Lecture Notes in Physics Edited by J. Ehlers, Menchen, K. Hepp, ZUrich R. Kippenhahn, MQnchen, H. A. Weidenmeller, Heidelberg and J. Zittartz, K61n Managing Editor: W. BeiglbSck, Heidelberg
110
David Park
Classical Dynamics and Its Quantum Analogues
Springer-Verlag Berlin Heidelberg New York 1979
Author David Park Department of Physics Williams College Williamstown, MA 01267/USA
ISBN 3-540-09565-9 Springer-Verlag Berlin Heidelberg N e w Y o r k ISBN 0-387-09565-9 Springer-Verlag New York Heidelberg Berlin Library of Congress Cataloging in Publication Data Park, David, 1919Classical dynamics and its quantum analogues. (Lecture notes in physics; v. 110) Bibliography: p. Includes index. I. Dynamics. 2. Mechanics.3. Quantum theory. I. Title, II. Series. QC133.P37 531'.11 79-23440 ISBN 3-540-09565-9 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin Heidelberg 1979 Printed in Germany Printing and binding: Bettz Offsetdruck, Hemsbach/Bergstr. 2]53/3140-543210
TABLE OF CONTENTS
I.
2.
RAYS
OF L I G H T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
i.i
Waves,
Rays,
1.2
Phase
Velocity
1.3
Dynamics
of a W a v e
Packet ..........................
1.4
Fermat's
Principle
of Least
1.5
Interlude
1.6
Optics
ORBITS
i
and O r b i t s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . and
on the
Group
1
Velocity ...................
Calculus
8 13
T i m e . . . . . . . . . . . . . . . . . . . 17
of V a r i a t i o n s . . . . . . . . . . . . 19
in a G r a v i t a t i o n a l
Field ....................
OF P A R T I C L E S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 58
2.1
Ehrenfest's
Theorems ...............................
38
2.2
Oscillators
and P e n d u l u m s . . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.3
Interlude
2.4
Driven
on E l l i p t i c
F u n c t i o n s . . . . . . . . . . . . . . . . . . . . 46
2.5
A Driven
2.6
Lagrange's
Equations ...............................
59
2.7
The
Pendulum, ...............................
64
2.8
Planets
and Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
2.9
Orbital
Oscillations
2.10
Orbital
Motion:
Oscillators ................................. Anharmonic
Double
Oscillator .....................
Other
2.12
Bohr
Integral
and
Orbits..77
2.13
The
Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Orbits
and Q u a n t u m
Principle
53
and S t a b i l i t y . . . . . . . . . . . . . . . . . 75
Vectorial
Hyperbolic 2.11
50
Mechanics:
of M a u p e r t u i s
and
82
D e g e n e r a c y ..... 83
its
Practical Utility..85
3.
4.
N-PARTICLE
SYSTEMS .......................................
3.1
Center-of-Mass
3.2
Two-Particle
3.3
Vibrating
3.4
Normal
3.5
The V i r i a l
HAMILTONIAN
89
Theorems ............................
89
Systems ...............................
96
Systems ..................................
97
Coordinates ................................ Theorem ................................
DYNAMICS ....................................
4.1
The
Canonical
4.2
Magnetic
4.3
Canonical
4.4
Infinitesimal
4.5
Generating
4.6
Deduction
Equations ...........................
Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transformations
.........................
Transformations
Finite
.....................
Transformations
106 iii iii i15 121 128
from
Infinitesimal of N e w
i01
Ones..135
Integrals ........................
139
IV
5.
4.7
Commutators
4.8
Gauge
and P o i s s o n
B r a c k e t s . . . . . . . . . . . . . . . . . . 142
Invariance ..................................
THE H A M I L T O N - J A C O B I
THEORY ..............................
5.1
The H a m i l t o n - J a c o b i
5.2
Step-by-Step
5.3
Interlude
on P l a n e t a r y
5.4
Action
the G e n e r a t o r
5.5
Jacobi's
5.6
Orbits
5.7
Coordinate
5.8
Curvilinear
5.9
Interlude
145 151
E q u a t i o n . . . . . . . . . . . . . . . . . . . . . . 151
Integration
of the H a m i l t e n - J a c o b i Equation..IS5
Motion
in G e n e r a l Relativity...159
6.
7.
8.
