Exploring Classical Mechanics. A Collection of 350+ Solved Problems for Students, Lecturers, and Researchers [2 ed.] 2020937520, 9780198853787, 9780198853794

Table of contents :
Title_Pages
Preface_to_the_second_English_edition
From_the_Preface_to_the_first_English_edition
Integration_of_onedimensional_equations_of_motion
Motion_of_a_particle_in_threedimensional_fields
Scattering_in_a_given_field_Collision_between_particles
Lagrangian_equations_of_motion_Conservation_laws
Small_oscillations_of_systems_with_one_degree_of_freedom
Small_oscillations_of_systems_with_several_degrees_of_freedom
Oscillations_of_linear_chains
Nonlinear_oscillations
Rigidbody_motion_Noninertial_coordinate_systems
The_Hamiltonian_equations_of_motion_Poisson_brackets
Canonical_transformations
The_HamiltonJacobi_equation
Adiabatic_invariants
Integration_of_onedimensional_equations_of_motions
Motion_of_a_particle_in_threedimensional_fields (1)
Scattering_in_a_given_field_Collision_between_particles (1)
Lagrangian_equations_of_motion_Conservation_laws (1)
Small_oscillations_of_systems_with_one_degree_of_freedom (1)
Small_oscillations_of_systems_with_several_degrees_of_freedom (1)
Oscillations_of_linear_chains (1)
Nonlinear_oscillations (1)
Rigidbody_motion_Noninertial_coordinate_systems (1)
The_Hamiltonian_equations_of_motion_Poisson_brackets (1)
Canonical_transformations (1)
The_HamiltonJacobi_equation (1)
Adiabatic_invariants (1)
Bibliography
Index

Citation preview

OUP CORRECTED PROOF – FINAL, 8/7/2020, SPi

EXPLORING CLASSICAL MECHANICS

OUP CORRECTED PROOF – FINAL, 8/7/2020, SPi

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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3

Great Clarendon Street, Oxford, OX2 6DP, United Kingdom Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries © Gleb L. Kotkin and Valeriy G. Serbo 2020 The moral rights of the authors have been asserted First Edition published in 2020 Impression: 1 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above You must not circulate this work in any other form and you must impose this same condition on any acquirer Published in the United States of America by Oxford University Press 198 Madison Avenue, New York, NY 10016, United States of America British Library Cataloguing in Publication Data Data available 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 Printed and bound by CPI Group (UK) Ltd, Croydon, CR0 4YY

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Preface to the second English edition

This book was written by the working physicists for students of physics faculties of universities. The first English edition of this book under the title Collection of Problems in Classical Mechanics was published by Pergamon Press in 1971 with the invaluable help by the translation editor D. ter Haar. This second English publication is based on the fourth Russian edition of 2010 and includes new problems from among those used in teaching at the physics faculty of Novosibirsk State University as well as the problems added in the publications in Spanish and French. As a result, this book contains 357 problems instead 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 previous editions, 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; and d – 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.

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From the Preface to the first English edition

This collection is meant for physics students. Its contents correspond roughly to the mechanics 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 students studying mechanics, but also to people who already know it. We follow the order in which the material is presented by Landau and Lifshitz, except that we start using the Lagrangian equations in § 4. The problems in §§ 1–3 can be solved using the Newtonian equations of motion together with the energy, linear momentum and angular momentum conservation laws. As a rule, the solution of a problem is not finished with obtaining the required formulae. 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 conditions of the problem are varied. We have, therefore, suggested further problems at the end of several solutions. A large portion of the problems were chosen for the practical classes with students from the physics faculty of the Novosibirsk State University for a course on theoretical mechanics given by Yu. I. Kulakov. We want especially to emphasize his role in the choice 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 grateful to 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 edition of our book.

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§1 Integration of one-dimensional equations 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) = − U20 (Fig. 1b); cosh αx c) U(x) = U0 tan2 αx (Fig. 1c).

(a)

(b)

(c)

U

U U x

x

x

Figure 1

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

U E

1.3. Give an approximate description of the motion of a particle 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.

a

x

Figure 2

1.4. Determine how the period of a particle moving in the field in Fig. 3 tends to infinity as 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

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4

[1.5

Exploring Classical Mechanics U

U

E Um

Um E

a

x

Figure 3

x1 b

a

c

x2

x

Figure 4

1.5. a) Estimate the period of the particle motion in the field U(x) (Fig. 4), when its energy 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 to x + dx. c) Determine during which part of the period the particle has a momentum mx˙ in the interval from p to p + dp. d) In the plane x, p = mx˙ represent qualitatively lines E(x, p) = const for the cases E < Um , E = U m , E > U m . 1.6. A particle of mass m moves along a circle of radius l in a vertical plane under the influence of the field of gravity (mathematical pendulum). Describe its motion for the case when its kinetic energy E in the lowest point is equal to 2mgl. Estimate the period of revolution of the pendulum for the case when E − 2mgl2mgl. 1.7. Describe the motion of a mathematical pendulum for an arbitrary value of the energy. Hint: The time dependence of the angle the pendulum makes with the vertical can be expressed 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 does not 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 the turning point. 1.9. Find the change in the motion of a particle caused by a small change δU(x) in the field U(x) in the following cases: a) U(x) = 12 mω2 x2 , δU(x) = 13 mαx3 ; b) U(x) = 12 mω2 x2 , δU(x) = 14 mβx4 . 1.10. Determine the change in the period of a finite orbit of a particle caused by the change in the field U(x) by a small amount δU(x).

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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 adding to the field U(x) a small term δU(x) in the following cases: a) U(x) = 12 mω2 x2 (a harmonic oscillator), δU(x) = 14 mβx4 ; b) U(x) = 12 mω2 x2 , δU(x) = 13 mαx3 ; c) U(x) = A(e−2αx − 2e−αx ), δU(x) = −Veαx (V  A). U0 with the energy E > U0 . Find the cosh2 αx particle delay time at the motion from x = −∞ to x = +∞ in comparison with the free motion time with the same energy. 1.12. The particle moves in the field U(x) =

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§2 Motion of a particle in three-dimensional fields γ 2.1. Describe qualitatively the motion of a particle in the field U(r) = − αr − 3 for r 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  −V , U(r) = 0,

when r < R, when r > R

(Fig. 5, “spherical rectangular potential well”) for different values of the angular momentum and of the energy. 2.3. Determine the trajectory of a particle in the field U(r) = α + β . Give an expression for the change in the direction of r r2 velocity when the particle is scattered as a function of angular momentum and energy.

U R

r

−V

Figure 5

β 2.4. Determine the trajectory of a particle in the field U(r) = αr − 2 . Find the time it r takes the particle to fall to the centre of the field from a distance r. How many revolutions around the centre will the particle then make? β 2.5. Determine the trajectory of a particle in the field U(r) = − αr + 2 . Find the r angle ϕ between the direction of radius vector at two successive passages through the pericentre (i.e., when r = rmin ); also find the period of the radial oscillations, Tr . Under what conditions will the orbit be a closed one? β 2.6. Determine the trajectory of a particle in the field U(r) = − αr − 2 . A field of this r kind arises in the motion of a relativistic particle in the Coulomb field in the special theory of relativity; see [7], § 42.1 for details.

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

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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 in the field U(r) for the following cases: −r

a) U(r) = − αer

; b) U(r) = −Ve−

2 r2

.

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 time to 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 number of 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 field U(r) = −α/r. The initial velocity of the particle is zero. Treat the orbit as a degenerate ellipse. 2.11. One particle of mass m moves along the x-axis from a long distance with velocity v towards the origin O of the coordinate system. Another particle of the same mass moves towards the origin O along the y-axis from a long distance with the same velocity magnitude. If the particles didn’t interact, the second would pass through point O in time τ after the first one. However, they repulse from each other, and potential energy of interaction is U(r) = α/r, where r is the distance between particles. Find the minimum distance between the particles. 2.12. Two particles with masses m1 and m2 move with velocities v1 and v2 from long −→ distances 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 time τ earlier than particle 2 would pass the point B. However, there is the force of attraction between the particles, which is given by the potential energy U(r) = −β/r 2 . 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 from infinity with an impact parameter ρ and an initial velocity v and the other one initially at rest. The masses of the particles are m1 and m2 , and the interaction law is U(r) = α/r n . 2.14. Determine in the centre of a mass system the finite orbits of two particles of masses 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 particle flying along the axis is turned back. 2.16. Find the inaccessible region of space for a beam of particles flying along the z-axis with a velocity v and being scattered by a field U(r) = α/r.

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8

Exploring Classical Mechanics

[2.17

2.17. Find the inaccessible region of space for particles flying with a velocity v from a point 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. A body, whose mass is negligible in comparison with the mass of the spacecraft, is thrown from the spacecraft with relative velosity v, directed to the centre of the Earth. Find the orbit 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 oscillations of a particle moving in the central field U(r) when this field is changed by a small amount δU(r). 2.21. Show that the orbit of a particle in the field α U(r) = − e−r/D r 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/r 1+ε , when |ε|  1. 2.23. Find the angular velocity of the orbit precession of a particle in the field U(r) = 12 mω2 r 2 +

β r4

for β  mω2 a6 , mω2 b6 , where a and b are parameters of the unperturbed trajectory:  r cos ϕ 2 a

+

 r sin ϕ 2 b

= 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 minimum distances of a particle from the z-axis are a and b, where a  l.

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2.30]

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

9

2.25. Study the motion of the Earth–Moon system in the field of the Sun. Assume that the mass of the Moon is 81 times less than the mass of the Earth, and the distance from the Earth to the Moon (r = 380 thousand km) is a lot less than the average distances to the Sun (R = 150 million km). a) For simplicity, taking that the plane of the Moon orbit coincides with the plane of the Earth orbit, show that the potential energy of the Earth–Moon system in the field of the Sun, averaged over a month, has the form U(R) = −

β α − 3, R R

where R is the distance from the Sun to the centre of mass of the Earth–Moon system. 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 the Earth’s orbit. Determine the related average velocity of precession for the Moon’s orbital plane. 2.26. Determine the angular velocity of the orbit precession of a particle in the field U(r) = − αr + δU(r) if the orbit eccentricity e is much less than 1, assuming δU(r) = δU(a) + (r − a)δU  (a) + 12 (r − a)2 δU  (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 field U(r) = − αr + 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) = γ γ − αr + 3 , assuming 3 to be a small correction to the Coulomb field. r r 2.29. Show that the problem of the motion of two charged particles in a uniform electric field E can be reduced to the problem of the motion of the centre of mass and that of the motion of a particle in a given field. 2.30. Under what conditions can the problem of the motion of two charged particles in a constant uniform magnetic field B be separated into the problem of the centre of mass motion and the relative motion problem? Take the vector potential of the magnetic field in the form A(ri ) = 12 [B, ri ] , i = 1, 2.

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10

Exploring Classical Mechanics

[2.31

2.31. Express the kinetic energy, the linear momentum, and the angular momentum of a system of N particles in terms of the Jacobi coordinates: m1 r1 + . . . + mn rn − rn+1 m1 + . . . + mn m1 r1 + . . . + mN rN = . m1 + . . . + mN

ξn = ξN

(n = 1 . . . N − 1),

2.32. A particle with a velocity v at infinity collides with another particle of the same mass m which is at rest. Their interaction potential energy is U(r) = α/r n and the collision is a central one. Find the point where the first particle comes to rest. 2.33. Prove that MB +

e [r, B]2 , 2c

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 magnetic field B(r) = gr/r 3 (the field of the magnetic monopole). Such form has a field of a thin long solenoid outside its end at distances which are large 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 charged particle moving in the field of a magnetic dipole μ, in the plane perpendicular to the vector μ. Take the vector potential of magnetic dipole in the form A(r) = [μ, r]/r 3 . 2.36. a) Give a qualitative description of the motion of a charged particle in the field U(r) = 12 mλr 2 , where r is a distance from the z-axis, for the case where there is a constant uniform 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 field U(r) = α/r 2 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 perpendicular to a constant uniform magnetic field B. Find the orbit of the particle. Study the case 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 uniform magnetic field B for the case when their orbits lie in the same plane which is perpendicular to B and where we may consider their interaction energy U(r) = e2 /r to be a small perturbation. 2.39. Show that the quantity F [v, M] −

α Fr + 12 [F, r]2 r

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.

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2.45]

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

11

2.40. Study the effect of a small extra term δU(r) = −Fr, where F = const, added to the Coulomb field on the finite orbit of a particle. a) Find the average rate of change of the angular momentum, averaged over one period. b) Find the time-dependence of the angular momentum, the size, and the orientation of 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 moving in the field U(r) = −α/r under influence of weak constant uniform electric E and magnetic B fields. a) Consider the limiting case when the magnetic field is perpendicular to the orbit plane 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 field U(r) = −α/r under the influence of a small perturbation δU(r, θ) = −βr 2 (3 cos2 θ − 1). Only consider the case when the orbit plane passes through the z-axis. This problem is a simplified model of the satellite motion in the Earth field taking into account the gravitational 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 plane of the Earth orbit, find the systematic change of the Moon orbit under the influence of the perturbation δU(r, χ) = − 12 m2 r 2 (3 cos2 χ − 1), where m is the Moon mass,  is the angular velocity of the Earth around the Sun, χ is the angle between Earth to Sun and Earth to Moon directions. 2.44. Find the systematic displacement of the finite orbit of a charged particle moving in the field U(r) = −α/r and in the field of the magnetic dipole μ, if the effect of the latter may be considered to be a small perturbation. Take the vector potential in the form A(r) = [μ, r]/r 3 . 2.45. Find the average precession rate of the orbit of a particle moving in the field U(r) = −α/r under the influence of a small additional “friction force” F = β v¨ (such 2 2 q , where q is the charge of form has the force of radiation damping; in this case, β = 3 3 c the particle and c is the velocity of light; see [2], § 75).

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§3 Scattering in a given field. Collision between particles

3.1. Find the differential cross-section for the scattering of particles with initial velocity parallel to the z-axes by smooth elastic surfaces of revolution ρ(z) for the following cases: z , 0  z  π a; a) ρ = b sin a n b) ρ = Az , 0 < n < 1; 2

c) ρ = b − az ,

a2  z < ∞. b

3.2. Find the surface of revolution which is such that the differential cross-section for elastic 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, 0,

when r < a, when r > a.

3.4. Find the cross-section for the process where a particle falls towards the centre of the field U(r) when U(r) is given by: γ β β a) U(r) = αr − 2 , b) U(r) = 2 − 4 . r r r 3.5. Calculate the cross-section for particles to hit a small sphere of radius R placed at the centre of the field U(r) for the cases: γ β a) U(r) = − αn , n  2; b) U(r) = 2 − 4 . r r r 3.6. A uniform beam of meteors with velocity v∞ flies towards the planet. What fraction of meteorites that fell on the planet falls on the part invisible from the beam side? Take a planet as a uniform ball of radius R and mass m0 .

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

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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): α α when r < R, r − R, 0, when r > R; 1 2 2 mω (r − R2 ), when r < R, b) U(r) = 2 0, when r > R.

a) U(r) =

3.8. Find the differential cross-section for the scattering of fast particles (E  V ) by the field U(r) for the following cases:   V 1 − (r 2 /a2 ), when r < a, a) U(r) = 0, when r > a;  V ln (r/a), when r < a, U(r) = b) 0, when r > a;  a2  U(r) = V ln 1 + 2 ; r   2  V 1− r2 , when r < R, U(r) = R 0, when r > R.

c)

d)

3.9. Find the differential cross-section for small angle scattering in the field β U(r) = 4 − α2 . r r 3.10. Find the differential cross-section for the scattering of particles by the field U(r) = −α/r 2 . 3.11. Find the differential cross-section for the scattering of fast particles (E  V ) by the following fields U(r): a) U(r) = Ve−

2 r2

; b) U(r) =

V . 1 +  2r2

Study in detail the limiting cases when the deflection angle is close to its minimum or maximum value. 3.12. A beam of particles with their velocities initially parallel to the z-axis is scattered by the fixed ellipsoid x2 a2

+

y2 b2

+

z2 c2

= 1.

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

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14

[3.13

Exploring Classical Mechanics

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 following fields U(r) (a is a constant vector): a) U(r) = ar ; b) U(r) = ar . r2 r3 3.14. Find the change in the differential cross-section for scattering of particles by the field U(r) when U(r) is varied by a small amount δU(r) for the following cases: β a) U(r) = αr , δU(r) = 2 ; r γ b) U(r) = αr , δU(r) = 3 ; r γ β c) U(r) = 2 , δU(r) = 3 . r r 3.15. Find the differential cross-section as function of the energy acquired by fast particles (E  V1, 2 ) due to their scattering in the field U(r, t) = (V1 + V2 sin ωt) e−

2 r2

.

3.16. A particle with velocity V decays into two identical particles. Find the distribution of the secondary particles over the angle of divergence, that is, the angle between the directions at which two secondary particles fly off. The decay is isotropic in the centre of 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 the angular distribution in the centre of mass system is dN 3 = sin2 θ0 d0 , N 8π where θ0 is the angle between the velocity V of the original particle and the direction in which one of secondary particle flies off in the centre of mass system. The velocity of the 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 after a moving particle of mass m1 has collided with a particle of mass m2 at rest.

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3.23]

§3. Scattering in a given field

15

3.20. Find the differential cross-section for the scattering of inelastic smooth spheres by similar ones at rest. 3.21. Find the change in intensity of a beam of particles travelling through a volume filled 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 dV when 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.

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§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 the point x = a in a time τ . Find the law of motion of particle, assuming it to be of the form x(t) = At 2 + Bt + C, and determine the parameters A, B, and C such that the action is a minimum. 4.2. A particle moves in the xy-plane in the field  U(x, y) =

0, V,

when x < 0, when x > 0,

and travels in a time τ from the point (−a, 0) to the point (a, a). Find its position as a function of time, assuming that it satisfies the equations x1, 2 (t) = A1, 2 t + B1, 2 , y1, 2 (t) = C1, 2 t + 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 motion under 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 Lagrangian equations 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 )?

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

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4.10]

§4. Lagrangian equations of motion

17

4.5. Write down the Lagrangian function and the equations of motion for a particle moving in a field U(x), introducing a “local time” τ = t − λx. 4.6. How does the Lagrangian function  L = − 1−

 dx 2 dt

transform when we change to the coordinates q and “time” τ through the equations x = q ch λ + τ sh λ t = q sh λ + τ ch λ 4.7. How do the energy and generalized momenta change under the following coordinate transformation qi = fi (Q1 , . . . , Qs , t)

i = 1, . . . , s?

4.8. How do the energy and generalized momenta which are conjugate to (a) the spherical polar and (b) the Cartesian coordinates transform under a change to a coordinate 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 reference which is moving with a constant velocity V? Take the Lagrangian function L in the moving frame of reference in either of two forms: a) L1 = L(r + Vt, r˙  + V, t), where L(r, r˙ , t) is the Lagrangian function in the original frame of reference; b) L2 = 12 mv2 − U(r + Vt, t). Here L2 differs from L1 by the total derivative with respect to the time of the function Vmr + 12 m V2 t. 4.10. Smooth rod OA of length l rotates around a point O in a horizontal plane with a constant angular velocity  (Fig. 6). The bead is fixed on the rod at a distance a from the point O. The bead is released, and after a while, it is slipping off the rod. Find its velocity at the moment when the bead is slipping?

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18

[4.11

Exploring Classical Mechanics y y´ m A

b

Ω

a

O

Figure 6

ϕ

Ωt

x´ x

Figure 7

4.11. At the end of the rod of length a, rotating with constant angular velocity  in the plane of the figure, is pivotally attached another freely rotating rod of length b, at the end of which there is a particle of mass m (Fig. 7). Find the Lagrangian function L(ϕ, ϕ, ˙ t) of a system and the frequency of small oscillations of the second rod. 4.12. Consider an infinitesimal transformation of the coordinates and the time of the form qi = 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 of the order of ε, inclusive), 

t2  t2  dq  dq  L q, , t dt = L q ,  , t  dt  . dt dt t1

t1

(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  ∂L i

∂ q˙i

(q˙i h − fi ) − Lh

is an integral of motion. 4.13. Generalize the theorem of the preceding problem to the case when under the transformation of the coordinates and the time the action changes in the following way: 

t2  t2  dq  dq  dF(q , t  )  L q, , t dt = L q ,  , t  + ε dt , dt dt dt  t1

t1

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

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4.20]

§4. Lagrangian equations of motion

19

4.14. Find the integrals of motion if the action remains invariant under a) b) c) d) e)

a translation; a rotation; a shift in the origin of the time; a screw shift; 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) = α n U(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 homogeneous function; e) an electromagnetic field rotating with a constant angular velocity  around the z-axis. 4.16. Find the time of the fall of the particle in the centre of the field U(r) = −(ar)2 /r 4 . 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 magnetic field 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]/r 3 , 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:

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20

[4.21

Exploring Classical Mechanics

a) L(x, x) ˙ = e−x

2 −x˙2

+ 2xe ˙ −x

2

x˙

e−y dy; 2

0

b) L(x, x, ˙ t) = 12 eαt (x˙2 − ω2 x2 ). 4.21. a) Write down the components of the acceleration vector for a particle in the spherical coordinate system. b) Find the components of the acceleration in the orthogonal coordinate qi , if the line element is given by the equation ds2 = h21 dq21 + h22 dq22 + h23 dq23 , where hi (q1 , q2 , q3 ) are the Lam´e coefficients. 4.22. Write down the equations of motion of a point particle using arbitrary coordinates 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=

m 2 eg (˙r + r 2 θ˙ 2 + r 2 ϕ˙ 2 sin2 θ) − ϕ˙ cos θ, 2 c

where r, θ, and ϕ are the spherical coordinates, describes the motion of a charged particle in the magnetic field B(r) = gr/r 3 (see problem 2.34). Find the integrals of motion. 4.24. Verify that one can use the Lagrangian functions L q˙21 L1 = − Uq1 , 2

q2 L2 = − 2 + Uq2 , 2C

(a) I

(b) L

U

B

B

A

A

U

C

q2

Figure 8

to get the correct “equations of motion” for q1 and q2 and the correct energies. Here q˙1 = I is the current flowing through the inductance L in the direction from A to B (Fig. 8a), q2 the charge on 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 the preceding problem to find the Lagrangian functions and the Lagrangian equations of motion for the circuits of Fig. 9a–9c.

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4.28]

21

§4. Lagrangian equations of motion (a) L

(c) L1

(b) U

L

C

C

C1

L2 C

C2

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 a pendulum 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 capacitor plate may be ignored; b) a core suspended from a spring with elastic constant k inside a solenoid with inductance L (x) which is a given function of the displacement x of a core (Fig. 10b).

(a)

(b) k

C(ϕ)

x

L

B B

C

ϕ

L(x)

a

C

A

m

D

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 angles to the AB axis. The inductance of the frame is L and the mass of the side CD is m; the masses 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 the equations 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.

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22

[4.29

Exploring Classical Mechanics

4.29. A particle moves in the field of gravity along a straight line which is rotating uniformly in a vertical plane. Write down the equations of motion and determine the moment of the forces of constraint. 4.30. One can describe the influence of constraints and friction on the motion of a system by introducing generalized constraint and friction forces Ri into the equations of motion: d ∂L ∂L − = Ri . dt ∂ q˙i ∂qi 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 motion invariant under a transformation to new generalized coordinates qi = qi (Q1 , . . . , Qs , t)? 4.31. Let the constraint equations be q˙β =

s 

bβn q˙n ,

β = 1, . . . , r,

n=r+1

while the Lagrangian function L(qr+1 , . . . , qs , q˙1 , . . . , q˙s , t) and the coefficients bβn do not depend on coordinates qβ . Show that the equations of motion can be written in the form r s   d ∂ L +  ∂L  ∂bβm − ∂bβn q˙ = 0, L − ∂ ∂ q˙β ∂qn ∂qm m dt ∂ q˙n ∂qn β=1

m=r+1

where  L(qr+1 , . . . , qs , q˙r+1 , . . . , q˙s , t) is the function obtained from L by using the constraint equations to eliminate the velocity q˙1 , . . . , q˙r . 4.32. A continuous string can be thought of as the limiting case of a system of N particles (Fig. 12) which are connected by an elastic thread, in the limit as N → ∞, a → 0, Na = const. The Lagrangian function for a discrete system is L(q1 , q2 , . . . , qN , q˙1 , q˙2 , . . . , q˙N , t) =

N+1  n=1

Ln (qn , qn − qn−1 , q˙n , t),

q qn

a 2a

na

Figure 12

x

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4.35]

§4. Lagrangian equations of motion

23

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 the Lagrangian equations of motion of a discrete system. b) Obtain an expression for the energy of a continuous system as the limiting case of the expression for the energy of a discrete system. Hint: Introduce the coordinate x of a point on the string together with the expression obtained as a result of taking the limit as a → 0, n = x/a → ∞: qn (t) − qn−1 (t) ∂q = lim , ∂x a  Ln (qn , qn − qn−1 , q˙n , t) ∂q ∂q  L x, q, , , t = lim . ∂x ∂t a q(x, t) = lim qn (t),

4.33. A charged particle moves in the potential field U(r) and a constant magnetic field B(r), where U(r) and B(r) are homogeneous functions of the coordinates of degrees k and n, respectively, that is, U(αr) = α k U(r), B(αr) = α n B(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 uniform constant magnetic field B. The potential energy U of the system is a homogeneous function of the coordinates, U(αr1 , . . . , αrs ) = α k U(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 according to the law Uik = U(xi − xk ), where xi is the coordinate of ith particle. Prove that besides the obvious integrals of motion P = m(x˙1 + x˙2 + x˙3 ), E = 12 m(x˙21 + x˙22 + x˙23 ) + U12 + U23 + U31 there is an additional integral of motion [23] A = mx˙1 x˙2 x˙3 − x˙1 U23 − x˙2 U31 − x˙3 U21 ,

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24

Exploring Classical Mechanics

[4.36

when the function U(x) has the following form: g2 , x2 g 2 a2 . b) U(x) = sinh2 ax

a) U(x) =

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 = x˙i (t = −∞) be such that v3 > v2 > v1 . Find vi = x˙i (t = +∞).

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§5 Small oscillations of systems with one degree of freedom

5.1. Find frequency ω of the small oscillations for particles moving in the following fields U(x): a) U(x) = V cos αx − Fx; b) U(x) = V (α 2 x2 − sin2 αx). 5.2. Find the frequency of the small oscillations for the system depicted in Fig. 13. The system 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 and in 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 first point charge. 5.4. Describe the motion along a curve close to a circle for a point particle in the central field U(r) = −α/r n (0 < n < 2). 5.5. Find the frequency of the small oscillations of a spherical pendulum (a particle of mass m suspended a string of length l) if the angle of deflection from the vertical, θ oscillates about the value θ0 . Ω

a

θ

A a

m

m R a

ϕ

a m

Figure 13

q m

q

Figure 14

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

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26

[5.6

Exploring Classical Mechanics

5.6. Find the correction to the frequency of the small oscillations of a diatomic molecule due to its angular momentum M. 5.7. Determine the free oscillations of the system shown in Fig. 15 for the case when the particle moves: A

k

m

k

B

2l

a) along the straight line AB; b) at a right angle to AB.

Figure 15

How does the frequency depend on the tension of the springs in the equilibrium position? 5.8. Find the free oscillations of the system (Fig. 16) in a uniform field of gravity for the case when the particle can only move vertically. 5.9. Find the stable small oscillations of a pendulum when its point of suspension moves uniformly along a circle of radius a with an angular velocity  (Fig. 17). The pendulum length is l (l  a). Ω a k 2l

m

l

R

k

L m

Figure 16

C U

Figure 17

Figure 18

5.10. Find the stable oscillations in the voltage across a capacitor and the current in a circuit with an e.m.f. U(t) = U0 sin ωt (Fig. 18). 5.11. Determine the law of motion of an oscillator with friction which initially is at rest and which is acted upon by a force F(t) = F cos γ t. 5.12. Determine the energy E acquired by an oscillator under the action of a force 2 F(t) = Fe−(t/τ ) 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¨ − μ2 x =

1 F(t); m

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5.19]

§5. Small oscillations of systems with one degree of freedom

27

b) of an oscillator with friction x¨ + 2λx˙ + ω02 x =

1 F(t). m

5.14. Find the differential cross-section for an isotropic oscillator to be exited to the energy ε by a fast particle (E  V ) if the interaction between the two particles is through the field U(r) = Ve−

2 r2

.

The energy of the oscillator is zero initially. 5.15. An oscillator can oscillate only along the z-axis. Find the differential cross-section for the oscillator to be excited to an energy ε by a fast particle E  V , if the interaction 2 2 between the particles is through the potential energy U(r) = Ve− r . The particle moves along 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 harmonic oscillator. Find the energy E(+∞) gained by the oscillator during the total time the force acts, 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) =

1 2 1 2

F0 eλt , F0

when t < 0,

(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 force F(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 discontinuity at t = 0 and at t = τ . F

F

τ

2τ

Figure 19

t

τ

2τ

t

Figure 20

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

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28

Exploring Classical Mechanics

[5.20

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 given by 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 stable mode for the case when F(t) = f1 cos ωt + f2 cos 2ωt. b) Repeat the calculation for the case when ∞ 

F(t) =

an einωt , a−n = a∗n .

n=−∞

c) Find the average over a long time interval of the work done by the force F(t) = f1 cos ω1 t + f2 cos ω2 t 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 with the force F(x, t) = f (t − x/V ), where x is the displacement of the oscillator from the equilibrium position and V is the wave velocity. Assuming x is small enough, find the connection 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 assume f (±∞)=0.

