CLASSICAL MECHANICS

Course by E. Kogan

Contents VIII. HAMILTON MECHANICS 26

A. From Velocities to Momenta 26
I. POTENTIAL WELL 2 B. Hamiltonians and Hamilton Equations 27
A. Newton Mechanics 2 C. Ignorable Coordinates 28
B. Simple Pendulum 3 D. Heavy Symmetric Top: General Solution 29
E. Stability of a Vertical Top 29
II. CENTRAL FIELD 4 F. Hamiltonian and Energy 30
A. Energy 4
B. Angular Momentum 5 IX. GRAVITATIONAL FIELDS 31
C. Kepler Problem 6
A. Gravitational Potential 31
D. Planetary Motion 7
B. The Tides 31
E. Interplanetary Travels 8
F. Laplace-Runge-Lentz Vector 10 C. Multipole Expansion of Potential 32
D. Precession of Equinoxes and Satellite Orbits 33
III. LAGRANGE MECHANICS 10 E. The Shape of the Earth 34
A. Generalized Coordinates and Lagrange’s
Equations 10 X. VARIATIONAL CALCULUS AND
B. Charged Particle in an Electromagnetic VARIATIONAL PRINCIPLES 34
Field 12 A. Euler-Lagrange Equation 34
C. Holonomic Constraints 12 B. The Brachistochrone Problem 35
D. The Two-Body Problem 13 C. Variational Problems with Constraints 36
D. Hamilton’s Principle 37
IV. SCATTERING THEORY 13
E. Jacobi’s Principle 38
A. Differential Scattering Cross-Section 13
F. Fermat’s Principle 38
B. Rutherford Scattering 14
C. Scattering in Arbitrary Potential 15
XI. CANONICAL TRANSFORMATIONS 39
D. Reaction Cross-Section 15
E. Elastic Collisions 16 A. Canonical Transformations 39
F. CM and LS Cross-sections 17 B. Generating Function 39
C. Application of Canonical Transformations 40
V. ROTATING FRAMES 17 D. Symmetries and Conservation Laws:
A. Laboratory System and Rotating System 17 Lagrange’s Mechanics 41
B. Inertial Forces 17 E. Symmetries and Conservation Laws:
C. Freely Falling Body 18 Hamilton’s Mechanics 41
D. Larmor Effect 19 F. Poisson Brackets and Canonical
Transformations 42
VI. RIGID BODY: NEWTON
MECHANICS 19 XII. HAMILTON-JACOBI MECHANICS 43
A. Basic Principles 19 A. The Action as a Function of the Coordinates43
B. Space System and Body System 19 B. Hamilton-Jacobi Equation 43
C. Tensor of Inertia 20
C. Conservative System 44
D. Rotation about a Fixed Axis 21
D. Hamilton-Jacobi Theory and Canonical
E. Euler’s Angles 21
Transformations 44
F. Free Motion of a Top (Space System) 22
E. Separation of Ignorable Variables 45
G. Euler’s Equations (Body System) 22
H. Effect of Small Force 23 F. Separation of Variables: General Case 45

VII. RIGID BODY: LAGRANGE XIII. ACTION-ANGLE VARIABLES 46

MECHANICS 24 A. Action-Angle Variables in a System of One
A. Kinetic Energy of a Rigid Body 24 Degree of Freedom 46
B. The Lagrangian of a Heavy Symmetric Top 25 B. Adiabatic Invariance 46
C. Steady Precession of a Heavy Symmetric C. Action-Angle Variables in Separable
Top 25 Systems 47
 2

I. POTENTIAL WELL We thus obtained differential equation of the first order,

which can be solved by separating the variables
A. Newton Mechanics √ ∫
m dx
t(x) = ± √ .
To describe the position of a point particle we intro-
2 E − V (x)
duce the radius-vector r. The motion of a particle is
The energy chosen fixes one of the initial conditions. If
governed by the Newton’s equation of motion
the second initial condition is given as x(0) = x0 , the
solution of the Eq. (3) is
mr̈ = F(r, ṙ, t). (1)
√ ∫ x
m dx′
The function F depends upon specific mechanical sys- t(x) = √ . (4)
tem we are dealing with. For example, for the case of 2 x0 E − V (x′ )
projectile
Consider, for example, the following problem
F(r, ṙ, t) = g, Problem A.2 * Suppose the earth’s orbital motion
about the sun suddenly stops. What fraction of a year
for the case of harmonic oscillator (a period of rotation of the earth about the sun) would
elapse before the earth falls into the sun?
F(r, ṙ, t) = −kr.
Using Eq. (4) we obtain
In these particular cases Eq. (1) can be easily solved √ ∫
analytically; the solution for the projectile is: ME R dx
T = √
2 0 − GMS ME
+ GMS ME
gt2 R x
r(t) = r(0) + v(0)t + , √ ∫ R √
2 R x
= dx .
the solution for the harmonic oscillator is: 2GMS 0 R−x

v(0) Making substitution

r(t) = r(0) cos(ωt) + sin(ωt), √
ω x 2Rdy
√ =y dx = ,
where ω = k/m. In general case Eq. (1) can not be R−x (y 2 + 1)2
solved analytically.
we obtain
In one dimension Eq. (1) takes the form
√ ∫ ∞
R3 2y 2
mẍ = F (x, ẋ, t), (2) T = dy .
2MS G 0 (y 2 + 1)2
In general case Eq. (2) can not be solved analytically.
Integrating by parts we obtain
Consider a particular case
∫ ∞ ∫ ∞
2y 2 1 π
mẍ = F (x). (3) dx 2 2
= dy 2 = .
0 (y + 1) 0 y +1 2
The solution of Eq. (3) (for arbitrary function F ) is based Taking into account that
on our ability to find an integral of motion - energy.
√
m 2 R3
ẋ + V (x) = E, 1 year = 2π ,
2 MS G
∫
where V (x) = − F (x)dx is the potential energy func- we obtain
tion.
1 year
T = ≈ 2 months.
Problem A.1 The potential energy of a particle is 25/2

1 Problem A.3 * The potential energy function of a par-

V (x) = k(a2 − x2 ) for |x| < a, ticle of mass m is
2
V (x) = 0 for |x| ≥ a, 1
V = − c(x2 − a2 )2 ,
2
where a and k are constants, a > 0. Describe the possible
types of motion for both cases k > 0 and k < 0. where a and c are positive constants.
 3

V(x) Problem A.5 * A particle of mass m moves in the re-

gion x > 0 under the force

E
F = −mω 2 (x − a4 /x3 ), (6)

where ω and a are constants. Sketch the potential energy

x1 x2 x function. Find the position of equilibrium and the period
of small oscillations about it. Show that the period of the
oscillations for arbitrary energy E is independent of E.
FIG. 1: Potential well; x1 and x2 are turning points [To do the integration, transform to the variable y = x2 ,
then add a constant to ”complete the square”, and finally
use a trigonometric substitution.]
(a) Sketch this function, and describe the possible types
of motion in three cases: E > 0; E < − 21 ca4 , and From Eq. (6) we obtain
− 12 ca4 < E < 0. ( )
(b) The particle starts from rest at x = −a, and is mω 2 a4
V (x) = x2 + 2 .
given a small push to start it moving to the right. What is 2 x
its velocity when it reaches the point x? Given that t = 0
is the instant when it reaches x = 0, find its position as Hence
a function of time. (Assume that the push has negligible √ ∫
dx
effect on its energy.) τ (E) = 2m √ ( )
E− mω 2
2 x2 + a4
x2
For finite motion in a potential well we can find the ∫ ∫ e
period as the function of energy for arbitrary V (x). 1 dy 1 dz π
= √ = √ = ,
√ ∫ x2 (E) ω 2E
− y 2 − a4 ω −e e −z
2 2 ω
dx mω 2 y
τ (E) = 2m √ (5)
x1 (E) E − V (x)
Problem A.6 Determine the period of oscillations, as
where the turning points x1 and x2 satisfy the equation a function of the energy, when a particle of mass m
moves in field for which the potential energy is a) U =
V (x1,2 ) = E. −U0 / cosh2 αx, where −U0 < E < 0; b) U = U0 tan2 αx.

For harmonic oscillator V (x) = kx2 /2, and

√ ∫ x0 B. Simple Pendulum
dx
τ (E) = 2m √
−x0 E − kx2 /2 Galileo began experimenting with pendulums after he
∫ x0
2 dx 2π saw a chandelier swinging in the Pisa Cathedral. He no-
= √ = , ticed each swing of the chandelier took the same amount
ω0 −x0 x0 − x2
2 ω0
of time, even though each swing was shorter than the pre-
where E = kx20 /2. vious swing. This led to Galileo’s discovery of the Law
Small oscillations near stable equilibrium position (in of the Pendulum.
any potential well) are harmonic oscillations, because we Motion of a simple pendulum is described by two co-
can expand potential energy ordinates x and y with the relation between them
1
V (x) = V (0) + xV ′ (0) + x2 V ′′ (0) + . . . x2 + y 2 = ℓ2 ,
2
and keep only quadratic terms which is called constraint. We can introduce a single
coordinate θ and express x and y, and hence kinetic and
1 2
kx , k = V ′′ (0).
V (x) = potential energy, through θ
2
Problem A.4 The potential energy function of a parti- 1 2 2 θ
T = ml θ̇ , V = mgl(1 − cos θ) = 2mgl sin2 .
cle of mass m is 2 2
cx From the integral of motion
V = 2 ,
x + a2
E = T + V,
where a and c are positive constants. Sketch the curve of
V as a function of x. Find the position of stable equilib- for the case E ≤ 2mgl we obtain, substituting for the
rium, and the period of small oscillations about it. Given energy E the amplitude θ0 ,
that the particle starts from this point with velocity v, find ( )
the values of v for which it (a) oscillates, (b) escapes to g 2 θ0 2 θ
2
θ̇ = 4 sin − sin . (7)
−∞, and (c) escapes to +∞. l 2 2
 4

x 3

2
θ l

τ/τs
m 1
y
FIG. 2: 0
0 1 2 3
θ0
For θ0 ≪ 1 we
√ obtain harmonic oscillations with the FIG. 3: Period of oscillations of a simple pendulum as a func-
frequency ω = l/g. For θ0 = 0 tion of an amplitude θ0 .
√ ∫
1 l dθ
t= . (8) Integrating term by term we get
2 g cos θ2
[ ( )2 ( )2 ]
π 1 1 · 3
The integral can be easily calculated K(k) = 1+ k2 + k4 + . . .
2 2 2·4
∫ ∫ (θ) ∫ ( )
dθ d 2 cos θ2 dx 1+x
=2 =2 = ln The first two terms of the expansion of the period with
cos θ2 cos2 θ2 1 − x2 1−x
( ) [ ( )]
respect to θ0 are
1 + sin θ2 π−θ √ (
= ln = 2 ln cot . (9) )
1 − sin 2
θ 4 ℓ 1 2
τ = 2π 1 + θ0 + · · · .
g 16
For arbitrary amplitude after substitution x =
sin θ2 / sin θ20 we obtain Eq. (7) in the form Problem B.1 Convince yourself that for θ0 → π the pe-
riod diverges ∝ − ln(π − θ0 ).
g
ẋ2 = (1 − x2 )(1 − k 2 x2 ),
l Problem B.2 ** The pendulum comprises a bob of mass
m supported by a light rigid rod of length l. Find the equa-
where k = sin(θ0 /2). For the period we obtain tion of motion for small displacements from the position
√ ∫ of unstable equilibrium. Show that if it is released from
ℓ 1 dx rest at a small angle to the upward vertical, then the time
τ =4 √ .
g 0 (1 − x2 )(1 − k 2 x2 ) taken for the angular displacement to increase by a factor
of 10 will be approximately (l/g)1/2 ln 20. Evaluate this
Making substitution ξ = sin x, we get time for a pendulum of length 1m, and find the angular
√ ∫ √ velocity of the pendulum when it reached the downward
ℓ π/2 dξ ℓ vertical.
τ =4 √ =4 K(k), (10)
g 0 1 − k sin ξ
2 2 g
II. CENTRAL FIELD
Integral in Eq. (10) is called the complete elliptic in-
tegral of the first kind - a well known special function.
A. Energy
We can present the integral (10) as a series with respect
to parameter k. Using the binomial series A particle is said to be in a force field (time dependent
in general case) if F = F(r, t). The force field is potential
α α(α − 1) 2 if we can introduce V (r, t), such that
(1 + x)α = 1 + x+ x + ...,
1! 2!
F = −∇V ⇐⇒ ∇ × F = 0.
we obtain
Well known examples:
1 1
√ = 1 + k 2 sin2 ξ
1 − k 2 sin ξ
2 2 1 2
F = −kr =⇒ V =kr
1·3 4 4 1·3·5 6 6 2
+ k sin ξ + k sin ξ + . . . . F = mg =⇒ V = −mg · r.
2·4 2·4·6
 5

For time independent potential field from Eq. (1) we get B. Angular Momentum
the energy conservation law
T + V = E = const, From Eq. (1) we obtain

where J̇ = G.
m
T = ṙ2 (11) where
2
is the kinetic energy. In Cartesian coordinates Eq. (11) J = mr × ṙ; G=r×F (18)
is
m( 2 )
are the angular momentum and the momentum of force
T = ẋ + ẏ 2 + ż 2 . (12)
2 F about the origin respectively. For central force G = 0.
Problem A.1 * A particle in a plane has potential en- Hence angular momentum is an integral of motion
ergy (in polars)
J = const. (19)
V = −k ln r + cr + gr cos θ,
Problem B.1 Let in the problem A.3 the point of ob-
where k, c and g are positive constants. Find the po-
servation was the perihelion. Find the angle at which the
sitions of equilibrium (a) if c > g, and (b) if c < g.
earth’s orbit, and the comet’s orbits cross.
Determine whether the equilibrium is stable.
A central field Conservation of direction of angular momentum means
r that the motion is coplanar and we can introduce in the
F = F (r) (13) plane of motion polars r, θ. In these coordinates Eqs.(12)
r
and (18) can be expressed as
is always conservative: it is described by V (r), found
from the equation 1 ( 2 )
m ṙ + r2 θ̇2 + V (r) = E (20)
dV 2
= −F (r) (14) mr2 θ̇ = J. (21)
dr
For gravitation field created by a point mass M We can interpret Eq. (21) as the second law of Kepler
GM m
V =− . (15) dS J
r = = const, (22)
dt 2m
Escape velocity from the earth can be found from the
equation which is thus valid for any central field.
1 GME m Substituting θ̇ = J/mr2 in the first equation, we get
E= mv 2 − = 0.
2 e R 1 2
mṙ + U (r) = E, (23)
It is convenient to take into account, that 2
GME where
= g. (16)
R2
Hence we get J2
U (r) = V (r) + .
2mr2
vescape = (2Rg)1/2 = 11.2 km s−1 . (17)
For the motion of a material point in a central field the
Problem A.2 ** Show that the minimal velocity Ves angle θ varies monotonically while the distance from the
(relative to the earth) allowing to the rocket sent from center of the field varies in the same way as r varies in
the earth to escape from the solar system is the one-dimensional problem with the effective potential
√ √ energy U (r). Minimum of the effective potential corre-
2
Ves = vescape 2 ( 2 − 1)2 = 16.7 km s−1 ,
+ vE spond to stable circular orbit, maximum and point of
inflection correspond to unstable circular orbits.
where vescape is given by Eq. (17).
Problem A.3 A comet is observed at a distance from Problem B.2 (a) A particle of mass m moves in a po-
the sun equal to half the radius of the earth’s orbit (as- tential
sumed circular), and its velocity at that point is twice k
the orbital velocity of the earth (the orbital velocity of the V (r) = −
earth is 30 km/sec). Find its velocity when it crosses r
the earth’s orbit. Whether the comet subsequently escape How many circular orbits are possible for a given value
from the solar system? of angular momentum? Are the orbits stable?
 6

* (b) The same for a potential C. Kepler Problem

c
V (r) = − λ
r Consider potential
where λ ̸= 0, 2 and the constant c is positive (negative) if k
λ is positive (negative). V (r) = ± . (26)
** (c) The same for a screened Coulomb potential r

e−αr In this case Eq. (24) takes the form

V (r) = −k ,
r 1 2 k J2
where k and α are positive constants. mṙ ± + = E. (27)
2 r 2mr2
[Hint: If you introduce dimensionless radius R = αr
and another dimensionless parameter A = J 2 α/mk, the Eq. (27) can be rewritten as
effective potential energy can be presented as ( )2
[ ] 1 dr 1 2mk 1 2mE
A e−R + ± 2 = , (28)
U (r) = αk − . r2 dθ r2 J r J2
2R2 R
Thus the problem is reduced to analysis of critical points by introducing new variable u = 1/r and the notation
of the function which depends upon a single parameter
A.] J2
l= (29)
mk
To find the equation of the trajectory we substitute
dr dr J can be rewritten in the form
ṙ ≡ θ̇ = , ( )2
dθ dθ mr2 du 2 2m
into Eq. (23). We obtain + u2 ± u = 2 E,
dθ l J
( )2
1 dr 2m 2mE or
+ 2 U (r) = . (24)
r2 dθ J J2 ( )2
dw e2
Problem B.3 A particle moving in a central field de- + w2 = , (30)
scribes the spiral orbit r = r0 ekθ . Show that the force law dθ l2
is inverse-cube and that θ varies logarithmically with t.
where w = u ± 1/l and
Substituting equation of the orbit into Eq. (21) we obtain
2EJ 2
dθ e2 = 1 + . (31)
mr02 e2kθ = J. mk 2
dt
Logarithmic dependence of θ with t is obvious. Substi- The solution of Eq. (30) is
tuting equation of the orbit into Eq. (24) we obtain e
w= cos θ,
k2 1 2m 2mE l
+ 2 + 2 V (r) =
r2 r J J2
which is equivalent to
Inverse-square dependence of potential energy with r
and, hence, inverse-cube dependence of force with r is r[e cos θ ∓ 1] = l. (32)
obvious.
The upper sign corresponds to repulsion and the lower
Problem B.4 ** A particle moves in two dimensions
to attraction.
under the influence of a central force determined by the
potential Problem C.1 Show that Eq. (32) can be obtained by
V (r) = αrp + βrq . directly calculating integral (25) for the potential (26).
Find the powers p and q which make it possible to achieve Problem C.2 a) Show that Eq. (32) for the case e < 1
a spiral orbit of the form r = cθ2 , with c a constant. (E < 0) can be presented in Cartesian coordinates as
Problem B.5 ** Find the central force which results in
x2 y2
the following orbit for a particle: + = 1,
a2 b2
r = a(1 + cos θ)
and hence describes an ellipse (Kepler’s first law); a is
The solution of Eq. (24) is called the semi-major axis and b - the semi-minor axis of
√ ∫
1 J/r2 dr the ellipse. Show that the surface of the ellipse is
θ= √ . (25)
2m E − U (r) S = πab.
 7

Problem C.3 (a) The semi-major axis of Jupiter’s or-

bit is 5.20 a.u. Find its orbital period in years.
(b) The orbit of an asteroid extends from the Earth’s
to that of Jupiter, just touching both. Find its orbital
period. (Everywhere in this Section treat the planetary
orbits as circular and coplanar.)