ACTION
AND
as
Generalization
of a C o n t a c t T r a n s f o r m a t i o n .... 164 ...........................
166
Integrals ..............................
171
Systems ................................
177
and
Coordinates ...........................
on C l a s s i c a l
Optics .....................
PHASE ........................................
6.1
The Old Q u a n t u m
6.2
Hydrogen
6.3
The A d i a b a t i c
6.4
Connections
6.5
Heisenberg's
6.6
Matter
6.7
Construction
Atom
Theory ............................
in the O l d
Quantum
Quantum
Quantum
Waves ...................................... Function ...................
THEORY OF P E R T U R B A T I O N S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . and P e r i o d i c
192
202
M e c h a n i c s . . . . . . . . . . . . . . . . 206
Mechanics ....................
of a W a v e
192
T h e o r y ........... 200
Theorem .............................
with
180 185
7.1
Secular
7.2
Perturbations
7.3
Adiabatic
7.4
Degenerate
7.5
Canonical
Perturbation
7.6
Newtonian
Precession ..............................
209 219 222 229
P e r t u r b a t i o n s . . . . . . . . . . . . . . . . 229
in Q u a n t u m
M e c h a n i c s . . . . . . . . . . . . . . . . 232
Perturbations ...........................
236
States ............ .....................
240
Appendix:
Time A v e r a g e
THE M O T I O N
OF A R I G I D
Theory .....................
of r n in K e p l e r
248
O r b i t s .......... 255
BODY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Velocity
245
Angular
8.2
The
8.3
Dynamics
8.4
Euler's
8.5
The
8.6
Quantum
8.7
Spinors ...........................................
289
8.8
Particles
298
Inertia
and M o m e n t u m . . . . . . . . . . . . . . . . . . . . .
257
8.1
Tensor ................................
in a R o t a t i n g
Coordinate
S y s t e m .......... 262
Equations .................................
Precession
of the E q u i n o x e s . . . . . . . . . . . . . . . . . . .
Mechanics
with
of a R i g i d
257 260
Body .................
Spin ...............................
268 275 283
9.
CONTINUOUS 9.1
SYSTEMS ......................................
Stretched
Strings .................................
9.2
Four M o d e s
9.3
Example:
A Plucked
9.4
Practical
Use
of V a r i a t i o n
9.5
More
One
Dimension ...........................
9.6
Waves
9.7
The M a t t e r
9.8
Classical
Than
of D e s c r i p t i o n . . . . . . . . . . . . . . . . . . . . . . . . . String ........................
302 304 311
P r i n c i p l e s . . . . . . . . . . . . . 313
in S p a c e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Field .................................. and Q u a n t u m
302
Descriptions
316 322 327
of N a t u r e ...... 332
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
336
INTRODUCTION The short Heroic age of Physics that started in 1925 was one of the rare occasions when deep considerations application in carrying out physical microphysics
of philosophy found
calculations.
In many parts of
the calculations have now become relatively straight-
forward even if not easy, but most physicists
seem to agree that some
difficulties of philosophical principle remain to be resolved.
This
situation has produced a great effect on the way people think and write about quantum mechanics,
a sort of gingerly approach to the
fundamentals and a tendency to emphasize what 50 years ago was new in the new theory at the expense of continuity with what had come before it.
Nowadays
someone who looks into the subject is more likely
to be struck by unexpected similarities between quantum and classical mechanics
than by contrasts he has been lead to expect,
and the main
purpose of these notes is to bring out the similarities.
It is
intended as a text in classical mechanics for readers who know a little classical and quantum mechanics but not very much of either, to teach them more of each and to emphasize the continuity between them. atom,
Most undergraduates for example,
I know are more familiar with the hydrogen
than with its planetary counterpart,
and since I
believe that historically and conceptually classical dynamics backbone of physics,
is the
I have treated planetary motion in some detail.
Since the relation between wave optics and ray optics is much the same as that between wave mechanics and particle orbits,
it has been
possible to do quite a lot of optics at the same time. Classical mechanics,
quantum mechanics,
and optics--the scope is
so vast that there has to be a principle of selection. mentioned those parts of elementary quantum mechanics the most fundamental interest and utility, of classical mechanics subjects
I have that are of
and explained those parts
that relate to them and illuminate
them.
