m i l t o n i a n we use
(4.16)
to express
E in terms of p:
H = 2-~ (~p -
eA~)2 + e@
I[4.18)
118
Here
p contains A though the energy does not.
in the historical
development
that it turned out to have a visible
(potential)
part.
field is rxp
(and not,
and an invisible
as one might perhaps
Show that the relativistic
Problem 4.9.
to (4.18)
(kinetic)
energy was
We can see now that the same is true of momentum.
Show that the angular m o m e n t u m of a particle
Problem 4.8.
magnetic
One of the difficulties
of the concept of mechanical
hamiltonian
think,
in a r×(p - eA).
corresponding
is H = c[(p
- eA) 2 + m c 2 ] ½ + e~'
(4.19)
An Electron Lens
It is not easy to find calculations by the use of Hamilton's example.
Fig.
equations.
4.2 shows a simplified
&
that are materially
The following
shortened
seems to be an
electron lens made of a single
I
0
Figure 4.2
circular
loop of wire.
use cylindrical
Loop of current serving as an electron lens,
Since the electron follows a spiral path, we
coordinates,
p, 8, z.
ring is given by an intractable the axis is approximated by z
Jackson,
1975, p. 177.
The vector potential
inside the
expression 2 which for points near
119
the axis
is a p p r o x i m a t e d by
p A@ = Iza a
p
z2]~/2
[(a + p)2 +
= I~a 2
(4.20)
(a2 + 2 2 ) ½
The l a g r a n g i a n is
p2~ L = ½ m(p 2 + p2~2
+ ~2)
+A (a 2 + z 2)
(see P r o b l e m 4.11),
and the h a m i l t o n i a n
A = I~a2e
h
is AP 2
H = I
(pp2 + pz 2) + ~ 1
Here 0 is a cyclic coordinate
[Pe
(4.21)
From ~ = dH/dpo
we have
+ 22)
and p@ = const.
2
3/2]
-
Ap e
[Po - (a2 + 22)3/2 ] This shows how far the h a m i l t o n i a n m o m e n t u m can be from one's idea.
intuitive
If you ask the e l e c t r o n why it is spiralling around the axis it
will answer that
it is conserving
its angular momentum.
Let us assume that far away at the object the beam originates the axis.
It then has 0 = 0 and so p@ = 0.
on
During the subsequent
mot ion, A = _ inca 2
+ Z2) 3/2
Since the m a g n e t i c field does no work, constant,
the electron's speed remains
and since the rays stay close to the axis this means
that z
is very nearly constant. To study the orbit, we have
dpp
=
_
dt and w i t h ~ = u -~ const,
dpp dz Assume
~H
~p
=_
A2p,
mC¢ 2 + z2) 3
this gives = dpp d__~t= _ A2p dt dz mu(a 2 + zZ) 3
that while passing through the ring the electron's
constant at b.
Then the total change in pp is f co AZb I dz App = - mu I (a2 + 22)3 _co
-
3~ A2b 8 muaS
p is about
120
From Fig. that
4.2,
tan# 0 = (pp/mU)initia 1 a n d tans i
tan# 0 + tan# i
and since tan¢ 0 = b/Zo,
tan¢i = b/Z i
=
_
(-pp/mu)final
=
so
,
&PPmu
this is
/_ I _ 3~ A2 3~ (eI~] iO + £i 8 m2u2a s - 8a - m u This is the equation
for a thin lens of focal length
8a
mu
f = ~
2
(e---~--~)
(4.22)
When the lens forms a real image it is not simply electrons
inverted,
for the
all spiral at a rate
de
de dt
A
S-; = S-? 2 ;
so that the total
rotation
=
=
In terms of the focal
+ z2)3~2
is
dz
2A
a 2 + z~3/2
mua 2
A
e
mu(~2
- m--J
length,
this is
e =-
. ~a)%
4t~f
If for example f = 3a, then O = -61 °. Note particularly differential
equations
that we have got these results but just by evaluating
first step in the direction we shall take in Chapter give a systematic way of reducing more or less difficult canonical
equations
developments,
Problem
4.10.
This
is a
5, which will
a large class of calculations
integrations.
is that
not by solving
integrals.
But the principal
they form the basis
to
merit of the
for further
formal
as the rest of this chapter will show. Use the first form of A e in (4.20)
and discuss
the
aberration produced.
Problem
4.11.
cylindrical
The construction
coordinates
Verify the formulas
Problem
4.12.
of L and H involve procedures
instead of the cartesian
with
ones used heretofore.
given in the text.
Show that
if a magnetic
field 8 is uniform and in the
121
z direction,
a possible A
x
= -%
By
Express
the h a m i l t o n i a n
Problem
4.13.
observed
(but not unique) A
To find the equations
the kinetic
hamiltonian
Problem
of Larmor's
Canonical
phenomena
z = z'
H =
p,2 _
in terms
and show that the
~'PO
the results
+
V(r)
(4.24)
of the last two problems, charged particles
of the same system without rotating
compare
in a magnetic
a magnetic
system of reference.
field
(This is the
Transformations
If a h a m i l t o n i a n
dynamical
order of the remaining
problem contains
are constants differential
to symmetries, are used.
important
cyclic coordinates
and can be used to lower the
equations.
Cyclic coordinates
but they only show up when appropriate
(In the Kepler problem,
leads to the conservation
use polar coordinates therefore
dynamical
theorem.)
their conjugate momenta
symmetry
(4.23)
0
Show that when written
is unchanged,
of a system of identical
field with the behavior
coordinates
=
y = x'sin~t + W'cos~t,
lagrangian.
Comparing
4.14.
correspond
governing
energy
but viewed from a suitably
4.3
z
becomes
the behavior
subject
A
let
- y'sin~t,
and find the transformed of momenta
= ½ Bx
y
is
in this case using the angular momentum.
on a merry-go-round,
x = x'cos~t
vector potential
for example,
of angular momentum,
but one has to
in order to get the cyclic coordinate.)
to be able to study the symmetries
ian and to capitalize
circular
on them by shifting
It is
of a hamilton-
to appropriate
coordinates
when one is found. With this motivation, canonical
equations.
the fact, m e n t i o n e d from Hamilton's
let us see how to change variables
The place
to start is Lagrange's
above but not yet used,
variational
6
in the
equations
and
that they are derivable
principle
L[q(t),
q(t),
t]dt
= 0
(4.25)
to provided
that the variations
end points.
The integral
in the 6q's are made to vanish at the
is known as Hamilton's
principal
functionjW,
122 and
(for fixed lower limit)
W(q,t)
it is a function
of the point q at
which all the paths arrive at time t.
In terms of coordinates
momenta,
is
the varied
integral
6
in (4.25)
IJt 0{pnqn•
- Hip(t), q(t), t]}dt
where we allow time-dependent difficult
to do so.
because whereas to vary
p's
interactions
We cannot at once
in varying
and
in H because
set it
(4.25) we set ~
and q's independently.
(4.26)
equal
to
= (6q)',
it is no more zero, however,
in (4.26) we wish
We must therefore proceed more
slowly. It will be useful requirement
that
in later work if we drop for a moment
the 6q's vanish at the limits of
perform a general
infinitesimal
variation.
integration
Denoting
constant)
the momenta,
the energy
and
the integral
before by W, we vary the path of integration with respect coordinates,
the as
to the
(which need not be considered
and the time at which a given point in the path is reached.
The first term of (4.26)
~[Pndqn ,
is
but we keep the time variable
for a moment:
6 I Pnqndt
= I [qn6Pn + Pn6(qn)]dt = (6qf, so
We have already noted that 6(5 )
that this is
I [qn6Pn + Pn(6qn)'] dt = I (~Pndqn + Pnd6qn) The second term of this integral as a change
dqn
in
(defined as
Returning
to
fPndqn that qndt) over
arises
from changing
and interpreted
the widths of the strips
which the integration
(4.26) we write
6W =
fPn6dqn
can be written
is carried
out.
the second term in the same way,
(6Pndqn
+ Pnd6qn
-
~Hdt
-
so
that
Hd6t)
~t 0
Integrating
t h e s e c o n d and f o u r t h terms by p a r t s
gives
6W = [pn6qn - H6t] tto + I~6Pndqnto Now suppose
6qndPn - 6Hdt + 6tdH)
that the path being varied
is what we shall call a
123
natural path,
one that is physically
the canonical
equations.
~H dt ~Pn
dqn = while by
possible
and therefore
satisfies
Since
~H--~--dt dPn = - ~qn
and
(4.8a) ~H ds = ~ d t
the last integral
is
it
_Jto~(3H 6Pn + --~qn~H6q n + ~3H 6t - ~H1dt = 0 Thus only the integrated part remains,
6W = [pn~qn if the variation
(4.271
H6t] t to
is away from a natural path, whether
or not the new
path is natural. Hamilton's lagrangian
principle
is stationary
that the variations selves
(4.251 asserts
vanish at the endpoints
are not varied.
and we have justified provided
of the
of q(t) provided
and the endpoints
them-
Under these restrictions 6w =
in (4.261,
that the integral
for arbitrary variations
the more general
0
(4.z81
form of the principle
expressed
in which the p's and q's may be varied independently
that the same restrictions
are observed.
Transformations Now let us adopt a new set of coordinates them to be functions
and momenta,
supposing
of the old ones and the time, pn(p, q, t) and
Qn(p, q, t). We assume that these relations are soluble to give pn(P, Q, tl and qn(P, Q, t). Further, we require that the equations of motion in P and Q shall again be canonical; that is, that when the changes
of variable have been made we again have
][PnQn - K(P, Q, t)]dt = 0 where K is the new hamiltonian of integration. preserve
called canonical
and 6Qn, like ~qn' vanish at the limits
Transformation
the canonical
(4.291
of this very general kind which
form of the equations
transformations.
of motion are properly
Today they are usually known as
124 contact
transformations,
somehwat
different
the subset consisting from changing
a term introduced by Hamilton
long ago in a
context, 3 and we shall use the term.
Q(q) and P(p,q)
of transformations
only the spatial
By contrast,
coordinates
that arise
are called point transforma-
tions. The final requirement natural path in (p,q) natural path. initial
for a contact
should,
(P,Q)
into
This will be so if, for corresponding
conditions,
same for all values
the integrals
is that
transformation
when translated in (4.26)
and
choices
(4.29)
of t O and tl, that is, if the integrands of a function whose variations
the limits of integration.
Since the variations
El(q,
of the
are exactly
most by the time derivative
such a function is the time derivative
there,
every
again be a the
differ at
vanish at
of q and Q vanish
of a suitable
function
Q, t):
6 I t~1(q, Q, t)dt = 6[FI (q' Q, t)] t = 0
(4.30)
to to Writing
out the time derivative
of F (q, Q, t), we have aF 1
Pn~n - H(p,q,t) which
is satisfied
= PnCn - K(P,Q,t) identically
provided
aF 1 .
(4.31)
~i (q' Q" t) ,
~qn
Bt
that
~FI (q" Q" t) Pn = -
aF 1
- --Bqn qn - ~Qn Qn
P
-
n
(4.32)
~Qn
and
~F1(q,~,t) H = K +
(We can also introduce (4.31) by different
9t
(4.33)
scale factors by multiplying
constants,
but what results
the two sides of
is only changes
in the
units of measurement.) There are other forms possible (4.30)
could solve the resulting
in a function F(q,
equations Pn = Pn (P'Q)
and reduce F to a function
To get a new transformation,
3
provided
it satisfies
it can be a function of any two of the variables p, q, P, Q,
but not more than two because
pn(p,Q)
for F:
For a condensed explanation (1923), p. 547.
Q, P), for example, we and qn = qn (P'Q)
for
of q and Q. replace FI(q,Q ) by a new function
of why the term is used,
see Sommerfeld
125
F2(q,P ) = Fl(q,Q )
where P is considered
as a function
will vanish at the endpoints. Pn~n
- H(p,q,t)
Then
= Pn$n
PnQn
of q and Q so t h a t (4.31)
- K(P,Q,t)
@F2 ~ BE2 - ~Pn " n - @qn qn As before,
this is satisfied provided aF 2
Pn
(q, P ) Qn
~F 2
"
H
=
K
+
(4.34)
at
n
In a similar way,
we can form two more functions
and generate
the
from
~F 3 (p, Q, t) aQ n "
Pn
• - QnPn
that
BP
@qn
transformations
aE 2 , ~t - PnQn
aF 2 (q, P ) "
=
its variation
is
aF 3 (p, Q, t) ap n ,
qn
H = K +
aF 3 (p, Q, t) at
(4.35)
and BF 4 (p,P, t) qn =
3F 4 (p,P,t)
@Pn
Example
I.
•
Qn =
3P n
BF 4 (p,P, t) ,
H = K +
Bt
Let
(4.3?)
F3(P,Q,t ) = PnQn
Then
(4.35)
gives
Pn = Pn" so t h a t as
it
this
is
sounds;
Example
2.
(4.36)
the
identity
we s h a l l
H=K
Qn = qn" transformation.
encounter
it
(This
is
not
as u s e l e s s
later.)
Let F l(q,Q)
= qnQn
We find Pn = qn"
This
is not very interesting
and coordinates
are merely
Qn = - Pn
computationally,
but it shows
that momenta
labels which we bestow on certain variables
126 for historical reasons.
Nothing is changed if we switch the labels
around. Point transformations
to new coordinates Q(q)
are often conven-
iently generated by F 3(p,Q)
since from
=
(4.38)
Pmfm(Q)
(4.35), ~fm
qn Example
3.
=
fn (Q)"
Pn
=
Pm ~Qn
To transform from C a r t e s i a n to polar coordinates
in a plane,
let x = Rcos@,
Px = m~,
y = Rsin@,
py = m~
SO that
F3(Px,Py,R,@ ) = R(PxCOS@
+ pySin@)
and along w i t h x and y we get
PR = Px c°sO + pysin@ = (XPx + ypy)/R PO = R(-PxSin@ yielding
the expressions
Problem 4.15.
+ pyCOS@) = Xpy - YPx
for radial and angular m o m e n t u m at once.
Find the t r a n s f o r m a t i o n from rectangular
to spherical
polar coordinates.
Problem
4.16.
Find the generator F3(P,X ) that caries out a rotation of
plane cartesian axes through an angle a: X = xcosa - ysina
P
= PxCOSa - pySina x
Y = xsinm + ycos~
The usual point t r a n s f o r m a t i o n s as are the infinitesimal
(4.39)
Py = PxSin~ + pyCOSm
are expressed in terms of F3's ,
transformations
to be discussed
in the next
section, but we shall encounter other forms in subsequent chapters. In particular, we shall be interested in canonical that leave the h a m i l t o n i a n invariant, as it was of p,q.
transformations
i.e., the same function of P,Q
This is what defines invariance.
127
Problem
4.17.
The hamiltonian of a simple harmonic oscillator ~(p,x)
= ~-~ + ½ 2m
is
~x 2
Carry out the contact transformation
FI(X~X ) = _ ½ m~x2cotX Find the new coordinate, oscillator
Problem
momentum,
~z = k/m and hamiltonian,
(4.40) and solve the
in the new variables.
4.18.
Find the F representing a transformation to a system of
coordinates rotating around the z axis, and verify that the hamiltonian in the rotating system is (4.24).
Problem
4.28.
coordinates
A point transformation
in terms of the new ones.
gives the new coordinates
Problem
4.80.
functions conditions
in the form (4.38) gives the old Find the transformation that
in terms of the old.
Show by comparing second derivatives
that a transformation
of the generating
is canonical if it satisfies certain
among which are
( n)q=
-
(%~m n) ~
Find the rest of these conditions.
Problem
4.81.
Show that the transformation Q = q½cos2p q =
p2 + Q2
P = q½sin2p
tan2p
=
P/Q
is canonical.
Problem
4.82.
generates
Find by integrating
(4.32) the form of Fl(q,Q)
that
the transformation of the preceding problem.
(Answer: FI(q,Q ) = ½ [ Q ( q _ Q 2 ) ½
+ qsin-1(Qq-½)]
Note that in all the examples given except that of Problem 4.18 the transformations
are independent of t and therefore, by
its later analogs, K(P,Q) = H(p,q).
(4.33) and
In the exception we are looking
at a system from a moving frame of reference and would not expect the energy to be the same.
128 4.4
Infinitesimal
Transformations
There are many transformations sequence
of infinitesimal
familiar
examples,
transformations
steps.
Rotations
and we shall encounter
are discontinuous,
polar coordinates.
Infinitesimal
interest because of their formal considerations invariance
that can be carried out in a and translations
others
in quantum mechanics,
and closeness
and because
to
are of special to analogous
the criterion
under such transformations
of the h a m i l t o n i a n
Some
that from cartesian
for example
transformations simplicity
are
less familiar.
for
is very easily
stated. Infinitesimal
coordinate
cases of infinitesimal displacement
transformations
canonical
of an object to the right
of coordinates
for rotations,
(or a displacement
to a
of the origin
£
the infinitesimal
X = x - ey,
f(x,y)
A function
and simple
Corresponding
to the left) we have
x+X=x+ whereas
are familiar
transformations.
and by infinitesimal
(4.41a)
form of <4.39) Y = y + sx
(4.42a)
is transformed
by infinitesimal
f(Z,Y)
+ ~f 3x
rotation
f(X,Y)
is
= f(x,y)
displacement
into
(4.41b)
into
--- f ( x , y )
(4.42b)
+ C (x ~f~y - y ~ )
Quantum M e c h a n i c s The application elementary.
of these formulas
The last two equations
in quantum mechanics
can each be written
is
in the form
f(x,z)
= (1 + ~ ) f ( x , y )
(4.43a)
f(x,y)
= (1 -
(4.43b)
with the inverse
where D is the appropriate
~(x,y)
is some dynamical
satisfies
a relation
~)f(z,z)
differential
variable
operator.
Suppose
in quantum mechanics
of the form aCx, y)f(x,y)
=
gCx,y)
~
that
and that
it
129
Operate with
cD)
(I +
on this equation
(1 + s D ) a ( x , y ) f ( x , y ) = By (4.43b),
(1 + e D ) g ( x , y )
this is (1 + e D ) a ( x , v ) ( 1
Thus the operator playing (x,v)
- sD)f(X,Y)
= g(X,Y)
the same role in (X,Y) that ~(x,y)
did in
is (1
which we may call A ( X , Y ) ,
+ eD)a(x,y)(1
[D,~]
-
(4.44)
sD)
and to first order this is
A(X,Z) where
= g(X,Y)
is the commutator
= ~(x,y)
D~-~D.
+ ~[D,~]
(4.45)
If D commutes with ~, then
A(X,Y) = ~(x,y) and we say that ~ is invariant under the transformation. If ~ happens
to be the hamiltonian
defines
a dynamical
factors
of i to make D hermitian
of measurement,
variable
we write
(4.43a)
f(X,Z)
operator
that satisfies
of some system, a conservation
and ~ to introduce
then law.
Adding
conventional
units
as
= (1 + ~
(4.46)
G)f(x,y)
where = - i~D
In contrasting designate
classical
operators. =
-
(4.47)
and quantum mechanics we generally use hats to
The two examples
given above give
^ Px
(displacement
i~
l~
=
in x)
(4.48)
and
= Xpy - YPx respectively,
which are recognized
angular momentum. invariance
The situation system is supposed wave function,
the quantity
are closely
$, called the generators
same as the operators
is especially
simple
which depends
(4.49)
of linear and is conserved.
Thus
linked in quantum mechanics. of the transformations,
giving the conserved
to be specified
in xy plane)
as the operators
If H is invariant,
and conservation
The operators
(rotation
dynamical
in quantum mechanics,
as completely
are the
variables. because
as possible by its
only on the coordinates
and time, f + I
a
130
variables
in general,
in classical variables. classical
while
theory require
the more closely specified coordinates,
Thus we may expect
momenta,
and
that the corresponding
systems
imagined
time, 2f + 1 formalism
in
dynamics will be a little more complicated.
Unitary Transformations In the two examples generator
given above
in (4.48) and
G turned out to be a hermitian
this was to be expected. ~' = (I + ic~-IG)~ translated
If ~(x,y)
describes
make no difference
operator.
describes
another
or rotated with respect
identical
the
Let us see why
a certain system
to the first.
to the n o r m a l i z a t i o n
(4.49),
system,
then
infinitesimally
Clearly
this can
of the wave function;
we must
have
I = ; ~'*~'(dx) =I~I + iE~-IG)¢(I + iEh-IG)~(dx) and only if, for all ~,
will we have
= [ ~*~(dx) = I Transformations
of the form
and their defining fundamental
property
eigenvalues, physics
hermitian
motion. laws:
operators
is that they have real with analogues
relation:
by hermitian
corresponds seen,
baryon number,
charge and so on.
a unitary
dynamical
leaves
particles,
electron number,
the hamiltonian of the
statement.
Physics,
is rich in conservation
muon number,
Each of these corresponds
by a
and vice
is a constant
is the converse
of elementary
theory and so provides
represented
transformation
variable
in classical
Thus we have a
variable
if the transformation
interest nowadays
the physics
operators.
to every dynamical
the corresponding
Of more
underlying
A
variables
As we have
especially
are called unitary,
of hermitian
operator
invariant,
(4.46) with G hermitian
is that they preserve normalization.
and so all dynamical
are represented
reciprocal versa.
property
to first order
charge,
hyper-
to some invariance
a clue as to its eventual
of the
structure.
131 Classical
Mechanics
An infinitesimal
transformation
is defined as one which differs
only infinitesimally from the identity
transformation.
the identity is F 3 = P n Q n (it is also F 2 = the same results). We write F3
where
e is infinitesimal,
function of p and q.
P
n
= Pn
e
-
noting
From
~G , 3Qn
PnQn
=
i,
that leads to
- eG(p,Q)
that in the limit e ÷ O, G becomes
(4.35), we have
= -Qn
qn
In Example
, but
-qnPn
~G J ?Pn
-
a
to order e
H(p,q)
=
(4.50)
K(P,Q)
Again to order e, this gives
APn
so that,
= P
Pn
n
3G
= - e
9qn"
Aqn
=
3G
e
(4.51)
9Pn
for any function a ( p , q ) ,
Aa
=
(4.52)
(a, G )
~
where we define
Ca, G) : This combination S. D. Poisson Obviously,
~a
~G
~Pn
~Pn
3qn
of derivatives
was
introduced
by comparison with the foregoing [D,~], although
G = - i~D
expresses
3G
~qn
that
into dynamics
significance,
by saying
generated by G means
invariance
it corresponds
one usually
that
[G,~] corresponds example,
formulas,
by
of a and G.
since it is not D but the hermitian
has physical
the correspondence
As a special
(4 53)
in 1809 and is called the Poisson bracket
to the commutator operator
3a
to i h ( G , a )
of H ( p , q )
under
(4.54)
the transformation
that
(G,H)
=
(4.55)
0
The result of Problem 4.24 shows that if this is so, the numerical value
Problem
of G remains
4.23.
dynamical
constant.
Verify
variables
the correspondence
familiar
(4.54) using some pairs
from classical
of
and quantum mechanics:
132
Pi (linear momentum) n i and pj. Problem 4.24.
and xj, L i (angular momentum)
and Lj, L i and xj,
In quantum mechanics,
d i ^ dt = ~ <[H,~]> for any time-independent operator ~.
dynamical
variable
Show that for any classical =
Show that
a represented
dynamical
by an
variable
a(p,q),
(4.57)
(a,s)
in agreement with the correspondence
Problem 4.25.
(4.56)
(4.54).
(4.57) contains
Hamilton's
equations
as
special cases.
Problem 4.26. (4.53)
Show that the canonical
coordinates
and momenta
in
satisfy
(qm,Pn)
(pm,Pn) = (qm,qn) = O, Problem 4.27.
If a dynamical
=
(4..58)
6mn
a(p,q) is re-expressed in terms
variable
of P and Q we call it A, A(P(p,q),
Express
Solution
Q(p,q))
(4.59)
= a(p,q)
9A/~P and 3A/~Q in terms of the generator of the transformation. Writing
~A BPn and similarly
~a ~Pm ~Pm ~Pn
-
@a 9qm ~qm ~Pn
+
for @A/@Qn, we need Bqm/~P n and 3pm/BPn.
to order c,
Pm = Pm + e
~G
~qm
"
qm = Qm - ~
@G
3Pm
so that, again to order E,
BPm -
6
~Pn
mn
+
e
@2G Bpn~qm"
-
~qm BPn
=
-
e
~2G ~Pm@Pn
-
and BA
~Pn
_
Ba
BPn
+
e
~a
~2G
(~P---m ~Pn~qm
_
~a
_~G
~qm BPn@Pm ~
From
(4.50),
133
or
(4.60a)
t @Pn
~Pn Finish the problem by showing that A
@a_.~_ =
cr @G
Bqn
Problem
4.28.
(4.60b)
a)
~ Bqn"
What transformation,
for an N-particle
system,
is
generated by the time-dependent generator G = Pt - M~ where P i s
the total
momentum,
M is
the t o t a l
mass, and [
locates
t h e c e n t e r o f mass?
Special
Transformations
Up till now we have not specified any generator G(p,q). Let us try some of the dynamical variables we have encountered so far.
i.
G = Px"
We have na
=
E
so that as in quantum mechanics,
Ca, Px)
@a
=
~
~'-~
p generates the linear displacement,
(4.41a). 2.
G = P8 = xpy
YPx ~a ~a__a_ ~a @a Aa = ~(x ~-~y + Px @py - y ~ - Py --)Bpx
corresponding
to a rotation through an angle e.
This
also is
as i n
quantum mechanics. 3.
s = H(p,q)
Aa = ECa, H} = e~
by (4.57).
(4.61)
Thus Aa is the change in a during a short time g, and the
natural motion of a system can be regarded as the continuous unfolding of a canonical transformation generated at each instant by H. corresponds
to Schr~dinger's
This
equation, which in a similar notation is
A~(q,t)
= - i ~ e H ~ (q, t )
(4.62)
in which H is the operator that takes us from one moment to the next.
134
Problem 4.29. generate
Because
the vector A of
an infinitesimal
the Kepler problem
invariant.
find the transformations will
is constant,
that leaves
Choose a single component,
it generates
it must
the hamiltonian say A
x,y,Z,px,Py , and Pz"
in
of
and z You
them of a more general kind than any of the forms dis-
find
cussed earlier, now continue changes
(2.91)
transformation
since the new coordinates
and prove
of variable,
complexities
the invariance
involve
the momenta.
of the hamiltonian
If you
under these
you will have some idea of the mathematical
that lie beneath
our relatively
simple
formulas.
Properties of Poisson Brackets The Poisson brackets
of dynamical
variables
satisfy
the following
relations: (a,b)
(a,b
+ c)
(a, bc)
=
=
in (4.63c) derive
is that if the variables
of
Time-independent
the Poisson bracket
(c,(a,b))
as well
identity. by Poisson
in 1809,
to P,Q by a canonical
(a,b) remains unchanged.
but that it leads to canonical
transformations
are simplest.
equations
Start with
(a,b) of two arbitrary dynamical variables
as functions
of p and q.
~j
Make a canonical
transformation
it only in b:
~Pi + 3P j ~ i ) - ~a__q_ ~b ~P~) . 9Pi (gb ~Qj ~Pi + ~Pj
= (a, Pj) ~~b. + J
(a, Qj) ~~b.
Now let b be the hamiltonian By Hamilton's
easy to
the last takes a little work
from p and q to P and Q, but for the moment make
Ca, b) - 9qi
(the order of factors
and all are trivially
(a,b), established value of
(4.63d)
= 0
theorem we must use the fact that not only is
there a change of variables
expressed
+
p,q are transformed
the numerical
To prove Poisson's motion.
(b,(c,a))
for commutators)
property
(4.63c)
+ b(a,c)
It is called Jacobi's
A remarkable
(4.63b)
(a,c)
that for Poisson brackets
(Problem 4.30).
transformation
+
+
apply to commutators
is correct
except
(a,b)
(a,b)c
(a,(b,c))
All these formulas
(4.63a)
= -(b,a)
H; in the new variables
equations,
(a,H) =(a, Pj)Qj - (a, Qn)P j =
(4.64) it is
K(P,Q).
of
135
But a is arbitrary,
so that if it is now considered as a function of
P and Q we must have ~a
(~,Pj) = ~Qj , Putting these relations
into
in w r i t i n g
theorem.
~a ~b DPj ~Qj
It results
(4.65)
in a convenience
of notation:
(a,b) we need not specify what independent variables
being used, as well,
~p.j
(4.64) gives
~a ~b (a,b) = ~Qj ~pj which is Poisson's
~a
(a, Qj)
and of course
it leads to computational
are
simplifications
as will be seen in Chapter 7, since some coordinates
are
m u c h simpler than others. As an exercise for the reader, Poisson's
Problem 4.31 asks for a proof of
theorem valid for infinitesimal
easily be c o n s t r u c t e d from
(4.60).
Problem 4.30.
Prove Jacobi's
Problem 4.31.
identity.
Show that for any
A(P,Q) and B(P,Q) defined as in (4.59),
the Poisson bracket is invariant under
Problem 4.32.
Show
ical v a r i a b l e s P
n
infinitesimal
transformations.
that if Pn and qn are transformed to new canonQn'
and
(Pm'Pn) = (Qm, Qn) = O, These relations
transformations, w h i c h can
(QmPn) = ~mn
(4.66)
can be taken as the d e f i n i t i o n of a contact transfor-
mation.
Problem 4.33. to
(4.65)
4.5
and
State and prove relations
analogous
(4.66).
G e n e r a t i n ~ Finite T r a n s f o r m a t i o n s If we iterate an i n f i n i t e s i m a l
one.
in quantum mechanics
from Infinitesimal Ones
t r a n s f o r m a t i o n we obtain a finite
How to do this is e s p e c i a l l y obvious
in quantum mechanics,
in
which
f ( x ) = (1 + i~ Suppose we w i s h
to make e cover a finite range X by iteration.
divide the range into n steps, with n eventually which
e = y/n.
Then
f(X)
Lira =
(1 + ix ~)nf(x) n~
infinite,
We
in each of
136
The limit is the exponential,
and even though ~ is an operator we can
still write
f(x)
=
exp
(/_z ~)f(x)
(4.67a)
._~ ~2 + "" .)f(x) 2~ 2
(4.67b)
or
(I + i--X- G
f(X)=
-
Classical contact t r a n s f o r m a t i o n s way,
are iterated
in much the same
though since the identity from elementary calculus
have to do a little more work.
We have, c o r r e s p o n d i n g
isn't there we to an infinites-
imal p a r a m e t e r
Ae(P,Q)
= a(p,q)
- e(G(p,q),
(4.68)
a(p,q))
Let us assume that the finite t r a n s f o r m a t i o n is of the form
Ay
=
a
-
y(G,a)
+ X2Y2(G, (Gja)) + X3Y3(G,(G,(G,a)) ) + ...
where the k's are to be determined. increment of y, we can write
(4.69)
With e as an infinitesimal
(4.52) as
dA~ = -
(G, Xy)
(4.70)
Applying t h i s to (4.69) gives - (G,a) + 2X2~{(G, (G,a))
+ 3X3X2(G,(G, (G,a))) 2(S,(G,(~,a)))
and c o m p a r i s o n of coefficients
= k 2 = ~, so that
.
gives
I
k3
(~I) n
- ~,
...,
kn =
n!
(4.71)
the finite t r a n s f o r m a t i o n can be w r i t t e n as
AX
=
a +
~
E i
(_~)n (~n "a)
(4.72)
n!
(Gn, a ) is the n-times iterated Poisson bracket.
where
Problem
4.34.
Generalize
transformations, analogous
Problem like
+
+ ...
to
4.35.
(4.57).
the infinitesimal
formula
(4.45) to finite
and find an e x p a n s i o n in m u l t i p l e commutators
(4.72). Equation
(4.72) gives the formal s o l u t i o n of equations
It is clumsy but it works.
Show it works by using it
137 to c o n s t r u c t
the s o l u t i o n of a simple harmonic oscillator that starts
at t = 0 w i t h x = a, p = 0.
Problem 4.36. by G = x n ?
What i n f i n i t e s i m a l contact t r a n s f o r m a t i o n is generated
By G = x n"9
Problem 4.32.
The i n f i n i t e s i m a l rotations
(4.42a) can be w r i t t e n in
matrix form as 1 -e
(~) where
= (e
x
i)%)
1 is the unit matrix.
= [i + ¢(I
0 -I
x
0)]% )
Iterate this to find the t r a n s f o r m a t i o n
that gives a finite rotation.
The Parameter
of the Transformation (4.72) with a = Pn or qn
If we apply the finite t r a n s f o r m a t i o n we generate
t r a n s f o r m e d variables
Pn (Y,P,q),
Qn (X,P,q)
w h i c h we shall use to define Pn and Qn as functions of y for fixed initial values Pn and qn c o r r e s p o n d i n g
to y = 0.
From either of
these we can solve for y to get y :
The r e l a t i o n parameter
to P + d P
~(P)
or
~ :
y(Q)
(4.73)
(4.52) gives the change in any function A(P,Q)
is varied.
and Q to Q + dQ.
dA
(4.74)
(A,G)
Let A be either of the functions y defined in (4.73). (y,G)
a parameter
that
P
With ¢ = dY we then have
dy -
Here y is
as the
Let y increase to y + dy and correspondingly,
enters
Then
= 1
(4.75)
our description
see that G is a c a n o n i c a l l y conjugate parameter. sarily c o n j u g a t e c o o r d i n a t e and momentum,
of
the system,
We
They are not neces-
though that is what they
may sometimes be, as in the following example will show.
Example
4.
We have seen above that G = xpy
of the axes in the xy plane.
YPx generates rotations
If the angle of r o t a t i o n is Y, we find
138 PX = Px c°sY - pysiny
X = xcosx - ysiny
(4.76) Py = pxsiny + pyCOSX
Y = xsinx + ycosy
These can be solved for y by writing
X + iZ = eiY(x + iy),
them as
PX + iPy = eiY(p x + ipy)
so that y = -iln(X + iY)
+ const.
Y = -iln[P X + iPy) + const.
or
In terms of the transformed variables
G = XPy
G retains
its form,
YPx
(4.78)
and we can now verify at once, using either of the forms not both!)
that
(4.75)
(4.77)
(4.77)
(but
is satisfied.
Problem
4.38.
Derive
Problem
4.39.
Carry out the differentiations
eq.
(4.78). that verify
(4.75)
in
the above example. Now in (4.74),
let A = H.
leaves H invariant,
If G generates
a transformation
that
we have dH d-~ = (R,a) = o
where
the derivative
on the left means
dH dx T~us H is independent and Q.
We have previously
in terms of P
cannot call it a
it is not in general one of the coordinates
argued that corresponding
nate there is a conserved m o m e n t u m
canonically
turn the argument
there is a conserved
~X
N is a cyclic parameter--we
because
last few paragraphs
8Qn
of y, however y may be expressed
In other words,
cyclic coordinate Q.
~H ~Pn 9H 8P n ~Y + ~Qn
to any cyclic coordi-
conjugate
to it.
The
the other way and show that if
dynamical quantity one can construct
a canonically
conjugate parameter w h i c h is cyclic. These results
clarify the relation between ~ and t.
already seen that the natural the hamiltonian,
We have
evolution of a system is generated by
with time as parameter.
Thus t and H are canonically
139
conjugate. sense of
But they are canonically conjugate as parameters
(4.75) and not as coordinate and momentum.
point of Hamilton's
equations
is to find p and q as functions of t, it
makes no sense to take t as a coordinate. ed carefully,
in the
Since the ultimate
This point should be absorb-
as it has caused a certain amount of confusion in the
literature. 4.6
D e d u c t i o n of New Integrals It is trivial in quantum m e c h a n i c s
that if H commutes with
dynamical variables a and b, it commutes with their commutator, the consequences are profound.
but
In classical mechanics Jacobi's
identity yields
d__ (a,b)
dt
so t h a t
if
important fore out
a and b a r e result,
of lowering
it
order
b))
then
gives
=
(a, Cs, b))
(a,b)
is
also.
a way o f f i n d i n g
of the
equations
- (b,C~,a)) This
is
integrals,
of motion,
a very and t h e r e -
without
carrying
any i n t e g r a t i o n s . One c a n ,
brackets, because
but
initial
of course, after and f o r
instant t o .
Isotropic
k e e p on f i n d i n g
a certain
a system with
integration,
The
constant,
since the
=-(~,Ca,
point
f degrees
integrals
nothing
o f f r e e d o m has
an e n e r g y - c o n s e r v i n g
by e v a l u a t i n g
new w i l l
be f o u n d .
at most
Poisson This
2f constants
s y s t e m one o f t h e m i s
is of
the
An example will show how this comes about.
Oscillator
A soluble example of the consequences
of Jacobi's
identity is a
simple harmonic oscillator
in two dimensions with x and y forces equal,
Fig.
is
4.3.
The H a m i l t o n i a n
1 (px 2 + py 2 ) + ½ k(x 2 + y2) H = ~-~
(4.79)
X
Figure 4.3
Isotropic oscillator with orbit arbitrarily oriented.
140
and from the circular
symmetry we know that the angular m o m e n t u m
L = xpy is constant.
To find another
the hamiltonian
integral,
of two uncoupled
Since each conserves
(Px 2
It is now a simple matter
-
(4.801
we note that H is the same as
oscillators
its energy,
B = 2-~
YPx
of the same frequency.
the difference
py2)
k(x 2
+ ½
to find a fourth
-
in energy
is constant:
y2)
(4.81)
integral by Jacobi's
2 C = (L,B) = ~ (pxPy + mkxy) Continuing,
theorem,
(4.82)
we find
(C,L) = 4B
so that no further new integrals
4kL
(C,B)
and
(4.85)
/17
can be found in this way.
fact found more than the 2f - 1 = 5 independent
integrals
We have expected,
in and
one can verify that H 2 =
B 2 +
%
Thus only three of the four integrals The integral
(4.82)
represents
C2 +
(4.84)
0~2L 2
are independent.
physical
property of the oscillator
that does not depend only on the energies of its component motions. To see what it is, write
the motions
x = asin~(t where
B is a constant
- tO),
Unlike H and B it depends
case the center of force
on the difference
(4.82)
is
in phase.
integral
(or perhelion).
is at the center
is ambigous,
tion will be better.
Then
- to) + 8]
2abm~2cos6
p r o b l e m has a vectorial
center of force to the aphelion of the vector
y = bsin[~(t
phase difference. C =
The Kepler
in z and y as
that points
from the
Since in the present
of the orbit,
the direction
and this warns us that some other descrip-
A glance at the integrals H, B, and C shows
~hat they define a constant
second-rank
tensor
1 Aij = ~-~ pip j + ½ k x . x .J
(4.8s)
141
in terms of w h i c h
(4.84) becomes
AxxAyy
- Axy
2
~2 = T
L2
~2
=
k
(4.86a)
while A
+ A
xx
yy
= H
To relate the tensor to the orbit,
(4.86b)
choose the x axis to coincide
w i t h the major axis of the ellipse and let t o = 0.
Then B = ~/2 and
the oscillator's m o t i o n is given by x = acos~t,
(a>b)
y = bsin~t
and we readily find that A
xx
= ½ ka 2,
A
yy
= ~ kb 2,
The axes chosen reduce A.. to diagonal pal axes of the tensor.
A
xy
= 0
form and are called the princi-
E v a l u a t i o n of H and L gives
= ½ k ( a 2 + b2),
L = kma2b 2
whence a2 = ~I [H + (H 2
-
~2L2)½]
b ~ =Z
When L = 0, b = 0 and the oscillations are linear. m o t i o n is in a circle,
Problem
4.40.
When eL = H the
and no larger value for L is possible.
Find the e c c e n t r i c i t y of the orbit in terms of H and L
and in terms of the components simple form w h e n p r i n c i p a l Problem
4.41.
ian coordinates Problem
4.42.
of A... Show that it reduces to a ~j axes are used.
Calculate the infinitesimal
transformations
of cartes-
and momenta generated by H, B, and C. Calculate the infinitesimal
transformations
of coordin-
ates and m o m e n t a generated by A... ~J Problem
4.45.
components Problem
4.44.
mechanics Problem
Find the P o i s s o n brackets
connecting the various
of A... ~J Solve the isotropic harmonic oscillator
in quantum
and give the d e g e n e r a c y of each level.
4.45.
degeneracies.
Do the same in polar coordinates and again note the
142
Problem 4.46.
Except for scale factors,
the Poisson bracket relations
(4.82) and
(4.83) are the same as those of the components of angular
momentum.
This suggests a way to find the energy levels of the
oscillator
in quantum mechanics.
quantum-mechanical
terms,
Transcribe
the entire d i s c u s s i o n into
find the commutators, which will again
resemble those of angular momentum,
and, noting
(4.84), use the appara-
tus of a n g u l a r - m o m e n t u m theory to find the energy levels and their degeneracies.
Be sure your results are the same as those of the last
two problems.
Problem 4.47. the trace of same.
Show that if the coordinate
A
axes are rotated by
(4.76),
(tr A = Axx + Ayy) and the determinant of A remain the
The trace occurs
in ( 4 . 8 6 ) .
What is the determinant?
Problem 4.48. For the Kepler p r o b l e m in the xy plane, the components A x and Ay of Laplace's vector integral are constant. Does (Ax, Ay) give a new
4.7
integral?
Commutators and Poisson Brackets shared by commutators
and
Poisson brackets that the two have a p r o f o u n d formal parallelism,
It is clear from the list of identities
and
yet it is not easy to see why dynamical v a r i a b l e s
they are so much alike.
In fact,
a and b in the Poisson bracket are functions
n o n c o m m u t i n g p's and q's,
it is not even obvious how derivatives
if the of should
be taken. With commuting variables,
~l'
"''' ~K the ordinary partial deriva-
tive is defined by aa Lim 1 a~. = e÷0 ~ [a(~l . . . . With
noncommuting
variables
~i + e . . . .
we d e f i n e
a more
~K) - a(a)] general
derivative
(4.87) s
involving an arbitrary function o(~), aa {c} = Lira 1 [a(al, aa. e+ e Bloore calls
"""
ai + ec . . . .
this a Fr&chet derivative.
~K ) - a(a)]
If c commutes with all the
~'s, then
aa.
s
F.J.Bloore, J. Phys. A. Math.,
(4.88)
aa.
Nucl.
Gen.
6 L7
(1973).
143 Evidently,
the Frechet
derivative
BCa+b) {c} Ba {c} + 3b De. - Be. ~
satisfies ~
{c},
We shall now prove a remarkable
theorem
~a.Ba{ [ a i , b ] } The proof
is by induction.
We write
the usual rules
that
~a
{c}
involving
commutators:
(summed)
= [a,b] the general
sum of pieces
each of which is of the general
and we assume
it to have been proved
another
{c} = y
@e.
(4.89a)
function a(~)
as the
form y a 3 ~ 2 ~ 2 ~ l ~ 3 ~ l a 4
for one such piece,
a'.
....
Now add
factor of e. on the left. J
(eda ') Be.
{[ai,b]} = Lira I
~+0 i (~j + e[aj,b] - ~j)
= [aj,b]a' + aj[a',b] Thus if (4.89a)
holds
the theorem holds is complete
trivially
and the theorem
To establish replacing
for a', it holds
a' + aj @a'
~
{[e~,b]}
= [aja',b]
for
aja'
Now set a' = ~k;
as in the steps just taken.
The induction
is proved.
the relation with Poisson brackets we rewrite
(4.89a)
i by j, b by ai, and a by b:
~Bb a.
{
[=j"ai]}
=
[b,ai]
=
[~i "b]
J or
aa.~b { [ a i , ~ j ] }
(4.89b)
J Now substitute
this into the left side of (4.89a):
@a.