Problem C.4 The parameters of the orbit of a satellite

are: the period is 102.2 minutes, the distance to the sur-
face of the Earth in the perigee is 210 km, in the apogee
– 1548 km. Find the mass of the Earth. Consider the
FIG. 4: In 1602, Kepler believed that the orbit of Mars was Earth as a ball with radius 6371 km. (The gravitation
oval; he later discovered that it was an ellipse with the Sun constant G = 6.67 · 10−11 N· m2 · kg−2 .)
at one focus. In fact, Kepler introduced the word ”focus” and
published his discovery in 1609. In 1705 Halley showed that Problem C.5 The eccentricity of the earth’s orbit is e =
the comet now named after him moved in an elliptical orbit 0.0167. If the orbit is divided in two by the minor axis,
around the Sun. show that the times spent in the two halves of the orbit
are ( 12 ± e/π) years, and evaluate the difference in days.
(Hint: The area required is half that of the ellipse ± a
Show that in this case
triangle.)
b2
1 − e2 = . (33) Problem C.6 * If the earth’s orbit is divided in two by
a2
the latus rectum, show that the difference in time spent
Show that this case is possible only for attractive inter- in the two halves, in years, is
action.
b) Show that Eq. (32) for the case e > 1 (E > 0) can be (2/π)[e(1 − e2 )1/2 + arcsin e],
presented in Cartesian coordinates as
and hence for small e about twice as large as the dif-
x2 y2 ference in Problem C.6. (Hint: Use Cartesian co-
− = 1,
a2 b2 ordinates to evaluate the required area. The identity
and hence describes an ellipse. Show that in this case π/2 − arcsin(1 − e2 )1/2 = arcsin e may be useful.) ** For
those who enjoy challenging integrals, it is interesting to
b2 solve the problem, using polar coordinates.
e2 − 1 = .
a2
Problem C.7 What is the trajectory of the comet in the
c) Show that in both cases problem A.3?
k
a= , (34)
2|E| D. Planetary Motion
J
b= √ . (35)
2m|E| For elliptic motion the solution of Eq. (23) is
√ ∫
d) What kind of curve describes Eq. (32) for e = 1 (E = m rdr
t= √
0)? 2|E| −r2 + |E|
kr
− 2m|E|
J2

Using the Kepler’s second law (22) and the formula for √ ∫
ma rdr
the surface of ellipse we obtain for the period of revolu- = √ .
tion: k a e − (r − a)2
2 2

√
S πab m 3/2 After natural substitution
τ= = J = 2π a ;
dS/dt k
2m r = a(1 − e cos ψ) (36)
thus we get Kepler’s third law
the integral takes the form
( τ )2 a3
m √ ∫
= a3 = , ma3 τ
2π k GM t= (1 − e cos ψ)dψ = (ψ − e sin ψ).
k 2π
that is the period of revolution around an elliptical orbit
depends only on the semi-major axis, and the squares of This is traditionally presented as
the revolution periods have the same ratio as the cubes
of their major semi-axes. µ = ψ − e sin ψ, (37)
 8

Q Attempts to solve Eq. (37) were numerous. Lagrange

obtained the solution as infinite series in powers of e
∑∞
en dn−1
ψ =µ+ n−1
sinn µ.
n=1
n! dµ

The series converges only if e < 0.66 . . . . For comets e

P may be too large for the solution to be of any use.

ψ θ Problem D.3 Solving Eq. (37) by iteration check up

explicitly the first three terms of Lagrange solution.

C S A Solution by Bessel starts from noticing that ψ − µ is a

periodic function of µ which vanishes when µ is an integer
multiple of π. Hence we can write
∞
∑
ψ−µ= An sin nµ.
n=1

∑∞
dψ
−1= nAn cos nµ.
dµ n=1
FIG. 5:
∫ π ( )
dψ πm
where µ = 2πt/τ is just the time in the units of the period − 1 cos mµdµ = Am .
0 dµ 2
(times 2π).
From Eqs. (36) and (32) we express the angle θ Since ψ = 0 when µ = 0 and ψ = π when µ = π we can
through ψ as change the variable to give
cos ψ − e ∫ π
cos θ = . (38) πm
1 − e cos ψ Am = cos mµdψ.
2 0

Calculating 1−cos θ and 1+cos θ from Eq. (38) we obtain Using Eq. (37) we get for the Fourier series coefficients
its alternative form ∫ π
√ 2
θ 1+e ψ Am = cos m(ψ − e sin ψ)dψ.
tan = tan . πm 0
2 1−e 2
Introducing Bessel functions
Problem D.1 Find Cartesian coordinates x and y as ∫
functions of ψ. 1 π
Jn (x) = cos(nψ − x sin ψ)dψ,
π 0
Alternatively, we can define ψ geometrically, as it is
shown on Fig. 5 we can present the series for the eccentric anomaly as
∑∞
Area of the circular sector ACQ 2Jn (ne)
ψ = 2π . ψ =µ+ sin nµ
Area of the auxiliary circle n
n=1
In fact, from geometric similarity we obtain
which converges for any value of e.
Area of the elliptic sector ASP
µ = 2π
Area of the ellipse
E. Interplanetary Travels
Area of the circular sector ASQ
= 2π
Area of the circle
Consider a body (planet, asteroid, comet, spacecraft)
S(ACQ) − S(CQS) a ψ − 12 ea2 sin ψ
1 2
on an elliptical orbit around the Sun with the semi-major
= 2π = 2 1 2 .
Area of the circle 2a axis a. As an application of Eq. (34), let us find velocity
of the body in the arbitrary point of the trajectory. If
Thus we reproduce Eq. (37). The angle θ is called in the point is on the distance r from the center, the energy
astronomy the true anomaly, the angle ψ is called the conservation law is
eccentric anomaly, and µ is called the mean anomaly.
mv 2 GMS m GMS m
Problem D.2 Obtain Eqs. (36) and (38) from Fig. 5. − =− .
2 r 2a
 9

Hence
( )
2 1
v 2 = GMS − .
r a

It is convenient to measure parameters of an orbit and ve-

locity in Earth orbit radius RE (considering Earth orbit
as circular) and orbital velocity
√
GMS
vE = = 30 km/s. (39)
RE
We obtain
√
2RE RE
v = vE − . (40)
r a
Problem E.1 * A spacecraft is to travel from the Earth
to Jupiter along an elliptical orbit which just touches each
FIG. 6: Voyager was a two-spacecraft mission to the outer
of the planetary orbits. Find the relative velocity of the planets: Jupiter, Saturn, Uranus, and Neptune. Instruments
spacecraft with respect to the Earth just after launching, aboard each Voyager studied the structure, composition, and
and that relative to Jupiter when it nears that planet. dynamics of the planets’ atmospheres, their magnetic fields
Where in its orbit must Jupiter be at the time of launch, and radiation belts, and the surfaces and atmospheres of their
relative to the earth? Where will the Earth be when it satellites. Voyager 1, launched September 5, 1977, passed
arrives? (In this problem ignore both the gravitational Jupiter in March 1979, Saturn in November 1980, and con-
attraction of the earth and the Jupiter.) tinued on a trajectory that took it out of the Solar System
above the plane of the planets. Voyager 2, launched August
Eq. (40) allows us to find velocity of a spacecraft rel- 20, 1977, reached Jupiter in July, 1979 and Saturn in August
ative to a planet vrel = vs − vP at their encounter. The 1981. Following a gravity assist by massive Saturn, Voyager 2
continued on to explore Uranus in January 1986 and Neptune
velocity of a spacecraft relative to the Sun after the en-
in August 1989. Pluto could not be included in Voyager’s tour
counter is because its orbital position was out of range of the spacecraft’s
path. The alignment of the outer planets that made such a
vs′ = vP + vrel n, (41) ”grand tour” possible will not occur again until the year 2157.
where n is the unit vector which shows the direction of
the velocity of a spacecraft relative to the planet after
(a) On reaching the vicinity of Jupiter, the spacecraft
the encounter.
is swung around the planet by its gravitational attraction.
The following three Problems should be solved after
Consider the encounter in the frame of reference, where
studying Section IV
Jupiter is at rest. What is the magnitude and direction
Problem E.2 A comet starts at infinity, goes around a of the velocity before scattering? What is the magnitude
star of mass M and goes out to infinity. The impact pa- after scattering? If the scattering angle in this frame is
rameter at infinity is b. The radius of closest approach is 90o , what must be the impact parameter? What is the
R. What is the speed of of the comet at closest approach? distance of closest approach, in terms of Jupiter radius?
( MJ = 318ME , Rj = 11RE .)
The problem is solved by writing down two conservation (b) If the maneuver is in the orbital plane, so that the
laws: of angular momentum and of energy final velocity of the spacecraft relative to Jupiter is radi-
ally away from the sun, what is its velocity in magnitude
v∞ b = vR and direction relative to the sun? Use the energy equation
2
v∞ v2 GM to determine the spacecraft’s farthest distance from the
= − .
2 2 R sun during the subsequent flight (its aphelion distance)
in astronomical units. Find also its new orbital period.
After obvious algebra one gets When it returns, what will be its perihelion distance?
√
2GM b2 Problem E.4 ** The orbit of an asteroid extends from
v= .
R(b2 − R2 ) the earth’s to that of Jupiter , just touching both. Suppose
that the asteroid approaches the earth with an impact pa-
Problem E.3 ** A spacecraft is to travel from the Earth rameter 5RE (where RE =earth’s radius), moving in the
to Jupiter along an elliptical orbit which just touches each same direction and overtaking it. Find the angle through
of the planetary orbits. which the asteroid is scattered, in the frame in which the
 10

Multiplying Eq. (42) by r we get

r · p × J − mkr = r · A, (43)

or
A J2
r· +r = .
mk mk
In polar coordinates with the origin in the center we ob-
tain the equation of conic
[ ]
A
r cos(θ − θ0 ) + 1 = l,
mk

where θ0 (the direction to the pericenter) is given by the

direction of A, and eccentricity e = A/mk is given by
the modulus of A.
Problem F.1 * Suppose the earth’s orbital velocity sud-
denly drops by 50%. What will be the parameters of the
new orbit?

FIG. 7: Plot of Voyager 2’s heliocentric velocity against its

distance from the sun, illustrating the use of gravity assist to III. LAGRANGE MECHANICS
accelerate the spacecraft by Jupiter, Saturn and Uranus. To
observe Triton, Voyager 2 passed over Neptune’s north pole
resulting in an acceleration out of the plane of the ecliptic and A. Generalized Coordinates and Lagrange’s
reduced velocity away from the sun Equations

Let us return to Eq. (1).

earth is at rest. What is its new velocity v relative to the
sun? Find the asteroid’s new orbital period. mr̈ = F(r, ṙ, t). (44)

Problem E.5 Look carefully at Fig. 7. To what extent A way to solve it involves projection of this equation on
the numbers you can extract from this Figure agree with the Cartesian axes, thus obtaining the system
what is written in this Section?
mẍi = Fi , (45)

where x1,2,3 are Cartesian coordinates (functions F1,2,3

F. Laplace-Runge-Lentz Vector are some functions of the coordinates and time). We are
now posing a more ambitious aim: to develop a general
Consider an alternative way to solve the Kepler prob- way to reduce Eq. (44) to the system of differential equa-
lem based on the existence of the integral of motion which tions for arbitrary coordinates q1 , q2 , q3 .
is called the Laplace-Runge-Lentz vector. In fact, from Using
the Newton equation
r = r(q1 , q2 , q3 , t), (46)
kr
ṗ = − 3 we introduce new basis vectors
r
we obtain ei = ∂r/∂qi . (47)
[ ]
kmr×(r × ṙ) ṙ r(r · ṙ) Problem A.1 Write down basis vectors for polar coor-
ṗ × J = − = km − .
r3 r r3 dinates system.
Taking into account that J =const, we get Multiplying Eq. (44) by the new basis vectors we obtain
( )
d d mkr ∂r ∂r
(p × J) = . mr̈ =F· ≡ Qi . (48)
dt dt r ∂qi ∂q i

Hence, there exists an integral of motion Simple algebra gives

( ) ( )
mkr ∂r d ∂r d ∂r
p×J− ≡ A = const. (42) mr̈ = mṙ − mṙ . (49)
r ∂qi dt ∂qi dt ∂qi
 11

Differentiating Eq. (46) we obtain In Cartesian coordinates dl2 = dx2 + dy 2 + dz 2 , hence

m
∂r ∂r T = (ẋ2 + ẏ 2 + ż 2 ).
ṙ = q̇i + . 2
∂qi ∂t
In polar coordinates dl2 = dr2 + r2 dθ2 , hence
Hence
1
∂r ∂ ṙ T = m(ṙ2 + r2 θ̇2 ).
= . 2
∂qi ∂ q̇i In cylindrical coordinates dl2 = dz 2 + dρ2 + ρ2 dφ2 , hence
Also, changing the order of differentiation we get 1
T = m(ż 2 + ρ̇2 + ρ2 φ̇2 ).
( ) 2
d ∂r ∂ ṙ
= . In spherical coordinates dl2 = dr2 + r2 dθ2 + r2 sin2 θdφ2 ,
dt ∂qi ∂qi
hence
Thus the lhs of Eq. (48) can be presented as 1
( ( )) ( ) T = m(ṙ2 + r2 θ̇2 + r2 sin2 θφ̇2 ).
2
∂r d ∂ 1 2 ∂ 1 2
mr̈ = mṙ − mṙ Consider particle of mass m and potential energy V (r)
∂qi dt ∂ q˙i 2 ∂qi 2
( ) in 2d. The Lagrangian in Cartesian coordinates is
d ∂T ∂T
= − . m
dt ∂ q˙i ∂qi T = (ẋ2 + ẏ 2 ) − V (x, y).
2
Thus instead of Eq. (45) we obtain Lagrange’s equations are
( ) ∂V ∂V
d ∂T ∂T mẍ + =0 mÿ + = 0.
− = Qi . (50) ∂x ∂y
dt ∂ q˙i ∂qi
Thus we reproduced Newton’s equations.
If the system is conservative, that is there exists potential In polars the Lagrangian takes the form
energy
m
L = (ṙ2 + r2 θ̇2 ) − V (r, θ).
V = V (r, t) (51) 2
Lagrange’s equations are
such that
∂V
∂V (r, t) mr̈ − mrθ̇2 + =0 (55)
F(r, t) = − , (52) ( )
∂r
∂r d ∂V
mr2 θ̇ + = 0. (56)
the rhs of Eq. (50) can be presented as dt ∂θ
Consider the case of central force:
∂V (r, t) ∂r ∂V (q, t)
− · =− , ∂V
∂r ∂q i ∂qi = 0.
∂θ
and we obtain Lagrange’s equations Eq. (56) can be integrated immediately
( )
d ∂L(q̇, q, t) ∂L(q̇, q, t) mr2 θ̇ = J. (57)
− = 0, (53)
dt ∂ q˙i ∂qi Substituting this into Eq. (55) we can integrate it to
where obtain
1 ( 2 )
L = T − V. (54) m ṙ + r2 θ̇2 + V (r) = E. (58)
2
Pay attention that not only the forces, but vectors in gen- Problem A.2 Consider 3D harmonic oscillator, that is
eral disappeared completely from Lagrange’s mechanics. a particle of mass m in the potential field V (r) = kr2 /2.
Both potential energy V , determined as a function of q (a) Find the Lagrangian in Cartesian coordinates. Write
by Eqs. (51), (46), and kinetic energy T are scalar func- down Lagrange’s equations and solve them.
tions; the latter can be presented as *(b) Find the trajectory of the particle.
**(c) Do the same as in (a) and (b) using polar coordi-
m dl2 nates. Formulate Kepler’s laws for harmonic oscillator.
T = ,
2 dt2 Problem A.3 ** During his work outside the Space Sta-
which allows us to write down kinetic energy explicitly tion astronaut threw hammer towards the earth. Describe
in arbitrary coordinate system if we know how to write the motion of the hammer with respect to the Station,
down Pythagoras theorem in this system. taking the velocity of the throw as 10 m/sec.
 12

B. Charged Particle in an Electromagnetic Field ω

Lagrange’s equations in the form of Eq. (53), are valid

not only for conservative systems. In particular, New-
ton’s equation for a particle in EM field

mr̈ = e(E + ṙ × B) (59)

θ
follows from the Lagrangian
1 2
L= mṙ + eṙ · A(r, t) − eϕ(r, t). (60)
2 Z
where the connection between fields and potentials is
FIG. 8:
∂A
E = −∇ϕ − , B = ∇ × A.
∂t
Exactly repeating the derivation of the previous Subsec-
In fact, tion, we obtain Lagrangian of simple pendulum
∂L 1 2 2
= mṙ + eA. L= ml θ̇ + mgl cos θ.
∂ ṙ 2
( ) The Lagrange’s equation is
d ∂L ∂A
= mr̈ + e + e(ṙ · ∇)A.
dt ∂ ṙ ∂t ml2 θ̈ + mgl sin θ = 0.

Problem C.1 A bead slides without friction on a

∂L parabolic wire of shape y = ax2 with the force of grav-
= e∇ (ṙ · A) − e∇ϕ.
∂r ity in the negative y direction. Write down Lagrangian
in terms of x. Find the equation of motion and then
Hence the Lagrange’s equation is
simplify it for the case of small oscillations.
[ ]
∂A
mr̈ = e −∇ϕ − + e [∇ (ṙ · A) − (ṙ · ∇) · A] . The same approach works when the particle has sev-
∂t eral degrees of freedom or/and the constraints are time
dependent. Consider a bead of mass m constrained to
If we take into account that
move without friction on a hoop of radius l rotating with
∇ (ṙ · A) − (ṙ · ∇) · A = ṙ × [∇ × A], constant angular velocity ω around a vertical diameter of
the hoop. The kinetic energy in spherical coordinates
1 ( )
we obtain Eq. (59).
T = m ṙ2 + r2 θ̇2 + r2 sin2 θφ̇2 , (62)
Problem B.1 (a) Write down explicitly Lagrangian 2
(60) in Cartesian coordinates. when the constraint r = l and φ = ωt are taken into
account, is
C. Holonomic Constraints ml2 2 mω 2
T = θ̇ + sin2 θ.
2 2
Consider a simple pendulum. The equation of motion
is Hence, the Lagrangian is

ml2 2
mr̈ = mg + f . (61) L= θ̇ − U (θ), (63)
2
The nature realizes the constraint by applying the force ( 2
)
fi , which we do not know a priori. However, we’ll be able where U (θ) = −mgl cos θ + lω 2g sin2
θ . Equilibrium
to formulate the problem in such a way, that the force points are found from equation
of constraints disappears. To do it, we introduce new [ ]
variable θ. Second, we multiply Eq. (61) by a new basis d lω 2
∂r U (θ) = 0 −→ sin θ 1 − cos θ = 0.
vector e = ∂θ . The force of constraint is perpendicular dθ g
to this direction
This equation always have solutions θ = 0, π. In addi-
f · e = 0. tion, for lω 2 /g > 1, it has solutions θ = cos−1 (g/lω 2 ).
 13

2 It is obvious, that if

1
L(q̇1 , q1 ; q̇2 , q2 ) = L1 (q̇1 , q1 ) + L2 (q̇2 , q2 ),

then the equations for q1 and q2 are decoupled.