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§6 Small oscillations of systems with several degrees of freedom

In problems 6.1–6.22, we study free and forced oscillations of simple systems (with two or three degrees of freedom) using common methods. In problems 6.19– 6.22, systems with the degenerate frequencies are presented. For more complex systems, it is helpful to use orthogonality of eigen-oscillations and the symmetry properties of the system. The corresponding theorems are given in problems 6.23 and 6.40 (see also [7], § 24 and § 25), whereas their application are illustrated, for instance, in problems 6.27–6.43 and 6.41–6.47, 6.49 (see also problems related 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 the perturbation theory. The general form of the perturbation theory for small oscillations is given in problem 6.35, whereas some specific examples can be found in problems 6.36, 6.38, 6.42, 6.43, 6.52 b. It is useful to note that the perturbation theory in quantum mechanics 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 particles move only vertically. Find the normal coordinates and express the Lagrangian function in terms of these coordinates. 6.2. Find the stable oscillations of the system described in the preceding problem, if the point of suspension moves vertically according to equation a(t), where a) a(t) = a cos γ t,   b) a(t) = a τt − 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

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

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30

[6.5

Exploring Classical Mechanics

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

2l

k

3m m

A m l

k m

Figure 21

l

m B

y

m

Figure 22

Figure 23

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

1 2

  x˙2 + y˙2 − ω12 x2 − ω22 y2 .

What is the trajectory of a point with Cartesian coordinate (x, y)? 6.6. Find the normal coordinates of the systems with the following Lagrangian functions: a) L = b) L =

1 2 1 2



 x˙2 + y˙2 − ω12 x2 − ω22 y2 + αxy;   m1 x˙2 + m2 y˙2 − x2 − y2 + β x˙y. ˙ (a) L1

L2

(b) L1 C2

C1

C1

L2 L

C2

Figure 24

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

B k1 m1 k2 m2 k3

Figure 25

A

B k

m k1 m

k

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 oscillations of the system.

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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 only along the straight line AB a) if at t = 0 one of the particles moves with velocity v while the second particle is at rest and the displacements from the equilibrium position of both particles are zero; b) if at t = 0 one of the particles is displaced from its equilibrium position over a distance a, while the other is at rest at its equilibrium position and both particles are at rest. 6.10. Determine the flux of energy flow from one particle to the other in the preceding problem. 6.11. Find the free oscillations of the system of Fig. 26 if each of the particles is acted upon 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 the initial moment the upper pendulum is vertical, while the bottom pendulum is displaced by an angle β  1, and their velocities are zero. The pendulum masses are M and m, respectively (M  m). (a)

(b)

A

k

k

A

k

k

1 m

3 m

l M l m

Figure 27

m

m k k

2 m

k

Figure 28

6.13. Find the stable oscillations of the system of Fig. 26 if the point A moves according to the relation a cos γ 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 relation a cos γ t. Study how the amplitude of the oscillations depends on the frequency of the applied 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 ring according to the law a cos γ t. 1 In this and similar problems, the ring is assumed to be smooth and fixed.

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32

[6.16

Exploring Classical Mechanics

6.16. Find the stable oscillations of the system of Fig. 26 when point A moves according to the relation a cos γ t. The particles are acted upon by the frictional forces proportional to their velocities. 6.17. Find the motion of the system of Fig. 25 when in the initial moment the particle are at rest in the equilibrium positions, and point A moves according to the relation a cos γ 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 field U(r, t) = −F(t)r, where the vector F(t) lies in the plane of the figure, for the following cases: a) F(t) = F0 cos γ t, b) the vector F(t) rotates with constant absolute amplitude magnitude with a frequency γ . 6.19. Find the eigen-oscillations of three identical particles connected by identical springs and moving along the ring (Fig. 30). C k A

k1

k

m 1

k

m B k1

m 2

3 m

k D

k

Figure 29

Figure 30

m

m 1

2k

3k

2m

3m

k m 2 k

k 4 m

6k

3 m

Figure 31

Figure 32

k

Determine the normal coordinates which reduce the Lagrangian function to a diagonal form. 6.20. Find the free oscillations of the system considered in the preceding problem, if at t = 0 one of the particles is displaced from its equilibrium position. The initial velocities of all particles are zero.

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6.24]

§6. Small oscillation of systems with several degrees of freedom

33

6.21. Find the eigen-oscillations of the system of three particles which can move along the 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=

1 (mij x˙i x˙j − kij xi xj ). 2 i, j

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

(l)

xi (t) = Ai cos(ωl t + ϕl ). Prove that the amplitudes corresponding to the oscillations with different frequencies ωl and ωs satisfy the relation 

(s)

(l)

Ai mij Aj =



i, j

(s)

(l)

Ai kij Aj = 0.

i, j

6.24. Consider a system described by the Lagrangian function L=

1 (mij x˙i x˙j − kij xi xj ) 2 i, j

which also has the different eigen-frequencies 1 < 2 < < . . . < N . The linear relationship 

ai xi = 0,

ai = const

i

is imposed upon the system. Prove that all eigen-frequencies ωl of the system with such relation lie between l : 1  ω1  2  ω2  . . .  ωN−1  N . 1 Since the product x˙ x˙ is symmetric with respect to the replacement of i ↔ j, then the matrix m can always i j ij be chosen as the symmetric one, mij = mji . The same holds for the matrix kij = kji .

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[6.25

Exploring Classical Mechanics

6.25. We can write the stable oscillations of the system described by the Lagrangian function L=

 1 (mij x˙i x˙j − kij xi xj ) + xi fi cos γ t, 2 i, j

i

 (l) in the form xi (t) = l λ(l) Ai cos γ 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 fi A(s) i = 0 for the sth normal oscillation. 6.26. A system of particles connected by springs can perform small oscillations. One of them (the particle A) be acted upon by the force F(t) = F0 cos γ t along the xdirection, while the other (the particle B) performs the stable oscillations, during which the projection of its displacement on the x -direction has the form xB = C cos γ t. Show that when force F acts upon the particle B along the x -axis, there will be the oscillations xA = C cos γ 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 those of particles 2 and 4 are M. 6.28. Find the eigen-oscillations of the system of four particles of Fig. 34; the particles move along the ring.

k

m 1

M 2 k

3 m

Figure 33

2m k

k

4 M

m 1

k

k

4

k 3 m

2 m

k

Figure 34

6.29. a) Find the normal oscillations of the system of Fig. 35. All particles and springs are identical. The tension in the springs at equilibrium is f = kl, where l is the equilibrium distance between the particles. Hint: Several of the eigen-vibrations are obvious. The determination of the others can be simplified by using the relations of problem 6.23. b) Find the eigen-oscillations of the system of four identical particles of Fig. 32 for the case where the mass of particle 5 is equal to zero and the springs in this place are

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6.34]

§6. Small oscillation of systems with several degrees of freedom

35

connected. The elasticity coefficients and tensions at equilibrium of all the springs are the 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 occur in this system? m1 C k m 5 3

2 A

1 4

k3

k2

m2

m3

B

D

k1

Figure 35

Figure 36

6.31. Which of the eigen-vibrations of the system of Fig. 30 remain practically unchanged 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 case a) and b) if initially the particles 1 and 3 are displaced over equal distances in opposite directions 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-vibrations are not uniquely determined. Even a small change in the mass of particles or a small change 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 as the 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 = 0 displaced by the same amount from their equilibrium positions in opposite direction so

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Exploring Classical Mechanics

[6.35

as to decrease their distance apart. Initially all velocities are equal zero. Describe the free oscillations of the system. 6.35. Consider a system described by the Lagrangian function L0 =

1 (mij x˙i x˙j − kij xi xj ). 2 i, j

Now consider the new system which has been obtained by the small change in the original Lagrangian function: δL =

1 (δmij x˙i x˙j − δkij xi xj ). 2 i, j

Find the small change of the eigen-frequencies of the new system if all eigen-frequencies of the original system are non-degenerate. 6.36. Find the small changes in the eigen-frequencies of the system of Fig. 34 when a small 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 the potential field U(r) = 12 m(ω12 x2 + ω22 y2 + ω32 z2 ) and in a uniform constant magnetic field B which is parallel to the z-axis. Consider in particular 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 the potential field U(r) = 12 m(ω12 x2 + ω22 y2 + ω32 z2 ) and in a weak constant uniform magnetic field B = (Bx , 0, Bz ), considering the effect of the magnetic field to be a small perturbation.

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6.42]

37

§6. Small oscillation of systems with several degrees of freedom

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

L C ⊕B

l m

6.40. Suppose that a system, making small oscillations (and hence its Lagrangian function L(x, x) ˙ as well), does not change its form with respect to the replacement xi →



Sij xj ;

x˙i →

j



Sij x˙j ,

Figure 37

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

j

Here the constant coefficients Sij = Sji satisfy the condition1 

Sij Sjk = δik .

j

Prove that a) if the eigen-oscillation xi = Ai cos(ωt + ϕ) is not degenerate, then the amplitude Ai is either symmetric or anti-symmetric with respect to the given transformation,   that is, either Sij Aj = +Ai or sij Aj = −Aj ; j

j

b) if the frequency is degenerate, one can choose the eigen-oscillation to be either symmetric or anti-symmetric; c) if the system is acted upon by an external force which is symmetric (or antisymmetric) with respect to this transformation, then the anti-symmetric (or the symmetric) eigen-oscillations is not excited. (This is one example of the so-called selection rules.) 6.41. Using symmetry considerations, find the eigen-oscillations of the system of Fig. 38.

k m

6.42. Determine the free oscillations of the system of Fig. 35 when at t = 0 particles 1 and 4 are displaced over equal distances in the horizontal direction towards each other in such a way that the distance between them decreases. At t = 0 the velocities of all 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).

m

k

2 3

1

m k

k 4

5

M

M k

Figure 38

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.

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38

[6.43

Exploring Classical Mechanics

6.43. Find corrections to the frequencies of the eigen-oscillations of the systems of four particles on the ring (Fig. 32). Such corrections arise from the small changes in the particles’ 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 for the system of Fig. 39. All the masses of the particles and the springs are identical. B k m A

1

2

4

3

l C l B m l 3 m D

A

B

Figure 39

3l

3l

A m

m E

Figure 40

6.45. Find the eigen-oscillations of the “scales” of Fig. 40. The suspension of a rigid frame BCD is implemented with a short flexible thread allowing any √ turns of the frame around 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. The masses of the rods and threads can be neglected. 6.46. Find the eigen-oscillations of the system of eight particles which are attached to the fixed frame by springs (Fig. 41). The elasticity coefficients k, the tensions f and the lengths l of springs are identical. 6.47. The frame shown in Fig. 41 oscillates along the direction AA according to the equation a cos γ t. At what values of the frequency γ will the resonant amplification of the oscillations be possible? B m

M

M

m

A

A m

m M

M

B

Figure 41

H

C

C

Figure 42

H

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6.54]

§6. Small oscillation of systems with several degrees of freedom

39

6.48. Find the eigen-oscillations of a linear symmetric molecule of acetylene C2 H2 (Fig. 42), assuming that the potential energy of the molecule depends on the distance between neighbouring atoms as well as on angles HCC. 6.49. Two identical particles are attached to a fixed frame by springs (Fig. 43). The system is symmetric with respect to the CF-axis. What information about the eigenoscillations of the system can be obtained if the elasticities and the tensions of the springs are unknown? 6.50. Classify the eigen-oscillations of a molecule of ethylene C2 H4 by their symmetry properties with respect to axes AB and CD (Fig. 44). In the equilibrium position all atoms of the molecule are located in the same plane. C

B A

F

Figure 43

1

D

C

E

3

H

2

H

H

A O C

C B

4 D

5

H

6

Figure 44

6.51. Determine the eigen-oscillations in a plane of a molecule which has the shape of an equilateral triangle. Assume that the potential energy depends only on the distances between atoms and that all atoms are the same. The angular momentum is equal to zero up 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 centre and atoms B in its vertices (the molecule of boron chloride BCl3 is an example of such a model). a) Using symmetry considerations, determine the degree of degeneracy of the eigenfrequencies of the molecule. b) Consider the oscillations which leave the molecule as an equilateral triangle, and the oscillations which output atoms of the plane. Determine how the frequencies of such oscillations will change when one of the atoms B (its mass m) is replaced by 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 various frequencies for the case of a “molecule” consisting of four identical “atoms” which has the form of a regular tetrahedron at equilibrium. 6.54. A molecule of methane CH4 has the form of a regular tetrahedron with fourth hydrogen atoms in vertices and one carbon atom in its centre.

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Exploring Classical Mechanics

[6.54

a) Determine the degree of degeneracy of the eigen-frequencies of the molecule. b) How many eigen-frequencies will experience resonance amplification when this molecule is under the influence of a uniform alternating electric field? (Such are, for instance, the electromagnetic waves of the infrared range whose wavelength is several orders of magnitude greater than the size of the molecule.) Note that the hydrogen and carbon atoms have charges of opposite signs. How does the amplitude of the oscillations of the carbon atom depend on the orientation of the molecule with respect to the electric field?

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§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 radio engineering.

7.1. Determine the eigen-vibrations of a system of N identical particles with masses m connected 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 suspended from each other (Fig. 47). The masses of all pendulums are the same; the lengths are equal 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 which can move along the ring (Fig. 48). All particles and the elastic constants of the springs are the same. Let the motion be that of a wave travelling along the ring. Check whether the 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 line for the following cases (consider the hint for problem 7.1):

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

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[7.6

Exploring Classical Mechanics

Nl m (N–1)l A

m

m

2l m m l

Figure 47

Figure 48

a) 2N particles, alternating with masses m and M connected by springs of elastic constant k (Fig. 49); b) 2N particles with masses m connected by springs with alternating elastic constants k and K (Fig. 50); m

M

m

M

A

k

K

Figure 49

k

K

K

k

Figure 50

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

K

k

k

K

Figure 51

7.6. a) Find the stable oscillations of the system described in problem 7.1 if the point A moves according to a cos γ t (Fig. 45). b) Let the left point, at which the spring of the same system is fixed, moves according to x0 = a cos γ t. Under what law should move the point A to make the oscillation represent 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 system of Fig. 49.

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7.10]

43

§7. Oscillations of linear chains

7.8. Determine the eigen-vibrations of a system of N particles which move along a straight line for the following cases: a) mi = m = mN , i = 1, 2, . . . , N − 1; the elastic constants of all the springs are the same (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 the cases when kN+1  k and kN+1  k. m

m mN

m

k

k

Figure 52

k

k

k

kN+1

Figure 53

7.9. a) N pendulums are connected by springs and move only in the vertical plane passing through the horizontal suspension line (Fig. 54). Find the eigen-vibrations of the system, if all the pendulums and the springs are the same and, in the position of equilibrium, the length of the spring is equal to the distance between the suspension points of neighbouring pendulums. b) Find the forced oscillations of the system of Fig. 54 for the case when the last particle is acted upon by the driving force F(t) = F sin γ t which is directed parallel to the suspension line. c) 2N identical pendulums are connected by identical springs and move only in vertical planes which are perpendicular to the circular line of suspension (Fig. 55). The distance between neighbouring suspension points is a. The length of each of the springs in the unstretched state is b. Study how the stability of small oscillations near the vertical line depends on the parameter b − a. Consider the radius of the circular suspension line R to be large enough so that small quantities l/R, a/R, b/R can be neglected. R

l

l k m

m

Figure 54

Figure 55

7.10. a) An AC voltage source U cos γ 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 of a travelling wave (i.e. in this case the voltage on each of the capacitors will differ from that of the neighbour’s only by a definite phase shift)? b) Do the same for the artificial line of Fig. 57.

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44

[7.11

Exploring Classical Mechanics L U

L

L

C

L C Z

C

Figure 56

L1 U

L2

L1

C

C

L2 C Z

Figure 57

7.11. Consider the elastic rod to be the limiting case of the system of N particles of Fig. 45 in the limit N → ∞, a → 0, where m and a are, respectively, the mass of particles and the distance between neighbouring particles at equilibrium, while Nm and Na kept constant. Write down the equations of motion for the oscillations of the rod as the limiting case of the equations of motion of the discrete system. Hint: Introduce the coordinate of a point of the rod at equilibrium ξ = na and consider the following quantities x(ξ , t) = lim xn (t),

∂x = lim xn (t) − xn−1 (t) ∂ξ a

with the limit a → 0. 7.12. Write down the equations of motion for the oscillations of the rod in the preceding problem taking into account the first non-vanishing correction due to a finite distance a between the neighbouring particles.

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§8 Non-linear oscillations

8.1. Determine the distortion in the oscillations of a harmonic oscillator which is caused by the presence of anharmonic terms in the potential energy for the following cases: a) δU(x) = 14 mβx4 ; b) δU(x) = 13 mαx3 . 8.2. Determine the distortion in the oscillations of a harmonic oscillator which is caused by the presence of anharmonic term δT = 12 mγ xx˙2 in the kinetic energy. 8.3. Determine the anharmonic corrections to the oscillations of a pendulum whose point 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 cos ω1 t + f2 cos ω2 t acting upon it, taking anharmonic corrections into account for the case when δU(x) = 1 3 3 mαx . 8.5. A pendulum consists of a particle of mass m in the field of gravity g suspended on a spring with the elasticity constant k (Fig. 58). The length of the spring in the free state is l0 . Find the anharmonic corrections to the oscillations of the pendulum. Use the Cartesian coordinates of the displacement of the particle from the equilibrium position.

k l

m y x

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

Figure 58

x¨ + 2λx˙ + ω02 x + βx3 = f cos ωt 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).

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

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[8.7

8.7. a) Determine the amplitude and phase of the stable vibration of an oscillation under conditions of parametric resonance: x¨ + 2λx˙ + ω02 (1 + h cos 2ωt)x + βx3 = 0 provided |ω − ω0 |  ω0 ,

h  1,

βx2  ω02 .

b) Determine the amplitude of the third harmonic in the stable vibration. 8.8. Determine the vibrations of the oscillator x¨ + ω02 (1 + h cos 2ωt)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. ω ω1 ω2 τ 2τ 3τ 4τ

t

Figure 59

8.10. Determine how amplitudes will change over time for the weakly coupled oscillators whose Lagrangian function is L = 12 m(x˙2 + y˙2 − ω2 x2 − 4ω2 y2 + 2αx2 y). 8.11. Find the frequency of the small free oscillations of a pendulum  whose point of suspension 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) = at a distance r  a;

α α − |r − a cos ωt| |r + a cos ωt|

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47

§8. Non-linear oscillations

b) a harmonic oscillator moving in the field U(r) = 2α

ar cos ωt. r3

8.13. Determine the motion of a fast particle entering the field U(r) = a(x2 − y2 ) sin kz at a small angle to the z-axis (k2 E  a). 8.14. Determine velocity of the orbit centre’s displacement for a charged particle in the weakly inhomogeneous magnetic field Bx = By = 0,

Bz = B(x),

ε=

B (x) r  1, B(x)

r=

mvc , eB(x)

where r is the orbit radius. 8.15. The following problem represents a mechanical model of phase transitions of the second kind. The iron ball of mass m can oscillate along the y-axis. It is connected to the spring whose potential energy has the form1 U(y) = −Cy2 + By4 . Using an electromagnet one can excite vibrations of the ball according to the equation y(t) = y0 cos γ t, where γ is considerably 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 the parameter T = y20 .

1 Such is, for example, the potential energy of the system shown in Fig. 15 where a ball can move only in the direction of the y-axis perpendicular to the AB line, and the length of the unstretched springs l0 is larger then l. Therein, if |y|  l, we have C = k(l0 − l)/l, B = kl0 /(4l 3 ). 2 Under the action of high-frequency force f (t) = −(y γ 2 /m) cos γ t, the amplitude of the forced vibration 0 tends to y0 in limit γ → ∞, and the corrections to the amplitude and the higher harmonics are small ∼ 1/γ 2 .

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§9 Rigid-body motion. Non-inertial coordinate systems

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

R ϕ A ψ

R O

9.2. Masses m and M are located at the vertices of a square with side lengths 2a (Fig. 61a). Find the components of the inertia tensor a) relative to the x-, y-, and z-axes; Figure 60 b) relative to the the x and y -axes which are the diagonals of the square and the z-axis. (a)

m y a M y O M

(b)

y

ax

(c) y m

m x 4a

m

M

2b M O' 2a

m x

m

2a

 2m x

Figure 61

9.3. Find the principal axes and the principal moments of inertia for the following systems: 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 2a and 4a (Fig. 61c).

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

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9.10]

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§9. Rigid-body motion

9.4. Give an expression for the moment of inertia In with respect to an axis parallel to a unit vector n and passing through the centre of mass of the body in terms of the components of the inertia tensor. x2

A

R x1

Figure 62

r

Figure 63

9.5. The cloakroom number tag is made of uniform thin plate. The tag is shaped as a disc of radius R with a hole in the shape of circle of radius R/2. The centre of this hole is at a distance R/2 from the tag’s edge (Fig. 62). Find the frequency of small oscillations of 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 of which 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 = (3xi xk − r 2 δik )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 on on 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 tidal forces 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 when it becomes equal to the month? Consider for simplicity that the Earth’s axis of rotation is 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 is 81 times greater than the mass of the Moon m; the distance from the Earth to the Moon equals R = 380 thousand km. 9.10. Two identical uniform balls, rotating with equal angular velocities ω, slowly approach closer and then rigidly connect with each other. Determine the motion of the newly formed body. Find which part of the initial kinetic energy has been converted into heat. Prior to connecting, the angular velocities of the balls were directed:

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[9.11

Exploring Classical Mechanics C D

B

A C

v D

B

A

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 horizontal plane at velocity v. At the moment when it touches another motionless ball, both balls become 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 with the plane. 9.12. A uniform rectangular parallelepiped has two spherical hinges attached to its opposite 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 impact parameters ρ1 and ρ2 is incident upon a uniform ellipsoid of rotation (with semi-axes a = b and c) (Fig. 66) and sticks to it. Describe the motion of the ellipsoid assuming its mass 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 a horizontal plane (Fig. 67). Study the motion of the gyrocompass at latitude α. The angular velocity of the Earth’s rotation equals .

z v O

ρ1

y ρ2

x

Figure 66

Figure 67

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Ω

θ O

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 angular velocity  (assume velocity of precession to be small). Find the angular velocity of the top when slipping of the disc vanishes. There were no nutations at the time of the touchdown. 9.16. An isotropic ellipsoid of revolution of mass m moves in the gravitational field produced by a fixed point of mass M. Use the spherical coordinates of the centre of mass and the Euler angles as generalized coordinates and determine the Lagrangian function of the system. Assume the size of the ellipsoid to be small compared to the distance from the centre of the field. Hint: The potential energy of the system is approximately equal to U(R) = mϕ(R) +

3 1  ∂ 2 ϕ(R) Dik , 6 ∂Xi ∂Xk i, k=1

where R = (X1 , X2 , X3 ) is the radius vector of the centre of the ellipsoid, Dik is the mass quadrupole tensor (see problem 9.7), and ϕ(R) = −GM/R is the potential of the gravitational field (cf. [2], § 42). 9.17. Determine the angular velocity of precession for the Earth’s axis caused by the attraction forces of the Sun and the Moon. For simplicity, consider the Earth as a uniform c ellipsoid of rotation whose equatorial axis a is greater than its polar axis c, so that a − a ≈ 1 300 . Assume the Earth’s and the Moon’s orbits to be the circumferences lying in the same plane. The tilt of the Earth’s axis to the plane of the Earth and Moon’s orbits equals 67◦ . 9.18. Write down the equations of motion for the components of the angular momentum along moving coordinate axes. These axes coincide with the principal axes of the inertia tensor. Integrate these equations for the case of the free motion of a symmetric top. 9.19. Use the Euler equations to study the stability of rotations around the principal axes of 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 of a vertical cylinder of radius b without slipping. Find the motion of the ball.

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9.21. a) A plane disc, symmetric around its axis, rolls in the field of gravity over a smooth horizontal 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 to an inclined plane the disc is placed upon. The plane makes a small angle α with the horizontal 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 ball which is slipping without friction on a horizontal plane in the field of gravity. The mass distribution of the ball is symmetrical with respect to the axis passing through the geometric 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 slipping over a horizontal plane. A

9.23. A particle is dropped from a height of h with zero initial velocity. Find its displacement from the vertical in the direction of the East and the South. For simplicity, consider the Earth as a uniform ball of radius R. 9.24. A vessel partially filled with gradually solidifying epoxy resin, is rotating in the field of gravity with an angular velocity ω2 around the AB-axis, which in turn rotates around the fixed CD-axis with an angular velocity ω1 (Fig. 69). What form will it take when the surface of the resin completely solidifies?2

ω2 C

D ω1

B

Figure 69

9.25. A particle moves in a central field U(r). Find the equation for its trajectory and describe its motion in a coordinate system which rotates uniformly with an angular velocity  parallel to its angular momentum M. 9.26. Find the small oscillations of the particle with mass m which is fastened to a frame by springs with elastic constants k1 and k2 . The frame rotates in its own plane with an angular 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 of contact neither slips nor rotates. The energy loss due to rolling friction can be neglected. 2 Problem by V. S. Kuzmin and M. P. Perelroizen.

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9.29]

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§9. Rigid-body motion Ω z k1

k2

γ

k1

x k2

Figure 70

Figure 71

9.27. A smooth paraboloid z=

x2 y2 + 2a 2b

rotates in the field of gravity around the vertical z-axis with an angular velocity ω. At what value 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, the elasticity constants k and the tension f (when the frame is at rest). The frame rotates with an angular velocity γ around the z-axis, which is shifted by a distance a from the centre of the frame (Fig. 71). Determine the equilibrium distance of the particle from the axis of rotation and investigate 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. Consider the frame of reference where the stars are motionless. Find the points at which a light particle in this system is also motionless. Investigate the stability of these “equilibrium positions”. (Ignore the points lying on a straight line connecting the stars.)

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§10 The Hamiltonian equations of motion. Poisson brackets

10.1. Let the Hamiltonian function H of a system of particles be invariant under an infinitesimal translation (rotation). Prove the linear (angular) momentum conservation law. 10.2. Use the Euler angles θ, ϕ, and ψ to find the Hamiltonian function of a symmetric top when there is no external field acting on the top. 10.3. Determine the Hamiltonian function of an anharmonic oscillator, if the Lagrangian function is given by the equation L=

1 2

  x˙2 − ω2 x2 − αx3 + βxx˙2 .

10.4. Describe the motion of a particle with the Hamiltonian function H(x, p) = 12 (p2 + ω02 x2 ) + 14 λ(p2 + ω02 x2 )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 c|p| form H(r, p) = (a beam of light). n(r) Find the trajectory if n(r) = ax. 10.7. Find the Lagrangian function for the case when the Hamiltonian function is c|p| p2 . a) H(r, p) = 2m − pa (a = const); b) H(r, p) = n(r) 10.8. Describe the motion of a charged particle in a uniform constant magnetic field B by solving the Hamiltonian equations of motion; take the vector potential in the form Ax = Az = 0,

Ay = xB.

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

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10.9. Study qualitatively the motion of a charged particle in a non-uniform magnetic field described by the vector potential A(r) = (0, hx2 , 0). Compare the result with the drift 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 one particle in a given potential field [24]. In problems 10.11 to 10.14 we are dealing with the motion of electrons in a metal or semiconductor. Electrons in a solid form a system of particles which interact both with themselves and with the ions which form the crystal lattice. Their motion is described by quantum mechanics. In solid state theory one is often able to reduce the problem of many interacting particles which form the solid to the problem of the motion of separate free particles (the so-called quasi-particles: electrons or holes, depending on the sign of their 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) =

  1  −4p2 ± 1.2p4 + 17(p2x p2y + p2x p2z + p2y p2z ) , 2m

where the coordinate axes are chosen to coincide with the crystal symmetry and m is the electron mass. In many cases it turns out that it is possible to consider the motion of the quasiparticles using classical mechanics (e.g. [16]). The function ε(p) is periodic with a period equal to the period of the so-called reciprocal lattice.1 Otherwise one can assume that ε(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 equal to that of the reciprocal lattice, multiplied by 2π h¯ ; for instance, for a simple cubic lattice with 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 − ce A 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  e  H(P, r) = ε P − A + eϕ c 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   h¯ ε(px , py , pz ) = ε px + 2π a , py , pz , where h¯ is the Planck constant.

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10.13. a) Determine the integrals of motion of an electron moving in a solid in uniform constant 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 field onto a plane at right angles to B in coordinate space can be obtained by rotation and change of scale of the orbit in momentum space. 10.14. Express the period of revolution of an electron in a uniform constant magnetic field 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, r n }, {p, (ar)2 }, where xi , pi , and Mi are the Cartesian components of the vectors, while a and b are the constant vectors. 10.16. Evaluate the Poisson brackets {Ai , Aj }, where A1 = 14 (x2 + p2x − y2 − p2y ), A3 = 12 (xpy − ypx ),

A2 = 12 (xy + px py ),

A4 = x2 + y2 + p2x + p2y .

10.17. Evaluate the Poisson brackets {Mi , jk }, { jk , il }, where ik = xi xk + pi pk . 10.18. Show that the Poisson bracket {Mz , ϕ} = 0, where ϕ is an arbitrary scalar function of 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 along the z-axis and f is the vector function of the coordinates and momenta of a particle, that is, f = r ϕ1 + p ϕ2 + [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 vector functions of r and p while a is the constant vector. 10.20. Evaluate the Poisson bracket {Mζ , Mξ }, where Mζ and Mξ are the components of the angular moment along the Cartesian ζ - and ξ -axes which are fixed in a rotating rigid body. 10.21. Write down the equations of motion for the components Mα of the angular momentum along the axes fixed in a freely rotating rigid body. The Hamiltonian function is of the form  H = 12 (I −1 )αβ Mα Mβ . α, β

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

M2 − γ MB, 2I

where I is the moment of inertia of the ball and γ is a gyromagnetic ratio. Write down the equations of motion for the vector M and find the law of its motion in the following cases: a) B = (0, 0, B0 ); b) B = (B1 cos ωt, B1 sin ωt, 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 vi are 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 momenta of 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 +

t t2 {H, f } + {H, {H, f }} + . . . , 1! 2!

where f = f (q(0), p(0)) and H = H(q(0), p(0)) is the Hamiltonian function. (Assume that the 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 , . . .)} = ∂ {f , ϕ}; ∂ϕ b) { f (q1 , p1 , q2 , p2 ), (ϕ(q1 , p1 , q2 , p2 ), q3 , p3 , . . .)} = ∂ {f , ϕ}; ∂ϕ  ∂ {f , ϕi }. c) { f (q, p), (ϕ1 (q, p), ϕ2 (q, p) . . .)} = i ∂ϕi 10.26. a) The Hamiltonian function depends on the variables q1 and p1 only through the 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.