The
in optics are pretty well determined by those choices.
There are no insects of the kind that walk on a weightless rod, no nonholonomic constraints,
no nonconservative
But the standard methods are h e r e - Lagrangian, Hamilton-Jacobi
together with their relations
spinning
forces even.
Hamiltonian and to quantum mechanics,
and considerable discussion of methods of exact and approximate solution and the role of invariance. theory by seeing how it works, in general relativity,
Believing that one learns a
I have included several calculations
including the deflection and retardation of
light by the sun and the precession of planetary perihelia.
Since
the relativistic precession must in practice be disentangled from the
VL11
precession produced by interaction with other planets, this as an example of classical perturbation quantum analogues, two particles, systems
but Chapters
of particles,
of continua.
including solid bodies,
analogy with the classical Inevitably, and made excuses selection taste.
Guided by the
deals with one or
3, 8, and 9 refer to the motions
The quantum mechanics
also a discussion
theory.
most of the classical mechanics
I have used
and to the mechanics
of rigid bodies
theory and
contrasts
of the roles of spinors
is treated
are noted.
in
There is
in the two theories.
I have discussed parts of physics
that appeal
for omitting parts that I find boring.
is finally not a matter
of
to me
Thus the
of logic so much as of personal
I trust there will be a few who share it.
The discussion Parts of elementary in many texts.
of classical mechanics
is fairly self-contained.
quantum mechanics not explained here are available
To avoid appearance
usually given references
of self-promotion
to my own book
I have not
(Park 1974) I , but all are to
be found there. I should mention a debt of gratitude once anyhow: mechanics problems
that many will recognize
to the book in w h i c h Max Born
(1927)
at
set classical
in order so that he could make a final assault on the of the old quantum theory.
But the assault was never made,
for the situation changed even as the book appeared, us more of the old mechanics
it sought to supersede
and so it tells than of the theory
it was intended to advance. There are a number of historical fewer than there should be.
tists who dared to think cosmically, after Newton's truth.
and in a little over a century
by people who aspired to nothing
They united mathematical
beauty with philosophical Perhaps
precision,
thanks to Karen Brownsword
they were right. for undertaking
difficult typing job at a time when she had more think about. David Park Williamstown,
i
out of
less than eternal
that there were no limits to the future of a civilization
which could produce such work. Special
though probably
death the science of dynamics was constructed
marble and bronze, and believed
references,
Kepler and Newton were the first scien-
The bibliography
is at the back of the notes.
a long and
important things
Massachusetts
to
CHAPTER
i.i
Waves,
I
RAVS OF LIGHT
Rays, and Orbits
Following the historical order of ideas is usually a good way of introducing people to physics, but the historical order is not the logical order.
If it were,
there would be no room for discovery,
since each new development would merely follow deductively from the preceding ones.
New discoveries usually involve radically new ideas
not implicit in what went before,
and deductive connections, when
they can be made at all, must be made backwards. the formulas of relativistic mechanics
You cannot derive
from those of Newton.
can derive Newtonian formulas from relativistic
You
ones by taking a
low-velocity limit, but even this is not a derivation of Newtonian physics,
since that physics takes place against a background of
absolute space and time which Einstein abandoned. contains a lesson:
This example
We may be able to derive the equations of an
earlier theory from those of a later one, but to recapture the theory itself requires
imagination and scholarship.
In this book we are going to establish mathematical between classical and quantum mechanics, optics.
The approach will be to consider classical mechanics as a
limiting case of quantum mechanics, case of wave optics. where necessary, anyhow.
connections
between ray optics and wave
and ray optics as a limiting
The conceptual background will be discussed
but the reader will be fairly familiar with it
One result of this approach will be the revelation that
classical and quantum theory are not so different conceptually as one thinks at first exposure.
Quantum mechanics
content and classical mechanics
is not:
is probabilistic
in
no amount of mathematical
fiddling will cross that gap, but it is often more conceptual than practical. The other transition,
from waves to rays,
is similar in many
ways except that it has largely escaped the rhetoric of twentiethcentury physics.
Of course the question of quanta is introduced
here if we claim that rays are the orbits of photons.
We are then
in the situation that light waves are classical and rays are quantum, while the orbits of particles are classical and their waves are quantum.
