OUPCORRECTEDPROOF – FINAL,8/7/2020,SPi EXPLORING CLASSICAL MECHANICS · OUPCORRECTEDPROOF –...

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E X P L O R I N G C L A S S I C A L M E C H A N I C S


Exploring Classical Mechanics

A Collection of 350+ Solved Problems for Students,Lecturers, and Researchers

Gleb L. Kotkin

Valeriy G. Serbo

Novosibirsk State University, Russia

Second revised and enlarged English edition

1


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Library of Congress Control Number: 2020937520

ISBN 978-0-19-885378-7 (hbk.)ISBN 978-0-19-885379-4 (pbk.)

DOI: 10.1093/oso/9780198853787.001.0001

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Preface to the second English edition

This book was written by the working physicists for students of physics faculties ofuniversities.

The first English edition of this book under the title Collection of Problems in ClassicalMechanics was published by Pergamon Press in 1971 with the invaluable help by thetranslation editor D. ter Haar. This second English publication is based on the fourthRussian edition of 2010 and includes new problems from among those used in teachingat the physics faculty of Novosibirsk State University as well as the problems added inthe publications in Spanish and French. As a result, this book contains 357 problemsinstead of the 289 problems that appeared in the first English edition.

We are grateful to A.V.Mikhailov for useful discussions of some new problems,to Z.K.Silagadze for numerous indications of misprints and inaccuracies in previouseditions, and to O.V. Karpushina for an invaluable help in preparation of this manuscript.

In this edition, the main notations are:m, e, r, p, and M = [r,p] – mass, charge, radius vector, momentum, and angular

momentum of a particle, respectively;L, H, E, and U – Lagrangian function, Hamiltonian function, energy, and potential

energy of a system, respectively;E and B – electric and magnetic field intensities, respectively;ϕ and A – scalar and vector potentials, respectively, of the electromagnetic field;c – velocity of light; andd� – solid angle element.For problems about the motion of particles in electromagnetic fields, we use Gaussian

units, and in problems on electrical circuits, SI units.


From the Preface to the first English edition

This collection is meant for physics students. Its contents correspond roughly to themechanics course in the textbooks by Landau and Lifshitz [1], Goldstein [4], or ter Haar[6]. We hope that the reading of this collection will give pleasure not only to studentsstudying mechanics, but also to people who already know it. We follow the order inwhich the material is presented by Landau and Lifshitz, except that we start using theLagrangian equations in § 4. The problems in §§ 1–3 can be solved using the Newtonianequations of motion together with the energy, linear momentum and angular momentumconservation laws.

As a rule, the solution of a problem is not finished with obtaining the requiredformulae. It is necessary to analyse the result, and this is by no means the “mechanical”part of the solution. It is also very useful to investigate what happens if the conditionsof the problem are varied. We have, therefore, suggested further problems at the end ofseveral solutions.

A large portion of the problems were chosen for the practical classes with studentsfrom the physics faculty of the Novosibirsk State University for a course on theoreticalmechanics given by Yu. I. Kulakov. We want especially to emphasize his role in thechoice and critical discussion of a large number of problems. We owe a great debt to I. F.Ginzburg for useful advice and hints which we took into account. We are very gratefulto V. D. Krivchenkov whose active interest helped us to persevere until the end.

We are extremely grateful to D. ter Haar for his help in organizing an English editionof our book.


§1

Integration of one-dimensionalequations of motion

1.1. Describe the motion of a particle in the following potential fields U(x):

a) U(x) = A(e−2αx − 2e−αx) (Morse potential, Fig. 1a);

b) U(x) = − U0cosh2 αx

(Fig. 1b);

c) U(x) = U0 tan2 αx (Fig. 1c).

U

(a) (b) (c)

U

U

x x x

Figure 1

U

E

ax

Figure 2

1.2. Describe the motion of a particle in the fieldU(x)= − Ax4 for the case when its energy is equal to zero.

1.3. Give an approximate description of the motion of aparticle in the field U(x) near the tuning point x = a (Fig. 2).

Hint: Use a Taylor expansion of U(x) near the point x = a.Consider the cases U ′(a) �= 0 and U ′(a) = 0, U ′′(a) �= 0.

1.4. Determine how the period of a particle moving in the field in Fig. 3 tends to infinityas its energy E approaches Um.

Exploring Classical Mechanics: A Collection of 350+ Solved Problems for Students, Lecturers, and Researchers. First Edition.Gleb L. Kotkin and Valeriy G. Serbo, Oxford University Press (2020). © Gleb L. Kotkin and Valeriy G. Serbo 2020.DOI: 10.1093/oso/9780198853787.001.0001


4 Exploring Classical Mechanics [1.5

U

E

ax

Um

U

EUm

x1 x2x

b a c

Figure 3 Figure 4

1.5. a) Estimate the period of the particle motion in the field U(x) (Fig. 4), when itsenergy is close to Um (i.e., E − Um � Um − Umin).

b) Determine during which part of the period the particle is in the interval from x tox + dx.

c) Determine during which part of the period the particle has a momentum mx in theinterval from p to p + dp.

d) In the plane x, p = mx represent qualitatively lines E(x, p) = const for the casesE < Um, E = Um, E > Um.

1.6. A particle of mass m moves along a circle of radius l in a vertical plane under theinfluence of the field of gravity (mathematical pendulum). Describe its motion for thecase when its kinetic energy E in the lowest point is equal to 2mgl.

Estimate the period of revolution of the pendulum for the case whenE − 2mgl�2mgl.

1.7. Describe the motion of a mathematical pendulum for an arbitrary valueof the energy.

Hint: The time dependence of the angle the pendulum makes with the vertical can beexpressed in terms of elliptic functions (e.g. see [1], § 37).

1.8. Determine the change in the motion of a particle moving along a section which doesnot contain turning points when the field U(x) is changed by a small amount δU(x).

Consider the applicability of the results obtained for the case of a section near theturning point.

1.9. Find the change in the motion of a particle caused by a small change δU(x) in thefield U(x) in the following cases:

a) U(x) = 12mω2x2, δU(x) = 1

3mαx3;

b) U(x) = 12mω2x2, δU(x) = 1

4mβx4.

1.10. Determine the change in the period of a finite orbit of a particle caused by thechange in the field U(x) by a small amount δU(x).


1.12] §1. Integration of one-dimensional equations of motion 5

1.11. Find the change in the period of a particle moving in a field U(x) caused by addingto the field U(x) a small term δU(x) in the following cases:

a) U(x) = 12mω2x2 (a harmonic oscillator), δU(x) = 1

4mβx4;

b) U(x) = 12mω2x2, δU(x) = 1

3mαx3;

c) U(x) = A(e−2αx − 2e−αx), δU(x) = −Veαx (V � A).

1.12. The particle moves in the field U(x) = U0cosh2 αx

with the energy E > U0. Find the

particle delay time at the motion from x = −∞ to x = +∞ in comparison with the freemotion time with the same energy.


§2

Motion of a particle inthree-dimensional fields

2.1. Describe qualitatively the motion of a particle in the field U(r) = −αr − γ

r3 for

different values of the angular momentum and of the energy.

2.2. Find the trajectories and the laws of motion of a particle in the field

U(r) ={−V , when r < R,

0, when r > R

U

R r

−V

Figure 5

(Fig. 5, “spherical rectangular potential well”) for different valuesof the angular momentum and of the energy.

2.3. Determine the trajectory of a particle in the field U(r) =αr + β

r2 . Give an expression for the change in the direction of

velocity when the particle is scattered as a function of angularmomentum and energy.

2.4. Determine the trajectory of a particle in the field U(r) = αr − β

r2 . Find the time it

takes the particle to fall to the centre of the field from a distance r. How many revolutionsaround the centre will the particle then make?

2.5. Determine the trajectory of a particle in the field U(r) = −αr + β

r2 . Find the

angle �ϕ between the direction of radius vector at two successive passages through thepericentre (i.e., when r = rmin); also find the period of the radial oscillations, Tr . Underwhat conditions will the orbit be a closed one?

2.6. Determine the trajectory of a particle in the field U(r) = −αr − β

r2 . A field of this

kind arises in the motion of a relativistic particle in the Coulomb field in the special theoryof relativity; see [7], § 42.1 for details.



2.16] §2. Motion of a particle in three-dimensional fields 7

2.7. For what values of the angular momentum M is it possible to have finite orbits inthe field U(r) for the following cases:

a) U(r) = −αe−�r

r ; b) U(r) = −Ve−�2r2.

2.8. A particle falls from a finite distances towards the centre of the field U(r) = −αr−n.Will it make a finite number of revolutions around the centre? Will it take a finite timeto fall towards the centre? Find the equation of the orbit for small r.

2.9. A particle in the field U(r) flies off to infinity from a distance r �= 0. Is the numberof revolutions around the centre made by the particle finite for the following cases?

a) U(r) = αr−n b) U(r) = −αr−n

2.10. How long will it take a particle to fall from a distance R to the centre of the fieldU(r) = −α/r. The initial velocity of the particle is zero. Treat the orbit as a degenerateellipse.

2.11. One particle of mass m moves along the x-axis from a long distance with velocityv towards the origin O of the coordinate system. Another particle of the same massmoves towards the origin O along the y-axis from a long distance with the same velocitymagnitude. If the particles didn’t interact, the second would pass through point O intime τ after the first one. However, they repulse from each other, and potential energy ofinteraction is U(r) = α/r, where r is the distance between particles. Find the minimumdistance between the particles.

2.12. Two particles with masses m1 and m2 move with velocities v1 and v2 from longdistances along the crossing lines, the distance between which is equal to

−→AB = ρ. If

particles didn’t interact, particle 1 would pass the point of minimum distance A at timeτ earlier than particle 2 would pass the point B. However, there is the force of attractionbetween the particles, which is given by the potential energy U(r) = −β/r2.

a) At what relation between ρ and τ will particles collide?

b) At what distance from the point A will such a collision occur?

2.13. Determine the minimal distance between the particles, the one approaching frominfinity with an impact parameter ρ and an initial velocity v and the other one initially atrest. The masses of the particles are m1 and m2, and the interaction law is U(r) = α/rn.

2.14. Determine in the centre of a mass system the finite orbits of two particles ofmasses m1 and m2, and an interaction law U(r) = −α/r.

2.15. Determine the position of the focus of a beam of particles close to the beam axis,when the particles are scattered in a central field U(r) under the assumption that a particleflying along the axis is turned back.

2.16. Find the inaccessible region of space for a beam of particles flying along the z-axiswith a velocity v and being scattered by a field U(r) = α/r.



2.17. Find the inaccessible region of space for particles flying with a velocity v from apoint A in all directions and moving in a potential field U(r) = −α/r.

2.18. Use the integral of motion A = [v,M] − α rr (the Laplace vector – see [1], §15 and

[7], § 3.3) to find the orbit of a particle moving in the field U(r) = −α/r.

2.19. The spacecraft is moving in a circular orbit of the radius R around the Earth. Abody, whose mass is negligible in comparison with the mass of the spacecraft, is thrownfrom the spacecraft with relative velosity v, directed to the centre of the Earth. Find theorbit of the body.

Hint: To find the orbit of the body, try using the Laplace vector.(This problem is formulated based on the real incident: during a spacewalk, the

cosmonaut A. Leonov threw the plug from the camera in the direction of the Earth –see [8], § 8.)

2.20. Determine in quadratures the change of the period T of radial oscillationsof a particle moving in the central field U(r) when this field is changed by a smallamount δU(r).

2.21. Show that the orbit of a particle in the field

U(r) = −α

re−r/D

is a lowly precession ellipse when rmax � D. Find the angular velocity of precession.

2.22. Find the precessional velocity of the orbit in the field U(r) = −α/r1+ε,when |ε| � 1.

2.23. Find the angular velocity of the orbit precession of a particle in the field

U(r) = 12 mω2r2 + β

r4

for β � mω2a6, mω2b6, where a and b are parameters of the unperturbed trajectory:

( r cosϕ

a

)2 +( r sinϕ

b

)2 = 1.

2.24. The particle slides on the surface of a smooth paraboloid of revolution whose axis(the z-axis) is directed straight up:

z = x2 + y2

2l.

Find the angular velocity of the orbit precession. The maximum and the minimumdistances of a particle from the z-axis are a and b, where a � l.



2.25. Study the motion of the Earth–Moon system in the field of the Sun. Assume thatthe mass of the Moon is 81 times less than the mass of the Earth, and the distance fromthe Earth to the Moon (r = 380 thousand km) is a lot less than the average distancesto the Sun (R = 150 million km).

a) For simplicity, taking that the plane of the Moon orbit coincides with the plane ofthe Earth orbit, show that the potential energy of the Earth–Moon system in thefield of the Sun, averaged over a month, has the form

U(R) = −α

R− β

R3 ,

where R is the distance from the Sun to the centre of mass of the Earth–Moonsystem. Determine the precession of perihelion for a 100-year period.

b) The plane of the Moon’s orbit makes an angle of θ = 5◦ with the plane of theEarth’s orbit. Determine the related average velocity of precession for the Moon’sorbital plane.

2.26. Determine the angular velocity of the orbit precession of a particle in the fieldU(r) = −α

r + δU(r) if the orbit eccentricity e is much less than 1, assuming

δU(r) = δU(a)+ (r − a)δU ′(a)+ 12 (r − a)2δU ′′(a),

where a = 12 (rmax + rmin) is the average orbit radius.

2.27. Determine the angular velocity of the orbit precession of a particle in the fieldU(r) = −α

r + δU(r) (δU(r) is a small correction) up to second order in δU(r) inclusively.

2.28. Find the equation of motion of the orbit of a particle moving in the field U(r) =−α

r + γ

r3 , assuming γ

r3 to be a small correction to the Coulomb field.

2.29. Show that the problem of the motion of two charged particles in a uniform electricfield E can be reduced to the problem of the motion of the centre of mass and that ofthe motion of a particle in a given field.

2.30. Under what conditions can the problem of the motion of two charged particles ina constant uniform magnetic field B be separated into the problem of the centre of massmotion and the relative motion problem?

Take the vector potential of the magnetic field in the form

A(ri) = 12 [B,ri] , i = 1, 2.



2.31. Express the kinetic energy, the linear momentum, and the angular momentum ofa system of N particles in terms of the Jacobi coordinates:

ξn = m1r1 + . . . + mnrn

m1 + . . . + mn− rn+1 (n = 1 . . .N − 1),

ξN = m1r1 + . . . + mNrN

m1 + . . . + mN.

2.32. A particle with a velocity v at infinity collides with another particle of the samemass m which is at rest. Their interaction potential energy is U(r) = α/rn and the collisionis a central one. Find the point where the first particle comes to rest.

2.33. Prove that

MB + e2c

[r,B]2,

is the integral of motion for a charged particle in a uniform constant magnetic field B.Here M = m[r,v], and c is the velocity of light.

2.34. Find the trajectory and the law of motion of a charged particle in the magneticfield B(r) = gr/r3 (the field of the magnetic monopole).

Such form has a field of a thin long solenoid outside its end at distances which arelarge compared to the solenoid’s diameter, but small compared to its length.

2.35. Give a qualitative description of the motion and the shape of the orbit of a chargedparticle moving in the field of a magnetic dipole μ, in the plane perpendicular to thevector μ. Take the vector potential of magnetic dipole in the form A(r) = [μ,r]/r3.

2.36. a) Give a qualitative description of the motion of a charged particle in the fieldU(r) = 1

2mλr2, where r is a distance from the z-axis, for the case where there is a constantuniform magnetic field B parallel to the z-axis present.

b) Find the law of motion and the orbit of a charged particle moving in the fieldU(r) = α/r2 in a plane perpendicular to a constant uniform magnetic field B.

2.37. A charged particle moves in the Coulomb field U(r) = −α/r in a plane perpen-dicular to a constant uniform magnetic field B. Find the orbit of the particle. Study thecase when B is small and the case when the field U(r) is a small perturbation.

2.38. Describe the motion of two identical charged particles in a constant uniformmagnetic field B for the case when their orbits lie in the same plane which is perpen-dicular to B and where we may consider their interaction energy U(r) = e2/r to be asmall perturbation.

2.39. Show that the quantity

F[v,M] − α

rFr + 1

2 [F,r]2

is a constant of motion in the field U(r) = −αr − Fr where F = const.

Give the meaning of this integral of motion when F is small.



2.40. Study the effect of a small extra term δU(r) = −Fr, where F = const, added tothe Coulomb field on the finite orbit of a particle.

a) Find the average rate of change of the angular momentum, averaged over oneperiod.

b) Find the time-dependence of the angular momentum, the size, and the orientationof the orbit for the case when the force F lies in the orbital plane.

c) Do the same as under b) for the case when the orientation of the force is arbitrary.

Hint: Write down the equations of motion for the vectors M = m[r,v] and A =[v,M] − α r/r averaged over one period and solve them.

2.41. Find the systematic displacement of a finite orbit of a charged particle movingin the field U(r) = −α/r under influence of weak constant uniform electric E and mag-netic B fields.

a) Consider the limiting case when the magnetic field is perpendicular to the orbitplane and the electric field is in this plane.

b) Consider the general case.

2.42. Find the systematic change of the elliptic orbit of a particle in the fieldU(r) = −α/r under the influence of a small perturbation

δU(r,θ) = −βr2(3cos2 θ − 1).

Only consider the case when the orbit plane passes through the z-axis. This problemis a simplified model of the satellite motion in the Earth field taking into account thegravitational field of the Moon near the Earth space.

2.43. Taking that the orbit of the Moon in the Earth field is an ellipse lying in the planeof the Earth orbit, find the systematic change of the Moon orbit under the influence ofthe perturbation

δU(r, χ) = −12m�2r2(3cos2 χ − 1),

where m is the Moon mass, � is the angular velocity of the Earth around the Sun, χ isthe angle between Earth to Sun and Earth to Moon directions.

2.44. Find the systematic displacement of the finite orbit of a charged particle movingin the field U(r) = −α/r and in the field of the magnetic dipole μ, if the effect of thelatter may be considered to be a small perturbation. Take the vector potential in the formA(r) = [μ,r]/r3.

2.45. Find the average precession rate of the orbit of a particle moving in the fieldU(r) = −α/r under the influence of a small additional “friction force” F = βv (such

form has the force of radiation damping; in this case, β = 23

q2

c3 , where q is the charge of

the particle and c is the velocity of light; see [2], § 75).


§3

Scattering in a given field. Collisionbetween particles

3.1. Find the differential cross-section for the scattering of particles with initial velocityparallel to the z-axes by smooth elastic surfaces of revolution ρ(z) for the following cases:

a) ρ = bsin za , 0 � z � πa;

b) ρ = Azn, 0 < n < 1;

c) ρ = b − a2

z , a2

b � z < ∞.

3.2. Find the surface of revolution which is such that the differential cross-section forelastic scattering by this surface coincides with the Rutherford scattering cross-section.

3.3. Find the differential cross-section for the scattering of particles by spherical“potential barrier”:

U(r) ={

V , when r < a,0, when r > a.

3.4. Find the cross-section for the process where a particle falls towards the centre ofthe field U(r) when U(r) is given by:

a) U(r) = αr − β

r2 , b) U(r) = β

r2 − γ

r4 .

3.5. Calculate the cross-section for particles to hit a small sphere of radius R placed atthe centre of the field U(r) for the cases:

a) U(r) = − αrn , n � 2; b) U(r) = β

r2 − γ

r4 .

3.6. A uniform beam of meteors with velocity v∞ flies towards the planet. What fractionof meteorites that fell on the planet falls on the part invisible from the beam side? Takea planet as a uniform ball of radius R and mass m0.



3.12] §3. Scattering in a given field 13

3.7. Find the differential cross-section for the scattering of particles by the field U(r):

a) U(r) ={α

r − αR , when r < R,

0, when r > R;

b) U(r) ={1

2 mω2(r2 − R2), when r < R,0, when r > R.

3.8. Find the differential cross-section for the scattering of fast particles (E � V ) by thefield U(r) for the following cases:

a) U(r) ={

V√

1 − (r2/a2), when r < a,0, when r > a;

b) U(r) ={

V ln(r/a), when r < a,0, when r > a;

c) U(r) = V ln(1 + a2

r2

);

d) U(r) ={

V(1 − r2

R2

), when r < R,

0, when r > R.

3.9. Find the differential cross-section for small angle scattering in the field

U(r)= β

r4 − α

r2 .

3.10. Find the differential cross-section for the scattering of particles by the field U(r) =−α/r2.

3.11. Find the differential cross-section for the scattering of fast particles (E � V ) bythe following fields U(r):

a) U(r) = Ve−�2r2; b) U(r) = V

1 + �2r2 .

Study in detail the limiting cases when the deflection angle is close to its minimum ormaximum value.

3.12. A beam of particles with their velocities initially parallel to the z-axis is scatteredby the fixed ellipsoid

x2

a2 + y2

b2 + z2

c2 = 1.

Find the differential scattering cross-section for the following cases:



a) the ellipsoid is smooth and the scattering elastic;

b) the ellipsoid is smooth and the scattering inelastic;

c) the ellipsoid is rough and the scattering elastic.

3.13. Find the differential cross-section for small-angle scattering by the followingfields U(r) (a is a constant vector):

a) U(r) = arr2 ; b) U(r) = ar

r3 .

3.14. Find the change in the differential cross-section for scattering of particles bythe field U(r) when U(r) is varied by a small amount δU(r) for the following cases:

a) U(r) = αr , δU(r) = β

r2 ;

b) U(r) = αr , δU(r) = γ

r3 ;

c) U(r) = β

r2 , δU(r) = γ

r3 .

3.15. Find the differential cross-section as function of the energy acquired by fastparticles (E � V1,2) due to their scattering in the field

U(r, t) = (V1 + V2 sinωt)e−�2r2.

3.16. A particle with velocity V decays into two identical particles. Find the distributionof the secondary particles over the angle of divergence, that is, the angle between thedirections at which two secondary particles fly off. The decay is isotropic in the centreof mass system and velocity of the secondary particles is v0 in that system.

3.17. Find the energy distribution of secondary particles in the laboratory system, if theangular distribution in the centre of mass system is

dNN

= 38π

sin2 θ0 d�0,

where θ0 is the angle between the velocity V of the original particle and the direction inwhich one of secondary particle flies off in the centre of mass system. The velocity ofthe secondary particles in the centre of mass system is v0.

3.18. An electron moving at infinity with velocity v collides with another electron at rest;the impact parameter is ρ. Find the velocities of the two electrons after the collision.

3.19. Find the range of possible values for the angle between the velocity directions aftera moving particle of mass m1 has collided with a particle of mass m2 at rest.



3.20. Find the differential cross-section for the scattering of inelastic smooth spheres bysimilar ones at rest.

3.21. Find the change in intensity of a beam of particles travelling through a volumefilled with absorbing centres; their density n cm−3, and the absorption cross-section σ .

3.22. Find the number of reactions occurring during a time dt in a volume element dVwhen two beams of particles with velocities v1, v2 and densities n1, n2, respectively,collide. The reaction cross-section is σ .

3.23. A particle of mass M moves in a volume filled with particles of mass m (m � M)which is at rest initially. The differential cross-section for scattering of M by m is dσ =f (θ)d�, and the collisions are assumed to be elastic.

a) Find the “friction force” acting on the particle M;

b) Find the average of the square of the angle over which the particle M is deflected.


§4

Lagrangian equations of motion.Conservation laws

4.1. A particle, moving in the field U(x) = −Fx, travels from the point x = 0 to thepoint x = a in a time τ . Find the law of motion of particle, assuming it to be of the formx(t) = At2 + Bt + C, and determine the parameters A, B, and C such that the action is aminimum.

4.2. A particle moves in the xy-plane in the field

U(x, y) ={

0, when x < 0,V , when x > 0,

and travels in a time τ from the point (−a, 0) to the point (a, a). Find its position as afunction of time, assuming that it satisfies the equations

x1,2(t) = A1,2t + B1,2,

y1,2(t) = C1,2t + D1,2.

The indices 1 and 2 refer, respectively, to the left-hand (x < 0) and right-hand (x > 0)

half-planes.

4.3. Prove by direct calculations the invariance of the Lagrangian equations of motionunder the coordinate transformation

qi = qi(Q1, Q2, . . . , Qs, t), i = 1, 2, . . . , s.

4.4. What is the change in the Lagrangian function in order that the Lagrangianequations of motion retain their form under the transformation to new coordinates and“time”:

qi = qi(Q1, Q2, . . . ,Qs, τ), i = 1, 2, . . . , s, t = t(Q1, Q2, . . . ,Qs, τ)?



4.10] §4. Lagrangian equations of motion 17

4.5. Write down the Lagrangian function and the equations of motion for a particlemoving in a field U(x), introducing a “local time” τ = t − λx.

4.6. How does the Lagrangian function

L = −√

1 −(dx

dt

)2

transform when we change to the coordinates q and “time” τ through the equations

x = qchλ + τ shλ

t = qshλ + τ chλ

4.7. How do the energy and generalized momenta change under the following coordinatetransformation

qi = fi(Q1, . . . , Qs, t) i = 1, . . . , s?

4.8. How do the energy and generalized momenta which are conjugate to (a) thespherical polar and (b) the Cartesian coordinates transform under a change to acoordinate system which is rotating around the z-axis

a) ϕ = ϕ′ + �t, r = r′

b) x = x′ cos�t − y′ sin�t, y = x′ sin�t + y′ cos�t

4.9. How do the energy and momenta change when we change to a frame of referencewhich is moving with a constant velocity V? Take the Lagrangian function L′ in themoving frame of reference in either of two forms:

a) L′1 = L(r′ + Vt, r′ + V, t), where L(r, r, t) is the Lagrangian function in the origi-

nal frame of reference;

b) L′2 = 1

2 mv′2 − U(r′ + Vt, t). Here L′2 differs from L′

1 by the total derivative withrespect to the time of the function

Vmr′ + 12 mV2t.

4.10. Smooth rod OA of length l rotates around a point O in a horizontal plane with aconstant angular velocity � (Fig. 6). The bead is fixed on the rod at a distance a fromthe point O. The bead is released, and after a while, it is slipping off the rod. Find itsvelocity at the moment when the bead is slipping?



OΩ

A

y

m

Ωt

ϕ

x

x´b

a

y´

Figure 6 Figure 7

4.11. At the end of the rod of length a, rotating with constant angular velocity � in theplane of the figure, is pivotally attached another freely rotating rod of length b, at the endof which there is a particle of mass m (Fig. 7). Find the Lagrangian function L(ϕ, ϕ, t) ofa system and the frequency of small oscillations of the second rod.

4.12. Consider an infinitesimal transformation of the coordinates and the time of theform

q′i = qi + εfi(q, t),

t′ = t + εh(q, t), ε → 0.

Let that the action is invariant under this transformation (with an accuracy to terms ofthe order of ε, inclusive),

t2∫t1

L(q,

dqdt

, t)

dt =t′2∫

t′1

L(q′, dq′

dt′, t′

)dt′.

(We emphasize that in the left and right sides of this equation is the same function L,but from different arguments.) Prove that the quantity

∑i

∂L∂ qi

(qih − fi)− Lh

is an integral of motion.

4.13. Generalize the theorem of the preceding problem to the case when under thetransformation of the coordinates and the time the action changes in the following way:

t2∫t1

L(q,

dqdt

, t)

dt =t′2∫

t′1

{L(q′, dq′

dt′, t′

)+ ε

dF(q′, t′)dt′

}dt′,

where F(q, t) is an arbitrary function of coordinates and time.



4.14. Find the integrals of motion if the action remains invariant under

a) a translation;

b) a rotation;

c) a shift in the origin of the time;

d) a screw shift;

e) the transformation of problem 4.6.

4.15. Find the integrals of motion for a particle moving in

a) a uniform field U(r) = −Fr;

b) a field U(r), where U(r) is a homogeneous function,

U(αr) = αnU(r);

specify, for what value of n the similarity transformation leaves the action invariant;

c) the field of a travelling wave U(r, t) = U(r − Vt), where V is a constant vector;

d) a magnetic field specified by the vector potential A(r), where A(r) is a homoge-neous function;

e) an electromagnetic field rotating with a constant angular velocity � around thez-axis.

4.16. Find the time of the fall of the particle in the centre of the field U(r) = −(ar)2/r4.At the initial moment, the coordinates and velocity of the particle are equal to r0 and v0.

4.17. Find the integral of motion corresponding to the Galilean transformations.Hint: Use the result of problem 4.13.

4.18. Find the integrals of motion of a particle moving in a uniform constant magneticfield B if the vector potential is given in the form:

a) A(r) = 12 [B,r]; b) Ax = Az = 0, Ay = xB.

4.19. Find the integrals of motion for a particle moving in

a) the field of magnetic dipole specified by the vector potential A(r) = [μ,r]/r3,where μ = const;

b) the field given by the vector potential Aϕ = μ/r, Ar = Az = 0.

4.20. Find the equations of motion of a system with the following Lagrangian function:



a) L(x, x) = e−x2−x2 + 2xe−x2x∫

0e−y2

dy;

b) L(x, x, t) = 12 eαt(x2 − ω2x2).

4.21.

a) Write down the components of the acceleration vector for a particle in the sphericalcoordinate system.

b) Find the components of the acceleration in the orthogonal coordinate qi, if the lineelement is given by the equation

ds2 = h21 dq2

1 + h22 dq2

2 + h23 dq2

3,

where hi(q1, q2, q3) are the Lame coefficients.

4.22. Write down the equations of motion of a point particle using arbitrary coordi-nates qi which are connected with the Cartesian coordinates xi by the relations:

a) xi = xi(q1, q2, q3), i = 1, 2, 3;

b) xi = xi(q1, q2, q3, t), i = 1, 2, 3.

4.23. Show that the Lagrangian function [25]

L = m2

(r2 + r2θ2 + r2ϕ2 sin2 θ)− egc

ϕ cosθ ,

where r,θ , and ϕ are the spherical coordinates, describes the motion of a charged particlein the magnetic field B(r) = gr/r3 (see problem 2.34). Find the integrals of motion.

U Uq2

CA

(a) (b)

A

B BI L

Figure 8

4.24. Verify that one can use the Lagrangian functions

L1 = L q21

2− Uq1, L2 = − q2

2

2C+ Uq2,

to get the correct “equations of motion” for q1 and q2and the correct energies. Here q1 = I is the currentflowing through the inductance L in the direction from A to B (Fig. 8a), q2 the chargeon the upper plate of the capacitor (Fig. 8b), and U the voltage between A and B(U = ϕB − ϕA).

4.25. Use the additivity property of the Lagrangian functions and the results of thepreceding problem to find the Lagrangian functions and the Lagrangian equations ofmotion for the circuits of Fig. 9a–9c.



L U C L C

L1 L2

CC1 C2

(a) (b) (c)

Figure 9

4.26. Find the Lagrangian functions for the following systems:

a) a circuit with a variable capacitor, the movable plate of which is connected to apendulum of mass m (Fig. 10), and the capacitor of which is a known function C(ϕ)

of the angle ϕ the pendulum makes with the vertical. The mass of the capacitorplate may be ignored;

b) a core suspended from a spring with elastic constant k inside a solenoid withinductance L (x) which is a given function of the displacement x of a core(Fig. 10b).

m

C(ϕ) L

ϕ

(a)

L(x)

xk

C

(b)

a

B

B

A

D

C

Figure 10 Figure 11

4.27. A perfectly conducting square frame can rotate around a fixed side AB = a(Fig. 11). The frame is placed in a constant uniform magnetic field B at right anglesto the AB axis. The inductance of the frame is L and the mass of the side CD is m; themasses of the other sides may be neglect.

Describe qualitatively the motion of the frame.

4.28. Use the method of the undetermined Lagrangian multipliers to obtain theequations of motion of a particle in the field of gravity when it is constrained to move

a) along a parabola z = ax2 in a vertical plane;

b) along a circle of radius r = l in a vertical plane.

Determine the forces of constraint.



4.29. A particle moves in the field of gravity along a straight line which is rotatinguniformly in a vertical plane. Write down the equations of motion and determine themoment of the forces of constraint.

4.30. One can describe the influence of constraints and friction on the motion of asystem by introducing generalized constraint and friction forces Ri into the equations ofmotion:

ddt

∂L∂ qi

− ∂L∂qi

= Ri.

a) How does the energy of the system vary with time?

b) What is the transformation of the forces Ri which lives the equations of motioninvariant under a transformation to new generalized coordinates

qi = qi(Q1, . . . , Qs, t)?

4.31. Let the constraint equations be

qβ =s∑

n=r+1

bβnqn, β = 1, . . . , r,

while the Lagrangian function L(qr+1, . . . , qs, q1, . . . , qs, t) and the coefficients bβn donot depend on coordinates qβ .

Show that the equations of motion can be written in the form

ddt

∂L∂ qn

− ∂L∂qn

+r∑

β=1

∂L∂ qβ

s∑m=r+1

(∂bβm

∂qn− ∂bβn

∂qm

)qm = 0,

where L(qr+1, . . . , qs, qr+1, . . . , qs, t) is the function obtained from L by using the con-straint equations to eliminate the velocity q1, . . . , qr .

q

xnaa 2a

qn

Figure 12

4.32. A continuous string can be thought of as the lim-iting case of a system of N particles (Fig. 12) which areconnected by an elastic thread, in the limit as N → ∞,a → 0, Na = const. The Lagrangian function for a discretesystem is

L(q1, q2, . . . , qN , q1, q2, . . . , qN , t)

=N+1∑n=1

Ln(qn, qn − qn−1, qn, t),



where qn is the displacement of the nth particle from its equilibrium position.

a) Obtain the equation of motion for a continuous system as the limit case of theLagrangian equations of motion of a discrete system.

b) Obtain an expression for the energy of a continuous system as the limiting case ofthe expression for the energy of a discrete system.

Hint: Introduce the coordinate x of a point on the string together with the expressionobtained as a result of taking the limit as a → 0, n = x/a → ∞:

q(x, t) = limqn(t),∂q∂x

= limqn(t)− qn−1(t)

a,

L(x, q,

∂q∂x

,∂q∂t

, t)

= limLn(qn, qn − qn−1, qn, t)

a.

4.33. A charged particle moves in the potential field U(r) and a constant magneticfield B(r), where U(r) and B(r) are homogeneous functions of the coordinates ofdegrees k and n, respectively, that is,

U(αr) = αkU(r), B(αr) = αnB(r).

Develop for this system the similarity principle, determining for what value of n it holds.

4.34. Generalize the virial theorem for a system of charged particles in a uniformconstant magnetic field B. The potential energy U of the system is a homogeneousfunction of the coordinates,

U(αr1, . . . , αrs) = αkU(r1, . . . , rs),

and the system moves in a bounded region of space with velocities which remain finite.

4.35. Three identical particles move along the same line and interact pairwise accordingto the law Uik = U(xi − xk), where xi is the coordinate of ith particle. Prove that besidesthe obvious integrals of motion

P = m(x1 + x2 + x3),

E = 12 m(x2

1 + x22 + x2

3)+ U12 + U23 + U31

there is an additional integral of motion [23]

A = mx1x2x3 − x1U23 − x2U31 − x3U21,



when the function U(x) has the following form:

a) U(x) = g2

x2 ,

b) U(x) = g2a2

sinh2 ax.

4.36. Consider the collision of the three particles described in the preceding problem.Assume x1 > x2 > x3, let the distances between the particles be infinitely large at t →−∞, and let their velocities vi = xi (t = −∞) be such that v3 > v2 > v1. Find v′

i = xi(t = +∞).


§5

Small oscillations of systems withone degree of freedom

5.1. Find frequency ω of the small oscillations for particles moving in the following fieldsU(x):

a) U(x) = V cosαx − Fx; b) U(x) = V (α2x2 − sin2 αx).

5.2. Find the frequency of the small oscillations for the system depicted in Fig. 13. Thesystem rotates with the angular velocity � in the field of gravity around a vertical axis.

5.3. A point charge q of mass m moves along a circle of radius R in a vertical plane andin the field of gravity. Another charge q is fixed at the lowest point of the circle (Fig. 14).Find the equilibrium position and the frequency of the small oscillations for the firstpoint charge.

5.4. Describe the motion along a curve close to a circle for a point particle in the centralfield U(r) = −α/rn (0 < n < 2).

5.5. Find the frequency of the small oscillations of a spherical pendulum (a particleof mass m suspended a string of length l) if the angle of deflection from the vertical, θ

oscillates about the value θ0.

a aθ

m

m

m

a a

Ω

A

R

qϕ

qm

Figure 13 Figure 14




5.6. Find the correction to the frequency of the small oscillations of a diatomic moleculedue to its angular momentum M.

A Bk km

2l

Figure 15

5.7. Determine the free oscillations of the system shown in Fig. 15for the case when the particle moves:

a) along the straight line AB;

b) at a right angle to AB.

How does the frequency depend on the tension of the springs in the equilibriumposition?

5.8. Find the free oscillations of the system (Fig. 16) in a uniform field of gravity for thecase when the particle can only move vertically.

5.9. Find the stable small oscillations of a pendulum when its point of suspension movesuniformly along a circle of radius a with an angular velocity � (Fig. 17). The pendulumlength is l (l � a).

k

k

m2l

a

l

m

Ω

LU

R

C

Figure 16 Figure 17 Figure 18

5.10. Find the stable oscillations in the voltage across a capacitor and the current ina circuit with an e.m.f. U(t) = U0 sinωt (Fig. 18).

5.11. Determine the law of motion of an oscillator with friction which initially is at restand which is acted upon by a force F(t) = F cosγ t.

5.12. Determine the energy E acquired by an oscillator under the action of a forceF(t) = Fe−(t/τ)2

during the total time its acts

a) if the oscillator was at rest at t = −∞;

b) if the amplitude of the oscillation at t = −∞ was a.

5.13. Describe the motion under the action of a force F(t)

a) of an unstable system described by the equation

x − μ2x = 1m

F(t);


5.19] §5. Small oscillations of systems with one degree of freedom 27

b) of an oscillator with friction

x + 2λx + ω20x = 1

mF(t).

5.14. Find the differential cross-section for an isotropic oscillator to be exited to theenergy ε by a fast particle (E � V ) if the interaction between the two particles is throughthe field

U(r) = Ve−2r2.

The energy of the oscillator is zero initially.

5.15. An oscillator can oscillate only along the z-axis. Find the differential cross-sectionfor the oscillator to be excited to an energy ε by a fast particle E � V , if the interactionbetween the particles is through the potential energy U(r) = Ve−2r2

. The particle movesalong the z-axis with velocity v∞, and the initial energy of the oscillator is ε0.

5.16. A force F(t), for which F(−∞) = 0, F(+∞) = F0, acts upon an harmonicoscillator. Find the energy E(+∞) gained by the oscillator during the total time the forceacts, and the amplitude of the oscillator at t → +∞, if it were at rest at t → −∞.

5.17. Find the energy acquired by an oscillator under the action of the force

F(t) ={

12 F0 eλt, when t < 0,12 F0 (2 − e−λt), when t > 0.

At t → −∞ the energy of the oscillator was E0.

5.18. Estimate the change in the amplitude of the vibrations of an oscillator when a forceF(t) is switched on slowly over a period of τ so that ωτ � 1. Assume F(t) = 0 for t < 0,F(t) = F0 for t > τ , while F(k) ∼ F0/τ

k (k = 0, 1, . . . , n + 1) for 0 < t < τ and F(s)(0) =F(s)(τ ) = 0 (s = 1, 2, . . . , n − 1), while the nth derivative of the force has a discontinuityat t = 0 and at t = τ .

F

tτ 2τ

F

t2ττ

Figure 19 Figure 20

5.19. Find the stable oscillations of an oscillator which is acted upon by a periodicforce F(t) in following cases:



a) F(t) = F · (t/τ − n), when nτ � t < (n + 1)τ (Fig. 19);

b) F(t) = F · (1 − e−λt′), t′ = t − nτ , when nτ � t < (n + 1)τ (Fig. 20).

c) Find the stable current through the circuit of Fig. 16 in which there is a.m.f. U(t) =(t/τ − n)V for nτ � t < (n + 1)τ . The internal resistance of the battery is zero.

5.20. An oscillator with eigen-frequency ω0 and with a friction force acting upon it givenby ffr = −2λmx has an additional force F(t) acting upon it.

a) Find the average work A done by F(t) when the oscillator is vibrating in a stablemode for the case when F(t) = f1 cosωt + f2 cos2ωt.

b) Repeat the calculation for the case when

F(t) =∞∑

n=−∞aneinωt, a−n = a∗

n.

c) Find the average over a long time interval of the work done by the force F(t) =f1 cosω1t + f2 cosω2t when the oscillator performs stable vibrations.

d) Find the total work done by the force

F(t) =∫ ∞

−∞ψ(ω)eiωt dω, ψ(−ω)=ψ∗(ω)

for the case where the oscillator was at rest at t → −∞.

5.21. A harmonic oscillator is in the field of the travelling wave that acts upon it withthe force F(x, t) = f (t − x/V ), where x is the displacement of the oscillator from theequilibrium position and V is the wave velocity. Assuming x is small enough, find theconnection between the energy �E and the momentum

�p =∫ ∞

−∞F(x(t), t)dt,

transferred to the oscillator. Consider the approximation quadratic in F only and assumef (±∞)=0.


§6

Small oscillations of systems withseveral degrees of freedom

In problems 6.1–6.22, we study free and forced oscillations of simple systems (with twoor three degrees of freedom) using common methods. In problems 6.19– 6.22, systemswith the degenerate frequencies are presented.

For more complex systems, it is helpful to use orthogonality of eigen-oscillationsand the symmetry properties of the system. The corresponding theorems are givenin problems 6.23 and 6.40 (see also [7], § 24 and § 25), whereas their application areillustrated, for instance, in problems 6.27–6.43 and 6.41–6.47, 6.49 (see also problemsrelated to vibrations of molecules in 6.48, 6.50–6.54).

How small changes in the system can influence its motion will be studied using theperturbation theory. The general form of the perturbation theory for small oscillationsis given in problem 6.35, whereas some specific examples can be found in problems6.36, 6.38, 6.42, 6.43, 6.52 b. It is useful to note that the perturbation theory in quantummechanics is constructed in a similar way.

Oscillations of systems in which gyroscopic forces act are studied in problems 6.37–6.39 (see also problems 9.27–9.30 and [7], § 23).

6.1. Find the free oscillations of the system of Fig. 21 for the case when the particlesmove only vertically. Find the normal coordinates and express the Lagrangian functionin terms of these coordinates.

6.2. Find the stable oscillations of the system described in the preceding problem, if thepoint of suspension moves vertically according to equation a(t), where

a) a(t) = acosγ t,

b) a(t) = a(

tτ − n

)for nτ � t < (n + 1)τ .

6.3. Find the free small oscillations of the coplanar double pendulum (Fig. 22).

6.4. Find the normal oscillations for the double pendulum (Fig. 23) with the angle 60◦.The angle is between the planes of vibrations of the upper particle with a mass of 3m




and the lower particle with mass m. The length of each rod is l, the masses of the rodsare negligible.

k

k

m

m B

A

x

m

m l

2l

y

l

l

3m

m


6.5. Find the free oscillations of the system described by the Lagrangian function

L = 12

(x2 + y2 − ω2

1x2 − ω22y2

).

What is the trajectory of a point with Cartesian coordinate (x,y)?

6.6. Find the normal coordinates of the systems with the following Lagrangianfunctions:

a) L = 12

(x2 + y2 − ω2

1x2 − ω22y2

) + αxy;

b) L = 12

(m1x2 + m2y2 − x2 − y2

) + βxy.

L1 L2 L1 L2

LC1 C1

(a) (b)

C2 C2

Figure 24

6.7. Find the eigen-oscillations of a system of coupled circuits: a) Fig. 24a; b) Fig. 24b.

Ak1 k2 k3m1 m2

B A Bk1k kmm

Figure 25 Figure 26

6.8. Find the eigen-oscillations of a system of particles which are connected by springs(Fig. 25). The particles can move only along the straight line AB. Find the free oscillationsof the system.


6.15] §6. Small oscillation of systems with several degrees of freedom 31

6.9. Find the free oscillations of the system of Fig. 26 where the particles can move onlyalong the straight line AB

a) if at t = 0 one of the particles moves with velocity v while the second particle isat rest and the displacements from the equilibrium position of both particles arezero;

b) if at t = 0 one of the particles is displaced from its equilibrium position over adistance a, while the other is at rest at its equilibrium position and both particlesare at rest.

6.10. Determine the flux of energy flow from one particle to the other in the precedingproblem.

6.11. Find the free oscillations of the system of Fig. 26 if each of the particles is actedupon by a frictional force which is proportional to the particle’s velocity.

6.12. Find the free small oscillations of the coplanar double pendulum of Fig. 27 if in theinitial moment the upper pendulum is vertical, while the bottom pendulum is displacedby an angle β � 1, and their velocities are zero. The pendulum masses are M and m,respectively (M � m).

l

l

m

M

A A

k k k

k k

k

(a) (b)

k

m m

m

m

1

2

3 m

Figure 27 Figure 28

6.13. Find the stable oscillations of the system of Fig. 26 if the point A moves accordingto the relation acosγ t in the direction of a straight line AB.

6.14. Find the stable oscillations of a system of two particles which move along the ring1

(Fig. 28a) for the case when the point A moves along the ring according to the relationacosγ t. Study how the amplitude of the oscillations depends on the frequency of theapplied force.

6.15. Three particles with mass m each, connected by springs, can move along the ring(Fig. 28b). Find the stable oscillations of the system if point A moves along the ringaccording to the law acosγ t.

1 In this and similar problems, the ring is assumed to be smooth and fixed.



6.16. Find the stable oscillations of the system of Fig. 26 when point A moves accordingto the relation acosγ t. The particles are acted upon by the frictional forces proportionalto their velocities.

6.17. Find the motion of the system of Fig. 25 when in the initial moment the particle areat rest in the equilibrium positions, and point A moves according to the relation acosγ t.The masses of all particles are equal (m1 = m2 = m).

6.18. Determine the stable oscillations of the particle of Fig. 29 moving in a variable fieldU(r, t) = −F(t)r, where the vector F(t) lies in the plane of the figure, for the followingcases:

a) F(t) = F0 cosγ t,

b) the vector F(t) rotates with constant absolute amplitude magnitude with a fre-quency γ .

6.19. Find the eigen-oscillations of three identical particles connected by identicalsprings and moving along the ring (Fig. 30).

C

A B

D

k

k

k1

k1

m

m

1

2 3m m

k

k

k

Figure 29 Figure 30

m

3k2k

3m2m

6k

m1

3

2 4

m

mm

kk

k k

Figure 31 Figure 32

Determine the normal coordinates which reduce the Lagrangian function to adiagonal form.

6.20. Find the free oscillations of the system considered in the preceding problem, if att = 0 one of the particles is displaced from its equilibrium position. The initial velocitiesof all particles are zero.



6.21. Find the eigen-oscillations of the system of three particles which can move alongthe ring (Fig. 31).

6.22. Find the normal coordinates of the system of four identical particles on the ring(Fig. 32).

6.23. Consider a system described by the Lagrangian function1

L = 12

∑i, j

(mij xi xj − kijxixj).

Let their eigen-oscillations be described by the following equation:

x(l)i (t) = A(l)

i cos(ωl t + ϕl).

Prove that the amplitudes corresponding to the oscillations with different frequen-cies ωl and ωs satisfy the relation

∑i, j

A(s)i mijA

(l)j =

∑i, j

A(s)i kijA

(l)j = 0.

6.24. Consider a system described by the Lagrangian function

L = 12

∑i, j

(mij xi xj − kijxixj)

which also has the different eigen-frequencies �1 < �2 < < .. . < �N . The linearrelationship

∑i

aixi = 0, ai = const

is imposed upon the system. Prove that all eigen-frequencies ωl of the system with suchrelation lie between �l :

�1 � ω1 � �2 � ω2 � . . . � ωN−1 � �N .

1 Since the product xi xj is symmetric with respect to the replacement of i ↔ j, then the matrix mij can alwaysbe chosen as the symmetric one, mij = mji . The same holds for the matrix kij = kji .



6.25. We can write the stable oscillations of the system described by the Lagrangianfunction

L = 12

∑i, j

(mij xi xj − kijxixj)+∑

i

xifi cosγ t,

in the form xi(t) = ∑l λ

(l)A(l)i cosγ t (see problem 6.23.) (Why?)

Express the coefficients λ(l) in terms of the fi and the A(l)i .

Study the γ -dependence of the λ(l).Show that λ(s) = 0, if

∑i fiA

(s)i = 0 for the sth normal oscillation.

6.26. A system of particles connected by springs can perform small oscillations. Oneof them (the particle A) be acted upon by the force F(t) = F0 cosγ t along the x-direction, while the other (the particle B) performs the stable oscillations, during whichthe projection of its displacement on the x′-direction has the form x′

B = C cosγ t.Show that when force F acts upon the particle B along the x′-axis, there will be the

oscillations xA = C cosγ t of the particle A (the reciprocity theorem).

6.27. Find the eigen-oscillations of the system of four particles moving along the ring(Fig. 33). All springs are identical, and the masses of particles 1 and 3 are m, while thoseof particles 2 and 4 are M.

6.28. Find the eigen-oscillations of the system of four particles of Fig. 34; the particlesmove along the ring.

1

3

2 4

m

m

M M

k

k

k

k

k

k k

m

m m

2m

4

2

1 3

k

Figure 33 Figure 34

6.29. a) Find the normal oscillations of the system of Fig. 35. All particles and springsare identical. The tension in the springs at equilibrium is f = kl, where l is the equilibriumdistance between the particles.

Hint: Several of the eigen-vibrations are obvious. The determination of the others canbe simplified by using the relations of problem 6.23.

b) Find the eigen-oscillations of the system of four identical particles of Fig. 32 forthe case where the mass of particle 5 is equal to zero and the springs in this place are



connected. The elasticity coefficients and tensions at equilibrium of all the springs arethe same as before.

6.30. Three particles with different masses mi (i = 1, 2, 3) move along the ring (Fig. 36).At what values of the elasticity coefficients ki will the degeneracy of frequencies occurin this system?

C

D

A B

km2

1 35

4

m1

m2 m3

k1

k3 k2

Figure 35 Figure 36

6.31. Which of the eigen-vibrations of the system of Fig. 30 remain practicallyunchanged when the following small changes are made in the system:

a) the elasticity coefficient of spring 1–2 is changed by a small amount δk;

b) a small mass δm is added to particle 3;

c) a small mass δm1 is added to particle 1 and δm2 to particle 2?

6.32. Describe the free oscillations of the system of the preceding problem for the casea) and b) if initially the particles 1 and 3 are displaced over equal distances in oppositedirections so as to decrease their mutual distance. All velocity are initially zero.

6.33. The system of Fig. 32 has degenerate frequencies; therefore, its eigen-vibrationsare not uniquely determined. Even a small change in the mass of particles or a smallchange in the elasticity of the springs can lead to the removal of degeneration.

Find the eigen-vibrations of the system of Fig. 32 which are practically the same asthe eigen-vibrations of the system which is obtained

a) by adding identical masses to the first and the second particles;

b) by changing the elasticity coefficients of the springs 1–2 and 3–4;

c) by adding an extra mass to the first particle.

6.34. The particles 1 and 3 of the system described in the problem 6.35b are at time t = 0displaced by the same amount from their equilibrium positions in opposite direction so



as to decrease their distance apart. Initially all velocities are equal zero. Describe the freeoscillations of the system.

6.35. Consider a system described by the Lagrangian function

L0 = 12

∑i, j

(mij xi xj − kijxixj).

Now consider the new system which has been obtained by the small change in the originalLagrangian function:

δL = 12

∑i, j

(δmij xi xj − δkijxixj).

Find the small change of the eigen-frequencies of the new system if all eigen-frequenciesof the original system are non-degenerate.

6.36. Find the small changes in the eigen-frequencies of the system of Fig. 34 when asmall mass δm is added to the first particle so that ε = δm/m � 1.

6.37. Determine the free oscillations of an anisotropic charged oscillator moving in thepotential field

U(r) = 12 m(ω2

1x2 + ω22y2 + ω2

3z2)

and in a uniform constant magnetic field B which is parallel to the z-axis. Consider inparticular the following limit cases:

a) |ωB| � |ω1 − ω2|,b) |ωB| � ω1,2,

c) ω1 = ω2 � |ωB|

(here ωB = eB/(mc)).

6.38. Determine the free oscillations of an anisotropic harmonic oscillator moving in thepotential field

U(r) = 12 m(ω2

1x2 + ω22y2 + ω2

3z2)

and in a weak constant uniform magnetic field B = (Bx, 0, Bz), considering the effect ofthe magnetic field to be a small perturbation.



m

C

l ⊕B

L

Figure 37

6.39. A mathematical pendulum is part of an electric chains(Fig. 37). A constant uniform magnetic field B is applied atright angles to the plane of the figure. Find the eigen-vibrationsof the system.

6.40. Suppose that a system, making small oscillations (andhence its Lagrangian function L(x, x) as well), does not changeits form with respect to the replacement

xi →∑

j

Sijxj ; xi →∑

j

Sij xj , i, j = 1, 2, . . . , N.

Here the constant coefficients Sij = Sji satisfy the condition1

∑j

SijSjk = δik.

Prove that

a) if the eigen-oscillation xi = Ai cos(ωt + ϕ) is not degenerate, then the amplitudeAi is either symmetric or anti-symmetric with respect to the given transformation,that is, either

∑j

SijAj = +Ai or∑j

sijAj = −Aj ;

b) if the frequency is degenerate, one can choose the eigen-oscillation to be eithersymmetric or anti-symmetric;

c) if the system is acted upon by an external force which is symmetric (or anti-symmetric) with respect to this transformation, then the anti-symmetric (or thesymmetric) eigen-oscillations is not excited. (This is one example of the so-calledselection rules.)

m

2

4 5

3 1mm

M M

k

kk

k

k

Figure 38

6.41. Using symmetry considerations, find the eigen-oscillationsof the system of Fig. 38.

6.42. Determine the free oscillations of the system of Fig. 35 whenat t = 0 particles 1 and 4 are displaced over equal distances inthe horizontal direction towards each other in such a way thatthe distance between them decreases. At t = 0 the velocities ofall particles equal zero. The tension of the springs is f = kl1;l − l1 � l, where l is the equilibrium distance between the particles(cf. problem 6.32).

1 This condition means that double application of such a replacement returns the system to its original state.For example, this is valid for reflections with respect to a plane of the system symmetry or rotations at 180◦relative to the symmetry axis.



6.43. Find corrections to the frequencies of the eigen-oscillations of the systems of fourparticles on the ring (Fig. 32). Such corrections arise from the small changes in theparticles’ masses: by δm1 for the particle 1 and by δm2 for the particle 2.

6.44. Using symmetry considerations, determine the vectors of the eigen-oscillations forthe system of Fig. 39. All the masses of the particles and the springs are identical.

B

B

AA

km

1 2

4 3

B DC

mm

m m EA

l l

l3

3l 3l

Figure 39 Figure 40

6.45. Find the eigen-oscillations of the “scales” of Fig. 40. The suspension of a rigidframe BCD is implemented with a short flexible thread allowing any turns of the framearound the point C. The length of the rods BC = CL = l, BD = l

√3, the length of the

threads AB = DE = 3l. The identical particles are fixed at points A, B, D, and E. Themasses of the rods and threads can be neglected.

6.46. Find the eigen-oscillations of the system of eight particles which are attached tothe fixed frame by springs (Fig. 41). The elasticity coefficients k, the tensions f and thelengths l of springs are identical.

6.47. The frame shown in Fig. 41 oscillates along the direction AA according to theequation acosγ t. At what values of the frequency γ will the resonant amplification ofthe oscillations be possible?

B

B

A

m

m

m

m

M M

M M

A

H HC C

Figure 41 Figure 42



6.48. Find the eigen-oscillations of a linear symmetric molecule of acetylene C2H2(Fig. 42), assuming that the potential energy of the molecule depends on the distancebetween neighbouring atoms as well as on angles HCC.

6.49. Two identical particles are attached to a fixed frame by springs (Fig. 43). Thesystem is symmetric with respect to the CF-axis. What information about the eigen-oscillations of the system can be obtained if the elasticities and the tensions of the springsare unknown?

6.50. Classify the eigen-oscillations of a molecule of ethylene C2H4 by their symmetryproperties with respect to axes AB and CD (Fig. 44). In the equilibrium position allatoms of the molecule are located in the same plane.

B

A E

D

C

F

H

CC

C

1

2

3 6

5

4H

H H

A

B

O D

Figure 43 Figure 44

6.51. Determine the eigen-oscillations in a plane of a molecule which has the shape ofan equilateral triangle. Assume that the potential energy depends only on the distancesbetween atoms and that all atoms are the same. The angular momentum is equal to zeroup to terms of first order in the amplitudes of the oscillations.

6.52. A molecule AB3 has the shape of an equilateral triangle with atom A in its centreand atoms B in its vertices (the molecule of boron chloride BCl3 is an example of sucha model).

a) Using symmetry considerations, determine the degree of degeneracy of the eigen-frequencies of the molecule.

b) Consider the oscillations which leave the molecule as an equilateral triangle, andthe oscillations which output atoms of the plane. Determine how the frequenciesof such oscillations will change when one of the atoms B (its mass m) is replacedby an isotope whose mass m + δm is close to m. The mass of atom A is mA.

6.53. Use symmetry arguments to determine the degree of degeneracy of variousfrequencies for the case of a “molecule” consisting of four identical “atoms” which hasthe form of a regular tetrahedron at equilibrium.

6.54. A molecule of methane CH4 has the form of a regular tetrahedron with fourthhydrogen atoms in vertices and one carbon atom in its centre.



a) Determine the degree of degeneracy of the eigen-frequencies of the molecule.

b) How many eigen-frequencies will experience resonance amplification when thismolecule is under the influence of a uniform alternating electric field? (Such are,for instance, the electromagnetic waves of the infrared range whose wavelength isseveral orders of magnitude greater than the size of the molecule.) Note that thehydrogen and carbon atoms have charges of opposite signs.

How does the amplitude of the oscillations of the carbon atom depend on theorientation of the molecule with respect to the electric field?


§7

Oscillations of linear chains

Chain of particles connected by springs are the simplest models used in theory of solids(e.g. see, [16]). The electrical analogues of such lines are r.f. lines employed in in radioengineering.

7.1. Determine the eigen-vibrations of a system of N identical particles with masses mconnected by identical springs with elastic constant k and moving along a straight line(Fig. 45).

Hint: Express the eigen-vibrations in the form of a superposition of travelling waves.

7.2. Repeat this for the system of Fig. 46 with one free end.

m k A

A

Figure 45 Figure 46

7.3. Determine the eigen-vibrations of the system of N plane pendulums suspendedfrom each other (Fig. 47). The masses of all pendulums are the same; the lengths areequal to Nl, (N−1)l, . . . , 2l, l, counting from the top.

7.4. Find the free oscillations of N particles which are connected by springs and whichcan move along the ring (Fig. 48). All particles and the elastic constants of the springsare the same. Let the motion be that of a wave travelling along the ring. Check whetherthe energy flux equals the product of the linear energy density and the group velocity.

7.5. Determine the free oscillations of a system of particles moving along a straight linefor the following cases (consider the hint for problem 7.1):




Nl

m

m

m

mm l

2l

(N–1)l

A

Figure 47 Figure 48

a) 2N particles, alternating with masses m and M connected by springs of elasticconstant k (Fig. 49);

b) 2N particles with masses m connected by springs with alternating elastic constantsk and K (Fig. 50);

AMM mm k k kK K K

Figure 49 Figure 50

c) 2N + 1 particles of mass m connected by springs with alternating elastic constantsk and K (Fig. 51).

k k kK K

Figure 51

7.6. a) Find the stable oscillations of the system described in problem 7.1 if the point Amoves according to acosγ t (Fig. 45).

b) Let the left point, at which the spring of the same system is fixed, moves accordingto x0 = acosγ t. Under what law should move the point A to make the oscillationrepresent a wave travelling along the chain towards the point A?

What will be in this case the flux of energy along the chain?c) Consider the same question as in a) for the system of Fig. 46.

7.7. The same questions as in a) and b) of the preceding problem, but for the systemof Fig. 49.


7.10] §7. Oscillations of linear chains 43

7.8. Determine the eigen-vibrations of a system of N particles which move along astraight line for the following cases:

a) mi = m �= mN , i = 1, 2, . . . , N − 1; the elastic constants of all the springs are thesame (Fig. 52). Study the cases when mN � m and mN � m;

b) ki = k �= kN+1, i = 1, 2, . . . , N; all the masses are the same (Fig. 53). Study thecases when kN+1 � k and kN+1 � k.

m m m mNk k k k k kN+1

Figure 52 Figure 53

7.9. a) N pendulums are connected by springs and move only in the vertical planepassing through the horizontal suspension line (Fig. 54). Find the eigen-vibrations ofthe system, if all the pendulums and the springs are the same and, in the position ofequilibrium, the length of the spring is equal to the distance between the suspensionpoints of neighbouring pendulums.

b) Find the forced oscillations of the system of Fig. 54 for the case when the lastparticle is acted upon by the driving force F(t) = F sinγ t which is directed parallel to thesuspension line.

c) 2N identical pendulums are connected by identical springs and move onlyin vertical planes which are perpendicular to the circular line of suspension (Fig. 55). Thedistance between neighbouring suspension points is a. The length of each of the springsin the unstretched state is b. Study how the stability of small oscillations near the verticalline depends on the parameter b − a. Consider the radius of the circular suspension line Rto be large enough so that small quantities l/R, a/R, b/R can be neglected.

mk

l

R

m

l

Figure 54 Figure 55

7.10. a) An AC voltage source U cosγ t is connected to one end of the artificial line(Fig. 56). Which chain Z with resistance R and inductance L0 (or do we need capacity?)must be connected to the other end of the line to produce the oscillations in the form ofa travelling wave (i.e. in this case the voltage on each of the capacitors will differ fromthat of the neighbour’s only by a definite phase shift)?

b) Do the same for the artificial line of Fig. 57.



L L L L

U C C C Z

Figure 56

L1 L2 L2L1

U C C C Z

Figure 57

7.11. Consider the elastic rod to be the limiting case of the system of N particles ofFig. 45 in the limit N → ∞, a → 0, where m and a are, respectively, the mass of particlesand the distance between neighbouring particles at equilibrium, while Nm and Na keptconstant. Write down the equations of motion for the oscillations of the rod as the limitingcase of the equations of motion of the discrete system.

Hint:Introduce the coordinate of a point of the rod at equilibrium ξ = na and considerthe following quantities

x(ξ , t) = limxn(t),∂x∂ξ

= limxn(t)− xn−1(t)

a

with the limit a → 0.

7.12. Write down the equations of motion for the oscillations of the rod in the precedingproblem taking into account the first non-vanishing correction due to a finite distance abetween the neighbouring particles.


§8

Non-linear oscillations

8.1. Determine the distortion in the oscillations of a harmonic oscillator which is causedby the presence of anharmonic terms in the potential energy for the following cases:

a) δU(x) = 14 mβx4; b) δU(x) = 1

3 mαx3.

8.2. Determine the distortion in the oscillations of a harmonic oscillator which is causedby the presence of anharmonic term δT = 1

2 mγ xx2 in the kinetic energy.

8.3. Determine the anharmonic corrections to the oscillations of a pendulum whosepoint of suspension moves along a circle (Fig. 17; l � a).

8.4. Determine the oscillations of a harmonic oscillator when there is a force

f1 cosω1t + f2 cosω2t

acting upon it, taking anharmonic corrections into account for the case when δU(x) =13 mαx3.

lm

k

y

x

Figure 58

8.5. A pendulum consists of a particle of mass m in the field of gravity gsuspended on a spring with the elasticity constant k (Fig. 58). The lengthof the spring in the free state is l0. Find the anharmonic corrections to theoscillations of the pendulum.

Use the Cartesian coordinates of the displacement of the particle fromthe equilibrium position.

8.6. Find the amplitude of the stable oscillations of an anharmonicoscillator which satisfies the equation of motion

x + 2λx + ω20x + βx3 = f cosωt

a) in the resonance region, |ω − ω0| � ω0;b) in the region where there is resonance with the tripled frequency of the force,

|3ω − ω0| � ω0 (frequency tripling).




8.7. a) Determine the amplitude and phase of the stable vibration of an oscillationunder conditions of parametric resonance:

x + 2λx + ω20(1 + hcos2ωt)x + βx3 = 0

provided

h � 1, |ω − ω0| � ω0, βx2 � ω20.

b) Determine the amplitude of the third harmonic in the stable vibration.

8.8. Determine the vibrations of the oscillator

x + ω20(1 + hcos2ωt)x = 0

provided

h � 1, |ω − ω0| � ω0

a) in the region where instability for parametric resonance occurs;b) close to the region of instability.

8.9. Let the frequency of a harmonic oscillator ω(t) change as is indicated in Fig. 59.Find the region where instability against parametric resonance occurs.

t

ω1

ω

ω2

2τ 3τ 4ττ

Figure 59

8.10. Determine how amplitudes will change over time for the weakly coupled oscillatorswhose Lagrangian function is

L = 12 m(x2 + y2 − ω2x2 − 4ω2y2 + 2αx2y).

8.11. Find the frequency of the small free oscillations of a pendulum whose point ofsuspension performs vertical oscillations with a high frequency γ (γ � √

g/l).

8.12. Find the effective potential energy for the following cases:

a) a particles of mass m moves in the field

U(r) = α

|r − acosωt| − α

|r + acosωt|at a distance r � a;


8.15] §8. Non-linear oscillations 47

b) a harmonic oscillator moving in the field

U(r) = 2αarr3 cosωt.

8.13. Determine the motion of a fast particle entering the field U(r) = a(x2 − y2)sinkzat a small angle to the z-axis (k2E � a).

8.14. Determine velocity of the orbit centre’s displacement for a charged particle in theweakly inhomogeneous magnetic field

Bx = By = 0, Bz = B(x), ε = B′(x)

B(x)r � 1, r = mvc

eB(x),

where r is the orbit radius.

8.15. The following problem represents a mechanical model of phase transitions of thesecond kind.

The iron ball of mass m can oscillate along the y-axis. It is connected to the springwhose potential energy has the form1 U(y) = −Cy2 + By4. Using an electromagnet onecan excite vibrations of the ball according to the equation y(t) = y0 cosγ t, where γ isconsiderably greater than the frequency of the eigen-vibration of the ball.2

Find how the frequency of small eigen-oscillations of the ball depends on theparameter T = y2

0.

1 Such is, for example, the potential energy of the system shown in Fig. 15 where a ball can move only inthe direction of the y-axis perpendicular to the AB line, and the length of the unstretched springs l0 is largerthen l. Therein, if |y| � l, we have

C = k(l0 − l)/l, B = kl0/(4l3).2 Under the action of high-frequency force f (t) = −(y0γ 2/m)cosγ t, the amplitude of the forced vibration

tends to y0 in limit γ → ∞, and the corrections to the amplitude and the higher harmonics are small ∼ 1/γ 2.


§9

Rigid-body motion. Non-inertialcoordinate systems

Rϕ

ψR

O

A

Figure 60

9.1. Find the normal oscillations of a uniform ring of radiusR, suspended on a string of length R in the gravity field g(Fig. 60). Consider only small oscillations in the plane of thering.

9.2. Masses m and M are located at the vertices of a squarewith side lengths 2a (Fig. 61a). Find the components of theinertia tensor

a) relative to the x-, y-, and z-axes;b) relative to the the x′ and y′-axes which are the diagonals of the square and the z-axis.

m

m

m

m

(a)

(b)

(c)

M

M

M

m

m

4a

2a2a

2b

2m

y y′

y′

x′

y′

x′x′

a

a x

M

O

O'

Figure 61

9.3. Find the principal axes and the principal moments of inertia for the followingsystems:

a) masses m and M at the vertices of a rectangle with side lengths 2a and 2b(Fig. 61b).

b) masses m and 2m at the vertices of a right-angled triangle with side lengths 2aand 4a (Fig. 61c).



9.10] §9. Rigid-body motion 49

9.4. Give an expression for the moment of inertia In with respect to an axis parallelto a unit vector n and passing through the centre of mass of the body in terms of thecomponents of the inertia tensor.

Ax2

x1

Rr

Figure 62 Figure 63

9.5. The cloakroom number tag is made of uniform thin plate. The tag is shaped as adisc of radius R with a hole in the shape of circle of radius R/2. The centre of this hole isat a distance R/2 from the tag’s edge (Fig. 62). Find the frequency of small oscillationsof the tag, suspended in the gravity field g and able to rotate around point A.

9.6. Determine the principal moments of inertia of a uniform ball of radius R inside ofwhich there is a cavity in the form of a ball of radius r (Fig. 63).

9.7. Express the components of the mass quadrupole moment tensor,

Dik =∫

(3xixk − r2δik)ρ dV

where ρ is the density, in terms of the components Iik of the inertia tensor.

9.8. Determine the frequency of small vibrations of a uniform hemisphere which lies onon a smooth horizontal surface in the field of gravity.

9.9. The period of the Earth’s rotation around its axis increases due to the action of tidalforces and may become equal to the period of circulation of the Moon around the Earth,that is, to the month. What will be the period of the Earth’s rotation at the moment whenit becomes equal to the month? Consider for simplicity that the Earth’s axis of rotationis perpendicular to the plane of the Earth and the Moon orbits. For numerical estimates,consider the Earth as a uniform ball of radius a = 6.4 thousand km and mass M which is81 times greater than the mass of the Moon m; the distance from the Earth to the Moonequals R = 380 thousand km.

9.10. Two identical uniform balls, rotating with equal angular velocities ω, slowlyapproach closer and then rigidly connect with each other. Determine the motion of thenewly formed body. Find which part of the initial kinetic energy has been converted intoheat. Prior to connecting, the angular velocities of the balls were directed:



v

A

A′

B′

C′

D′

D B

C

Figure 64 Figure 65

a) perpendicular to the line of the balls’ centres and parallel to each other;b) one along the line of the balls’ centres, and the other, perpendicular to this line.

9.11. A uniform ball of radius r and mass m rolls, without slipping, on a horizontalplane at velocity v. At the moment when it touches another motionless ball, both ballsbecome rigidly connected (Fig. 64). The plane is smooth; therefore, upon connecting,the balls freely slide on it. What are the forces with which the balls act upon the plane?The acceleration of gravity is large enough so that the balls are in constant contact withthe plane.

9.12. A uniform rectangular parallelepiped has two spherical hinges attached to itsopposite vertices A and C′. The parallelepiped rotates freely around the diagonal AC′with an angular velocity � (Fig. 65). Find the forces acting upon the hinges.

9.13. A particle moving parallel to the Oy-axis with velocity v and with impactparameters ρ1 and ρ2 is incident upon a uniform ellipsoid of rotation (with semi-axesa = b and c) (Fig. 66) and sticks to it. Describe the motion of the ellipsoid assuming itsmass to be much larger than that of the incident particle.

9.14. A gyrocompass is a rapidly rotating disc whose spinning axis is confined to ahorizontal plane (Fig. 67). Study the motion of the gyrocompass at latitude α. Theangular velocity of the Earth’s rotation equals �.

z

v

ρ1

ρ2

O

x

y

Figure 66 Figure 67



O

Ω

θ

Figure 68

9.15. A top with a fixed fulcrum O touches the horizontal plane with the edge of its disc(Fig. 68). Before the touchdown, the top was spinning around its axis with an angularvelocity � (assume velocity of precession to be small).

Find the angular velocity of the top when slipping of the disc vanishes. There were nonutations at the time of the touchdown.

9.16. An isotropic ellipsoid of revolution of mass m moves in the gravitational fieldproduced by a fixed point of mass M. Use the spherical coordinates of the centre of massand the Euler angles as generalized coordinates and determine the Lagrangian functionof the system. Assume the size of the ellipsoid to be small compared to the distance fromthe centre of the field.

Hint: The potential energy of the system is approximately equal to

U(R) = mϕ(R)+ 16

3∑i,k=1

Dik∂2ϕ(R)

∂Xi∂Xk,

where R = (X1, X2, X3) is the radius vector of the centre of the ellipsoid, Dik is themass quadrupole tensor (see problem 9.7), and ϕ(R) = −GM/R is the potential of thegravitational field (cf. [2], § 42).

