The Cambridge Handbook of Physics Formulas

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The Cambridge Handbook of Physics Formulas

GRAHAM WOAN Department of Physics & Astronomy University of Glasgow

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PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE

The Pitt Building, Trumpington Street, Cambridge, United Kingdom CAMBRIDGE UNIVERSITY PRESS

The Edinburgh Building, Cambridge CB2 2RU, UK http://www.cup.cam.ac.uk 40 West 20th Street, New York, NY 10011-4211, USA http://www.cup.org 10 Stamford Road, Oakleigh, Melbourne 3166, Australia ´ 13, 28014 Madrid, Spain Ruiz de Alarcon c Cambridge University Press 2000
This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2000 Printed in the United States of America Typeface Times Roman 10/12 pt.

System LATEX 2ε [tb]

A catalog record for this book is available from the British Library. Library of Congress Cataloging in Publication Data Woan, Graham, 1963– The Cambridge handbook of physics formulas / Graham Woan. p.

cm.

ISBN 0-521-57349-1. – ISBN 0-521-57507-9 (pbk.) 1. Physics – Formulas. QC61.W67

I. Title.

1999

530 .02 12 – dc21

ISBN 0 521 57349 1 hardback ISBN 0 521 57507 9 paperback

99-15228 CIP

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Contents

Preface

page vii

How to use this book

1

1

3

Units, constants, and conversions 1.1 Introduction, 3 • 1.2 SI units, 4 • 1.3 Physical constants, 6 • 1.4 Converting between units, 10 • 1.5 Dimensions, 16 • 1.6 Miscellaneous, 18

2

Mathematics

19

2.1 Notation, 19 • 2.2 Vectors and matrices, 20 • 2.3 Series, summations, and progressions, 27 • 2.4 Complex variables, 30 • 2.5 Trigonometric and hyperbolic formulas, 32 • 2.6 Mensuration, 35 • 2.7 Differentiation, 40 • 2.8 Integration, 44 • 2.9 Special functions and polynomials, 46 • 2.10 Roots of quadratic and cubic equations, 50 • 2.11 Fourier series and transforms, 52 • 2.12 Laplace transforms, 55 • 2.13 Probability and statistics, 57 • 2.14 Numerical methods, 60

3

Dynamics and mechanics

63

3.1 Introduction, 63 • 3.2 Frames of reference, 64 • 3.3 Gravitation, 66 • 3.4 Particle motion, 68 • 3.5 Rigid body dynamics, 74 • 3.6 Oscillating systems, 78 • 3.7 Generalised dynamics, 79 • 3.8 Elasticity, 80 • 3.9 Fluid

dynamics, 84

4

Quantum physics

89

4.1 Introduction, 89 • 4.2 Quantum definitions, 90 • 4.3 Wave mechanics, 92 • 4.4 Hydrogenic atoms, 95 • 4.5 Angular momentum, 98 • 4.6 Perturbation theory, 102 • 4.7 High energy and nuclear physics, 103

5

Thermodynamics 5.1 Introduction, 105 • 5.2 Classical thermodynamics, 106 • 5.3 Gas laws, 110 • 5.4 Kinetic theory, 112 • 5.5 Statistical thermodynamics, 114 • 5.6 Fluctuations and noise, 116 • 5.7 Radiation processes, 118

105

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Solid state physics

123

6.1 Introduction, 123 • 6.2 Periodic table, 124 • 6.3 Crystalline structure, 126 • 6.4 Lattice dynamics, 129 • 6.5 Electrons in solids, 132

7

Electromagnetism

135

7.1 Introduction, 135 • 7.2 Static fields, 136 • 7.3 Electromagnetic fields (general), 139 • 7.4 Fields associated with media, 142 • 7.5 Force, torque, and energy, 145 • 7.6 LCR circuits, 147 • 7.7 Transmission lines and waveguides, 150 • 7.8 Waves in and out of media, 152 • 7.9 Plasma physics, 156

8

Optics

161

8.1 Introduction, 161 • 8.2 Interference, 162 • 8.3 Fraunhofer diffraction, 164 • 8.4 Fresnel diffraction, 166 • 8.5 Geometrical optics, 168 • 8.6 Polarisation, 170 • 8.7 Coherence (scalar theory), 172 • 8.8 Line

radiation, 173

9

Astrophysics

175

9.1 Introduction, 175 • 9.2 Solar system data, 176 • 9.3 Coordinate transformations (astronomical), 177 • 9.4 Observational astrophysics, 179 • 9.5 Stellar evolution, 181 • 9.6 Cosmology, 184

Index

187

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Chapter 3 Dynamics and mechanics