~
{[ai,aj]}}
=
[a,b]
(4.90)
J This identity will give us a relation between brackets. and ql'
Divide
the dynamical
variables
commutators
and Poisson
a i into two sets, Pl'
"'" Pf
... qf, with [qm,Pn ] = i~mn
(4 91)
144
As ~i and aj go through all the p's and q's we get
3.aBqn{8BP~ {i~}}
BPnBa'-q--{ @k
+
[a,b]
{_i~}} =
or by (4.88),
[a,b] where the derivatives
i~ ~a
3b
~a {~b } @Pn @qn
(4.92)
of b are the usual ones defined
in (4.87).
The
quantity on the right is a version of the Poisson bracket which is appropriate version
to noncommuting
if the variables
variables
commute.
and which reduces
similarity between Poisson brackets
and commutators.
If we were to overlook all questions side of (4.92), derivative,
the Frgchet derivative
of commutation
would become
which would amount to putting
there are additional
relations.
[a,b]
on the right
the ordinary
an = sign into
terms left over proportional
the omitted commutation
to the e-number
This is the origin of the formal
(4.54).
to ~, ~2,
But
... from
We have therefore
= i~(a,b)
(4.93)
+ O C~ ~)
From a formal point of view, nothing more can be done to establish the ally,
correspondence
between classical
the correspondence
and quantum theories.
(4.54) can be obtained only by keeping
commutators
(4.91)
cummutators
in going from
procedure.
Since there is no universal
order the numbers
in going from
to (4.93).
physics
have taken pains to symmetrize
Problem
recipe
that
tells
should be written
the quantity A in (2.107)
we recommended
cummutes with the hamiltonian. scription
This is not a consistent in what if they are
It is for this reason that although we in Problem 2.70
verify that it is in fact the correct requirement
all all
the omitted terms of the order of ~ must be
supplied by other arguments. make it hermitian,
(4.90) to (4.92) and ignoring
(4.92)
of classical
to become operators,
Specific-
of hermiticity
into operators
Except
that
and therefore
the reader
form by seeing whether in the simplest
it
cases the
is not enough to make sure that the trans-
is correct.
C a l c u l a t e [ p 2 , x 2 ] by t a k i n g F r ~ c h e t d e r i v a t i v e s and a l s o by e v a l u a t i n g t h e commutator d i r e c t l y . Compare the r e s u l t s w i t h 4.48.
i~(p2,x2).
145 4.8
Gauge Invariance It is perhaps
surprising
that the Hamiltonian
in an electromagnetic
particle
than the fields. particles,
The fields
whereas
field involve
are defined
the potentials
tional definition
them or the forces they exert. particles, well,
potentials
defined by the sources When one studies
pattern,
is affected by the existence
understand,
that produce
the laws of motion in
for example,
of a potential
rather than fields
though in classical
The potentials
opera-
as we
fields govern the orbits of
through which the beams pass is entirely rence of potentials
7.8), and indeed,
govern not only orbits but phase relations
so that an interference
phases,
do not have any such obvious
one finds that whereas
quantum mechanics
for a rather
in terms of forces on charged
(but see Park 1974, Sec.
shall see, are not even uniquely
equations
the potentials
Thus the occur-
in quantum mechanics it remains
on
even if the region
field-free.
mechanics
A and ¢ determine
as
which depends
is easy to
somewhat
the fields according
opaque.
to the
relations: ~A
B = VxA
Now introduce
new potentials
A'
where ~(~,t)
and E determined
according
to
@' = @ - ~
(differentiable) are unchanged;
The latitude
but it introduces mation
of definition
potentials
into something
The classical
as we shall show,
(4.95) with arbitrary
of a simple static electric
field at all,
equation
by the forces
is often convenient
some complications,
of A and @ via
Such a change
The values
they
for computation, for transfor-
I can convert even the
and/or magnetic
field,
or no
time-dependent and elaborate.
of motion, m~ = e(E + v x B )
is gauge-invariant Schr~dinger's
because
of 8
this is what is meant by
saying that A and @ are not uniquely determined exert.
(4.95)
function.
is called a gauge transformation. by (4.94)
(4.94)
~X
= A + VX,
is an arbitrary
of the potentials
E = -V@ --~-~.
it depends
only on the fields,
(4.96)
but in
equation
iH ~
1 (p - eA)2 + e@]~ = [~m
(4.97)
146
the potentials occur explicitly.
The equation can, however, be made
gauge-invariant like (4.96) if we introduce a change in the (unobservable) phase of the wave function connected with the gauge transformation (4.95): 9'
:
e~P~,
e
p
= ~ X(x,t)
- ~)e~P~
(4.98)
Then, in (4.97), i~
Similarly
~~ t
-
e~'~'
= i~ ~t e ~
-
e(~
=
e ip
-
(~
=
eiP[i~ ~
[i~
~
~p
TT+
~h
e¢ - ~ T T ) ~ ]
e~]
-
(Problem 4.49) (p
-
eA')~'
=
e~P(~
-
(4.99)
eA)~
from which follows by simple steps the gauge-invarian~e of (4.97) (Problem 4.50).
Pauli has called the transformations
(4.98) and
(4.95) gauge transformations of the first and second kinds respectively.
It is clear that the physical results of the theory should be
independent of the choice of potentials, but how this is to be achieved raises some interesting questions•
Problem 4.49.
Prove (4.99).
Problem 4.50.
Prove that the complete Schr~dinger equation (4.97) is
gauge invariant.
Problem 4.51. Prove that the eigenvalues of the equation H~ n = En~ n depend on the gauge chosen but that those of (H
-
e¢)~n
:
en~ n
are gauge-invariant.
Perturbation
Theory
Let us start with a system bound together by its own internal forces and described by a hamiltonian ^
~o : #-#-~m+
eV
(4.lOO)
147
written
in one-particle
are not restricted immersed
form for simplicity
in an external
electromagnetic
and that we wish to study the effects to be considered
wave equation
in cartesian
=
act on A,@ but not on W.
-
+
are considered
to split H into an unperturbed
~, this is not a gauge-invariant
The
The
+
(4.101)
equation. as a perturbation,
it is
part H 0 plus a perturbing
but since it is ~ - eA that
potential,
A,@,
coordinates,
it is a gauge-invariant
If the external potentials natural
this system is
field B,E with potentials
=
and, as we have seen,
our considerations
of this field on the system.
gauge transformations is now,
though
Suppose now that
to one particle.
is gauge-invariant
separation.
and not
The fact does not mean
that we will get wrong results, since nothing has been falsified or omitted,
but it affects
the approximations
of interpretation.
difficulty called @n"
Perturbation
Suppose
theory calls
we make and it leads to a
the eigenfunctions for an expansion
of H 0 are
of the form
(4.102)
= X On(t)@ n with the o's determined by an approximation o's have a physical
procedure.
Further,
IOnl 2 is the probability
interpretation:
the system will be found in the nth quantum state of the
measured
unperturbed
system.
Now if a gauge transformation
is carried out and ~ becomes
the o's will be entirely different, their physical @n in (4.102)
interpretation
~n (See Problem
cannot be kept.
4.51.)
equation.
of gauge.
Such an equation
=en¢n"
only if @n satisis
(4.103)
~ = ~ - e¢
that the phase of @n will vary in the
This means
On the other hand,
into the "unperturbed"
accomplished
the perturbing
equation,
potential
by A is
and we do not seem to have
much.
For a complete solve the auxiliary
and general
discussion
problem of calculating
sufficient
approximation
Often this
is not necessary.
we can choose
that
We can keep it only if
same way as that of ~ and the o's in (4.102) will be unaffected now built
e~,
and it would at first appear
also changes by e c~, and this happens
fies a gauge-invariant
changes
the
that if
it would now be necessary the solutions
of
(4.103)
and then using them for the expansion If the external magnetic
to
(4.102).
field is zero
a gauge in which A = 0; in fact we normally
to
do this
148 without
thinking
the effects
about it, putting the entire interaction
of the magnetic
field are small,
can achieve
the same simplification
by choosing
numerically
small and then omitting
it.
system is immersed interaction, -eE.r,
in electromagnetic
to good approximation,
the potential
into e~.
Thus
a gauge in which A is
(Problem 4.61)
radiation
if the
one can put the entire
into e@, which comes out to be
energy of a particle
in an electrostatic
That this special choice of gauge is the only one consistent the interpretation
If
even though not zero, we
of ICn[ 2 as a probability
field. with
was pointed out long ago
by Willis Lamb. ~ Problem
4.52.
mechanics equation
Show that when a dynamical variable
involves potentials of motion
which depend explicitly
i
dt
4.53.
mechanics
on time,
~
: <~[~'~]
+ ii
Show that the corresponding
(4.104)
>
relation
in classical
is dn
~
d'-{ = CQ,H) + @--~
Problem
4.54.
complete
Show that in classical
hamiltonian
which depends
and quantum mechanics,
is simple,
gauge-invariant,
and physically
reasonable
in in both
and quantum mechanics.
4.56.
Show that the operator
~' : eiV~e - i v has the same matrix elements transformed
state
(4.98)
(and expectation
Show that p' = p - eVl.
Problem
4.57.
Just as a canonical
leaves the Poisson brackets transformation
(4.106) values)
that the corresponding
state.
s
the
in (4.101) has a time rate of change
Show that the time rate of change of h defined
4.55.
(4.103)
Problem
H defined
(4.105)
on the gauge.
Problem
classical
its
is d<~>
Problem
~ in quantum
of the form
transformation
unchanged (4.106)
(Poisson's
to new p's and q's theorem),
leaves invariant
W.E.Lamb, Jr., Phys. Rev. 85, 259 (1952). of Phys. 101, 62 (1976).
in the gauge-
~ had in the original
show that a
the commutation
See also K,H. Tang, Ann.
149
relations
of the transformed
if ~ is a function
Problem
4.58.
formation
operators.
of operators
Consider
(Note that this is still valid
as well as coordinates.)
in hamiltonian
mechanics
the canonical
trans-
generated by
F3 (P,Q, t ) = PnQn -el (Q, t ) Find the new m o m e n t u m tion between
transformations
Problem
4.59.
and the new h a m i l t o n i a n
this transformation (4.95)
involving
Show that
A = asin(k.r the magnetic
on the rela-
and the gauge
the field variables.
if the external
- ~t) ,
field is a plane wave with
@ = bsin(k-r
field B is automatically
ality of E requires
and comment
of particle variables
- ~t)
transverse,
(4.107)
while
the transvers-
that k.a
b ---~T~ Show that B is now p e r p e n d i c u l a r
Problem
4.60.
Find the X that transforms A' = a'sin(k.r
where
the new amplitude
Problem
4.81.
to E.
- ~t),
i.e.,
in the preceding
(Note that A" is longitudinal depends
- ~t),
@"= -E(t).r
potential
amounts
to setting
Problem
4.62.
original
Note also it isn't,
For evaluating matrix
in which
The magnitude
of light a thousand
elements,
the wave function differs of A" is therefore
than that of A' by a factor of kr, and the "optical with the w a v e l e n g t h
k is small.
and @"is gauge-invariant.
on time.)
r gives the size of the region from zero.
problem,
generated by X = -A',r and show
that though @"may look like an electrostatic
significantly
that the spatial deriva-
are
A" = - (a'.r)kcos(k-r
since E in general
(4.108)
to k.
field implies
Carry out the gauge transformation
that the new potentials
¢' = 0
a' is p e r p e n d i c u l a r
A weak magnetic
tives of A are small,
this to
smaller
approximation,"
times the radius
of an atom,
this equal to zero.
Show that A" gives
the same magnetic
A. How does it happen that an A" which
field as the
in the situation
150
mentioned above is hundreds
of times smaller than A describes
the same
field? Problem
4.63.
Show that in addition to energy, linear momentum,
angular momentum,
and
the equations for an n-particle system have a
further integral C =
n ~--I miri
n - t i~= =1
and comment on the physical meaning of the constancy of C. Problem
4.64.
Show that
if the forces between particles
in an
n-particle system were inverse-cube~
there would be two more general
integrals of the equations of motion
(Wintner 1947, who says that
Jacobi knew them) D = 2Ht -
n ~I i=
ri'pi
a nd F = Ht 2 -
n ~ i=1
(tri'P i - ½ miri 2)
CHAPTER 5 The principal thing
THE HAMILTON-JACOBI THEORY
object of classical
is at time t; that is, to find a set of
object of quantum mechanics physics,
but you can calculate
Hamiltonian
dynamics
differential
find
equations.
since ~ has a value
everything
mathematical
descriptions
are so wide
partial
familiar
equation
equation
of Lagrange
matical
is known.
necessary
the subtler question Further, to problems
In Chapter
is known,
answers
5.1
The Hamilton-Jacobi
sketched
as in Figure
W is evaluated
difference
and a study of the
above as the solution
in the sense that if a wave function values,
etc.,
the desired answers Classical mechanics
is then easy. theory;
is more specific;
the
We shall find that exactly soluble problems and quantum theory,
in Chapter
but
is much
7.
E~uation
5.1.
qt space of Lagrange, and paths can be We have already
seen in (4.27)
that if the
for a natural path and then compared with
for a slightly different path running between end points
when the
should be easy to find.
have to be made quantum mechanics
We are once more in the function
if the eikonal
will do much to illuminate
These matters will be discussed
the value
assumptions,
about equally hard in classical
that when approximations easier.
differential
(1.31), which was
the same thing in classical
a catch.
1.3),
came out of the
relation.
of expectation
are more detailed.
are usually
equation
function was characterized
is solved,
is, however,
equations
to
(Sec.
will help to explain the mathe-
for the arguments
We have a right to expect equation
theorems
comes a partial
the two theories,
the calculation
the two
6 we shall see how to recon-
under suitable
in quantum mechanics,
it solves
between
It is now our task to show that out
and Hamilton
of their physical
the wave
and
~(q,t) that are the same as the
These arguments
relation between
assumptions
eikonal
for
that is the same as the eikonal
struct quantum mechanics,
The equation
that they seem to belong
differential
dynamics.
from wave optics.
equation
Lagrangian
The differences
but in Ehrenfest's
ordinary
of Newtonian
of the dynamics derived
of ideas,
differential
equations
is in the sense of classical
for every q and every t. equation.
for example,
qn(t). The principal ~(q,t). From
all there is to know.
differential
universes
every-
q(t) and ~(t) or p(t) by means of ordinary In ~(q,t), q is in no sense a function of t,
is a partial different
is to find where
is to find a wave function
this you cannot calculate where
There
dynamics
in a slightly different
time,
slightly
the change
in Y
152 is given by t
6W
=
[pn6qn
-
(4.27)
H6t] t o
If the variations
6q and 6t are zero at the initial point A, then W is
8
&t 0 Figure 5.1
t
Adjacent paths in qt space leading from a common initial point.
defined by this relation point.
Varying
as a function of q and t at the arrival
the q's and t one at a time gives @W
3qn
and substituting
~W
Pn"
~t
the first of these ~W S(~,
This
is the Hamilton-Jacobi
Cartesian occurs
coordinates
of a de Broglie wave;
3W + // = 0
equation. ~
form.
different here
called it the principal
(5.1)
into the second gives
(5.2)
We have seen it before,
for a single particle,
in a more general
meaning was entirely
q,t)
- H(p,q,t)
The equation in Chapter
i.
in (1.31).
Here
in
it
is the same but its There W was the phase
it is the action integral.
(Hamilton
function.)
i This equation was first given by Hamilton, Phil. Trans. Roy. Soc. (1834), II, p. 247, a year before the paper containing what are now called Hamilton's equations. Hamilton's discussion, however, was somewhat more involved than that given here, and it was clarified and generalized by Jacobi in his lectures in the winter of 1842-43 (see Jacobi 1869). For somewhat condensed discussions of the theory's development see Dugas 1955 and Lanczos 1970.
153
Now we must given later,
start integrating.
the energy
If, as in the examples
is constant,
to be
we can do one integration
in (5.1)
immediately:
W(q,t)
S(q)
=
where E is the constant numerical called the c h a r a c t e r i s t i c
Hamilton
-
(5.2)
dependent
and time-independent
and
(5.4)
have solved for S(q). (4.27)
and
to)E
-
(5.3)
value of the energy,
=
and S, which
satisfies
function,
~S H(~-I, q) o~ Equations
(t
(5.4)
E
are the classical SchrDdinger
analogues equations.
How do we find useful
results
of the timeSuppose we from it?
From
(5.3),
6S = 6W + 6[(t - to)E ] = pn6qn
E6t + E6t + (t - to)6E
or
(s.s)
~S = pn6qn + (t - to)6E so that S may conveniently (The t r a n s f o r m a t i o n familiar
in thermodynamics,
already encountered Sec.
be regarded
from W(q,t)
4.1.)
From
as a function of q and E.
to S(q,E)
is of a type especially
called a Legendre
transformation.
in going from L(q,~)
an example
to H(p,q)
We have in
(5.5), we have t - t O = 9S ~--E "
Pn = 9S ~qn
(5.6a,b)
The first of these shows the rather curious way in which on~ deduces the temporal behavior
of a system from the timeless
given by solutions
(5.4).
(1.32).
Example
of
description
The second we have already seen in
How the procedure works can be seen from a simple
I.
Simple Harmonic
Oscillator
2m
so
p
=
~~s -E =
[~m(E - ½ k x 2 ) ] ~
example.
154
Thus
S = (mk)½ I [-~
- x2]~dx
(5.7) = ½ (ink) ½
This has been written sketched
in Fig.
looking at S.
{x(
- x2) ½ + -~- s i n -
[(
)½x]} + c o n s t .
down to show how the solution
6.2b.
Actually
Differentiating
looks;
one is not normally
under the integral
it is
interested
in
sign gives
dx
~E~-~s = t -
t o =
(~) ~ (s
-
~
kx2) ½ X
= (~)½ s i n -1 (2E/k)
giving
t(x,E).
The inverse
X = (~½ As promised, integral
=
k/m
point of the solution was finding
(5.6b)
is familiar.
form in (I.Ii),
with a quantum
~2
Finding x(t) from a knowledge
that gives S.
part of the phase
is
sin~o(t - t o )
the critical
The relation Cartesian
%
We encountered
the
of S was trivial.
it first in
in which s = ~-IS is taken to be the spatial
of a de Broglie wave,
and the momentum
associated
is p = ~k = VS
The meaning
of (5.6a)
also becomes
clear if we think of ~ - I W as the
W is a function W(E,q,t).
entire phase of the wave.
a localized wave packet we superpose waves belonging values
of E as in (1.20)
and Fig.
In order to make to a range of
1.2:
I
ei~-iW(E,q,t)A(E)dE
where A(E) has its maximum at E. condition
for constructive ~W
a~ -
As in the earlier discussion,
interference o
(~
is =
~)
the
155
and with
(5.3)
realized
that ~-iw
trivially. mechanics assumption
this is the same as
What allows
is a phase
(5.6a).
Thus as soon as it is
the relations
is remarkable
(5.6) come out almost
is that the structure
these relations
of Newtonian
to be derived without making the
of a wave.
Problem 5.1.
Figure
around a cylinder.
5.2 shows a pendulum made by wrapping
a string
Find the period.
I
Figure 5.2
5.2
Step-by-Step
Integration
The main computational constants
of the motion
To illustrate Problem 5.1.
of the Hamilton-Jacobi
strategy
in dynamics
When E is constant,
s
=
w
is the use of
in order to simplify calculations.
see here how to use constant momenta coordinates.
Equation
+
E(t
-
and their conjugate
the integral
to)
=
ftI
5
+
expression
We shall
cyclic for S is
~)dt
t
= it (pnqn - H + E)dt to or s
=
IiPndqn ~0
along a natural path on which p and q both depend on t and H has
(s.8)
156
the c o n s t a n t accordingly
v a l u e E.
If n o w ql'
constant,
'''' qk
we can c a r r y
are c y c l i c
and PI'
out the c o r r e s p o n d i n g
"''' Pk
integrations
at once,
S(q)
Call the i n t e g r a l
and w i t h
=
E n=1
(qn
- qno)Pn
f
J n=k+l Pndqn
+
It is
Sk(q).
qo
Sk = S -
k E n=1
(t - t o ) 6 E
-
(qn
(5.5) we have
k ~S k =
~ n=1
pn6qn
+
n=k+l
pn6qn
+
(t - t )6E o
S k is a f u n c t i o n
together with
k
~ n=1
f
Clearly,
(5.9)
qno)Pn
-
pn~qn
k Z n=1
of the e n e r g y
the v a r i a b l e s
qk+l'
...
-
E n=l
(qn
qno)~Pn
Cq n - q n o ) 6 P n
and p
qf.
, "'"
Pk'
all c o n s t a n t s ,
It s a t i s f i e s
3S k (qn - qno )
n = 1,
..
(5.10a)
k
~Pn ~S k ~qn - Pn
~S k ~--~-- =
Example 2.
t
n = k+1 . . . .
-
f
to
(5.i0c)
The K e p l e r P r o b l e m
W i t h ql = 8, q2 = r, 2 Pr
H =
2m
(5.1Oh)
L 2
+ 2-~-~ - ~r
P8 = L = const.
~SI =
sI :
I
T21½dr
[~m(E + ~) - V ~ -
157 Everything
flows
from this
integral.
To find the orbit,
we have
dr
@ - @
. . . .
o
~S t
-
t
[compare w i t h the tables.
and
first
r2]L2
dr =
(5.11b)
m
[2m(E
~E
(2.72) The
r2[2m(E + ~)
1
= o
(5.11a)
L
8L
(2.70)].
does n o t
+
Both
~r) - L2]½r 2J integrals
involve
can be f o u n d
t and is the e q u a t i o n
in all for the
orbit, e - e
w i t h e as in result.
= sin_ I r - a(1 - e 2)
= sin_ I r - L 2 / m y O
er
(2.74b),
The s e c o n d
~(t
er
and this, w h e n
equation
- t
solved
) = sin -I r
- a
[e 2 _ ( r
a~
orbital
is u s e f u l
to r e a d the S o l a r
of course, have
be s o l v e d
rederived
Example
3.
Here
to p r e d i c t
Kepler
Problem
the r e l a t i o n
-
where E T
includes
formulas
in w h i c h E did n o t
the rest
velocity
the p o s i t i o n s
results
System
as in
(2.81).
This
like a c l o c k but
form.
less
It cannot,
At any rate, we
in a few lines.
energy
Relativity and m o m e n t u m
V(r)] 2 = c 2 p 2 + energy mc 2 .
is t a k e n from
(mc2) 2
(4.19)
(5.12)
For c o m p a r i s o n w i t h e a r l i e r
i n c l u d e mc 2, we shall w r i t e E T
o r b i t s E is a g a i n n e g a t i v e . )
p2 = 2m[E(l
/
of the p l a n e t s .
in c l o s e d
in S p e c i a l
between
[E T
(For p l a n e t a r y
angular
to give r(t)
our e a r l i e r
- a} 2 ]½ a
~
w h e r e ~ is the a v e r a g e if one w a n t s
is our e a r l i e r
gives
0
so if one w a n t s
for r,
= E
+ m c 2.
This gives
E E + 2--~-~T) - (1 + m-~2)V( r)]
1 + ~2
V(r) 2
(S.13a)
158 By comparison,
the n o n r e l a t i v i s t i c version is
p2 = 2m[E - V ( r ) ] so that in special r e l a t i v i t y
(5.13b)
it is as if the energy had been increased
to = E(1 + E/2mc 2)
Es.r. and V(r)
(5.14)
had been changed to
(1 + mczE--~r)V ( r )
Vs.r.(r)=
The rest of the w o r k proceeds
2mc21 V ( r ) 2
as in Example
(5.15)
W i t h Y(r)
2.
= -V/r, we
have
p r 2 = 2re[E(1
E r + 2--~Ec2) + Z~] - L2
-
r2X 2 / c 2
y'
E = (1 + m--~-f) Y
(5.16)
The orbit is
dr O
-
=L
@
r2[2m(Es.r.
0
The integral
+ y ' r - 1 ) _ (L 2 - y 2 / c 2 ) r - 2 ] ½
is no more difficult than that in C5.11a), L2
--
and gives
.~2d--2
y(m + E c - 2 ) r =
(5.Z7a) 1 -
es.r.
Sin[(1
-
-L-2Y o - ~2 ) ½ ( O
-
@0 ) ]
where E e
2 = I + s.r.
These are of the same form as
(L 2 - y2c-2) s.r. (m + Ea-2)2¥ 2
(5.17b)
(2.74), w i t h only slight differences
the dependence of orbital and dynamical p a r a m e t e r s
except for the
occurence of
sin[(/
y2 )%(8
L2c 2
- @ )] o
-~ s i n [ ( /
-
y2 2 - - ~ - ~ ) (8 - @o)
W i t h this, r comes b a c k to a previous value after one cycle when 2
(1 - ~ 5 2 - ~ c 2 ) ( e has increased by 2~.
- eo)
That is, the increase
in @ is
in
159
2
AO : 27(I + ~ )
and the orbit precesses
forward through an angle
~y2
27
[email protected]. = L 2 c 2 in each revolution.
In planetary
6es.r.
If the relativistic out of special
Starting
5.2.
equation,
-E
e2
dynamics,
with E = - G M m / 2 a ,
and Newton's
from
the nuclear
correction.
(5.19)
interest because
Problem
5.3.
is 6 times as great.
(5.12), write down a relativistic and exhibit
Show that there are no stable (These results
wave the
is states when
are at most of methodo-
they ignore the effects of spin.)
A ball of unit mass rolls under the action of gravity
(take g = i) inside a cup whose axis is vertical surface
in this way
law, this would be the preces-
The true precession
charge z > 137/2.
logical
this is
= (I -TGM2)ea c z
derive the energy levels of hydrogen,
relativistic
(5.18)
me2
theory of gravity were put together
relativity
sion of planetary perihelia.
Problem
~ 1 -
is of the form z = f [ r ) .
and whose inner
Show that the hamiltonian
of the
system is
H =
Problem
5.4.
Suppose
Hamilton-Jacobi
5.3
Interlude
½ ( pr2 + 1 + fr 2
the surface
on Planetary Motion
Einstein calculated
fr
is a paraboloid, the motion
=
~¢~r
f = ½ Kr 2
Use the
for small values of r.
in General Relativity
corTection
from the general
of planetary perihelia.
roughly analogous
) + f
r2
method to analyze
The first observable precession
po a
to Newtonian
dynamics
that
theory of relativity was the The calculation proceeds
to our earlier treatment
in a way
of light in Sec. 1.6.
There
we set the speed of light equal to c by setting d82 = 0 and then determined
the path by Fermat's principle.
In (4.11) we saw that the
160
variational
principle
for the motion of a free particle was written
in terms of the invariant gravitational considered invariant
ds.
In general
force introduced
to be moving
as such; a particle
to (4.11)
with ds as already given in Chapter
from
expression.
=
(I - 2 l r ) - I d r
The equivalent
because
of the negative
geometry
geodesics
let the radial coordinate
r
(1.54)
condition
in two dimensions
(5.20)
are
is the great
are harder to visualize the d82 between two
negative,
or zero.
(1.54) leads to an intractable
ducing a new radial coordinate
transforms
are usually more convenient
sign in the metric:
points may be positive,
Solution using
are those
of space and time
surface that is the shortest path between two
In space-time
Instead,
expressions
the coordinates
A familiar example
circle on the earth's
(1.54a)
2 - r2de 2
(1.54b)
Curves that satisfy the extremal
points.
neighboring
(S.20)
GM/c 2
(1.54) by assigning
called geodesics.
is
: 0
in some other way, and some assignments than others.
The only
i,
- 2k/r)dt 2 -
k
derivable
in its orbit is
in flat space-time
6 I d8
or any equivalent
there is no
freely in a curved space-time.
law that reduces
d8 z = c 2 ( 1
relativity
into what
in (1.54) be called r'.
r according ,
integral
= r(1
(Prob. 5.5). Intro-
to (5.21)
+
is called the isotropic
form, which will
be easier to handle: d82 h(r)
= (
= h(r)c2dt
I - h/2r ] 2 I + k/2r" "
6 I mc{c
the unnecessary -
[he 2
We also abbreviate
=
(I
+
)
(5.22b)
theory will become as clear constants
of (4.10) and write
k(r 2 + r2@2)]½}dt
in which the integrand now functions I.
(5.22a)
~ k(r)
So that the relation with nonrelativistic as possible we introduce
r2d@ 2)
2 - k(r)(dr2+
as a Lagrangian
the square root by R:
= 0
(5.23)
and will be called
161
I = mc(c
R2
- R),
=
~°2
_
~(~2
+
r2~2)
(5.24)
The m o m e n t a formed from I are ko Pr
•
kc
= -R- m r ,
= L (const.)
mr2~
PO = T
W i t h these we define a H a m i l t o n i a n in the usual way H = ~
R
L2 (pr 2 + ~-f) + mcR - m o 2 = E (const.)
E x p r e s s i n g R in terms of the m o m e n t a gives R2
R2 = ~ o 2
L2
~(Pr2
+~)
from w h i c h we can solve for R and w r i t e H as 2
H
=
[hc2(m2c
L2]]½
Pr 2 + ~
+
kr2~ J
mc 2 =
(5.25)
E
or 2
(~ + mc2)2
=
Pr + -V-
hc2(m~o2
L2
~-~
+
From this, E pr 2 = 2m~ ¼ (i + 2-~) w h i c h is to be compared w i t h
+ m~c~k(¼ - :) - L~ a
(5.16).
(s26)
In the first term of (5.26) we
expand k h -I t o the first power of k and in the second, where the c o e f f i c i e n t is m u c h larger, we keep two powers: -1< h = l + r i.x.~ +
..
1 - 1)
k (~
""
=
2k61
r -
+
3k
~r
+
""
")
and find E
pr 2 = 2m[E(1
This
can be written
in terms
e x p r e s s e d in the style of
Vg.r" The o n l y the
significant
correction
predicted
is
4E
L2
of a general
relativity
+ $--~-E) + (1 + m-~y) ~] -
difference
perihelion
2/°2
(5.27)
potential
(5.15),
= ( / + ~4E -~)V(r)
6 times
r 26Y
from (5.16)
as great
precession
3~ V ( r ) 2 - me---
is
as that therefore
is
in the
of special
(5.27a) last
term,
relativity.
where The
162
6~GM
[email protected].=
(1
-
e2)ac
(S.28)
z
We have already given an estimate of this precession
for a nearly
circular orbit in (2.88). The experimental situation
is best for Mercury,
for which a is
small and e = 0.2, a relatively
large value.
is 43.03 seconds of arc per century.
Here the predicted value
The observed value must be
picked out from among a number of much larger effects. 2 In fact, the total precession observed is about 5600", contributed mostly by the precession planets
(Sec.
residual 41", 3
of the equinoxes 7.6).
discrepancy
were concerned
1.01±0.02
times the theoretical Problem
metric
value computed value.
that in principle
to be further improved.
The present
a small planet
smaller than Mercury's.
from astronomical
data ~
is
Radar observations~, 5 yield allow the Newtonian
"observed"
calculations
value is 1.005±0.02
value.
Set up the same calculation
5.5.
there was a
theory of about
enough about it that they had
with an orbital radius
times the theoretical
more exact distances
and Newtonian
the law of gravity or inventing
(1979) "observed"
of other
paper was published,
between observation
(to be called Vulcan) The best
8.3) and the influence
When Einstein's
and astronomers
considered modifying
(Sec.
using the Schwarzschild
(1.54), and show that it leads to a more difficult
Show that the substitution
(5.21)
integral.
leads once more to the results
of
this section. Problem
Attempting
5.6.
to make a field theory of gravity by rough
analogy with the electrostatic
field, show that in this theory the
energy density of the gravitational
field is negative.
that E < 0 in (5.27) and that all energy produces we can understand Interpreting
qualitatively
the correction
the corrections
Problem
5.7.
Evaluate
field,
to E and y in (5.27).
in y as a correction
of the sun, M@, how large is the correction?
Remembering
a gravitational
to the effective mass
(MO = 1.99 × 1033 gm.)
the planetary precession
by a perturbation
calculation. 2 G.M.Clemence,
Revs. Mod. Phys. 19, 361 (1947).
3 The generally accepted value was that of S. Newcomb, Wash. Astron. Papers 6, 108 (1898), who gave 41.24" ±2.09". The effect had first been established as 36" by U.J.J. Leverrier in 1859. I.I. Shapiro
in Hegyi 1973.
s I.I. Shapiro, G.H. Pettengill, M.E. Ash, R.P. Ingalls, and R.B. Dyce, Phys. Rev. Lett. 28, 1594 (1972).
D.B. Campbell,
163
The precession
Solution:
the expression
comes only from the last term in (5.27),
for the characteristic
S =
[2m(E
+ ~)
- ~
=
I [2m(E
+ ~)
-
function
+ r~]½dr
~]½dr
~ -
+ ~
2
6X2
r212m(E
EI
~2 ~)Y" - ~L2]½ +
This is to be integrated between the two turning points, the orbit.
Although
and the integrand integral
the square root vanishes
of the second integral
nonetheless
converges.
=
-~[
The integral
my
+
z, -
can be done by complex
Integration
-
~S
~
~-E-E] +
Problem
5.8.
precession
"".
2~ (I +
=
E ~-g-T)
is 6e
from which follows
over a complete
angle is @ =
Thus the precession
all around
at the turning points
(_2ms)½
and the corresponding
=
~c
(s.29)
~-£
the usual result.
Show that if the perturbing
energy is C/r 3, the
is 6@ = 6 ~ C m 2 y L~
(We shall use this result later, Fermat's
...
is infinite there, the
methods (Born 1927) or taken from tables. cycle of r gives s
so
is effectively
in Sec.
(s.3o) 8.5).
Principle
In Sec. 1.6 Fermat's deflection
principle
of least time was used to find the
of light by the sun's gravity.
this principle
governs
We can now easily show that
the orbits of particles with vanishing
rest
mass. Taking the lagrangian
of the particle L = mc(c
- R)
to be the integrand
in (5.23), (5.31)
164
with R as in (5.24), we find by the usual procedure
a conserved
E = mc2(hcR -I - I) Solve this for R and use it to rewrite mo
L along the natural
(5.32)
(5.31):
2
E + mc 2[E + (I - h)mc ~]
path.
Now let m approach
zero keeping E constant:
L = mo z ÷ 0 which
is constant
(Figure
energy
(5,33)
along the natural path.
Compare
a natural path
2.15) with another path which runs between the same two points
but requires
a time which differs by St from that of the natural path.
By (4.27), p.
123 , with
~qn = O, ~W = -E6t
But with
mc ~ << E.
(5.34)
(5.33), we know that ~W is of the order of mc2~t, Thus
6t must be zero:
the varied path,
6t, takes the same time as the natural stationary with respect
to deviations
one.
with
to first order
in
That is, the time is
from the natural path,
and this
is Fermat's principle. 5.4
Action as the Generator If a system's
principle Sec.
for S.
of a Contact Transformation
energy is conserved, Starting with
there is a variational
(5.8), we vary S as we varied W in
S.l: q
~S
= ~ (~Pndqn
+ Pn6dqn )
J
qo
= lq(~Pndq n - ~qndPn ) + [pn~qn ~ q qo qo
q
=
l
q ~H
(~n
~H
Cs.3s)
~Pn + --~qn ~qn) dt + [pn~qn]
qo
qo
6~
165
If we consider only variations unchanged,
and if 6q vanishes
then ~S = 0.
in which the numerical
value of H is
as before at the limits of integration,
Because the time at the endpoints
special case of the Maupertuis
principle
is not varied,
(2.102)
this
is of little practical
use, but we can see from it that S is part of the theory of contact transformations. S can be varied
(for constant
energy)
by varying
or the final q's; it is a function of q and qo" from (5.35),
Comparison with an El(q, Q).
~S ~qon = - P o n "
(4.32)
transformation
from the initial
set Po' qo to the final set p,q,
we can understand
2f constants
The equation
f first-order
f constants. answer
generates
the We
the infinitesimal S
form of this transformation. an important point.
of integration
for their complete
for S, H (~
contains
of
of motion for a system with f degrees of freedom are of
2f, and require
solution.
S generates
of a system as the time goes from t to t + dr.
the integrated
The equations
the definition
from an old set of
Correspondingly,
seen that the hamiltonian
With this interpretation order
derivatives,
(5.36a,b,c)
satisfies
the transformation
p, q to a new set P, Q.
have previously
~S ~ = 0 ~t
shows that S(q,qo)
F 1 generates
transformation
generates
Its partial
are
~S ~qn = Pn"
coordinates
either the initial
,q) = E
(5.4)
derivatives and a complete
solution contains
How are we to find the other f constants?
is suggested by the case we have considered:
only
The general
suppose that the
f constants of the complete solution are q o l . . . q o f. Then C4.32) states that the path of the system is defined by [5.36b,0], and this brings in the additional a solution. Problem
5.9.
calculated Problem
Integrate
the last few paragraphs 5.11.
gravitational
are needed to specify
between S as
I, p. 153, and the F 1 in Problem 4.22.
a free particle moving
Problem
Pon that
Analyze the source of the resemblance
in Example
5.10.
f constants,
the Hamilton-Jacobi
in 2 dimensions,
equation for S describing
and show how the program of
is carried out in this case.
Do the same for the motion of a projectile field.
in a uniform
166
Problem
5.12.
Give the formal modification
to the theory developed
above that makes use of S k rather than S when there are k constants of the motion. Problem
5.13.
Apply
the results
motion of an isotropic potential
Jacobi's
problem to the
oscillator with
= ½ k(Xl2 + x22 )
leave the subject where Hamilton
But in solving a partial
other ways
(5.37)
Generalization
The last few paragraphs in 1834.
harmonic
energy V(Xl,X2)
5.5
of the previous
two-dimensional
to introduce
of the coordinates. us to use constants
constants
differential of integration
We need a more general of integration
by Jacobi in his lectures
equation
left it
there are
than as initial values
approach
as they occur.
that will
enable
This was provided
in the winter of 1842-1843
in K~nigsberg
(now Kaliningrad). The point is to generate possible. 6
as many constants
Suppose we can find a contact
forms p, q into new coordinates transforms
that trans-
and momenta and at the same time
H into a number E independent
all the new momenta will be constants depend on time in a specially
of the motion as
transformation
of the new coordinates.
and the new coordinates
simple way.
The trick,
Then
will
of course,
is
to find the transformation w h i c h does this. Following
the pattern of F 2 in (4.34), we wish to find a transforma-
tion function which we shall call p's into new constant properties
to be determined.
to H but depending transformation
Pn =
that will
H is to be transformed
only on the constant
equations
and the equations
-S(q,a)
transform
the old
~'s and the old q's into new coordinates
~ with
to a new E, equal
~'s and not on the ~'s.
The
are
~s(q,a) ~qn
"
~n =
~
~a
of motion for the new canonical
an" = - 9E(a)~¢n = O,
Sn : 9E__E_~_ =~n
~n(e)
(5.38a,b) n variables
are
(const.)
(5.39)
6 We restrict ourselves to situations in w h i c h ~H/~t = O, since they are of principal interest to physicists. For an excellent discussion of the general case as well as the one presented here, see Goldstein 1950, Ch. 9.
167 Both equations
of (5.39)
can be integrated
over again that the a's are constant ~n
=
~n t
at once:
the first states
and the second gives +
(5.40)
~n
The constants
~ and B are known as the first and second integrals
the equations
of motion respectively.
A Special
Choice
of
a's
A simple way in which the above conditions choose one of the constants
can be fulfilled
(5.39)
is to
~, say ~i' to be the energy, E(oO = c~1
Then
of
(5.41)
gives C~l =
1
,
~°2 = ~3 = "'" = 0
and with B ° = -to, ¢i = t - to,
¢n = 6n"
so that all the coordinates are constants. mined?
n = 2, S . . . .
except the first one,
Now how is the transformation
We start from H ( p , q )
and use
f
(5.42)
like all the momenta,
function S to be deter-
(5.38) and
(5.41) to get
aS H(~-~I" q) = ~i This
is again the time-independent
in a more general solution
context,
are not merely
constants
ered two ways quantities
in which constants
of the coordinates
they occur.
in the but any
So far, we have encount-
can occur in S: In Sec,
equation for S, but
of integration
5.9
as conserved we shall encounter
a
as the constants w h i c h arise in the separation of variables
of a partial
differential
are f + i in number: equation.
however
and as initial values.
third way:
Hamilton-Jacobi
since the constants
the initial values
of integration
(5.43)
~i
equation.
However
plus f others
they arise,
the constants
that come from integrating
But since S occurs only differentiated
in the equation,
the one
of the constants
is a trivial additive one that carries no information
about dynamics.
If we agree to ignore this one, a complete
depends
on f constants
~i'
Once S has been found, we use (5.38) aS(q,~)
a~ 1
as (q, ~)
= t - t
o"
solution
"'' ~f"
a~ n
=
6n
and (5.42) n = 2,
to see that 3 ....
f
(5.44a, b)
168
The
first
is already
familiar
... f do not involve they are solved
from
(5.6).
time and describe
The equations
for n = 2,
the shape of the orbits when
to give
q : qCe,B) Time is introduced relation
between
for q(e,8,t-to)
equations possible belong
Example
only By
q's, e's,
(5.44a),
(s.4s)
which
and t - t o .
is in general
Solving
may be difficult,
if the e's are the f independent
to a general
4:
Charged
If the field (relativistic)
of the f equations
Particle
in a Uniform
is in the x direction,
set of
but it is in principle
constants
solution
hamiltonian
a single
the whole
of integration
that
(5.44).
Electric
Field is -eEx and the
the potential
is
= (c2p 2 + m % 4 ) ½ -eEx = e I Since py and Pz are cyclic
they are constants
e3' so that
e 2 and
px 2 = (e I + eEx) 2c -2 _ e22 - u32 - m2c 2 and the general
S(q,e)
formula
for S satisfying
(5.38a)
is
: f [(el + eEx) Zc-2 - e22 - e3Z - ra2c2]½dx + e2Y + e3z
To integrate
this let
CL1 + eEx = CV Then
SCq, c~)
Differentiating c
t - t o =7~;
1
= -~c J (V2 "_ (~2 2 with respect
-
_
ra2c2)½dv + e2Y + e3z
by putting
m~c2)½
-
(v°Z
in initial
e 1 = c C P z o 2 + c~22 + e 3 2
-
~2
2
~32
values:
+ r/t2c2) ½ _ eEz 0
or
V02 = c-2(Ul
(5.46)
to e I gives
[Cv2 - e22 - e32
This can be simplified
e3 2
+ eExo)2
= pz02
+ ~22 + ~32 + m2c z
_ m2c2)%]
169 and using
this relation
twice gives
t - t o = c e1E {[ (~i + eEx) 2 - (~i + e E X o ) 2
1
{[2~leE( x
= tee
Untangled,
- Xo)
+ e2E2 (x 2
+ c 2 P¢o 2]% _ C P x o }
- Xo2) +
c2P¢02] ½
-CPxo}
this is + e E ( x 2 - x o 2)
2~l (X - XO)
from which x(t)
=
can be further
c2[2pxo(
t
- to)
disengaged.
+ eE(t
-
to)2 ]
(5.47)
To see what the expression
means, let eE = 0 and write ~i = mrC2' Pxo = m r V x o with m r the relativistic mass. Then (5.47) is just x - x o = Vxo(t
- to)
Other special
cases
which is not surprising. be similarly
and approximations
derived.
To continue
with the program, ~S
~"a2
62"
From these we readily
find
so that the projection
return
~S
~3
to (5.44):
63
Y - 62
~2
z-
~3
63
of the trajectory
(constants)
on the yz plane
line through the point (82,83) with slope ~2/e3. (Problem 5.15) gives y and z as functions of x. Problem
Discuss
5.14.