9.17. Determine the angular velocity of precession for the Earth’s axis caused by theattraction forces of the Sun and the Moon. For simplicity, consider the Earth as a uniformellipsoid of rotation whose equatorial axis a is greater than its polar axis c, so that a − c

a ≈1

300 . Assume the Earth’s and the Moon’s orbits to be the circumferences lying in thesame plane. The tilt of the Earth’s axis to the plane of the Earth and Moon’s orbitsequals 67◦.

9.18. Write down the equations of motion for the components of the angular momen-tum along moving coordinate axes. These axes coincide with the principal axes ofthe inertia tensor. Integrate these equations for the case of the free motion of asymmetric top.

9.19.Use the Euler equations to study the stability of rotations around the principal axesof the moment of inertia tensor of an asymmetric top.

9.20. A uniform ball of radius a moves in the field of gravity along the inner surface ofa vertical cylinder of radius b without slipping. Find the motion of the ball.



9.21. a) A plane disc, symmetric around its axis, rolls in the field of gravity over a smoothhorizontal plane without friction. Find its motion in the form of quadratures.

Answer in detail the following questions:Under what conditions does the angle of inclination of the disc to the plane remain

constant?If the disc rolls in such a way that its axis has a fixed (horizontal) direction in space,

at what angular velocity will the rotation around this axis be stable?b) A disc rolls without slipping over a horizontal surface. Find the equations of motion

for this case and answer the same questions as in a).c) Do the same for a disc which rolls without slipping over the horizontal plane and

without rotation around the vertical axis.1

d) A disc rotates without slipping around its diameter which is at right angle toan inclined plane the disc is placed upon. The plane makes a small angle α with thehorizontal plane. Find the displacement of the disc over a long time period.

9.22. a) Find in terms of quadratures the law of motion of an inhomogeneous ballwhich is slipping without friction on a horizontal plane in the field of gravity. Themass distribution of the ball is symmetrical with respect to the axis passing through thegeometric centre and the centre of mass of the ball.

b) Find the equations of motion of the ball described in 9.22a if it rolls without slippingover a horizontal plane.

C

A

B

D

ω1

ω2

Figure 69

9.23. A particle is dropped from a height of h with zeroinitial velocity. Find its displacement from the vertical in thedirection of the East and the South. For simplicity, considerthe Earth as a uniform ball of radius R.

9.24. A vessel partially filled with gradually solidifyingepoxy resin, is rotating in the field of gravity with an angularvelocity ω2 around the AB-axis, which in turn rotatesaround the fixed CD-axis with an angular velocity ω1(Fig. 69). What form will it take when the surface of theresin completely solidifies?2

9.25. A particle moves in a central field U(r). Find the equation for its trajectory anddescribe its motion in a coordinate system which rotates uniformly with an angularvelocity � parallel to its angular momentum M.

9.26. Find the small oscillations of the particle with mass m which is fastened to a frameby springs with elastic constants k1 and k2. The frame rotates in its own plane with anangular velocity � (Fig. 70). The particle moves in the plane of the frame.

1 This means that the cohesion of the disc to the plane at the “point” of contact is so firm that the area ofcontact neither slips nor rotates. The energy loss due to rolling friction can be neglected.

2 Problem by V.S.Kuzmin and M.P.Perelroizen.



k1 k1

k2

Ω

k2

z

x

γ

Figure 70 Figure 71

9.27. A smooth paraboloid

z = x2

2a+ y2

2b

rotates in the field of gravity around the vertical z-axis with an angular velocity ω. At whatvalue of ω will the lower position of a particle inside the paraboloid become unstable?The acceleration of gravity is g = (0, 0, −g).

9.28. A particle of mass m is fixed inside a frame by the strings of the lengths l, theelasticity constants k and the tension f (when the frame is at rest). The frame rotateswith an angular velocity γ around the z-axis, which is shifted by a distance a from thecentre of the frame (Fig. 71).

Determine the equilibrium distance of the particle from the axis of rotation andinvestigate the stability of this equilibrium positions.

Consider the following cases:

a) the particle can only move along the springs;

b) any displacements of the particle are possible.

9.29. Two stars move along the circular orbits around their centre of mass. Considerthe frame of reference where the stars are motionless. Find the points at which a lightparticle in this system is also motionless. Investigate the stability of these “equilibriumpositions”. (Ignore the points lying on a straight line connecting the stars.)


§10

The Hamiltonian equationsof motion. Poisson brackets

10.1. Let the Hamiltonian function H of a system of particles be invariant underan infinitesimal translation (rotation). Prove the linear (angular) momentum conserva-tion law.

10.2. Use the Euler angles θ , ϕ, and ψ to find the Hamiltonian function of a symmetrictop when there is no external field acting on the top.

10.3. Determine the Hamiltonian function of an anharmonic oscillator, if the Lagrangianfunction is given by the equation

L = 12

(x2 − ω2x2

)− αx3 + βxx2.

10.4. Describe the motion of a particle with the Hamiltonian function

H(x, p) = 12 (p2 + ω2

0x2)+ 14 λ(p2 + ω2

0x2)2.

10.5. Do the same for H(x, p) = A√

p − xF.

10.6. Find the equations of motion for the case when the Hamiltonian function is of the

form H(r, p) = c|p|n(r) (a beam of light).

Find the trajectory if n(r) = ax.

10.7. Find the Lagrangian function for the case when the Hamiltonian function is

a) H(r, p) = p2

2m − pa (a = const); b) H(r, p) = c|p|n(r) .

10.8. Describe the motion of a charged particle in a uniform constant magnetic field Bby solving the Hamiltonian equations of motion; take the vector potential in the form

Ax = Az = 0, Ay = xB.



10.12] §10. The Hamiltonian equations of motion 55

10.9. Study qualitatively the motion of a charged particle in a non-uniform magneticfield described by the vector potential A(r) = (0, hx2, 0). Compare the result with thedrift approximation.

10.10. Show that the problem of motion of two particles with opposite charges (e and −e)in a uniform constant magnetic field is reducible to the problem of the motion of oneparticle in a given potential field [24].

In problems 10.11 to 10.14 we are dealing with the motion of electrons in a metal orsemiconductor. Electrons in a solid form a system of particles which interact both withthemselves and with the ions which form the crystal lattice. Their motion is describedby quantum mechanics. In solid state theory one is often able to reduce the problem ofmany interacting particles which form the solid to the problem of the motion of separatefree particles (the so-called quasi-particles: electrons or holes, depending on the sign oftheir charge) for which, however, the momentum dependence ε(p) (“dispersion law”)of the energy is complicated. For instance, for “holes” in silicon crystal

ε(p) = 12m

[−4p2 ±

√1.2p4 + 17(p2

xp2y + p2

xp2z + p2

yp2z)

],

where the coordinate axes are chosen to coincide with the crystal symmetry and m is theelectron mass.

In many cases it turns out that it is possible to consider the motion of the quasi-particles using classical mechanics (e.g. [16]). The function ε(p) is periodic with a periodequal to the period of the so-called reciprocal lattice.1 Otherwise one can assume thatε(p) has an arbitrary form.

10.11. It is well known that ε(p) is a periodic function of p with a period which is equalto that of the reciprocal lattice, multiplied by 2π h; for instance, for a simple cubic latticewith lattice constant a, the period of ε(p) is equal to 2π h/a).

Describe the motion of an electron in a uniform electric field E.Hint to problems 10.12–10.14: In these problems it is convenient to introduce the

kinematic momentum p = P − ecA besides the generalized momentum P (here A is the

vector potential of the magnetic field).

10.12. Obtain the equation of motion using the Hamiltonian function

H(P, r) = ε(P − e

cA

)+ eϕ

where the electron charge e is taken to be negative, e < 0.

1 For instance, for a crystal the lattice of which has a smallest period a in the x-direction, we have

ε(px, py, pz) = ε(px + 2π h

a , py, pz

), where h is the Planck constant.



10.13. a) Determine the integrals of motion of an electron moving in a solid in uniformconstant magnetic field B. What does the “orbit” in momentum space look like?

b) Prove that the projection of the electron orbit in a uniform constant magnetic fieldonto a plane at right angles to B in coordinate space can be obtained by rotation andchange of scale of the orbit in momentum space.

10.14. Express the period of revolution of an electron in a uniform constant magneticfield B in terms of the area of S(E, pB) of the section cut off by the plane pb = pB/B =const of the surface ε(p) = E in momentum space.

10.15. Evaluate the Poisson brackets:

a) {Mi, xj}, {Mi, pj}, {Mi, Mj};b) {ap, br}, {aM, br}, {aM, bM};c) {M, rp}, {p, rn}, {p, (ar)2},

where xi, pi, and Mi are the Cartesian components of the vectors, while a and bare the constant vectors.

10.16. Evaluate the Poisson brackets {Ai, Aj}, where

A1 = 14 (x2 + p2

x − y2 − p2y), A2 = 1

2 (xy + pxpy),

A3 = 12 (xpy − ypx), A4 = x2 + y2 + p2

x + p2y .

10.17. Evaluate the Poisson brackets {Mi, jk}, {jk, il}, where ik = xixk + pipk.

10.18. Show that the Poisson bracket {Mz, ϕ} = 0, where ϕ is an arbitrary scalar functionof the coordinates and momenta of a particle, ϕ = ϕ(r2, p2, rp).

Show also that the Poisson bracket {Mz, f } = [n, f ], where n is the unit vector alongthe z-axis and f is the vector function of the coordinates and momenta of a particle, thatis, f = rϕ1 + pϕ2 + [r, p]ϕ3 and ϕi = ϕi(r2, p2, rp).

10.19. Evaluate the Poisson brackets {f , aM}, {fM, lM}, where f and l are the vectorfunctions of r and p while a is the constant vector.

10.20. Evaluate the Poisson bracket {Mζ , Mξ }, where Mζ and Mξ are the componentsof the angular moment along the Cartesian ζ - and ξ-axes which are fixed in a rotatingrigid body.

10.21. Write down the equations of motion for the components Mα of the angularmomentum along the axes fixed in a freely rotating rigid body. The Hamiltonian functionis of the form

H = 12

∑α,β

(I−1)αβMαMβ .



10.22. In this problem we consider the model of the electron and nuclear paramagneticresonance (see [16], Ch. IX). The Hamiltonian function of the magnetized ball ina uniform magnetic field B has the form

H = M2

2I− γ MB,

where I is the moment of inertia of the ball and γ is a gyromagnetic ratio. Write downthe equations of motion for the vector M and find the law of its motion in the followingcases:

a) B = (0, 0, B0);

b) B = (B1 cosωt, B1 sinωt, B0) and M = (0, 0, M0) in the initial time.

10.23. Evaluate the Poisson brackets {vi, vj} for a particle in a magnetic field. Here viare the components of the particle velocity in the Cartesian coordinates.

10.24. Prove that the value of any function f (q(t), p(t)) of the coordinates and momentaof a system at time t can be expressed in terms of the values of the q and p at t = 0 as

f (q(t), p(t)) = f + t1! {H, f }+ t2

2! {H, {H, f }}+ . . . ,

where f = f (q(0), p(0)) and H = H(q(0), p(0)) is the Hamiltonian function. (Assume thatthe series converges.)

Apply this formula to evaluate q(t), p(t), q2(t), and p2(t) for the following cases:

a) a particle moving in a uniform field of force;

b) a harmonic oscillator.

10.25. Prove the equalities

a) { f (q1, p1), �(ϕ(q1, p1), q2, p2, . . .)} = ∂�

∂ϕ{f , ϕ};

b) { f (q1, p1, q2, p2), �(ϕ(q1, p1, q2, p2), q3, p3, . . .)} = ∂�

∂ϕ{f ,ϕ};

c) { f (q, p), �(ϕ1(q, p), ϕ2(q, p) . . .)} = ∑i

∂�

∂ϕi{f , ϕi}.

10.26. a) The Hamiltonian function depends on the variables q1 and p1 only throughthe function f (q1, p1)

H = H( f (q1, p1), q2, p2, . . . , qN , pN).

Prove that the function f (q1, p1) is an integral of motion.



b) Find the integrals of motion for a particle in the field U(r) = arr3 (use the spherical

coordinates).

10.27. A particle in the field U(r) = −α/r is known to have the integral of motion(see [1], §15 and [7], §3.3)

A = [v, M] − αrr.

a) Evaluate the Poisson brackets {Ai, Aj}, {Ai, Mj},b) Evaluate the Poisson brackets

{H, J1,2}, {J1i, J2j}, {J1i, J1j}, {J2i, J2j}

for the finite motion ( E < 0), if the vectors J1,2 are

J1,2 = 12

(M ±

√m

−2EA

).

Compare these Poisson brackets with the Poisson brackets for the components ofthe angular momentum M.

Express the Hamiltonian function H in terms of J1 and J2.


§11

Canonical transformations

11.1. Determine the canonical transformation defined by the following generatingfunctions:

a) F(q, Q, t) = 12mω(t)q2 cotQ.

Write down the equations of motion for a harmonic oscillator with frequency ω(t) interms of the variables Q and P.

b) F(q, Q, t) = 12mω

[q − F(t)

mω2

]2cotQ.

Write down the equations of motion for a harmonic oscillator which is acted upon byan external force F(t) in terms of the variables Q and P.

11.2. Determine the generating function �(p, Q) which produces the same canonicaltransformation as the generating function F(q, P) = q2eP .

11.3. What is the condition that a function �(q, P) can be used as a generating functionfor a canonical transformation?

Consider in particular the example

�(q, P) = q2 + P2.

11.4. Prove that a rotation in (q, p) space is a canonical transformation for a system withone degree of freedom.

11.5. Consider the small oscillations of an anharmonic oscillator with the Hamiltonianfunction

H = 12

(p2 + ω2x2

)+ αx3 + βxp2

under the assumption that |αx| � ω2, |βx| � 1.




Find the parameters a and b for the canonical transformation produced by generatingfunction � = xP + ax2P + bP3 such that the new Hamiltonian function does not containany anharmonic terms up to first-order terms in αω−2Q and βQ. Determine x(t).

11.6. Determine the parameters a and b for the canonical transformation produced bythe generating function

� = xP + ax3P + bxP3,

in such a way that that small oscillation of an anharmonic oscillator described by theHamiltonian function

H = 12

(p2 + ω2

0x2)

+ βx4

can be reduced to harmonic oscillations in terms of the new variables Q and P. Neglectterms of second order in βω−2Q2 in the new Hamiltonian function.

11.7. Small oscillations of the coupled oscillators are described by the Hamiltonianfunction

H = 12m

[p2

1 + (mω1x)2 + p22 + (mω2y)2

]− mαx2y.

Neglecting the terms of the second degree in α, find the parameters a,b, and c for thecanonical transformation produced by the generating function

�(x,y,PX ,PY ) = xPX + yPY + axyPX + bP2XPY + cx2PY

such that in the new variables the system is reduced to two independent oscillators. Findx(t) and y(t) with the first-order corrections in α.

11.8. Prove that the following transformation is canonical:

x = X cosλ + PY

mωsinλ, y = Y cosλ + PX

mωsinλ,

py = −mωX sinλ + PY cosλ, px = −mωY sinλ + PX cosλ.

Determine the new Hamiltonian function, H ′(P, Q), if the old Hamiltonian function is

H(p, q) = p2x + p2

y

2m+ 1

2 mω2(x2 + y2)

(cf. with problem 11.18). Describe the motion of a two-dimensional oscillator atY = PY = 0.


11.15] §11. Canonical transformations 61

11.9. Use the transformation of the preceding problem to reduce the Hamiltonianfunction of an isotropic harmonic oscillator in a magnetic field described by the vectorpotential A(r) = (0, xB, 0), to a sum of squares and determine its motion.

11.10. Use the canonical transformation to diagonalize the Hamiltonian function of ananisotropic charged harmonic oscillator with potential energy

U(r) = 12 m(ω2

1x2 + ω22y2 + ω2

3z2),

which is situated in a uniform constant magnetic field determined by the vector potentialA(r) = (0, xB, 0).

11.11. Apply the canonical transformation of problem 11.8 to pairs of normal coordi-nates corresponding to standing waves in the system of particles on the ring consideredin problem 7.4 to obtain the coordinates corresponding to the travelling waves.

11.12. Prove that the following transformation is a canonical one:

x = 1√mω

(√

2P1 sinQ1 + P2),

px = 12

√mω(

√2P1 cosQ1 − Q2),

y = 1√mω

(√

2P1 cosQ1 + Q2),

py = 12

√mω(−√

2P1 sinQ1 + P2)

Find the Hamiltonian equations of motions for a particle in a magnetic field describedby the vector potential A(r) =

(−1

2yB, 12xB, 0

)in terms of the new variables introduced

through the previously described transformation with ω = eBmc .

11.13. What is the meaning of the canonical transformation produced by the generatingfunction �(q, P) = αqP?

11.14. Prove that a gauge transformation of the potentials of the electromagnetic field isa canonical transformation for coordinates and momenta of charged particles, and findthe corresponding generating function.

11.15. It is well known that replacing the Lagrangian function L(q, q, t) by

L′(q, q, t) = L(q, q, t)+ dF(q, t)dt

,

where F(q, t) is an arbitrary function of the coordinates and time lives the Lagrangianequations of motion invariant. Prove that this transformation is a canonical one and findits generating function.



11.16. Find the generating function for the canonical transformation which consist inchanging q(t) and p(t) to

Q(t) = q(t + τ) and P(t) = p(t + τ),

where τ is constant for the following cases:

a) a free particle,

b) motions in a uniform field of force, U(q) = −Fq;

c) a harmonic oscillator.

11.17. Discuss the physical meaning of the canonical transformations produced by thefollowing generating functions:

a) �(r, P) = rP + δaP;

b) �(r, P) = rP + δϕ[r,P];

c) �(q, P, t) = qP + δτH(q, P, t);

d) �(r, P) = rP + δα(r2 + P2),

where r is the Cartesian radius vector while δa, δϕ, δτ , and δα are infinitesimalparameters.

11.18. Prove that the canonical transformation produced by the generating function

�(x, y, PX , PY ) = xPX + yPY + ε(xy + PXPY ),

where ε → 0 is a rotation in phase space.

11.19. Write down the generating function for infinitesimal canonical transformationscorresponding to

(a) a screw motion;

b) a Galilean transformation;

c) a change to a rotating system of reference.

11.20. A canonical transformation is produced by the generating function �(q, P) =qP + λW (q, P), where λ → 0.

Determine up to first-order terms the change in value of an arbitrary function f (q, p)

when we change arguments:

δf (q, p) = f (Q, P)− f (q, p).



11.21. For the case when the Hamiltonian function is

H(r, p) = p2

2m+ ar

r3 (a = const),

determine the Poisson bracket {H, rp} and use the result to obtain an integral of theequations of motion. Here, it is convenient to use the results of the preceding problemand problem 11.13.

11.22. Determine how the r and p dependence of M, p2, pr, and H(r, p, t) changeunder the canonical transformations of problem 11.17.

11.23. Show that if

{W1(q, p), W2(q, p)} = 0,

the result of applying two successive infinitesimal canonical transformations which areproduced by the generating functions

�i(q, P) = qP + λiWi(q, P), λi → 0, i = 1, 2,

will be independent of the order in which they are taken, up to and including second-order terms.

11.24. Determine the canonical transformation which is the result of N successiveinfinitesimal transformations produced by the generating function

�(q, P) = qP + λ

NW (q, P), λ = const, N → ∞;

a) W (r, P) = [r, P]a, a = const; b) W (x, y, PX , PY ) = Ai,

where Ai is defined in problem 10.16.Hint: Construct—and solve for the different concrete forms of W —differential

equations which Q(λ) and P(λ) must satisfy.

11.25.

a) What is the change with time in the volume, the volume in momentum space, andthe volume in phase-space which are occupied by a group of particles which movefreely along the x-axis? At t = 0 the particle coordinates are lying in the intervalx0 < x < x0 + �x0, and their momenta in the range p0 < p < p0 + �p0.

b) Do the same for particles which move along the x-axis between two walls.Collisions with the walls are absolutely elastic. The particles do not interact withone another.

c) Do the same for a group of harmonic oscillators.



d) Do the same for a group of harmonic oscillators with friction.

e) Do the same for a group of anharmonic oscillators.

f) We shall describe the particle distribution in phase space at time t by thedistribution function w(x, p, t) which is such that w(x, p, t)dxdp is the numberof particles with coordinates in the interval from x to x + dx and momenta in therange from p up to p + dp. Determine the distribution functions of a group of freeparticles and of a group of harmonic oscillators, if at t = 0

w(x, p, 0) = 12π�p0�x0

exp{− (x − X0)2

2(�x0)2− (p − P0)2

2(�p0)2

}.

11.26. For the variable

a = mωx + ip√2mω

eiωt,

a) find the Poisson bracket {a∗, a}. Express the Hamiltonian function of a harmonicoscillator

H0 = p2

2m+ 1

2 mω2x2

in terms of variables a and a∗.

b) Prove that Q = a and P = ia∗ are the canonical variables. Find a new Hamiltonianfunction H ′

0(Q, P).

c) For an oscillator with the anharmonic addition to potential energy δU = 14 mβx4,

average the Hamiltonian function H ′(Q, P) over the period of fast oscillations2π/ω.

Use the averaged Hamiltonian function to determine the slow changes invariables Q and P.

d) Study the variation of the amplitude of oscillations for a harmonic oscillator underthe action of the nonlinear resonant force

H = p2

2m+ 1

2 mω2x2 + m2ω2αx4 cos4ωt.

11.27. Study the variation of the amplitude of oscillations for the system of threeharmonic oscillators with the weak nonlinear coupling

H = 12m

(p2x + p2

y + p2z)+ 1

2 m(ω21x2 + ω2

2y2 + ω23z2 + αxyz),



when |ω1 − ω2 − ω3| � ω1, |αx| � ω21. Consider in more detail the cases when in the

initial moment |y| � |x|, z = 0, y = z = 0.Use the same method as in the preceding problem.

11.28. The Hamiltonian function of an anharmonic oscillator with the parametricexcitation has the form

H = p2

2m+ 1

2 mω2(1 + hcos2γ t)x2 + 14 mβx4.

Introduce the canonical variables

a = mωx + ip√2mω

eiγ t, P = ia∗.

a) Determine the new Hamiltonian function H ′(a, P, t) and average it over the periodof fast oscillations 2π/γ .

b) Study the change in an amplitude of oscillations in the resonance region |γ − ω| �hω, h � 1 when in the initial moment quantity a is close to zero.

11.29. Check that the transformation

x = Qcosγ t + 1mωP sinγ t,

p = −mωQsinγ t + P cosγ t

is canonical. Determine the new Hamiltonian function H ′(Q, P, t) for an oscillator withthe parametric excitation:

H = p2

2m+ 1

2 mω2x2(1 − hcos2γ t).

b) Average H ′(Q, P, t) by the period 2π/γ and describe qualitatively the motion of apoint on the phase plane QP. Take h � 1, ε = 1 − γ /ω � 1.

11.30. Consider the motion of two weakly coupled oscillators

H = H0 + V , H0 = 12m

(p2x + p2

y)+ 12 m(ω2

1x2 + ω22y2),

V = mβxysin(ω1 − ω2)t, β � ω21 ∼ ω2

2.

In the plane of the variable x, px/(mω1) change the variable to the coordinate system X ,PX/(mω1) which rotates clockwise with an angular velocity ω1. Make a similar change



for the variables y, py/(mω2). Prove that the variables X , Y , and PX , PY are the canonicalvariables.

Determine the new Hamiltonian function H ′(X , Y , PX , PY , t) and average it over thetime tav such that

1ω1,2

� tav � ω1,2

β,

For the period t � ω1,2/β, study the change over time of the amplitudes of oscillationsfor the variables x and y.

b) Do the same for V = mβxysin(ω1 + ω2)t.


§12

The Hamilton–Jacobi equation

12.1. Describe the motion of a particle moving in the field U(r) by using the Hamilton–Jacobi equation for the following cases:

a) U(r) = −Fx;

b) U(r) = 12

(mω2

1x2 + mω22y2

).

12.2. Describe the motion of a particle which is scattered in the field U(r) = ar/r3.Express the equation of the trajectory in terms of quadratures; express it analyticallyfor the case when Eρ2 � a, where ρ is the impact parameter. Before the scattering thevelocity of the particle is parallel to the vector (−a).

12.3. Find the cross-section for the small-angle scattering of particles with velocitiesbefore the scattering anti-parallel to the z-axis for the cases where the scattering is causedby the following fields U(r):

a) U(r) = a cosθ

r2;

b) U(r) = b cos2 θ

r2;

c) U(r) = b(θ)

r2.

12.4. Find the cross-section for a particle to fall into the centre of one of the followingfields U(r):

a) U(r) = arr3

; b) U(r) = arr3

+ λr ;

c) U(r) = arr3

− γ

r4; d) U(r) = b(θ)

r2.

Assuming that all directions of a are equally probable, evaluate the averages of thecross-section obtained.




12.5. Find the cross-section for particles to hit a sphere of radius R placed at the centreof the field U(r) = ar/r3.

12.6. Describe the motion of particles which are scattered by and fall towards the centreof the field U(r) for the following cases:

a) U(r) = a cos θ

r2;

b) U(r) = −a(1 + sin θ)

r2.

The particle velocity before scattering is parallel to the z-axis.Give an expression for the trajectories in terms of quadratures and analytically for the

case when Eρ2 � a.For the first field, find also an analytic expression for the trajectory of a particle falling

into the centre at Eρ2 � a.

12.7. Describe the motion of a particle falling towards the centre of the field U(r) =ar/r3 for the case when at infinity the particle moved along the straight line y = ρ,x = −z tanα, where ρ is the impact parameter. The vector a is along the z-axis andthe initial spherical coordinates of the particle are r = ∞, θ = π − α, ϕ = 0. Express the

trajectory in terms of quadratures and analytically for the case when α2 <2Eρ2

a � 1.

12.8. a) Determine in terms of quadratures the finite orbit of a particle moving in the

field U(r) = acosθ

r2− α

r for the case when Mz = 0, where the z-axis is taken along the

direction of a.b) Do the same for the field U(r) = acosθ

r2+ γ

r4.

12.9. Under what condition is the orbit in the preceding problem a closed one?

12.10. Describe qualitatively the motion of a particle in the field

U(r) = arr3

− αr .

12.11. Find the values of the angular momentum Mz for which the orbits in the followingfields U(r) are finite:

a) U(r) = γ

r4− bcos2 θ

r2;

b) U(r) = bcos2 θ

r2− α

r .

Describe the orbits in both cases.


12.15] §12. The Hamilton–Jacobi equation 69

12.12. Describe the motion in terms of parabolic coordinates for a particle moving inthe field U(r) for the following cases:

a) U(r) = −αr ;

b) U(r) = −αr − Fr.

In b), consider only finite orbits it terms of quadratures.

12.13. A particle starts from the origin at an angle α to the z-axis inside a smooth elasticellipsoid of revolution:

x2

a2+ y2

a2+ z2

c2= 1.

Find the region inside the ellipsoid which can not be reached by the particle.

12.14. A particle moves in a smooth horizontal plane in the area bounded by an elasticsmooth wall having the form of ellipse

x2

a2 + y2

b2 = 1.

The particle starts to move from a point on the x-axis at the angle α to it:

(x,y) = (x0,0), (x, y) = v(cosα,sinα).

Find the region inside the ellipse which can not be reached by the particle.Hint: Use the elliptic coordinates ζ ,ϕ, determined by the equations

x = c coshζ cosϕ, y = c sinhζ sinϕ,

0 ≤ ζ < ∞, −π ≤ ϕ < π ,

where c is the parameter of transformation, c2 = a2 − b2.

m

zO2 O1

r1r2

Figure 72

12.15. Describe in terms of quadratures the trajectory ofa particle moving in the field two Coulomb centres,

U(r) = αr1

− αr2

(Fig. 72). At infinity the velocity of the particle wasparallel to the axis O2O1z. Describe the motion of aparticle falling onto a “dipole” formed by these centres.



12.16. A short magnetic lens is produced by a field governed by the vector potential

Aϕ = 12 rB(z), Ar = Az = 0,

where B(z) is non-vanishing in the interval |z| < a. A beam of electrons close to the z-axis incident onto the lens from point (0, 0, z0). Find the point (0, 0, z1) onto which thebeam is focused. Assume that z0,1 � a.

Hint: Find the solution of the Hamilton–Jacobi equation in the form of expansion ofS(r, ϕ, z, t) in powers of r:

S(r, ϕ, z, t) = −Et + pϕϕ + f (z)+ rψ(z)+ 12 r2σ(z)+ . . .

12.17. A magnetic lens is produced by a field governed by the vector potential

Aϕ = 12 rB(z), Ar = Az = 0,

where

B(z) = B1 + 2z2

.

A beam of electrons close to the z-axis is incident from the point (0, 0, z0). Find thepoints where it will be focused.

Hint: Find a solution of the Hamilton–Jacobi equation as an expansion in r.

12.18. How can one find the action as a function of the coordinates and time, if thesolution of the Hamilton–Jacobi equation is known?

12.19. Formulate and prove the theorem about the integrating the equations of motionusing the complete solution of the equation

∂S∂t

+ H(−∂S

∂p, p, t

)= 0,

where H(q, p, t) is the Hamiltonian function (Hamilton–Jacobi equation in the momen-tum representation).

12.20. Use the Hamilton–Jacobi equation in the momentum representation to find thetrajectory and law of motion of a particle moving in a uniform field.


§13

Adiabatic invariants

13.1. A particle of mass m is suspended from a thread which passes through a small ringA (Fig. 73). Determine the average force exerted upon the ring by the thread when thependulum performs small oscillations. Find the change in the energy of the pendulumwhen the ring is slowly displaced vertically.

13.2. A particle moves in a rectangular potential well of width l. Consider the collisionsof the particle with the “wall” of the well to find how the energy of the particle will changewhen l is changed slowly.

A

B

x

y

ϕ

m

M

Figure 73 Figure 74

13.3. A ball of mass m moves between a heavy piston of mass M � m and the bottom ofthe cylinder in the gravitation field g (Fig. 74). Equilibrium distance from the bottom ofthe cylinder to the piston is equal to X0. Considering that the velocity of the ball is muchmore than the piston’s velocity, determine the law of motion of the piston, averaged overthe period of the ball’s motion. Find the frequency of small oscillations of the piston.Assume the collisions to be elastic. (This is a model of a “gas” consisting of a singlemolecule.)

13.4. A small ball is jumping up and down on an elastic plate in a lift. What is the changein the maximum height the ball reaches if the acceleration of the lift is slowly changed?How does the height vary if the plate is raised slowly?




13.5. How does the energy of a particle moving in the field U(x,α) change when theparameters of the field change slowly for the following cases:

a) U(x,α) = A(e−2αx − 2e−αx); b) U(x,α) = − U0

cosh2 αx;

c) U(x,α) = U0 tan2 αx; d) U(x,α) = α|x|n.

Hint: Use the formula T = 2π∂I∂E

([1], § 49).

13.6. A particle moves down an inclined plane AB (Fig. 75) and is reflected elasticallyby a wall at the point A. How does the maximum height the particle reaches change whenthe angle α changes slowly?

A

B

αα

AO

Figure 75 Figure 76

13.7. A pendulum is placed upon an inclined plane OA (Fig. 76). How does its amplitudechange, when the angle α is changes slowly?

13.8. Determine the adiabatic invariant for a mathematical pendulum for the case whenthe amplitude of the oscillations is not small.

13.9. Two elastic small balls of small radius and with masses m and M, respectively,(m � M) move along the straight line OA (Fig. 77). At O the particle m is reflectedelastically by a wall. Assume that the velocity of the lighter particle at t = 0 is much largerthat of the heavier particle and determine the motion of the heavier particle averaged overperiod of motion of lighter particle.

13.10. In this problem we consider a simplified model of an ion H+2 . Two particles of

mass M each and one of mass m (m � M), located between them, move along a straightline AB (Fig. 78). The lighter particle is attracted with the constant force f to each ofthe heavier particles and is reflected elastically at collisions with the them. Assume thevelocity of the lighter particle to be much higher than that of the heavier particles. Findthe frequency of small oscillations of the distance between the heavier particles averagingon the fast motion of the lighter one.


13.18] §13. Adiabatic invariants 73

Om M

A

A

M MmB

Figure 77 Figure 78

13.11. Use the method of successive approximations to solve the equations for P and Qof problem 11.1a for the case when the frequency changes slowly, |ω| � ω2, |ω| � |ω|ω,up to and including terms of first order in ω/ω2.

What is the advantage of the variables P, Q over p, q in this case?

13.12. Check that up to terms of first order in ω/ω2 the expression

q = 1√ω

ei∫

ωdt

satisfies the equation

q + ω2(t)q = 0.

13.13. A force F(t) acts upon a harmonic oscillator. Find the time-dependence of the

adiabatic invariant I = 12π

∮pdq.

13.14. Find a connection between the volume and the pressure of a “gas”consisting ofparticles which move parallel to the edges and inside an elastic cube, when the size of thecube changes slowly.

13.15. A particle moves inside an elastic parallelepiped. How does its energy change

a) if the size of the parallelepiped changes slowly;

b) if the parallelepiped rotates slowly?

13.16. A particle moves inside an elastic sphere, the radius of which changes slowly.How does its energy change, and how does the angle at which it hits the sphere change?

13.17. Determine the change in the energy and in the trajectory of a particle performinga finite motion in the field U(r) in the following cases:

a) U(r) = − γ

rn(0 < n < 2);

b) U(r) = arr3

+ γ

r4,

when the coefficient γ changes slowly.

13.18. Determine the change in the energy of a particle moving in the central field U(r)when a small extra field δU(r) is slowly “switched on”.



13.19. Find the time-dependence of the energy of a system two coupled harmonicoscillators with the Lagrangian function

L = 12 (x2 + y2 − ω2

1x2 − ω22y2)+ αxy

when ω1 changes slowly.What is the change in the orbit of the point (x, y)?

13.20. Let the coupling between the oscillators in the preceding problem be small:α � ω2

1,2. Prove that if we are far from degeneracy, when ω1 = ω2, the adiabatic invariantscalculated neglecting the coupling are conserved, but that they change rapidly when wepass slowly through the region of degeneracy.

13.21. In which range of frequency ω1(t) will the adiabatic invariants of the harmonicoscillators in problem 13.19 change sharply, if the coupling is δU(x, y) = βx2y?

13.22. Determine the shortest distance a particle approaches the edge of a dihedralangle α after being reflected elastically from its faces. The angle at which particle isincident on one of the faces at a distance of l from the edge is ϕ0.

Solve this problem by two methods: either by using a reflection method and solving itdirectly or by using adiabatic invariant assuming that α and ϕ0 are small.

13.23. Define the boundaries of the region in which a particle moves between two elastic

surfaces y = 0 and y = a cosh αx√cosh2αx

. The particle starts at the origin at the angle ϕ to the

y-axis in the plane xy (α, ϕ � 1). Also determine the period of the oscillations alongthe x-axis.

13.24. What is the change in the radius and the centre of the orbits of a charged particlemoving in a uniform magnetic field which is slowly changing its strength? Take the vectorpotential in one of the following two forms

a) A(r) = (0, xB, 0);

b) Ar = Az = 0, Aϕ = 12 rB.

13.25. Calculate the adiabatic invariants for a charged isotropic harmonic oscillatormoving in a uniform magnetic field. Take the vector potential in the form Ar = Az = 0,Aϕ = 1

2 rB(t).

13.26. a) Determine the adiabatic invariants for a charged anisotropic harmonicoscillator with the potential energy

U(r) = 12 m(ω2

1x2 + ω22y2 + ω2

3z2)

moving in a uniform magnetic field B parallel to the z-axis. Take the vector potential inthe form A(r) = (0, xB, 0).



b) Let initially B = 0 and let the trajectory of the oscillator fill the rectangle |x| � a,|y| � b. Find the motion of the oscillator after the magnetic field slowly increasesits strength up to a value which is so high that ωB = eB/(mc) � ω1,2.

c) Let the magnetic field be weak (ωB � ω1 − ω2 > 0) and let the oscillator originallyoscillate near along the x-axis. What happens to its motion when the value of ω1decreases slowly to reach a value ω′

1 < ω2 such that ωB � ω2 − ω′1?

13.27. A particle performs a finite motion in a plane at right angles to a magneticdipole μ. What is the change in the energy of the particle when the magnitude μ changesslowly?

13.28. Determine the period of the oscillations along the z-axis of an electron in amagnetic trap. The magnetic field of the trap is symmetric with respect to the z-axis,and we have

Bϕ = 0, Bz = Bz(z), Br = −12 rB′

z(z),

Consider the following cases:

a) Bz(z) = B0

(1 + λ tanh2 z

a

);

b) Bz(z) = B0

(1 + z2

a2

).

13.29. How does the energy and the period of oscillations of an electron moving in themagnetic trap of the preceding problem change when the field parameters B0, λ, and achange slowly?

13.30. Determine the change in the energy of a particle moving in a central field U(r)when a weak iniform magnetic field B is slowly “switched on”.

13.31. The number of single-valued integrals of motion is known to increase whenthe motion becomes degenerate. Find the integrals of motion when a particle movesin the field

U(x, y) = 12 mω2(x2 + 4y2).

13.32. Find the action and angle variables for the following systems:

a) a harmonic oscillator;

b) a particle moving in the field U(x) ={∞, when x < 0,

xF, when x > 0.



13.33. Use the generating function

S(x, P) =x∫

0

√2m|E − U(x)|dx,

to perform a canonical transformation for the case of a particle moving in the periodicfield

U(x) =⎧⎨⎩

0, when na < x <(n + 1

2

)a,

V , when(n + 1

2

)a < x < (n + 1)a,

n = 0, ±1, ±2, . . . ,

when E > V , and where E in the expression for S(x, P) can be derive as a function of Pfrom the equation

P =a∫

0

√2m|E − U(x)|dx.


§1

Integration of one-dimensionalequations of motions

1.1. a) The energy of the particle E is determined by the initial values x(0) and x(0).The motion follows from the energy conservation law

–A

U

xx1 x2 x3

E′

E′′

Figure 79

12mx2 + U(x) = E. (1)

When E � 0, the particle can move in the region x � x1:the motion is infinite (E = E′ in Fig. 79). When E < 0(E = E′′), the particle moves in the interval x2 � x � x3:the motion is finite. The turning points are determinedfrom (1) through U(xi) = E:

x1 = 1α ln

√A(A + E) − A

E, when E > 0,

x1 = − ln2α , when E = 0, (2)

x2,3 = 1α ln

A ∓ √A(A − |E|)|E| , when E < 0.

From (1) we obtain

t =√

m2

x∫x(0)

dx√E − U(x)

. (3)

whence

x(t) = 1α ln

A − √A(A − |E|)cos(αt

√2|E|/m + C)

|E| , when E < 0, (4)




x(t) = 1α ln

[12

+ Aα2

m (t + C)2]

, when E = 0, (5)

x(t) = 1α ln

√A(A + E)cosh(αt

√2E/m + C)− A

E, when E > 0. (6)

The C constants are defined by the initial values of x(0). For example, in (4), whenx(0) > 0

C = arccosA − |E|eαx(0)

√A − (A − |E|) .

Turning points (2) are also easily found from (4) to (6).

When E < 0 the motion is periodic according to (4) with a period T = πα

√2m|E| .

When E is close to the minimum value of U(x), Umin = U(0) = −A (that is, when

ε = A − |E|A � 1), the period

T ≈ T0

(1 − ε

2

), T0 = π

α

√2mA

is nearly independent of E. In this case we can write (4) in the form

x(t)=−1α

ln(1−ε)+1α

ln[1−√

ε cos(

2π

Tt+C

)]≈−

√ε

αcos

(2π

T0t+C

). (7)

The particle now performs harmonic oscillations near the point x = 0 with an amplitude√ε/α determined by the difference E − Umin, and with a frequency which is independent

of the energy. This kind of motion for an energy E close to Umin occurs in any field U(x),in which potential energy near the point of minimum x = a has a non-zero secondderivative U ′′(a) �= 0. (For more information, see § 5, and [7], § 19.)

When E � 0 the particle coming to the right reaches the turning point x1 (see (2)),turns back, and goes to infinity. Its velocity approaches the value

√2E/m from above.

b) x(t) = 1α arsinh

[√− |E| + U0

|E| sin(αt√

2|E|/m + C)

], when E < 0,

x(t) = ±1α arsinh

⎡⎣

√E + U0

Esinh(αt

√2E/m + C)

⎤⎦ , when E > 0,


1.3] §1. Integration of one-dimensional equations of motions 81

x(t) = ±1α arsinh(αt

√2U0/m + C), when E = 0;1

c) x(t) = 1α

arcsin

[√E

E + U0sin

(αt

√2(U0 + E)/m + C

)].

Explain why in some of the formulae two signs occur?

1.2.

x(t) = x0

1 ± tx0√

2A/m, x0 = x(0).

The sign in the denominator is the opposite of that of x(0). Let for definiteness

x(0) > 0. When x(0) > 0, the particle reaches infinity after a time√

m/(2Ax20). Of course,

for real system the particle only reaches finite, though long, distance, corresponding tothe distance over which the specified field U(x) has the given form.

When x(0) < 0, the particle asymptotically approaches the point x = 0.

1.3. Near the turning point U(x) ≈ E − (x − a)F, where F = −U ′(a), that is, the motionof the particle can be considered to occur under the action of a uniform constant force F.Assuming that x(0) = a, we get

x(t) = a + Ft2

2m.

The further away from the point x = a we go, the less accurate this formula becomes.It takes a time τ ∝ s to traverse a short path length s if it is far from a turning point.

However, if this path length is at the turning point, τ ≈ √2ms/|F|, that is, τ ∝ √

s.

U

ε

xa

Figure 80

If U ′(a) = 0 (see Fig. 80), one must extend theexpansion of U(x) to the next term:

U(x) = E + 12U ′′(a)(x − a)2.

In this case

x(t) = a + se±λt with s = x(0)− a,λ2 = −U ′′(a)

m,

where m is the particle mass, and where the sign in theindex is determined by the direction of the velocity at t = 0. It takes an infinite time toreach the turning point.

1 arsinhx = ln(x +√

x2 + 1).



1.4. If U ′′(a) �= 0, then T ∝ lnε, where ε = Um − E. If

U ′′(a) = . . . = U (n−1)(a) = 0, U (n)(a) �= 0,

we have t ∝ ε−(n−2)/(2n).

1.5. a) When ε = E − Um is small, the particle moves most slowly near the point x = a.Therefore, the entire period of movement T can be estimated by the time T1 of the (backand forth) passage of the small interval δ in the neighbourhood of this point a − δ < x <

a + δ:

T1 = √2m

a+δ∫a−δ

dx√E − U(x)

≈ T .

In the neighbourhood of x = a, we present U(x) in the form

U(x) = Um − 12k(x − a)2,

where k = −U ′′(a). If ε is small enough, you can choose δ such that velocity v on theboundary of the interval is a lot more then the minimum one (at x = a),

12mv2 ∼ 1

2kδ2 ε,

and at the same time to be δ � L = x2 − x1, i. e.

√ε

k� δ � L = x2 − x1.

As a result,

T1 = 2

√mk

ln2kδ2

ε. (1)

The time T2 spent by a particle along the intervals x1 < x < a − δ and a + δ < x2satisfies the relation

T2 � Lv

∼√

mk

lδ

.

When ε decreases, T1 increases so that, for sufficiently small ε, T2 � T1, and we can use(1) to estimate the period of the motion. This formula is asymptotically exact. As ε → 0,its relative error tends to zero as 1/ lnε. But with the same logarithmic accuracy we canreplace δ by L in (1) and omit the multiplier 2 under the sign of logarithm:



T = 2

√mk

lnkL2

ε. (2)

If U ′′(a) = 0, but U (4) = −K �= 0, we have

T = 4(

6m2

εK

)1/4 ∞∫0

dx√1 + x4

= 11.6(

m2

εK

)1/4

and the relative error tends to zero as ε1/4 at ε → 0.

x1 x2x a xx+dx

w

Figure 81

b) If we observe the motion of the particle over atime which is much larger compared to the period T ,the probability to find the particle between x andx + dx is

w(x)dx = 2dtT

=√

2mdx

T√

E − U(x),

where 2dt is the a entire time which the particle spends on an interval dx over the period.The dependence of the probability density w(x) on x is shown in Fig. 81.

The probability w(x)dx is represented by the shaded area (the total area under thecurve is unity). For sufficiently small ε the area under the central maximum gives themain contribution T1/T to the total area under the curve. Despite the fact that w(x) → ∞as x → x1,2, the contribution from the regions near the turning points is relatively small.

c)

w(p)dp = 1T

∑k

∣∣∣∣dtkdp

∣∣∣∣dp = 1T

∑k

dp∣∣∣∣dU(xk)

dx

∣∣∣∣,

where xk = xk(p) is the various roots of the equation p2

2m + U(x) = E.The graph w(p) is shown in Fig. 82, where

p1 = √2m(E − Um), p2 = √

2m[E − U(c)], p3 = √2m[E − U(b)].



–p3 –p2 –p1 p1 p2 p3 p

w~

Figure 82

p

x1

2

2

3

3

4

Figure 83

d) The lines E(x, p) = const (the phase trajectories of theparticle) are shown in Fig. 83, where curves are numberedin ascending order of energies. For U(c) � E < Um, thephase trajectory 2 is double-connected. The arrows indi-cate the direction of the motion of the point representingthe particle state.

1.6. We put the value of the potential energy at the lowestpoint equal to zero. If E = 2mgl we have

ϕ(t) = −π + 4arctan(

e±t√

g/l tanϕ(0)+ π

4

),

where ϕ is the angle between the pendulum and the vertical. The sign in the index is thesame as that of ϕ(0). The pendulum asymptotically approaches its upper position.

In the case 0 < E − 2mgl � 2mgl the pendulum rotates, slowly crossing its upperposition. One can estimate the period of rotation applying the result (2) of the precedingproblem:

T =√

lg

lnε0

E − 2mgl; ε0 = 4π2mgl.

1.7. Once again, we put the value of the potential energy at the lowest point equal tozero, then the energy is equal to

E = 12ml2ϕ2 + mgl(1 − cosϕ).

Let at the moment t0 the angle ϕ(t0) = 0 and for definiteness ϕ(t0) > 0. By introducingk = √

E/2mgl, we have

t = 12

√lg

ϕ∫0

dϕ√k2 − sin2 ϕ

2

+ t0. (1)



At k < 1 the pendulum oscillates within the interval −ϕm � ϕ � ϕm and k =sin ϕm

2 . The substitution sinξ = 1k sin ϕ

2 is reduced the integral (1) to the form1

t =√

lg F(ξ , k)+ t0.

Hence

ϕ = 2arcsin[ksn(u, k)], u = (t − t0)

√gl.

The period of oscillations is

T = 4

√lg

K(

sinϕm

2

).

In the extreme cases (cf. with the problem 1.4)

T = 2π

√lg

(1 + ϕ2

m

16

), when ϕm � 1,

T = 4

√lg ln 8

π − ϕm, when π − ϕm � 1.

If k > 1, the pendulum does not oscillate but rotates and from (1) we obtain

t = 1k

√lg

F(

ϕ

2,

1k

)+ t0, ϕ = 2Arcsinsn

(u,

1k

), u = k(t − t0)

√gl.

1 Functions

F(ξ , k) =∫ ξ

0

dα√1 − k2 sin2 α

, E(ξ , k) =∫ ξ

0

√1 − k2 sin2 α dα

are the so-called incomplete elliptic integrals of the first and second kind, while functions

K(k) = F(π

2, k

), E(k) = E

(π

2, k

)

are the so-called complete elliptic integrals of the first and second kind, respectively. If u = F(ξ , k), then ξ isexpressed in one of the elliptic Jacobi functions, namely by the elliptic sine: sinξ = sn(u, k).

Here is also the formulae for the two limiting cases:

K(k) = π

2

(1 + k2

4

), when k � 1; K(k) = 1

2ln 16

1 − k2, when 1 − k � 1.

Tables and formulae of these functions can be found, for instance, in [11].



The rotation period of the pendulum is

T = 2k

√lg

K(

1k

).

In particular, if E − 2mgl � 2mgl we get

T =√

lg

lnε0

E − 2mgl,

where ε0 = 32mgl. This result differs from the rather crude estimation made in the pre-ceding problem in the value of ε0, that is in a quantity which does not depend onE − 2mgl.

1.8. The motion in the field U(x)+ δU(x) is governed by the equality

t =√

m2

∫ x

b

dx√E − U(x)− δU(x)

, (1)

where we have assumed that at t = 0, x = b. Expanding the integrand in (1) in powersof δU(x), we obtain

t = t0(x)+ δt(x), (2)

where

t0(x) =√

m2

∫ x

b

dx√E − U(x)

, (3)

δt(x) = 12

√m2

∫ x

b

δU(x)dx[E − U(x)]3/2 . (4)

Let the orbit for the case when δU(x) = 0, which is determined by the equationt = t0(x), be x = x0(t). We then have from (2)

x = x0(t − δt(x)), (5)

where in small correction δt(x) we substitute for x the function x = x0(t). Expanding (5)in terms of δt, we finally obtain

x = x0(t)− x′0(t)δt(x0(t)). (6)



Near a turning point x = x1 one has to be careful, as the expansion (2) becomesinapplicable since the correction is δt(x) → ∞ as x → x1.

It is remarkable, however, that formula (6) turns out to be correct up to the turningpoint if

|δU ′(x)| � |F|, F = −U ′(x1). (7)

This is because although with the approach to a turning point δt increases, the depen-dence x(t) near the extremum turns out to be weak.

It is obvious that close to x1 the unperturbed motion has the form

x0(t) = x1 + F2m

(t − t1)2. (8)

Adding δU shifts the turning point on δx1, according to equation

U(x1 + δx1)+ δU(x1 + δx1) = E,

hence δx1 = δU(x1)/F. Taking into account the perturbation δU similarly to (8) we have

x(t) = x1 + δx1 + F2m

(t − t1 − δt1)2 (9)

(because of (7), the correction to F is neglected). Make sure that the calculation of theformula (6) results in (9).

Divide the integration region in (4) into two parts: from b to c and from c to x, where clies near x1. In the second area we can put δU = δU(x1) and U(x) = E − (x − x1)F. Then

δt =√

mδU(x1)√2F3(x − x1)

+ δt0, (10)

δt0 = 12

√m2

∫ c

b

δU(x)dx(E − U)3/2 −

√mδU(x1)√F3(c − x1)

.

Substituting (10) and (8) into (6) and neglecting (δt0)2, we obtain (9) with δt1 = δt0.

1.9. a) Here we can use the results of the preceding problem. For the unperturbedmotion with b = 0 and x1 = a we have

x0(t) = asinωt, E = 12mω2a2.



In this case |δU/U | � ε = αaω2 � 1. For the correction we have

δt(x) = α

3ω3

(√a2 − x2 + a2

√a2 − x2

− 2a)

,

δt(x0(t)) = ε

3ω

(cosωt + 1

cosωt− 2

).

Substituting these expressions into (6) of the preceding problem, we get

x(t) = a sin ωt − 13 εa(cos2 ωt + 1 − 2cosωt).

Up to and including first-order terms in ε, we have

x(t) = asin(ωt + 2

3 ε)

− 12 εa − 1

6 εacos2ωt

(cf. problem 8.1 b).b) Acting in the same way as previously described, we obtain

x(t) = asinωt + εa(

32 ωt cosωt − 9

8 sinωt − 18 sin3ωt

), ε = βa2

4ω2 � 1. (1)

This result has a relative accuracy ∼ ε2 in one period, and after 1/ε periods theformula (1) becomes completely inapplicable. Taking into account the periodic characterof the motion, we can extend the result (1) over a larger period of time. With accuracyup to and including first-order terms of ε, the formula (1) is transformed into a clearlyperiodic form

x(t) = a(1 − 9

8 ε)

sin[ω

(1 + 3

2 ε)

t]− 1

8 εasin[3ω

(1 + 3

2 ε)

t]

. (2)

The corrections, which were not taken into account in (1), have lead to the frequencychange of the order of ε2ω, so that (2) has the relative accuracy ∼ ε during 1/ε periods(cf. problem 8.1 and, for more detail, see [7], § 29.1).

1.10. The change in the period is

δT = √2m

⎡⎢⎣

x2+δx2∫x1+δx1

dx√E − U(x)− δU(x)

−x2∫

x1

√E − U(x)dx

⎤⎥⎦ . (1)

One can not expand the integrand (1) in terms of δU(x) since the requirements of thetheorem about differentiation improper integrals with respect to a parameter are violated,



since the resulting integral diverges. However, we can expand the integrand in termsof δU(x) up to terms of fist order in δU(x) if we rewrite δT in the form

δT = 2√

2m ∂

∂E

⎡⎢⎣

x2+δx2∫x1+δx1

√E − U(x)− δU(x)dx −

x2∫x1

√E − U(x)dx

⎤⎥⎦ , (2)

whence

δT = −√2m ∂

∂E

x2∫x1

δU(x)dx√E − U(x)

= − ∂

∂E(T〈δU〉), (3)

where

〈δU〉 = 1T

T∫0

δU[x(t)]dt (4)

is time-average value of δU .The time spent near the turning points contributes little to the period, provided,

of course, that U ′(x1,2) �= 0 (cf. problem 1.3). Equation (3) can therefore give a goodapproximation.

We note that sometimes even small extra term δU(x) may strongly affect the particlemotion (e.g. see, problem 1.11 b, c).

Higher-order terms in the expansion of δT in terms of δU can be obtained by similarmeans:

T = √2m

∞∑n=0

(−1)n

n!∂n

∂En

x2∫x1

[δU(x)]n dx√E − U(x)

. (5)

The formal expression (5) may be an asymptotic or even a convergent series.

1.11. a) According to (5) of the preceding problem the correction to the period2π/ω is

δT = −3πβE2mω5 .

This correction is small for sufficiently small E.

U

E

U

x

Um

U+δU

Figure 84

b) In Fig. 84 we have given the potentialenergies U(x) and U(x)+ δU(x). When E > Um =mω6/(6α2) the extra term clearly means that theparticle can move to infinity. When E is close to Um,



the period of oscillation increases without bound (as | ln(Um − E)|; see problem 1.4) sothat one cannot determine the change in period just by a few terms in series (5) of thepreceding problem.

When E � Um the correction to the period is

δT = 5π

18ω

EUm

.

c)

δT = 3πAV√

m√2α|E|5/2

;

the formula is applicable when |E| |Um| ≈ √8AV , (E < 0).

1.12. The particle delay time is

τ =+∞∫

−∞

(1v

− 1v0

)dx = 1

αv0ln

EE − U0

,

where v =√

2m |E − U(x)|, v0 =

√2Em (cf. problem 1.1b).


§2

Motion of a particle inthree-dimensional fields

2.1. To study the orbits of the particle we use the energy and angular momentum

Ox

r

rϕr

rϕ

˙˙

˙

Figure 85

conservation laws:

mr2

2+ U(r) = E, (1)

m[r, r] = M. (2)

It follows from (2) that the orbit lies in a plane. Introducingthe polar coordinates in that plane (see Fig. 85) we get

12 mr2 + 1

2 mr2ϕ2 + U(r) = E, (3)

mr2ϕ = M. (4)

Using (4) to eliminate ϕ from (3), we find

mr2

2+ Ueff(r) = E, (5)

where

Ueff(r) = U(r)+ M2

2mr2 .

The radial motion can thus be considered as one-dimensional motion in the field Ueff(r).For a qualitative discussion of the character of the orbits we use curves of

Ueff(r) = −α

r− γ

r3 + M2

2mr2 (6)

for different values of M (Fig. 86).




Ueff

Ueff Ueffr1 r2

Umax

Umin

a

(a) (b) (c)

bc d

r1

r2 r r r

Figure 86

When 12αγ m2 < M4, then Ueff has two extrema at

r1,2 = M2 ∓ √M4 − 12αγ m2

2mα.

The maximum value of Ueff(r1) = Umax is positive when M4 > 16αγ m2 (Fig. 86a)and negatively when 12αγ m2 < M4 < 16αγ m2 (Fig. 86b); in both these cases we haveUeff(r2) = Umin < 0.

If M4 < 12αγ m2, then the function Ueff(r) is monotone (Fig. 86c).Let us consider the case a) in somewhat more detail. If E > Umax the particle coming

from infinity falls in the centre of the field. The quantity of ϕ then increases accordingto (4). This is all we need to sketch the orbit (see Fig. 87a).

O

(a) (b)

x

O

br1

a

Figure 87 Figure 88

At large distances such that γ

r3 � αr , the main term in U(r) is the term (−α/r) and

the orbit differs little from the hyperbola. (See problem 2.8 on the form of a trajectoryat r → 0.)

If the energy E is close to Umax the particle pass very slowly through the range ofvalues of r which are close to r1. At the same time the radius vector turns round with anangular velocity ϕ ≈ M

mr21

so that the particle may perform many revolutions around the

centre before it passes through this range of values (Fig. 87b).



If E = Umax the particle approaches the point of r = r1 (cf. problem 1.3). The orbitis a spiral approaching the circle of radius r1 and centre O (Fig. 88 curve a). If, on theother hand, the particle starts in the region r < r1 and is moving away from the centre, theorbit also approaches this circle, but this time from the inside (Fig. 88 curve b). Finally,motion along the circle r = r1 is possible when E = Umax. However, the motion alongthis circle is unstable one, as any change in the values of E or M will bring onto the orbitwhich moves away from the circle.

If 0 < E < Umax a particle coming from infinity will be reflected from the potentialbarrier Ueff(r) and again move off to infinity. Examples of such orbits are given in Fig. 89(curves a and b). If the energy is close to Umax the particle makes many revolutionsaround the centre before the radial velocity r changes sign. The closer the energy isto zero (for a fixed value of M this corresponds to larger values of the impact parameter)the less twisted the orbit of the particle.

When E < Umax the case is also possible of the particle falling towards the centre ofthe field, when it is moving in the region r < a. The orbit for this case is given in Fig. 90.

O

ba

B

A

C

OO

D

A

C

B

c

d


When Umin < E < 0 the particle can also perform radial oscillations in region c � r � d(Fig. 91). If the energy is close to zero, the amplitude of the radial oscillations will be large;their period can also become large and the radius vector may perform several revolutionsduring one radial oscillation. When the energy close to Umin the orbit close to a circlewith radius r2 while the angle over which the radius vector turns during one period ofthe radial oscillation depends on the quantities α, γ , M (compare problem 5.4). WhenE = Umin the particle moves along of the previously mentioned circle.

One can analyse the motion of the particle in a similar way for the other cases.What are the particular features of the orbit when M4 = 12αγ m2?Let us now consider the orbits in more detail. One can obtain the equation for the

orbits from the (4) and (5). From (5) we get

r = drdt

= ±√

2m

[E − Ueff(r)], (7)

or

t = ±√

m2

∫dr√

E − Ueff(r)+ C. (8)

Using (4) to eliminate dt from (7), we find the equation of the orbit



ϕ = ± M√2m

∫dr

r2√

E − Ueff(r)+ C1. (9)

Let us consider the case M4 > 12αγ m2. If the particle moves towards the centre, wemust take the lower sign in (7) and, of course, in (8). If at t = 0, r = r0, we can rewrite (8)in the form

t = −√

m2

r∫r0

dr√E − Ueff(r)

. (10)

Equality (10) is an implicit equation for r as the function of t. If the orbit passesthrough the point r = r0, ϕ = ϕ0, the equation of the orbit becomes

ϕ = −M

r∫r0

drr2|pr| + ϕ0, (11)

where we have once again taken the lower sign and where

|pr| = √2m[E − Ueff(r)].

In particular, if the velocity of the particle at infinity is makes an angle ψ with the x-axis,we must put r0 = ∞, ϕ0 = π − ψ .

If E > Umax, (10) and (11) completely determine law of motion and the particle orbit.However, if 0 < E < Umax, these equations correspond only to the section AB of the

orbit (see Fig. 89, curve a). At the point B the radial component of the velocity r vanishesand then changes sign. The section BC of the orbit is thus described by (9) with the uppersign, and we must redetermine the constant. It is helpful to write (9) in the form

ϕ = M

r∫rmin

drr2|pr| + C1. (12)

As long as C1 is not determined, we can choose the lower limit of the integral arbitrarily.According to (12), we now have

C1 = ϕ(rmin). (13)

Determining ϕ(rmin) from (11), we get the following equation for the section BC of theorbit:

ϕ =⎛⎝

r∫rmin

−rmin∫r0

⎞⎠ M dr

r2|pr| + ϕ0. (14)



Similarly we can determine the time dependence of r along the section BC

t =⎛⎝

r∫rmin

−rmin∫r0

⎞⎠

√m2

dr√E − Ueff(r)

. (15)

If Umin < E < 0, a < r0 < b, r(0) < 0, ϕ|t=0 = ϕ0, (11) describes the section AB of theorbit in Fig. 91. The section BC is described by the equation

ϕ = M

r∫a

drr2|pr| + ϕ1, (16)

where the angle ϕ1 can be obtained by putting r = a in (11). The equation for thesection CD is

ϕ = −M

r∫b

drr2|pr| + ϕ2, (17)

where ϕ2 is defined from (16) with r = b, and so on. Substituting the values ϕ1 and ϕ2in (16) and (17), we get the equations for BC and CD in the form

ϕ = M

⎛⎝

r∫a

−a∫

r0

⎞⎠ dr

r2|pr| + ϕ0, (18)

ϕ = M

⎛⎝

r∫b

+b∫

a

−a∫

r0

⎞⎠ dr

r2|pr| +ϕ0 = M

⎛⎝−

r∫a

+2

b∫a

−a∫

r0

⎞⎠ dr

r2|pr| + ϕ0. (19)

One verify easily that the equation that section of the orbit which corresponds to thenth radial oscillation, taken the section AB to be the first, has the form1

1 Equation (20) for the orbit can be written in the form

cosγ (ϕ +α) =⎛⎝γ M

r∫a

dr

r2|pr |

⎞⎠ ,

where

π

γ= M

b∫a

dr

r2|pr | , α = M

a∫r0

dr

r2|pr | − ϕ0.



ϕ = M

⎛⎝±

r∫a

+2(n − 1)

b∫a

−a∫

r0

⎞⎠ dr

r2|pr| + ϕ0. (20)

In the equations given here we have assumed that the angle ϕ changes continuouslyand we have not introduced the limitation 0 � ϕ < 2π . There is an infinity of values of ϕ

corresponding to a given value of r, corresponding to different values of n and differentsigns in (20): ϕ is a many-valued function of r; on the other hand, r is a single-valuefunction of ϕ.

All other cases can be treated in a similar fashion.

2.2. Outside the sphere of radius R the particle moves with a speed√

2E/mand inside with a speed

√2(E + V )/m. Depending on the values of E and M, we get

different kinds of orbits.When M2

2mR2 − V < E < M2

2mR2 , the particle can either move inside the sphere and

be reflected at its surface (Fig. 92a), or, provided E > 0, we can have an infinite orbit

(which can be a straight line, see Fig. 93b). When M2

2mR2 < E, we can have a reflected

orbit (Fig. 92b).

What is the shape of the orbit when E = M2

2mR2 − V ?

(a) (b)

Figure 92

2.3. To determine the equation of the orbit we use the equations

ϕ =∫

M dr

r2√

2m(E − Ueff), Ueff = U(r)+ M2

2mr2 (1)



and we then get1

r = pecosγ (ϕ − ψ)− 1

, (2)

where

p = 2α

(β + M2

2m

), e =

√1 + 4E

α2

(β + M2

2m

), γ =

√1 + 2mβ

M2 , (3)

E > 0, and ψ is an arbitrary constant.The orbit is the curve which is obtained from a hyperbola by reducing the polar angles

by a factor γ (Fig. 93). The constant ψ determines the orientation of the orbit.The direction of the asymptotes is determined by the condition r → ∞, or

e cos(ϕ1,2 − ψ) = 1. The direction of the velocity before and after the scattering areat the angle

π − (ϕ1 − ϕ2) = π − 2γ

arccos1e

= π − 2γ

arctan

√4Eα2

(β + M2

2m

).

Ox

ϕ1ϕ2

ψ

E

Ueff

rmr

Figure 93 Figure 94

2.4. It is useful first to study the character of the orbit using a graph Ueff(r). Thisgraph is given in Fig. 94 for the case where β < M2/2m. In that case all orbits areinfinite and lie in the region r � rm when E > 0. The equation of the orbits are thesame as in problem 2.3, (2), except that in equations (3) we must substitute β for (−β).

1 If we write the integral for ϕ in the form

MM

ϕ =∫

M dr

r2

√2m

(E − M2

2mr2 − αr

) ,

whereM2 = M2 + 2mβ, the integral is reduced to the corresponding integral in the Kepler problem(see [1], § 15 and [7], § 3.4).



The main difference with the orbits found in problem 2.3 occurs because now γ < 1.An example of an orbit is shown in Fig. 95. (The point of the inflection A is determinedby the condition dU(r)/dr = 0, i. e. r = 2β/α.)

A

O

Ueff

Umax

E

a br

Figure 95 Figure 96

The form of Ueff(r) for the case β > M2/2m is shown in Fig. 96.If

E > Umax = α2

4(β − M2/(2m)),

the particle flying from infinity will fall into the centre of the field. The equation of theorbit can be obtained from the equation of problem 2.3. We must then replace β by (−β)

and ψ by ψ + π/2γ , and also use the formulae

sin ix = i sinh x,√−x = i

√x.

As a result, we get

r = p′

e′ sinhγ ′ (ϕ − ψ)+ 1, (1)

p′ = 2α

(β − M2

2m

), e′ =

√4Eα2

(β − M2

2m

)− 1, γ ′ =

√2mβ

M2 − 1. (2)

The orbit for this case is shown in Fig. 97a. We note that as r → 0, ϕ → ∞. This meansthat a particle coming from infinity and falling into the centre of the field makes an infinitenumber of revolutions around the centre.

O

(a) (b)

O

Figure 97



If E < Umax, we see from Fig. 96 that the particle can either move in the region b �r < ∞ (scattering), or in the region 0 < r � a (falling into the centre.) We obtain theequation for the orbit, using the equation cos ix = coshx, and for the second case againthe substitution of ψ + π/γ for ψ :

r = p′

1 ∓ e′′ coshγ ′(ϕ − ψ), e′′ =

√1 − 4E

α2

(β − M2

2m

). (3)

For the case E = Umax, it is not possible to use (2) of the problem 2.3, as we haveassumed in its derivation that e = 0, and we must start again from the integral (1). Wefind

r = p′

1 + cexp(−γ ′ϕ),

that is

r = p′

1 ± exp[−γ ′(ϕ − ψ)]or r = p′

depending on the initial value of r. The orbit is either a spiral starting at infinity andapproaching asymptotically the circle with radius r = p′, or a spiral starting near thecentre and approaching the same circle asymptotically, or it is the circle itself (Fig. 97b).

Finally, for the case β = M2/(2m), it is also simplest to start again from the originalintegral. In that case, scattering occurs and the equation for the orbit is

r = α/E1 − mα2(ϕ − ψ)2/(2M2E)

.

The time it takes the particle to fall into the centre of the field is found from theequation

t =√

m2

r∫0

dr√E − Ueff

.

For instance, when the orbit has the form given by (2), the time to fall into the centrefrom a distance r is given by

t = 1E

√m2

(√Er2 − αr + β − M2/(2m) −

√β − M2/(2m)

)++ α

2E

√m2E

(arsinh

(2Er/α)− 1e′

+ arsinh1e′

).



O

Figure 98

2.5. The equation of the orbit is

r = p1 + ecosγ (ϕ − ψ)

,

where p, e, and γ are defined in problem 2.3. When E < 0, the orbitis the finite one, and1

Tr = πα√

m√2|E|3/2

, �ϕ = 2π

γ.

The orbit is closed if γ is the rational number. In Fig. 98, to show the orbit for γ ≈ 5.

2.6. When β < M2/(2m) we have

r = p1 − ecos γ (ϕ − ψ)

, p = 2α

(M2

2m− β

),

γ =√

1 − 2mβ

M2 , e =√

1 + 4Eα2

(M2

2m− β

);

if E < 0, then �ϕ = 2π/γ and Tr will be the same as in the preceding problem.When β > M2/(2m) we have (using the notations of the problem 2.4)

r = p′

e′ sinhγ ′(ϕ − ψ)− 1, if E > Umax,

r = p′

e′′ coshγ ′ (ϕ − ψ)− 1, if E < Umax.

2.7. a) A finite orbit is possible if the function Ueff(r) has a minimum. The equationU ′

eff(r) = 0 can be reduced to the form f (x) = M2/(αm), where f (x) = x(x + 1)e−x,x = r. Using a graph of f (x) one sees easily that this equation has real roots only whenm2/(αm) is less than the maximum value of f (x), for x > 0. This maximum value

is equal to (2 + √5)exp

[−1

2 (1 + √5)

]≈ 0.84. A finite orbit is thus possible, provided

M2 < 0.84αm/.b) A finite orbit is possible, provided M2 < 8mV/(e22).

2.8. In the equation for the orbit (see (1) of the problem 2.3), we can neglect for smallvalues of r the quantity E, when n = 2, and also the term M2/(2mr2), when n > 2. We

1 The period Tr is the same as in the field U0 = −α/r. To determine Tr it is sufficient to note that adding aterm β/r2 to the field U0 has the same effect on the radial motion as increasing M. However, the period Tr isindependent of M in the Coulomb field U0.



obtain then (see Fig. 99)

ϕ = − M ln(r/r0)√2mα − M2

+ ϕ0, when n = 2,

ϕ = 2Mr−1+n/2√

2mα(n − 2)+ C, when n > 2.

n=2 n>2

Figure 99

It turns out that the number of revolutions is infinite onlywhen n = 2.

The time it takes the particle to fall into the center isfinite as the radial velocity increases when the center isapproached.

2.9. The number of revolutions of the particle around thecentre is infinite only in the case b) E = 0, n = 2.

2.10. The time it takes the particle to fall into the centre isπ

√mR3/(8α).

2.11. As it is known, the two-body problem is reduced to two other problems: the motionof the centre of mass and the relative motion of a particle with reduced mass (in thiscase equal to m/2) in the field U(r). We consider only the relative motion. Introducingthe relative distance r = r1 − r2 and taking into account that when time t is close tothe initial time t∞ → −∞ one can assume that r1 = (vt,0,0) and r2 = (0,v(t − τ),0).Using these expressions, we find the initial velocity v∞ = (v,−v,0), the energy E =12 (m/2)v2∞ = 1

2mv2, the angular momentum M = (m/2)[r,v∞] = −(0,0,Eτ), and theeffective potential energy

Ueff (r) = α

r+ M2

mr2 .

After that, the minimum distance which is equal to

rmin = α

mv2

[1 +

√1 + 1

2

(mv3τ/α

)2]

,

can be found from the equation Ueff (rmin) = E.

2.12. The problem is reduced to two others: the motion of the centre of mass

R = m1r1 + m2r2

m1 + m2(1)

with constant velocity R and the relative motion of particles with reduced massm = m1m2/(m1 + m2) in the field U(r), where the relative distance is r = r1 − r2.



a) Let’s take the point A to be the origin of the coordinate system, and the originof the time will be the moment when particle 1 would fly through this point if therewere no particle interaction. Under this condition, the position of the particles wouldbe, respectively, r1 = v1t and r2 = v2(t − τ)+ ρ, where ρ ⊥ v1,2. These expressions canbe used to determine all of the following: the law of motion of the centre of mass; theenergy E = 1

2mv2; the angular momentum M = m[r,v] = −m([v1,v2]τ + [ρ,v]); andthe effective potential energy

Ueff (r) = − β

r2 + M2

2mr2 = −2mβ − M2

2mr2

in the relative motion problem (here v = v1 − v2). Note that from the expressions for Eand M, the time t vanishes as expected. It can be seen from the equation

E = 12mr2 + Ueff (r) (2)

that the particle falls to the centre under the condition

2mβ > M2 = m2([v1,v2]2τ2 + ρ2v2

).