3.1

Introduction

Unusually in physics, there is no pithy phrase that sums up the study of dynamics (the way in which forces produce motion), kinematics (the motion of matter), mechanics (the study of the forces and the motion they produce), and statics (the way forces combine to produce equilibrium). We will take the phrase dynamics and mechanics to encompass all the above, although it clearly does not! To some extent this is because the equations governing the motion of matter include some of our oldest insights into the physical world and are consequentially steeped in tradition. One of the more delightful, or for some annoying, facets of this is the occasional use of arcane vocabulary in the description of motion. The epitome must be what Goldstein1 calls “the jabberwockian sounding statement” the polhode rolls without slipping on the herpolhode lying in the invariable plane, describing “Poinsot’s construction” – a method of visualising the free motion of a spinning rigid body. Despite this, dynamics and mechanics, including fluid mechanics, is arguably the most practically applicable of all the branches of physics. Moreover, and in common with electromagnetism, the study of dynamics and mechanics has spawned a good deal of mathematical apparatus that has found uses in other fields. Most notably, the ideas behind the generalised dynamics of Lagrange and Hamilton lie behind much of quantum mechanics.

1 H.

Goldstein, Classical Mechanics, 2nd ed., 1980, Addison-Wesley.

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64

3.2

Dynamics and mechanics

Frames of reference

Galilean transformations Time and positiona

r = r  + vt t = t

(3.1) (3.2)

Velocity

u = u + v

(3.3)

Momentum

p = p  + mv

(3.4)

r,r v t,t u,u

velocity in frames S and S 

p,p 

particle momentum in frames S and S  particle mass

m

Angular momentum

J = J + mr × v + v× p t

(3.5)

Kinetic energy

1 T = T  + mu · v + mv 2 2

(3.6)

a Frames







position in frames S and S  velocity of S  in S time in S and S 

J ,J 

angular momentum in frames S and S 

T ,T 

kinetic energy in frames S and S 

γ v

Lorentz factor velocity of S  in S

c

speed of light

S

m

S r

r

vt

coincide at t = 0.

Lorentz (spacetime) transformationsa Lorentz factor

−1/2  v2 γ = 1− 2 c

(3.7)

Time and position x = γ(x + vt ); x = γ(x − vt) (3.8)   y =y (3.9) y=y ; z = z; z = z (3.10) ! ! v " v " (3.11) t = γ t + 2 x ; t = γ t − 2 x c c Differential dX = (cdt,−dx,−dy,−dz) four-vectorb (3.12)

 S S

x,x

t,t

X

x-position in frames S and S  (similarly for y and z) time in frames S and S

v  x x

spacetime four-vector

a For frames S and S  coincident at t = 0 in relative motion along x. See page 141 for the transformations of electromagnetic quantities. b Covariant components, using the (1,−1,−1,−1) signature.

Velocity transformationsa Velocity ux + v ux = ; 1 + ux v/c2 uy ; uy = γ(1 + ux v/c2 ) uz uz = ; γ(1 + ux v/c2 ) a For

ux − v 1 − ux v/c2 uy uy = γ(1 − ux v/c2 ) uz uz = γ(1 − ux v/c2 )

ux =

γ

Lorentz factor = [1 − (v/c)2 ]−1/2

v c

velocity of S  in S speed of light

ui ,ui

particle velocity components in frames S and S 

(3.13) (3.14) (3.15)

frames S and S  coincident at t = 0 in relative motion along x.

 S S

u v  x x

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65

3.2 Frames of reference

Momentum and energy transformationsa Momentum and energy px = γ(px + vE  /c2 ); py = py ; pz = pz ; E = γ(E  + vpx ); 2

px = γ(px − vE/c2 ) py = py pz = pz E  = γ(E − vpx ) 2 2

E −p c =E −p c 2

2 2

Four-vectorb a For

(3.16) (3.17) (3.18) (3.19)

γ

Lorentz factor = [1 − (v/c)2 ]−1/2

v c

velocity of S  in S speed of light

px ,px E,E 

x components of momentum in S and S  (sim. for y and z) energy in S and S  (rest) mass total momentum in S momentum four-vector

= m20 c4

(3.20)

m0 p

P = (E/c,−px ,−py ,−pz )

(3.21)

P

 S S

v  x x

frames S and S  coincident at t = 0 in relative motion along x. components, using the (1,−1,−1,−1) signature.

b Covariant

Propagation of lighta ! " v ν = γ 1 + cosα ν c

Doppler effect

(3.22)



cosθ + v/c 1 + (v/c)cosθ cosθ − v/c cosθ = 1 − (v/c)cosθ

cosθ = Aberration

b

Relativistic beamingc

P (θ) =

(3.23) (3.24)

sinθ 2γ 2 [1 − (v/c)cosθ]2

(3.25)