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b) Find the integrals of motion for a particle in the field U(r) = ar (use the spherical r3 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) r A = [v, M] − α . r 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

 1 m J1, 2 = M± A . 2 −2E Compare these Poisson brackets with the Poisson brackets for the components of the angular momentum M. Express the Hamiltonian function H in terms of J1 and J2 .

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§11 Canonical transformations

11.1. Determine the canonical transformation defined by the following generating functions: a) F(q, Q, t) = 12 mω(t)q2 cot Q. Write down the equations of motion for a harmonic oscillator with frequency ω(t) in terms of the variables Q and P.  2 b) F(q, Q, t) = 12 mω q − F(t)2 cot Q. mω Write down the equations of motion for a harmonic oscillator which is acted upon by an external force F(t) in terms of the variables Q and P. 11.2. Determine the generating function (p, Q) which produces the same canonical transformation as the generating function F(q, P) = q2 eP . 11.3. What is the condition that a function (q, P) can be used as a generating function for a canonical transformation? Consider in particular the example (q, P) = q2 + P 2 . 11.4. Prove that a rotation in (q, p) space is a canonical transformation for a system with one degree of freedom. 11.5. Consider the small oscillations of an anharmonic oscillator with the Hamiltonian function   H = 12 p2 + ω2 x2 + αx3 + βxp2 under the assumption that |αx|  ω2 , |βx|  1.

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

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Find the parameters a and b for the canonical transformation produced by generating function  = xP + ax2 P + bP 3 such that the new Hamiltonian function does not contain any anharmonic terms up to first-order terms in αω−2 Q and βQ. Determine x(t). 11.6. Determine the parameters a and b for the canonical transformation produced by the generating function  = xP + ax3 P + bxP 3 , in such a way that that small oscillation of an anharmonic oscillator described by the Hamiltonian function   H = 12 p2 + ω02 x2 + βx4 can be reduced to harmonic oscillations in terms of the new variables Q and P. Neglect terms of second order in βω−2 Q2 in the new Hamiltonian function. 11.7. Small oscillations of the coupled oscillators are described by the Hamiltonian function H=

 1  2 p1 + (mω1 x)2 + p22 + (mω2 y)2 − mαx2 y. 2m

Neglecting the terms of the second degree in α, find the parameters a, b, and c for the canonical transformation produced by the generating function 2 (x, y, PX , PY ) = xPX + yPY + axyPX + bPX PY + cx2 PY

such that in the new variables the system is reduced to two independent oscillators. Find x(t) and y(t) with the first-order corrections in α. 11.8. Prove that the following transformation is canonical: PY sin λ, mω py = −mωX sin λ + PY cos λ, x = X cos λ +

PX sin λ, mω px = −mωY sin λ + PX cos λ. y = Y cos λ +

Determine the new Hamiltonian function, H  (P, Q), if the old Hamiltonian function is H(p, q) =

p2x + p2y 2m

+ 12 mω2 (x2 + y2 )

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

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11.9. Use the transformation of the preceding problem to reduce the Hamiltonian function of an isotropic harmonic oscillator in a magnetic field described by the vector potential 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 an anisotropic charged harmonic oscillator with potential energy U(r) = 12 m(ω12 x2 + ω22 y2 + ω32 z2 ), which is situated in a uniform constant magnetic field determined by the vector potential A(r) = (0, xB, 0). 11.11. Apply the canonical transformation of problem 11.8 to pairs of normal coordinates corresponding to standing waves in the system of particles on the ring considered in problem 7.4 to obtain the coordinates corresponding to the travelling waves. 11.12. Prove that the following transformation is a canonical one: 1  x= √ ( 2P1 sin Q1 + P2 ), mω  √ px = 12 mω( 2P1 cos Q1 − Q2 ), 1  y= √ ( 2P1 cos Q1 + Q2 ), mω  √ py = 12 mω(− 2P1 sin Q1 + P2 ) Find the Hamiltonian equations  of motions for  a particle in a magnetic field described 1 1 by the vector potential A(r) = − 2 yB, 2 xB, 0 in terms of the new variables introduced eB . through the previously described transformation with ω = mc 11.13. What is the meaning of the canonical transformation produced by the generating function (q, P) = αqP? 11.14. Prove that a gauge transformation of the potentials of the electromagnetic field is a canonical transformation for coordinates and momenta of charged particles, and find the 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 Lagrangian equations of motion invariant. Prove that this transformation is a canonical one and find its generating function.

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11.16. Find the generating function for the canonical transformation which consist in changing 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 the following generating functions: a) b) c) d)

(r, P) = rP + δaP; (r, P) = rP + δϕ[r, P]; (q, P, t) = qP + δτ H(q, P, t); (r, P) = rP + δα(r2 + P2 ),

where r is the Cartesian radius vector while δa, δϕ, δτ , and δα are infinitesimal parameters. 11.18. Prove that the canonical transformation produced by the generating function (x, y, PX , PY ) = xPX + yPY + ε(xy + PX PY ), where ε → 0 is a rotation in phase space. 11.19. Write down the generating function for infinitesimal canonical transformations corresponding 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).

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11.21. For the case when the Hamiltonian function is H(r, p) =

p2 ar + 3 2m r

(a = const),

determine the Poisson bracket {H, rp} and use the result to obtain an integral of the equations of motion. Here, it is convenient to use the results of the preceding problem and problem 11.13. 11.22. Determine how the r and p dependence of M, p2 , pr, and H(r, p, t) change under 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 are produced by the generating functions i (q, P) = qP + λi Wi (q, P),

λi → 0,

i = 1, 2,

will be independent of the order in which they are taken, up to and including secondorder terms. 11.24. Determine the canonical transformation which is the result of N successive infinitesimal transformations produced by the generating function λ W (q, P), λ = const, N → ∞; N a) W (r, P) = [r, P] a, a = const; b) W (x, y, PX , PY ) = Ai , (q, P) = qP +

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, and the volume in phase-space which are occupied by a group of particles which move freely along the x-axis? At t = 0 the particle coordinates are lying in the interval x0 < 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 with one another. c) Do the same for a group of harmonic oscillators.

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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 the distribution function w(x, p, t) which is such that w(x, p, t) dx dp is the number of particles with coordinates in the interval from x to x + dx and momenta in the range from p up to p + dp. Determine the distribution functions of a group of free particles and of a group of harmonic oscillators, if at t = 0 w(x, p, 0) =

 (x − X )2 (p − P )2  0 0 1 . exp − − 2 2π p0 x0 2(x0 ) 2(p0 )2

11.26. For the variable mωx + ip iωt a= √ e , 2mω

a) find the Poisson bracket {a∗ , a}. Express the Hamiltonian function of a harmonic oscillator H0 =

p2 + 1 mω2 x2 2m 2

in terms of variables a and a∗ . b) Prove that Q = a and P = ia∗ are the canonical variables. Find a new Hamiltonian function H0 (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 oscillations 2π/ω. Use the averaged Hamiltonian function to determine the slow changes in variables Q and P. d) Study the variation of the amplitude of oscillations for a harmonic oscillator under the action of the nonlinear resonant force H=

p2 + 1 mω2 x2 + m2 ω2 αx4 cos 4ωt. 2m 2

11.27. Study the variation of the amplitude of oscillations for the system of three harmonic oscillators with the weak nonlinear coupling H = 1 (p2x + p2y + p2z ) + 12 m(ω12 x2 + ω22 y2 + ω32 z2 + αxyz), 2m

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11.30]

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when |ω1 − ω2 − ω3 |  ω1 , |αx|  ω12 . 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 parametric excitation has the form H=

p2 + 1 mω2 (1 + h cos 2γ t)x2 + 14 mβx4 . 2m 2

Introduce the canonical variables mωx + ip iγ t a= √ e , 2mω

P = ia∗ .

a) Determine the new Hamiltonian function H  (a, P, t) and average it over the period of 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 1 P sin γ t, x = Q cos γ t + mω p = −mωQ sin γ t + P cos γ t is canonical. Determine the new Hamiltonian function H  (Q, P, t) for an oscillator with the parametric excitation: H=

p2 + 1 mω2 x2 (1 − h cos 2γ t). 2m 2

b) Average H  (Q, P, t) by the period 2π/γ and describe qualitatively the motion of a point on the phase plane Q P. Take h  1, ε = 1 − γ /ω  1. 11.30. Consider the motion of two weakly coupled oscillators H0 = 1 (p2x + p2y ) + 12 m(ω12 x2 + ω22 y2 ), 2m V = mβxy sin(ω1 − ω2 )t, β  ω12 ∼ ω22 .

H = H0 + V ,

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

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for the variables y, py /(mω2 ). Prove that the variables X, Y , and PX , PY are the canonical variables. Determine the new Hamiltonian function H  (X, Y , PX , PY , t) and average it over the time tav such that ω1, 2 1 ω1, 2  tav  β , For the period t  ω1, 2 /β, study the change over time of the amplitudes of oscillations for the variables x and y. b) Do the same for V = mβxy sin(ω1 + ω2 )t.

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§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 mω12 x2 + mω22 y2 . 12.2. Describe the motion of a particle which is scattered in the field U(r) = ar/r 3 . Express the equation of the trajectory in terms of quadratures; express it analytically for the case when Eρ 2  a, where ρ is the impact parameter. Before the scattering the velocity of the particle is parallel to the vector (−a). 12.3. Find the cross-section for the small-angle scattering of particles with velocities before the scattering anti-parallel to the z-axis for the cases where the scattering is caused by the following fields U(r): θ; a) U(r) = a cos 2 r

2 b) U(r) = b cos2 θ ;

c) U(r) =

r b(θ ) r2

.

12.4. Find the cross-section for a particle to fall into the centre of one of the following fields U(r): a) U(r) = ar ; r3

b) U(r) = ar + λr ; r3 γ b(θ ) c) U(r) = ar − ; d) U(r) = 2 . r3 r4 r Assuming that all directions of a are equally probable, evaluate the averages of the cross-section obtained.

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

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12.5. Find the cross-section for particles to hit a sphere of radius R placed at the centre of the field U(r) = ar/r 3 . 12.6. Describe the motion of particles which are scattered by and fall towards the centre of the field U(r) for the following cases: θ; a) U(r) = a cos 2

b) U(r) = −

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 Eρ 2  a. For the first field, find also an analytic expression for the trajectory of a particle falling into the centre at Eρ 2  a. 12.7. Describe the motion of a particle falling towards the centre of the field U(r) = ar/r 3 for the case when at infinity the particle moved along the straight line y = ρ, x = −z tan α, where ρ is the impact parameter. The vector a is along the z-axis and the 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 α 2
0) and let the oscillator originally oscillate near along the x-axis. What happens to its motion when the value of ω1 decreases slowly to reach a value ω1 < ω2 such that ωB  ω2 − ω1 ? 13.27. A particle performs a finite motion in a plane at right angles to a magnetic dipole μ. What is the change in the energy of the particle when the magnitude μ changes slowly? 13.28. Determine the period of the oscillations along the z-axis of an electron in a magnetic 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 rBz (z),

Consider the following cases:   a) Bz (z) = B0 1 + λ tanh2 az ;  2 b) Bz (z) = B0 1 + z2 . a

13.29. How does the energy and the period of oscillations of an electron moving in the magnetic trap of the preceding problem change when the field parameters B0 , λ, and a change 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 when the motion becomes degenerate. Find the integrals of motion when a particle moves in 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, x F, when x > 0.

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[13.33

13.33. Use the generating function x  2m|E − U(x)| dx, S(x, P) = 0

to perform a canonical transformation for the case of a particle moving in the periodic field   ⎧ ⎨0, when na < x < n + 1 a, 2   U(x) = n = 0, ±1, ±2, . . . , ⎩V , when n + 1 a < x < (n + 1)a, 2

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

a  2m|E − U(x)| dx. 0

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§1 Integration of one-dimensional equations 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 E

x1

U 2 1 2 mx˙ + U(x) = E.

x2

x

x3

E –A

Figure 79

(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 determined from (1) through U(xi ) = E:

√ A(A + E) − A , E x1 = − lnα2 , √ A ∓ A(A − |E|) 1 , x2, 3 = α ln |E| 1 ln x1 = α

when

E > 0,

when

E = 0,

when

E < 0.

(2)

From (1) we obtain  t=

m 2

x x(0)

dx . √ E − U(x)

(3)

whence 1 ln A − x(t) = α

√ √ A(A − |E|) cos(αt 2|E|/m + C) , |E|

when E < 0,

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

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  2 Aα 1 1 2 x(t) = α ln + m (t + C) , 2 √ √ 1 ln A(A + E) cosh(αt 2E/m + C) − A , x(t) = α E

when

E = 0,

(5)

when

E > 0.

(6)

The C constants are defined by the initial values of x(0). For example, in (4), when x(0) ˙ >0 A − |E|eαx(0) C = arccos √ . A − (A − |E|) Turning points (2) are also easily found from (4) to (6).

 π When E < 0 the motion is periodic according to (4) with a period T = α 2m |E| . When E is close to the minimum value of U(x), Umin = U(0) = −A (that is, when A − |E|  1), the period ε= A   ε π 2m T ≈ T0 1 − , T0 = 2 α A is nearly independent of E. In this case we can write (4) in the form √     √ 1 1 2π 2π ε x(t)=− ln(1−ε)+ ln 1− ε cos t+C ≈− cos t +C . α α T α T0

(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 second derivative 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.



− |E| + U0 sin(αt 2|E|/m + C) , |E| ⎤ ⎡ E + U 0 1 x(t) = ± α arsinh ⎣ sinh(αt 2E/m + C)⎦ , E

1 b) x(t) = α arsinh

when

E < 0,

when

E > 0,

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1.3]

81

§1. Integration of one-dimensional equations of motions

1 arsinh(αt 2U /m + C), x(t) = ± α 0 1 c) x(t) = arcsin α



when

E = 0; 1

  E sin αt 2(U0 + E)/m + C . E + U0

Explain why in some of the formulae two signs occur? 1.2. x(t) =

x0 , x0 = x(0). 1 ± tx0 2A/m

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 to the 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 motion of 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 +

Ft 2 . 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. If U  (a) = 0 (see Fig. 80), one must extend the expansion of U(x) to the next term: U U(x) = E + 12 U  (a)(x − a)2 .

ε

In this case x(t) = a + se±λt with s = x(0) − a, λ2 = −

U  (a) , m

a

x

Figure 80

where m is the particle mass, and where the sign in the index is determined by the direction of the velocity at t = 0. It takes an infinite time to reach the turning point. 1 arsinh x = ln(x +

x2 + 1).

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[1.4

1.4. If U  (a) = 0, then T ∝ ln ε, 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 (back and forth) passage of the small interval δ in the neighbourhood of this point a − δ < x < a + δ:  √ dx T1 = 2m ≈ T. √ E − U(x) a+δ

a−δ

In the neighbourhood of x = a, we present U(x) in the form U(x) = Um − 12 k(x − a)2 , where k = −U  (a). If ε is small enough, you can choose δ such that velocity v on the boundary of the interval is a lot more then the minimum one (at x = a), 2 1 2 mv

∼ 12 kδ 2 ε,

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

ε  δ  L = x2 − x1 . k

As a result,  m 2kδ 2 ln . T1 = 2 k ε

(1)

The time T2 spent by a particle along the intervals x1 < x < a − δ and a + δ < x2 satisfies the relation  ml L . T2  ∼ v kδ 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/ ln ε. But with the same logarithmic accuracy we can replace δ by L in (1) and omit the multiplier 2 under the sign of logarithm:

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1.5]

83

§1. Integration of one-dimensional equations of motions

 m kL2 ln . T =2 k ε

(2)

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

6m2 T =4 εK

1/4 ∞ 0



m2 = 11.6 √ εK 1 + x4 dx

and the relative error tends to zero as ε1/4 at ε → 0. b) If we observe the motion of the particle over a time which is much larger compared to the period T , the probability to find the particle between x and x + dx is

1/4

w

x1 x x+dx

√ dt 2m dx w(x) dx = 2 = √ , T T E − U(x)

a

x2

x

Figure 81

where 2 dt 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 the curve is unity). For sufficiently small ε the area under the central maximum gives the main 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)     dtk   dp 1   dp = 1  ,  w(p) dp =    dU(xk )  T T dp  k k   dx 

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

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[1.6

Exploring Classical Mechanics ~ w

–p3 –p2

–p1

p1

p2

p3 p

Figure 82

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

p 4 3 2 1

3 2

x

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

±t

ϕ(t) = −π + 4 arctan e

√

g/l

ϕ(0) + π tan 4



Figure 83

,

where ϕ is the angle between the pendulum and the vertical. The sign in the index is the same 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 upper position. One can estimate the period of rotation applying the result (2) of the preceding problem:  T=

l ε0 ln ; g E − 2mgl

ε0 = 4π 2 mgl.

1.7. Once again, we put the value of the potential energy at the lowest point equal to zero, then the energy is equal to E = 12 ml 2 ϕ˙ 2 + mgl(1 − cos ϕ). Let at the moment t0 the angle ϕ(t0 ) = 0 and for definiteness ϕ(t ˙ 0 ) > 0. By introducing k = E/2mgl, we have 1 t= 2

 ϕ l dϕ  + t0 . g 2 − sin2 ϕ k 2 0

(1)

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1.7]

§1. Integration of one-dimensional equations of motions

85

At k < 1 the pendulum oscillates within the interval −ϕm  ϕ  ϕm and k = ϕ ϕ sin 2m . The substitution sin ξ = 1 sin 2 is reduced the integral (1) to the form1 k  t = gl F(ξ , k) + t0 . Hence  g ϕ = 2 arcsin[k sn(u, k)], u = (t − t0 ) . l The period of oscillations is  T =4

l  ϕm  K sin . g 2

In the extreme cases (cf. with the problem 1.4)    2 ϕm l T = 2π g 1 + , 16  T = 4 gl ln 8 , π − ϕm

when ϕm  1, when π − ϕm  1.

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

 ϕ 1 l F , + t0 , g 2 k

1 Functions

F(ξ , k) =



1 ϕ = 2Arcsin sn u, , k

 ξ

dα , 0 1 − k2 sin2 α

E(ξ , k) =

 g u = k(t − t0 ) . l

 ξ 1 − k2 sin2 α dα 0

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 is expressed in one of the elliptic Jacobi functions, namely by the elliptic sine: sin ξ = sn(u, k). Here is also the formulae for the two limiting cases:  K(k) = π 2

2 1+ k 4

 ,

when k  1;

K(k) = 1 ln 16 , 2 1 − k2

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

when 1 − k  1.

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[1.8

The rotation period of the pendulum is  2 T= k

 l 1 K . g k

In particular, if E − 2mgl  2mgl we get  l ε0 ln , g E − 2mgl

T=

where ε0 = 32mgl. This result differs from the rather crude estimation made in the preceding problem in the value of ε0 , that is in a quantity which does not depend on E − 2mgl. 1.8. The motion in the field U(x) + δU(x) is governed by the equality  t=

m 2



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 powers of δU(x), we obtain t = t0 (x) + δt(x),

(2)

where  t0 (x) = δt(x) =

1 2

m 2 



x b

m 2



dx , √ E − U(x) x

b

δU(x) dx . [E − U(x)]3/2

(3)

(4)

Let the orbit for the case when δU(x) = 0, which is determined by the equation t = 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) − x0 (t)δt(x0 (t)).

(6)

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§1. Integration of one-dimensional equations of motions

87

Near a turning point x = x1 one has to be careful, as the expansion (2) becomes inapplicable since the correction is δt(x) → ∞ as x → x1 . It is remarkable, however, that formula (6) turns out to be correct up to the turning point if |δU  (x)|  |F|,

F = −U  (x1 ).

(7)

This is because although with the approach to a turning point δt increases, the dependence 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 +

F (t − t1 )2 . 2m

(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 +

F (t − t1 − δt1 )2 2m

(9)

(because of (7), the correction to F is neglected). Make sure that the calculation of the formula (6) results in (9). Divide the integration region in (4) into two parts: from b to c and from c to x, where c lies near x1 . In the second area we can put δU = δU(x1 ) and U(x) = E − (x − x1 )F. Then √ m δU(x1 ) δt = + δt0 , 2F 3 (x − x1 )   √ 1 m c δU(x) dx m δU(x1 ) δt0 = − . 2 2 b (E − U)3/2 F 3 (c − x1 )

(10)

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 unperturbed motion with b = 0 and x1 = a we have x0 (t) = a sin ωt,

E = 12 mω2 a2 .

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[1.10

In this case |δU/U|  ε = αa2  1. For the correction we have ω  α 2 a2 2+ √ a − x − 2a , 3ω3 a2 − x2  ε 1 δt(x0 (t)) = cos ωt + −2 . 3ω cos ωt δt(x) =

Substituting these expressions into (6) of the preceding problem, we get x(t) = a sin ωt − 13 εa (cos2 ωt + 1 − 2 cos ωt). Up to and including first-order terms in ε, we have   x(t) = a sin ωt + 23 ε − 12 εa − 16 εa cos 2ωt (cf. problem 8.1 b). b) Acting in the same way as previously described, we obtain x(t) = a sin ωt + εa



3 2

 βa2 ωt cos ωt − 98 sin ωt − 18 sin 3ωt , ε =  1. 4ω2

(1)

This result has a relative accuracy ∼ ε2 in one period, and after 1/ε periods the formula (1) becomes completely inapplicable. Taking into account the periodic character of the motion, we can extend the result (1) over a larger period of time. With accuracy up to and including first-order terms of ε, the formula (1) is transformed into a clearly periodic form           x(t) = a 1 − 98 ε sin ω 1 + 32 ε t − 18 εa sin 3 ω 1 + 32 ε t .

(2)

The corrections, which were not taken into account in (1), have lead to the frequency change of the order of ε2 ω, 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 ⎥ − E − U(x)dx⎦ .

(1)

x1

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

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1.11]

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§1. Integration of one-dimensional equations of motions

since the resulting integral diverges. However, we can expand the integrand in terms of δU(x) up to terms of fist order in δU(x) if we rewrite δT in the form ⎡ ⎤ x2+δx2 x2 √ ⎢ ⎥ δT = 2 2m ∂ ⎣ E − U(x) − δU(x) dx − E − U(x) dx⎦ , ∂E x1 +δx1

(2)

x1

whence  √ δU(x) dx ∂ = − ∂ (T δU), δT = − 2m √ ∂E ∂E E − U(x) x2

(3)

x1

where 1

δU = T

T δU[x(t)] dt

(4)

0

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 good approximation. We note that sometimes even small extra term δU(x) may strongly affect the particle motion (e.g. see, problem 1.11 b, c). Higher-order terms in the expansion of δT in terms of δU can be obtained by similar means:  ∞ √  (−1)n ∂ n [δU(x)]n dx . T = 2m √ n! ∂E n E − U(x) n=0 x2

(5)

x1

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 period 2π/ω is 3πβE δT = − . 2mω5 This correction is small for sufficiently small E. b) In Fig. 84 we have given the potential energies U(x) and U(x) + δU(x). When E > Um = mω6 /(6α 2 ) the extra term clearly means that the particle can move to infinity. When E is close to Um ,

U U+δU Um

U

E x

Figure 84

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[1.12

the period of oscillation increases without bound (as | ln(Um − E)|; see problem 1.4) so that one cannot determine the change in period just by a few terms in series (5) of the preceding problem. When E  Um the correction to the period is δT =

5π E . 18ω Um

c) √ 3π AV m δT = √ ; 2 α|E|5/2 √ the formula is applicable when |E| |Um | ≈ 8AV , (E < 0). 1.12. The particle delay time is +∞ τ= −∞

where v =



2 m |E − U(x)|, v0 =

1 1 − v v0



dx =

1 E ln , αv0 E − U0

2E (cf. problem 1.1b). m

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§2 Motion of a particle in three-dimensional fields

2.1. To study the orbits of the particle we use the energy and angular momentum conservation laws: rϕ˙ r˙

m˙r2

+ U(r) = E, 2 m[r, r˙ ] = M.

(1) (2)

It follows from (2) that the orbit lies in a plane. Introducing the polar coordinates in that plane (see Fig. 85) we get 1 2

r ϕ O

r˙ x

Figure 85

m˙r 2 + 12 mr 2 ϕ˙ 2 + U(r) = E, mr 2 ϕ˙ = M.

(3) (4)

Using (4) to eliminate ϕ˙ from (3), we find m˙r 2 + Ueff (r) = E, 2

(5)

where Ueff (r) = U(r) +

M2 . 2mr 2

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 γ M2 − 3+ r r 2mr 2

for different values of M (Fig. 86).

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

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(b)

Ueff

(c)

Umax c r2 d a r1 b

r

Ueff r1 r2

Ueff r

r

Umin

Figure 86

When 12αγ m2 < M 4 , then Ueff has two extrema at  M 2 ∓ M 4 − 12αγ m2 . r1, 2 = 2mα The maximum value of Ueff (r1 ) = Umax is positive when M 4 > 16αγ m2 (Fig. 86a) and negatively when 12αγ m2 < M 4 < 16αγ m2 (Fig. 86b); in both these cases we have Ueff (r2 ) = Umin < 0. If M 4 < 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 according to (4). This is all we need to sketch the orbit (see Fig. 87a).

r1 (a)

b O

(b)

O

a

x

Figure 87

Figure 88

γ  αr , the main term in U(r) is the term (−α/r) and r3 the orbit differs little from the hyperbola. (See problem 2.8 on the form of a trajectory at r → 0.) If the energy E is close to Umax the particle pass very slowly through the range of values of r which are close to r1 . At the same time the radius vector turns round with an angular velocity ϕ˙ ≈ M2 so that the particle may perform many revolutions around the mr1 centre before it passes through this range of values (Fig. 87b). At large distances such that

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If E = Umax the particle approaches the point of r = r1 (cf. problem 1.3). The orbit is a spiral approaching the circle of radius r1 and centre O (Fig. 88 curve a). If, on the other hand, the particle starts in the region r < r1 and is moving away from the centre, the orbit 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 along this circle is unstable one, as any change in the values of E or M will bring onto the orbit which moves away from the circle. If 0 < E < Umax a particle coming from infinity will be reflected from the potential barrier 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 revolutions around the centre before the radial velocity r˙ changes sign. The closer the energy is to 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 of the field, when it is moving in the region r < a. The orbit for this case is given in Fig. 90. A

b a

C

D

O O

B

A

c O B

d

C

Figure 89

Figure 90

Figure 91

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 revolutions during one radial oscillation. When the energy close to Umin the orbit close to a circle with radius r2 while the angle over which the radius vector turns during one period of the radial oscillation depends on the quantities α, γ , M (compare problem 5.4). When E = 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 M 4 = 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  dr 2 r˙ = =± [E − Ueff (r)], (7) dt m or



m t=± 2



dr + C. √ E − Ueff (r)

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

(8)

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M ϕ = ±√ 2m



dr + C1 . √ r 2 E − Ueff (r)

[2.1

(9)

Let us consider the case M 4 > 12αγ m2 . If the particle moves towards the centre, we must take the lower sign in (7) and, of course, in (8). If at t = 0, r = r0 , we can rewrite (8) in the form  r m dr . (10) t=− √ 2 E − Ueff (r) r0

Equality (10) is an implicit equation for r as the function of t. If the orbit passes through the point r = r0 , ϕ = ϕ0 , the equation of the orbit becomes r ϕ = −M r0

dr + ϕ0 , r 2 |pr |

(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˙ vanishes and then changes sign. The section BC of the orbit is thus described by (9) with the upper sign, and we must redetermine the constant. It is helpful to write (9) in the form r ϕ=M rmin

dr r 2 |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 the orbit: ⎛ r ⎞  rmin ⎠ M dr + ϕ0 . ϕ=⎝ − (14) r 2 |pr | rmin

r0

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

95

Similarly we can determine the time dependence of r along the section BC ⎛ t=⎝

r

rmin

⎞ rmin  ⎠ m √ dr − . 2 E − Ueff (r)

(15)

r0

If Umin < E < 0, a < r0 < b, r˙(0) < 0, ϕ|t=0 = ϕ0 , (11) describes the section AB of the orbit in Fig. 91. The section BC is described by the equation r ϕ=M a

dr + ϕ1 , r 2 |pr |

(16)

where the angle ϕ1 can be obtained by putting r = a in (11). The equation for the section CD is r ϕ = −M

dr r 2 |pr |

b

+ ϕ2 ,

(17)

where ϕ2 is defined from (16) with r = b, and so on. Substituting the values ϕ1 and ϕ2 in (16) and (17), we get the equations for BC and CD in the form ⎛ ϕ = M⎝ ⎛ ϕ = M⎝

r

b +

b

a −

a

r0

r − a

⎞

a r0

⎞ ⎠ dr + ϕ0 , r 2 |pr | ⎛

⎠ dr + ϕ0 = M ⎝− r 2 |pr |

r

(18)

b +2

a

a −

a

r0

⎞ ⎠ dr + ϕ0 . r 2 |pr |

(19)

One verify easily that the equation that section of the orbit which corresponds to the nth 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

⎛ cos γ (ϕ + α) = ⎝γ M

r a

where π =M γ

b a

dr , α=M r 2 |pr |

a r0

⎞ dr ⎠ , r 2 |pr | dr − ϕ0 . r 2 |pr |

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⎛ ϕ = M ⎝±

r

b

a

+2(n − 1) a

− a

r0

⎞ ⎠ dr + ϕ0 . r 2 |pr |

(20)

In the equations given here we have assumed that the angle ϕ changes continuously and 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 different signs in (20): ϕ is a many-valued function of r; on the other hand, r is a single-value function 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/m and inside with a speed 2(E + V )/m. Depending on the values of E and M, we get different kinds of orbits. 2 2 When M 2 − V < E < M 2 , the particle can either move inside the sphere and 2mR 2mR be reflected at its surface (Fig. 92a), or, provided E > 0, we can have an infinite orbit 2 (which can be a straight line, see Fig. 93b). When M 2 < E, we can have a reflected 2mR orbit (Fig. 92b). 2 What is the shape of the orbit when E = M 2 − V ? 2mR

(a)

(b)

Figure 92

2.3. To determine the equation of the orbit we use the equations  ϕ=

M dr , √ 2 r 2m(E − Ueff )

Ueff = U(r) +

M2 2mr 2

(1)

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97

p , e cos γ (ϕ − ψ) − 1

(2)

and we then get1 r= where

2 M2 p= β+ , α 2m

 e=



4E M2 1+ 2 β + , 2m α

 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 are at the angle 

2 M2 1 2 4E π − (ϕ1 − ϕ2 ) = π − arccos = π − arctan β + . γ e γ 2m α2

Ueff ϕ1

ϕ2

E

ψ

x

O

Figure 93

rm

r

Figure 94

2.4. It is useful first to study the character of the orbit using a graph Ueff (r). This graph is given in Fig. 94 for the case where β < M 2 /2m. In that case all orbits are infinite and lie in the region r  rm when E > 0. The equation of the orbits are the same 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

 M ϕ= M

dr M

, 2 r 2 2m E − M 2 − αr 2mr 

= M 2 + 2mβ, the integral is reduced to the corresponding integral in the Kepler problem where M (see [1], § 15 and [7], § 3.4). 2

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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 determined by the condition dU(r)/dr = 0, i. e. r = 2β/α.) Ueff A

Umax E

O

a

Figure 95

r

b

Figure 96

The form of Ueff (r) for the case β > M 2 /2m is shown in Fig. 96. If E > Umax =

α2 , 4(β − M 2 /(2m))

the particle flying from infinity will fall into the centre of the field. The equation of the orbit 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 p , e sinh γ  (ϕ − ψ) + 1 

M2 4E  β− e = − 1, 2m α2

r=

M2 2 β− , p = α 2m 

(1)  

γ =

2mβ − 1. M2

(2)

The orbit for this case is shown in Fig. 97a. We note that as r → 0, ϕ → ∞. This means that a particle coming from infinity and falling into the centre of the field makes an infinite number of revolutions around the centre. (a)

(b) O

O

Figure 97

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99

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 the equation for the orbit, using the equation cos ix = cosh x, and for the second case again the substitution of ψ + π/γ for ψ: p , r= 1 ∓ e cosh γ  (ϕ − ψ)

 

e =



4E M2 1− 2 β − . 2m α

(3)

For the case E = Umax , it is not possible to use (2) of the problem 2.3, as we have assumed in its derivation that e = 0, and we must start again from the integral (1). We find r=

p , 1 + c exp(−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 and approaching asymptotically the circle with radius r = p , or a spiral starting near the centre and approaching the same circle asymptotically, or it is the circle itself (Fig. 97b). Finally, for the case β = M 2 /(2m), it is also simplest to start again from the original integral. In that case, scattering occurs and the equation for the orbit is r=

α/E 1 − mα 2 (ϕ − ψ)2 /(2M 2 E)

.