This answers the first question.b) To find the point of impact, we find its time considering relative motion and then

determine where at that moment the centre of mass is.From (2), we find the time of collision

t = t∞ −0∫

r∞

dr√2[E − Ueff(r)]/m

.

Notice the we took into account that the radial velocity r is negative during the processof particles falling on each other (such process corresponds to the movement from theinitial, very large distance r∞ to the point of incidence r = 0). For very large values ofr, one can ignore the particle interaction and take the initial time equal to t∞ = −r∞/v.In the end, the time of collision of particles is equal to

t = 1v

(√r2∞ + a2 − a − r∞

), a2 = 2mβ − M2

(mv)2 .

Substituting√

r2∞ + a2 ≈ r∞ + a2

2 r∞ and discarding the term containing a2/(2vr∞), weget t = −a/v and finally find the distance where the particles will collide,

R(t) = − (m1v1 + m2v2)a + m2v2vτ

(m1 + m2)v.



2.13. The relative motion is characterized by the angular momentum M = mvρ and theenergy E = 1

2mv2, where m = m1m2/(m1 + m2) is the reduced mass. The distance to bedetermined follows from the condition Ueff(rmin) = E. Solutions can be obtained easilyfor n = 1, 2, and 4.

O

O S

ρϕ0

Figure 100 Figure 101

2.14. The particle orbit is:m

m1,2

pr1,2

= 1 ± ecosϕ,

where

m = m1m2

m1 + m2, p = M2

mα, e =

√1 + 2EM2

mα2 ;

E and M are the total energy and angular momentum of the system. The particles movein similar conical sections with a common focus, and their radius vectors are at anymoment in the opposite directions (Fig. 100).

2.15. One sees easily from Fig. 101 that

OS = ρ(cotϕ0 − cot2ϕ0),

where

ϕ0 = ρ

∞∫rmin

dr

r2

√1 − U(r)

E − ρ2

r2

.

At ρ → 0

ϕ0 = ρ

∞∫rmin

drr2

√1 − U/E

− 2ρ3E3/2 ∂

∂E

∞∫rmin

dr

r4√

E − U+ . . .



(cf. problem 2.27), so that

OS =⎛⎝2

∞∫rmin

drr2

√1 − U/E

⎞⎠

−1

+O(ρ2) . . .

The point of S is the virtual focus of the beam of scattered particles so that up to andincluding terms of first order in ρ the position of the points where the asymptote of theorbit intersects the axis of the beam is independent on ρ.


pr

= ecos(ϕ − ϕ0)− 1,

where p = M2

mα , e =√

1 + 2EM2

mα2 , while ϕ0 is determined from the condition that ϕ → π ,

as r → ∞, so that cosϕ0 = −1/e. The region which can not be reached by the particlesis bounded by the envelope of the family of orbits.

To find it, we differentiate the equation for the orbit,

M2

mαr+ 1 + cosϕ − M

α

√2Em

sinϕ = 0 (1)

with respect to the parameter M

2Mmr

−√

2Em

sinϕ = 0, (2)

and eliminate M from (1) and (2). The result is

2α

E r= 1 − cosϕ.

Oz

Figure 102

The inaccessible region is thus

r <2α

E(1 − cosϕ)

which is bounded by a paraboloid of rotation (Fig. 102).

2.17. ρ > 2aδ

1 − δ2 − (1 − δ)2 cosϕ, where δ = mav2

2α, OA = a.



2.18. We take the scalar product of the equation

[v,M] − αrr

= A

with r. Denoting by ϕ the angle between R and A, we get

M2

m− αr = Ar cosϕ,

or

pr

= 1 + ecosϕ,

where

p = M2

mα, e = |A|

α.

We note that the vector A is directed from the centre of the field to the point r = rmin.

2.19. In a field U(r) = −α/r, the body, first as a part of the spacecraft, flew in a circleof radius R at a velocity V . If the separation of the body from the spacecraft occurs atR = (0,R) and velocity V = (V ,0), the Laplace vector

A0 = [V,M0] − αRR

= mV 2R − αRR

is zero and the velocity of the body before separation was V = √α/(mR), the angular

momentum M0 = mRV = √mαR.

Obviously, the orbit of the body lies in the same plane as the orbit of the spacecraft,and the velocity of the body at the first moment after separation from the spacecraft isV + v. The angular momentum of the body has not changed, m[R,V + v] = M0, sincev ‖ (−R), but the Laplace vector is no longer zero

A = [V + v,M0] − αRR

= [v,M0] = mRvV = αvV

V 2 .

As it is known, the Laplace vector is directed towards the pericentre (the point of theorbit with the smallest radius) and is αe, where e is the eccentricity. From here we getthe value e = v/V . Assuming that e < 1, we find that the orbit of the body is an ellipse,whose parameter is p = M2/(mα) = R, and whose pericentre lies on the x-axis. Thus, theequation of the body orbit in polar coordinates r,ϕ has the form

r = R1 + (v/V ) cosϕ

.



2.20. δT = −∂δI∂E

, where

δI = T〈δU〉 = √2m

r2∫r1

δU(r)dr√E − Ueff(r)

(cf. problem 1.10). Similarly, we can write for the change in angular distances between

successive pericentre passages (r = rmin) in the form δ�ϕ = ∂δI∂M

(cf. [1], § 15, problem 3;

§ 49).

2.21. The field U(r) differs little from the Coulomb field U0(r) = −α/r in the regionr � D. Therefore, a finite orbit which is close to an ellipse with a parameter p and aneccentricity e determined by the integrals of motion E and M will retain its shape, butwill change its orientation. The velocity of rotation � of the ellipse is determined by thedisplacement of the pericentre over a period � = δ�ϕ/T0 which we can evaluate usingthe equations from the preceding problem with δU = α

D − αr2D2 while T0 is the period

in the Coulomb field.1 The result of the calculation is � = M/(2mD2). We can write theequation of the orbit in the form

pr

= 1 + ecosγ ϕ, γ = 1 − �T0

2π. (1)

The deviation of the curve (1) from the actual orbit is of first order in δU ; that is, duringone period (1) describes the orbit with the same degree of accuracy as the equationfor the fixed ellipse. However, (1) retains the same accuracy during many periods. It istherefore just this equation which can be called the “correct zero approximation”.

In other words, only secular first-order effects have been taken into account in (1).

2.22. The field U(r) = −α/r1+ε differs little from the Coulomb field so that the orbit ofthe particle in this field will be a slowly precessing ellipse. Expanding U(r) in terms of ε,we can write it in the form

U(r) = U0(r)+ δU ,

1 It is convenient to change to an integration over ξ , where (see [1], § 15)

r = α

2|E| (1 − ecosξ).



where

U0(r) = − α

r, δU = εα

rln

rR

, α = αR−ε

and R is a constant which characterizes the size of the orbit.We can evaluate the shift of pericentre for the period

δ�ϕ = ∂

∂M

T∫0

δU dt

(see problem 2.20) by making the substitution

r = − α

2E(1 − ecosξ), t = T

2π(ξ − esinξ),

where

e =√

1 + 2EM2

mα2 , T = πα

√m

2|E|3

(see [1], § 15). The result is

� = δ�ϕ

T= εα

2π

∂

∂M

2π∫0

lnα(1 − ecosξ)

2|E|R dξ =

= −ε|E|π

∂e∂M

2π∫0

cosξ dξ

1 − ecosξ= 2π

T1 − √

1 − e2

e2 ε.

At ε > 0, the direction of the orbit precession coincides with the direction of motion ofa particle over orbit, and at ε < 0 the precession is opposite to it.

2.23. � = −32

βmω

a2 + b2

a4b4M|M| (see also [7], § 4).

2.24. The Lagrangian function reads

L = m2

(x2 + y2

)− mg

2l

(x2 + y2

)+ m

2x2 + y2

l2

(x2 + y2

)=

= m2

(r2 + r2ϕ2

)− mg

2lr2 + mr2

2l2r2.

If we neglect the last term, then it will be convenient to consider the particle motionin the Cartesian coordinates. It leads to harmonic oscillations with frequency ω = √

g/l



along the x- and y-axes, that is, to the elliptic orbit. The exact equation of the trajectoryis convenient to obtain in the cylindrical coordinates using the integrals of motion Eand M = (0, 0, M):

ϕ =∫

M√

1 + r2/l2 dr

r2√

2m[E − Ueff(r)], Ueff(r) = mgr2

2l+ M2

2mr2 .

Expansion

√1 + r2

l2= 1 + r2

2l2+ . . .

leads to the equation

ϕ =∫

M dr

r2√

2m(E − Ueff)+ M

2ml2

∫dr√

2m(E − Ueff)= ϕ0(r)+ � t,

where ϕ0(r) corresponds to the motion over an ellipse, and

� = M2ml2

= ab2l2

ω

determines the angular velocity of the precession.

m0

R1

m1

m2

r1

r2

R2

R

O

Figure 103

2.25. The Lagrangian function of the Earth–Moon system is

L = 12 m1R2

1 + 12 m2R2

2 + Gm0m1

R1+ Gm0m2

R2+ Gm1m2

|R1 − R2| ,

where R1 and R2 are radius vectors of the Earth and the Moonin the heliocentric coordinate system, m1, m2 are their masses,m0 is the mass of the Sun, and G is the gravitational constant.We further introduce the coordinates of the centre of mass of theEarth and the Moon (the point O on Fig. 103) and their relativedistance

R = m1R1 + m2R2

m1 + m2, r = R2 − R1,



then

R1,2 = R + r1,2, r1,2 = ∓ mm1,2

r,

m = m1m2

m1 + m2≈ m2.

Using the expansion

1Ri

= 1R

(1 + 2Rri

R2 + r2i

R2

)−1/2

=

= 1R

− Rri

R3 + 12R3

[3

(Rri)2

R2 − r2i

]+ . . .

and taking into account that

m1r1 + m2r2 = 0, m1r21 � m2r2

2 ≈ m2r2,

we obtain

L = L1 + L2 − δU , (1)

L1 = m1 + m2

2R2 + Gm0(m1 + m2)

R,

L2 = m2

r2 + Gm1m2

r, δU = −Gm0m

2R3

[3

(Rr)2

R2 − r2]

.

If we neglect the term δU , the problems of motion of the centre of mass O and ofrelative motion are separated and reduced to the Kepler problem (for simplicity, we willtalk further about the motions of the Earth around the Sun and the Moon around theEarth).

a) In the problem of the motion of the Earth around the Sun, the smallness of δU ischaracterized by a ratio

δU(Gm0m1/R)

∼ m2r2

m1R2 ∼ 10−7.

Considering the orbit of the Moon as a circle of radius r lying in the plane of the Earth’sorbit, one has

δU = −2β

R3

(3cos2 χ − 1

), β = 1

4 Gm0m2r2, (2)



where χ is the angle between vectors R and r. This angle changes to 2π in 29.5 days(a so-called synodic month which is a time interval between the two sequent newMoons). Averaging expression (2) over that interval leads to the following replacementcos2 χ → 1

2 . As a result,

δU = − β

R3 .

The perihelion displacement per year (see [1], problem 3 to § 15) is

δϕ = 6πβ

α1R2 = 3πm2

2m1

( rR

)2, α1 = Gm0m1.

The perihelion shift per century equals

100δϕ = 7.7′′.

It should be noted that the total displacement of the Earth’s perihelion for one centuryequal to 1158′′ is mainly due to the influence of the Jupiter and the Venus.

It is worth noting that the relativistic correction can be estimated as

∼ 100 · 2π

(Vc

)2

∼ 1′′, V ≈ 30 km/s

(see [2] § 39, § 101 and [7], § 41.2).b) Studying the motion of the Moon around the Earth, we can also consider δU as a

small correction to L2:

δU(Gm1m2/r)

∼ �2

ω2 ∼ 10−2, �2 = Gm0

R3 , ω2 = Gm1

r3 .

Here the period 2π/ω = 27.3 days is the star (or sidereal) month and the period 2π/�

is 1 year. The correction δU leads to various distortions of the the Moon’s orbit as theripple of eccentricity, the perihelion shift (cf. problem 2.43) and so on. We will consideronly one of them, neglecting their mutual influence and assuming the unperturbed orbitof the Moon as a circle; we also take R = const.

We introduce the geocentric coordinate system Oxyz with the x-axis directed along theline of intersection of the plane of the Moon’s and the Earth’s orbits with (the line of lunarnodes), the y-axis lying in the plane of the Earth’s orbit and the z-axis perpendicular tothis plane and directed to the northern celestial hemisphere (Fig. 104). The coordinatesof the Sun in this system are

−R = R(cosϕ, sinϕ, 0), ϕ = �t + ϕ0,



ϕ

θψ

z

r

–R

m2

m1

m0

x

y

Figure 104

and the coordinates of the Moon are

r = r(cosψ , cosθ sinψ , sinθ sinψ), ψ = ωt + ψ0,

where ϕ0 and ψ0 are determined by the moments of passage of the Sun and the Moonthrough the x-axis.

Averaging quantity

(Rr)2 = R2r2(cosϕ cosψ + cosθ sinϕ sinψ)2

over one month leads to the replacements

cos2 ψ → 12 , sin2 ψ → 1

2 , sinψ cosψ → 0,

and to equations

〈(Rr)2〉month = 12 R2r2

(cos2 ϕ + cos2 θ sin2 ϕ

), (3)

while averaging over one year leads to these replacements

cos2 ϕ → 12 , sin2 ϕ → 1

2 .

It gives

〈(Rr)2〉 = 14 R2r2

(1 + cos2 θ

).

As a result,

〈δU〉 = −Gm0m8R3 r2

(3cos2 θ − 1

).



The rotation of the orbital plane can be traced by considering the angular momentumM = m[r, r] perpendicular to this plane. The equation of its motion is M = K, where

K = −[r,

∂δU∂r

]

is the moment of the forces acting upon the Moon. Because the change of the angle θ isthe rotation around the x-axis,

Kx = −∂δU∂θ

,

that is,

〈K〉 =(

−3Gm0m4R3 r2 sinθ cosθ , 0, 0

).

Since the moment of forces 〈K〉 is perpendicular to the vector M, it leads only to itsrotation. The x- and y-axes in this case also rotate; therefore, the angular velocity �prof precessions of the vector M is directed along the z-axis. It can be found from thecondition

[�pr,M] = 〈K〉,

which gives

(�pr

)z = −〈Kx〉

My= 〈Kx〉

Mωr2 sinθ= −3�2

4ωcosθ ≈ − �

17.9.

Thus we found that the precession period of the Moon’s orbit is 2π/�pr = 17.9 years.This motion is called a retreat of the lunar nodes. The correct meaning of this period is18.6 years. Given the crudeness of our approximations, our result could be consideredas a good agreement with the correct one.

This period determines the repetition period of the solar and lunar eclipses cycles.That period (or, more precisely, its triple value containing integer 19765 days) wasalready known to the priests of the ancient Babylon.

Note that by limiting the averaging over a month; that is, using (3), we would discoverthe unevenness of the precession within a year:

�pr →(1 − 1

2 cos2ϕ)�pr.



2.26. Similarly to problem 2.21, we have

� = ∂

∂M〈δU〉 = a

2∂e2

∂M

[δU ′(a)+ a

2δU ′′(a)

]=

= −√

amα

[δU ′(a)+ a

2δU ′′(a)

] M|M| .

If we take into account the next terms of the expansion δU over (r − a), that will give thecontribution of � e2� into �.

2.27. The perihelion displacement for the period

�ϕ =r2+δr2∫

r1+δr1

M drr2

√2m(E − Ueff − δU)

can be presented as

�ϕ = −43

√2m

∂2

∂M∂E

r2+δr2∫r1+δr1

(E − Ueff − δU)3/2dr.

Up to the second order in δU , we have

�ϕ = 2π + δ1ϕ + δ2ϕ,

δ1ϕ = t ∂

∂M〈δU〉,

δ2ϕ = − ∂2

2∂M∂E[T〈(δU)2〉],

where 〈f 〉 is the value of f (r), averaged over the period T of the unperturbed motion(cf. problem 2.20). The angular velocity of the precession is

� = δ1ϕ + δ2ϕ

T + δT= δ1ϕ

T+ δ2ϕ

T− δ1ϕ δT

T2 ,

δT = − ∂

∂E(T〈δU〉).



2.28. If we expand the integrand in the equation for the orbit,

ϕ =∫

M dr

r2

√2m

(E − M2

2mr2 + αr − γ

r3

) , (1)

in terms of δU = γ /r3, we can rewrite this equation in the form

ϕ = ϕ0(r)+ δϕ(r), (2)

where

ϕ0(r) =∫

M dr

r2

√2m

(E − M2

2mr2 + αr

) , (3)

δϕ(r) =∫

γ M dr

2r5

√2m

(E − M2

2mr2 + αr

)3. (4)

If we neglect the correction δϕ(r) in (2), we obviously get the equation of the orbit ina Coulomb field (see [1], § 15)

pr

= 1 + ecosϕ, (5)

where

p = M2

mα, e =

√1 + 2EM2

mα2 .

If we take into account the correction δϕ(r) in (2), we get, instead of (5),

pr

= 1 + ecos(ϕ − δϕ(r)). (6)

In the right-hand side of (6) we can expand in terms of δϕ(r) and use in δϕ(r) for rthe relation r = r0(ϕ) which follows from (5). The result is

pr

= 1 + ecosϕ + eδϕ(r0(ϕ))sinϕ. (7)



Integrating (4), we find1

δϕ(r0(ϕ)) =

= m2αγ

M4

{−3ϕ − 2e2 + 1

esinϕ − 1 + ecosϕ

e2 sinϕ[2e + (e2 + 1)cosϕ]

}. (8)

Substituting (8) into (7) gives the result which is accurate up to and including first-

order terms in ζ = m2αγ

M4 :

pr

= 1 + ecos[(1 + 3ζ )ϕ] − ζ(2e2+1)sin2 ϕ−

− ζ1 + ecosϕ

e[2e + (e2 + 1)cosϕ] (9a)

or

p′

r= 1 + e′ cosλϕ + f cos2ϕ,

λ = 1 + 3ζ ,

p′ = p[1 + 3

2 ζ(e2 + 2)]

,

e′ = e(

1 + ζ3e4 − 2

2e2

),

f = 12 ζ e2.

(9b)

1 We can rewrite (4) in the form

δϕ = ∂

∂M

∫ √2mγ dr

2r3√

E − M2

2mr2 + αr

,

and use (5) to change the integration into one over ϕ:

δϕ = ∂

∂Mm2γα

M3

∫(1 + ecosϕ)dϕ =

= −3m2γα

M4 (ϕ + esinϕ) + m2γα

M3∂e∂M

sinϕ + m2γα

M3∂ϕ

∂M(1 + ecosϕ). (8′)

From (5) we find

2M

mα2r= ∂e

∂Mcosϕ − e

∂ϕ

∂Msinϕ, ∂e

∂M= e2 − 1

eM

and substituting this into (8′), we obtain (8).



In the vicinity ϕ = 0 and ϕ = π , the expression (2) can no longer be applied as δϕ

increases infinitely. Equation (9) for the orbit, however, valid also in those regions (cf.problem 1.8).

In the case of an infinite orbit (E � 0), (9) is the solution of the problem. If E < 0,(9) remains the equation for the orbit only during a few revolutions,1 namely, as longas 3ζϕ � 1. Taking only the secular part δϕ = −3ζϕ in (8) into account, we obtain theequation

pr

= 1 + ecosλϕ, λ = 1 + 3ζ , (10)

which describes the path over a long section of the orbit; that is, a “corrected” zeroapproximation (as distinct from (5); compare problem 2.21). It is obviously easy totransform (9) also in such a way that it describes the orbit over a long stretch with anaccuracy up to and including first-order terms, and we have

p′

r= 1 + e′ cosλϕ + f cos2λϕ. (11)

2.29. It is sufficient to prove that the Lagrangian function can be split into two partswhen expressed in terms of the centre of mass coordinates

R = m1r1 + m2r2

m1 + m2,

and the relative coordinates, r = r2 − r1:

L = L1(R, R)+ L2(r, r),

L1(R, R) = 12 (m1 + m2)R2 + (e1 + e2)ER,

L2(r, r) = 12 mr2 − e1e2

r− m

(e1

m1− e2

m2

)Er.

The function L1(R, R) determines the motion of the centre of mass which is the sameas the motion of a particle of mass m1 + m2 and charge e1 + e2 in uniform field E. Therelative motion, determined by L2(r, r), is the same as the motion of a particle withmass m = m1m2/(m1 + m2) (the reduce mass) in the Coulomb field as well as in uniformfield E.

One could, of course, have obtained the same result starting from the equations ofmotion of a particles.

2.30. The Lagrangian function

1 In particular, the radius vector r must be a periodic function of ϕ.



L = 12

(m1r2

1 + m2r22

)− e1e2

|r1 − r2| + e1

cA(r1)r1 + e2

cA(r2)r2

can be split into two parts depending only on R, R and r, r (using the notations ofproblem 2.29, c is the velocity of light), if e1/m1 = e2/m2. In this case we get:

L = 12 (m1 + m2)R2 + e1 + e2

cA(R)R + 1

2 m r2 − e1e2

r+ e1m2

2 + e2m21

c(m1 + m2)2 A(r) r.

2.31. T = 12

∑Nn=1 μnξ

2n, p = μN ξN , M = ∑N

n=1 μn[ξn, ξn], where

1μn

= 1∑nk=1 mk

+ 1mn+1

(n = 1, . . . , N − 1), μN =N∑

k=1

mk.

2.32. Let the coordinates of the incident particle and of the particle initially at rest be x1and x2, respectively, and let at t = 0, x1 = −R, x2 = 0. The centre of mass of the systemmoves accordingly to the equation X = −1

2R + 12vt. The relative motion (x = x2 − x1)

follows from the equation law

t = −√

m4

∫ x

R

dx√mv2

4 − αxn

.

The first particle thus has the following position:

x1 = X − 12 x = −1

2 R + 12

R∫x

dx√1 − 4α/

(mv2xn

) − 12 x.

The distance between the particles decreases until it becomes xmin =(

4α

mv2

)1/n

and then

increases again. When it is again equal to R, the first particle has come to the point

x1f = xmin

[∫ R/xmin

1

(1√

1 − z−n− 1

)dz − 1

]. (1)

The point where the first particle comes to rest is the limit of x1f , as R → ∞.If n � 1, then x1f → ∞, as R → ∞; that is, both particles go to infinity after the

collision.



2.34. The equation of motion (c is the velocity of light)

mv = egcr3 [v,r] (1)

has the integrals of motion

12 mv2 = E, (2)

m[r,v] − egc

rr

= J. (3)

Multiplying (1) by r, we obtain rv = 0 or d2

dt2r2

2 − v2 = 0, where

r2 = r20 + v2(t − t0)2. (4)

Multiplying (3) by r/r, we obtain

− egc

= J cosθ , (5)

where θ is the polar angle in the spherical coordinates with the z-axis parallel to thevector J. Projection (3) onto the z-axis

mr2ϕ sin2 θ − egc

cosθ = J

together with (5) gives

ϕ = Jmr2 . (6)

At t = t0, we have v = r0ϕ sinθ , besides, from (6) we get J = mr0vsinθ

, and the result is(taking into account (5))

tanθ = −mr0vceg

. (7)

Integrating the equation ϕ = r0vr2(t) sinθ

, we obtain

ϕ = ϕ0 + 1sinθ

arctanv(t − t0)

r0. (8)

The particle thus moves with a constant velocity v on the surface of the cone θ = const.By introducing

ψ = (ϕ − ϕ0)sinθ , (9)



we rewrite (8) as

tanψ = v(t − t0)

r0. (10)

Equations (4), (9), and (10) can be interpreted as follows: the motion of a point on thecone involute is uniform and straightforward, while r and ψ are the polar coordinates inthe plane of the involute.

2.35. The motion of a charged particle in an electromagnetic field is determined by theLagrangian function

L = 12 mv2 − eϕ + e

cvA, (1)

where ϕ and A are the scalar and vector potentials, and c is the velocity of light (see [7],§ 10 and [2], § 16–17).

Using cylindrical coordinates, we have

L = m2

(r2 + r2ϕ2 + z2)+ ec

μr2ϕ

(r2 + z2)3/2 . (2)

We see from the equation of motion for z

mz + 3ec

μr2ϕz(r2 + z2)5/2 = 0 (3)

that for z = 0 the component of the force along the z-axis vanishes. The orbit thus liesin the plane z = 0 when z(0) = z(0) = 0.

As ϕ is a cyclic coordinate, we have

∂L∂ϕ

= mr2ϕ + eμcr

= pϕ = const. (4)

From this equation it follows that the projection of the momentum Mz = m[r,v]z = mr2ϕ

is not saved, but the sum of Mz + eμcr = pϕ is retained. For the case of the infinite motion,

pϕ is the value of Mz at r → ∞. Moreover, the energy conservation law is satisfied (since∂L/∂t = 0):

12m(r2 + r2ϕ2) = E. (5)

Using (4) to eliminate ϕ from (5) we get

12mr2 + Ueff(r) = E, (6)



where

Ueff(r) = 12mr2

(pϕ − eμ

cr

)2. (7)

The radial motion thus takes place as in one-dimensional field Ueff(r).We have drawn Ueff(r) given by (7) for the cases pϕ < 0 and for the case pϕ > 0 in

Figs. 105a and 105b, respectively.

E

(a) (b)

E2

E1

r1 r2

Um

ra b

r

UeffUeff

Figure 105

In the case when pϕ < 0 the orbit is always infinite. To give a qualitative descriptionof the orbit, it is useful to use (4) to write

ϕ = −|pϕ |mr2 − eμ

mcr3 . (8)

The velocity with which the radius vector of the particle turns round has the samedirection all the time and increases when the particle approaches the dipole. Curve 1in Fig. 106 shows such an orbit. The orbit is symmetric with respect to straight lineconnecting the centre of the field with the point r = rmin.

When pϕ > 0, scattering can occur for any energy E, but if

E < Um = c2p4ϕ

32me2μ2

(see the line E = E1 in Fig. 105b), there is also the possibility of finite orbit. From theequation

ϕ = pϕ

mr2 − eμmcr3

it follows that ϕ > 0 when r > r1 = eμcpϕ

and ϕ < 0 when r < r1. At r = r1 there is a“sticking point” in ϕ.

A particle with an energy E > Um (see the line E = E2 in Fig. 93b) is scattered whilein two points (where r = r1) its velocity is parallel to its radius vector (curve 2 in Fig. 106).



O 1

2

3

Oa

AB C

b

r1


At E < Um there can occur scattering without sticking points for ϕ (curve 3 ofFig. 106) or a finite orbit in an annulus a � r � b (Fig. 107). In the latter case the particlecan perform both a direct (section AB) and “counter” (section BC) motion as far as ϕ

is concerned.

2.36. a) It is convenient to use cylindrical coordinates and to take the vector potentialin the form

Aϕ = 12 rB, Az = Ar = 0.

In the z-direction the motion is uniform, while in a plane perpendicular to the z-axiswe have a finite motion. In Fig. 108 we show the projection of the orbit on this plane.The orbits a, b, and c correspond, respectively, to the cases1 pϕ > 0 and energy valuesE⊥ = E − 1

2 mv2z in the regions

U1 < E⊥ < U2, E⊥ = U2, E⊥ > U2,

where

U1 = (� − �)pϕ , U2 = λpϕ

2�, � = eB

2mc, � =

√�2 + λ.

An orbit for the case when pϕ < 0 is given in Fig. 108d, while Fig. 108e depicts theorbit for the case where pϕ = 0.

The orbit for the particle in the xy-plane can easily be determined once we knowthe motion of a free isotropic oscillator of frequency � (see [1], § 23, problem 3 and[7], § 4)2

r2 = a2 cos2 �t + b2 sin2 �t, ϕ = −�t + arctan(

ba

tan�t)

.

1 To fix our ideas we assume B > 0; pϕ is the generalized momentum corresponding to the coordinate ϕ.2 The branches of arctan

(ba tan�t

)should be chosen such that the angle ϕ is a continuous function of t.



Here the minimum (b) and maximum (a) radii are determined by the energy E =12 m�2 (a2 + b2)− pϕ� and the angular momentum pϕ = m�ab; the origins for t and ϕ

are chosen that ϕ(0) = 0, r(0) = a.Note that the period of radial oscillations, T = π/� is independent of E and pϕ . The

angle over which the radius vector turns during this period is

�ϕ = π(±1 − �/�) for pϕ ≷ 0,

where the sign is the same as that for pϕ , and

�ϕ = −π�/� for pϕ = 0.

The angle �ϕ does not depend on E.1

What will be the motion when λ < 0?It is interesting to compare the motion of the particle in this problem with the motion

of the charged particle in crossed electric and magnetic fields (see [2], § 22).


ϕ =√

m2

∫ pϕ

mr2 − �

√E − Ueff(r)

dr, (1)

where � = eB2mc , c is the velocity of light and

Ueff(r) = −α

r+ 1

2 mr2(�− pϕ

mr2

)2.

By looking at a graphs of Ueff(r) one can study the character of the motion qualitatively.One must then pay attention to the fact that ϕ changes sign when r passes through the

value r0 =√

pϕ

m�. The result is orbits such as the ones in Fig. 108a–e.2 The different

orbits occur when

a) pϕ > 0, Umin < E < U0, where Umin is the minimum value of Ueff(r) and U0 =Ueff(r0);

b) pϕ > 0, E = U0;

c) pϕ > 0, E > U0;

d) pϕ < 0;

e) pϕ = 0.

1 Another way to solve this problem is given in problem 6.37.2 As we are only interested in a qualitative study, using the shape of Ueff(r), we can use the same approximate

treatment as in problem 2.36. Of course, the exact form of the orbits is different in the two problems.



(a)

(d) (e)

(b) (c)

Δϕ

Figure 108

In the latter case, the particle falls into the centre at the first loop.Let us consider in somewhat more detail two limiting cases.Equation (1) can be written in the form

ϕ = pϕ√2m

∫dr

r2

√E + pϕ�+ α

r − p2ϕ

2mr2 − 12 m�2r2

− �t. (2)

We must thus describe the effect of the magnetic field as a change in the energy E′ =E + pϕ� and an extra term in the field U(r) = −α

r which is δU(r) = 12 m�2r2, each of

which leads to a precession of the orbit and an additional precession with an angularvelocity −�. If the magnetic field B is sufficiently small, the term δU(r) can be consideredto be a small correction to

U0(r) = p2ϕ

2mr2 − α

R,

provided the following condition is satisfied,

δU(r) � |U0(r)|, (3)

for the whole range of the motion.The precession velocity caused by δU can be found from

�′ = δ�ϕ

T= 1

T∂

∂pϕ(T〈δU〉), (4)

where the averaging of δU is over the motion of the particle in field U0 with an energy E′and an angular momentum pϕ , while T is the period of the motion (cf. problems 2.20



and 2.21). If we perform the calculation,1 we get

�′ = −3�2pϕ

4|E′| ; (5)

we can assume δU to be a small correction if, apart from (3), the condition δ�ϕ � 2π

is also satisfied, or

�2pϕα√

m|E′|−5/2 � 1. (6)

It is, of course, impossible to consider δU to be a small correction when E′ � 0, as in thatcase neglecting δU may introduce a qualitatively change the character of the motion.

The quantity �′ may turn out to be either small or large compared to �. The sign �′is the opposite to that of pϕ ; that is, the direction of this velocity is the opposite of that ofthe motion of the particle in its orbit. The direction of the velocity � is determined bythe magnetic field.

The orbit is thus an ellipse precessing with the angular velocity

�prec = −�+ �′. (7)

To be more exact, the orbit is a fixed ellipse if the system of reference rotates with anangular velocity �prec, because it is possible that �T � 1.

It is interesting to compare the results with the Larmor theorem (see [7], § 17.3, [2],§ 45, and problem 9.26).

Is there a case when although E is positive we can consider the field δU(r) to be asmall correction?

Let us now consider the case when the field U(r) = −α/r can be considered to be asmall correction. If U(r) were not present, the motion would be along a circle. Its radius aand the distance b from of its centre from the centre of the field can be express in termsof the maximum and minimum distances of the particle from the centre of the field

r21,2 = E + pϕ�± √

(E + 2pϕ�)E

m�2 . (8)

There are now two possibilities (see Fig. 109), depending on how the circle is placedwith respect to the origin. If pϕ� < 0, case 109a occurs, and if pϕ� > 0, case 109b occurs.In both cases, we have

b2 = E + 2pϕ�

2m�2 , a2 = E2m�2 . (9)

1 To evaluate 〈δU〉, it is convenient to use the variables used in problems 2.21 and 2.22; since the periodin the field U0 is independent of pϕ , we can take T from under the differentiation sign in (4).



Taking U(r) into account leads to a systematic displacement of this circle (the so-called drift), while its radius and the distance from the origin of the fields which aredetermined by a and b do not change; that is, its centre moves along a circle of radius b.The angular velocity of the displacement of the centre of the circle is

γ = ∂

∂pϕ〈U〉,

where the averaging of U(r) is over the uniform motion along the circle. Let us restrictourselves to the case when a � b. In that case, we may simple assume that

〈U(r)〉 = −α

b, (10)

so that

γ = α

2m�b3 .

We note that in this case the linear drift velocity is equal to c|E|/B, where e|E| = α/b2 isthe force acting upon the particle at a distance b (cf. [2], § 22).

2.38. The problem of the motion of two identical charged particles in a uniformmagnetic field can be reduced to the problem on the motion of the centre of mass andthe problem of the relative motion (see problem 2.30).

For the centre of mass coordinates, we have

X = Rcosωt, Y = −Rsinωt, (1)

where ω = eB/(mc).The relative motion is the same as that of a motion of a particle of mass m/2

and charge e/2 in the field U(r) = e2/r and in a uniform magnetic field B. This motion issimilar to the one considered in the preceding problem (we need only change m to m/2,e to e/2 and α to −e2 in formulae). Let us restrict ourselves to the case where the radius aof the orbit is small compared to the distance b from the centre of the fields (Fig. 109b).We can easily find the frequency of the radial oscillations to a higher degree of accuracythan we was done in preceding problem, by expanding

Ueff(r) = e2

r+ p2

ϕ

mr2 + 1162mω2r2 − 1

2 pϕω

in a series in the vicinity of the minimum where r = b (see [1], § 21). From the condition



A A

OO

b

(a) (b)

b

a aθ θ

Figure 109

U ′eff(b) = 0, we find

pϕ =√

m2

16ω2b4 − m

2e2b ≈ m

2b2

(ω

2− γ

), γ = 2e2

mωb3 , (2)

and we thus get finally for ωr =√

2U ′′eff(b)m the result ωr = ω − 1

2 γ , and for the separationof the particles

r = b + acos(ωr t + α). (3)

To find ϕ(t), we apply the principle of conservation of generalized momentum to pϕ =12 mr2

(ϕ + 1

2ω)

. Using (2) and (3), we get

ϕ(t) = −γ t − ab

sin(ωr t + α)+ ϕ0. (4)

Using (3) and (4), we get for the relative coordinates

x = r cosϕ = bcos(γ t − ϕ0)+ acos(ωt + 1

2 γ t + β)

,

y = r sinϕ = −bsin(γ t − ϕ0)− asin(ωt + 1

2 γ t + β)

,(5)

where β = α − ϕ0. The first terms here correspond to the motion of the centre of thecircle with a drift velocity bγ , and the second terms to the motion along this circle with anangular velocity ω + 1

2γ .

2

1




Coordinates of the particles, x1,2 = X ± 12 x, y1,2 = Y ± 1

2 y can be written in the form

x1,2 = ±12 bcos(γ t − ϕ0)+ ρ1,2 cos(ωt + ψ1,2),

y1,2 = ∓12 bsin(γ t − ϕ0)− ρ1,2 sin(ωt + ψ1,2),

(6)

where

ρ1,2 =√

R2 + 14 a2 ± aRcos

(12 γ t + β

),

tanψ1,2 =±asin

(12 γ t + β

)

2R ± acos(

12 γ t + β

) .

The centres of the circles along which the particles move thus rotate around theorigin with an angular velocity γ (a drift velocity bγ /2) while their radii oscillate witha frequency γ /2 (Fig. 110).

Another limiting case when a � b (distance of the centre of the orbits to the originsmall compared to the radii of the orbits (Fig. 111)) may give a clear insight intothe mechanism of energy “exchange”. The work carried out by the the force of theinteraction on the second particle is clearly positive, while the work carried on the firstparticle is negative, when taken over many periods.

2.39. One can easily prove that the given quantity is a constant by using the equationsof motion and writing this quantity in the form

AF + 12 [F,r]2,

where

A = [v,M] − αrr

is the Laplace vector (see [1], § 15 and [7], § 14). For small values of F the orbit willbe close to an ellipse, with its semi-major axis along the direction of the vector A andwith an eccentricity e = |A|/α. In this case AF ≈ const, or ecosψ ≈ const, where ψ isthe angle between A and F.

2.40. When there is a small extra term δU(r) in the potential energy, the quantitiescharacterizing the motion of the particle, such as the angular momentum, pericentrumposition and so on, change, although they do not change their values appreciable over ashot time interval (a few periods of the unperturbed motion). However, these changesmay add up over an extended time, so that some of the quantities may happen to changeby large amounts.



In particular, the orbit remains elliptical for a shot time interval. Its semi-major axis,a = α

2|E| , which is determined by the energy, does not change over a long time, while

the eccentricity e =√

1 − M2

mαa and the orientation of the orbit are both liable to secular

changes.

xA

F<r>

y

Oψ

Figure 112

a) The change in the angular momentum is determined bythe equation

M = [r,F]. (1)

Average this equation over one period, we obtain

〈M〉 = [〈r〉,F], (2)

where

〈r〉 = 1T

∫ T

0r(t)dt. (3)

For averaging, we use a coordinate system with the z-axis parallel to M and the x-axisparallel to A (Fig. 112). Here

A = [v,M] − αrr

is an additional integral of motion in the Kepler problem. Recall that the vector A isdirected from the centre of the field to the pericentre, and |A| = αe. Clearly, the vector−〈r〉 is parallel to the x-axis.

Making the substitution

x = a(cosξ − e), t = T2π

(ξ − esinξ),

we get

〈x〉 = a2π

2π∫0

(cosξ − e)(1 − cosξ)dξ = −3ae2

. (4)

Therefore

〈r〉 = −3ae2

A|A| = − 3a

2αA. (5)



b) If the force F is at right angles to M, it is obvious from symmetry considerationthat the orbit lies in a plane and the vector M retains its direction—apart possibly fromthe sign.

Omitting the averaging sign, we can write (2) and (5) in the form

M = 32 aeF sinψ , (6)

where ψ is the angle between A and F. Using the fact that

ecosψ = e0 = const (7)

(see problem 2.39), and eliminating e and ψ from (6) we find

M = 32 aF

√1 − e2

0 − M2

maα. (8)

Integrating (8), we get

M = M0 cos(�t + β), (9)

where

� = 3F2

√a

mα, M0 =

√maα(1 − e2

0),

as well as

e =√

1 − (1 − e20)cos2(�t + β).

F

Figure 113

The orbit is thus an ellipse which oscillates about the directionof F and an eccentricity which changes in tune with the oscillations(Fig. 113). The direction of motion along the ellipse also changes,together with the sign of M. The period 2π/� of the oscillationof the ellipse is much longer than the period T of the rotation ofthe particle along the ellipse.

c) In the general case, we consider also the change in thevector A. Using the equations of motion, we easily get

A = 1m

[F,M] + [v, [r,F]]. (10)



For the averaging in (10), we use the equations

〈xx〉 =⟨

ddt

x2

2

⟩= 0,

〈yy〉 =⟨

ddt

y2

2

⟩= 0,

〈xy〉 + 〈yx〉 =⟨

ddt

xy⟩= 0,

〈xy〉 − 〈yx〉 = Mm

.

(11)

As a result, we get

〈A〉 = 32m

[F,M]. (12)

We have thus for M and A, averaged over one period (in the following we omit theaveraging sign), the following set of equations

A = 32m

[F,M],

M = 3a2α

[F,A].(13)

The components of these vectors along the direction of F are conserved:

MF = const, AF = const. (14)

(The same result could also easily be obtained from other considerations). For thetransverse component M,

M⊥ = M − F(MF)

F2 , (15)

we obtain from (13)

M⊥ + �2M⊥ = 0. (16)

In a coordinate system OX1X2X3 with the X3-axis parallel to F we have for thesolution of (13):

M1 = B1 cos�t + C1 sin�t,

M2 = B2 cos�t + C2 sin�t.(17)



We then obtain from (13):

A1 = − 3F2m�

(B2 sin�t − C2 cos�t),

A2 = 3F2m�

(B1 sin�t − C1 cos�t).(18)

As we should expect, the constants B1,2 and C1,2 are determined by the initial values ofthe vectors M and A.

X3

A

M Fρ

σ

μ

Figure 114

The end point of the vector M describes an ellipse with itscentre on the X3-axis and lying in a plane μ which is parallelto the OX1X2-plane (Fig. 114). The end of the vector A alsodescribes an ellipse with centre on the X3-axis and lying in aplane σ which is parallel to μ; this second ellipse is similar tothe first one, but rotated over an angle π/2: A remains thusat right angles to M all the time. The plane of the orbit isperpendicular to M, and the vector A determines the directionto the pericentre of the orbit.

The plane of the orbit thus rotates (precesses) around F.The angle between the plane of the orbit ρ and F oscillatesabout its mean value. The eccentricity and the angle betweenthe projection of F upon the plane ρ and the direction tothe pericentre also oscillate about their mean values. All these motions occur with afrequency �.

We should bear in mind that we have neglected those corrections in F which are ofthe first order but which do not lead to secular effects. Our solution is valid for a timeinterval of the order of several periods of the orbital precession.

The reader should checked whether the next approximations will lead to a qualitativechange in the character of the motion (e.g., to a possible departure of the particle toinfinity). The exact solution of the problem of the motion of a particle in the fieldU = −α

r − Fr, which can be given in parabolic coordinates (see problem 12.12b), showsthat such effects do not take place if E < 0 and if F is sufficiently small.

It should be emphasized that the appearance of secular changes of the orbit underthe action of infinitesimal constant perturbation is connected with a degeneracy of theunperturbed motion.

In [8], § 7.3 one can find a solution of this problem using the canonical perturbationtheory.

2.41. According to the Larmore’s theorem (see [7], § 17.3 and [2], § 45), the orbit of aparticle in a uniform magnetic field B rotates around the centre of the Coulomb field with

the angular velocity � = − qB2mc , where q and m are the charge and mass of a particle and

c is the velocity of light. In this case, the angular momentum M and the Laplace vector Aratate with the same angular velocity:



M1 = [�,M], A1 = [�,A]. (1)

In the preceding problem (see (13)), the averaged velocity change of the vectors Mand A under the influence of the constant force F = qE has been determined:

M2 = 3a2α

[F,A], A2 = 32m

[F,M]. (2)

The averaged velocity change of the vectors M and A under the influence of bothfields is the following sum:

M = M1 + M2, A = A1 + A2. (3)

a) Let us direct the x-axis along the electric field and the z-axis along the magneticone. Then the equations (3) take the form

Ax = −�Ay, Ay = �Ax − 3F2m

M, M = 3aF2α

Ay.

The solution of this system is

Ax = �

ωC cos(ωt + β)+ C1,

Ay = C sin(ωt + β),

M = −3aF2αω

C cos(ωt + β)+ 2m�

3FC1,

where ω = √�2 + 9aF2/(4mα) and constants C, β, and C1 are determined by the initial

values of A and M.Thus, the end of the vector A moves along an ellipse with the axes which are parallel to

the axes x and y (Fig. 115) and with the centre on the x-axis. The orbit wiggles (or rotateswith �C > ωC1), and its eccentricity changes periodically. When 9aF2C > 4mα�ωC1,the elliptical orbit periodically turns into the straight-line segment.

O x

y

A

Figure 115



b) Let us denote

�F = 32

F√

amα

, N = A√

maα

,

then we can write (3) as

M = [�,M] + [�F ,N], N = [�,N] + [�F ,M].

Adding and subtracting these equations, we find

J1,2 = [ω1,2 ,J1,2],

where

J1,2 = 12 (M ± N), ω1,2 = �± �F .

Thus, the vectors Ji, i = 1, 2 rotate with constant angular velocities ωi

Ji(t) = Ji(0)cosωi t +[ωi

ωi,Ji(0)

]sinωi t + ωi

Ji(0)ωi

ω2i

(1 − cosωi t),

and vectors

M = J1 + J2, A =√

α

ma(J1 − J2)

are completely determined by their initial values.We will not analyse this answer in detail, only note that ω1 = ω2 when the electric

and magnetic fields are mutually orthogonal.

2.42. Let the xz-plane be where the particle moves. The equations for the Laplacevector A (see (10) in problem 2.40 where one must substitute F = −∂δU(r)/∂r) can bepresented in the form

Ax = 2β(5xzz − 2xz2) = −6β

mzM + 2β

ddt

(xz2),

Az = −2β(4xz − zx2) = −4β

mxM − 2β

ddt

(x2z),

where M = m(zx − xz) is a slowly changing function of time.



After averaging these equations, we obtain

Ax = −6βMm

〈z〉 = 9βamα

AzM,

Az = −4βMm

〈x〉 = 6βamα

AxM,

where a is a large semi-axis of an ellipse. Hence

Ax

Az= 3Az

2Ax;

that is, in contrast to problem 2.40b, the end of the vector A varies not along a straightline but along a hyperbola

A2x = 3

2 A2z + const.

The time dependence of A can be found from the equation

t = mα

9βa

∫dAx

MAz,

in which M and Az must be expressed as the functions of Ax.For example, for the case when at the initial moment Az(0) = 0, Ax(0) = αe0 (here e0

is the initial eccentricity value), we have

Az = α

√23 (u2 − e2

0), M =√

53 mαa(c2 − u2),

where

u = Ax

α, c =

√3 + 2e2

0

5.

Over time, the orbit slowly rotates in the xz-plane and turns into a segment whose anglewith the x-axis is

ψ = −arctan(√

2/3k)

, k =√

3 − 3e20

3 + 2e20

,



during the time

τ = 13β

√mα

10a3

c∫e0

du√(u2 − e2

0)(c2 − u2)

.

The integral in the previous expression is reducible to the complete elliptic integral ofthe first kind (see footnote in the solution of problem 1.7)

τ = Tα

6π√

10cβa3K(k),

where T is the period of unperturbed motion.Let us use our model to estimate the lifetime of a satellite whose initial orbit has a

small eccentricity e0 ∼ 0.1 and is close to a circle of radius a ∼ 40000 km. In this case,the period of the satellite T is about one day, and the time of the fall on the Earth (i.e. thetime at which its minimum distance becomes equal to the radius of the Earth) is of theorder of τ . The estimate is valid for the relation: α/β ∼ (m1/m2)R3, where m1 and m2 arethe respective masses of the Earth and the Moon, and R is the distance from the Earthto the Moon. From here we get the estimate

τ ∼ T22

T, T2 = 2π√

Gm1/R3,

where T2 is the period of the Moon’s rotation and G is the gravitational constant. Thus,the time during which such a satellite will fall to the Earth, is about a few years. However,the correct fall time is much less due to the presence of the air resistance even at highaltitudes.

2.43. Here we will use the geocentric coordinate system. Let’s assume the strait linebetween the Earth and the Sun as the z-axis, and let the x-axis be in the plane of theEarth’s orbit. This coordinate system rotates around the y-axis with an angular velocity�. In this frame of reference, δU is evidently time-independent, so that the integral ofmotion is the energy

E = 12 mv2 − α

r+ δU − 1

2 m�2r2,

where α = Gm1m, G is the gravitational constant and m1 is the Earth’s mass. The force

δF = −∂δU∂r



and the force of inertia

Fi = m�2r + 2m[v,�]

leads to the distortion of the elliptical orbit of the Moon, while the major semi-axis of theellipse a = α/(2|E|) stays almost unchanged.

The rate of change of the Laplace vector A consists of two terms: A1 and A2 thatcorrespond to δF and Fi (cf. problem 2.41). The term A1 has already been foundin problem 2.41a:

A1x = 92 �ζAz, A1z = 3�ζAx,

where ζ = M�a/α. Since the eccentricity of the Moon’s orbit is small e = 0.055, we have

M ≈ ma2ω, α ≈ mω2a3, ζ ≈ �

ωω ≈ 29.5

365≈ 1

12,

where ω is the angular velocity of the Moon around the Earth in the given (rotating)coordinate system.

The force of inertia leads to rotation of the vector A with an angular velocity −� (notethat the orbit would be fixed in 0x0yz0 frame of reference in the absence of the term δU):

A2x = −�Az, A2z = �Ax.

Thus1

Ax = −�(1 − 9

2 ζ)

Az, Az = �(1 + 3ζ )Ax. (1)

Integration of (1) gives

Ax = Bcos(�′t + φ), Az = B(1 + 15

4 ζ)

sin(�′t + φ), (2)

where �′ = �(1 − 34 ζ ), B, and φ are constants. In the 0xyz-frame of reference, the vector

A rotates around the y-axis with an average angular velocity −�′. In the system 0x0yz0,it rotates with the angular velocity (so-called precession)

�prec = �− �′ = 34 ζ�.

1 Note that according to (1)AA = AxAx + AzAz = 7.5�ζAxAz.

Using the relations A = αe, e2 = 1 − M2/(maα), we can estimate ζ = �aM/α ∼ 7.5e2ζ2� � �. The value ofζ in (1) can be considered constant.



The small changes of |A| defined by (2) correspond to the small ripples of the orbit’seccentricity.

2.44. The Lagrangian function of the system (q is the particle charge and c is the velocityof light)

L = 12 mv2 + α

r+ q

c[μ,r]

r3 v

is the same—apart from the notations—as the one considered in [2] (problem 2 to§ 105).

The equations of motion for the angular momentum M and the Laplace vector A arethe following

M = qmcr3 [M,μ],

A = qmcr3 [A,μ] + 3q(M,μ)

m2cr5 [M,r].

When averaging over the period T of unperturbed motion, these equations describe thesystematic change of vectors M and A:

〈M〉 = [�′,M], �′ = − qcma3(1 − e2)3/2 μ, (1)

〈A〉 = [�,A], � = �′ − 3M(�′M)

M2 , (2)

where a and e are again the semi-major axis and the eccentricity of the unperturbedellipse. Equation (1) can also be rewritten as

〈M〉 = [�,M], (3)

since the vector �− �′ is parallel to M.From (2) and (3), it is seen that the ellipse, along which the particle moves, precesses

“as a whole” with an angular frequency �. Other interpretation can also be givenbased on (1) and (2): in the coordinate system, rotating with the frequency �′, thevector M as well as the plane of motion of a particle are stationary, while the vector Aand the perihelion of the orbit revolve with the constant frequency � − �′ around thedirection M.

We also indicate that it is helpful to perform the averaging of 1/r3 and r/r5 usingthe angle ϕ instead of the variable t:

⟨1r3

⟩= 1

T

T∫0

dtr3(t)

= mTM

2π∫0

dϕ

r(ϕ)= m

TMp

2π∫0

(1 + ecosϕ)dϕ = 2πmTMp

,



⟨ rr5

⟩= βA, β = 1

A2T

T∫0

Arr5 dt = m

ATM

2π∫0

cosϕ

r2(ϕ)dϕ = 2πme

TMp2A.

2.45. When averaging the equation for the Laplace vector (see (10) of problem 2.40)

A = 1m

[F,M] + [v, [r,F]]

we take into account that 〈F〉 = 0 and that, according to the unperturbed equations ofmotion,

mv = ddt

mv = ddt

(−α

rr3

)= −αv

r3 + 3αr(rv)

R5 .

It gives (see the preceding problem)

〈A〉 = −αβ

m2

⟨[v,M]

r3

⟩= αβ

m2

⟨Ar3 + αr

r4

⟩= − 3αβ

2m2a3(1 − e2)3/2 A.

The rate of change of the vector A is directed in the direction opposite to the vector Aitself. Vector A is directed to pericentre of the orbit and its magnitude is A = αe. Thus,the additional force does not cause precession of the orbit, but leads to the reduction ofthe eccentricity.

It can also be shown (see [2], § 75, problem 1) that a particle will fall on the centreover a finite time due to the energy loss.


§3

Scattering in a given field. Collisionbetween particles

z

θ

Figure 116

3.1. a) Fig. 116 shows that the angle of deflection of a particle θ

is twice the angle of the slope of the tangent to the surface ofrevolution at the point of collisions. We have therefore

tan(θ/2) = dρ

dz= b

acos

za

.

Hence we have

ρ2 = b2 − a2tan2(θ/2)

and thus

dσ = π |dρ2| = πa2tan(θ/2)dθ

cos2(θ/2)= a2 d�

4cos4 θ2

.

The possible deflection of the particle lie within the range of angles from zero, as ρ → b,to θm = 2arctan(b/a), as ρ → 0.

As a result,

dσ

d�=

⎧⎪⎨⎪⎩

a2

4cos4(θ/2), when 0 < θ < θm,

0, when θm < θ .

b) dσd�

= A2/(1−n)

2(1 − n)sinθ cos2(θ/2)[ncot (θ/2)](1+n)/(1−n).




c)dσ

d�=

⎧⎪⎨⎪⎩

√2a

(b − a

√tg(θ/2)

)4sin3/2 θ

√cos(θ/2)

, when 0 < θ < θm

0, when θ > θm

where θm = 2arctan(b2/a2

).

3.2. It is the paraboloid of revolution ρ2 = αz/E. The reader should check whetherthe trajectories of the particles scattered by the field U(r) = −α/r and by the paraboloidapproach one another as r → ∞.

3.3. At E > V

dσ

d�=

⎧⎪⎪⎨⎪⎪⎩

a2n2

4cos θ2

(ncos θ

2 − 1)(

n − cos θ2

)(1 + n2 − 2ncos θ

2

)2 + a2

4, when 0 < θ < θm,

0, when θm < θ < π ,

where

n = √1 − (V/E), θm = 2arccosn.

Unlike the case of scattering on a potential well (see [1], § 19, problem 2), here it isnecessary to take into account the possibility of the elastic reflection at impact parametersan < ρ < a.

3.4. a)

σ =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

π

(β

E− α2

4E2

), when E >

α2

4β,

0, when E <α2

4β.

How does the cross-section change when α changes sign?

b)

σ =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

π

(2

√γ

E− β

E

), when E >

β2

4γ,

0, when E <β2

4γ.

3.5. a) We first consider the motion of a particle in the field U(r) = −α/rn. Thebehaviour of

Ueff(r) = Eρ2

r2 − α

rn



for different values of the impact parameters ρ is shown in Fig. 117.

E

Ueff

r0

12 3

rminr

Figure 117

For large values of ρ (curve 1), the particles are scattered, and the minimum distancefrom the centre of the field rmin(ρ), which is determined by the equation Ueff(rmin) = E,decreases with the decreasing ρ up to the value r0, achieved when ρ = ρ0 (curve 2). Ateven smaller ρ, the particle falls into the centre (curve 3).

The values r0 and ρ0 are determined by the following equations

Ueff(r0) = E, U ′eff(r0) = 0

and equal to

r0 =[(n − 2)α

2E

]1/n

, ρ0 =√

nn − 2

[(n − 2)α

2E

]1/n

.

If R > r0, then all particles for which rmin � R will hit the sphere and the cross-sectionwill be

σ = πρ2(rmin = R) = πR2(1 + α

ERn

).

If R < r0, only those particles which fall into the centre of the field will hit the sphere,and in that case

σ = πρ20 = πn

n − 2

[(n − 2)α

2E

]2/n

.

b) If γ > ER4 the cross-section is

σ = π

(2

√γ

E− β

E

)



provided 2√

γ E > β; if γ < ER4 then

σ = πR2(

1 + γ

ER4 − β

ER2

)

provided 2√

γ E > β or provided both 2√

γ E < β and R < R1 or R > R2, otherwiseσ = 0; here

R21,2 = β

2E∓

√β2

4E2 − γ

E.

3.6. Let us take the origin of the coordinate system in the centre of the planet direct thex-axis parallel to v∞ and the y-axis parallel to the impact parameter ρ. The motion ofmeteorites occurs in the field

U(r) = −α

r= −Gm0m

r,

where G is the gravitational constant. The orbit of the meteorite in the polar coordinates is

r(ϕ) = p1 + ecos(ϕ − ϕA)

, p = M2

mα= ρ2

ξR, e =

√1 +

(ρ

ξR

)2

, ξ = Gm0

Rv2∞,

where ϕA is the polar angle of point of the meteorites’ smallest distance from the origin,coinciding with the direction of the Laplace vector

A = [v,M] − αrr

= α e(

1,ρ

ξR

)= α e(cosϕA,sinϕA).

The polar coordinates of the meteorites falling on the planet satisfy the equationr(ϕ) = R at ϕA < |ϕ| < π . In this case, the impact parameter of such meteorites is lessthan the maximum ρmax as defined by the equation rmin = r(ϕa) = R, from which weobtain ρmax = R

√1 + 2ξ . The meteorite falling on the invisible part of the planet satisfies

the equation r(ϕ) = R at ϕA < |ϕ| < π/2 with impact parameters from the intervalρmin < ρ < ρmax. The corresponding impact parameter ρmin is determined by theequation r(ϕ = π/2) = R, from which we obtain the equation

ρ2min/(ξR)

1 + ρmin/(ξR)= R.

Hence

ρmin = 12 R

(1 + √

1 + 4ξ)

.



Thus, the fraction of particles falling on the invisible part of the planet is equal to

δ = πρ2max − πρ2

min

πρ2max

= 12

−√

1 + 4ξ

2(1 + 2ξ).

At high energy of the flying particles, ξ � 1, this fraction is small (ξ2). With decreasingthe energy of flying particles, the value of δ increases as well, reaching the value of δ = 1/2at ξ � 1.

3.7. a) dσd�

= R2(1 + λ)

4(

1 + λsin2 θ

2

)2 , where λ = 4RE(RE + α)

α2 .

How can one explain the result which one obtains when α + 2RE = 0?b) Let us introduce the Cartesian coordinates in the plane of the orbit of the particle

with the x-axis directed along the axis of the beam and the y-axis directed along theimpact parameter. The motion of the particle in the region r � R is a harmonic oscillationalong each of the coordinates, and in the initial moment of this oscillation (for t = 0)y0 = ρ, y0 = 0, x0 = −√

R2 − ρ2, x0 = v, thus

x = −√

R2 − ρ2 cosωt + vω

sinωt, y = ρ cosωt. (1)

The moment of release of the particle from action of the forces is determined by theequation

x2 + y2 = R2, (2)

and the angle θ between the particle velocity at this moment and the x-axis by theequation

sinθ = yv

= −ρω

vsinωt. (3)

Substituting (1) and (3) into (2), we obtain

ρ4 − ρ2R2(1 + λsin2 θ)+ 14 R4(1 + λ)2 sin2 θ = 0,

where λ = v2/(R2ω2). As a result,

ρ21,2 = R2

2(1 + λsin2 θ ∓

√cos2 θ − λ2 sin2 θ cos2 θ),



and the cross-section

dσ = π(|dρ21 | + |dρ2

2 |) = πd(ρ21 − ρ2

2) = R2(1 + λ2 cos2θ)d�

2√

1 − λ2 sin2 θ.

If λ > 1, the scattering is only possible at angles smaller than

θm = arcsinR2ω2

v2 ,

and at θ → θm, the differential cross-section dσ/d� increases indefinitely. Suchbehaviour of the cross-section is called rainbow scattering (see [12], chapter 5, § 5).A similar type of cross-section behaviour leads to formation rainbows in the scatteringof light by drops of water.

See also problems 3.9 and 3.11 as examples of the rainbow scattering.

3.8. a)dσ

d�=

⎧⎨⎩

a2

θ2m

, when θ ≤ θm = π V2E ,

0, when θ > θm.

b) dσ

d�=

{πa2

4θmθsin(πθ/θm), when θ ≤ θm = π V

2E ,

0, when θ > θm.

c)dσ

d�=

⎧⎪⎨⎪⎩

a2θ2m(θm − θ)

θ3(2θm − θ)2 , when θ � θm = πVE

,

0, when θ > θm.

d) When calculating the scattering angle (see [1], § 20 and [7], § 6.2)

θ =∞∫

−∞

Fydx2E

= 2VE

ρ√

R2 − ρ2

R2

we take into account that the force is

Fy =

⎧⎪⎨⎪⎩

2VyR2 = 2Vρ

R2 , when −√

R2 − ρ2 < x <

√R2 − ρ2,

0, when |x| >

√R2 − ρ2.



From here we find

ρ21,2 = 1

2R2(1 ∓

√1 − θ2/θ2

m),

where θm = V/E, so thus

dσ = π(|dρ21 | + |dρ2

2 |) = πd(ρ21 − ρ2

2).

Finally,

dσ

d�=

⎧⎪⎨⎪⎩

R2

2θm√

θ2m − θ2

, when 0 < θ < θm,

0, when θ > θm.

Compare this result with the result of problem 3.7b, which was obtained for the field thatis different from the given by a sign.

θm

ρ12 ρ2

2 ρ32 ρ2

θ

dσ

θmθ

dΩ


3.9. One can easily evaluate the angle of deflection of a particle, using a general formula,1

and we have

θ =∣∣∣∣ 3πβ

4Eρ4 − πα

2Eρ2

∣∣∣∣ . (1)

The function θ(ρ2) is shown in Fig. 118. From (1) we find

ρ21 = πα

4Eθ

(√1 + θ

θm− 1

), ρ2

2,3 = πα

4Eθ

(1 ∓

√1 − θ

θm

),

where θm = πα2

12Eβ.

1 It is simplest to use both terms of (1) from [1], § 20, problem 2.



For the cross-section, we have

dσ = π(|dρ21 | + |dρ2

2 | + |dρ23 |) = πd(−ρ2

1 + ρ22 − ρ2

3) == πα

8Eθ3

(1 + θ/(2θm)√

1 + θ/θm+ 2 − θ/θm√

1 − θ/θm− 1

)d�. (2)

This result is valid provided each of the terms in (1) is much less than unity. An estimateshows that the condition θ � 1 is sufficient for this. Equation (2) is obtained when θ < θm.If θm � 1 and θm < θ � 1, we have for the cross-section

dσ = π |dρ21 | = πα

8Eθ3

(1 + θ/(2θm)√

1 + θ/θm− 1

)d�.

Fig. 119 shows how dσ/d� depends on θ . The differential cross-section becomesinfinite both as θ →0 and as θ →θm. The total cross-section for scattering into a rangeof angles adjoining θ = 0 is infinite as the small-angle scattering corresponds to largeimpact parameters.

The total cross-section for scattering of particles into a range of anglesθm − δ < θ < θm,

θm∫θm−δ

2πdσ

d�θ dθ = π2αδ1/2

2Eθ3/2m

,

is finite and tends to zero as δ → 0.What is the relation between the number of scattered particles reaching a counter and

its size, if the counter is located at the angle θm?

3.10. The velocity of the particle after scattering is at an angle

θ = π − π√1 − a2/ρ2

, a2 = α

E. (1)

to its original direction. A counter registers particles scattered over an angle |θ | < π

together with those particles which have made several revolution around the scatteringcentre (Fig. 120a). The observed angle of deflection χ lies in the range 0 < χ < π andsatisfies the relation

−θ = 2π l ± χ , (2)



ρ(χ,1,–)

ρ(χ,1,+)

ρ(χ,1,+)ρ(χ,1,–) ρ (χ,0,+)

ρ(χ,0,+)

zχ

(a) (b)

a

θ,χ

θ

ρ

π

–π

–2π

–3π

–4π

χχ

Figure 120

where l = 0, 1, . . . corresponds to the upper sign in (2) and l = 1, 2, . . . to the lower signin (2) (Fig. 120b). From (1) and (2), we have

ρ2(χ , l, ±) = a2 + πa2

2

(1

2π l ± χ− 1

2π l + 2π ± χ

).

The cross-section is

dσ = π

∞∑l=0

|dρ2(χ , l, +)| + π

∞∑l=1

|dρ2(χ , l, −)|.

Using the fact that

dρ2(χ , l, +)

dχ< 0,

dρ2(χ , l, −)

dχ> 0,

we find that

dσ = πd[−

∞∑l=0

ρ2(χ , l, +)+∞∑

l=1

ρ2(χ , l, −)

]=

= 12 π2a2d

(1

2π − χ− 1

χ

)= πa2(2π2 − 2πχ + χ2)

2χ2(2π − χ)2 sinχd�.

The cross-section dσ/d� turns out to infinity as χ → π . This leads to the fact thatthe cross-section for scattering in a small finite solid angle

�� =∫ π

π−χ0

2π sinχ dχ = πχ20



is equal to

�σ = a2

2π

∫ π

π−χ0

d�

χ= a2χ0 = a2

√��

π.

The appearance of such features of the differential cross-section is because the angle ofdeviation equal to π is reached at the values of the impact parameter ρ(π , l, ±), which arenonzero. In a solid angle ��, proportional to the square of a small value �χ ≡ χ0, flyingparticles fall before scattering through sites 2πρ�ρ, areas which are proportional to thefirst degree �ρ ∝ χ0. This feature of scattering is called radiance (see [12], Ch. 5, § 5).

The same feature is present in scattering by an angle where χ = 0, but in this case itis masked by an infinite cross-section due to scattering of particles with the arbitrarilylarge impact parameters. Radiance in the forward scattering could occur, for example,under a constraint the diameter of the projectile to the centre of the beam of particles.

3.11. a) One sees easily that the condition E � V means that the angle over which theparticle is deflected during the scattering is small. The change in momentum is (see[7], § 6.2)

�p = − ∂

∂ρ

∞∫−∞

U(|ρ + vt|)dt = 2√

πVv

xe−x2,

where x = �ρ. The angle of deflection is

θ = �pp

= √π

VE

xe−x2. (1)

We cannot solve this equation for x in analytic form. However, from the graph of thefunction xe−x2

(Fig. 121), we see that (1) has two roots when θ < θm = VE

√π2e .

xe–x2

x1

Eθ

x2x

1√2

√

1√2e

V π

dσdΩ

θθm


Using (1) and the relation

dθ = √π

VE

(1 − 2x2)e−x2dx,



we can write the expression for the cross-section

dσ = π(|dρ21 | + |dρ2

2 |) = 2π

�2 (x1 dx1 − x2 dx2)

in the form

dσ

d�= 1

�2θ2

(x2

2

2x22 − 1

+ x21

1 − 2x21

).

When θ � θm, then x1 � 1 and x2 � 1, so that

dσ

d�= 1

2�2θ2 .

When θm − θ � θm, we can solve (1) by expanding xe−x2in a series near the maximum.

We then get

x1,2 = 1 ∓ √1 − θ/θm√

2,

dσ

d�= 1

2�2θ2m√

1 − θ/θm.

Fig. 122 shows dσ/d� as function of θ . The singularity at θ = θm is integrable(cf. problem 3.9).

Discuss whether the presence of the singularity in the cross-section at θ = θm isconnected with the approximations made in the solution of the problem?

b) dσd�

= 1�2θ2

(x1 + x2

11 − 2x1

+ x2 + x22

2x2 − 1

), where

x1,2

(1 + x1,2)3 =(

2EθπV

)2.

When θ � θm = π

3√

3VE we have

dσ

d�= πV

4�2Eθ3 .

When θm − θ � θm we have

dσ

d�=

√3

2√

2�2θ2m√

1 − θ/θm.

3.12. a) A particle with a velocity v before the collision will have a velocityv′ = v − 2n(nv) after the collision, where n is a unit vector which is normal to the surfaceof the ellipsoid. Using the relations1

1 We know from differential geometry that

n ∝ grad

(x2

a2 + y2

b2 + z2

c2 − 1

),

while the value of N is determined by the relation n2 = 1.



v = v(0, 0, 1), n = 1N

( xa2 ,

yb2 ,

zc2

),

we get

v′ = v(

− 2xzN2a2c2 , − 2yz

N2b2c2 , 1 − 2z2

N2c4

). (1)

Introducing spherical coordinates with the z-axis along v we can write

v′ = v(sinθ cosϕ, sinθ sinϕ, cosθ),

and from (1) we then have

tanϕ = a2yb2x

, cosθ = 1 − 2z2

N2c4 , sin2 θ =(

2zN2c2

)2 (x2

a4 + y2

b4

). (2)

For the cross section, we have

dσ = dxdy =∣∣∣∣ ∂(x, y)∂(θ , ϕ)

∣∣∣∣dθ dϕ,

where the dependence of x and y on θ and ϕ follows from (2) and from the equation forthe ellipsoid.

To evaluate the Jacobian, it is helpful to introduce an auxiliary variable u such that

x = a2ucosϕ, y = b2usinϕ.

From (1) and (2), it then follows that

sinθ = 2zuN2c2 , 1 − cosθ = 2z2

N2c4 , tan(θ/2) = zuc2

and from the equation of the ellipsoid, we find

u−2 = a2 cos2 ϕ + b2 sin2 ϕ + c2 tan2(θ/2).

Moreover, we have

∂(x, y)∂(θ , ϕ)

= ∂(x, y)∂(u, ϕ)

∂(u, ϕ)

∂(θ , ϕ)= a2b2

2∂u2

∂θ.



We thus get, finally,

dσ

d�= a2b2c2

4cos4 θ2

(a2 cos2 ϕ + b2 sin2 ϕ + c2tan2 θ

2

)2 .

What is the limit which we must take to obtain from this result the cross-section forscattering by a paraboloid?

b) dσd�

= a2b2c2

cos3 θ(a2 cos2 ϕ + b2 sin2 ϕ + c2tan2θ)2 .

c) dσd�

= a2b2c2 cosθ

sin4 θ(a2 cos2 ϕ + b2 sin2 ϕ + c2 cot2 θ

)2 .

3.13. a) The change in the scattering momentum is (see [7], § 6.2)

�p = − ∂

∂ρ

∞∫−∞

U(ρ + vt)dt = − ∂

∂ρ

π(aρ)

vρ. (1)

Let the z-axis be parallel to v and the y-axis be perpendicular to a. We then have

�px = −πax

v

ρ2y

ρ3 , �py = πax

vρxρy

ρ3 . (2)

The direction of the velocity after scattering can be characterized by two spherical angles

tanϕ = �py

�px, θ = �p

p. (3)

It is clear from (2) that scattering only occurs when

12 π < ϕ < 3

2 π .

From (2) and (3), we find

ρx = ±πax

2Esinϕ cosϕ

θ, ρy = ∓πax

2Ecos2 ϕ

θ. (4)

For the cross-section, we find

dσ =∑

dρx dρy =∑∣∣∣∣∣

∂(ρx, ρy)

∂(θ , ϕ)

∣∣∣∣∣ dθ dϕ =(πax

E

)2 cos2 ϕ

2θ4 d�. (5)



The summation in (5) addresses the two possible values of ρ (see (4)).

b) dσd�

= |a⊥|2Eθ3 , where a⊥ is the component of a perpendicular to v∞. The cross-

section turns out to be symmetrical with respect to v∞ (although the field is by no meanssymmetric with respect to this directions.)

3.14. The change in the angle of deflection of the particle is given by the equation (seeproblem 2.20)

δθ(ρ) = − 1E

∂

∂ρ

∞∫rmin

δU(r)dr√1 − ρ2

r2 − U(r)E

. (1)

We then find from the equation θ = θ0(ρ)+ δθ(ρ):

ρ = ρ0(θ)− δθ(ρ0(θ))dρ0(θ)

dθ

(cf. problem 1.8). The function θ0(ρ) and hence the function ρ0(θ) are the expressionsobtained when δU(r) = 0. We then get for the cross-section

dσ

dθ= π

∣∣∣∣dρ2

dθ

∣∣∣∣ = π

∣∣∣∣∣dρ2

0(θ)

d(θ)− d

dθ[2ρ0(θ)δθ(ρ0(θ)]

dρ0(θ)

dθ

∣∣∣∣∣ =

= dσ0

dθ− d

dθ

[δθ(ρ0(θ))

dσ0

dθ

].

a) δ dσdθ

= −πβE

ddθ

{π − θ − sinθ

sinθ

};

b) δ dσdθ

= −2πγα

ddθ

{32 (π − θ)[1 + tan2(θ/2)] − 3 tan(θ/2)− sinθ

};

c) δ dσdθ

= − 4γ

π√

βEddθ

{(π − θ)2√θ(2π − θ)

}.