0 For the Kepler problem
U/mgl

1 2 Gm1 m2 1 GM µ
−1 L2 = µṙ + = µṙ2 + ,
2 r 2 r
and the period of a satellite motion s determined by sum
−2
of masses rather than by the mass of the heavier body.
E.g. for the earth-moon system m1 /m2 = 1/81.3. If we
−3 were to compute the period of the moon’s orbit from the
−π 0
θ π Kepler’s law by comparing it with the period of small
earth satellite without taking the effect into account we
would get the period 4 hours too long.
FIG. 9: Effective potential energy of a heavy particle on a
rotating ring as a function of θ for lω 2 /g = .5 (solid line) and Problem D.1 A double star is formed of two compo-
lω 2 /g = 4 (dashed line). nents, each with mass equal to that of the sun. The dis-
tance between them is 1 a. u. [Astronomical unit (a.u.)
is the semi-major axis of the earth’s orbit ≈ 1.5×108 km.]
For lω 2 /g < 1 the system has a stable equilibrium θ = 0. What is the orbital period?
For lω 2 /g > 1 this equilibrium becomes unstable, but
there appear two new stable equilibrium positions, cor- The mass determining the motion of the double star is
responding to cos θ = g/lω 2 . twice the mass of the sun. Because the period of motion
is inversely proportional to the square root√of mass, the
Problem C.2 A particle of mass m is constrained to orbital period of the double star is 1 year/ 2.
move in a plane which rotates with constant angular ve-
locity ω. Find and solve the equations of motion of the
particle, including the force of gravity: a) when the axis IV. SCATTERING THEORY
of rotation is vertical; b) when the axis of rotation is hor-
izontal. A. Differential Scattering Cross-Section

D. The Two-Body Problem Consider a beam of particles with the flux (a number
of particles crossing unit area normal to the beam in unit
time) fin impinging on a target. Some of the particles
Consider two particles with the interaction energy
are scattered and we can measure the flux of scattered
V (r1 − r2 ).
particles fout (it will in general depend upon where we
1 1 put the detector, that is upon scattering angle θ. We
L= m1 ṙ21 + m2 ṙ22 − V (r1 − r2 ). define the differential scattering cross-section
2 2
We define the position of center of mass R and the rela- dσ fout L2
(Θ) = ,
tive position r dΩ fin
m 1 r1 + m 2 r2 where L is the distance from the target to the detector.
R= ; r = r1 − r2 . (64)
m1 + m2 It is really the ratio between the element of surface in the
plain perpendicular to the incoming flux to the element of
Thus solid angle that the particles which traversed through the
m2 r m1 r abovementioned element of surface were scattered into.
r1 = R + ; r2 = R − , We can calculate differential scattering cross-section if we
m1 + m2 m1 + m2
know connection between impact parameter and scatter-
and we get ing angle.
1 1 dσ = b|db|dφ; dΩ = sin ΘdΘdφ.
L= M Ṙ2 + µṙ2 − V (r) = L1 + L2 ,
2 2
Hence we get
where
 
m1 m2 dσ b(Θ)  db(Θ) 
(Θ) = . (66)
sin Θ  dΘ 
M = m1 + m2 ; µ = . (65)
m1 + m2 dΩ
 14

FIG. 10: Mapping of dσ to dΩ.

B. Rutherford Scattering
θ
i In 1908 Hans Geiger and Ernest Marsden, working
with Ernest Rutherford of the Physical Laboratories at
b R the University of Manchester, measured the angular dis-
tribution of alpha particles scattered from a thin gold
foil.
From Eq. (32) follows, that the directions in which r
becomes infinite are:

Notice that θ = ± cos−1 (1/e) in the repulsive case

θ = π ± cos−1 (1/e) in the attractive case. (71)
J
b= . (67)
mv∞
In both cases the scattering angle Θ is
Consider scattering of point particles by a hard sphere.
For an elastic scattering Θ = π − 2 cos−1 (1/e). (72)
π−Θ
i= . (68) Eq. (72) can be rewritten in the form
2
Hence e2 − 1 = cot2 (Θ/2). (73)
b = R sin i = R cos(Θ/2), (69)
Finally, taking into account Eq. (33), we obtain connec-
and, hence, tion between impact parameter and scattering angle
dσ 1
(Θ) = R2 . (70) b2 = a2 cot2 (Θ/2), (74)
dΩ 4
Pay attention that and the differential scattering cross-section in the form
∫ ∫
dσ dσ
dΩ = 2π (Θ) sin ΘdΘ = πR2 . dσ a2
dΩ dΩ = 4 . (75)
dΩ 4 sin (Θ/2)
Problem A.1 Consider molecules of radius R1 moving
toward the right with identical velocities scattering from For the scattering of an α-particle on a nucleus with
dust particles of radius R2 that are at rest. Consider both atomic number Z we have a = Zq 2 /E, where q is the
as hard spheres and find the differential cross sections. proton charge.
 15

Problem C.2 Find scattering cross-section for potential

V (r) = −α/r2 (α > 0).

Problem C.3 * Find the differential scattering cross

section in the potential
{ α
r − R for r < R .
α
V (r) =
0 for r > R

Problem C.4 * Show that for repulsive scattering Eq.

(76) for the angle of scattering as a function of the impact
parameter, b, can be rewritten as
∫ 1
ρdρ
Θ(b) = π − 4b √ [ ] ,
2 1−
E − b (1 − ρ )
0 V 2 2 2
rm

by changing the variable of integration to some func-

tion ρ(r). Because of the definite limits of integration,
this formula has advantages for numerical calculations
of Θ(b).
C. Scattering in Arbitrary Potential
Problem C.5 ** Apply the formulation of the preceding
problem to compute numerically Θ(b) and the differential
After introducing the impact parameter b = J/mv∞
cross section dσ/dΩ for the repulsive potential
from Eq. (25) we obtain
  V0
  V =
 ∫ ∞  1+r
 dr 
Θ(b) = π − 2b √ [ ]
,
 (76)
 rm V (r)  and for a total energy E = 1.2V0 . Does the scattering
 r r 1− E −b 
2 2
exhibit a rainbow?

where rm nullifies the denominator of Eq. (76). Equiva-

lently, after making change of variable x = b/r D. Reaction Cross-Section
 
 ∫ xmax  We define the reaction total cross-section as the cross-
 dx 
Θ(b) = π − 2 √ 
 (77) sectional area crossed by all particles participating in a
 0 1 − x2 − V (b/x)  given reaction. The mean free path
E

where xmax nullifies the denominator. λ = 1/nσ,

Problem C.1 * Put V = k/r in Eq. (77), calculate where n is the concentration of scatterers. Consider a
integral and show that thus way you reproduce Rutherford beam of particles with flux f impinging normally on a
formula (75). wall. The attenuation of the beam is described by an
equation
Consider potential V (r) = α/r2 (α > 0). From the
df (x)
Eq. (77) we get = −f (x)/λ.
dx
[ ( ]
α )−1/2 The solution is
Θ(b) = π 1 − 1 + 2 , (78)
b E
f (x) = f e−x/λ ,
or, equivalently,
where λ is the mean free path.
2 α (π − Θ)2
b = . (79) Problem D.1 A beam of particles strikes a wall con-
E Θ(2π − Θ)
taining 2 × 1029 atoms m−3 . Each atom behaves like a
Hence sphere of radius 3 × 10−15 m. Find the thickness of wall
that exactly half the particles will penetrate without scat-
dσ π2 α π−Θ tering. What thickness would be needed to stop all but
= . (80)
dΩ E sin Θ Θ2 (2π − Θ)2 one particle in 106 ?
 16

q1
q*
1 p1 2 θ1
1 θ* 2
θ2
q2 p* -p *
-q *
FIG. 11: Collision as it looks in the LS.

Problem D.2 * A particle of mass m is acted on by an

FIG. 12: Collision as it looks in the CM system; from the
attractive force whose potential is given by U ∝ r−4 . Find
energy conservation law we get |p∗ | = |q ∗ |.
the total cross section for capture of the particle coming
from infinity with an initial velocity v∞ .
Problem D.3 ** A star of mass M and radius R is
moving with velocity v through a cloud of particles of den-
sity ρ. If all the particles which collide with the star are
trapped by it, show that the mass of the star will increase
at a rate
q1 q2
dM
(
2 2GM R
) q*
= πρv R +
dt v2
Given that M = 1031 kg and R = 108 km, find how the
effective cross-section compares with the geometric cross-
θ1 θ* θ2
section πR2 for velocities of 1000 km s−1 , 100 km s−1 m 1 p* p*
and 10 km s−1 . m2
[Hint: Consider the problem in the reference frame
moving with the star. Take into account that the dis-
tance of closest approach rmin for a hyperbolic orbit is
l
p1
rmin = e∓1 .]
FIG. 13: Eqs. (81) summarized in a vector diagram.
E. Elastic Collisions

Consider particle with mass m1 impinging on a par- Eq. (82) looks especially simple for equal masses. In this
ticle at rest with mass m2 . The scattering problem is case we get
most easily solved in the CM system. As it was shown is
Subsection III D, we should consider the scattering of the
particle of reduced mass µ = m1 m2 /(m1 + m2 ) with the Θ1 = Θ∗ /2. (84)
same as in the Laboratory system (LS) initial velocity
v∞ and impact parameter b. After finding scattering an-
gle in the CM system Θ∗ , we can calculate the scattering
Problem E.1 A proton is elastically scattered through
Θ1 and recoil Θ2 angles in the LS. The connection be-
an angle of 56o by a nucleus, which recoils at an angle of
tween the angles is obtained from the connection between
600 . Find the atomic mass of the nucleus and the fraction
the momenta in the CM and LS. Because the LS moves
of the kinetic energy transferred to it.
relative to the CM system with the velocity p∗ /m2 , the
connection between the momenta is
m1 ∗ m1 ∗
p1 = p∗ + p , q1 = q∗ + p , q2 = −q∗ + p∗(81)
.
m2 m2 Problem E.2 * Two identical charged particles, each of
mass m and charge e, are initially far apart. One of
Hence we get
the particles is at rest at the origin, and the other is
sin Θ∗ approaching it with the velocity v along the line y = b,
tan Θ1 = , (82) z = 0, where b = 2e2 /mv 2 . Find the scattering angle in
m1 /m2 + cos Θ∗
1 the CM frame, and the directions in which two particles
Θ2 = (π − Θ∗ ). (83) emerge in the LS.
2
 17

F. CM and LS Cross-sections Consider first rotations about the OZ axis. From simple
geometrical reasoning follows that
Analysis of the scattering problem in the CM system x′1 = x1 cos Φ − x2 sin Φ
gives
x′2 = x1 sin Φ + x2 cos Φ
dσ x′3 = x3 .
(Θ∗ ).
dΩ∗
Considering general case and presenting radius-vector in
To find the LS differential cross-sections we use the rela- the form
tion
r = n(n · r) + [r − n(n · r)], (88)
dΩ∗ sin Θ∗ dΘ∗
= where n is a unit vector in the direction of OZ axis, we
dΩ sin ΘdΘ
obtain that the coordinates of the point after the rotation
Thus for the scattering of the first particle are connected with its coordinates before the rotation by
the Equation
dσ1 sin Θ∗ dΘ∗ dσ
(Θ1 ) = (Θ∗ ), (85) r′ = n(n · r) + [r − n(n · r)] cos Φ + (n × r) sin Φ. (89)
dΩ sin Θ1 dΘ1 dΩ∗
Consider rotation during an infinitesimal time dt. In
where Θ∗ is given by Eq. (82), and for the recoil of the this case the angle of rotation is proportional to dt (with
second particle some coefficient of proportionality ω) and Eq. (89) takes
the form
dσ2 2 sin(π − 2Θ2 ) dσ
(Θ2 ) = (π − 2Θ2 )
dΩ sin Θ2 dΩ∗ ω × r]dt,
dr = [⃗ (90)
dσ
= 4 cos Θ2 ∗ (π − 2Θ2 ) where we have introduced the vector of angular velocity
dΩ
ω
⃗ = nω. Being presented as
Consider Coulomb scattering of two identical parti-
ω × r],
ṙLS = [⃗ (91)
cles, the first initially moving and the second at rest.
In the CM frame of reference the solution is given by Eq. (93) is a manifestation of a general rule, which gives
Eq. (75). Because the particles are identical we should the derivative, in the LS, of any vector constant in the
sum up cross-sections for the scattered particles and the RS. If the vector in addition changes in the RS, the law of
recoiled particles. transformation of an arbitrary vector b can be presented
as
dσ dσ1 dσ2
(Θ) = (Θ) + (Θ). (86)
dΩ dΩ dΩ ω × b].
ḃLS = ḃ + [⃗ (92)

After some algebra we get Returning to the problem of transformation of radius-

( ) vector, let the RS moves in addition translationally with
dσ 1 1 the velocity V relative to the LS. Then according to the
= a2 cos Θ + . (87)
dΩ sin4 Θ cos4 Θ law of addition of velocities we get
⃗ × r,
vLS = V + v + ω (93)
V. ROTATING FRAMES where v is the velocity of a point relative to the LS, and
v is the velocity of the same point relative to the RS.
A. Laboratory System and Rotating System Differentiating Eq. (93) we obtain

⃗ × [⃗
v̇LS = V̇ + v̇ + ω ω × r] + 2⃗ ⃗˙ × r.
ω×v+ω
A rotation is a rigid body movement which keeps a
point fixed. This definition applies to rotations within
both two and three dimensions (in a plane and in space, B. Inertial Forces
respectively.) All rigid body movements are rotations,
translations, or combinations of the two. Euler’s rota-
tion theorem says that a rotation in three-dimensional The second law of Newton mr̈LS = F, in the RS is
space keeps an entire line fixed. From Euler theorem fol- presented as
lows that a space rotation can be specified by the axis of mr̈ = F − mV̇ + mω 2 r⊥ + 2mv × ω ⃗˙ ,
⃗ + mr × ω
rotation, defined by unit vector n = (n1 , n2 , n3 ), and the
angle of rotation Φ. In this section we’ll talk about ro- where r⊥ = r − ω ω · r)/ω 2 is the part of r perpendicular
⃗ (⃗
tations of the frame of reference of the Rotating System ⃗ . The term mω 2 r⊥ is called the centrifugal force; the
to ω
(RS) with respect to inertial Laboratory System (LS). term 2mṙ × ω⃗ is the Coriolis force.
 18

ω By roughly sketching the effective potential energy curve,

show that there are three such positions.
N *** (b) Investigate the stability of the Lagrangian
points considering small deviations of a third body from
one of this points. Given that M1 > M2 , show that the
points are stable, provided that ratio M1 /M2 is slightly
R less than 25.
(There are several examples in the solar system of La-
grangian orbits, including the so-called Trojan asteroids
S that lead and trail Jupiter. The two stable points of the
Earth-Moon system have also been suggested as suitable
locations for future large space habitats.)
FIG. 14: Skyhook
Equilibrium positions in the rotating system are those
for which the sum of all forces acting on the test body is
Problem B.1 * The science fiction writer R. A. Hein- equal to zero.
line describes a ”skyhook” satellite that consists of a long
GM1 (r − r1 ) GM2 (r − r2 )
rope placed in orbit at the equator, aligned along a ra- − − + ω 2 (r − R) = 0, (94)
dius from the center of the earth, and moving so that the |r − r1 |3 |r − r2 |3
rope appears suspended in space above a fixed point on the
where
equator. The bottom of the rope hangs free just above the
surface of the earth (radius R). Assuming that the rope M 1 r1 + M 2 r2
has uniform mass per unit length (and that the rope is R=
M1 + M2
strong enough to resist breaking!), find the length of the
rope. is the COM radius-vector, and the rotation angular ve-
locity is found from the equation
Problem B.2 Find the centrifugal acceleration at the
equator of the planet Jupiter, and of the sun. In each G(M1 + M2 )
= ω 2 a,
case, express your answer also as a fraction of the sur- a2
face gravity. (The rotation periods are 10 hours and 27
days, the radii 7.1 × 104 km and 7.0 × 105 km, and the where a is the distance between M1 and M2 . Substituting
masses 1.9 × 1027 kg and 2.0 × 1030 kg.) in Eq. (94) we obtain
[ ] [ ]
1 1 1 1
Problem B.3 A bird of mass 2 kg is flying at 10m/s in M1 (r − r1 ) 3 − = M2 (r − r2 ) − 3 .
latitude 60o N, heading due east. Find the horizontal and a |r − r1 |3 |r − r2 |3 a
vertical components of the Coriolis force on it.
Non-collinear equilibrium positions correspond to both
Problem B.4 * An orbiting space station may be made expressions in the braces equal to zero.
to rotate to provide an artificial gravity. Given that the
radius is 25 m, find the rotation period required to produce
an apparent gravity equal to 0.7g. A man whose mass is C. Freely Falling Body
75 kg runs around the station in one direction and then
in the other (i.e. on a circle around the inside of the Newton equation for free fall in rotating frame is
cylindrical wall) at 5 m s−1 : find his apparent weight in
each case. What effects will he experience if he climbs a v̇ = g + 2v × ω.
ladder to a ”higher” level, i.e. closer to the axis, climbing
at 1 m s−1 ? The first two terms of expansion of the solution with
respect to ω are
Problem B.5 ** (a) Two bodies of masses M1 and M2
are moving in circular orbits about their center of mass. v = gt + g × ωt2 .
The restricted three-body problem concerns the motion of
Integrating the solution we obtain
a third small body of mass m (≪ M1 or M2 ) in their
gravitational field (e.g. a spacecraft in the vicinity of the gt2 t3
earth-moon system). Considering the forces acting on r=h+ +g×ω . (95)
2 3
the third body in a rotating frame in which M1 and M2
are fixed, show that there are two equilibrium positions in Let us chose our axes so that i is east, j north, and k up.
each of which the three bodies form an equilateral trian- Then
gle (the so called Lagrangian points), and several equilib-
rium positions on the line of of centers of M1 and M2 . ω
⃗ = (0, ω sin θ, ω cos θ),
 19

ω ⃗ = −(e/2m)B we get
If we chose ω

k ( e )2
r̈ = − r + B × (B × r)
j k mr3 2m
θ If we assume that the magnetic field is sufficiently weak
the last term drops out also.
North

Problem D.1 ** A beam of particles of charge q and

velocity v is emitted from a point source, roughly parallel
with a magnetic field B, but with a small angular disper-
sion. Show that the effect of the field is to focus the beam
to a point at a distance z = 2πmv/|q|B from the source.
Calculate the focal distance for electrons of kinetic en-
ergy 500 eV in a magnetic field of 0.01 T. (Charge of
where θ is the colatitude. And hence Eq. (95) in coordi- electron = −1.6 × 10−19 C, mass = 9.1 × 10−31 kg, 1 eV
nate notation is = 1.6 × 10−19 J.)
1 1
x = ωgt3 sin θ, y = 0, z = h − gt2 .
3 2
VI. RIGID BODY: NEWTON MECHANICS
So the deviation of the body freely falling from the height
h is
( )3/2 A. Basic Principles
1 2h
x = ωg sin θ.
3 g Consider arbitrary system of point particles. The
equation of motion for the particle α is
Problem C.1 * Solve the problem of particle falling
from height h above the equator by using an inertial ∑
frame, and verify that the answer agrees with that found mα r̈α = F(e)
α + Fαβ , (96)
β
using rotating frame. [Hint: Use equations of motion in
polar coordinates]. where Fαβ is the force exerted on the particle α by parti-
Problem C.2 * A body is thrown vertically upward to cle β Summing up the equations and taking into account
a height h at a attitude λ in the Northern Hemisphere. that according to the third law of Newton Fαβ = −Fβα ,
Where will it strike the ground? we obtain

Problem C.3 * A projectile is launched due north from Ṗ = F, (97)

a point in colatitude θ at an angle 14 π to the horizontal, ∑
and aimed at a target whose distance is y (small compared where P = mα ṙα is the total angular momentum and
∑ (e)
to R). Show that if no allowance is made for the effects F = Fα is the total force.
of the Coriolis force, the projectile will miss its target by Multiplying Eqs. (96) by rα , summing up and taking
a distance into account that the forces Fαβ and Fβα act along the
( 3 )1/2 ( ) same line, we obtain
2y 1
x=ω cos θ − sin θ .
g 3 J̇ = G, (98)
Evaluate this distance if θ = 60o and y = 40 km. Why is ∑
where J = mrα × ṙα is the total angular momentum
it that the deviation is to the east near the north pole, but ∑ (e)
to the west both at the equator and near the south pole? and G = rα × Fα is the total torque. Eqs. (97) and
(98) hold for any system of particles. For the case of a
rigid body they in principle solve the problem.
D. Larmor Effect

B. Space System and Body System

Consider motion of charged particle in magnetic field.
d2 r k dr The Space System (SS) is the frame of reference which
m 2
= − 3r + e × B
dt r dt does not rotate with respect to the inertial LS; the origin
In the rotating frame of the SS is frozen into some chosen point in the body.
If the rotation is about the fixed pivot we can chose the
r̈ + 2ω × ṙ + ω × (ω × r) = origin of the SS in the pivot; as a result the space system
k e is inertial, and, in addition, the torque of the forces from
− 3 r + (ṙ + ω × r) × B the pivot is zero. If there is no fixed pivot, the space
mr m
 20

system, whichever origin we chose, moves translationaly Problem C.1 *Prove the following relations
together with the chosen point and is, in general, non- ∑
inertial. Newton’s equations of motion in such frame of Tr I = 2 mr2
reference are different from Eq. (96); they take the form ∑ 1
∑ mxi xk = (Tr I) δik − Iik
mr̈α = F(e) Fαβ − mα a, 2
α + (99)
β Choosing a suitable set of axes we can always reduce
tensor of inertia to a diagonal form
where we have added fictitious inertial forces (a is the
acceleration of the frame of reference). However, if the  
I1 0 0
origin is chosen in the center of mass of the body, after I =  0 I2 0  ,
multiplying Eqs. (99) by ri and summing up we again 0 0 I3
obtain Eq. (98). Because forces of gravitation are also
proportional to masses, choosing origin in the center of and Eq. (101) to
mass we also nullify their torque. Also, Eq. (97) can be
rewritten as J1 = I1 ω1 , J2 = I2 ω2 , J3 = I3 ω3 . (104)
R̈CM = F, Such axes are called the principal axes, and the corre-
sponding diagonal elements I1 , I2 and I3 – the principal
where RCM is the radius-vector of the center of mass.
moments of inertia. If I1 = I2 = I3 the top is called
Problem B.1 * Calculate acceleration of a cylinder spherical. If I1 = I2 ̸= I3 the top is called symmetric.
rolling without slipping down an inclined plane making For a body with cylindrical symmetry, choosing axis
angle θ to the horizontal. OZ on the direction of the symmetry axis we obtain the
tensor of inertia in the diagonal form, with I1 = I2 . If
Body System (BS) is the frame of reference with the the body is in addition flat (z = 0 for all points), we get
axes frozen into the body. I3 = 2I1 .