The time it takes the particle to fall into the centre of the field is found from the equation  t=

m 2

r 0

dr . √ E − Ueff

For instance, when the orbit has the form given by (2), the time to fall into the centre from a distance r is given by 1 t= E



   m Er 2 − αr + β − M 2 /(2m) − β − M 2 /(2m) + 2 

(2Er/α) − 1 α m 1 arsinh + + arsinh . 2E 2E e e

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2.5. The equation of the orbit is r=

p , 1 + e cos γ (ϕ − ψ)

O

where p, e, and γ are defined in problem 2.3. When E < 0, the orbit is the finite one, and1 √ πα m Tr = √ , 2|E|3/2

ϕ =

Figure 98

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 < M 2 /(2m) we have p , 1 − e cos γ (ϕ − ψ)  2mβ γ = 1− , M2 r=



2 M2 −β , α 2m 

4E M 2 e = 1+ 2 −β ; 2m α

p=

if E < 0, then ϕ = 2π/ γ and Tr will be the same as in the preceding problem. When β > M 2 /(2m) we have (using the notations of the problem 2.4) r= r=

p e sinh γ  (ϕ − ψ) − 1

,

p

e cosh γ  (ϕ − ψ) − 1

if E > Umax , ,

if E < Umax .

2.7. a) A finite orbit is possible if the function Ueff (r) has a minimum. The equation  (r) = 0 can be reduced to the form f (x) = M 2 /(αm), where f (x) = x(x + 1)e−x , Ueff x = r. Using a graph of f (x) one sees easily that this equation has real roots only when m2 /(αm) is less than the value of f (x), for x > 0. This maximum value  maximum √  √ is equal to (2 + 5) exp − 12 (1 + 5) ≈ 0.84. A finite orbit is thus possible, provided M 2 < 0.84αm/ . b) A finite orbit is possible, provided M 2 < 8mV /(e2 2 ). 2.8. In the equation for the orbit (see (1) of the problem 2.3), we can neglect for small values of r the quantity E, when n = 2, and also the term M 2 /(2mr 2 ), when n > 2. We 1 The period T is the same as in the field U = −α/r. To determine T it is sufficient to note that adding a r r 0 term β/r 2 to the field U0 has the same effect on the radial motion as increasing M. However, the period Tr is independent of M in the Coulomb field U0 .

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

obtain then (see Fig. 99) M ln(r/r0 ) + ϕ0 , ϕ = −√ 2mα − M 2 2Mr −1+n/2 ϕ=√ + C, 2mα(n − 2)

when n = 2, when n > 2.

It turns out that the number of revolutions is infinite only when n = 2. The time it takes the particle to fall into the center is finite as the radial velocity increases when the center is approached. 2.9. The number of revolutions of the particle around the centre is infinite only in the case b) E = 0, n = 2.

n=2

n>2

Figure 99

2.10.  The time it takes the particle to fall into the centre is π mR3 /(8α). 2.11. As it is known, the two-body problem is reduced to two other problems: the motion of the centre of mass and the relative motion of a particle with reduced mass (in this case equal to m/2) in the field U(r). We consider only the relative motion. Introducing the relative distance r = r1 − r2 and taking into account that when time t is close to the 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 = 1 1 2 2 2 (m/2)v∞ = 2 mv , the angular momentum M = (m/2)[r, v∞ ] = −(0, 0, Eτ ), and the effective potential energy Ueff (r) =

α M2 + 2. r mr

After that, the minimum distance which is equal to   

2 α 1 3 rmin = 1 + 1 + 2 mv τ/α , mv2 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=

m1 r1 + m2 r2 m1 + m2

(1)

˙ and the relative motion of particles with reduced mass with constant velocity R m = m1 m2 /(m1 + m2 ) in the field U(r), where the relative distance is r = r1 − r2 .

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[2.12

a) Let’s take the point A to be the origin of the coordinate system, and the origin of the time will be the moment when particle 1 would fly through this point if there were no particle interaction. Under this condition, the position of the particles would be, respectively, r1 = v1 t and r2 = v2 (t − τ ) + ρ, where ρ ⊥ v1,2 . These expressions can be used to determine all of the following: the law of motion of the centre of mass; the energy E = 12 mv2 ; the angular momentum M = m[r, v] = −m ([v1 , v2 ]τ + [ρ, v]); and the effective potential energy Ueff (r) = −

β M2 2mβ − M 2 + = − r 2 2mr 2 2mr 2

in the relative motion problem (here v = v1 − v2 ). Note that from the expressions for E and M, the time t vanishes as expected. It can be seen from the equation E = 12 m˙r 2 + Ueff (r)

(2)

that the particle falls to the centre under the condition   2mβ > M 2 = m2 [v1 , v2 ]2 τ 2 + ρ 2 v2 . 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 0 t = t∞ − r∞

dr  . 2[E − Ueff (r)]/m

Notice the we took into account that the radial velocity r˙ is negative during the process of particles falling on each other (such process corresponds to the movement from the initial, very large distance r∞ to the point of incidence r = 0). For very large values of r, 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 1 t= v

 2mβ − M 2 2 2 r∞ + a − a − r∞ , a2 = . (mv)2

 2 2 2 + a2 ≈ r + a Substituting r∞ ∞ 2 r∞ and discarding the term containing a /(2vr∞ ), we get t = −a/v and finally find the distance where the particles will collide, R(t) = −

(m1 v1 + m2 v2 )a + m2 v2 vτ . (m1 + m2 )v

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103

2.13. The relative motion is characterized by the angular momentum M = mvρ and the energy E = 12 mv2 , where m = m1 m2 /(m1 + m2 ) is the reduced mass. The distance to be determined follows from the condition Ueff (rmin ) = E. Solutions can be obtained easily for n = 1, 2, and 4.

O

ϕ0 O

ρ S

Figure 100

Figure 101

2.14. The particle orbit is: m p = 1 ± e cos ϕ, m1, 2 r1, 2 where m1 m2 m= , m1 + m2

M2 p= , mα

 e=

1+

2EM 2 ; mα 2

E and M are the total energy and angular momentum of the system. The particles move in similar conical sections with a common focus, and their radius vectors are at any moment in the opposite directions (Fig. 100). 2.15. One sees easily from Fig. 101 that OS = ρ(cot ϕ0 − cot 2ϕ0 ), where ∞ ϕ0 = ρ rmin

 r2

dr

ρ2 1 − U(r) E − r2

.

At ρ → 0 ∞ ϕ0 = ρ rmin

dr − 2ρ 3 E 3/2 ∂ √ ∂E r 2 1 − U/E

∞ rmin

dr + ... √ r4 E − U

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(cf. problem 2.27), so that ⎛ OS = ⎝2

∞

rmin

⎞−1 dr ⎠ + O(ρ 2 ) . . . √ r 2 1 − U/E

The point of S is the virtual focus of the beam of scattered particles so that up to and including terms of first order in ρ the position of the points where the asymptote of the orbit intersects the axis of the beam is independent on ρ. 2.16. The equation of the orbit is p = e cos(ϕ − ϕ0 ) − 1, r  2 2 M where p = mα , e = 1 + 2EM2 , while ϕ0 is determined from the condition that ϕ → π , mα as r → ∞, so that cos ϕ0 = −1/e. The region which can not be reached by the particles is bounded by the envelope of the family of orbits. To find it, we differentiate the equation for the orbit, M M2 + 1 + cos ϕ − mαr α



2E sin ϕ = 0 m

(1)

with respect to the parameter M 2M − mr



2E sin ϕ = 0, m

(2)

and eliminate M from (1) and (2). The result is 2α = 1 − cos ϕ. Er The inaccessible region is thus r


2 2aδ , where δ = mav 2 2α , OA = a. 1 − δ − (1 − δ) cos ϕ 2

Figure 102

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105

2.18. We take the scalar product of the equation [v, M] −

αr =A r

with r. Denoting by ϕ the angle between R and A, we get M2 − αr = Ar cos ϕ, m or p = 1 + e cos ϕ, r where p=

|A| M2 , e= . mα 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 circle of radius R at a velocity V . If the separation of the body from the spacecraft occurs at R = (0, R) and velocity V = (V , 0), the Laplace vector A0 = [V, M0 ] − α

R R = mV 2 R − α R R

√ 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 is V + v. The angular momentum of the body has not changed, m[R, V + v] = M0 , since v  (−R), but the Laplace vector is no longer zero A = [V + v, M0 ] − α

R V = [v, M0 ] = mRvV = αv 2 . R V

As it is known, the Laplace vector is directed towards the pericentre (the point of the orbit with the smallest radius) and is αe, where e is the eccentricity. From here we get the value e = v/V . Assuming that e < 1, we find that the orbit of the body is an ellipse, whose parameter is p = M 2 /(mα) = R, and whose pericentre lies on the x-axis. Thus, the equation of the body orbit in polar coordinates r, ϕ has the form r=

R . 1 + (v/V ) cos ϕ

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[2.20

2.20. δT = − ∂δI , where ∂E  √ δU(r) dr δI = T δU = 2m √ E − Ueff (r) r2

r1

(cf. problem 1.10). Similarly, we can write for the change in angular distances between successive pericentre passages (r = rmin ) in the form δϕ = ∂δI (cf. [1], § 15, problem 3; ∂M § 49). 2.21. The field U(r) differs little from the Coulomb field U0 (r) = −α/r in the region r  D. Therefore, a finite orbit which is close to an ellipse with a parameter p and an eccentricity e determined by the integrals of motion E and M will retain its shape, but will change its orientation. The velocity of rotation  of the ellipse is determined by the displacement of the pericentre over a period  = δϕ/T0 which we can evaluate using α − αr while T is the period the equations from the preceding problem with δU = D 0 2D2 1 2 in the Coulomb field. The result of the calculation is  = M/(2mD ). We can write the equation of the orbit in the form p = 1 + e cos γ ϕ, r

γ = 1−

T0 . 2π

(1)

The deviation of the curve (1) from the actual orbit is of first order in δU; that is, during one period (1) describes the orbit with the same degree of accuracy as the equation for the fixed ellipse. However, (1) retains the same accuracy during many periods. It is therefore 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/r 1+ε differs little from the Coulomb field so that the orbit of the 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 (1 − e cos ξ ). 2|E|

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107

where α U0 (r) = − , r

δU =

ε α r ln , r R

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 T ∂ δϕ = δU dt ∂M 0

(see problem 2.20) by making the substitution r=−

α (1 − e cos ξ ), 2E

t=

T (ξ − e sin ξ ), 2π

where  e=

1+

2EM 2 , m α2

 m T = π α 2|E|3

(see [1], § 15). The result is ε α ∂ δϕ = = T 2π ∂M

2π α (1 − e cos ξ ) ln dξ = 2|E|R 0

ε|E| ∂e =− π ∂M

2π 0

√ cos ξ dξ 2π 1 − 1 − e2 = ε. 1 − e cos ξ T e2

At ε > 0, the direction of the orbit precession coincides with the direction of motion of a particle over orbit, and at ε < 0 the precession is opposite to it. 3 β a2 + b2 M (see also [7], § 4). 2.23.  = − 2 mω a4 b4 |M| 2.24. The Lagrangian function reads  mg   m x2 + y2   m 2 x˙ + y˙2 − x2 + y2 + x˙2 + y˙2 = 2 2 2l 2 l  mg 2 m 2 mr = r˙ + r 2 ϕ˙ 2 − r 2 + 2 r˙2 . 2 2l 2l

L=

If we neglect the last term, then it will be convenient to consider the particle motion  in the Cartesian coordinates. It leads to harmonic oscillations with frequency ω = g/l

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along the x- and y-axes, that is, to the elliptic orbit. The exact equation of the trajectory is convenient to obtain in the cylindrical coordinates using the integrals of motion E and M = (0, 0, M):  ϕ=

 M 1 + r 2 /l 2 dr , √ r 2 2m[E − Ueff (r)]

Ueff (r) =

M2 mgr 2 + . 2l 2mr 2

Expansion  1+

r2 r2 = 1 + 2 + ... 2 l 2l

leads to the equation  ϕ=

M M dr + √ 2 2 2ml r 2m(E − Ueff )



dr = ϕ0 (r) +  t, √ 2m(E − Ueff )

where ϕ0 (r) corresponds to the motion over an ellipse, and =

M ab = 2ω 2 2ml 2l

determines the angular velocity of the precession. 2.25. The Lagrangian function of the Earth–Moon system is m2 r2

˙ 2 + 1 m2 R ˙ 2 + Gm0 m1 + Gm0 m2 + Gm1 m2 , L = 12 m1 R 1 2 2 R1 R2 |R1 − R2 | where R1 and R2 are radius vectors of the Earth and the Moon in 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 the Earth and the Moon (the point O on Fig. 103) and their relative distance m1 R1 + m2 R2 R= , m1 + m2

O r1 m1

R2 R R1

m0

r = R2 − R1 ,

Figure 103

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

109

then R1, 2 = R + r1, 2 , m=

r1, 2 = ∓

m r, m1, 2

m1 m2 ≈ m2 . m1 + m2

Using the expansion  −1/2 ri2 1 1 2Rri = = 1+ 2 + 2 Ri R R R   1 Rri (Rri )2 1 2 = − 3 + 3 − ri + . . . R R 2R3 R2 and taking into account that m1 r1 + m2 r2 = 0,

m1 r12  m2 r22 ≈ m2 r 2 ,

we obtain L = L1 + L2 − δU, m1 + m2 ˙ 2 Gm0 (m1 + m2 ) R + , L1 = 2 R   m 2 Gm1 m2 Gm0 m (Rr)2 2 3 2 −r . , δU = − L2 = r˙ + 2 r 2R3 R

(1)

If we neglect the term δU, the problems of motion of the centre of mass O and of relative motion are separated and reduced to the Kepler problem (for simplicity, we will talk further about the motions of the Earth around the Sun and the Moon around the Earth). a) In the problem of the motion of the Earth around the Sun, the smallness of δU is characterized by a ratio m2 r 2 δU ∼ ∼ 10−7 . (Gm0 m1 /R) m1 R2 Considering the orbit of the Moon as a circle of radius r lying in the plane of the Earth’s orbit, one has δU = −

 2β  2 3 cos χ − 1 , R3

β = 14 Gm0 m2 r 2 ,

(2)

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[2.25

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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 new Moons). Averaging expression (2) over that interval leads to the following replacement cos2 χ → 12 . As a result, δU = −

β . R3

The perihelion displacement per year (see [1], problem 3 to § 15) is δϕ =

3π m2  r 2 6πβ = , 2m1 R α1 R2

α1 = Gm0 m1 .

The perihelion shift per century equals 100 δϕ = 7.7 . It should be noted that the total displacement of the Earth’s perihelion for one century equal 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

V c

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 2 ∼ 2 ∼ 10−2 , (Gm1 m2 /r) ω

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 the ripple of eccentricity, the perihelion shift (cf. problem 2.43) and so on. We will consider only one of them, neglecting their mutual influence and assuming the unperturbed orbit of the Moon as a circle; we also take R = const. We introduce the geocentric coordinate system Oxyz with the x-axis directed along the line of intersection of the plane of the Moon’s and the Earth’s orbits with (the line of lunar nodes), the y-axis lying in the plane of the Earth’s orbit and the z-axis perpendicular to this plane and directed to the northern celestial hemisphere (Fig. 104). The coordinates of the Sun in this system are −R = R(cos ϕ, sin ϕ, 0),

ϕ = t + ϕ0 ,

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

111

z m2 r m1

ψ θ ϕ

y

–R m0

x

Figure 104

and the coordinates of the Moon are r = r(cos ψ, cos θ sin ψ, sin θ sin ψ),

ψ = ωt + ψ0 ,

where ϕ0 and ψ0 are determined by the moments of passage of the Sun and the Moon through the x-axis. Averaging quantity (Rr)2 = R2 r 2 (cos ϕ cos ψ + cos θ sin ϕ sin ψ)2 over one month leads to the replacements cos2 ψ → 12 ,

sin2 ψ → 12 ,

sin ψ cos ψ → 0,

and to equations   (Rr)2 month = 12 R2 r 2 cos2 ϕ + cos2 θ sin2 ϕ , while averaging over one year leads to these replacements cos2 ϕ → 12 ,

sin2 ϕ → 12 .

It gives   (Rr)2 = 14 R2 r 2 1 + cos2 θ . As a result, δU = −

 Gm0 m 2  2 r θ − 1 . 3 cos 8R3

(3)

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[2.25

The rotation of the orbital plane can be traced by considering the angular momentum ˙ = K, where M = m[r, r˙ ] perpendicular to this plane. The equation of its motion is M 

∂δU K = − r, ∂r



is the moment of the forces acting upon the Moon. Because the change of the angle θ is the rotation around the x-axis, Kx = −

∂δU , ∂θ

that is,

3Gm0 m 2 r sin θ cos θ, 0, 0 . K = − 4R3 Since the moment of forces K is perpendicular to the vector M, it leads only to its rotation. The x- and y-axes in this case also rotate; therefore, the angular velocity pr of precessions of the vector M is directed along the z-axis. It can be found from the condition [pr , M] = K , which gives

pr



=− z

 Kx Kx 32 = − cos θ ≈ − . = 2 My 4ω 17.9 Mωr sin θ

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 is 18.6 years. Given the crudeness of our approximations, our result could be considered as 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 19 765 days) was already known to the priests of the ancient Babylon. Note that by limiting the averaging over a month; that is, using (3), we would discover the unevenness of the precession within a year:   pr → 1 − 12 cos 2ϕ pr .

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2.26. Similarly to problem 2.21, we have  a 2 a  = ∂ δU = ∂e δU  (a) + δU  (a) = ∂M 2 ∂M 2   M a   a =− δU (a) + δU  (a) . mα 2 |M| If we take into account the next terms of the expansion δU over (r − a), that will give the contribution of  e2  into . 2.27. The perihelion displacement for the period r2+δr2

ϕ = r1 +δr1

r2

M dr √ 2m (E − Ueff − δU)

can be presented as 4√ ∂2 ϕ = − 2m 3 ∂M∂E

r2+δr2

(E − Ueff − δU)3/2 dr. r1 +δr1

Up to the second order in δU, we have ϕ = 2π + δ1 ϕ + δ2 ϕ, δ1 ϕ = t ∂ δU , ∂M ∂2 δ2 ϕ = − [T (δU)2 ], 2∂M∂E 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 δ1 ϕ + δ 2 ϕ δ1 ϕ δ2 ϕ δ1 ϕ δT , = + − T + δT T T T2 δT = − ∂ (T δU ). ∂E =

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2.28. If we expand the integrand in the equation for the orbit,  ϕ=

 r2

M dr



2 γ 2m E − M 2 + αr − 3 2mr r

,

(1)

in terms of δU = γ /r 3 , we can rewrite this equation in the form ϕ = ϕ0 (r) + δϕ(r),

(2)

where  ϕ0 (r) = 





M dr 2

r 2 2m E − M 2 + αr 2mr

δϕ(r) =



γ M dr

2

2r 5 2m E − M 2 + αr 2mr

,

(3)

3 .

(4)

If we neglect the correction δϕ(r) in (2), we obviously get the equation of the orbit in a Coulomb field (see [1], § 15) p = 1 + e cos ϕ, r

(5)

where M2 p= , mα

 e=

1+

2EM 2 . mα 2

If we take into account the correction δϕ(r) in (2), we get, instead of (5), p = 1 + e cos(ϕ − δϕ(r)). r

(6)

In the right-hand side of (6) we can expand in terms of δϕ(r) and use in δϕ(r) for r the relation r = r0 (ϕ) which follows from (5). The result is p = 1 + e cos ϕ + eδϕ(r0 (ϕ)) sin ϕ. r

(7)

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

115

Integrating (4), we find1 δϕ(r0 (ϕ)) = m2 αγ = M4



 2e2 + 1 1 + e cos ϕ 2 −3ϕ − sin ϕ − 2 [2e + (e + 1) cos ϕ] . (8) e e sin ϕ

Substituting (8) into (7) gives the result which is accurate up to and including firstm2 αγ : order terms in ζ = M4 p = 1 + e cos[(1 + 3ζ )ϕ] − ζ (2e2 +1) sin2 ϕ− r 1 + e cos ϕ −ζ [2e + (e2 + 1) cos ϕ] e

(9a)

or p = 1 + e cos λϕ + f cos 2ϕ, r λ = 1 + 3ζ ,   p = p 1 + 32 ζ (e2 + 2) ,

3e4 − 2 e = e 1 + ζ , 2e2

(9b)

f = 12 ζ e2 .

1 We can rewrite (4) in the form

δϕ = ∂ ∂M



√ 2mγ dr ,  2 2r 3 E − M 2 + αr 2mr

and use (5) to change the integration into one over ϕ: m2 γ α δϕ = ∂ ∂M M 3

 (1 + e cos ϕ) dϕ = =−

3m2 γ α m2 γ α ∂e m2 γ α ∂ϕ (ϕ + e sin ϕ) + sin ϕ + (1 + e cos ϕ). 4 3 M M ∂M M 3 ∂M

From (5) we find 2M ∂ϕ = ∂e cos ϕ − e sin ϕ, ∂M ∂M mα 2 r and substituting this into (8 ), we obtain (8).

2 ∂e = e − 1 ∂M eM

(8 )

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[2.29

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 long as 3ζ ϕ  1. Taking only the secular part δϕ = −3ζ ϕ in (8) into account, we obtain the equation p = 1 + e cos λϕ, r

λ = 1 + 3ζ ,

(10)

which describes the path over a long section of the orbit; that is, a “corrected” zero approximation (as distinct from (5); compare problem 2.21). It is obviously easy to transform (9) also in such a way that it describes the orbit over a long stretch with an accuracy up to and including first-order terms, and we have p = 1 + e cos λϕ + f cos 2λϕ. r

(11)

2.29. It is sufficient to prove that the Lagrangian function can be split into two parts when expressed in terms of the centre of mass coordinates R=

m1 r1 + m2 r2 , m1 + m2

and the relative coordinates, r = r2 − r1 : ˙ + L2 (r, r˙ ), L = L1 (R, R) ˙ = 1 (m1 + m2 )R ˙ 2 + (e1 + e2 )ER, L1 (R, R) 2

e1 e2 e2 e1 L2 (r, r˙ ) = 12 m˙r2 − − Er. −m r m1 m2 ˙ determines the motion of the centre of mass which is the same The function L1 (R, R) as the motion of a particle of mass m1 + m2 and charge e1 + e2 in uniform field E. The relative motion, determined by L2 (r, r˙ ), is the same as the motion of a particle with mass m = m1 m2 /(m1 + m2 ) (the reduce mass) in the Coulomb field as well as in uniform field E. One could, of course, have obtained the same result starting from the equations of motion of a particles. 2.30. The Lagrangian function 1 In particular, the radius vector r must be a periodic function of ϕ.

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2.32]

117

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

L=

1 2



 m1 r˙ 21 + m2 r˙ 22 −

e1 e2 e1 e2 + A(r1 )˙r1 + A(r2 )˙r2 |r1 − r2 | c c

˙ and r, r˙ (using the notations of can be split into two parts depending only on R, R problem 2.29, c is the velocity of light), if e1 /m1 = e2 /m2 . In this case we get: e m2 + e2 m21 ˙ + 1 m r˙ 2 − e1 e2 + 1 2 ˙ 2 + e1 + e2 A(R)R L = 12 (m1 + m2 )R A(r) r˙ . 2 c r c(m1 + m2 )2 2.31. T =

1 2

N

˙2 n=1 μn ξ n ,

p = μN ξ˙ N , M =

1 1 1 = n + μn mn+1 k=1 mk

N

˙

n=1 μn [ξ n , ξ n ],

(n = 1, . . . , N − 1),

where

μN =

N 

mk .

k=1

2.32. Let the coordinates of the incident particle and of the particle initially at rest be x1 and x2 , respectively, and let at t = 0, x1 = −R, x2 = 0. The centre of mass of the system moves accordingly to the equation X = − 12 R + 12 vt. The relative motion (x = x2 − x1 ) follows from the equation law 

m t=− 4



x



R

dx

.

mv2 − α 4 xn

The first particle thus has the following position: R x1 = X −

1 2

x = − 12

R+

1 2 x

dx 1 

 − 2 x. 2 n 1 − 4α/ mv x

1/n 4α The distance between the particles decreases until it becomes xmin = and then mv2 increases again. When it is again equal to R, the first particle has come to the point 

R/xmin

x1f = xmin 1

 1 − 1 dz − 1 . √ 1 − z−n

(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.

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[2.34

2.34. The equation of motion (c is the velocity of light) mv˙ =

eg [v, r] cr 3

(1)

has the integrals of motion mv2 = E, eg r m[r, v] − = J. c r 1 2

(2) (3)

2 2 Multiplying (1) by r, we obtain rv˙ = 0 or d 2 r2 − v2 = 0, where dt

r 2 = r02 + v2 (t − t0 )2 .

(4)

Multiplying (3) by r/r, we obtain −

eg = J cos θ, c

(5)

where θ is the polar angle in the spherical coordinates with the z-axis parallel to the vector J. Projection (3) onto the z-axis mr 2 ϕ˙ sin2 θ −

eg cos θ = J c

together with (5) gives ϕ˙ =

J . mr 2

(6)

At t = t0 , we have v = r0 ϕ˙ sin θ, besides, from (6) we get J = mr0 v , and the result is sin θ (taking into account (5)) tan θ = − Integrating the equation ϕ˙ =

mr0 vc . eg

(7)

r0 v , we obtain r 2 (t) sin θ

ϕ = ϕ0 +

1 v(t − t0 ) . arctan sin θ r0

(8)

The particle thus moves with a constant velocity v on the surface of the cone θ = const. By introducing ψ = (ϕ − ϕ0 ) sin θ,

(9)

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

119

we rewrite (8) as tan ψ =

v(t − t0 ) . r0

(10)

Equations (4), (9), and (10) can be interpreted as follows: the motion of a point on the cone involute is uniform and straightforward, while r and ψ are the polar coordinates in the plane of the involute. 2.35. The motion of a charged particle in an electromagnetic field is determined by the Lagrangian function e L = 12 mv2 − eϕ + vA, c

(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=

e m 2 μr 2 ϕ˙ . (˙r + r 2 ϕ˙ 2 + z˙2 ) + 2 c (r 2 + z2 )3/2

(2)

We see from the equation of motion for z m¨z +

3e μr 2 ϕz ˙ =0 2 2 c (r + z )5/2

(3)

that for z = 0 the component of the force along the z-axis vanishes. The orbit thus lies in the plane z = 0 when z(0) = z(0) ˙ = 0. As ϕ is a cyclic coordinate, we have ∂L = mr 2 ϕ˙ + eμ = p = const. ϕ ∂ ϕ˙ cr

(4)

From this equation it follows that the projection of the momentum Mz = m[r, v]z = mr 2 ϕ˙ eμ is not saved, but the sum of Mz + cr = 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): 1 r 2 + r 2 ϕ˙ 2 ) = E. 2 m(˙

(5)

Using (4) to eliminate ϕ˙ from (5) we get 1 r 2 + Ueff (r) = E, 2 m˙

(6)

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where Ueff (r) =

1  eμ 2 − . p ϕ cr 2mr 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. (a)

(b) U eff

Ueff

E2

E

Um E1

r

a r1 b r2

r

Figure 105

In the case when pϕ < 0 the orbit is always infinite. To give a qualitative description of the orbit, it is useful to use (4) to write ϕ˙ = −

|pϕ | eμ − . 2 mr mcr 3

(8)

The velocity with which the radius vector of the particle turns round has the same direction all the time and increases when the particle approaches the dipole. Curve 1 in Fig. 106 shows such an orbit. The orbit is symmetric with respect to straight line connecting the centre of the field with the point r = rmin . When pϕ > 0, scattering can occur for any energy E, but if E < Um =

c2 p4ϕ 32me2 μ2

(see the line E = E1 in Fig. 105b), there is also the possibility of finite orbit. From the equation ϕ˙ =

pϕ eμ − 2 mr mcr 3

eμ it follows that ϕ˙ > 0 when r > r1 = cp and ϕ˙ < 0 when r < r1 . At r = r1 there is a ϕ “sticking point” in ϕ. A particle with an energy E > Um (see the line E = E2 in Fig. 93b) is scattered while in two points (where r = r1 ) its velocity is parallel to its radius vector (curve 2 in Fig. 106).