3.15. The energy acquired by the particle,

ε = (p + �p)2

2m− p2

2m≈ v�p,

is to the first order determined solely by the change of the longitudinal component ofthe momentum. As we assume that the deflection of the particle is small, we can (afterdifferentiating) put r = ρ + v(t − τ) in the expression for the force



F = −∂U(r, t)∂r

,

acting upon the particle. Here ρ is the impact parameter and τ the time when the particleis at its smallest distance from the centre. Therefore, we have

ε = v∫ ∞

−∞F(t)dt = εme−�2ρ2 cosϕ ,

εm = √π V2

ω

�ve−ω2/(4�2v2), ϕ = ωτ ,

and the scattering cross-section for particles with a given value of τ when cosϕ > 0(cosϕ < 0) is

dσ

dε=

⎧⎨⎩

π

�2|ε| , when 0 < ε < εm cosϕ (0 > ε > εm cosϕ),

0, when |ε| > εm|cosϕ|.

In the incident beam, there are particles with different values of τ . If we average thecross-section over the phase ϕ, for ε > 0 using, for example, the formula

⟨dσ

dε

⟩= 1

2π

α∫−α

dσ

dεdϕ, α = arccos

ε

εm,

we get⟨dσ

dε

⟩=

⎧⎨⎩

1�2|ε| arccos

|ε|εm

, when |ε| < εm,

0, when |ε| > εm.

3.16. The distribution of decay particles is

dNN

= λ2 sinθ dθ

cos3 θ√

1 − λ2tan2θ, λ = V 2 − v2

0

2Vv0

in the angle interval 0 � θ � arctan (1/λ) if V > v0 and in the angle interval(π − arctan(1/|λ|) � θ � π if V < v0.

3.17. The distribution of decay particles by energies T = 12 mv2 has the form

dNN

= 6(Tmax − T)(T − Tmin)

(Tmax − Tmin)3 dT , Tmin � T � Tmax,



where

Tmin = 12 m(v0 − V )2, Tmax = 1

2 m(v0 + V )2.

3.18.

tanθ1 = cotθ2 = α/(Eρ), E = 12 mv2,

v1 = Eρv√α2 + E2ρ2

, v2 = αv√α2 + E2ρ2

.

3.19.

12 π � θ � π , when m1 < m2,

θ = 12 π , when m1 = m2,

0 � θ � 12 π , when m1 > m2.

ρ

θυ0

υ′0 υ′

χ

Figure 123

3.20. The velocity component normal to the sphere’s surfaceat the point of impact becomes zero in the centre-of-masssystem while the tangential component v′

0 is conserved (seeFig. 123). The scattering cross-section as function of the angleof deflection χ of the particle in the centre-of-mass system is

dσ = π |dρ2| = 4a2π |d cos2 χ | = 4a2 cosχ d�.

To transform to the laboratory system, we use the equation

tgθ = v′0 sinχ

v′0 cosχ + v0

= sinχ cosχ

1 + cos2 χ

to find the equation

cos2 χ1,2 = 32 cos2 θ − 1 ± 1

2 cosθ√

9cos2 θ − 8.

Taking into account that there are two possible connections between χ and θ , we obtain

dσ = 4πa2(|d cos2 χ1| + |d cos2 χ2|) =

= 4πa2d(cos2 χ2 − cos2 χ1) = 4a2 5 − 9sin2 θ√1 − 9sin2 θ

d�,

where 0 < θ < arcsin(1/3).



If the spheres impinging upon those at rest are identical so that there are no meansof distinguishing between the two after the collisions, we must add to the cross-sectionwhich we have just obtained the cross-section

dσ ′ = 4a2 cosθ d� (0 < θ < 12π),

which refers to the spheres which were originally at rest flying off at an angle θ

3.21. When passing the distance x intensity is

I(x) = I(0)e−nσx.

3.22. dN = σn1n2|v1 − v2|dV dt.

3.23. a) Ffr = 2πmv2n∫ π

0f (θ)(1 − cosθ)sinθ dθ ;1

b) 〈�2〉 = 2π(

mM

)2nl

∫ π

0f (θ)sin3 θ dθ ;

where l is the path travelled by a particle of mass M; v, its velocity; and n, theconcentration of light particles.

1 The quantity∫

(1 − cosθ) dσd�

d� is called the transport cross-section—as distinct from the total cross-

section∫

dσd�

d�.


§4

Lagrangian equations of motion.Conservation laws

4.1. Assuming that t = 0 when the particle is at x = 0, we find C = 0, and from thecondition that x = a when t = τ , we find that B = a

τ − aτ . Substituting the function

x(t) = At2 +(aτ

− Aτ)

t

into the action, we find

S =τ∫

0

L(x, x)dt =τ∫

0

(12 mx2 + Fx

)dt = mA2τ3

6+ ma2

2τ− FAτ3

6+ Faτ

2.

From the condition that the action be a minimum, ∂S∂A

= 0, it follows that A = F2m . It is

clear that in the present case the law of motion

x(t) = Ft2

2m+

(aτ

− Fτ

2m

)t

is exact. However, the only thing the solution given here allows us to state is that this lawis in some sense the best one among all possible laws of this kind.

In order that we can be certain that this law of motion gives a smaller value for S thanany other x(t), that is, that it is the true law, we must verify that it satisfies the Lagrangianequation of motion.




4.2.x(t) =

⎧⎨⎩

vxt − a, when 0 < t < t0,√v2

x − 2V/m(t − τ)+ a, when t0 < t < τ ,

y(t) = at/τ ,

where vx = [2aV/(mτ)]1/3 and t0 = a/vx.

4.3. From the relation

L(Q, Q, t) = L(q(Q, t), q(Q, Q, t), t) = L(

q,∂q∂Q

Q + ∂q∂t

, t)

we get

ddt

∂L∂Q

= ∂q∂Q

ddt

∂L∂ q

+ ∂L∂ q

ddt

∂q∂Q

,

∂L∂Q

= ∂L∂q

∂q∂Q

+ ∂L∂ q

∂ q∂Q

.

Bearing in mind that∂ q∂Q

= ddt

∂q∂Q

, we obtain

ddt

∂L∂Q

− ∂L∂Q

= ∂q∂Q

(ddt

∂L∂ q

− ∂L∂q

). (1)

The validity of the equation ddt

∂L∂ q

− ∂L∂q

= 0 thus leads to the validity of the analogous

equation

ddt

∂L∂Q

− ∂L∂Q

= 0.

If there are several degrees of freedom, instead of (1) we get the following equations:

ddt

∂L∂Qi

− ∂L∂Qi

=∑

k

∂qk

∂Qi

(ddt

∂L∂ qk

− ∂L∂qk

).

4.4. L(Q, dQ

dτ, τ

)= L

(q, dq

dt , t)

dtdτ

.



Here

q = q(Q, τ),dqdt

= dqdτ

dτ

dt,

dqdτ

= ∂q∂Q

dQdτ

+ ∂q∂τ

,dtdτ

= ∂t∂Q

dQdτ

+ ∂t∂τ

.

4.5. L = 12m x2

1 + λx − (1 + λx)U(x), x = dxdτ

.

4.6. L = −√

1 −(

dqdτ

)2.

The problem is a purely formal one. However, both this Lagrangian function and thetransformation consider (“improved” by the introduction of dimensional factors) havea simple physical meaning in the theory of relativity (e.g. see [2], § 4 and § 8).

4.7. The generalized momentum Pl = ∂L∂Ql

and the energy

E′ =∑

l

PlQl − L

transform as

Pl =∑

k

∂fk∂Ql

pk, E′ = E −∑

k

pk∂fk∂t

.

4.8. Applying the formulae of the preceding problem, we obtain

a) p′r = mr′ = pr , p′

ϕ = mr′2(ϕ′ + �) = pϕ ,E′ = E − �pϕ ;

b){

p′x = px cos�t + py sin�t,

p′y = −px sin�t + py cos�t.

(1)

It follows from (1) that p′ = p, while (1) also gives the rules for the law of transformationof vector components when we are changing to a coordinate system which is rotated overan angle �t. We emphasize that p′ �= mv′ (cf. [1], § 39).

4.9. a) E′ = E − Vp, p′ = p;b) E′ = E − Vp + 1

2 mV 2, p′ = p − mV.



The two expressions for the energy differ by a constant. Usually, we employ the secondformula, as it agrees with the definition of the energy in the theory of relativity.

4.10. In the frame of reference rotating together with the rod, the law of conservationof energy is correct (taking into account the centrifugal energy):

−12 m�2a2 = 1

2 mv2r − 1

2 m�2l2,

where vr is the velocity directed along the rod at the moment of the bead slipping off.From here we get v2

r = �2(l2 − a2). In the laboratory reference of frame, the velocity vtdirected across the rod should be added (vt = �l) to this velocity vr . Thus, the value ofthe total velocity is

v =√

v2r + v2

t = �√

2l2 − a2.

The velocity of the slipping bead can be changed significantly if the end of the rod isbent in the direction towards or against the motion.

4.11. In a rotating frame of reference x′y′ (Fig. 7), the Lagrangian function is equal to(see [7], § 17.2)

L(r′,v′, t) = 12 m

(v′2 + �2r′2) + mv′[�,r′].

We direct the x′-axis along rod a and introduce the angle ϕ, which determines thedirection of rod b with respect to the direction of rod a (see Fig. 7). Substituting thevector r′ = (a + bcosϕ,bsinϕ) in the Lagrangian function, we obtain

L(ϕ, ϕ, t) = 12 mb2ϕ2 + mab�2 cosϕ

(after removing 12 m�2

(a2 + b2

)+ m�ϕb(acosϕ + b), representing the total derivative ofthe function of ϕ with respect to the time). The Lagrangian equation

bϕ = −a�2 sinϕ

is identical to that of the pendulum of length b in the field of gravity g = a�2, so thefrequency of small oscillations is ω = �

√a/b.

If we add a uniform constant magnetic field B, perpendicular to the plane of rotation,then the addition to the Lagrangian function will be

�L(ϕ, ϕ, t) = mab�ωB cosϕ, ωB = eBmc

,



where e is the particle charge and c is the velocity of light. (This result occurs if we choosethe vector potential A = 1

2 [B,r] and remove the terms representing the total derivativeof the function of ϕ with respect to time.)

Thus, the effect of the magnetic field is reduced to a simple substitution of �2

for �(�+ ωB) in the Lagrangian function without magnetic field. In particular, for�+ ωB = 0, the oscillations of the angle ϕ will be completely “turned off”.

q

A

B

B ′

A′ −δq−δt

t

Figure 124

4.12. Let qi = gi(t) describe the motion of the system (trajec-tory AB in Fig. 124). As the form of the action is invariantwhen we change to the variables q′

i, t′, the motion is alsodescribes by the equations q′

i = gi(t′). If we express theseequations in terms of the variables qi, t, we have up to firstorder in ε:

qi(t − δt) = gi(t)− δqi,

with

δqi = εfi(g(t), t), δt = εh(g(t), t)

(trajectory A′B′ in Fig. 124).Small changes in the coordinates and time of the beginning and the end of the motion

when we change from the trajectory AB to the trajectory A′B′ lead to the following changein the action:

SA′B′ − SAB =[

∂S∂t

(−δt)+∑

i

∂S∂qi

(−δqi)

]B

A

.

Here we have (see [1], § 43)

∂S∂t

=L −∑

i

∂L∂ qi

qi = − E(t), ∂S∂qi

= ∂L∂ qi

=pi(t).

On the other hand, according to the conditions of the problem SAB = SA′B′ , so that

E(tA)εh(qA, tA)−∑

i

pi(tA)εfi(qA, tA) =

= E(tB)εh(qB, tB)−∑

i

pi(tB)εfi(qB, tB),



or

Eh −∑

i

pifi = const.

The proved theorem is, in essence, a united derivation of the various conservationlaws. The importance of this theorem increases if we take into account the fact that thesame theorem takes also place in the field theory (Noether’s theorem, see [13] and [14]).

4.13.∑

i

∂L∂ qi

(qih − fi)− Lh − F = const.

4.14. a) The linear omentum;

b) the angular momentum;

c) the energy;

d) Mz + h2π

pz = const, where h is the pitch of the screw;

e) Ex − pxt = const: the integral of motion of the centre of inertia of the system(see [2], § 14).

4.15. a) The potential energy is U(r) = −Fr, and this energy and at the same timethe action remain unchanged under transformation in direction at right angles to F orunder rotating around axis parallel to F. Integral of motion are thus the linear momentumcomponents at right angles to F and the angular momentum component parallel to F.As the Lagrangian function is independent of the time, the energy is an integral of themotion.

The statement that different points in some region are “equal” means that the valueof the potential energy (but not of the force!) is the same in all those points.

b) The similarity transformation r′ = αr leaves the form of the action invariant if thetime is transformed as t′ = βt simultaneously. The contribution of the kinetic energy tothe action

∫m2

v′2 dt′ = α2

β

∫m2

v2 dt

remains unchanged at β = α2, and the contribution of the potential energy

−∫

U(r′)dt′ = −αnβ

∫U(r)dt = −αn+2

∫U(r)dt

remains unchanged when n = −2. To use the theorem, formulated in problem 4.12, wewrite the infinitesimal similarity transformation as α = 1 + ε:

r′ = (1 + ε)r, t′ = (1 + 2ε)t, ε → 0,



so thus f = R, h = 2t and the integral of motion reads

∂L∂v

(vh − f )− Lh = 2Et − mvr = C. (1)

From (1) we can find r(t) taken into account that rv = 12

dr2

dt :

r2 = 2Em

t2 − 2m

Ct + C1. (2)

If E < 0, the particle falls into the centre (in this case, r → ∞). If E > 0, it is helpfulto introduce other constants τ , B instead of C, C1 and write (2) as

r2 = 2Em

(t − τ)2 + B.

At B > 0 the dependence r(t) is the same as for the free motion of particles with thevelocity v0 = √

2E/m and with the impact parameter ρ = √B. At B < 0, the particle falls

into the centre.The fields for which the conditions of this problem are satisfied are given, for example,

in problems 12.6, 12.7 and in [1], problem 2 to § 15.

c) E − Vp = const;

d) rp − 2Et = const, where p = mv + ecA(r), if A(αr) = α−1A(r);

e) E − pϕ� = const.

4.16. One can use the results of problem 4.15b (see (2), taking into account the initialconditions)

r2(t) = 2Em

t2 + 2r0v0t + r20. (1)

The fall time is determined from the quadratic equation r2(t) = 0, which gives two roots

t1,2 = 12E

(−mr0v0 ± √

β)

, β = (mr0v0)2 − 2mEr2

0.

From (1) we can see that the fall must happen at E < 0; its time is determined bythe positive root t2. As for the case when E > 0, the fall is possible only at r0v0 < 0 andβ > 0. In this case, one should choose the smallest t2 among the two positive roots.



Please note that in this problem the value M2 − 2m(ar/r)2 is an integral of motionwhich coincides with the (−β) (see problem 10.26). Therefore, the equation for theradial motion

E = 12 mr2 − β

r2

leads to the following fall time

t = −m∫ 0

r0

dr√2mE − β/r2

= 12E

(√β + 2mEr2

0 − √β

),

same as t2.

4.17. mr − pt = const (cf. problem 4.14e).Is this integral of motion the eighth independent integral for a closed system (apart

from E, M, and p)?

4.18. a) Let the z-axis be parallel to B. A translation along the z-axis or a rotation aroundit lives the form of A, and hence also the form of the action, invariant. We have thus thefollowing integrals of motion:

pz = ∂L∂ z

= mz and Mz = xpy − ypx = m(xy − yx)+ eB2c

(x2 + y2).

Moreover, the energy

E = 12 m(x2 + y2 + z2)

is an integral of motion.b) E = 1

2 m(x2 + y2 + z2), p′y = my + eBx/c, p′

z = mz (cf. problem 10.8).Symmetry considerations allow us to determine various integrals of motion depending

on the choice of the vector potential for a given field B. However, the quantities E, pz = p′z,

Mz, and p′y are all integrals of motion, independent of the choice of A.

4.19. a) E = 12 m(r2 + r2ϕ2 + z2), pϕ = mr2ϕ + eμ

cr2

(r2 + z2)3/2 , where we use the cylin-

drical coordinates and the z-axis is taken parallel to the vector μ (cf. problem 2.35).b) From the symmetry properties of the given field we can obtain the following integral

of motion:

pz = mz, Mz ≡ pϕ = mr2ϕ + eμc

, E = 12 m(r2 + r2ϕ2 + z2).



However, the motion in this “field” is free-particle motion. Indeed, the Lagrangianfunction,

L = 12 mv2 + eμ

cϕ,

differs from the Lagrangian function of a free particle only by the total derivative withrespect to time of the function eμϕ/c (of course, in this case we have B = rotA = 0).

We note that in the case when μ is a function of time, pz and Mz remain integrals ofmotion.

4.20. a) x + x = 0. The same equation could have obtained from the Lagrangian functionL1(x, x) = x2 − x2. It is well known that if two Lagrangian functions differ by a totalderivative of a function of the coordinates and the time, they lead to the same Lagrangianequations of motion. The reverse statement is incorrect.

b) x + αx + ω2x = 0.

4.21. a) In the spherical coordinates the Lagrangian equations of motion for a particlemoving in a field U are

m(r − rϕ2 sin2 θ − rθ2)+ ∂U∂r

= 0,

m(r2θ + 2rrθ − r2ϕ2 sinθ cosθ)+ ∂U∂θ

= 0,

m(r2ϕ sin2 θ + 2rrϕ sin2 θ + 2r2θ ϕ sinθ cosθ)+ ∂U∂ϕ

= 0

We can easily write them in the form

m(v)i = (F)i,

where the components of the force are the components of (−gradU):

Fr = −∂U∂r

, Fθ = −1r

∂U∂θ

, Fϕ = − 1r sinθ

∂U∂ϕ

.

Hence we have

(v)r = r − rϕ2 sin2 θ − rθ2,

(v)θ = rθ + 2rθ − rϕ2 cosθ sinθ ,

(v)ϕ = rϕ sinθ + 2rϕ sinθ + 2rθ ϕ cosθ .

b) (v)i = 12hi

(ddt

∂

∂ qi− ∂

∂qi

)ds2

dt2= hiqi +

3∑k=1

(2qi qk

∂hi

∂qk− q2

khkhi

∂hk

∂qi

).



4.22. a) The Lagrangian function

L = 12 m

3∑i,k=1

gikqi qk − U(q1, q2, q3),

with

gik =3∑

l=1

∂xl

∂qi

∂xl

∂qk,

leads to the equations

m3∑

k=1

gskqk + m3∑

k, l=1

�s,kl qkql = −∂U∂qs

(s = 1, 2, 3), (1)

where the

�s,kl = 12

(∂gsk

∂ql+ ∂gls

∂qk− ∂gkl

∂qs

).

are the so-called Christoffel symbols of the first kind.b) Using the notation q4(x, t) = t, we can proceed as under 4.22a and obtain the same

formulae, merely replacing∑3

1 by∑4

1.What is the meaning of the terms in (1) which contain �1,k4 (k = 1, 2, 3, 4) if the ql

are Cartesian coordinates in a rotating frame of reference (see problem 4.8)?

4.23. Since the proposed Lagrangian function differs from the Lagrangian function ofthe free-moving particles only by the term

L1 = − egc

ϕ cosθ

the components of the force in the spherical coordinates (see problem 4.21a) have theform

Fr = 0,

Fθ = 1r

∂L1

∂θ= egϕ sinθ

cr,

Fϕ = − 1r sinθ

ddt

∂L1

∂ϕ= egθ

cr

and coincide with the components of the Lorentz force



ec[v,B] = eg

cr3 [v,r].

Since ∂L/∂t = 0, the integral of motion is the energy E = 12 mv2. Similarly, since

∂L/∂ϕ = 0, the integral of motion is the generalized momentum

pϕ = mr2ϕ sin2 θ − egc

cosθ = Jz.

In the given Lagrangian function, the direction of the z-axis can be chosen arbitrary; thisleads to the conservation of the vector

J = m[r,v] − egc

rr.

Besides, the action corresponding to this Lagrangian function is invariant under thesimilarity transformation. Therefore, the integral of motion is

prr − 2Et = mrr − 2Et

(cf. problem 4.15b).

4.24. The equations of motion are

−L I = ϕA − ϕB,q2

C= ϕB − ϕA.

We shall assume that the potential source is a capacitor with a very large capacitance C0and that its charge at the moment when q1 = 0 is Q. The energy of the system includingthe potential source and the inductance is

E0 = (Q + q1)2

2C0+ 1

2 L q21.

Shifting the zero of the energy and considering the limit Q/C0 → U as C0 → ∞, we get

E = E0 − Q2

2C0= Uq1 + 1

2 L q21.

This is the form of the energy which leads to the Lagrangian function given in theproblem.

Similar to this, the energy of a particle of mass m in a uniform force field −F(t) is12 mx2 + Fx.

The same Lagrangian function may be obtained from that of an electromagneticfield including interactions between an electromagnetic field and charges (see [3], § 27and § 28):



L = 18π

∫(E2 − B2)dV + 1

c

∫AjdV −

∫ϕρ dV (1)

(using Gaussian units).Generally speaking, the electromagnetic field is a system with an infinite number of

degrees of freedom. However, the fields inside the capacitor and inductance are specifiedby the charge q2 or the current q1. Using equations

crotB = 4π j, divE = 4πρ

(and given that the fields are concentrated inside a limited volume), we obtain

∫ϕρ dV = 1

4π

∫ϕ divEdV =

= 14π

∫div(ϕE)dV − 1

4π

∫Egradϕ dV = 1

4π

∫E2 dV ,

and similarly

1c

∫AjdV = 1

4π

∫B2 dV ,

so that

L = 18π

∫ (B2 − E2

)dV .

Therefore, the Lagrangian function can be expressed in terms of the electric field energyinside a capacitor as

18π

∫E2 dV = q2

2

2C

and in term of the magnetic field energy inside the inductance as

18π

∫B2 dV = 1

2 L q21

(see [3], § 2 and § 33).

4.25. a) L = 12 L q2

1 − q22

2C + U(q2 − q1);



b) L = 12 L q2 − q2

2C ;

c) L = 12 L1q2

1 + 12 L2q2

2 − q21

2C1− q2

22C2

− (q1 + q2)2

2C .

4.26. a) L = 12 ml2ϕ2 + 1

2 L q2 + mgl cosϕ − q2

2C(ϕ);

b) L = 12 mx2 + 1

2 L (x)q2 − 12 kx2 + mgx − q2

2C .

4.27. Let ϕ be the angle of rotation of the frame around the AB-axis, such that ϕ = 0gives the direction of the magnetic field; let q be the current in the frame (the positivedirection is the one from A to D). The Lagrangian function for the system is

L = 12 ma2ϕ2 + 1

2 L q2 + Ba2qsin vfi.

The integrals of motion are the energy,

E = 12 ma2ϕ2 + 1

2 L q2, (1)

and the momentum conjugate to the cyclic coordinate q which is associated with the totalmagnetic flux through the frame,

∂L∂ q

= L q + Ba2 sinϕ = �0.

The current through the frame is thus uniquely determined by its position:

q = �0 − Ba2 sinϕ

L.

Substituting this value q in (1) we obtain

E = 12 ma2ϕ2 + Ueff(ϕ), Ueff(ϕ) =

(�0 − Ba2 sinϕ

)2

2L. (2)

The problem of system’s motion is thus reduced to a one-dimensional one.

Ueff

Um

Umax

ϕ−π

2 2 2π π 3π

Figure 125



Let us consider the case 0 < �0 < Ba2 in more detail. The function Ueff(ϕ) for thiscase is given in Fig. 125. One sees that when E > Umax = (�0 + Ba2)2/(2L ) the framerotates and ϕ is a periodic time function with period

T = √2ma

π/2∫−π/2

dϕ√E − Ueff(ϕ)

.

When Umax > E > (�0 − Ba2)2/(2L ) = Um, the frame performs periodic oscillationswithin the angular interval ϕ1 < ϕ < π − ϕ1, where

ϕ1 = arcsin�0 − √

2L EBa2 ;

the period tends to infinity as E → Umax (see problem 1.5). When 0 < E < Um one canhave oscillations either in the interval ϕ1 < ϕ < ϕ2 or in the interval π − ϕ2 < ϕ < π − ϕ1,where

ϕ2 = arcsin�0 + √

2L EBa2 .

How will the character of the frame’s rotation change if one assume it to have a smallresistance?

4.28. a) The equations of motion for the system can be obtained using a Lagrangianfunction with an additional term responsible for the constraints (see [5], § 2.4)

L∗ = 12 m(x2 + z2)− mgz + λ(z − ax2),

where λ is a time-dependent Lagrangian multiplier. The equation of motion

mx = −2 lmax, (1)

mz + mg = λ (2)

together with the equation of constraints z = ax2 completely determine the motion of theparticle.

On the right-hand side of (1) and (2) there are the components of the reaction forcesalong the two axes: Rx = −2λax and Rz = λ. They can be rewritten in terms of thecoordinate and velocity of the particle with the equation of constraints as

Rx = −2axRz, Rz = (2ax2 + g)m1 + 4a2x2 .



b) mr − mg cosϕ − mrϕ2 = λ,

mr2ϕ + 2mrrϕ + mgr sinϕ = 0,

r = l.

The reaction force,

λ = −mg cosϕ − mlϕ2,

lies along r.

4.29.

L∗ = 12 m(r2ϕ2 + r2)+ mgr cosϕ + λ(ϕ − �t);

λ = 2mrr�+ mgr sin�t is the generalized force corresponding to the coordinate ϕ (themoment of the torque force).

4.30. a) dEdt = −∂L

∂t+

s∑i=1

qiRi.

b) The law of the transformation of the left-hand sides of the equations of motion isgiven in problem 4.3:

ddt

∂L∂Qi

− ∂L∂Qi

=∑

k

∂qk

∂Qi

(ddt

∂L∂ qk

− ∂L∂qk

).

The same law of transformation must be for the right-hand sides:

Ri =∑

k

∂qk

∂QiRk. (1)

If the law of transformation of coordinates is not included evidently the time, thevelocities are transformed according to the law

qi =∑

k

∂qi

∂QkQk,

which is reverse to (1).In other words, the components of the force Rk form a covariant vector, while the

velocity components form a contravariant vector in s-dimensional space (see [2], § 83).Therefore one can find the reaction forces Ri in terms of any generalized coordinates if

the constraint and friction forces are known in Cartesian coordinates. In particular, if the

friction forces are given in terms of a dissipative function, Ri = −∂F∂ qi

, the transformation

of F is reduced to a change of variables.



4.31. The equations stated in this problem can be obtained by eliminating the λβ fromthe following equations:

ddt

∂L∂ qβ

− ∂L∂qβ

= λβ , β = 1, . . . , r,

ddt

∂L∂ qn

− ∂L∂qn

= −r∑

β=1

λβbβn, n = r + 1, . . . , ., s.

The following relations must be taken into account:

∂L∂ qn

= ∂L∂ qn

+r∑

β=1

∂L∂ qβ

bβn,

∂L∂qn

= ∂L∂qn

+r∑

β=1

s∑m=r+1

∂L∂ qβ

∂bβm

∂qnqm.

Thus, the equations of motion for a system with non-holonomous constraints differ fromthe Lagrangian equations of motion although the constraints equations permit one toeliminate certain coordinates and velocities from the Lagrangian function.

4.32. a) Taking into account that qn is occur both in Ln and in Ln+1, we obtain theLagrangian equations of motion

ddt

∂Ln

∂qn= ∂Ln

∂qn+ ∂Ln

∂(�qn)− ∂Ln+1

∂(�qn+1)(�qn = qn − qn−1), (1)

whence we find as a → 0

1a

∂Ln

∂ qn→ ∂L

∂(∂q/∂t),

1a

∂Ln

∂qn→ ∂L

∂q,

1a

[∂Ln

∂(�qn)− ∂Ln+1

∂(�qn+1)

]→ − ∂

∂x∂L

∂(∂q/∂x),

so that the equations (1) are reduced to the following equations:

∂

∂t∂L

∂(∂q/∂t)+ ∂

∂x∂L

∂(∂q/∂x)= ∂L

∂q. (2)

Here the derivatives ∂/∂t and ∂/∂x should be applied to the functions q(x, t) and itsderivatives.



The system of N ordinary differential equations (1) thus transforms to one equationin the partial derivatives (2). For a continuous system the variable x indicates a fixedpoint on the string.

We do not consider the physical consequences of the (2) since systems with an infinitenumber of degrees of freedom are the subject of field theory, but not of mechanics(see [5], Ch. 11; [2], § 26 and § 32).

b) E =∫ {

∂L∂(∂q/∂t)

∂q∂t

− L

}dx.

4.33. The Lagrangian function is

L = 12 mv2 − U(r)+ e

cA(r)v,

where A(r) is the vector potential of the magnetic field, B = rotA (of course, A(r) canalways be taken as a homogeneous function of the coordinates of degree n + 1). If as aresult of the similarity transformation,

r → αr, t → α1−k/2 t,

the transformation of the vector potential is the same as that for the velocity, that is,n = −1 + k/2, we have L → αkL. Therefore, the equations of motion remain invariantafter this transformation and the principle of mechanical similarity holds (see [1], § 10).

It is clear that the principle of mechanical similarity remains valid for a magnetic fieldif it is constant in space and if its value changes by a factor α−1+k/2 under a similaritytransformation (e.g. see, problems 2.34– 2.37 and 6.37).

4.34. The kinetic energy of the system is T = ∑a

12 mav2

a, so that

2T =∑

a

∂T∂va

va = ddt

(∑a

mavara

)−

∑a

ramava.

The first term ddt

(∑a mavara

), which is the total time derivative with respect of time

of a bounded function becomes zero after averaging over large time interval (see [1], § 10and [7], § 7). Substituting

mava = − ∂U∂ra

+ ea

c[va,B]

into the second term and averaging over the time we obtain

⟨2T + B

∑a

ea

c[ra,va]

⟩= k〈U〉,



where the pointed brackets 〈〉 indicate the time averaging. In particular, if the magneticfield B is homogeneous and constant, then

2 〈T〉 = k〈U〉 − 2〈μ〉B,

where

〈μ〉 =⟨

12c

∑a

ea[ra,va]

⟩

is the average magnetic moment of the particle system. If eama

= em , then μ = e

2mcM,where M is the angular momentum of the system.

4.35. a) Write dAdt in the form of two terms

dAdt

= [m(x1x2x3 + x1x2x3 + x1x2x3)− x1U23 − x2U13 − x3U12]−− (x1U23 + x2U13 + x3U12). (1)

Using the equations of motion

mx1 = F12 + F13, mx2 = F21 + F23, mx3 = F31 + F32, (2)

Fik = −Fki = −∂Uik

∂xi= 2g2

(xi − xk)3 (3)

and introducing the relative distances

x1 − x2 = x, x2 − x3 = y, x1 − x3 = z,

write the second term of (1) as

−2g4

m

[(1x3 + 1

z3

)1y2 +

(− 1

x3 + 1y3

)1z2 −

(1z3 + 1

y3

)1x2

]. (4)

Combine the terms with the same powers of z, we rewrite (4) as

2g4(x − y)m

(1

x3y3 − x2 + xy + y2

z2x3y3 − x + yz3x2y2

).

Substituting z = x + y, it is easy to prove that this expression becomes zero.The first term of (1) is zero for arbitrary forces. To show this, it is sufficient to use the

equations of motion in form (2) and substitute



U ik = ∂Uik

∂(xi − xk)(xi − xk) = −(xi − xk)Fik.

Finally, note that in this field the similarity transformation does not change anaction, and, as a consequence, the is the fourth (additional) integral of motion (seeproblem 4.15b)

m(x1x1 + x2x2 + x3x3)− 2Et. (5)

4.36. When two particles approache each other, their interaction energy becomesinfinite. As a result, the particles can not pass “one through the other” and the orderof their location on the line is conserved.

When two particles of equal mass but with arbitrary interaction energy (only allowingthe expulsion of particles) collide, these particles simply exchange velocities. (This isfollows from the laws of conservation of energy and momentum.) If the collisions ofthree particles occur one after another, so that during the convergence of two particlesthe third particle is far away from them, it will simply lead to the exchange of velocities.Thus the collisions will end when the fastest particle is ahead and the slowest one isbehind, in other words, in this case

v′1 = v3, v′

2 = v2, v′3 = v1. (1)

In general, when all three particles simultaneously converge, the values of the velocitieswill not be conserved.

The more surprising is that for the forces specified in the preceding problem, theresult (1) is conserved. This can be shown using all three integrals of motion: P, E, andA. Taking into account that the functions Uik → 0 as t → ±∞ and comparing P, E, andA at t → +∞ and at t → −∞, we obtain three equations:

v′1 + v′

2 + v′3 = v1 + v2 + v3,

(v′1)2 + (v′

2)2 + (v′

3)2 = v2

1 + v22 + v2

3,

v′1v′

2v′3 = v1v2v3.

(2)

Solving this system with respect to v′i, we get, generally speaking, six different solutions.

However, all these solutions can be guessed.It is easy to verify that the solution (1) satisfies the system (2). Furthermore, since the

equations (2) are obviously symmetric with respect to all six possible permutations ofthe particles, it is clear that the remaining roots of the set (2) can be obtained by simplepermutations from (1).

After that, it is easily to prove that only the result (1) can be realized at t → +∞, sinceany other options imply the possibility of the further collisions of the particles (due toinequalities v3 > v2 > v1).


§5

Small oscillations of systems withone degree of freedom

5.1. a) ω2 = Vα2

m

√1 −

(F

Vα

)2; a minimum in U(x) occurs when F < Vα;

b) ω2 = 8π3

Vα4

m x20

(�(3/4)�(1/4)

)2, where the amplitude x0 is determined from the

equality

E = 12 mx2 + 1

3 Vα4x4 = 13 Vα4x4

0.

5.2. The Lagrangian function of the system is (see [1], § 5, problem 4)

L = ma2[θ2(1 + 2sin2 θ)+ �2 sin2 θ + 2�20 cosθ],

where we have introduce �20 = 2g/a.

When � > �0 the potential energy of the system,

U(θ) = −ma2(�2 sin2 θ + 2�20 cosθ),

has a minimum when cosθ0 = �20/�2. Expanding U(θ) in the neighbourhood of θ0, and

in the kinetic energy putting

1 + 2sin2 θ = 1 + 2sin2 θ0 = 3 − 2(

�0

�

)4

≡ M2ma2 ,

we get

L = 12 mx2 − 1

2 kx2,




where k = U ′′(θ0), x = θ − θ0. Hence we get

ω2 = kM

= �2 �4 − �40

3�4 − 2�40

(� > �0).

When � � �0, the oscillation frequency is proportional to the angular velocity of rota-tion, ω = �/

√3 and θ0 = π/2; when � → �0, small oscillations occur with a frequency

ω → 0 and θ0 → 0.If � < �0, we can consider oscillations near θ0 = 0 for elastic collisions of the lateral

particles

ω2 = �20 − �2 (� < �0).

If � = �0, the potential energy U has a minimum at θ0 = 0 and in the neighbourhoodof it, we can write

U = ma2�20

(−2 + 1

4 θ4)

,

that is, the oscillations are non-linear in an essential way. Retaining also in the kineticenergy terms up to the fourth order, we get

2π

ω= T = 4

θm∫0

√1 + 2θ2dθ√

�20(θ4

m − θ4)

.

Here θm is the amplitude of the oscillations (see [1], § 11, problem 2a).

5.3. The Lagrangian function of the system is

L = 12 mR2ϕ2 − U(ϕ), U(ϕ) = 2mgR

[sin2 ϕ

2+ 2s3

sin(ϕ/2)

],

where s =(

q2

8mgR2

)1/3

.

When s < 1, the stable equilibrium position ϕ0 is determined by the condition

sin(ϕ0/2) = s, and ω2 = 3gR (1 − s2).

When s > 1, the point A is a position of stable equilibrium and ω2 = gR (s3 − 1).

When s = 1, the potential energy U(ϕ) has a minimum at ϕ0 = π and in the neigh-bourhood of it

U(ϕ) = 332

mgR(ϕ − π)4 + 6mgR,



that is, the oscillations are non-linear in an essential way. As a result, we get

2π

ω= T = 16

√R3g

ϕm∫0

dα√ϕ4

m − α4),

where ϕm is the amplitude of the oscillations (see [1], § 11, problem 2a).

5.4. r = r0 + acosω(t − t0),

ϕ = ϕ0 + �(t − t0)− 2a�

r0ωsin[ω(t − t0)],

where r0, ϕ0, a, and t0 are integration constants (a � r0), and

� =√

nα

mr−(n+2)/20 , ω = �

√2 − n.

5.5. At the point θ = θ0 the effective potential energy

Ueff(θ) = M2z

2ml2 sin2 θ− mgl cosθ

(see [1], § 14, problem 1) has a minimum, so that U ′eff(θ0) = 0. We thus obtain

M2z = m2l3g sin4 θ0

cosθ0,

and the frequency of small oscillations is

ω(θ0) =√

U ′′eff(θ0)

ml2=

√gl

1 + 3cos2 θ0

cosθ0.

For this calculation to be the applicable the condition

12 U ′′

eff(θ0)(θ)2 � 16 |U ′′′

eff(θ0)|(θ)3,

with θ the oscillation amplitude, must be satisfied. If θ0 ∼ 1, it is satisfied for θ � 1.If, however, θ0 � 1, we have U ′′′

eff(θ0) ∝ 1/θ0, and the oscillations in θ can be consideredto be small only when θ � θ0. The result obtained, ω = 2

√g/l, is nevertheless valid

also for θ ∼ θ0 when the oscillations in θ are no longer harmonic. Indeed, in this casesmall harmonic oscillations with frequency

√g/l occur along the x- and y-axes, that is,



the pendulum moves along an ellipse executing two oscillations in the angle θ for eachrevolution (see [1], § 23, problem 3 and [7], § 4).

5.6. The effective potential energy for radial oscillations of the molecule is

Ueff(r) = 12 mω2

0(r − r0)2 + M2

2mr2 ,

where r is the distance between the atoms and m the reduced mass. The extra term whichis assumed to be a small correction leads to a small shift in the equilibrium position

δr0 = M2

m2ω20r3

0

.

We determine the shift in the frequency by expanding Ueff(r) in a series near the pointr0 + δr0:

Ueff(r) = 12 mω2

0(r − r0 − δr0)2 + M2

2mr20

+ 3M2

2mr40

(r − r0 − δr0)2.

We get from this a correction to the frequency

δω = 3M2

2m2ω0r40

= 3�2

2ω0,

where � = M/(mr20) is the average angular velocity of the molecule rotation.

5.7. a) We get for the displacement from the equilibrium position:

x = x0 cosωt + x0

ωsinωt =

√x2

0 + x20

ω2 cos(ωt + ϕ),

tanϕ = − x0

ωx0, ω2 = 2k

m.

b) Let the tension in each spring be f . For small displacements |y| � √fl/k, where

l is the separation between the points where the springs are fixed, the oscillations areharmonic y = Acos(ωt + ϕ), and ω2 = 2f /(ml).

When f = kl, the oscillation frequency is equal to that in part 5.7a. In the case ofsprings in which there is no tension (f = 0) the oscillations are non-linear, the restoringforce is F = −ky3/l2, and the frequency (cf. problem 5.1b) is

ω =√

π�(3/4)

�(1/4)

√2km

ym

l,

where ym is the amplitude of the oscillations.



If the particle can move in the xy-plane, its motion – for the case when f = 0 and x andy are small – consists of harmonic oscillations along the x- and y-axes with frequenciesω2

x = 2k/m and ω2y = 2f /(ml), respectively (see problem 6.3).

5.8. Let y be the coordinate of the particle reckoned from the upper suspension point,and 2l the distance between the two suspension points. The Lagrangian function of thesystem,

L = 12 my2 − k(y − l)2 + mgy = 1

2 my2 − k(y − l − mg

2k

)2 + const,

describes a harmonic oscillator with frequency ω2 = 2k/m and equilibrium position y0 =l + mg/(2k), so that y = y0 + Acos(ωt + ϕ).

If we take the displacement from the equilibrium position as coordinate, we caneliminate the gravity field from the Lagrangian function.

5.9. For the angle between the pendulum and the vertical we have

ϕ = a�2

g − l�2 sin�t,∣∣∣∣ a�2

g − l�2

∣∣∣∣ � 1

(see also problem 8.3).It is also possible the oscillations of the pendulum near the direction of the radius

vector

ϕ = �t + ga�2 sin�t, �2 � g

a.

5.10. The result for the current in circuit

I = dqdt

= U0 sin(ωt − ϕ)√R2 + (ωL − 1/(ωC))2

, tanϕ = ωL − 1/(ωC)

R

can be obtained by solving the Lagrangian equations of motion for q. The Lagrangianfunction of the system is

L = 12 L q2 − q2

2C+ qU

(see problem 4.25), and the dissipative function is equal to 12Rq2 (see [3], § 48).

5.11. The general solution of the equation of motion (see [1], § 26)

x + 2λx + ω20x = F

mcosγ t



under such conditions that ω2 = ω20 − λ2 > 0, has the form

x(t) = e−λt(acosωt + bsinωt)+

+ F[(ω20 − γ 2)cosγ t + 2λγ sinγ t]

m[(ω20 − γ 2)2 + 4λ2γ 2]

,

where a and b are constants to be determined from the initial conditions. If we put x(0) =x(0) = 0, we find finally

x(t) = F

m[(ω20 − γ 2)2 + 4λ2γ 2]

[(ω2

0 − γ 2)(cosγ t − e−λt cosωt)+

+2λγ

(sinγ t − ω2

0 + γ 2

2γωe−λt sinωt

)]. (1)

F x(a)

(b)x

mω0|ε|

2π

|ε|

t

t

Figure 126

Let us study this solution in the region near theresonance, γ = ω0 + ε, |ε| � ω0. If there is no fric-tion at all, that is, λ = 0, the motion of the oscillatornear resonance will show beats:

x = Fmω0ε

sin(εt/2) · sinω0t, (2)

and the amplitude and the frequency of the beatsare determined by how near resonance we are(Fig. 126a). However, when γ = ω0, that is, at exactresonance, we get by letting ε → 0:

x = F2mω0

t sinω0t, (3)

that is, we get oscillations with an amplitude a(t) which increases indefinitely a(t) =Ft/(2mω0) (Fig. 126b).

When there is even a small amount of friction (λ � ω0) the character of the motionchanges qualitatively. From (1) we get easily in the case when λ � |ε| instead of (2)

x(t) = F2mω0ε

√1 − 2e−λt cosεt + e−2λt cos(ω0t + ϕ1(t)). (4)

Here ϕ1(t) is the phase of the oscillations which changes slowly with time. The amplitudeof the oscillation oscillates slowly with a frequency |ε| about the value F/(2mω0|ε|),gradually approaching that value (Fig. 127a). It is remarkable that during the transientstage the amplitude may reach twice the value of the amplitude of the steady-stateoscillations.



When |ε| � λ � ω0,

x = F2mω0λ

(1 − e−λt)sinω0t. (5)

x

x

x

t

t

t

F

F

F

2mω0|ε|

2√2mω0|ε|

2mω0 λ

2π

|ε|

(a)

(b)

(c)

Figure 127

In that case we have the transient process with a smoothly increasing amplitude whichasymptotically approaches the value F/(2mω0λ) which is determined by the frictioncoefficient λ (Fig. 127b). Finally, if the quantities ε and λ are of the same orderof magnitude, |ε| ∼ λ � ω0, we get oscillations of the amplitude around the valueF/(2

√2mω0|ε|) which corresponds to the steady-state oscillations which are reached

very slowly (see Fig. 127c for the case ε ≈ λ).The system thus proceeds to steady-state oscillations for these three cases (Fig. 127)

over a time of the order of t ∼ 1/λ, as is clear from (1).

A

xO

B E

Emax

τm τ


One can use a vector diagram (see Fig. 128) to study qualitative how the oscillationsproceeds to a steady state (transient process) when λ � ω0. The forced oscillation isdepicted by the component of the vector

−→OA, which rotates with an angular velocity γ ,



onto the x-axis. The vector of free oscillations,−→AB, rotates with an angular velocity ω,

and its length decreases as e−λt. In the initial moment−→AB + −→

OA ≈ 0.What is the nature of the transient process, if x(0) = 0, x(0) = 0?

5.12. a) The energy acquired by the oscillator,

E = πF2

2mτ2 exp

[−1

2 (ωτ)2]

depends on how fast the force is switched on (i.e., on the parameter ωτ). For aninstantaneous impact (ωτ � 1) or a very slow switching on, (ωτ � 1), the energy transferis small, while the maximum energy transfer, Emax = πF2/(mω2e), is reached whenτm = √

2/ω (Fig. 129).b) If x → acos(ωt + ϕ), as t → −∞,1 we have

E = E(+∞)− E(−∞) =

= πF2

2mτ2e−(ωτ)2/2 − √

πaωτFe−(ωτ)2/4 sinϕ.

Depending on the value of ϕ, the oscillator gains or loses energy. This change in energyis similar to the absorption or stimulated emission of light by an atom.

When we average over the phase ϕ we get the same result as in point 5.12a.

5.13. a) x(t) = 12μ

[ξ1(t)− ξ2(t)], where

ξ1,2 = e±μt

⎡⎣

t∫0

1m

F(τ )e∓μτ dτ + x0 ± μx0

⎤⎦ .

b) x(t) = 1ω Im

{eiωt−λt

[t∫

0

1mF(τ )eλτ−iωτ dτ + x0 + (iω + λ)x0

]},

where ω =√

ω20 − λ2.

5.14. The force

F(t) = − ∂

∂rU(|r − r0(t)|), (1)

1 The meaning of ϕ is that of the “impact phase”, that is, the phase which the oscillator would have had att = 0, if there were no force acting upon it.



is acting upon the oscillator; here r(t) is the deflection of the oscillator and r0(t) theradius vector of the impinging particle. Assuming that the particle is little deflected, wecan put r0(t) = ρ + vt with ρ being the impact parameter (vectors ρ and v are mutuallyorthogonal). Assuming also that the amplitude of the vibrations of the oscillator is smallwe can put r = 0 in (1) (after differentiating) and then

F(t) = −2�2V (ρ + vt) · exp(−�2ρ2 − �2v2t2).

The oscillations along ρ and v are independent and the corresponding energy excitationsare equal to

12m

∣∣∣∣∣∣+∞∫

−∞Fρ(t)e−iωtdt

∣∣∣∣∣∣2

and1

2m

∣∣∣∣∣∣+∞∫

−∞Fv(t)e−iωtdt

∣∣∣∣∣∣2

, (2)

where Fv and Fρ are the components of F in the directions of v and ρ. The total energyexcitation of the oscillator is

ε = ε1

(1 + x

a

)e−x, ε1 = πV 2

2Eae−a, (3)

where

E = 12 mv2, a = 1

2

( ω

�v

)2, x = 2(�ρ)2.

The cross-section for exciting the oscillator to an energy between ε and ε + dε is

dσ = π∑

k

|dρ2k | = π

2�2

dε

ε

∑k

∣∣∣∣ a + xk(ε)

1 − a − xk(ε)

∣∣∣∣ , (4)

where the xk are the roots of (3).For a further consideration, it is most helpful to solve (3) graphically, in the same way

as was done in problem 3.11a. When ε � ε1, we get dσ = π

2�2dεε (in (4) we assume

that xk(ε) � 1, xk � a). For large ε the result depends on the value of a. If a > 1, we canonly have ε < ε1 (see Fig. 130a; for the cross-section see Fig. 130a). However, if a < 1

we can have ε < ε2 = πV 2

2Ee (see Fig. 131b) and at ε = ε1 the function dσ/dε will have adiscontinuity while for ε2 − ε � ε2 it has an integrable singularity,



ε ε

ε1 ε1

ε2

x−a −a−a+1 1−a x

(a) (b)

Figure 130

dσ dσ

dε dε

ε1 ε1 ε2ε ε

(a) (b)

Figure 131

dσ = π√2�2

dε

ε2

1√1 − ε/ε2

(see Fig. 131b).

5.15. If the oscillator has an “impact phase” ϕ (see problem 5.12b), we get, by repeatingthe calculations of the preceding problem, for the energy of the oscillator the expression(ε0 is the initial energy of the oscillator)

ε = ε1e−2(�ρ)2 + 2√

ε1ε0e−(�ρ)2cosϕ + ε0, (1)

where

ε1 = π

4E

(Vω

�v

)2

exp[− ω2

2(�v)2

].

When cosϕ > 0, we have ε > ε0 for all ρ, while we can also have values ρ1,2 such thatε < ε0, if cosϕ < 0. Solving (1) for ρ2, we find

ρ2 = 1�2 ln

√ε1/ε0

−cosϕ +√

(ε/ε0)− sin2 ϕ

, when ε > ε0,

ρ21,2 = 1

�2 ln√

ε1

√ε0|cosϕ| ±

√ε − ε0 sin2 ϕ

, when cosϕ < 0

and ε0 > ε > εmin = ε0 sin2 ϕ.



Hence we have

dσ = π

∣∣∣∣dρ2

dε

∣∣∣∣dε = π

2�2

dε

ε − ε0 sin2 ϕ − cosϕ ·√

εε0 − ε20 sin2 ϕ

, (2)

when ε > ε0, and

dσ = πd(−ρ21 + ρ2

2) = 1�2

ε0π |cosϕ|dε

(ε0 − ε)

√εε0 − ε2

0 sin2 ϕ

(3)

when εmin < ε < ε0 and cosϕ < 0.Averaging over all possible phases ϕ for a given ε, we obtain

⟨dσ

dε

⟩= π

2�2|ε0 − ε| (4)

(Fig. 132). The averaging is performed, using the formulae

⟨dσ

dε

⟩= 1

2π

2π∫0

dσ

dεdϕ,

dσ

dε

ε0ε

Figure 132

when ε > ε0 and

⟨dσ

dε

⟩= 1

2π

π+α∫π−α

dσ

dεdϕ,

when ε < ε0. Here α = arcsin√

ε/ε0.The singularity of the cross-sections (2), (3), and (4), as ε → ε0, is connected with

the fact that the oscillator is excited for any, however large, ρ.What is the case of the additional singularity in (3) and why does it not appear in (4)?

5.16. We have the Lagrangian function

L = 12 mx2 − 1

2 mω2x2 + xF(t).

We then have for the energy of the system

E(t) = 12 m(Reξ)2 + 1

2 m(Imξ)2 − F(t)ω

Imξ = 12 m

∣∣∣∣ξ − iF(t)mω

∣∣∣∣2

− F2(t)2mω2



where

ξ = x + iωx = eiωt

t∫−∞

e−iωτ 1m

F(τ )dτ (1)

(see [1], § 22). Although the expression for the energy has a well-definite limit ast → ∞, the integral defining ξ(t) has no limit as t → ∞ (since F(τ ) → F0 as τ → ∞).Integrating (1) by parts, we obtain

ξ(t) = iF(t)mω

− ieiωt

mω

t∫−∞

F ′(τ )e−iωτ dτ , (2)

where F ′(τ ) → 0, as τ → ∞ and the integral converges as t → ∞. It is clear from (2) thatas t → ∞ the motion of the oscillator in this case is a harmonic oscillation (the secondterm in (2)), around a new equilibrium position x0 = F0/(mω2) (the first term in (2)).The energy transferred to the oscillator is in accordance with this given by

E(+∞) = F20

2mω2 + 12mω2

∣∣∣∣∣∣∞∫

−∞F ′(t)e−iωtdt

∣∣∣∣∣∣2

.

5.17.

E = E(+∞)− E(−∞) = F20

2mω2 + λ4F20

2mω2(λ2 + ω2)2 − aλ2F0 cosϕ

λ2 + om2 ,

E0 = 12mω2a2 and ϕ is the “impact phase” (see footnote to problem 5.12b).

5.18. If in the formula (see [1], § 22)

ξ(τ ) = ξ(0)eiωτ + eiωτ

τ∫0

F(t)m

e−iωtdt

we integrate by parts n times, we get

ξ(τ ) = ξ(0)eiωτ + iF0

mω+ F(n)(+0)eiωτ − F(n)(τ − 0)

m(iω)(n+1)+

+ eiωτ

m(iω)(n+1)

τ∫0

F(n+1)(t)e−iωtdt.



Here |ξ(0)| = a0ω, where a0 is the amplitude of the oscillations at the moment theforce is switched on. The penultimate term in this formula is of order of magnitudeF0mω(ωτ)−n, while the last tem is, generally speaking, considerable smaller – providedF(n+1)(t) changes smoothly. The square of the amplitude of the oscillations

1ω2

∣∣∣∣ξ − iF0

mω

∣∣∣∣2

is for t > τ of the order of magnitude

(a0 + F0

mω2

1(ωτ)n

)2

.

Thus, if the force F(t) is switched on slowly and smoothly, the energy transferred isvery small.

5.19. a) During the time interval 0 � t � τ the oscillations have the form

x = Ftmω2τ

+ Bsinωt + C cosωt.

The oscillations will be stable if

x(τ ) = x(0), x(τ ) = x(0).

These conditions lead to the following set of equations

Fmω2 + Bsinωτ + C(cosωτ − 1) = 0,

B(cosωτ − 1)− C sinωτ = 0,(1)

which determines the constants B and C. Thus we have for 0 � t � τ

x(t) = Fmω2

[tτ

− sin(ωt − ωτ/2)

2sin(ωτ/2)

]. (2)

However, if t lies in the interval nτ � t � (n + 1)τ (where n is an integer), we mustreplaced t on the right-hand side of (2) by t′ = t − nτ (0 � t′ � τ).



When ωτ is close to an integer number times 2π the second term in (2) turns out tobe very large – a case which is close to resonance. When ωτ = 2π l (l an integer), therecan be no stable oscillations – the set (1) is inconsistent.1

b) x(t) = 1ω Im

[iFmω + F

m(λ + iω)e−λt + Aeiωt

], when 0 � t � τ ; here

A = − Fm(λ + iω)

1 − e−λτ

1 − eiωτ;

while for nτ � t � (n + 1)τ we must replace on the right-hand side t by t′ = t − nτ .c) When ω0 = (L C)−1/2 > λ = R/(2L ) the stable current is

I(t) = − 1√ω2

0 − λ2

VL

Im(

eαt

1 − eατ+ 1

ατ

), α = −λ + i

√ω2

0 − λ2 (1)

for 0 � t � τ . When n � tτ � n + 1, we must in the formula for the current replace t by

t′ = t − nτ .Can one use (1) to obtain an expression for the stable current when ω0 < λ or when

ω0 = λ?

5.20. a) A = 1T

∫ T

0F(t)x(t)dt =

= λω2

m

[f 21

(ω2 − ω20)2 + 4λ2ω2

+ 4f 22

(ω20 − 4ω2)2 + 16λ2ω2

],

that is, the two harmonics in the force both transfer energy, independently of one another(the period T = 2π/ω).

b) A = 4λω2

m

∞∑n=1

|an|2n2

(ω20 − n2ω2)2 + 4λ2ω2n2 .

c) 〈A〉 = λm

[f 21 ω2

1(ω2

0 − ω21)2 + 4λ2ω2

1+ f 2

2 ω22

(ω20 − ω2

2)2 + 4λ2ω22

].

1 If we write the force as a Fourier series

F(t) = 12 F −

∞∑l=1

Fπ l

sin(2π lt/τ),

we see that each harmonic term in the force which is acting can cause a resonance build-up. When τ = 2π l/ωwe have for sufficiently large t (how large?)

x(t) ∼ Ft2πmωl

cosωt.



When we average over a long time interval, T � 2π/ω1,2, it turns out that each of thetwo forces f1 cosω1t and f2 cosω2t acts independently on the oscillator. This is becauseonly the squares of trigonometric functions have non-vanishing averages,

1T

T∫0

sin2 ω1t dt = 12

+ 14ω1T

(1 − sin2ω1T) → 12

,

as T → ∞, while the average values of cross-products such as sinω1t · cosω1t,sinω1t · cosω2t, and so on vanish: for instance,

1T

T∫0

sinω1t · cosω2t dt = 1 − cos(ω1 − ω2)t2(ω1 − ω2)T

+ 1 − cos(ω1 + ω2)T2(ω1 + ω2)T

→ 0,

as T → ∞.d) The displacement of the oscillator is

x =∞∫

−∞

ψ(ω)eiωtdω

ω20 − ω2 + 2iλω

,

so that the total work done by the force F(t) is equal to

A =∞∫

−∞x(t)F(t)dt = 8πλ

m

∞∫0

ω2|ψ(ω)|2(ω2

0 − ω2)2 + 4λ2ω2dω. (1)

To prove this equality, one uses the inverse Fourier transform

∞∫−∞

F(t)eiωtdt = 2πψ∗(ω).

When λ � ω0 the main contribution to the integral (1) comes from the vicinity of theeigen-frequency of the oscillator ω = ω0. We have thus

A ≈ 4π |ψ(ω0)|2ω0

m

⎡⎣λ

∞∫0

dω2

(ω20 − ω2)2 + 4λ2ω2

0

⎤⎦ .

The factor inside the square brackets, which can easily be evaluated, is independent ofλ, when λ → 0:



A = |2πψ(ω0)|22m

(cf. [1], (22.12)).

5.21. Taking into account that the displacement of an oscillator x is a small value of thefirst order in F, we get

E =∞∫

−∞F(x, t)x dt ≈

∞∫−∞

f (t)x dt,

P =∞∫

−∞F(x, t)dt ≈

∞∫−∞

[f (t)− f (t)

xV

]dt.

Integrating the second term by parts, we find

P =∞∫

−∞f (t)dt + E

V.

In particular, when

∞∫−∞

f (t)dt = 0, we have E = Vp.

The condition of smallness of x can be demonstrated on the example of the actionof wave groups f (t) = fe−|t|/τ cosγ t on an oscillator. A small parameter in expansion ofF(x, t) is x/λ where λ = 2πV/γ is a characteristic wave length, that is,

|x|λ

= f γ2πm|ω2 − γ 2| � 1.


§6

Small oscillations of systems withseveral degrees of freedom

6.1. Let xi be the displacement of the ith particle from the equilibrium position(i = 1, 2). The Lagrangian function of the system is

L = 12 m

(x2

1 + x22

)− 1

2 k[x21 + (x1 − x2)2]. (1)

The equations of motion

mx1 + k(2x1 − x2) = 0, mx2 + k(x2 − x1) = 0

can be reduced to a set of algebraic equations through the substitution xi = Aicos(ωt + ϕ):

(−mω2 + 2k)A1 − kax2 = 0, −kA1 + (−mω2 + k)A2 = 0. (2)

This set has a non-trivial solution only when its determinant vanishes:

(−mω2 + 2k)(−mω2 + k)− k2 = 0. (3)

From here we get the eigen-frequencies

ω21,2 = 3 ∓ √

52

km

.

Of the two equations (2), only one is independent, because of (3). Substituting thevalues ω1 and ω2 in (2), we obtain the ratios of the amplitudes

A1 = 2√5 + 1

A2 ≡ A for ω = ω1,

A1 = − 2√5 − 1

A2 ≡ B for ω = ω2.




The eigen-vibrations of the system are thus

x1 = Acos(ω1t + ϕ1)+ Bcos(ω2t + ϕ2),

x2 =√

5 + 12

Acos(ω1t + ϕ1)−√

5 − 12

Bcos(ω2t + ϕ2).(4)

The constant A, B, ϕ1, and ϕ2 are determined by the initial conditions.The free oscillations (4) completely describe the motion of the system. However, when

solving various problems, such as, for instance, problems with extra force (see prob-lems 6.2b and 6.25), or when developing the perturbation theory (see problem 6.35), orwhen quantizing the system, it is more convenient to use the normal coordinates. This isbecause the normal coordinates qi, defined by the equalities

x1 = q1 + q2,

x2 =√

5 + 12

q1 −√

5 − 12

q2,(5)

reduce the Lagrangian function (1) to a sum of squares:

L = 5 + √5

4m(q2

1 − ω21q2

1)+ 5 − √5

4m(q2

2 − ω22q2

2), (6)

while the equations of motion become one-dimensional:

qi + ω2i qi = 0, i = 1, 2.

This is similar how we consider the problem of the motion of two interacting particles.The latter can be reduced to the problems of the centre-of-mass motion and the motionof a particle with the reduce mass in the given field of force.

Finally, we note that a more general case of a system of N particles with one suspensionpoint is considered in problem 7.2.

6.2. The Lagrangian function of the system is (cf. the preceding problem)

L = 12 m

(x2

1 + x22

)− 1

2 k[(x1 − a(t))2 + (x1 − x2)2

].

If we discard the term −12ka2(t), which is the total derivative with respect to the time,

the function L can be rewritten as

L = L0 + �L, �L = x1ka(t), (1)


6.2] §6. Small oscillations of systems with several degrees of freedom 193

where L0 is the Lagrangian function of a system with the fixed suspension point (see (1)from the preceding problem). This way of writing the Lagrangian function is moreconvenient as it enables us to immediately write the external force “vector”:

(Fx1

Fx2

)=

(ka(t)

0

).

a) The equations of motion

mx1 + k(2x1 − x2) = kacosγ t,

mx2 + k(x2 − x1) = 0,(2)

can be reduce to a inhomogeneous linear set of two equations to determine A and Bthrough the substitution1

x1 = Acosγ t, x2 = Bcosγ t.

A,B

Akm√

B

ω1

ω1 ω2

ω2γ

γ

|A|

(a)

(b)

Figure 133

Thus, we get

A = ak(−mγ 2 + k)

m2(γ 2 − ω21)(γ 2 − ω2

2),

B = ak2

m2(γ 2 − ω21)(γ 2 − ω2

2),

where ω21,2 = 3 ∓ √

52

km are the eigen-frequencies of the

system. The frequency dependence of the amplitudes Aand B is shown in Fig. 133a.

When the frequency goes through the resonance valuesγ = ω1,2, the amplitude A and B change sign; this corre-sponds to a change of π in the phase of the oscillations.At the frequency γ = √

k/m the oscillations of the upperparticle are completely damped: A = 0.

In Fig. 133b, we have shown qualitatively how |A|depends on the frequency of the applied force when there is friction present.

1 The general solution of the set of equations in (2) is a superposition of free and forced oscillations. As soonas there is even a smallest amount of friction present, the free oscillations are damped, so that after a long timeinterval the solution of (2) is independent of the initial conditions and consist of the forced oscillations (3).



At what frequencies γ will the oscillations of the upper particle be damped, if wesuspend from the lower particle another particle on the same spring?

b) Introducing the normal coordinates q1,2 (see formula (5) from the precedingproblem), we can write the Lagrangian function (1) in the form

L = L1(q1, q1)+ L2(q2, q2),

L1,2 = 5 ± √5

4m(q2

1,2 − ω21,2q2

1,2)+ q1,2ka(t)

(cf. (6) of the preceding problem).The problem is thus reduced to finding the stable oscillations of two independent

harmonic oscillators on each of which a sawtoothed force acts (see problem 5.19a).Certainly, in point 6.2a, it was possible to solve the problem by using normal

coordinates (cf. problem 6.25).

6.3. First, we introduce a system of Cartesian coordinates with the centre in the point ofsuspension and the y-axis directed down along the vertical. As generalized coordinates,we choose the coordinates x1 and x2 of the points A and B. In the expression for thepotential energy U = −mgy1 − mgy2 we substitute y1(x1) and y2(x1, x2) and take intoaccount all terms up to second order of magnitude in x1,2/l:

y1 =√

4l2 − x21 ≈ 2l − x2

1

4l,

y2 = y1 +√

l2 − (x2 − x1)2 ≈ 3l − x21

4l− (x2 − x1)2

2l,

while in the expression for the kinetic energy

T = 12 m(x2

1 + y21 + x2

2 + y22)

we substitute y1 and y2 up to the first order:

y1 = −x1

2lx1 ≈ 0, y2 = y1 − (x2 − x1)

l(x2 − x1) ≈ 0.

After that, the Lagrangian function

L = 12 m(x2

1 + x22)− mg

2lx2

1 − mg2l

(x2 − x1)2 + 5mgl

coincides with the Lagrangian function of the system considered in problem 6.1, if wetake k = mg/l and discard the nonessential constant 5mgl. Therefore, the functions x1(t)and x2(t) found in problem 6.1 are also valid for the double pendulum.

If the pendulum suspension point moves according to the law x0 = a(t) � l, it is easilydemonstate that we return to the Lagrangian function considered in problem 6.2.



6.4. Let ϕ and ψ be the deflection angles of the upper and lower particles from thevertical. The normal oscillations are ψ = 2ϕ with the frequency

√4g/(5l) and ψ = −2ϕ

with the frequency√

4g/(3l).

6.5. The motion is described by the equations

x = acos(ω1t + ϕ), y = bcos(ω2t + ψ).

yb

a−a

−b

x

Figure 134

Constant a, b, ϕ, and ψ are determined by theinitial conditions. The orbit lies inside the rectangle(Fig. 134):

−a � x � a, −b � y � b.

Generally speaking, the orbit “fills” the wholerectangle. More precisely, if the ratio ω1/ω2 is irra-tional, the orbit comes arbitrary close to any pointinside this rectangle. The motion of the point is notperiodic in this case, although the motion of its components along the two coordinateaxis is. However, if the ratio ω1/ω2 is rational (lω1 = nω2 with l and n integers), the orbitis a closed curve, a so-called Lissajous figure. The motion in this case is periodic, andthe period is 2πn/ω1.

y

x

Q1

Q2

ϕ

Figure 135

6.6. a) The transition to normal coordinates isfor this system simply a rotation in the (x, y)-plane(Fig. 135):

x = Q1 cosϕ − Q2 sinϕ, y = Q1 sinϕ + Q2 cosϕ.(1)

Indeed, the kinetic energy does not change its formunder the rotation while the coefficient of Q1Q2 inthe potential energy, which is equal to

−12

(ω2

1 − ω22

)sin2ϕ − α cos2ϕ,

can be made to vanish, if we determine the parameter ϕ from the condition

cot2ϕ = ω22 − ω2

1

2α.

The dependence of ϕ on ω1 is shown in Fig. 136; the region where ϕ changes fromϕ = 0 to ϕ = π/2 has a width of the order of α/ω2 .



When the coupling is weak, α � |ω21 − ω2

2|, the normal oscillations are localized,that is, when ω1 < ω2, we have ϕ ≈ 0 and x ≈ Q1, y ≈ Q2, while for ω1 > ω2, ϕ ≈ π/2,and x ≈ −Q2, y ≈ Q1.

When |ω21 − ω2

2| � α the normal oscillations no longer localized:

ϕ ≈ π

4, x ≈ 1√

2(Q1 − Q2), y ≈ 1√

2(Q1 + Q2)

(see [1], § 23, problem 1).

π

π

ϕ

2

4

ω2ω1

Ω1,2Ω2

Ω1

ω1ω2

ω2


The normal frequencies,

21,2 = 1

2

[ω2

1 + ω22 ∓

√(ω2

2 − ω21)2 + 4α2

], (2)

lie outside the range of partial frequencies1; that is, 1 < ω1 and 2 > ω2 (to fix the ideaswe assume ω1 < ω2). Relations of this kind for systems with many degrees of freedomare known as “Rayleigh theorem” (see [15] and problem 6.24).

Fig. 137 shows how 1,2 depends on ω1. It is clear from this figure that the normalfrequencies 1,2 differ little from the partial frequencies ω1,2 (just as the normalcoordinates Q1 and Q2 differ little from coordinates x and y) when α is small, everywhere,except the degeneracy region, where

|ω21 − ω2

2| � α.

When ω1 becomes sufficiently small, one of the normal frequencies becomes imaginary:the system ceases to be stable.

In terms of the coordinates Q1 and Q2 the law of motion and the orbits are the sameas in the preceding problem.

1 We follow Mandelstam [9] in calling those frequency which are obtained from the original system whenx ≡ 0 (or y ≡ 0) partial frequencies.



b) The normal coordinates can in this case be obtained from the results of thepreceding problem, simply by replaced ω2

1,2 → m1,2 and α → −β, while the normalfrequencies are the reciprocal of the normal frequencies 1,2 of the preceding problem.Why is this the case?

Can one see from the form of the Lagrangian function that the normal frequenciesare independent of the sign α (or β) without finding 1,2 explicitly?

6.7. a) The Lagrangian function of the system is (see problem 4.23)

L = 12

(L1q2

1 + L2q22

)− 1

2

[q2

1

C1+ q2

2

C2+ (q1 + q2)

2

C

],

where q1 and q2 are the charges on the upper plates of the capacitors C1 and C2. Intro-ducing new variables

√L1q1 = x and

√L2q2 = y, we obtain the Lagrangian function of

the problem 6.6a with the parameters

ω21 = 1

L1

(1C

+ 1C1

), ω2

2 = 1L2

(1C

+ 1C2

), α = 1

C√

L1L2.

b) By the change of variables q1 = √C1x and q2 = √

C2y we can change theLagrangian function of the this system to the Lagrangian function of problem 6.6bwith the parameters

m1,2 = (L + L1,2)C1,2, β = L√

C1C2.

Can these systems become unstable?

6.8. Let x1 and x2 be the displacements of the particles m1 and m2 from their equilibriumpositions. By the change of variables

√m1x1 = x and

√m2x2 = y, we obtain for the system

the same Lagrangian function as in problem 6.6a.We can obtain the answer to the problem in various limiting cases without solving it.

For instance, if all ki = k and m1 � m2, we can have a normal vibration with a very lowfrequency 2

1 = 3k/(2m2), x1 = x2/2 (the particle m1 is, so to say, part of the spring,while the particle m2 vibrates between springs of stiffness k/2 on the left and k on theright) and one with a very high frequency 2

2 = 2k/m1 (when the particle m2 is almost atrests). One can find the amplitude of the oscillations of the second particle considering itsmotion as being under the influence of an extra force kx1 with a high frequency (see [1],formula (22.4)): x2 = − m1

2m2x1.

It is of interest to consider in a similar way the cases

a) m1 = m2, k1 = k2 � k3;

b) all stiffnesses are different, but of the same order of magnitude, and m1 � m2;

c) k2 � k1 = k3, and the masses m1 and m2 of the same order of magnitude.



6.9. a) x1,2 = v2ω1

sinω1t ± v2ω2

sinω2t, when k1 � k, the oscillations have the form ofbeats:

x1 = vω

cosεt · sinωt, x2 = − vω

sinεt · cosωt.

b) x1,2 = 12a(cosω1t ± cosω2t); when k1 � k, we have

x1 = acosεt · cosωt, x2 = asinεt · sinωt.

Everywhere in this problem ω21 = k

m and ω22 = 2k1 + k

m , ε = k12kω1, ω = 1

2 (ω1 + ω2).

6.10. The energy transferred from the first to the second particle during the time dt isequal to the work done by the force F = k1(x1 − x2):

dE = k1(x1 − x2)dx2 = k1(x1 − x2)x2 dt,

and the energy flux is dE/dt = k1(x1 − x2)x2. In the limiting case when k1 � k inproblem 6.9a, the flux of energy averaged over the fast oscillations is equal to

12 εmv2 sin2εt.

6.11. The equations of motion

mx1 + k1(x1 − x2)+ kx1 + αx1 = 0,

mx2 + k1(x2 − x1)+ kx2 + αx2 = 0

split into two equations for the normal coordinates, when we use the substitutionx1,2 = (q1 ± q2)/

√2:

q1 + ω21q1 + 2λq1 = 0,

q2 + ω22q2 + 2λq2 = 0,

where ω21 = k/m, ω2

2 = (k + 2k1)/m, 2λ = α/m.We have thus, when λ < ω1,2 (see [1], § 25)

x1,2 = e−λt[acos(γ1t + ϕ1)± bcos(γ2t + ϕ2)],

where γ1,2 =√

ω21,2 − λ2.

The characteristic equation for the system of Fig. 25 is no longer biquadratic whenthere is friction present, but of the fourth degree, and it is therefore considerable morecomplicated to find the eigen-vibrations.