(5.47) when
Pxo = 0.
relativistic
problem
nonrelativistic (It will b e g o o d
In the plane z = 0, find y(x)
Problem
5.16.
If the constants
7
P.M.
invariant
limits
of
to solve the non-
a n generate
[see Eq.
[4,55)),
in the above example. transformations
Am. J. of Phys.
57, 1161
which leave
what transformations
6 n generate? 7
Campbell,
further work
from the beginning.)
5.15.
the hamiltonian
is a straight
A little
and r e l a t i v i s t i c practice
Problem
constants
can
(1968),
do the
170
Solution:
Write
pn[q,a )
(5.38a,b)
and (5.40) as
DS(q,a)Dqn ,
=
where the time dependence
Sn(q,a )
that
(symmetries
DS(q~a)Dan
~n(a)t
-
of q is such as to make 6n constant.
(4.57) and (4.55) the constant E invariant
=
a's generate
transformations
of H), but ~ contains
By
which leave
the time explicitly,
so
(4.57) becomes
d8 n =
(Sn, H)
D8 ~--~
+
dt
and constancy of 8 does not imply that ~ is a symmetry of H, as generating
an infinitesimal
transformation
Take B
to new coordinates
n
and
momenta with
6qm
e =
The second relation
6pm
DS n (q,a) ~Pm
0
6pm = -c
DSn (q,a) ~qm
is
D
-e =
DE Da
Dqm
D - -~ Da n
DS
= -e
n Dqm
9Pm(q'a) Da n
Whenever we change a n infinitesimally, ~Pm = (DPm/Dan)~an" so that here, ~a = -e. The 8's change the ~'s by varying the momenta while keeping n
the coordinates
fixed,
first integrals
generate
generate
changes
and so they will in general symmetries
change H also.
Thus,
of H, while second integrals
in the numerical values of the first integrals.
Integral Surfaces Let us look at the integration procedure. that f = 3.
There will be three constants,
additive ones that arise from the integration.
as ( q , a ) Da 2
each define an algebraic whose coordinates
_- ~ 2 '
--
Dc~3
The two equations
(5.48)
B3
are ql' q2' q3 they define two surfaces.
intersect
Finally,
the last equation,
point.
for simplicity
relation between ql' q2' and q3; in the space
surfaces moves
Suppose
a I plus two other non-
in a line which is the system's (5.44a),
as time passes and intersects This gives the configuration
defines
These
orbit Fig.
5.3.
a third surface which
the orbital
line in a moving
of the system at time t.
171
Figure 5.3 Problem 5.17.
The orbit is the intersection of two integral surfaces. Construct the three surfaces mentioned above for a
particle moving freely Problem 5.18.
in space.
Construct the three surfaces for the path of a projectile
in a uniform gravitational
5.6
field.
Orbits and Integrals If one wants to know what a system will do after having been set
in motion in a certain way, there is nothing to do except to solve the equations.
But there are other kinds of questions one can ask that do
not require everything to be solved:
is the orbit closed or does it
fill a certain region, and in the latter case what region does it fill? What is the character of orbits in phase space?
Are there any regions
of phase space or configuration space into which a system is forbidden to move?
Questions of this kind that require generalizations
about
classes of solutions are often of more interest than the solutions themselves,
and they are sometimes relatively easy to ask and answer
in terms of the integral surfaces. In the following paragraphs we shall see the kind of arguments that can be based on the construction of integral surfaces without making any attempt to solve Hamilton-Jacobi considerations
to follow are geometrical
equations.
Because the
in nature and we are going to
illustrate them with pictures, we are obliged to treat systems with only a few degrees of freedom.
The methods,
however,
are general.
Motion Under a Central Force Suppose we wish to study the general properties an attractive central potential V(r).
of motion under
The particle's position and
motion are given by a point in the phase space whose coordinates are, for example, r, 8, Pr' and p@.
Since this is too many for convenience,
172
we agree to ignore the angle e
by considering
that three-dimensional
slice of the four-dimensional
space which corresponds
addition,
P8 is a constant.
for central
forces,
to 8 = 0.
In
To see what this
allows us to say about the orbit, write
(s.49)
E = % (pr 2 + p92/r 2) + V(r) and PO = r [ 2 E In the space whose coordinates surface; defines
are r, Pr' and p@ this defines
a typical one is shown in Fig. another
surfaces
such surface,
intersect,
Fig.
(s.so)
- V(r) - prZ] ~;
a plane.
5.4b, defines
5.4a.
a
The c o n d i t i o n p @ = const.
The line along which the the values of r and Pr
Pr
r Figure 5.4
(a) Integral surface corresponding to (5.50). (b) Intersection of the integral surface with the plane P8 = const. (Only the half of the figure with Pe > 0 is shown.)
compatible with the given values of E and Pg" If we now represent
the system in the space r, Pr with P8 fixed
and 8 = 0, we arrive at the kind of orbital Figure
5.5, in which each successive
Figure 5.5
representation
revolution
shown in
of the particle brings
In this orbit in rOp space the locus of points at which the path pierces the plane 0 =r 0 is a closed curve.
173
it through the plane the contour a position
8 = 0 at a certain point,
determined
by the construction
to distinguish
between orbits
and these points
of Fig.
5.4.
lie on
We are now in
that are closed and those
that are not. Two-Dimensional
Isotropic
The h a m i l t o n i a n (neglecting
Harmonic
Oscillator
of this system is w r i t t e n
unimportant
constants)
H = %
[ p ~ + po2/r
w i t h P8 equal to a constant, system is shown in Fig.
5.6a.
L.
in polar coordinates
as 2]
E
(S.Sl)
contour
in r, Pr for this
+ ½ r2 :
A typical
But there are other integrals
of this
r
Figure 5.6
system
(a) (b)
Contour in r, Pr determined by (5.51) for fixed values of E and PS" Intersection of the previous contour with that defined by (5.53).
(Sec. 4.6).
One of them is B = ½ Cpx 2 - p y 2
which
in polar coordinates
+ m2 _ y ~ )
is 2
B : ½ (pr 2 + r 2 - Pr82)cos 28 - PrP8 2-~
sin28
(s.sz)
174
Let us assume that B ~ 0 and look at the plane defined by @ = 0. B
0, take @ = ~/2.)
(5.53)
pr 2 + r 2 - L 2 / r 2 = 2B
at the same time as (5.51). The allowed values points
correspond
Fig.
directions
these intersect;
the two
around the orbit.
There is
and ~e now see graphically how
Iprl ,
of the second integral
forces
the orbit to be closed.
What role does the third integral
5.17.
curves.
for every value of @; thus for each @ there is
only one possible r and one
Problem
5.6b shows a pair of typical
of r and Pr occur where to opposite
one of these diagrams the existence
(If
Then the path in (r,Pr) must satisfy
(4.82) play in fixing
the orbit just discussed? Poincar@
Tubes
If an orbit is not closed but angular momentum we can still draw conclusions path pierces
from the r, Pr plot,
the r, Pr plane with
lying on the closed contour which, the given constant values fixed.
What will
circumstances
@ = 0 it pierces as in Fig.
of E and L.
the new contour
can any two contours
is still conserved, for each time the it at a point
5.4, is determined by
Suppose we change E, leaving L
look like?
Only u n d e r very special
intersect,
since if they did there
would be two orbits which at a certain moment had the same values of r, r, L, and @.
With r and L the same,
two orbits would have to coincide two contours some moment
can cross
can give rise to two different
point.
corresponding
If we continue
enables Fig.
the system's
The contour of Fig. the space r, Pr' which spirals
cannot intersect.
Fig.
5.7a
and no smaller E is possible.
construction
given by Poincar~ which to be visualized:
the orbit by including
the angle
return to
@ (Fig.
5.7b),
5.7a is now seen as a cross section of a tube in
@, and the particle's
orbit in this space is a path
around the tube but never leaves
for the exceptional
This that do
to decrease E, the contour will narrow to a
dynamical behavior
5.5 and complete
Contours
at
to three values of E for a central
The orbit is now a circle,
There is a geometrical
subsequent motions.
balanced on its point.
not contain such points of instability shows contours
The only way in which
is if exactly the same initial conditions
is the situation of a pencil
potential.
@ would be the same and the
throughout.
points
never touch each other~
at which two orbits
in this diagram,
it.
Further,
intersect,
except
the tubes
in which p~ is fixed,
the tubes
175
c o r r e s p o n d i n g to smaller E lie entirely inside those c o r r e s p o n d i n g to larger E.
Figure 5.7
(a) Contours in r, Pr corresponding to different values of E for an attractive central potential. (b) Poincar~ tubes whose cross sections are the contours shown in (a).
An Angle-Dependent Potential The next case to be d i s c u s s e d is one in which the orbit is not re-entrant.
There is no integral c o r r e s p o n d i n g to B above and because
@ is not a cyclic v a r i a b l e The h a m i l t o n i a n
the angular m o m e n t u m is no longer conserved.
is I
H
=
-~ (Pr
2
+
P8 2
I r2 1 ) + ~ + 7
r3c°s3e
(5.54)
r~ The form is chosen here because it has r e c e i v e d considerable d i s c u s s i o n since it was first studied n u m e r i c a l l y by H 6 n o n and Heiles 8 a number of years
ago.
because (Sec.
The H a m i l t o n - J a c o b i
e q u a t i o n cannot be solved exactly
it is impossible to carry out a separation of the variables
5.9),
and so we resort to the computer.
The b e h a v i o r of this simple system shows c o n s i d e r a b l e variety, d e p e n d i n g on the value of the energy,
Since the potential
is not
symmetric b e t w e e n @ : 0 and e = ~, we set x : rsin@, y = rcos@ plot y and
py
and
instead of r and Pr' getting a contour c o r r e s p o n d i n g to
O = z (negative y) at the same as that c o r r e s p o n d i n g to 8 = 0 [positive y].
Figure
5.8a shows ii0 successive passages through the plane.
again, the points lie on contours.
Here
Clearly there is an integral of the
m o t i o n w h i c h is r e s p o n s i b l e for the contour in the same way as L was in the situations
in w h i c h it is conserved.
In this case the exact
analytic form of the c o n s e r v e d q u a n t i t y is u n k n o w n and one m u s t M . H ~ n o n and C.Heiles, Astron. J. 69, 73 (1964); see also M.H~non, Quart. J. Appl. Math. 27, 291 (1969); E . M . M c M i l l a n in Brittin and 0dabasi 1971; and J.Ford, A d v a n c e s in Chem. Phys. 24, 187 (1973).
178 laboriously construct it by approximation. ~ For comparison, Fig. 8b shows the corresponding motion when the angular dependence in [5.64)
+ + + + ÷
+ +
÷÷ ÷
+ +
+
+ + +
÷
~xxXX
+
+
<
t
+
+
X
I
I
XXXX
X
X XXXX I
X XXXX
X )¢XXX
X X X X ~x
t
t
X XxxxX
~
I
I
X X
+
4'
+ +
t
÷,
+ ÷ ÷ ÷
+ +
+ +
Figure 5.8a
+÷**
Computed points in an r, pm plot using the potential (5.54).
+++÷+÷÷+%
/
+
XXXx x
%
X
÷
X
~,,:
;~.
÷
+*. +÷~*+
xxxxxxx~x
g
~ +,
Figure 5.8b
9
The same omitting the factor of cos38 in the potential, from the plane; +: entering plane.
F.G.Gustavson,
Astron. J. 71, 670 ~966].
X
x:
emerging
177
1
is omitted and the potential
1
is ~ r 2 + ~ r s, and the points lie on a
contour determined by the angular momentum But the system has an unexpected
integral.
feature.
Fig.
5.9 represents
the same system started on the same point in the y py plane with a
+
X + +
+
+÷
+ X +
+
4.
X +
Pu
X
+
t" X
+
X
++X
4.
K
+
X
K
4.
X
+
4. 4+ +
X
X
)k X
xx
)K
X
x
X
I
I
I
X
I
4"
+
xXx
~ X
4.
XXX
X
X
+
X
4. +
4.
4. 4"
4. 4.
X
X
X
X X
4.
X÷ X K
X K
+
+
X 4'
4"
X X
Figure 5.9
energy.
What has happened
surfaces
are complex
This produces
R
+
The same plot as in Fig. 5.8= but at a slightly higher energy.
slightly greater here.
4. +
4. + + 4 .
,~
There is no sign of the effect of an integral is apparently
a strong instability
in the observed motion, probably
amplified by the effect of round-off 5.7
Coordinate
errors
in the computer.
Systems
The power of the methods evident
that parts of the integral
in form, with many folds and self-intersections.
in simple examples,
of solution we have developed
where almost every constant
fairly obvious way from a cyclic coordinate. been the most familiar ones,
straight
is not
arises
These coordinates
lines and angles.
But a
in a have
178
number of calculations where the meaning
must be carried out in less ordinary
of cyclicity
is not intuitively
section we shall see why certain calculations special
coordinate
systems
types of coordinates
clear.
require
and then go on to discuss
in three-dimensional
coordinates,
In this
the use of the different
space in which solutions
are feasible.
The Anisotropic Oscillator The two-dimensional Cartesian
coordinates
anisotropic
oscillator
by the hamiltonian
i___
s = 2m ~ x 2 + py2) + ~1 (klX2 + and the Hamilton-Jacobi
equation
~S 2 (~-~)
~[
+
3S 2] (~-j)
This can be solved by separation processes factors
each depending
a sum, as here. Assuming
k2y2)
(5.55)
is +
1 (klZ2 + k2Y2 ) = a~
of variables.
that occur by this name:
equation or any other
is defined in
separation
on a single variable
(5.56)
There are two into a product
of
as with Schr~dinger's
that starts with a laplacian, and separation into
(We shall see later how the two processes
are related.)
that
s(=,y) + st(x) + s2@) we find
[
1 (dS1) 2 + -~ klX 2] + [ ~m (d s 2)2 dz dy
and the only way this can be satisfied a constant.
This gives
IDS121 --2m ( ) dx
+ -2 klX2 = c~2"
Since
IDS221 ) dy
--2m (
1 + "~ k2Y 2] = o~I
is for each bracketed
term to be
2 + -~ k2y
= aI - a2
(5.57)
(dSl/dX) 2 and (dS2/dY)2 cannot become negative, the values of x
and y are bounded:
-Ax<X
-Ay
(5. SScz)
where
A x = (2(~2/ki)½,
A~[2(~ 1 - a2)/k2 ]½
(5.58b)
179
H o w one knows motion
that
oscillates
these
limits
between
are a c t u a l l y
them is d i s c u s s e d
reached, in Sec.
and that the
2.2 u n d e r A Note
on A r c - S i n e s . The
solutions
SCx, y) = where
of
[2m(a 2
the i n t e g r a l s
(5.57) _ -~ 1
can be w r i t t e n
klX 2) ]½dx +
I [2m((~ 1 - a 2 - ~
m a y be c o n s i d e r e d
a g r e e d not to d i s t i n g u i s h
additive
down from
as i n d e f i n i t e
constants.
(5.7):
k2y2)]½dy
b e c a u s e we have
To find the orbits we
have
½klX2)½-
(0. 2 -
8c~ 2
(C~l - a 2
_
2)
=
S2
½ k2Y
or
/ _ sin_1 x ~i Ax What
curves
introduce
I sin_1 _ y _ = B2 ~2 Ay
this s t r a n g e
formula
the time v a r i a b l e
~2 =
represents
as a p a r a m e t e r
~SCx,y) ~al
becomes by
= 2_~ sin_ I ~K_ =
~2
Ay
t
k/m
(5.59)
clear
if we
(5.44a), -
t
o
so that sin-1AY
= ~2 (t - to)
y = Ay sin~2(t
or
- to)
(5.60a ,b)
Y Putting
the first of these
into
(5.59)
x = AxSin[ml(t Equations Fig.5.10
(5.60a,c)
are the r e c i p e
is an example.
The
gives (s.60o)
- t o ) + 82]
for d r a w i n g
curves w i l l
Lissajous
ordinarily
figures;
fill up
Ch) Figure 5.10
Lissajous figures defined by (5.69) for different frequency ratios. (a) If ~i and ~2 are incommensurable the curve finally fills the allowed region. (b) Here ~I = ~2 and the allowed region is not filled.
180 the entire allowed area defined by (5.58). is clear why cartesian variables, expressed
coordinates
If these are the orbits,
were right for separating
since in no other system are the boundaries in terms of one variable (Fig.
the
of this region
at a time.
But suppose the two frequencies now reentrant
it
happen to be equal.
The orbit is
5.10 ) and it no longer fills the available
In this case the problem can be separated
in polar coordinates
space. also.
(Problem 4.43 does it for quantum mechanics.) In Sec.
Problem 5.32.
generator
5.6 we showed that S can be regarded as the
of the contact transformation
initial state to its present state. a 2 of the foregoing
discussion
so that their Lagrangian
L Separate
=
=
are coupled together
+
~2)
_
~
~Cx 2
-
y~)
_
~2xu
new coordinates
u),
figures
+
y =
~-%Cv
in u,v that express
+
u)
the solution of this
Sketch them in x,y and note how the boundaries
figures are expressed 5,8
harmonic oscillators
by introducing
~-%(v
Sketch the Lissajous equation.
that define the
form of the S that generates
is
(~2
~
the variables x
Curvilinear
of the
in each system.
Coordinates
The simple examples curvilinear
~I and
and verify that it does so.
Two identical
Problem 5.33.
the constants
in terms of the numbers
initial state, write down the explicit this transformation
that takes a system from its
Expressing
coordinates
of the last section suggest how the use of should be regarded:
if all the Lissajous
figures generated by a certain orbit can be bounded by coordinate lines
(or surfaces)
be separated. following
in a certain set of coordinates,
If not, the calculation
example in plane parabolic
introduction
the variables
may be very difficult.
coordinates
will serve as an
to this approach and will show how one can identify
classes of potential
functions
for which curvilinear
coordinates
useful. Example
5.
Introduce variables
can
The
a new way of covering
~ and n defined by
the xy plane by means of the
are
181
x = ( ~ n ] ½, If we eliminate
y = @ (~ -
~,
~ > 0
(5,61)
first ~ and then n, we find X2
Y which,
n)
-
~2
X2
2n
__ ,.~2
2~
Y :
(5.62)
for c o n s t a n t ~ and ~, are two sets of confocal parabolas all of
shose intersections
are at right angles,
%
Fig.
5.11.
• X
o Figure 5.11 Parabolic coordinates. We can also easily show that
~=r
+y,
(5.63)
~=r-y
A routine c a l c u l a t i o n gives the l a g r a n g i a n as
L :
I
m(~ + n)
~62
(=~- +
"~)
-
v(~,n)
The m o m e n t a are
p~ :
Tm(~
+ n)
,
Pn : Y m ( ~
+ n)
(5.64)
and 2 ~p, 2 + npq2 H : ~ ~ + n + v(~,n)
(s.6s)
182
The Hamilton-Jacobi
equation
2
1
in these variables
~S
~S 2
m ~ + ~ [ ~ C ~ ]~ + n ( ~ ] Suppose
now that
V(~,n)
is
]+ v(~,n) = %
is of the form
vC~,n)
fl (~] + f2 (q)
(S.66]
=
+ n Then
[~
+ n ( ~ ) 2] + fi({)
3~
and S separates
additively
+ f2(n)
into SI($)
- ~i(~ + n) :
+ S2(n)
0
where
@S
2
~(~-T-)2 + fl (~) - % ~ 2 n(~--~-] aS 2 + f2Cn] Note that the constant
~2 has entered
= ~2
- ~i n = -~2
not via a cyclic
conservation
of energy but via the procedure
in a partial
differential
equation.
(5.67)
coordinate or the
for separating
Such constants
variables
are called constants
of separation. Since
(~SI/~)2
and
al$ + a 2 - f i ( $ )
C~$2/~)2 ,> O,
are not negative,
°~lq - ~2 - f2 (q)
depending
on the f's and the values
establish
limits
vanishes
for ~ and n.
is a turning
oscillates
between
of integration exactly
determine
regions
of the ~n plane.
depending
If it vanishes
the two turning
9 0
expressions
twice,
points.
have
C5.68)
even though they may not
they at least restrict
The equations
on the left
the particle
The role of the constants
in these expressions:
an orbit,
must
of ~i and ~2' these may or may not
Where one of the
point.
is clear
we
(5.67)
it to certain
are easy or difficult
on the forms of fl and f2' but let us see what has already
been accomplished.
In solving
separation
~2 which
constant
the equations
we have
is a new constant
introduced
of the motion,
a
given by
either of two forms 2
~2 = ~ CP~
2
+ f l (~)
- ~i ~ (5.69)
= - ~ nPq z -
f2(n)
+ ~1 n
183
A special case is Kepler motion,
in which ~+
n
This is of the form (5.66] with fl : f2 = - Y Let us symmetrize the expressions
(5.70)
(5.69) by writing ~2 as half the
sum 1
2
(5.65]
this
~2 : ~ [~p~ With
~i
equal
to
H in
~2 = ~ + n
np~ 2) - % ~ z ( ~ reduces
-
- n)
to
Pn
-
my(~ -
n)]
(5.71)
To see what this has to do with anything else we know, let us calculate the component of the vector integral A in (2.91] that lies along the axis of the parabolas, A Y = ~1 (mpxpy _ ypx2)
+ 2Lit
[5.72)
On substituting the expressions that transform this into parabolic coordinates we find
[5.73)
Ay = -~2/m
(There is no need to ask about Ax, since from (2.94) A has only one independent component.] It should be noted that the constant which was derived in Chapter 2 by an ingenious de~ice appears here in the course of a routine application of the separation of variables in parabolic coordinates. Of course, the right set of coordinates had to be chosen, but there is not much ingenuity in this, since as we shall see there are only eleven kinds of coordinates in which the separation is possible, and most of these correspond to potentials so esoteric in form that they are unlikely to find physical application.
Further, we have succeeded
in generalizing our earlier result, since the functions fl and f2' instead of being given by (5.70), are now arbitrary. Finally, we may if we wish perform an actual calculation.
The characteristic function
is [
- @2
fl($)
f
~2
f2"(N)
184
through which the constants
el and u2 will make their appearance
in
the solution. Find fl and f2 corresponding
Problem 5.34.
electric
field,
and evaluate
the integral
to a hydrogen atom in an
~2 in terms of Cartesian
coordinates. Problem 5.35.
Carry out the change of variables
Problem 5.36.
Show that the sets of coordinate
all intersect
each other at right angles.
for the separation Problem 5.3?.
particle
Find the physical
lines given by (5.62)
Why is this fact important
of variables?
Set up the equations
attracted
leading to (5.65).
of motion in 3 dimensions
for a
to a fixed center by an aribtrary potential significance
show that when the coordinates
of the two separation
V(r).
constants,
and
are suitably oriented with respect
to
the initial motion our earlier results are reproduced. Problem 5.38.
parabolic
Solve the plane Kepler problem in two dimensions
in
coordinates.
Problem 5.39.
With the major axis of a keplerian
axis, show that the formula for the ellipse 1 + e
÷
n
1 - e
=
ellipse along the y
in parabolic
coordinates
is
(5.74)
2a
Figure 5.12 A Keplerian ellipse in parabolic coordinates. The point representing the particle oscillates back and forth along the llne. Problem 5.40.
Show from
(5.68) that ~ + ~ ~ 4a in the Kepler problem,
and show that this is compatible Problem 5.41.
Discuss
with
(5.74).
the motion of a particle under an inverse-
square force in terms of intersecting
surfaces
done in Sec.
5.6, but here use the ~, p~ plane.
what happens
if a uniform electric
in phase space as was You may wish to show
field is also present.
185
The conditions under which it is possible to separate variables in the Hamilton-Jacobi equation have been much studied, notably by St~ckel in the 1890's.
For a perfectly general physical system, there are
infinitely many kinds of coordinates in which separation is possible. For a particle in three-dimensional Euclidean space it was finally shown in 1934 by Eisenhart *° that there are exactly ii real coordinate systems
(i.e., systemsin which the coordinates are real numbers) that
satisfy St~ckel's criteria.
They are discussed in Morse
and Feshbach
1953, Ch. 5, with diagrams in which certain readers will see the coordinates in three dimensions. The potentials for which each system allows variables to be separated are the same in classical and quantum mechanics and are given in most economical form in Eisenhart's paper of 1948. *Q The variety of problems for which variables can be separated is extremely small, though fortunately for students and teachers it contains some important cases.
The two-body problem is equivalent to
the one-body problem, but when we leave that behind almost nothing is exactly soluble any more except for very special choices of initial conditions.
We have to resort to approximate procedures,
except for certain qualitative considerations, theory is not specially advantageous. 11 approximately,
and here,
the Hamilton-Jacobi
If equations have to be solved
in letters or by machine, ordinary differential equations
are easier to deal with, and most practical calculations start from Hamilton's 5.9
equations.
We shall discuss a few simple methods in Chap. 7.
Interlude on Classical Optics Although it is not possible to create situations in which the
index of refraction varies as freely as potentials do in dynamics, the theory developed in this chapter has its close counterpart in optics. Only the derivations are different, since we start from a wave theory. Let w(r,t) be the phase at a certain point in a monochromatic light ray.
If one goes to a neighboring point r + ~r the phase changes
according to (i.i0) and (i. II) by K.6r.
If one evaluates it also at
a slightly different time the phase changes by m~t. ~w ~
analogously to (4.27). ,0 87
L.P.Eisenhart,
(1948).
K-6r
-
~6t
Thus (5.75)
As before, we introduce a time-independent
Phys. Rev. 45, 427 ~1954].
See also Phys. Rev. 74,
,i An introduction to Hamilton-Jacobi perturbation theory is given by T.L.Perrell, Am. J. Phys. 39, 622 (1971).
186 p h a s e by a L a g r a n g e t r a n s f o r m a t i o n
which
= w +
=
K.~r
toCt
-
t o)
(5.763
-
to)6~
(5.773
satisfies
6s so
e
+ (t
that 8__f8 t - t o = 8to
K -- Vs, From the definition
of K we have by (l.S)
[~co n(to, r ) ] 2
(783 2 The equation integrate
(5.78)
is the same as
=
0
(5.79)
(1.8) but now we know more about how to
it.
Let us consider medium of uniform propagation,
first the propagation
n.
Taking
of a signal
the x axis parallel
through a
to the direction of
we have
('~[r~')ds2 = [~.con(to) ] 2
(5.80)
or
d8 -d-~ =
where k = ton/c as u s u a l .
k,
8
"
kx
(5.78)
From t
=
t
~8 dk ~to - ~
o
-
-
to~)
x
or =
= 2sF~
(t
=
v g. ( t
"
to)
(s.813
where v
is the group velocity, This is in analogy with the particle g velocities given by the analogous procedures in dynamics. We shall see in Sec.
6.5 why it comes out this way.
A more interesting see Fig. (5,79)
5.13.
With
case is one treated before in Example 1.3,
n(to,z)
we can separate variables
and integrate
to give
(s.823 The orbit is determined by
187 ~8
[
~i
=
x
-
~i
dz
(0
J
'
[k~(~,z)
~s ~2
=
~22]½
~i 2 -
~
'
= -f n(~,z)
81
dz
= y - ~2 J [ k 2 ( ~ , z )
~12 - ~22] ½ = 8 2
-
whence x - ~i
~1 and the p r o p a g a t i o n
y - ~2
=
[
a2
dz
= J [k2C~'z)
(s. 83)
- a12 - ~2 21%
of a signal along the beam proceeds
at a rate
given by
dk k "~'~ dz t -- t
0
(s. 84)
=
(k ~ - ~I ~
~2) ~
X Figure 5.13
Bending of a ray of light as it enters the atmosphere above a plane earth.
Problem 5.42.
Integrate
(5.79)
Problem 5.43.
According
to Example
waves
to get
(5.82).
I.i, the propagation
in shallow water of depth h < g/m2
of water
(i.e., kh < i) satisfies ~2 h
where h is some function of x and beach,
discuss
y.
Assuming
a uniformly
shelving
the m o t i o n of a large isolated wave as it approaches
straight beach.
Problem
5.44.
Show that
(5.83]
is equivalent
to Snell's
law for a
a
188
horizontally Problem
4.45.
layered medium. Show that where
Figure 5.14
and n = n(r), horizontally
there is circular
Problem
5.46.
= cos~,
(Fig.
5.4)
Variables in the Smith-Helmholtz relation (5.86).
the invariant
corresponding
layered medium
to Snell's
law for a
is nrsin X
The constant
symmetry
=
const.
(5.86)
is known as the S m i t h - H e l m h o l t z
invariant.
Show that in terms of the direction cosines
the Smith-Helmholtz
invariant
n(ay
- 8x)
(Note that if y >> x this reduces
a = sin~,
(5.86) is
= const.
to Snell's
law for a horizontally
layered medium.) Problem
5.47.
according
Assuming
that the atmospheric
n(r)
find
the
index of refraction varies
to
difference
whose apparent
between
zenith
= I
the
distance
~o e'~(r-R)
+
real
R
and apparent
is
eo,
Fig.
1
(8~S
2
S.15a.
Solution
The eikonal
equation is ~S
(T~)
2
and with S = S r + S 8 this is
+--r2
:
n2
(r)
earth
directions
radius of a star
189
as e so
as =
"
~
C~2]#
-
"~
(n2
=
r--2"
so that
i
r
S =
a 2 ] ½dr
(n 2 - -#~j
+ ~O
(5.87)
R
Figure 5.15
The p a t h
For calculating the bending of a ray of light as it enters the atmosphere above a spherical earth.
of the
ray
is g i v e n
i
by
r
-a
= B, or
as/aa
dr r
(s.88)
=
2(n2
~2 ~ + e
-
r2J
R
To i n t r o d u c e
@
we n o t e
that
from
Fig.
5.15
(b),
0 p
d° I
tan@ o -- R ~
(s.89)
r=R
By (s. 88)
dO dr
n
=
R2(no2 r=R
_ R2 ~-i~
0
=
n(R)
=
1
+
v
0
190
and this,
in
(5.89),
gives = noRsin@ o
To i n t e g r a t e
(5.88), we w r i t e
the i n t e g r a l
dr 2
_
~9o ~ 2 ~
-
r 2j
the
by R in
is
denominator
only of
a few kilometers
the
second
this
dr
thick
integral
we c a n r e p l a c e
and
r
find
[I - e - p ( r - R ) ]
- /__ sin_ I a__ ~o r ~R 2 (1 - ~ - ~2) 3/2
In (5.88),
in 9o as
R
atmosphere
the
to first order
~2 312 ~-~)
_
R Since
(5.90)
is ~)
sin_1~_ + _ r
o
~ R ~ (I
Now we can d e t e r m i n e
the c o n s t a n t
sin-1 or
by
[:
5Y, 3/2 ~-~) 2
-
a =
-
8.
e -~6r-R)]
W h e n r = R,
0
=
S
@ = 0:
a = Rsin8
8,
(5.90)
szn~ = n sin@ O
which
+
the r e a d e r w i l l
readily
solve
O
to give (5.91)
8 = 8o + ~ o t a n @ ° + "'" W h e n r : ~,
the ray has
the a s y m p t o t i c OVV 0 pR 2
In the first t e r m
(I
-
a 2 3/2 + @
O
(5.91)
@ :
=8
~--~-)
it is a c c u r a t e
a = Rsin0 o, and w i t h
direction
enough
to r e p l a c e
65.90)
by
we f i n d
8o = ~ o t a n O o
(1
-
1 0
or,
finally,
0
- 0 ° = ~o(1
-
~
9o tan30o ) tan0 ° - p-~
(s.92)
191
if @
is not too close to ~/2. The first term comes from Snell's o the second is the effect of the curvature of the earth's surface. angle,
though small,
accurate
is of great importance for astronomers,
formulas have b e e n developed.
law; The
and very
The m e t h o d used here fails
w h e n @o is very near ~/2, and a more delicate analysis is required. A value for Vo under typical conditions of Prob.
is given by the formula
i.ii as 2.77 × 10 -4, and a c o r r e s p o n d i n g value for U is
0.14 km -I
W i t h these @~ - @o = 57.1" tan@ ° - 0.064" tan3@ °
At @
= 30 ° the d e f l e c t i o n is 32.9"; o or 9.28'.
at 85 ° it has increased to 557",
Problem
5.48.
Find the term in v 2 that corrects o
Problem
5.49.
Calculate the apparent change in the shape of the sun's
disc as it sinks towards the horizon.
(That it often looks different
from this can be a t t r i b u t e d to inhomogenieties a clear evening,
(5.92).
in the atmosphere.)
the apparent height is about 3/4 the width.
In
CHAPTER 6
It is a relatively mechanics
ACTION AND PHASE
easy matter to derive formulas
from those of quantum mechanics.
about probabilities
into statements
of classical
One translates
about measurable
one finds ways to eliminate ~, either by considering
statements
quantities,
which phase changes much more rapidly than other parameters or by taking averages
in a suitable way
To invent quantum mechanics
(Sec.
phase.
was much more difficult,
This is the WKB approximation
Hamilton-Jacobi
to introduce
from this approximation
equation of Newtonian physics, It is necessary
hypotheses
1.3)
for ways
law.
of quantum mechanics
little new physics:
furnished by the theory are estimates penetrabilities.
(Sec.
It was
the ~ that allows ~-IW to be considered
and turns out finally to produce for the phase resulting
in
2.1).
had to be found in which to introduce ~ into dynamical not enough to introduce
and
situations
as a
(Sec.
Equation
i.i)
[1.6)
is nothing but the and the only new results
of energy levels and barrier
at some point to change something,
that do not belong
to Newtonian physics,
in
order to invent a new dynamics. Bohr's
theory simply grafted a quantum principle
onto a Newtonian original laws.
theory of motion.
and widely underrated,
Heisenberg
in 1925 was the first to do that.
of Bohr's
he rewrote
it in terms of matrices,
Jacobi dynamics, Heisenberg's
(Sec.
SchrBdinger
making 6.5).
theory.
Starting
from an
(Sec.
of Heisenberg
waves with Hamilton-
as he did so, to make a
9.7) which later turned out to be equivalent And much later,
6.4)
changes as he did so, and Independently
united de Broglie's
but again making changes (Sec.
though very
theory in terms of Fourier amplitudes
arrived at quantum mechanics
wave mechanics
involving
ideas,
did not point toward new dynamical
expression
and a year later,
De Broglie's
to
in 1948 Feynman showed that if
one takes ~-lW for the phase of a wave and makes careful use of Hamilton's
principle
and the principle
arrive at quantum mechanics arbitrary
hypotheses
(Sec.
by a path which
is particularly
the reader's
but it may deepen his understanding
the old and new theories
and, in addition,
in which new discoveries
are made.
6.1
one can free of
6.7).
This chapter will not advance dynamics,
of superposition,
knowledge
of classical
of the relation between
show something
of the ways
The Old Quantum Theor Z Quantum phenomena were not first noticed at any one instant,
a good place to begin is Wien's observation
(1895)
that
but
the spectral
193
distribution of black-body radiation has a maximum at a wavelength m
inversely proportional
to theJ absolute temperature of the radiation, g = W (const.) m
and it became clear t h a t the value of the constant of proportionality is about 0.294 cm K, independent of the size, shape, and material of the enclosure: Boltzmann's Nowadays
it is a constant of nature.
Introducing a factor of
k allows the constant to be expressed in mechanical units.
one writes Wien's law in terms of the frequency: kW
kW
kT = T m- =
7
~m
The value of k W / e t h a t could be found from information then available was about i0 -2~ erg sec.
It was
(or, in hindsight,
should have been)
clear that a successful theory of thermal radiation must contain a universal
constant of this size
and units.
(give or take a dimensionless
factor)
The constant was introduced into mechanics by Planck in
his famous hypothesis
that the energy of an oscillator of frequency
is quantized in s t e p s
separated by an interval hw.
Although the hypothesis worked,
it was hard to justify)
as 1906 Planck was thinking about expressing
as early
it in another way, I
He
had noticed that the orbit in phase space of the point representing an oscillator
is an ellipse: p2
q2
2mE + 2E/k = I
and t h a t the area of the ellipse corresponding
to a given energy is
A = 2~(m/k) ½ E = E / ~
As one goes from one energy level to the next, E increases by hv and A increases by ~A=h
This is a recipe for quantizing
the oscillator which puts into
evidence the quantity h independently of any physical property of the oscillator,
and Planck accordingly felt that it might be of fundamental
significance.
He and other workers
imposing the same
thereupon tried the effect of
condition on rotators
and other systems with one
! See for example
Planck
1932,
Pt. 4, Ch.
III.
194
degree of freedom. can be written
The area enclosed by a closed orbit in phase space
as f f d p d x
convenient notation
integrated
over the orbit,
but a more
is ~pdx
=
nh
the orbit
n =
(6.1)
2 ....
integrated
around
Figure 6.1.
The integral (6.1) around a closed orbit in phase space gives the area of the orbit, since the doubly shaded area is covered twice in opposite directions.
The alternative when Bohr's circular
(Fig.
1,
6.1).
forms of quantization
theory appeared.
were discussed until
Here the basic quantum hypothesis
orbits was that an electron's
constant
1913, for
angular momentum
satisfied Po
= nh/2~
n
=
1,
This does not follow from the quantization in fact,
of energy in equal steps;
it gives rise to energies
En
But
(6.2)
2 ....
=
2~2me ~ naha
n
=
1,
(6.3)
2 ....
(6.2) does follow from an action principle
JPodO = 2~po = nh The two laws appeared general principle. integral
to be examples
In several
around a complete J
= ~Pmdqm
degrees
(6.4)
in one dimension
of freedom,
of a
with the phase
cycle denoted by J, it is =
nh
(summed over m)
(6.5a)
195
which was appealing under
a contact
quantization By (4.32), momenta
because
it can easily be shown that J is invariant
transformation
has nothing
omitting
and that therefore
the recipe
to do with the particular
time variation,
for
coordinates
g in two systems
used.
of coordinates
and
satisfies
J
~Pmqmdt
=
~PmQm dt
=
dt
-
dt
But since values,
in a complete
cycle both q and Q come back to their
the last term vanishes
J = ~Pmdqm is
proved.
B u t i's n o t
quantizing
J,
since
o f f r e e d o m s u c h as it
does not
recipe 2 is
lead to
and Pvdqv
~PmdQm
(summed)
to
Newtonian mechanics
correct
a hydrogen
=
the
=
suspended.
of
(6.5b)
w i t h more t h a n
an e l l i p t i c a l
levels
at
(6.5a)
nvh,
is not summed over v.
to identify products
systems
energy
terms
~Pvdqv
quantize to
atom w i t h
to quantized
Jv
future
=
when a p p l i e d
separate
all.
and w r i t e
nv
=
I,
orbit
How the procedure works
J = ZJv,
2 ....
(6.6)
The ball moves
in each segment continuous
is
based on classical physics.
= ± (2mE) ½
the function
Integrating
Phil. Mag.
(x
-
walls
at x = 0
Xo)
segments
in Fig.
6.2a.
oscillator
across
so as to make
S
(The characteristic is shown for comparison
in
the box and back gives = 2(2mE)
to the quantum condition,
W. Wilson,
it encounters
Joining
for the harmonic
J
2
to
1 will be
of (dS/dxJ 2 = 2mE are of the form
of the path.
gives
(5.7)
6.2b.)
According
is
can be seen from a calculation
freely except where
The solutions S
Fig.
v in the
the summation convention
A Ball Between Two Walls
and x = £.
function
6.8)
where
to see that in the limit of high quantum numbers,(6.6)
closely connected with arguments
Example I.
(Prob.
We shall use the subscript
in which
by
one d e g r e e
The c o r r e c t
be carried out in the next section, but following Example instructive
initial
and the invariance
29, 795
½
the ball's
(1915).
energy
in its nth state
196
will be
E
-n
n2h 2 8m~2
n = O,
I,
2....
5 5
.... X 0
Figure 6.2. Characteristic function S(x) for (a) a particle bouncing between two walls; (b) a harmonic oscillator.
Problem
6.1.
What is the q u a n t u m - m e c h a n i c a l version of this result?
How do they differ?
The
Correspondence
The
Limit
following argument is an e l e m e n t a r y part of Bohr's theory.
Consider circular orbits with quantum number n for convenience; it is easily shown
(Prob.
6.4)
then
the orbital frequency of an
that
electron in the nth orbit is
~n
me
=
(6.7)
n3~ 3
The energy of the state n is (6.8)
me
E =
2n2~ 2
and the frequency of light emitted in a transition from state n to state n - ~ is ~n,n-a
= ~-I(E n _ En_a )
me ~
= ~
1
((n
-
1
~2
"~)
197 ( n_~ * ~ -2c~ Y ÷ ....
me
~17 )
or
mn,n-~
= ~n
+ "'"
Thus in the limit of large quantum numbers, starting
transitions
of the orbital
~ = I,
2,
the frequencies
from state n are just the various
frequency,
and the results
(6.9)
...
emitted in
Fourier harmonics
of quantum theory are in
harmony with what one would expect by classical
reasoning.
the formula above could have been obtained by writing
Note that
Taylor's
series
as
(n_
but since n varies
~
: V~ -
~ TffT~
in integer
That the nonsensical
+ : - - +n2
n-T
+
"'"
steps the derivative makes no sense.
procedure
leads to the same result as the careful
one used above will be important
in simplifying
the developments
to
follow. Derivations
o f the Q u a n t u m
It is instructive the quantum condition energy.
Supposean
Conditions
to turn the foregoing (6.6)
argument around and derive
from the simple notion of a quantum of
oscillating
system with one degree of freedom is
in its nth quantum state, where n is a large number and we are in the domain of the correspondence (-An) quanta at its frequency
principle.
Let the system emit a few
of oscillation,
v, so that the energy
E(n) decreases by
Denoting
differentiation
in the schematic way just explained,
we can
write this as dn
1
where T is the period of oscillation From
T
of the oscillator
in state n.
(5.6) we can write f in terms of J, the value of S on
integration
around a complete
cycle: T = dd 2-E
(6 i 0 )
198
Thus dJ dE
which
integrates
essentially
as we can verify a complete
dn d--E
+
const.
nh
result
from the value
(6.11)
(6.1)
for a harmonic
of S given in (5.7).
oscillator,
Integration
over
cycle gives
the quantum
2~(m/k)
condition E
The constant, and not
(6.5),
½
= E/~
is =
as we now know,
The same argument (6.6),
=
Planck's
J =
so t h a t
h
to J
This gives
_
nh~
const.
+
is ½hr.
explains
why for a multiply
is the correct
periodic
rule for quantization.
Such a
system can emit quanta with
frequencies
each of its periodicities.
We then have for the vth degree of freedom AE v =
If as in all the examples then J reduces before general
case,
we have given the variables and the argument
For a nonseparable
EinStein 3
proposed
system,
equal
cycles
to n h .
convenient
of each of the various
The anisotropic
though
tion has already of y, the phase ra 2 J
= r8
been done. integral
example
condition:
discussed because
If we integrate
as
is of course
the
Integrate
number
of
and set this
on page
178 is a
most of the calcula-
over r cycles
of x and s
is
-
nh
r,
s,
n
=
O,
1,
2,
...
(6.12)
~2
(Here a2 and a I - ~2 are the energies x and y.)
goes through
8(a I - a2) +
~i
to
are separable,
an integral
periodicities,
oscillator
too elementary
which
a new quantum
over any closed path in q space which makes complete
corresponding
&nvhVv(nv)
to a sum as in (6.5)
for each Jr"
and energies
system
The only way in which
associated
this relation
with the motions
can be satisfied
in
for
all integral values of r and 8 is if a2/Vl and (al a2)/~ 2 are each integral multiples of h. Thus each of the oscillator's degrees of
3
A.Einstein,
Verh.