Problem C.2 Find the tensor of inertia

C. Tensor of Inertia (a) of a uniform hollow sphere of mass M and outer and
inner radii a and b with respect to its center;
Consider rotation of the body in the SS frame of ref- (b) * of a homogeneous cylinder of mass M whose height
erence. If the body is rotating with the angular velocity is h and whose radius is R with respect to its center.
ω, the velocity of each point of the body is determined (c) ** of a homogeneous cone of mass M whose height is
by the radius-vector r of this point (see Eq. (93)): h and whose base has a radius R with respect to its apex.
⃗ × r.
vSS = ω For a homogeneous ellipsoid of mass M and semi-major
axes a, b and c, the inertia tensor in the principal axes of
The connection between angular momentum and angular
the ellipsoid is
velocity is
 2 
∑ ∑ 2
J= mr × (⃗
ω × r) = ⃗ − r(r · ω
m[r2 ω ⃗ )], (100) 1  b +c 0 0
.
Iik = M 0 a2 + c 2 0 (105)
5 0 0 2
a +b 2
where the sum is over all particles comprising the body.
This refers also to Eqs. (102) and (103). In tensor nota- CM
The tensor of inertia for the center of mass Iik gives
tion Eq. (100) is
opportunity to calculate tensor of inertia about any other
′
Ji = Iik ωk , (101) point Iik . Taking into account that x′i = xi − Xi , where
r is the radius-vector relative to the center of mass, r′ is
where the radius-vector relative to the new center, and R is the
∑ ( ) radius-vector of the new center relative to the center of
Iik = m r2 δik − xi xk (102) mass and M is the mass of the body, we obtain
∑
is the moment of inertia tensor or simply tensor of inertia ′
Iik = m [(xl − Xl )(xl − Xl )δik − (xi − Xi )(xk − Xk )]
∫ ∫ ∫ 
ρ(y∫2 + z 2 )dV ∫ − ρxydV − ∫ ρxzdV CM
= Iik + M (Xl Xl δik − Xi Xk ). (106)
I =  − ∫ ρxydV ρ(x∫2 + z 2 )dV ∫ − ρyzdV .
− ρxzdV − ρyzdV ρ(x2 + y 2 )dV Problem C.3 Find the principal moments of inertia of
(103) a uniform solid cube of mass m and edge length 2a (a)
(Because typically the distribution of mass is continuous with respect to the mid-point of an edge, and (b) with
we substituted integrals for sums.) respect to a vertex.
 21

D. Rotation about a Fixed Axis Z

Consider a simple particular case – rotation about a
x3 ϕ
fixed axis. In this case there is a singe coordinate – angle
φ of the rotation about the axis. If we introduce the unit ψ
vector determining the direction of the axis n, then the x2
angular velocity
ωi = φ̇ni . θ
Choosing the axis of rotation as z and 3 -axis, from 3
equations (98) we need only its one component: Y
I33 φ̈ = G · n. (107) ϕ x1
The projection of the torque can be presented as
θ ψ

G · n = [r × F] · n = r · [F × n] = rmin F sin θ, X N

where rmin is the distance between the line of action of

FIG. 15: Euler’s angles
the force and the axis of rotation, and θ is the angle
between the direction of the force and the direction of
the axis of rotation. The Eq. (106) has the form
Let us consider physical pendulum. Then
[ CM ]
G = −mgl sin φ, I = ni nk Iik + M (Xl Xl δik − Xi Xk )
[ ]
2
where l is the distance between the axis of rotation and = I CM + M R2 − (n · R)
the center of mass. So we get: = I CM + M (n × R)2 .
I φ̈ = −mgl sin φ.
For small oscillations E. Euler’s Angles
√
I
T = 2π (108) There are many ways to chose coordinates which de-
mgl termine the orientation of the body. We’ll consider the
most often used set – Euler’s angles. Let X, Y, Z are the
Problem D.1 Find, using Eq. (108) the period of small
axes of the SS, and x1 , x2 , x3 are the axes of the BS. We
oscillations for a simple pendulum of length L
get
Problem D.2 A thin rod of length l can rotate in a ver-
tical plane about a horizontal axle passing at a distance ω
⃗ = φ̇eZ + ψ̇e3 + θ̇eN ,
x from the center of the rod. Find the period of small
where eZ , e3 , eN are the unit vectors in the directions
oscillations of the rod. At what distance from the cen-
of Z, x3 and N respectively. Collecting the components
ter must a thin rod be suspended to obtain the maximum
along each axis in body system, we have
frequency? What does the value ωmax of this frequency
equal? ω1 = φ̇ sin θ sin ψ + θ̇ cos ψ
Problem D.3 A uniform plank of mass m is suspended ω2 = φ̇ sin θ cos ψ − θ̇ sin ψ
horizontally by two vertical ropes, one at each end. If ω3 = φ̇ cos θ + ψ̇. (109)
one is severed, find the tension, immediately afterwards,
in the remaining rope. In space system, we have
Problem D.4 * (a) A simple pendulum supported by a ωX = ψ̇ sin θ sin φ + θ̇ cos φ
light rigid rod of length l is released from rest with the rod ωY = −ψ̇ sin θ cos φ + θ̇ sin φ
horizontal. Find the reaction at the pivot as a function
of the angle of inclination. ωZ = φ̇ + ψ̇ cos θ. (110)

Problem D.5 ** A door is constructed of a thin homo- Problem E.1 ** A thin uniform disk of radius a and
geneous slab of material: it has a width of 1m. If the mass m is rotating with a uniform angular velocity ω
door is open through 90◦ , it is found that on release it about a fixed axis passing through its center but inclined
closes itself in 2 s. Assume that the hinges are friction- at an angle α to the axis of symmetry. Find the direc-
less, and show that the line of hinges must make an angle tion and magnitude of the torque which is exerted at the
of approximately 3◦ with the vertical. bearings.
 22

F. Free Motion of a Top (Space System) ω3

For G = 0 from Eq. (98) we obtain J = const. Let us
take the Z-axis of the fixed system of co-ordinates in the

ω1
direction of the constant angular momentum J of the top.
Projecting J at the axes of the BS, we can write down
Eq. (104) as

I1 ω1 = J sin θ sin ψ medallion rotates twice as fast as the wobble rate–two to

I2 ω2 = J sin θ cos ψ one.” (Richard P. Feynman ”Surely You’re Joking Mr.
Feynman!”)
I3 ω3 = J cos θ. (111) Analyze the problem. Do you agree with the observa-
tion or think that the memory failed the great physicist?
Using Eq. (109), we obtain
( ) Problem F.3 * A coin in a horizontal plane is tossed
I1 φ̇ sin θ sin ψ + θ̇ cos ψ = J sin θ sin ψ into the air with angular velocity components ω1 about
( ) a diameter through the coin and ω3 about the principal
I2 φ̇ sin θ cos ψ − θ̇ sin ψ = J sin θ cos ψ axis perpendicular to the coin. If ω3 were equal to zero,
( ) the coin would simply spin around its diameter. For ω3
I3 φ̇ cos θ + ψ̇ = J cos θ. (112) nonzero, the coin will precess. What is the minimal value
of ω1 /ω3 for which the wobble is such that the same face
Eq. (112) can be solved for θ̇, φ̇, ψ̇. The solution for the of the coin is always exposed to an observer looking from
case of a symmetric top I1 = I2 ̸= I3 , above? With a little practice, this is a clever way to ar-
range the outcome of a coin flip!
θ̇ = 0
J
φ̇ = G. Euler’s Equations (Body System)
I1
J(I1 − I3 )
ψ̇ = cos θ, Alternatively we can analyze free motion of a rigid
I1 I3
body in the BS. Applying the law of transformation given
means that by Eq. (92) to Eq. (98), in the BS we obtain
J̇ + ω × J = G. (114)
θ = const
J Projecting this Equation on the axes and using Eq. (104)
φ̇ = = const
I1 we obtain
J
ω3 = cos θ = const. (113) I1 ω̇1 + (I3 − I2 )ω2 ω3 = G1 (115)
I3
I2 ω̇2 + (I1 − I3 )ω1 ω3 = G2 (116)
From Eq. (113) we obtain the picture of the free motion I3 ω̇3 + (I2 − I1 )ω1 ω2 = G3 . (117)
of a symmetric top. It rotates with a constant angular
velocity ω3 about it’s figure axis x3 , which by itself pre- Consider free motion G1 = G2 = G3 = 0 of a symmet-
cesses around the direction of angular momentum, main- ric top (I1 = I2 ) Eq. (117) is solved immediately
taining a constant angle θ to it. The rate of precession φ̇ ω3 = const. (118)
is proportional to ω3 .
After substituting the solution in Eqs. (115) and (116)
Problem F.1 * A thin square plate is thrown into the we obtain linear system
air so that it rotates freely under zero torque. The period
of the rotation of the plate about the symmetry axis of I1 ω̇1 + (I3 − I1 )ω3 ω2 = 0
the plate is 1s. If the symmetry axis makes an angle 45◦ I1 ω̇2 + (I1 − I3 )ω3 ω1 = 0,
with the direction of the angular momentum of the plate,
with the obvious solution
find the period of the precession of the axis about this
direction. ω1 = A sin Ω̃t
ω2 = A cos Ω̃t, (119)
Problem F.2 ”...I was in the cafeteria and some guy,
fooling around, throws a plate in the air. As the plate where Ω̃ = ω3 (I1 − I3 )/I1 . Substituting Eq. (119) into
went up in the air I saw it wobble, and I noticed the Eq. (111), we obtain
red medallion of Cornell on the plate going around. It
was pretty obvious to me that the medallion went faster I1 A sin Ω̃t = J sin θ sin ψ
than the wobbling. ... when the angle is very slight, the I1 A cos Ω̃t = J sin θ cos ψ,
 23

which immediately gives

θ = const (120)
Z x3
ω3 (I1 − I3 )
ψ = t. (121)
I1
CM
Eqs. (118), (120) and (121) reproduce the solution ob-
tained previously using the Space System.
θ
Problem G.1 A Frisbee is thrown into the air in such a
r
way, that it has a definite wobble. The air friction exerts
a frictional torque −c⃗ω on the rotation of the Frisbee.
Mg
(a) * Show that the component of ω ⃗ in the direction of
the symmetry axis decreases exponentially with time.
(b) ** Show that the angle α between the symmetry axis
⃗ 0 and leave in Eqs. (115)-(117) only terms linear with
ω
and the angular velocity vector ω decreases with time if
respect to small deviations of ω
⃗ from (0, 0, Ω). Thus Eq.
the moment of inertia about the symmetry axis is larger
(117) in this approximation is
than the moment about the axes normal to the symmetry
axis, which is the case for a flat-type object. Thus the ω̇3 = 0, (126)
degree of wobble steadily diminishes if there is air friction.
giving ω3 = Ω = const. Hence, Eqs. (115),(116) can be
The Euler equations are
written as
I1 ω̇1 + (I3 − I1 )ω2 ω3 = −cω1 (122)
I1 ω̇1 + (I3 − I2 )Ωω2 = 0 (127)
I1 ω̇2 + (I1 − I3 )ω1 ω3 = −cω2 (123)
I2 ω̇2 + (I1 − I3 )Ωω1 = 0. (128)
I3 ω̇3 = −cω3 . (124)
After we differentiate Eq. (127) with respect to time and
Solving Eq. (124) we obtain
substitute ω̇2 from Eq. (128) we obtain
ω3 = Ae−ct/I3
ω̈1 + Bω1 = 0, (129)
For α we obtain
where
ω3
cos α = √ 2 .
ω1 + ω22 + ω32 (I1 − I3 )(I2 − I3 )Ω2
B= .
I1 I2
To get ω12 + ω22 we multiply Eq. (122) by ω1 , Eq. (123) √
by ω2 and add up the results. Thus we obtain The two √ independent solutions of√Eq. (129) are sin( √ B)
and cos( B) if B > 0, and exp( −B) and exp(− −B)
I1 (ω1 ω̇1 + ω2 ω̇2 ) = −c(ω12 + ω22 ). if B < 0. Hence If I3 > I1 , I2 or I3 < I1 , I2 , then B > 0.
The solution is If I2 > I3 > I1 , then B < 0. So the free rotation of the
asymmetric top about the principal axis with the mini-
ω12 + ω22 = B 2 e−2ct/I1 . mal or maximal moment of inertia is stable, the rotation
about the principal axis with the intermediate moment
Hence of inertia is unstable.
A Problem G.2 A solid rectangular box, of dimensions
cos α = √ .
2
B e −2(1/I1 −1/I3 )t + A2 100mm×60mm×20mm, is spinning freely with angular
velocity 240 r.p.m. Determine the frequency of small os-
Taking into account that I3 > I1 , we see that limt→∞ α =
cillations of the axis if the axis of rotation is (a) the
0.
longest, and (b) the shortest axis.
The general solution of Eqs. (115)-(117) for the free
motion of the asymmetric top is rather complicated. We
notice however, that there are three trivial particular so- H. Effect of Small Force
lutions of these equations - rotation of about one of a
principal axes, say
Consider heavy symmetric top rotating about the fixed
⃗ 0 = (0, 0, Ω).
ω (125) point, at a distance r from the CM. Eq. (98), in this case
takes the form
In this case ω
⃗ remains constant also in the LS. To analyze
the stability of such a solution, we can consider ω close to J̇ = n × rM g, (130)
 24

where n is the unit vector in the direction of the figure Problem H.2 * A gyroscope consisting of a uniform
axis. The solution of Eq. (130) is rather complicated; solid sphere of radius 0.1 m is spinning at 3000 r.p.m.
later we’ll obtain general solution of the problem in the about a horizontal axis. Due to faulty construction, the
framework of Hamilton mechanics. Now we consider the fixed point is not precisely at the center, but 20 µm away
case of small force and in addition assume that the body from it along the axis. Find the time taken for the axis
was initially freely rotating about the figure axis. (This to move through 1o .
case is in fact approximate gyroscope theory.) Let us
make an approximation Problem H.3 * A toy gyroscope has a mass of 100 g and
is made in the form of a uniform disk of radius a = 2
J = I3 ω3 n, cm fastened to a light spindle, the center of the disk being
where n is the unit vector in the direction of the figure 2 cm from the pivot. If the gyroscope is set spinning at
axis. Thus Eq. (130) takes the form a rate of 20 revolutions per second, find the period for
steady horizontal precession.
d
(I3 ω3 n) = n × rM g. (131)
dt
Writing down Eq. (131) as VII. RIGID BODY: LAGRANGE MECHANICS

I3 ω̇3 n + I3 ω3 ṅ = n × rM g, (132) A. Kinetic Energy of a Rigid Body

we see that it’s solution is ω3 = const, and vector n which
rotates about the vertical direction with the angular ve- The position of an arbitrary point in a body is given
locity by equation
M gr rLS = R + r, (135)
φ̇ = . (133)
I3 ω3
The meaning and the range of applicability of this ap- where R is the radius vector of the origin of the BS.
proximation will be clarified in the next Section. Differentiating Eq. (135) and using Eq. (93) we obtain

Problem H.1 A heavy top consists of a uniform, solid dR

vLS = ⃗ × r.
+ω (136)
cone of height 100 mm and base radius 20 mm. It is dt
spinning with its vertex fixed at 7200 r.p.m. Find the
period(s) of the steady precession, assuming the angle be- To characterize the general motion of a rigid body we
tween the cone axis and the vertical to be small. need two vectors: R and angular velocity ω⃗.
Kinetic energy can be written in the form
Let us calculate the momemts of inertia of the cone.
∫ h ∫ az/h ∫ h ∑1 ( )2 ( )2
M a4 dR 1 dR
I3 = 2π dz rdrρr2 = 2π dzz 4 T = m +ω⃗ ×r = M
0 0 πha2 /3 4h4 0 2 dt 2 dt
dR 1 ∑
3M a2 +M · (⃗
ω × rCM ) +
2
ω × r) , (137)
m (⃗
= dt 2
10
∫ h ∫ az/h ∫ 2π
I1 = dz rdr ρ(z 2 + r2 sin2 θ) where rCM is the radius vector of the center of mass.
0 0 0 If the body is rotating about a fixed pivot we take the
[ 2 ]∫ h the origin in the pivot, the first and the second terms in
M a a4 3M (4h2 + a2 )
= 2π 2 2
+ 4
dzz 4 = ; Eq.(137) disappear. Otherwise we take the the origin in
πha /3 2h 8h 0 20
the center of mass and the second term disappears.
The distance between the vertex and the center of gravity The third term is
is
1∑ 1∑
∫ h ∫ az/h Trot = ω × r)2 =
m(⃗ m [ω 2 r2 − (⃗
2
ω · r) ]
1 3h 2 2
r =
πha2 /3
2π dz rdrz =
4
. 1∑
0 0 = m [ωi ωi xk xk − ωi xi ωj xj ]
2
The solution (using Eq. (133)) is 1∑ 1
= m [xk xk δij − xi xj ]ωi ωj = Iik ωi ωk(138)
.
2πI3 ω3 4πa2 ω3 2 2
T = = ≈ .76s. (134)
M gr 5gh Choosing the principal axes as the axes of the body
The period of fast precession is system we obtain Eq. (138) as