6.12. The Lagrangian function of the double pendulum is

L = 12 Mx2

1 + 12 mx2

2 − Mg2l

x21 − mg

2l

[x2

1 + (x2 − x1)2]

,

where x1 and x2 are the displacements of M and m points from the vertical passingthrough the suspension point (cf. problem 6.3). The replacements

x1 = x√M

, x2 = y√m

reduce the Lagrangian function to the form considered in problem 6.6a, with

ω21 − ω2

2 = 2mgMl

� α = gl

√mM

.

In this case we have

x = 1√2

(Q1 − Q2), y = 1√2

(Q1 + Q2),

where Qi = ai cosi t + bi sini t,

1,2 =√

gl

∓ γ

2, γ =

√mgMl

.

Taking into account the initial conditions Q1,2(0) = lβ√

m/2, Q1,2(0) = 0, we get

x1 = lβ

√mM

sinγ t sin

√glt,

x2 = lβ cosγ t cos

√glt.

Thus, the pendulums oscillate “in turn”, and the amplitude of the upper pendulumis

√M/m times smaller than that of the lower one.

6.13.

x1 = ak(k1 + k − mγ 2)

m2(γ 2 − ω21)(γ 2 − ω2

2)cosγ t,

x2 = akk1

m2(γ 2 − ω21)(γ 2 − ω2

2)cosγ t,

where ω21 = k/m, ω2

2 = (k + 2k1)/m.



6.14. x1 = x2 = akk − mγ 2 cosγ t, where the xi is the displacement along the ring from

the equilibrium positions of ith particle. Resonance is possible only at one of the normalfrequencies when γ 2 = k/m (see problem 6.25).

6.15. Let xi be the displacement of ith particle along the ring from the equilibriumposition and then

x1 = x3 = ak(ω22 − γ 2)

m(γ 2 − ω21)(γ 2 − ω2

3)cosγ t,

x2 = 2ak2

m2(γ 2 − ω21)(γ 2 − ω2

3)cosγ t,

where the eigen-frequencies ωi are

ω21,3 =

(2 ∓ √

2) k

m, ω2

2 = 2km

.

Note that when γ = ω2 we have the displacement x1 = x3 = 0, and x2 = −acosγ t.Why is the number of the resonances in system less than the number of the normal

frequencies?

6.16. The equations of motion are (cf. problem 6.13)

mx1 + αx1 + kx1 + k1(x1 − x2) = kaReeiγ t,

mx2 + αx2 + kx2 + k1(x2 − x1) = 0.

We look for solution of these equations in the form

x1 = Re(Aeiγ t

), x2 = Re

(BeIγ t

).

For A and B we get

(−mγ 2 + 2imλγ + k + k1)A − k1B = ka,

−k1A + (−mγ 2 + 2imλγ + k + k1)B = 0



with 2mλ = α, whence

A = ak(k + k1 − mγ 2 + 2iλmγ )

m2(γ 2 − 2iλγ − ω21)(γ 2 − 2iλγ − ω2

2),

B = akk1

m2(γ 2 − 2iλγ − ω21)(γ 2 − 2iλγ − ω2

2),

x1 =ak

√(γ 2 − 1

2ω21 − 1

2ω22

)2 + 4λ2γ 2 cos(γ t + ϕ1 + ϕ2 + ψ)

m√

[(γ 2 − ω21)2 + 4λ2γ 2][(γ 2 − ω2

2)2 + 4λ2γ 2],

x2 = akk1 cos(γ t + ϕ1 + ϕ2)

m2√

[(γ 2 − ω21)2 + 4λ2γ 2][(γ 2 − ω2

2)2 + 4λ2γ 2],

ω21 = k

m, ω2

2 = k + 2k1

m, tanϕ1,2 = 2λγ

γ 2 − ω21,2

, tanψ = 4λγ

ω21 + ω2

2 − 2γ 2.

There is a phase difference ψ between the oscillations of the two particles; theoscillations of the first particle are never completely damped. As function of a frequencyof the applied force, γ , the amplitudes of the oscillations have either one or two maxima,depending on the ratio of the parameters ω1, ω2, and λ.

6.17. If x and y are the displacements from the equilibrium position of the first and thesecond particle, respectively, we have for the Lagrangian function of the system

L = 12 m

(x2 + y2 − k1 + k2

mx2 − k2 + k3

my2 + 2k2

mxy

)+ k1axcosγ t;

y

Q1

Q2

F

F1

F2

xϕ

Figure 138

it differs from the Lagrangian function consid-ered in problem 6.6a only in the term xk1acosγ t,corresponding to a force k1acosγ t acting uponthe first particle. We shall use here the notationsof problem 6.6a. The partial frequency ω1,2 =√

(k1,3 + k2)/m corresponds to the eigen-frequencyof a system which we obtain by fixing the second(first) particle, that is, by putting y = 0 (respec-tively, x = 0). When we change to the normal coor-dinates Q1 and Q2, the Lagrangian function becomes

L = 12 m

(Q2

1 − 21Q2

1 + Q22 − 2

2Q22

)+ (F1Q1 + F2Q2)cosγ t,



where F1 = k1acosϕ and F2 = −k2asinϕ are the components of the amplitude of theforce F = k1a onto the normal coordinates Q1 and Q2 (Fig. 138). For the coordi-nates Q1,2, we get the equations of motion of the harmonic oscillators with frequen-cies 1,2 under the action of the applied forces F1,2 cosγ t. The initial conditions areQi(0) = Qi(0) = 0. We get

Q1,2 = F1,2(cosγ t − cos1,2t)

m(21,2 − γ 2)

.

It is interesting to consider resonance at the second eigen-frequency for the systemin the weak coupling approximation,

k2

k3 − k1≡ ε � 1

(to fix the ideas, we have assume that k1 < k3). If we put γ = 2(1 + ε1), we have

Q1 = k1ak1 − k3

(cosω2t − cosω1t),

Q2 = − k1aε

mω22ε1

sin(ε1

ω2

2t)

sinω2t when |ε1| � 1,

Q2 = − k1a2mω2

εt sinω2t when ε1 = 0.

We see thus that for even the smallest coupling the amplitude Q2 can become large oreven steadily increases with time, although the rate of change will then be small. Sincethe angle of rotation is small (sinϕ = ε), the displacements are x = Q1 − εQ2 and y = Q2.

What is the rate of change of the amplitude of the vibrations for resonance near thefirst frequency γ = 1?

How will the character of the vibrations change if a small friction force acts on bothparticles, proportional to the velocity (cf. problem 5.11)?

6.18.

a) x = F0 cosϕ

m(ω21 − γ 2)

cosγ t, y = F0 sinϕ

m(ω22 − γ 2)

cosγ t,

where ω21 = 2k1/m, ω2

2 = 2k/m, ϕ is the angle between the vector F0 and the AB-axisand x and y are the displacements from the equilibrium position along AB and CD. Theparticle oscillates along a strait line through the centre.

It is interesting that when γ 2 = ω21 sin2 ϕ + ω2

2 cos2 ϕ, this strait line is at right angles tothe vector F0. In this case the work done by the acting force is equal to zero. Therefore, itseems that even the smallest amount of friction must lead to a damping of the oscillations.Explain this situation.



b) x = Fm(ω2

1 − γ 2)cosγ t, y = F

m(ω22 − γ 2)

sinγ t.

The trajectory is an ellipse with semi-axes

a = F

m|ω21 − γ 2| , b = F

m|ω22 − γ 2| .

If the quantities (ω21 − γ 2) and (ω2

2 − γ 2) have opposite signs, the particle movesclockwise along the ellipse, while the force vector rotates counterclockwise.

How does the picture given here of the motion of the particle change when the tensionin the springs in the equilibrium position differs from zero?

6.19. Let xi be the displacement of the ith particle along the ring from the equilibriumposition. The three particles can rotate on the ring with a constant angular velocity:

x1 = x2 = x3 = Ct + C1 = q1(t), ω1 = 0. (1)

The oscillations where the particles 1 and 2 move towards one another with equalamplitude,

x1 = −x2 = Acos(ω2t + α) = q2(t), x3 = 0, ω2 = √3k/m, (2)

have the same frequency as the oscillations where the particles 2 and 3 move towardsone another:

x1 = 0, x2 = −x3 = Bcos(ω3t + β) = q3(t), ω3 = ω2. (3)

We introduce the “displacement vector”

r =⎛⎝x1

x2x3

⎞⎠ ,

and we can then write the oscillations (1)–(3) as three vectors (see Fig. 139),

r1 =⎛⎝1

11

⎞⎠q1, r2 =

⎛⎝ 1

−10

⎞⎠q2, r3 =

⎛⎝ 0

1−1

⎞⎠q3.

Any linear superposition of the r2 and r3 vectors is again an oscillation with thefrequency ω2. Thus, in a space with Cartesian coordinates x1, x2, and x3, a set of solutionsthat represent the oscillations with twice degenerate frequency ω2 = ω3 defines a plane



r1 r2 r3 r3′

Figure 139

which passes through the vectors r2 and r3.1 As easy to see from (4), both of thesevectors—and, hence, all vectors lying in this plane—are orthogonal to the vector r1 (thegeneral orthogonality condition seen in problem 6.23).

The Lagrangian function of the system is

L = 12 m

(x2

1 + x22 + x2

3

)− 1

2 k[(x1 − x2)2 + (x2 − x3)2 + (x3 − x1)

2]

. (5)

The normal coordinates must simultaneously diagonalize two quadratic forms—thekinetic energy and the potential energy. The kinetic energy in (5) is already proportionalto the sum of the squares of the velocities. Therefore, the transformation from xito the normal coordinates which does not changing its form must be an orthogonaltransformation. Hence, the vectors of the corresponding normal oscillations must bemutually orthogonal. The vectors ri are independent, but not mutually orthogonal:r1r2 = r1r3 = 0, but r2r3 = 0. To obtain the normal coordinates, we must choose twomutually orthogonal vectors in the plane of vectors r2 and r3. These can be, for example,the vector r2 and the orthogonal vector eq3 (the unit vector e has been found from thecondition er1 = er2 = 0). As a result, a set of normalized vectors2

r′1 = r1√

3, r′

2 = r2√2

, r′3 = eq3 = 1√

6

⎛⎝ 1

1−2

⎞⎠q3 (6)

allow us to define the normal coordinates:

x1 = 1√3

q1 + 1√2

q2 + 1√6

q3,

x2 = 1√3

q1 − 1√2

q2 + 1√6

q3,

x3 = 1√3

q1 − 2√6

q3,

(7)

1 Note that the linear combination in this plane of the form αr1(t) +βr2(t) is an oscillation either along astrait line (when α = β, β +π) or along an ellipse (when α = β).

2 The factors 1/√

3 and 1/√

2 are introduced so that the vectors ri are normalized by the condition rirk =δikq2

i . Under this condition, the transformation (7) is orthogonal.



which reduce the Lagrangian function (5) to the form

L = 12 m(q2

1 + q22 − ω2

2q22 + q2

3 − ω23q2

3). (8)

Of course, any coordinates obtained from q2, q3 by an orthogonal transformation (i.e.,by a simply rotation around r1) are also the normal coordinates.

6.20. The initial conditions for the displacements xi along the ring are

x1(0) = a, x2(0) = x3(0) = xi(0) = 0.

We have thus for the normal coordinates qi (see formula (7) of the preceding problem)the initial conditions:

q1(0) = a√3

, q2(0) = a√2

, q3(0) = a√6

, qi(0) = 0.

Therefore, we find

q1 = a√3

, q2 = a√2

cosω2t, q3 = a√6

cosω3t,

and taking into account that ω2 = ω3, we get finally

x1 = a3

+ 2a3

cosω2t, x2 = x3 = a3

− a3

cosω2t.

6.21. We use the notations of problem 6.19. The Lagrangian function of the system is

L = 12 m

(x2

1 + 2x22 + 3x2

3

)− 1

2 k[2(x1 − x2)2 + 6(x2 − x3)2 + 3(x3 − x1)2

]. (1)

The equations of motion are reduced to the system of three algebraic equations bysubstitution xi = Ai cos(ωt + ϕ):

(−mω2 + 5k)A1 − 2kA2 − 3kA3 = 0,

−2kA1 + (−2mω2 + 8k)A2 − 6kA3 = 0,

−3kA1 − 6kA2 + (−3mω2 + 9k)A3 = 0.

(2)

This system has a non-trivial solution when its determinant is equal to zero:

ω2(mω2 − 6k)2 = 0.



From here we find the eigen-frequencies of the system:

ω1 = 0, ω2 = ω3 = √6k/m.

The value of ω1 = 0 corresponds to the evident solution—the rotation along the ringwith the constant angular velocity

r1 =⎛⎝1

11

⎞⎠q1, q1(t) = Ct + C1. (3)

For matching frequencies ω2 = ω3, only one equation in set (2) is independent:

A1 + 2A2 + 3A3 = 0. (4)

Any set of values of Ai that satisfies the condition (4) gives us the oscillations with thefrequency ω2. In particular, it is possible to choose such oscillations where either the first,second, or the third particle is at rest:

r2 =⎛⎝ 0

3−2

⎞⎠q2, r3 =

⎛⎝ 3

0−1

⎞⎠q3, r4 =

⎛⎝ 2

−10

⎞⎠q4. (5)

qi = Ci cos(ω2t + ϕi), i = 2, 3, 4.

According to (4), any linear combination of vectors (5) is orthogonal to the vector

⎛⎝1

23

⎞⎠ .

It is easy to verify that the set of vectors

r1, r2, r′3 =

⎛⎝ 5

−1−1

⎞⎠q3 (6)

allows us, as in problem 6.19, to determine the normal coordinates which lead theLagrangian function (1) to the diagonal form. The vectors in (6) satisfy not the ordinarycondition of orthogonality (as in problem 6.19), but the condition of the “weightorthogonality” (see problem 6.23).



6.22. The vectors of the normal oscillations are

r1 = 1√2

⎛⎜⎜⎝

10

−10

⎞⎟⎟⎠q1, r2 = 1√

2

⎛⎜⎜⎝

010

−1

⎞⎟⎟⎠q2,

r3 = 12

⎛⎜⎜⎝

1−1

1−1

⎞⎟⎟⎠q3, r4 = 1

2

⎛⎜⎜⎝

1111

⎞⎟⎟⎠q4,

(1)

ql = Al cos(ωl t + ϕl), l = 1, 2, 3; q4 = A4t + A5,

ω1 = ω2 = √2k/m, ω3 = 2

√k/m.

The Lagrangian function of the system reads

L = 12 m

(q2

1 + q22 + q2

3 + q24 − ω2

1q21 − ω2

2q22 − ω2

3q23

).

This is, of course, not the only possible choice. Any vectors obtained from the onesgiven here through a rotation in the plane determined by the vectors r1 and r2 will alsobe vectors of the normal oscillations. For instance,

r′1 = 1

2

⎛⎜⎜⎝

11

−1−1

⎞⎟⎟⎠q′

1, r′2 = 1

2

⎛⎜⎜⎝

1−1−1

1

⎞⎟⎟⎠q′

2, r′3 = r3, r′

4 = r4 (2)

(rotation over π/4). However, the vectors r1, r′2, r3, r4 will not reduced the Lagrangian

function to a sum of squares, although they also are independent.

6.23. The amplitude of the normal oscillations satisfy the equations

−ω2l

∑j

mijA(l)j +

∑j

kijA(l)j = 0, (1)

−ω2s

∑j

mijA(s)j +

∑j

kijA(s)j = 0. (2)



Multiply (1) by A(s)i and (2) by A(l)

i and summing both equations over i, we get insteadof (1) and (2)

−ω2l

∑ij

mijA(l)j A(s)

i +∑

ij

kijA(l)j A(s)

i = 0, (3)

−ω2s

∑ij

mijA(s)j A(l)

i +∑

ij

kijA(s)j A(l)

i = 0. (4)

Subtracting (4) from (3) and using the fact that mij=mji and kij=kji, we obtain

(ω2s − ω2

l )∑

ij

mijA(s)i A(l)

j = 0;

that is, when ωs = ωl ,

∑ij

mijA(s)i A(l)

j = 0, (5)

and at the same time from (3) we obtain

∑ij

kijA(s)i A(l)

j = 0. (6)

It is helpful to use the terminology from linear algebra. We shall call the set ofamplitudes of a given oscillation the amplitude vector A(l) = (A(l)

1 , A(l)2 , . . . , A(l)

n ). Therelations (5) and (6) which we have just prove mean that the amplitudes A(s) and A(l)

are mutually orthogonal, provided that the scalar product is defined by means of themetric tensors mij or kij .

In the case of degeneracy (if ωs = ωl), the amplitude A(s) and A(l) do not necessarysatisfy (5) and (6). However, one can in that case always choose—indeed, in several was—such amplitudes that they satisfy (5) and (6) and also reduce the Lagrangian functionto a sum of squares.

6.24. Introducing the normal coordinates

xi =∑

l

A(l)i ql ,

we transform the constraint equations to the form

∑l

blql = 0, bl =∑

i

aiA(l)i .



Equations of motion with an indefinite Lagrangian multiplier λ

Ml(ql + 2l ql) = blλ

can be solved by putting

ql = Cl cos(ωt + ϕ), λ = �cos(ωt + ϕ).

Expressing Cl from

Ml(2l − ω2)Cl = bl�,

and substituting it into the constraint equations, we obtain for new frequencies

�∑

l

b2l

Ml(2l − ω2)

= 0.

y

ω12

Ω12 Ω2

2 Ω32

ω22

ω2

Figure 140

To study this equation, it is helpful to use the graph(Fig. 140)

y(ω2) =∑

l

b2l

Ml(2l − ω2)

= 0.

Note that the function y(ω2) changes the sign bypassing through the infinite value when ω2 = 2

l .After that, the location of the roots ωl becomes quite clear. If any of the coefficients bl iszero, then the corresponding normal oscillation (and its frequency) do not change whenone applies a constraint.

The fact considered in this problem allows for a simple geometric interpretation(see [8], § 24).

6.25. Substituting xj = ∑l λ

(l)a(l)j cosγ t into the equations of motion,

∑j

mij xj +∑

j

kijxj = fi cosγ t,

we obtain the following equations to determine the coefficients λ(l):

−γ 2∑I,j

mijλ(l)A(l)

j +∑j,l

kijλ(l)A(l)

j = fi. (2)



The simplest way to solve these equations is by using the orthogonality relations (5) and(6) of problem 6.23.

To do this, we multiply the equations in (2) by A(s)i and sum over i; then we get

λ(s) = Fs

Ms(ω2s − γ 2)

,

where

Fs =∑

i

A(s)i fi, Ms =

∑i, j

mijA(s)i A(s)

j , Ks =∑i, j

kijA(s)i A(s)

j .

The quantity ωs = √Ks/Ms is the sth eigen-frequency of the system, in accordance

with (4) of problem 6.23.1 The γ -dependence of the λ(s) shows resonances.For the normal coordinates qs, introduced by the the equations

xi =∑

s

A(s)i qs(t), (3)

instead of the equations (1), we have the following equations of motion:

Msqs + Ksqs = Fs cosγ t. (4)

Hence, if the force vector fi is orthogonal to the amplitude vector of the sth normaloscillation,

∑i

A(s)i fi = 0,

the corresponding normal coordinate satisfies the equation for free oscillations, and thereis no resonance for ω = ωs. Note that the work done by the applied force is in this caseequal to zero:

∑i

fi dxi =∑

i

fiA(s)i dqs = 0.

1 If some of the eigen-frequencies are degenerate, we shall assume that the amplitudes of the eigen-oscillations corresponding to them are chosen in such a way that they satisfy the orthogonality relations (5)and (6) of problem 6.23.



Consider the case where the force vector is parallel to the amplitude vector of one ofthe normal oscillation,

fi

A(s)i

= const (i = 1, 2, . . . , N).

Can such a force excite other normal vibrations?

6.26. The stable oscillations can be presented in the form (see preceding problem)

xi =∑

j

βij fj ,

where

βij =∑

l

A(l)i A(l)

j

Ml(ω2l − γ 2)

.

The reciprocity theorem reflects that βij = βji.How will the formulation of this theorem change if the coordinates xi and xj have

different dimensions (e.g. for the case of an electromechanical system)?

6.27. The normal oscillations are

⎛⎜⎜⎝

1111

⎞⎟⎟⎠q1,

⎛⎜⎜⎝

10

−10

⎞⎟⎟⎠q2,

⎛⎜⎜⎝

010

−1

⎞⎟⎟⎠q3,

⎛⎜⎜⎝

1−m/M

1−m/M

⎞⎟⎟⎠q4,

where

q1 = At + B, qi = Ai cos(ωi t + αi), i = 2, 3, 4;

ω22 = 2k

m, ω2

3 = 2kM

, ω24 = 2k(M + m)

mM.

The first three oscillations are easily guessed, and the last one can be obtained using theorthogonality conditions to the first three. Since the particle masses are different, theorthogonality condition of the two normal oscillations with the amplitudes A and B havethe form

mA1B1 + MA2B2 + mA3B3 + MA4B4 = 0

(see problem 6.23).



6.28. Let xi be the displacement of the ith particle along the ring. We can easily guesstwo normal vibrations:

r1 =

⎛⎜⎜⎝

1111

⎞⎟⎟⎠q1(t), r2 =

⎛⎜⎜⎝

10

−10

⎞⎟⎟⎠q2(t), (1)

q1(t) = C1t + C2, q2(t) = A2 cos(ω2t + ϕ2), ω2 = √2k/m.

The two other vectors must be orthogonal to the vectors (1) in the metric determinedby the coefficients of the quadratic form of the kinetic energy (see problem 6.23); thatis, they must have the form

r =

⎛⎜⎜⎝

aba

−a − 12b

⎞⎟⎟⎠q(t). (2)

Substituting (2) into the equations of motion for the first and the second particles,

mx1 + k(2x1 − x4 − x2) = 0, mx2 + k(2x2 − x1 − x3) = 0,

we get two equations to determine the values of a, b and frequencies:

(−mω2 + 3k)a − 12 kb = 0,

−2ka + (−mω2 + 2k)b = 0.(3)

Solving (3), we find

ω23,4 = 5 ∓ √

52

km

, b3,4 = (1 ± √5)a3,4

or

r3,4 =

⎛⎜⎜⎝

11 ± √

51

−12 (3 ± √

5)

⎞⎟⎟⎠q3,4(t),

q3,4 = A3,4 cos(ω3,4t + ϕ3,4).



6.29. a) Let xi, yi, and zi be the displacements of the ith particle from its equilibriumposition. The Lagrangian function of the system has the form (see problem 5.7)

L = L1(x, x)+ L1(y, y)+ L1(z, z),

L1(x, x) = 12 m

(x2

1 + x22 + x2

3 + x24 + x2

5

)− 1

2 k[x2

1 + (x1 − x5)2++ (x5 − x3)

2 + x23 + x2

2 + (x2 − x5)2 + (x5 − x4)2 + x2

4

];

therefore, the oscillations in the x-, y-, and z-directions proceed independently. We caneasily guess three of the normal oscillations in the x-direction:

r1 =

⎛⎜⎜⎜⎜⎝

10

−100

⎞⎟⎟⎟⎟⎠q1, r2 =

⎛⎜⎜⎜⎜⎝

010

−10

⎞⎟⎟⎟⎟⎠q2, r3 =

⎛⎜⎜⎜⎜⎝

1−1

1−1

0

⎞⎟⎟⎟⎟⎠q3, (1)

qi = Ai cos(ωi t + ϕi), ω1 = ω2 = ω3 = √2k/m.

The other two normal oscillations must be orthogonal to the vectors (1) and thereforehave the form1

r4,5 =

⎛⎜⎜⎜⎜⎝

aaaad

⎞⎟⎟⎟⎟⎠q4,5.

Substituting this vector into the equations of motion for the first and fifth particle,

mx1 + k(2x1 − x5) = 0,

mx5 + k(4x5 − x1 − x2 − x3 − x4) = 0,

we obtain two equations to determine the unknown parameters a, d, and the frequenciesω4,5:

(−ω2m + 2k)a − kd = 0,

−4ka + (−mω2 + 4k)d = 0.(3)

1 Let r4,5 = (a, b, c, e, d) ≡ r; then the orthogonality conditions (r, r1) = (r, r2) = (r, r3) = 0 lead to theequations a = b = c = e.



Solving (3), we find ω4,5 =√

(3 ∓ √5)k/m and d4,5 = (−1 ± √

5)a4,5, and finally

r4,5 =

⎛⎜⎜⎜⎜⎝

1111

−1 ± √5

⎞⎟⎟⎟⎟⎠q4,5. (4)

The results for oscillations along the y- and z-axes are the same as for those alongthe x-axis. Thus, there are altogether three different frequencies in the system: ω1 =√

2k/m which is ninefold degenerate and two frequencies ω4,5 =√

(3 ∓ √5)k/m which

are threefold degenerate (see problem 6.42 about the lifting of the degeneracy).b) The oscillations along the z-axis can easily be guessed:

r1 =

⎛⎜⎜⎝

10

−10

⎞⎟⎟⎠q1, r2 =

⎛⎜⎜⎝

010

−1

⎞⎟⎟⎠q2, r3,4 =

⎛⎜⎜⎝

1∓1

1∓1

⎞⎟⎟⎠q3,4,

qi = Ai cos(ωi t + ϕi), ω1 = ω2 = ω3 =√

2fml

, ω4 =√

fml

,

where f is the tension in the springs, and l the length of one of the spring at equilibrium.If f = kl, then the oscillations in the x- or y-directions will have the same form as in

the z-direction if we put r = (x1, x2, x3, x4) (or r = (y2, y1, y4, y3)). If, however, f = kl,the degeneration is lifted. Two of the normal oscillations with frequencies ω1 = √

2k/mand ω2 = √

2f /(ml) are the same as r1 and r2. The two other must have the form

⎛⎜⎜⎝

abab

⎞⎟⎟⎠cos(ωt + ϕ)

because of the orthogonality condition. To find them, it is sufficient to consider theequations of motion of two particles:

mx1 + k(2x1 − x5) = 0, mx2 + fl(2x2 − x5) = 0.

Here

x5 = k(x1 + x3)+ (f /l)(x2 + x4)

2k + 2f /l



is the coordinate of the point where the springs are joined together which is determinedfrom the condition that the potential energy be a minimum for given values of x1,2,3,4.

Solving these equations, we obtain

ω23 = f + kl

ml, b3 = − f

kla3;

ω24 = 2kf

m(f + kl), b4 = kl

fa4.

6.30. In this case, the result can be obtained by a simple generalization of the results ofthe problem 6.21 without an explicit calculation of the eigen-frequencies ωi.

Let ω1 = 0 and the degeneracy ω2 = ω3 be in the system. The frequency ω1 = 0corresponds to the rotation of the particles along the ring

r1 =⎛⎝1

11

⎞⎠(Ct + C1).

Due to the degeneracy of the frequency ω2 any vector

r1 =⎛⎝A1

A2A3

⎞⎠cos(ωt + ϕ), (1)

which satisfies the condition

m1A1 + m2A2 + m3A3 = 0, (2)

is a normal oscillation with the frequency ω = ω2. (The equality (2) is the orthogonalitycondition for the vector r1 in the metric defined by the coefficients of the quadratic formof the kinetic energy—see problem 6.23). In particular, one can chose such a normaloscillation (1) that the first particle will be at rest:

A1 = 0, m2A2 + m3A3 = 0. (3)

If we substitute (3) in the equations of motion

(m1ω2 − k2 − k3)A1 + k3A2 + k2A3 = 0,

k3A1 + (m2ω2 − k1 − k3)A2 + k1A3 = 0,

k2A1 + k1A2 + (m3ω2 − k1 − k2)A3 = 0,



we see that they can have a non-zero solution only when

k3A2 + k2A3 = 0. (4)

Comparing (3) and (4), we find that m2k2 = m3k3. Repeating the similar reasoning forthe cases where the second or the third particle is at rest, we find that the coefficients kimust satisfy the condition

m1k1 = m2k2 = m3k3 (5)

for the case of the frequency degeneracy. On the other hand, it is clear from the previousconsideration that (5) is a sufficient condition for the degeneracy of the frequencies.Indeed, if the condition (5) is satisfied, then there are three different normal oscillationswith the frequencies other than zero. Of these three oscillations only two are linearlyindependent due to (2). From here it follows that these three oscillations have the samefrequency.

Thus, (5) is the necessary and sufficient condition for the degeneracy of thefrequencies.

6.31. It is helpful to use for the solution the method explained in problem 6.28.a) The normal oscillations are

r1 =⎛⎝1

11

⎞⎠(C1t + C2), r2 =

⎛⎝ 1

−10

⎞⎠A2 cos(ω2t + ϕ2),

r3 =⎛⎝ 1

1−2

⎞⎠A3 cos(ω3t + ϕ3),

ω22 = (3 + 2ε)k

m, ω2

3 = 3km

, ε = δkk

;

(1)

for small ε, they are close to the oscillations (6) of problem 6.19: the amplitudesof the oscillation are the same, but all frequencies are different. Therefore, althoughin problem 6.19 any superposition of the vectors r′

2 and r′3 again gave normal oscillations,

now the choice of the vectors r2 and r3 is completely unambiguous.b) The normal oscillations

r1 =⎛⎝1

11

⎞⎠(C1t + C2), r2 =

⎛⎝ 1

−10

⎞⎠A2 cos(ω2t + ϕ2), (2)



r3 =⎛⎜⎝

11

− 21 + ε

⎞⎟⎠A3 cos(ω3t + ϕ3),

ω22 = 3k

m, ω2

3 = 3 + ε

1 + ε

km

, ε = δmm

(2)

are for small ε close to the oscillations (6) of problem 6.19. If the extra mass had beenadded to particle 2, the normal oscillations

r1 =⎛⎝1

11

⎞⎠(C1t + C2), r2 =

⎛⎝ 1

0−1

⎞⎠a2 cos(ω2t + ϕ2),

r3 =⎛⎜⎝

1− 2

1 + ε1

⎞⎟⎠A3 cos(ω3t + ϕ3),

would be close to the superposition of the normal oscillations (6) of problem 6.19.

c) r1 =⎛⎝1

11

⎞⎠(C1t + C2),

r2,3 =⎧⎨⎩a2,3

⎛⎝ 1

0−1 − ε1

⎞⎠+ b2,3

⎛⎝ 0

1−1 − ε2

⎞⎠⎫⎬⎭cos(ω2,3t + ϕ2,3),

where

b2,3

a2,3= 1

ε2

(ε2 − ε1 ±

√ε2

1 + ε22 − ε1ε2

),

ω22,3 ≈ k

m

(3 − ε1 − ε2 ∓

√ε2

1 + ε22 − ε1ε2

), εi = δmi

m.

6.32. a) We expand the initial displacement

r(0) =⎛⎝ a

0−a

⎞⎠

in terms of the vectors ri (see (1) of the preceding problem) taken at t = 0:

r(0) = r1(0)+ r2(0)+ r3(0). (1)



We do the same for the initial velocities:

r(0) = r1(0)+ r2(0)+ r3(0). (2)

From the set of equations in (1) and (2), we find for the constants the following values:A2 = A3 = a/2, C1 = C2 = ϕ2 = ϕ3 = 0, or

r = a2

⎛⎝ cosω2t + cosω3t

−cosω2t + cosω3t−2cosω3t

⎞⎠ ≈ a

⎛⎝cos(εω3t/6)cosω3t

sin(εω3t/6)sinω3t−cosω3t

⎞⎠ .

The motion of the particles 1 and 2 thus shows beats with a frequency which isdetermined by the perturbation δk, while the particle 3 performs a simple oscillationwith frequency ω3. We emphasize that even a very small perturbation δk leads to secularchanges which become appreciable after sufficiently long times (cf. problem 2.40).

6.33. a), b)

r1 =

⎛⎜⎜⎝

1−1−1

1

⎞⎟⎟⎠q1(t), r2 =

⎛⎜⎜⎝

11

−1−1

⎞⎟⎟⎠q2(t), r3,4 =

⎛⎜⎜⎝

1∓1

1∓1

⎞⎟⎟⎠q3,4(t);

c) the same as in problem 6.22, (1).

6.34. x1,2 = −x3,4 = ±12 acos

√(2k + 2δk)/m t + 1

2 acos√

2k/m t;

the oscillations of the particles show beats (cf. problem 6.32).

6.35. We can expect that the changes of the frequencies and of the vectors of the normaloscillations will be small, so we use the method of successive approximations. It is helpfulto use the normal coordinates of the original system (see problem 6.25)

xi =∑

l

A(l)i ql .

In this case, δL takes the form

δL = 12

∑l,s

(δMlsql qs − δKlsqlqs), (1)

where

δMls =∑i, j

δmijA(l)i A(s)

j , δKls =∑i, j

δkijA(l)i A(s)

j ; (2)



the equations of motion are

Ml(ql + ω2l ql) = −

∑s

(δMlsqs + δKlsqs). (3)

Assuming that only the oscillation qn is excited in zero approximation, we can leave inthe right-hand side of (3) only the term s = n.

To determine the correction to the frequency ωn, it is enough to write the singleequation (with l = n):

(Mn + δMnn)qn + (Mnnω2n + δKnn)qn = 0,

where

(ωn + δωn)2 = Mnnω

2n + δKnn

Mn + δMnn,

as a result,

δωn = δKnn

2ωnMn− ωn δMnn

2Mn. (4)

The equations with l = n allow us to find corrections to the vector of the normaloscillation. In this case, the right-hand sides of the equations can be considered as thegiven forces with the frequency ωn. The excitation of the oscillations ql , as we haveexpected, is weak, as these “forces” are small.

We can also get the following approximation for the ωn and the vectors of the normaloscillations (e.g. see [14], Ch. 1, § 5).

It is worth noting that the value δMnn in (2) represents a supplement to the doubledkinetic energy of the system provided that the velocities xi = A(n)

i . From this, it follows, inparticular, that with increasing of particles’ masses, the quantity δMnn � 0 and, accordingto (4), δωn � 0. In the similar way, it is easy to see that if one increases the stiffness ofthe springs, the eigen-frequencies can only increase (see [8], § 24 and [15]).

It is important to understand what will change when we look for a correction to thedegenerate frequency (let ωp = ωn). In this case, the “force” in the right-hand sides ofthe equations (3) becomes resonant. Therefore, the coordinate qp(t) increases with time,and it must be taken into account in the right-hand sides of the equations (3). Therefore,in this case, one must use the equations (3) together with l = n, p, leaving in the right-hand sides only the terms with s = n, p

Mn(qn + ω2nqn) = −δMnnqn − δMnpqp − δKnnqn − δKnpqp,

Mp(qp + ω2pqp) = −δMpnqn − δMppqp − δKpnqn − δKppqp.

(5)



It is clear that this can be applied to the case ωp ≈ ωn as well.Thus, to determine the corrections to all the eigen-frequencies, including the degen-

erate ones, one can drop all terms in the �L (see (1)) which contain products of thenormal coordinates related to the different frequencies of the original system.

6.36. We use the notions and the results of problems 6.28 and 6.35. It is obvious thatδω1 = 0. For other frequencies, we have the correction

δωn = −ωn

2

∑i, j

(rn)iδmij(rn)j

∑i, j

(rn)imij(rn)j

, n = 2, 3, 4. (1)

The matrix mij of the kinetic energy is diagonal; moreover,

m11 = m22 = m33 = m, m44 = 2m. (2)

The matrix δmij has a single non-zero element;

δm11 = �m. (3)

Substituting expressions (2) and (3) into (1), as well as the components of the vectorsof the normal oscillations rn, which have been found in problem 6.28, we obtain

δω2 = −14 εω2, δω3,4 = −3 ∓ √

540

εω3,4.

6.37. We choose the vector potential to be

A(r) = 12 B(−y, x, 0);

the Lagrangian function is then

L = 12 m

(x2 + y2 + z2

)− 1

2 m(ω2

1x2 + ω22y2 + ω2

3z2)

+ 12 mωB(xy − yx),

where ωB = eBmc . For x and y, we get

x + ω21x − ωBy = 0,

y + ω22y + ωBx = 0.

It is helpful to look for oscillations in the form

x = Re(Aeit), y = Re(Beit).



The set of equations

(ω21 − 2)A − iωBB = 0,

iωBA + (ω22 − 2)B = 0

leads to the oscillations

x = Re(Akeikt

)= ak cos(kt + ϕk),

y = Re

(Ak

−iωBk

ω22 − 2

k

eikt

)= ak

ωBk

ω22 − 2

k

sin(kt + ϕk),

Ak = akeiϕk , k = 1, 2

with frequency

21,2 = 1

2

[ω2

1 + ω22 + ω2

B ±√(

ω21 + ω2

2 + ω2B

)2 − 4ω21ω2

2

],

for which

12 = ω1ω2

is true.To fix our ideas, let ω1 > ω2, ωB > 0. The first of the oscillations which we have found

is then a clockwise the motion along an ellipse with its major axis along the x-axis, whilethe second oscillation is a counterclockwise motion along an ellipse with its major axisalong the y-axis.

The motion along the z-axis turns out to be a harmonic motion which is independentof the magnetic field,

z = a3 cos(ω3t + ϕ3).

The free motion of the oscillator is a superposition of the oscillations obtained. We callthese oscillations normal oscilliations, just generalizing the concept of normal oscillations:the motion in the x- and y-directions occur with the same frequency, but with a shift inphase. It is impossible to reduce the Lagrangian function to diagonal form using onlya linear transformation of the coordinates as the transition to normal coordinates in thepresent case is connected with a canonical transformation (see problem 11.8–11.10).

a) If the magnetic field is weak, ωB � ω1 − ω2, the ellipses of the normal oscillationsare strongly elongated, and the frequencies

1,2 ≈ ω1,2 ± ω2Bω1,2

2(ω21 − ω2

2)



are close to ω1,2. The orbit of the oscillator without a magnetic field fills a rectangle withsides parallel to the coordinate axes (see problem 6.5); the influence of a weak magneticfield is merely to slightly deform the region filled by the orbit. The Larmor theorem (see[7], § 17.3) is not applicable as the field U(r) is not symmetric with respect to the z-axis.

b) In a strong magnetic field, ωB � ω1,2, the normal oscillation with frequency1 ≈ ωB takes place along circle, and the normal oscillation with frequency 2 ≈ω1ω2/ωB along an ellipse with axes which are in the x- and y-directions and which standin the ratio ω2/ω1. The motion is thus along the circle with a centre which moves slowlyalong the ellipse.

It is well-known that if a charged particle moves in a strong uniform magnetic fieldin a plane at right angles to the field, the occurrence of a weak, quasi-uniform field U(r)

(i.e. such that the force F = −∂U∂r

changes little within the circular orbit) leads to a slow

displacement (drift) of the centre of the orbits in a direction at right angles to F (i.e.along the equipotential lines of U(r)) (see [2], § 22). Note that in our case a similar driftoccurs also in a strongly inhomogeneous oscillator field.

c) If ω1 = ω2, the normal oscillations in the (x, y)-plane correspond to motions along

a circles in opposite senses with frequencies 1,2 = ω ± 12 ωB, where ω =

√ω2

1 + 14 ω2

B.

In a system rotating with the angular velocity (−12 ωB) the frequencies of both motions

thus turns out to be equal to ω. Such motions are the normal oscillations of an isotropicoscillator with frequency ω. Indeed, the sum and difference of such oscillations with equalamplitudes,

(cos ωt

−sin ωt

)±

(cos ωtsin ωt

)

are linear oscillations in the x- or y-directions (if we neglect the motion in the directionof the magnetic field).

If the magnetic field is weak, ωB � ω1, we have ω ≈ ω1, and the whole effect of thefield on the motion of the oscillator reduces to a rotation (“precession”) around the z-axis with a frequency (−1

2 ωB) (Larmor theorem, compare [7], § 17.3 and [2], § 45). If,however, ωB � ω1, there is no longer any obvious use for the rotating system which wehave employed.

6.38. We can solve the equations of motion,

x + ω21x = ωzy,

y + ω22y = −ωzx + ωxz,

z + ω23z = −ωxy,

with

ωx = eBx

mc, ωz = eBz

mc



by the method of successive approximations. We look for the coordinates in the formx = x(1) + x(2), y = y(1) + y(2), z = z(1) + z(2), where x(2), y(2), and z(2) are small com-pared to x(1), y(1), and z(1). In first approximation we neglect small terms in the right-hand sides of the equations:

x(1) = Acos(ω1t + α),

y(1) = Bcos(ω2t + β),

z(1) = C cos(ω3t + γ ).

We get

x(2) = −ωzω2Bsin(ω2t + β)

ω21 − ω2

2

,

y(2) = ω1ωzAsin(ω1t + α)

ω22 − ω2

1

− ωxω3C sin(ω3t + γ )

ω22 − ω2

3

,

z(2) = ωxω2Bsin(ω2t + β)

ω23 − ω2

2

.

(2)

The corrections turns out to be small, provided |ωz| � |ω1 − ω2| and |ωx| � |ω2 − ω3|.The normal oscillations are oscillations along ellipses which are strongly alongated alongthe coordinate axes.

If, however, for example, |ωz| � |ω1 − ω2|, |ωx| � |ω2 − ω3|, according to (2) x(2)

and y(2) are no longer small. This is connected with the fact that the frequencies ofthe “forces” ωzy(1) and −ωzx(1) in (1) turn out to lie close to the eigen-frequencies ofthe oscillator. In that case we must retain the resonant terms in the first approximationequations:

x(1) + ω21x(1) − ωzy(1) = 0,

y(1) + ω22x(1) + ωzx(1) = 0,

z(1) + ω23z(1) = 0;

(3)

that is, it is necessary to take the effect of Bz on the motion exactly into account. Weconsider the set (3) in problem 6.37. For the second-order corrections, we have theequations

x(2) + ω21x(2) = 0,

y(2) + ω22y(2) = ωxz(1),

z(2) + ω23z(2) = −ωxy(1).



For the sake of simplicity, we restrict ourselves to the case ω1 = ω2 ≡ ω � ωz and wehave then the following normal oscillations:

⎛⎝x

yz

⎞⎠ = Re

⎧⎪⎨⎪⎩A1

⎛⎜⎝

1i

ωωxω2

3 − ω2

⎞⎟⎠ei(ω+ωz/2)t+

+ A2

⎛⎜⎝

1−i

−ωωxω2

3 − ω2

⎞⎟⎠ei(ω−ωz/2)t + A3

⎛⎜⎝

0iωxω3

ω23 − ω2

1

⎞⎟⎠eiω3t

⎫⎪⎬⎪⎭ .

(4)

In the approximation used, the normal oscillations (4) with frequencies ω ± 12 ωz, take

place along a circle in planes which make angles ∓ωxω/(ω23 − ω2) with the (x, y)-plane,

while the oscillation with frequency ω3 is along an ellipse in the (y, z)-plane which isstrongly elongated in the z-direction.

6.39. We assume that the oscillations of the pendulum are small oscillations: we reckonthe angle ϕ counterclockwise from the vertical, and as the second coordinate we choosethe charge q on the right-hand plate. When the pendulum is deflected over an angle ϕ, themagnetic flux through the circuit is equal to � = const − 1

2Bl2ϕ, so that the Lagrangianfunction of the system is (see problem 4.24)

L = 12

(ml2ϕ2 + L q2 − mglϕ2 − q2

C− Bl2ϕq

).

If we introduce the coordinates x = lϕ and y = √L /mq, the Lagrangian function of

our system differs from the one considered in problem 6.37 (with parameters ω21 = g

l ,

ω22 = 1

L C , ωB = − Bl2√

mLand z = 0) only by a total derivative with respect to the time:

12mωB

d(xy)dt . The equations of motion of problem 6.37 and their solutions are thus also

valid in the present case.

6.40. Let

r = Acos(ωt + ϕ), A = (A1, A2, . . . , AN) (1)

be any normal oscillation. Since the replacement

xi →∑

j

Sijxj



does not change the form of the Lagrangian function, a normal oscillation of the form

Sr = SAcos(ωt + ϕ), (SA)i =∑

j

SijAj (2)

must be valid together with (1). Here S is a matrix with elements Sij , which have theproperties:

ST = S, SS = E, (3)

where E is a unit matrix and ST —the transposed matrix S.a) If the given frequency ω is not degenerate, the solution (2) can differ from (1)

except for a common factor: Sr = cr, moreover,

SSr = c Sr = c2 r. (4)

Since SS = E, we obtain from (4) that r = c2r, or c2 = 1 and c = ±1. Therefore, for anon-degenerate frequency, we get

or Sr = +r, or Sr = −r.

b) If the frequency ω is degenerate, then oscillations (1) and (2) may not coincide.But their sum and difference

r ± Sr = (A ± SA)cos(ωt + ϕ)

are the normal oscillations with the same frequency and have all necessary properties ofsymmetry.

c) The addition to the Lagrangian function has the form �L = ∑i

fixi, where

f = (f1, f2, . . . , fN)

is the external force acting upon the system.Let the force f be symmetric, while the normal oscillation ra (1) be anti-symmetric

with respect to transformation S; that is,

Sf = +f , Sra = −ra. (5)

This force does not affect the oscillation ra if the vectors f and ra are mutually orthogonal(see problem 6.25):

(f , ra) = 0. (6)



It follows from (5) that

(Sf , Sra) = −(f , ra). (7)

On the other hand, the left-hand side of (7) can be rewritten as

(Sf , Sra) = (f , ST Sra) = +(f , ra) (8)

due to (3). Then, comparing (7) and (8), we obtain (6).Do parts a)–c) of the problem remain unchanged if we do not require in advance the

condition ST = S?

6.41. Let xi be the displacement of the ith particle along the ring from the equilibriumposition; for definiteness, we consider a counterclockwise displacement as positive. Thesystem is clearly symmetric with respect to rotation over an angle 180◦ around the AB-axis which passes through the equilibrium position of the second particle and the centreof the ring. Therefore, the Lagrangian function of the system

L = 12m

(x2

1 + x22 + x2

3

)+ 1

2M(x2

4 + x25

)− 1

2k

[4∑

i=1

(xi − xi+1)2 + (x5 − x1)2

]

does not change its form under the replacement corresponding to such a rotation

x2 → −x2 x1 → x3, x3 → −x1, x4 → −x5, x5 → x4. (1)

Using the symmetry considerations (see the preceding problem) and the orthogonal-ity conditions, it is easy to reduce this problem with five degrees of freedom to the twoindependent problems with two degrees of freedom each.

Indeed, the vectors of the normal oscillation, which are symmetric and anti-symmetricwith respect to the transformation (1), have the form

rs =

⎛⎜⎜⎜⎜⎝

a0

−ab

−b

⎞⎟⎟⎟⎟⎠cos(ωst + ϕs), ra =

⎛⎜⎜⎜⎜⎝

cdcff

⎞⎟⎟⎟⎟⎠cos(ωat + ϕa).



In addition, one anti-symmetric “oscillation” can be easily guessed—it is a rotation of allthe particles around the ring

ra1 =

⎛⎜⎜⎜⎜⎝

11111

⎞⎟⎟⎟⎟⎠(Ct + C1), ωa1 = 0.

Two other (besides ra1) anti-symmetric oscillations must be orthogonal to ra1 with themetric tensor determined by the coefficients of the kinetic energy, that is,

m(2c + d)+ 2Mf = 0. (2)

As a result, the oscillations ra and rs have only two undefined coefficients. To determinethem, it is enough to use only two equations of motion out of five, for example, those forthe first and fifth particles:

mx1 + k(2x1 − x2 − x5) = 0,

Mx5 + k(2x5 − x4 − x1) = 0.(3)

Substituting the explicit form of rs, we find for the two symmetrical oscillations

b1,2 =(m

kω2

s1,2 − 2)

a1,2,

ω2s1,2 = k

2mM

(2M + 3m ∓

√4(M − m)2 + 5m2

).

Similarly, substituting in (3) the vector ra and taking into account (2), we find

c2,3 =(

1 − Mk

ω2a2,3

)f2,3, d2,3 = −2c2,3 − 2M

mf2,3,

ω2a2,3 = k

2mM

(4M + m ∓

√12

(4M − m)2 + 12

m2

).

6.42. The considered system is close to the one studied in problem 6.29a; the Lagrangianfunction in our problem differs by a small quantity

δL = δL1(x, x)+ δL2(y, y)+ δL3(z, z),

δL1(x, x) = 12 εk

[x2

2 + (x2 − x5)2 + (x4 − x5)2 + x24

],



where ε = (l − l1)/l � 1. The oscillations in x, y, and z occur independently; in thefollowing we are only interested in the x-oscillations.

To determine the corrections to the frequencies, it is helpful to use the method ofsuccessive approximations (see problem 6.35). The frequencies ω3,4 are non-degenerate;therefore, we can directly use (4) of problem 6.35 to these oscillations.

The frequency ω1 of the original problem (6.29a) is threefold degenerate; therefore,it may seem that to determine the corrections to the frequencies and vectors of thenormal oscillation we will have to consider the set of equations of (5) from problem 6.35.However, the symmetry properties of the system enable us to indicate the vectors of thenormal oscillations of the original system, which change a little when we add δL. Theseare the vectors (1) from problem 6.29, because only they have the certain symmetryproperties. Namely, the oscillation r3 is symmetric with respect to the AB-axis and anti-symmetric with respect to the CD-axis, and the oscillation r1 is symmetric and r2, anti-symmetric with respect to both axes. Corrections to the frequencies of these oscillationscan also be calculated using (4) from problem 6.35.

Substituting x1 = −x3 = 1, x2 = x4 = x5 = 0, we find

δK11 = −2δL1 = 0;

therefore, δω1 = 0. Similarly,

δK22 = −2δL1(x1 = x3 = x5 = 0, x2 = −x4 = 1) = −4εk,

M2 = 2L1(x1 = x3 = x5 = 0, x2 = −x4 = 1, xi = 0) = 2m,

and δω2 = −ε√

k/(2m);

δK33 = −4εk, M3 = 4m, δω3 = −12 ε

√k/(2m).

Representing the vector of the initial displacements

r(0) =

⎛⎜⎜⎜⎜⎝

a00

−a0

⎞⎟⎟⎟⎟⎠

and the vector of the initial velocities r(0) = 0 as r(0) = ∑iri(0) and r(0) = ∑

i ri(0),respectively, we find that

A1 = A2 = A3 = 12 a, A4 = A5 = ϕi = 0.



Thus, in the given approximation, the fourth and fifth normal oscillations are not excited,and the oscillations of the particles

x1,3 = 12 a(±cosω1t + cosω3t),

x2,4 = 12 a(±cosω2t − cosω3t), x5 = 0

are beats (see problem 6.32).

6.43. In this problem, it is helpful to use the method of successive approximations (seeproblem 6.35). The change in mass leads to the correction to the Lagrangian functionof the form

δL = 12 (δm1 x2

1 + δm2 x22).

This correction must be expressed in terms of the normal coordinates of the originalsystem (see problem 6.22). In this case, the coefficient in front of the product of thegeneralized velocities q1q2 corresponding to the degenerate frequencies is zero. Otherproducts ql qs (for ωl = ωs) can be omitted as this issue has been noted in problem 6.35.We get

δL1 = 14 δm1 q2

1 + 14 δm2 q2

2 + 18 (δm1 + δm2)(q2

3 + q24).

The Lagrangian function L + δL1, as well as the Lagrangian function of the originalsystem, is reduced to the terms, each of which contains only one of the coordinates ql .As a result, the coordinates ql remain the normal ones, and to calculate the correctionsto the frequencies, one can use (4) of problem 6.35:

δω1 = −14 ε1ω1, δω2 = −1

4 ε2ω3, δω3 = −18 (ε1 + ε2)ω3,

δω4 = 0, εi = δmi/m.

All eigen-frequencies of the system become different; the ambiguity in choice of thevectors of the normal oscillations disappears since with the accuracy up to εi these vectorsare the vectors (1) from problem 6.22.

It is noteworthy that when δm1 = δm2, the frequencies ω1 + δω1 and ω2 + δω2coincide with each other again (up to the second-order corrections |δω1 − δω2| ∼ ε2

1ω1).In this case, the Lagrangian function L + δL1 once more leads to the ambiguous choiceof the vectors of the normal oscillations. However, in the exact solution of the problemat δm1 = δm2, the vectors of the normal oscillations have the form

⎛⎜⎜⎝

11

−1 − ε

−1 − ε

⎞⎟⎟⎠ ,

⎛⎜⎜⎝

1−1

12 (3 ∓ √

4 + 4ε + 9ε2)

−12 (3 ∓ √

4 + 4ε + 9ε2)

⎞⎟⎟⎠ ,

⎛⎜⎜⎝

1111

⎞⎟⎟⎠



and at small ε they are close to the vectors (2) of problem 6.22 (up to the normalizationfactors, omitted here). The sharp change in the form of the normal oscillations occursin a very narrow range of masses |δm1 − δm2| � ε2m (cf. problem 6.6a). To determinethe vectors of the normal oscillations in this range of values of δm1 and δm2, it would bepossible to use the next approximation in the method of successive approximations.

6.44. Obviously, the motion of particles in the direction of the AA- and BB-axes isindependent. We will consider only the motion along the AA-axis.

For the first and fourth particles, we assume the displacements to be positive to theleft, while for the second and third, positive to the right. According to the results ofproblem 6.40, the normal oscillations r = (x1, x2, x3, x4) can be chosen either symmetricor anti-symmetric with respect to the AA- and BB-axes. The oscillations which aresymmetric with respect to the AA-axis have x1 = x4, x2 = x3. If they are also symmetricwith respect to the BB-axis, then x1 = x2, x3 = x4, so for this twofold symmetricoscillation we have rss = (1, 1, 1, 1)qss.

For the oscillations which are symmetric with respect to the AA-axis and anti-symmetric with respect to the BB-axis, we have x1 = x4, x2 = x3 and x1 = −x2, x4 = −x3,so that

rsa = (1, −1, −1, 1)qsa.

Similarly, we find

ras = (1, 1, −1, −1)qas, raa = (1, −1, 1, −1)qaa.

The vectors of the normal oscillations in the vertical direction can be found in the sameway.

The oscillation frequencies can be found by substituting the vectors obtained in theequations of motion.

In the presence of degeneration, there are a lot of normal oscillations (except for theones as previously described) which do not have the previously described symmetryproperties. It is easy to figure out, for example, that the frequencies of the ss and aaoscillations coincide, ωss = ωaa = 2

√k/m, if the tension of the springs is not arbitrarily,

but equals kl (where l is the length of each of the springs in the equilibrium position). Inthat case, however, any superposition of vectors raa and rss will be the normal oscillations,for example, the vector (1, 0, 1, 0)qss.

Similarly, we can find the normal oscillations along the BB-axis.

6.45. Taking into account the symmetry properties allows us to reduce this systemwith seven degrees of freedom to a few simple properties with not more than twodegrees of freedom each. Indeed, due to the symmetry of the system with respect toa plane perpendicular to the plane of the scales, all the normal oscillations can be choseneither symmetric or anti-symmetric with respect to this plane. In addition, the normal



oscillations are divided into those which displace the particles from the plane of the scalesand those which do not. We will consider only the latter ones.

Let α, β, and γ be the deviation angles from the vertical of the centre of theframe, threads BA, and DE, respectively. Apart from the obvious symmetric oscillationsα = 0, β = −γ with frequency

√g/(3l), there are two anti-symmetric oscillations for

which β = γ . Since the contributions of different normal oscillations to the Lagrangianfunction are additive, then to find the anti-symmetric oscillations, it will suffice to knowonly term

La = ml2(2α2 + 9β2 + 3αβ)− mgl(α2 + 3β2),

corresponding to this type of oscillations. As a result, we obtain two anti-symmetricoscillations: α = β = γ with frequency

√2g/(7l) and α = −3β = −3γ with frequency√

2g/(3l).To describe the oscillations that displace the particles from the plane of the scales, we

use the Cartesian coordinates of the particle displacements from the equilibrium positionalong of the x-axis, which in turn is directed perpendicular to the equilibrium plane ofthe scales. It is obvious that the symmetric oscillations

xA = xE, xB = xD

coincide with the oscillations of the coplanar double pendulum1; therefore,

ω2s =

(7 ∓ √

37) g

3l, xA = (6 ± √

37)xB.

Of the two anti-symmetric oscillations

xA = −xE, xB = −xD,

one is obvious: it is a rotation of the scales around the vertical axis: xA = xB. The otheranti-symmetric oscillation is orthogonal to the one we found, and therefore, we havexA = −xB = xD = −xE . But this oscillation does not shift the centre of each thread inthe first approximation, so the frequency of such oscillations coincides with the frequencyof the pendulum of the length 3l/2; that is, it equals

√2g/(3l).

6.46. Obviously, the motions of particles along the AA- and BB-axes and in the directionperpendicular to the plane of the frame are independent. As an example, we will considerthe oscillations along the AA-axis.

1 See [1], § 23, problem 2 with parameters

m1 = m2 = 2m, l1 = 12 l, l2 = 3l, ϕ1 = 2xB

l, ϕ2 = xA − xB

3l.



It is helpful to place the components of the oscillation vector in the table similar inshape to the frame. For particles on the left from the axis-BB, we take the displacementsto the left as positive, while for the particles located on the right we take the displacementsto the right as positive. The oscillations ss symmetric with respect to both AA- and BB-axes have the form

(x X X xx X X x

).

The oscillations x, X are reduced to the oscillations of the system considered in prob-lem 6.8 with m1 = m, m2 = M, where we must also put k1 = k + k′, k2 = k, k3 = 2k + k′,k′ = f /l. Thus, there are two ss oscillations.

The sa-oscillations, which are symmetric with respect to the AA-axis and anti-symmetric with respect to the BB-axis, have the form

(x X −X −xx X −X −x

),

where, in this case, k3 = k′, k1 = k + k′.Similarly, we have for as-and aa-oscillations:

(x X X x

−x −X −X −x

), k1 = k + 3k′, k3 = 2k + 3k′,

(x X −X −x

−x −X X x

), k1 = k + 3k′, k3 = 3k′.

The remaining sixteen normal oscillations can be found in the similar manner.

6.47. In the notation of the preceding problem, the force vector is

(−k − k′, −k′, k′, k + k′−k − k′, −k′, k′, k + k′

)acosγ t.

It is orthogonal to all the normal oscillations, except for symmetric–anti-symmetric (sa).Therefore, the resonance only occurs for two frequencies γ = ωsa

1,2, where

(ωsa1,2)

2 = 12mM

{k(2M + m)+ f

l(M + m)±

±√[

k(2M − m)+ fl(M − m)

]2

+ 4k2Mm

⎫⎬⎭ .



6.48. The linear molecule C2H2 has only seven normal oscillations: two flexuraloscillations in two mutually perpendicular planes and three longitudinal oscillations (e.g.see [1], § 24).

The problem of determining the two symmetric longitudinal oscillations x1 = −x4,x2 = −x3 is reduced to problem 6.8 with parameters k2 = kHC, k3 = 2kCC, k1 = 0,where kHC and kCC are stiffness of the HC and CC bonds. The anti-symmetriclongitudinal oscillation x1 = x4, x2 = x3, mHx1 + mCx2 = 0 has the frequency

ωa1 =√

kHC(mH + mC)

mHmC

(the same as the “molecule” CH). Note that the longitudinal displacement of themolecule as a whole, x1 = x2 = x3 = x4, can be considered as the second anti-symmetricoscillation, and the condition of its orthogonality to the first anti-symmetric oscillationcoincides with the requirement that the total linear momentum of the molecule is zero.

The transverse symmetric oscillation reads y1 = y4, y2 = y3,

mHy1 + mCy2 = 0, ω2s3 = kHCC(mH + mC)

l2HCmHmC.

Here kHCC is the stiffness of the molecule at the flexure; that is, the potential energyincreases by the term kHCCδ2/2 when the bond HCC bends over an angle δ.

The transverse anti-symmetric oscillation reads y1 = −y4, y2 = −y3,

mClCCy2 + mH(lCC + 2lHC)y1 = 0,

ω2a2 = kHCC[mCl2CC + mH(2lHC + lCC)2]

mHmCl2HCl2CC

,

where lHC and lCC are the equilibrium distances between the H–C and C–C atoms,respectively. Note that the relation between the displacements of the atoms of thisoscillation can be found from the requirement that the total angular momentum of themolecule is zero (or due to orthogonality to the vector of the rotation of a molecule asa whole y1 = −y4, y2 = −y3, y2(2lHC + lCC) = y1lCC, which can be considered as thesecond anti-symmetric oscillation).

6.49. We denote the displacements of the particle from the equilibrium position inthe direction BD through x1 and x2, and in the direction CF—through y1 and y2. Therotation around the axis of symmetry CF over the angle 180◦ leads to the replacements:

x1 � −x2, y1 � y2;



that is, the transformation S, which is the rotation of the oscillation vector r =(x1, y1, x2, y2), has the form

Sr = (−x2, y2, −x1, y1).

Due to the symmetry of the system, the normal oscillations of two types are possible init: rs – symmetric with respect to the CF-axis (for which Srs = rs) and anti-symmetric ra(for which Sra = −ra). For the symmetric oscillations, we have (−x2, y2, −x1, y1) =(x1, y1, x2, y2), from where

rs = (x, y, −x, y).

Analogously,

ra = (x, y, x, −y).

For the anti-symmetric oscillations, the rotation around the axis of symmetry is equiva-lent to the change of the oscillation phase by π .

Two symmetric and two anti-symmetric oscillations are possible; to determine thefrequency and vectors of each pair of oscillations, it is sufficient to use only two equationsof motion.

We can obtain some more information about the type of normal oscillations withoutknowing stiffness of the springs. The condition of orthogonality of the oscillation vectorsrs1 and rs2 can be reduced to equality

2m(xs1xs2 + ys1ys2) = 0,

which shows that the vectors of the displacements of, for example, particle 1 at each ofthe two symmetrical oscillations are mutually orthogonal (Fig. 141a). The same appliesto the antisymmetric oscillations (Fig. 141b).

s1

a1

a1

a2

a2s1

s2 s2

(a) (b)

Figure 141

The direction in which the particle displaces at each of the normal oscillationscannot be determined without knowing stiffness of the springs. Indeed, if stiffnessand tension springs AC and CE are small, then the particle displacements under thenormal oscillations are directed almost along or across the springs BD. On the contrary,



if the stiffness of springs BD is low, the normal oscillations occur almost along or acrossthe springs AC and CE.

Of course, if the normal oscillation is degenerate, one can choose other vectors ofthe normal oscillations which do not possess the symmetry properties. For example, ifwe “switch off” spring BD, each of the particles will be able to oscillate along or acrosssprings AC and CE.

6.50. We denote the displacement of each atom from its equilibrium position in the OD-directions through xi, in the OA-directions through yi, and perpendicular to the plane ofthe molecule through zi.

The oscillations of the molecule C2H4 are divided into those which leave the moleculeplanar and those which output atoms of the plane of the molecule. We will start with thelatter.

It is helpful to present the vector of displacement of the atoms in the form

r =⎛⎝z1 z4

z2 z5z3 z6

⎞⎠ .

For the oscillation symmetric with respect to the AB-axis, we have

z4 = z1, z5 = z2, z6 = z3.

If, in addition, this oscillation is anti-symmetric with respect to the CD-axis, we get

z3 = −z1, z2 = −z2, z5 = −z5, z6 = −z4.

As a result,

rsa =⎛⎝ 1 1

0 0−1 −1

⎞⎠qsa.

This “oscillation” turns out to be, in fact, the rotation around the CD-axis.Similarly, the symmetric oscillations with respect to both axes are

rss =⎛⎝z1 z1

z2 z2z1 z1

⎞⎠ .



One of them, in particular, is

rss,1 =⎛⎝1 1

1 11 1

⎞⎠qss,1.

This is a translational motion. The amplitudes of the other oscillation can be determinedby taking into account its orthogonality to the vector rss,1:

4mz1 + 2Mz2 = 0,

where m and M are the masses of the hydrogen and carbon atoms, respectively:

rss,2 =⎛⎝1 1

−2mM −2m

M1 1

⎞⎠qss,2.

The antisymmetric oscillation with respect to both axes,

raa =⎛⎝ 1 −1

0 0−1 1

⎞⎠qaa,

is a torsional oscillation around the CD-axis.Finally,

ras,i =⎛⎝1 −1

ai −ai1 −1

⎞⎠qas,i.

Here for the rotation around the AB-axis (the oscillation rsa,1), the factor a1 = lOC/lOH isthe ratio of the distances of the carbon atoms and hydrogen atoms from the AB-axis in theequilibrium position; for the flexural oscillation (ras,2), it is

a2 = −2mM

lOH

lOC.

The similar approach can also be applied to the oscillations which do not output atomsfrom the plane of the molecule.



We can present the corresponding displacement vectors in the form

r =⎛⎝x1 y1 x4 y4

x2 y2 x5 y5x3 y3 x6 y6

⎞⎠ .

The general form of the as-oscillations is

ras =⎛⎝x1 y1 x1 −y1

x2 0 x2 0x1 −y1 x1 y1

⎞⎠ .

One of the as-“oscillations” is the translational motion of the molecule as a whole alongthe x-axis (x1 = x2, y1 = 0). The other two as-oscillations are shown in Fig. 142a and b.

as

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

as sa

sa

ss ss ss

aa aa

Figure 142

To visualize the kind of displacements of atoms under these vibrations, note that thedistance between the carbon atoms does not change: the C–C link “does not work”. Ifwe neglect the interaction between the relatively distant atoms (e.g., 1–4, 1–5 in Fig. 43)and also stiffness of the angles of the form 1–2–5, then the considered oscillations willcoincide with those which are symmetric with respect to the CD-axis oscillations of two“molecules” H2C occurring in anti-phase (cf. problem 6.48; oscillations of molecules ofthe form A2B are considered in [1], § 24, problem 2).

The general form of sa-oscillations is

rsa =⎛⎝ x1 y1 −x1 y1

0 y2 0 y2−x1 y1 x1 y1

⎞⎠ .

In addition to the translational motion in the direction of the y-axis, there are twoother oscillations (Fig. 142c and d). One of them (Fig. 142c) can be presented as the



oscillations of the two “molecules” H2C which are antisymmetric with respect to the CD-axis and occur in phase. The other (Fig. 142d) is the rotations of the “ molecules” H2Cin the opposite directions. If we completely neglect the constraints of the distant atomsand the stiffness of the angles 1–2–5, then the frequency of this oscillation will be zero.

Among aa-oscillations, there is a rotation of the molecule as a whole in the xy-plane, the anti-symmetric oscillations of the “molecules” H2C in anti-phase (Fig. 142e)and their rotations in one direction (Fig. 142f).

Three ss-totally symmetric oscillations are also possible as shown in Fig. 142g–i.

6.51. The Lagrangian function is

L = 12 m(u2

1 + u22 + u2

3)− 12 k

{(|r10 − r20 + u1 − u2| − l)2+

+(|r20 − r30 + u2 − u3| − l)2 − (|r30 − r10 + u3 − u1| − l)2}

,

where l = |r10 − r20| = |r20 − r30| = |r30 − r10|; ua is the displacement of the ath atomfrom its equilibrium position, which is determined by the radius vector ra0. Since |ua| �l, we have

L = 12 m(u2

1 + u22 + u2

3)−−1

2 k{(e12(u1 − u2))2 + (e23(u2 − u3))2 + (e31(u3 − u1))

2}

,(1)

where

e12 = r10 − r20

l, e23 = r20 − r30

l, e31 = r30 − r10

l.

In the system of reference where the total momentum equals zero, m(u1 + u2 + u3) = 0,the condition

u1 + u2 + u3 = 0 (2)

is satisfied. Moreover, we impose upon ua the condition

[r10,u1] + [r20,u2] + [r30,u3] = 0, (3)

which is equivalent to the requirment that the angular moment of the molecule,

M = m∑

a

[ra0 + ua, ua], (4)

vanishes up to and including terms of first order in ua.



x1

x2x3

y3

y1

y2

u1

r10

r20 r30

Figure 143

It is helpful for the description of the motion tointroduce for each atom its own Cartesian system ofcoordinates (Fig. 143), thus retaining symmetry inthe description of the system. Equations (2) and (3)in terms of these coordinates read1

y1 +(

−12

y2 +√

32

x2

)+

(−1

2y3 +

√3

2x3

)= 0,

(5)

y2 +(

−12

y3 +√

32

x3

)+

(−1

2y1 −

√3

2x1

)= 0,

(6)

and hence

y1 = x3 − x2√3

, y2 = x1 − x3√3

, y3 = x2 − x1√3

,

and

L = 5m6

(x2

1 + x22 + x2

3

)− m

3(x1x2 + x2x3 + x3x1)− 3

2k(x2

1 + x22 + x2

3

).

One normal oscillation (totally symmetric) is obvious:

x(1)1 = x(1)

2 = x(1)3 = 1√

3q1. (7)

The two other oscillations are orthogonal to the first oscillation which leads to thecondition2

x(s)1 + x(s)

2 + x(s)3 = 0, s = 2, 3. (8)

One of these oscillations is symmetric with respect to the x1-axis: x(2)2 = x(2)

3 ; the other is

antisymmetric: x(3)1 = 0, x(3)

2 = −x(3)3 .

1 For instance, multiplying both sides of the equality (2) by e23, we get (5). Note that the vector e23 in thedifferent coordinate systems has the coordinates

e23 = (0, 1)1 =(√

32

, −12

)2

=(

−√

32

, −12

)3

,

while ua = (xa, ya)a. Equality (6) follows from (5) through a cyclic permutation of the indices.2 The metric kij is used. In (7) and (9) the factors before q are chosen so that x(l)2

1 + x(l)22 + x(l)2

3 = q2l .



Taking into account (8), we have

x(2)1 = −2x(2)

2 = −2x(2)3 =

√23

q2,

x(3)2 = −x(3)

3 = 1√2

q3.(9)

The replacement,

x1 = 1√3

q1 +√

23

q2,

x2 = 1√3

q1 − 1√6

q2 + 1√2

q3,

x3 = 1√3

q1 − 1√6

q2 − 1√2

q3,

leads to the Lagrangian function of the form

L = 12 m(q2

1 + 2q22 + 2q2

3)− 32 k(q2

1 + q22 + q2

3). (10)

The normal oscillations corresponding to these coordinates are shown in Fig. 144. Theirfrequencies are

Figure 144

ω1 =√

3km

, ω2 = ω3 =√

3k2m

.

The form of the Lagrangian function (10) is retained under the rotation in the q2, q3-plane.

The angular momentum with the account of quadratic in ua terms

|M| = m

∣∣∣∣∣∑

a

[ua, ua]

∣∣∣∣∣ = m|q2q3 − q3q2|

can be non-zero if there is a phase difference between the oscillations q2 and q3.It is interesting to analyse what changes are introduced in this picture if we consider

now the possibility that the potential energy may depend on the angles between the bonds.



It is clear that such a dependence will not affect the frequency of the q1 oscillations.The frequencies of the q2 and q3 oscillations will be changed, but a twofold degeneracywill remain. Indeed, together with a q-oscillation, we can also have another oscillationobtained from the original q-oscillation by a rotation over 2π/3. Its frequency must bethe same as that of the original oscillation. On the other hand, the q-oscillation will differ(only the q1-oscillation remains the same under a rotation over 2π/3). Thus, we find twoindependent oscillations with the same frequency. The normal coordinates must in thiscase satisfy only one condition: they must be orthogonal to q1; in particular, q2 and q3remain normal coordinates.

6.52. a) We introduce the coordinates of atoms B in the same way as in the precedingproblem; for atom A—the coordinates will be x4, y4, z4 with the axes parallel to x1, y1, z1and with the beginning in the centre of the triangle.

There are four degrees of freedom of motions which output atoms from the xy-plane.Three of them correspond to the translational motion along the z-axis and rotationsaround the x4- and y4-axes, and one to the oscillation (for which, obviously, z1 = z2 = z3,mAz4 + m(z1 + z2 + z3) = 0). The frequency of this oscillation ω1 is not degenerate; it isonly by chance that it could coincide with the frequency of any other oscillations.

Let us consider the oscillation of atoms in the xy-plane symmetric with respect tothe x4-axis. The general form of such oscillation is

y1 = 0, x2 = x3, y2 = −y3, y4 = 0.