Therefore building a bridge between waves and rays will
be anti-historical
either for light or for matter, whichever way we
progress. One of the useful by-products
of this study will be to see
whether there are any useful concepts that can be isolated and
distinguished
in the dichotomy classical-quantum.
We shall discuss
the question later in more detail, but for the moment
it may be
instructive to consider de Broglie's relation ~p = h connecting the wavelength and momentum of a beam of electrons. "classical
The
limit" of such a relation is the one in which h is
replaced by zero. but only that,
This does not mean that h is treated as a variable,
in ordinary units,
either % or p is much larger than h.
There are, in the limit, two possibilities.
Either % is large and p
is effectively zero or p is large and ~ is effectively
zero.
That is,
either the theory contains the word wavelength but not the word momentum or the other way around. are perfectly familiar: mechanics.
In the quantum theory one sometimes speaks of momentum
and w a v e l e n g t h To b e g i n point the
in the at
use of Planck's just
the
varying
index
constant.
us r e c a l l
that
w a v e s and l i g h t era;
they
Further,
from a historical
rays
both
can be d i s c u s s e d
as we s h a l l
see
from SchrSdinger
construction
of light
rays
in
waves
belong without
(1.S), is
to the
the
formally
i n a medium o f
of refraction. condition
for
o f t h e wave s h a l l
and any l e n s e s the
light
classical
same as t h e
wave-length
let
of Newtonian orbits
A necessary
a nd i f
least,
of the
construction
same s e n t e n c e .
our development,
of view,
physics
Both kinds of theory without h
examples are classical optics and classical
or holes
it
any s u c h
construction
be " s m a l l ; "
goes through
index of refraction
varies,
that
is,
is the
that
a r e many w a v e l e n g t h s it
varies
the
beam i t s e l f
relatively
wide, little
o v e r one w a v e l e n g t h :
I~'~'l
> I,
v¢
--
(g/~)~
Long waves move faster, groups
are visible,
a pool and watch Problem
1.8
Problem
1.9
one can produce
depends
-~
½(g/~) ~
and since in water both phase waves a splash by dropping through
of a given frequency,
and wave
a stone into
the group.
show qualitatively
how
on the depth.
When sound passes
electromagnetic
vg
the phase waves passing
For waves
the wavelength
,
down a pipe of square cross section or
waves pass through a square waveguide
of side a, the
propagation
in the lowest mode is governed by the relation
and Lorrain
1962)
4Corson
13
~2
Find v 0 and
Vg and
= 02(k2
particle
of relativity
The relation between
1.10
272 a2 )
note the simple relation between them.
any conflict with the results Problem
+
in special relativity
Is there
theory?
energy and momentum of a free
is
E 2 = C 2 p 2 + (md2) 2 where m is the rest mass.
With E = ~
and p = ~k,
same questions
as
in the preceding problem. Problem
Careful m e a s u r e m e n t
1.11
of w a v e l e n g t h 299707.2
± 0.8 km s -I n =
1 +
m e a s u r e d value
a speed of light equal to
The index of refraction
(27259.9
w h e n k is given in ~m. one forgets
using an interrupted beam of light
X = 0.589 ~m in air yields
~-2
+ 153.50
What correction
in order to find c?
+
of air is l -~) x 10 -e
1.317
must be applied to the
What correction
that group velocity was measured
is obtained
and corrects
if
only for
phase velocity? Problem
A slow electron moves with velocity
1.1~
region enclosed by a metallic K as in (1.5),
1.3
Dynamics
I00 ms -I through the
shell at a potential
of 5 x 106V.
With
find v~ and Vg.
of a Wave Packet
At this point in the discussion, particle mechanics
someone who knew nothing
would have enough information
ing it.
In one dimension,
centered
at x, and ask how p changes with time.
because p = m V g .
we can apply
dV
dz
m
to a wave packet
d p_ =
m
dt
p dx dt = -~ 2-~ vg = -2-~x
In more interesting
parallel
to the momentum,
argument
is instructive
we have
dV
(1.13)
of
to begin reconstructSince E is constant,
dV
cases, when the force is not
to think a little harder but the
and illustrates
the importance
of the phase
from left to right in Fig.
1.4 w h i c h is
velocity. Consider a beam moving deflected by some lateral of constant further
inhomogeneity
s to stay in phase,
than the inside,
in K(r).