2πI1 π(4h2 + a2 ) I1 ω12 I2 ω22 I3 ω32

T = = ≈ .17s. Trot = + + . (139)
I3 ω3 a2 ω 3 2 2 2
 25

θ θ
a a
M M
o

2m m
m
B. The Lagrangian of a Heavy Symmetric Top
Problem A.1 A uniform block of mass m and dimen-
sions a×2a×3a spins about a long diagonal with angular
The Lagrangian of a heavy symmetric top according
velocity ω. Using a coordinate system with origin at the
to Eq. (139) is
center of the block,
a) Find the kinetic energy I1 2 I3
b) Find the angular velocity vector and the angular mo- L= (ω1 + ω22 ) + ω32 − M gr cos θ.
2 2
mentum vector about the origin.
In the present approach instead of solving Eqs. (97) Using Eq. (109) and taking into account that ω12 + ω22 =
and (98), which was done in Newton mechanics we solve φ̇2 sin2 θ + θ̇2 , we obtain
Lagrange equations for the Lagrangian
I1 ( 2 2 ) I (
3
)2
L = T − V. (140) L= φ̇ sin θ + θ̇2 + φ̇ cos θ + ψ̇
2 2
For the rotation about a fixed axis −M gr cos θ. (142)
I
Trot = φ̇2 . (141) Problem B.1 ** A simple gyrocompass consists of a gy-
2 roscope spinning about it’s axis with angular velocity ω3 .
Problem A.2 * Solve Problem B.1 from Section VI in The moment of inertia about this axis is I3 , that about a
the framework of Lagrange mechanics. transverse axis is I1 .The gyroscope is free to rotate in a
Problem A.3 A uniform solid cylindrical drum of mass horizontal plane. If the gyro is placed at the earth’s equa-
M and radius a is free to rotate about a horizontal axis. tor, the angular velocity of of the earth being Ω, show that
(a) Masses m and 2m are joined by a light inextensible the axis of the gyro will oscillate about the north-south di-
cable which runs without slipping over a drum. Using the rection; and for small amplitudes of oscillations find this
angular position of the pulley as generalized co-ordinate, period. Remember that ω3 ≫ Ω is an excellent approxi-
write down the Lagrangian function, and the Lagrange mation.
equation. Find the acceleration of the masses.
(b)* A cable is wound on the drum, and carries on its
free end a mass m. Find the Lagrangian function for the C. Steady Precession of a Heavy Symmetric Top
system if the cable is elastic with elastic potential energy
1 2
2 kx , where x is an extension of the cable. Show that
If coordinates q1 , . . . , qs do not appear in Lagrangian
the motion of the mass m is a uniform acceleration with explicitly, that is if
a superimposed oscillation of angular frequency given by
ω 2 = k(M + 2m)/M m. Find the amplitude of this os- L = L(q̇1 , . . . , q̇n ; qs+1 , . . . , qn ), (143)
cillation if the system is released from rest with the cable
unextended. these coordinates are called ignorable. Steady motion
is characterized by all non-ignorable coordinates being
Problem A.4 * A uniform bar of mass M and length constant:
2l is suspended from one end by a spring of forth con-
stant k. The bar can swing freely only in one vertical q̇i = 0, i = s + 1, . . . , n.
plane, and the spring is constrained to move only in the
vertical direction. Set up the equations of motion in the Lagrange equations
Lagrangian formulation.
( )
d ∂L ∂L
Problem A.5 ** A uniform plank of length 2a is placed − = 0, i = 1, . . . , n
with one end on a smooth horizontal floor and the other dt ∂ q̇i ∂qi
against a smooth vertical wall. Write down the La-
grangian function, using angle θ of inclination to the can be written explicitly as
horizontal as a generalized co-ordinate. Given that the
∑n
∂2L ∑n
∂2L ∂L
plank is initially at rest at an inclination of 60◦ , find the q̈j + q̇j − = 0, i = 1, . . . , n.
angle at which it looses the contact with the wall. j=1
∂ q̇i ∂ q̇ j j=1
∂ q̇ i ∂q j ∂q i
 26

In the second sum for j = 1, . . . , s we get ∂L/∂qj = 0, Problem C.3 * Find the Lagrangian function for a
and for j = s + 1, . . . , n we get q̇j = 0. Hence we have symmetric top whose pivot is free to slide on a smooth
horizontal table, in terms of the generalized co-ordinates
∑s
∂2L ∂L X, Y, φ, θ, ψ, and the principal moments about the center
q̈j − = 0, i = 1, . . . , n. (144)
∂ q̇i ∂ q̇j ∂qi of mass. (Note that Z is related to θ.) Show that the hor-
j=1
izontal motion of the center of mass may be completely
Consider i = 1, . . . , s. In this case ∂L/∂qi = 0, Hence we separated from the rotational motion. Are the preces-
obtain sional angular velocities greater or less than in the case
of a fixed pivot? Show that steady precession at a given
∑s
∂2L
q̈j = 0, i = 1, . . . , s, value of θ can occur for a smaller value of ω3 than in the
j=1
∂ q̇i ∂ q̇j case of a fixed pivot.

which means that The Lagrangian is

q̈i = 0, i = 1, . . . , s, 1( 2 ) I∗ ( )
L = Ẋ + Ẏ 2 + r2 sin2 θ̇2 + 1 φ̇2 sin2 θ + θ̇2
2 2
which means that all the velocities referring to the ignor- I3 ( )2
able coordinates are constant. For i = s + 1, . . . , n from + φ̇ cos θ + ψ̇ − M gr cos θ, (149)
2
Eq. (144) we obtain
where I1∗ is the component of the inertia tensor, calcu-
∂L lated when the origin is chosen in the center of mass.
= 0, i = s + 1, . . . , n,
∂qi When we considered the motion of the top with the fixed
which serves as connections between the appropriate ve- pivot, we were choosing the origin in the fixed pivot and
locities. obtained I1 . One can see that I1∗ < I1 . For steady pre-
Looking at the Lagrangian Eq. (142) we see that the cession we recover Eq. (145) (with the modified I1 )
co-ordinates ψ and φ are ignorable. Hence the steady
I1∗ cos θφ̇2 − I3 ω3 φ̇ + M gr = 0.
precession of a heavy symmetric top is described by the
equation The result for the fast precession
∂L [ ]
= sin θ I1 cos θφ̇2 − I3 ω3 φ̇ + M gr = 0. (145) I3 ω3
∂θ φ̇1 =
I1∗
cos θ
The solutions for the precession rate are
√ shows that the precession rate is largen than in the case
I3 ω3 ± I32 ω32 − 4I1 M gr cos θ of fixed pivot. In the approximation which was used to
φ̇ = . (146)
2I1 cos θ obtain Eq. (148), the rate of the precession does not
change. In the next approximation we obtain Eq. (148)
For the case of fast top: I32 ω32 ≫ I1 M gr cos θ we obtain in the form
fast precession
M gr I1∗ M 2 g 2 r2 cos θ
I3 ω3 φ̇2 = + ,
φ̇1 = , (147) I3 ω3 I33 ω32
I1 cos θ
and slow precession and hence the precession rate is less than for the case of
fixed pivot.
M gr
φ̇2 = . (148) Problem C.4 ** Obtain equation
I3 ω3
Now, in particular, we understand that the approximate I1 cos θφ̇2 − I3 ω3 φ̇ + M gr = 0
gyroscope theory describes steady precession of a fast
top, and find the limits of applicability of this approxi- from Euler’s equations of motion, by imposing the re-
mation. quirement that the motion be a steady precession.

Problem C.1 Return to Problem H.1 from Section VI.

Solve it using exact Equation (146). Compare with VIII. HAMILTON MECHANICS
the solutions based on approximate Equations (147) and
(148). Comment on the results. A. From Velocities to Momenta
Problem C.2 Analyze steady motion of a particle in a
central field, described by Lagrangian The form of the Lagrange’s equations
( )
m 2 d ∂L ∂L
L= (ṙ + r2 θ̇2 ) − V (r). = (150)
2 dt ∂ q̇i ∂qi
 27

suggests the idea to to make a transformation from ve- Hence the Hamiltonian is
locities q̇i to momenta ( )
1 2 p2θ
H= pr + 2 + V (r), (155)
∂L 2m r
pi = . (151)
∂ q̇i
Problem B.1 Show that from Lagrangian
The transformation at least simplifies the lhs of Eqs. m 2
(150); say, first equation becomes L= ṙ − V (r)
2
( )
∂L follows Hamiltonian, which can be written as
ṗ1 = .
∂q1 q2 ,...,q̇1 ,q̇2 ,... 1 ( 2 )
H = px + p2y + p2z + V (x, y, z),
Since we introduced momenta instead ( of ) velocities 2m ( )
∂L
we need to express partial derivative ∂q 1 2
p2φ 2
1
q2 ,...,q̇1 ,q̇2 ,... H = pr + 2 + pz + V (r, φ, z),
through the derivative (. . . )q2 ,...,,p1 ,p2 ,... . This can be 2m r
( )
achieved by a transformation of the Lagrangian, that is 1 p2θ p2φ
2
by introducing instead of the Lagrangian the Hamilton H = pr + 2 + 2 2 + V (r, θ, φ),
function 2m r r sin θ
∑
H(p1 , . . . , pn ; q1 , . . . , qn ) = pi q̇i − L, (152) in Cartesian, cylindrical and spherical systems respec-
i tively.

where in the lhs q̇i should be expressed through pi using Problem B.2 Find the Hamiltonian for the anharmonic
Eq. (151). (Such simultaneous transformation of the oscillator, described by the Lagrangian
(part of) independent variables and of the function is
ẋ2 ω 2 x2
called in mathematics Legendre’s dual transformation.) L= + − αx3 + βxẋ2
For the differential of the Hamiltonian we obtain 2 2
∑ ∑ ∑ ∂L ∑ ∂L Problem B.3 Show that the Hamiltonian function for a
δH = pi δ q̇i + q̇i δpi − δ q̇i − δpi . particle of charge q in an electromagnetic field is
i i i
∂ q̇i i
∂pi
H = (p − qA)2 /2m + qϕ.
Using Eq. (151) we obtain
Problem B.4 Given a Hamiltonian function H(p, q, t),
∑ ∑ ∂L
δH = q̇i δpi − δpi . how does one obtain the corresponding Lagrangian?
i i
∂pi
Problem B.5 Solve the Problem 62 from Section III us-
Hence we obtain instead of n second order differential ing Hamilton’s equations.
equations by Lagrange 2n first order differential equa-
Problem B.6 Find the law of motion of the particle
tions by Hamilton
whose Hamiltonian is
∂H ( 2 )2
q̇i = p2 ω02 x2 p ω02 x2
∂pi H(x, p) = + +λ +
2 2 2 2
∂H
ṗi = − . (153) Hamilton’s equations are
∂qi
[ ( 2 )]
dp p ω 2 x2
= −ω02 x 1 + 2λ + 0
B. Hamiltonians and Hamilton Equations dt 2 2
[ ( 2 2 2
)]
dx p ω x
= p 1 + 2λ + 0 .
If the Lagrangian is dt 2 2
m 2 Because the Hamiltonian is integral of motion we can
L= ẋ − V (x),
2 introduce the constant
the momentum is p = mẋ, and the Hamiltonian is p2 ω 2 x2
+ 0 = Ẽ. (156)
p2 2 2
H= + V (x).
2m Thus the Hamilton’s equations become
In 2D in polars the Lagrangian of a particle in a central dp [ ]
field is = −ω02 x 1 + 2λẼ
dt
m dx [ ]
L = (ṙ2 + r2 θ̇2 ) − V (r). (154) = p 1 + 2λẼ .
2 dt
 28

The solution is
[ ]
x = A sin ω0 (1 + 2λẼ)t + ϕ
g θ R
[ ]
p = Aω0 cos ω0 (1 + 2λẼ)t + ϕ .
m
From Eq. (156) we obtain

ω02 A2 Problem B.10 A small particle of mass m slides with-

= Ẽ. out friction on the inside of a hemi-spherical bowl, of
2
radius R, that has its axis parallel to the gravitational
Problem B.7 * The point of suspension of a simple pen- field g. Use the polar angle θ and the azimuthal angle
dulum of length l and mass m is constrained to move on a φ to describe the location of the particle (which is to be
parabola z = ax2 in the vertical plane. Derive a Hamilto- treated as a point particle).
nian governing the motion of the pendulum and its point a) Write the Lagrangian for the motion.
of suspension. b) Determine formulas for the generalized momenta pθ
and pφ .
Problem B.8 Find the Hamiltonian for the system de- c) Write the Hamiltonian for the motion.
scribed in Problem A.4 Section VII and obtain Hamilton’s d) Develop Hamilton’s equations for the motion.
equations of motion. e). Combine the equations so as to produce one second
order differential equation for θ as a function of time.
Problem B.9 A mass m is suspended by a spring of neg- f ) If θ = θ0 and θ̇ = 0, independent of time, calculate
ligible mass, force constant k, and unstretched length b the velocity (magnitude and direction).
from a point which has a constant upward acceleration g) If at t = 0, θ = θ0 , θ̇ = 0 and φ̇ = 0, calculate the
a0 . Gravity g is acting vertically downward. Find the maximum speed at later times.
Lagrangian function and from it the equation of motion
in the vertical direction. Determine also the Hamilto-
nian function, and write Hamilton’s equations. What is C. Ignorable Coordinates
the period of the motion?
The first advantage of Hamilton’s mechanics is mani-
If x is the extension of the spring, the Lagrangian is fested in treatment of ignorable coordinates. If the co-
ordinates q1 , . . . , qs are ignorable, they do not appear in
( )
m kx2 at2 Hamiltonian either. In this case the latter takes the form
L= (a0 t − ẋ)2 − − mg −x .
2 2 2 H = H(p1 , . . . , pn ; qs+1 , . . . , qn ). (157)
The Lagrange equation of motion Hence for i = 1, . . . , s

m(ẍ − a0 ) = −kx + mg, pi = const = αi .

And we get the problem with n − s degrees of freedom,
can be presented as
described by Hamiltonian
mẍ = −kx + m(g + a0 ). H = H(α1 , . . . , αs , ps+1 , . . . , pn ; qs+1 , . . . , qn ).

Obviously, the period of motion is For the Hamiltonian (155) the coordinate θ is ignor-
able, pθ = const = J, and the problem is reduced to a
√
m one dimensional one, with the Hamiltonian
T = 2π .
k p2r J2
H= + + V (r). (158)
The Hamiltonian is 2m 2mr2
( ) The Hamiltonian (158) is by itself integral of motion (see
p2 kx2 at2 Eqs (170) and (278)):
H= + pa0 t + + mg −x .
2m 2 2
p2r J2
+ + V (r) = E. (159)
The Hamilton equations of motion are 2m 2mr2
p From Hamilton’s Equation
ẋ = + a0 t
m ∂H pr
ṗ = −kx + g. ṙ = =
∂pr m
 29

we obtain explicit connection between ṙ and pθ and fi- 500

nally obtain Eq. (159) in the well-known form
400
mṙ2 J2
+ + V (r) = E. 300

U/Mgr
2 2mr2
Problem C.1 A simple pendulum of mass m and length 200
l hangs from a trolley of mass M running on smooth
horizontal rails. The pendulum swings in a plane paral- 100
lel to the rails. Using the position x of the trolley and
the angle of inclination θ of the pendulum as generalized 0
0 1 2 3
co-ordinates, find the Hamiltonian and show that x is ig- θ
norable.
FIG. 16: Effective potential for p2φ /2M grI1 = 100,
p2ψ /2M grI1 = 1.
D. Heavy Symmetric Top: General Solution

pψ = I3 ω3 ) are integrals of motion. In fact, the existence

The transition from the Lagrangian to Hamilton can be
of these integrals of motion can be inferred directly from
presented elegantly using matrix. For a heavy symmetric
Eq. (130), which shows that the projection of the torque
top the Lagrangian in this language is language
of the force of gravity both on the axis OZ, and on the
1 axis x3 is equal to zero.
L= Kij q̇i q̇j − V (q1 , q2 , q3 ), Thus Hamiltonian (161) describes a one-dimensional
2
problem: a particle in a potential well. The shape of the
where (q1 , q2 , q3 ) = (θ, φ, ψ), V (q1 , q2 , q3 ) = M gr cos θ well is determined by the integrals of motion pφ and pψ ,
and and, ignoring the θ-independent term is
 
I1 0 0 (pφ − pψ cos θ)2
K =  0 I1 sin2 θ + I3 cos2 θ I3 cos θ  U (θ) = M gr cos θ +
2I1 sin2 θ
. (162)
0 I3 cos θ I3
Ruler was the first to find the solution of the problem of
The momenta are heavy symmetric top in terms of elliptic functions.
∂L After finding θ(t) we can find φ(t) and ψ(t) by inte-
pi = = Kij q̇j . (160) grating Hamilton’s Equations
∂qi
∂H pφ − pψ cos θ
In matrix notation Eq. (160) is φ̇ =
=
∂pφ I1 sin2 θ
p = K q̇ ⇐⇒ q̇ = K −1 p, ∂H pφ − pψ cos θ pψ
ψ̇ = =− 2 cos θ + . (163)
∂pψ I1 sin θ I3
and the Hamiltonian is
1 Looking at Eq. (163) we recognize the difference between
H = pT q̇ − L = pT K −1 p − pT K −1 KK −1 p + V (q) velocity and momentum. Thus, in spite of the fact that
2 pφ and pψ are integrals of motion in the present problem,
1 T −1
= p K p + V (q), the velocities φ̇ and ψ̇ do change with time (except for
2
the case of steady procession).
where
 1

I1 0 0 E. Stability of a Vertical Top
 − I1cos 
K −1 =  0 I1 sin2 θ
1 θ
sin2 θ ,
2
0 − I1cos θ
sin2 θ
1 cos θ
I3 + I1 sin2 θ
Consider a heavy symmetric top which initially stands
vertically, that is θ(0) = 0 and θ̇(0) = 0. The question
or explicitly, substituting p = (pθ , pφ , pψ )T , is what will happen with this top further? From the
Hamiltonian (161) it is clear that pφ = pψ = I3 ω3 and
p2θ (pφ − pψ cos θ)2 p2ψ the potential energy has the form
H= + + + M gr cos θ. (161)
2I1 2I1 sin2 θ 2I3 I 2 ω2 θ
U (θ) = M gr cos θ + 3 3 tan2
The co-ordinates ψ and φ are ignorable. Hence pφ (the 2I1 2
( )
projection of angular momentum on the axis OZ), and 2 θ
= M gr cos θ + x tan , (164)
pψ (the projection of angular momentum on the axis x3 ; 2
 30