The displacement vector contains four independent parameters: x1, x2, y2, x4; that is,such motions have four degrees of freedom. One of them corresponds to the translationalmotion of the molecule along the x4-axis, another one to totally symmetric oscillation

x1 = x2 = x3, y1 = y2 = y3 = x4 = y4 = 0

with the frequency ω2 (Fig. 145a), and the other two—to the oscillations with thefrequencies ω3 and ω4 which break symmetry of the molecule (Fig. 145b and c).

−

(a)

(d)

(b) (c)

Figure 145

Of the four remaining degrees of freedom, one is the translational motion along the y4-axis, and another is the rotation around the z4-axis. The last two are the oscillations thatcan be obtained from the previously discussed oscillations with the frequencies ω3 andω4 by a rotation on an angle 2π/3 around the z4-axis (cf. preceding problem).



As a result, the frequencies ω1 and ω2 are non-degenerate, while the frequencies ω3and ω4 are twofold degenerate.

Note that the vector of the oscillation which is antisymmetric with respect to the x4-axes can be obtained from the vector of the symmetric oscillation. For this, it is sufficientto take a certain superposition of the oscillations obtained from the latter by the rotationaround the z4-axis over angles ±2π/3; namely, it must be their difference (see Fig. 145d).

b) The changes in the eigen-frequencies can be determined using the perturbationtheory (see problem 6.35):

δω = −ω

2δMM

.

For a totally symmetric oscillation, we choose x1 as a normal coordinate. Then quanti-ties M and δM are defined as the coefficients in the expressions for the kinetic energyand the corrections to it, respectively:

32 mx2

1 = 12 Mx2

1, 12 δmx2

1 = 12 δMx2

1,

so that

δω2 = −ω2δm6m

.

For the oscillations along the z-axis, the kinetic energy and the correction to it are equalto, respectively,

32 m(1 + 3m/mA) z2

1 and 12 δmz2

1,

so that

δω1 = −12 ω1

mAδm3m(3m + mA)

.

For example, when we replace one atom of chloride with the atomic weight of 35 by theisotope with the atomic weight of 37 in the molecule of boron chloride, it reduces thefrequency ω1 and ω2 by 0.1% and 1%, respectively.

6.53. Let the oscillation for which the molecule retains its shape (Fig. 146a) have afrequency ω1.

The frequency ω2 of the oscillation which retains its form under rotations around ODover 2π/3 (Fig. 146b) is, in general, different from ω1. One can obtain anotherdisplacement of the atoms by taken a reflection in the plane BCO; we obtain oscillationswhich differ from the second oscillation only in that atoms A and D change roles. Thefrequency of that oscillation ω3 = ω2. Similarly, in the reflection in plane AOC, the roles of



D D D

B B

OC C C

B

AA A

(a) (b) (c)

Figure 146

atoms B and D are changed, while the frequency remains the same, ω4 = ω2. This fourthoscillation cannot be reduced to a superposition of previous oscillations as, in contrastto those, it is not symmetric with respect to plane AOD.

The oscillation which is symmetric with respect to planes AOB and DOC (Fig. 147c)has a frequency ω5 which is different from ω1 and ω2. A rotation over an angle 2π/3around OD which results in a cyclic permutation of A, B, and C, leads to an oscillationwhich is symmetric with respect to planes COA and DOB and its frequency ω6 = ω5.

The molecule has thus three eigen-frequencies which are, respectively, non-degerenate, twofold degenerate, and threefold degenerate.

In conclusion, we note that the molecules considered in problems 6.51 and 6.53 are,clearly, not to be found in nature. However, a similar approach can also be used for realmolecules.

6.54. a) When a totally symmetric and twofold-degenerate oscillations, considered in thepreceding problem (Fig. 146a and b) occur, the carbon atom remains at rest. There areextra two threefold degenerate frequencies. The corresponding oscillations are similar tothe ones shown in Fig. 147b. The only difference is that the carbon atom oscillates eitherin the same direction as the atom D or in the opposite direction.

b) The addition to the Lagrangian function describing the action of the electricfields E(t) is

δL = E(t)∑

a

eaua,

where ea is the charge and ua, displacement of ath atom.In any oscillation m

∑4a=1 ua + mCu5 = 0, so that

5∑a=1

eaua = −e1

(mC

m+ 4

)u5.

For the totally symmetric and twofold-degenerate oscillations, are quantity∑

eaua = 0and these oscillations are not excited. For the oscillation of Fig. 147b, in contrast,∑

eaua = 0 and the similar oscillations are excited.As a result, resonance is possible on two frequencies.



The vector E can be decomposed into three terms Ej which are parallel to axes ofsymmetry of each of the three oscillations of the molecule with degenerate frequency.Each of the terms Ej will lead to the oscillation of the carbon atom with an ampli-tude proportional to Ej , and with the same proportionality coefficient �. Therefore,u5 = �E and |u5| does not depend on the orientation of the molecule. It may be shownthat the oscillation amplitudes of the hydrogen atoms |u1,2,3,4| do also not depend on theorientation of the molecule and that they are parallel to the vector E.


§7

Oscillations of linear chains

7.1. The Lagrangian function of the system is

L(x, x) = 12 m

N∑n=1

x2n − 1

2 k

[x2

1 +N∑

n=2

(xn − xn−1)2 + x2N

], (1)

where the xn is the displacement of the nth particle from its equilibrium position. We alsointroduce the coordinate of the equilibrium position of the nth particle, Xn = na, wherea is the equilibrium length of one spring. The Lagrangian equations of motion

mx1 + k(2x1 − x2) = 0,mxn + k(2xn − xn−1 − xn+1) = 0, n = 2, 3, . . . , N − 1,mxN + k(2xN − xN−1) = 0

(2)

are equivalent to the set

mxn + k(2xn − xn−1 − xn+1) = 0, n = 1, 2, . . . , N (3)

with the additional condition

x0 = xN+1 ≡ 0. (4)

We expect from physical considerations that the normal oscillations of the system willbe standing waves. It is, however, helpful to consider

xn = Aei(ωt±nϕ). (5)

The set of N the equations in (3) then reduces to the equation,

ω2 = 4km

sin2 ϕ

2, (6)




which determines the relationship between the frequency ω and the difference in thephase of neighbouring particles ϕ. The meaning of the substitution (5) consist ofthe choice for xn of a solution in the form of a travelling wave with a wave vectorK = ϕ/a such that nϕ = naK = KXn. Equation (6) thus establishes the relation betweenthe frequency and the wave vector.

The conditions of (4) can be satisfied by taking a superposition of the waves travellingin two directions,

xn = Aei(ωt−nϕ) + Bei(ωt+nϕ).

The condition x0 = 0 gives A = −B, or xn = 2iBsin(nϕ)eiωt, that is, a standing wave.From the condition at the other end, xN+1 = 0, we determine the spectrum of possiblefrequencies.

The equation sin(N + 1)ϕ = 0 has N independent solutions

ϕs = π sN + 1

, s = 1, 2, . . . , N. (7)

In fact, s = 0 and s = N + 1 give vanishing solutions; for s = N + l, the phase ϕN+l =−ϕN−l+2 + 2π ; that is, the solution corresponding to s = N + l can be expressed interms of the solutions corresponding for s = N − l + 2. From (6) and (7), we find Ndifferent frequencies:

ωs = 2

√km

sinϕs

2= 2

√km

sinπ s

2(N + 1), s = 1, 2, . . . , N. (8)

1 2 Ns

ωs

Figure 147

The different frequencies are shown in Fig. 147 by the discretepoints on the sine curve. The vector of the normal oscillationscorresponding to the sth frequency is

rs =

⎛⎜⎜⎝

x1x2. . .

xN

⎞⎟⎟⎠ =

√2

N + 1

⎛⎜⎜⎝

sinϕssin2ϕs

. . .

sinNϕx

⎞⎟⎟⎠qs(t), (9)

where

qs(t) = Re(2iBseIωst) = Cs cos(ωst + αs)

is the sth normal coordinate, while the factor

√2

N + 1=

[ N∑n=1

sin2 nϕs

]−1/2



is introduced to get the normalization: (rs, r′s) = δss′q2

s . The general solution is a super-position of all normal oscillations:

xn =N∑

s=1

√2

N + 1qs(t)sinnϕs.

The matrix, given the transition from the xn to the qs,

Uns =√

2N + 1

sinπns

N + 1,

is an orthogonal matrix which reduces the Lagrangian function to a sum of squarescorresponding a set of N different oscillators:

L =N∑

s=1

Ls(qs, qs), L(qs, qs) = 12 m(q2

s − ω2s q2

s ).

7.2. The equations of motion for the given system are the same as the equations (3) ofthe previous problem but now with the additional conditions x0 = 0, xN = xN+1. Thus,we get

ϕs = (2s − 1)π

2N + 1, s = 1, 2, . . . , N,

ωs = 2

√km

sin(2s − 1)π

2(2N + 1),

xn =∑

s

sinnϕs · As cos(ωst + αs).

For the particular case when N = 2, see problem 6.1.

7.3. We use displacements of each of the particles from the vertical as the generalizedcoordinates (cf. problem 6.3). In such variables, the given problem is completelyreduceble to problem 7.2 with k/m = g/l.

7.4. The equations of motion are the same as the equations (3) of problem 7.1 but nowwith the additional conditions x0 = xN and xN+1 = x1. Therefore,

ϕs = 2π sN

, s = 0, 1, . . . , N − 1,

ωs = 2

√km

sinsπN

;



the frequencies ωs and ωN−s are the same, and the corresponding wave vectors

Ks = ϕs

a= 2π − ϕN−s

a= −KN−s + 2π

a

determine the travelling waves moving in opposite directions. The frequency ω0 = 0corresponds to the motion of all particles along the ring with a constant velocity. In thissystem, oscillations of the form

x(s)n = ReAsei(ωst−nϕs)

are possible, that is, waves which travel along the ring. The previously mentioned twofolddegeneracy of the frequencies corresponds to waves moving in opposite directions. Thepresence of two such waves with equal amplitudes gives a standing wave:

x(s)n ± x(N−s)

n = 2|As|{

cosnϕs cos(ωst + αs),sinnϕs sin(ωst + als).

(2)

This is also a normal oscillations (all particles move either in phase or in anti-phase).In terms of the appropriate normal coordinates,

xn =R∑

s=1

(qs1 cosnϕs + qs2 sinnϕs)+ q0, R = 12 (N − 1), (N : odd), (3)

the Lagrangian function is reduced to diagonal form:

L = 12 Nm

{q2

0 + 12

R∑s=1

[q2

s1 + q2s2 − ω2

s (q2s1 + q2

s2)]}

. (4)

(If the number of particles is even, (3) and (4) must be somewhat changed as thefrequency ωN/2 is non-degenerate; (1) defines a standing wave for s = N/2.)

It is interesting to note that rotations in the qs1, qs2-planes,

qs1 = q′s1 cosβs − q′

s2 sinβs,

qs2 = q′s1 sinβs + q′

s2 cosβs,

which leave the Lagrangian function (4) invariant, correspond to a displacement of anodes of the standing waves:

xn = q0 +R∑

s=1

[q′s1 cos(nϕs − βs)+ q′

s2 sin(nϕs − βs)].



The average energy flux along the ring is for the travelling waves (1) given by (cf.problem 6.10)1

Sav = ω

2π

2π/ω∫0

k(xn−1 − xn)xn dt = 12 k|A|2ω sinϕ,

while the group velocity is

vgr = dω

dK=

√km

acosϕ

2,

where a is the equilibrium length of one spring, while K = ϕ/a is the wave vector. Theenergy is

E = 12 m

N∑n=1

x2n + 1

2 kN∑

n=1

(xn − xn−1)2 = 2N|A|2ksin2 ϕ

2= 1

2 Nmω2|a|2,

and we have thus

ENa

vgr = Sav.

7.5. a) The equations of motion are

mx2n−1 + k(2x2n−1 − x2n−2 − x2n) = 0,

Mx2n + k(2x2n − x2n−1 − x2n+1) = 0,(1)

here x0 = x2N+1 = 0, n = 1, 2, . . . , N.We shall look for a solution in the form of travelling waves with different amplitudes

x2n−1 = Aei[ωt±(2n−1)ϕ],

x2n = Bei[ωt±2nϕ].(2)

To determine A and B, we get a set of homogeneous equations,

(−mω2 + 2k)A − k(e−iϕ + eiϕ)B = 0,

−k(e−iϕ + eiϕ)A + (−Mω2 + 2k)B = 0,(3)

1 In the following, we drop the index s for the sake of simplicity. The calculation of flux Sav and of theenergy E is very helpful carried out using complex variables (cf. [2], § 48).



which have a non-trivial solution only if its determinant vanishes. This condition deter-mines the relations between the frequency and the difference in phase of neighbouringparticles:

ω2(∓) = k

μ

⎛⎝1 ∓

√1 − 4μ2

mMsin2 ϕ

⎞⎠ , μ = mM

m + M. (4)

2kμ√

1 2 3 Ns

ω

ω(+)

ω(−)

2k√m

2k√M

Figure 148

The boundary conditions are satisfied only by well-defined linear combinations of the travelling waves (2),namely,

x2n−1 = As sin(2n − 1)ϕs cos(ωst + αs),

x2n = Bs sin2nϕs cos(ωst + αs),

for which ϕs = sπ/(2N + 1). Since ϕ2N+1−s = π − ϕs, weget different frequencies, having chosen one of the twosigns in (4) only for s = 1, 2, . . . , N. These are indicated(for this case, when M > m) in Fig. 148 by discrete dots on the two different branches;the lower one, ω(−), is usually called the acoustic, and the upper one, ω(+), is called theoptical branch.

The general solution has the form

x2n−1 =N∑

s=1

sin(2n − 1)ϕs[A(+)s cos(ω(+)st + αs)+A(−)s cos(ω(−)st+βs)],

x2n =N∑

s=1

sin2nϕs[B(+)s cos(ω(+)st + αs)+ B(−)s cos(ω(−)st + βs)],

where the A(±)s and B(±)s are according to (3) connected by the relation

B(±)s = 2k − mω2(±)s

2kcosϕsA(±)s.

It is remarkable that B(−)s and A(−)s corresponding to the acoustic frequencies havethe same signs, while B(+)s and A(+)s for the optical frequencies have the opposite signs(i.e., neighbouring particles with masses m and M move in antiphase). The distributionof the amplitudes of the oscillations for the case N = 8, s = 2 is shown in Fig. 149 wherethe ordinate gives the numbers of the particles, and the abscissa gives the amplitudes(149a gives an optical and 149b, an acoustic mode).

How can one get from the results obtained here to the limiting case m = M (seeproblem 7.1)?



xl

l

l

xl

(a)

(b)

Figure 149

b) The normal oscillations are

x(s)2n = As sin2nϕs cos(ωst + αs),

x(s)2n−1 = As

K sin2nϕs + ksin(2n − 2)ϕs

k + K − mω2s

cos(ωst + αs),

where

ω2s = 1

m

[K + k ∓

√(K − k)2 + 4Kkcos2 ϕs

]

and ϕs is determined from the equation

tan(2N + 1)ϕs = −K − kK + k

tanϕs, s = 1, 2, . . . , N,

0 < ϕs < π/2.

Fig. 150a shows the optical and acoustic branches of the frequencies for the case K > k.What happens in the limiting case K = k?

c) We have ϕs = π2(N + 1)

; for s = 1, 2. . .N, we get 2N normal vibrations and normal

frequencies which have the same form as under point 7.5b (Fig. 150b).



2(K+k)√ m2(K+k)√ m

2K√ m

2K√ m

2k√ m

K+k√ m2k√ m

1 2 3 1 2 3N N N+1ss

ω ω

ω(+) ω(+)

ω(−) ω(−)

(a) (b)

Figure 150

There is one more, (2N + 1), normal oscillation x2n = 0 and Kx2n−1 = −kx2n+1, thefrequency of which, ω2

0 = (K + k)/m, lies in the forbidden band between the opticaland acoustic branches. The distribution of the amplitudes of this oscillation is shown inFig. 151 where the ordinate and abscissa show, respectively, the numbers of the particlesand their amplitudes of the oscillations. The particles with an even numbers are notmoving while neighbouring particles with odd numbers are moving in antiphase withthe exponentially damped amplitudes, when we move away from the left-hand end ofthe chain.

xl

1 32l

Figure 151

7.6. a) We look for a solution of the equations of motion,

mxn + k(2xn − xn−1 − xn+1) = 0, n = 1, 2, . . . , N, (1)

with the boundary conditions x0 = 0 and xN+1 = acosγ t in the form of standing wavesxn = Asinnϕ cosγ t, so that the first boundary condition is immediately satisfied. Fromthe second boundary condition, we find constant A = a/sin(N + 1)ϕ, while from (1) weget for the “wave vector” ϕ of the standing wave

sin2 ϕ

2= mγ 2

4k.



When γ 2 < 4k/m, the stable oscillations

xn = asinnϕ

sin(N + 1)ϕcosγ t (2)

have a larger amplitude when the denominator sin(N + 1)ϕ is close to zero. But thisis just the condition which determined the spectrum of the eigen-frequencies ωs (seeproblem 7.1); that is, we have then a near-resonance situation, γ ≈ ωs. When

γ � ω1 = 2

√km

sinπ

2(N + 1),

the oscillations in (2) correspond to a slow extension and compression of all springs asa whole:

xn = an

N + 1cosγ t.

If γ 2 > 4k/m, we change ϕ to π − iψ in (2) and get

xn = (−1)N+1+nasinhnψ

sinh(N + 1)ψcosγ t,

where

cosh2 ψ

2= mγ 2

4k.

The oscillations are (exponentially when nψ � 1) damped towards the left-hand end ofthe chain. The reasonableness of this result is particularly clear when γ 2 � 4k/m whenthe frequency of the applied force lies appreciable above the limit of the spectrum ofthe normal frequencies. In that case, the particle on the extreme right oscillates with asmall amplitude in anti-phase with the applied force while the (N − 1)st particle is infirst approximation at rest. Then we can consider the motion of the (N − 1)st particle asa forced oscillation caused by an applied force of high frequency arising from the Nstparticle, and so on.

We note that a similar damping of the wave occurs in the phenomenon of completeinternal reflection (e.g., when short wavelength radio-waves are reflected by the iono-sphere).

What is the form of the stable oscillation when γ 2 = 4k/m?b) To find the normal oscillations we have consider, first, the solution in the form of

the travelling wave (see problem 7.1). If the wave travels to the right, to the point A, thenthe displacements are



xn = Re(Aei(γ t−nϕ)

),

where ϕ is determined by the equation

γ 2 = 4km

sin2 ϕ

2.

In the given problem x0 = acosγ t, so we find A = a. Therefore,

xN+1 = Re(Aei(γ t−(N+1)ϕ)

)= acos(γ t − (N + 1)ϕ).

The energy flux from left to right is equal to the work (averaged over period) ofthe spring, which connects the (n − 1)th and nth particles, over the nth particle (cf.problem 6.10):

〈k(xn−1 − xn)xn〉 = 12 γ ka2 sinϕ = 1

4 mγ 2a2√

(4k/m)− γ 2.

When γ > 2√

k/m, there will be no travel wave.

c) xn = acos

(N − n + 1

2

)ϕ

cos(N + 1

2

)ϕ

cosγ t,

sin2 ϕ

2= mγ 2

4k, when γ 2 <

4km

,

xn = (−1)nasinh

(N − n + 1

2

)ψ

sinh(N + 1

2

)ψ

cosγ t,

ch2 ψ

2= mγ 2

4k, when γ 2 >

4km

.

7.7. If the frequency of the applied force lies in the range of the acoustic eigen-frequencies, 0 < γ 2 < 2k/M, or in the region of the optical eigen-frequencies, 2k/m <

γ 2 < 2k/μ (see problem 7.5a), the stable oscillations are

x2n−1 = asin(2n − 1)ϕ

sin(2N + 1)ϕcosγ t,

x2n = ±√

2k − mγ 2

2k − Mγ 2 asin2nϕ

sin(2N + 1)ϕcosγ t,



where

cos2 ϕ = (2k − Mγ 2)(2k − mγ 2)

4k2 ,

while the upper (lower) sign corresponds to the frequency γ lying in the acoustic(optical) frequencies.

For frequencies lying in the “forbidden band” 2k/M < γ 2 < 2k/m, we have

x2n−1 = (−1)N+n+1acosh(2n − 1)ψ

cosh(2N + 1)ψcosγ t,

x2n = (−1)N+n+1a

√2k − mγ 2

Mγ 2 − 2ksinh2nψ

cosh(2N + 1)ψcosγ t,

sinh2 ψ = (2k − mγ 2)(Mγ 2 − 2k)

4k2 ,

while for frequencies γ 2 > 2k/μ, which lie above the limit of the optical branch (this isalso the “forbidden band”),

x2n−1 = asinh(2n − 1)χ

sinh(2N + 1)χcosγ t,

x2n = −a

√2k − mγ 2

2k − Mγ 2

sinh2nχ

sinh(2N + 1)χcosγ t,

cosh2 χ = (Mγ 2 − 2k)(mγ 2 − 2k)

4k2 ,

the oscillations are damped exponentially towards the left-hand side end of the chain.

7.8. a) We look for the solution for the equations of motion

mxn + k(2xn − xn−1 − xn+1) = 0, n = 1, 2, . . . , N − 1, (1)

mNxN + k(2xN − xN−1) = 0 (2)

(with the boundary condition x0 = 0), in the form of standing waves:

xn = Asinnϕ cos(ωt + α), n = 1, 2, . . . , N − 1,

xN = Bcos(ωt + α).(3)

From (1), we get the relation

ω2 = 4km

sin2 ϕ

2. (4)



Using (3) and (4), we get from (1) and (2) the set of equations

AsinNϕ − B = 0.

−Asin(N − 1)ϕ +(−2mN

msin2 ϕ

2+ 2

)B = 0.

Hence, B = AsinNϕ, while the parameter ϕ is determined as the solution of thetranscendental equation

sinNϕ

(4mN

msin2 ϕ

2− 2 + cosϕ

)= cosNϕ sinϕ. (5)

When mN � m, we have, apart from the obvious normal oscillations,

x(s)n = As sinnϕs cos(ωst + αs), n = 1, 2, . . . , N,

tanNϕs ≈ m2mN

cotϕs

2, s = 1, 2, . . . , N − 1,

when the particle mN is practically stationary (sinNϕs � 1), as well as a normal oscillationwith the amplitudes of the particles which are linearly decreasing to the left-hand side ofthe chain:

x(N)n = B

nN

cos(ωNt + αN), ω2N = k

mN

(1 + 1

N

).

The particle mN then oscillates between springs of stiffness k (to the right) and k/N (tothe left). That (5) has such a solution can be seen as follows. Assuming ϕ to be smalland retaining only the main terms, we get from (5): ϕ2 = m

mN

(1 + 1

N

)completely in

accordance with the assumption made.When mN � m, we have the usual oscillations characteristic for a system of (N − 1)

particles with a spring of stiffness k/2 at the right-hand end (the parameter ϕs and the

frequency ωs are determined from the equation tanNϕ = − sinϕ2 − cosϕ

). Apart from those

oscillations, there is also a normal oscillation with the amplitudes of the particles whichdecrease to the left-hand side of the chain:

x(N)n = (−1)N+nB

sinhnψ

sinhNψcos(ωNt + αN),

cosh2 ψ

2= m

2mN� 1, ω2

N = 2kmN

.

Formally, the value of the parameter ψ can be obtained from (5) using the substi-tution ϕ = π − iψ and assuming ψ to be large. This normal oscillation can in thefirst approximation be considered to be the simple oscillation of the small mass mN



particle while the other particles are at rest, while we afterwards consider the motionof the other particles as forced oscillations under the action of high-frequency forcekxN = kBcos(ωNt + αN), acting upon the right-hand side of the chain of N − 1 identicalparticles (cf. problem 7.6a).

Note that in this problem the oscillations of an impurity atom in a crystal are modelled.If the mass of the impurity atom is significantly different from the mass of atoms formingthe crystal, the frequency of oscillations, in which the greatest amplitude has an impurityatom, can occur in the forbidden band, and such oscillations are localized near this atom.

b) When kN+1 � k, the solution is the same as the solution of problem 7.2. WhenkN+1 � k, there are normal oscillations for which Nth particle is practically at rest:

x(s)n = As sinnϕs cos(ωst + αs), ω2

s = 4km

sin2 ϕs

2,

ϕs ≈ sπN

, s = 1, 2, . . . , N − 1.

The parameter ϕs is defined from

(2sin2 ϕ

2− kN+1

k

)sinNϕ = cosNϕ sinϕ, (6)

which in the considered approximation has the form

tanNϕ = − kkN+1

sinϕ.

Equation (6) has yet one more solution which we can obtain by putting ϕ = π − iψand assuming ψ to be large. In this case, we have

x(N)n = (−1)N+nBN

sinhnψ

sinhNψcos(ωNt + αN),

cosh2 ψ

2= kN+1

4k, ω2

N = kN+1

m;

that is, the amplitudes of the particles of this oscillation decrease towards the left-handside of the chain.

How can we obtain this last oscillation using the results of the problem 7.6?

7.9. Let ϕn be the displacement of nth pendulum from the vertical.a) The sth normal oscillation is

ϕn = As cos(n − 12 )ψs · cos(ωst + αs),



where the frequency spectrum (Fig. 152) begins with the value ω0 = √g/l:

ω2s = g

l+ 4k

msin2 ψs

2,

ψs = π sN

, s = 0, 1, 2, . . . , N − 1.

ω02

ωs2

0 1 2 3 N−1 s

Figure 152

b) In the region of the eigen-frequency of the system

ω0 < γ <

√ω2

0 + 4k/m,

the forced oscillations are

ϕn = F cos[(n − 12 )ψ]

2kl sinNψ sin(ψ/2)sinγ t;

γ 2 = gl

+ 4km

sin2 ψ

2, 0 < ψ < π .

When γ → ωs, the resonances arise because sinNψ → sinNψs → 0.In the region of low frequencies γ < ω0, all pendulums oscillate in the same phase:

ϕn =F cosh

(n − 1

2

)χ

2kl sinhNχ sinh(χ/2)sinγ t; γ 2 = g

l− 4k

msinh2 χ

2> 0.

If, at the same time, the stiffness of the springs is small

k

m(ω2

0 − γ 2) = ε � 1,

then the oscillation amplitudes rapidly decrease to the left-hand side

ϕn = ϕNεN−n.

In the high-frequency region γ >

√ω2

0 + 4k/m, the neighbouring pendulums oscil-late in anti-phase

ϕn =(−1)N−n+1F sinh

(n − 1

2

)χ

2kl sinhNχ cosh(χ/2)sinγ t, γ 2 = g

l+ 4k

mch2 χ

2.



At a very high frequency mγ 2/k � 1, the oscillation amplitudes are also decrease rapidlyto the left-hand side:

ϕn =(

− kmγ 2

)N−n

ϕN .

c) It is clear that in the linear approximation at b − a = 0 all pendulums oscillateindependently with the frequency ω0 = √

g/l.With the growth of the parameter b − a, the springs first decrease the restoring force

of gravity and then begins to “push apart” the neighbouring pendulums, causing theinstability of the small oscillations of the pendulums near the vertical.


L = 12 ml2

2N∑n=1

ϕ2n − U , U =

2N∑n=1

(−mgl cosϕn + 1

2 kr2n

),

where elongation of the nth spring is (we assume l|�n| � a)

rn =√

a2 + 4l2 sin2(�n/2) − b ≈ a − b + l2

2a�2

n − l2(a2 + 3l2)24a3 �4

n,

�n = ϕn − ϕn+1.

With the accuracy up to terms of the ϕ4n order of magnitude inclusively, we have

U = 12 mgl

∑n

(ϕ2

n − α�2n − 1

12 ϕ4n + β�4

n

)+ const,

α = (b − a)klamg

, β = 112

α + kbl3

4mga3 .

(1)

ω02

ωs2

0 1 2 3 N s

Figure 153

The equations of motion in the linear in ϕn approximation

ϕn + gl[ϕn − α(2ϕn − ϕn+1 − ϕn−1)] = 0

have solutions in the form of the travelling waves

ϕn = Aei(ωt±nψ) (2)

with the frequencies (see Fig. 153)

ω2s = g

l

(1 − 4α sin2 ψs

2

), ψs = π s

N, s = 0, 1, . . . , N,



where the frequencies ω0 and ωN are non-generate, while the rest frequencies are twofolddegenerate.

It can be seen that when

4α − 1 > 0, or b − a >mga4kl

, (3)

the oscillations are unstable; that is, some ω2s become negative. The frequency ωN is

the first to vanish, which corresponds to ψN = π . The normal oscillation is of the type of“accordion”, in which the neighbouring particles oscillate in anti-phase: ϕn = −ϕn−1. It isnatural, therefore, to look for a new equilibrium position ϕn0 in the form of “accordion”:

ϕ10 = −ϕ20 = ϕ30 = −ϕ40 = . . . = −ϕ2N0 = ϕ. (4)

The value ϕ can be found from the equilibrium condition ∂U∂ϕn

= 0 or

ϕn − α(2ϕn − ϕn+1 − ϕn−1)− 16 ϕ3

n + 2β(ϕn − ϕn+1)3 + 2β(ϕn − ϕn−1)3 = 0. (5)

It gives

ϕ = ±√

6(4α − 1)

192β − 1.

Let us now consider the small oscillations near the new equilibrium position (4). Wewill introduce the small displacements

xn = ϕn − ϕn0,

and then the potential energy (1) will be equal to the expression

U = 12 mgl

∑n

[(1 − 1

2 ϕ2)x2

n − (α − 24βϕ2)(xn − xn+1)2]+ const (6)

up to terms x2n inclusively. By comparing (1) and (6), it is easy to see that xn has a solution

in the same form as ϕn (2) with frequencies

ω2s = g

l

[1 − 1

2 ϕ2 − 4(α − 24βϕ2)sin2(ψs/2)]

.

However, now for the small ϕ2 < α24β

< 12 , all frequencies ω2

s are positive (see (3)):

ω2s >

gl

[1 − 1

2 ϕ2 − 4(α − 24βϕ2)]

= 2gl

(4α − 1) > 0;

that is, the small oscillations near the new equilibrium position (4) are stable.



Thus, with the increase of the parameter α, the original configuration of the verticalpendulum is replaced by the “accordion”. This change in symmetry of the system issimilar to the change of symmetry of the thermodynamic systems due to the phasetransitions of the second kind. In this case, the external parameters such as temperature,magnetic field, and so on will be analogous to the α (e.g., [20]).

Of course, (5) can have other non-zero solutions, except the one found in (4). Forexample, this equation is satisfied by the value ϕn = √

6, which, however, is not physical,as it corresponds to the large deflection angles, while the decomposition (1) itself is validonly for small ϕ.

7.10. a) Let we denote the current in nth coil as qn. The Lagrangian function is

L = 12

N∑n=1

[L q2

n − 1C

(qn − qn+1)2]

+ 12 L0q2

N+1 + Uq1 cosγ t

(the current through the Z chain is qN+1). The resistance R can be introduced into equa-tions of motion using the dissipative function

F = 12 Rq2

N+1.

The equations of motion are

L q1 + 1C

(q1 − q2) = U cosγ t, (1)

L qn + 1C

(2qn − qn−1 − qn+1) = 0, n = 2, 3, . . . , N, (2)

L0qN+1 + 1C

(qN+1 − qN) = −RqN+1. (3)

We are looking for a solution in the form

qn = Re{Aeiγ t−inϕ},

and we can consider, without losing generality, that −π � ϕ � π . From (2) and (3), weobtain

γ 2 = 4L C

sin2(ϕ/2), (4)

−γ 2L0 + 1C

(1 − eiϕ) = −iγ R. (5)

Hence,

R = sinϕ

γ C, L0 = 1 − cosϕ

Cγ 2 = 12 L .



Since R > 0, there must be ϕ > 0; that is, the wave travels in the direction of the L Rchains. The amplitude can then be determined from the equations (1).

For γ 2 > L C/4, the propagation of travelling waves in the artificial line is impossible(cf. problem 7.6a).

b) The equations of motion

L1q2n−1 + 1C

(2q2n−1 − q2n−2 − q2n) = 0, n = 2, 3, . . . , N,

L2q2n + 1C

(2q2n − q2n−1 − q2n+1) = 0, n = 1, 2, . . . , N, (6)

coincide, up to notations, with (1) in problem 7.5; besides,

L1qn + 1C

(q1 − q2) = U cosγ t, (7)

L0q2N+1 + 1C

(q2N+1 − q2N) = −Rq2N+1. (8)

We are looking for a solution in the form

q2n−1 = Aeiγ t−i(2n−1)ϕ,

q2n = Beiγ t−i2nϕ.(9)

Without losing generality, we can take −π � ϕ � π . From (6), we get

(1 − γ 2/γ 21 )A − cosϕ · B = 0,

cosϕ · A − (1 − γ 2/γ 22 )B = 0,

γ 21,2 = 2

L1,2C,

(10)

and

cos2 ϕ = (1 − γ 2/γ 21 )(1 − γ 2/γ 2

2 ).

Let, for example, γ1 < γ2. The condition 0 � cos2 ϕ � 1 is satisfied when 0 � γ � γ1,

which is the region of “acoustic” waves (cf. problem 7.5) and when γ2 � γ �√

γ 21 + γ 2

2 ,which is the region of “optic” waves. Outside these regions, the propagation of travellingwaves is impossible (cf. problem 7.7).

From (8), we obtain

R + iγL0 = iγ C

− iγ C

BA

eiϕ = iγ C

(1 − B

Acosϕ

)+ B

Asinϕ

γ C(11)



with the condition (B/A)sinϕ > 0. In the region γ � γ1, the amplitudes A and B have

the same signs, so that ϕ > 0. In the region γ2 � γ �√

γ 21 + γ 2

2 , on the contrary, B/A < 0and ϕ < 0. Substituting the values of B/A, cosϕ and sinϕ in (11), we obtain as a result

R =√

L1 + L2

2C

(1 − L1L2C

L1 + L2· γ 2

2

)2 − L1Cγ 2

2 − L2Cγ 2 , L0 = 12 L1.

The negative value of ϕ in the region of “optical” oscillations means that the phasevelocity of the travelling wave is directed from the Z chain to the voltage source. Thegroup velocity, on the contrary, has the opposite direction (e.g. see, Fig. 148 where vgr ∝dω(+)

ds < 0; cf. problem 7.5). The group velocity is the velocity of the wave packet. If thewave packet moves along the chain, then the oscillation energy is localized in the regionwhere the wave packet is at the moment, so naturally that the group velocity determinesthe flow of energy.

7.11. The equation for the oscillations of the discrete system (see (3) of problem 7.1)can be written in the form

xn − kaam

(xn+1 − xn

a− xn − xn−1

a

)1a

= 0. (1)

The quantity ma = Nm

Na becomes in the limit the linear density of the rod ρ. The relativeextension of the section a; that is, the quantity(xn − xn−1)/a is proportional to the forceacting upon it,

F = kaxn − xn−1

a,

and ka thus becomes in the limit the elastic modulus � of the rod. Equations (1) in thelimit thus becomes the wave equation

∂2x(ξ , t)∂t2

− v2 ∂2x(ξ , t)∂ξ2 = 0, (2)

where v = √�/ρ is the phase velocity of the wave.

Instead of a set of N coupled ordinary differential equations, we have obtained onepartial differential equation (cf. problem 4.32).

Note that in our derivation we had to make the important assumption that thefunction xn(t) tends to a well-defined limit x(ξ , t) which is a sufficiently smooth function.



7.12. If a is small, we can approximate the displacement as

xn = x(ξ , t),

xn±1 = x(ξ ± a, t) = x(ξ , t)± a∂x(ξ , t)

∂ξ+ a2

2∂2x(ξ , t)

∂ξ2 ±

± a3

6∂3x(ξ , t)

∂ξ3 + . . .

Therefore, (2) of problem 7.11 changes to

∂2x∂t2

− �

ρ

∂2x∂ξ2 − �a2

12ρ

∂4x∂ξ4 = 0. (1)

While each of the equations (1) of problem 7.11 contains the displacements of thethree neighbouring points (long-range interaction), (1) here contains the displacementx in a given point ξ (short-range interaction). The last term

− �a2

12ρ

∂4x∂ξ4

in (1) corresponds to approximate account of the small difference between the systemconsidered and a continuous one. This leads, in particular, to the fact that the phasevelocity appears to be dependent on the wavelength.

Substituting x = acos(ωt − kξ) in the equation, we obtain

v = ω

k=

√�

ρ

(1 − a2k2

12

).

Such phenomena, which are due to the long-range interaction, are called spatialdispersion.


§8

Non-linear oscillations

8.1. a) We solve the equation of motion

x + ω20x = −βx3 (1)

using the method of successive approximations (cf. problem 6.35):

x = x0 + δx = Aeiω0t + A∗e−iω0t + δx, (2)

where A(t) is the complex amplitude

A = 12 aeiϕ.

The “force”

−βx30 = −βA3e3iω0t − 3βA2A∗eiω0t − 3βAA∗2e−iω0t − βA∗3e−3iω0t

contains the resonant terms

−3βA2A∗eiω0t − 3βAA∗2e−iω0t = −3β|A|2x,

which are more convenient to attach to the term ω20x in left-hand side of (1).

This results in replacement

ω20 → ω2 = ω2

0 + 3β|A|2.

For δx, we obtain

δx + ω2δx = −β(A3e3iωt + comp. conj.),




from where

δx = βA3

8ω2 e3iωt + comp. conj.

As a result,

x = acos(ωt + ϕ)+ βa3

32ω2 cos(3ωt + 3ϕ),

ω = ω0 + 3βa2

8ω0

(cf. [1], § 28 and [7], § 29.1). Fig. 154 depicts the function x(t).

x

x

t

t

β >0

β <0

x +

t

αa2

2ω02


When β > 0, there is a “limitation” of the oscillations; when β < 0, the maximabecome sharper. These properties of the oscillations, as well as the sign of the correctionsto the frequency, can easily be considered by looking at the graph of U(x). For othersolution methods, see problems 1.9 and 11.26d.

b) Solving the problem in the same way as in point 8.1a, we obtain

δx = αA2

3ω20

e2iω0t − 2α|A|2ω2

0

+ αA∗2

3ω20

e−2iω0t;

that is

x=acos(ω0t + ϕ)− αa2

2ω20

+ αa2

6ω20

cos(2ω0t + 2ϕ).

The distortion of the oscillations is non-symmetric (Fig. 155).



In the following approximation one must take into account the term −2αx0δx in the“force” −α(x0 + δx)2 which contains the resonant terms

−2α2

3ω20

A2A∗eiω0t − 2α2

3ω20

AA∗2e−iω0t + 4α2|A|2ω2

0

x = 10α2

3ω20

|A|2x.

This results to the replacement

ω0 → ω0 − 5α2a2

12ω30

.

8.2. x = acosωt − 14 γ a2 cos2ωt + 1

4 γ a2, ω = ω0 + 116 γ 2a2ω0.

8.3. ϕ = a�2

g − l�2 cos�t + a2�4

2(g − l�2)(g − 4l�2)sin2�t (the notation is that of

problem 5.9).

8.4. x = x(0) + x(1) + . . .,

x(0) = f1 cosω1t

m(ω20 − ω2

1)+ f2 cosω2t

m(ω20 − ω2

2),

x(1) = − αf 21

2m2ω20(ω2

0 − ω21)2

− αf 22

2m2ω20(ω2

0 − ω22)2

−

− αf 21 cos2ω1t

2m2(ω20 − 4ω2

1)(ω20 − ω2

1)2− αf 2

2 cos2ω2t

2m2(ω20 − 4ω2

2)(ω20 − ω2

2)2−

− αf1f2 cos(ω1 − ω2)t

m2[ω20 − (ω1 − ω2)2](ω2

0 − ω21)(ω2

0 − ω22)

−

− αf1f2 cos(ω1 + ω2)t

m2[ω20 − (ω1 + ω2)2](ω2

0 − ω21)(ω2

0 − ω22)

.

What combinational frequencies will occur when we take into account ananharmonic correction of the form δU = 1

4 mβx4?

8.5. Let x and y be the displacements of the pendulum from the equilibrium positionin the horizontal and vertical directions. We expand the Lagrangian function

L = 12 m(x2 + y2)− 1

2 k(√

(l − y)2 + x2 − l0

)2

− mgy

in series of small x and y up to the third order inclusively

L = 12 m(x2 + y2 − ω2

1x2 − ω22y2 + 2αx2y)+ . . . ,



where

ω21 = g

l, l = l0 + mg

k, ω2

2 = km

, α = kl02ml2

.

Using the same method as in problem 8.1, we obtain

x = acos(ω1t + ϕ1)− αab2ω2(2ω1 + ω2)

cos(ω+t + ϕ+)+

+ αab2ω2(2ω1 − ω2)

cos(ω−t + ϕ−),

y = bcos(ω2t + ϕ2)+ αa2

2ω22

+ αa2

2(ω22 − 4ω2

1)cos(2ω1t + 2ϕ1),

where ω± = ω1 ± ω2 and ϕ± = ϕ1 ± ϕ2.The solutions obtained are valid as long as the frequency ω2 is not close to 2ω1. When

ω2 = 2ω1, the anharmonic corrections cease to be small and can lead to the significantpumping of the energy from the x- to y-oscillations and back. This case is considered inproblem 8.10.

8.6. a) We look for a solution in the form

x = Aeiωt + A∗e−iωt.

Equating the coefficients of eiωt, we get

(ω20 − ω2 + 2iλω + 3β|A|2)a = 1

2 f ,

and, hence,

[(ω20 − ω2 + 3β|A|2)2 + 4λ2ω2]|A|2 = 1

4 f 2.

A study of this equation which is cubic in |A|2 can be done in the same way as the studyof the analogous equation (29.4) in [1].

This equation is square in ω2, so that the dependence of |A|2 on ω2 can be easilyrepresented graphically (see [7], § 30).

8.7. a) We look for a solution of the oscillations equation

x + 2λx + ω20(1 + hcos2ωt)x + βx3 = 0 (1)



in the form

x = Aeiωt + A∗e−iωt, (2)

and we retain only the terms containing e±iωt.1 Putting the coefficients of e±iωt equal tozero, we find

12 hω2

0A +(ω2

0 − ω2 − 2iωλ + 3β|A|2)

A∗ = 0,

12 hω2

0A∗ +(ω2

0 − ω2 + 2iωλ + 3β|A|2)

A = 0.(3)

We can only have a non-vanishing A if∣∣∣∣12 hω2

0 ω20 − ω2 − 2iωλ + 3β|A|2

ω20 − ω2 + 2iωλ + 3β|A|2 1

2 hω20

∣∣∣∣ = 0. (4)

Hence,

|A|2 = 13β

[ω2 − ω2

0 ±√(

12 hω2

0

)2 − (2ωλ)2

]. (5)

|A|2

A

B

C

D Eω2

Figure 156

From (3), we get2

sin2ϕ = ImAA∗ = − 4λ

hω0,

cos2ϕ = ∓ 2

hω20

√(12 hω2

0

)2 − (2ωλ)2.(6)

Thus,

x = acos(ωt + ϕ), (7)

where A = 12 aeiϕ .

Fig. 156 shows how |A|2 depends on ω2 (to fix the ideas, we assume β > 0). In somefrequency ranges two options or three (including zero values) different amplitudes ofstable oscillations are possible.

The amplitudes corresponding to the sections AD and CD are not realized in actualcases since those oscillations are unstable (for a proof of this for the section AD, seethe next problem; for a study of the stability of the oscillations along sections ABC, CD,and DE, see [10]).

1 Assume that the terms in e±3iωt are appreciably smaller and will be compensated by the contribution tox from the third harmonic, as will become clear in the following discussion.

2 Equations (6) determine the phase apart from a term nπ . There is no sense in determining the phase withgreater precision as a change in the phase by π corresponds simply to a shift in the time origin.



b) When we take the third harmonic into account, x has the form

x = Aeiωt + A∗e−iωt + Be3iωt + B∗e−3iωt. (8)

We assume that |B| � |A|, which will be confirmed by the results. Substituting (8)into (1), we split off the terms containing e3iωt; we then drop the product of B withsmall parameters. We find out that

B =(

116 h + 1

8 βA2ω−2)

A (9)

and, indeed, |B| � |A|.Therefore,

x = acos(ωt + ϕ)+ bcos(3ωt + ψ),

where b = 2|B|, ψ = argB.It is easy to notice that the fifth harmonic turns out to be smaller than second order

(∼ h2A), the seventh ∼ h3A, and so on. The even harmonics do not occur. This is thebasis of the method used to evaluate the amplitudes.

8.8. a) We look for a solution to the equation of motion in the form

x(t) = a(t)cosωt + b(t)sinωt, (1)

where a(t) and b(t) are slowly changing functions of the time. To determine a(t) and b(t),we get the following set of equations (cf. [1], § 27)

a +(ω − ω0 + 1

4 hω0

)b = 0,

b −(ω − ω0 − 1

4 hω0

)a = 0.

(2)

If |ω1 − ω0| < 14 hω0, its solution is

a(t) = α1(C1e−st + C2est),

b(t) = α2(C1e−st − C2est),(3)

where

s = 14

√(hω0)2 − 16(ω − ω0)2, α1,2 = √

hω0 ± 4(ω − ω0).

Hence,

x = C3est cos(ωt + ϕ)+ C4e−st cos(ωt − ϕ), (4)

where tanϕ = α2/α1 (Fig. 157).



The oscillations thus increase, generally speaking, without limit. The rate of theirincrease which is characterized by the quantity s is, indeed, small. In actual cases,the increase in the amplitude of the oscillations is cut off, for instance, if the influenceof the anharmonic terms becomes important (see problem 8.7) or the reaction of theoscillations on the device which periodically changes its frequency becomes important.

It is useful to draw attention to the analogy between the results obtained and theparticular solution of the problem of the normal oscillations of a chain of particlesconnected by springs of different stiffness (problem 7.5b). An inhomogeneity withperiod 2a along the chain leads to a build-up along the chain of the amplitude of thestable oscillations, when the “wavelength” is equal to 4a (Fig. 151) in a similar way thata periodic change with time of the frequency of an oscillator will lead to an increase inthe amplitudes with time.

An even more complete analogy can be observed in problem 7.7. The region ofinstability with respect to parametric resonance corresponds to the forbidden zone inthe spectrum of oscillations of the chain.

Similar equations are obtained in quantum mechanics in the problem of the motionof a particle in a periodic field. In that problem, we also met with “forbidden bands” and“surface states”.

b) If |ω − ω0| > 14 hω0, we have

x = Cβ1 sin(�t + ψ)cosωt − Cβ2 cos(�t + ψ)sinωt,

where � = 14

√16(ω − ω0)2 − (hω0)2,

β1,2 ={ √

4(ω − ω0)± hω0, when ω > ω0,±√

4(ω0 − ω)∓ hω0, when ω < ω0.

The oscillations are beats:

x = C√

4|ω − ω0| ∓ hω0 cos(2�t + 2ψ)cos(ωt + θ), when ω ≷ ω0,

where θ is a slowly varying phase (see Fig. 158). If the frequency approaches the limitof the instability region, the depth of the modulation of the oscillations approaches thetotal amplitude and their period increases without limit.

x

t

x

t


What is the form of the oscillations when |ω − ω0| = 14 hω0?



8.9. Let x = eiω1t, when 0 < t < τ . We then have in the interval τ < t < 2τ ,

x = aeiω2t + be−iω2t

where a and b are determined from the “matching” condition for t = τ :

x(τ − 0) = x(τ + 0), x(τ − 0) = x(τ + 0).

Hence, we have

a = ω1 + ω2

2ω2ei(ω1−ω2)τ ,

b = ω2 − ω1

2ω2ei(ω1+ω2)τ .

Similarly, we find that for 2τ < t < 3τ

x = αeiω1t + βe−iω1t,

where

α = e−iω1τ(

cosω2τ + iω2

1 + ω22

2ω1ω2sinω2τ

),

β = i sinω2τ e3iω1τ ω21 − ω2

2

2ω1ω2.

It is clear that the oscillation in the form

Aeiω1t + Be−iω1t (1)

for 0 < t < τ , after a period of 2τ will become

A(αeiω1t + βe−iω1t)+ B(α∗e−iω1t + β∗eiω1t) == (αA + β∗B)eiω1t + (βA + α∗B)e−iω1t.

We now look for such a linear combination (1) that it retains its form—apart froma multiplying factor—after a period 2τ :

(αA + β∗B)eiω1t + (βA + α∗B)e−iω1t = μ(Aeiω1t′ + Be−iω1t′),

where t′ = t − 2τ ,



αA + β∗B = μe−2iω1τ A,

βa + α∗B = μe2iω1τ B.(2)

Set (2) has a non-trivial solution, provided

(α − μe−2iω1τ )(α∗ − μe2iω1τ )− ββ∗ = 0,

whence

μ1,2 = γ ±√

γ 2 − 1,

where

γ = Re(αe2iω1τ ) = cosω1τ cosω2τ − ω21 + ω2

2

2ω1ω2sinω1τ sinω2τ .

After n periods, the oscillation

x1,2 = A1,2(eiω1t + λ1,2e−iω1t), 0 < t < τ ,

λ1,2 = μ1,2e−2iω1τ − α

β∗

has changed to

x1,2(t) = μn1,2A1,2(eiω1t′ + λ1,2e−iω1t′), 0 < t′ = t − 2nτ < τ .

Any oscillation is a superposition of oscillations such as x1,2; in particular, the realoscillation (which is the only one which has a direct physical meaning)

x(t) = Aeiω1t + A∗e−iω1t, 0 < t < τ ,

0

1

2

3

4

5

1 2 3 4 5 6

(ω22 −ω1

2)τ2

2π2

(ω12 +ω2

2)τ2

2π2

Figure 159



is the sum of x1(t)+ x2(t) with

A1 = A∗ − λ2Aλ1 − λ2

, A2 = λ1A − a∗

λ1 − λ2.

If γ < 1, |μ1,2| = 1 and the oscillations x1,2(t) (and at the same time x(t)) remainbounded.

If, however, γ > 1, we have μ1 > 1, and the amplitude of the oscillations increaseswithout bound. This is the case of the onset of parametric resonance. We can easilyverify that if the frequency difference is small, |ω1 − ω2| � ω1, this condition is satisfiedif the frequencies lie close to nπ/τ :

|(ω1 + ω2)τ − 2πn| <(ω1 − ω2)

2τ

ω1 + ω2.

We show in Fig. 159 (taken from [17]) the regions of instability against parametricresonance.


x + ω2x − 2αxy = 0,

y + 4ω2y − αx2 = 0.

We look for a solution in the form

x = Aeiωt + A∗e−iωt + δx,

y = Be2iωt + B∗e−2iωt + δy,

assuming that A and B are the slowly varying amplitudes of the oscillations, while morerapidly oscillating terms δx and δy can be neglected: |A| � ω|A| � ω2|A|, |B| � ω|B| �ω2|B|, δx ∼ δy � |A|.

Leaving only the terms with eiωt (e2iωt, respectively) and neglecting |A|, |B|, we get

ωA + iαBA∗ = 0,

4ωB + iαA2 = 0.(1)

Clearly, from (1) follows

|A|2 + 4|B|2 = C = const (2)

(this is the law of conservation of energy) and

A∗2B + A2B∗ = D = const. (3)



Using (1), we find

ωddt

|A|2 = −iα(A∗2B − A2B∗). (4)

Squaring (4) and taking into account (2) and (3), we get( d

dt|A|2

)2 = −α2

ω2

[(A∗2B + A2B∗)2 − 4|A|4|B|2

]=

= α2

ω2

[|A|4(C − |A|2)− D2

].

(5)

This equation is similar to the law of conservation of energy for the problem aboutthe one-dimensional motion of a particle with a coordinate |A|2. It is helpful tostudy this equation using the graph of the “potential energy” U(|A|2) = (|A|2 − C)|A|4(see Fig. 160).

U

−D2

|A|2

Figure 160

Fig. 160 shows that the amplitude |A| experiencesoscillations which lead to the beats. The dependenceof the amplitudes |A| and |B| on the time can beexpressed in the elliptic functions (we will not do thishere).

Note that, in this case, not only the depth ofbeating, but also the period will depend on the initialamplitudes and phases. This is in direct contrast tothe vibrations of oscillators with linear coupling (see problem 6.9).

This problem is relevant, for instance, to the coupling of the longitudinal and flexuraloscillations of a molecule CO2 (the so-called Fermi resonance; see [18]) and to the doublingand division of the light frequency in nonlinear optics (see [19]).

8.11. ω2 = a2γ 2

2l2+ g

l . When a2γ 2

2l2>

gl , a second stable equilibrium position appears

in the form the straight upward vertical; the oscillation frequency around it is equal to

ω2 = a2γ 2

2l2− g

l (see [1], § 30, problem 1).

8.12. a)

Ueff = α2

mω2

[a2

r6 + 3(ar)2

r8

], (1)

Notice that the dependence Ueff ∝ r−6 is characteristic of intermolecular forces. If wesubstitute in (1) the values1 α ∼ e2 ∼ (5 · 10−10 ESU)2, a ∼ 10−8 cm, and ω ∼ 1016 s−1,which are typical for atoms, and as a mass we choose the electron mass, m ∼ 10−27 g,

1 In the CI system: α ∼ e2

4πε0∼ (10−19 K)2

10−11 (F/m) · 10, m ∼ 10−30 kg, a ∼ 10−10 m, Ueff ∼ 10−18 (a/r)6 J.



we get Ueff ∼ 10−40 erg · cm6/r6, which is close to the correct value for van der Waalsinteraction, as far as order of magnitude is concerned. This result may serve as anindication of the physical nature of this interaction. A complete calculation of the vander Waals forces is only possible using quantum mechanics.

b)

Ueff = α2

m(ω2 − ω20)

[a2

r6 + 3(ar)2

r8

],

where ω0 is the eigen-frequency of the oscillator.

8.13. The motion along the z-axis is nearly uniform, z = vt. In the xy-plane, the particleis acted upon by a fast oscillating force

fx = 2Axsinkvt, fy = 2Aysinkvt.

The corresponding effective potential energy is Ueff = 12m�2(x2 + y2), where

� = √2A/(mkv).

According to the initial conditions, we have for the frequency of the force oscillationkv � so that the force is, indeed, fast oscillating. Thus, in the xy-plane, the particleperforms a harmonic oscillation with frequency � around the z-axis.

This problem illustrates the principle of strong focus of particle beams in accelerators.


mx = ec

B(x)y,

my = − ec

B(x)x.

We look for the law of motion in the form

x = X + ξ , y = Y + η, (1)

where the terms ξ and η describe the fast motion over the almost circular orbit, whileX and Y are the slow displacements of its centre (cf. [1], § 30). Substituting (1) into theequations of motion, we expand B(X + ξ) in powers of ξ :

X + ξ = ωY + ωη + emc

B′(X)ξ(Y + η),

Y + η = −ωX − ωξ − emc

B′(X)ξ(x + ξ ),



and then consider separately the fast oscillating and slowly changing terms. For theoscillating terms, we obtain

ξ = ωη, η = −ωξ , ω = emc

B(X),

where

ξ = r cosωt, η = −r sinωt.

For the slowly changing terms, we have

X = ωY + emc

B′(X)〈ξ η〉,Y = −ωX − e

mcB′(X)〈ξ ξ 〉,

(2)

where

〈ξ η〉 = −r2ω〈cos2 ωt〉 = −12 r2ω, 〈ξ ξ 〉 = 0.

Since X , Y ∼ εωx, εωY , the left parts in (2) can be put equal to zero.As a result,

Y = er2

2mcB′(X) = 1

2 εv, x = 0.

The rate of the displacement of the orbit centre (the drift velocity) in a more general caseis considered in [2], § 22, problem 3 and in [10], § 25.

8.15. The equation of motion of the ball is

my = −dU(y)dy

+ f (t).

The proper motion of the ball under the action of a spring is described by the “low-frequency” displacement x = y − y0 cosγ t, for which

mx = −dU(x + y0 cosγ t)dx

.

If we average this proper motion over a period of 2π/γ of the high-frequency motionusing

〈cos2n+1 γ t〉 = 0, 〈cos2 γ t〉 = 12 ,



we get the effective force and the corresponding effective potential energy

Ueff(x) = Ax2 + Bx4, A = −C + 3By20.

The graph of the function Ueff(x) is shown in Fig. 161.

Ueff

A>0

A<0−x0 x0

x

Figure 161

When A > 0 or

T = y20 > Tc = C

3B,

the ball oscillates near the point x = 0 with the frequency

ω = √2A/m ∝ √

T − Tc.

When A < 0 or T < Tc, the minima of Ueff(x) are locatedat the points

±x0 = ±√−A

2B,

and the ball oscillates near one of the points with the frequency

ω =√−4A

m∝ √

Tc − T .

The emerging picture is very close to the picture of the phase transitions of thesecond kind, described by the phenomenological theory of Landau [20]. The fast forcedoscillations are analogous to the thermal motion (corresponding to the optical modesof oscillations of the system which is not connected with the transition), and the valueT = y2

0 is the analog of temperature. When T is large, the system oscillates aroundthe equilibrium position x = 0. In this case, there is the symmetry with respect to thereplacement of x → −x. When the temperature decreases reaching a value T < Tc, theball begins to oscillate around one of the new equilibrium positions: x0 or −x0. In thiscase, the symmetry x → −x obviously vanishes. Moreover, the value Tc is an analog of thetemperature of the phase transition of the second kind. In the neighbourhood of T = Tc,the value x0 is small, x0 ∝ √

Tc − T , and frequency ω of the eigen-oscillations is small.


§9

Rigid-body motion. Non-inertialcoordinate systems

9.1. Let us introduce the generalized coordinates as follows: the angle of the stringdeflection from the vertical, ϕ, and the angle between the vertical and the radius of thering connecting the point of its attachment to the centre of the ring, ψ . The kinetic energyof the ring is the sum of the energy of its rotation around the ring centre 1

2 mR2ψ2 and theenergy of the translational motion 1

2 m(Rϕ + Rψ)2 (since the angles ϕ and ψ are small,we can consider velocities of the suspension point and the centre of the ring as parallel).The potential energy is determined by the position of the ring centre

U = mgR[(1 − cosϕ)+ (1 − cosψ)] ≈ 12 mgR(ϕ2 + ψ2).

Using Lagrangian function

L = 12 mR2

[(ϕ + ψ)2 + ψ2 − g

R

(ϕ2 + ψ2

)],

we get the equations of motion

ϕ + ψ + gR

ϕ = 0, ϕ + 2ψ + gR

ϕ = 0.

The solution of these equations has the form

(ϕ

ψ

)=

(−1 + √5

2

)A1 cos(ω1t + χ1)+

(−1 − √5

2

)A2 cos(ω2t + χ2),

where the normal frequencies are equal to

ω1,2 =√

3 ± √5

2gR

.




9.2. a)

⎛⎝2a2(m + M) 2a2(m − M) 0

2a2(m − M) 2a2(m + M) 00 0 4a2(m + M)

⎞⎠ ,

b)

⎛⎝4a2M 0 0

0 4a2m 00 0 4a2(m + M)

⎞⎠ .

9.3. For both cases, one of the principal axes is the axis perpendicular to the plane ofthe figure going through the centre of mass of the figure (z-axis). The principal x-axis isorientated at an angle ϕ with respect to O′x′ of each of two figures. The principal y-axisis perpendicular to the x-axis. Both these axes pass through the centre of mass of thefigures.

a) The coordinates of the centre of mass in the O′x′y′ system (x′ = b, y′ = a) are

Izz = 2(a2 + b2)(M + m),

Ixx = (a2 + b2)(M + m)∓√

(b2 − a2)2(M + m)2 + 4a2b2(M − m)2,

Iyy = (a2 + b2)(M + m)±√

(b2 − a2)2(M + m)2 + 4a2b2(M − m)2,

when a ≷ b, ϕ = 12 arctan 2ab(M − m)

(a2 − b2)(M + m).

x

x′′

x′

y′′y′

O′m

m

y

2mO

ϕ

Figure 162

b) The coordinates of the centre of mass O in the O′x′y′z′system (Fig. 162) are x′ = y′ = a, z′ = 0. In the Ox′′y′′z′′ coordi-nate system with the axes which are parallel to the axes x′y′z′, theinertia tensor is

I ′′ik = 4ma2

⎛⎝3 1 0

1 1 00 0 4

⎞⎠ .

In the transition to the Oxyz system, which is rotated over an angleϕ around z′′-axis, the coordinates are converted to:

x = x′′ cosϕ + y′′ sinϕ, y = −x′′ sinϕ + y′′ cosϕ, z = z′′,



and the components of the inertia tensor, as products of the coordinates, are

Ixx = I ′′xx cos2 ϕ + 2I ′′

xy sinϕ cosϕ + I ′′yy sin2 ϕ =

= 4ma2(3cos2 ϕ + sin2ϕ + sin2 ϕ),

Iyy = 4ma2(cos2 ϕ − sin2ϕ + 3sin2 ϕ),

Izz = 16ma2,

Ixy = 4ma2(−sin2ϕ + cos2ϕ),

Ixz = Iyz = 0.

We take the angle ϕ such that the condition Ixy = 0 will satisfy; for instance,ϕ = π/8. Then

Ixx = 4ma2(2 + √2), Iyy = 4ma2(2 − √

2).

9.4. In =∑i,k

Iiknink.

9.5. First, we express the kinetic and potential energy for the solid disc and then makethe corrections by subtracting the contribution of the hole.

Moments of inertia of the disc relative to its centre are

Id3 = 1

2 mR2, Id1 = 1

4 mR2,

where m is the mass of the disc. With respect to the suspension point A, the moments ofinertia are

IdAi = Id

i + mR2, IdA3 = 3

2 mR2, IdA1 = 5

4 mR2.

The contribution of the hole can be obtained by making the substitutionm → −m/4, R → R/2. As a result, the moments of inertia for the tag with respectto the suspension point are equal to

I3 = 4532

mR2, I1 = 4564

mR2.

Kinetic energy of the tag at oscillations in the plane of the disc or across are 12 I3ϕ

2 and12 I1ϕ

2, respectively.Potential energy of the disc at an angle ϕ (in the plane of the disc or across) is Ud(ϕ) =

mgR(1 − cosϕ). After the same substitutions (that is, taking the hole into account), thepotential energy of the tag equals U(ϕ) = 7

8 mgR(1 − cosϕ) ≈ 716 mgRϕ2.



As a result, the frequency of small oscillations for the motion in the plane of the tag is

ω3 =√

2845

gR

,

while for the motion across the plane of the tag, it is

ω1 =√

5675

gR

.

9.6. The centre of mass is a point on the axis of symmetry at a distance

(R − r)r3

R3 − r3

to the left of the centre of the ball. The body is a symmetric top. The moment of inertiawith respect to the axis of symmetry is

I3 = mR3 − r3

25

(R5 − r5),

and with respect to any two perpendicular axes passing through the centre of mass, themoment of inertia is

I1 = I2 = mR3 − r3

[25

(R5 − r5)− (R − r)2r3R3

(R3 − r3)

].

9.7. Dik = (∑n Inn

)δik − 3Iik (see [2], § 99).

9.8. The centre of mass lies on the axis of the hemisphere at a distance 38R from the

centre of the ball, where R is the radius of the ball. The moment of inertia around anyaxis perpendicular to the axis of symmetry is

I = 25 mR2 − m

(38 R

)2 = 83320 mR2,

where m is the mass of the hemisphere. The centre of mass can only move along thevertical. Let ϕ be the angle over which the hemisphere is turned, and z the height of thecentre of mass above the plane so that

z = R − 38 Rcosϕ.


L = 12 Iϕ2 + 1

2 mz2 − mgz,



or when ϕ is small

L = 12 Iϕ2 − 3

16 mgRϕ2.

The frequency of the small oscillations is thus given by

ω =√

12083

gR

.

9.9. The current distances from the centre of mass of the Earth–Moon system to theEarth and the Moon are equal to m

M + mR and MM + mR, respectively, and the angular

momentum of the system is

m(

MRM + m

)2

�M + M(

mRM + m

)2

�M + I�E = J�M + I�E,

J = MmM + m

R2, (1)

where �M and �E are the angular velocities of the Moon around the Earth and theEarth around its own axis (�E/�M ≈ 28). In (1), we considered the Moon as a materialpoint, and for the Earth we took into account the rotation around the centre of mass andaround its own axis (with the moment of inertia I = 2

5Ma2).At the moment when a day becomes equal to a month, the angular velocity of the

Earth’s rotation ω will coincide with the angular velocity of the Moon. Simultaneouslythe distance from the Earth to the Moon (according to the Kepler’s third law) will becomeequal to R(�M/ω)2/3, and the angular momentum

[J(

�M

ω

)4/3

+ I

]ω. (2)

From (1) and (2), we find for the dimensionless variable x = ω/�E the equation

x(1 + k − x)3 = k3 �M

�E, (3)

where

k = J�M

I�E= 5

2

(Ra

)2 mM + m

�M

�E≈ 3.8.

Equation (3), or x(4.8 − x)3 = 2.01, has two real roots: x1 ≈ 1/55 and x2 ≈ 4.The first root corresponds to the future, and the second root, to the past. Accordingly,in the first case, the month will be equal to 55 current days, whereas in the second case,



the month equalled 6 hours. The distance from the Earth to the Moon will be 1.6R, andit used to be 2.6a (in this model!).

For more realistic (than the one considered here) models of evolution of the Earth–Moon system see, for instance, [21], Ch. 2.

9.10. a) The body rotates with the angular velocity 27 ω; the fraction 5

7 of the initial kineticenergy transfers into the heat.

b) The line of the centres rotates with the angular velocity of√

27 ω around the direction

of the angular momentum M, which has an angle of 45◦ with respect to this line. Thebody rotates around a line of the centres with the angular velocity 5

14 ω; the fraction 1928

of the initial kinetic energy transfers into the heat.

9.11. f1,2 = m(

g ± v2

14r

).

9.12. It is helpful to use the moving coordinate system with the origin at centre of massand axes x1, x2, x3 parallel to the edges AB = a, AD = b, AA′ = c of the parallelepiped.In this system, the angular velocity is

� = �n, where n = l/l, l = (a, b, c),

and the angular momentum of the parallelepiped

M = (I1�1, I2�2, I3�3),

where

I1 = 23 m(b2 + c2) = 2

3 m(l2 − a2), I2 = 23 m(l2 − b2), I3 = 2

3 m(l2 − c2)

are its principal moments of inertia. Thus,

M = 23 m�l2 n − 2�m

3l· (a3,b3,c3).

The vector M is fixed with respect to the system x1x2x3; that is, it rotates in the laboratorysystem with the angular velocity �, and therefore, M = [�,M].

Let the forces acting upon the parallelepiped at the points A and C′ be equal to−f and f (we do not take into account gravity). The moment of these forces K = [l, f ].The equation of motion M = K leads to the equality

�[n ,M] = l[n , f ],



which allows us to determine the component f⊥ of the force f perpendicular to thevector n:

f⊥ = �

lM⊥ = �

l{M − n(Mn)} =

= 2m�2

3l4(a4 + b4 + c4) · (a, b, c)− 2m�2

3l2· (a3, b3, c3).

The component of the force f parallel to the diagonal AC′ cannot be found in themodel which considers the parallelepiped (and hinges) as a non-deformable rigid body.It is easy to see that if we act upon the parallelepiped at the points A and C′ by forcesNn and −Nn, we will not affect its motion.

As a result, the forces applied to the A and C′ hinges are equal to f and −f ,

f = −2m�2

3l2(a3, b3, c3)+ N ′(a, b, c),

where N ′ is an indefinable value. We have introduced

N ′ = N − 2m�2

l5(a4, b4, c4).

In a laboratory system, the vector f⊥ rotates with the angular velocity �.

9.13. The moment of inertia of the ellipsoid with respect to the axis of symmetry isI3 = 2

5Ma2; with respect to any perpendicular to that axis and passing through the centreof mass, we have I1 = 1

5M(a2 + c2), where M is the mass of the ellipsoid.The impinging particle of mass m M transfers to the ellipsoid a momentum

p = (px,py,pz) = mv(0,−1,0) and an angular momentum M = mv(ρ1,0,−ρ2).In the system of reference moving with the velocity p/(M + m), we find (cf. [1],

§ 33 and [7], § 47) that the ellipsoid will rotate around the c-semi-axis with anangular velocity

�3 = Mz

I3= −5mvρ2

2Ma2 ,

while at the same time it is precessing around the direction of M with an angular velocity

� = |M|I1

=5mv

√ρ2

1 + ρ22

M(a2 + c2).

9.14. We denote the angular velocity of rotation of the disc around its axis as ψ

and the angle between this axis and the direction on the north as ϕ. The angularvelocity of the disc in the inertial system is ω = �+ ϕ + ψ , while its projections are



ω3 = ψ + �cosα cosϕ on the disc axis, ω1 = ϕ + �sinα on the vertical, and ω2 =�cosα sinϕ on the horizontal axis perpendicular to the disc axis. The Lagrangianfunction equals the kinetic energy (we take into account that I1 = I2):

L = 12 I1(ϕ + �sinα)2 + 1

2 I1�2 cos2 α sin2 ϕ + 1

2 I3(ψ + �cosα cosϕ)2.

It is helpful to study the motion using the integrals of motion pψ and E = pψψ + pϕϕ − L:

pψ = I3(ψ + �cosα cosϕ),

E = 12 I1(ϕ

2 − �2 sin2 α)− 12 I1�

2 cos2 α sin2 ϕ++1

2 I3(ψ2 − �2 cos2 α cos2 ϕ).

Excluding ψ , we find

E = 12 I1ϕ2 + Ueff(ϕ)+ const,

where

Ueff(ϕ) = −pψ�cosα cosϕ + 12 I1�2 cos2 α cos2 ϕ.

Let us consider in detail the important case of ψ � only. Then pψ ≈ I3ψ and

Ueff(ϕ) ≈ −pψ�cosα cosϕ.

The function Ueff(ϕ) has a minimum at ϕ = 0 (i.e., in the north direction). The axis ofthe gyrocompass oscillates around this direction. For small oscillations, we have

Ueff(ϕ) = 12 I3ψ�cosα · ϕ2 + const,

and the oscillation frequency of the axis is

√I3

I1�ψ cosα.

For example, for a gyroscope making about ten thousand turns per minute, the oscillationperiod equals approximately half a minute

(for I3

I1cosα ∼ 1

).

How can we take the moment of inertia of the gyrocompass’ frame into account?

9.15. It is helpful to use the moving coordinate system with the vertical z-axis andthe x-axis which passes through the following points: the O and the point of contact ofthe disc with the table surface. This system of reference rotates around the z-axis with



the angular velocity ϕ. Here we have the equation for the angular momentum projectionson the moving axes

(dMdt

)i+ [ϕ, M]i = Ki, (1)

where M is the angular momentum of the top with respect to the stationary point O,and K is the moment of the forces acting upon the top (see [1], § 36 and [7], § 45). Theprojection of (1) onto the x-axis is

Mx − ϕMy = Kx.

It is evident that My = 0 and Kx = 0; therefore, Mx = const.Let I1 = I2 �= I3 be the principal moments of inertia of the top with respect to point O.

In the initial moment, we have

M = �I3, Mx = �I3 cosθ .

When the slipping stops, the x-axis becomes the instant axis of rotation; that is, theangular velocity of the top ω will be directed along the x-axes, and Mx = I3ωcos2 θ +I1ω sin2 θ . Thus,

ω = �I3 cosθ

I3 cos2 θ + I1 sin2 θ.

Note that ϕ = −ωtanθ when the slipping has stopped.

9.16. Let a = b �= c be the semi-axes of the ellipsoid, R, �, and , the sphericalcoordinates of the centre of mass of the ellipsoid, and θ , ϕ, and ψ the Euler angles, andlet the x3-axis of the moving system of reference be along the c-semi-axis. The kineticenergy of the body is (see [1], § 35)

T = 12 m(R2 + R2�2 + R2 2 sin2 �)++ 1

2 I1(ϕ2 sin2 θ + θ2)+ 1

2 I3(ϕ cosθ + ψ)2,(1)

where I1 = I2 = 15 m(a2 + c2) and I3 = 2

5 ma2 are the moments of inertia of the ellipsoidwith respect to the x1-, x2-, and x3-axes.