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121

3 B

2

C

a O

O

1

A

r1 b

Figure 106

Figure 107

At E < Um there can occur scattering without sticking points for ϕ (curve 3 of Fig. 106) or a finite orbit in an annulus a  r  b (Fig. 107). In the latter case the particle can 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 potential in the form Aϕ = 12 rB, Az = Ar = 0. In the z-direction the motion is uniform, while in a plane perpendicular to the z-axis we 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 values E⊥ = E − 12 mv2z in the regions U 1 < E⊥ < U 2 , E⊥ = U 2 , E⊥ > U 2 , where − )pϕ , U2 = U1 = (

 λpϕ eB = 2 + λ. , = ,  2 2mc

An orbit for the case when pϕ < 0 is given in Fig. 108d, while Fig. 108e depicts the orbit for the case where pϕ = 0. The orbit for the particle in the xy-plane can easily be determined once we know (see [1], § 23, problem 3 and the motion of a free isotropic oscillator of frequency  2 [7], § 4) t + b2 sin2  t, r 2 = a2 cos2 

ϕ = −t + arctan

b t . tan  a

1 To fix our ideas we assume B > 0; p is the generalized momentum corresponding to the coordinate ϕ.   ϕ 2 The branches of arctan b tan  t should be chosen such that the angle ϕ is a continuous function of t.

a

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[2.37

Here the minimum (b) and maximum (a) radii are determined by the energy E = 1 2 2 2 2 m (a + b ) − pϕ  and the angular momentum pϕ = mab; the origins for t and ϕ are chosen that ϕ(0) = 0, r(0) = a. is independent of E and pϕ . The Note that the period of radial oscillations, T = π/ angle over which the radius vector turns during this period is ) for pϕ ≷ 0, ϕ = π(±1 − / where the sign is the same as that for pϕ , and for pϕ = 0. ϕ = −π / 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). 2.37. The equation of the orbit is  ϕ=

m 2



pϕ 2 − mr dr, √ E − Ueff (r)

(1)

eB , c is the velocity of light and where  = 2mc Ueff (r) = −

pϕ 2 α 1 2 + 2 mr  − 2 . r mr

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 pϕ value r0 = 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 U (r), we can use the same approximate eff

treatment as in problem 2.36. Of course, the exact form of the orbits is different in the two problems.

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2.37]

§2. Motion of a particle in three-dimensional fields (a)

(b)

123

(c)

Δϕ

(d)

(e)

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  pϕ dr − t. ϕ=√  2 2m p r 2 E + pϕ  + αr − ϕ 2 − 12 m2 r 2 2mr

(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) = − αr which is δU(r) = 12 m2 r 2 , each of which leads to a precession of the orbit and an additional precession with an angular velocity −. If the magnetic field B is sufficiently small, the term δU(r) can be considered to be a small correction to U0 (r) =

p2ϕ 2mr 2

−

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  =

δϕ 1 ∂ (T δU ), = T T ∂pϕ

(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

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[2.37

and 2.21). If we perform the calculation,1 we get  = −

32 pϕ ; 4|E  |

(5)

we can assume δU to be a small correction if, apart from (3), the condition δϕ  2π is also satisfied, or √ 2 pϕ α m|E  |−5/2  1.

(6)

It is, of course, impossible to consider δU to be a small correction when E   0, as in that case 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 of the motion of the particle in its orbit. The direction of the velocity  is determined by the 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 an angular 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 a small correction? Let us now consider the case when the field U(r) = −α/r can be considered to be a small correction. If U(r) were not present, the motion would be along a circle. Its radius a and the distance b from of its centre from the centre of the field can be express in terms of the maximum and minimum distances of the particle from the centre of the field 2 r1,2 =

 E + pϕ  ± (E + 2pϕ )E . m2

(8)

There are now two possibilities (see Fig. 109), depending on how the circle is placed with 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  , 2m2

a2 =

E . 2m2

(9)

1 To evaluate δU , it is convenient to use the variables used in problems 2.21 and 2.22; since the period in the field U0 is independent of pϕ , we can take T from under the differentiation sign in (4).

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2.38]

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125

Taking U(r) into account leads to a systematic displacement of this circle (the socalled drift), while its radius and the distance from the origin of the fields which are determined 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 = ∂ U , ∂pϕ where the averaging of U(r) is over the uniform motion along the circle. Let us restrict ourselves to the case when a  b. In that case, we may simple assume that α U(r) = − , b

(10)

so that γ=

α . 2mb3

We note that in this case the linear drift velocity is equal to c|E|/B, where e|E| = α/b2 is the 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 uniform magnetic field can be reduced to the problem on the motion of the centre of mass and the problem of the relative motion (see problem 2.30). For the centre of mass coordinates, we have X = R cos ωt,

Y = −R sin ωt,

(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 is similar 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 a of 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 accuracy than we was done in preceding problem, by expanding Ueff (r) =

p2ϕ e2 1 + 2 + 16 2 mω2 r 2 − 12 pϕ ω r mr

in a series in the vicinity of the minimum where r = b (see [1], § 21). From the condition

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126

[2.38

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(b)

A a b θ O

b

A θ

a

O

Figure 109  (b) = 0, we find Ueff

 pϕ =

 m2 2 4 m 2 m ω ω b − e b ≈ b2 −γ , 16 2 2 2 

and we thus get finally for ωr = of the particles

γ=

2e2 , mωb3

(2)

 2Ueff (b) the result ωr = ω − 12 γ , and for the separation m

r = b + a cos(ωr t + α).

(3)

To find  ϕ(t), we  apply the principle of conservation of generalized momentum to pϕ = 1 1 2 ϕ˙ + 2 ω . Using (2) and (3), we get 2 mr ϕ(t) = −γ t −

a sin(ωr t + α) + ϕ0 . b

(4)

Using (3) and (4), we get for the relative coordinates   x = r cos ϕ = b cos(γ t − ϕ0 ) + a cos ωt + 12 γ t + β ,   y = r sin ϕ = −b sin(γ t − ϕ0 ) − a sin ωt + 12 γ t + β ,

(5)

where β = α − ϕ0 . The first terms here correspond to the motion of the centre of the circle with a drift velocity bγ , and the second terms to the motion along this circle with an angular velocity ω + 12 γ .

2

1

Figure 110

Figure 111

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

127

Coordinates of the particles, x1, 2 = X ± 12 x, y1, 2 = Y ± 12 y can be written in the form x1, 2 = ± 12 b cos(γ t − ϕ0 ) + ρ1, 2 cos(ωt + ψ1, 2 ), y1, 2 = ∓ 12 b sin(γ t − ϕ0 ) − ρ1, 2 sin(ωt + ψ1, 2 ),

(6)

where 

  R2 + 14 a2 ± aR cos 12 γ t + β ,   ±a sin 12 γ t + β  . tan ψ1, 2 = 2R ± a cos 12 γ t + β ρ1, 2 =

The centres of the circles along which the particles move thus rotate around the origin with an angular velocity γ (a drift velocity bγ /2) while their radii oscillate with a frequency γ /2 (Fig. 110). Another limiting case when a  b (distance of the centre of the orbits to the origin small compared to the radii of the orbits (Fig. 111)) may give a clear insight into the mechanism of energy “exchange”. The work carried out by the the force of the interaction on the second particle is clearly positive, while the work carried on the first particle is negative, when taken over many periods. 2.39. One can easily prove that the given quantity is a constant by using the equations of motion and writing this quantity in the form AF + 12 [F, r]2 , where A = [v, M] −

αr r

is the Laplace vector (see [1], § 15 and [7], § 14). For small values of F the orbit will be close to an ellipse, with its semi-major axis along the direction of the vector A and with an eccentricity e = |A|/α. In this case AF ≈ const, or e cos ψ ≈ const, where ψ is the angle between A and F. 2.40. When there is a small extra term δU(r) in the potential energy, the quantities characterizing the motion of the particle, such as the angular momentum, pericentrum position and so on, change, although they do not change their values appreciable over a shot time interval (a few periods of the unperturbed motion). However, these changes may add up over an extended time, so that some of the quantities may happen to change by large amounts.

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In particular, the orbit remains elliptical for a shot time interval. Its semi-major axis, α , which is determined by the energy, does not change over a long time, while a = 2|E| 

M 2 and the orientation of the orbit are both liable to secular the eccentricity e = 1 − mαa changes. a) The change in the angular momentum is determined by the equation

˙ = [r, F]. M

(1)

F

O ψ A

Average this equation over one period, we obtain

x y

˙ = [ r , F], M

(2) Figure 112

where 1 r = T



T

r(t) dt.

(3)

0

For averaging, we use a coordinate system with the z-axis parallel to M and the x-axis parallel to A (Fig. 112). Here A = [v, M] −

αr r

is an additional integral of motion in the Kepler problem. Recall that the vector A is directed 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(cos ξ − e), t =

T (ξ − e sin ξ ), 2π

we get

x =

a 2π

2π 3ae . (cos ξ − e)(1 − cos ξ ) dξ = − 2

(4)

0

Therefore r = −

3a 3ae A = − A. 2 |A| 2α

(5)

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

b) If the force F is at right angles to M, it is obvious from symmetry consideration that the orbit lies in a plane and the vector M retains its direction—apart possibly from the sign. Omitting the averaging sign, we can write (2) and (5) in the form ˙ = 3 aeF sin ψ, M 2

(6)

where ψ is the angle between A and F. Using the fact that e cos ψ = e0 = const

(7)

(see problem 2.39), and eliminating e and ψ from (6) we find  ˙ = 3 aF M 2

1 − e20 −

M2 . maα

(8)

Integrating (8), we get M = M0 cos(t + β),

(9)

where =

3F 2



a , mα

M0 =

 maα(1 − e20 ),

as well as e=



1 − (1 − e20 ) cos2 (t + β).

The orbit is thus an ellipse which oscillates about the direction of 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 oscillation of the ellipse is much longer than the period T of the rotation of the particle along the ellipse. c) In the general case, we consider also the change in the vector A. Using the equations of motion, we easily get ˙ = 1 [F, M] + [v, [r, F]]. A m

(10)

F

Figure 113

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[2.40

For the averaging in (10), we use the equations 

 d x2 xx ˙ = = 0, dt 2   d y2 = 0, yy ˙ = dt 2   d xy = 0, xy ˙ + yx ˙ = dt M xy ˙ − yx ˙ = . m

(11)

As a result, we get ˙ = A

3 [F, M]. 2m

(12)

We have thus for M and A, averaged over one period (in the following we omit the averaging sign), the following set of equations ˙ = 3 [F, M], A 2m 3a ˙ [F, A]. M= 2α

(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 the transverse component M, M⊥ = M −

F(MF) , F2

(15)

we obtain from (13) ¨ ⊥ + 2 M⊥ = 0. M

(16)

In a coordinate system OX1 X2 X3 with the X3 -axis parallel to F we have for the solution of (13): M1 = B1 cos t + C1 sin t, M2 = B2 cos t + C2 sin t.

(17)

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

131

We then obtain from (13): 3F (B2 sin t − C2 cos t), 2m 3F A2 = (B1 sin t − C1 cos t). 2m A1 = −

(18)

As we should expect, the constants B1, 2 and C1, 2 are determined by the initial values of the vectors M and A. The end point of the vector M describes an ellipse with its X3 centre on the X3 -axis and lying in a plane μ which is parallel μ to the OX1 X2 -plane (Fig. 114). The end of the vector A also describes an ellipse with centre on the X3 -axis and lying in a plane σ which is parallel to μ; this second ellipse is similar to the first one, but rotated over an angle π/2: A remains thus M F at right angles to M all the time. The plane of the orbit is ρ perpendicular to M, and the vector A determines the direction to the pericentre of the orbit. A σ The plane of the orbit thus rotates (precesses) around F. The angle between the plane of the orbit ρ and F oscillates Figure 114 about its mean value. The eccentricity and the angle between the projection of F upon the plane ρ and the direction to the pericentre also oscillate about their mean values. All these motions occur with a frequency . We should bear in mind that we have neglected those corrections in F which are of the first order but which do not lead to secular effects. Our solution is valid for a time interval of the order of several periods of the orbital precession. The reader should checked whether the next approximations will lead to a qualitative change in the character of the motion (e.g., to a possible departure of the particle to infinity). The exact solution of the problem of the motion of a particle in the field U = − αr − Fr, which can be given in parabolic coordinates (see problem 12.12b), shows that 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 under the action of infinitesimal constant perturbation is connected with a degeneracy of the unperturbed motion. In [8], § 7.3 one can find a solution of this problem using the canonical perturbation theory. 2.41. According to the Larmore’s theorem (see [7], § 17.3 and [2], § 45), the orbit of a particle in a uniform magnetic field B rotates around the centre of the Coulomb field with qB the angular velocity  = − 2mc , 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 A ratate with the same angular velocity:

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˙ 1 = [, M], M

˙ 1 = [, A]. A

[2.41

(1)

In the preceding problem (see (13)), the averaged velocity change of the vectors M and A under the influence of the constant force F = qE has been determined: ˙ 2 = 3a [F, A], M 2α

˙ 2 = 3 [F, M]. A 2m

(2)

The averaged velocity change of the vectors M and A under the influence of both fields is the following sum: ˙ =M ˙ 1+M ˙ 2, M

˙ =A ˙ 1+A ˙ 2. A

(3)

a) Let us direct the x-axis along the electric field and the z-axis along the magnetic one. Then the equations (3) take the form A˙x = −Ay ,

3F M, A˙y = Ax − 2m

˙ = 3aF Ay . M 2α

The solution of this system is  C cos(ωt + β) + C1 , ω Ay = C sin(ωt + β), 2m 3aF C cos(ωt + β) + C1 , M=− 2αω 3F

Ax =

 where ω = 2 + 9aF 2 /(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 rotates with C > ωC1 ), and its eccentricity changes periodically. When 9aF 2 C > 4mαωC1 , the elliptical orbit periodically turns into the straight-line segment. y A O

Figure 115

x

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

133

b) Let us denote 3 F = F 2



 a ma , N=A , mα α

then we can write (3) as ˙ = [, M] + [F , N], N ˙ = [, N] + [F , M]. M Adding and subtracting these equations, we find J˙ 1, 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) cos ωi t +

 ωi Ji (0)ωi , Ji (0) sin ωi t + ωi (1 − cos ωi t), ωi ωi2

and vectors  M = J1 + J2 , A =

α (J1 − J2 ) ma

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 Laplace vector A (see (10) in problem 2.40 where one must substitute F = −∂δU(r)/∂r) can be presented in the form 6β d ˙ 2 ) = − zM + 2β (xz2 ), A˙x = 2β(5xzz˙ − 2xz m dt 4β d 2 A˙z = −2β(4xz ˙ − zx ˙ ) = − xM − 2β (x2 z), m dt where M = m(zx˙ − xz) ˙ is a slowly changing function of time.

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[2.42

After averaging these equations, we obtain βa M Az M, A˙x = −6β z = 9 m mα βa M Ax M, A˙z = −4β x = 6 m mα where a is a large semi-axis of an ellipse. Hence 3Az A˙x = ; ˙ 2Ax Az that is, in contrast to problem 2.40b, the end of the vector A varies not along a straight line but along a hyperbola A2x = 32 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



2 3

(u2 − e20 ),

M=



5 3

mαa(c2 − u2 ),

where  Ax u= , α

c=

3 + 2e20 . 5

Over time, the orbit slowly rotates in the xz-plane and turns into a segment whose angle with the x-axis is    3 − 3e20 ψ = −arctan 2/3 k , k = , 3 + 2e20

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

135

during the time 1 τ= 3β



mα 10a3

c e0

du  . (u2 − e20 )(c2 − u2 )

The integral in the previous expression is reducible to the complete elliptic integral of the first kind (see footnote in the solution of problem 1.7) τ=

Tα K(k), √ 6π 10 cβa3

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 ∼ 40 000 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. the time at which its minimum distance becomes equal to the radius of the Earth) is of the order of τ . The estimate is valid for the relation: α/β ∼ (m1 /m2 ) R3 , where m1 and m2 are the respective masses of the Earth and the Moon, and R is the distance from the Earth to the Moon. From here we get the estimate τ∼

T22 , T

2π T2 =  , 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 high altitudes. 2.43. Here we will use the geocentric coordinate system. Let’s assume the strait line between the Earth and the Sun as the z-axis, and let the x-axis be in the plane of the Earth’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 of motion is the energy E = 12 mv2 −

α + δU − 12 m2 r 2 , r

where α = Gm1 m, G is the gravitational constant and m1 is the Earth’s mass. The force δF = −

∂δU ∂r

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[2.43

and the force of inertia Fi = m2 r + 2m[v, ] leads to the distortion of the elliptical orbit of the Moon, while the major semi-axis of the ellipse a = α/(2|E|) stays almost unchanged. ˙ 1 and A ˙ 2 that The rate of change of the Laplace vector A consists of two terms: A ˙ 1 has already been found correspond to δF and Fi (cf. problem 2.41). The term A in problem 2.41a: A˙1x = 92 ζ Az ,

A˙1z = 3ζ Ax ,

where ζ = Ma/α. Since the eccentricity of the Moon’s orbit is small e = 0.055, we have M ≈ ma2 ω,

α ≈ mω2 a3 ,

ζ≈

 29.5 1 ω≈ ≈ , ω 365 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 − (note that the orbit would be fixed in 0x0 yz0 frame of reference in the absence of the term δU): A˙2x = −Az ,

A˙2z = Ax .

Thus1   A˙x = − 1 − 92 ζ Az ,

A˙z = (1 + 3ζ )Ax .

(1)

Integration of (1) gives Ax = B cos( t + φ),

  Az = B 1 + 15 ζ sin( t + φ), 4

(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 0x0 yz0 , it rotates with the angular velocity (so-called precession) prec =  −  = 34 ζ . 1 Note that according to (1)

AA˙ = Ax A˙x + Az A˙z = 7.5 ζ Ax Az . 2 ˙ Using the relations A = αe, e = 1 − M 2 /(maα), we can estimate ζ˙ = aM/α ∼ 7.5 e2 ζ 2   . The value of ζ in (1) can be considered constant.

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137

The small changes of |A| defined by (2) correspond to the small ripples of the orbit’s eccentricity. 2.44. The Lagrangian function of the system (q is the particle charge and c is the velocity of light) L = 12 mv2 +

α q [μ, r] + v r c r3

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 are the following q [M, μ], mcr 3 . ˙ = q [A, μ] + 3q(M, μ) [M, r] A mcr 3 m2 cr 5

˙ = M

When averaging over the period T of unperturbed motion, these equations describe the systematic change of vectors M and A: ˙ = [ , M], M ˙ = [, A], A

q μ, cma3 (1 − e2 )3/2 ( M)  =  − 3M , M2

 = −

(1) (2)

where a and e are again the semi-major axis and the eccentricity of the unperturbed ellipse. 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 given based on (1) and (2): in the coordinate system, rotating with the frequency  , the vector M as well as the plane of motion of a particle are stationary, while the vector A and the perihelion of the orbit revolve with the constant frequency  −  around the direction M. We also indicate that it is helpful to perform the averaging of 1/r 3 and r/r 5 using the angle ϕ instead of the variable t: 

 T 2π 2π 1 1 m m dt dϕ 2π m = = = (1 + e cos ϕ) dϕ = , T TMp r3 r 3 (t) TM r(ϕ) TMp 0

0

0

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r = βA, r5

1 β= 2 A T

T 0

Ar m dt = ATM r5

2π

[2.45

2π me cos ϕ dϕ = . r 2 (ϕ) TMp2 A

0

2.45. When averaging the equation for the Laplace vector (see (10) of problem 2.40) ˙ = 1 [F, M] + [v, [r, F]] A m we take into account that F = 0 and that, according to the unperturbed equations of motion, m¨v =

d d  r αv 3αr(rv) mv˙ = −α 3 = − 3 + . dt dt r r R5

It gives (see the preceding problem)     αβ [v, M] αβ A αr 3αβ ˙ A = − 2 = 2 3 + 4 =− 2 3 A. 3 m r m r r 2m a (1 − e2 )3/2 The rate of change of the vector A is directed in the direction opposite to the vector A itself. 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 of the eccentricity. It can also be shown (see [2], § 75, problem 1) that a particle will fall on the centre over a finite time due to the energy loss.

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§3 Scattering in a given field. Collision between particles

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 of revolution at the point of collisions. We have therefore tan (θ/2) =

θ

b z dρ = cos . dz a a

z

Figure 116

Hence we have ρ 2 = b2 − a2 tan2 (θ/2) and thus dσ = π |dρ 2 | = π a2 tan (θ/2)

dθ a2 d = . cos2 (θ/2) 4 cos4 2θ

The possible deflection of the particle lie within the range of angles from zero, as ρ → b, to θm = 2 arctan(b/a), as ρ → 0. As a result, ⎧ ⎪ ⎨

a2 , dσ = 4 cos4 (θ/2) d ⎪ ⎩0,

when 0 < θ < θm , when θm < θ.

A2/(1−n) b) dσ = [n cot (θ/2)](1+n)/(1−n) . d 2(1 − n) sin θ cos2 (θ/2)

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

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Exploring Classical Mechanics

 ⎧ √   ⎪ 2a b − a tg(θ/2) ⎨ dσ  , when 0 < θ < θm = 3/2 4 sin θ cos(θ/2) d ⎪ ⎩ 0, when θ > θm

c)



where θm = 2 arctan b2 /a2 . 3.2. It is the paraboloid of revolution ρ 2 = αz/E. The reader should check whether the trajectories of the particles scattered by the field U(r) = −α/r and by the paraboloid approach one another as r → ∞. 3.3. At E > V ⎧



⎪ a2 n2 n cos 2θ − 1 n − cos 2θ a2 ⎪ ⎨ + , when 0 < θ < θm , dσ

2 4 = 4 cos 2θ 1 + n2 − 2n cos 2θ d ⎪ ⎪ ⎩ 0, when θm < θ < π , where n=

 1 − (V /E),

θm = 2 arccos n.

Unlike the case of scattering on a potential well (see [1], § 19, problem 2), here it is necessary to take into account the possibility of the elastic reflection at impact parameters a n < ρ < a. ⎧  3.4. a) α2 β α2 ⎪ ⎪ ⎪ − , π , when E > ⎨ E 4E 2 4β σ= ⎪ α2 ⎪ ⎪ when E < . ⎩0, 4β How does the cross-section change when α changes sign? b)

⎧  γ β ⎪ ⎪ ⎪π 2 − , ⎨ E E σ= ⎪ ⎪ ⎪ ⎩0,

when E >

β2 , 4γ

when E
r0 , then all particles for which rmin  R will hit the sphere and the cross-section will be  α  σ = πρ 2 (rmin = R) = π R2 1 + . ERn If R < r0 , only those particles which fall into the centre of the field will hit the sphere, and in that case   π n (n − 2)α 2/n σ = πρ02 = . n−2 2E b) If γ > ER4 the cross-section is  β γ − σ =π 2 E E

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√ provided 2 γ E > β; if γ < ER4 then  σ = πR

2

γ β 1+ − ER4 ER2



√ √ provided 2 γ E > β or provided both 2 γ E < β and R < R1 or R > R2 , otherwise σ = 0; here  β R21,2 = ∓ 2E

β2 4E

2

−

γ . E

3.6. Let us take the origin of the coordinate system in the centre of the planet direct the x-axis parallel to v∞ and the y-axis parallel to the impact parameter ρ. The motion of meteorites occurs in the field U(r) = −

Gm0 m α =− , r r

where G is the gravitational constant. The orbit of the meteorite in the polar coordinates is p M2 ρ2 r(ϕ) = , p= = , e= 1 + e cos(ϕ − ϕA ) mα ξR

 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  ρ r = α e (cos ϕA , sin ϕA ). A = [v, M] − α = α e 1, r ξR The polar coordinates of the meteorites falling on the planet satisfy the equation r(ϕ) = R at ϕA < |ϕ| < π . In this case, the impact parameter of such meteorites is less than the maximum √ ρmax as defined by the equation rmin = r(ϕa ) = R, from which we obtain 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 interval ρmin < ρ < ρmax . The corresponding impact parameter ρmin is determined by the equation r(ϕ = π/2) = R, from which we obtain the equation 2 /(ξ R) ρmin = R. 1 + ρmin /(ξ R)

Hence ρmin =

1 2

   R 1 + 1 + 4ξ .

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3.7]

§3. Scattering in a given field

143

Thus, the fraction of particles falling on the invisible part of the planet is equal to √ 2 − πρ 2 πρmax 1 1 + 4ξ min δ= . = − 2 2 2(1 + 2ξ ) πρmax At high energy of the flying particles, ξ  1, this fraction is small (ξ 2 ). With decreasing the energy of flying particles, the value of δ increases as well, reaching the value of δ = 1/2 at ξ  1. 3.7. a) dσ = d

4RE(RE + α) . R2 (1 + λ) 2 , where λ = α2 θ 4 1 + λ sin2 2 

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 the impact parameter. The motion of the particle in the region r  R is a harmonic oscillation along each of the coordinates, and in the initial moment of this oscillation (for t = 0)  y0 = ρ, y˙0 = 0, x0 = − R2 − ρ 2 , x˙0 = v, thus  v x = − R2 − ρ 2 cos ωt + sin ωt, ω

y = ρ cos ωt.

(1)

The moment of release of the particle from action of the forces is determined by the equation x2 + y2 = R2 ,

(2)

and the angle θ between the particle velocity at this moment and the x-axis by the equation sin θ =

y˙ ρω =− sin ωt. v v

Substituting (1) and (3) into (2), we obtain ρ 4 − ρ 2 R2 (1 + λ sin2 θ) + 14 R4 (1 + λ)2 sin2 θ = 0, where λ = v2 /(R2 ω2 ). As a result, 2 ρ1,2

 R2 2 = (1 + λ sin θ ∓ cos2 θ − λ2 sin2 θ cos2 θ), 2

(3)

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[3.8

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and the cross-section dσ = π(|dρ12 | + |dρ22 |) = π d(ρ12 − ρ22 ) =

R2 (1 + λ2 cos 2θ)d  . 2 1 − λ2 sin2 θ

If λ > 1, the scattering is only possible at angles smaller than θm = arcsin

R2 ω 2 v2

,

and at θ → θm , the differential cross-section dσ /d increases indefinitely. Such behaviour 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 scattering of light by drops of water. See also problems 3.9 and 3.11 as examples of the rainbow scattering. ⎧ ⎨ a2 , when θ ≤ θ = π V , dσ m 2E = θ2 d ⎩ 0,m when θ > θ . m

3.8. a)

b)

c)

dσ = d



π a2 V 4θm θ sin(π θ/θm ), when θ ≤ θm = π 2E , 0, when θ > θm .

⎧ 2 2 ⎪ ⎨ a θm (θm − θ) , dσ = θ 3 (2θm − θ)2 d ⎪ ⎩0,

when θ  θm = π

V , E

when θ > θm .

d) When calculating the scattering angle (see [1], § 20 and [7], § 6.2) ∞ θ= −∞

 Fy dx V ρ R2 − ρ 2 =2 2E E R2

we take into account that the force is ⎧ 2Vy 2V ρ ⎪ ⎨ = 2 , 2 R Fy = R ⎪ ⎩0,

  when − R2 − ρ 2 < x < R2 − ρ 2 ,  when |x| > R2 − ρ 2 .

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3.9]

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§3. Scattering in a given field

From here we find 2 ρ1, 2=

 1 2 R (1 ∓ 1 − θ 2 /θm2 ), 2

where θm = V /E, so thus dσ = π(|dρ12 | + |dρ22 |) = π d(ρ12 − ρ22 ). Finally, ⎧ ⎪ ⎨

R2  , dσ = 2θm θm2 − θ 2 d ⎪ ⎩ 0,

when 0 < θ < θm , when θ > θm .

Compare this result with the result of problem 3.7b, which was obtained for the field that is different from the given by a sign. dσ dΩ

θ θm ρ12

ρ22 ρ32

ρ2 θm

Figure 118

θ

Figure 119

3.9. One can easily evaluate the angle of deflection of a particle, using a general formula,1 and we have    3πβ π α  . (1) − θ =  4Eρ 4 2Eρ 2  The function θ(ρ 2 ) is shown in Fig. 118. From (1) we find ρ12

πα = 4Eθ



 θ 1+ −1 , θm

2 ρ2, 3

πα = 4Eθ

π α2 . where θm = 12Eβ 1 It is simplest to use both terms of (1) from [1], § 20, problem 2.