ν

frequency received in S

ν α γ

frequency emitted in S  arrival angle in S Lorentz factor = [1 − (v/c)2 ]−1/2 velocity of S  in S speed of light

v c

S y

c α x

S S  y y

θ,θ emission angle of light in S and S 

v θ c x x

P (θ) angular distribution of photons in S

frames S and S  coincident at t = 0 in relative motion along x. travelling in the opposite sense has a propagation angle of π + θ radians.& c Angular distribution of photons from a source, isotropic and stationary in S  . π P (θ) dθ = 1. 0 a For

b Light

Four-vectorsa Covariant and contravariant components

x0 = x0

x1 = −x1

x2 = −x2

x3 = −x3

(3.26)

Scalar product

xi yi = x0 y0 + x1 y1 + x2 y2 + x3 y3

(3.27)

xi

xi ,x i four-vector components in frames S and S 

Lorentz transformations x0 = γ[x + (v/c)x ]; 1

0

1

1

0

x = γ[x + (v/c)x ]; 2

2

x =x ;

xi

covariant vector components contravariant components

x = γ[x0 − (v/c)x1 ] 0

1

x = γ[x − (v/c)x ] 3

x =x

1

3

0

(3.28)

v

Lorentz factor = [1 − (v/c)2 ]−1/2 velocity of S  in S

c

speed of light

γ

(3.29) (3.30)

frames S and S  , coincident at t = 0 in relative motion along the (1) direction. Note that the (1,−1,−1,−1) signature used here is common in special relativity, whereas (−1,1,1,1) is often used in connection with general relativity (page 67). a For

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Dynamics and mechanics

Rotating frames Vector transformation



dA dt





dA dt

+ ω× A

(3.31)

Acceleration

˙v = ˙v  + 2ω× v  + ω× (ω× r  )

(3.32)

Coriolis force

F cor = −2mω× v 

(3.33)

F cen = −mω× (ω× r  )

(3.34)

= +mω 2 r ⊥

(3.35)

Centrifugal force

Motion relative to Earth Foucault’s penduluma a The

3.3

=

A



S

S

stationary frame rotating frame angular velocity of S  in S  ˙v ,˙v accelerations in S and S  v  velocity in S  r  position in S  F cor coriolis force m particle mass

(3.36) ˙ sinλ m¨ y = Fy − 2mωe x ˙ cosλ m¨z = Fz − mg + 2mωe x

(3.37) (3.38)

Ωf = −ωe sinλ

(3.39)

ω

λ z

nongravitational force latitude local vertical axis

y x

northerly axis easterly axis

Ωf

pendulum’s rate of turn

F cen r ⊥

F cen centrifugal force r ⊥ perpendicular to particle from rotation axis Fi

y sinλ − ˙z cosλ) m¨ x = Fx + 2mωe (˙

any vector

S S ω

m

r ωe y

z x

λ

ωe Earth’s spin rate sign is such as to make the rotation clockwise in the northern hemisphere.

Gravitation

Newtonian gravitation Newton’s law of gravitation

Newtonian field equationsa

Fields from an isolated uniform sphere, mass M, r from the centre a The

Gm1 m2 rˆ 12 F1= 2 r12

(3.40)

g = −∇φ

(3.41)

∇2 φ = −∇ · g = 4πGρ

(3.42)

 GM  − 2 rˆ (r > a) r g(r) =  − GMr rˆ (r < a) 3  a GM  − (r > a) r φ(r) =   GM (r 2 − 3a2 ) (r < a) 2a3

gravitational force on a mass m is mg.

(3.43)

m1,2 F1 r 12 ˆ

masses force on m1 (= −F 2 ) vector from m1 to m2 unit vector

G g φ ρ

constant of gravitation gravitational field strength gravitational potential mass density

r M a

vector from sphere centre mass of sphere radius of sphere

M (3.44)

a

r

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67

3.3 Gravitation

General relativitya Line element

Christoffel symbols and covariant differentiation

ds2 = gµν dxµ dxν = −dτ2 1 Γαβγ = g αδ (gδβ,γ + gδγ,β − gβγ,δ ) 2 φ;γ = φ,γ ≡ ∂φ/∂xγ Aα;γ = Aα,γ + Γαβγ Aβ Bα;γ = Bα,γ − Γβαγ Bβ

(3.45)

(3.46) (3.47) (3.48) (3.49)

= Γαµγ Γµβδ − Γαµδ Γµβγ + Γαβδ,γ − Γαβγ,δ Bµ;α;β − Bµ;β;α = R γµαβ Bγ

(3.50)

Rαβγδ = −Rαβδγ ; Rβαγδ = −Rαβγδ Rαβγδ + Rαδβγ + Rαγδβ = 0

(3.52) (3.53)

Dv µ =0 Dλ

(3.54)

ds

invariant interval

dτ gµν dxµ Γαβγ

proper time interval metric tensor differential of xµ Christoffel symbols

,α ;α φ Aα

partial diff. w.r.t. xα covariant diff. w.r.t. xα scalar contravariant vector

Bα

covariant vector

R αβγδ

Riemann tensor

vµ

tangent vector (= dxµ /dλ) affine parameter (e.g., τ for material particles)