Deutsche
physikalische
Gesellschaft
19, 82 (1917).
199
freedom must be quantized separately. is that it is separable.
The trouble with this example
There is no soluble nonseparable Hamilton-
Jacobi equation known to the author. Problem
6.2.
Planck's formula
for the spectral distribution in black-
body radiation is
dI
=
8~ahl-S
l
exp(ha/~kT)
- 1
Locate the maximum of this curve, make a rough calculation of the constant W in (6,1) and, using a modern value of k, see how good a value of h is implicit in the Problem V(x)
6.3.
I
= ~
nineteenth-century measurement.
Find how the period of an oscillator obeying the law
Kx ~ depends
on the energy and amplitude of oscillation.
The
action integral J can be evaluated with the help of the integral formula 1
r2
(6.131
(1 - x~)~dx 0
which may be derived with profit and enjoyment from the integral for the beta function. tm-l(l
-
t)n-ldt
=
rF(m (m) +r (
= B (m, n)
0
and some of the identities satisfied by the gamma function F(n). ProbLem 6.4. Find the energy levels of the oscillator in the last problem using the quantum condition in phase space, and show that the
calculation is almost exactly equivalent to a solution by the WKB approximation. Problem
6.5.
Derive (6.8) for a circular orbit.
Problem
6.6.
The potential Y(r) binding two quarks together is usually
supposed to increase with distance. V(r)
= g
Supposing that In
r
o
show that the quantum conditions gives energy levels En = 2-~ In ~ for s states
nh
(zero angular momentum) where p is the reduced mass of
the two quarks.
The integral is not as difficult as it looks at first.
200 What is the difference
in energy between the nth state and the ground
state? 6.2
H y d r o g e n Atom i n t h e O l d Quantum T h e o r y We c a n u s e t h e r e c i p e
and t h e i r lies
degeneracies.
i n t h e xy p l a n e ,
to
(6.6) If
1 the total
For an a r b i t r a r y
the
2
angular
orientation
(Pr
is
Pa 2 *
energy levels
a r e c h o s e n so t h a t
then the hamiltonian H = ~
w h e r e Pa i s
find
coordinates
easily
ez
(6.14)
momentum, w h i c h i s of the axes,
the orbit
f o u n d to be
)
r2
of hydrogen
around the z axis.
one f i n d s
in spherical
coordinates 1
p82
H = -~
where
p 2
~2
(pr 2 +
+ ~
e2
)
- -~-=
p is the reduced mass of electron and proton.
cyclic,
pc is constant
(6.15)
E
Since
~ is
and we can write s = st(v)
+
So(O) + p ~
46.16)
Then ~)S0 2 p 2 (TO--) + sln2@=¢ - -- ~@
(const.)
46,17)
and ~S r
(~) If we write
2
c~0 e2 + - ~ = 2m(E + T - )
(6.17] as PO 2 + ~
and compare total
(6.18~
(6.15)
with
(6.14),
: ~O
we r e c o g n i z e
s 0 as t h e s q u a r e o f
the
a n g u l a r momentum,
J¢ : . ~ p ¢ d ¢
: rich
(6.19a)
~D
C6.19b)
2
=
J~--
-
[~(~
sln20J e2 + T)
~
- -~2]
½
h
dr = n r
(6.19o)
201
These are (Prob.
6.7)
(6.20)
Jr
=
-(J@ + Jq5) + ~e2(-2-~EE)½
Thus
p~ = n ~ h / 2 ~
~0½ = p~ + n o h / 2 ~
It was customary in Bohr's theory to write ~@
½
= (n~ + n o ) h / 2 ~
as £h12n and n# as m,
so that £ = m +n o Since m can be positive or negative while no, n r > 0 we have -£ < m < £ as is given also by quantum mechanics,
(6.21)
although in the newer theory
the square of the orbital angular momentum is written as Z(£ + 1)~ 2. From (6.20) the energy is given by
E = -
2w2Pe~ 2w2Pe~ 2wZpe~ ~J~ + Jo + Jr )2 = - h2(n~ + n o + nr) 2 = - h2(£ + nr)
(6 22)
Evidently there are a number of sets of integers n. corresponding any value of E except the lowest; degenerate.
to
these levels are said to be
Problem 6.10 shows that if n~ + n@ + n r = n, the
degeneracy of the nth level is n 2. After the elliptic orbits of hydrogen had been quantized in this way, Sommerfeld
(1923) used the same methods to quantize the orbits
of the relativisitc theory.
His results agreed with experiment, yet
we now know that this was a coincidence, (but then unknown)
for he omitted the important
effect of electron spin.
The same formula was
correctly re-derived from quantum mechanics by Dirac in 1927.
Problem
6.7.
Carry
out the integrations of (6.19b,c). ~
In both,
it is
The necessary indefinite integrals are in Dwight's Tables, but the best way to carry out closed integrals of this kind is by the use of complex variables. It is not only faster when you know how to do it; it also enables more difficult integrals to be done. For the relevant techniques see Sommerfeld 1923, Note 5; Born 1927, App.2; Goldstein 195@, Sec. 9.7.
202 necessary
to think carefully what the limits of integration
start on J@ take cos@ = V as the variable integrand will then involve
separately.
Show that if the quantum rule J = nh is adopted, with J
6,8.
(6.Sa), 6.9.
Problem
To
1 + __r__9)
I
and the two terms can be integrated
given by
are. The
(5 - ~z)-1, which can be written as
I
Problem
of integration.
a continuous
Evaluate
applied electric the integrand
spectrum results.
the energy shift in a hydrogen
field F.
The action integral
in powers of F.
see Sommerfeld
atom due to an
is difficult;
This is a long and difficult
1923, Note Ii; Born 1927, App.
expand calculation;
2; and Goldstein
1950 for
support. Problem
6.10.
Derive
(6.1S) from the corresponding
lagrangian.
Problem
6.51.
Carry out the separation of variables
Problem
6.12.
Show that if n~ + n@ + n r = n, the degeneracy
that yields (6.16). of the
th level is n 2 6.3
The Adiabatic
Theorem
There is a remarkable sought to establish
argument by which Bohr and Ehrenfest s
the universality
of the quantization
procedure
(6.6) by showing that if it works for one class of potential fewer)
dimensions
it works for all.
originally
due to Boltzmann,
mechanical
transformability
Suppose
The argument
that in a cyclic system
that describe
that the change in the action integral
change the parameters Haar 1961).
The discussion
We imagine parameters) s
a(t)
we make a
The theorem
J is then very slow,
in
as much as we want, without
amount,
provided
that we
slowly enough.
A general proof is rather difficult make it a proof,
theorem.
the system that is slow
in any way.
the sense that we can change the parameters changing J by more than any preassigned
of
called the adiabatic
(orbiting or oscillating)
and does not cause the system to resonate
in 3 (or
on a theorem,
which Bohr called the principle and Ehrenfest
change in some of the parameters states
depends
(Saletan and Cromer 1971, ter
to follow omits the epsilons
that would
but is intended to show why the theorem is true. that the hamiltonian
H
contains
a parameter
(or
which has been varying slowly since an initial
N.Bohr, Kgl. Danske Vid. Selsk. Skr., nat.-math. 1 (1918); reprinted in van der Waerden 1967.
Afd.,
time
8. Raekke
IV.
203 to .
We assume
that it is possible
such that although a(t) time,
changes
to define a time interval
only by a small amount
the system goes through many cycles.
hamiltonian
changes
a since t
changes, the
at a rate
equations,
(6.23)
OH p + ~OH q + ~~H a• = ~-~ ~H a• = ~-~
H(p, q , a ( t ) ) by Hamilton's
As a(t)
t o to t
Sa during this
so that the change
in H due to the change
in
is
OH = it OH ~dt
o
J If a has changed
slowly during
the interval,
this expression by its time average
it
6H = |
Oa
t Now we calculate integration
adt
adt
= t
=
Oa
(6.24)
0
in J, the value of S
is over a single
in
over a cycle,
0
the change
OH/Oa
we can replace
=
SPn~ndt
when the
cycle
6.I = ~ (pn6qn + qn~Pn)dt
= [pn~qn] The integrated The
integral,
+ ~ (-Pn6qn + qnOPn )dt
part vanishes by Hamilton's
very nearly on integration equations
~j = ~(O~n
(compare
the derivation
of
in a.
(2.65)).
The total change
is
~H Bpn} d t Oqn + ~OP n
the changes w h i c h have accumulated change
over a cycle.
In this relation,
(6.25)
6q and 0p are
since t o as a consequence
of the
in H is
6H = ~H ~pnOPn ~H Oq----nOqn + ~ + ~-~ Oa so t h a t r OH Oa) d t OJ = ~ (OH - 7a
- f coa
)oadt
(6.26)
204
by (6.24).
Now integrate
again as in deriving
~-H ~6adt
(6.24)
0
(6.27)
so that 6J is not of the order of 6a, as one might have expected, smaller.
If one wants
a, the analysis
to know how J actually
changes
but
as a function of
is much more complicated. 6
Not every slow change of potential hump in the potential
of Fig.
leaves J constant.
6.3 is slowly raised,
I f the
a particle
becomes
E
Figure 6.3
Potential which violates the adiabatic theorem if the hump is raised.
trapped on one side of the well or the other and J changes discontinuously.
The theorem does not apply here because
at the moment the
trapping occurs a small change in a makes a very large change in p and q.
Thus
(6.27] cannot be used and the argument
of Bohr and Ehrenfest was that if one has a system
The argument such as a harmonic
oscillator (6.6)
in i, 2, or 3 dimensions
quantization
rule
transformed,
by slowly changing
entirely different
quantization
in which the
is known to work, one can imagine
system,
But the action integral points analogous
fails.
the shape of the potential,
say an atom or even something
once equal to nh remains
to that of Fig.
it being into an
like H2 +
so, if no singular
6.3 are encountered,
and so the
rule is universal.
There is a further argument
that makes the adiabatic
plausible basis for a theory of quantization. system is slowly distorted
Suppose
in the way we have described.
will slowly change as the external
theorem a
a physical
forces do their work.
Its energy This means
that there will not be a sudden quantum jump from one value of n to another,
s
for this would involve a discontinuous
change
in energy or in
P.O.Vandervoort, Annals of Phys. 12, 436 (1961); A.A.Slutskin, Soy. Phys. JETP 18, 676 (1964). Calculations of this kind are important in the theory of a plasma confined by a slowly varying magnetic field. See Chandrasekhar 1960, p. 48.
205 some other parameter which ought to change slowly and continuously. Thus the quantum numbers with
should be adiabatic variables,
in accordance
(6.5a).
Example
2.
A ball is set to bouncing
which slowly changes
onthe
its acceleration.
hard floor of an elevator
How does the duration of one
bounce vary? If the ball starts upward at a speed Vo, its trajectory
is given
by y : rot - ½ gt 2
where g is the acceleration
of gravity plus that of the elevator.
bounce lasts for a time T = 2Vo/g.
[
The phase integral
One
is
Tm
j =
2dt = 2 3 mVo
-0
Thus,
= ~I
mg2T 3 :
const.
g
T ~ g "2/3.
Example
$.
A particle
is gradually
circles
strengthened.
in a uniform magnetic
How do the radius
field B which
and kinetic energy of the
orbit change with B? With B constant,
the orbit is determined by the balance of forces,
m r m 2 = er~B,
so that the frequency value.
By [4.16),
m = eB/m
(6.28)
is determined hy B but the radius can have any
the phase integral
is
J : ~p.d~ = ~(mv + e A ) . d £ where A i s velocity
the vector
v is
Cnearly)
potential constant
second t e r m i s s i m p l i f i e d
of the field.
In the first
i n m a g n i t u d e and p a r a l l e l
by S t o k e s ' s J = ~mvd£
theorem:
+ e I VxA.dS
= 2~mvr + ~r2eB
With v = mr = e B r / m by [6.28), this is J = 3~r2eB = 3eft = const.
term, to d~.
the The
206
where # is the total flux through the orbit. varies
as B -#, and multiplying
If J is constant,
= m2 shows that the kinetic energy varies as B. methods
for heating
a plasma is to immerse
is then slowly increased. since a magnetic particle's
Although
6.13.
oscillating varies,
energy,
remember field.
that a changing magnetic
1 is suspended a mass m
frequency,
field is slowly established
How does the electron's
Problem
6.15.
moved.
How do the frequency and kinetic £?
Problem
It has been hypothesized
1973, p. 45; Reines year.
Discuss
energy change? i,
p. 195, are slowly
energy of the ball vary with
by Dirac and others
the effect of changes
in G on planetary
i how the phase integral S arises
the solution of Schr~dinger's
of the form A(x]
exp[is(x)].
is pursued,
one encounters
an almost exact integration (6.1)
is possible
level.
since p involves
fails.
(Merzbacher
Here 1970, Ch.
in the allowed and forbidden
is replaced by
~pdx = (n + #)h which,
equation by an
the question what to do if
7), and on using this to connect solutions one finds that
in the
This is the WKB approximation.
there are the turning points where the approximation
regions
orbits and on
with Quantum Mechanics
of approximating
If the analysis however
(Mehra
1972, p. 56) that certain of the fundamental
We have seen in Chap. expression
to
to the moon, which can now be measured very accurately.
Connections
process
perpendicular
of nature are really changing by about i part in i0 I° per
the distance
6.4
If
a circular orbit in a Bohr
The walls of the box of Example
the separation
constants
As the acceleration
and energy vary?
in the nth quantum state, how does n change?
A magnetic
6.16.
k.
amplitude
the orbital plane of an electron executing hydrogen atom.
field
It is this that does the work.
at the end of a spring of stiffness
how do the spring's
6.14.
field which
it might at first be argued that
In the elevator of Example
it was initially Problem
One of the standard
it in a magnetic
field does no work it cannot posibly change the
kinetic
gives rise to an electric Problem
r
the two sides of the equation by (eB/m) 2
E, is an implicit
(6.29) equation for the nth energy
The wave function between the turning points a and b is
approximated
hy
207
~(=) =
Nk-@cos[
z
(6.30)
a ~ x _< b
and that in the forbidden regions by analogous This is the point of closest contact between
exponential
expressions.
the old quantum theory
and quantum mechanics. But the WKB theory does more than enable us to derive and correct the equations
of the old theory;
it helps us to understand how the new
one is related to what went before. expectation
Quantum mechanics
value of a dynamical variable
the corresponding
operator
a(p,x)
states that the
is given in terms of
~ by
= I~*(x)~(p~x)~Cx)dx
(6.31)
Let us see how this formula works in the especially which a is a function of x only.
We consider n to be large, but
will later return to the question of small n. the forbidden
regions
simple case in
contributes
negligibly
The wave function to the integral,
in
which
we write as
I
k-1(=)a(x)cos~[
a
so
kC=')dx' - I
(6.32a)
=
a
Ibk-1 (=)cos~ [
~]dx
[~k C x ' ) d = '
1~]~x
a
using
(6.30), where ~kC=)
If n is large,
= pCx)
the (cosines)ain
:
{2m[~-
vC=)]} ~
(6.32a) oscillate many times and can be
replaced by their average value of 1/2:
Ib [~l¢(x)]-la(x)dx Ja -- "'b v~ Let us interpret ~ik as the classical momentum:
(a) = I
J
m-l ~dx dt
208 and finally,
change the variable
of integration
from x to t:
Itb a[x(t)]dt ta
where ~ is mechanics averages the
the
Thus e x p e c t a t i o n
correspond,
in the theory.
first-order
to a rule
that
a small
in energy
by the
values the
system.
to time
which says
that
state
see
corresponds
system produces
of the perturbing
We s h a l l
from the adiabatic
rule
by a perturbing
unperturbed
of a classical
time average
in quantum
quantum numbers,
produced
in the
perturbation
in the unperturbed
this rule follows
of large
In particular,
change given
evaluated
limit
perturbation
A
H1 i s
(6.33a)
= ~
time average.
in classical
hamiltonian energy
=
an
energy
in Chap.
7 that
theorem.
Problem 6.17. Suppose that a is a dynamical variable containing p and x. Is (6.33a) still true? (Consider first ~ itself, then polynomials containing
terms like x2~m~ 3 ...)
Problem 6.78. out above,
Using
(6.12) and a calculation
similar to that carried
show that for large m, the dipole matrix elements
which define allowed optical
transitions
are given in classical
terms
by
cos(~amt)x(t)dt
a = m - n
(6.33B)
~0
where the time T covers a whole number of cycles, frequency
in the mth state,
It follows
and a << m.
from the result of the last problem that the matrix
elements of x correspond
to the Fourier components
variable,
and we see how to interpret
vanishes,
the quantum transition
corresponds
either.
proportional
of the classical
the fact that if
is not allowed.
to the same frequency,
at this frequency, physics
~m is the classical
The Fourier component
and if the system is not vibrating
one would not expect any radiation
Furthermore,
in classical
since the intensity of radiation
is
on the one hand to l
square of the amplitude Fourier amplitude,
of the electromagnetic
we see that intensities
are related by our formula.
wave coupled to the
as well as selection rules
209
The
Correspondence
Principle
At high quantum numbers, theories
are supposed
the results of classical
to correspond.
and quantum
This is an obvious
requirement
of any version of a quantum theory and of course it emerges equations
of the two theories.
approximation that there
is reasonably
is a one-to-one
semi-classical
methods
What is not trivial
accurate
useful theory.
between states deduced by
and those of the exact theory.
though heuristic
given above,
theory
(Sec.
although until
argument.
rule relates
to quantum mechanics
remember
the connection
between
S = ~¢.
If we traverse
the configuration
AS = J = nh,
principle,
of quantum
6.5) explained why it is true Bohr preferred
to call it the correspondence How Einstein's
One is not led
and it furnished a
guide in the early development
It is known as the correspondence
Heisenherg's
is that the WKB
at low values of n also, and
correspondence
to expect this by the arguments
from the
is clear if we
action and phase found in Chap.
space in any closed loop,
whence A¢ = ~ n
We see that the quantum-mechanical space in all its dimensions, condition
of periodicity
natural generalization freedom.
(6.34)
"wave" is a wave in configuration
however many they are, that
the wave must satisfy,
and that
of the usual quantum condition
from a moving boat the apparent
is the
(6.34) is the
to f degrees of
else.
The wavelength
If one studies ocean waves
frequency will change with the boat's
speed and direction but the wavelength
is a property of the water and
of a "matter wave," on the other hand,
is given by the momentum
of the matter,
motion of the observer.
The wave function
carrier of phase and amplitude,
and that depends partly on the in quantum mechanics
are described.
In the next
section we shall see that though phase and amplitude quantum mechanics, Heisenberg's
is a
which are the two crucial parameters
in terms of which natural processes
6.5
(6.34)
There really isn't any "matter wave" in the same sense that
there is a sound wave or an ocean wave.
nothing
I:
are essential
waves are not. Quantum Mechanics
After Hamilton had shown that there is an affinity between the equations
of Newtonian mechanics
and those of wave motion,
century elapsed before evidence was discovered relationship
might have a counterpart
almost a
that this formal
in nature,
It is not
to
210
derogating
from SchrSdinger's
discovery
to say that if he had not
followed his own path to his wave equation
(Sec.
would soon have derived it in another way. general
formulation
of quantum mechanics
9.7) someone
The first correct
the foundering
The successes
and
did not stem from the idea of
waves at all, but rather from a series of brilliant to correct
else
quantum mechanics
guesses about how
of Bohr and Sommerfeld.
of this theory were always more noted than its failures,
but by 1925 it was clear that the theory was giving wrong answers-the excited states of helium, calculate
by perturbation
for example,
theory because
should have been easy to
their energies
differ only
slightly from those of hydrogen but Born and Heisenberg ~ were unable to get agreement with experiment. possible,
using the separation
Similarly,
of variables
of Sec.
the electronic motion in the ion H2 + exactly effects
of nuclear motion)
but here again,
and so determine
experiment
Copenhagen,
was Bohr's
though Heisenberg
the right path,
2.12, to quantize
(neglecting
the small
the ion's binding
energy, 8
disagreed.
The locus of the discussions quantum mechanics"
it should have been
that finally gave birth to the "new
Institute
for Theoretical
Physics at
was quite alone when he finally found
in June 1925, on the North Sea island of Helgoland.
It is unnecessary
to try to follow the full history here--much
of it
is given in van der Waerden 1967 and it is full of false starts and good guesses,
but it may be instructive
the old quantum theory was modified
to see the formal way in which
into the new.
Heisenberg 9 started with the claim that an atomic theory should involve observable
quantities
and that it is therefore pointless,
example,
to talk about the orbit of an electron
modified
also the most characteristic
theory,
around a nucleus,
~mn
He
and singular result of Bohr's
that except in the limit of large quantum numbers,
frequency
for
the
of light emitted from an atom in a transition
from
state m to state n is not the same as the frequency of any of the electronic motions
mmn
inside the atom.
The trouble
is, of course,
that
depends on the final state as well as the initial one:
~mn and Heisenberg's
Ann.
~-~ (Em - En)
C6.35)
first step was to assume that all dynamical
7 M.Born and W. Heisenberg, 8 W.Pauli,
=
Z,Physik
quantities
16, 229 (1923).
d. Physik 68, 177 (1922).
9 W.Heisenberg, Z. Physik 33, van der Waerden 1967.
879 (1925).
Translated
and annotated
in
211
have the same kind of dual dependence:
a coordinate,
for example, was
to occur in the theory as a quantity
Xmn(t) = Xmnei~mnt though
its precise relation
The corresponding
momentum
to experiment
remained
to be specified.
is •
Pmn where v is the particle's
(6.36a)
=
°
~Xmn
--
(6.36b)
~mnXmn
mass.
What these two-index quantities mean was not clear, but fluctuating
quantities
for radiation manifest another
like the electric dipole moments themselves
during transitions
(m ~ n), while constant quantities
single state
(m = n).
this kind is suggested, can be understood us examine
such as energy refer to a
How to calculate with two-index but only suggested,
limit of large quantum numbers,
[see
quantities
of
by the fact that in the
the different
as Fourier harmonics
responsible
from one state to
spectral
frequencies
(6,9) and (6.33b)].
Let
this relation more closely.
Fourier Amplitudes Suppose which,
a one-dimensional
according
to Bohr's
Then its coordinate
system is in its mth quantum state,
theory,
can be represented
x re(t) = where the
the
state
Xm(t)
is
X~ a r e m
m.
a set
They
it oscillates
are
of
2
constant
not
all
with frequency
in
a m-
by the Fourier series
xae ia~mt
(6.37)
m
Fourier
independent
amplitudes because
appropriate the
to
coordinate
real:
x m*(t)
= Xm(t)
or
Za*e -ia~mt m
= ~ X a ei~mt m
= ~ X -ae -ia~mt m
so t h a t x~* = Xm-~
(6.38)
212 The correspondence (6.36).
principle
suggests
as calculated by classical given by
(6.35).
frequency
~
n
considerations
Classically,
and its Fourier
a system
according
a neighboring
to quantum
should be the same as that
radiates
theory,
(Note that here and subsequently,
aE an n
than infinitesimal
quantities,
to an
to a ratio of finite rather
as already remarked
In the limit of large n, the two frequencies like
n to
(6.39)
the "derivative" with respect
in integer steps refers
~ in expressions
from state
is
index which varies
from this that
at the natural
in a transition
~_2 (E n - E n . ~ ) : ~ - i
=
to
frequency
~ : I, ~, 3, ...
state n - e, the frequency
mqu
(6.37)
the radiated
harmonics:
mcl : ~ n whereas
how to relate
In the limit of large quantum numbers,
(6.37)
following
are equal. corresponds
(6,9).)
We learn to m - n in
(6.34) . Quantum
Amplitudes
Calculations
are performed w i t h functions
momenta.
To see how functions m a y be formed,
represent
x m 2 in a series
analogous
Zm2(t ) = Let 8 : Y - ~.
~
and
to (6.37).
We write
X~X~ei(~+~)~m t
Then x 2(t)
=
m
SO t h a t
of coordinates
let us see how to
the Fourier
~
amplitude for
(6.40)
x~xY-~eiY~mt mm
~,Y
Xm2 i s
S X~X7-~
(6.41)
mm
We have suggested above that the Fourier amplitudes the quantum amplitudes
in (6.36],
and that
X e corresponds m Now what corresponds
to
(6.41)?
to X
Combining
together w i t h their time dependence
correspond
in some sense, m3m-~
(6.42)
the quantum amplitudes
according
to C6.40] gives
to
213
i~-1(Em-Em_e)t X
Z Xm,m_~e a corresponding
to (6.41).
m,m_y+
e
i~-*(Em-Em_y+~)t
This does not have the required time
dependence (6.36a), but if we change it to
Z X c~
ei~ - * (Ern-Em_ ~) t x
ei~ -I (Em_~-Em_ Y) t
or, more succinctly, = ~ x
a
(m2)m'm-Y then
the
terms
agreement (6.40).
Then
with
Em_ ~ c a n c e l (6.36a)
The n o t a t i o n
(6.43)
as w i t h
k runs
over
Heisenberg's merely
first
condition
(6.39) the
invention. the
recognize
it.
The Quantum
let
quantity
m - ~ = k and m - y = n.
for
to to
of the
Note that
introducing
this
this into
agree with
for
This
point
is
he h a d s t o p p e d
a new n o t a t i o n ; the
new p h y s i c s .
multiplication
he h a d n e v e r
recipe
system.
at
quantum theory
matrix
(6.44)
E XmkXkn k
so as t o make i t
Apparently
according
a nd p r o c e e d e d
old
he was
rule
=
quantum states
an e q u a t i o n
recognize that
corresponding
c a n be s i m p l i f i e d :
all
transcribing
changing
the
is
(x 2 ]ran where
(6.43)
(x~)m,m. Y h a s a t i m e d e p e n d e n c e i n
and
as w e l l
m,m-~ x m-a,m-y
The r e a d e r
but Heisenberg
heard
by
frequency
of matrices.
will did not
He n o t e d
(XY)mn ~ (YX)m n
forming products,
draw m o m e n t o u s c o n c l u s i o n s .
Condition
In the old quantum theory,
Planck's
constant made its appearance
via the action integral,
Pvdqv = mvh (See.
6.1),
w h e r e t h e mv a r e
find how ~ occurs Again,
v = I, integers.
in a calculation
the calculation
2. . . .
Heisenberg's
f
C6.45) next
task
was t o
5ased on the two-index quantities.
by Fourier amplitudes
pro(t) = Z P~e i ~
mt
furnished a guide.
Let
214
Then in the mth state of our o n e - d i m e n s i o n a l
system,
2~m-1
J
Jp
m =
The integral
1
• mXmd t
t .... ~ iB~ t._ ~mXm e mav
~a £ ~ m ~m e
= a~S] °
is zero except for the terms in which B = -e:
Jm = ~ - - Z m
or J
The b a s i c increases
in
assumption steps
o f h, J
m
of Bohr's
(6.46)
~ aPeX -a
a
/nm
method of quantization
~ aPeX-a m r~
J
m
(6.41) was trans-
(6.43), -2~i
in which the exponential with r e s p e c t
that
= mh + const.
Transcribe it in terms of quantum amplitudes just as scribed into
is
so t h a t
= -2~i
m
= -2~i
to
2 aPm,m_aXm_a,m
(6.47)
= mh
Now d i f f e r e n t i a t e
time factors now cancel.
m! -2~i
with " d i f f e r e n t i a t i o n "
Z a ~m e
as in
(Pm,m-~Xm-a,m)
(6.39).
= h
H e i s e n b e r g now constructs the
integer relation of w h i c h this is the quantum limit. O~ ~'~ f ( r n )
= f(m
so that in the limit of small ~,
e with ~ = h/2~.
(Pm+a,mXm,m+e
+ a)
(6.48) becomes - Pm,m_eXm_e,m)
This is Heisenberg's
(XmkPk m
or in matrix notation,
He writes
- f(m)
= £~
second invention.
in the first term and m - e = k in the second.
k
(6.48)
- PmkXkm)
= i~
Then
(6.49) Let m + e = k
215 (xp
-
[6.50)
PX]mm = i ~
This is the first time that either the imaginary i or n o n - c o m m u t i n g dynamical quantities, what Dirac later called q-numbers, had ever occurred in physics
in any essential way.
A Simple Application The a p p l i c a t i o n of these ideas to a harmonic oscillator
is so
simple that it is not an adequate test of the theory, and even in his first paper H e i s e n b e r g w e n t on to consider the more difficult p r o b l e m of the energy levels of an anharmonic oscillator, but the simpler problem contains : mass and
some, at least, of the essential
Pmn = ~Xmn" i~
=
or
(6.50)
i~Z CXnk~knXkn k =
because
~kn = -~nk"
ideas.
With
is
-
mnkXnkXkn )
2 ~ ~knXnkXkn
The great s i m p l i f i c a t i o n here is that the
o s c i l l a t o r is harmonic;
its m o t i o n is at the frequency
there are no higher harmonics.
~n,n-1
= ~ and
Thus the sum has only two n o n v a n i s h i n g
terms: =
2~(Xn, n+lXn+l, n
-
Xn, n-lXn_l,n)
(~6. Sl)
Heisenberg assumes, i n analogy w i t h C6.38]tthat X
=
X
mn
so that
nm
(.6,51) is
C6.s2) In this r e l a t i o n let m = 0, c o r r e s p o n d i n g there
is no lower state,
Xm,m_ I
to the ground state,
Since
= 0 and
IXo,112 = We can now use
(6.52) to find all the IXm,m+:[ 2 =
Note t h a t Xmm = 0:
IXm,m+712.
(m +
:) 2~
C6.55)
the o s c i l l a t o r ' s displacement x has, of course, no
zero-frequency Fourier component, and the time-dependence of x is
216
manifest a part.
only during a radiative
Now let us calculate
transition,
the energy.
in which Xmn(m # n) plays
We assume the formula from
classical physics ,~ = ~ ~ ( x 2 + ~ 2 x 2 )
and transcribe
it as
Hmn The nonvanishing
H
mn
=
1J[ (X2) mn + ~)2(X2)mn ]
~
have n = m±2 and n = m.
Hm,m+ 2 = ½ P(Xm, m+iXm+l,m+2
We find
+ ~2Xm, m+lXm+l,m+2)
= ½ ~ [ ( . i ~ ) X m , m+l(_i~)Xm+l,m+ 2 + ~2 Xm,m+lXm+l,m+2 ~
and s i m i l a r l y ,
~Mcm-2. : O.
H mm = ~ ~[~m,m+iXm+1,m
The diagonal matrix
+ xm~milXM-l~m
= 0
element is
+ ~2(xm, m+lXm+ljm + Xm, m_iXm_1,m)]
or
By
this is
(6.53)I ,
or ~mm = (m + ~ ) ~ Thus the energy is represented diagonal matrix.
by the elements
(6.55) of a time-independent
The extra half in the energy was not entirely new.
It had been conjectured
earlier as a necessary
correction
to the
Planck formula but this was the first time it had emerged from a mathematical
scheme.
The unfamiliar Xmn.
Clearly,
oscillations. amplitude
quantity
in all this is the quantum amplitude
it has something Let
the
energy
to do with the amplitude
of state
A m as one w o u l d c a l c u l a t e
E m = ½ (2m + I ) ~
= ½p~
it
m b e (m + ½ ) ~ . classically
Am
is
of the Then t h e
g i v e n by
217
This is something
like
(6.53) but not the same.
thing about the calculation
of (6.5S)
has it been assumed that the oscillator The
idea
of a trajectory
And the remarkable
is that nowhere, is executing
even implicitly,
sinusoidal motion.
does not occur.
Equation of Motion Suppose that some dynamical variable f is represented by a matrix
fmn = Fmnei~mnt Then its time derivative
is
fmn = immnFmnei~mnt i = ~ (Hmm - Hnn)fmn i = ~ (Hmmfmn - fmnHnn ) If H is a diagonal matrix,
this is i
(6.56)
finn = ~ (Hf - fH)mn Heisenberg's dynamical
coordinates standing 1973.)
theory was put together out of a belief that the
variables
of atomic physics
and momenta
represent
in the classical
of the correspondence
sense,
relations.
something
and out of an under-
(See Heisenberg
He was also able to assume that, transcribed
quantum amplitudes,
the hamiltonian
would still be of use. classical triumphs
hamiltonians
functions
It is the micraculous
from classical physics
clothed in new mathematics
particles
may be much greater,
classical
physics will tell us what to write down.
The difficulties
And the theory was ill-adapted
variable
to computation.
failed to derive the Balmer formula; mechanics
of
original paper it is not at all clear what the
That was clarified gradually
of hard work.
and properties
for there is no reason to think that
quantum amplitudes mean or how a dynamical
months
of
that has led to the
of the future theory that will explain the structure
states at once.
in Mehra
into his new
effectiveness
of atomic theory in the last half-century.
In Heisenberg's
other than
Heisenberg
Pauli accomplished
In the next year Schr~dinger
and shortly afterward
can refer to two
in the next few years. tried and
it in several
invented wave
showed that it is mathematically
218 equivalent with matrix mechanics where the two overlap. for physical
systems
having a continuous for example--but erative motions
that can be described by coordinates range of variables--the
situations
involving
of many particles
They overlap and momenta
r, O, # of atomic orbitals
spin and isospin or the coop-
are usually treated by matrix
methods, which are in principle more general.
Problem 6.1~.
Assuming
that Heisenberg's
quantum amplitude
the matrix element <mlxln> , verify that the equations
Xmn
involving
is
this
quantity derived above are in fact correct. I°
Problem
6.20.
Show that the nondiagonal
are zero for the harmonic
oscillator.
Problem
constructed
6.21.
Heisenberg
as the expression argument,
the commutation
of the old quantum condition
new quantum amplitudes. quantum mechanics
(6.45)
In order to appreciate
start with the commutation that you need,
(px - xP)mn
elements
relation,
(m ~ n)
relation
(6.50)
in terms of his
the subtlety of the plus anything else from
and derive the old quantum condition
from it. Shortly after Heisengerg's to generalize
(6.S0)
to include
paper, Born and Jordan 11
showed how
the off-diagonal
as well,
elements
[xp - px)mn = i ~ m n The theory was then complete could solve or approximate of degrees of freedom. and detailed structure
in that if one worked hard enough one
any dynamical
It is natural
of Poisson brackets.
fundamental
discussion,
Jacobi
i0
to ask whether the more complex
This turns out to be the case.
Pauli 12 has shown that if one defines
structure of Poisson brackets
identity and postulates
all of classical
dynamics
Having established
actually
system with a finite number
of classical mechanics can be constructed
a knowledge algebraic
'
xx
M.Born and P.Jordan,
12
W.Pauli,
In a
the
an equation of motion of the form (4.57~
can be derived.
this
from
by means of (4.63) with the
we can verify that X_~ and X_ m ~ are
t h e same i n t h e c o r r e s p o n d e n c e Nuov.
(6.57)
Zeits.
cim. I0, 1176
.
.
limit.
f/l
See '~1~.~'2).
f. Physik 34, 858 (1925) (19S3), Sec.
2.
219
6.6
M a t t e r Waves In 1923 de Broglie :3 p r o p o s e d the existence of m a t t e r waves,
derived the r e l a t i o n b e t w e e n w a v e l e n g t h and m o m e n t u m by a relativistic argument,
and p r o p o s e d experiments by which the hypothesis
checked.
We shall not follow his arguments here, but will instead
could be
show some connections with what we have already done. The p r o p a g a t i o n constant k of a matter wave is given by
k = ~-~p = {~m~-2[E-
and the p h a s e change a l o n g an e l e m e n t of p a t h d~ = ~ - 1 ( p ' d r
so t h a t at
(6.58)
v(r)]} ½
- Edt)
dr a n d a t i m e d t
is
E = ~
t h e p h a s e c h a n g e i n g o i n g from a p o i n t
r ° at
t o to a point
r
t is r, t
A¢
i
= ~-I
(p.k
- E) dt
roJ t o
integrated natural
a l o n g the n a t u r a l
path,
path
we c a n r e p l a c e
connecting
E by H ( p , r )
r ° and r.
Since
it
is a
and
rj t
A¢ = ~-1
= ~-lW(r,t)
L(r,r)dt
(6.59)
Jro to where the n o t a t i o n a function
reflects
only of the
approximation,
the
fact
end p o i n t
a de B r o g l i e
that
of the
W is
(for
a given
integration.
wave o r i g i n a t i n g
at
ro,
r ° and to)
Thus, i n t h i s t o i s g i v e n by a
"wave f u n c t i o n " ~(r,t) where a i s
= a(r,t)exp
some s u i t a b l y
phase-integral but here if
W is In
mechanics provides (6.5a)
under contact used.
general
still:
complexity is
(6.34)
that
described
L. de B r o g l i e ,
This
though there
(6.60)
- Et]
is nothing
f r o m t h e wave e q u a t i o n
mechanics plus
no ~ , that
transformations. In
i [SCr)
de B r o g l i e ' s
but the i n S e c . 1.1~ hypothesis.
must be such a c o n s t a n t
as a p h a s e .
we h a v e n o t e d
to a complete oscillation dinates
derived
from c l a s s i c a l
t o be c o n s i d e r e d
= a(r,t)exp
chosen amplitude.
approximation
derived
Classical
i W(r,t)
the phase
integral
J is
Thus t h e a d v a n c e i n p h a s e
of the
system is
independent
we h a v e t h e s t a t e m e n t
t h e same a p p l i e s
177,
corresponding
of the coor-
of a hypothesis
more
to a system of arbitrary
by a n y n u m b e r o f v a r i a b l e s , Comptes r e n d u s
invariant
107,
148
not just (1923).
the three
220
that occur in (6.60).
In its full generality,(6.35)
in any number of canonical
coordinates.
describes
a wave
The wave is a wave in q
space, which only for a single point particle
coincides with the x, y,
z of ordinary space. Stationary
Phase
Let us suppose
it has been possible
equation and determine
the principal
to solve the Hamilton-Jacobi
function W(q,e,t).
From what
has been said above it might appear that W/~ is the phase of the wave function of the corresponding corresponds constant, consider
quantum system, but in fact the phase
to a new characteristic
which we shall call U.
function,
first the time-independent
(6.61)
in Sec.
5.7.
to the phase of a t i m e - i n d e p e n d e n t
is Seen if we consider distribution
to
- 8n~ n
where the 8Vs and e's are those introduced corresponds
from W by a
case, for which we define
U = S(q,~)
quantity
differing
It will be most convenient
a wave packet characterized
That this wave function
by a narrow
of the ~'s,
~(q) = ; A(~)ei~-lUdf~ where tion
df~ = dalda2...da f
and A ( a )
for
centered at ~n as in Fig. 6.4.
Figure 6.4.
is the condition
for stationary
function ~(q) will differ appreciably of the phase with respect ~U
~(~ n
or
each a n represents
The condition
a distribu-
for constructive
Distribution in A (~n) that forms a wave packet.
interference derivative
(6.62)
o~ n
=
"~ n
=
0
phase, Sec.
1.2.
The wave
from zero only where the to each ~n is zero.
n
=
1,
...
f
This is
221
~S "~ as in (5.57~.
We see that
satisfies Jacobi's
n
Sn
(~ = ~ ' = n n
(6.62) represents
a wave packet whose center
equations.
In classical dynamics, with
(5.57) satisfied,
a variation in U is
given by @S
6u
=
~qn ~qn
~S + ~
~en - S n ~ n
~nSSn
-
or
6U = ~BS n
6qn - an~8 n
Thus U is conveniently regarded as a function of q and 8 with BU The difference
~S
~U
in meaning between U and S is clear when we remember
that the e's correspond to conserved dynamical variables and the 8's to initial positions
in q space.
Thus S, with given values of the e's,
allows one to calculate families of orbits corresponding to different initial positions,
whereas
U, with given initial positions,
to calculate families of orbits corresponding the e's.
allows one
to different values of
It is the interference between different members of these
a-families
that creates the wave packets.
We now see why the speed of a light signal in (5.91) came out to he the group velocity, and Wheeler
and it is for this reason that Misner, Thorne,
(1973) refer to the Hamilton-Jacobi
"condition for constructive
equation as the
But note that what has
interference."
been done here is in no sense a mere extension of the Hamilton-Jacobi theory or an alternative way of expressing the argument, introduced an idea that
is entirely new to mechanics,
of superposing waves to form a wave packet.
for we have
the possibility
Though this is a funda-
mental feature of classical wave theories it makes no sense at all within the framework of classical mechanics. Hamilton-Jacobi mechanics
The assimilation of
into a wave theory depends on two entirely
new ideas, superposition of wave amplitudes
and introduction of a
universal constant having the same units as S. of Hamilton's
From the publication
first paper "On a general method of expressing the paths
of light,
and of the planets, by the coefficients
function"
(1833) to de Broglie's
tentative
because it still lacked a wave equation,
of a characteristic
hypothesis,
took 90 years.
rather useless From the
222
W r i g h t brothers
Problem 6.22.
to the moon landing Show, by adapting
that it is impossible corresponds and 8n"
standard methods
and definitions,
to form a wave function of the form
simultaneously
to arbitrarily
What "indeterminacy
Problem 6.33.
took 66 years.
principle"
How does one change
that
(6.62)
narrow distributions
of
n
results?
(6.62)
so as to produce
a time-
dependent wave function? 6.7
Construction
of a Wave Function
The difficulty with a wave function
like
only to the natural paths of the particles that waves
diffract
(6.64)
is that it refers
described,
whereas we know
and do not follow the natural paths
following
argument,
principle
of superposition
functions
out of action integrals
exactly.
The
due to Feynman, I~ shows how one can use the to construct
true quantum-mechanical
evaluated
along paths
wave
that are not
natural. To find the wave amplitude
at r in Fig.
6.5, Huyghens'
principle
A
[ Figure 6.5
r
The amplitude at r is found by coherent from the slits.
states that one must superpose r via the different
slits,
superposition
of contributions
the rays that have travelled from r ° to
each with its proper phase.
Assuming
that
we know the wave function to the left of the grating, we construct the amplitude the slits,
at r at time t as the sum of amplitudes
contributed
by
the ith one being i ~Wi(r,t)
C6.63)
aiwbi(r)e
*~ R.P.Feynman,
Revs. Modern PhFs.
20, 367
(1948).
223
amplitude
where a i is the slit width,
at slit i from the source at ro, w is the
and
!W.(r,~ wbi(r)e ~ * is the amplitude,
together with its proper phase, produced
time t by a wave of unit amplitude arriving at the slit. b. involves
the inverse-square
do not know it a priori. of course,
Hamilton's
law and some obliquity
at r at The factor
factor,
It must come out of the theory.
principal
function
evaluated
but we
Wi(r,t ) is,
along the natural
path from the slit to r. The role of the barrier was to allow us to discuss tion to the amplitude different grating, of Fig.
points
at r made by Huyghens
along the plane of the grating.
all points of the plane 6.6.
the contribu-
wavelets originating
contribute,
If we take away the
and we have the situation
at r is not just that contributed
The amplitude
at
natural path from r ° to r, which is a straight
by the
line.
.A !
r
X
'A Figure 6.6
Every point of the plane can be regarded as a sllt.
Finally we recognize
that there is nothing
plane A-A or this particular consider
set of broken-line
all imaginable paths from r
special
about the
paths.
We therefore
and r that occupy the same time O
interval.
Fig. 6.7 shows the natural path together with some of the
varied paths.
We have introduced
path is curved.
a force field, so that the natural
We must evaluate W along each path,
calculate e iW/~,"
and find some way of summing over the different paths. W differ for different paths; because ~ is small, widely,
The values of
the W/~ differ
and the terms of the sum tend to cancel each other.
Now however we make use of the stationary property slight variations The situation
of W:
for
away from the natural path, it changes very little,
is the same as in Sec.l,2 and Sec. 6.6, where we found
the motion of a wave packet by the principle
of stationary phase.