 θ 1∓ 1− , θm

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[3.10

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For the cross-section, we have dσ = π(|dρ12 | + |dρ22 | + |dρ32 |) = π d(−ρ12 + ρ22 − ρ32 ) =  2 − θ/θm πα 1 + θ/(2θm ) + − 1 d. = √ √ 8Eθ 3 1 + θ/θm 1 − θ/θm

(2)

This result is valid provided each of the terms in (1) is much less than unity. An estimate shows 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 |dρ12 | =

πα 8Eθ 3



1 + θ/(2θm ) − 1 d. √ 1 + θ/θm

Fig. 119 shows how dσ /d depends on θ. The differential cross-section becomes infinite both as θ → 0 and as θ → θm . The total cross-section for scattering into a range of angles adjoining θ = 0 is infinite as the small-angle scattering corresponds to large impact parameters. The total cross-section for scattering of particles into a range of angles θm − δ < θ < θm , θm 2π θm −δ

dσ π 2 αδ 1/2 θ dθ = , 3/2 d 2Eθm

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 π , θ =π− 1 − a2 /ρ 2

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 scattering centre (Fig. 120a). The observed angle of deflection χ lies in the range 0 < χ < π and satisfies the relation −θ = 2π l ± χ,

(2)

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3.10]

147

§3. Scattering in a given field (a)

(b)

ρ(χ,0,+)

ρ(χ,1,+)

π χ

z

χ

ρ(χ,1,–)

θ,χ χ a

ρ(χ,1,+)ρ(χ,1,–)

ρ

ρ (χ,0,+)

θ

–π –2π –3π –4π

Figure 120

where l = 0, 1, . . . corresponds to the upper sign in (2) and l = 1, 2, . . . to the lower sign in (2) (Fig. 120b). From (1) and (2), we have ρ 2 (χ, l, ±) = a2 +

π a2 2



1 1 − 2π l ± χ 2π l + 2π ± χ

.

The cross-section is dσ = π

∞ 

|dρ 2 (χ, l, +)| + π

l=0

∞ 

|dρ 2 (χ, l, −)|.

l=1

Using the fact that dρ 2 (χ, l, +) < 0, dχ

dρ 2 (χ, l, −) > 0, dχ

we find that    ∞ ∞  ρ 2 (χ, l, +) + ρ 2 (χ, l, −) = dσ = π d − l=0



= 12 π 2 a2 d

l=1

1 1 − 2π − χ χ



=

π a2 (2π 2 − 2π χ + χ 2 ) d. 2χ 2 (2π − χ)2 sin χ

The cross-section dσ /d turns out to infinity as χ → π . This leads to the fact that the cross-section for scattering in a small finite solid angle   =

π π−χ0

2π sin χ dχ = π χ02

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[3.11

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is equal to a2 σ = 2π



d 2 2  = a χ0 = a . χ π

π π−χ0

The appearance of such features of the differential cross-section is because the angle of deviation equal to π is reached at the values of the impact parameter ρ(π , l, ±), which are nonzero. In a solid angle , proportional to the square of a small value χ ≡ χ0 , flying particles fall before scattering through sites 2πρρ, areas which are proportional to the first 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 it is masked by an infinite cross-section due to scattering of particles with the arbitrarily large 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 the particle is deflected during the scattering is small. The change in momentum is (see [7], § 6.2) ∂ p = − ∂ρ

∞ −∞

√ 2 π V −x2 xe , U(|ρ + vt|) dt = v

where x = ρ. The angle of deflection is θ=

p √ V −x2 = π xe . p E

(1)

We cannot solve this equation for x in analytic form. However, from thegraph of the 2 π. function xe−x (Fig. 121), we see that (1) has two roots when θ < θm = VE 2e dσ dΩ 2

xe –x 1 √2e Eθ V√ π

x1

1 √2

x2

Figure 121

x θm

Figure 122

Using (1) and the relation dθ =

√ V 2 π (1 − 2x2 )e−x dx, E

θ

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§3. Scattering in a given field

we can write the expression for the cross-section dσ = π(|dρ12 | + |dρ22 |) =

2π (x1 dx1 − x2 dx2 ) 2

in the form dσ 1 = 2 2 d  θ



x22 2x22 − 1

+



x21 1 − 2x21

.

When θ  θm , then x1  1 and x2  1, so that 1 dσ = . d 2 2 θ 2 When θm − θ  θm , we can solve (1) by expanding xe−x in a series near the maximum. We then get 2

x1, 2 =

1∓

√ 1 − θ/θm , √ 2

dσ 1 . = √ 2 2 d 2 θm 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 is connected with the approximations made in the solution of the problem?  x1 + x21 x2 + x22 x1, 2 1 2Eθ 2 . dσ = 2 2 1 − 2x + 2x − 1 , where = b) πV d  θ 1 2 (1 + x1, 2 )3 π V we have When θ  θm = √ 3 3E πV dσ = . d 4 2 Eθ 3 When θm − θ  θm we have

√ 3 dσ . = √ √ d 2 2 2 θm2 1 − θ/θm

3.12. a) A particle with a velocity v before the collision will have a velocity v = v − 2n (nv) after the collision, where n is a unit vector which is normal to the surface of the ellipsoid. Using the relations1 1 We know from differential geometry that

n ∝ grad



 x2 y2 z2 + + − 1 , a2 b2 c2

while the value of N is determined by the relation n2 = 1.

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[3.12

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v = v (0, 0, 1),

n=

1 x y z , , , N a2 b2 c2

we get  2xz 2yz 2z2 v = v − 2 2 2 , − 2 2 2 , 1 − 2 4 . N a c N b c N c

(1)

Introducing spherical coordinates with the z-axis along v we can write v = v (sin θ cos ϕ, sin θ sin ϕ, cos θ), and from (1) we then have tan ϕ =

a2 y , b2 x

cos θ = 1 −

2z2 , N 2 c4

 sin2 θ =

2z N 2 c2

2 

x2 y2 + . a4 b4

(2)

For the cross section, we have    ∂(x, y)    dθ dϕ, dσ = dx dy =  ∂(θ, ϕ)  where the dependence of x and y on θ and ϕ follows from (2) and from the equation for the ellipsoid. To evaluate the Jacobian, it is helpful to introduce an auxiliary variable u such that x = a2 u cos ϕ,

y = b2 u sin ϕ.

From (1) and (2), it then follows that sin θ =

2zu , N 2 c2

1 − cos θ =

2z2 , N 2 c4

tan (θ/2) =

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) ∂(x, y) ∂(u, ϕ) a2 b2 ∂u2 = = . ∂(θ, ϕ) ∂(u, ϕ) ∂(θ, ϕ) 2 ∂θ

z uc2

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3.13]

§3. Scattering in a given field

151

We thus get, finally, dσ = d

a2 b2 c2  2 . 4 cos4 2θ a2 cos2 ϕ + b2 sin2 ϕ + c2 tan2 2θ

What is the limit which we must take to obtain from this result the cross-section for scattering by a paraboloid? a2 b2 c2 . b) dσ = 3 2 2 d cos θ(a cos ϕ + b2 sin2 ϕ + c2 tan2 θ)2 c) dσ = d



a2 b2 c2 cos θ

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

∞ −∞

π(aρ) U(ρ + vt)dt = − ∂ . ∂ρ vρ

(1)

Let the z-axis be parallel to v and the y-axis be perpendicular to a. We then have px = −

π ax ρy2 , v ρ3

py =

π ax ρ x ρ y . v ρ3

(2)

The direction of the velocity after scattering can be characterized by two spherical angles tan ϕ =

py , px

θ=

p . p

(3)

It is clear from (2) that scattering only occurs when 1 2

π < ϕ < 32 π .

From (2) and (3), we find ρx = ±

π ax sin ϕ cos ϕ , 2E θ

ρy = ∓

π ax cos2 ϕ . 2E θ

(4)

For the cross-section, we find dσ =



   π a 2 cos2 ϕ   ∂(ρx , ρy )  x dρx dρy = d.   dθ dϕ =  ∂(θ , ϕ)  E 2θ 4

(5)

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The summation in (5) addresses the two possible values of ρ (see (4)). |a | b) dσ = ⊥ 3 , where a⊥ is the component of a perpendicular to v∞ . The crossd 2Eθ section turns out to be symmetrical with respect to v∞ (although the field is by no means symmetric with respect to this directions.) 3.14. The change in the angle of deflection of the particle is given by the equation (see problem 2.20) 1 δθ(ρ) = − ∂ E ∂ρ

∞

rmin

δU(r) dr

ρ2 1 − 2 − U(r) E r

.

(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 expressions obtained when δU(r) = 0. We then get for the cross-section    2  dρ 2 (θ )   dρ  d dρ (θ ) dσ 0   =π 0 = π  − (θ )δθ (ρ (θ )] [2ρ = 0 0   dθ dθ d(θ) dθ dθ    dσ0 dσ0 d = − δθ(ρ0 (θ )) . dθ dθ dθ   πβ a) δ dσ = − E d π − θ − sin θ ; sin θ dθ dθ   2π γ b) δ dσ = − α d 32 (π − θ)[1 + tan2 (θ/2)] − 3 tan(θ/2) − sin θ ; dθ dθ   4γ d (π − θ)2 . c) δ dσ = −  dθ π βE dθ θ(2π − θ) 3.15. The energy acquired by the particle, ε=

(p + p)2 p2 − ≈ vp, 2m 2m

is to the first order determined solely by the change of the longitudinal component of the momentum. As we assume that the deflection of the particle is small, we can (after differentiating) put r = ρ + v(t − τ ) in the expression for the force

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F=−

∂U(r, t) , ∂r

acting upon the particle. Here ρ is the impact parameter and τ the time when the particle is at its smallest distance from the centre. Therefore, we have  ε=v εm =

∞ −∞

F(t)dt = εm e−

2 ρ 2 cos ϕ

√ ω −ω2 /(4 2 v2 ) π V2 e , v

,

ϕ = ωτ ,

and the scattering cross-section for particles with a given value of τ when cos ϕ > 0 (cos ϕ < 0) is ⎧ π dσ ⎨  2 |ε| , = dε ⎩ 0,

when 0 < ε < εm cos ϕ

(0 > ε > εm cos ϕ),

when |ε| > εm | cos ϕ|.

In the incident beam, there are particles with different values of τ . If we average the cross-section over the phase ϕ, for ε > 0 using, for example, the formula 

 α dσ 1 dσ = dϕ, dε 2π dε

α = arccos

−α

we get 

⎧ ⎨ 1 arccos |ε| , dσ εm =  2 |ε| ⎩ dε 0, 

ε , εm

when |ε| < εm , when |ε| > εm .

3.16. The distribution of decay particles is dN λ2 sin θ dθ = , √ N cos3 θ 1 − λ2 tan2 θ

λ=

V 2 − v20 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 dN 6(Tmax − T )(T − Tmin ) dT , = N (Tmax − Tmin )3

Tmin  T  Tmax ,

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[3.18

where Tmin = 12 m(v0 − V )2 ,

Tmax = 12 m(v0 + V )2 .

tan θ1 = cot θ2 = α/(Eρ),

E = 12 mv2 ,

3.18.

Eρv v1 =  , 2 α + E2ρ2

αv v2 =  . 2 α + E2ρ2

3.19. 1 2

π  θ  π , when m1 < m2 , θ = 12 π , when m1 = m2 ,

0  θ  12 π , when m1 > m2 . υ0 χ ρ

θ

υ0

υ

3.20. The velocity component normal to the sphere’s surface at the point of impact becomes zero in the centre-of-mass system while the tangential component v0 is conserved (see Fig. 123). The scattering cross-section as function of the angle of deflection χ of the particle in the centre-of-mass system is dσ = π |dρ 2 | = 4a2 π |d cos2 χ| = 4a2 cos χ d.

Figure 123

To transform to the laboratory system, we use the equation tg θ =

v0 sin χ sin χ cos χ =  v0 cos χ + v0 1 + cos2 χ

to find the equation cos2 χ1, 2 =

3 2

 cos2 θ − 1 ± 12 cos θ 9 cos2 θ − 8.

Taking into account that there are two possible connections between χ and θ, we obtain dσ = 4π a2 (|d cos2 χ1 | + |d cos2 χ2 |) = 5 − 9 sin2 θ = 4π a2 d(cos2 χ2 − cos2 χ1 ) = 4a2  d, 1 − 9 sin2 θ where 0 < θ < arcsin(1/3).

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3.23]

§3. Scattering in a given field

155

If the spheres impinging upon those at rest are identical so that there are no means of distinguishing between the two after the collisions, we must add to the cross-section which we have just obtained the cross-section dσ  = 4a2 cos θ d (0 < θ < 12 π ), 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−nσ x . 3.22. dN = σ n1 n2 |v1 − v2 | dV dt.  π 2 3.23. a) Ffr = 2π mv n f (θ)(1 − cos θ) sin θ dθ;1 0  2 π m nl b) 2  = 2π M f (θ) sin3 θ dθ; 0

where l is the path travelled by a particle of mass M; v, its velocity; and n, the concentration of light particles.

 1 The quantity



section

dσ d. d

(1 − cos θ ) dσ d is called the transport cross-section—as distinct from the total crossd

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§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 the condition that x = a when t = τ , we find that B = τa − aτ . Substituting the function x(t) = At 2 +

a τ

 − Aτ t

into the action, we find τ S=

L(x, x) ˙ dt = 0

τ 

1 2

 mA2 τ 3 ma2 FAτ 3 Faτ mx˙2 + Fx dt = + − + . 6 2τ 6 2

0

F . It is From the condition that the action be a minimum, ∂S = 0, it follows that A = 2m ∂A clear that in the present case the law of motion   Ft 2 a Fτ x(t) = + − t 2m τ 2m is exact. However, the only thing the solution given here allows us to state is that this law is 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 than any other x(t), that is, that it is the true law, we must verify that it satisfies the Lagrangian equation of motion.

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

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4.4]

4.2.

§4. Lagrangian equations of motion

⎧ ⎨vx t − a, x(t) =

⎩ v2 − 2V /m (t − τ ) + a, x

157

when 0 < t < t0 , when t0 < t < τ ,

y(t) = at/τ , where vx = [2aV /(mτ )]1/3 and t0 = a/vx . 4.3. From the relation   ∂q ˙ ∂q ˙ ˙  L(Q, Q, t) = L(q(Q, t), q(Q, ˙ Q, t), t) = L q, Q+ , t ∂t ∂Q we get d ∂ L = ∂q d ∂L + ∂L d ∂q , ˙ ∂ q˙ dt ∂Q ∂Q dt ∂ q˙ dt ∂ Q ∂ L = ∂L ∂q + ∂L ∂ q˙ . ∂q ∂Q ∂ q˙ ∂Q ∂Q Bearing in mind that

∂q ∂ q˙ = d , we obtain dt ∂Q ∂Q d ∂ L − ∂ L = ∂q ˙ ∂Q ∂Q dt ∂ Q



 d ∂L ∂L − . ∂q dt ∂ q˙

(1)

The validity of the equation d ∂L − ∂L = 0 thus leads to the validity of the analogous dt ∂ q˙ ∂q equation d ∂ L = 0. L − ∂ ˙ ∂Q dt ∂ Q If there are several degrees of freedom, instead of (1) we get the following equations: d ∂ L − ∂ L = ∂qk ˙ i ∂Qi ∂Qi dt ∂ Q k

    dq dQ , τ = L q, , t dt . 4.4.  L Q, dτ dt dτ



 d ∂L − ∂L . dt ∂ q˙k ∂qk

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Here dq = dt dq ∂q dQ ∂q = + , ∂τ ∂Q dτ dτ q = q(Q, τ ),

x˙2 − (1 + λx) 4.5.  L = 12 m 1 + ˙ U(x), λx˙

dq dτ , dτ dt dt dQ ∂t = ∂t + . ∂τ ∂Q dτ dτ

x˙ = dx . dτ

 2 dq  . 4.6. L = − 1 − dτ The problem is a purely formal one. However, both this Lagrangian function and the transformation consider (“improved” by the introduction of dimensional factors) have a simple physical meaning in the theory of relativity (e.g. see [2], § 4 and § 8). 4.7. The generalized momentum Pl = ∂L and the energy ˙l ∂Q E =



˙l −L Pl Q

l

transform as Pl =

∂fk pk , ∂Ql

E = E −

k

k

pk

∂fk . ∂t

4.8. Applying the formulae of the preceding problem, we obtain a) pr = m˙r  = pr , pϕ = mr  2 (ϕ˙  + ) = pϕ , E  = E − pϕ ;  b) p = px cos t + py sin t, x

py = −px sin t + py cos t.

(1)

It follows from (1) that p = p, while (1) also gives the rules for the law of transformation of vector components when we are changing to a coordinate system which is rotated over an angle t. We emphasize that p = mv (cf. [1], § 39). 4.9. a) E  = E − Vp, p = p; b) E  = E − Vp + 12 mV 2 , p = p − mV.

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4.11]

§4. Lagrangian equations of motion

159

The two expressions for the energy differ by a constant. Usually, we employ the second formula, 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 conservation of energy is correct (taking into account the centrifugal energy): − 12 m2 a2 = 12 mv2r − 12 m2 l 2 , where vr is the velocity directed along the rod at the moment of the bead slipping off. From here we get v2r = 2 (l 2 − a2 ). In the laboratory reference of frame, the velocity vt directed across the rod should be added (vt = l) to this velocity vr . Thus, the value of the total velocity is v=

 v2r + v2t =  2l 2 − a2 .

The velocity of the slipping bead can be changed significantly if the end of the rod is bent 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 v2 + 2 r 2 + mv [, r ]. We direct the x -axis along rod a and introduce the angle ϕ, which determines the direction of rod b with respect to the direction of rod a (see Fig. 7). Substituting the vector r = (a + b cos ϕ, b sin ϕ) in the Lagrangian function, we obtain L(ϕ, ϕ, ˙ t) = 12 mb2 ϕ˙ 2 + mab2 cos ϕ   (after removing 12 m2 a2 + b2 + mϕb ˙ (a cos ϕ + b), representing the total derivative of the function of ϕ with respect to the time). The Lagrangian equation bϕ¨ = −a2 sin ϕ is identical to that of the pendulum oflength b in the field of gravity g = a2 , so the frequency 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) = mabωB cos ϕ, ωB =

eB , mc

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[4.12

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where e is the particle charge and c is the velocity of light. (This result occurs if we choose the vector potential A = 12 [B, r] and remove the terms representing the total derivative of 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

4.12. Let qi = gi (t) describe the motion of the system (trajectory AB in Fig. 124). As the form of the action is invariant when we change to the variables qi , t  , the motion is also describes by the equations qi = gi (t  ). If we express these equations in terms of the variables qi , t, we have up to first order in ε:

B B A

−δq

−δt

A t

qi (t − δt) = gi (t) − δqi ,

Figure 124

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 change in the action: 

B ∂S ∂S (−δqi ) . (−δt) + SA B − SAB = ∂t ∂qi i

A

Here we have (see [1], § 43) ∂S = L − ∂L q˙ = − E(t), ∂S = ∂L = p (t). i ∂t ∂ q˙i i ∂qi ∂ q˙i i

On the other hand, according to the conditions of the problem SAB = SA B , so that E(tA )εh(qA , tA ) −



pi (tA )εfi (qA , tA ) =

i

= E(tB )εh(qB , tB ) −

i

pi (tB )εfi (qB , tB ),

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4.15]

161

§4. Lagrangian equations of motion

or Eh −



pi fi = const.

i

The proved theorem is, in essence, a united derivation of the various conservation laws. The importance of this theorem increases if we take into account the fact that the same theorem takes also place in the field theory (Noether’s theorem, see [13] and [14]). 4.13.

∂L i

∂ q˙i

(q˙i h − fi ) − Lh − F = const.

4.14. a) The linear omentum; b) the angular momentum; c) the energy; h p = const, where h is the pitch of the screw; d) Mz + 2π z e) Ex − px t = 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 time the action remain unchanged under transformation in direction at right angles to F or under rotating around axis parallel to F. Integral of motion are thus the linear momentum components 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 the motion. The statement that different points in some region are “equal” means that the value of 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 the time is transformed as t  = βt simultaneously. The contribution of the kinetic energy to the action 

m 2  α2 v dt = 2 β



m 2 v dt 2

remains unchanged at β = α 2 , and the contribution of the potential energy  −

U(r )dt  = −α n β



 U(r)dt = −α n+2

U(r)dt

remains unchanged when n = −2. To use the theorem, formulated in problem 4.12, we write the infinitesimal similarity transformation as α = 1 + ε: r = (1 + ε)r,

t  = (1 + 2ε)t,

ε → 0,

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[4.16

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so thus f = R, h = 2t and the integral of motion reads ∂L (vh − f ) − Lh = 2Et − mvr = C. ∂v From (1) we can find r(t) taken into account that rv = r2 =

1 2

(1)

dr 2 : dt

2E 2 2 t − Ct + C1 . m m

(2)

If E < 0, the particle falls into the centre (in this case, r˙ → ∞). If E > 0, it is helpful to introduce other constants τ , B instead of C, C1 and write (2) as r2 =

2E (t − τ )2 + B. m

At B > 0 the √ dependence r(t) is the same as for the free√motion of particles with the velocity 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 + ce A(r), if A(αr) = α −1 A(r); e) E − pϕ  = const. 4.16. One can use the results of problem 4.15b (see (2), taking into account the initial conditions) r 2 (t) =

2E 2 t + 2 r0 v0 t + r02 . m

(1)

The fall time is determined from the quadratic equation r 2 (t) = 0, which gives two roots t1,2 =

  1  −mr0 v0 ± β , β = (mr0 v0 )2 − 2mEr02 . 2E

From (1) we can see that the fall must happen at E < 0; its time is determined by the positive root t2 . As for the case when E > 0, the fall is possible only at r0 v0 < 0 and β > 0. In this case, one should choose the smallest t2 among the two positive roots.

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4.19]

§4. Lagrangian equations of motion

163

Please note that in this problem the value M 2 − 2m(ar/r)2 is an integral of motion which coincides with the (−β) (see problem 10.26). Therefore, the equation for the radial motion E = 12 m r˙2 −

β r2

leads to the following fall time  t = −m

0 r0



dr 2mE − β/r 2

1 = 2E

 

 2 β + 2mEr0 − 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 around it lives the form of A, and hence also the form of the action, invariant. We have thus the following integrals of motion: eB pz = ∂L = mz˙ and Mz = xpy − ypx = m(xy˙ − yx) ˙ + (x2 + y2 ). ∂ z˙ 2c Moreover, the energy E = 12 m(x˙2 + y˙2 + z˙2 ) is an integral of motion. b) E = 12 m(x˙2 + y˙2 + z˙2 ), py = my˙ + eBx/c, pz = 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 = pz , Mz , and py are all integrals of motion, independent of the choice of A. 2 eμ 4.19. a) E = 12 m(˙r 2 + r 2 ϕ˙ 2 + z˙2 ), pϕ = mr 2 ϕ˙ + c 2 r 2 3/2 , where we use the cylin(r + z ) 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 = mr 2 ϕ˙ +

eμ , c

E = 12 m(˙r 2 + r 2 ϕ˙ 2 + z˙2 ).

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164

[4.20

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However, the motion in this “field” is free-particle motion. Indeed, the Lagrangian function, L = 12 mv2 +

eμ ϕ, ˙ c

differs from the Lagrangian function of a free particle only by the total derivative with respect to time of the function eμϕ/c (of course, in this case we have B = rot A = 0). We note that in the case when μ is a function of time, pz and Mz remain integrals of motion. 4.20. a) x¨ + x = 0. The same equation could have obtained from the Lagrangian function ˙ = x˙2 − x2 . It is well known that if two Lagrangian functions differ by a total L1 (x, x) derivative of a function of the coordinates and the time, they lead to the same Lagrangian equations of motion. The reverse statement is incorrect. b) x¨ + α x˙ + ω2 x = 0. 4.21. a) In the spherical coordinates the Lagrangian equations of motion for a particle moving in a field U are m(¨r − r ϕ˙ 2 sin2 θ − r θ˙ 2 ) + ∂U = 0, ∂r m(r 2 θ¨ + 2r r˙θ˙ − r 2 ϕ˙ 2 sin θ cos θ) + ∂U = 0, ∂θ 2 2 2 2˙ m(r ϕ¨ sin θ + 2r r˙ϕ˙ sin θ + 2r θ ϕ˙ sin θ cos θ) + ∂U = 0 ∂ϕ We can easily write them in the form ˙ i = (F)i , m(v) where the components of the force are the components of (− grad U): Fr = − ∂U , ∂r

1 Fθ = − ∂U , r ∂θ

Fϕ = −

1 ∂U . r sin θ ∂ϕ

Hence we have ˙ r = r¨ − r ϕ˙ 2 sin2 θ − r θ˙ 2 , (v) ˙ θ = r θ¨ + 2˙r θ˙ − r ϕ˙ 2 cos θ sin θ, (v) ˙ ϕ = r ϕ¨ sin θ + 2˙r ϕ˙ sin θ + 2r θ˙ ϕ˙ cos θ . (v) ˙ i= 1 b) (v) 2hi



d ∂ − ∂ dt ∂ q˙i ∂qi



  3 ds2 = h q¨ +  2q˙ q˙ ∂hi − q˙2 hk ∂hk . i i i k k hi ∂q ∂qk dt 2 i k=1

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4.23]

§4. Lagrangian equations of motion

165

4.22. a) The Lagrangian function 3

L = 12 m

gik q˙i q˙k − U(q1 , q2 , q3 ),

i, k=1

with gik =

3 ∂xl ∂xl , ∂qi ∂qk l=1

leads to the equations

m

3 k=1

gsk q¨ k + m

3 k, l=1

s, kl q˙k q˙l = − ∂U ∂qs

(s = 1, 2, 3),

(1)

where the  s, kl =

1 2

 ∂gsk ∂gls ∂gkl + − . ∂ql ∂qk ∂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 31 by 41 . 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 of the free-moving particles only by the term L1 = −

eg ϕ˙ cos θ c

the components of the force in the spherical coordinates (see problem 4.21a) have the form Fr = 0, 1 ∂L1 eg ϕ˙ sin θ = , r ∂θ cr 1 d ∂L1 eg θ˙ Fϕ = − = r sin θ dt ∂ ϕ˙ cr Fθ =

and coincide with the components of the Lorentz force

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Exploring Classical Mechanics

[4.24

e eg [v, B] = 3 [v, r]. c cr 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 = mr 2 ϕ˙ sin2 θ −

eg cos θ = Jz . c

In the given Lagrangian function, the direction of the z-axis can be chosen arbitrary; this leads to the conservation of the vector J = m[r, v] −

eg r . c r

Besides, the action corresponding to this Lagrangian function is invariant under the similarity transformation. Therefore, the integral of motion is pr r − 2Et = m˙r r − 2Et (cf. problem 4.15b). 4.24. The equations of motion are −L I˙ = ϕA − ϕB ,

q2 = ϕB − ϕA . C

We shall assume that the potential source is a capacitor with a very large capacitance C0 and that its charge at the moment when q1 = 0 is Q. The energy of the system including the potential source and the inductance is E0 =

(Q + q1 )2 1 + 2 L q˙21 . 2C0

Shifting the zero of the energy and considering the limit Q/C0 → U as C0 → ∞, we get E = E0 −

Q2 = Uq1 + 12 L q˙21 . 2C0

This is the form of the energy which leads to the Lagrangian function given in the problem. Similar to this, the energy of a particle of mass m in a uniform force field −F(t) is 1 2 m 2 x˙ + Fx. The same Lagrangian function may be obtained from that of an electromagnetic field including interactions between an electromagnetic field and charges (see [3], § 27 and § 28):

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4.25]

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§4. Lagrangian equations of motion

L=

1 8π

 (E2 − B2 ) dV +

1 c



 Aj dV −

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 specified by the charge q2 or the current q˙1 . Using equations c rot B = 4π j,

div E = 4πρ

(and given that the fields are concentrated inside a limited volume), we obtain  ϕρ dV =

1 4π

 ϕ div E dV = =

1 4π

 div(ϕE) dV −

1 4π

 E grad ϕ dV =

1 4π

 E2 dV ,

and similarly 1 c



1 Aj dV = 4π

 B2 dV ,

so that L=

1 8π

   B2 − E2 dV .

Therefore, the Lagrangian function can be expressed in terms of the electric field energy inside a capacitor as 1 8π

 E2 dV =

q22 2C

and in term of the magnetic field energy inside the inductance as 1 8π

 B2 dV = 12 L q˙21

(see [3], § 2 and § 33). 4.25.

a) L = 12 L q˙21 −

q22 + U(q2 − q1 ); 2C

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168 b) L = 12 L q˙2 −

q2 ; 2C

c) L = 12 L1 q˙21 + 12 L2 q˙22 − 4.26.

[4.26

Exploring Classical Mechanics

q21 q2 (q + q2 )2 − 2 − 1 . 2C1 2C2 2C

a) L = 12 ml 2 ϕ˙ 2 + 12 L q˙2 + mgl cos ϕ −

q2 ; 2C(ϕ)

b) L = 12 mx˙2 + 12 L (x)q˙2 − 12 kx2 + mgx −

q2 . 2C

4.27. Let ϕ be the angle of rotation of the frame around the AB-axis, such that ϕ = 0 gives the direction of the magnetic field; let q˙ be the current in the frame (the positive direction is the one from A to D). The Lagrangian function for the system is L = 12 ma2 ϕ˙ 2 + 12 L q˙2 + Ba2 q˙ sin vfi. The integrals of motion are the energy, E = 12 ma2 ϕ˙ 2 + 12 L q˙2 ,

(1)

and the momentum conjugate to the cyclic coordinate q which is associated with the total magnetic flux through the frame, ∂L = L q˙ + Ba2 sin ϕ =  . 0 ∂ q˙ The current through the frame is thus uniquely determined by its position: q˙ =

0 − Ba2 sin ϕ . L

Substituting this value q˙ in (1) we obtain 

E = ma ϕ˙ + Ueff (ϕ), 1 2

2 2

0 − Ba2 sin ϕ Ueff (ϕ) = 2L

2 .