R αβγδ Riemann tensor

Geodesic equation

where

DAµ dAµ ≡ + Γµαβ Aα v β Dλ dλ

(3.51)

(3.55)

λ

Geodesic deviation

D2 ξ µ = −R µαβγ v α ξ β v γ Dλ2

(3.56)

ξµ

geodesic deviation

Ricci tensor

Rαβ ≡ R σασβ = g σδ Rδασβ = Rβα

(3.57)

Rαβ

Ricci tensor

Einstein tensor

Gµν = R µν −

(3.58)

Gµν R

Einstein tensor Ricci scalar (= g µν Rµν )

Einstein’s field equations

Gµν = 8πT µν

(3.59)

T µν p

stress-energy tensor pressure (in rest frame)

Perfect fluid

T µν = (p + ρ)uµ uν + pg µν

(3.60)

ρ uν

density (in rest frame) fluid four-velocity

Schwarzschild solution (exterior)

1 µν g R 2

    2M 2M −1 2 dr ds2 = − 1 − dt2 + 1 − r r + r 2 (dθ 2 + sin2 θ dφ2 )

(3.61)

spherically symmetric mass (see Section 9.5) (r,θ,φ) spherical polar coords. t time M

Kerr solution (outside a spinning black hole) ∆ − a2 sin2 θ 2 2Mr sin2 θ dt − 2a dt dφ 2 2 2 2 (r 2 + a2 )2 − a2 ∆sin2 θ 2 2 dr + 2 dθ2 sin θ dφ + + 2 ∆ ds2 = −

J

angular momentum (along z)

a ≡ J/M ∆ ≡ r2 − 2Mr + a2 (3.62) 2  ≡ r2 + a2 cos2 θ a General relativity conventionally uses “geometrized units” in which G = 1 and c = 1. Thus, 1kg = 7.425 × 10−28 m etc. Contravariant indices are written as superscripts and covariant indices as subscripts. Note also that ds2 means (ds)2 etc.

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Dynamics and mechanics

3.4

Particle motion

Dynamics definitionsa Newtonian force

F = m¨r = p˙

(3.63)

F m

force mass of particle

r

particle position vector

Momentum

p = m˙r

(3.64)

p

momentum

Kinetic energy

1 T = mv 2 2

(3.65)

T v

kinetic energy particle velocity

Angular momentum

J = r× p

(3.66)

J

angular momentum

Couple (or torque)

G = r× F

(3.67)

G

couple

(3.68)

R0 mi ri

position vector of centre of mass mass of ith particle position vector of ith particle

Centre of mass (ensemble of N particles) a In

N R0 =

mi r i i=1 N i=1 mi

the Newtonian limit, v  c, assuming m is constant.

Relativistic dynamicsa Lorentz factor

−1/2  v2 γ = 1− 2 c

(3.69)

γ v c

Lorentz factor particle velocity speed of light

Momentum

p = γm0 v

(3.70)

p m0

relativistic momentum particle (rest) mass

Force

F=

F

force on particle

t

time

Rest energy

Er = m0 c2

(3.72)

Er

particle rest energy

Kinetic energy

T = m0 c2 (γ − 1)

(3.73)

T

relativistic kinetic energy

E = γm0 c2

(3.74) E

total energy (= Er + T )

Total energy

dp dt

2 2

= (p c

(3.71)

+ m20 c4 )1/2

(3.75)

a It

is now common to regard mass as a Lorentz invariant property and to drop the term “rest mass.” The symbol m0 is used here to avoid confusion with the idea of “relativistic mass” (= γm0 ) used by some authors.

Constant acceleration v = u + at 2

2

v = u + 2as 1 s = ut + at2 2 u+v t s= 2

(3.76) (3.77) (3.78) (3.79)

u

initial velocity

v t s a

final velocity time distance travelled acceleration

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3.4 Particle motion

Reduced mass (of two interacting bodies) r m2

m1

centre of mass

r2

r1

m1 m2 m1 + m2 m2 r r1 = m1 + m2 −m1 r r2 = m1 + m2

(3.80)

µ mi

reduced mass interacting masses

(3.81)

ri

position vectors from centre of mass

(3.82)

r |r|

r = r1 − r2 distance between masses

Moment of inertia

I = µ|r|2

(3.83)

I

moment of inertia

Total angular momentum

J = µr×˙r

(3.84)

J

angular momentum

Lagrangian

1 L = µ|˙r |2 − U(|r|) 2

(3.85)

L

Lagrangian

U

potential energy of interaction

µ=

Reduced mass Distances from centre of mass

Ballisticsa Velocity

v = v0 cosα xˆ + (v0 sinα − gt) yˆ (3.86) v 2 = v02 − 2gy

(3.87)

Trajectory

gx2 y = xtanα − 2 2v0 cos2 α

Maximum height

h=

v02 sin2 α 2g

(3.89)

Horizontal range

l=

v02 sin2α g

(3.90)

a Ignoring

constant.