224
Only varied paths in the immediate neighborhood of the natural path contribute
appreciably to the sum. Is
X
Figure 6.7
Every path contributes to the amplitude at r.
It is difficult to carry out the sum in order to evaluate wave functions;
the problem is one of weighting and lies in the mathematical
domain of measure theory.
In spite of considerable
effort,
few calcu-
lations have been done that could not have been done otherwise,
but in
his first paper Feynman at least shewed that wave functions so constructed will be those with which we are already familiar. Consider the simplest case of one-dimensional motion, drawn in xt space in Fig. 6.8(a)
i.
The last infinitesimal
x, i~
segment of the path is
.
x, t t
L-E; X
Figure 6.8
(~) Path in one dimension represented in Lagranglan qt space. Magnified view of the end of the path.
I
(b)
shown in Cb), where the trajectories have drawn close together and only a time ~ remains before they arrive at (x,t). 15
If the wave
That wave functions might be constructed in this way was suggested long ago by Dirac, Phys. Z. Sowj~tunion 5, 64 C1933), reprinted in Schwinger 1958.
225
amplitude at p o i n t x, and time t is given by some ~(x', t -s) we can construct the amplitude at (x,t) range dx'
starting in the
at x',
: I ~(x',
~(x,t) where dm'corresponds natural
out of contributions
to w in
one contribute,
The i n t e r v a l s
(6.63).
we t a k e
are
t - E)b(x')ei~-lWdz'
now a l l
Since
b outside
only paths the
infinitesimal.
close
to
the
integral. We w r i t e
k as
(x - z ' ) e -1 s o t h a t
=½
n
(~
x') ~
-
m ~ _
- V(z)
and (x
W = L ¢ = ½ m
E
x')
2 -
(6.64)
mV(x)
It is c o n v e n i e n t to introduce ---- X
-
dx' -- -d~
Xtj
so that
~(x,t)
Since
only
O(x,t)
values
= b
I
~ exp ~i "2-~--rm~ a eV(x)]~(x
of x'
= be - i ~ ' 2 V ( x )
near x matter,
exp(
~
~
- ~,
t - E)d~
we e x p a n d i n p o w e r s
[¢(z, t - e) - ~ ~ ( x ~
~x 2
o f 6,
t
+ ... ] d~
Since
I
~e-aX2dx
=
(ITa- 1
__
(i~a-1)~
3~
46.65)
we have
I
~eiax2dx
_co
(6.66)
226
while
I
I
• =x2e~aX~dx
~ xe i aX z dx = O,
-
oo
i ~-da
"
(i~a_1
)½
(6.67)
Thus
~(x,t) = b . ~ . Expand each side in powers
~(x,t)
= b . ~ .
We get agreement
ie~
[~(x,t-~) +
~2~(x~t-c) ~x 2
of e, keeping only e°and
...] +
e:
×
in the limit
e ~ 0 if
b = (2-~e) ½
(6.68)
The first power of e then gives
which may be familiar. The mathematical Feynman,
method is not familiar.
"One feels," wrote
"as Cavalieri must have felt calculating
pyramid before the invention of calculus."
the volume of a
Wiener had done something
along the same lines in the 1920's in connection with the mathematical theory of Brownian motion, 16 but the subject was, and remains, unknown to physicists? 7 The method's Feynman and Hibbs Insights
Gained
possibilities
are explored
in
1965.
from
Huyghens'
Principle
What has been done here is very important, far the most direct route from the equations the Schr6dinger
largely
equation.
for it furnishes
by
of classical physics
to
There are certain cases in which this
z6 N.Wiener, J.Math. and Phys. 2, 131 Revs. Modern Phys. 33, 79 (1961).
(1923).
See also S.G.Brush,
1~ A clear and detailed review, with extensive literature references, is D.I.Blochintsev and B.M.Barbashov, Soy. Phys. Uspekhi 15, 193 (1972). See also J.B.Keller and D.W.McLaughlin, Am. Math. Monthly 82, 451 (1975).
227
directness
is a great advantage,
prescriptions unambiguous equation
for "quantizing"
results.
An example
An equation of motion, interpretive
apparatus, density,
however,
(1965).
commutation Feynman's
The mathematics
The interpretation
apparatus
of
are derived in Yourgrau
is a momentum,
generalized
and
and Mandelstam
and the
(1968).
to include the dynamics is, so that physics
royal road of hamiltonian mechanics.
theory leads in a straight
the
I~I 2 as a
are given by Feynman and
any more than Schr6dinger's
needs Heisenberg's
One needs an
is done somewhat more generally
theory is not readily
spin and isospin,
is not a theory.
the proof that <-iN3/~x>
relations
of Schrodinger's
and one needs to be able to generalize
other parts of the interpretive Hibbs
theory do not yield
is the establishment
laws to new situations.
probability
in cases in which the usual
,8
in a curved space.
dynamical
notably
a classical
of
still
But Feynman's
line from the Hamilton-Jacobi
theory to
the wave equation and shows where and why ~ comes in, and with it we have succeeded classical
Finally, In Sec.
the argument of Sec. I.i, in which
there is an interesting
2.6 we expressed
Hamilton's initial
in reversing
theory was derived from the quantum theory.
principle,
question of physical
the equations
which states that of all paths connecting
and final states in qt space, that actually
one which minimizes
the action integral.
as a purely mathematical
prescription
a question of the general
followed
is the
This can of course be taken of motion,
sense one inevitably
encounters
form "How does the system know which path
gives the least action if it tries only one of them?" leaves
given
for finding equations
but if one tries to give it a physical
dynamics
content.
of motion in the form of
this question unanswered,
Classical
but in the light of this
discussion we can say that it tries all of them, and that the natural path emerges as a result of the property Problem
Show by considering
6.24.
that the integral of (6.66) be justified?
Problem 6.25.
Derive by Huyghens'
free particle moving
the area under successive
converges.
the derivation
of stationary phase.
What about
principle
(6.67)?
loops, How can
the wave function of a
in a unit circle starting with an arbitrary
(X,to). S o l u t i o n . 18
There is
natural paths leading ,8
a trick
here,
for
t h e sum o v e r a l l
possible
to x includes not only paths in which it moves a
For a recent discussion ( 1 9 7 2 ) ; 24, 980 ( 1 9 7 3 ) .
see K.-S. Cheng, J. Math.
Phys. 13, 1723
228
short distance
from x' to x b u t
also those in which it travels a whole
number of times around the circle in going from x' to x. corresponding
to n revolutions
and ~(z,t)
is
= 2 mX e
Wn
= b
~ n=--eo
The action
(x + 2 ~ n
12~e
i~-IWn(x)
- x') ~
~(xv,t-e)dx
'
0
Now show, assuming
that ~(x',t-e)
is single-valued
on the unit circle,
that
*(x,t)
and complete include the W
= b
I ~
the calculation. n
for n~
O?
im
exp[2--~ (x - x ' ) a ] ~ ( x ' , t - e ) d x
'
What goes wrong if one forgets to
CHAPTER 7
THEORY OF PERTURBATIONS
The number of calculations either
in classical
or in quantum dynamics,
the number of useful results part of physics, imations,
that can be carried out exactly,
so obtained
engineering,
and dynamical
and the worst difficulties
problem are no longer separable. corresponding
problem
but the resulting
is comparatively
is even smaller. astronomy
involves
arise when the variables
approxof a
What one hopes to do is to find a
in which not only are the variables
differential
small and
The greater
equations
separable
can be solved exactly,
and
then move from the solved problem to the unsolved one by a systematic procedure
of approximation,
The procedures in somewhat
usually
involving
analogous
fashions,
very far since what is observable
theories.
An astronomer,
for example,
in the sky and has less interest an
electron's
in series.
can be carried out
but there is no use in pursuing
comparison
physics
expansion
in quantum and in classical mechanics
position
wishes
differs
the
in the two
to calculate positions
in knowing energies,
whereas
in atomic
is not an observable while its energy
is. Because astronomers' cal theory of perturbed
need for precision
the most eminent mathematicians Because alone,
and our discussion
ous situations 7.1
subject we shall mostly leave it
of the classical
theory of perturbations
to examples which throw some light on analog-
in quantum mechanics.
Secular and Periodic
Perturbations
To someone who has encountered mechanics,
the classi-
in history have had a hand in it.
it is a very specialized
will mostly be confined
is very great,
orbits has been carried very far, and many of
perturbation
theory in quantum
it might seem quite obvious how to proceed
in classical
mechanics.
Let us say that we wish to calculate
the perturbation
the Earth's
orbit by the action of Mars.
that the perturbing
force or energy carries
Assume
of
a factor e and write the Earth's position as
r(t) = ro(t ) + erl(t ) + eZr2(t ) + ... put it back into the equations of the same order in e. perturbation
of motion,
and gather together
In this procedure
terms are small,
that
lerll
not so, since even a small perturbation,
terms
it is assumed that the
~< Irol , but this is just if allowed to run long enough,
will finally bring the Earth to a point very far from where it would otherwise have been.
230
To see what happens let us use perturbation of a harmonic changes
oscillator
the stiffness
(if the amplitude
if one disregards
this warning
theory on a trivial problem,
by the addition of another spring that
k to k + e and the frequency
is fixed)
the oscillator's
x(t) = asin(m
asin~t + ...
increase without
ment remains bounded.
It is spurious
limit, whereas
of course
because
the displace-
To get a finite result we must sum an infinite
number of terms of the perturbation can be done in another simple Anharmonic
(7.2)
+ ...
is called a secular perturbation.
Ix - xol
Then
becomes (7.1)
= Xo(t ) + ¢a~itcoset
it has
to ~ + e~ I.
displacement
+ eml)t
= asin~t + E ~ I ~
This result
and proceeds,
the p e r t u r b a t i o n
series
in ¢.
Let us see how this
example.
Oscillator
The example
is a harmonic
oscillator
force so that the equation of motion
perturbed by an anharmonic
is
+ ~02x
= cx 3
(7.3)
If we try to solve this by letting X = X 0 + 6X 1 + £2X 2 + ... we see how the secular perturbation 2
..
2
2
""
(7.4)
arises: 2
3
2
XO + ~oXo + E(Xl + ~OXl ) + ~ (X2 + ~oX2 ) = £(XO + 3gxoXl + " ' ' ) whence Xo + ~ 0 2 X o
= O
xl + ~02xl = x03 x2 + ~oZX2 = 3 X o 2 X l etc.
The first equation has a solution x 0 = asin~ot
Then the second equation
is
(7.s)
231 ~i + ~02Xl = a3sin3~ot
this equation without knowing
If we encountered judge
= % a3(3sinmot
it to be the equation of a harmonic
force having a Fourier component The resonance would transfer tionally
to t (Prob.
oscillator
an ultimately
amount of energy
would ultimately
increase propor-
7.1).
that the frequency
(7.1)]
driven by a
divergent
To get out of the difficulty we suppose example
its context we would
in resonance with the oscillator.
and its amplitude
to the oscillator
- sin3eot)
changes
[as in the primitive
from ~0 to m, where
~02 = ~2 + el I + e212 + ... and choose l I so that the secular perturbation
vanishes.
(7.6) The terms
through e 2 are given by
Xo + el41 + e2x2 + (~2 + et I + e 2 1 2 ) ~ 0 + ex 1 + e2x2 ) = e(Xo + eXl)3 Suppose we require
that 2(0) = o
x(O) = a, The
zeroth-order
(7.7)
equation
of w h i c h the solution
(7.7)
from
that satisfies
(7.8)
is
C7.8)
is
X 0 = aCOS~t The next equation
is
~i + ~2Xl
= -llXo
+ X03 + % a3(3cos~t
= -llacos~t and the resonant
secular
term cancels X1
The remaining
=
if
$ a~
(7.9)
equation Xl + 0°2Xl = % a3cos3oJt
has a solution
+ cos3~t)
(7.10)
232
xI
= ~
G
3
[cosmt
-
cos3mt)
(7.11)
which when combined with x o is consistent with the boundary (7.8).
condition
The frequency we have found is
= (~02 - e~l)#= ~0(I -
S~l 3~a 2) ) = ~0(I 2~02 8~02
(7.12)
and the solution so far is £a 2
xCt)
The secular
= act
~
~a 3
+ 3--~-~T,cosmt - ~
term is gone and only periodic perturbations
see the origin of the word anharmonic: the amplitude
of swing.
series because
of this kind occur in quantum mechanics;
examples are in the theory of many-body interacting
fields.
We
in e, but in fact if we were to expand
of e it would be an infinite
Difficulties
remain.
the frequency now depends on
In one sense we have stopped after the first
term of the series expansion x(t) in powers
e.
(7.13)
cossmt
m contains notable
systems and in the theory of
The method used here is there called "summing
an infinite number of diagrams,"
and is necessary
if finite results
are to be obtained. Problem ? . 1 . frequency m.
mr
Find
the
Problem 7.2.
(7.11)
A series
L-C c i r c u i t
= (LC) -½ i s
i(t)
current
driven when
Find the general
(no by
resistance)
a sinusoidal
m < mr,
m = mr,
with
a natural
voltage and
of
frequency
m > mr -
solution of (7.i0) and verify that
is the right solution to choose.
Problem ?.3.
Look again at (2.27) and verify that the results of
Sec. 2.3 agree with those obtained here. Problem
?.4.
Problem ?.S.
Find the next terms in e in the example Discuss
is very slightly 7.2
the motion
damped with a force proportional
Perturbations
the calculations
we ask are easier:
if it
to ~.
a wave function,
are easier because
though the oscillator
physics may end up far from where perturbation,
oscillator
in Quantum Mechanics
In quantum mechanics questions
of the anharmonic
above.
the
in classical
it would have been with no
in a nondegenerate
little changed by a small perturbation.
state at least,
We expect an expansion
in
is
233
powers of e to go through, perturbation
correctly.
and it does,
if we handle the secular
It is easy to arrange things so that the
danger of secular perturbations
is not in evidence;
this is what is
usually done, but we shall be a little more naYve here. The problem
to be solved is i~ ~
where,
= (H 0 + eV)~
(7.14)
in general H 0 and V are both operators,
moment t h a t
V is independent
been possible
of time.
and we assume for the
We assume also that
to solve the unperturbed
it has
problem to find the stationary
states, ~ n (°) i~ ~ Ot
(o) = HO~ n
(0)
(7.15)
= En~ n
~n (0) = ¢n e - i K - I E n t
We identify
the state sought by saying that it is to be that state
which in the absence of the perturbation the unperturbed
stationary
Let us assume t h a t
would coincide with one of
say ~a (0).
states,
the spatial part of ~ can be represented
by
a sum
~a + where c
that
n
Cn¢ n
(7.16)
is of order e, but that the time dependence,
of ~a'
i s changed by t h e a d d i t i o n =
(7.14)
instead of being
o f an e n e r g y At:
, . .ei~ -I + AE) t (Ca + I Cn@n) (Ea
where the prime indicates this into
E
n~a
that the term n = a is omitted.
(7.17) Putting
gives
(E a + A E ) ( ~ a + E ' C n ¢ n) = E a ¢ a + Z ' C n E n ¢ n + eve a + eE'CnV¢ n
or AECa
Now multiply
+ 2'(E a + At - En)Cn¢ n =cV¢
by Cm* and integrate.
At = ~<~Ivl~>
a + eE'CnV$ n
(7.18)
If m = a,
÷ ~Z'c <~IVln> n
(7.19)
234
where the matrix element < a I V l n >
is defined as
(7.20)
TO o r d e r ~, perturbation
(7.19) gives the familiar of energy, AE
To c o n t i n u e , we r e t u r n and i n t e g r a t i n g w i t h m ~ a : (E a +
Cm = E
AE
e
=
to
- Em)C m
(7.18),
=
formula for the first-
+
O(e 2 )
this
£ <mlVIa>
+
time multiplying
by Cm*
eZ'Cn <m[Vln>
e - E (<mlVla> + Z ' C n <m]VIn> ) + AE m
a
order
(7.21)
m ~ a
All these relations are exact, and they form the basis for an iterative evaluation of AE and cm .
For example, using the first term of (7.21)
in (7.19) gives
AE
=
e
+
¢2
~, n
E
a
+
(7.22)
"'"
n
and further terms may be written by inspection.
The only difficulty
with the application of this formula is that it contains AE implicitly and must be solved for it, but this expansion is simpler to write down than the usual expansion in E that contains AE only on the left-hand side.
It was first given by L~on Brillouin in 1933. As in the discussion of classical perturbation theory,
not really an expansion in e, since the expressions
this is
(7.17) for ~ and
(7.22) for AE contain all powers of e, but it achieves a formal simplicity beyond that of the more conventional Schr~dinger perturbation theory precisely because it goes to the root of the problem of secular perturbations.
Although the problem solved in this way can
be dealt with otherwise in other forms of perturbation theory, most forms of the quantum theory of fields require the correction to the frequency to 5e made explicitly as we have done here.
The process by
which this is done is known as renormalization.
Calculation
of
Perturbed
Wave
Functions
If we use the first term of (7.21) to calculate a "first-order" perturbation to the wave function, we have
235
Cm
=
~
+
AE
-
<mlVla>
E
a
0(e 2)
+
(7.23a)
m
and
<mIvl=> + AE
¢ = ~a + A¢,
A¢ = e E' E m
Sometimes
- E
a
only a few of the matrix elements
~m
(7.23b)
m
are different
<mlVla>
zero and the sum is easy, but sometimes
it has many terms and,
addition,
in the form of an integral
it may involve
a contribution
from a range of energies this reason the solution H^ 0
- Ea
E
in w h i c h
m
it is often convenient of a differential
the spectrum
equation.
Multiplying
- AF)A~, = -~
z' <mlVl=>% : -~(Z ~ l v l a > m
to
by
of V~ a in terms of ~m' so with
in (7.22),
(~1 o
The quantity
Cm -
m
The sum is just the expansion --
(7.23b)
For
of A¢
AE gives
(#o - z
AE
is continuous.
to reduce the determination
from
in
Ea -
on the right
involves
In the ordinary o n e - p a r t i c l e
= -e(V-
(7.24)
a
only the unperturbed wave function.
Schr6dinger
equation,
where
2 V2 + V o ( r ) ~^o = - 4~-~
it is useful
to write
A{ = E X ( ~ ) ¢ a ( r ) A short calculation
-
then shows
~2 ~-~ [ v 2 x
that X satisfies
VCa'VX + 2 - - ]
- <=lvl=>
(7.25)
x = -v + <=lvl=>
Ca
T h i s may be r e a s o n a b l y Problem
7.6.
Show that
easy t o s o l v e , (7.19)
is, in an obvious notation,
A~ = ~ ( < a I v i a >
Problem
first-
?.7.
Find the first-order
and second-order
oscillator p e r t u r b e d
perturbed
+
perturbed wave function energies
by an interaction
~x ~.
and the
of a simple harmonic
236
Problem
Using the method just explained,
7.8.
find the lowest-order
correction to the ground-state energy and wave function of a hydrogen atom in an electric field F. Problem
Show that if AE is smaller than all the energy differ-
?.9.
ences E n
Ea,
the implicit equation
(7.22) can be solved for AE to
give, to third order, an expansion for AE:
< IvIn>_
m n
+
a
(Em - E a ) ( E n - E )
m
(Em _ Ea) 2
]
...
(7.26)
This series in e is the original perturbation expansion given by SchrSdinger in his earliest papers.
The second p a r t of the third-
order term is called the wave function renormalization
term.
Such
terms appear in increasing profusion when the series is carried further (though,
7.3
in practice,
nobody does).
Adiabatic Perturbations In this section we shall compare the classical and quantum
theories of a perturbed bound state, tive result.
in physical content and quantita-
The physical content of the quantum perturbation theory
is clear, especially if the unperturbed state is nondegenerate:
a
bound state is defined by a potential energy and the usual boundary conditions
at infinity.
One tries to alter the known solution of a
problem similar to the one in hand in such a way that the eigenvalue equation is satisfied approximately and the boundary conditions exactly.
If the unperturbed state is degenerate,
set of stabilized eigenfunctions
one determines
a
in a way which is explained in texts
on elementary quantum mechanics and it is these that are altered by the perturbing potential. is not so clear, met,
The physical sense of the classical theory
for there is no criterion for a stationary state to be
and any number of states of roughly the same energy could be
taken as the effect of perturbing a given unperturbed state.
We need
a criterion by which "the" perturbed state corresponding to a given unperturbed one may be defined. The criterion if found in the adiabatic principle:
let us
consider the calculation not as a certain expansion in powers of e but as the a t t e m p t miraculous,
to find out what happens if by some means, physical or
the perturbation is established very slowly.
The adiabatic
237
theorem (Sec.
6,3 ) t h e n s t a t e s
that
S = J~pndqn : ~ p n q n d t remains constant in generalized
in value.
coordinates
It follows
Suppose that
initially
addition
2~[E(t)
- V(t)]dt
t h e e n e r g y was Eo, a c o n s t a n t ,
Then g r a d u a l l y
of a perturbing
of energy
that this is
S :
e n e r g y was V0 .
from the definition
the potential
Vl(t ).
Then t h e c o n s t a n c y
Vo :
Eo
~-
~o -
and the potential
energy is
c h a n g e d by t h e
of S implies
that
Vz
or
E: As long as V 1 is changing, that V 1 stops changing
the constant
EO + VZ
E varies during a cycle, but if we imagine
when it reaches
the desired value, ~ becomes
energy of the system, E = Eo + F1
(7.27)
This very much resembles
the first-order
for we have seen in Sec.
6.4 that expectation
mechanics theless,
correspond
to time averages
of energy is obtained
the unperturbed
ing the perturbation? quantum-mechanical
mechanics.
if the average
If the perturbing ly modify our earlier
Never-
V is computed using Is the perturbed
also the result of adiabatically
establish-
This can be answered when we have found the
description
Adiabatic Transitions
in quantum
limit, while the first-order
rather than the perturbed motion.
energy in quantum mechanics
(7.17)
in quantum mechanics,
values
in classical
(7.27) is exact in the adiabatic
perturbation
replace
formula
of adiabatic behavior.
in Quantum Mechanics e n e r g y V d e p e n d s on t , treatment.
we s h a l l
have to slight-
S i n c e 5E now d e p e n d s on t ,
we
by
= (Ca + X ' C n ( t ) ¢ n ) e - i ~ - l j ( E a + b E ) d t Substitution
into the wave equation gives
(E a + AE)(Ca + Z'CnCn) + i~E'enCn = EaCa + Z'CnEnCn + eV¢a + eZ)CnVCn
238
or AE¢ a + E ' (E a + AE
- En)Cn~ n + i~E'Cn~ n =
~V~a
+
g~ I cnV~ ~
From this we get as before
AE(t)
=
e
+ eZ'c
n
(t)
(7.28)
>
while if m ~ a, i~c m +
Em)a m = E <mIV(t) la> + Z'an<mlV(t) ln> (7.293
(E a + h E ( t )
If we calculate Cm only to the first order, i~c m
+
(E a
t =
-~;
V(t)
= eYt~,
(7.30)
~<mlV(t)[a>
Em)C m =
Now let us suppose that V(t)
(7.29) becomes
has been turned on continuously from
where y is eventually
to become very small.
The
solution of (7.30) is
e re(t) -- - ~ i
m
which with y÷O is exactly
AE(t)
That is, nothing
=
(7.23a)
e
-
to first order.
~ 2r.'
(to second order)
The energy shift is
m
-
E
adiabatic and stationary state perturbation
function of time.
in the sense of time average and expectation value,
only to the first
The expectation value
is taken with respect to unperturbed wave functions,
the time average in (7.27) motion of the system.
is with respect to the actual perturbed
The analogy is pursued in Problem 7.13; let us
see how good it is from a numerical point of view.
Example
1.
The
for the perturbed energy correspond,
order in e, but this is not surprising. < a l V ( t ) la>
The
theories give the same
This is the adiabatic theorem in quantum mechanics.
classical and quantum formulas
(7.32)
a
is changed in our earlier result
except that AE has beocme a slowly-varying answer.
i-~
a
Anharmonic
Oscillator
The hamiltonian corresponding
H = p~+ ½ m~o2X~ + V 2m
to (7.3)
is V =
gmx ~
239
Classical mechanics An u n p e r t u r b e d motion has
Ec
Xo = ac°s~0t" The first-order
perturbation
Ec( 1 ) the bar denotes
= ½ mco 2a2
0
of energy, which
over a cycle of the unperturbed motion,
where
(05
% ~mx-T= ~s
=
(i)
3
ema
(7.33)
~
In terms of Ec(°)
a time average.
Ec
is the average of V
is
e
this
is
(0)2
(7.34)
= 8 mO~o~ Ec
Quantum mechanics E
(1)
(7.3s)
=
qn
or
Eqn
(1)
3 e ~2
8 m o~02 [(n + ½) 2 + %]
Since Eqn(O) = (n + ½)~U~O, this
Eqn The classical
(15
3
is
C
- 8 m~04 [Eqn
and quantum results
(052
+ Eqo
correspond
the zero-point
energy E (o) qo
Problem 7.10.
Calculate (7.33) and (7.3S).
Problem 7.11.
Calculate
result
in classical
(0)2]
(7 36)
at high energies where
can be ignored.
and compare the perturbed
and quantum
energies
theory w h e n a harmonic
placed in a uniform field of force.
that
oscillator
(This is a second-order
is
calcula-
tion.)
Problem 7.12.
Solve the preceding p r o b l e m
new coordinate
$ = x + ~ and choosing ~ appropriately.
Problem 7.13.
Let , be the exact solution of =
s~
and ~(o) be an e i g e n f u n c t i o n ^
~o ¢
=
of
exactly by introducing
Ho +
o
(o)
E(0) =
¢
(05
v
a
240
¢ so that 1~ *~(dx) = 7.
normalized
E = E (°)
Show t h a t
(7.37)
+ (¢(°)~V~(dx) 4
exactly.
If ~ is the state that results
tion is established close)
7.4
adiabatically,
from ~(0) when the perturba-
this is the closest
(but not very
quantum analog of (7.27).
Degenerate
States
Everybody knows that in quantum mechanics, applied to a degenerate splits. classical
state the degenerate
(We shall see below what the criterion mechanics
the situation
energy we read frequency. apparatus was developed
for "usually" the same,
is.)
In
if for
(see for example Corben and Stehle 1960)
for
but it lies beyond the cutoff of this
quantum mechanics,
is of narrow interest, A Classical
is essentially
is no longer current.
place of elementary
is
In the old quantum theory an extensive
dealing with this situation, book and moreover
when a perturbation
energy level usually
Since the subject
is a common-
while in classical mechanics
it
our discussion here will be very short.
Oscillator
We start with a nondegenerate
oscillator
in two dimensions,
for
which pl ~ + p2 2 H0 =
(7.38a)
+ ½ m(~ol2Xl 2 + t022x22 ]
2ra To this precessing restoring
system
(Fig.
7.1) we add a symmetrical
but nonlinear
force V = % e ( x 2 + y2) 2
(7.38b)
Y'1
Figure 7.1
Orbit of a harmonic oscillator in 2 dimensions. ) Degenerate case, e I = ~2"
~) el # ~2"
241
Writing
the unperturbed
oscillation
in parametric
x = asin0~l% ,
form (5.70) as
y = bsin(~2t + @)
we note that E (0) = ½m(~012a + 0322b)
independent E ~)
(7.39)
of ¢, but that E (I) involves ~:
= % e[a~sin--T~-~it + b~sin~(~2t
In the first two terms, done carefully.
sin~t
=
+ @) +
2aab2sin2mltsin2(m2t + ~)]
3/8, but the last average must be
We have to average
%{ COS2[(C~ 1 + ~02)t + £~] - 2COS[((O 1 + (~2)t + £~]COS[(~ 1
+ COS2[(~ 1 If ~ zero.
~2)t
~2)t
- ~]}
~ ~2' the first and last terms give ½ and the middle If
- ~]*
term gives
~i = ~2 the last term gives cos2@, while the other two are
unchanged.
Thus
E (z)
= % ~ [ ~3 (a" + b ~) + ~1 a262 (1 + ~1 c o s 2 ~ ) ]
(7.40a)
where
1
6 =
~i ~ ~2
(7.40b)
~i = ~2
These formal expressions different
reveal that the degenerate
from the nondegenerate
nondegenerate,
the perturbed
the unperturbed turbed motion
In the degenerate ellipse,
of the ellipse
filling curve of the nondegenerate orientation.
The existence
ing to the same unperturbed splitting
of degenerate
state is
only on the amplitudes
and hence on the energies
is a stationary
spatial orientation
case is essentially
If the unperturbed
energy depends
oscillations
the x I and x 2 directions.
one.
case,
of
associated with
in which the unper-
E (I) depends on @ but not on the
(Problem
7.14).
For the space-
motion there is no such thing as
of different perturbed energy is the classical
levels by a perturbation
energies
correspond-
analog of the
in quantum mechanics.
242
We may ask how the discontinuity continuous
analysis.
an idealization;
It arises,
in (7.40) arises out of a
of course,
only because we have made
we are willing to wait an infinite
take the time average.
time in order to
In fact, if ~2 - ~i = 2~/(I year),
say, it is
only a matter of convenience whether we say that the system is degenerate.
In this case the ellipse precesses
and whether we say it is stationary emphasize.
Whether we say that two energy levels are degenerate
depends not on the vanishing practical Problem
very slowly in space,
depends on what we wish to
possibility
of the difference between them but on the
of observing
it.
Show that in the above example
7.14.
equal and the semidiameters
of the resulting
if the frequencies
are
ellipse are A and B, the
shift in energy is E(1)
Degeneracy
=
-I~ e(3A ~ + 2A2B 2 + 3B ~)
~n Q u a n t u m M e c h a n i e e
If the unperturbed
system has several
states corresponding
same energy, we must choose the right linear conbination the unperturbed
state from which to start the perturbation
That is, if the erthogonal
time-independent
degenerate
to the
of them as expansion.
unperturbed
states are ~i "'" @d' where d is the degree of degeneracy,
we form
the linear combination d ~(o) corresponding
to the energy E (0)
~o +
~¢:
=
E i=I
(7.41)
ci~ i
Then from
(~(o) + ~(~))(~(o)
+ ~(1))
+ (v - E ( i ) ) z c i ~
i = o
we find, to first order, (H o
(The ¢ previously here.)
Multiply
E(°))~ (i)
used to distinguish
orders of perturbation
this equation by any ~j~ and integrate.
ZoiC<j]V]i>
- E(l} 6ij)
= 0
j :
1 ....
is omitted
This gives d
or, written out,
(
- E(i))o i + < l l v l ~ > c 2
+
+
....
o
243
+ C<21vlz>
The condition
etc.
- ~x~)~2
+ <mlvls>~3
that there exist a nontrivial
is that the determinant
of the coefficients
I
<51vii>
- F(x~aid
states.
Examples
I = o
hydrogen--so oscillator
d different values of
linear combinations
of unperturbed
eigenfunctions.
are found in all books on quantum
is usually the same example,
the linear Stark effect in
we shall not work it out in detail here.
discussed
above,
state are still degenerate levels
(7.42)
are known as stabilized
of this procedure
mechanics--it
solution for the o's
and unknown E (I) , is an algebraic
to d different
These combinations
are split.
Almost Degenerate
In the perturbed
the ground state and the first excited under the perturbation,
but higher energy
(Problem 7.15) Levels
It is interesting the separation
o
vanish,
equation of dth order which in general produces E (I) corresponding
+ ....
to note what happens
of the unperturbed
in quantum mechanics
levels becomes very small.
as
We
shall show by the example of a two-level
system that the transition
from the nondegenerate
case is a continuous
to the degenerate
one.
Suppose we have a system that has a pair of levels close together: H0~I and by "close"
= El0) ~i "
(0) ~2
H0~2 = E2
is meant that the difference
~2~(°)
with the energy shift produced by the perturbation. other states of the system have energies shall show that under the perturbation not degenerate,
become mixed.
unperturbed
far away from these.
V the two states,
We
even though is
~¢
the solution by a linear combination
of the two
functions, = el~ 1 + c2~ 2
Substituting
is comparable
We assume that
The equation to be satisfied
(B o + v ) ~ : and we approximate
" ~i~(°)
into the equation and multiplying
by #2" and integrating
(7.43) first by ~i* and then
gives a pair of equations
244
E + ~i ~(o) )Cl + V12c2 = 0
(Vll
(7.44) V21c 1 + (V22 - E + E 2( 0 ) ) # 2 = 0 where we have abbreviated equation
The resulting
quadratic
for E has the solutions
_(o)) + E = %(E z(o) + ~2 + _ _(o) (o))z + ½{Vii + V22 - [(Vii V22 + ~i - E2 + 41V1212] %} with equally In Fig.
(7.45)
long formulas
for e I and a 2. _(o) and ~2 ~(o) 7.2 we have assigned numerical values to ~i
and have taken V as proportional
to a parameter
'
h which is allowed to
ICJI
0
°
0
0 (b)
Figure 7.2
Energy levels of (7.45) and coefficients of (7.43) as functions of the strength of the perturbation. (b) Plot of the energy levels in (a) on and expanded scale.
vary from 0 to i. coefficients comparable
The energy levels
fell and
Thus in classical
and the plot of the
to the splitting
of the unperturbed
levels
mixing of states. theory from a practical
quantum theory by calculation,
Problem
do not cross,
It21 shows that as the matrix elements become
in magnitude
there is considerable
levels
l-
point of view and in
the situation when there are degenerate
is only a point on a continuum. ?.15.
Study t h e q u a n t u m - m e c h a n i c a l d e g e n e r a c i e s o f t h e u n p e r -
t u r b e d and perturbed states of the oscillator defined by (7.38). Problem
7.16.
What are the conditions
necessary
if the splitting of
245
the perturbed what value 7.5
levels
is
of X does
Canonical
basis,
formal
basis.
part the
occur?
Perturbation
The c l a s s i c a l formal
t o go t h r o u g h
this
theory
but
of perturbations
calculations
tend
We a s s u m e as b e f o r e
that
the
V(p,
q,
part
We a s s u m e I t h a t these
these
8,
as
constant
perturbed
functions
of time,
constant,
for
hamiltonian
as
V(C,
this,
t).
a very
First,
consists
handsome
the
of a soluble
a complete
qn(a,
soluble
q,
for
the
t)
8,
solution
t)
of
(7.46)
constants,
a n an d
i = 1....
dynamical
system,
and t h e
quantities
8n.
Let
the p's
that
V(p,
(7.47)
they must satisfy
and q ' s defined
HO + V) = (C i ,
suppose
2f
HO) = 0
C's, still
are by
no l o n g e r
(7.47),are
the
same
no l o n g e r
V)
q,
(7.48)
t)
has been w r i t t e n
via
(7.45)
Then ~C.
( c i , v) =
so t h a t
long.
We w r i t e
t),
are
Ci = (Ci" To e v a l u a t e
c a n be g i v e n
t o be v e r y
t).
6 i = (C i , In the
At
constants
c i = Ci(P, Considered
7.2?
p r o b l e m as
pn(a,
us c a l l
in Fig.
t h e minimum s p l i t t i n g ?
Theory
Ho an d a s m a l l unperturbed
a minimum as
What i s
(7.48)
~V
~C.
~
~C.
,.,J -
~V
~C.
~
~Pn DCj D4nJ,
~qn ~Cj ~Pn
is
6¢ = (c¢, cj) aV ~c.
(7.49)
3
where i it
is
an d j
exact
independent evaluate
of the
to
2f.
by P o i s s o n ) s
choice
The m e r i t theorem,
of coordinates
of this the
expression
Poisson
an d momenta
brackets p, q u s e d
is
that are
to
them.
Equation is
run from i
an d t h a t
if
Ci
i
See Sec.
(7.48)
has
a precise
a time-independent
5.7.
counterpart
dynamical
variable
in quantum mechanics: that
is
constant
in the
246 unperturbed
system,
then in the perturbed
d
system, A
= ~i < [ ( H ^ o + ^V ) ,
: ~i < [ ~ , c i ] > _
~i]>
(7
50)
dt
The foregoing for calculation, pertaining
equations
are all exact and consequently
useless
since the C's are defined in terms of quantities
to the perturbed
system.
But if we write
C. = C! O) + C! I )
where the C! O) are calculated
+
(7.51)
...
from the unperturbed
system,
(7.49)
becomes ~v (o)
b! l )
=
and higher approximations
(c~° ~ , "
c! °)) ~
(7
+
j
CO! o) J can be obtained similarly. .
.
.
In order to see how the formalism works out in practice, apply it first to a particularly Example
2.
simple and obvious
5z)
.
we
situation.
Perturbed Harmonic Oscillator.
Let
1 H o = ~--~(p 2 +
and suppose that the perturbation
m2~2x2
consists
) of attaching
C7.53) an extra
spring to it: V = ½ kx 2 The solution of the unperturbed
X
=
C7.54)
problem is of the form
U- (1° ) s i n m t
+
C 2( ° ) c o s ~ t
(7.ss) p whence the constants
=
mm(C O)cosmt
corresponding
u~ 2( O ) c o s ~ t )
-
to this zeroth-order
approximation
are
C1( 0 )
=
xsin~t
+
J?-- c o s ~ t mo~
(7.56)
C2(o) = x c o s ~ t
- J2_ mm s i n ~ t
There is only one Poisson bracket,
J
c2
m~
247 and it is identically
constant.
V (°)
With
= ½ kCC~°)sin~t
(°)cos~t)2
+ C2
we find for the first-approximation • (1) C1
~o) =
(C
~v(O)
^(o)) ,
U2
mk~
@O~°)
+ C 2(O)cosmt)cosmt
(C °)sin~t
@v(O) k
i
"
Let us assume
~2
that the perturbation
Then for t > 0, we integrate C(Z) 1 = - ~ k
C~I) = ~ k
the perturbed
cosmt)
~(o)
This shows
the secular
represents method
part of a cycle,
the perturbation
is cumbersome
this problem
Perturbation 1918,
Brouwer
(I
of the oscillator
to
phase by
in k it correctly
(Problem efficient
simplicity
is the basis 1961).
perturbations
of celestial
7.17).
The
approach
makes
to
it valuable
Its principal
difficulty
in the solar system
arising
from resonances.
orbits
the Sun exactly
that when Jupiter times, since
there actually
and the resonance
of energy between
mechanics
is in the
are secular perturba-
produces
It happens, 5 times,
for
Saturn
a much larger
the two than would normally
the force
(Plummer
such as we have just encountered;
tions
so, it is small,
will not continue
situations.
and Clemence
it 2.0134
(7.57)
but to first order
since
exchange
0.
sin~t)
the oscillator's
is the more difficult
orbits
=
C (0) + C (I)
sin~t
and obviously
has changed
2, but its formal
theory
of secular
k 2rao-----T
kt 2m~
this
example,
t
cosset)]
-
by substituting
x(t)
and we have seen a more
in Chapter
in more complicated
avoidance
perturbation
after the perturbation
an appreciable
+ u. ( 2o )
~(O)(cos~t + u2
- uI
be valid
at
in (7.55), we find
= C(°)(sinmtl +
xCt)
established
+ 2C °)cot + 02~(O)sin2~t]
~t - C °)sin2~t
If now we evaluate for the C£°l's
was suddenly
to find
[u~(0) I (] - cos2(0t)
[2~(O)u 1
+ c~(0) 2 cos~t)sinmt
(C °)sin~t
mto
B~ 1
occur.
is weak and the periodicity
Even of the
248
resonance
is 880 years.
To obtain the accuracy
required
or the space program requires very long calculations, are mostly simple in principle, will
in astronomy
but the methods
as the example in the following
section
show.
Problem
2.17.
Carry out the calculations
leading to (7.57) and show
how the result is related to the exact solution. Problem
2.18.
Find a way to eliminate
the secular t e r m s
in evaluating
the C i(i) , and verify the result by an exact calculation. 7.6
Newtonian Precession The general-relativity
precession
precession
of the orbit of Mercury
largest part is accounted
is only a small part of the
as viewed from the Earth.
for by the precession
system used, which is tied to the slowly-changing Earth's
axis and will be evaluated
be discussed here,
is contributed
and the part a t t r i b u t a b l e
in Section 8.5.
essentially
direction
Another part,
to general
relativity
is only the residue
from the observed value.
two coordinates
and two momenta,
but it is convenient
to give a and the two components
A, since by (2.93), A = ye.
The direction
A
e.
For this
of the Laplacian
integral
of A is given by @a'
7.3, and @ao is the value of @a at the time t = 0.
F i g u r e 7.3
to
We first define the orbit by the magnitude
and direction of its major axis a, and its e c c e n t r i c i t y
Fig.
to
the m o t i o n of a planet we need four numbers,
choose them as follows. is suffices
of the
by the action of the other planets,
after the others have been subtracted To characterize
The
of the coordinate
In the
_
~ l g l e s and c o o r d i n a t e s f o r c a l c u l a t i n g Mercury by t h e a t t r a c t i o n of. J u p i t e r .
the perturbation
of t h e o r b i t
of
249
absence
8a would remain
of perturbations
to calculate
its rate of change.
constant
at @ao; we are going
To locate the planet
in the orbit
thus defined we assume that at some time t = t o the planet was at aphelion, initial
and measure
direction.
its subsequent
The planet's
displacement
position
relative
to that
and motion can be calculated
from the four numbers
a, To calculate
A x,
Ay,
the Poisson brackets
to
(7.58)
in (7.52) we need a set of
coordinates
and momenta.
convenience
and does not affect the result,
is only a matter
of
but the calculation
is
Which ones we choose
very short if we remember
the view of the Hamilton-Jacobi
that was taken in Sec. 5.5, in which the characteristic taken to be the generator coordinates mass
of a contact
transformation
8n and momenta en with constant values.
is set equal
to i, they are
(Problem
C~l = E = - -Y-2a
~2
~z
82 =
function
is
to a set of If the planetary
7.21)
L
=
equation
[ya(1
=
- e2)] ½
(7.59)
In
=
to
-
the unperturbed
evaluate
OaO
system the components
of A are constant;
we
them at time to:
A x = -yesinfl2, and the other variables
a
=
-
are found from
We can now calculate
(Ax, Ay
(7.59):
2 e2 = I + ~
2al,Y
the required
)
=
(7.60)
Ay = yecos82
8Ax 8Ay 8fl----28a 2
~i~2
brackets
3A~
~A
z
I
= --X~ (Am2 + Ay
(7.61)
almost at sight:
x _
382 ~ 2
2)
2&l&2
(7.62)
or
(Az, Ay)
= ~
[¥a(I
-
e~)] ~
(7.63a)
and similarly,
(a,
to)--
2a2
Y
(7.63b)
250
(Ax"
to)
=
a ( 1 - e 2) A ye2 x
(7.63a)
(Ay,
t o)
=
a(1~e2- e 2) A y
(7.63d)
(a, Ax)
=
The largest perturbation Jupiter.
Figure
(a, Ay)
=
0
(7.63e)
of the orbit of Mercury
7.4 shows the situation.
is created by
We assume that Jupiter(~+)
Figure 7.4 Angles and coordinates for calculating the perturbation of the orbit of Mercury by the attraction of Jupiter. moves
in a circular orbit of radius a 2 with an angular
while Mercury perturbing
(~) moves
in an ellipse
characterized
frequency
~2'
by a I and e I .
The
energy is
V
=
- ~6
6
In terms of the orbital variables
=
(7.64)
Gin.,,
this contains
1 I
= [r12
_ ~rlalc°se
= ~ az nZ__O
+ a22] ½
Pn(C°Se)
e = 8z
82
(7.65)
251
where t h e Pn'S a r e t h e L e g e n d r e p o l y n o m i a l s , 1 t h e e x p r e s s i o n f o r ~ in powers o f r l / a Z : = 1,
Po(COSO)
and so forth.