3 Thus the effective potential coincides with that given by

Eq. (164)
2 ( )
θ
U (θ) = M gr cos θ + x tan2
U/(Mgr)
,
1 2
where x = I32 ω32 /(2M grI1∗ ). Hence the boundary be-
0 tween stable and unstable equilibrium I32 ω32 = 4M grI1∗
corresponds to smaller ω3 .
−1
0 1 2 3
θ F. Hamiltonian and Energy
FIG. 17: U (θ) of Eq. (164) as a function of θ for x = 2.1
(solid line) and x = .1 (dashed line).
Consider Lagrangian
1∑
L=T −V = mα ṙ2α − V (r, t). (167)
where x = I32 ω32 /(2M grI1 ). Expanding effective poten- 2 α
tial energy near θ = 0
[ ] Introducing new variables
θ2 θ2
U (θ) ≈ M gr 1 − +x , (165) rα = rα (q1 , q2 , . . . qn , t), (168)
2 4
we obtain
we see, that the the top with θ = 0 is in stable equilibrium ( )
for x > 2 and in unstable equilibrium for x < 2. In the 1∑ ∑ ∂rα ∂rα
latter case appears a new stable equilibrium which can T = mα q̇j q̇k
2 α
∂qj ∂qk
be found from the equation jk
( ) ( )2
[ ] ∑ ∑ ∂rα ∂rα 1∑ ∂rα
′ x + mα q̇j + mα
U (θ) = M gr sin θ −1 + = 0. (166) j α
∂t ∂qj 2 α ∂t
2 cos4 θ2
= T2 + T1 + T0 .
Problem E.1 * A pencil is set spinning in an upright In matrix notation
position. How fast must the spin be in order that the
pencil will remain in an upright position? Assume that 1 T
L= q̇ K q̇ + NT q̇ + T0 − V (q).
the pencil is a uniform cylinder of length a and diameter 2
b. Find the value of the spin in revolutions per second Hence
for a = 20 cm and b = 1 cm.
p = K q̇ + N ⇐⇒ q̇ = K −1 (p − N),
Problem E.2 ** To prove that the effective potential
function U (θ) of the symmetric top has only a single min- and
imum, show that the equation U (θ) = E can be written 1
as a cubic equation f (z) = 0 in the variable z = cos θ, H = pT q̇ − L = (p − N)T K −1 (p − N) − T0 + V (q).
2
with three roots in general. Show, however, that f (z) has
Hence
the same sign at both z = ±1, and hence that there are
either two roots or none between these points: foe every H = T2 − T0 + V.
E there are at most two values of θ for which U (θ) = E.
In a natural system time does not appear among the
Problem E.3 ** Obtain the Hamiltonian function for arguments in Eq. (168). In such system T = T2 , and
the pivot with freely sliding pivot, described in Problem Hamiltonian is just the energy of the system.
C.3 Section VII. Find whether the minimum angular Differentiating Lagrangian L(q̇, q, t) with respect to
velocity required for stable vertical rotation is greater or time, we obtain
less than in the case of a fixed pivot. Can you explain ( )
this result physically? dL ∑ ∂L ∂L ∂L
= q̇j + q̈j + . (169)
dt j
∂qj ∂ q̇j ∂t
From the Lagrangian (149) we obtain
Using Euler-Lagrange equations, we can rewrite Eq.
p2 + p2Y p2θ (169) as
H= X +
2M 2(I1∗ + M R2 sin2 θ)  
p2ψ dL d ∑ ∂L  ∂L
(pφ − pψ cos θ)2 = q̇j + .
+ + + M gr cos θ. dt dt ∂ q̇j ∂t
2I1∗ sin2 θ 2I3 j
 31

Hence Problem A.1 Prove that inside a homogeneous sphere

of radius a
dH ∂L
=− . GM R
dt ∂t g(R) = −
( a3)
If L does not depend explicitly on t we obtain a conser- 3 R2
vation theorem Φ(R) = −GM − 3 .
2a 2a
H = const. (170) Problem A.2 The density of a spherical planet of radius
a with a molten core of radius 12 a is given by ρ for 12 a <
For a natural conservative system this is just the energy
r < a and 5ρ for r < 21 a, where ρ is a constant. Find:
conservation law:
a) the force per unit mass inside or outside the planet,
b) the gravitational potential for any distance from the
T + V = E = const.
planet’s center.

Problem A.3 * Find the r dependence of the mass den-

IX. GRAVITATIONAL FIELDS sity ρ(r) of a planet for which the gravitational force has
constant magnitude throughout its interior.
A. Gravitational Potential
Problem A.4 If a narrow tunnel were dug through the
earth along a diameter, show that the motion of a particle
Generalizing Eq. (15) for the case of gravitation field in the tunnel would be simple harmonic. Compare the
created by several point masses and introducing gravita- period to the orbital period of a satellite in a circular
tion potential orbit close to the earth. Assume that the density of the
earth is uniform and neglect the earth’s rotation. Show
V (R)
Φ(R) = , that all the results remain valid for a straight tunnel dug
m between any two points on the earth surface.
we obtain
Problem A.5 ** A diffuse spherical cloud of gas of den-
∑ Mj sity ρ is initially at rest, and starts to collapse under its
Φ(R) = −G . own gravitational attraction. Find the radial velocity of a
j
|R − rj |
particle which starts at a distance a from the center when
it reaches the distance r. Hence, neglecting other forces,
Free fall acceleration in such field is show that every particle will reach the center at the same
instant, and that the time taken is (3π/32ρG)1/2 . Eval-
g(R) = −∇Φ(R) uate this time in years if ρ = 10−19 kg/m3 . (Assume that
particles do not overtake those that start nearer the cen-
If potential is created by a continuous mass distribu- ter. The substitution r = a sin2 θ may be used to perform
tion gravitation potential can be found from the equation the integration.)
∫
ρ(r) 3
Φ(R) = −G d r. (171)
|R − r| B. The Tides

Alternatively, we can solve field equation in the differen-

The potential at the earth surface is
tial form
Φ(ρ, θ) = ΦE + ΦM + ΦCF ,
∇2 Φ = 4πGρ,

or in the integral form GmE

ΦE = − ,
∫ ρ
n · gdS = −4πGM,
S GmM
ΦM = − √ 2
where M is the total mass inside the surface S. RM + ρ2 − 2rρ cos θ
[ ]
Thus the field outside any spherically symmetric mass 1 ρ ρ2 ( )
distribution is the same as if all the mass was concen- = −GmM + 2 cos θ + 3 3 cos 2
θ − 1 ,
RM RM 2RM
trated in the center
GM R ω2 2
g=− . ΦCF = − (x + ρ2 − 2xρ cos θ),
R3 2
 32

ρ a
η
θ
r
E E b
x M
RM

C. Multipole Expansion of Potential

where

G(mE + MM ) We have
ω2 = 3
RM 1 1
= √ .
RM mM |R − r| R 1 − 2 R·r r2
x= R2 + R2
mE + mM
Hence expanding with respect to inverse powers of 1/R
Because the surface of the water should be equipotential we obtain
we get
1 1 R · r 3(R · r)2 − R2 r2
= + + + ....
Φ(a + h, θ) = const, |R − r| R R3 2R5
Hence
where a is the radius of the Earth. Expanding and ne- GM
glecting terms quadratic and higher in h and terms of Φ(R) = Φ0 (R) + Φ1 (R) + Φ2 (R) + · · · = −
R
order [(mE + mM )/mE ](a/r)3 we obtain ∫
GM R · r 3(R · r)2 − R2 r2 3
− − G ρ(r) d r + . . . .(173)
mE h mM a2 ( ) R3 2R5
2
− 3 3 cos2 θ − 1 = const.
a 2RM If we chose origin in the center of mass, then Φ1 (R) = 0.
Taking into account that
Hence
(R · r)2 = Xi Xj xi xj ,
3mM a4 cos2 θ and the results of Problem (C.1) from Section VI, the
h(θ) = 3 + h0 . (172)
2mE RM term Φ2 can be presented as
( )
G 3Xi Iij Xj G
We have mM /mE = 1/81.3 and a/RM = 1/60.3. Thus Φ2 (r) = − T r I .= (3IR − T r I),
using a = 6370 km we find ∆h = 0.36 m. 2R3 R2 2R3
Eq. (172) is equally applicable for the tide induced by where IR is the moment of inertia about the direction of
the Sun, provided we take appropriate mass and distance R (Eq. (??)).
from the earth. For the sun mS /mE = 3.33 × 105 and Let the directional cosines of the radius-vector R in
a/rS = 4.26 × 10−5 . The sun tide ∆h = 0.16 m. the system of the principal axes of the body be α, β, γ.
Then
Problem B.1 * The distance between the earth and the IR = I1 cos2 α + I2 cos2 β + I3 cos2 γ.
moon is gradually increasing (because of tidal friction).
Estimate the height of the tides when the moon was 10 If the gravitating body is a symmetric top,
earth radii away.
IR = I1 + (I3 − I1 ) cos2 γ.

Problem B.2 ** Two small identical uniform spheres of Hence

density ρ and radius r are orbiting the earth in a circular 3IR − T r I = 3I1 + 3(I3 − I1 ) cos2 γ − 2I1 − I3
orbit of radius b. Given that the spheres are just touching,
with their centers in line with earth’s center, and that the = (I3 − I1 )(3 cos2 γ − 1).
only force between them is gravitational, show that they Finally we get
will be pulled apart by the earth’s tidal force if b is less
than bc = 2(ρe /ρ)1/3 a, where ρE is the mean density 3(R · k)2 − R2
Φ2 (R) = G(I3 − I1 ) ,
of the earth. (This is an illustration of the existence of 2R5
the Roche limit,within which small planetoids will be torn where k is a unit vector in the direction of the symmetry
apart by tidal forces.) axis.
 33

z where mS is the mass of the Sun. The equation for the

steady precession is

N θ ∂L
= 0,
∂θ
where the Lagrangian is
k
I1 ( 2 2 ) I (
3
)2
y L= φ̇ sin θ + θ̇2 + φ̇ cos θ + ψ̇ − V (θ).
λ 2 2
Differentiating and ignoring the term with φ̇2 for slow
x precession we obtain

FIG. 18: Diagram of figure axis of the earth relative to orbit 1 ∂V 3GmS
φ̇ = − = − 3 , ϵI cos θ. (174)
of mass point. I3 ω3 sin θ ∂θ 2RS ω3
where RS is the distance between the Sun and the Earth.
Problem C.1 ** Find the gravitational potential at It is convenient to present Eq. (174) in the form
large distances of a thin circular loop of radius a and
Tprec 2T0
mass m, up to terms of order R−3 . =− , (175)
T0 3T3 ϵI cos θ
The earth is a spheroid (ellipsoid of revolution) with where T0 is one year
semi-axes, a, a, c and very small oblateness ϵ:
( )2
2π GmS
a ≈ c ≈ 6400km T0
=
RS3
, (176)
a−c 1
ϵ= ≈ .
a 300 T3 is one day, and θ = 23◦ 27′ . Thus for the period of
precession we obtain
If we consider the earth as homogeneous we can use the
results of Eq. (105) and get (taking into account that 2 · 365.25
ϵ ≪ 1) Tprec = = 81000 [years]. (177)
3 · 0.00327 · cos 23◦ 27′
2 If we take into account also the influence of the moon
I1 ≈ I3 ≈ M a2 we should divide the result of Eq. (177) by 1 +
5
I3 − I1 mM RS3 /mS RM3
= 3.2, where RM is the distance between
ϵI ≡ ≈ ϵ, the Moon and the Earth. Taking mM /mS = 3.7 × 10−8
I3
and RS /RM = 3.9 × 102 we obtain
In reality the earth is strongly nonhomogeneous and
Tprec = 26000 years, (178)
1
I1 ≈ I3 ≈ M a 2 which is observed by astronomers.
3 The radius vector of the satellite is determined by the
ϵI ≈ .00327.
angle between the normal to the orbit plane and the earth
axis θ and the azimuthal angle describing the position of
the satellite at the orbit λ. In this case averaging gives
D. Precession of Equinoxes and Satellite Orbits
an opportunity to consider satellite as a spinning ring,
V = ms Φ, and I3 = ms r2 and ω3 refer in this case to the
Consider motion of the Sun around the earth; the fig- ring (ms is the mass of the satellite). Consider a satellite
ure axis of the earth is in the xz plane. For the potential in a circular orbit of radius r in a plane inclined to the
we obtain equator at an angle θ. We get for the precession rate
G(I3 − I1 ) φ̇ 3 I3 − I1
Φ2 (R) = [3 sin2 θ cos2 λ − 1], =− cos θ, (179)
2R3 ω3 2 M R2
where λ is ecliptic longitude. Averaging with respect to where ω32 = GM
a3 , and M is the mass of the earth. We get
time, for a circular orbit we obtain
[ ] φ̇ 1 ( a )2
G(I3 − I1 ) 3 ≈ − ϵI cos θ. (180)
Φ̄2 (θ) = sin2
θ − 1 . ω3 2 R
2R3 2
A plane of a close satellite orbit with the inclination to
the equator, say 30o , precesses completely around 2π in
V = mS Φ̄2 (θ), about 700 orbits of the satellite.
 34

E. The Shape of the Earth is minimal (or maximal). If y(x) is such a function, then
for any function η(x), such that η(a) = η(b) = 0,
To find the equilibrium shape of the earth we should
d
equilibrate the potentials at the pole and at the I[y + ϵη]|ϵ=0 = 0. (187)
equator, including the centrifugal potential: Φcent = dϵ
− 12 ω 2 r2 sin2 θ. In other words
GM GM a J2 2
GM GM J2 1 ∫ b [ ]
− =− − − ω 2 a2 , (181) ∂L ∂L
c
+
c3 a 2a 2 η + ′ η ′ dx = 0. (188)
a ∂y ∂y
where
Integrating by parts we get
I3 − I1 ∫ x2 [ ( )]
J2 = . (182) ∂L d ∂L
M a2 − ηdx = 0. (189)
Keeping only the terms up to the fist order with respect x1 ∂y dx ∂y ′
to ϵ we get
Hence we obtain Euler-Lagrange equation
2
ω a ( )
= 2ϵ − 3J2 , (183) ∂L d ∂L
g0 − = 0. (190)
∂y dx ∂y ′
where g0 = GM/a2 . For constant density (J2 = 52 ϵ) this
gives ϵ = 1/230. In two particular cases we cam immediately obtain the
first integral of the Euler-Lagrange equation:
Problem E.1 The rotation period of Jupiter is approx- 1. f does not depend on y.
imately 10 hours. Its mass and radius are 318 ME and In this case we get
11.0RE respectively (E=earth). Calculate approximately
its oblateness, neglecting the variation of density. (The ∂L
= C. (191)
observe value is about 1/15.) ∂y ′

If from Eq. (191) we can explicitly find

X. VARIATIONAL CALCULUS AND
VARIATIONAL PRINCIPLES y ′ = Φ(x), (192)

then it can be solved in quadratures

A. Euler-Lagrange Equation
∫
y = Φ(x)dx. (193)
Variational calculus is a branch of mathematics that
is a sort of generalization of calculus. It seeks to find
the path, curve, surface, etc., for which a given function 2. L does not depend on x
has a stationary value (which, in physical problems, is In this case we get
usually a minimum or maximum). ( )
∂L d ∂L
Consider a problem of finding a curve on the plane, y′ − y′ = 0. (194)
which connects two points (x1 , y1 ) and (x2 , y2 ) and has ∂y dx ∂y ′
a minimal length. We should minimize
∫ x2 √ ∂L
l= dx2 + dy 2 . (184) y′ − L = C. (195)
x1
∂y ′

In other words among all functions y(x), such that If from Eq. (195) we can explicitly find
y(x1 ) = y1 , y(x2 ) = y2 we should find the function, for
which integral y ′ = Φ(y), (196)
∫ x2 √
then the equation can be solved in quadratures
l= 1 + y ′2 dx (185)
x1 ∫
dy
x= . (197)
is minimal. Φ(y)
Variational calculus deals with the problem of finding
function y(x), with given values y(a) and y(b), such, that The problem of geodesic on a plain corresponds to ei-
functional ther one of the particular cases, mentioned above. We
∫ b get
I= L(y ′ , y, x)dx = I[y(x)] (186)
a
y ′ = C =⇒ y = Cx + b. (198)
 35

So the geodesic between the points (x1 , y1 ) and (x2 , y2 ) The geodesic is a great circle, that is the intersection of
is a sphere and a plain passing through the center of the
sphere.
y2 − y1 y1 x2 − x1 y2
y= x+ (199) If we were looking for φ as a function of θ, we would
x2 − x1 x2 − x1 get
To find geodesic on a sphere ∫ √
l= 1 + sin2 θφ′2 dθ.
r = 1, (200)

we should minimize From Eq. (191) we obtain

∫ √
dθ2 + sin2 θdφ2 . sin2 θφ′
l= (201) √ = C,
1 + sin2 θφ′2
Let us look for θ as a function of φ. We get
which exactly coincide with Eq. (204).
∫ φ2 √
l= θ′2 + sin2 θdφ. (202) Problem A.2 (a) Find the geodesic on a plane using
φ1
polar coordinates.
From Eq. (195) we obtain (b)* Show that the geodesic on the surface of a right cir-
cular cylinder is a segment of a helix.
θ′2 √ (c)* Find the shortest path between the (x, y, z) points
√ − θ′2 + sin2 θ = C, (203)
θ′2 + sin2 θ √ −1, 0) and (0, 1, 0) on the conical surface z = 1 −
(0,
x2 + y 2 . What is the length of the path? Note: this is
or the shortest mountain path around a volcano.
1
θ′2 = sin4 θ − sin2 θ. (204) Problem A.3 ** Write out and solve the Euler equa-
C2 tions to make the following integrals stationary. Change
Making the change of variable w = cot θ, we get the independent variable, if needed, to make the Euler
equation simpler.
1 ∫ x2 √
w′2 + w2 = − 1 = A2 . (205)
C2 I1 = 1 + y 2 y ′2 dx
x1
The solution is obvious ∫ x2
I2 = (y 2 + y ′2 )dx
cot θ = A cos(φ − α). (206) x1
∫ x2
yy ′2
If the geodesic connects two points θ1 , φ1 and θ2 , φ2 , the I3 = dx
x1 1 + yy ′
coefficients A and α are found from the system ∫ √
t2
y 2 + y ′2
cot θ1 = A cos(φ1 − α) I4 = dx.
t1 y
cot θ2 = A cos(φ2 − α). (207)

Problem A.1 ** Two cities - Tel-Aviv, Israel and San- B. The Brachistochrone Problem
Diego, CA - have the same latitude 32o N, but, different
longitudes: Tel-Aviv is 34o E and San-Diego is 117o W.
Find the shape of the curve down which a bead sliding
Do you think that the shortest path between them is along
from rest and accelerated by gravity will slip (without
the 32o latitude line? If not, what is the maximal distance
friction ) from one point to another in the least time. The
between the path and the latitude line?
term derives from the Greek (brachistos) ”the shortest”
Eqs. (200) and (206) can be presented as and (chronos) ”time, delay.” We should make stationary
the integral
{
r=1 ∫ √
cos θ = A cos α sin θ cos φ + A sin α sin θ sin φ
.(208) 1 1 + y ′2
T =√ √ dx
2g y
Multiplying the second equation by r, and rewriting it in
Cartesian coordinates, we obtain Eq. (208) in the form (we have chosen y = 0 at the initial point, and downward
{ direction of the axis y). Eq. (195) gives
r=1
. (209)
z = ax + by y(1 + y ′2 ) = C.
 36