(3.88)

v0

initial velocity

v α g

velocity at t elevation angle gravitational acceleration

ˆ t

unit vector time

h

maximum height

l

range

the curvature and rotation of the Earth and frictional losses. g is assumed

yˆ v0 h

α l

xˆ

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Dynamics and mechanics

Rocketry 

Escape velocitya

2GM vesc = r

Specific impulse

Isp =

Exhaust velocity (into a vacuum) Rocket equation (g = 0) Multistage rocket

2γRTc u= (γ − 1)µ 

∆v =

constant of gravitation mass of central body central body radius specific impulse

(3.92)

u g R γ

effective exhaust velocity acceleration due to gravity molar gas constant ratio of heat capacities

(3.93)

Tc µ ∆v Mi Mf

combustion temperature effective molecular mass of exhaust gas rocket velocity increment pre-burn rocket mass post-burn rocket mass

M

mass ratio

1/2

Mi ∆v = uln Mf N 

escape velocity

(3.91)

u g



vesc G M r Isp

1/2

 ≡ ulnM

(3.94)

ui lnMi

N

number of stages

(3.95)

Mi ui

mass ratio for ith burn exhaust velocity of ith burn

(3.96)

t

burn time

θ

rocket zenith angle

∆vah

velocity increment, a to h

∆vhb ra rb

velocity increment, h to b radius of inner orbit radius of outer orbit

i=1

In a constant gravitational field

∆v = ulnM − gtcosθ 

Hohmann cotangential transferb

GM ∆vah = ra  ∆vhb =

GM rb

1/2 

2rb ra + rb

1/2 

 1−

1/2

2ra ra + rb

−1 (3.97) 1/2

transfer ellipse, h

a

b

(3.98) a From

the surface of a spherically symmetric, nonrotating body, mass M. between coplanar, circular orbits a and b, via ellipse h with a minimal expenditure of energy.

b Transfer

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3.4 Particle motion

Gravitationally bound orbital motiona U(r) potential energy

GMm α ≡− r r

Potential energy of interaction

U(r) = −

Total energy

J2 α α E =− + =− r 2mr 2 2a

Virial theorem (1/r potential)

E = U /2 = − T U = −2 T

Orbital equation (Kepler’s 1st law) Rate of sweeping area (Kepler’s 2nd law)

r0 = 1 + ecosφ , r a(1 − e2 ) r= 1 + ecosφ

Semi-major axis

a=

Semi-minor axis

b=

Eccentricityb Semi-latusrectum Pericentre Apocentre Phase Period (Kepler’s 3rd law)

or

G M m α

constant of gravitation central mass orbiting mass ( M) positive constant

(3.100)

E J

total energy (constant) total angular momentum (constant)

(3.101) (3.102)

T ·

kinetic energy mean value

(3.103)

r0 r e

semi-latus-rectum distance of m from M eccentricity

(3.99)

(3.104)

J dA = = constant dt 2m

(3.105)

A

area swept out by radius vector (total area = πab)

r0 α = 2 1−e 2|E|

(3.106)

a

semi-major axis

b

semi-minor axis

2a

J r0 = (3.107) (1 − e2 )1/2 (2m|E|)1/2  1/2  1/2 2EJ 2 b2 e= 1+ = 1 − (3.108) mα2 a2

J 2 b2 = = a(1 − e2 ) mα a r0 = a(1 − e) rmin = 1+e r0 = a(1 + e) rmax = 1−e r0 =

(3.109)

r0

r φ M

ae

2b rmax

(3.110)

rmin pericentre distance

(3.111)

rmax apocentre distance

(J/r) − (mα/J) (3.112) (2mE + m2 α2 /J 2 )1/2  1/2 ! m "1/2 m = 2πa3/2 P = πα 3 2|E| α (3.113) cosφ =

m

A

φ

orbital phase

P

orbital period

rmin

an inverse-square law of attraction between two isolated bodies in the nonrelativistic limit. If m is not  M, all explicit references to m in Equations (3.100) to (3.113) should be replaced by the reduced mass, µ = Mm/(M +m), and r taken as the body separation. The distance of mass m from the centre of mass is then rµ/m (see earlier table on Reduced mass). Other orbital dimensions scale similarly. b Note that if the total energy, E, is < 0 then e < 1 and the orbit is an ellipse (a circle if e = 0). If E = 0, then e = 1 and the orbit is a parabola. If E > 0 then e > 1 and the orbit becomes a hyperbola (see Rutherford scattering on next page). a For