Pl(COS@]
= cosO,
P2(COS@)
We shall need only these
Since both @ and r I are periodic only interested simplify
in the cumulative
everything
the sum
by taking
calculated
: ½ (3C0S2@
three
I)
terms.
functions
effect
by e x p a n d i n g
of t, and since we are
after a long time, we can
time averages.
In the first term of
(7.65) we have
rl(t)cos@(t ) = rl(t ) (cos@icos@ 2 + sin@isin@2) in w~ichrlCOS@ 1 and rlsin@ 1 are periodic cos@ 2 and sine 2 have period incommensurable, the entire argument,
T 2. Since T 1 and T 2 are different and term averages to zero, while by the same
P2 (cos@)can he replaced --
:
by 3
P2 = ~ (~ -
To c o m p l e t e
the
second
term
we n e e d
of this kind are quite easily to this chapter
Thus effectively,
(7.66)
time
average
details
r12 .
Averages
are in the appendix
(1 + ~3 e l 2 ) a l
=
energy
(7.67)
2
is
(1 + 3 el2)(
al ) 2 ]
is given by the time derivative
rate of change
of
is
the perturbing
The precession
I
= ~
the
[I + :
the average
:)
calculated;
and the result rl2
in t with period T 1 and
of A.
We find for
of A X
-T-A x = (A x, ~) by
(7.63e). T-
Ax
With =
=
(7.61),
36a12 , 8a 2
this is
3 6a12
- 8 Y2a23
-
3 @a12 4 y2a23
(A x, e~)
(Ax" Ax2 + Ay
(A x, Ay ) Ay
2)
4 Ya23
el 2
y
252
One revolution of Mercury takes a time 2~a 1 Y
and in this time, on the average, A x changes by 7--
3
A x = AxTI
~
= - i 2~ ~
aI 3 (-~2} (I - e l 2 ) # A y
Let us choose axes so that at the moment of interest A lies along the y axis.
Then the forward precession
AAx radians A
In
one
Y century,
(Earth)
parameters
x 360x602
A 27 arc sec. Y makes 415.2 revolutions.
Mercury
The r e m a i n i n g
are m = mo
al a2
-
AA
=
in one revolution is
-
=
Jl 4.5m e 329390m0=
57.91x10ekm 278.3xlO6km
9 . 5 4 8 x I 0 -~
=
@
=
Earth
7 . 4 4 1 x 1 0 -2
e I = 0.2056
and we find for the precession
in one century attributable P
to Jupiter
= 155"
The precession of Mercury's orbit caused by Jupiter and all the other planets
is calculated as 531" per century.
equinoxes
The precession of the
once in about 25780 years adds 5026" to this.
value is 5600".
The observed
The difference 5600"
(531"
+ 5026")
= 43"
agrees with the value calculated in Sec. 5.3. The foregoing calculation bation.
is an example of a first-order pertur-
Higher orders are, as one might expect, more difficult and
are described in works on celestial mechanics. Problem
7.19.
Show that the constants
in (7.59) are those which arise
from solving the Kepler problem in two dimensions. Problem
7.20
Calculate the brackets
in (7.63).
Problem
2.21.
The eccentricity of Jupiter's orbit is e 2 =
0.056.
Refine the calculation done above to see whether this makes any difference.
253
Problem 7.22. expansion
Problem 7.25. momenta,
Find the effect of the next nonvanishing
Calculate (Ax, Ay) using Cartesian coordinates and
and show that (7.63a) results.
Problem 7.24.
The interaction with other planets not only causes a
planet's orbit to precess; in its orbit. average,
term of the
(7.65).
it also alters the position of the planet
By how many seconds of arc in one revolution,
on the
does Jupiter change the orbital position of Mercury?
Problem 7.25.
As seen from the earth,
the moon travels in an
approximately elliptical path in a plane
(Fig.
7.5), slightly inclined
Z
/x Figure 7.5
Coordinates for Problem 7.25.
to the plane of the earth's orbit around the sun (the plane of the ecliptic).
Owing to the action of the sun, the plane of the moon's
orbit wobbles--i.e., revolution
the angular momentum £ precesses
in 18.61 y.
at a rate of one
Assuming that the earth orbits the sun in a
circle of radius R, calculate this rate. Take as variable the vector r running from the earth to the moon and show that, averaged over a year,
the perturbing energy provided by
interaction with the sun is
= 2R GMp [ s ( ~ R r )2
rz -~-¢]
(7.68)
where M is the sun's mass and ~ is the reduced mass of earth and moon. Now, calculate the rate at which £ precesses.
The answer is
254
T __~=! 3 m~
m@ + m M
(R)3 ~ sec~
where T
is the period of the precession, T is one month, and the P other variables are obvious. Numerical values are m
+ m
• R
=
6.057xlO27g
M
=
1.971×10~3g
D =
1.496xlOZScm
a~
a =
5°8t43"
=
3.8155×I01°cm
CHAPTER 7 APPENDIX Time Avera g e of Start w i t h
rn
in K e p l e r Orbits
(2.70) or
(5.11),
E = energy
t =
J
dr
2 (~
+
Zr)
]%
L
This can be r e w r i t t e n more neatly ties e x p r e s s e d by (2.75a,b]
-E = -Y-" ~a "
~
z
=
(1 -
and
e
L = angular momentum y = G/en
in terms of the basic orbital proper-
(2.77),
2) mT -2E2 = (1 -
eZ)yma,
axis of the orbit,
T is the orbital period.
We find
T
I
rdr
[e2a 2-
(a
-
e is the eccentricity,
r)2] ½
a t once from a t a b l e ,
reached
variable2
i n 1609.
but our goal
introduced
i n t e r m s o f ~,
(7.69)
=
a(1
-
ecos~)
C7.70)
becomes
t
esin~)
(7.71)
J
w h i c h is known as Kepler's equation. We w i s h to evaluate
rn
I fr(t)ndt
From ( 7 . 7 1 ) ,
at
= 2-~ ( I
- ecos~)d~
so that
2
is
by K e p l e r
Let r
and,
~ first
and
(7.69)
T h i s c a n o f c o u r s e be i n t e g r a t e d by t h e u s e o f an a n g u l a r
= 2n(~)½a 3/~ Y
where a is the semi-major
t = 2Ta
T
Kepler's name for this v a r i a b l e was the eccentric anomaly, E. have changed the letter to avoid confusion with the energy.
We
256
n
r n
where the integral
=
~--~/(:
- ecos¢)n+ld~
a
is from 0 to 2~.
Positive values
of n are easy;
negative values are calculated by contour integration or by expanding the integrand in powers of e.
A few values are (r/a) n
3
Se2 + ~3 e ~
1+
3e2
1
e2
I
-I
1
-2
(:
-.5
-
e2)-%
CHAPTER 8
THE MOTION OF A RIGID BODY
In the narrow world of the exact sciences there are several kinds of rigid body.
There are nuclei, generally not spherical and
precessing in the atomic field or some applied field.
There are
molecules, whose shapes may be very complex, and there are stars and planets interacting through gravitation. field of engineering,
In the closely adjacent
and especially space engineering,
there is an
immense variety of oscillating and rotating devices, some of which, such as the gyroscopic stabilizers used on space craft, are extremely sophisticated. In every case the rigidity is an idealization.
Nuclear and
molecular spectra show series of lines associated with rigid-body motions among many other series produced by motions in which the system changes its shape.
Stars and planets are far from rigid, and
the designers of real machinery are always conscious of the elastic properties of the materials they use.
Nevertheless there are
phenomena characteristic of rigid-body motion in general, and in this chapter we shall study some of them. The discussion will be limited by a somewhat arbitrary restriction. involved in inertial guidance
Except for a few
(Sec. 8.4), man-made devices generally
involve supports of some kind, fixed axes or restrictive gimbals by which forces are exerted from outside to constrain the system, molecules,
Nuclei,
and astronomical bodies float in the fields that influence
their motion.
We shall consider only this latter class, skipping
with regret (except for Problem 8.20) the theory of the spinning top, an object with one fixed point. mathematically
For 200 years the top was analyzed
in order to confirm the insights of childhood; more
recently students have been introduced to tops in order to confirm the insights of the blackboard.
Instead, riding the crest of
astrology, we shall consider the precession of the equinoxes. 8.1
An~ular Velocit Z and Momentum The velocity of a point revolving around a fixed axis can be
very simply described.
Choose a point 0 on the axis to be the origin
of coordinates and locate P by the vector r o.
Define the angular-
velocity vector m directed along the axis according to the right-hand rule, with the length equal to the numerical angular velocity m.
Then
the velocity of P is given, in magnitude and direction, by v
:
~oxr °
(8.1)
258
regardless
of
the
location
of
0.
Note
that
the
components
of
m are
V
P
Figure 8.1
Relation between linear velocity and angular velocity.
not introduced
as time derivatives
this important
point in Section 8.5.
If the axis
of anything.
itself is moving,
with velocity V,
so that the point
In all
the
the
cases
the
of mass
V is
zero
With particle
we s h a l l
center
center
mass,
this P.
consider,
of mass, is
and
at
the
If
remains
let
(8.2)
o axis
of
a n d we may c h o o s e
rest.
(8.1)
choice, It
0 is in motion
(8.1) becomes v = V +~xr
through
We shall return to
us
now 0 i s valid
evaluate
chosen
even the
rotation
passes
coordinates
if
as
the
~ changes
angular
so
that
center its
of
direction.
momentum of
the
is
i = mroXV = mroX(~Xr o) or
(8.5)
1 = m[ro2C0 - ( a ~ . r o ) r o ]
Note that convenient
1 is not in general to introduce
~i
=
parallel
vector
m(ro2~i
indices
to ~.
It
will
explicitly
- roiroj~ j)
in
the
sequel
and write
(summed)
or £i = m(ro2~ij The fundamental
hypothesis
- roiroj)~j
about a rigid body
C8.4) (never really
be
259
encountered
in nature)
relative positions. Thus
is that all parts
This requires
the angular m o m e n t u m
of it maintain
of a rigid body is given by
Li = Ioij~j"
Ioij = 2m(ro2~ij - roiroj)
summed over all the masses, with the components mass.
The formula
looks nice but
can be remedied by introducing The coordinates
it is useless,
a new coordinate
we have used so far,
(8.1)
thus far hold equally well of vectors
of vectors
fixed in a rotating from fixed axes.
All
system. and directions
of the
about their rates of change.
in a rotating
Of course,
coordinate
thinking
to be careful,
system appears
We shall see presently
system,
for the
them are unchanged
if we start
we shall have coordinate
by
the relations we have derived
and the angles between
adopt such a system. of change
This
v is the rate of change of ro, but it was introduced
and not as a time derivative.
lengths
of time.
fixed in space and centered
to, v, and ~; we have said nothing
To be sure,
to each
for as the body moves
on O, have only served to specify the magnitudes vectors
~8.5)
~k appropriate
and both Iij and mj are functions
all the r's change,
the same
that they all share the same ~.
if we
about the rates
since a vector
to move when viewed
how to take this into
account. We choose axes center of mass unchanged:
fixed in the rigid body, with the origin at the
as before.
v represents
axes, but the components axes.
The position
of the vectors
velocity with respect
of v are evaluated with respect
of a particle with respect
given by the fixed vector
L i = Iijw j where the components
The interpretation a particle's
r, and
is to fixed
to the moving
to the moving
axes
is
(8.5) becomes
Iij = Zm(r26ij - rirj)
I.. ~J are now constant.
This will greatly
(8.6) simplify
the later analysis. The same quantities energy,
T.
occur in an evaluation
of the body's
We have T = ½ Zm(~oxr) 2 = ~ /:m[a~2r 2 -
= ½ Zm(r2~ij - r i r j ) ~
j
(~o.r)2]
kinetic
260
That is, __- ½
_r..~.o~.
(8.7)
zj z j
or
T = ½ L.~
This recalls a particle,
(8.8)
the fact that one can write T = ½ p.v for the motion of but here there
is the difference
that in general
L and
are not parallel. 8.2
The Inertia Tensor An object like lij , bearing more than one index,
tensor ment
if it has the right transformation
is that it should transform
like a product
it is in our case formed from the product is satisfied
automatically.
a linear operator
of vectors
Considered
The require-
of vectors,
A tensor can be regarded
and as a thing.
is called a
properties.
this criterion in two ways:
the vector ~ into the vector L by rotating
its length.
Considered
ties
it represents
of a solid object just as other physical
ted by various vectors components
and scalars.
Written
as
as a linear operator,
I transforms
as a thing,
and since
~ and changing
the inertial proper-
properties
are represen-
out in detail,
the
are
I Zm(y + z 2) - Zmxy - Zmxz 1 Iij = -Zmxy Zm(x2 + z 2) - Zmyz
(8.9)
-Emxz - Zmyz Em(x2 + y2) It is customary
to label the rows of such an array by the first index
and the columns by the second, but here it does not matter,
since I is
symmetrical, x..
zj
=
z..
(8.10)
jz
Clearly also,
Iii This diagonal also used) single
sum,
~summed)
= 2Zmr 2
called the trace of the tensor
in invariant under rotations
component
of
Iij
(8.11) (the German
of the coordinate
has already been encountered
Spur
system.
in (3.15)
is A
and the
261
formula
(3.16)
is a restricted
form of (8.6).
of I.. ~J are called coefficients products of inertia. It is the products Suppose
of inertia
they were all zero.
II
Iij =
The diagonal
elements
of inertia and the off-diagonal that make calculations
ones
difficult.
Then we would have
I2
0
0 1
Ii = III, etc.
(8.12a)
13
and
L
Ilmx~ + I2~y ] + I3ez~
(8.12b)
z = ~(zaO~x~ + _ h % ~ + z 3 % ~)
(8.12o)
=
This can always be achieved by the proper choice of axes with respect to the body in which they are fixed. e.g.,
Goldstein
details
1950, Konopinski
The fact is proved in many books, Rather than reproduce
the
here, we remark only that in most cases the body has axes of
symmetry and one merely needs following
8.2.
to align the axes with them.
The
example and problems will show how this occurs.
ExampZe I. Fig.
1969.
A uniform
To evaluate
slab in the shape of an isosceles
triangle,
Zmxy, divide the slab in strips like s I.
For
>
Figure 8.2
Isosceles-triangular slab of uniform thinkness.
each value of y the strip gives equal contributions negative x, so I12 = 0. Since positive vanish.
from positive
and
For I13 and 123, sum first along the z axis.
and negative values of z contribute
equally,
Thus even though the slab is not symmetrical
these also
with respect~to
262
y, its symmetry with respect diagonalize
to the other two axes suffices
to
the inertia tensor. Prove that for any inertia tensor reduced to diagonal
Problem 8.1.
form, I l + 12 ~ 13 ,
Problem 8.~.
A section of pipe w i t h inside and outside
r 2 rolls down an inclined plane. the two limiting
bar.
Find the moment
What
system
calculate
of inertia
tensor of a triangular
in a Rotating
but this,
ary axes.
coordinates
of course,
flat
Find a new
refers
It is not hard to express
rotating bodies
to motion with respect it with respect
on a few related
a be rigidly attached
xyz, which
dimensions
and
System
equation governing
is
to station-
to rotating
axes,
topics.
to a coordinate
system S,
is rotating with angular velocity ~ relative
a fixed system So, coordinates stationary
(0,i,0).
8.2 with literal
Coordinate
but before doing so we digress Let a vector
and
of inertia.
The basic differential (3.11),
h hangs
is its period of oscillation?
the slab in Fig.
its coefficients
8.3 D y n a m i c s
and give
in which the tensor will be diagonal.
Provide
Problem 8.5.
radii r I and
the acceleration,
~ and height
slab bounded by the points (a,0,O), (i,0,0) coordinate
(8.13)
pipe and a solid cylinder.
shop sign of width
from a horizontal
Problem 8.4.
Calculate
cases of a thin-walled
A wooden
Problem 8.3.
suspended
1 3 + I 1 ~ 12
12 + 13 > I 1 ,
XoYoZ o.
in S, moves with respect ~o = ~xa
If a is moving with respect
to
The tip of the vector,
to S o with a velocity (a fixed in S)
(8.14a)
to S at a rate ~, then this velocity
is
added to that just given, ~e = ~xa + ~
(a moving
This relation will presently be used to transform rigid bodies. the vector
To
see some of its further
r is the position vector
to the fixed axes the particle's
the equations
implications,
of a particle.
velocity
is
(8.14b)
in S)
suppose
for that
Then with respect
263
Write
~×r
+
~
(8.15)
v
=
~xr
+
v
t~.loj
v with respect
the absolute
to moving
axes.
again in
to fixed axes is
acceleration, :
~
0
~o is a hybrid,
fixed axes.
= & O0
(8.14b]
0
a
xr
+ ~ox~"
0
+
0
0
the rate of change of v (not vo) with respect
So that everything
a
to fixed
Now take the derivative
The rate of change of v ° with respect
a
axes use
0
v ° is the velocity of the particle with respect
the same way.
where
=
this as
Once more: axes;
ro
three times:
:
Lcoxm)xr
=
~xr
+
+
~xr
+
~x~mxr)
mx(mxr)
+
2~×v
+
mxv
+
mxv
+
v
+
(8.17a)
0
This can also be w r i t t e n a
=
gxr
+
as
(m.r)m
-
m2r
+
2mxv
+
(8.17b)
0
The first, corrects
third,
and last terms should be obvious.
the third when r is not p e r p e n d i c u l a r
in elementary physics, Like the centrifugal pays
for looking
~x~xr,
Fig.
to the room,
With respect
(approximately)
straight,
curve towards
of dynamics rotating
to describe
frame,
accelerations. apparent
however
the motions
they must contain
the marble's
on it,
path will be
it leaves on the turntable
it is started. of particles
forces
acceleration.
Put a piece of
and roll an inked marble
but the tracks
the left,
is
it is part of the price one
at the world from a merry-go-round. turntable
will
The second
to ~ as it usually
and the fourth term is the Coriolis
acceleration
paper onto a p h o n o g r a p h 8.3.
to
on the right will refer to moving
If we write
as seen in the
corresponding
If the real force on the particle
laws
to these
is F = ma ° the
force is
mv
=
F
-
m~×r
-
The third term is the centrifugal
m~xm×r
-
2mmxv
(8.18)
force and the last is the Coriolis
264
force.
Figure 8.3
(a) Path of a marble rolling on a rotating turntable as viewed from the room. (~) As viewed by an observer rotating with the turntable.
The Coriolis
force is most noticeable
southern hemisphere
the situation
If there is a high-pressure deflected towards
the northern hemisphere
region produces the relations
the direction
away from it is
counterclockwise
a clockwise
are opposite.
of highs
in the
as shown in Fig.8.3.
region at P, air flowing
the left and ends up circulating
around P; a low-pressure can often deduce
in the atmosphere;
is qualitatively
circulation.
In
Knowing
one
and lows from shifts
them, in the
wind as they go past.
Foucault's Pendulum Another
illustration
of the Coriolis
long pendulum hung carefully Fig.
8.4 shows a suitable
the earth's
set of coordinates,
= 0,
S is the earth's
small displacements,
~
y
= ficos/,
with respect
angular velocity
~
to w h i c h
= fisinl
(8.19)
and I is the latitude.
For
z
the force on the p e n d u l u m
F x = - ~ x£,
and omitting
of a
angular velocity has the components
x where
force is the behavior
so that it can swing in any direction.
the negligible
is
Py = -£Y
centrifugal
force gives
265
m~
= F -
(8.zo)
2m~xv
or
m~ = Fy
- 2mCcozx
- oJx~ )
LO Z
~K~ra 8.4
Since the pendulum's
Coordinates for Foucault~s pendulum.
vertical m o t i o n
is negligible,
these are
= - ~ x + 2~sinX~
=
To solve these neatly, a complex
2nsinX~
the xy
plane
as a complex plane with
coordinate
The equation
+ jy
=
~
j2
= -1
for E is = - ao2E-
phase
_
consider
x
This
_
is solved by setting of m o t i o n
locates points
is a V(-l)
2jasink~
~. = A e i v t
~o 2
where
g/~
the i that gives the time
that has nothing
on the plane tangent
=
to do with the j that
to the earth.
two square roots of -i; i and j may designate
There
are, of course,
the same one or
266
different ones. ently.
The point is that the designations
are made independ-
We find
~2 = ~0 2 + 2£j~%~
~h = ~sinl
which has two solutions
v1
= ij~ l +- (COo2 + ~12) ½
2
since we are neglecting
terms in ~2.
The general solution is
= Aei~l t + Be i~2t
= e-J~k t (Ae{wo t + Be-i~o t)
The pendulum must be carefully released from rest:
~(o)
--
A + B,
~(0)
=
o
from which f~X
A = ½ (1 - ij - ~ o ) ~ ( 0 ) , and
~(t) = e - J ~ t ~ ( O ) ( c o s ~ o t
The second term in the parentheses
+ j --~ sin~ot)
is very small.
Essentially,
(8.21)
the
pendulum swings at a frequency ~o in a plane that rotates clockwise with the angular velocity ~ .
There is no use mounting such a
pendulum on the equator. Foucault tried his pendulum first on a modest scale in 1851 and then installed a heavy iron sphere suspended by a 67-m wire from the dome of the Pantheon in Paris.
It created something of a sensation
as people realized that this was the first direct dynamical proof of the Copernican hypothesis sky,
that it is the earth that turns and not the
(Today one says more carefully that a description that takes
the sky as a reference system requires than one centered on the earth.
simpler equations of motion
To say that one turns and the other
267
does not is not a p h y s i c a l
Hamiltonian
Formulation
The H a m i l t o n i a n By (4.52)
statement.)
the change
in a rotating system contains
a kinematical
term.
in a d y n a m i c a l variable a c o r r e s p o n d i n g to an
i n f i n i t e s i m a l r o t a t i o n of coordinates Aa c
=
around the ith axis is
A@i (a,L i )
or, dividing by At,
ac = ~i(a'Li ) and this gives &
= (a,~o.L) + 0
w h e n we include the change & w i t h respect to the rotating axes.
This
is the change in a w i t h respect to fixed axes, given by the h a m i l t o n i a n as u s u a l ,
O
=
(a, s)
Thus the change in a w i t h respect to rotating axes can be w r i t t e n as
=
(~, ~eff)
in w h i c h the effective h a m i l t o n i a n
is
Her f = H - e.t This is useful where,
as in the example of the Foucault pendulum,
angular v e l o c i t y ~ is a p a r a m e t e r of the rotating system. we shall consider in Sec.
(8.22)
situations
Presently
in w h i c h it is a dynamical variable,
8.6 we shall see how to write down the H a m i l t o n i a n
the
and
for that
case.
Problem 8.6
If the M i s s i s s i p p i River is about 1 km wide at latitude
45 ° and the current flows at about 7 m/s, how m u c h higher
is the water
level on the west bank than on the east?
Problem 8.7
In 1833
mine shaft in Saxony.
(Dugas 1955)
F. Reich studied free fall in a
The shaft was 188 m deep and in 106 observations
he found that a test object fell an average 28 mm east of the point directly below w h e r e it was dropped. theory?
Is this in accordance with
268
Problem
v.
A ball is thrown vertically
8.8
Where will it strike
Problem
upward with initial velocity
the ground?
What is the shape of the curves described by (8.21)?
8.9
Show the function of the small second term. Problem
Use the hamiltonian
8.10
(8.22)
to derive
the equations
for
the Foucault pendulum. L a r m o r ts T h e o r e m
If an atom is immersed without changing theories
in a magnetic
its shape;
field it begins
to turn
this is the basis of semi-classical
of the Zeeman effect.
The general
statement
is Larmor's
theorem : If a system of identical each other,
charged particles,
and with a central mass,
its motion becomes of a constant
(except for centrifugal
rotation
interacting
is placed in a magnetic effects)
among field N,
the superposition
around the central mass and the accelerations
produced by the other forces.
The precessional
angular velocity
e B = = 2-mThe proof is a direct application on a particle
Then in the rotating
system, m~
is satisfied
theorem is applicable the stationary
(8 23)
of (8.18).
Let the total force
be given by F
and if (8.23)
is
f
=
+
(8.20) gives =
r
+
(8.24)
evxB
for each particle
vx(2m~+eB)
only nonmagnetic
in practice
central mass,
+
...
forces remain.
The
only to atoms because except
for
all the particles must have the same
ratio e/m.
8.4
Euler's Equations After this digression
we can q u i c k l y f o r m u l a t e t h e e q u a t i o n s
the angular velocity of a rigid body subject The equation of motion mass so that ~ = 0.
to arbitrary
torques.
is (3.11) with the origin at the center of
In the notation of Sec. [
= M O
8.3,
for
269
With respect
to axes fixed in the body, +
and if the axes are oriented
~xL
:
this is
(8.2s)
M
so that I is diagonal,
this is
Ii~x + (13 - I2)~y~z = M1 (8.26)
I2~y + (II - I3)~X~ z = M2 I3~z + (12 - Ii)~x~y = M3 These equations,
which sum up the theory of rigid bodies
ably symmetrical
and appealing
the comment tribus
"Summa totius theoriae motus corporum rigidorum
formulis
satis simplicibus
As we have mentioned
before,
in 8.1 as kinematical If Euler's
equations
object with torques object's
vectors,
applied,
8.5, but if the torques
equations
of M depend on the
further development,
the
in this case is to note that of energy and angular
L is constant with respect
and to specify the individual
angles which we are not ready to discuss,
body-fixed
coordinates,
L 2 is constant
(Prob.
to space-
components
but even in the
8.11).
We thus have
Izmx2 + I2my2 + I3~z2 = 2T (const.)
(8.27a)
I12~x2
(8.27b)
To find the general
+ 122~y2 + I32~z 2 = L 2 (const .)
solution
of the Euler equations we solve
for ~y and ~z' say, in terms of ~z and substitute first Euler equation,
z
which we postpone
are zero we can integrate
contain the conservation
The angular momentum
fixed axes, of course,
(2.19).
of anything. of a spinning
as they stand.
momentum. requires
the behavior
To specify what angular velocities
requires
The first step of the integration Euler's
~¢ were introduced
not as time derivatives
the components
his
A word of caution.
the angular velocities
angular orientation.
equations
continetur. ''1
are used to analyze
have to do with angles to Sec.
in a remark-
form, were given by Euler in 1760 with
The details
which yields
an elliptic
are left for Prob.
integral
of the form
8.12.
The whole theory of the motion of rigid bodies
these three rather simple formulas.
(8.27)
them into the
is contained
in
270
Asymmetric
Top
An ordinary blackboard three different becomes what
top,
will detect
into the air it
and with a little practice
in the classroom.
so that it spins nearly
around one of the axes of
w h i c h we shall name the third axis.
observed to precess
of an object with
If thrown
results may be demonstrated
Toss the eraser
about a fixed direction
The axis will be
in space,
that if the long axis is the one chosen,
is in the opposite axis
is an example
of inertia.
is known as an asymmetric
the following symmetry,
eraser
coefficients
is chosen,
direction
from the spin, whereas
the precession
is the other way,
and sharp eyes the precession
if the shortest
The following
analysis will show why. By the conventions momentum
of the experiment,
around one of the axes,
say axis
the initial
angular
3, is much greater
around the other two, so that ~z >> ~ x and ~ y , and Euler's
than that
third
equation becomes z3~ z ~ o
Thus ~z is approximately equations
constant;
(8.28)
call
it ~, and the other two
are Ii~ x +
(I 3 - I2)~O~y
= 0
(8.29) 12~y
These
equations
+
(I 1 - I 3 ) ~
x =
0
can be solved by the substitutions x
and the condition
= ae ~t ,
~
for solubility
X2
=
y
= beXt
(8.30)
turns out to be
(13-12) (Ii-13)
~z
(8.31)
IiI 2
The four possible
situations
are
I3
~ If,
12
X 2 ~< 0
13
> I 1 , 12
X 2 ..< 0
I 1
<
13
< I2,
12
< 13
<
I 1
12 > 0
271 If 12 < 0 the m o t i o n is periodic, while is unstable
and the axis wanders
the solution is left for Prob.
Symmetric
if 12 > 0 the initial m o t i o n
through large angles.
The rest of
8.12.
Top
If two of the moments of inertia are equal s i t u a t i o n simplifies. ~3 = ~, a constant.
(or nearly so),
Setting 12 = I 1 in ( 8 . 2 6 ) g i v e s
the
~3 = 0, or
The remaining two equations are
13 - I 1 $o + - - ~ x ii
0
=
y
(8.32) 13
-
I 1
-
~
y Introduce
ii
= 0 x
the a b b r e v i a t i o n 13 - I 1 P
11 (There is no occasion for
and solve these by setting ~x + iv y = ~.
both an i and a j because the m o t i o n is pure precession.)
- i~p~
=
This gives
0
and ~(t)
=
~(O)e i~pt
(8.33)
This gives the p r e c e s s i o n of the a n g u l a r - v e l o c i t y vector with respect to the b o d y - f i x e d coordinates. Something of the sort is seen in the motion of the earth, Fig. 8.5. Z
Figure 8.5 Careful m e a s u r e m e n t s
Wobbling of the earth on its axis.
have shown t h a t
the earth's axis of rotation
272 wanders
counterclockwise
around the North Pole in an irregular path
never more than about Sm from the Pole. 13
Since
I 1
-
1
(8.34) II
(we shall see in Sec. earth's rotation
8.5 how this value is determined),
is counterclockwise
of the precession
is correct.
than the 300 days predicted explained by the irregular discrepancy
~04
as seen from the pole,
The mean period
from (8.34). loading
The irregularities tides,
Then by (8.14a),
according
are
and the
lack of rigidity.
the precession
is in the same direction.
Let k be a unit vector along the axis of symmetry, z axis.
the sign
is about 50% longer
of atmospheric
in the period by the earth's
As seen from outside,
and since the
k moves with respect
which we call the
to space-fixed
axes
to ~:0 --- ~xk,
k II ,z
or
Thus
~0 ~ k o x + 6kO v = ~Y - 6 ~ x = - i ~ ( t ) and integration
using
(8.33) gives
<0 =
_
~(0)
(8.3s)
eimpt
P
The sense of the precession according Problem
is the same as that of ~ or opposite
as 13 > 11 o r 13 < I 1 , 8.11.
Show that Euler's
Fig.
8.6
equations
imply the constancy
of
g and L 2 as defined in ( 8 . 2 7 ) . Problem
8.12.
the Jacobian
Complete elliptic
to assign the labels The solution Problem
8.13.
the solution of Euler's
functions
sn, cn, and dn.
Suppose
that
in the reduction
been very nearly but not quite equal to I I.
in terms of
It will be convenient
i, 2, 3 to the axes by assuming
is given in Landau and Lifschitz
have been different?
equations that
L 2 > 213T.
(1969). leading
to (8.32) 12 had
How would the results
273 Problem 8.14.
Show
that
for
a symmetrical
L where
k is as b e f o r e
of s y m m e t r y . plane.
Thus
Draw
=
I1(~
a body-fixed
~,
diagrams
unit
the
for
> 0 and m
m
(8.36)
vector
space-fixed
P about
(Ii = 12),
k~p)
+
k, and
top
pointing vector
along
L all
the
lie
axis
in a
< 0 to s h o w h o w m and
k precess
P
L.
Figure 8.6
Precession of short and long symmetrical tops.
The Gyrocompass In earth
1852,
with
demonstrated form, its
Fig.
axis the
in
year
after
the
turn
11 = 12 i n
wheel.
If
space
device
around
to
with
consists the
the
we l e t
he had demonstrated
occurred
a smaller
8.7,
can
Since in
the
a pendulum , it
that
a gyroscope. of
spin
in
plane
xy
11 a n d I z r e m a i n
constant
with
and t o make them c o n s t a n t
reason ing
that
for
introducing
the
we c a n w r i t e
orientation down t h e
the of
as
we n e e d n o t the
inertia
z axis,
could
In its
a gyroscope
aligned
the
rotation it
of
the
be
simplest
suspended
so t h a t
vertical.
gyroscope, it
the
Foucault
rotating the
components
axes of
long axes xyz
the
as in
= ~sink
the
+ $
the
coefficients
axis
of
x and y axes spin
was t h e
first
changes vector
change,
X
the
fix the
as
~ that
is
only
place. the
of
Remember-
earth
rotates,
describes
the
274
=
~cos~sin~
=
~cos~cos~
Y Z
where ~ describes
(8.37)
the turning of the gyroscope's
axis.
7¢
Figure 8.7
Foucault gyroscope suspended above the earth's surface.
To find the equations we note that
if I 1 = I2,
of motion
(8.6)
Lx = I1~x,
in the non-spinning
becomes
Ly = Ii~y ,
where s is the spin angular velocity. fixed coordinates
is given as before by
I1~x + (13 - /1)~y~z z1~y + (zl
With the gyroscope e corresponds
suspended
to thousands
coordinates
Lz = I3(~z + 8) The transition (8.14b),
+ 13smy
z3)mx~ z
as in Fig.
from moving
and we find
=
M
x
z~8~ x = My
(8.38)
8.7, M
of revolutions
to
-- M -- O, and since x y per minute while co Z
275 corresponds constant.
to
less
With
than one revolution per day, we can take 8 to be
(8.37),
the vanishing
of M x then gives
Ii~ + (13 - ll)~2cos2lsin~cos~
The middle
term is negligible,
+ 138ficoslsin~ = 0
and if ~ is small the rest reduces
I3 $ + (Ti-~cosX)O
Equilibrium direction
= o
is when ~ = 0 and the gyroscope
it oscillates
to
(8.39)
points north;
around this
with a period
p = ~
I1
(
½
)
(8.40a)
/38flcosX As the ready recokoner will at once agree, 2~
x
366.25
=
= 365.25
x
24
×
7.292
×
1 0 -s
(8.40b)
S -I
602
and with s of the order of 102s -I, P is of the order of i minute. Foucault's gyrocompass
demonstration
to damp the oscillations, disturbances
that
largely replaced
apparatus
has been developed
that
result
never worked very well.
to keep it spinning, from changes
the magnetic
The
from it is equipped with devices of speed
compass where
and to compensate and course;
it has
accurate bearings
are
needed. Problem
Using
A gyroscope
8.15.
the non-spinning
the direction
of the gyroscope's
usual phenomena Problem
brium
8.16
spinning
coordinate
of the gyroscope
In what direction
around the y axis has I 1 = 12 < 13.
system evaluate axis
the moment
M when
is changed and show that the
are explained. does the gyrocompass
if it is being carried north at a velocity v?
point
at equili-
Give a numerical
estimate.
8.5
The Precession
of the Equinoxes
It has been known Since Hipparchus before
that,
that the North Pole traces out a circle
to put it another way, ecliptic
(~125 BC), and possibly
the equinoxes move backwards
at a rate of about
50 arc seconds per year.
long
in the sky, or around Fig.
the 8.8 shows
276
the plane of the ecliptic,
containing
the earth's orbit,
the equatorial
plane along the line g~.
line of nodes.
The sun moves clockwise
of the ecliptic.
At ~, the ascending
Astronomers
around the earth in the plane
node,
it passes
~
Figure 8.8 equatorial
plane;
at ~, the descending
of Pisces.
enters Aquarius
The precession
node,
it passes below it.
The
towards a point in the Zodiacal
The Age of Aquarius
is a gyroscopic Fig.
8.9.
is to begin when it
effect that arises because
The calculation
the
of the precessional
The sun exerts a torque on a spheriodal earth because the near side is attracted more strongly than the far side.
velocity will be done in four parts: deriving
ec|£pf£c
(Prob. 8.19).
earth is not a sphere,
Figure 8.9
above the
Precession of the equinoxes.
ascending node points at present constellation
intersecting
call this the
the potential
with the spheroidal
function
earth,
deriving
the equations
for the interaction
solving
the equations,
of motion,
of another body
and evaluating
the
effects produced by the sun and the moon.
Euler's Angles We have first to establish and angles. orientation
the relations
between angular velocity
The best way to do this is due to Euler:
define the
of the rotating body by means of angles and then express
277
the components
of ~ in terms of them and their derivatives.
8.10 shows an appropriate
construction
in two stages.
Fig.
To get from the
zo
Figure 8.10
The Euler angles are generated by rotating the ~ tipping the z axis.
space-fixed
axes to axes oriented with respect
but not spinning with it, first rotate with the line of nodes;
then rotate
the desired
of the z axis.
changing, of nodes
inclination
and if the body's
angular
is 9, the components
plane and then
to the spinning body,
the xy plane
so t h a t z coincides
the Vz plane so as to achieve If the angles
orientation
0 and ¢ are
relative
of ~ in the moving
to the line
frame can be written
down by inspection,
x = Csin0
:8.41)
V z
(If two of the
coefficients
of inertia
necessary
to fix the axes
designate
a rotation of the entire
axis;
see for example
in the body,
Goldstein
We can now proceed
are not equal,
so that it is
the third angle @ has to
coordinate
system around the z
1951.)
in either
of two ways:
find differential
equations
for 8, ~, and @ in terms of the moment M by Euler's
equations
in the form
(8.26),
or write down the Lagrangian
system and follow the route of Lagrange, Jacobi
to a solution.
Hamilton,
We shall use Lagrange's
for the
or Hamilton
equations.
and
278
The kinetic energy of the earth's rotation is
T which
½ [ll(~0X2 + ~0ya) + /3O~Z2 ]
=
is T = ½
The potential
[ZI(@2 + sinZ@$ 2) + Z3($COS@ + ~)2]
(8.42)
energy depends on the situation; we must next find one
to describe the i n t e r a c t i o n between the sun and the earth.
The Interaotion Potential Let us first evaluate
the c o n t r i b u t i o n of the sun.
is considered at rest at the origin,
If the earth
the sun moves around it in a
path which we shall take to be a circle of radius R traversed at a constant angular v e l o c i t y travels clockwise.)
x 0
=
(The negative
sign is because
it
is given by
Rcosnt,
In the m o v i n g coordinate these
-n.
Its p o s i t i o n
Yo
=
-
(8.43)
Rsinnt
system the sun's coordinates
are related to
by
x = XoCOS~ + y0sin~ y = (-x0sin~ + YO cos~)cos@ z
(8.44)
(x0sin¢ - YO cos~)sin@
=
(These are best derived by carrying out in succession the two rotations
shown in Fig.
8.10.)
With
(8.43), these take an obvious
form, x
=
Ecos(~
+
nt)
y = -Rsin(~ + nt)cose
(8.45)
z = Rsin(~ + nt)sine
We have now to calculate
the interaction p o t e n t i a l w h e n the sun is
at this point in the sky, Fig.
8.11.
The distance from the sun to the
element of mass dm is =
[(r-r')2] ½,
r.r = R 2
279
and
the p o t e n t i a l
is V =
- GM
[ d_.~m
J integrated perform
over
the
Legendre's
2_ = Z [~,2
spheroid
of the
expansion
2 _ 2r'.r
+ R2]½
:
earth.
Since
r' << R
, we
can
in L:
1 ~
r'.r + -7-
[1
+
3(r'.r) 2 - r ' Z R 2 2R"
+
"'']
M
/ Figure 8.11
The
Coordinates for calculating the attraction of a mass M by a spheriodal earth,
interaction
v(r)
:
potential
f J
- z..~ |o(x ',y',z')[1 + ~ R
2R~
r T2
- ~
Assume sense
is
that that
X' , y T , or
the p(x')
(x'
+
2X2
+ y
12~2
...]dxVdyVdz
earth's
density
= p(-x'),
(x'x + y'y + z'z) +
etc.
Z' go out and we h a v e
+ Z
,2Z2
+
2x'y
exy + ~y ,Z!
yZ
+
~xVz
v
is s y m m e t r i c a l l y Then left
all
terms
distributed
in the
containing
a single
Txz)
280 v(r)
=
___J aM
{:
p(=',y',z')
R
+ ~
term of the i n t e g r a l
the rest are c l o s e l y
r ' 2 ) x 2 + (3y '2 - r ' 2 ) y 2
-
m
related
dx'dy'dz'
is the p o i n t - m a s s
Vo = - R
while
'~
( 3 z '2 - r ' 2 ) z a ] }
+
The first
[(Sx
=
potential
f
Pd3r'
to the c o e f f i c i e n t s
of inertia.
From r I1 =
~
(r '2
_
x'2)Pd3r
+ 13 =
2
I r'2pd3r'
'
J and I I +
we r e a d i l y
deduce
that
J which,
Thus
v(r)=
or,
in terms
V(r)
R = - G~m
precession,
-
V(r)
aMm R
12
+ 13
Similar
ann u a l m o t i o n
GMm R
2I 1
results
hold
for a n a l o g o u s
+
is v e r y fast w i t h r e s p e c t this
(8.46)
+ nt) sin2@]
expression
over
to the
time.
the sun by a ring of the same mass
sin2(¢
GM ~R3
...
angles,
2R'[ (13 - II)[ 1 - 3sin2(¢ G~
Since
-
~M 2R, s (/3 - II)( x2 + y2. _ 2z 2)
and we can a v e r a g e
= -
' =
f i n a l l y we get
to r e p l a c i n g
the earth.)
- r,2)d3r
is I 3 - I i.
of the E u l e r
The sun's
amounts
p(3x,Z
since I I = 12,
expressions.
12
+ nt)
averages
(-r3 - I 1 ) ( 1
(This
encircling
to i/2, we have
3 - ~ sin2e)
(8.47)
281
S o l u t i o n of the Equations
The lagrangian (8.42)
and
that describes
the earth's
rotation
is formed
from
(8.46):
GM
+ ~
where we have omitted coordinates
(I3
the irrelevant
¢ and ~ are cyclic,
3
- Ii](l
first
- ~
(8.48)
sin28) The
term of (8.47).
so that we have at once two constant
momenta
~L
~--~ = (llsin2e
+ 13cos2B)$
+ 13~cos
8 : p@
(const.]
(8.49) ~--~= BL I3($cose + ~) = p~
where we have written more
rapid
(const)
~ for the daily rotational
than $ by a factor
of 365 x 26000,
P¢ = 13~cos8,
with the observed motion,
may once have been,
analogous
so the approximation Neglecting
and if the precession angular
constant.
in which whatever
to those of (8.33),
3GM
+ ~
(I 3 - /l)sinecos8
is uniform with
8 constant,
velocity 3GM I3 - II c o s e
to Kepler's
is there
are now very small,
to ~, we find that the equation
2R 3
According
This
wobbles
is good enough.
$ compared
I18 + 13nSsinS
precessional
Since ~ is
P% = I3~
from which we can see that 8 is very nearly consistent
rate ~.
these are very nearly
third
13
law,
R~
T2
for 8 is
= 0
this gives
for the
282
where T is one year.
Thus
$
where the negative the ecliptic.
I3 I__________ ! = 13
=
-
67[2 ( ~ T ) 2
13 - I] i3
COS0
sign means that the precession
The relevant
1 304
data, noting
0 =
(8.40b),
23o27 '
=
is backwards
along
are
7.29
x 1 0 - s s -I
"
and, for the sun's motion,
~f = 2~ x 366.
This gives a precessional
period of about 80,000 years, much too long. saved when we realize
that
the moon,
The calculation
the sun, is also much closer to the earth and actually more to the precession.
is
though much less massive
When its contribution
than
contributes
is included
(Prob.
8.17)
we find a value near the observed value
Pobs=
Problem
8.17.
Calculate
~obs
(8.$0)
25780y
the contribution
of the moon to the precession
of the equinoxes. Problem
8.18.
magazines,
the
According
March 21 to April 19. May 17 (depending that
whatever
of precession, 8.19.
8.20.
in Aries
arbitrary
and
from 17 to
of boundaries).
take no account of
when were they written down?