The problem of system’s motion is thus reduced to a one-dimensional one. Umax

Ueff Um

−π 2

π 2

π

Figure 125

3π 2

ϕ

(2)

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4.28]

§4. Lagrangian equations of motion

169

Let us consider the case 0 < 0 < Ba2 in more detail. The function Ueff (ϕ) for this case is given in Fig. 125. One sees that when E > Umax = (0 + Ba2 )2 /(2L ) the frame rotates and ϕ˙ is a periodic time function with period T=

π/2

√

2ma −π/2

dϕ . √ E − Ueff (ϕ)

When Umax > E > (0 − Ba2 )2 /(2L ) = Um , the frame performs periodic oscillations within the angular interval ϕ1 < ϕ < π − ϕ1 , where √ 0 − 2L E ϕ1 = arcsin ; Ba2 the period tends to infinity as E → Umax (see problem 1.5). When 0 < E < Um one can have oscillations either in the interval ϕ1 < ϕ < ϕ2 or in the interval π − ϕ2 < ϕ < π − ϕ1 , where √ 0 + 2L E . ϕ2 = arcsin Ba2 How will the character of the frame’s rotation change if one assume it to have a small resistance? 4.28. a) The equations of motion for the system can be obtained using a Lagrangian function with an additional term responsible for the constraints (see [5], § 2.4) L∗ = 12 m(x˙2 + z˙2 ) − mgz + λ(z − ax2 ), where λ is a time-dependent Lagrangian multiplier. The equation of motion m¨x = −2 lmax, m¨z + mg = λ

(1) (2)

together with the equation of constraints z = ax2 completely determine the motion of the particle. On the right-hand side of (1) and (2) there are the components of the reaction forces along the two axes: Rx = −2λax and Rz = λ. They can be rewritten in terms of the coordinate and velocity of the particle with the equation of constraints as Rx = −2axRz ,

Rz =

(2ax˙2 + g)m . 1 + 4a2 x2

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[4.29

m¨r − mg cos ϕ − mr ϕ˙ 2 = λ,

b)

mr 2 ϕ¨ + 2mr r˙ϕ˙ + mgr sin ϕ = 0, r = l. The reaction force, λ = −mg cos ϕ − ml ϕ˙ 2 , lies along r. 4.29. L∗ = 12 m(r 2 ϕ˙ 2 + r˙2 ) + mgr cos ϕ + λ(ϕ − t); λ = 2mr r˙ + mgr sin t is the generalized force corresponding to the coordinate ϕ (the moment of the torque force). q˙i Ri . 4.30. a) dE = − ∂L + dt ∂t s

i=1

b) The law of the transformation of the left-hand sides of the equations of motion is given in problem 4.3:   d ∂ L = ∂qk d ∂L − ∂L . L − ∂ ˙ i ∂Qi ∂Qi dt ∂ q˙k ∂qk dt ∂ Q k The same law of transformation must be for the right-hand sides:  Ri =

∂qk Rk . ∂Qi

(1)

k

If the law of transformation of coordinates is not included evidently the time, the velocities are transformed according to the law q˙i =

∂qi ˙ k, Q ∂Qk k

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 , the transformation ∂ q˙i of F is reduced to a change of variables.

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4.32]

§4. Lagrangian equations of motion

171

4.31. The equations stated in this problem can be obtained by eliminating the λβ from the following equations: d ∂L − ∂L = λβ , dt ∂ q˙β ∂qβ

β = 1, . . . , r,

d ∂L − ∂L = − λβ bβn , dt ∂ q˙n ∂qn r

n = r + 1, . . . , ., s.

β=1

The following relations must be taken into account: ∂ L = ∂L + ∂L b , ∂ q˙n ∂ q˙n ∂ q˙β βn r

∂ L = ∂L + ∂qn ∂qn

β=1 r

s ∂L ∂bβm q˙ . ∂ q˙β ∂qn m

β=1 m=r+1

Thus, the equations of motion for a system with non-holonomous constraints differ from the Lagrangian equations of motion although the constraints equations permit one to eliminate 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 the Lagrangian equations of motion d ∂Ln ∂Ln+1 ∂Ln ∂Ln = + − ∂qn ∂(qn ) ∂(qn+1 ) dt ∂qn

(qn = qn − qn−1 ),

(1)

whence we find as a → 0 1 ∂Ln → ∂L , ∂(∂q/∂t) a ∂ q˙n

1 ∂Ln → ∂L , ∂q a ∂qn

  1 ∂Ln+1 ∂Ln ∂L , − →− ∂ ∂x ∂(∂q/∂x) a ∂(qn ) ∂(qn+1 ) so that the equations (1) are reduced to the following equations: ∂L ∂ ∂L + ∂ = ∂L . ∂t ∂(∂q/∂t) ∂x ∂(∂q/∂x) ∂q

(2)

Here the derivatives ∂/∂t and ∂/∂x should be applied to the functions q(x, t) and its derivatives.

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[4.33

The system of N ordinary differential equations (1) thus transforms to one equation in the partial derivatives (2). For a continuous system the variable x indicates a fixed point on the string. We do not consider the physical consequences of the (2) since systems with an infinite number of degrees of freedom are the subject of field theory, but not of mechanics (see [5], Ch. 11; [2], § 26 and § 32).    ∂ L ∂q − L dx. b) E = ∂(∂q/∂t) ∂t 4.33. The Lagrangian function is e L = 12 mv2 − U(r) + A(r)v, c where A(r) is the vector potential of the magnetic field, B = rot A (of course, A(r) can always be taken as a homogeneous function of the coordinates of degree n + 1). If as a result of the similarity transformation, r → αr,

t → α 1−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 → α k L. Therefore, the equations of motion remain invariant after 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 field if it is constant in space and if its value changes by a factor α −1+k/2 under a similarity transformation (e.g. see, problems 2.34– 2.37 and 6.37).  4.34. The kinetic energy of the system is T = a 21 ma v2a , so that   ∂T d va = ma va ra − ra ma v˙ a . 2T = ∂va dt a a a   The first term d ma va ra , which is the total time derivative with respect of time a dt of a bounded function becomes zero after averaging over large time interval (see [1], § 10 and [7], § 7). Substituting ea ma v˙ a = − ∂U + [va , B] ∂ra c into the second term and averaging over the time we obtain  2T + B

ea a

c

 [ra , va ] = k U ,

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4.35]

§4. Lagrangian equations of motion

173

where the pointed brackets indicate the time averaging. In particular, if the magnetic field B is homogeneous and constant, then 2 T = k U − 2 μ B, where 

1 ea [ra , va ] μ = 2c a



ea = e , then μ = e M, is the average magnetic moment of the particle system. If m m 2mc a where M is the angular momentum of the system. 4.35. a) Write dA in the form of two terms dt dA = [m(¨x1 x˙2 x˙3 + x˙1 x¨ 2 x˙3 + x˙1 x˙2 x¨ 3 ) − x˙1 U˙ 23 − x˙2 U˙ 13 − x˙3 U˙ 12 ]− dt − (¨x1 U23 + x¨ 2 U13 + x¨ 3 U12 ).

(1)

Using the equations of motion m¨x1 = F12 + F13 ,

m¨x2 = F21 + F23 ,

Fik = −Fki = −

m¨x3 = F31 + F32 ,

2g 2 ∂Uik = ∂xi (xi − xk )3

(2) (3)

and introducing the relative distances x1 − x2 = x,

x2 − x3 = y,

x1 − x3 = z,

write the second term of (1) as 2g 4 − m



1 1 + 3 3 x z



     1 1 1 1 1 1 1 + − 3+ 3 − 3+ 3 . y2 x y z2 z y x2

(4)

Combine the terms with the same powers of z, we rewrite (4) as 2g 4 (x − y) m



 x2 + xy + y2 x+y 1 − − . x3 y3 z2 x3 y3 z3 x2 y2

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

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U˙ ik =

[4.36

∂Uik (x˙ − x˙k ) = −(x˙i − x˙k )Fik . ∂(xi − xk ) i

Finally, note that in this field the similarity transformation does not change an action, and, as a consequence, the is the fourth (additional) integral of motion (see problem 4.15b) m(x1 x˙1 + x2 x˙2 + x3 x˙3 ) − 2Et.

(5)

4.36. When two particles approache each other, their interaction energy becomes infinite. As a result, the particles can not pass “one through the other” and the order of their location on the line is conserved. When two particles of equal mass but with arbitrary interaction energy (only allowing the expulsion of particles) collide, these particles simply exchange velocities. (This is follows from the laws of conservation of energy and momentum.) If the collisions of three particles occur one after another, so that during the convergence of two particles the 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 is behind, in other words, in this case v1 = v3 ,

v2 = v2 ,

v3 = v1 .

(1)

In general, when all three particles simultaneously converge, the values of the velocities will not be conserved. The more surprising is that for the forces specified in the preceding problem, the result (1) is conserved. This can be shown using all three integrals of motion: P, E, and A. Taking into account that the functions Uik → 0 as t → ±∞ and comparing P, E, and A at t → +∞ and at t → −∞, we obtain three equations: v1 + v2 + v3 = v1 + v2 + v3 , (v1 )2 + (v2 )2 + (v3 )2 = v21 + v22 + v23 , v1 v2 v3 = v1 v2 v3 .

(2)

Solving this system with respect to vi , 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 of the particles, it is clear that the remaining roots of the set (2) can be obtained by simple permutations from (1). After that, it is easily to prove that only the result (1) can be realized at t → +∞, since any other options imply the possibility of the further collisions of the particles (due to inequalities v3 > v2 > v1 ).

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§5 Small oscillations of systems with one degree of freedom   2 2 V α 5.1. a) = m 1 − VFα ; a minimum in U(x) occurs when F < V α;  2 V α 4 x2 (3/4) , where the amplitude x is determined from the b) ω2 = 8π 0 3 m 0 (1/4) equality ω2

E = 12 mx˙2 + 13 V α 4 x4 = 13 V α 4 x40 . 5.2. The Lagrangian function of the system is (see [1], § 5, problem 4) L = ma2 [θ˙ 2 (1 + 2 sin2 θ) + 2 sin2 θ + 220 cos θ ], where we have introduce 20 = 2g/a. When  > 0 the potential energy of the system, U(θ) = −ma2 (2 sin2 θ + 220 cos θ), has a minimum when cos θ0 = 20 /2 . Expanding U(θ ) in the neighbourhood of θ0 , and in the kinetic energy putting  1 + 2 sin2 θ = 1 + 2 sin2 θ0 = 3 − 2

0 

4 ≡

M , 2ma2

we get L = 12 mx˙2 − 12 kx2 ,

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

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[5.3

where k = U  (θ0 ), x = θ − θ0 . Hence we get ω2 =

4 − 40 k = 2 M 34 − 240

( > 0 ).

When    √0 , the oscillation frequency is proportional to the angular velocity of rotation, 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 neighbourhood of it, we can write   U = ma2 20 −2 + 14 θ 4 , that is, the oscillations are non-linear in an essential way. Retaining also in the kinetic energy terms up to the fourth order, we get 2π =T =4 ω

θm √ 1 + 2θ 2 dθ  . 2 (θ 4 − θ 4 )  0 m 0

Here θm is the amplitude of the oscillations (see [1], § 11, problem 2a). 5.3. The Lagrangian function of the system is

ϕ 2s3 + , L = mR ϕ˙ − U(ϕ), U(ϕ) = 2mgR sin 2 sin(ϕ/2) 1 2

2 2

2



1/3 q2 . 8mgR2 When s < 1, the stable equilibrium position ϕ0 is determined by the condition 3g sin(ϕ0 /2) = s, and ω2 = R (1 − s2 ). g When s > 1, the point A is a position of stable equilibrium and ω2 = R (s3 − 1). When s = 1, the potential energy U(ϕ) has a minimum at ϕ0 = π and in the neighbourhood of it where s =

U(ϕ) =

3 mgR(ϕ − π )4 + 6mgR, 32

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5.5]

§5. Small oscillations of systems with one degree of freedom

177

that is, the oscillations are non-linear in an essential way. As a result, we get  ϕm 2π R dα = T = 16 , 4 − α4) ω 3g ϕm 0

where ϕm is the amplitude of the oscillations (see [1], § 11, problem 2a). 5.4.

r = r0 + a cos ω(t − t0 ), 2a sin[ω(t − t0 )], ϕ = ϕ0 + (t − t0 ) − r0 ω

where r0 , ϕ0 , a, and t0 are integration constants (a  r0 ), and  =

nα −(n+2)/2 r , m 0

√ ω =  2 − n.

5.5. At the point θ = θ0 the effective potential energy Ueff (θ) =

Mz2 2ml 2 sin2 θ

− mgl cos θ

 (θ ) = 0. We thus obtain (see [1], § 14, problem 1) has a minimum, so that Ueff 0

Mz2 =

m2 l 3 g sin4 θ0 , cos θ0

and the frequency of small oscillations is  ω(θ0 ) =

 (θ ) Ueff 0 = ml 2



g 1 + 3 cos2 θ0 . l cos θ0

For this calculation to be the applicable the condition 1 2

  Ueff (θ0 )( θ )2  16 |Ueff (θ0 )|( θ )3 ,

with θ the oscillation amplitude, must be satisfied. If θ0 ∼ 1, it is satisfied for θ  1.  (θ ) ∝ 1/θ , and the oscillations in θ can be considered If, however, θ0  1, we have Ueff 0 0 to 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 case small harmonic oscillations with frequency g/l occur along the x- and y-axes, that is,

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[5.6

the pendulum moves along an ellipse executing two oscillations in the angle θ for each revolution (see [1], § 23, problem 3 and [7], § 4). 5.6. The effective potential energy for radial oscillations of the molecule is Ueff (r) = 12 mω02 (r − r0 )2 +

M2 , 2mr 2

where r is the distance between the atoms and m the reduced mass. The extra term which is assumed to be a small correction leads to a small shift in the equilibrium position δr0 =

M2 m2 ω02 r03

.

We determine the shift in the frequency by expanding Ueff (r) in a series near the point r0 + δr0 : Ueff (r) = 12 mω02 (r − r0 − δr0 )2 +

M2 3M 2 + (r − r0 − δr0 )2 . 2mr02 2mr04

We get from this a correction to the frequency δω =

3M 2 32 = , 2ω0 2m2 ω0 r04

where  = M/(mr02 ) is the average angular velocity of the molecule rotation. 5.7. a) We get for the displacement from the equilibrium position:  x˙0 x = x0 cos ωt + sin ωt = ω 2k x˙0 tan ϕ = − , ω2 = . ωx0 m

x20 +

x˙20 cos(ωt + ϕ), ω2

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 are harmonic y = A cos(ωt + ϕ), and ω2 = 2f /(ml). When f = kl, the oscillation frequency is equal to that in part 5.7a. In the case of springs in which there is no tension (f = 0) the oscillations are non-linear, the restoring force is F = −ky3 /l 2 , and the frequency (cf. problem 5.1b) is  √ π (3/4) 2k ym ω= , (1/4) m l where ym is the amplitude of the oscillations.

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5.11]

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179

If the particle can move in the xy-plane, its motion – for the case when f = 0 and x and y are small – consists of harmonic oscillations along the x- and y-axes with frequencies ωx2 = 2k/m and ωy2 = 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 the system,  mg 2 L = 12 my˙2 − k(y − l)2 + mgy = 12 my˙2 − k y − l − + const, 2k describes a harmonic oscillator with frequency ω2 = 2k/m and equilibrium position y0 = l + mg/(2k), so that y = y0 + A cos(ωt + ϕ). If we take the displacement from the equilibrium position as coordinate, we can eliminate the gravity field from the Lagrangian function. 5.9. For the angle between the pendulum and the vertical we have

a2

g − l2  1

a2 ϕ= sin t, g − l2

(see also problem 8.3). It is also possible the oscillations of the pendulum near the direction of the radius vector ϕ = t +

g sin t, a2

g 2  . a

5.10. The result for the current in circuit I=

dq U0 sin(ωt − ϕ) = , 2 dt R + (ωL − 1/(ωC))2

tan ϕ =

ωL − 1/(ωC) R

can be obtained by solving the Lagrangian equations of motion for q. The Lagrangian function of the system is L = 12 L q˙2 −

q2 + qU 2C

(see problem 4.25), and the dissipative function is equal to 12 Rq˙2 (see [3], § 48). 5.11. The general solution of the equation of motion (see [1], § 26) x¨ + 2λx˙ + ω02 x =

F cos γ t m

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[5.11

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under such conditions that ω2 = ω02 − λ2 > 0, has the form x(t) = e−λt (a cos ωt + b sin ωt)+ +

F[(ω02 − γ 2 ) cos γ t + 2λγ sin γ t] m[(ω02 − γ 2 )2 + 4λ2 γ 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  F x(t) = (ω02 − γ 2 )(cos γ t − e−λt cos ωt)+ m[(ω02 − γ 2 )2 + 4λ2 γ 2 ] 

 ω02 + γ 2 −λt e sin ωt . (1) +2λγ sin γ t − 2γ ω Let us study this solution in the region near the resonance, γ = ω0 + ε, |ε|  ω0 . If there is no friction at all, that is, λ = 0, the motion of the oscillator near resonance will show beats: x=

F sin(εt/2) · sin ω0 t, mω0 ε

(2)

and the amplitude and the frequency of the beats are determined by how near resonance we are (Fig. 126a). However, when γ = ω0 , that is, at exact resonance, we get by letting ε → 0: x=

F t sin ω0 t, 2mω0

(a)

F x mω0|ε|

t 2π |ε|

(b)

x t

Figure 126

(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 motion changes qualitatively. From (1) we get easily in the case when λ  |ε| instead of (2) x(t) =

F 1 − 2e−λt cos εt + e−2λt cos(ω0 t + ϕ1 (t)). 2mω0 ε

(4)

Here ϕ1 (t) is the phase of the oscillations which changes slowly with time. The amplitude of 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 transient stage the amplitude may reach twice the value of the amplitude of the steady-state oscillations.

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5.11]

§5. Small oscillations of systems with one degree of freedom

181

When |ε|  λ  ω0 , x=

F (1 − e−λt ) sin ω0 t. 2mω0 λ

(a)

(5)

x

F 2mω0|ε|

t 2π |ε|

(b) F 2mω0 λ

x

t

(c)

x

F 2√2mω0|ε| t

Figure 127

In that case we have the transient process with a smoothly increasing amplitude which asymptotically approaches the value F/(2mω0 λ) which is determined by the friction coefficient λ (Fig. 127b). Finally, if the quantities ε and λ are of the same order of magnitude, |ε| ∼ λ  ω0 , we get oscillations of the amplitude around the value √ F/(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). E

B

Emax A x

O

Figure 128

τm

τ

Figure 129

One can use a vector diagram (see Fig. 128) to study qualitative how the oscillations proceeds to a steady state (transient process) when λ  ω0 . The forced oscillation is −→ depicted by the component of the vector OA, which rotates with an angular velocity γ ,

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[5.12

−→ 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,   π F2 2 τ exp − 12 (ωτ )2 2m

E=

depends on how fast the force is switched on (i.e., on the parameter ωτ ). For an instantaneous impact (ωτ  1) or a very slow switching on, (ωτ  1), the energy transfer 2 2 is small, √ while the maximum energy transfer, Emax = π F /(mω e), is reached when τm = 2/ω (Fig. 129). b) If x → a cos(ωt + ϕ), as t → −∞,1 we have E = E(+∞) − E(−∞) = =

π F 2 2 −(ωτ )2 /2 √ 2 − πaωτ Fe−(ωτ ) /4 sin ϕ. τ e 2m

Depending on the value of ϕ, the oscillator gains or loses energy. This change in energy is 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. 1 [ξ (t) − ξ (t)], where 5.13. a) x(t) = 2μ 1 2 ⎡

t

ξ1, 2 = e±μt ⎣

⎤ 1 F(τ )e∓μτ dτ + x˙0 ± μx0 ⎦ . m

0



  t  1 1 λτ −iωτ dτ + x˙ + (iω + λ)x , b) x(t) = ω Im eiωt−λt 0 0 m F(τ )e 0  where ω = ω02 − λ2 . 5.14. The force F(t) = − ∂ U(|r − r0 (t)|), ∂r

(1)

1 The meaning of ϕ is that of the “impact phase”, that is, the phase which the oscillator would have had at t = 0, if there were no force acting upon it.

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5.14]

§5. Small oscillations of systems with one degree of freedom

183

is acting upon the oscillator; here r(t) is the deflection of the oscillator and r0 (t) the radius vector of the impinging particle. Assuming that the particle is little deflected, we can put r0 (t) = ρ + vt with ρ being the impact parameter (vectors ρ and v are mutually orthogonal). Assuming also that the amplitude of the vibrations of the oscillator is small we can put r = 0 in (1) (after differentiating) and then F(t) = −2 2 V (ρ + vt) · exp(− 2 ρ 2 −  2 v2 t 2 ). The oscillations along ρ and v are independent and the corresponding energy excitations are equal to

+∞

2



1 −iωt F (t)e dt ρ

2m

and

−∞

+∞

2



1 −iωt F (t)e dt v

, 2m

(2)

−∞

where Fv and Fρ are the components of F in the directions of v and ρ. The total energy excitation of the oscillator is  x  −x ε = ε1 1 + e , a

ε1 =

π V 2 −a ae , 2E

(3)

where E = 12 mv2 ,

a=

1 2

 ω 2 , v

x = 2(ρ)2 .

The cross-section for exciting the oscillator to an energy between ε and ε + dε is dσ = π

 k

|dρk2 | =

π dε 

a + xk (ε)

1 − a − x (ε) , 2 2 ε k

(4)

k

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 dε (in (4) we assume 2 ε that xk (ε)  1, xk  a). For large ε the result depends on the value of a. If a > 1, we can only have ε < ε1 (see Fig. 130a; for the cross-section see Fig. 130a). However, if a < 1 V 2 (see Fig. 131b) and at ε = ε the function dσ/dε will have a we can have ε < ε2 = π2Ee 1 discontinuity while for ε2 − ε  ε2 it has an integrable singularity,

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184

[5.15

Exploring Classical Mechanics (a)

(b)

ε

ε2

ε1

−a

ε

ε1

x

−a+1

−a

x

1−a

Figure 130 (a)

(b) dσ dε

dσ dε

ε1

ε

ε2 ε

ε1

Figure 131

π dε 1 dσ = √ √ 2 2 ε2 1 − ε/ε2 (see Fig. 131b). 5.15. If the oscillator has an “impact phase” ϕ (see problem 5.12b), we get, by repeating the calculations of the preceding problem, for the energy of the oscillator the expression (ε0 is the initial energy of the oscillator) √ 2 2 ε = ε1 e−2(ρ) + 2 ε1 ε0 e−(ρ) cos ϕ + ε0 ,

(1)

where ε1 =

π 4E



Vω v

2

  ω2 exp − . 2(v)2

When cos ϕ > 0, we have ε > ε0 for all ρ, while we can also have values ρ1, 2 such that ε < ε0 , if cos ϕ < 0. Solving (1) for ρ 2 , we find √ ε1 /ε0  , − cos ϕ + (ε/ε0 ) − sin2 ϕ √ ε1 1 2  , ρ1, = ln 2 2 √  ε0 | cos ϕ| ± ε − ε0 sin2 ϕ 1 ρ = 2 ln  2

and ε0 > ε > εmin = ε0 sin2 ϕ.

when ε > ε0 ,

when cos ϕ < 0

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5.16]

185

§5. Small oscillations of systems with one degree of freedom

Hence we have

2

dρ dε

dε = π  , dσ = π

dε 2 2 ε − ε sin2 ϕ − cos ϕ · εε − ε2 sin2 ϕ 0 0 0

(2)

when ε > ε0 , and dσ = π d(−ρ12 + ρ22 ) =

1 ε0 π | cos ϕ| dε   2 (ε − ε) εε − ε2 sin2 ϕ 0 0 0

(3)

when εmin < ε < ε0 and cos ϕ < 0. Averaging over all possible phases ϕ for a given ε, we obtain 

 dσ π = dε 2 2 |ε0 − ε|

(4)

(Fig. 132). The averaging is performed, using the formulae 

 2π dσ 1 dσ = dϕ, dε 2π dε 0

when ε > ε0 and 

π+α   dσ 1 dσ = dϕ, dε 2π dε π−α

dσ dε

ε0

ε

√ Figure 132 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 mx˙2 − 12 mω2 x2 + xF(t). We then have for the energy of the system

F(t) iF(t)

2 F 2 (t) 1 Im ξ = 2 m ξ − E(t) = m(Re ξ ) + m(Im ξ ) − − ω mω 2mω2 1 2

2

1 2

2

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[5.17

where t ξ = x˙ + iωx = e

iωt

e−iωτ

−∞

1 F(τ ) dτ m

(1)

(see [1], § 22). Although the expression for the energy has a well-definite limit as t → ∞, the integral defining ξ(t) has no limit as t → ∞ (since F(τ ) → F0 as τ → ∞). Integrating (1) by parts, we obtain iF(t) ieiωt ξ(t) = − mω 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) that as t → ∞ the motion of the oscillator in this case is a harmonic oscillation (the second term 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

∞

2



F02 1

 −iωt E(+∞) = + F (t)e dt . 2mω2 2mω2

−∞

5.17. E = E(+∞) − E(−∞) =

F02 λ4 F02 aλ2 F0 cos ϕ + − 2 , 2 2 2 2 2 2mω 2mω (λ + ω ) λ + om2

E0 = 12 mω2 a2 and ϕ is the “impact phase” (see footnote to problem 5.12b). 5.18. If in the formula (see [1], § 22) τ ξ(τ ) = ξ(0)e

iωτ

+e

iωτ

F(t) −iωt dt e m

0

we integrate by parts n times, we get iF0 F (n) (+0)eiωτ − F (n) (τ − 0) + + mω m(iω)(n+1) τ eiωτ + F (n+1) (t)e−iωt dt. m(iω)(n+1)

ξ(τ ) = ξ(0)eiωτ +

0

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187

Here |ξ(0)| = a0 ω, where a0 is the amplitude of the oscillations at the moment the force is switched on. The penultimate term in this formula is of order of magnitude F0 −n mω (ωτ ) , while the last tem is, generally speaking, considerable smaller – provided (n+1) F (t) changes smoothly. The square of the amplitude of the oscillations

1

iF0

2 ξ − mω ω2 is for t > τ of the order of magnitude  a0 +

F0 1 2 mω (ωτ )n

2 .

Thus, if the force F(t) is switched on slowly and smoothly, the energy transferred is very small. 5.19. a) During the time interval 0  t  τ the oscillations have the form x=

Ft + B sin ωt + C cos ωt. mω2 τ

The oscillations will be stable if x(τ ) = x(0),

x(τ ˙ ) = x(0). ˙

These conditions lead to the following set of equations F + B sin ωτ + C(cos ωτ − 1) = 0, mω2 B(cos ωτ − 1) − C sin ωτ = 0,

(1)

which determines the constants B and C. Thus we have for 0  t  τ x(t) =

  F t sin(ωt − ωτ/2) − . 2 sin(ωτ/2) mω2 τ

(2)

However, if t lies in the interval nτ  t  (n + 1)τ (where n is an integer), we must replaced t on the right-hand side of (2) by t  = t − nτ (0  t   τ ).

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[5.20

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When ωτ is close to an integer number times 2π the second term in (2) turns out to be very large – a case which is close to resonance. When ωτ = 2π l (l an integer), there 1 can be no stable oscillations – the set (1) is inconsistent.   1 Im iF + F −λt iωt , when 0  t  τ ; here b) x(t) = ω mω m(λ + iω) e + Ae A=−

F 1 − e−λτ ; m(λ + iω) 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  αt  e 1 1 V Im + , I(t) = −  1 − eατ ατ ω02 − λ2 L

 α = −λ + i ω02 − λ2

(1)

for 0  t  τ . When n  τt  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=

1 T

T

F(t)x(t) ˙ dt =

0

λω2 = m



f12 (ω2 − ω02 )2 + 4λ2 ω2

+

4f22 (ω02 − 4ω2 )2 + 16λ2 ω2

,

that is, the two harmonics in the force both transfer energy, independently of one another (the period T = 2π /ω). ∞ 2 |an |2 n2 . b) A = 4λω m (ω02 − n2 ω2 )2 + 4λ2 ω2 n2 n=1   f12 ω12 f22 ω22 λ c) A = m + . (ω02 − ω12 )2 + 4λ2 ω12 (ω02 − ω22 )2 + 4λ2 ω22

1 If we write the force as a Fourier series

F(t) = 12 F −

∞  F sin(2π lt/τ ), πl l=1

we see that each harmonic term in the force which is acting can cause a resonance build-up. When τ = 2π l/ω we have for sufficiently large t (how large?) x(t) ∼

Ft cos ωt. 2π mωl

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189

When we average over a long time interval, T  2π /ω1, 2 , it turns out that each of the two forces f1 cos ω1 t and f2 cos ω2 t acts independently on the oscillator. This is because only the squares of trigonometric functions have non-vanishing averages, T

1 T

sin2 ω1 t dt =

1 1 1 + (1 − sin 2ω1 T ) → , 2 4ω1 T 2

0

as T → ∞, while the average values of cross-products such as sin ω1 t · cos ω1 t, sin ω1 t · cos ω2 t, and so on vanish: for instance, 1 T

T sin ω1 t · cos ω2 t dt = 0

1 − cos(ω1 − ω2 )t 1 − cos(ω1 + ω2 )T + → 0, 2(ω1 − ω2 )T 2(ω1 + ω2 )T

as T → ∞. d) The displacement of the oscillator is ∞ x=

ψ(ω)eiωt dω

ω2 − ω2 + 2iλω −∞ 0

,

so that the total work done by the force F(t) is equal to ∞ A= −∞

8π λ x(t)F(t) ˙ dt = m

∞ 0

ω2 |ψ(ω)|2 dω. (ω02 − ω2 )2 + 4λ2 ω2

(1)

To prove this equality, one uses the inverse Fourier transform ∞

F(t)eiωt dt = 2π ψ ∗ (ω).