The given potential energy for the interaction of the ellipsoid with the gravitationalcentre can be transformed into the form

U = −GmMR

− GMD4

· 3cos2 α − 1R3 , (2)

where D = 2(I1 − I3) and α is the angle between the radius vector R and the x3-axis.



The unit vector e3, which determines the direction of the x3-axis, has the components

e3 = (sinθ sinϕ, −sinθ cosϕ, cosθ).

Hence, we have

cosα = Re3

R= cosθ cos�+ sinθ sin�sin(ϕ − ). (3)

From (1), (2), and (3), we finally get the expression for the Lagrangian functionL = T − U .

9.17. Let us first consider the Sun’s influence alone without the Moon’s influence. Theorigin of the coordinate system will coincide with the Sun; the X3-axis will be directedperpendicular to the plane of the Earth’s orbit, and the x3-axis, to the north (see notationsof the preceding problem).

The angular velocity of the Earth’s axis precession ϕ is evidently small compared to thedaily ψ and the annual velocities of the Earth’s rotation. Therefore, in the Lagrangianfunction we save only the first-order terms in ϕ. Moreover, we put R, � = π/2, and

as the constants; we also average cos2 α over year: 〈cos2 α〉 = 12 sin2 θ . As a result, we

obtain

L = 12 I1θ

2 + 12 I3(ψ

2 + 2ψϕ cosθ)− 3GMm(a2 − c2)

20R3 sin2 θ .

Since pψ = I3(ψ + ϕ cosθ) and pϕ = I3ψ cosθ are conserved, it follows that ψ andθ = 23◦ are conserved as well (up to values of the order of ϕ). The equation of motionfor the angle θ

I1θ + I3ϕψ sinθ + 3GMm(a2 − c2)

10R3 sinθ cosθ = 0

(taking into account that θ ∼ θ2 ∼ ϕ2) leads to the equation

ϕ = − 3GM

4R3ψ

a2 − c2

a2 cosθ .

Substituting

a2 − c2

a2 ≈ 2(a − c)a

,GM

R3 = 2,



where 2π/ = 1year, we get

ϕ ≈ −32

a − ca

2

ψcosθ ≈ −16′′ per year.

The angular velocity of the precession caused by the Moon is obtained from (2) byreplacement the mass of the Sun M with the mass of the Moon and R with the distancefrom the Earth to the Moon. This angular velocity turns out to be −31′′ per year. Thetotal angular velocity is ϕ = −48′′ per year. The observed value ϕexp = −50.2′′ per year(see [21], Ch. 2).

Thus, the Earth’s axis rotates around the x3-axis with a period of about 26 thousandyears in the opposite direction with respect to the rotation of the Earth around the Sun(the so-called precedence of the equinoxes).

9.18.

M1 +( 1

I2− 1

I3

)M2M3 = K1,

M2 +( 1

I3− 1

I1

)M3M1 = K2,

M3 +( 1

I1− 1

I2

)M1M2 = K3.

When I1 = I2 and K1,2,3 = 0, we get

M1 = Bcos(ωt + α), M2 = Bsin(ωt + α), M3 = const,

where ω =(

1I1

− 1I3

)M3 (see [1] § 36; [7] § 47, and compare problem 10.21).

9.19. Let us consider the motion around an axis which lies close to the principal x1-axis.From the Euler equation (see [1], equation (36.5))

�1 + I3 − I1

I1�2�3 = 0,

we get �1 = const, neglecting terms proportional to �2,3/�1 1. The two otherequations become linear in �2 and �3. Assuming that

�2,3 ∝ est, (1)

we get for s the equation

s2 = (I1 − I3)(I2 − I1)

I2I3�2

1. (2)



When I2 < I1 < I3 or I3 < I1 < I2, (2) has real roots so that according to (1) therotation around the x1-axis is unstable.

If, however, I1 is either the largest or the smallest moment of inertia, (2) has imaginaryroots; that is, the change in �2 and �3 are in the nature of oscillations and rotation aroundthe x1-axis is stable.

9.20. The motion of the ball is determined by the equations

mv = mg + f , (1)

Iω = [a, f ], (2)

v + [ω, a] = 0, (3)

where m is the mass of the ball; I = 25 ma2, its moment of inertia; a, radius of the ball

directed to the point of its touch with by cylinder; f , the force applied upon the ball atthis point (i.e. the sum of the reaction and friction forces); v, the velocity; and ω, theangular velocity of the ball.

It is helpful to use the cylindrical coordinates with the z-axis directed along the axis ofthe cylinder. It is necessary to take into account that projection of the derivatives of anyvector A on the axes of the moving system of reference is determined by the equations

(A)ϕ = Aϕ + [ϕ, a]ϕ = Aϕ + ϕAr,

(A)r = Ar + [ϕ, A]r = Ar − ϕAϕ ,(4)

where r = b − a, ϕ, and z are the coordinates of the centre of the ball and ϕ = vϕ/r. Fromthe equations

mvϕ = fϕ , Iωz = afϕ , vϕ + aωz = 0

we obtain

vϕ = const, ωz = const, fϕ = 0;

while from the equations

mvz = −mg + fz, vz − aωϕ = 0,

I(ωϕ + ϕωr) = −afz, I(ωr − ϕωϕ) = 0

it follows that

(I + ma2)ωϕ + Iϕ2ωϕ = 0,



and hence

ωϕ = C cos(�t + α), � =√

II + ma2 ϕ =

√27

vϕ

r,

ωr = − 5gr2avϕ

+√

72

C sin(�t + α),

z = z0 + aC�

sin(�t + α).

Thus, the ball makes the harmonic oscillations in height, and the radial component ofthe angular velocity is varying at the same time with these oscillations.

9.21. a) We use as our generalized coordinates the X- and Y-coordinates of the centreof the disc and the Euler angles ϕ, θ , and ψ (see [1], § 35). We take the Z-axis to bevertical and the (moving) x3-axis along the axis of the disc. The angle ϕ is betweenthe intersection of the plane of the disc with the XY-plane and the X-axis (the line ofnodes1). The Lagrangian function is

L = 12 m(X2 + Y2 + a2θ2 cos2 θ)+ 1

2 I1(θ2 + ϕ2 sin2 θ)+

+12 I3(ϕ cosθ + ψ)2 − mgasinθ ,

where I1 = I2 and I3 are the principal moments of inertia of the disc and the x1-, x2-, andx3-axes are along its principal axes; a is the radius of the disc and m, its mass; the centreof the disc is at a distance Z = asinθ from the XY-plane. The generalized momenta

mX = pX , mY = pY ,

I1ϕ sin2 θ + I3 cosθ (ϕ cosθ + ψ) = pϕ ≡ MZ ,

I3(ϕ cosθ + ψ) = pψ ≡ M3

(1)

together with the energy are integrals of motion. In the coordinate system moving withthe constant velocity (X , Y , 0), the centre of the disc moves only vertically. From (1), wehave

ϕ = MZ − M3 cosθ

I1 sin2 θ, ψ = M3

I3− MZ − M3 cosθ

I1 sin2 θcosθ , (2)

1 In the following discussion, we will use the ξ- and η-axes lying in the horizontal plane: the ξ-axis is alongthe line of nodes and the η-axis is at right angles to the ξ-axis; that is, eη = [eZ ,eξ ].



and substituting this into the expression for energy, we obtain

E = 12

(I1 + ma2 cos2 θ)θ2 + M23

2I3+ (MZ − M3 cosθ)2

2I1 sin2 θ+ mgasinθ . (3)

The dependence θ(t) is determined from this equation in the form of quadratures, afterwhich one can get ϕ(t) and ψ(t) from (2). The dip angle of the disc oscillates, and theprecession velocity ϕ and the angular velocity of rotation around the axis of the disc ψ

change at the same time. Equations (2) and (3) are similar to those for the motion of aheavy symmetric top (e.g. see, equations (4)–(7) from [1], § 35, problem 1).

The rolling of the disc is stable if �23 > mgaI1/I2

3 . The rotation is stable if �2Z > mga/I1.

b) If the rolling disk is not slipping, in addition to the force of gravity mg and thevertical reaction force Q, there is also the horizontal friction force f acting on the disc.

It is helpful to write the equation of motion

M = [a, Q] + [a, f ]

and its components along the Z-, x3-, and ξ-axes1

MZ = fξ acosθ , M3 = fξ a, (4)

ddt

∂L∂θ

− ∂L∂θ

= fηasinθ . (5)

Here a is the vector from the centre of the disc to the point of its contact with the plane.If we take the components of the equation

mV = f + Q + mg

along the ξ- and η-axes (see equation (4) of the preceding problem), we have thefollowing equations for fξ and fη:

fξ = m(V)ξ = m(V ξ − ϕVη),

fη = m(V)η = m(V η + ϕVξ ).(6)

The condition of rolling without slipping V + [�, a] = 0 leads to

Vξ = −a(ψ + ϕ cosθ), Vη = −aθ sinθ . (7)

1 These equations are the Lagrangian equations for the Euler angles when there is no friction force.



Substituting (1), (6), and (7) in (4) and (5), we obtain a set of equations for the Eulerangles. The disc can move along the plane without slipping and without leaving the plane,provided

| f | � μm(g + Z), g + Z � 0,

where μ is the friction coefficient.If we put θ = 0, we get ϕ = ψ = 0. Moreover, the following relation exists between

θ , ϕ, and ψ :

I ′3ϕ(ϕ cosθ + ψ)sinθ − I1ϕ

2 sinθ cosθ + mgacosθ = 0 (8)

(from now on, we use I ′1,3 = I1,3 + ma2). The centre of the disc moves with a velocity,

which has a constant absolute magnitude

V = a|�3| = a|ψ + ϕ cosθ |,

along a circle with radius R = V/|ϕ|.Condition (8) can also be expressed as

I ′3RV 2 = |I1aV 2 cosθ − mga2R2 cotθ |.

In particular, if the mass of the disc is concentrated in its centre (I1 = I3 = 0), we getthe elementary relation: V 2 = gR|cotθ |.

Gyroscopic effects which appear when I1,3 differs from zero may turn out to beimportant. For example, when R a , we have 3

2V 2 = gR|cotθ | for a uniform disc(2I1 = I3 = 1

2ma2), but 2V 2 = gR|cotθ | for a hoop (2I1 = I3 = ma2).If the disc is rolling vertically, we have

θ = π/2, ϕ = 0, ψ = �3 = const. (9)

To study the stability of this motion, we put

θ = (π/2)− β, β 1, ϕ ∼ β ∼ β�3 �3, ϕ ∼ β ∼ �3 ϕ�3

into (1) and (4)–(7), retaining only the first-order terms. We then get

MZ = I1ϕ + I3�3β = const, �3 = const,

I ′1β − I ′

3�3ϕ − mgaβ = 0,(10)



and hence

I ′1β +

(I3I ′

3

I1�2

3 − mga)

β = I ′3

I1MZ�3.

If

�23 >

I1mgaI3I ′

3, (11)

small oscillations in β and ϕ occur with the angle θ differing from π/2:

β = I ′3�3MZ

I1I ′1ω

2 + β0 cos(ωt + δ),

ϕ = −mgaMZ

I1I ′1ω

2 − I3

I1�3β0 cos(ωt + δ),

where

ω2 = I3I ′3

I1I ′1�2

3 − mgaI ′1

.

The direction of motion of the disc also executes small vibrations; the disc’s motionis not about a straight line, but about the circle of radius a�3I2

1ω2/(mgaMZ).Therefore, if there are small deviations in the initial conditions from (9), we may have

either small oscillations near an “equilibrium” motion along a straight line—if Mz = 0,β0 �= 0—or a new “equilibrium” motion—if Mz �= 0, β0 = 0.

If the inequality (11) is not satisfied, the motion is not stable.One can say that the motion with θ = const can occur, if in the θ-, ϕ-, ψ-“space” the

representative point lies on the surface determined by (8). It can easily be seen that, ifcondition (11) holds, the motion is stable with respect to perturbations which removethe representative (θ-, ϕ-, ψ)-point from the surface (8) while it is neutral with respect toperturbation shifting the point along the surface. A similar situation holds for the stabilityof the motion of the disc along the smooth plane; we must merely replace I ′

1,3 by I1,3.Rotation of the disc around its vertical diameter is stable, if �2

Z > mga/I ′1.

c) If there can be no rotation around a vertical axis, the following condition must besatisfied:

�Z = ϕ + ψ cosθ = 0. (12)

In this case, there is an additional, “frictional torque” N, which is directed along thevertical, acting on the disc. Instead of (4), we then have



MZ = fξ acosθ + N, M3 = fξ a + N cosθ . (13)

The integration of the equations of motion can easily be reduce to quadratures (incontrast to the equations under problem 9.21b).

Motion with a constant angle of inclination is possible if conditions (8) is satisfiedas well as condition (12), which means that ϕ and ψ are completely determined by theangle θ . In this case,

R = a|sinθ tanθ |, V 2 = mga3 sinθ

I ′3 + I1 cot2 θ

.

Rolling of the disc in a vertical position is stable if �23 > mga/I ′

3.d) If there is a small inclination, the term δL = −mgαX must be added to the

Lagrangian function; the Y-axis lies in the plane of the disc and is horizontal, theX-axis lies in the plane at right angles to the Y-axis, pointing upwards, and the Z-axis is atright angles to the plane. In this case, an interesting effect can be observed: although theadditional force is directed against the X-axis, the disc moves in the opposite directionto the Y-axis; that is, the disc moves without losing height.

Substituting

θ = (π/2)− β, β 1, ψ ∼ β �Z , ψ ∼ β ∼ �Z ψ�Z

into in (1) and (4) to (7) and adding the contribution from δL, we get

I ′1β + (I1 − I ′

3)�2Zβ − I ′

3�Zψ − mgaβ = −mgaα cos�Zt,

I ′3ψ + (I3 + 2ma2)�Z β = mgaα sin�Zt,

and hence

ψ = −(

2 + 2ma2

I ′3

+ mga

I ′3�

2Z

)α sin�Zt, β = −2α cos�Zt.

Substituting (7) and θ , ϕ = �Zt, and ψ into

X = −Vξ sinϕ − Vη cosϕ, Y = Vξ cosϕ − Vη sinϕ

and averaging over one period of rotation, we find

〈X〉 = 0, 〈Y〉 =(

1 + ma2

I ′3

+ mga

2I ′3�

2Z

)αa�Z ;

that is, the disc is displaced without losing height.



9.22. a) The position of the ball is determined by the coordinates of its centre of massX , Y , Z and the Euler angles θ , ϕ, ψ (see [1], § 35), which determine the orientation ofthe principal axes of inertia. Let the Z-axis be vertical and the x3-axis be directed fromthe centre of mass to the geometric centre of the ball; let x3 = b be the position of thegeometrical centre of the ball and a be its radius. The analysis of the motion of the ballis analogous to the analysis in the preceding problem.

If MZ �= M3, there is a minimum Ueff(θ) when θ0 �= 0, π , and if the energy E =Ueff(θ0), we can have a steady rotation of the ball with a constant angle θ = θ0 andZ = a − bcosθ0. The instantaneous point of contact of ball and plane has the velocityv = (bϕ + aψ)sinθ0 along the direction of the line of nodes.

b) Let

Rc = (X + bsinθ sinϕ, Y − bsinθ cosϕ, a)

be the coordinates of the geometric centre of the ball, and let

�= (�X ,�Y ,�Z)=(θ cosϕ + ψ sinθ sinϕ, θ sinϕ − ψ sinθ cosϕ, ϕ + ψ cosθ)

be the angular rotational velocity of the ball, while a = (0, 0, −a) is the radius vector fromthe geometric centre of the ball to the point where the ball is in contact with the plane.The condition that the ball rolls without slipping

Rc + [�, a] = 0

is a non-holonomic constraint

X = θ (a − bcosθ)sinϕ − (aψ + bϕ)sinθ cosϕ,

Y = −θ (a − bcosθ)cosϕ − (aψ + bϕ)sinθ sinϕ.(4)

The equation of motion are

mX = λ1, mY = λ2, (5)

MZ = (λ1 cosϕ + λ2 sinϕ)bsinθ , (6)

m3 = (λ1 cosϕ + λ2 sinϕ)asinθ , (7)

ddt

∂L∂θ

− ∂L∂θ

= (−λ1 sinϕ + λ2 cosϕ)(a − bcosθ); (8)

they contain the friction forces and the torques produced by these forces on theright-hand sides. Using the constraint conditions (4), we can express the Lagrangianmultipliers λ1 and λ2 in terms of the Euler angles. Indeed, the components of the frictionforce along the line of nodes f‖ and at right angles to it, f⊥, are given by



f‖ = mθ ϕ(a − bcosθ)− mddt

[(aψ + bϕ)sinθ], (9)

f⊥ = −mddt

[θ (a − bcosθ)] − mϕ(aψ + bϕ)sinθ . (10)

Note that (6) and (7) are the same as (1) if one substitutes into (1) the frictionforce f‖ (9) instead of the dry friction force ∓f . The quantity Mza − M3b = C is theintegral of motion (2), as discussed before.

9.23. To solve the problem, it is necessary to take into account that the height of theparticle above the Earth, h, is small compared to the radius R of the Earth and that thecentrifugal acceleration, R�2 (where � is the angular velocity of the Earth), is smallcompared to the acceleration of free fall, g, towards the surface of the Earth. There arethus two small parameters in the problem:

ε1 = hR

� ε2 = R�2

g∼ 0.003.

Let R + r be the coordinate of the particle reckoned from the centre of the Earth. Theequation of motion has the following form if we take into account the dependence of gon the distance r from the surface of the Earth (up to the second order in ε2):

r = −GMR + r

|R + r|3 + 2[v, �] + [�, [R + r, �]], (1)

where G is the gravitational constant and M is the mass of the Earth. We expand thefirst term into series in small parameter r/R � ε1:

r = g + 2[v, �] + g1 + [�, [r, �], r(0) = h, v(0) = 0, (2)

g = −GMRR3 + [�, [R, �], g1 = GM

R2

(3R

(rR)

R3 − rR

)[1 + O(ε1)].

The Coriolis acceleration 2[v, �] ∼ gt� ∼ √ε1ε2g and g1 ∼ ε1g have the first order of

smallness, while [�, [r, �]] ∼ ε1ε2g has the second order.The vertical h is anti-parallel to the vector g and, it makes a small angle α =

ε2 sinλcosλ with the vector R (here λ is the northern [geocentric] latitude; i.e. the anglebetween the plane of the equator and the vector R). We choose the z-axis along thevertical upwards, the x-axis along the meridian towards to the south, and the y-axis alongthe latitude to the east, then

g = (0, 0, −g), R = R(sinα, 0, cosα), � = �(−cos(α + λ), 0, sin(α + λ).



In zero approximation, the particle moves with acceleration −g along the z-axis. In thefirst approximation, the Coriolis acceleration results in the deviation to the east, and themagnitude of the displacement is y ∼ √

ε1ε2 gt2 ∼ √ε1ε2 h. The acceleration g1 has the

component ∼ ε1g along z, and components along x and y only of the second order;therefore, in the first approximation, the effect of g1 will only lead to an increase of thetime of the fall from a height of h by ∼ ε1

√2h/g. The deviation to the south thus occurs

only in the second order. Then we write (2) in projections into the chosen axes andtake the terms of the zero, first order, and second order for the z-, y-, and x-component,respectively:

z = −g, y = −2z�cosλ,

x = 2y�sinλ + gR

(3zsinα − x)+ �2zsinλcosλ.

Using the method of successive approximations, we obtain

z = h − 12 gt2, y = 1

3 gt3�cosλ, x = 2ht2�2 sinλcosλ.

Substituting the time of the fall t = √2h/g, we find deviations to the east and south

y = 13

√8ε1ε2 hcosλ, x = 2ε1ε2hsin2λ.

9.24. We use the reference system, which is rotating with a vessel; the x-axis is directedalong AB, the origin of the coordinates is placed in a fixed point, that is, the intersectionof the AB- and CD-axes. The angular velocity of the system is

� = ω1 + ω2 = (ω2, ω1 cosω2t, −ω1 sinω2t).

Besides the force of gravity mg, the inertia forces act upon each particle of fluid of mass min this system of reference (see [1], § 39):

the Coriolis force 2m[v, �],

the centrifugal force ∂

∂r12 m[�, r]2,

and the force m[r, �], where � is the rate of change of the vector � in the inertialsystem.

When the resin hardens, the particle velocities (relative to the vessel) turn to zero, andthe Coriolis force vanishes. The other forces need to be averaged over the periods ofrotation:

〈mg〉 = −mg (〈cosω1t〉, 〈sinω1t sinω2t〉, 〈sinω1t cosω2t〉) = 0,

if ω1 �= ω2, and

〈m[r, �]〉 = m[r, 〈�〉] = 0,



since 〈�〉 = 〈[ω1, ω2]〉 = [〈ω1〉, ω2] = 0. Finally,

⟨∂

∂r12 m[�, r]2

⟩= 1

2 m∂U(r)

∂r,

where

U(r) =⟨[�, r]2

⟩=

⟨�2r2 − (�r)2

⟩=

=(ω2

1 + ω22

)r2 −

⟨(xω2 + yω1 cosω2t − zω1 sinω2t)2

⟩=

= ω21x2 +

(12 ω2

1 + ω22

)(y2 + z2).

(1)

The surface of the liquid will be located along the level U(r) = const. The resin willharden in the form of an ellipsoid of rotation.

What will change in this result when ω1 = ω2?When g = 0 and ω2 → 0, we get an obviously wrong answer from (1). Why?

9.25.

t =√

m2

∫dr√

E + M�− Ueff,

ϕ =√

m2

∫ (M

mr2 − �)

dr√

E + M�− Ueff,

where E is the energy, M is the angular momentum in the rotating system of refer-ence, and Ueff(r) = U(r)+ M2/(2mr2). Bear in mind that E = E0 − M� and M = M0,where E0 and M0 are the energy and angular momentum, respectively, in the inertialsystem.

It is interesting to note that the centrifugal potential energy −12 m�2r2 does not enter

into Ueff(r).

9.26. In the system of reference, which is fixed in the frame, the Lagrangian function ofthe problem is the same as the one considered in problem 6.37, with z = 0 and with theparameters

ωB = −2�, ω21,2 = 2

m

(k1,2 + f2,1

l

)− �2.

When ω21,2 > 0, the motion of the particles is the same as that of an anisotropic oscillator

in the magnetic field B = −2mc�/e. The orbit of the particle for the case ω1 = ω2 isshown in Fig. 109 of problem 2.36. In particular, if ω1 = ω2 = 0, the motion of theparticle is the same as the motion of a free particle in a magnetic field:



x = x0 + acosωBt, y = y0 − asinωBt;

that is, the particle moves uniformly along a circle of radius a and centre (x0, y0). Itis interesting to investigate what the motion of the particle is in the rest system, whichcorresponds to the latter case, especially for a = 0 or for x0 = y0 = 0.

If the centrifugal force is larger than the restoring force from the two springs, ω21,2 < 0,

then the particle still executes small oscillations. Although the potential energy has amaximum at x = y = 0, the equilibrium position is stable because of the Coriolis force.

If ω21 and ω2

2 have opposite signs, so that the point x = y = 0 is a saddle point of thepotential energy, the equilibrium position is unstable.

It is interesting to compare these results with those of problem 5.4. One sees thatin a system rotating with an angular velocity � the point r0, ϕ0 lies on the ridge of the“potential circus”; the potential energy

U(r) = − α

rn − 12

m�2r2

has a maximum in the direction of the radius vector, but it does not change in thedirection of the azimuth. In this case, one of the eigen-frequency is ω, but the otherone vanishes: the equilibrium position is neutral with respect to some perturbations (e.g.a changes in ϕ0).

9.27. The Lagrangian function of a particle, which is expressed in its coordinates andvelocity in the system of reference rotating with the paraboloid, has the form

L = m2

(v + [ω, r])2 + mgr =

= m2

[(x − ωy)2 + (y + ωx)2 + z2 − g

(x2

a+ y2

b

)].

The quantity z can be omitted for the small oscillations, and then the equations of motionare

x − 2ωy +( g

a− ω2

)x = 0,

y + 2ωx +(g

b− ω2

)y = 0.

(1)

We look for a solution in the form

x = Aei�t, y = Bei�t;

and for �2 we get

�4 −( g

a+ g

b+ 2ω2

)�2 +

( ga

− ω2)(g

b− ω2

)= 0.



It is easy to prove that its roots are real. However, when( g

a− ω2

)(gb

− ω2)

< 0

one of the roots becomes negative, �21 < 0, so that the corresponding motion

x = A1e|�1|t + A2e−|�1|t,y = B1e|�1|t + B2e−|�1|t

leads to the departure of the particle from the origin. This means that the lower positionof the particle becomes unstable. Assuming for definiteness that a > b, we obtain theinstability region:

ga

< ω2 <gb

.

Note that if ω2 > g/b, the motion is stable although the potential energy in the rotatingsystem

U(x,y) = −12 m

(ω2 − g

a

)x2 − 1

2 m(ω2 − g

b

)y2

is not a potential pit, but a potential hump. Stability in this case is provided by the actionof the Coriolis force (see also [7], § 23).

9.28. a) The potential energy (including the centrifugal one) is

U(x) = k(x − a)2 − 12 mγ 2x2.

The equilibrium position x0 is determined by the condition U ′(x0) = 0; it is equal to

x0 = 2ka2k − mγ 2 .

Note that when the rotation frequency γ becomes greater than the frequency of theeigen-oscillations of the particle

√2k/m, we have x0 < 0; at the same time, when mγ 2

2k, the equilibrium position is close to the axis of rotation.The sign of U ′′(x0) = 2k − mγ 2 defines the stability of the equilibrium position: when

mγ 2 < 2k, the equilibrium is stable; when mγ 2 > 2k, it is not.b) Obviously, the equilibrium position is the same as in the preceding point 9.28a. To

study the stability of the oscillations, we will consider the potential energy of the systemat the small displacements from the equilibrium position:

U(x, y, z) = U(x0, 0, 0)+ 12 k1(x − x0)2 + 1

2 k2y2 + 12 k3z2,



where

k1 = 2k − mγ 2,

k2 = f + k(x0 − a)

l + x0 − a+ f − k(x0 − a)

l − x0 + a− mγ 2 = k3 − mγ 2.

The sign of ∂2U(x0,0,z)

∂z2 |z=0 = k3 defines the stability of the equilibrium position

with respect to the small deviations along the z-axis. When k3 > 0, the z-oscillations arestable. When k3 < 0, they are unstable since the deviation of the particle from the point(x0, 0, 0) increases in time (until the interaction with the walls of the frame, which wedo not consider here, becomes significant). Upon comparing this results with problem9.27, we can conclude that the equilibrium position with respect to the small deviationsin the xy-plane is stable when k1k2 > 0, and it is unstable when k1k2 < 0.

Let us consider in detail the case of fast rotation of the frame when

mγ 2 2k. (1)

In that case,

k3 ≈ 2fl − ka2

l2 − a2 = 2k(l − l0)l − a2

l2 − a2 , (2)

where we introduce the length of a free string l0 in accordance with the equationf = k(l − l0). From (2), it follows that 0 < k3 < 2k when fl > ka2, while k3 < 0 whenfl < ka2. Besides, we find that in the limit (1) both k1 < 0 and k2 < 0. This means thatwhen mγ 2 2k and fl > ka2, the equilibrium position is stable, in contrast to the resultsof 9.28a.

my

m1 m2

r2

r1

r

xO

l

Figure 163

9.29. We use the Cartesian coordinates in the rotating systemof reference with the origin of this system places in the centreof mass. The system itself rotates with an angular velocity ω

around the z-axis, while the stars are located along the x-axis(Fig. 163).

Let the x coordinates of the stars be

a1,2 = ∓ m2,1

m1 + m2a,

where m1,2 are the masses of the stars and a is the distance between them. From equality

m1m2

m1 + m2aω2 = Gm1m2

a2



(where G is the gravitational constant), we obtain

ω2 = Gm1 + m2

a3 .

The potential energy of a body of mass m (including the centrifugal potentialenergy) is

U(x, y, z) = −Gmm1

r1− Gmm2

r2− 1

2 mω2(x2 + y2),

where r1,2 are the distances to the stars and r = (x, y, z) is the radius vector of the body.

The equilibrium position of the body is determined by the condition ∂U∂r

= 0, or

∂U∂x

= Gm

(m1

r31

+ m2

r32

− m1 + m2

a3

)x − Gm

(m1a1

r31

+ m2a2

r32

)= 0,

∂U∂y

= Gm

(m1

r31

+ m2

r32

− m1 + m2

a3

)y = 0,

∂U∂z

= Gm

(m1

r31

+ m2

r32

)z = 0.

Hence, z = 0 and (when y �= 0) r1 = r2 = a.Thus, the stars and the equilibrium point are at the vertices of a right triangle. There

are two such points (the so-called Lagrange points):

x0 − a1 = a2 − x0 = 12 a, y0 = ±

√3

2 a, z0 = 0.

The potential energy near the Lagrange points has the form

U(x0 + x1, y0 + y1, z0 + z1) == U(x0, y0, z0)− 3

8 mω2x21 − mαx1y1 − 9

8 mω2y21 + 1

2 mω2z21,

α = ±3√

34a3 G(m1 − m2).

The motion along the z-axis is obviously stable. The equations of motion in the xy-planeread

x1 − 34 ω2x1 − αy1 − 2ωy1 = 0,

y1 − 94 ω2y1 − αx1 + 2ωx1 = 0.



The substitutions x = Aei�t, y = Bei�t lead to the equation for �:

�4 − ω2�2 + 2716 ω4 − α2 = 0.

Its roots are real for 16α2 � 23ω4, that is, when

27(m1 − m2)2 � 23(m1 + m2)

2 .

This condition is satisfied if the mass of one of the stars (say, the first one) satisfiesthe condition m1/m2 � 1

2 (25 + 3√

69) ≈ 25 . In this case, the motion of bodies in theneighbourhood of Lagrange points is stable. Stability of motion is provided by theCoriolis force (cf. problem 9.28).

On the x-axis, there are three more points where ∂U∂r

= 0, but the motion near them

is unstable.Note that for the Sun–Jupiter system, asteroids are observed just near the Lagrange

points (for more detail see [5], § 3.12).


§10

The Hamiltonian equationsof motion. Poisson brackets

10.1. Let ε be a vector of infinitesimal displacement; we then have

ra → r′a = ra + ε, pa → p′

a = pa,

H(ra, pa) = H(r′a, p′

a).

Hence,∑

a∂H∂ra

= 0. Using the Hamiltonian equation, we get

P =∑

a

pa = −∑

a

∂H∂ra

= 0, P = const.

For an infinitesimal rotation δϕ, we obtain

ra → r′a = ra + [δϕ, ra], pa → p′

a = pa + [δϕ, pa],

H(ra, pa) = H(r′a, p′

a),∑

a

{∂H∂ra

[δϕ, ra] + ∂H∂pa

[δϕ, pa]}

= 0 =

=∑

a

{−pa [δϕ, ra] + r[δϕ, pa]} = −δϕ∑

a

ddt

[ra,pa],

or

M =∑

a

[ra,pa] = const.

10.2. H = p2θ

2I1+ (pϕ − pψ cosθ)2

2I1 sin2 θ+ p2

ψ

2I3.




10.3. H = p2

2(1 + 2βx)+ 1

2 ω2x2 + αx3. In particular, for small oscillations (|αx| �ω2, |βx| � 1)

H = 12 p2 + 1

2 ω2x2 + αx3 − βxp2 + 2β2x2p2 − . . . ,

and up to and including terms linear in α and β the extra term in the Hamiltonianfunction of the harmonic oscillator is connected with the extra term in the Lagrangianfunction through the equation δH = −δL (see [1], § 40).

10.4. x = acos(ωt + ϕ), p = −ω0asin(ωt + ϕ), where

ω = (1 + 2λE0)ω0, E0 = 12 ω2

0a2.

10.5. p = p0 + Ft, x = x0 + AF (

√p0 + Ft − √

p0 ). The given Hamiltonian functiondescribes the motion of a charged vortex ring in liquid helium in the presence of ahomogeneous electric field along the x-axis [26]. The characteristic features of suchmotion are the following: the momentum of the vortex increases with the time, whilevelocity of its motion drops x = a/(2

√p0 + Ft ).

10.6. r = cpnp , p = cp

n2∂n∂r

, p = |p|.The given Hamiltonian function describes the propagation of light in a transparent

medium with refractive index n(r) in the geometric optics approximation (see [3], § 65).The “particle” is a wave packet, and r(t) gives the law of its motion; r is its group velocity,while the vector p, which is perpendicular to the wave front, determines the wave vector.

When n(r) = ax, the trajectory is

x = C1 cosh(

yC1

+ C2

),

where the C1 and C2 are determined by the initial and final points of the trajectory,respectively.

10.7. a) L = 12m(v − a)2;

b) L = 0: such “particles” cannot be described using a Lagrangian function (see [2],§ 53).

10.8. The given vector potential determines a magnetic field B in the z-direction.The Hamiltonian function is

H(x, y, z, px, py, pz) = p2x + p2

z

2m+ 1

2m

(py − e

cBx

)2.



Since H depend neither on y nor on z, we have

py = const, pz = const.

If we present H in the form

H = p2x

2m+ 1

2 mω2(x − x0)2 + p2z

2m,

where

ω = eBmc

, x0 = cpy

eB,

we see that x and px are obtained from the same Hamiltonian function as a harmonicoscillator. Therefore, we have

x = acos(ωt + ϕ)+ x0, px = −mωasin(ωt + ϕ).

To determine y and z, we use the equations

y = ∂H∂py

= 1m

(py − e

cBx

)= −ωacos(ωt + ϕ), z = pz

m,

whence

y = −asin(ωt + ϕ)+ y0, z = pz

mt + z0.

The particle moves along a spiral with its axis parallel to B. The generalized momen-tum py determines the distance of that axis from the yz-plane.

x1 x

E

Ueff

−x1

y

xx1−x1




10.9. The magnetic field is directed along the z-axis and equals 2hx. The motion alongthe z-axis is uniform. Here we consider only the motion in the xy-plane. The Hamiltonianfunction

H = p2x

2m+ 1

2m

(py − eh

cx2

)2

depends neither on y nor on t. Therefore, the integrals of motion are the generalizedmomentum py and energy E:

py = my + ehc

x2,

E = 12 mx2 +Ueff(x), Ueff(x)= 1

2m

(py − eh

cx2

)2

.

When py � 0, the function Ueff(x) is shown in Fig. 164, while an approximate view ofthe orbits can be found in Fig. 165. Note that the velocity

y = −|py|m

− ehmc

x2

is negative everywhere and oscillates near the value py/m.

Ueff

Um

xx0−x0

Figure 166

When py > 0, the function

Ueff(x) = e2h2

2mc2 (x2 − x20)2, x0 =

√pyceh

is shown in Fig. 166. Now the velocity

y = ehmc

(x20 − x2)

has both positive and negative values for any value of E. The approximate view of theorbits is given in Fig. 167; Fig. 167a–e corresponds to the decreasing values of energies.

At high energies, E � Um = p2y/(2m), the amplitude of the oscillations along the x-axis

is large; the average over the period value 〈x2〉 is larger than x20. Therefore, the averaged

value

〈y〉 = ehmc

(x2

0 − 〈x2〉)

is negative (see Fig. 167a). When the energy decreases, the value 〈y〉 increases to zero(Fig. 167b) and then becomes positive (Fig. 167c). When energy E = Um, the particle,



y

y

y

y

y

x0

x0

x0

x0

x

E � Um

E � Um

EUm

x

x

x

x

(a) (d)

(e)

(b)

(c)

Figure 167

having in the initial moment x > x0 and x < 0, asymptotically approaches the y-axis(Fig. 167d).

Finally, when E < Um, the particle moves either in the region near (−x0) or inthe region near x0 (Fig. 167e). That 〈y〉 > 0 is easily demonstrable. When |x − x0| � x0,the particle moves in a circle, the centre of which slowly drifts along the y-axis. To find thevelocity of the drift, it is necessary to take into account the first anharmonic corrections(see [1], § 28 and [7], § 29)

x = x0 + acosωt − a2

4x0(3 − cos2ωt),

when calculating 〈x2〉. That leads to

〈y〉 = cE

2hx20e

(cf. problem 8.14).

10.10. Introducing the coordinates of the centre of mass R and the relative motion r (cf.problems 2.29 and 2.30), we present the Lagrangian function of the system in the form



L = 12 MR2 + e

2c[B, r]R + L1(r, r)+ e

2c[B, R] r, (1)

where M = m1 + m2,

L1(r, r) = 12 mr2 + e

2c[B′, r] r + e2

r,

B′ = m2 − m1

m2 + m1B

and m is the reduced mass.We rewrite the last term in (1) as

e2c

[B, R] r = e2c

[B, r]R + ddt

e2c

[B, R]r.

Omitting the total derivative with respect to time, we have

L = 12 M R2 + e

c[B, r]R + L1(r, r).

The Hamiltonian function of the system now has the form

H = 12M

(P − e

c[B, r]

)2 + 12m

(p − e

2c[B′, r]

)2 − e2

r.

This function has no evident dependence on R and t; therefore, the generalizedmomentum

P = ∂L∂R

= MR + ec

[B, r] (2)

and the total energy of the system

E = 12 MR2 + 1

2 mr2 − e2

r

are conserved. The last equation can be rewritten in the form

E = 12 mr2 + Ueff(r), Ueff(r) = − e2

r+ 1

2M

(P − e

c[B, r]

)2.



From here follows (with the assumption P = const) that a particle of mass m moves inthe field with the effective potential energy Ueff (r) and in the constant uniform magneticfield B′. If the z-axis is directed along B, then

Ueff(r) = − e2

r+ 1

2 mω2[(x − a)2 + (y − b)2] + const,

ω = eBc√

m1m2, b = − cPx

eB, a = cPy

eB.

After finding r(t), the law of motion of the centre of mass can be determined from (2):

R(t) = PtM

− eMc

[B,

∫ t

0r(t)dt

]+ R0.

10.11.

p = p0 + eEt, ε(p)− eEr = ε0,

(r − r0)eE = ε(p0 + eEt)− ε(p0).

Here r0, p0, and ε0 are the constants.

10.12.

p = eE + ec

[v, B] .1

10.13. a) ε(p) = E, pB = const, where pB is the component of the momentum alongthe magnetic field B. The trajectory in the momentum space is determined by theintersection of the two surfaces: ε(p) = E and pB = const.

b) From the equation of motion p = ec [r, B], it is clear that the projection of the

electron orbit on a plane perpendicular to the magnetic field B is obtained from the orbitin momentum space by a rotation over π/2 around B and by changing the scale by afactor c

eB .

10.14.

S =E∫

Emin

dE∮

dp|v⊥| , T = c

eB

∮dp

|v⊥| = ceB

∂S∂E

,

where v⊥ is the component of the vector ∂ε

∂p, which is orthogonal to B.

1 For detail about the motion of electrons in a metal (problems 10.10–10.14) see, for istance, [16].



10.15. a) −∑

k

eijkxk;1 −∑

k

eijkpk; −∑

k

eijkMk.

b) ab;

{aM, br} =⎧⎨⎩

∑i

aiMi,∑

j

bjxj

⎫⎬⎭ =

∑ij

aibj{Mi, xj} =

= −∑ijk

aibjeijkxk = −[a, b]r; −[a, b]M.

c) 0; nr rn−2; 2a(ar).

10.16. {Ai, Aj} = −∑

k

eijkAk, {Ai, A4} = 0, where i, j, and k take on the values 1, 2, and

3, respectively (cf. problem 10.15a).

10.17.

{Mi, jk} = −∑

l

eijl lk −∑

l

eikl lj ;

{ jk, il} = δijMlk + δikMlj + δjlMik + δklMij ,

where Mkl = pkxl − plxk.

10.18. When the system as a whole is rotated around the z-axis over an infinitesimalangle ε, the change δϕ in any function of the coordinates and momenta is in the firstorder in ε given by

δϕ = ϕ(x−εy, y + εx, z, px−εpy, py + εpx, pz)− ϕ(x, y, z, px, py, pz) =

= ε

(−∂ϕ

∂xy + ∂ϕ

∂yx − ∂ϕ

∂pxpy + ∂ϕ

∂pypx

)= ε{Mz, ϕ}.

If ϕ is a scalar, this change under rotation must vanish, and thus {ϕ, Mz} = 0. If ϕ = fx isthe component of a vector function, its change under rotation is δfx = −εfy, and thus

{Mz, fx} = −fy or {Mz, f } = [n, f ]

(cf. [1], § 42, problems 3 and 4).

1 eijk is the completely antisymmetric tensor:

e123 = e231 = e321 = 1, e132 = e321 = e213 = −1;

all other components of eijk vanish.



What is the value of the Poisson brackets {Mz, Txx}, where Txx is a component of atensor function?

10.19. [f , aM] = [f , a]; {fM, lM} = [f , l]M + ∑ik

MiMk{fi, lk}.

10.20. Substituting into the second formula of the preceding problem f = eζ andl = eξ , where eζ and eξ are unit vectors along the ζ - and ξ-axes of the moving system ofreference, we have

{Mζ , Mξ } = +Mη. (1)

This equation differs in the sign of the right-hand side from the analogous relation for thecomponents of the angular momentum along the axis of the fixed system of coordinates,

{Mz, Mx} = −My. (2)

yy, η

x x,ξ

M MM′

M′ξ

δMx

ε

εMξ δMξ

ξ′

η′(a) (b)

δMx −εMy δMξ +εMη

Figure 168

As was shown in problem 10.18 (see also [5], § 9.7), the Poisson brackets (2)characterizes the change of the component Mx when the coordinate system is rotatedas a whole over an infinitesimal angle ε (Fig. 168a)

δMx = ε{Mz, Mx} = −εMy.

The Poisson brackets in (1), which are equal to

{eζ M, eξ M} = {Mζ , eξ }M,

characterizes the change in the components of the fixed vector M along the eξ -axis whenthe moving coordinate system is rotated over an infinitesimal angle around the ζ -axis(Fig. 168b; in the figure the ξ , η, ζ -axis are the same as as x, y, z-axes before the rotation).

10.21. Mα = ∑βγ δ eαβγ (I−1)γ δMβMδ. In particular, if we choose the moving system of

coordinates such that the tensor of inertia Iαβ is diagonal, we obtain the Euler equation(see [1], § 36) using the relation Mα = Iα�α).



10.22. The equation of motion is

Mi = {H, Mi} = γ∑jk

eijkMjBk, or M = −γ [B, M];

that is, the vector M processes with an angular velocity −γ B.a) The vector M precesses around the direction of B:

Mx = Mx(0)cos(γ B0t)+ My(0)sin(γ B0t),

My = −Mx(0)sin(γ B0t)+ My(0)cos(γ B0t),

Mz = Mz(0).

b) The vector M rotates with an angular velocity −γ B, which in turn rotates aroundthe z-axis with an angular velocity ω. It is helpful to use the rotating system of coordinates,in which the vector B is fixed. In that system, the components of the angular velocity ofthe vector M are equal to

ω′x = −γ B1, ω′

y = 0, ω′z = −γ B0 − ω ≡ ε.

At a given initial condition, the components M in the rotating system of reference are

M′x = −a

ε

λM0(1 − cosλt),

M′y = aM0 sinλt,

M′z =

(ε2

λ2 + a2 cosλt)

M0,

where

λ =√

ε2 + γ 2B21, a = γ B1√

ε2 + γ 2B21

.

In the fixed system, we have

Mx = M′x cosωt − M′

y sinωt,

My = M′x sinωt + M′

y cosωt,

Mz = M′z.

When B1 � B0, the dependence of amplitudes Mx,y on ω is resonant: generally speaking,these amplitudes are small ∼ M0B1/B0, but for |ε| = |ω + γ B1| � γ B1, they increasedramatically, reaching values ∼ M0. In particular, for ω = −γ B0, we get



Mx = M0 sinγ B1t sinγ B0t,

My = M0 sinγ B1t cosγ B0t,

Mz = M0 cosγ B1t.

10.23. {vi, vj} = − em2c

∑k eijkBk.

10.24. a) p(t) = p + Ft, r(t) = r + ptm + Ft2

2m ;b) p(t) = pcosωt − mωqsinωt,

q(t) = qcosωt + pmω

sinωt.

Of course, these quantities can more simply be evaluated without using Poissonbrackets. However, this method can easily be taken over in quantum mechanics.

10.25. a) According to the preceding problem, we have

dfdt

= {h, f } = ∂H∂f

{ f , f } = 0.

b) The Hamiltonian function is

H = p2r

2m+ 1

2mr2 f (θ , ϕ, pθ , pϕ),

where

f (θ , ϕ, pθ , pϕ) = p2θ + p2

ϕ

sin2 θ+ 2macosθ .

Integrals of motion are E, pϕ , and, according to the preceding problem, f .

10.26. a)

{Ai, Aj} = 2Hm

3∑k=1

εijkMk,

{Ai, Mj} = −3∑

k=1

εijkAk;



b) {H, J1,2} = 0, {J1i, J2j} = 0,

{J1i, J1j} = −3∑

k=1

εijk J1k, {J2i, J2j} = −3∑

k=1

εijk J2k,

H = − mα2

4(J21 + J2

2

) .

The vectors J1 and J2 are the independent integrals of motion. Each has the samePoisson brackets for their components as an usual angular momentum. The presence oftwo such “momenta” is closely related to the so-called hidden symmetry of the hydrogenatom (see [22], Ch. I, § 5).


§11

Canonical transformations

11.1. a)q =

√2Pmω

sinQ, p = √2mωP cosQ,

Q = ω + ω

2ωsin2Q, p = −P

ω

ωcos2Q.

In this case, P and Q are action and angle variables. These variables are more helpfulthan p and q for solving the problem by the method of the perturbation theory, if thefrequency ω(t) changes slowly, |ω| � ω2 (see problem 13.11).

b)

q = Fmω2 +

√2Pmω

sinQ, p = √2mωP cosQ,

Q = ω − Fω

cosQ√2mωP

, P = − Fω

√2Pmω

sinQ.

11.2. �(p, Q) = −Q(

1 + ln p2

4Q

).

11.3. The function �(q1, q2, . . . , qs, P1, P2, . . . , Ps) determines a canonical transforma-tion, provided

det∂2�

∂qi∂Pk�= 0.

11.4. Let

Q = qcosα − psinα, P = qsinα + pcosα.

Then we have

{P, Q}p,q = −{q, p}p,q sin2 α + {p, q}p,q cos2 α = 1.




For a system with one degree of freedom, this is sufficient for the transformation to becanonical.

11.5. One sees easily (and subsequent calculations verify) that the canonical transfor-mation must be close to the identity transformation and that the terms ax2P and bP3

in the generating function are small. To solve the equations

p = P + 2axP, Q = x + ax2 + 3bP2,

which determine the canonical transformation, for x and p, we replace x by Q and p by Pin the small terms:

p = P + 2aQP, x = Q − aQ2 − 3bP2. (1)

We proceed in the similar fashion when we express the Hamiltonian function in termsof the new variables:

H ′(Q, P) = 12 (P2 + ω2Q2)+ αQ3 + βQP2 + 2aQP2 − aω2Q3−−3bω2QP2 + terms of fourth degree in Q and P.

Putting

α − aω2 = 0, β + 2a − 3bω2 = 0,

the third-order terms vanish. In the approximation indicated in the problem, we have

Q = Acosωt, P = −ωAsinωt,

and from (1) (cf. [1], § 28)

x = Acosωt − αA2

ω2 −(β + α

ω2

)A2 sin2 ωt.

11.6. Reducing the Hamiltonian function to the form considered in problem 10.4,we get

x = Q − 5β

8ω20

Q3 − 9β

8ω40

QP2,

where

Q = Acosωt, P = −ω0Asinωt, ω = ω0 + 3β

2ω0A2

(cf. [1], § 28).



11.7. Because the terms xPX + yPY in the generating function correspond to theidentical transformation, it is clear in advance that the parameters a,b, and c will beproportional to α. Therefore, the canonical transformation for px with required precisionhas the following form

px = ∂�

∂x= PX + ayPX + 2cxPY ≈ PX + aYPX + 2cXPY .

Acting similarly, we get

py ≈ PY + aXPX , x ≈ X − aXY − 2bPXPY , y ≈ Y − bP2X − cX2.

Substituting these expressions into a new Hamiltonian function

H ′ = H + ∂�

∂t= H

on the condition that this function corresponds (with the necessary accuracy) to the twoindependent oscillators,

H ′ = 12m

[P2

X + (mω1X)2 + P2Y + (mω2Y)2

],

we find

a = − 2α

4ω21 − ω2

2

, b = − 2α

(4ω21 − ω2

2)m2ω22

, c = − α (2ω21 − ω2

2)

(4ω21 − ω2

2)ω22

.

The new canonical variables correspond to the two free oscillators

X = Acos(ω1t + ϕ1), Y = B cos(ω2t + ϕ2),

PX = −mω1Asin(ω1t + ϕ1), PY = −mω2B sin(ω2t + ϕ2).

As a result, with the required accuracy, we find

x(t) = A cos(ω1t + ϕ1)− αAB2ω2(2ω1 + ω2)

cos(ω+t + ϕ+)+

+ αAB2ω2(2ω1 − ω2)

cos(ω−t + ϕ−),

y(t) = B cos(ω2t + ϕ2) + αA2

2ω22

− αA2

2(4ω21 − ω2

2)cos(2ω1t + 2ϕ1) ,

where ω± = ω1 ± ω2 and ϕ± = ϕ1 ± ϕ2.



The obtained solutions are valid until the frequency ω2 is close to 2ω1. For ω2 = 2ω1,anharmonic corrections stop being small and can lead to significant transferring energyfrom x to y oscillations and back (see problem 8.10).

11.8. H ′(P, Q) = H(P, Q); when X = Asin(ωt + ϕ), Y = 0 the oscillator perform amotion along an ellipse:

x = Acosλ sin(ωt + ϕ),

y = Asinλ cos(ωt + ϕ).

11.9. To make the notation less cumbersome, it is helpful for the time being to putm = ω = e = c = 1. One can easily reintroduce these factors in the final expressions. Thetransformation of problem 11.8 is a rotation in the xpy- and ypx-planes, which thereforeleaves the form of that part of the Hamiltonian function, which is equal to

12 (x2 + y2 + p2

x + p2y),

invariant. On the other hand, the correction due to the terms 12B2x2 − Bxpy is equal to

12 B2(X2 cos2 λ + P2

Y sin2 λ + 2XPY sinλcosλ)++ B(X2 − P2

Y )sinλcosλ − B(cos2 λ − sin2 λ)XPY .

The off-diagonal term XPY vanishes if we put

sin2 λ − cos2 λ + Bsinλcosλ = 0, that is, tan2λ = 2B

.

After a few simple transformations, the Hamiltonian function is reduced to the form

H = 12m

(P2

X + P2Y tan2 λ

)+ 1

2 mω2(X2cot2 λ + Y2

). (1)

The variables X and Y thus perform harmonic oscillations with frequencies whichare, respectively, equal to

ω1 = ω cotλ =√

ω2 +(

eB2mc

)2

+ eB2mc

,

ω2 = ω tanλ =√

ω2 +(

eB2mc

)2

− eB2mc



(cf. [2], § 21, problem). Each of the coordinates X and Y corresponds to a motionalong an ellipse; an arbitrary oscillation is a superposition of two such motions(cf. problems 6.37 and 11.8).

It is interesting to note that when B → 0 , λ = π/4 (and not λ = 0). This means thateven for very weak field B the “normal” oscillations is “circularly polarized”. On the otherhand, oscillations corresponding to the coordinates X or Y with λ = 0, which if therewere no field B would be linear polarized, slowly change their direction of polarization,as soon as there is a field B present.

If the magnetic field is variable, we must add to the Hamiltonian function (1) thepartial derivative with respect to the time of the generating function

� = −(

mωxy + PXPY

mω

)tanλ + xPX + yPY

cosλ

(expressing it in terms of X , Y , PX , and PY ; see also the footnote to problem 13.26).

11.10. Putting into the canonical transformation of the preceding problem

ω = ω2, tan2λ = 2ωBω2

ω2B + ω2

1 − ω22

, ωB = eBmc

,

we get

H ′ = 12m

(P2

X + �22

ω22

P2y + p2

z

)+ 1

2 m(�2

1X2 + ω22Y2 + ω2

3z2)

,

where �1,2 was defined in problem 6.37.

11.11. The transformation (λ = π/4)

qs1 = 1√2

(Xs + PYs

Nmωs

), qs2 = 1√

2

(Ys + PXs

Nmωs

)

leaves the form of the Hamiltonian function

H = p20

2Nm+

R∑s=1

[p2

s1 + p2s2

2Nm+ 1

2 Nmω2s (q2

s1 + q2s2)

]=

= p20

2Nm+

R∑s=1

[P2

Xs + P2Ys

2Nm+ 1

2 Nmω2s (X2

s + Y2s )

].



invariant (cf. problem 11.8). The oscillation corresponding to Xs = Acos(ωst + β) is

xn = A√2

sin(ωst + nϕs + β),

and the oscillation corresponding to Ys = Bcos(ωst + β) is

xn = B√2

sin(−ωst + nϕs − β) .

11.12. The new Hamiltonian function is H ′ = ωP1, and in the new variables, theequations of motion have the form

P1 = P2 = Q2 = 0, Q1 = ω.

What is the change in the Hamiltonian function H ′, if B depends on the time?

11.13. The transformation is p = αP, r = Q/α, which is a similarity transformation.

11.14. The gauge transformation

A′ = A + ∇f (r, t), ϕ′ = ϕ − 1c

∂f (r, t)∂t

can be written as a canonical transformation,

R = r, P = p + ec∇f , H ′ = H − e

c∂f∂t

,

if one uses the generating function

�(r, P, t) = rP − ec

f (r, t).

11.15. �(q, P) = qP − F(q, t).

11.16. b) Fτ (q, Q) = −12 Fτ(q + Q)− m

2τ(q − Q)2;

c) Fτ (q, Q) = mω2sinωτ

[2qQ − (q2 + Q2)cosωτ ].

11.17. a) Q = r + δa, P = p: a shift of the systems as a whole over δa (or a shift of thecoordinate system over −δa).

b) Up to and including first-order terms, we have

Q = r + [δϕ, r], P = p + [δϕ, p].

The transformation is a rotation of the coordinate system over an angle −δϕ.



c) Q(t) = q(t + δτ), P(t) = p(t + δτ), H ′(P, Q, t) = H(p, q, t + δτ). The transforma-tion is a shift in the time by δτ (cf. [1], § 45, [7] § 36.2).

d) Q = r + 2pδα, P = p − 2rδα.The transformation is a rotation over an angle 2δα in each of the xipi-planes

(i = 1, 2, 3) in phase space.

11.19. a) �(r, P) = rP + nPδa + n[r, P]δϕ, where δa is the displacement along thedirection of n while δϕ = 2π

h δa is the angle of rotation around n; (h is the pitch of thescrew);

b) �(r, P, t) = rP − VPt + mrV;c) �(r, P, t) = rP − t δ�[r, P].

11.20. δf (q, p) = λ{W , f }p,q. Indeed, substituting the values of the new variables,

P = p − λ∂W∂q

, Q = q + λ∂W∂P

,

into f (Q, P) and expanding the expression obtained in powers of λ, we get up to andincluding first-order terms

δf (q, p) = λ∂f∂q

∂W (q, p)

∂p− λ

∂f∂p

∂W (q, p)

∂q.

11.21. Putting � = rP + λrp in the preceding problem, we get a similarity transforma-tion with α = 1 + λ (see problem 11.13). The given Hamiltonian function is such that

H ′(P, Q) = α−2H(P, Q),

and therefore

λ{H, rp} = H ′ − H = −2λH(λ → 0).

On the other hand, {H, rp} = ddt (rp), and hence

rp − 2Et = const

(cf. problem 4.15b).

11.23. Let δ1q and δ1p be the changes in the coordinates and momenta connected withthe transformation defined by �1.1 Then we have

1 Let us as an example indicate the change in the momentum up to second-order terms:

δ1p = P − p = −λ1∂W1(q, P)

∂q= −λ1

∂W1(q, p)

∂q+ λ2

1∂2W1(q, p)

∂p∂q∂W1(q, p)

∂q.



f (q + δ1q, p + δ1p) = f (q, p)+ λ1{W1(q, p), f (q, p)}+ λ21ϕ1(q, p). (1)

We now apply to each of the terms of the right-hand side of (1) another transformation,defined by the function �2:

f (q + δ21q, p + δ21p) = f + λ2{W2, f }+ λ1{W1, f }++λ1λ2{W2, {W1, f }}+ λ2

1ϕ1 + λ22ϕ2.

(2)

The transformation of λ21ϕ1(q, p) gives a correction of higher than the second order. If

we apply these transformations in the reverse order, the result is

f (q + δ12q, p + δ12p) = f + λ1{W1, f }+ λ2{W2, f }++λ1λ2{W1, {W2, f }}+ λ2

1ϕ1 + λ22ϕ2

(3)

The difference between (2) and (3) lies only in the second-order terms which areproportional to λ1λ2. Subtracting (3) from (2), we obtain

λ1λ2({W2, {W1, f }}− {W1, {W2, f }}) = λ1λ2{ f , {W1, W2}}.

Therefore, we see that, in particular, shifts λW = δaP (see problem 11.17) commute,while this is not the case for rotations around different axes, λW = δϕ [r, P].

Is the statement, which is inverse of the one in the problem, correct?

11.24. A canonical transformation with a variable parameter λ can be considered to bea “motion” where λ plays the role of the time and W (q, p) plays role of the Hamiltonianfunction (cf. problem 11.1c). The equations of “motion” are

dQdλ

= ∂W (Q, P)

∂P,

dPdλ

= −∂W (Q, P)

∂Q.

One can also easily obtain these equations formally from the result of problem 11.20.a) The infinitesimal change in the coordinates and momenta under the given canonical

transformation has the form

δr = − λ

N{W , r} = λ

N{Ma, r} = −[n, r]δϕ,

δp = −[n, p]δϕ,

where

M = [r, p], n = aa

, δϕ = −λaN

.



This transformation is a rotation of the coordinate system over an angle δϕ around thedirection of n. If we take the z-axis along a, we finally obtain

X = xcosϕ + ysinϕ, Y = ycosϕ − xsinϕ, Z = z

and similar equations for the components of the momentum.b) The infinitesimal change in the coordinates and momenta under the canonical

transformation given by A1 has the form

δx = px δϕ, δy = −py δϕ, δpx = −xδϕ, δpy = yδϕ,

where δϕ = λ/(2N). This transformation represents a rotation over an angle +δϕ in thexpx-plane and over an angle −δϕ in the ypy-plane. Therefore, we have

X = xcosϕ + px sinϕ, Y = ycosϕ − py sinϕ,

PX = −xsinϕ + px cosϕ, PY = ysinϕ + py cosϕ.

Similarly, A2 and A3 represent the rotation over an angle ϕ or −ϕ in the xpy- and ypx-planes and in the xy- and pxpy-planes, respectively, while A4 represents the rotation overan angle 4ϕ in the xpx- and ypy-planes.

Not every rotation in phase space is the canonical transformation. For example,rotation in the xpy-plane is not a canonical transformation.

It is interesting to compare the motion of a two-dimensional isotropic harmonicoscillator (its Hamiltonian function is H = 1

2A4) and the motion of a particle in thexy-plane in the arbitrary potential field having an axial symmetry, U(x2 + y2). In bothcases, the integrals of motion are the angular momentum 2A3, the conservation of whichis due to invariance of the system with respect to rotations around the z-axis. For theoscillator, in addition, there are integrals of motion A1 and A2, the conservation ofwhich is connected with a “hidden” symmetry—with the invariance of the Hamiltonianfunction with respect to certain rotations in phase space. In this sense, the oscillator issimilar to a particle in three-dimensional central field, for which there are three integralof motion Mx,y,z.

The presence of additional integrals of motion for the two-dimensional oscillator leadsto fact that point (x, y, px, py) in phase space moves along a closed line, while for aparticle in the arbitrary field U(x2 + y2), the phase trajectory “fills” two-dimensionalsurface (see [1], § 52, [7] § 44).

11.25. a) The volume specified in momentum and in phase space does not change withtime, but in coordinate space the volume is spread out. Thus, if at t = 0 the state of thesystem is represented by the rectangle ABCD (Fig. 169), it goes over to the parallelogramA′B′C′D′ (AD = A′D′) after a time t, and the distance in the x-direction between the

points A′ and C′ is equal to �x = �x0 + �p0m t. As time goes on, this parallelogram

degenerates into a narrow, very long strip.



B

p

p0+Δp0

p0+Δp0x0+Δx0 x0+Δx0+x0+p0t/m

p0

x0

A D

B′

A′ D′

C′C

m

x

t

Figure 169

b) If there is a wall at the point x = L, the state of the system can no longer berepresented by the parallelogram A′B′C′D′, but must have the form shown in Fig. 170a.When time marches on the initial phase volume, ABCD, changes into a number ofvery narrow parallel strips which are almost uniformly distributed inside two rectangles0�x�L, p0�p�p0 +�p0 and 0�x�L, −p0 −�p0�p� − p0 (Fig. 171b).

p0+Δp0 p0+Δp0

−p0−Δp0 −p0−Δp0

p0 p0

−p0 −p0

p pB

A D

L L

C

x x

A′

B′ C′

D′

(a) (b)

Figure 170

c) The phase orbit of an oscillator with energy E and frequency ω is the ellipse x2

a2 +p2

b2 = 1 with semi-axes a =√

2Emω2 , b =

√2Em . All points of the specified phase volume

move along such ellipsis and return to their initial state after a period T = 2π/ω. Thedimension of the specified “volume” in coordinates space, �x, and in momentum space,�p, oscillate with frequency 2ω. In contrast to the case sub 11.25b there is no spreadingout of the specified phase volume into the whole of the available phase space.

d) For an oscillator with friction (friction force Ffr = −2mλx), we have

x = ae−λt cos(ωt + ϕ),

p = mx = −mae−λt[ω sin(ωt + ϕ)+ λcos(ωt + ϕ)],



and the oscillations are damped so that the phase orbit is a spiral,

x2

a2 +(

p + λmxmaω

)2

= e−2λt.

The specified phase volume decreases until it vanishes. The non-conservation of phasevolume here is connected with the fact that the system is not canonical: to describe itfully, we need know not only the Lagrangian function L = 1

2m(x2 − ω2

0x2)

but also thedissipative function F = 1

2mλx2 (see [1], § 25).If we choose for this system the “Lagrangian function” in the form

L′ = 12 me2λt

(x2 − ω2

0x2)

(cf. problem 4.20), the phase volume specified will be conserved for the appropriate

canonical variables x and p′ = ∂L′∂ x

; however, in this case the generalized momentum

p′ = mxe2λt will not have a simple physical meaning, as before.e) Since the period of the motion in this case depends on the energy, the phase space

volume is spread out with time, “filling” the whole of the available region of phase space(cf. sub 11.25b).

Let the initial specified region be

x0 < x < x0 + �x, p0 < p < p0 + �p.

One can easily estimate the time which is such that during that period the fastest particleshave made one more oscillation than the slowest ones:

τ ∼ T2

�T, �T ∼ dT

dE�E, �E ∼ p0�p

m+

∣∣∣∣dU(x0)

dx

∣∣∣∣�x.

f) Let there be N particles such that the points in phase space which representtheir state are at time t = 0 distributed with a density Nw(x0, p0, 0) and that they moveaccording to the equations

x = f (x0, p0, t),

p = ϕ(x0, p0, t).(1)

Here

f (x0, p0, t) = x0 + p0

mt, ϕ(x0, p0, t) = p0



for free movement and

f (x0, p0, t) = x0 cosωt + p0

mωsinωt,

ϕ(x0, p0, t) = −mωx0 sinωt + p0 cosωt

for harmonic oscillators. The number of particles in the specified region of phase space,all points of which move according to the same law, remains constant; in particular, foran infinitesimal phase volume dxdp, we have

Nw(x, p, t)dxdp = Nw(x0, p0, 0)dx0 dp0.

According to the Liouville theorem (see [1], § 46 and [7], § 40)∂(x,p)

∂(x0,p0)= 1, and

therefore

w(x, p, t) = w(x0, p0, 0). (2)

Using (1) to get expressions for x0 and p0,

x0 = f (x, p, −t), p0 = ϕ(x, p, −t)

and substituting this into (2), we obtain

w(x, p, t) = w( f (x, p, −t), ϕ(x, p, −t), 0),

or

w(x, p, t) = exp[−α(x − X)2 − β(x − X)(p − P)− γ (p − P)2]2π�p0�x0

,

where X = f (X0, P0, t), P = ϕ(X0, P0, t) while the coefficients α, β, γ for freeparticles are

α = 12(�x0)2 , β = − t

m(�x0)2 ,

γ = 12(�p0)2 + t2

2m2(�x0)2 ,



and for oscillators,

α = cos2 ωt2(�x0)2 + m2ω2 sin2 ωt

2(�p0)2 ,

γ = cos2 ωt2(�p0)2 + sin2 ωt

2m2ω2(�x0)2 ,

β = sinωt cosωt(

mω

(�p0)2 − 1mω(�x0)2

).

We show in Figs. 171 and 172 how regions in phase space in which

2π�x0�p0 w(x, p, t) � 12

p

p0

X0 X x

p

P

X x


(for free particles and for harmonic oscillators, respectively) move about. These regionsare ellipses which are deformed as time marches on.1 Their centres are displacedaccording to the same law (1) as the particles. In the case of free particles, this ellipseis spread out without limit, but in the case of the oscillators, it only pulsates. We notethat the distributions in coordinate and in momentum space are no longer independent:w(x, p, t) cannot be split into two factors of the form w1(x, t) · w2(p, t).

It is interesting to consider the coordinate distribution function (independent of thevalues of the momenta)

w(x, t) =∞∫

−∞w(x, p, t)dp

1 If the scales along the p- and x-axes in the phase space of the harmonic oscillators are chosen such thatmω = 1, the phase orbits are circles, and the specified region in phase space rotates around the origin withoutbeing deformed.



or the momentum distribution function

w(p, t) =∞∫

−∞w(x, p, t)dx.

These distributions turn out to be Gaussian with maxima at X and P, respectively:

w(x, t) = 1√2π�x

e− (x−X)2

2(�x)2 ,

w(p, t) = 1√2π�p

e− (p−P)2

2(�p)2 ,

where for free movement

(�x)2 = (�x0)2 + (�p0)2

m2 t2, (�p)2 = (�p0)2,

and for the oscillators

(�x)2 = (�x0)2 cos2 ωt + (�p0)2

m2ω2 sin2 ωt,

(�p)2 = (�p0)2 cos2 ωt + m2ω2(�x0)2 sin2 ωt.

11.26. a) {a∗, a} = −i, H0 = ωa∗a.b) The variables P and Q are canonical because {P, Q} = 1. The generating function

is determined from the equation

dF = p(x, Q)dx − P(x, Q)dQ

and equals

F(x, Q, t) = i2 mωx2 + i

2 Q2e−2iωt − i√

2mωxQe−iωt.

The new Hamiltonian function is, therefore,

H ′0(Q, P) = H0 + ∂F(x, Q, t)

∂t= 0.

c) We note that the expression −3Q2P2/(2m2ω2) is the only term in the expression

x4 =(

Qe−iωt − iPeiωt√

2mω

)4



which does not contain the time. Whence, the average Hamiltonian function is

〈H ′(Q, P)〉 = − 3β

8mω2 Q2P2.

In the following, we will omit the brackets 〈〉 denoting averaging over time. Now it isclear that −iQP = |Q0|2 = |a|2 is the integral of motion.

The Hamiltonian equations are

Q = −iεQ, P = iεP, ε = 3β|Q0|24mω2 ,

where

Q = Q0e−iεt, P = iQ∗0eiεt;

whence

x = 1√2mω

(Q0e−Iω′t + Q∗0eiω′t) = x0 cos(ω′t + ϕ).

The influence of the additive term δU is reduced to change of frequency

ω′ = ω + 3β|Q0|24mω2 = ω + 3βx2

0

8ω

d) The new Hamiltonian function

H ′(Q, P, t) = m2ω2α

(Qe−iωt − iPeiωt

√2mω

)4e4iωt + e−4iωt

2

after averaging is reduced to

〈H ′(Q, P)〉 = 18 α(Q4 + P4).

For the variable ξ = −iQP = |a|2, proportional to the square of the oscillation amplitude,the equation of motion is

ξ = {〈H ′〉, ξ} = −12 iα(P4 − Q4).

Taking into account that

14 α(Q4 + P4) = A = const,



we find

ξ2 = −4A2 + α2ξ4.

Thus, the variable ξ changes in the same way as the coordinate of the particle(with mass equal to one) in the field V (ξ) = −1

2α2ξ4 and with the energy −2A2

(cf. problem 1.2). Note that the amplitude increases to infinity for a finite time (theso-called explosive increase in amplitude).

Of course, the use of the average Hamiltonian function is correct only for |ξ | � ωξ ,that is, when ξ � ω/α (cf. problem 8.1).

11.27. We introduce the new variables:

a = mωx + ipx√2mω

eiωt, b = mω2y + ipy√2mω2

eiω2t, c = mω3z + ipz√2mω3

eiω3t,

(ω = ω2 + ω3) and canonical conjugate mometa ia∗, ib∗, ic∗. New Hamiltonian function,averaged over periods 2π/ω2,3, is

〈H ′〉 = ε|a|2 + η(a∗bc + ab∗c∗),

ε = ω1 − ω, η = α

4√

2mωω2ω3.

The equations of motion

a = −iεa − iηbc,

b = −iηac∗,

c = −iηab∗

have the integrals of motion1

〈H ′〉 = A, |a|2 + |b|2 = B, |a|2 + |c|2 = C.

The equation

ddt

|a|2 = iη(ab∗c∗ − a∗bc)

can be presented in the form more helpful for qualitative investigation:

ξ2 + V (ξ) = 0,

1 Integrals B and C are called integrals of Manley–Rowe.



where ξ = |a|2 and

V (ξ) = (A − εξ)2 − 4ξη2(B − ξ)(C − ξ).

In the initial moment c = 0, therefore a = εC, B > C and

V (ξ) = (C − ξ)2(ε2 − 4ξη2)− 4ξη2(B − C)(C − ξ).

The function V (ξ) for the cases ε2 < 4η2C and ε2 > 4η2C is shown in Fig. 173.In the first case, the variable ξ performs oscillations; that is, the system experiences

beats. The energy is periodically pumped back and forth between the x-oscillator andthe y and z-oscillators. In the second case (i.e., when there are small initial amplitudesand a large “detuning” ε), the y- and z-oscillations are not excited.

For more details, see [19].

11.28. a) 〈H ′〉 = ε|a|2 + μ|a|4 + η(a2 + a∗2), where

ε = ω − γ , μ = 3β

8mω2 , η = 18 hω.

V

V

C

ε2<4η2C

ε2>4η2C

ξ

ξ

(a)

(b)

Figure 173

b) The equations of motion

−a = i(ε + 2μ|a|2)a + 2iηa∗,

a∗ = i(ε + 2μ|a|2)a∗ + 2iηa

have a constant solution

a0 = 0, |a1|2 = 2η − ε

2μ.

When ξ = |a|2, we get

ξ = −2iη(a∗2 − a2).

Taking into account that

〈H ′〉 = εξ + μξ2 + η(a2 + a∗2) = C = const,

we obtain

ξ2 + V (ξ) = 0,

where

V (ξ) = 4η2[(a∗2 + a2)2 − 4|a|4] = 4(C − εξ − μξ2)2 − 16η2ξ2.



V

ξ

ξm

Figure 174

In the case considered, the value of C is small accord-ing to the initial conditions. In the resonance region|ε| < 2η, the graph of function V (ξ) (Fig. 174)demonstrates that ξ performs the oscillations inthe range from zero up to ξm ≈ 2|a1|2.

