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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
Ruiz de Alarcón 13, 28014 Madrid, Spain
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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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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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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66
Dynamics and mechanics
Rotating frames
Vector transformation
dA
dt
dA
dt
+ ω× A
(3.31)
Acceleration
v̇ = v̇ + 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
v̇,v̇ accelerations in S
and S v velocity in S r position in S F cor coriolis force
m particle mass
(3.36)
mÿ = Fy − 2mωe ẋsinλ
mz̈ = Fz − mg + 2mωe ẋcosλ
(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
mẍ = Fx + 2mωe (ẏ sinλ − ż cosλ)
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
r̂ 12
F1=
2
r12
(3.40)
g = −∇φ
(3.41)
∇2 φ = −∇ · g = 4πGρ
(3.42)

GM

− 2 r̂ (r > a)
r
g(r) =

− GMr r̂ (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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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 = mr̈ = ṗ
(3.63)
F
m
force
mass of particle
r
particle position vector
Momentum
p = mṙ
(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× ṙ
(3.84)
J
angular momentum
Lagrangian
1
L = µ|ṙ|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 x̂ + (v0 sinα − gt) ŷ
(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
ŷ
v0
h
α
l
x̂
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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
main
January 25, 2000
14:26
72
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.
main
January 25, 2000
14:26
73
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
main
January 25, 2000
14:26
74
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)
main
January 25, 2000
14:26
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.
main
January 25, 2000
14:26
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
main
January 25, 2000
14:26
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
G1 = I1 ω̇1 + (I3 − I2 )ω2 ω3
G2 = I2 ω̇2 + (I1 − I3 )ω3 ω1
G3 = I3 ω̇3 + (I2 − I1 )ω1 ω2
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
main
January 25, 2000
14:26
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