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Dynamics and mechanics

Rutherford scatteringa y trajectory for α < 0

b

x

scattering centre

a

χ

rmin trajectory for α > 0

Scattering potential energy

Scattering angle

Closest approach

Semi-axis Eccentricity Motion trajectoryb Scattering centrec

Rutherford scattering formulad a Nonrelativistic

rmin

a

(α<0)

(α>0)

α U(r) = − r # < 0 repulsive α > 0 attractive |α| χ tan = 2 2Eb   |α| χ α rmin = csc − 2E 2 |α| = a(e ± 1) |α| 2E  2 2 1/2 4E b χ e= +1 = csc α2 2 a=

4E 2 2 y 2 x − 2 =1 α2 b  2 1/2 α 2 x=± + b 4E 2 dσ 1 dN = dΩ n dΩ ! α "2 χ = csc4 4E 2

(3.114)

U(r) potential energy

(3.115)

r α

particle separation constant

χ E b

scattering angle total energy (> 0) impact parameter

(3.116) (3.117) (3.118)

rmin closest approach a hyperbola semi-axis e eccentricity

(3.119)

(3.120) (3.121)

x,y position with respect to hyperbola centre

(3.122) dσ dΩ

(3.123) (3.124)

differential scattering cross section n beam flux density dN number of particles scattered into dΩ Ω solid angle

treatment for an inverse-square force law and a fixed scattering centre. Similar scattering results from either an attractive or repulsive force. See also Conic sections on page 38. b The correct branch can be chosen by inspection. c Also the focal points of the hyperbola. d n is the number of particles per second passing through unit area perpendicular to the beam.

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3.4 Particle motion

Inelastic collisionsa m1

m2

v1

m1

v2

Before collision Coefficient of restitution

Loss of kinetic energyb

v2

After collision

v2 − v1 = (v1 − v2 )  = 1 if perfectly elastic

(3.125) (3.126)

 = 0 if perfectly inelastic

(3.127)

T −T = 1 − 2 T

 vi vi

coefficient of restitution pre-collision velocities post-collision velocities

T ,T 

total KE in zero momentum frame before and after collision

mi

particle masses

(3.128)

m1 − m2 (1 + )m2 v1 + v2 m1 + m2 m1 + m2 m2 − m1 (1 + )m1 v2 = v2 + v1 m1 + m2 m1 + m2

v1 = Final velocities

m2

v1

(3.129) (3.130)

the line of centres, v1 ,v2  c. zero momentum frame.

a Along b In

Oblique elastic collisionsa

θ

Before collision m1

Directions of motion

Relative separation angle

tanθ1 = θ2 = θ

a Collision

After collision

m1

m2 sin2θ m1 − m2 cos2θ

(m21 + m22 − 2m1 m2 cos2θ)1/2 v m1 + m2 2m1 v v2 = cosθ m1 + m2

v2

θ1 v1

v

  > π/2 if m1 < m2 θ1 + θ2 = π/2 if m1 = m2   < π/2 if m1 > m2 v1 =

Final velocities

m2

θ2 m2

θ

(3.131) (3.132)

θi mi

angle between centre line and incident velocity final trajectories sphere masses

(3.133)

(3.134)

v

(3.135)

vi

between two perfectly elastic spheres: m2 initially at rest, velocities  c.

incident velocity of m1 final velocities

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3.5

Dynamics and mechanics

Rigid body dynamics

Moment of inertia tensor Moment of inertia tensora &



Iij =

(r 2 δij − xi xj ) dm

(3.136)

& &  − xy dm − xz dm (y 2 + z 2 ) dm & & 2 &   (x + z 2 ) dm − yz dm  I =  − xy dm & & & 2 − xz dm − yz dm (x + y 2 ) dm

r δij

r2 = x2 + y 2 + z 2 Kronecker delta

moment of inertia tensor dm mass element

I

(3.137)

xi

position vector of dm

Iij

components of I

IijY

tensor with respect to centre of mass ai ,a position vector of centre of mass m mass of body

Y − ma1 a2 I12 = I12

(3.138)

Y + m(a22 + a23 ) I11 = I11

(3.139)

Iij = IijY + m(|a|2 δij − ai aj )

(3.140)

Angular momentum

J = Iω

(3.141)

J

angular momentum

ω

angular velocity

Rotational kinetic energy

1 1 T = ω · J = Iij ωi ωj 2 2

(3.142)

T

kinetic energy

Parallel axis theorem

ii are the moments of inertia of the body. Iij (i = j) are its products of inertia. The integrals are over the body volume.

aI

Principal axes



 0 0 I3

I

principal moment of inertia tensor principal moments of inertia angular momentum

Principal moment of inertia tensor

I1 I =  0 0

Angular momentum

J = (I1 ω1 ,I2 ω2 ,I3 ω3 )

(3.144)