Thus, the transition to the stable oscillationξ = |a1|2 (cf. problem 8.7) can be provided only bysome previously unaccounted for mechanism, suchas friction. This transition may be very long. Please note that this transition process hasthe character of beats even at zero “detuning”, ε = 0, in contrast to the transition processin the linear oscillations (see problem 5.11).

11.29. The average Hamiltonian function is

〈H ′(Q, P, t)〉 = H ′(Q, P) = 12 mω2

(ε − 1

4 h)

Q2 + 12m

(ε + 1

4 h)

P2.

Quantity√

Q2 + P2/(m2ω2) represents the amplitude of the oscillation. The variables Qand P change very little over the period 2π/γ , as can be easily seen from the Hamiltonianequations which contain small parameters ε and h.

On the Q, P-plane, a point representing the state of the system moves along the lineH ′(Q, P) = C = const. Families of such lines for area of the parametric resonance |ε| <

h/4 and its neighbourhood |ε| > h/4 are shown in Fig. 175a and 175b, respectively. In thefirst case, the amplitude ultimately increases without limit; in the second case, it performsthe bits (cf. problem 8.8).

P

P

QQ

(a) (b)

Figure 175

11.30. a) It is easy to check (cf. problem 11.4) that the given transformation is canonical.At V = 0, the motion of the x-oscillator is represented by the motion of the point

along the circumference in the x, px/(mω1)-plane with the frequency ω1. The radius ofthis circumference



a =√

x2 + p2x

m2ω21

matches the amplitude of the oscillations along the x-axis. In the X , PX/(mω1)-plane thiswill be a fixed point x = x(0), PX = px(0). Thus, the new variables are time-independentat V = 0 and, therefore, H ′

0 = 0.1

When V �= 0, these variables are time-dependent, but because the new Hamiltonianfunction H ′ = H ′

0 + V = V is small, the average motion in these variables is slow. Indeed,after averaging we have the new Hamiltonian function

〈H ′〉 = − β

4ω1ω2(ω1XPY − ω2YPX ) ,

and from the Hamiltonian equations

X = β

4ω1Y , Y = − β

4ω2X ,

we easily get

X = Acos(γ t + ϕ), Y = −√

ω1

ω2Asin(γ t + ϕ), γ = β

4√

ω1ω2� ω1,2.

Similarly, for the new momenta, we have

PX = mω1Bcos(γ t + ψ), PY = −m√

ω1ω2Bsin(γ t + ψ).

Thus, in the X ,PX/(mω1)-plane there occurs slow (with frequency γ ) motion alongan ellipse. This motion corresponds to the oscillations along the x-axis with slowlyvarying amplitude

a(t) =√

X2 + (PX/mω1)2 =√

A2 cos2(γ t + ϕ)+ B2 cos2(γ t + ψ),

that is, to the beats. Analogously, the oscillation amplitude along the y-axis is

b(t) =√

ω1

ω2

√A2 sin2(γ t + ϕ)+ B2 sin2(γ t + ψ).

1 From the Hamiltonian equations for new variables (e.g. X = ∂H ′0/∂PX = 0), it follows that H ′

0 isindependent of them. Therefore, H ′

0 = f (t), where f (t) is an arbitrary time function, which we can put equal tozero without losing generality.



This shows that the energy of x- and y-oscillators Ex = 12mω2

1a2(t) and Ey =12mω2

2b2(t), respectively, and their sum E = Ex + Ey are not conserved. However, thevalue which can be called the total number of quanta is conserved:

n = Ex

hω1+ Ey

hω2= mω1

2 hbarC2.

Here C = √A2 + B2, and h is the Planck constant.

In particular, at ϕ = ψ = 0 the beat amplitude reaches zero

x = X cosω1t + PX

mω1sinω1t = C cosγ t cos(ω1t + ϕ0),

y = −C√

ω1

ω2sinγ t cos(ω2t + ϕ0), tanϕ0 = −B

A,

and the energy oscillates with the frequency 2γ :

E = 12 mω1C2

(ω1 cos2 γ t + ω2 sin2 γ t

).

Note that even a weak coupling |V | � H0 = E leads to large changes in the energy�E ∼ E. Thus, when ϕ = ψ = 0 and ω1 � ω2, we have

�E = 12 mω1 (ω1 − ω2)C2 ∼ 〈E〉 = 1

4 mω1 (ω1 + ω2)C2.

Note that this problem coincides with problem 11.27 related to three interactingoscillators considered in the limit of the energy of the z-oscillator Ez � Ex,y so high thatthe beats of the x- and y-oscillators have almost not effect on its motion

z = z0 sinω3t, ω3 = ω1 − ω2.

In this case, β = 12αz0, and nh is the same as one of the integrals of Manley–Rowe,

namely, the integral B in notation of problem 11.27. The third oscillator plays the roleof a large energy reservoir with which x- and y-oscillators exchange energy.

b) New canonical variables increase exponentially with time



X = Aeγ t + Be−γ t, Y =√

ω1

ω2

(Aeγ t − Be−γ t) ,

PX = mω1(Deγ t + Fe−γ t) , PY = −m

√ω1ω2

(Deγ t − Fe−γ t) ,

that corresponds to the exponentially growing amplitude of the oscillations along the x-and y-axes. In this case, the difference in the number of quanta is conserved

Ex

hω1− Ey

hω2= 2mω1

h(AB + DF).


§12

The Hamilton–Jacobi equation

12.2. It is clear that the trajectory is a curve in a plane. If we use polar coordinates, wecan separate the variables in the Hamilton–Jacobi equation, provided we take the polaraxis z along a. The complete integral of the Hamilton–Jacobi equation is

S = −Et ±∫ √

β − 2macosθ dθ ±∫ √

2mE − β/r2 dr. (1)

To fix the signs in (1), we use the relations

pr = mr = ∂S∂r

= ±√

2mE − β

r2 , (2)

pθ = mr2θ = ∂S∂θ

= ±√β − 2macosθ . (3)

On the initial part of the trajectory r < 0, θ > 0 (we assume that the trajectory lies abovethe z-axis; see Fig. 176). We must thus take the upper sign in front of the first radical

in (1) and the lower sign in front of the second radical. ∂S∂β

= B is the equation of the

trajectory:

θ∫0

dθ√β − 2macosθ

+r∫

∞

dr

r2√

2mE − β/r2= B. (4)

The lower limits of the integrals can be chosen arbitrarily as long as the constant B is notdetermined. From our choice of lower limits and the condition that θ → 0 as r → ∞, itfollows that B = 0.

The constant β is an integral of motion of our problem, and from (3), we have

β = p2θ + 2macosθ .




M

N

L

Oz

K

θmθmax

Figure 176

It can be expressed in terms of the particle parameters when r → ∞ and θ → 0, that is,before the collision, when pθ = mvρ (ρ is the impact parameter):

β = 2m(Eρ2 + a), E = 12 .mv2

When r changes from ∞ to

rm =√

β

2mE=

√ρ2 + a

E,

which is determined by the condition pr = 0, θ changes from zero to a value θm which issuch that

θm∫0


+rm∫

∞

dr

r2√

2mE − β/r2= 0. (5)

A further increase in θ is accompanied by an increase in r; pr then changes sign. Theequation for the part LM of the orbit is

θ∫θm


−r∫

rm

dr

r2√

2mE − β/r2= 0; (6)

it is more helpful to use (5) and (6) and write it in the form:

θ∫0


−∞∫

rm

dr

r2√

2mE − β/r2−

r∫rm

dr

r2√

2mE − β/r2= 0. (7)



At r → ∞ the trajectory asymptotically approaches a straight line parallel to ON. Theangle θmax can be found from the equation1

θmax∫0


= 2

∞∫rm

dr

r2√

2mE − β/r2= π√

β. (8)

The equation ∂S∂E

= A determines r as function of t. If we choose A such that r(0) = rm,

we get

r =√

v2t2 + r2m =

√ρ2 + a

E+ v2t2. (9)

The integral over r in (4) and (7) can be evaluated elementarily, but those over θ reduceto elliptical integrals.

If Eρ2 � a, we can expand the integrand in (4) and (7) in powers of 2ma/β ≈a/(Eρ2). Up to and including first-order terms, we get

r sinθ = rm

(1 − a

2Eρ2 cosθ

)(10)

(see Fig. 177).

M

K

L

Oz

M′K′

Figure 177

In this approximation, the angle over which thevelocity of the particle is deflected after the scatteringis zero. This can be explained by the fact that theaction of the force along different sections of theorbit (which to first approximation is a straight lineK ′M′) is self-concealing.

12.3. a) To determine the angle over which theparticle is deflected, we must expand in (8) of the

preceding problem the powers of a/(Eρ2) up to the second order. We get the equation

θmax + maβ

sinθmax + 34

(maβ

)2 (θmax + 1

2sin2θmax

)= π . (1)

1 We draw attention to the following method to avoid the calculation of the integral over r in (8). This integralis independent on a and must therefore equal the left-hand side of (8) also when a = 0. But in that case, clearly,θmax = π , and the integral over θ is trivial.



Solving this equation up to order (ma/β)2, we find the angle of deflection1

χ = π − θmax = 34

π

(maβ

)2

= 3π

(a

4Eρ2

)2

. (2)

The scattering cross-section is

dσ = π |dρ2| =√

3πad

16Eχ5/2 . (3)

The dependence on χ which we have obtained is the same as for small-angles scatteringin the field γ /r4, which decreases much faster than U(r).

b) dσd

= πb8Eχ3 .

c) When Eρ2 � |b(θ)|, we have, instead of (10) of problems 12.2, the followingexpression for all θ :

θ + mβ

∫ θ

0b(θ)dθ + 3

2m2

β2

∫ θ

0b2(θ)dθ =

={

arcsin rmr , when 0 < θ < θm,

π − arcsin rmr , when θm < θ < θmax,

up to and including second-order terms. Here, the constant β is equal toβ = 2m[Eρ2 + b(0)] ≈ 2mEρ2 and rm ≈ ρ.

If

π∫0

b(θ)dθ = π〈b〉 = 0,

we can limit ourselves to the first approximation for which

χ = π − θmax = mβ

π〈b〉,

1 We look for θmax in the form θmax = θ0 + θ1 + θ2 + . . ., where θ1 ∼ (ma/β) θ0. In zeroth approximationwe get from (1) θ0 = π ; in first approximation

θ1 +(

maβ

)sinθ0 = 0,

where θ1 = 0; in second approximation

θ2 + maβ

θ1 cosθ0 + 34

(maβ

)2 (θ0 + 1

2sin2θ0

)= 0,

whence follows (2).



and the small-angle scattering cross-section,

dσ

d= π〈b〉

4Eχ3 ,

is the same as in the central field

U = 〈b〉r2 .

If, however, 〈b〉 = 0, we must take the second-order terms into account, and we get

dσ

d=

√3π

8Eχ5/2

√12 〈b2〉.

12.4. a) We can separate the variables in the Hamilton–Jcobi equation if we choose thespherical coordinates with the z-axis parallel to a. The canonical momenta are

pr = mr = −√

2mE − β/r2,

pθ = mr2θ = ±√

β − 2macosθ − p2ϕ/sin2 θ ,

pϕ = mr2ϕ sin2 θ = const.

(1)

One finds the constant

β = p2θ + p2

ϕ

sin2 θ+ 2macosθ

easily by noticing that

p2θ + p2

ϕ

sin2 θ= M2,

where M is the total angular momentum of the particle; it is helpful to evaluate M forr → ∞, θ → π − α (α is the angle between v∞ and a); that is, before the collision,

β = 2m(Eρ2 − acosα).

According to (1), the particle can fall into the centre when β < 0 or

ρ2 <aE

cosα. (2)



This is thus possible if α < π/2 and in that case that the cross-section is

σ = πaE

cosα.

Averaging over all possible directions of a gives

〈σ 〉 = 14π

π/2∫0

πaE

cosα · 2π sinα dα = πa4E

.

It is interesting that the area defined by condition (2) is a circle with centre on the axis ofthe particle beam, although the potential field is not symmetric with respect to that axis.

b)

σ =⎧⎨⎩

πaE

cosα − πλ2

4E2 , when 0 < α < αm = arccosλ2

4aE0, when αm < α < π .

〈σ 〉 = πa4E

(1 − λ2

4Ea

)2

.

c)

σ =

⎧⎪⎨⎪⎩

πaE

cosα + 2π

√γ

E, when 0 < α < αm = π − arccos

2√

γ Ea

,

0, when αm < α < π .

〈σ 〉 = π

(a

4E+

√γ

E+ γ

a

).

d)

σ = −πb(π − α)

E,

provided b(π − α) < 0.

12.5.

σ ={πR2 + πa

E cosα, when acosα > −ER2,0, when acosα < −ER2,

where α is the angle between v∞ and a.



12.6. a) We use the same notation as in problem 12.2. The equation for the initial partof the orbit (r → ∞, θ → π) is

π∫θ


=∞∫

r

dr

r2√

2mE − β/r2, (1)

with

β = 2m(Eρ2 − a). (2)

When β > 0, the angle θ decreases when r changes from ∞ to rm and then againincreasing to ∞. The equation for the part of the orbit after r had passed through theminimum distance to the centre is

π∫θ


= π

2√

β+

r∫rm

dr

r2√

2mE − β/r2. (3)

It is clear that when Eρ2 � a the orbit (1) and (3) are the same as (11) in problems 12.2.When β < 0, the particle can fall into the centre of the field (note that it follows

from (2) that only values β � −2ma are admissible.) In that case, r decreases monoton-ically from ∞ to 0. The angle θ decreases from π to the value θ1 for which pθ vanishes(section AB of the orbit; see Fig. 178). We then have β − 2macosθ1 = 0. After that, theangle increases until it reaches the value 2π − θ1 (section BC of the orbit)

∞∫r

dr

r2√

2mE − βr−2=

π∫θ1


+θ∫

θ1


. (4)

B

n=12

θ1

3

C

Oz

D

E

A

2π–θ

Figure 178

In point C, the momentum pθ again changes sign, and θ decreases until it reaches thevalue θ1 in point D; it then increases again, and so on.



The equation of the complete orbit can be written in the form∞∫

r

dr

r2√

2mE − βr−2= (−1)n

π∫θ


+ 2n

π∫θ1


(n = 0, 1, 2, . . .) (5)

A single value of θ (θ1 < θ < 2π − θ1) corresponds to an infinite number of valuesof r (n can take any non-negative integer value since the integral on the left-hand sideof (5) increases without bound as r → 0). Thus, the particle performs infinitely manyoscillations between the straight lines BD and CE before it falls into the centre.

In the case of small impact parameters Eρ2�a, π − θ � 1 so that we can writein (5) cosθ ≈ −1 + 1

2 (π − θ)2. The final result is1

θ = π − ρ

√2Ea

sin[

1√2

arsinh(

1r

√aE

)]. (6)

The law of motion r(t) is determined in the same way as in problem 12.2. When β > 0,the law is determined by (9) from problem 12.2. When β < 0, we have

r(t) = v√

t2 − τ2, v = √2E/m, −∞ < t < τ = −√|β|/(2E), (7)

and the particle falls into the centre at time τ .b) If β > 0 ( Eρ2 > a), we have

π∫θ

dθ√1 + 2ma

β(1 + sinθ)

={

arcsin rmr , when θm < θ < π ,

π − arcsin rmr , when θmin < θ < θm.

If β < 0 (Eρ2 < a), we have

⎛⎜⎝

π∫θ1

±θ∫

θ1

+2l

θ2∫θ1

⎞⎟⎠ dθ√

β + 2ma(1 + sinθ)=

∞∫r

dr

r2√

2mE − β/r2,

where l is the number of complete oscillations in angle (from θ1 to θ2 and back again)performed by the particle, and the (±) sign corresponds to counterclockwise (clockwise)motion (Fig. 179a), and

θ1 = −arcsinEρ2

a, θ2 = π + arcsin

Eρ2

a.

1 arsinhx = ln(x +√

1 + x2).



θ1

θ2

O z

(a) (b)

Figure 179

If β = 0 (Eρ2 = a), we have

r = ρ√2F(θ)

, F(θ) = lntan

[18 π + 1

4 (π − θ)]

tan(18 π)

, −12 π < θ < π .

The particle moves along the orbit of Fig. 179b, and we have r = −√2E/mt (t < 0; the

particle falls into the centre at t = 0).

12.7. The complete integral of the Hamilton–Jcobi equation is (see [1] § 48)

S = −Et + pϕϕ ±∫ √

β − 2macosθ − p2ϕ

sin2 θdθ −

∫ √2mE − β

r2 dr.

The generalized momenta are the same as in problem 12.4a. The particle can fall intothe centre if β = 2m(Eρ2 − acosα) < 0 (which is clearly satisfied if α2 < 2Eρ2/a � 1).

The equation for the orbit,

ϕ = ±∫ θ

π−α

pϕ dθ

sin2 θ

√β − 2macosθ − p2

ϕ

sin2 θ

, (1)

r0

r= ∓sinh

∫ θ

π−α

√|β|dθ

sin2 θ


ϕ

sin2 θ

, r20 = |β|

2mE (2)

can, in general, not be integrated to produce elementary functions. However, one caneasily describe the motion qualitatively if one notes that (1) which gives a relation betweenthe angles θ and ϕ is, apart from the notation, the same as the equation for the motionof a spherical pendulum (see [1], § 14, problem 1). The particle thus moves in such away that the point where its radius vector intersects the surface of a sphere of radius ldescribes the same curve as does a spherical pendulum of length l, energy β/(2ml2),and angular momentum pϕ in the field of gravity g = −a/(ml3). This curve is enclosedbetween two “parallel” circles on the sphere corresponding to θ = θ1 and θ = θ2.



If α2 < 2Eρ2/a � 1, one can easily integrate (1) and (2):

θ = π −√

ε + 12 α2 −

(ε − 1

2 α2)

cos(

2√

ma|β| arsinh

r0

r

),

θ = π − 2α√

ε√2ε + α2 + (2ε − α2)cos2ϕ

, ε = Eρ2

a.

(3)

It is clear from (3) that a particle when falling into the centre moves in the regionbetween two conical surfaces θ1 � θ � θ2 rotating around the z-axis, while one completerotation around the z-axis corresponds to two complete oscillations in the angle θ . In thisapproximation the orbit is closed for a spherical pendulum (it is an ellipse).

12.8. a) If the particle does not fall into the centre, the equation for a finite orbit is

pr

= 1 + ecos f (θ), (1)

where

p = β

mα, e =

√1 + 2Eβ

mα2 , f (θ) =∫ θ

θ1

dθ√1 − 2ma

βcosθ

,

while the constants E and β satisfy the inequalities −mα2/(2β) < E < 0 and β > 0.If 0 < β < 2ma, the orbit “fills” the region ABCDEF (Fig. 180),

r1 � r � r2, r1,2 = p1 ± e

, θ1 � θ � θ2, θ1 = arccosβ

2ma, θ2 = 2π − θ1;

that is, it approaches any point in this arbitrary closely.If β = 2ma,

f (θ) = √2 ln tan(θ/4)+ C1, (2)

and the orbit lies inside the ring r1 � r � r2 (Fig. 181).If β > 2ma, the orbit fills the ring r1 � r � r2. In particular, if β � 2ma, we have

f (θ) = θ + ζ sinθ + 34 ζ 2θ + 3

8 ζ 2 sin2θ + C2, (3)

where ζ = ma/β. This is a slightly deformed ellipse, the nature of the deformation beingdetermined by its orientation. Equation (3) is valid when arccosζ−2 � θ � arccos(−ζ−2).It is interesting to make a comparison with the results of problem 2.28.



C F A

E

D

B

zO


12.9. If the motion lies inside the ring r1 � r � r2, we have∫ θ1+2π

θ1

dθ√1 − 2ma

βcosθ

= 2πnl.

If the motion lies in the region r1 � r � r2, θ1 � θ � θ2, we have∫ θ2

θ1

dθ√1 − 2ma

βcosθ

= πnl

(n and l are integers.)

12.10. We can separate variables in the Hamilton–Jcobi equation, if we take the z-axisalong the vector a (see [1],§ 48, (48.9)). The radial motion,

t =√

m2

∫dr√

E + αr − β

2mr2

,

is, when β � 0, the same as the motion of a particle in a Coulomb field −α/r with angularmomentum β and energy E. When β < 0, the particle can fall into the centre. The

equations of the orbit are ∂S∂pϕ

= const, ∂S∂β

= const. The first of these,

ϕ = ±∫

pϕ dθ

sin2 θ


ϕ

sin2 θ

is the same as the equation for the orbit of a spherical pendulum with energy β/(2ml2)and angular momentum Mz = pϕ in the field of gravity g = −a/(ml3) (see [1], § 14,problem 1). The second equation connects r and θ . One can also use the analogy with aspherical pendulum for the analysis of that equation.



12.11. a) |Mz| <√

mb/2.b) A finite orbit is possible for any value of Mz.

12.12. b) The complete integral of the Hamilton–Jacobi equation is (see [1], § 48,problem 1)

S = −Et + pϕϕ +∫

pξ (ξ)dξ +∫

pη(η)dη,

where

pξ = ±√

12 m(E − Uξ (ξ)), Uξ (ξ) = p2

ϕ

2mξ2 − mα + β

mξ− 1

2 Fξ ,

pη = ±√

12 m(E − Uη(η)), Uη(η) = p2

ϕ

2mη2 − mα − β

mη+ 1

2 Fη.

The trajectory and the law of motion are determined by the equations

∂S∂β

= B, ∂S∂pϕ

= C, ∂S∂E

= A,

that is,

∫dξ

ξpξ (ξ)−

∫dη

ηpη(η)= 4B, ϕ − pϕ

4

∫dξ

ξ2pξ (ξ)− pϕ

4

∫dη

η2pη(η)= C,

−t + m4

∫dξ

pξ (ξ)+ m

4

∫dη

pη(η)= A.

(1)

Uξ Uη

Uξ max

Uη min

ξ1 ξ2ξ

E

a a

b

b

ηη1 η2

(a) (b)

Figure 182

When studying the character of the motion, we must determine the region admissiblefor the values of ξ and η for given values of E, pϕ , and β. Fig. 182 gives the shape of ofthe effective potential energies Uξ (ξ) and Uη(η).



A

B

C

zO D

Figure 183

If F = 0, and when −mα < β < mα (see curves182a) and E < 0, the motion in both ξ and η isfinite, but for E > 0, it is infinite. When a small forceF > 0 appears, the curve Uξ (ξ) shows a maximum(see curves 182b); when Uηmin < E < Uξ max, themotion is finite, as before. The motion is restrictedto the region ξ1 � ξ � ξ2, η1 � η � η2 (see Fig. 183)in the “ρz-plane”, while the ρz-plane itself rotatesaround the z-axis with an angular velocity ϕ. Theorbit fills the region of space formed by rotating

the figure ABCD around the z-axis (see also problem 2.40). When Uξ max < E, themotion is infinite.

When F increases, the quantity Uξ max decreases and Uηmin increases. WhenUξ max becomes less than Uηmin, finite motion becomes impossible (when β < −mα +32 (Fmp4

ϕ)1/3, there are no Uξ max extrema).

12.13. In elliptical coordinates, we have

ρ = σ

√(ξ2 − 1)(1 − η2), z = σξη, σ =

√c2 − a2.

The potential energy,

U ={∞, when ξ > ξ0 = c/σ ,

0, when ξ < ξ0,

depends only on ξ , and we can separate the variables in the Hamilton–Jacobi equation(see [1], § 48).

The complete integral is

S = −Et + pϕϕ ±∫ √

2mσ 2E + β − 2mσ 2A(ξ)

ξ2 − 1− p2

ϕ

(1 − ξ2)2 dξ ±

±∫ √

2mσ 2E − β

1 − η2 − p2ϕ

(1 − η2)2 dη,

(1)

where

A(ξ) = (ξ2 − η2)U(ξ) = U(ξ).



For a particle flying through the origin, pϕ = 0. From the previous discussion, itfollows that

pξ = ±√

2mσ 2E + β − 2mσ 2A(ξ)

ξ2 − 1= mσ 2(ξ2 − η2)

ξ2 − 1ξ , (2)

pη = ±√

2mσ 2E − β

1 − η2 = mσ 2(ξ2 − η2)

1 − η2 η. (3)

In the origin (η = 0, ξ = 1), we have

η = ±√

2mσ 2E − β

mσ 2 , ξ = 0, z = σ(ξη + ηξ ) = σ η

and from the condition√

2Em

cosα =√

2mσ 2E − β

mσ,

we find β = 2mσ 2E sin2 α.

z

ρ

Figure 184

The region of the inaccessible values of η is determined by the condition

2mσ 2E − β/(1 − η2) < 0 or |η| > |cosα|.

The motion thus takes place in the region |η| < |cosα|, 1 < ξ < ξ0 (see orbits in Fig. 184).

12.14. The transformation from the Cartesian coordinates to elliptical coordinates hasthe inverse transformation as the Jacobian of the transformation

∂(x,y)∂(ζ ,ϕ)

= c2 D, D = cosh2ζ − cos2 ϕ = sinh2ζ + sin2 ϕ



vanishes only when ζ = 0, ϕ = ±π . This transformation is similar to the transition fromthe Cartesian coordinates to the polar ones. For a given value of ζ (and for all values of ϕ),the corresponding points lie on an ellipse

x2

c2cosh2ζ+ y2

c2sinh2ζ= 1,

and for a given value ϕ (and for all values of ζ ), the corresponding points lie on the partof the hyperbola (half of one branch)

x2

c2 cos2ϕ− y2

c2 sin2ϕ= 1.

The potential energy in the elliptic coordinates has a simple form

U(ζ ) ={

0, when ζ < ζa,∞, when ζ > ζa,

where the parameter ζa is related to the ellipse axes: a = ccoshζa, b = csinhζa.Taking into account that

y = c(ζ coshζ cosϕ − ϕ sinhζ sinϕ), x = c(ζ sinhζ sinϕ + ϕ coshζ cosϕ),

x2 + y2 = c2 (ζ 2 + ϕ2)D,

we find the Lagrangian function of the system in the elliptic coordinates

L = 12 mc2 (ζ 2 + ϕ2)D − U(ζ ).

Next, we find the generalized momenta

pζ = mc2Dζ , pϕ = mc2Dϕ

and the Hamiltonian function

H(pζ ,pϕ ,ζ ,ϕ) = p2ζ + p2

ϕ

2mc2D+ U(ζ ).

The particle moves with a constant velocity, which changes direction after collisionswith the wall indicated by the ellipse. In the collision, the angle of incidence is equal tothe angle of reflection. Of course, the particle velocity and its direction of motion remainconstant between the walls. However, the question remains as to which areas may befilled by the trajectory of the particle and which will be inaccessible to it (at given initial



conditions). This is what we are going to find out using the Hamilton–Jacobi equation.We know in advance that these areas do not depend on the velocity value.

Since the Hamiltonian function does not have the evident dependence on time, wecan use the equation for an abbreviated action S(ζ ,ϕ, t) = −Et + S0(ζ ,ϕ):

12mc2D

(∂S0

∂ζ

)2

+ 12mc2D

(∂S0

∂ϕ

)2

+ U(ζ ) = E. (1)

Next we consider the region ζ < ζa only where U(ζ ) = 0. In this region the variables inthe Hamilton–Jacobi equation are separated:

S0(ζ ,ϕ) = S1(ζ )+ S2(ϕ).

Indeed, (1) with U(ζ ) = 0 can be represented as

(dS1(ζ )

dζ

)2

− 2mc2E cosh2 ζ = −(

dS2(ϕ)

dϕ

)2

+ 2mc2E cos2 ϕ.

The left-hand side of this equality is independent of ϕ, whereas the right-hand side isindependent of ζ ; therefore, each of them is a constant. Let’s define such constant as(−β). Thus,

S1 = ±∫ √

2mc2E cosh2 ζ − β dζ , S2 = ±∫ √

β − 2mc2E cos2 ϕ dϕ.

The motion occurs in the region where U = 0, so E > 0; it is also clear from the formS2 that β > 0. Thus, we obtain the complete integral of the Hamilton–Jacobie equationS(ζ ,ϕ, t) = −Et + S0(ζ ,ϕ,E,β), which depends on two arbitrary constants E and β. Bydifferentiating S with respect to β and equating the result to the constant C, we get theequation for trajectory in the form f (ζ ,ϕ,E,β,C) = 0. The equation ∂S/∂E = t0 allowsus to find the functions ζ(t) and ϕ(t). However, we already have a clear idea of thetrajectory (consisting of straight segments) and the law of motion. Generalized momentapζ and pϕ can be found by differentiating S with respect to ζ and ϕ:

pζ = ±√

2mc4E cosh2 ζ − β = mc2Dζ , (2)

pϕ = ±√

β − 2mc4E cos2 ϕ = mc2Dϕ. (3)

Constants E,β,C, t0 are determined by the initial values of coordinates ζ ,ϕ andvelocities ζ , ϕ.

If β > 2mc4E, then according to (3) pϕ will never equal zero at any value of ϕ. Thesign of ϕ is conserved; the motion always occurs clockwise (or counterclockwise).



In accordance with (2), the condition 2mc4E cosh2 ζm = β defines the minimum valueζm: ζm ≤ ζ ≤ ζa. When ζ reaches the value ζa, the sign in (2) changes from positive tonegative to correspond to the reflection from the wall. When ζ reaches the value ζm, thevelocity ζ becomes zero, and the negative sign in (2) is replaced by the positive one. Inthis case, the trajectory touches the ellipse which is limited to the area of motion (seeorbits in Fig. 185)

x2

c2cosh2ζm+ y2

c2sinh2ζm= 1.

x

y

x

y


If β < 2mc4E, then according to (3) the function ϕ becomes zero for ϕ = ϕmdetermined by the equation 2mc4E cos2 ϕm = β. In this case, the sign in (3) changes.The coordinate ζ changes in the range 0 ≤ ζ ≤ ζa; that is, it runs through all valid values.At the boundaries of this interval, the sign in (2) changes. The area of the particle motionis limited by both branches of the hyperbola (see orbits in Fig.186)

y2

c2 cos2ϕm− x2

c2 sin2ϕm= 1.

The initial values of Cartesian coordinates and velocity components allows us to findthe constants

E = 12 mv2, ϕ0 = 0, ccoshζ0 = x0,

tanα = vy

vx= ζ

ϕ= pζ

pϕ

=√

2mc4E cosh2 ζ0 − β

β,

β = 2mc4E cosh2 ζ0 tan2 α.

Each time ζ or ϕ changes sign, the constants C and t0 are updated, but E and β remainunchanged during the whole time of the particle motion.



Depending on whether the value cosh2 ζ0 tan2 α is larger or smaller the unit, theinaccessible area will be limited to an ellipse or a hyperbola. In the case when it equals aunit, there will be no inaccessible areas at all. It is also possible to have closed orbits.

12.15. One can separate the variables in the Hamilton–Jacobi equation using ellipticalcoordinates (see [1], § 48, problem 2 with α1 = −α2 = α). For a particle coming frominfinity along the z-axis, the constant β = −2mEρ2 + 4mασ , where ρ is the impactparameter.

O1

O2

Figure 187

At β < 0 the orbit is qualitatively the same as theorbit of a particle which is scattered in the field of apoint dipole (see problem 12.6a).

If β > 0, the particle “falls” onto the dipole(i.e., it passes in its motion through the sectionO1O2) and then goes off again to infinity. Moreover,if pη(η1) = 0 when η1 < 0, the particle moves inthe region bounded by the hyperbole η = η1 (seeFig. 187).

12.16. In the Hamilton–Jacobi equation,

∂S∂t

+ 12m

[(∂S∂z

)2

+(

∂S∂r

)2

+(

1r

∂S∂ϕ

− e2c

B(z)r)2

]= 0, (1)

we can separate the time and the angle ϕ:

S = −Et + pϕϕ + S(r, z). (2)

Considering in the following only orbits which intersect with the z-axis, we put pϕ = 0.It is not possible to separate the variables r and z, and we shall look for the integralapproximately in the form of an expansion in r:

S(r, z) = S0(z)+ rψ(z)+ 12 r2σ(z)+ . . . . (3)

Since the radial momentum,

pr = ∂S∂r

= ψ(z)+ rσ(z)+ . . . , (4)

for a particle flying along the z-axis (with r = 0) vanishes, for the particle beamconsidered we have ψ(z) = 0. Substituting (3) into (1) and comparing coefficients ofthe same powers of r, we obtain (cf. [2], § 56, problem 2)



S0(z) = pz, p = √2mE, (5)

pσ ′(z)+ σ 2 + e2

4c2 B2(z) = 0. (6)

Outside the lens (where |z| > a, B(z) = 0), we have from (6)

σ(z) = pz + C1

, when z < −a, (7)

σ(z) = pz + C2

, when z > a. (8)

The equation of the orbit,

∂S∂C1,2

= − pr2

2(z + C1,2)2 = B1,2 (9)

is an equation of strait lines intersecting the z-axis in the points −C1,21; that is, z0 = −C1

and z1 = −C2. From (6), we get

pσ(a)− pσ(−a)+a∫

−a

σ 2 dz + e2

4c2

a∫−a

B2(z)dz = 0. (10)

Since |z0,1| � a, it follows from (7) and (8) that

σ(±a) = − pz1,0

. (11)

Let us estimate∫ a−a σ 2 dz. According to (6), σ(z) is a monotonic function. Therefore,

we have

a∫−a

σ 2 dz � 2ap2

z21,0

� pσ(±a).

It thus follows from (10) that

1|z0| + 1

z1= e2

4c2p2

a∫−a

B2(z)dz = 1f

. (12)

The condition |z0,1| � a is, indeed, satisfied when a � cp/(eB).

1 When z is close to C1,2, σ → ∞ so that expansion (3) becomes inapplicable. However, the equations (9)for the orbits remain valid also for that region.



12.17. The whole of the calculations of the preceding problem up to (6) is applicablealso in this problem. The substitution σ = pf ′/f reduces (6) to the form

(1 + �2z2

)2f ′′(z)+ e2B2

4c2p2 f = 0,

and after that the substitution

�z = tanξ , −12 π < ξ < 1

2 π , f (z) = η(ξ)

cosξ

gives

η′′(ξ)+ λ2η(ξ) = 0,

where

λ2 = 1 + e2B2

4c2�2p2 .

Hence,

σ = 12 �psin2ξ + λ�pcos2 ξ cot(λξ + α),

and the equation for the orbit becomes

∂S∂α

= − �pr2λcos2 ξ

2sin2(λξ + α)= B,

or

r cosξ = B′ sin(λξ + α).

When r = 0,

λξn + α = πn,

and thus α = −λ arctan(�z0) so that the points where the beam is focused are given by

�zn = tan(arctan�z0 + nπ

λ

).

Depending on the magnitude of λ, there will be one or several points of focusing.



12.18.

S(q, q0, t, t0) = f (q, α(q, q0, t, t0), t)− f (q0, α(q, q0, t, t0), t0),

where f (q, α, t) is the complete integral of the Hamilton–Jacobi equation, while thefunction α(q, q0, t, t0) is determined by the equation (or set of equations for the caseof several degrees of freedom)

∂f (q, α, t)∂α

= ∂f (q0, α, t0)

∂α.


§13

Adiabatic invariants

13.1. E2l = const.Let us explain the answer. On the ring A there acts a force F determined by the tension

T in the thread. For the small angles ϕ of deflection of a pendulum, we get

Fx = mgϕ, Fy = 12 mgϕ2

(the y-axis is directed vertically upwards; the x-axis is in the plane of oscillations.) Sincethe length of the thread AB = l changes slowly, we can average the force over a period ofthe oscillation, ϕ = ϕ0 cosωt, ω = √

g/l, assuming the length of the thread to be constant.We get

Fx = 0, Fy = 14 mgϕ2

0.

l

2l2l

t

t

t

υ

υ

υl

(a)

(b)

(c)

Figure 188

When the ring is displaced over a distancedy = dl, the energy decreases by Fydy = 1

4mgϕ20dl.

Since E = 12mglϕ2

0 , we have

dE = − E2l

dl.

Hence, we have E2l = const.

13.2. After the particle has collided with both walls,its velocity v is changed by 2l. The condition that thechange is slow means |2l| � v.

We choose a time interval �t such that

2lv

� �t � l

|l| .




Such a �t exists because of the slowness condition. During such a time interval there arev�t/(2l) pairs of collision with the walls, and the velocity is changed by

�v = −vl�tl

.

Integrating, we obtain vl = const or El2 = const.It is interesting to study in somewhat more detail how the product vl changes. This is

easily done by studying the functions l(t) and v(t) (see Fig. 188a and b). In Fig. 188c, wehave drawn the function I(t) = v(t)l(t). The quantity vl oscillates around the practicallyconstant value 〈vl〉 while the amplitude of the oscillation has the relative value �I/I ∼ l/v.

The deviation of 〈vl〉 from a constant value is of higher order of smallness:

ddt

〈vl〉 ∼ l2.

13.3. The particle of low mass moves much faster than the piston. To determine thevalue of the particle velocity v, we assume that it moves between the slowly and smoothlyshifting walls (bottom of the vessel and the piston). Ignoring the effect of gravity on themotion of this very fast particle, we can use the conservation of the adiabatic invariantvX = const = C (see problem 13.2). To determine the law of motion of the piston, wecan use the law of energy conservation

12 MX2 + MgX + 1

2 mv2 = E

(here we have neglected the contribution of the potential energy of a light particle).Substituting v = C/x, we obtain the equation for the motion of the piston:

12 MX2 + Ueff (X) = E, Ueff (X) = MgX + mC2

X2 ,

Here Ueff (X) is the effective potential energy of the piston, averaged over the period of

motion of a light particle. The function Ueff (X) has at a minimum at X0 =(

2mC2

Mg

)1/3

.

The piston can oscillate near X = X0 with frequency

ω =√

U ′′eff (X0)

M=

√3gX0

.

This problem demonstrates the so-called adiabatic approximation. We consider themotion of a system consisting of bodies that differ strongly in mass where light particlesmove much faster than the heavy ones. In this case, the coordinates of heavy particles canbe considered as parameters defining the field in which the light particles are moving. The



change of these parameters is sufficiently slow and smooth so that adiabatic invariantsare conserved. Depending on the values of these parameters (i.e., on the coordinates ofheavy particles), the energy of the light particles changes, and the source of these changes(positive or negative) is the kinetic energy of heavy particles. When studying the motionof heavy particles, this energy of the light particles can be added to the potential energyof the heavy particles. Of course, this implies averaging over time the motion of the lightparticles.

Such an approximation is used in studying the motion of atoms inside both a moleculeand a crystal. Of course, in these problems the motion of particles, as well as lightelectrons and massive ions, is described not in terms of classical mechanics, but ratherin terms of quantum mechanics. However, the essence of the adiabatic approximation ismore helpful to demonstrate the approximation using a simple example from classicalmechanics. The same question is dealt with in problem 13.9 where an extremelysimplified model of a molecule is discussed.

13.4. If g(t) = g − a(t) were constant, the motion of the ball would be described by

z(t) = h − 12 2gt2 for − √

2h/g < t <√

2h/g).

A change in g(t) by �g leads to a change in the potential energy by mz�g, and over aperiod by m〈z〉�g, where 〈z〉 = 2

3 h is the time average of z.There is a slow change in the total energy, �(mgh), due to the changes in the potential

energy. Therefore m�g · 23h = �(mgh) or g�h + 1

3h�g = 0, whence h ∝ g−1/3.In this proof we have essentially followed by the same method which in the general

case can be applied to prove that∮

pdq is constant (see [1], § 49).Of course, we could directly have used the results of the general theory in this problem

(and in the two other preceding problems).If the plate is raised slowly (but g(t) = const), then h = const. This is clear when the

velocity of the plate is constant (it is sufficient to change to a system of reference fixedin the plate). If the velocity changes the result cannot change as it depends, according tothe general theory, only on the height of the support of the plate. It is assumed that therelative change in velocity during a time

√2h/g is small and the plate is risen smoothly.

13.5. a) E = −A(

1 − αI√2mA

)2

;

b) E = −U0

(1 − αI√

2mU0

);

c) E = αI√

2U0m + α2I2

2m ;

d) E =[√

π2m

I2 α1/n �

(1n + 3

2

)/�

(1n)]2n/(n+2)

.

13.6. h ∝ (sinα)2/3.

13.7. a ∝ (sinα)−1/4.



13.8. I = 8ml√

glπ

[E

(sin ϕ0

2

)− K

(sin ϕ0

2

)cos2 ϕ0

2

], where K(k) and E(k) are the

complete elliptic integrals of the first and second kind, respectively (see footnote in thesolution of problem 1.7).

13.9. Let the coordinates of the particles m and M, reckoned from the point O, be xand X . The motion of the light particle can be consider approximately as the motionbetween two walls, one of which is moving. As the condition

|x| |X | (1)

is satisfied, the average over one period of the product |x|X = C is conserved (seeproblem 13.2). Eliminating x from the energy conservation law,

12 mx2 + 1

2 Mx = E,

we find that the effect of the light particle upon the motion of the heavy particle isequivalent to the appearance of a potential energy

U(X) = mC2

2X2 .

The equation

12 MX2 + U(X) = E

leads to the law of motion:

X =√

mC2

2E+ 2E

M(t − τ)2.

The constants E, C, and τ can be determined from the initial values of X , X , and x (theyare independent of x(0)). This method of solving the problem becomes inapplicablewhen condition (1) is not satisfied.

13.10. We denote the coordinates of the heavy particles by X1,2 and the light particlesby x. When X1 < x < X2, the potential energy is

U = (x − X1)f + (X2 − x)f = (X2 − X1)f .

Therefore, the light particle moves freely between the heavy particles, and

|x|(X2 − X1) = C = const



(see the preceding problem). Taking this into account, we get the following equationfor the relative motion of the heavy particles (X = X2 − X1) from the law of energyconservation

14 MX2 + mC2

2X2 + fX = E.

Expanding

Ueff(X) = mC2

2X2 + fX

near the minimum X0 = (mC2/f )1/3, we find the frequency of the small oscillationsof an “ion”

ω2 = 2U ′′(X0)

M= 6f

MX0.

13.11. We expand the frequency in a series in t in the equations for P and Q,

Q = ω + ω

2ωsin2Q, P = −P

ω

ωcos2Q.

Q

p

q

t

t

t

(a)

(b)

(c)

Figure 189

Restricting ourselves to the first-order corrections,we get for P and Q the equations

Q = (ω0t + ϕ)+ 12 ω0t2 + ω0

2ω0

t∫0

sin2Q(t)dt, (1)

P = P0

(1 − ω0

ω0

t∫0

cos2Qdt)

, (2)

where ω0 and ω0 are the values of the frequencyand its derivative at time t0 = 0, and we have ω0 = ε2ω2

0with ε � 1.

The phase Q and the amplitude A = √2P/(mω0) of

the perturbed motion differ relatively little from theirunperturbed values Q0 = ω0t + ϕ and A0 = √

2P0/(mω0) even for time intervals whichare much longer than the period 2π/ω of the oscillations (see Fig. 189).



Thus, for a time t ∼ 1/(εω0), the second term in (1) is of order unity, and the thirdterm of order ε, and thus

Q = ω0t + ϕ + 12 ω0t2, P = p0.

However, this change in phase leads to the fact that the perturbed motion in terms of thevariables p and q,

q(t) = A0 cos(ω0t + ϕ + 1

2 ω0t2)

,

will differ appreciably from unperturbed motion,

q0(t) = A0 cos(ω0t + ϕ),

in such a way that

q(t)− q0(t) ∼ q0(t).

When one tries to construct a perturbation theory for the variables q and p, one obtainsfor the first-order corrections q1(t) the equation

q1 + ω20q1 = −2ω0ω0tA0 cos(ω0t + ϕ),

which has a resonant force which increases with time. The solution obtained in such atheory is thus applicable only for small time intervals of the order of a few periods of theoscillations

2π

ω0� 1

εω0.

13.13. We transform the Hamiltonian function of the system

H(x, pt) = p2

2m+ 1

2 mω2x2 − xF(t) = E(t) (1)

to the form

12 mω2

(x − F

mω2

)2

+ p2

2m− F2

2mω2 = E(t).



From this, it is clear that the orbit in phase space is an ellipse which is displaced alongthe x-axis over a distance F/(mω2) with semi-axes

a =√

2E

mω2 + F2

m2ω4 , b =√

2mE + F2

ω2 .

Apart from the factor 1/(2π), the adiabatic invariant is the area of this ellipse

I = 12 ab = E + F2/(2mω2)

ω. (2)

Here the meaning of E + F2/(2mω2) is that of the energy of oscillations near the displaceequilibrium position (cf. problem 5.16). Substituting the value of E from (1) into (2),we can present the result in the form

I = m2ω

∣∣∣∣x + iω(

x + Fmω2

)∣∣∣∣2

=

= m2ω

∣∣∣∣∣∣1m

t∫0

eiω(t−τ)F(τ )dτ + eiωt [x(0)+ iωx(0)] − iF(t)mω

∣∣∣∣∣∣2

.

Here we can use equations (22.9) and (22.10) from [1] for the quantity x + iωx.Integrating by parts, we get

I(t) = I(0)− x(0)

ω2

t∫0

F(t)sinωt dt−

− 1ω

[x(0)− F(0)

mω2

] t∫0

F(t)cosωt dt + 12mω3

∣∣∣∣∣∣t∫

0

F(τ )e−iωτ dτ

∣∣∣∣∣∣2

.

If the force changes slowly, I(t) will oscillates near I(0). If F(t) → const as t → ∞, thetotal change in the adiabatic invariant, I(∞)− I(0), can be very small (see problem 5.18).

13.14. PV 5/3 = const.

13.15. a) E = π2

2m

(I21

a2 + I22

b2 + I23

c2

), where a, b, and c are the lengths of the edges of the

parallelepiped, and Ik = const.(b) The absolute magnitude of the components of the velocity along each of the edges

are conserved.



13.16. In spherical coordinates, the variables separate. The angular momentum M isstrictly conserved. (Moreover, Mz is an adiabatic invariant corresponding to the angle ϕ).The adiabatic invariant for the radial motion is

Ir = 1π

R∫rmin

√2mE − M2

r2 dr. (1)

We can find the function E(R) without evaluating the integral (1). The substitution r =Rxgives

Ir = 1π

1∫xmin

√2mER2 − M2

x2 dx = Ir(ER2, M), (2)

whence ER2 = const.We get thus for the angle of incidence α the equation

sinα = rmin

R= M√

2mER= const.

13.17. a) E ∝ γ2

2−n ; b) E ∝ γ −1.

13.18. Equating the values of the adiabatic invariant before and after the switching onof the field,

rmax∫rmin

√E − M2

2mr2 − U(r)dr =rmax∫

rmin

√E + δE + M2

2mr2 − U − δU dr

we get

δE = 〈δU〉 = 2T

rmax∫rmin

δU dr√2m

(E − M2

2mr2 − U) .

13.19. E = I1�1 + I2�2 (in the notation of problem 6.6a). The orbit fills the rectangle

|Q1| �√

I1/�1, |Q2| � √I2/�2.

The conditions that the theory of adiabatic invariants is applicable is

|�i| � �2i , |�i| � �i|�i| (i = 1, 2).



Outside the region of degeneracy, these conditions reduce to the same conditions forω1(t). In the region of degeneracy, |ω2

1 − ω22| ∼ α, and the second condition is more

restrictive and gives |ω1| � α (the region of degeneracy is traversed during a time whichis considerable longer than the period of the beats).

13.20. When the coupling αxy is not present, the system splits into two independentoscillators with coordinates x and y. The corresponding adiabatic invariants are

Ix = Ex

ω1, Iy = Ey

ω2,

where Ex and Ey are the energies of these oscillators.When the coupling is taken into account the system consists of two independent

oscillators with coordinates Q1 and Q2. If the frequency changes slowly, the quantities

I1 = E1

�1, I2 = E2

�2.

are conserved.Outside the region of degeneracy the normal oscillations are strongly localized, and

when ω1 < ω2, then Q1 = x, Q2 = y, while when ω1 > ω2, then Q1 = +y, Q2 = −x. Thus,when ω1 < ω2, Ix = I1, Iy = I2, while when ω2 < ω1, then Ix = I2, Iy = I1 (Fig. 190).

I

I1 Iy

IxI2

ω1ω2

A

B C

L(t)

D

l


We shall illustrate this by the following example. Two pendulum, of which the length ofone can be changed slowly, are coupled by a spring with small stiffness (Fig. 191). Whenthe length l and L of the pendulums are appreciable different, the normal oscillations arepractically the same as the oscillations of one or the other pendulum.

Let initially L > l and the first pendulum AB oscillates with amplitude ϕ0, and thependulum CD with a very small amplitude. When L is decreased, the amplitude of theoscillations of the pendulum CD remains small until its length becomes almost equal to l.When L ≈ l, its amplitude increases (and when L = l both pendulums will oscillate withthe same amplitudes, ϕ0/

√2, in anti-phase). When L is decreased further, practically all

the energy transfers to the pendulum CD and its amplitude becomes ϕ1 = ϕ0 (l/L)3/4, asfor a separate pendulum.

If we traverse the degeneracy region relatively fast, ω1 α, such a transfer of energybetween the oscillators will not take place. Moreover, if ω1 � ω2

1, ω1 � ω1ω1, Ix and Iyare conserved.



13.21. From the equations of motion

x + ω21x + 2βxy = 0, (1)

y + ω22y + βx2 = 0, (2)

we see easily that the coupling between the oscillators leads to a large energy transferwhen 2ω1 ≈ ω2.

Let

x = a(t)cos(ω1t + ϕ), y = b(t)cos(ω2t + ψ).

If a b, the term

βx2 = 12 βa2 + 1

2 βa2 cos(2ω1 + 2ϕ)

in (2) will play the role of an applied force, leading to a resonance increase in y. If,however, a � b, the term

2βxy = 2βbxcos(ω2t + ψ)

in (1) leads to a parametric building-up of the oscillation in x. The detail of the study ofsystem (1) and (2) can be found in problem 8.10.

The region of resonance interaction is

|2ω1 − ω2| � βbω1

.

In general, a strong resonance interaction between the oscillators occurs whennω1 = lω2, where n and l are integers. However, the widths of the regions of frequenciesin which these resonances occur are extreamly small when n and l are not too small(see [1], § 29). We can therefore neglect their influence on the motion of the oscillatorsfor values of ω1 that are not too small (provided they are sufficiently small that we canuse the theory of adiabatic invariants).

y

y0(x)

x

Figure 192

13.22. Let the particle moving in the xy-plane at a smallangle to the y-axis (|x| � |y) be reflected from the x-axisand from the curve y0(x) (Fig. 192).

If we assume that we know how the particle moves in thex-direction, we can study the motion in the y-direction bytaken x(t) to be a slowly changing parameter. The adiabaticinvariant,

∮py dy = 2|py|y0(x) = 2πI,

will remain constant, and that equation determines the function py(x).



O EC Ax

BD

C′D′

E′

α

ϕ0

Figure 193

To determine x(t), we can use the energy conser-vation law

m2x2 + p2y(x) = 2mE.

The minimum distance xmin is determined by thecondition

p2y(xmin) = 2mE.

Substituting y0(x)=x tanα and 2πI =2√

2mEl tanα cosϕ0, we get

xmin = l cosϕ0.

The solution by means of the reflection method is clear from Fig. 193. This methodgives the exact solution applicable at any angle α and ϕ0 but it cannot be generalized tothe cases when y0(x) is not a straight line.

13.23. tanhαxmax = tanϕ, T = 2π

αv√

cos2ϕ(see the preceding problem).

13.24. a) The problem of the motion of a particle in a magnetic field reduces for thegiven choice of vector potential to the problem of the motion of a harmonic oscillator(see problem 10.8). The adiabatic invariant is

I = E − p2z/(2m)

ω∝ v2⊥

B∝ πa2B,

where v⊥ is the transverse to the magnetic field component of the particle velocity anda = cmv⊥/(eB) is the radius of the particle orbit (see [2], § 21).

The relation I ∝ πa2B can be interpreted simply: the radius of the orbit changes insuch a way that the magnetic field flux through the area circumscribed by it remainsconstant. The distance of the centre of the orbit from the yz-plane is equal to x0 =cpy/(eB), and decreases with increasing B.

The occurrence of a drift of the orbit is connected with the appearance of an electricfield,

E = −1c

∂A∂t

=(

0, −1c

xB, 0)

,

when the magnetic field changes (cf. [2], § 22). The electric field vectors E and the driftvelocity vdr are shown in Fig. 194 for different orbit positions of the particle.



b) The Hamiltonian function is in cylindrical coordinates

H = p2z

2m+ p2

r

2m+ 1

2mr2

(pϕ − eB

2cr2

)2

.

The quantity pz and pϕ are integrals of motion.The adiabatic invariant for the radial motion is

πIr =rmax∫

rmin

√2mE⊥ − 1

r2

(pϕ − eB

2cr2

)2

dr,

which after the substitution r = ζ/√

B becomes

ζmax∫ζmin

√2mE⊥

B− 1

ζ 2

(pϕ − e

2cζ 2

)2dζ = πIr

(pϕ ,

E⊥B

).1

Therefore, E⊥/B = const; that is, the energy of transverse motion, E⊥ = 12 mv2⊥, changes

in the same way as under sub 13.23a. The distance r0 of the centre of the orbit to theorigin is

r0 = rmax + rmin

2= ζmax + ζmin

2√

B∝ 1√

B.

When B increases, the centre of the orbit approaches the origin (Fig. 195).

yE

E

x

vdr

vdr

E

E

vdr vdr


1 It is interesting that Ir does not actually depend on pϕ when pϕ > 0. Indeed,∂Ir∂pϕ

is the change of the

angle �ϕ over the time of the one radial oscillation; besides, when pϕ > 0, the origin of coordinates lies outsidethe orbit (see Fig. 109b), and therefore �ϕ = 0.



When B changes, there occurs an electric field

(E)ϕ = − r2c

B, (E)r = (E)z = 0,

of which the field of lines are closed circles.In real conditions, a uniform magnetic field can exist only in a limited region of

space. The electrical field occurring when the magnetic is changed depends very stronglyon the shape of that region and the conditions at its boundaries (see [2], § 21). Forexample, the field considered under sub 13.23a could occur near a conducting planein which there was a current, while the field under sub 13.23b could be produced ina solenoid.1

The strong dependence of the nature of the particle motion on the weak field E evenin the case of extremely small B can be explained by the presence of degeneracy (whenB = const the periods in the two coordinates x and y, or r and ϕ are the same).

We note that the quantity E⊥/B turns out to be an adiabatic invariant in both cases.One can prove that this result is independent of the choice of the form of A (see [2], § 21and [10], § 25).

13.25. The adiabatic invariants are

Iz = Ez

ω, Iϕ = pϕ ,

Ir = 1π

ζmax∫ζmin

√Cζ 2 − p2

ϕ − m2ζ 4 dζ

ζ= Ir(pϕ , C), (1)

where

C = 2mE⊥ + eBpϕ/c√ω2 +

(eB

2mc

)2, Ez = 1

2 mv2z , E⊥ = 1

2 mv2⊥,

and v⊥ is the transverse to the magnetic field component of the particle velocity.Therefore

Ez ∝ ω, E⊥ + eB2mc

pϕ ∝√

ω2 +(

eB2mc

)2

. (2)

1 The change in the electric field E connected with the change in the choice of the form of A would notoccur if we had simultaneously changed the scalar potential by Bxy/(2c) (gauge transformation).



The given vector potential defines a magnetic field which is symmetric with respect tothe z-axis going through the centre of the oscillator. If we make a different choice for A,

Ax = Az = 0, Ay = xB(t), (3)

we get, in fact, a different physical problem. The Lagrangian functions for these twoproblems differ by

δL = ddt

( e2c

Bxy)

− e2c

Bxy; (4)

that is, the difference is very small if we drop in (4) the inessential total derivative withrespect to time.

In the preceding problem, where the motion was degenerate, just this extra term ledto a complete change in the direction and drift velocity of the orbit. In the present case,however, the motion of the oscillator is not degenerate when B = const and we canneglect the extra term δL (cf. problem 13.20). The relations (2) are thus valid also fora different choice of A. When one passes through the degeneracy region (B = 0), theequations (2) remains valid only when we choose the axially symmetric field given inthe problem. For instance, the behaviour of the oscillator in the field (3) when B passesthrough zero requires additional study.

13.26. a) Using a canonical transformation, one can reduce the Hamiltonian functionto a sum of the Hamiltonian functions for two independent oscillators (for X and Y ; seeproblem 11.10). For each of the oscillators the ratio of the energy to the frequency is anadiabatic invariant. We remind ourselves that the oscillations of each of them correspondto motion along an ellipse (see problem 6.37). In terms of the amplitude ak of theoscillations in the x-direction, for instance, the adiabatic invariants are equal to

Ik = ma2k

2�k

�4k − ω2

1ω22

�2k − ω2

2

(k = 1, 2).

When the parameters of the system are changing, we must also add to the newHamiltonian function the partial derivative with respect to the time of the generatingfunction which is equal to −λ[mω2XY + PXPY/(mω2)] (see problem 11.18). Thiscorrection term is small (λ � �k) and can be neglected provided the eigen-frequenciesare not the same (cf. problem 13.20). One must consider separately the degenerate casewhen ω1 = ω2 and a magnetic field which can vanish.

When ω1 = ω2, the different choice of the adiabatic invariants is possible (seepreceding problem).

b) To fix the ideas, let ω1 > ω2. The motion is along a circle of radius a√

ω1/(ωB)

with a frequency ωB, but the centre of the circle moves along an ellipse with semi-axesin the x- and y-direction and equal to



bω2√ω1ωB

and b√

ω1

ωB

with a frequency ω1ω2/ωB.c) The oscillation will proceed almost in the y-direction; its amplitude is increased by

a factor√

ω1/ω2 (cf. problem 13.20).

13.28. a) The motion of the particle in the xy-plane takes place under the action ofa magnetic field which is slowly varying as the particle moves along the z-axis. Theadiabatic invariant

I⊥ = E⊥[

eB(z)

mc

]−1

is then conserved (see problem 13.24). From the energy conservation law, we have

12 mz2 + I⊥

eB(z)

mc= E.

The particle moves in the z-direction as if it moved in a potential field

U(z) = I⊥eB(z)

mc.

The period of the oscillations is (cf. problem 2b from [1], § 11)

T = 2πa

v√

λsin2 α − cos2 α,

where α is the angle between the velocity v of a particle and the z-axis. Particles, forwhich tan2 α < 1/λ are not contained in the trap. The condition for the applicability of thetheory of adiabatic invariants consists in the requirement that the change in the magneticfield during one period of revolution of the particle be small. This gives mcλvz � aeB0.

As an example of a magnetic trap, we can observe the radiation belts of the Earth.

b) T = 2πavsinα

.

13.29. a) (λE⊥ − Ez)a2 = const, E⊥/B0 = const, E = E⊥ + Ez;b) E⊥/B0 = const, aEz

√B0 = const.

13.30. If we neglect in the Hamiltonian function

H = p2r

2m+ p2

θ

2mr2 + p2ϕ

2mr2 sin2 θ− eBpϕ

2cm+ e2B2r2 sin2 θ

8mc2 + U(r)


374 Exploring Classical Mechanics [13.31.

the term which is quadratic in B, we can separate the variables in the Hamilton–Jacobiequation.

The adiabatic invariants have the form

Iϕ = pϕ , Iθ = 1π

θ2∫θ1

√β − p2

ϕ

sin2 θdθ = Iθ (pϕ , β),

Ir = 1π

r2∫r1

√2m

[E + eBpϕ

2mc− U(r)

]− β

r2 dr = Ir

(E + eBpϕ

2mc, β

).

When B is slowly changed, the quantities

pϕ , β = p2θ + p2

ϕ

sin2 θ, and E + eBpϕ

2mc

thus remain constant.

13.31..

p2x

2m+ 1

2 mω2x2 = E1,p2

y

2m+ 2mω2y2 = E2,

(m2ω2x2 − p2x)y + xpxpy = A,

(m2ω2x2 − p2x)py − 4mωxpxy = m{E2, A}.

13.32. a) w = arctanmωxp , I = p2

2mω+ 1

2 mωx2.These variables are convenient, for instance, to develop perturbation theory (seeproblem 13.10).

b) Let initially the particle moves to the right from the point x = 0, and we shall chooseS such that S = 0 for x = 0. In that case,

S =x∫

0

|p|dx = πI − πa

[(Ia

)2/3

− Fx

]3/2

,

where

I = 1π

xm∫0

|p|dx = aE3/2, a = 2√

2m3πF

, xm = EF

, |p| = √2m(E − xF).



If the motion is to the left,

S =⎛⎝

xm∫0

−x∫

xm

⎞⎠ |p|dx = πI + πa

[(Ia

)2/3

− Fx

]3/2

and so on. For the nth oscillation,

S = (2n − 1)πI ∓ πa

[(Ia

)2/3

− Fx

]3/2

(the upper [lower]) sign corresponds to motion to the right [left]; Fig. 196).One can use the S(x, I) as a generating function to change to new canonical action

and angle variables (see [1], § 49). The new variables are connected with the old ones inthe following way:

x = 1π2F

(Ia

)2/3 {π2 − [(2n − 1)π − w]2

},

p = 32

aF(

Ia

)1/3

[(2n − 1)π − w],

where x is a periodic function of w, while w in multivalued function of x (Fig. 197).

S

xxm

x

2π 4π 6π w

xm


13.33. From the equation

P =a∫

0

√2m(E − U)dx

we find

E = P2

2ma2 + 12 V + mV 2a2

8P2 .



The abbreviated action is

S0=x∫

0

pdx=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

√2mE(x − na) + nP, when na < x <

(n + 1

2

)a,

√2m(E − V )

[x − (n + 1

2 )a]+

+ √2mE

(12 a

)+ nP

, when(n + 1

2

)a < x < (n + 1)a.

Eliminating E and introducing A = ma2V2P2 , we get the generating function for the

canonical transformation under consideration

S0(x, P) = Pa

·

⎧⎪⎨⎪⎩

(1 + A)(x − na) + na, when na < x <(n + 1

2

)a,

(1 − A)(x − na) + Aa + na, when(n + 1

2

)a < x < (n + 1)a.

From the equations

p = ∂S0

∂x, Q = ∂S0

∂P,

we get

p(P, Q) = Pa

·{

1 + A, when n < Q < n + Q0,

1 − A, when n + Q0 < Q < n + 1,

x(P, Q) = a ·

⎧⎪⎪⎨⎪⎪⎩

Q − nA1 − A

, when n < Q < n + Q0,

Q + (n + 1)A1 + A

, when n + Q0 < Q < n + 1,

where Q0 = 12 (1 − A). The variables P and Q are analogous to action and angle variables,

and the quantity aQ is the average particle velocity.


Bibliography

[1] L. D. Landau, E. M. Lifshitz (1976) Mechanics, Pergamon Press, Oxford.[2] L. D. Landau, E. M. Lifshitz (1980) Classical Theory of Fields, Butterworth-

Heinenann, Oxford.[3] L. D. Landau, E.M. Lifshitz (1984) Electrodynamics of Continous Media, Pergamon

Press, Oxford.[4] H. Goldstein (1960) Classical mechanics, Addison-Wesley, Reading, Mass.[5] H. Goldstein, C. Poole, J. Safko (2000) Classical Mechanics, Addison-Wesley,

Reading, Mass.[6] D. ter Haar (1964) Elements of Hamiltonian Mechanics, North Holland, Amsterdam.[7] G. L. Kotkin, V.G. Serbo, A. I. Chernykh (2010) Lectures on Analytical Mechanics,

Regular and Chaotic Dynamics, Moscow-Izhevsk.[8] V. I. Arnold (1989) Mathematical Methods of Classical Mechanics, Springer-Verlag,

Berlin.[9] L. I. Mandelstam (1972) Lectures on Theory of Oscillations, Nauka, Moscow.

[10] N.N.Bogolyubov, Yu. A. Metropolskii (1958) Asymptotic Methods in the Non-linearOscillations, Noordhoff, Groningen.

[11] M. Abramowitz, I. A. Stegun (1965) Handbook of Mathematical Functions, Dover,New York.

[12] N. F. Mott, H. S. W. Massey (1985) The Theory of Atomic Collisions, ClarendonPress, Oxford.

[13] N.N.Bogolyubov, D. V. Shirkov (1980) Introduction to the Theory of QuantizedFields, Wiley, New York.

[14] P. Courant, D. Hilbert (1989) Methods of Mathematical Physics, Wiley, New York.[15] J. V. Stratt (Lord Rayleigh) (1945) Theory of Sound, Vol. I., Chap. IV, Dover,

New York.[16] C. Kittel (1968) Introduction to Solid State Physics, Wiley, New York.[17] J.P.Den Hartog (1985) Mechanical Vibrations, Dover, New York.[18] E. Fermi (1971) Scientific Papers, Vol. I, p. 440, Nauka, Moscow.[19] N.Bloembergen (1965) Nonlinear Optics, Appendix I, §§ 5,6, W. A. Benjamin,

New York.[20] L. D. Landau, E.M. Lifshitz (1980) Statistical Physics, Chap. XIV, Pergamon Press,

Oxford.[21] F. D. Stacey (1969) Physics of the Earth, Wiley, New York.


378 Bibliography

[22] A. I. Bazh, Ya. B. Zeldovich, A. M. Perelomov (1971) Scattering, Reactions andDecays in Non-Relativistic Quantum Mechanics, Nauka, Moscow.

[23] P. Calogero, O. Ragnisco (1975) Lettere al Nuovo Cimento 13, 383.[24] M.S.Ryvkin (1975) Doklady Akademii Nauk, (USSR) 221, No 1.[25] V. V. Sokolov Nuclear Physics, (USSR) 23, 628 (1976); 26, 427 (1977).[26] F. Rafe (1974) Scientific American, 12.


Index

The numbers refer to the problems; for instance, 4.12 is problem 12 of § 4.

Acoustic branch 7.5, 7.7Adiabatic approximation 13.3,

13.9, 13.10Angle and action

variables 11.1, 13.32,13.33

Anomaly, eccentric 2.44

Barrier, spherical 3.9Beam focusing 2.15, 8.13,

12.16, 12.17Beats 5.11, 6.9, 6.34, 8.8

Christoffel symbol 4.22Circuits 4.24–4.26, 5.10, 5.19,

6.7Conservation law

angular momentum 4.14,10.1

energy 2.1, 4.14momentum 2.1, 4.14, 10.1

Constraints 4.28–4.31Continuous systems 4.32, 7.11,

7.12Curvilinear coordinates 4.21,

4.22

Degenerate frequencies 6.19,6.21, 6.22, 6.29, 6.30

Delay time 1.12Dipol, motion in 3.12, 10.26,

12.2–12.10Dissipative function 5.10,

11.25Distribution of decay

particles 3.16, 3.17Displacement of the Earth’s

perihelion 2.25Displacement from the

vertical 9.23Drift 2.27, 2.38, 6.37, 8.14,

10.9

Earth-Moon system 2.25, 9.9,9.17

Electric field, motion in 2.29,2.40, 4.15, 6.18, 10.11,10.24, 11.16, 12.1, 12.20

Electron and nuclearparamagneticresonance 10.23

Electron motion in periodiclattices 10.11, 10.13

Energy flux 6.10, 7.4, 7.6Euler angles 9.14, 9.16, 9.21,

9.22, 10.2, 13.15Euler equations 9.10, 11.7

Fermi resonance 8.10Frequency

division 8.10doubling 8.10tripling 8.6

Forbidden band 7.5, 7.7Friction 5.11, 5.13, 5.20,

11.25

Gauge transformation 11.14,13.24

Generating functionfor the canonical

variables 11.7–11.11for the identity

transformation 11.5for the gauge

transformation 11.14for the similarity

transformation 11.13Gravity, motion under 4.28,

4.29, 5.2, 5.8, 9.22Gyrocompass 9.14

Hidden symmetry of thehydrogen atom 10.26

Holes 10.12

Impact phase 5.12, 5.15Impurity atom in a crystal

7.8Inaccessible region 2.16, 12.13,

12.14Inertial tensor 9.2–9.4, 9.6Infinitesimal

transformations 10.1,10.18, 11.17–11.19, 11.24

Integrals ofManley–Rowe 11.27,11.30

Jacobi coordinates 2.24

Kepler problem 2.3, 2.44,10.26

Lagrangian multipliers 4.28,6.24

Lagrange points 9.29Laplace vector 2.19, 2.39,

2.41–2.45, 3.6Larmor theorem 2.37, 2.41,

6.37, 10.22Lifetime of a satellite 2.42Light, propagation of 10.6Liouville theorem 11.25Lissajous figures 6.5Localised oscillations 6.6

Magnetic dipole 2.35, 2.44,4.19, 10.22

Magnetic field, motion in 2.30,2.33, 2.36–2.38, 4.15, 4.18,4.19, 4.33, 4.34, 6.37–6.39,10.8, 10.12, 10.13, 10.14,10.23, 11.9, 11.12, 12.16,12.17, 13.24–13.30

Magnetic monopol 2.34,4.23

Magnetic trap 13.27, 13.28


380 Index

Moleculesdiatomic 5.6four-atomic 6.53triatomic 6.51, 9.26

Noether’s theorem 4.12Normal coordinates 6.1, 6.2,

6.4, 6.6, 6.11, 6.17, 6.19,6.21, 6.22, 6.25, 6.27,6.28, 6.30, 6.31, 6.37

Optical branch 7.5, 7.7Orthogonality of the normal

oscillations 6.23, 6.25,6.27–6.30

Oscillatoranharmonic 1.11, 8.1, 8.2,

8.4–8.7, 10.2, 11.5, 11.6,11.25

harmonic 1.9, 1.11,5.11–5.20, 6.37, 6.38, 8.8,8.9, 8.12, 11.8–11.10,11.16, 11.25, 12.1,13.11–13.13, 13.24–13.26,13.31, 13.32

Oscillators, coupled 11.15,13.19–13.21

Partial frequency 6.6, 6.17Pendulum 1.6, 1.7, 4.26, 5.5,

5.9, 6.39, 8.3, 8.9, 12.10,13.1, 13.7, 13.8

Pendulums, coupled 13.20Perturbation theory 1.8–1.10,

2.20, 2.28, 2.38, 2.40,3.14, 6.31, 6.33, 13.11,13.15, 13.16, 13.18

Phase-spacedistributions 11.25

Phase trajectories 1.5Potential

Coulomb 2.6, 2.21, 2.28,2.29, 2.37, 2.44, 12.15

Morse 1.1perturbed 1.8–1.10, 2.20Yukawa 2.7, 2.21

Principal axes 9.2, 9.3, 9.12,9.13, 9.19

Precession of orbit 2.1Probability in the coordinate

space 1.5Probability in the momentum

space 1.5

Quadrupole momenttensor 9.7, 9.16

Quasi-momentum 10.12Quasi-particles 10.12

Radial motion 2.1Radial oscillations 2.1Radiance 3.10Rayleigh’s theorem 6.6Reciprocity theorem 6.26Resonance 5.11, 5.19, 6.14,

6.15, 8.6, 13.21Resonance, parametric 8.8, 8.9Retreat of the lunar nodes 2.25Rotating frame, motion in 4.10,

4.11, 9.15, 9.22–9.28

Scattering § 3 5.14, 5.15,12.3–12.5

Rutherford 3.2

small angle 3.9, 3.11, 3.13,12.3

Solids, motion in 10.11–10.13Spatial dispersion 7.12Spiral orbit 2.1, 2.4Strong focus of particle beams

in accelerators 8.13Symmetry properties of the

normaloscillations 6.45–6.54

Total number of quanta 11.30Top

asymmetric 9.19symmetric 9.6, 9.15, 10.2

TransformationsGalilean 4.17, 11.19gauge 11.14, 13.24infinitesimal 11.23, 11.24similarity 4.15, 4.16, 4.33

Transport cross-section 3.23Turning point 1.1, 1.3, 1.8Two-body problem 2.11, 2.12,

2.29, 2.30, 2.38, 10.10

van der Waals forces 8.12Virial theorem 4.34Vortex ring in liquid

helium 10.5

Wave vector 7.1, 7.4Waves

standing 7.1travelling 7.1, 7.5, 7.10

Wellspherical 2.2square 13.2

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