C. Variational Problems with Constraints

Consider the problem of finding maximum (or mini-

mum) of the function
The solution is y = f (x1 , x2 , . . . xn )
∫ ∫ √
dy dy y
x= √ = √ . under constraints
C
−1 C −y
y
gi (x1 , x2 , . . . xn ) = 0; i = 1, . . . , m. (212)
After substitution
For example, think about finding minimum of function
φ C
y = C sin2 = (1 − cos φ). (210) y = x1 + x2
2 2
we obtain under constraint
∫ x21 + x22 = 1.
dφ sin2 φ2 cos φ2 C
x=C = (φ − sin φ) + B. (211)
cos φ2 2
To solve the problem we introduce the function
Eqs. (210) and (211) present the parametric representa- h(x1 , . . . xn ; λ1 , . . . , λm ) = f − λ1 g1 − · · · − λm gm ;
tion of the cycloid. The cycloid is the locus of a point
on the rim of a circle of radius a rolling along a straight the indefinite parameters λ1 , . . . , λm are called the La-
line. It was studied and named by Galileo. grange multipliers. The theorem says that the critical
The brachistochrone problem was one of the earliest points of function y under conditions (212) are the un-
problems posed in the calculus of variations. Newton was conditional critical points of the function h. That is they
challenged to solve the problem in 1696, and did so the are the solutions of the system
very next day. In fact, the solution, which is a segment of
a cycloid, was found by Leibniz, L’Hospital, Newton, and ∂f ∂g1 ∂gm
− λ1 − · · · − λm = 0; j = 1, . . . , n
the two Bernoullis. Johann Bernoulli solved the problem ∂xj ∂xj ∂xj
using the analogous one of considering the path of light gi = 0; i = 1, . . . , m.
(213)
refracted by transparent layers of varying density. Ac-
tually, Johann Bernoulli had originally found an incor- The schematic proof is as follows. If there were no
rect proof that the curve is a cycloid, and challenged his constraints, we would have find the critical point from
brother Jakob to find the required curve. When Jakob the equation
correctly did so, Johann tried to substitute the proof for
his own. dy = 0.
Consider a case of two points at the same height and
at a distance l. In this case, choosing the origin in the It means that the linear functional dy, defined on a
initial point we obtain from Eq. (210) φ ∈ [0, 2π], and tangent space with the coordinates (dx1 , dx2 , . . . dxn ) is
from Eq. (211) C = l/π, B = 0. equal to zero.
When constraints are taken into account, we demand
Problem B.1 Show that the time of motion in this case dy = 0 on the subspace of the initial tangent space, for
is which
√( ) ( )2
∫ 2π dx
2
dy √ dgi = 0; i = 1, . . . , m.
+ dφ
1 dφ 2πl
T =√ √ dφ = . That we are interested in the subspace annihilated by
2g 0 y g
linear functionals dg1 , dg2 , . . . dgm . Any other liner func-
tional which is zero on the subspace is a linear combina-
Problem B.2 ** Find the Euler Lagrange equation de-
tion of the above mentioned functionals:
scribing the brachistochrone curve for a particle moving
inside a spherical Earth of uniform mass density. Obtain
dy = λ1 dg1 + λ2 dg2 + · · · + λm dgm .
an expression for the time of travel along the brachis-
tochrone between two points on the surface of the Earth. Consider now the problem of finding the stationary of
How long would it take to go from New York to Los An- the functional
geles (assumed to be 3000 miles apart) along a brachis-
∫ b
tochrone tunnel (assuming no friction) and how far below
the surface would the deepest point of the tunnel be? F [y(x)] = L(y ′ , y, x)dx
a
 37

subject to a constraint
∫ b
G[y(x)] = g(y ′ , y, x)dx = C = const. (214)
a x
It is equivalent to finding the stationary value of the func- R
tional
∫ b ωt
H= (L − λg)dx. l
a θ
The solution depends upon λ, which is found from Eq. m
(214). y
For example, given two points x1 and x2 on the x axis,
and an arc length l > x2 − x1 , the shape of the curve
enclosing the largest area is given by the equation D. Hamilton’s Principle
∫ √
δ [y − λ 1 + y ′2 ]dx = 0. State of a system, defined by coordinates and time q, t
is a point, and it’s motion is a trajectory in a space-time.
−λy ′2 √ Hamilton’s Principle states that given initial (q0 , t0 ) and
√ − y + λ 1 + y ′2 = C. final (q1 , t1 ) points in space-time, the trajectory of the
1+y ′2
system with the Lagrangian L(q̇, q, t) is such that the
integral
λ2
y ′2 = − 1. ∫ t1
(C + y)2
δS = δ L(q̇, q, t)dt = 0. (216)
∫ √ t0
dy
x= √ = λ2 − (C + y)2 + B. (Action takes the least possible value among all others
λ2
(C+y)2 −1
trajectories with the same initial and final points.) La-
grange equations follow from this variational principle.
(x − B)2 + (C + y)2 = λ2 . Two Lagrangians L2 and L1 , connected by the relation
Problem C.1 A curve of given total length is drawn in d
a plane between the points (±a, 0). Find the shape that L2 (q̇, q, t) = L1 (q̇, q, t) + Φ(q, t),
dt
will enclose the largest possible area between the curve
and the x-axis. where Φ is some function of q and t (not q̇), are equivalent
Consider because the contribution from the term dΦ(q, t)/dt to the
∫ a uniform chain hanging from two fixed action does not depend upon the trajectory.
points.
∫ yds is stationary subject to the condition
ds =const. The shape of the chain is given by the Problem D.1 ** A pendulum clock is designed for use
equation on a gravity-free spacecraft.The mechanism is a simple
∫ √ pendulum (mass m at the end of a massless rod of length
δ (y − λ) 1 + y ′2 dx = 0.
l) hung from a pivot, about which it can swing in a plane.
To provide artificial gravity, the pivot is forced to rotate
From Eq. (195) we obtain
at a frequency ω in a circle of radius R in the same plane
y ′ (y − λ) as the pendulum arm. Show that the motions θ(t) of this
√ = C.
1 + y ′2 pendulum are identical to the motions θ(t) of a simple
pendulum in a uniform gravitational field of strength g =
The solution is
∫ ω 2 R.
dy y−λ
x= √( )2 = C cosh−1 + A,
C The coordinates of the mass are
y−λ
C −1
x = R sin ωt + l sin(ωt + θ)
or, equivalently, y = R cos ωt + l cos(ωt + θ).
x−A
y = λ + C cosh . (215) Hence the Lagrangian is
C
This curve is called catenary. The word catenary is de- m[ 2 2 ]
L= R ω + l2 (ω + θ̇)2 + 2Rlω cos θ(ω + θ̇) ,
rived from the Latin word for ”chain.” In 1669, Jungius 2
disproved Galileo’s claim that the curve of a chain hang-
which is equivalent to
ing under gravity would be a parabola. The equation was
obtained by Leibniz, Huygens, and Johann Bernoulli in m 2 2
L= l θ̇ + mRω 2 l cos θ.
1691 in response to a challenge by Jakob Bernoulli. 2
 38

E. Jacobi’s Principle

Consider

L = L(q̇1 , . . . , q̇n , q1 , . . . , qn , t)

Let qn be a cyclic variable. Then solving Equation

∂L
= pn FIG. 19: Snell’s law
∂ q̇n
for q̇n , we can obtain
Consider, for example, motion of a particle in a cen-
L = L(pn , q̇1 , . . . , q̇n−1 , q1 , . . . , qn−1 , t). tral field. In polar coordinates, and using angle θ as a
parameter, the abridged action is
Consider Hamilton’s principle, only now we are fixing ∫ √
only q1 , . . . , qn−1 (and pn ). In this case Eq. (216) gives S0 ∼ [E − V (r)](r2 + r′2 )dθ. (221)
∫ t1
δ L(q̇, q, t)dt = pn [δqn (t1 ) − δqn (t0 )], The first integral of the Euler-Lagrange equation is
t0
√ r′2 √
or E − V (r) √ − [E − V (r)](r2 + r′2 ) = C.
∫ t1 r2 + r′2
δ [L − pn q̇n ]dt = 0. (217) (222)
t0 Thus we obtain

Let the Lagrangian does not involve the time explicitly. [E − V (r)]r4 = C 2 (r2 + r′2 ). (223)
The action can be presented as
∫ τ1 Problem E.1 * Analyze the relation between Eq. (223)
q′ q′ and Eq. (24).
S= L( 1′ , . . . , n′ , q1 , . . . , qn )t′ dτ,
τ0 t t
The time does not appear in the Jacobi Principle. It
where τ is some parameter. Similar to Eq. (217) we determines the path of the point in the configuration
obtain variational principle in the form space, not the motion in time. Yet this last problem
∫ t1 ∫ t1 ∫ τ1 ∑ can be solved by integrating Equation
′ ′ ′
δ [Lt − Et ]dτ = δ [L − E]t dτ = δ pi dqi = 0. dl
t0 t0 τ0 i dt = √ . (224)
2m(E − V )
Among all the trajectories starting at q0 and ending at
q1 and satisfying energy conservation law, the actual tra-
jectory is such that the integral F. Fermat’s Principle
∫ ∑
S0 = pi dqi (218) To explain the law of rectilinear propagation of light
i
in an optically homogeneous media, the law of specular
reflection and Snell’s law
is minimal. S0 is an abridged action and the statement
itself is called Maupertuis Principle. Taking into account n1 sin θ1 = n2 sin θ2 , (225)
that
∑ ∑ √ √ where n is the refractive index, Fermat postulated that
pi dqi = pi q̇i dt = 2T dt = 2mT dl = [2m(E − V )]dl,the particles of light travel in a media with the velocity
i i
(219) v = c/n, (226)
we obtain, that given initial and final points in space,
among all the trajectories satisfying energy conservation where c is the velocity of light in vacuum, and their tra-
law, the actual trajectory is such that jectory is such that the time of travel between two points
∫ q1 √ q) and q1 is minimal. Thus we immediately obtain Fer-
mat’s Principle which determines the shape of light ray
S0 = [2m(E − V )]dl (220)
q0 in a region of variable refractive index
∫ q1
is minimal. The statement itself is called Jacobi’s Prin-
δ n(r)dl = 0. (227)
ciple. q0
 39

Consider, for example, the 2D problem with n = n(y). Lagrange Equations are invariant relative to such trans-
From the variational principle formation, that is Equations for the new variables
∫ √ ( )
δ n(y) 1 + y ′2 dx = 0 (228) d ∂L(Q̇, Q, t) ∂L(Q̇, Q, t)
− =0
dt ∂ Q̇i ∂Qi
we obtain the first integral of the Euler-Lagrange equa-
tion coincide with those for the old ones.
Hamilton’s Equations
n(y)
√ = C. (229) ∂H(p, q, t) ∂H(p, q, t)
1 + y ′2 q̇i = ; ṗi = − (234)
∂pi ∂qi
Taking into account that
are invariant relative to a transformation
1 dx
√ =√ = sin θ, (230) Qi = Qi (q1 , . . . qn ; p1 , . . . pn )
1 + y ′2 dx2 + dy 2
Pi = Pi (q1 , . . . qn ; p1 , . . . pn ), (235)
where θ is the angle between the light ray and the y axis
we see the geometrical meaning of Eq. (229): if there exists function H ′ (P, Q, t) such, that new vari-
ables satisfy equations
n(y) sin θ(y) = const. (231)
∂H ′ (P, Q, t) ∂H ′ (P, Q, t)
Q̇i = ; Ṗi = − . (236)
Let the interface y=const separates two regions with re- ∂Pi ∂Qi
fractive indices n1 and n2 . Then Eq. (231) is just Snell’s
law. In this case the transformation (235) is called canonical.
Transformation (235)is canonical if and only if there
Problem F.1 The refractive index n of a medium is a exist some functions H ′ and F such that
function only of the distance r from a fixed point O. ∑ ∑
* (a) Prove that the equation of a light ray, assumed to lie pi dqi − Hdt = Pi dQi − H ′ dt + dF. (237)
i i
in in a plane through O, traveling in the medium satisfies
(in polar coordinates) The reason lies in the fact that Hamilton equations are
( )2 Euler-Lagrange equations for the action considered as a
1 dr r2 n2 (r) functional in phase space
= −1
r2 dθ a2 n2 (a)
∫ t1 [∑ ]
where a is the distance of the ray from O at the point at S= pi q̇i − H(p, q, t) dt. (238)
which dr/dθ = 0. t0 i
** (b) If n = [1 + (α2 /r2 )]1/2 and the ray starts and ends
far from O, find the angle through which the ray is turned In fact, integrating by parts we obtain
if its minimum distance from O is a. ∫ t1 ∑ ( )
∂H ∂H
δS = δpi q̇i + pi δ q̇i − δpi − δqi dt
Comparing Jacobi and Fermat Principles we see that t0 i
∂pi ∂qi
the shape of light ray can be found from the solution of ∫ t1 ∑ [( ) ( ) ]
∂H ∂H
mechanical problem with the parameters = q̇i − δpi − ṗi + δqi dt
t0 i
∂pi ∂qi
V (r) ∼ −n(r)2 . t1
E = 0, (232) ∑ 

+ pi δqi  (239)
.

i t0
XI. CANONICAL TRANSFORMATIONS
Addition to the integrand of a full derivative does not
A. Canonical Transformations change the Euler-Lagrange equations.

Making arbitrary (invertible) point transformations

B. Generating Function
Qi = Qi (q1 , . . . qn ; t) (233)
From Eq. (237) follows that every function F generates
in a Lagrangian we obtain canonical transformation. In fact, let

L(Q̇, Q, t) = L(q̇(Q, t), q(Q, t), t). F = F1 (q, Q, t). (240)

 40

Substituting F in Eq. (237) we obtain From Eq. (249) we obtain

∂F1 H ′ = P. (251)
pi =
∂qi Hamilton equations now are
∂F1
Pi = − ∂H
∂Qi Ṗ = − =0
∂Q
∂F1
H′ = H + . (241) ∂H
∂t Q̇ = = 1, (252)
∂P
If we use generating function in the form and the solution is
∑
F = F2 (q, P, t) − Pi Qi , (242) P = const
i Q = t + Q0 .
it generates transformation defined by the equations Inverting Eq. (250) we obtain
√
∂F2 2P
pi = q= sin(ωQ).
∂qi k
∂F2
Qi = Thus returning to old variables we obtain a well known
∂Pi solution
∂F2 √
H′ = H + . (243) 2P
∂t q= sin(ωt + φ0 ). (253)
k
Notice, that to be able to solve Eq. (243) relative to P
and Q we should have If we make canonic transformation with the generating
function
( 2 )
∂ F2 ∫ √ ( )
det ̸= 0. (244) kq 2
∂qi ∂Pj F2 = 2m P − dq − P t, (254)
2
we get
C. Application of Canonical Transformations
H ′ = 0. (255)
Consider harmonic oscillator Problem C.1 * Can you regain from Eqs. (254) and
p 2
kq 2 (255) the solution (253)?
H= + . (245)
2m 2 Problem C.2 Can you solve the problem of harmonic
Let us make canonical transformation generated by the oscillator using canonical transformation generated by
function the function
∫ √ ( ) mωq 2
kq 2 F1 = cot Q?
F2 = 2m P − dq. (246) 2
2
Problem C.3 * (a) Check up that the function
From Eq. (243) we obtain ∑
F2 = fi (q1 , . . . qn ; t)Pi (256)
√ ( ) i
kq 2
p = 2m P − dq (247) generates point transformation. (b) What transformation
2
√ ∫ is generated by a function
m dq
Q = √ , (248) F2 = fi (q1 , . . . qn ; t)Pi + g(q1 , . . . qn ; t)?.
2 2
P − kq2
Problem C.4 ** Consider small oscillations of the an-
√ harmonic oscillator with the Hamiltonian
where ω = k/m. Thus the canonical transformation
generated by the function (246) is: P2 ω 2 x2
H= + + αx3 + βxp2 ,
2 2
p2 kq 2
P = + (249) where αx ≪ ω 2 , βx ≪ 1. In the generating function
2m 2 [ ] F2 = xP + ax2 P + bP 3 chose parameters a and b such,
√
−1 −1 k that new Hamiltonian up to the terms αω2 Q, βQ would
Q = ω sin q . (250)
2P not contain anharmonic terms, and find x(t).
 41

D. Symmetries and Conservation Laws: Lagrange’s Using Lagrange’s equations we obtain

Mechanics [ ( ) ]
d ∂L ∂Q
δL = δλ. (267)
In all our solutions integrals of motion played a very dt ∂ q̇ ∂λ λ=0
important role. Remind, that integral of motion is a
function of q, q̇ (and possibly t) which does not change in Now let the transformation does not change the La-
the process of motion. Any ignorable coordinate qi gives grangian, that is
us integral of motion
Q̇ Q
∂L L(Q̇, Q) = L(̸ q̇, ̸ q), (268)
= const. (257)
∂ q̇j or in other words
For example, Lagrangian of a particle in a central field,
being written in polars does not contain θ. So it has δL ≡ 0. (269)
integral of motion - angular momentum
Then from Eq. (267) we obtain integral of motion
dL ( )
= mr2 θ̇ = const. (258) ∂L(q̇, q, t) ∂Q(q)
dθ̇ = const. (270)
∂ q̇ ∂λ λ=0
However, the same Lagrangian written in Cartesian co-
ordinates, More generally
m √ ∑ ∂L ( ∂Qi )
L = (ẋ2 + ẏ 2 ) − V ( x2 + y 2 ), (259) = const. (271)
2 ∂ q̇i ∂λ λ=0
i
does not have ignorable variables. So the question is:
why angular momentum is the integral of motion for the Thus we have proved the Noether Theorem: If a La-
Lagrangian (259), but is not an integral of motion, for grangian is invariant under a family of transformations,
example, for the Lagrangian the system posses a constant of the motion.
m √ We can generalize Noether theorem to the case, when
L = (ẋ2 + ẏ 2 ) − V ( x4 + y 4 )? (260) under the transformation (264) the Lagrangian changes
2 according to the law
Consider some Lagrangian
d
δL = Φ(q, t)δλ, (272)
L = L(q̇, q, t) (261) dt
and a transformation, depending upon continuous pa- where Φ is some function. In this case the integral of
rameter motion is
∑ ∂L(q̇, q, t) ( ∂Qi (q) )
Q = Q(λ, q) (Q(0, q) ≡ q). (262) − Φ(q, t) = const. (273)
i
∂ q̇i ∂λ λ=0
For example, rotation through an angle φ about axis OZ
Problem D.1 * Show, that Lagrangian (259) is invari-
X = x cos φ − y sin φ
ant under rotations. Then using Noether theorem show
Y = x sin φ + y cos φ (263) that m(−ẋy + ẏx) = const.
We restrict ourselves by infinitesimal transformations
( ) E. Symmetries and Conservation Laws: Hamilton’s
∂Q
δq ≡ Q(δλ, q) − q = δλ. (264) Mechanics
∂λ λ=0

For the rotation In Hamilton’s Mechanics, integral of motion is a func-

tion of q, p (and possibly t) which does not change in
δx = −yδφ, δy = xδφ, δz = 0. (265) the process of motion. Let the system is described by a
Hamiltonian
Consider change of the Lagrangian(261) under the in-
finitesimal transformation H = H(p1 , . . . , pn ; q1 , . . . , qn ; t). (274)
[ ]
∂L ∂L
δL = δq + δ q̇ δλ Consider an arbitrary function F (p, q, t). We get
∂q ∂ q̇
[ ( ) ( ) ]
∂L ∂Q ∂L d ∂Q dF ∂F ∑ ∂F ∑ ∂F ∂F
= + δλ. (266) = + q̇j + ṗj = + [F, H](275)
∂q ∂λ λ=0 ∂ q̇ dt ∂λ λ=0 dt ∂t ∂qj ∂pj ∂t
 42