ωi components of ω along principal axes

Rotational kinetic energy

1 T = (I1 ω12 + I2 ω22 + I3 ω32 ) 2

(3.145)

T

Moment of inertia ellipsoida

T = T (ω1 ,ω2 ,ω3 ) ∂T (J is ⊥ ellipsoid surface) Ji = ∂ωi # ≥ I3 generally I1 + I2 = I3 flat lamina ⊥ to 3-axis

Perpendicular axis theorem

Symmetries

a The

0 I2 0

(3.143)

Ii J

I1 = I2 = I3

asymmetric top

I1 = I2 = I3 I1 = I2 = I3

symmetric top spherical top

ellipsoid is defined by the surface of constant T .

kinetic energy

(3.146) I3

(3.147) I1

I2

(3.148) lamina (3.149)

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75

3.5 Rigid body dynamics

Moments of inertiaa Thin rod, length l

Solid sphere, radius r Spherical shell, radius r

Solid cylinder, radius r, length l

Solid cuboid, sides a,b,c

I1 = I2 = I3  0

l

ml 2 12

(3.150)

2 I1 = I2 = I3 = mr 2 5

I1 = m(b2 + c2 )/12 I2 = m(c2 + a2 )/12

(3.156) (3.157)

I3 = m(a2 + b2 )/12

(3.158)

I3 I2

(3.153)

l I1 I2

r

I3

(3.155) I1

2

3 h m r2 + 20 4

3 I3 = mr 2 10

I1 r

(3.154)

I1 = I2 =

I2

I1

(3.152)

2 I1 = I2 = I3 = mr 2 3   m 2 l2 I1 = I2 = r + 4 3 1 I3 = mr 2 2



Solid circular cone, base radius r, height hb

I3

(3.151)

a

I3

I2

b

c

 (3.159)

h

I3 I 2 I r

(3.160)

1

2

2

Solid ellipsoid, semi-axes a,b,c

I1 = m(b + c )/5 I2 = m(c2 + a2 )/5 I3 = m(a2 + b2 )/5

(3.161) (3.162) (3.163)

Elliptical lamina, semi-axes a,b

I1 = mb2 /4 I2 = ma2 /4 I3 = m(a2 + b2 )/4

(3.164) (3.165) (3.166)

I3 a c b I2 I1 I2 b I3 a

I1

I2 2

I1 = I2 = mr /4 I3 = mr 2 /2

Disk, radius r

Triangular plate a With

c

m I3 = (a2 + b2 + c2 ) 36

(3.167) (3.168)

r

I1

I3 a

(3.169)

b I3

c

respect to principal axes for bodies of mass m and uniform density. The radius of gyration is defined as k = (I/m)1/2 . b Origin of axes is at the centre of mass (h/4 above the base). c Around an axis through the centre of mass and perpendicular to the plane of the plate.

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76

Dynamics and mechanics

Centres of mass Solid hemisphere, radius r

d = 3r/8 from sphere centre

(3.170)

Hemispherical shell, radius r

d = r/2 from sphere centre

(3.171)

Sector of disk, radius r, angle 2θ

2 sinθ d= r 3 θ

from disk centre

(3.172)

Arc of circle, radius r, angle 2θ

d=r

from circle centre

(3.173)

Arbitrary triangular lamina, height ha

d = h/3 perpendicular from base

(3.174)

Solid cone or pyramid, height h

d = h/4 perpendicular from base

(3.175)

Spherical cap, height h, sphere radius r

solid: d =

3 (2r − h)2 from sphere centre 4 3r − h shell: d = r − h/2 from sphere centre

Semi-elliptical lamina, height h ah

sinθ θ

d=

4h 3π

(3.176) (3.177)

from base

(3.178)

is the perpendicular distance between the base and apex of the triangle.

Pendulums Simple pendulum Conical pendulum Torsional penduluma

   l θ02 1 + + ··· P = 2π g 16

P period

(3.179)

g gravitational acceleration l length θ0 maximum angular displacement

(3.180)

α cone half-angle

(3.181)

I0 moment of inertia of bob C torsional rigidity of wire (see page 81)



 l cosα 1/2 g  1/2 lI0 P = 2π C P = 2π

Equal double pendulumc a Assuming

P  2π

l √

(2 ± 2)g

α

m

1 (ma2 + I1 cos2 γ1 P  2π mga 1/2 2 2 + I2 cos γ2 + I3 cos γ3 ) (3.182) 

m l



Compound pendulumb

l θ0

a distance of rotation axis from centre of mass m mass of body Ii principal moments of inertia γi angles between rotation axis and principal axes

1/2 (3.183)

the bob is supported parallel to a principal rotation axis. b I.e., an arbitrary triaxial rigid body. c For very small oscillations (two eigenmodes).