For the surfaces
the outside of the beam must move
14
dv~
where
v~(p)dt
v~(p + dp)dt
p
p + dp
p is the radius
v~(p)
+ dp dp
p + dp
of curvature
Solving
of the path.
this for
i/p gives 1 p
Since by
(1.18) v ¢
=
1
~v~
v~ ~p
~p
~/K with ~ constant, this is ~ P
Figure 1.4
~inv~
31nK 3P
Surfaces of constant s in a beam traversing a medium of varying index of refraction.
Now we can compare
the behavior
of light and matter:
Light K
Matter
c n (r)
=
ink
(1.26)
=
inn(r)
1 ~ inn(r) p =3p
K
+
const.
(1.27c)
=
2m ~E" g
i n k = ~1 l n [ E 1 1 3V/Op -p = 2 E-V
-
V(r)]
-
=
(1.27a,
½
Y(r)]
+
const.
1
3V/3p
-2 ½mY g 2
b)
15
so that 2
8V mv Fp---~=.q
(1.28)
P
The first is, as we shall see, a g e n e r a l i z a t i o n of Snell's law; the second is H u y g h e n s ' s Note that -~V/~p,
c e l e b r a t e d formula for centripetal
force
(1659).
the rate of increase of V in the direction p, is
exactly the force transverse
to the beam.
Note also how Planck's
constant got lost in taking the logarithmic derivative.
Example Solution:
Derive Snell's
1.2
Fig.
law for a h o r i z o n t a l l y
layered medium.
1.5 shows the s i t u a t i o n w h e n n depends only on z.
P
Figure 1.5
Since d@ = - d R ~ o ,
Geometry of a curved ray.
we have
1
d@
dlnn
dlnn
~z
or
dlnn dz
From the figure, ~
dlnn
=
dz/d~ - ~ ~z
-cos@, ~
d@ =
0
= sin@ and
+ coted@ = d l n n
+ din sine = 0
or
nsin@ = const.
(1.29)
16
This
is for a medium
with respect
to I.
we may suppose
that varies
continuously with z and not too fast
But I has been lost sight of in this discussion;
it as small as we like,
take place very rapidly,
though not,
the limiting
is Snell's
at a plane
form
(1.29)
interface,
in 1696 seems
derived empirically
case of a continuously
to his argument
in Section
case in which n varies develop
other methods
Problem
1.13
as it sets?
at the moment
varying medium.
We shall
For the still more general we shall have
to
The index of refraction
Assume
of air
that the earth is flat and that
the bottom of the sun's disc appears
to
The calculation will be done for a curved earth
1.14
Assuming
that the sun's angular diameter
is 30' and
the instant of sunset as in the preceding problem,
change in the visual
1.15
Adapt
show directly as occurs
Problem
(1.29) holds
5.47.
defining
Problem
Johann Bernoulli
that
below.
at ground level.
touch the horizon.
Problem
1.5
In
of light
By how large an angle is the center of the sun's disc
sunset occurs in Prob.
in 1621. ~
in more than one direction,
displaced by refraction is 1.000293
in n may
law for the refraction
to have been the first to recognize
in the more general return
and so the change
of course discontinuously.
that
shape of the sun's disc as it sets. the argument
(1.29) holds
at an air-glass
1.16
find the
of the preceding paragraph at a sharp discontinuity
so as to
in n such
interface.
What theorem in mechanics
corresponds
to the optical
law (1.29)?
Expression
of Classiea~
To conclude mechanics.
Mechangcs
in Terms of Phase
this section we forge another
The total
time-dependent
part is 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,
> 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:
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 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
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> = =
= O. Since the usual eigenfunctions tiation values satisfying The classical
(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
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 ~ 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
+
+
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
is defined as
(7.20)
= I ~a*gCndX
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
+ AE - E
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
as
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 -
)A¢
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 - - ]
-
(7.25)
x = -v +
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>_ + n
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)
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
<mlV(t)[a> E
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)
= = 3(4--k~o)~C2n2 + 2n + 1)
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 3
+
....
o
243
~x
+ C
The condition
etc.
- ~x~)~2
+ <mlvls>~3
that there exist a nontrivial
is that the determinant
of the coefficients
I and this, with given
- 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
by V... zj
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
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
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'