We have introduced Poisson brackets of two arbitrary The function G is called generator of the transformation.
functions of coordinates and momenta F and G as From Eq. (243) we obtain
∑n ( )
∂F ∂G ∂F ∂G ∂G(q, p, t)
[F, G] ≡ − . (276) δqi = δλ
∂qi ∂pi ∂pi ∂qi ∂pi
i=1
∂G(q, p, t)
If the function F does not depend explicitly upon time δpi = − δλ. (280)
∂qi
and
For example, translation in the direction OZ is generated
[F, H] = 0, (277) by the operator pz and rotation about the axis OZ (Eq.
(265)) is generated by the operator Jz = xpy − ypx .
then F is an integral of motion. From Eq. (275) we get Consider change of an arbitrary function of coordinates
dH ∂H and momenta F under transformation generated by G.
= . (278) From Eq. (280) we obtain
dt ∂t
∑n ( )
So if the Hamiltonian does not contain time explicitly, ∂F ∂F
δF = δqi + δpi
for example for a natural conservative system, it is an i=1
∂qi ∂pi
integral of motion. In this example the Hamiltonian has ∑n ( )
also the meaning of energy. ∂F ∂G ∂F ∂G
= − δλ ≡ [F, G] δλ.
Consider, however, spherical pendulum constrained to i=1
∂q i ∂p i ∂p i ∂qi
rotate about an axis. From the Lagrangian (63) we get
Let G is the transformation of symmetry
p2 1
H = θ 2 − ml2 ω 2 sin2 θ + mgl(1 − cos θ). (279) δH = [H, G] = 0,
2ml 2
Hamiltonian is not an energy, because the system is not and does not depend explicitly upon t. Then using Eq.
natural, but it is conserved. (275) and an obvious relation
[H, G] = − [G, H] ,
Problem E.1 For a one-dimensional system with the
Hamiltonian we obtain
2
p 1 dG
H= − 2 = 0,
2 2q dt
show that there is a constant of the motion D = pq
− Ht. that is G is conserved. For example, if Hamiltonian is
2
invariant with respect to spatial translations in the OZ
Problem E.2 * The Hamiltonian for two-dimensional direction, then pz = const, and if Hamiltonian is invari-
harmonic oscillator can be written as ant relative to rotations about OZ, then Jz = const.
1 2 1 2 Problem E.3 * Show that the Poisson brackets of the
H= (px + m2 ω 2 x2 ) + (p + m2 ω 2 y 2 ).
2m 2m y components of angular momentum are
Verify that the components of a symmetrical two- [Jx , Jy ] = Jz .
dimensional tensor A defined as
Interpret this result in terms of the transformation of one
1 component generated by another.
Aij = (pi pj + m2 ω 2 xi xj )
2m
are constants of motion.
F. Poisson Brackets and Canonical
The reason for the existence of an integral of motion Transformations
is the particular symmetry. The natural language for ex-
pressing a symmetry is the formalism of canonical trans- An approach to canonical transformations alternative
formations. The identity transformation is realized by to that based on generating functions uses Poison brack-
the function ets. Let us present Eq. (237) as
∑  
F2 = qi Pi . ∑ ∑ ∂Qj ∑ ∂Qj
i
pi − Pj  dqi − Pj dpi
i j
∂qi j
∂pi
Hence the infinitesimal canonical transformation is real-  
ized by the function ∑ ∂Qj
+ H ′ − H − Pj  dt = dF. (281)
F2 = qi Pi + G(q, P, t)δλ = qi pi + G(q, p, t)δλ. j
∂t
 43

The relations between mixed second derivatives give us Hence

the conditions for the canonicity of the transformation as
δS = p(t1 )δq1 − p(t0 )δq0 − H(t1 )δt1 + H(t0 )δt0 . (285)
[Qi , Qj ] = 0; [Pi , Pj ] = 0; [Qi , Pj ] = δij . (282)
From Eq. (285) we obtain
Evidently if the transformation is canonical, the inverse
transformation is also canonical. ∂S
= p (286)
Problem F.1 Use Eq. (282) to check up that transfor- ∂q
mation (249), (250) is canonical. ∂S
= −p0 . (287)
∂q0
Problem F.2 Prove that transformation
√ √ Eq. (287) being solved for q
p = 2mωP cos Q; q = 2P/mω sin Q.
q = q(t, t0 ; q0 , p0 )
is canonical. Use this transformation to solve the problem
of harmonic oscillator (245). gives the solution of the equations of motion. For exam-
ple, from Eq. (284) we obtain
Problem F.3 Prove that the transformation
m(x − x0 )
p2 p0 = .
Q1 = q1 cos α − sin α, P1 = βq2 sin α + p1 cos α, t − t0
β
p1
Q2 = q2 cos α − sin α, P2 = βq1 sin α + p2 cos α, Inverting we obtain
β
p0
where α and β are some constants, is canonical. x = x0 + (t − t0 ),
m
which is, of course, correct for free motion.
XII. HAMILTON-JACOBI MECHANICS
Problem A.1 ** Show that the function
A. The Action as a Function of the Coordinates
mω 2
S= (x + x20 ) cot[ω(t − t0 )] − mωxx0 csc[ω(t − t0 )]
2
Let us consider, for arbitrary Lagrangian L(q̇, q, t), ac-
tion S as a function of the initial point q0 and initial time is the action for the linear harmonic oscillator (245).
t0 and of the final point q and final time t Substitute this action into Eq. (287) and check up that
∫ t you regain the well known solution of the harmonic os-
S[q0 , t0 ; q, t] = L(q̇, q, t)dt. (283) cillator problem.
t0

(The integral is calculated for an actual trajectory.) For B. Hamilton-Jacobi Equation

p2
example, for Hamiltonian H = 2m we get
Consider arbitrary Hamiltonian
m (x − x0 )2
S= . (284)
2 t − t0 H = H (q1 , . . . , qn , p1 , . . . , pn , t) . (288)
Consider variation of the action with the variations of
From Eq. (285) we get
of the initial and final points. We get
∫ t1 +δt1 ∫ t1 ∂S
= −H, (289)
δS = L(q̇ + δ q̇, q + δq)dt − L(q̇, q)dt ∂t
t0 +δt0 t0
= L(t1 )δt1 − L(t0 )δt0 or, explicitly, using Eq. (286)
∫ t1 [ ( )] 1
∂L d ∂L ∂L  ( )
+ − δqdt + δq . ∂S ∂S ∂S
t0 ∂q dt ∂ q̇ ∂ q̇ 0 ∂t
+ H q1 , . . . , qn ,
∂q1
,...,
∂qn
,t = 0 (290)

By Lagrange’s equations the quantities in the square

brackets vanish; also the variations of the final and initial Problem B.1 (a) Show that action (284) satisfies Eq.
p2
coordinates are (290) for Hamiltonian H = 2m .
*(b) The same for the action and Hamiltonian from the
δq1 = δq(t1 ) + q̇(t1 )δt1 , δq0 = δq(t0 ) + q̇(t0 )δt0 . Problem A.1.
 44

The idea of Hamilton was to use Eq. (290) with the C. Conservative System
appropriate initial (t → t0 ) conditions to find action and,
hence, the solution of the problem. For conservative system we may always look for S in
Jacobi understood that we can use any complete solu- the form
tion of the equation (290), that is a solution in the form:
S (q1 , . . . , qn , t) = W (q1 , . . . , qn ) − Et, (296)
S = S (q1 , . . . , qn , α1 , . . . , αn , t) .
where W (q1 , . . . , qn , α1 , . . . , αn−1 , E) is the solution of
where α1 , . . . , αn are arbitrary constants. Note that there the equation
are infinitely many such solutions, and action is just one
of them. Any such solution is called Hamilton’s Principal ( )
∂W ∂W
Function. H q1 , . . . , qn , ,..., = E, (297)
∂q1 ∂qn
Let us define functions qi (t) and pi (t) in the following
way: qi (t) are the solutions of the equations and is called Hamilton’s Characteristic Function. Eq.
(291) takes the form
∂S
= βi , i = 1, . . . , n (291)
∂αi ∂W ∂W
= βi , (i = 1, . . . , n − 1); − t = βn . (298)
where β1 , . . . , βn are arbitrary constants, and pi (t) are ∂αi ∂E
defined by simple differentiation p2
For example, for Hamiltonian H = 2m + V (x) we look
∂S for S in the form
pi = , i = 1, . . . , n. (292)
∂qi
S = W (x) − Et, (299)
Note, that to be able to solve Eq. (291) we should have
( 2 ) where W is the solution of the equation
∂ S ( )2
det ̸= 0.
∂αi ∂xj 1 dW
+ V (x) = E. (300)
2m dx
Below we prove that such defined q and p satisfy Hamil-
ton equations We get
∫ √
∂H ∂H
q̇j = ; ṗj = − . (293) W = 2m (E − V (x))dx. (301)
∂pj ∂pj

Using the fact that, due to Eq. (292), H(q, p, t) can be Substituting this into Eq. (296) and differentiating with
presented as H(q, α, t) we obtain. respect to E we obtain
( ) √ ∫
d ∂S ∂ ∂ ∂S m dx
0=
dt ∂αi
=
∂t
+ q̇j
∂qj ∂αi t= √ . (302)
2 (E − V (x))
∂2S ∂pj ∂H ∂pj
= + q̇j =− + q̇j Problem C.1 ** (a) Find explicitly the Hamilton’s
∂αi ∂t ∂αi ∂αi ∂αi
( ) Principal Function of the type (301) for linear harmonic
∂H ∂pj ∂pj ∂H ∂pj
=− + q̇j = q̇j − . (294) oscillator.
∂pj ∂αi ∂αi ∂pj ∂αi
(b) Looking at the action from the Problem A.1 of Sec-
Similarly tion X, can you suggest the Hamilton’s Principal Func-
( ) tion for the harmonic oscillator not of the type (296)?
dpi ∂ ∂ ∂S Check up that this function generates correct solution of
= + q̇j the problem.
dt ∂t ∂qj ∂qi
( ) ( ) ( )
∂H ∂H ∂pj
=− +
∂qi α ∂pj q ∂qi α D. Hamilton-Jacobi Theory and Canonical
( ) ( ) ( ) Transformations
∂H ∂H ∂pj
=− −
∂qi p ∂pj q ∂qi α
( ) ( ) ( ) Consider arbitrary Hamiltonian H (q, p, t) and make
∂H ∂pj ∂H
+ =− . (295) the canonical transformation with the generating func-
∂pj q ∂qi α ∂qi p tion
( )
Hence solutions of Eqs. (291) and (292) satisfy Eqs. P1 Pn
F2 (q, P, t) = S q1 , . . . , qn ; ̸ α1 , . . . , ̸ αn , t .
(293).
 45

where S is the solution of Eq. (290). We obtain Differentiating with respect to E we get
√ ∫
H ′ = 0. m dr
t= √ . (312)
2 E − V (r) − 2mr
J2
The solution of the equations of motion 2

Pi = αi , Qi = βi Differentiating with respect to J we get

being inserted into Eq. (243) gives Eqs. (291),(292). ∫ J
√ r 2
If we make canonical transformation with the generat- θ= dr. (313)
ing function 2m[E − V (r)] − J2
r2
( Pn−1 Pn
)
P1
F2 (q, P ) = W q1 , . . . , qn ; ̸ α1 , . . . , ̸ αn−1 , ̸ E , (303) Eqs. (312) and (313) coincide with those we obtained in
Section II.

where W is the solution of Eq. (297), we get Problem E.1 * Solve the problem of the motion of a
point projectile in a vertical plane, using the Hamilton
H ′ = Pn . (304) Jacobi method. Find both the equation of the trajectory
and the dependence of the coordinates on time, assuming
The equations of motion have the form
the projectile is fired off at time t = 0 from the origin with
Ṗi = 0 the velocity v0 , making an angle α with the horizontal.
{
0 i = 1, . . . , n − 1
Q̇i = . (305)
1 i=n
F. Separation of Variables: General Case
The solution
Let Hamilton has the form
Pi = αi
{
βi i = 1, . . . , n − 1 H = H (f (q1 , p1 ) , q2 , . . . qn , p2 , . . . pn , t) . (314)
Qi = (306)
t + βn i = n
Then we can seek solution of Hamilton-Jacobi equation
being inserted in Eq. (243) gives us Eq. (298). in the form

S = S ′ (q2 , . . . , qn , t) + S1 (q1 ), (315)

E. Separation of Ignorable Variables
where S1 is the solution of the equation
To solve Kepler problem we should find W (r, θ) satis- ( )
∂S1
fying f q1 , = α1 (316)
∂q1
[( )2 ( )2 ]
1 ∂W 1 ∂W and S ′ is the solution of the equation
+ 2 + V (r) = E. (307)
2m ∂r r ∂θ
( )
∂S ′ ∂S ′ ∂S ′
θ is cyclic variable, so + H α1 , q2 , . . . , qn , ,..., , t = 0. (317)
∂t ∂q2 ∂qn
W = Jθ + Wr (r), (308) Thus we effectively have one degree of freedom less. If
we can get
where for Wr we get
[( )2 ] S = W1 (q1 ) + · · · + Wn (qn ) − Et, (318)
1 dWr J2
+ 2 + V (r) = E. (309)
2m dr r the system is called separable.
In spherical coordinates, for example, the variables are
So separable if
∫ √
J2 b(θ) c(φ)
Wr = 2m[E − V (r)] − 2 dr (310) V (r, θ, φ) = a(r) + + 2 2 . (319)
r r2 r sin θ
and Physically interesting case corresponds to
∫ √
J2 b(θ)
S = −Et + Jθ + 2m[E − V (r)] − 2 dr. (311) V (r, θ, φ) = a(r) + . (320)
r r2
 46

(say, a gravitational field of two closely separated point If

masses). In this case the Hamilton-Jacobi equation takes
the form p2
H= + V (q), (330)
( )2 [( )2 2m
1 ∂W 1 ∂W
+ a(r) + + 2mb(θ) then
2m ∂r 2mr2 ∂θ
∫
( )2 ] (2m)1/2 q2
1 ∂W I= [E − V (q)]1/2 dq. (331)
+ 2 = E.(321) π q1
sin θ ∂φ
For a harmonic oscillator (V = mω 2 q 2 /2),
We seek the solution in the form
H = ωI. (332)
W = Jφ φ + W1 (r) + W2 (θ), (322)
Problem A.1 * A particle of mass m moves in a one-
obtaining for the functions W1 and W2 equations
dimensional potential
( )2 ( )
dW2 Jφ2
+ 2mb(θ) + = α2 1 λ
dθ sin2 θ V (q) = kq 2 + 2 ,
( )2 2 q
dW1 α2
+ 2ma(r) + 2 = 2mE (323) where k, λ, q > 0. Sketch the potential and the phase
dr r
portrait. Show that the energy E and action I are related
by
XIII. ACTION-ANGLE VARIABLES
E = (kλ)1/2 + 2I(k/m)1/2
A. Action-Angle Variables in a System of One and that the period is then independent on the amplitude.
Degree of Freedom
Problem A.2 * A particle of mass m is constrained to
Consider conservative system of one degree of freedom. move under the action of gravity in the vertical xz-plane
After we have found the Hamilton’s characteristic func- on a smooth cycloid curve given parametrically by
tion W (q, E), we can take the generating function F not
only in the form (303) x = l(θ + sin θ), z = l(1 − cos θ).

F2 = W (q, P ), (324) Show that a suitable Hamiltonian is

but also in a more general form form p2θ

H= + mgl(1 − cos θ).
4ml2 (1 + cos θ)
F2 = W (q, E(P )), (325)
Use action/angle variables to show that the frequency of
where E(P ) is any function we like. After the transfor- oscillation of the particle is independent of its amplitude,
mation we get Hamiltonian i.e. it is the same for all initial conditions with |θ| < π.
(The substitution s = sin 12 θ can be useful.)
H = E(P ). (326)

Action-angle variables correspond to specific choice of

E(P ), given by the equation B. Adiabatic Invariance
I ∫∫
1 1 Consider a mechanical system executing a finite mo-
P (E) = pdq ≡ dpdq (327)
2π 2π tion in one dimension; λ is a slowly changing parameter.
That is we have
(Chosen in such a way momentum is called action and
is traditionally designated as I; we also consider finite H = H(q, p, λ(t)), (333)
motion). The coordinate corresponding to such a mo-
mentum is called angle where
∂W dλ
ϕ= . (328) T ≪ λ. (334)
∂I dt
The period of motion T (I) is given by the equation Averaging equation
∂I dE ∂H ∂H dλ
T = 2π . (329) = = (335)
∂E dt ∂t ∂λ dt
 47

with respect to period of motion, we obtain Problem B.2 ** For the small oscillations of a pendu-
( ∫ ) H ∂H dq lum whose length l is varying very slowly show that the
dE ∂H dλ 1 T ∂H dλ ∂λ ∂H dλ maximum angular displacement θmax ∝ l−3/4 . Hence
= = dt = H dq∂p . show that the maximum sideways displacement from the
dt ∂λ dt T 0 ∂λ dt dt
∂H
∂p vertical is ∝ l1/4 and that the maximum acceleration is
(336) ∝ l−3/4 . (Note that this last result implies, as l decreases
Putting p = p(q, E, λ) and differentiating with respect to slowly, an increasing risk for spaghetti eaters from sauce
λ equation detachment!)
H(q, p, λ) = E (337)
Problem B.3 ** A superball is bouncing vertically up
we have and down. It has a velocity v0 when it strikes the ground.
The acceleration due to gravity is slowly reduced by 10%
∂H ∂H ∂p
+ =0 (338) during a very long period of time. Assuming that the
∂λ ∂p ∂λ collisions of the ball with the ground are elastic, find the
or corresponding change in v0 .
∂H
∂p
∂λ
∂H
=− . (339) Problem B.4 ** A child of mass m on a swing rises
∂p
∂λ her center of mass by a small distance b every time the
swing passes the vertical position, and lowers her mass by
Hence we get the same amount at each extremal position. Assuming
H ∂p small oscillations, calculate the work done by the child
dE dλ ∂λ dq
=− H ∂p
(340) per period of oscillation. Show that the energy of the
dt dt ∂E dq swing grows exponentially according to dE/dt = αE and
or determine the constant α.
I ( )
∂p dE ∂p dλ
+ dq = 0, (341)
∂E dt ∂λ dt C. Action-Angle Variables in Separable Systems
which can be rewritten in the form
Suppose the problem is completely separable, that is
dI
= 0 =⇒ I = const. (342)
dt W = W1 (q1 , α1 . . . , αn−1 , E)+· · ·+Wn (qn , α1 . . . , αn−1 , E).
And the energy changes according to the law (348)
Let us introduce new momenta I1 , . . . , In by the gener-
E = H(I, λ). (343) ating function
Consider, for example, small oscillations of simple pen- F = W (q1 . . . , qn ; I1 . . . , In ), (349)
dulum.
ml2 2 p2 where W (I) is obtained from W (α) using equations
L= θ̇ − mglθ2 =⇒ H = θ 2 + mglθ2 . (344)
2 2ml I I
1 1 dWk
We have Ik = pk dqk = dqk = Ik (α1 . . . , αn−1 , E).
2π 2π dqk
√ (350)
g
E= I. (345) After the canonical transformation we obtain
l
So, when l changes adiabatically, H(I1 . . . , In ) = E(I1 . . . , In ). (351)
E ∼ l−1/2 . (346) The connection between new coordinates ϕi and old co-
ordinates qi is
Taking into account, that E = mglθ02 , we obtain
∂W
θ0 ∼ l−3/4 . (347) ϕi = . (352)
∂Ii
Problem B.1 * (a) A particle moves in periodic mo-
tion in one dimension under the influence of a potential During one period
V (x) = F |x|, where F is a constant. Using action-angle I I
variables find the period of the motion as a function of ∂ϕi ∂ ∂W
∆ϕi = dqk = dqk
the particle’s energy. ∂qk ∂Ii ∂qk
I
** (b) Suppose that the parameter F is slowly varied ∂ ∂Ik
= pk dqk = 2π = 2πδik (353)
from an initial value. What happens to the energy of the ∂Ii ∂Ii
particle? The amplitude of oscillation? The period?