Find a chart of the sky t h a t
Calculate
in newspapers
from about April
assignment
the tables used by astrologers
and estimate when the Age of Aquarius Problem
published
the first sign of the zodiac,
It is actually
on a somewhat
Assuming
Problem
to the horoscopes
sun is in Aries,
shows the vernal
equinox
will begin.
the rate of uniform precession
of the top
shown in ~ig. 8.12, The top's nutation can he calculated by the equations derived here. It involves either elliptic integrals or their trigonometric example Goldstein Problem
8.21.
explanation
approximations
and is not very difficult
(see for
1950), but we shall not pursue the question.
If one did not believe
for the precession
relatiyity
theory,
of planetary perihelia
a possible
is that the sun
283
is slightly oblate. value to Mercury
Assuming
an oblateness
(43.1" per century,
that gives the correct
e = 0.2056, T = 88 days), use
(8.46) with @ = ½~ (and the oblateness now that of the sun) to find the precession to be expected for the earth e = 0.0167).
The result of Prob.
(S.0 ± 1.2" obs.,
5.9 will be useful.
l
Figure 8.12 Spinning top. 8.6
Quantum Mechanics of a Rigid Body Although the quantum theory of rigid-body motion is a hybrid of
quantum dynamical structure,
laws and purely classical
it nevertheless provides
nuclear spectra,
ideas of rigidity and
insight into both molecular and
and thus has some contact with reality.
In analyzing
it we shall also have occasion to develop more clearly some ideas about rotating
coordinates
for the subsequent
that will be needed
in quantum mechanics
sections.
In the work thus far we have started from Euler's or Lagrange's equations of motion,
but quantum mechanics
from a hamiltonian.
That given in (8.22) will not do here because it
applies
starts most naturally
to situations where the angular velocity ~ is given; here it
is a dynamical variable and we must start again. however, we pause to study the commutation
momentum
relations for angular
as expressed in a rotating system of
The Commutation
Relations
Before doing so,
coordinates.
for Angular Momentum
In order to explain the origin of the angular-momentum operator we have derived in Sec. 4.4 a special case of it: gives infinitesimal y and z.
angular displacements
Quantum mechanically,
in the coordinate picture, components
the operator that
applied to functions
we were dealing with orbital quantities
and the commutation relations
of h follow directly
of x,
(Prob.
for the
8.22) from those for the
284 linear momentum. For the derivation to come we need a more direct and general approach which is provided by the theory of this chapter. be an arbitrary vector fixed in a rotating coordinate
Let a
system.
Then
in a time 6t its tip will move a distance 6a 0 = ~ x a 6 t
with respect to the fixed frame.
(8.51)
It will be convenient
for the
moment to consider the variation of the scalar quantity A
=
(8.52)
x.a
where the I. are a set of arbitrary non-transforming constants. z Further, we introduce the infinitesimal angular displacements during the time 6t:
50.
=
m.6t
i
=
x,
Y,
(8.53)
z
Then by the rule for permuting dots and crosses, Problem 8.23,
6Ao = X . 6 O x a = - 6 O . X x a In quantum mechanics, variable,
suppose
(8.54)
that ~ is some vectorial dynamical
in general a function of coordinates, momenta,
and A = k.8.
spins, etc.,
In the form of (4.45) we express the change in A in
terms of commutation with an operator J defined by
6~
=
i6o.[3,~]
(8 5 s )
The geometrical meaning of a rotation requires that (8.54) hold in quantum mechanics
also, so that since 68 is arbitrary ^
[~,
(x.~)]
= ix×~
(8.56)
This can be written in component form if we introduce the totally antisymmetric tensor eij k = ±l:
cijk
= -ejik
= -~ikj,
e123
= I
(8.57)
That is, eijk = +l when the sequence ijk can be obtained from 123 by an even number of permutations write any vector product
and -I otherwise.
Using this we can
in component form as (AxB) i
= eijkAjB k
(8.58)
285 and ( 8 . 5 6 )
becomes ^
~j[Ji,aj] : iEijkhj~ k With ~. arbitrary, we deduce that J ^
.
[Ji,aj] This
is
the familiar
cyclic
^
: ~ijkak
(8.59)
relation
[Jl,a2] = ia 3
etc.
(8.60)
^
and i f ,
as a s p e c i a l
case, ~ = J, ^
it
gives
^
iJ 3
[Ji,J2] =
etc.
(8.61)
A familiar example of operators that satisfy this relation is the components of orbital angular momentum, Prob. 8.22.
Real angular
momentum contains a factor of ~ which the reader can insert where necessary. The theory of rigid-body motion is expressed in terms of rotating coordinate frames.
In such a frame, the rotation can be described in
a way perfectly symmetrical to what we have just done with respect to fixed frames:
consider the apparent change $ in the rotating frame
of a vector that is fixed in space. o
=
0
By (8.14b), =
~xa+a
or = -~xa
(8.62)
Let K be the operator in the rotating frame that generates angular displacements in the same way that J does in the fixed frame. The derivation goes through exactly as before except for the negative sign; we find that [Ki,~j]:-
i~ijkak
(8.63)
and
[ki,kj] = -i~ijkK k k represents angular momentum in the rotating frame just as represents it in the fixed frame.
(8.64)
286
Problem 8.22.
and using
Defining
the operator
the commutation
commutation
practice,
as
i~6ij
=
for the factor of ~ the components
relations
Problem 8.23
angular momentum
relations
[~i,Pj] show that except
of orbital
of L satisfy
the
(8.61).
Verify
(8.58)
from the definition
use these relations ax[bxe]
to derive =
(a.c)b
(8.57)
and, for
the vector formulas -
(a'b)c
and a.[bxc]
Problem 8.24.
=
To see the dynamical
[axb].c
content
of (8.61),
consider
a
A
rotating
system with a magnetic moment
gyromagnetic
ratio.
In classical
given by M = yL, where y is the
physics
exert a torque on it equal to -BxM.
a magnetic
field B will
In quantum mechanics
and compare
it with the classical
Problem 8.25.
formula.
Solve the equations
set up in the preceding
that is, solve the equation of motion find the eigenfunctions
for L in classical
and eigenvalues
problem;
physics
in quantum physics.
the relation between the two solutions which may seem, to be so different
the inter-
Find d
action is described by a term -B.M in the hamiltonian.
and
Explain
at first sight,
as to be almost unrelated.
The Equations of Motion Although scopic
rigidity
is even more of an idealization
scale that it is with laboratory
body in terms of the classical
apparatus,
on the micro-
we define a rigid
hamiltonian
K22 x32)
+ v
H = 2__ii +
I2 +
I3
(8. ss)
287
where
the potential
is expressed
in terms of the Euler angles.
The
justification
for this form is that it leads to the right equations
motion
8.26).
(Prob.
(12 = Ii)
We shall consider
in force-free motion.
transcribed
from
(8.65),
here only a symmetric
For this,
of
top
the quantum-mechanical
can be written as
(8.66)
The eigenfunctions
and
eigenvalues
those of $3 and $2; the negative makes no difference
of K~ and ~2 correspond
~k,m k
to
sign in the c o m m u t a t i o n relation
to the eigenvalue
write down the eigenvalues
closely
spectrum and we can at once
of H as
= 1 ~211_!. k ( k + l ) ~ Zl
+ (I__ 1 ~] z3 - Y~'l)m~
(8.67a)
where
k = O,
Because
1,
2....
the classical
is essentially not expect
and
-k
theory from which
<. m k
(8.67b)
<, k
these expressions
originate
a theory of the orbital motion of particles,
to find half-odd values
of k, and in experiments
one does none are
found. R o t a t i o n a l States of Nualei
If we try to apply the abstract situations
Suppose nucleus
in the real world,
for example
the others. nucleus
But what does
Thus neither
well
the individual
any differently
in the restricted
consists
it mean to a particle Nothing:
a rotating
nucleons
at rest. spectra.
of particles,
each
provided by
to say that the spherically
symmetri-
from one that does not rotate.
nor the nucleus made up of them
in a spherical
have no rotational
above to
symmetric
symmetric potential
is indistinguishable
than in the same nucleus shells)
is a spherically
The nucleus
in a spherically
as a whole rotates?
cal potential behaves
that the rotator
with I 1 = 12 = 13.
one of which moves
formalism developed
we at once run into complications.
nucleus
Spherical
imagined
nuclei
to be rotating
(those with closed
The same applies when I 1 = 12
sense that the symmetrical
object does not rotate
288
about its axis of symmetry
and the possible
states 2 are therefore
only
those with m k = 0: ~2
Ek, mk Rotational
effects
least evident) numbers
are most evident
in the rotation
of neutrons
of k are even.
2i I k(k + l) (i.e.,
and protons.
For such nuclei,
symmetry
through the nucleus
contains
an even number of particles.
the resulting wave even in k. shows
divides
function
a set of rotational
compared w i t h theoretical given.
effects
the allowed values
it into halves
of
each of which
Thus if the halves are exchanged
is invariant
of sign takes place
under the interchange;
the odd levels
do not occur.
Table
levels of U 238 obtained by Coulomb values
are
nuclei with even
if we note that the plane
an even number of changes
Experimentally,
level be correctly
single-nucleon
spectra of spheroidal
This can be understood
particle by particle,
(8.67c)
adjusted by requiring
The discrepancy
and
that is, 8.1
excitation
that the first
is a measure
of the depart-
ure from rigid rotation. Table
8.1 ~
Rotation Levels of U 238
k
Eob s (KEY)
Eth (Key)
0
0
0
2
Problem
Problem
-+ 2
149
+ 1
6
309
+ 3
313
+ 1
8
523
+ 4
536
+- 2
787
+ 6
819
+ 4
12
1098
+ 8
1162
+- 5
Use
(4.57)
hamiltonian
8.27.
transcribing
(44.7+.2)
148
i0
8.26.
classical
44.7+.2
4
to derive Euler's
equations
Show that the quantum-mechanical (8.65)
(8.26)
from the
(8.65). hamiltonian
formed by
does not commute with k 3 if all I's are different,
2 The reader may already have encountered an example of this situation in considering the specific heat of a diatomic gas, to which molecular rotation contributes two degrees of freedom and not the three that one w o u l d expect if the molecule could rotate about its axis. See Andrews 1975, Ch. ii. 3
F.S.Stephens,
3, 435 (1959).
Jr.,
R.M.Diamond,
and I.Perlman,
Phys.
Rev. Letters
289
and that therefore energy eigenfunctions cannot be taken to be eigenfunctions
Problem
8.28.
r = roA~3,
The usual expression
where r 0 = 1.2fm.
from the data in Table the nucleus 8.7
of this asymmetrical rotator
of ~2 and 23. 4 for the radius of a nucleus is
Compare the moment of inertia calculated
8.1 with that estimated on the assumption that
is roughly a sphere of radius r.
Spinors We have seen in Chapter 4 one way to introduce a change of
coordinate axes.
If it is an infinitesimal
rotation around the third
axis, for example, we write fCxl,x2) corresponding an angle (8.51),
e. in
to
a clockwise
which
if
corresponding as
~ 6t
= e,
z
of
describe
rotations
a geometric
the
a clockwise
object
the
ex2,x2
on a vector
= ~
we h a v e
x
y
= 0,
rotation.
In
the
telling
called
of
the
what
a spinor
axes (not
(8.68) through
the
axes)
is
6a o = 0
We c a n
following
+ ~xi)
coordinate
operation
rotation
vector. by
of
-
6ay ° = eax,
a counterclockwise
rotation
~
= -eay,
to
) = f(xl
rotation
The c o r r e s p o n d i n g
8axo
tion
ie ~3)fCxl,x2 + -~--
÷ (1
describe
basis
or
do
to
that
can be used
a rota-
a clockwise
discussion,
they
such
we s h a l l
vectors
and, to
later,
represent
to a
vector.
Orthogonal
Transformations
A linear transformation of a vector can be written in the form a'.
=
c..a.
~j
J
(summed)
(8.69a)
or equivalently in matrix notation a' = ca A rotation in any number of dimensions
(8.69b) is defined as a transformation
that leaves the length of a vector invariant:
ai'a i' = aia i = c i j a j c i k a k How to determine the energy levels in this case is shown in Landau and Lifschitz 1958, Sec. i01. A brilliant and very complete discussion is Casimir 1931.
290
and if this
cijcik Introducing write
a i we must have
is to hold for any vector
the notation
T to
8jk
=
denote
the
(8.70a) transpose
of a matrix,
we c a n
this as T
o jiOik = 8jk or,
in matrix
notation, T
c c = 1 from w h i c h
it follows
by d e f i n i t i o n 0
Transformations
so r e s t r i c t e d
Problem 8.29. tion is that
Show
that
T
that
=
O-
that
(8.70a,d)
.
another
the m a t r i x w h i c h
Problem 8.30. orthogonal,
In order
and
{~5
property
=
rotates
to see why
by it into two orthogonal
(8.51)
~o
orthogonal
transformations.
of an o r t h o g o n a l
transforma-
(8.70d)
6ik
axes
in the
xy plane satisfies
the t r a n s f o r m a t i o n
is called
An
vectors.
infinitesimal
rotation
in 3 dimensions
is defined
by
(8.53), a ÷ a t = a + 8Oxa
Writing
70c)
that two o r t h o g o n a l vectors a and b are t r a n s f o r m e d
show
Problem 8.31.
i
are called
cijckj and show
(8.70b)
(8.71a)
aid = Bid + Yij' or e = 1 + y, show that (8.71a)
is r e p r e s e n t e d
by the m a t r i x
y =
In order
to describe
v e c t o r we shall
introduce
Co 03 8@ 3
0
802
801
the
-8 1
infinitesimal
three m a t r i c e s
rotations
Ji by w r i t i n g
(8.71b)
of a space (8.71)
as
291
6a
o
= -i(80.3)a
(8.72a)
with
J1
=
0
-i
i
J2
:
0
0
0
0
0
J3
=
0
0
0
0
(8.72b)
Problem 8.32.
Show that these matrices
Problem 8.33.
Show that the matrices
relations
(8.61).
The negative (8.68)
from (8.71b).
J. satisfy the commutation
It is for this reason that we have called them J.
sign in (8.72a)
rotates
can be extracted
reflects
the basis clockwise,
the fact that whereas
J in (8.72)
rotates
~ in
a vector
clockwise. Problem 8.34.
(8.61),
Since the matrices
they represent
angular momentum?
an angular momentum.
That is, what is j2?
(You may need to consult the eigenvalues
J. satisfy the commutation
relations
What is the value of this
Find the eigenvalues
a book such as Arfken
of J3.
1970 to see how to find
of a matrix.)
Spinors
It has happened devices
that were
several
times in physics
first introduced
have turned out to contain profound physics. theory
is an example.
Another
example
by which Klein was able to simplify the top
(Sommerfeld
that mathematical
in order to simplify
The Hamilton-Jacobi
is the Cayley-Klein
some calculations
1952, Klein 1896, Whittaker
the theory of rigid-body show how they arose, will be similar
of spin ~.
motion
It is not necessary
a embedded
in it.
1950).
earlier:
to repeat
but we shall
and what they have to do with spin.
rotation of the rigid body by telling what happens vector
in the theory of
turned out to be the
in the new variables,
to that we have followed
parameters,
1937, Goldstein
In the hands of Pauli and Dirac these parameters key to a theory of particles
calculations
The approach
to describe
the
to an arbitrary
292
A spinor to represent
is a vector with an ordinary
the following way.
2 complex components which can be used
3-dimensional
vector with real components
The spinor u that represents
a vector
in
a is defined
by the relations aI
2ReUl*U2,
a2 = 2 I m U l *
from which it follows
at once that aia ~•
Thus any transformation complex
length
=
~=
(l~xl
Here the dagger represents
lull
=
the
(8.75)
conjugate,
conjugate
defined as the
and the transpose:
for
e,
The vector independent
=
c
*T
(8.76)
a is overspecified by u because whereas
real components,
u has
2 complex ones,
The nature of the redundancy
spinor u (~=i , 2) represents represents
a rotation preserves
1~21 = = ~ %
+
the hermitian
cf
numbers.
(8.74)
u, defined as
formed by taking the complex
any matrix
I~21=3 =
= +
which represents
lul 2 of the spinor I~1 ~ =
matrix
=
exactly
the same vector.
defined by 4 real
is clear from
a certain vector
a has 3
(8.73):
if a
a, then the spinor
e iX U
s
We shall remove this redundancy
in a moment.
Transformations
of u
An arbitrary
linear transformation u+u'
where
the matrix A depends
of u has the form
= au
(8.77)
on four complex numbers, (8.78)
The a d j o i n t
transforms
by t h e
adjoint
matrix,
u%' = utAt and so if (8.75)
is to be invariant we must have u t , u , = utAtAu
293
This
is satisfied
for all u provided
that
ATA = I Such transformations form
(8.78)
(8.79)
are called unitary,
the elements
and if A is written
are thereby restricted
(Prob.
in the
8.55) by the
relations I~I = +
I~I ~ =
But the A so defined is A' = eiCA where compare Prob.
lal ~ = =,
+ya~--
and we have already
of a.
It follows
zero is the obvious
and it is these transformations that satisfies
A
but rather
=
that
8.37)
and
Thus we make A i: (8.81)
to rotations
of a.
~ ) ,c~,2 + ,~,2 = 1
(8,82)
a ~
the general properties
of such an A, we
transformations.
Transformations
The most general matrix unity that satisfies
in the infinitesimal
neighborhood
of
(8.82) has the form A =
e2 + iE1
I I + ie 3 e2 +
where
{but
[8.81) has the form
shall look at its infinitesimal Infinitesimal
CProb.
det A = I that correspond
( -8a ~
than investigate
(8.79)
i.e., with d e t e r m i n a n t
AiA = 1,
2x2 matrix
seen
so
that X take some
one to choose.
not only unitary but also unimodular;
Any
from
of A is of the form e iX with X real
value;
(8.80)
since if A is unitary,
so we can fix the phase of A simply by requiring particular
0
that A' and A both give rise to exactly the same
of the components
the determinant
~s*
is still too general,
# is any real phase,
8.56)
transformation
Isl ~ +
ze I
] - ie3
(8.83)
J
the ~. are real infinitesimals and quantities of order ~2 are z We w i s h to see what transformations of a are induced by
neglected.
such transformations (8.75)
of u.
It is found
(Prob.
8.58) by means
of
that they are a
÷
a'
=
a
+
2~:xa
C8.84)
294
where e is the vector whose components are ¢i" Comparison with (8.71a) shows that these are the rotations we have already seen, with
¢i The g e n e r a l
form
(8.83)
can be written
unimodular
(8.85)
(8.8s)
as
1 0 -i 0 ) + g2( i 0 )
0 A = I - i[¢i(i and this, with
= ½ a°i
I 0 + ¢3( 0 _•)]
allows us to write the infinitesimal unitary
t r a n s f o r m a t i o n s of u in a form exactly analogous
to that of
(8.72a), 6u
where the m a t r i c e s
ol
0 = (1
a. are the Pauli m a t r i c e s
I 0 )"
0 -i ~2 = ( i 0 )"
These matrices have the p r o p e r t y that called hermitian.
(8.86)
= -i(6@.½~)u
~i = ~ti; such
matrices
(8.87)
are
They generate rotations of a via the spinor u just
as the matrices J generate
them directly.
If we look at the finite t r a n s f o r m a t i o n s infinitesimal
1 0 = (0-1 )
o3
c o r r e s p o n d i n g to the
(8.83) we encounter a remarkable
feature of spinors.
Let
the rotation be around the third axis so that ¢i = e2 = O and
U
As in Sec.
-~
I I+ig 0
0) E3
U
1-ie
=
½
~03
3
4.5, the finite t r a n s f o r m a t i o n is formed by iteration,
Lira
1+2 N
~,~o
o
O
A(O3) 1 - =i
or
e-~ s
and there are similar formulas Prob.
8.41.
for the other two e l e m e n t a r y rotations,
If the angle of r o t a t i o n is 2~, so that a returns
original orientation,
(8.88)
to its
the components of u are m u l t i p l i e d by -I, and
it is not until a has rotated through 4~ that u returns to its
295
original value.
It is natural
oddity has any counterpart suprisingly,
is yes.
to ask whether
this mathematical
in the physical world.
A simple demonstration
The answer, perhaps
popularized
by Dirac shows
that it is indeed possible
to perform an experiment which tells
whether
2~ has been made but is unable to detect a
a rotation
rotation
through
through 4~.
Attach three
2-ft pieces
piece of cardboard hand now rotates the triangle
of string to the corners
and give them to three hands to hold.
the cardboard
through
A fourth
4~ around a horizontal
is held still it is now possible
strings without
of a triangular
to disentangle
letting go the ends so that no evidence
axis.
of the
rotation remains,
but if the rotation
be disentangled.
That this fact is indeed a physical manifestation
spinor transformations topological
is through 2z the strings cannot
argument. 5
out. s
An iron foil is magnetized strips.
wave functions determined
effectively
8.35.
Problem
8.36.
and we may
8.37.
directions
along
and the change in phase can be
pattern.
through
has been carried
through the foil have their The results
show clearly the
2z.
Show that the most general (8.80). Show that if A yields
A' = eitA yields Problem
in opposite
rotated,
by the interference
restricted by
Such an experiment
Slow neutrons passing
sign change under rotation Problem
not topology,
there is any dynamical phenomenon which could detect a
rotation of 2~ but not one of 4z. parallel
of
can be shown by a subtle but straightforward
The subject of this book is dynamics, ask whether
If
the
form of a unitary A is C8.78)
a certain transformation
of a,
the same transformation.
Show from the properties
the determinant o f A,
of determinants
that if D is
(8.79) requires
[Di 2 = I
whence
D = e ~X
for some real X. Problem
8.38.
Show that if A given by (8.83)
ding transformation Problem
s
8.39.
E.D.Bolker,
acts on u, the correspon-
of a is (8.84).
Show that the matrices
½~i' with a~ as in (8.87),
Am. Math. Monthly 80, 977 (1973).
A.G.Klein and G.I.Opat, Letters 37, 238 (1976).
Phys, Rev. DII,
523 [1975);
Phys. Rev.
296
satisfy the commutation earlier matrices
Problem 8.40.
=
the last relation
Problem 8.41. other way,
for angular momentum
Show that the Pauli matrices
aiaj where
relations
like the
J. in (8.72a).
Eijkak
satisfy
and
ai 2 = 1
is true for each value of i.
With the derivation of
(8.88)
as a model or in some
find the matrices A corresponding
around the first and second coordinate
to finite rotations
of a
axes.
Spinors and Vectors In (8.73) we showed how to find the components of u.
The m o t i v a t i o n was
ad hoc.
of a given those
(8.74), but the substitution
Let us find a heater
form
for
(8.73)
looks somewhat
and also solve the
inverse p r o b l e m of finding u given a. Since the components
of a are bilinear
functions
of u and u t, there must exist three matrices for every u, the corresponding
a + a + 6exa, This
gives
u
(Prob.
(8.89)
= utMu
the unknown matrices
rotation of coordinates
M i with the property that
a is given by a
We can identity
of the components
if we carry out the infinitesimal
given by + u
-
½ i(~e.~)u,
ut + u t
+ ½ iut(6e.a)
8.42)
gijk~@jak
= ½
i6ejut[~j,Mi]u
(8.89a)
and since this is to be true for all 8@j, the use of (8.89) gives
~ijkUtMk u = ½ Since this,
in turn,
iut[ajaMi]u
is to be true for all u, M must satisfy
(8.9o)
½ [Mi,a j] = i~ijkMk It follows
from the result of Prob.
~i this relation
is satisfied,
the only solution. and a m u l t i p l i c a t i v e
8.39 that if M. is proportional
and a little thought
[Mi contains four numbers,
(8.90)
factor remains undetermined.]
to
shows that this is is three relations, It follows
from
297
direct c a l c u l a t i o n
(Prob:
8.43) that u%o.u
with
which
compact from
we h a v e
expression
solved for
the
the
(8.91)
= a.
problem
set
relations
in
(8.73)
(8.89), with
by giving
which
a is
a
found
u. Our next
task
is
to
find
u given
a.
From
(8.73)
and
(8.74)
we
have a I
a which
is
satisfied
ia 2
-
=
2UlU2
w i t h U c h o s e n so that +
I.~1 ~
U1
a 3
by u I = p ( a 1 - i a 2 ),
I.~l ~
~
=
(8.74)
I~1~% ~
u2
~(a
=
-
a 3)
is also satisfied:
+ ==~ + a~ _ = ~
+ =~)
=
~1.1~=~
- =~) :
so that °
i a 2 ) e ~'×
(a 1 U 1
=
,
[2Ca
-
U 2
[ ½(a - a3)]½e ix
=
a3)] ½
w h e r e X is an a r i b t r a r y phase common to both. more t r a n s p a r e n t form if we express angles
(8.9z)
These formulas take a
a in terms of the usual polar
@ and ¢: a 1 = asin@cos¢,
a 2 = asinesin¢,
a 3 = acose
whence a I -
We find
(Prob.
ia 2 =
asin@e
-i~
8.45)
u I =
a½cos½Se
-i~/2,
w h e n X has b e e n set equal to
u2 =
a½sin½@e
¢/2 for symmetry.
i~/2
(8.93)
Note again that
changes sign w h e n ¢ = 2~ and returns to its original value w h e n
u ¢=
4~.
Spinors are of interest today b e c a u s e they enter the c o n s t r u c t i o n of q u a n t u m - m e c h a n i c a l wave functions. way,
of course,
functions
as products.
They can enter them in a trivial
For example,
of h y d r o g e n c o r r e s p o n d i n g
quantum number n are of the form
the three p - s t a t e wave
to a given value of the radial
298 ½
I::>
= 2-
11o>
= X3fn(r)
I:-:>
(xi + ix2]fn(r)
= 2-½(x I - ~x2)fn(r)
If v is the spinor that represents
= 2½Vl*V2fn(r)
I:o>
=
But this and formulas
Iv21bfn(r)
~ -
into other terms. containing
an arbitrary product
Ivll 2 +
spin
Iv212
transcriptions
(fermions),
of the
We get something new when
a single spinor as a factor or,
of spinor components.
It turns
the spins of
and it is this fact, to be
that gives spinors
the discussion
given here.
Problem
8.42.
Derive Eq.
Problem
8.43.
Show by direct calculation
their names and motivates
(8.89a). that
of a using the spinor components
Problem
8.44.
Derive
the formulas
68.92).
Problem
8.45.
Derive the formulas
(8.93].
8.8
r =
(8.94b)
of this kind are needed to describe
of half-odd
discussed briefly below,
component
,
like it are only trivial
wave functions
out that expressions particles
(Ivll
= ~½vlv2*fn(r)
vectorial wave functions more generally,
the vector ×, these are
11:>
I:-I>
we construct
(8.94a)
(8.91) holds for each
of (8.92).
Particles with Spin The angular-momentum
the angular dependence in (8.94), describes
for example, a particle
certain directional
properties
of particles
of their wave functions.
are associated with We have already seen
how the vectorial wave function ~ = rf(r)
of spin i. properties;
A spinor is a geometric
object with
let us see how it can be combined
into
a wave function. The three wave functions the components
in (8.94a)
of the 3-component
are linear combinations
vectorial wave function rf(r),
of
299
chosen so as to diagonalize
J3"
The wave function
= uf(r)
is
an a n a l o g o u s
angular
(8.9s)
2-component object.
The o p e r a t o r
momentum when t h e wave f u n c t i o n
is
that
represents
a 2-component spinor
S = ½ ~o and the expectation a certain vector from
conjugate
is (8.96)
values of the components
spinor u and its hermitian
the
u%,
of S are formed using the
If the spinor corresponds
a, then we find the expectation
to
value of S at once
(8.91): <S> =
Thus although physical
space
It follows
in spin space
½ha
the spinor has only 2 components,
it represents,
as before,
from the principles
a quantity
in this way has an intrinsic
(Since f(r)
is spherically
symmetric,
that the object
angular momentum equal to 1/2.
there
is in this case no orbital
To see this let the basis vectors
be ~ and 6, so that an arbitrary u is w r i t t e n components
in
in 3 dimensions,
of quantum mechanics
represented
angular momentum.)
(8.97)
of the u-space
in terms of its (complex)
as
= Ul~ + U26 In this basis
the basis
vectors have components
1 a = (0)
These are
the
two i n d e p e n d e n t
0 6 = (1)
and eigenfunctions
Saa = ½ ~ a , and t h u s
they represent
o r down.
In the state
values
S 3 = ½~ a n d -½~
the states described
of S3:
S36 = - ½ ~ 6 in which the by ( 8 . 9 3 ) ,
spin
is
certainly
the probabilities
a r e cos2½@ a n d s i n 2 ½ e r e s p e c t i v e l y .
is obvious mathematically;
what iS n o t o b v i o u s m a t h e m a t i c a l l y
nature
to play with
has g i v e n us t h i n g s
formalism. such as e ,
It has been found that p, n , p ,
such as A, Z*, E*,
correspond
a number of the
and ~- have spin 3/2.
of spin
Higher
this
is
that
particles
several others
It is a relatively
relativistic
O, 1/2, and i.
All
to the
familiar
A, I, and ~ have spin 1/2, while
matter to find the wave equations, for particles
that
up
of the
simple
and nonrelativistic
spins have been an
300
unusually stubborn problem which seems finally to have been solved in recent years. Nothing in the foregoing argument provides any information as to the dynamics of the systems described;
requires further hypotheses.
that
we got the classical hamiltonian for rigid-body motion by putting the body together out of point particles each of which obeys Newton's laws. Clearly no such simple c o n s t r u c t i o n
is possible here.
It is assumed
in the modern theory that the hamiltonian contains no part corresponding to a particle's energy of spin; such a part, if it did exist, would ordinarily be constant and not participate in any dynamical exchanges of energy.
Only in an a t t e m p t
to treat transitions between
the states S(S = ½, m = 1190 MeV) and Z * ( S = 3/2, m = 1383 MeV), for example, would such a thing come into play, and nothing at all simple has come out a t t e m p t s
to make a theory.
(Readers with anti-
quarian i n t e r e s t s may consult Pauli 1946). The interaction between spin and an external electromagnetic field is introduced through the hypothesis known as minimal electromagnetic coupling, as follows: write the field-free nonrelativistic hamiltonian in the form ~o
(o'~) ~
= it= 2m
2m
C8 98)
where the equivalence of the two forms follows from the algebraic properties mentioned in Prob. 8.40.
Then use the rule (4.18) that
includes a magnetic field by the substitution p + p
-
eA
and finally, add the scalar potential e V .
[8.99)
On simplifying [ ~ . C p - e A ) ]
2
we find =
(P"eA)2 2m
-
~ S.B m
+ eV
C8 lOO)
in which the middle term represents the coupling of a particle with gyromagnetic ratio e/m to the magnetic field. We see that the electron's magnetic interaction emerges out of the hypothesis C8.98). It should be noted t h a t we have merely expressed one hypothesis in terms of another, 8 though the step [8.99) can be justified by fundamental arguments
(Park 1967, p. 188).
7
W.J. Hurley, Phys. Rev. D4, 3605 (1971).
8
J.Kalckar, Nuovo cim. 8A, 759 C1972),
301
Historically, "derivation" wards,
(8.100) was written down by Pauli in 1927.
given here made its way into the literature
and its only merit
electron's
gyromagnetic
in which the electron
is that it gives a way of understanding
ratio without becoming
is pictured
as a spinning
The real value of these considerations from rotational
transformations
in four dimensions. relativistic
well as the existence
in all of physics.
8.46.
Derive
(8.100)
from
Problem
8.47.
Show from (8.100)
(8.98)
gyrocompass?
is now one of the The matrices
now
ratio emerges,
as
and (8.99).
that d<S>/dt belongs
Can a hydrogen nucleus
which may
of positrons.
equation of motion as classical physics 8.48.
of charge.
This was done by Dirac in
but the same gyromagnetic
Problem
Problem
distribution
wave equation
and properties
the
in arguments
to Lorentz transformations,
1928 and the resulting
most solidly based relations
involved
lies in their generalization
be viewed as rotations
have 4 rows and columns,
The
long after-
to the same
gives for the same situation.
in a hydrocarbon
serve as a
CHAPTER 9
CONTINUOUS SYSTEMS
If we consider the motions of a continuous
s y s t e m - a stretched
string or the air in a room-- it might at first seem that we would have to deal with the infinite limit of an N-particle complexities would arise. to understand everything not so ambitious.
the individual molecules are doing, but we are
The right approach is to forget all about discrete-
ness and deal from the beginning with continuous coordinates.
system and that vast
This would of course be true if we wanted
This does not always work.
functions of the
If for example we want to
study the elastic vibrations
of a crystal we must start from the inter-
molecular
of this kind go back to Cauchy in 1822
forces.
Arguments
and are of a much higher order of difficulty than any to be attempted here.
We shall consider a few examples of mechanical vibrations--
stretched strings,
vibrating membranes,
sound w a v e s - and then an
example of a field that can be treated in much the same w a y - the matter field ~. 9.1
Stretched Strings The simple one-dimensional
and vibrating transversely
system of a string fastened at the ends
is a good introduction to the entire sub-
ject, since the extension to more than one dimension is only a matter of detail.
In the continuum limit the string has a number f of degrees
of freedom that becomes infinite,
and its configuration can be speci-
fied by giving the displacement u(t) corresponding along the string.
This suggests
to every point x
that f is not only infinite but that
it has the power of the continuum-- like the points in a line, the degrees of freedom cannot even be counted. kind of generality,
But we do not need that
for the string is continuous
and every point moves
very much like the other points in its immediate neighborhood; restriction will greatly simplify the analysis.
this
We label the points
on the string by the variable x and specify the displacement by the continuous
function u(x,t)
(Fig.
9.1a) which,
in the case considered
here, must vanish at the two ends: u(O,t)
Another way to characterize
=
u(Z,t)
=
(9.1)
0
the string is to express u(x,t) as a
Fourier decomposition, u(x,t)
- , , imKx = ~ am[~Je
K =
~
where the value of K is explained in Fig. 9.1b:
(9.2) the longest wave-
303
length, u(x,t)
corresponding
to m = +i,
is 2£.
Because
the displacement
is a real number, we have co
u(x,t)
= u*(x,t)
co
= 7~ a m
*e - i m K x
E a
=
--oo
so that the coefficients
-m
*e i m K t
_oo
will satisfy
(9.3)
a m*(t ) = am(t )
t~
t
•
,
\
%__-
x=O
Figure 9.1
×=L
In the Fourier r e p r e s e n t a t i o n infinity,
This results
the various (9.3)
coordinates
it by treating
move
independently
a more general
if u(x,0)
special
and &(x,#)
u(z,t) will remain real at all subsequent It will be instructive
on
as a The condi-
since one prefers
to have
of each other, but we can in which
to take on complex values, cases by the imposition are real,
for example,
times.
to study the problem
x descr.
Fourier descr.
lagrangian
1
3
hamiltonian
2
4
which we shall take up in order.
of u(x,t)
class of problem,
is at liberty
out the desired
of the points
the analysis.
of an inconvenience,
is not used and u(x,t) conditions:
and not t h a t
simplify
coordinates
but it is a countable
from the assumed continuity
is something
and then selecting boundary
in number,
the infinity of the integers
(9.5)
overcome
zl
the am(t ) are the generalized
They are infinite
function of x, and it will greatly tion
t
(a) String secured at the ends. (b) Representation of its normal modes as odd harmonics of a wave twice as long.
of the system. a line.
o
in four ways:
of then
304 9.2
1)
Four Modes of Description
Lagrange's Equations,
x Description
Let T be the tension of the cord, a shape given by uCx,t).
and let it be pulled aside into
The old length was £.
The new length is
~9.4)
~' = I de = Io [I + u'2Cx, t)]~dx =Io[I + ½ u'2Cx, t)]dx where u' = ~u/gx. the displacement
We have made the approximation is small.
of assuming
that
This is called the linear approximation,
since it leads to a linear equation of motion.
Problem 9.2 shows
that things become much more complicated when nonlinear terms are included. The slightly displaced
string is stretched by an amount
£'-
~
=
#
u'adx 0
and the work required
to do this is the potential
v=~ z If p is the linear mass density, see Prob.
9.22),
energy,
u ,2dx
(9.Sa)
o which we shall assume constant
(but
then the kinetic energy is obviously
T = ~
~2dx
C9.5b)
o where & = ~u/~t. The Lagrangian
of the string is then
o This is the sum of contributions the string.
S :
Ldt = Jt 0
where L,
the lagrangian
dx 0
density, L =
In g e n e r a l
from each element of length along
The action is given by
Csee P r o b l e m 9 . 3 ) ,
dtLCu',~)
C9.7a)
to
is
% Cp; 2 -
(9.7b)
~u '2)
L is a function
LCu,u',~ ) .
S is also
305
the sum of the actions belonging string,
each of which moves
exerted
by the adjacent
to successive
in response
forces
that the lagrangian
can be written
in terms of the system's
is a little
long
forces
is the potential
is true but not quite obvious.
show generally energies
dx of the
to time-dependent
That ½ Tu'2dx
elements.
energy produced by these
elements
of a continuous
mechanical
total kinetic
(Yourgrau and M a n d e l s t a m
To system
and potential
1968,
Chap.
8) and
The reason is that it is of no
we shall not pursue the matter here.
general use to be able to do so, for the main utility of variational methods
today is in the theory of fields, where
whole field at once; with the string,
principle
Problem
as can be done
of m o t i o n of the string are derived from varying S.
6u vanishes
at t O and t I by the terms of Hamilton's
Hamilton's
principle
and two independent
the string
~-u =
In our particular
aL
~u
-
~ ax
is not free to
6S = 0 is thus a problem
variables,
~L au,
in one
which has been shown in
1.21 to lead to the Euler-Lagrange ~L
motion
the
elements.
and at x = 0 and x = £ because
move there. dependent
one does not put it together,
out of individual
The equation The v a r i a t i o n
one considers
equation
~ aL ~-~ = o
-
case L is independent
(9.8~
of u, and the equation of
is TU" - p~
= 0
or u"
This
-
is d'Alembert's
~2
u(x,t)
it.
~/p
(9.9)
equation and its solutions
for any doubly differentiable
satisfies
v2 =
~. -- 0
= f(x
- vt)
+ g(x
That the sum of two solutions
the result of linearity,
and the velocity
is again a solution
is independent
does not change of amplitude
at the speed v.
and frequency.
of a system are the motions resulting
In
as the wave progresses
of each part is periodic with the same frequency. is m, the normal modes
is
and f and g represent waves of arbitrary
the w a v e f o r m
The normal modes
general,
+ vt)
w a v e f o r m m o v i n g to right and left respectively this a p p r o x i m a t i o n
are extremely
function of the form
in which the motion If the frequency
from (9.9) are the solutions 2 u" + ~-f u = 0
of (9.10)
306
specified by the appropriate boundary conditions.
Problem
For many field calculations boundary conditions
9.1.
like
(9.1) are inappropriate because they introduce reflected waves. alternative
is the periodic boundary condition u(Z) = u(0).
An
This
changes the value of K in (9.2) and also the conditions on ~u in Hamilton's principle. Problem
9.2.
Show how the theory developed above is changed.
Find the wave equation that results if the wave amplitude
is great enough so that the first three terms in the expansion
(9.4)
must be used. Problem
9.3.
A stretched string is attached at intervals to a
straight bar by springs of unstressed
Figure 9.2
length d, Fig. 9.2.
Find the
To illustrate Problem 9.3.
wave equation and give a solution.
Assume the springs are close
enough together so that their effect is essentially continuous. Problem
9.4.
The normal-mode vibrations
of a string are those in
which each part of it moves sinusoidally with the same frequency. Find the normal modes and their frequencies for the stretched string and also for the modification shown in Figure 9.2. Problem
9.5.
A bent board stores potential energy because the fibres
on one side of the median plane are stretched and those on the other side are compressed.
Assuming
that the fibres obey Hooke's law, show
that the stored energy is given by an expression of the form
V = o
u"2dx 0
Find Lagrange's diving board.
equation for the free oscillations
of a uniform
The boundary conditions are u = u' = 0 at the left end
307
and u"
= 0 at
frequency
of
the the
right
because
no
lowest
mode of
oscillation.
FT
string
9.6.
applied
there.
Find the
To illustrate Problem 9.5.
Find the equation
fixed at the ends
is
T
Figure 9.3
Problem
torque
for longitudinal
if a static
vibrations
tension f produces
of a
a stretch
ef£ in a string of length £.
2)
Hamilton's
Equations,
x Description
We define the m o m e n t u m
~
conjugate
to the variable u by
aL and the h a m i l t o n i a n
density H by H = z~
The
equation
of
motion
for
z
is aL
for
arbitrary
~H = "n'6~ + ~
T h u s we c a n
regard
variations
~
in
u,
aL 6"~ ~L Tu au - a-~ - ~,
a function =
where
the functional
while
(9.13) gives
(9.11)
~L
(9.12)
ax a u '
- aL
fl a s
L
(9.8),
{ = g~uwhile
-
of
~,
6u' = uS~ -
~,
aH aH aH - ~ + ax au
derivative
and u',
u,
= -
and
~L
6u -
u',
~H %-~u'
~L
gg,
and
6u'
(9.12)
(9.13)
becomes
(9.14a)
is of the same form as that in (1.46),
_- aH
6H
(9
14b)
308
These are Hamilton's
equations
for a one-dimensional
continuous
system.
We can easily verify that H is the energy density by seeing that its linear integral H is conserved.
we
a(u,u',~,w')
For any density
with
have •
d-T =
plus terms evaluated (9.14)
•
~a ~ + ~a
C-~-~ u + ~ d ' u ' + 5-4-
~)&
~-,
at the boundaries
which will normally vanish.
By
this is dA
6a ~H
d-T =
~a ~H
( ~ ~ - ~ ~-~)dx
(9.15)
o The resemblance the conservation
to [4.57)
defining
Poisson brackets
of energy follows
In the present case,
is obvious
and
at once.
from (9.7b), ~r =
pu
(9.16)
and
~2
T u,2
(9.17a)
while the total energy is
I~
I'£ 0
(9.17b)
Tu,2)d x
0
as one could have written with much less trouble from
Problem 9.7.
Find Hamilton's
equations
normal mode of the system of Fig.
Problem
g.8.
Find Hamilton's
(9.5).
and solve them for the lowest
9.2.
equations
for a nonuniform
cord in which
p is a function of x.
3) Lagrange's Equations, Fourier Description In this description am(t) and the velocities mentioned,
the coordinates
are the Fourier amplitudes
are their time derivatives.
we shall not impose the conditions
but allow them to follow from the boundary complex it is not obvious how to minimize
(9.3)
As already on the coordinates
conditions. or maximize
But if u is a complex S:
309
is -21 smaller
or larger than i?
The solution
is to make S automatic-
ally real by defining
L=~ noting
that
if,
the old one.
as we a n t i c i p a t e ,
Putting
u
into
is
real,
(9.18)
C9.18) the
new
L is
m ~°°
The integral
the
same
as
gives
da ~m = dtm
i ~S co C~ ama*m' - -Emint ,l(2ama,m,)ei(m-m')KXdx
L =
to
(9.2)
~T I~,I ~
I~1 ~
m t=_oo
of e
i(m-m')Kz
vanishes
unless m' = m, and it is then equal
~:
z
c = ~
5"~
(0 I~m I ~ -
m2Z2
laml 2)
We could take Rea m and Ima m as the coordinates, to take a
m
and a ~ ( P r o b .
9.12).
m
6L ~a ~ m
The
(gig)
but it is more natural
equation
~3L -~'-~-
d aL 2"%~-~ "-~= o
m
m
gives
~" or, with
d ~
~ maK2a m
p • -~ a m )
(9.9),
am + (mKv) 2am for each m, and ~L/~a m = 0 gives of the continuous simple harmonic coordinates.