l

I0 a

I1

I3 I2

l

m

l m

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77

3.5 Rigid body dynamics

Tops and gyroscopes J herpolhode

3 ω

space cone

invariable plane

J3 polhode

Ωp body cone

moment of inertia ellipsoid

θ support point

a

2 gyroscope

prolate symmetric top

a

Euler’s equations

Free symmetric topb (I3 < I2 = I1 ) Free asymmetric topc

Steady gyroscopic precession

˙ 1 + (I3 − I2 )ω2 ω3 G1 = I1 ω ˙ 2 + (I1 − I3 )ω3 ω1 G2 = I2 ω ˙ 3 + (I2 − I1 )ω1 ω2 G3 = I3 ω I1 − I3 ω3 I1 J Ωs = I1

Ωb =

Ω2b =

(I1 − I3 )(I2 − I3 ) 2 ω3 I1 I2

Ω2p I1 cosθ − Ωp J3 + mga = 0

# Mga/J3 (slow) Ωp   J3 /(I1 cosθ) (fast)

mg

(3.184) (3.185) (3.186)

Gi

external couple (= 0 for free rotation)

Ii ωi

principal moments of inertia angular velocity of rotation

(3.187)

Ωb

body frequency

(3.188)

Ωs J

space frequency total angular momentum

Ωp θ J3

precession angular velocity angle from vertical angular momentum around symmetry axis mass

(3.189)

(3.190) (3.191)

m g a

gravitational acceleration distance of centre of mass from support point moment of inertia about support point

Gyroscopic stability

J32 ≥ 4I1 mgacosθ

(3.192)

Gyroscopic limit (“sleeping top”)

J32  I1 mga

(3.193)

Nutation rate

Ωn = J3 /I1

(3.194)

Ωn

nutation angular velocity

Gyroscope released from rest

Ωp =

(3.195)

t

time

a Components

mga (1 − cosΩn t) J3

I1

are with respect to the principal axes, rotating with the body. body frequency is the angular velocity (with respect to principal axes) of ω around the 3-axis. The space frequency is the angular velocity of the 3-axis around J , i.e., the angular velocity at which the body cone moves around the space cone. c J close to 3-axis. If Ω2 < 0, the body tumbles. b b The

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78

3.6

Dynamics and mechanics

Oscillating systems

Free oscillations x t

oscillating variable time

γ

damping factor (per unit mass)

ω0

undamped angular frequency

(3.197)

A

amplitude constant

where ω = (ω02 − γ 2 )1/2

(3.198)

φ ω

phase constant angular eigenfrequency

x = e−γt (A1 + A2 t)

(3.199)

Ai

amplitude constants

x = e−γt (A1 eqt + A2 e−qt )

(3.200)

− ω02 )1/2

(3.201)

2

Differential equation

dx dx + ω02 x = 0 + 2γ dt2 dt x = Ae−γt cos(ωt + φ)

Underdamped solution (γ < ω0 ) Critically damped solution (γ = ω0 ) Overdamped solution (γ > ω0 )

where q = (γ

2

an 2πγ = an+1 ω  π ω0 if  Q= 2γ ∆

Logarithmic decrementa

(3.202)

∆ = ln

Quality factor

(3.196)

 Q1

(3.203)

∆

logarithmic decrement

an

nth displacement maximum

Q

quality factor

a The decrement is usually the ratio of successive displacement maxima but is sometimes taken as the ratio of successive displacement extrema, reducing ∆ by a factor of 2. Logarithms are sometimes taken to base 10, introducing a further factor of log10 e.

Forced oscillations Differential equation

Steadystate solutiona

d2 x dx + ω02 x = F0 eiωf t + 2γ dt2 dt

(3.204)

x = Aei(ωf t−φ) ,

(3.205)

where

A = F0 [(ω02 − ωf2 )2 + (2γωf )2 ]−1/2 F0 /(2ω0 ) [(ω0 − ωf )2 + γ 2 ]1/2 2γωf tanφ = 2 ω0 − ωf2 

(γ  ωf )

(3.206) (3.207) (3.208)

x t γ

oscillating variable time damping factor (per unit mass)

ω0 F0

undamped angular frequency force amplitude (per unit mass) forcing angular frequency amplitude phase lag of response behind driving force

ωf A φ

Amplitude resonanceb

2 ωar = ω02 − 2γ 2

(3.209)

ωar amplitude resonant forcing angular frequency

Velocity resonancec

ωvr = ω0

(3.210)

ωvr velocity resonant forcing angular frequency

Quality factor

Q=

(3.211)

Q

quality factor

Impedance

Z = 2γ + i

(3.212)

Z

impedance (per unit mass)

a Excluding

ω0 2γ ωf2 − ω02 ωf

the free oscillation terms. frequency for maximum displacement. c Forcing frequency for maximum velocity. Note φ = π/2 at this frequency. b Forcing