= Am e im~°t
where A m and B m are complex conditions problem
conjugate.
The problem
to that of a set of independent
is, the am(t ) are the normal-mode is
+ B m e -im~°t constants
and the conditions
(9.20)
~o = Kv
(9.21)
to be determined by the initial
at the boundaries.
We have solved the
in Fourier variables w i t h about as much labor as it took to
state it in terms of x. initial
That
The solution of (9.20)
am(t)
0
the complex
cord is thus reduced
oscillators.
=
conditions.
To complete
the solution,
we must put in the
Suppose we are given the initial
configuration
310
and the initial velocity u(x,O)
u(x,@)
of the cord at each point.
We
have
u(x,o)
= ~
(A m + B m ) e i m K x
(9.22a)
m(A m - B m ) e i m K x
(9.22b)
oo
u(x,O)
= ico o
Z
By Fourier analysis we can find the coefficients Bm) and then solve for A m and B m.
m(A m
real, then a m obeys in detail in Sec. 4)
Hamilton's
(9.3)
If u(x,O)
9.14).
and u(x,O)
are
We shall work out an example
9.3.
Equations,
The momentum
(Prob.
A m + B m and
Fourier
conjugate
to a
Description
is m
2z_p_~ ,
at Pm = ~ - - = m
and, of course, of this,
Pm
the momentum
m
conjugate
The hamiltonian
to a
is the complex conjugate m
is easily seen to be
co
H : Hamilton's
equations
2 C-~
z
%T IPm 12 + -T-
laml 2)
m2K2
lead at once to (9.20),
(9.23)
and the energy comes out
as
E when a
=
z2~
is given by (9.21).
z
m2Clnm 12 +
IBml =)
The time-dependent
(9.24)
terms have cancelled
m
and E is
constant.
Problem
9.9.
Derive the hamiltonian
Problem
9.10.
Show that Hamilton's
Lagrange's Problem
equations
8.11.
9.12.
Evaluate
E with particular
and Ima . m
m
are equivalent
to
attention
to the cancella-
terms.
Show that the equations
a m and am* i n d e p e n d e n t l y
Rea
equations
for the system just discussed.
tion of t i m e - d e p e n d e n t Problem
(9.23).
are equivalent
obtained
from (9.19) by varying
to those obtained by varying
Problem
9.14.
Fourier
amplitudes
9.3
Show that if u(x,o)
Example:
derived from
and u(x,O)
(9.22)
are real,
satisfy
then the
(9.3) automatically.
A Plucked String
In this section we shall work out an example of the general considerations
given above and propose
Fig.
the initial position
9.4 shows
0
I/Z
Figure 9.4 and released
u(x,O) =
Since u(x,O)
a few corollary problems.
of a string pulled to one side
String displaced at the middle.
from rest at t = 0.
~2hZ-ix
(O<x<~,/2)
L 2hg-~(z-x)
(z/2<x<~)
= 0, the time dependence
,
~(x,O)
of the normal
=
0
coordinates
(9.2s)
is
given by c o s i n e s , and u(x,t)
=
Z
A eimKxcosmKvt
(9.26)
m
The constants conditions.
can now be chosen so as to satisfy the boundary In order that u(x,t)
must go together
= 0 at x = 0 and £, the exponentials
to form sines,
u(x,t)
= Z B sinmKxcosmKvt 1
m
With this, co
u(x,O)
and
B
m
is readily
= Z B sinmKx 1 m
found from
2 B m = -~
u (x, O) sinmKxdx 0
K = ~/i
(9.27)
312
B m -- ~
Thus B
m
= 0 for m even,
8h
sin
m~T
and m-1
Sm = ( - I ) ~
8h
odd)
(m
~T2m~
(9.28)
The motion of the string is given by
u(x,t)
= ~8h
I
(sinT~xcosKvt
The wavelengths
are 2 ~ / f ,
from an elementary
~
2~/3K,
construction.
harmonics
depend on the squares
therefore
vary as I, 1/34 . . . . .
understood Fig.
by sketching
1
sin3Kxcos3fvt
...
+ ~ T sinSKxcos5Kvt
, or 2Z, 2£/3,
The intensities of the Fourier
...
- ...)
(9.29)
, as is clear
of the various
components
of u and
Why the even terms vanish can be
some of the even
modes
and comparing
them with
9.4.
Euler's
Sums
The fourier application.
expansion
just used has an interesting mathematical
With 0 5 x 5 £ / 2 ,
n
the expansion
is
Z J _ . L sin % 1 sin n ~ x odd ~ 2 n 2 £
-
2hx £
1
Let x
=
~/2:
n
8h ~2n2
odd
sin
2 n~
- - =
2
1
2
n
odd
8h - - = ~2n2
h
1
so that 7T 2
=
So
2 n=1,3,5,...
-n-2 = - -8
(9.30)
Let oo
S
Then
=
2 n=l
1 n2
a
Se =
X n--£ n = 2 , 4, 6, . . .
(9 . 31)
313
S=S
I But also S e = ~ S
(Prob.
o
These
=
e
4 Thus S = ~ S O and
9.15).
~2 S
+S
72
- -
6
"
SO
sums are not especially
were obtained by Euler
~
- -
8
~2 S
"
easy to derive
=
e
(9.3z)
- -
24
in other ways.
They
in 1737.
Problem
9.15.
1 Show that S e = ~ S.
Problem
9.16.
Evaluate
Problem
8.17.
What
Z n -4. 1
is the harmonic
content
of the sound produced when
a string
is held at the point x = x o and released
x
where n is an integer?
o
=
£/n
Problem
plucked.
Assuming
the h a r m o n i c Problem
A piano
9.18.
string
a piano
dissonant with the note. order not to excite
not
string at x = xo, find
of the sound produced.
Show that the ninth harmonic
9.19.
What if
is set in motion by being struck,
that the hammer hits
content
from rest?
it?
Where
of a given note is
should the hammer
Are the m a n u f a c t u r e r s
strike
the string
of grand pianos
in
aware
of this fact? Problem
An electron
9.20.
well with infinite properly
normalized.
Practical We have
computations
shown in Sec.
can do something finding what
measured
Use of V a r i a t i o n
easier
quite
2.13 that Maupertuis's
similar.
In the double-integral
9.4,
that if the electron's (n~/Z) 2 ( ~ 2 / 2 m ) ?
Principles
Consider
variational
principle makes
and in continuum mechanics for example
mode of a string.
it is, and that will be useful
-iat
into a potential
it will turn out to be
in point mechanics,
the lowest vibrational
let u ~ e
introduced
What is the p r o b a b i l i t y
energy is subsequently
9.4
is m i r a c u l o u s l y
sides with a wave function of the form of Fig.
in assessing
some we
the problem of
(We know of course the approximation.)
principle
with ~ to be determined.
The integration
over t is now
314
trivial,
Suppose
and we have
the string stretches
from x = -i to x = +i, and let ~ = p~o2/T
be the unknown quantity that gives w. t1
6 ~
Then
(;~I.l ~ - l.'l~)d= = 0
(9.34)
J-I with
6u = 0 at the ends.
The calculus
of variations
(Prob. 9.23)
leads
to the equations we have already solved,
but it will be instructive
pretend for a while
the problem
that we cannot
To solve the problem single antinode, parameters.
approximately,
satisfying
We calculate
parameters make
let us invent a curve with a
the boundary
=
Suppose
conditions
for example
- x ~)(c o
(1
+
in (9.34)
16
8
s = C-i-~ x - ~)°o and the equations
@S/Be
z
°
and containing
free
of the
that (9.35)
e l x 2)
This is zero at x = +i and is symmetrical of the integral
to
exactly.
the action and then find what values
it a minimum. u
solve
right and left.
The value
is 16
88 )o
+ C-~-~5 ~ - -~)°o°~ + C ~ = 0 and
~S/~o
I
=
0
2
( ~ ; ~ - 5)°0 + (7 ;~ - I ) ~ i
~ - ~-~
2
give =
0
(9.36) (2~, - 7 ) % + (~ ; ~ - : I ) o I = o In order for these vanish,
to have any solution
and this gives
the eigenvalue ~2
Whose smallest
root
(lowest l
The exact value
=
_
28h
+
63
frequency) 14
-
at all,
1/]--~ =
=
2.46744
TF 2
The error is about ~0~%.
-~- =
0
is
is =
their determinant
must
condition
2. 4 6 7 4 0
C9.37)
315
Notes:
They don't always
a)
of a trial interest
in connection
16
s =
8
(-ff x -
~)o °
2
one,
our choice
approximations
is of
to ~.i
- x ~) is not too bad:
Co(I ~S
,
Clearly
so much so that the result
with algebraic
Even an approximation
b)
come out this well.
function was a happy
16
~ =
8
0 = 2o o ( - i Y ~ - ~ ) "
~ = 2.5
0
We h a v e r e a l l y
c)
normalization extra
unknown i s
The g r e a t
d)
o n l y f o u n d one p a r a m e t e r
is undetermined, generally
but
simpler.
accuracy with which the
n o t matched by a c o r r e s p o n d i n g substitute
The e x a c t
~ into
we f i n d
Uva r
Co(1
solution
-
1.2207x
an e r r o r
fractional
example, with
root
of that
Problem
9.21.
- 1.2337x
Problem
9.22.
of the
Problem
the 9.23.
variations Problem
equation
i
given
of a String
of the
Quart.
for
9.7 that
the
of the
/(6-~0%) =
square
~%.
(9.34) was not a proof. writing
down a solution
that it satisfies
secured
(9,34).
at x = ±i is given by
(-l~x~+l)
the solution
string
(9.38)
l o w e s t n o r m a l mode. of
obtained is rather
of motion has been studied
S. Ramanujan,
case,
= XoCOS2 ~x4
to the results
A piano
~ + ...)
= 0.7071
in Sec.
on t, and showing
The aensity
Show t h a t
(½)
is of the order
the wave equation,
periodically
frequency
see
In this
The demonstration
leads
9.24.
We s h a l l
eigenvalue.
l(x) Estimate
~)
2 + 0.2537x
Uexac t
in the eigenfunction
Fill the gap by deriving that depends
2 + 0.2207x
c o = 1,
o f a b o u t ~% .
error
I f we
is
Uvar(½) = 0.7086, with
has been determined
in the eigenfunction.
the approximation
Uexac t = c o c o s ~ x = c o ( 1 At x = ½, f o r
eigenvalue
accuracy
(9.36) =
of u, namely el/C ° since
the homogeneous approach with the
J. Math.
(9.34)
of
earlier. stiff.
in Prob.
45, 350
by t h e c a l c u l u s
How t o f i n d 9.5.
(1914).
its
F i n d t h e phase and
is
316 group velocities for the propagation carry the problem point where results.
of determining
a difficult
Require
of waves
down such a string and
its normal-mode
transcendental
equation
frequencies
for the frequencies
that u and u' vanish at the ends.
four dynamical methods
to the
Any or all of the
developed above may be used.
Approximate
the
frequency of the nth normal mode to the first order in the stiffness parameter
c.
Problem 9.25.
Assuming
carry out a variational
a reasonable calculation
form for the lowest of the piano
string's
eigenfunction, fundamental
frequency.
9.5
More Than One Dimension The equation
same argument
for the vibration
of a drumhead
as that used to derive
the drumhead's
vertical
the equation be u~x,y,t~.
displacement
follows
from the
for the string.
Let
Then if we consider
an element of area dxdy to be drawn on it when u = 0, the x direction by a factor 1 + ½ ~u/~x) ~
will be stretched
ly for the y direction,
as in ~9.4), and similar-
so that the area will
change to
~u 2][ 1 + ½ L'y;y .~u) 2] dxdy [1 + ½ (~-d) and t h e
increase
in area
is
thus 8u 2
SA=
If will
the
drumhead
is
in
be p r o p o r t i o n a l
V = ~
Problem 9.26. deriving
~[C~)
isotropic to
[(~x )
this
tension
the
potential
energy
stored
quantity,
+ (~y)2]dxdy =
Let the drumhead's
its equation
~u ~
%-F) ls=@
+
~
(9.39)
(Vu) 2dxdy
area density be s, and finish
of motion
~2U ~x2
~2U + ~-~--
I
~2U
v 2 ~t 2 -
o
(9.40)
How d o e s v 2 d e p e n d on P and o?
Changes of Variables Most drumheads
are round,
given most conveniently
and the boundary condition
if u is considered
as a function
is therefore of r and @,
317 with u(a,@)
where a is the radius operator
of the drum. formula
to make the transformation of Vu are the spatial directions,
du/~x
The variables
in the laplacian
in the variational
integral.
Choose the directions
of increasing r and
au
du
d--7 "
(Vu)e
=
~
1 du =
de
r
becomes
and the entire action
1 (~u) 2]rdrd e
integral
[ { ~d (~-~) dU 2
s =
The general problem
u r = au/dr,
is
- ~[
(~_uu)2 1 (-~) au dr + 7-~
2]}rdrded t
is of the form
6 ff[
L(U, Ur, U e ) r d r d @ d t
u e = du/d@.
Carrying
= 0
out the variation
6ue)drdOdt
= 0
~ ur) - -~ (r ~-~O)]6udrdedt +
+
uis continuous
infinite,
gives
the second and third terms by parts:
~u - ~
If
The components
directions:
II
Integrate
or one may
but it is also very convenient
rates of change of u in two orthogonal
(VU)r = (9.39)
in tables,
and du/dy.
@ as two new orthogonal
where
(9.41)
in (9.40) may be changed by direct calculation,
look up the necessary
Thus
= 0
constant,
line vanishes.
ff
[r ~aL Ur
r=R
6u]r = O dedt +
fl
~t
[r ~ u e
~U]°
e=o
drdt
= 0
a t r = 0, and 6u = 0 at r = R, where R may be or a function
of 8, the first term in the second
(It may be necessary
to think for a moment to see why
318 this is so.) The last term vanishes because @ = 0, and we have left 6(rL)
~u
as the Euler-Lagrange equation. to denote partial
0
(9.42)
In the present case, using subscripts
differentiation,
L and
=
@ = 2~ is the same line as
=TUt°2
_ ~2 CUr2 + ~1 u@2)
(9.43)
(9.42) gives 1
1
(9.44)
C~Utt - ~(Urr + r Ur + -~T u@@) = 0 The expression dimensions,
in parentheses
and the method
any kind of coordinate coordinates xl, x2,
is capable
... qy, given as functions
the
volume element
integral
to
If the new
of the Cartesian
(x)
becomes =
Jdq1...dqr
is
[
S =
/[u(q) ....
Jdql...dqrdt
and this gives at once the Euler-Lagrange taken as the lagrangian. they include
described by Maxwell's
Of course,
(9.45)
equation
for u when JL is
there are many other systems
than
here that are described by a laplacian
the drumhead considered operator;
generalization
= ~--'(-~
dx1...dx r The a c t i o n
in two
find the Jacobian J
so t h a t
of immediate
system in any number of dimensions.
are ql, q2,
... mr,
is indeed the laplacian operator
continuous mechanical and Schr~dinger's
systems
equations.
as well as fields We shall discuss
the latter separately below.
Wave Equation
in Polar Coordinates
The wave equation 1
in the surface 1
is
1
Urr + ~ ur + -~T u80 - -~T u t t = 0
v 2
= ~/o
(9.46)
S i n c e an a r b i t r a r y waveform can be formed o f normal-mode v i b r a t i o n s
Fourier superposition,
it is sufficient
to consider vibrations
at
by
319
angular frequency m, for w h i c h 1
1
42
Urr + ~ u r + 7~-u00 + 4~u = 0 (This is sometimes called H e l m h o l t z ' s one of the ii v a r i e t i e s 5.10).
equation.)
Polar coordinates
for w h i c h the wave equation is separable
are
(Sec.
If we set
u(r,O) then
(9.47)
= m2/V
R(r)@(O)
=
(9.47) becomes
(R"
"~ ,7. R! + 42R) @ + r~ RG" = 0
+
or r2 R
(R" + / R + 12R) = r @
Since the left side is i n d e p e n d e n t of @ and the right side is independent of r, b o t h must be constant. Set the constant equal to m 2. (9"
The @ equation is -m2(9
=
whence @ = A sinm@ + B cosm@ m m
(9.48)
and c o n t i n u i t y on traversing a complete circle requires that m be real and an integer.
The r e q u a t i o n is now
R"
+
This is Bessel's e q u a t i o n solutions,
!r R' + (42
m2 -
(Arfken 1970).
7£ )R
=
0
(9.49)
It has two classes of
roughly analogous to the sine and cosine solutions of
(9.10),
R = CmJm(lr) The Neumann functions Nm, kind,
are all infinite
in 1764.
(9.50)
also called Bessel functions of the second
at r = 0 and so have no place here.
functions of the first kind, Jm' w e r e Euler,
+ DmNm(4r )
(predictably)
The Bessel
first studied by
They are r e p r e s e n t e d by the series
Jm(X) = (½x)m m
[I
~ - I(m+1)
~½x)~ + 1.2(m+l)(m+2) . . . . ]
(9.51)
320 and for x > > m
are given a s y m p t o t i c a l l y by
~m ( x )
(~) %os (~ -
~
Jm 's
Figure 9 • S shows the first few extensively tabulated
7
-
~-)
(9.52)
as a function of
(Abramowitz and Stegun 1964)
x
They are
and available in
computer subroutines.
Figure 9.5
Three Bessel Junctions Jm(X).
The solution we have found is of the form
u ( r , O ) -- Jm(hr)(Amsinm@ + BmCOSm@) and the complete s o l u t i o n m u l t i p l i e s t.
this by a sinusoidal
function of
Since drums are u s u a l l y struck into motion, we shall look at
solutions
u(r,@,t)
that satisfy
u(r,O,o)
= o,
u(r,O,o)
(9.S5)
= Vo(r,O)
so that
u(r,O,t)
=
First, we must find the frequencies ~m" the r e q u i r e m e n t that n u m b e r Of zeros; the nth zero of by
(9 . s 4 )
2 Jm(Xr) (AmSinm@ + BmCOSmO)sinmmt m=O
Jm(~a)
= 0.
For each
They are d e t e r m i n e d by
m, Jm(X)
has an infinite
the first few are listed in Table 9.1.
Jm(x).
Then
Jm(~a)
= 0 implies that
Let x
ha = Xmn
mn
be
or,
(9.47) mn
V a
= - x
mn
(9.55)
B2'1
These
are the frequencies
not in the ratios
present
of small
not in any harmonic
in the sound of a drum.
integers
(or any integers)
They are
and therefore
are
series. Table
Zeros of Jm (x)
9.1
Numbers in parentheses are calculated from (9.52). first
second
third
2.40
5.52
8.65
11.79
fourth (11.78)
3.83
7.02
10.17
13.32
(13.35)
5.14
B.42
11.62
14.80
(14.92)
6.38
8.76
13.02
16.22
(16.49)
To satisfy the initial condition we write ~(r,@,0) =
Z ~mnJm(hmna) ~mnSinm@ + BmnCOSm@ ) = Vo(r,@ )
(9.56)
mj n where we have
introduced
extra indices
corresponding
to the various
zeros of each J . For the stretched string we solved the corresponding m (9.22) by Fourier inversion. The same is possible here
equations because
the Helmholtz
equation
equation with eigenvalues different,
The eigenvalues
and the differential
eigenfunctions
operator
are all orthogonal.
that they form a complete determined
(9.47) has the form of an eigenvalue
kmn"
set.
from a knowledge
are all real and
is hermitian;
A more advanced
It follows
of v(r,@)
therefore
the
investigation
that the A
shows
and B
can be
What
frequen-
mn mn and the problem, from the formal
point of view, is solved.
Problem 9.27.
A round drum is hit in the exact center.
cies are heard? approximate there be?
formula
numerical
values
for the rest.
Problem 9.28.
for the first few and an
Roughly how many frequencies
You may w i s h to experiment
of different
beaten?
Give
beating
will
a drum with implements
sizes. What
frequencies will be present when a square
Are they h a r m o n i c a l l y
related?
drum is
How would a square drum
sound compared with a round one?
Problem 9.29. solution 5.52008.
Estimate
of Bessel's
the two lowest
equation.
zeros of Jo(x)
The correct values
by a variational
are 2.40483
and
322 9.6
Waves in Space In this section we shall consider sound waves as a prototype of
other kinds.
The mathematical
as those in aerodynamics,
Propagation
functions to be encountered are the same
electromagnetic
theory,
and quantum theory.
of Sound
The lagrangian for sound waves can be found by evaluating potential and kinetic energies of an elastic medium
the
(we shall speak of
air, but only in order to be specific).
This argument is more
complicated than the underlying physics,
however
(Goldstein 1950, Sec.
11.3), and we shall follow a shorter path. Let the constant background pressure and density of the air be Po and
0o, the sound wave producing small fluctuations so that p and p
vary periodically. may be called x.
Consider a plane wave moving in a direction that Fig. 9.6 shows an element of volume.
"1
Its mass is
F+ P ' I
x ×l+dx Figure 9.6
Element of volume of a gas with applied pressures.
pAdx, which differs from
PoAdXby
terms of higher order,
force on it acting to the right is -Adp. ~v
PoAd x
and the net
Newton's second law is
x = -Adp ~t
or
~V
,==_~
~t
Po ~x
(9.57)
Where we use partial derivatives because at each point variables such as p, p, and the fluid velocity v are functions of both x and t. vectorial
form (9.57)
In
is ~._xv = _ !
~t
Po
vp
(9. s8)
323
The motion of the air must satisfy the equation of continuity which states that air is not created or destroyed, ~P + v . Cpv)
o
=
@t
Again to first order in the variations,
this is
~-~@t + Po v'v = 0 Finally,
the increments
(9.59)
in pressure and density are related by the
bulk modulus of the gas, B: ~P = i
Po
We can use t h i s
to eliminate
t h e wave e q u a t i o n continuity
is
for
p ~or p)
the pressure.
found from
(9.59)
(9.60)
Ap
B
from the e q u a t i o n s , I n terms o f p,
L e t us f i n d
the e q u a t i o n o f
and ( 9 . 6 0 ) :
v.v=_iaP
B ~t
To e l i m i n a t e
v,
take the time derivative
of this
and compare i t
with
the divergence of (9.58):
a-T v . v
=
5
=0
-
~
at2
=
-
p--j
VZP
or
This
is
d'Alembert's
B/no
equation for
the pressure,
(961)
and p s a t i s f i e s
the
same e q u a t i o n . To f i n d that
it
B we need some h y p o t h e s i s
obeyed B o y l e ' s
temperature
a b o u t t h e gas.
l a w and d e r i v e d B = Po"
Newton assumed
T h i s assumes t h a t
remains constant as the sound wave goes through.
waves in the ordinary frequency range the variations
the
But for
are far too rapid
for this and we do better to assume that the compressions and rarefactions take place adiabatically.
The adiabatic law is pV Y = const, or p = KpY
where y is the ratio of the specific heat at constant pressure to that at constant volume,
equal to 1.403 for air at 0°C and one atmosphere.
Thus Ap = K y p Y - : A p
and dividing the last two equations gives
324
Ap
_
Ap
po- -Co for departures
from the background
speed of sound V
At a pressure
is (ypo/Po)½
=
of 76 cm of mercury
Po
=
1"0132x10S
(9.62)
= YPo
B
and the computed
Thus
values.
at 0°C,
nm-2'
PO
=
1.2929
kg m -3
and V = 331. 6 ms -I
in agreement
with observation.
The lagrangian
density L
where,
as before,
that it leads Solution
to the d'Alembert
according
(Sec.
5.10),
Cartesian
9.32]
(9.61].
topic,
for there are many kinds
conditions
treatment
knowledge
of quantum mechanics
coordinates
are easy,
in each
1932,
to show the
on the one hand and
on the other.
of the
to appropriate
of
and for a
1970, Bateman
a few lines
of the drumhead
can be separated
and each one leads
imposed,
(Arfken
Here we shall only sketch
The variables
(Prob.
Equation
to the earlier
to the reader's
63]
(9
one easily verifies
we refer to other books
1949).
is
- V1{ ~2 ]
equation
to the boundary
discussion
Sommerfeld relation
, since
is a long and complex
solutions general
½[(Vp]2
=
p = @p/Bt
o f the Wave
This
for sound waves
ii St[ckel
systems
sets of functions.
but there are few rectangular
boxes
in
nature. Here we shall discuss only spherical polar coordinates. The wave equation in these coordinates can be derived by changing variables in the lagrangian
9r2
~ ~r
If we consider
integral
~
~
only a single
(Sec.
(sin@
9.5), Prob.
9.40),
and we find
+ r2sin2@ 2@ 2- v-[ ~t2 =
frequency
to the last term becomes
(m/v]Zp.
325
We begin by separating
the radial
p(r,e,¢)
variable
from the angular
ones:
= R(r)~(e,¢)
where
~-~IR=O
7 a (sin@ ~-Y) + 1 ~21~ r 2 ~ " 4- r2 R'+ X2R)Y + [s-~ "~ 30 sin2@ - -
If we divide by RY the variable constant
of separation
r is separated
X - - -v
out, and introducing
a
K gives R"
F2 R ~ + (h2
+
-
K~ - ) R
=
O
(9.66)
and
7 3 sin@ TO If the quantum-mechanical angular momentum,
BY)
(sine
7
3e
+ si~
operator
omitting
-
so that
(9.66)
for the square the result
1 3 sine n-----~ si 3--~
is the eigenvalue
We know that the eigenvalues
There
is
-£,
....
a second
by the functions
set
of
are however magnetic
essential
theory,
p m
=
the
of
kind,
because
that the
P£m(cosO)eim¢
(9.69)
associated
(9.66) Q m.
Legendre
in which
These
they become
in many calculations
notably
(9.68)
harmonics
are
solutions
of the second
use in quantum mechanics
(9.67)
lore of quantum mechanics
are the spherical
+£ a n d t h e
3~
~
£ = 0 , 1 , 2 ....
Y~m(e,O) m :
coordinates
is
K are
and it is part of the standard
where
l
3-'@-
of the orbital into polar
= K~'(o.,¢)
K = Z(Z+I)
y(@,@)
3
(9.66)
equation
~2:z(e,¢)
functions
+ KY = 0
the ~, is transformed
(see any book on quantum mechanics), ~2 =
32/"
~
p m
replaced
are almost never of
infinite
at e = 0.
in acoustics
those of the design
is
functions.
They
and electro-
of loudspeakers
and
antennas. With
(9.68) we return
to the radial wave equation
almost but not quite of the form of Bessel's ~(r)
=
~(r)
(Xr) ½
equation,
[9.65).
It is
but if we write
326
it changes to z"
+
-I
z '
r
+ 2FX
~ ]
z =
(9.70)
0
This is Bessel's equation of order ~ + ½, and it has two solutions,
J~+½(Ir), which is finite at the origin, and N£+½(Ir), which is singular there.
The radial function R is therefore of the form
R(r) = j ~ ( t r )
n~(tr)
or
(9.71)
where
~+~(~r)
JR+½(Xr) j~(kr) = are called expressible
(hr) ½ ,
n~(kr)
(~r) ½
s p h e r i c a l B e s s e l and Neumann f u n c t i o n s . i n t e r m s o f s i n e s and c o s i n e s :
Jo (x)
no(x)
= sin Xx
(9.72)
They are all
COSX X
(9.73)
J l (X) - sinXx2
cOSXx
nl(x)
cosx X2
sinx .~
These functions are appropriate to the solution of standing-wave problems;
incoming and outgoing waves are formed from linear combina-
tions such as
i lix Jo(X ) ± ino(X) = x ~ e etc.
(9.74)
They are called Hankel functions. These few remarks are enough to show why most of the mathematical apparatus of the quantum theory of angular momentum lay ready to hand when the theory was first invented. The angular part of the solutions of the wave equation in polar coordinates is standard, and only the radial equations, when quantum mechanics came along, were somewhat unfamiliar.
Problem 9.30.
Derive the laplacian operator for spherical coordinates
by changing variables in the v~riational integral.
Problem 9.31.
It is possible to compute the average molecular
velocity in a gas from its bulk properties by kinetic theory, and the sound velocity from (9.61).
ProbZem 9.32.
How do they compare?
Show that (9.63) is the lagrangian density for sound
waves and find the corresponding hamiltonian density.
Find the
327 numerical
factor that makes this an energy density. Starting
Problem
9.33.
derived
in the previous
acoustic
from the general problem,
energy transmitted
formula for energy density
find a general
it to a plane wave and show that the result
is reasonable.
not know how to do this, look at the analogous current density Problem
in elementary Assume
9.34.
that when sound waves
next to the surface
and formulate
the appropriate discuss
derivation
waveguides
wave equation
Using the result of Prob. What happens to this except
require
a little more thought.)
Problem
9.35.
Solve the preceding
cross section, 9.7
9.33,
and boundary
conditions.
evaluate
that will
the sound energy
(The theory of electromagnetic the boundary
that
problem
conditions
for a pipe of circular
radius a.
The Matter Field In his first paper on wave mechanics,
1926, 2 SchrSdinger Hamilton-Jacobi
made a strange proposal.
action is something
ing to avoid introducing proposed
submitted on 27 January Recognizing
that the
like a phase, yet obviously wish-
complex numbers
into physical
quantities,
he
to write S in the form
where K is a constant Jacobi equation
S = KZn@
(9.75)
to be determined.
In terms of ~, the Hamilton-
is x
a~L
H(xi" ~ ~x. In c a r t e s i a n
coordinates, (v~)~
(9.76)
ahead and solves
Annalen
) = E
_ 2___m [E
Ka
- V(r)]~
2 =
0
of classical physics;
it ones gets nothing new.
d. Physik
(9.76)
is
But this is merely a transcription
2
the
(this is not quite true),
if the sound frequency presented
is lower than t h i s limit? is similar
of the
the motion of sound waves down a pipe of rectan-
passing down the pipe. to the pipe
Specialize (If you do
reach a solid surface
do not move
gular cross section a x b and find the lowest frequency propagate.
for the
quantum mechanics.)
air molecules Using these,
expression
per second through a unit area.
79, 361 (1926).
Schr~dinger
if one goes proposed
328
instead a physical principle:
~
is to be chosen so as to make the
volume integral of the quantity on the left a minimum and not necessarily
zero.
We have therefore a variational problem of the form
I {(V~)2 - K22m [E - V ( r ) ] ~ 2 } d v which yields the differential
eigenvalues of this
v(r)]~
equation,
= o
(9.78)
as e v e r y b o d y knows, c o i n c i d e
V(r) = .e2/r and K = ~.
with the Balmer levels when
In a note added in proof, SchrSdinger tion:
(9.77)
equation ~m [ ~ _
The d i s c r e t e
= 0
suggests a better formula-
Write the integral to be varied as
[KaT(r. ~-~) + V ( r ) , 2 ] d v
(9.79a)
where T is the classical expression for kinetic energy, subject to the condition that ~ remain normalized
to unity,
f~2dv This is an equivalent,
9.36.
reluctance
(9.79b)
but neater formulation in terms of what is
called an isoperimetric problem,
Problem
= 1
to be discussed next.
The strange form of (9.75) reflects Schr6dinger's
to introduce complex numbers
instead with ~ = exp(iS/K),
into mechanics.
Starting
show how he could have phrased his argument
so as to introduce a complex ~.
Isoperimetric
Problems
There is a class of problems prototype
in the calculus of variations whose
is to be found in the Aeneid: ~
is given has the largest area?" answer is trivially a circle. problems of the kind:
"What figure whose perimeter
It is called Dido's problem,
and the
It is a special case of a class of
What function y(x) maximizes
or minimizes
the
integral
S =
I
f(y,y',x)
dx
(9.80a)
x0
where f is given,
3
Aeneid,
the variations vanish at the endpoints,
I, lines 367-8.
and another
329
integral,
T =
g(y,yV,x) dx
(9.80b)
X
is at the same time required
to have a given numerical value?
On varying y(x), we are requiring tionary, while is a number
~T = 0 because
to be chosen,
that ~S = 0 because S is sta-
T is fixed.
Let S' = S + XT, where
called a Lagrange multiplier.
Then
6S' = 0 is a necessary depending
condition
on the value
Problem 9 . 3 7 . be m a x i m i z e d
are satisfied
a nonrotating
The area to
[r 2 + r'CO)2]½dO located
equation.
A uniform chain of length ~ is suspended between
(Xo, yo ) and
Problem 9.39. planetary
is solved.
of a circle w i t h its point arbitrarily
the Euler-Lagrange
two points
takes on its given value,
use polar coordinates.
d~=Jo
[r2dO2 + dr2]
Show that the equation
Problem 9.38.
to find a
I o2~ ½r2d@, while the perimeter is
is
P =
satisfies
C9.80b)
and the problem
In Dido's problem,
a y(x,X)
If now it is possible
value of I such that T calculated by all the conditions
(9.81)
on y(x), and its solution yields
of X chosen.
X
Somewhere planet)
radii
(xt,yl).
What
is the equation
in the universe
suspended
in height.
the
for its curve?
there may be an antenna
from towers
(above
far apart and several
Find the formula for the curve of the
suspended wire.
SchrBdinger's
Variational Principle
In Schr~dinger's
s'
where
revised version,
=
[~-~ ( v ~ ) ~ + v ( r ) ~
X is the Lagrange multiplier.
(9.77) and so of course
the results
a little clearer
that the desired
integral
exists
to i.
(9.79b)
the integral
This
X~]dv
(9,8Z)
is of the same form as
are the same, but it now becomes
functions
at all--since
That is, a ~ corresponding
2 +
to be varied is
~ are those for which
if it exists
to a stationary
the
it can be set equal state must be
330
normalizable. SchrSdinger's
second paper ~ explored
and the Hamilton-Jacobi the time-dependent of mechanics
equation,
Schr6dinger
in Chapter i.
new discovery,
similar
but having no logical
The variational
principle
a very general property of eigenvalue Suppose K represents function belonging
principle
problems
any hermitian
to the eigenvalue
for (9.82) was a for a typical
connection with it,
~ does not refer to any mechanical
system represented.
(9.78)
equation with which we began the study
His variational
in form to that given in C9.63)
spatial wave phenomenon, since Schr6dinger's
the relation between
and introduced what is now called
property of the
does, however,
reflect
as is easily seen.
operator
and f is an eigen-
~:
kf -- ~f Now write
(9.83)
K as ~^
K and imagine
~z_ZLL~__Cq_ ~
that f is varied slightly
The quantity
< given by (9.84)
have a value nearby.
Varying
(9 84)
= if~fdq
from the exact eigenfunction.
is no longer an eigenvalue,
but it will
f and < in the equation
< I f*fdq = ; f*Kfdq gives
6~ I f*fdq + ~ I (f*6f + f6f*)dq = I ~f*Kfdq + I f*K6fdq and by the use of (9.83) and the hermiticity 6< Thus if
6f
f f q = 0
6~ = 0
this
to (9.85)
e, ~< is of the
fact has already been noted in connection
(9.34). We can also prove very easily
energy by variational ~n be the exact let
or
is of the order of some small parameter
order of e 2 or smaller; with
of K this reduces
some
that
(but unknown)
eigenfunctions
varied function ~ be expanded
=
Annalen
d. Physik
in calculating
a
ground-state
means we always obtain too high a value.
79, 489
in
Z en~ n n=O
(1926).
of the hamiltonian
terms
of them:
Let H, and
331
where if ~ and the @n are normalized, z I%12 n:O
= I.
The expectation value of H in the state ~ is
Z m,n
I Cm*~m*HCn~ndq
If E o is the ground-state
=
Z n=O
Icnl2E
n
energy then
< H > : Z I%I E n
Z I%I E o
or
so that E
Problem
O
>
(9.86)
E 0
is a lower bound on
9.40.
Knowing the form of the wave function for the ground
state of a simple harmonic oscillator, make at least two variational to that of an anharmonic oscillator with VCx)
approximations
Which approximation
Example
I
The Helium
Examples
= -~ Kx ~.
is the best?
Atom
of the use of Schr~dinger's variational principle
for
finding stationary states are given in every text on quantum mechanics, but it may be useful to outline one such calculation,
omitting computa-
tional details which at first sight obscure the simplicity of what is being done. ed in Sec.
The system is a quantum~mechanical
The problem is to estimate function of a helium atom. completely
analog of that discuss-
7.6.
intractable
energy and wave
It is a three-body problem and therefore
in classical mechanics because the details of
motion are so complicated. the details;
the ground-state
But in quantum mechanics we do not follow
the indeterminacy principle deprives them of meaning.
Instead, we treat it as a system with a very simple behavior. The idea is to assume that the wave function for each electron in the ground state is not very different from that of an electron in a hydrogen-like charge be Ce.
atom of suitable but unknown nuclear charge.
Let the
We shall solve the problem by varying the parameter
~.
The normalized wave function for two electrons in the ground state is
332
(rl'r2)
Ta 3
=
The complete H a m i l t o n i a n
=
where
-
a = ~ m:~2 e2
for the two-electron
2-m (Vz 2 + V2
Z is the nuclear
operator
e-a(rl+r2)
charge,
Ze 2
Ze 2
r I
r 2
(9.87)
system is e2
+
and the expectation
(9.88)
value of this
in the assumed state is found to be
~ )~] me ~ 2R2
W = 2[~ 2 - 2(Z where m e ~ / 2 ~ 2 = 1 Rydberg
= 13.606
eV.
(9.89)
W goes through a minimum when
is chosen equal to Z-5/16; we explain this p i c t u r e s q u e l y that each electron
is shielded
5/16 of the time.
The energy
from the nucleus is then
E = Wmi n = - 2 ( Z For helium,
This
one begins
eV.
{9.90)
The experimental
is perhaps
comparison with what we could achieve but if one compares motion,
- /~)2Ry
this is -5.695 Ry = -77.49
-78.98 eV; the error is 1.9%.
by saying
by the other electron
not impressive
value
is
by
in simple vibration problems,
it instead with the three-body problem of orbital to appreciate
the simplifications
brought by
quantum mechanics.
Problem
9.41.
Carry out the integrations
Problem
9.42.
Estimate
simple pendulum function
to derive
(9.90).
the two lowest quantum energy states of a
variationally.
satisfy
necessary
What criterion must a variational
if it is to yield an estimate
wave
of the first excited
state? Problem
9.43.
Show that the lagrangian
density of the complex matter
field is
~2 L : ~ Problem 9.44.
{~*~ - ~*~) + ~
Iv~l 2 + V(r) l ~ [ 2
(9.91)
Show by c a l c u l u s o f v a r i a t i o n s t h a t the c o n d i t i o n
^
C l a s s i c a l and Quantum D e s c r i p t i o n s o f Nature As f a r as anyone knows, quantum mechanics i s r i g h t ,
and one might
suppose t h a t c l a s s i c a l mechanics i s t h e r e f o r e "wrong," or at best some approximation to the c o r r e c t t h e o r y .
C l a s s i c a l physics does not
333
contain h and in the course of deriving equal to zero.
example de Broglie's
relation,
and let h approach of units,
zero.
procedure.
set h
Take for
as already noted in Chapter I, kp
changes
it one must somewhere
But this is not an unambiguous
=
h
If we are not to become
involved in trivial
this can be done in two ways: p
fixed
X
fixed
or
p÷O
k÷O The first leads, as we have shown several the second, mechanics without
to a classical
and wave optics
anybody
limiting invoked
theory of waves.
except Hamilton
to support
It was possible
for
to exist side by side for more than a century realizing
cases of the same fundamental
them could be perceived
times, to classical mechanics;
that they might be different
theory.
The relation between
only by a mathematician,
since the experiments
them had little to do with each other and were
ordinarily
described
in quite different
connection
of classical physics with quantum physics we must stop
worrying
about which one is right,
remark:
the question physics
languages.
To understand
the
and remember Niels Bohr's profound
can answer
is not what nature is, but
rather what can be said about it. Classical mechanics talking
about nature,
and quantum mechanics
and classical mechanics
terms used in quantum mechanics. warns us that the language take it too literally,
for there are limits of experiments
are embodied in ordinary
in the indeterminacy language without
Bohr's principle confusing physical
transgressing
of complementarity
quantities
apparatus
and how to express physics
them is the subject of
(Park 1974, p. 58).
It is very
to a massive
since the "physical
to think of them as belonging more
than to nature as it "really
volume,
What is a particle? object,
following
situated
of
can never be directly known to us in is."
Much of this book has been concerned with the dynamics particles.
for
These limits
in terms of "the perturbation
any way and hence it is appropriate to our conceptual
if we
to its applicability
by the act of measurement,"
about to be perturbed
is deceptive
in microphysics.
relations,
to explain indeterminacy
quantities"
for
At the same time, quantum mechanics
of classical mechanics
explaining
the results
are languages
provides many of the
In classical physics
at a point or occupying
an exactly definable
trajectory
of
the word refers a very small
in space and time
334
that is determined by external fields. mechanics
To the student of quantum
the concept of particle may seem weakened and vague, but
that is because he is trying to find a way to fit a classical concept into another theory in which its role is quite different. essence,
"In its
it is not a material particle in space and time but,
in a
way, only a symbol on whose introduction the laws of nature assume an especially simple form."
(Heisenberg 1952, p. 62.)
If one understands
this one can stop worrying about which slit a photon passes through in an experiment on diffraction. The laws of nature themselves
are clear and economical descrip-
tions from which one can predict what will happen in experiments have not yet been performed. these descriptions
that
By what sometimes seems a miracle,
are sometimes expressed in such general and
universal mathematical
language that we can feel as Galileo did when
he wrote that in contemplating them we are perceiving God himself perceives
it.
But these descriptions
the contemplation of man.
the world as
are made by and for
They derive from experience,
logical or necessary way, and thus they are not unique; have value for different purposes.
but not in a different ones
This is especially evident when
two theories overlap, when they offer a choice between different ways of describing
the same set of facts.
the subjects discussed in this book. those of classical optics
Such is often the situation with Three languages are used here:
(phase, wavelength, ...); classical mechanics
~osition, momentum, mass, ...); and quantum mechanics, which uses a larger and more mathematical vocabulary.
And, of course,
in the
classical mechanics of elastic media as described in this chapter there are also terms of the classical wave theory from which, by analogy, the wave terminology of optics and quantum physics are derived. does not derive one language
One
from another, but one can make accurate
translations when the subject is one which can be discussed adequately in both, and it often helps our thought to do so. Classical mechanics
talks about a world in which particles are
things which can be described in an objective manner. mechanics
typically answers
the question:
physical system using a certain experimental procedure, probability
Quantum
"If I measure a certain
£hat I will obtain a certain answer?"
what is the
There is a long
and imposing history of debate on how the word "probability" as used here should best be defined.
(See for example Jammer 1966, 1974).
Even knowing this, one often says for example that the ground-state energy of a hydrogen atom "is" -13.6 eV.
We know enough about physics
to know that one will rarely if ever get into trouble saying such a
335
thing, but the language
is not that of quantum mechanics
derived from it; it is simply borrowed,
for convenience,
and cannot be from classical
physics. It is useful
in physics
their classical meanings
really need quantum mechanics situation,
words
to use words
for their elucidation.
from the old languages
but rather as metaphors
like wave and particle
in
in order to help explain phenomena which
or parables.
Here,
in a new
are not to be used literally
In quantum theory the wave is
not a wave in any physical medium but only in a certain mathematical function,
and a particle
that these are metaphors
is not a thing.
have exact and literal meanings. that new theories
are formulated
It is by this linguistic process and progress
of this book has had an opportunity formal relationships
We must always remember
expressed by words which in other contexts is made~ and the reader
to perceive not only the close
between the equations
of the old theories
those of the new, but also the vastly different they describe
the world.
languages
and
in which
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