−∞

When λ  ω0 the main contribution to the integral (1) comes from the vicinity of the eigen-frequency of the oscillator ω = ω0 . We have thus ⎤ ⎡ ∞  dω2 4π |ψ(ω0 )|2 ω0 ⎣ ⎦. A≈ λ m (ω02 − ω2 )2 + 4λ2 ω02 0

The factor inside the square brackets, which can easily be evaluated, is independent of λ, when λ → 0:

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A=

[5.21

|2π ψ(ω0 )|2 2m

(cf. [1], (22.12)). 5.21. Taking into account that the displacement of an oscillator x is a small value of the first order in F, we get ∞ E =

∞ F(x, t)x˙ dt ≈

−∞ ∞

P = −∞

f (t)x˙ dt,

−∞

∞  x f (t) − f˙(t) dt. F(x, t) dt ≈ V −∞

Integrating the second term by parts, we find ∞ P = −∞

f (t) dt +

E . V

∞ f (t) dt = 0, we have E = V p.

In particular, when −∞

The condition of smallness of x can be demonstrated on the example of the action of wave groups f (t) = fe−|t|/τ cos γ t on an oscillator. A small parameter in expansion of F(x, t) is x/λ where λ = 2π V /γ is a characteristic wave length, that is, |x| fγ =  1. λ 2π m|ω2 − γ 2 |

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§6 Small oscillations of systems with several 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 x˙21 + x˙22 − 12 k[x21 + (x1 − x2 )2 ].

(1)

The equations of motion m¨x1 + k(2x1 − x2 ) = 0,

m¨x2 + k(x2 − x1 ) = 0

can be reduced to a set of algebraic equations through the substitution xi = Ai cos(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 2 ω1, 2=

√ 3∓ 5 k . 2 m

Of the two equations (2), only one is independent, because of (3). Substituting the values ω1 and ω2 in (2), we obtain the ratios of the amplitudes 2 A2 ≡ A √ 5+1 2 A2 ≡ B A1 = − √ 5−1

A1 =

for ω = ω1 , for ω = ω2 .

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

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[6.2

The eigen-vibrations of the system are thus x1 = A cos(ω1 t + ϕ1 ) + B cos(ω2 t + ϕ2 ), √ √ 5+1 5−1 A cos(ω1 t + ϕ1 ) − B cos(ω2 t + ϕ2 ). x2 = 2 2

(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 problems 6.2b and 6.25), or when developing the perturbation theory (see problem 6.35), or when quantizing the system, it is more convenient to use the normal coordinates. This is because the normal coordinates qi , defined by the equalities x1 = q1 + q2 , √ √ 5+1 5−1 x2 = q1 − q2 , 2 2

(5)

reduce the Lagrangian function (1) to a sum of squares: L=

√ √ 5− 5 5+ 5 m(q˙21 − ω12 q21 ) + m(q˙22 − ω22 q22 ), 4 4

(6)

while the equations of motion become one-dimensional: q¨ i + ωi2 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 motion of 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 suspension point is considered in problem 7.2. 6.2. The Lagrangian function of the system is (cf. the preceding problem)     L = 12 m x˙21 + x˙22 − 12 k (x1 − a(t))2 + (x1 − x2 )2 . If we discard the term − 12 ka2 (t), which is the total derivative with respect to the time, the function L can be rewritten as L = L0 + L,

L = x1 ka(t),

(1)

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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 more convenient as it enables us to immediately write the external force “vector”: 

F x1 F x2



 =

 ka(t) . 0

a) The equations of motion m¨x1 + k(2x1 − x2 ) = ka cos γ t, m¨x2 + k(x2 − x1 ) = 0,

(2)

can be reduce to a inhomogeneous linear set of two equations to determine A and B through the substitution1 x1 = A cos γ t, Thus, we get

x2 = B cos γ t.

(a) A,B

A=

ak(−mγ 2 + k) , m2 (γ 2 − ω12 )(γ 2 − ω22 )

B=

ak2 , m2 (γ 2 − ω12 )(γ 2 − ω22 )

√ mk ω1

A

B γ

ω2

√ 2 = 3 ∓ 5 k are the eigen-frequencies of the (b) where ω1, 2 2 m |A| system. The frequency dependence of the amplitudes A and B is shown in Fig. 133a. When the frequency goes through the resonance values γ = ω1, 2 , the amplitude A and B change sign; this corresponds to a change ofπ in the phase of the oscillations. ω1 ω2 At the frequency γ = k/m the oscillations of the upper Figure 133 particle 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.

γ

1 The general solution of the set of equations in (2) is a superposition of free and forced oscillations. As soon as there is even a smallest amount of friction present, the free oscillations are damped, so that after a long time interval the solution of (2) is independent of the initial conditions and consist of the forced oscillations (3).

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[6.3

At what frequencies γ will the oscillations of the upper particle be damped, if we suspend from the lower particle another particle on the same spring? b) Introducing the normal coordinates q1, 2 (see formula (5) from the preceding problem), we can write the Lagrangian function (1) in the form L = L1 (q1 , q˙1 ) + L2 (q2 , q˙2 ), √ 5± 5 2 2 L1, 2 = m(q˙21, 2 − ω1, 2 q1, 2 ) + q1, 2 ka(t) 4 (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 of suspension 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 the potential energy U = −mgy1 − mgy2 we substitute y1 (x1 ) and y2 (x1 , x2 ) and take into account all terms up to second order of magnitude in x1, 2 /l: y1 =



4l 2 − x21 ≈ 2l −

x21 , 4l

 x2 (x2 − x1 )2 y2 = y1 + l 2 − (x2 − x1 )2 ≈ 3l − 1 − , 4l 2l while in the expression for the kinetic energy T = 12 m(x˙21 + y˙21 + x˙22 + y˙22 ) we substitute y˙1 and y˙2 up to the first order: y˙1 = −

x1 x˙1 ≈ 0, 2l

y˙2 = y˙1 −

(x2 − x1 ) (x˙2 − x˙1 ) ≈ 0. l

After that, the Lagrangian function L = 12 m(x˙21 + x˙22 ) −

mg 2 mg x − (x2 − x1 )2 + 5mgl 2l 1 2l

coincides with the Lagrangian function of the system considered in problem 6.1, if we take 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 easily demonstate that we return to the Lagrangian function considered in problem 6.2.

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6.4. Let ϕ and ψ be the deflection angles of the upper and lower  particles from the vertical. The normaloscillations 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 = a cos(ω1 t + ϕ),

y = b cos(ω2 t + ψ).

Constant a, b, ϕ, and ψ are determined by the initial conditions. The orbit lies inside the rectangle (Fig. 134): −a  x  a,

y b −a

−b  y  b.

a x

Generally speaking, the orbit “fills” the whole −b rectangle. More precisely, if the ratio ω1 /ω2 is irrational, the orbit comes arbitrary close to any point Figure 134 inside this rectangle. The motion of the point is not periodic in this case, although the motion of its components along the two coordinate axis is. However, if the ratio ω1 /ω2 is rational (lω1 = nω2 with l and n integers), the orbit is a closed curve, a so-called Lissajous figure. The motion in this case is periodic, and the period is 2π n/ω1 . 6.6. a) The transition to normal coordinates is for this system simply a rotation in the (x, y)-plane (Fig. 135): x = Q1 cos ϕ − Q2 sin ϕ,

y = Q1 sin ϕ + Q2 cos ϕ. (1)

Indeed, the kinetic energy does not change its form under the rotation while the coefficient of Q1 Q2 in the potential energy, which is equal to

y Q2 Q1 ϕ x

Figure 135

  − 12 ω12 − ω22 sin 2ϕ − α cos 2ϕ, can be made to vanish, if we determine the parameter ϕ from the condition cot 2ϕ =

ω22 − ω12 . 2α

The dependence of ϕ on ω1 is shown in Fig. 136; the region where ϕ changes from ϕ = 0 to ϕ = π/2 has a width of the order of α/ω2 .

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[6.6

Exploring Classical Mechanics

When the coupling is weak, α  |ω12 − ω22 |, 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 |ω12 − ω22 |  α the normal oscillations no longer localized: ϕ≈

1 π 1 , x ≈ √ (Q1 − Q2 ), y ≈ √ (Q1 + Q2 ) 4 2 2

(see [1], § 23, problem 1).

π 2

Ω1,2

ϕ

Ω2

ω2

π 4 ω1

ω2

Ω1

ω2

Figure 136

ω1

Figure 137

The normal frequencies,

21, 2

=

1 2

ω12 + ω22 ∓

 2 2 2 2 (ω2 − ω1 ) + 4α ,

(2)

lie outside the range of partial frequencies1 ; that is, 1 < ω1 and 2 > ω2 (to fix the ideas we assume ω1 < ω2 ). Relations of this kind for systems with many degrees of freedom are 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 normal frequencies 1, 2 differ little from the partial frequencies ω1, 2 (just as the normal coordinates Q1 and Q2 differ little from coordinates x and y) when α is small, everywhere, except the degeneracy region, where |ω12 − ω22 |  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 same as in the preceding problem. 1 We follow Mandelstam [9] in calling those frequency which are obtained from the original system when x ≡ 0 (or y ≡ 0) partial frequencies.

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197

b) The normal coordinates can in this case be obtained from the results of the 2 →m preceding problem, simply by replaced ω1, 1, 2 and α → −β, while the normal 2 frequencies 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 frequencies are 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=

1 2

  L1 q˙21 + L2 q˙22 − 12



q22 q21 (q1 + q2 )2 + + , C1 C2 C

where q1 and q2 are the √ charges on the√upper plates of the capacitors C1 and C2 . Introducing new variables L1 q1 = x and L2 q2 = y, we obtain the Lagrangian function of the problem 6.6a with the parameters ω12

1 = L1



 1 1 + , C C1

ω22

1 = L2



 1 1 + , C C2

α=

1 . √ C L1 L2

√ √ b) By the change of variables q1 = C1 x and q2 = C2 y we can change the Lagrangian function of the this system to the Lagrangian function of problem 6.6b with the parameters m1, 2 = (L + L1, 2 )C1, 2 ,

β =L

 C1 C2 .

Can these systems become unstable? 6.8. Let x1 and x2 be the displacements of the particles m1 and m2 from their equilibrium √ √ positions. By the change of variables m1 x1 = x and m2 x2 = 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 low frequency 21 = 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 the right) and one with a very high frequency 22 = 2k/m1 (when the particle m2 is almost at rests). One can find the amplitude of the oscillations of the second particle considering its motion as being under the influence of an extra force kx1 with a high frequency (see [1], m1 x . formula (22.4)): x2 = − 2m 1 2 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.

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[6.9

v sin ω t ± v sin ω t, when k  k, the oscillations have the form of 6.9. a) x1, 2 = 2ω 1 2 1 2ω2 1 beats: x1 =

v cos εt · sin ωt, ω

x2 = −

v sin εt · cos ωt. ω

b) x1, 2 = 12 a(cos ω1 t ± cos ω2 t); when k1  k, we have x1 = a cos εt · cos ωt,

x2 = a sin εt · sin ωt.

k and ω2 = 2k1 + k , ε = k1 ω , ω = 1 (ω + ω ). Everywhere in this problem ω12 = m 1 2 2 m 2 2k 1 6.10. The energy transferred from the first to the second particle during the time dt is equal to the work done by the force F = k1 (x1 − x2 ): dE = k1 (x1 − x2 ) dx2 = k1 (x1 − x2 )x˙2 dt, and the energy flux is dE/dt = k1 (x1 − x2 )x˙2 . In the limiting case when k1  k in problem 6.9a, the flux of energy averaged over the fast oscillations is equal to 1 2

εmv2 sin 2εt.

6.11. The equations of motion m¨x1 + k1 (x1 − x2 ) + kx1 + α x˙1 = 0, m¨x2 + k1 (x2 − x1 ) + kx2 + α x˙2 = 0 split into two equations for the normal coordinates, when we use the substitution √ x1, 2 = (q1 ± q2 )/ 2: q¨ 1 + ω12 q1 + 2λq˙1 = 0, q¨ 2 + ω22 q2 + 2λq˙2 = 0, where ω12 = k/m, ω22 = (k + 2k1 )/m, 2λ = α/m. We have thus, when λ < ω1, 2 (see [1], § 25) x1, 2 = e−λt [a cos(γ1 t + ϕ1 ) ± b cos(γ2 t + ϕ2 )], 2 − λ2 . where γ1, 2 = ω1, 2 The characteristic equation for the system of Fig. 25 is no longer biquadratic when there is friction present, but of the fourth degree, and it is therefore considerable more complicated to find the eigen-vibrations.

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199

6.12. The Lagrangian function of the double pendulum is  Mg 2 mg  2 x1 − x1 + (x2 − x1 )2 , 2l 2l

L = 12 M x˙21 + 12 mx˙22 −

where x1 and x2 are the displacements of M and m points from the vertical passing through the suspension point (cf. problem 6.3). The replacements x x1 = √ , M

y x2 = √ m

reduce the Lagrangian function to the form considered in problem 6.6a, with ω12 − ω22 =

2mg g α= Ml l



m . M

In this case we have 1 x = √ (Q1 − Q2 ), 2

1 y = √ (Q1 + Q2 ), 2

where Qi = ai cos i t + bi sin i t,  1, 2 =

g γ ∓ , l 2

 γ=

mg . Ml

√ ˙ 1, 2 (0) = 0, we get Taking into account the initial conditions Q1, 2 (0) = lβ m/2, Q 

 m g x1 = lβ sin γ t sin t, M l  g t. x2 = lβ cos γ t cos l Thus, √ the pendulums oscillate “in turn”, and the amplitude of the upper pendulum is M/m times smaller than that of the lower one. 6.13. ak(k1 + k − mγ 2 ) cos γ t, m2 (γ 2 − ω12 )(γ 2 − ω22 ) akk1 x2 = cos γ t, m2 (γ 2 − ω12 )(γ 2 − ω22 )

x1 =

where ω12 = k/m, ω22 = (k + 2k1 )/m.

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[6.14

Exploring Classical Mechanics

ak cos γ t, where the x is the displacement along the ring from i k − mγ 2 the equilibrium positions of ith particle. Resonance is possible only at one of the normal frequencies when γ 2 = k/m (see problem 6.25). 6.14. x1 = x2 =

6.15. Let xi be the displacement of ith particle along the ring from the equilibrium position and then x1 = x3 = x2 =

ak(ω22 − γ 2 ) m(γ 2 − ω12 )(γ 2 − ω32 )

cos γ t,

2ak2 cos γ t, m2 (γ 2 − ω12 )(γ 2 − ω32 )

where the eigen-frequencies ωi are  √ k 2 = 2 ∓ , ω1, 2 3 m

ω22 =

2k . m

Note that when γ = ω2 we have the displacement x1 = x3 = 0, and x2 = −a cos γ 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) m¨x1 + α x˙1 + kx1 + k1 (x1 − x2 ) = ka Re eiγ t , m¨x2 + α x˙2 + 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 − k1 B = ka, −k1 A + (−mγ 2 + 2imλγ + k + k1 )B = 0

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6.17]

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with 2mλ = α, whence A= B=

ak(k + k1 − mγ 2 + 2iλmγ ) , m2 (γ 2 − 2iλγ − ω12 )(γ 2 − 2iλγ − ω22 ) m2 (γ 2 − 2iλγ

akk1 , − ω12 )(γ 2 − 2iλγ − ω22 )

 2 ak γ 2 − 12 ω12 − 12 ω22 + 4λ2 γ 2 cos(γ t + ϕ1 + ϕ2 + ψ) , x1 = m [(γ 2 − ω12 )2 + 4λ2 γ 2 ][(γ 2 − ω22 )2 + 4λ2 γ 2 ] x2 =

akk1 cos(γ t + ϕ1 + ϕ2 ) , m2 [(γ 2 − ω12 )2 + 4λ2 γ 2 ][(γ 2 − ω22 )2 + 4λ2 γ 2 ]

ω12 =

k k + 2k1 2λγ 4λγ , ω22 = , tan ϕ1,2 = , tan ψ = 2 . 2 2 m m γ − ω1,2 ω1 + ω22 − 2γ 2

There is a phase difference ψ between the oscillations of the two particles; the oscillations of the first particle are never completely damped. As function of a frequency of 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 the second particle, respectively, we have for the Lagrangian function of the system   k1 + k2 2 k2 + k3 2 2k2 L = 12 m x˙2 + y˙2 − x − y + xy + k1 ax cos γ t; m m m it differs from the Lagrangian function considered in problem 6.6a only in the term xk1 a cos γ t, corresponding to a force k1 a cos γ t acting upon the first particle. We shall use here the notations of  problem 6.6a. The partial frequency ω1, 2 = (k1, 3 + k2 )/m corresponds to the eigen-frequency of a system which we obtain by fixing the second (first) particle, that is, by putting y = 0 (respectively, x = 0). When we change to the normal coordinates Q1 and Q2 , the Lagrangian function becomes

y Q2 Q1

F1

ϕ F

F2

Figure 138

  ˙ 2 − 2 Q2 + Q ˙ 2 − 2 Q2 + (F1 Q1 + F2 Q2 ) cos γ t, L = 12 m Q 1 1 1 2 2 2

x

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[6.18

where F1 = k1 a cos ϕ and F2 = −k2 a sin ϕ are the components of the amplitude of the force F = k1 a onto the normal coordinates Q1 and Q2 (Fig. 138). For the coordinates Q1, 2 , we get the equations of motion of the harmonic oscillators with frequencies 1, 2 under the action of the applied forces F1, 2 cos γ t. The initial conditions are ˙ i (0) = 0. We get Qi (0) = Q Q1, 2 =

F1, 2 (cos γ t − cos 1, 2 t) . m( 21, 2 − γ 2 )

It is interesting to consider resonance at the second eigen-frequency for the system in the weak coupling approximation, k2 ≡ε1 k3 − k1 (to fix the ideas, we have assume that k1 < k3 ). If we put γ = 2 (1 + ε1 ), we have k1 a (cos ω2 t − cos ω1 t), k1 − k3  ω  k1 aε 2 Q2 = − sin ε1 t sin ω2 t 2 mω22 ε1 k1 a Q2 = − εt sin ω2 t 2mω2

Q1 =

when |ε1 |  1, when ε1 = 0.

We see thus that for even the smallest coupling the amplitude Q2 can become large or even steadily increases with time, although the rate of change will then be small. Since the angle of rotation is small (sin ϕ = 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 the first frequency γ = 1 ? How will the character of the vibrations change if a small friction force acts on both particles, proportional to the velocity (cf. problem 5.11)? 6.18. a) x =

F0 cos ϕ cos γ t, m(ω12 − γ 2 )

y=

F0 sin ϕ cos γ t, m(ω22 − γ 2 )

where ω12 = 2k1 /m, ω22 = 2k/m, ϕ is the angle between the vector F0 and the AB-axis and x and y are the displacements from the equilibrium position along AB and CD. The particle oscillates along a strait line through the centre. It is interesting that when γ 2 = ω12 sin2 ϕ + ω22 cos2 ϕ, this strait line is at right angles to the vector F0 . In this case the work done by the acting force is equal to zero. Therefore, it seems that even the smallest amount of friction must lead to a damping of the oscillations. Explain this situation.

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§6. Small oscillations of systems with several degrees of freedom

203

F F cos γ t, y = sin γ t. m(ω12 − γ 2 ) m(ω22 − γ 2 ) The trajectory is an ellipse with semi-axes

b) x =

a=

F m|ω12 − γ 2 |

,

b=

F m|ω22 − γ 2 |

.

If the quantities (ω12 − γ 2 ) and (ω22 − γ 2 ) have opposite signs, the particle moves clockwise along the ellipse, while the force vector rotates counterclockwise. How does the picture given here of the motion of the particle change when the tension in 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 equilibrium position. 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 equal amplitude, x1 = −x2 = A cos(ω2 t + α) = q2 (t),

x3 = 0,

ω2 =

 3k/m,

(2)

have the same frequency as the oscillations where the particles 2 and 3 move towards one another: x1 = 0,

x2 = −x3 = B cos(ω3 t + β) = q3 (t),

ω3 = ω2 .

(3)

We introduce the “displacement vector” ⎛ ⎞ x1 r = ⎝x2 ⎠ , x3 and we can then write the oscillations (1)–(3) as three vectors (see Fig. 139), ⎛ ⎞ 1 r1 = ⎝1⎠ q1 , 1

⎛

⎞ 1 r2 = ⎝−1⎠ q2 , 0

⎛

⎞ 0 r3 = ⎝ 1⎠ q3 . −1

Any linear superposition of the r2 and r3 vectors is again an oscillation with the frequency ω2 . Thus, in a space with Cartesian coordinates x1 , x2 , and x3 , a set of solutions that represent the oscillations with twice degenerate frequency ω2 = ω3 defines a plane

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[6.19

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r1

r2

r3

r3

Figure 139

which passes through the vectors r2 and r3 .1 As easy to see from (4), both of these vectors—and, hence, all vectors lying in this plane—are orthogonal to the vector r1 (the general orthogonality condition seen in problem 6.23). The Lagrangian function of the system is     L = 12 m x˙21 + x˙22 + x˙23 − 12 k (x1 − x2 )2 + (x2 − x3 )2 + (x3 − x1 )2 .

(5)

The normal coordinates must simultaneously diagonalize two quadratic forms—the kinetic energy and the potential energy. The kinetic energy in (5) is already proportional to the sum of the squares of the velocities. Therefore, the transformation from xi to the normal coordinates which does not changing its form must be an orthogonal transformation. Hence, the vectors of the corresponding normal oscillations must be mutually orthogonal. The vectors ri are independent, but not mutually orthogonal: r1 r2 = r1 r3 = 0, but r2 r3 = 0. To obtain the normal coordinates, we must choose two mutually 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 the condition er1 = er2 = 0). As a result, a set of normalized vectors2 r1 r 1 = √ , 3

r2 r 2 = √ , 2

⎛ ⎞ 1 1 r 3 = eq3 = √ ⎝ 1⎠ q3 6 −2

(6)

allow us to define the normal coordinates: 1 1 1 x1 = √ q1 + √ q2 + √ q3 , 3 6 2 1 1 1 x2 = √ q1 − √ q2 + √ q3 , 3 6 2 1 2 x3 = √ q1 − √ q3 , 3 6

(7)

1 Note that the linear combination in this plane of the form αr (t) + βr (t) is an oscillation either along a 1 2 strait 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 r are normalized by the condition r r = i i k δik q2i . Under this condition, the transformation (7) is orthogonal.

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§6. Small oscillations of systems with several degrees of freedom

205

which reduce the Lagrangian function (5) to the form L = 12 m(q˙21 + q˙22 − ω22 q22 + q˙23 − ω32 q23 ).

(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) = x˙i (0) = 0.

We have thus for the normal coordinates qi (see formula (7) of the preceding problem) the initial conditions: a q1 (0) = √ , 3

a q2 (0) = √ , 2

a q3 (0) = √ , 6

q˙i (0) = 0.

Therefore, we find a q1 = √ , 3

a q2 = √ cos ω2 t, 2

a q3 = √ cos ω3 t, 6

and taking into account that ω2 = ω3 , we get finally x1 =

a 2a + cos ω2 t, 3 3

x2 = x3 =

a a − cos ω2 t. 3 3

6.21. We use the notations of problem 6.19. The Lagrangian function of the system is     L = 12 m x˙21 + 2x˙22 + 3x˙23 − 12 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 by substitution xi = Ai cos(ωt + ϕ): (−mω2 + 5k)A1 − 2kA2 − 3kA3 = 0, −2kA1 + (−2mω2 + 8k)A2 − 6kA3 = 0, −3kA1 − 6kA2 + (−3mω + 9k)A3 = 0. 2

This system has a non-trivial solution when its determinant is equal to zero: ω2 (mω2 − 6k)2 = 0.

(2)

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[6.21

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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 ring with the constant angular velocity ⎛ ⎞ 1 r1 = ⎝1⎠ q1 , 1

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 the frequency ω2 . In particular, it is possible to choose such oscillations where either the first, second, or the third particle is at rest: ⎛

⎞ 0 r2 = ⎝ 3⎠ q2 , −2

⎛

⎞ 3 r3 = ⎝ 0⎠ q3 , −1

qi = Ci cos(ω2 t + ϕi ),

⎛

⎞ 2 r4 = ⎝−1⎠ q4 . 0

(5)

i = 2, 3, 4.

According to (4), any linear combination of vectors (5) is orthogonal to the vector ⎛ ⎞ 1 ⎝2⎠ . 3 It is easy to verify that the set of vectors ⎛

r1 ,

r2 ,

⎞ 5 r 3 = ⎝−1⎠ q3 −1

(6)

allows us, as in problem 6.19, to determine the normal coordinates which lead the Lagrangian function (1) to the diagonal form. The vectors in (6) satisfy not the ordinary condition of orthogonality (as in problem 6.19), but the condition of the “weight orthogonality” (see problem 6.23).

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207

6.22. The vectors of the normal oscillations are ⎛

⎞ 1 1 ⎜ 0⎟ ⎟ q1 , r1 = √ ⎜ 2 ⎝−1⎠ 0 ⎛ ⎞ 1 ⎟ 1⎜ −1 ⎟q , r3 = ⎜ ⎝ 1⎠ 3 2 −1

⎛

⎞ 0 1 ⎜ 1⎟ ⎟ q2 , r2 = √ ⎜ 2 ⎝ 0⎠ −1 ⎛ ⎞ 1 1⎜ 1⎟ ⎟ q4 , r4 = ⎜ ⎝ 2 1⎠ 1

(1)

ql = Al cos(ωl t + ϕl ), l = 1, 2, 3; q4 = A4 t + A5 ,   ω1 = ω2 = 2k/m, ω3 = 2 k/m. The Lagrangian function of the system reads   L = 12 m q˙21 + q˙22 + q˙23 + q˙24 − ω12 q21 − ω22 q22 − ω32 q23 . This is, of course, not the only possible choice. Any vectors obtained from the ones given here through a rotation in the plane determined by the vectors r1 and r2 will also be vectors of the normal oscillations. For instance, ⎛

⎞ 1 1 ⎜ 1⎟ ⎟ q , r 1 = ⎜ 2 ⎝−1⎠ 1 −1

⎛

⎞ 1 1 ⎜−1⎟ ⎟ q , r 2 = ⎜ 2 ⎝−1⎠ 2 1

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 −ωl2



(l)

mij Aj +

j

−ωs2

 j



kij Aj = 0,

(l)

(1)

(s)

(2)

j

(s)

mij Aj +

 j

kij Aj = 0.

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[6.24

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(l) Multiply (1) by A(s) i and (2) by Ai and summing both equations over i, we get instead of (1) and (2)

−ωl2



(l)

(s)

mij Aj Ai +

ij

−ωs2





(l)

(s)

(3)

(s)

(l)

(4)

kij Aj Ai = 0,

ij (s)

(l)

mij Aj Ai +

ij



kij Aj Ai = 0.

ij

Subtracting (4) from (3) and using the fact that mij = mji and kij = kji , we obtain (ωs2 − ωl2 )



(s)

(l)

mij Ai Aj = 0;

ij

that is, when ωs = ωl , 

(s)

(l)

(5)

(s)

(l)

(6)

mij Ai Aj = 0,

ij

and at the same time from (3) we obtain 

kij Ai Aj = 0.

ij

It is helpful to use the terminology from linear algebra. We shall call the set of (l) (l) (l) amplitudes of a given oscillation the amplitude vector A(l) = (A1 , A2 , . . . , An ). The relations (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 the metric tensors mij or kij . In the case of degeneracy (if ωs = ωl ), the amplitude A(s) and A(l) do not necessary satisfy (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 function to a sum of squares. 6.24. Introducing the normal coordinates xi =



(l)

Ai ql ,

l

we transform the constraint equations to the form  l

bl ql = 0,

bl =

 i

(l)

ai Ai .

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6.25]

209

§6. Small oscillations of systems with several degrees of freedom

Equations of motion with an indefinite Lagrangian multiplier λ Ml (¨ql + 2l ql ) = bl λ 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 



b2l

l

Ml ( 2l − ω2 )

= 0.

To study this equation, it is helpful to use the graph (Fig. 140) y(ω ) = 2



b2l

l

Ml ( 2l − ω2 )

y ω12

= 0.

Ω12

Ω22

ω22 Ω23

ω2

Figure 140

Note that the function y(ω2 ) changes the sign by passing through the infinite value when ω2 = 2l . After that, the location of the roots ωl becomes quite clear. If any of the coefficients bl is zero, then the corresponding normal oscillation (and its frequency) do not change when one applies a constraint. The fact considered in this problem allows for a simple geometric interpretation (see [8], § 24). 6.25. Substituting xj =



lλ

(l) a(l) cos γ t j



mij x¨ j +

j

into the equations of motion,



kij xj = fi cos γ t,

j

we obtain the following equations to determine the coefficients λ(l) : −γ 2

 I,j

(l)

mij λ(l) Aj +

 j,l

(l)

kij λ(l) Aj = fi .

(2)

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[6.25

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The simplest way to solve these equations is by using the orthogonality relations (5) and (6) of problem 6.23. (s) To do this, we multiply the equations in (2) by Ai and sum over i; then we get Fs , Ms (ωs2 − γ 2 )

λ(s) =

where Fs =



A(s) i fi ,

Ms =

i



(s) mij A(s) i Aj ,

Ks =

i, j



(s) kij A(s) i Aj .

i, 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)

Ai qs (t),

(3)

s

instead of the equations (1), we have the following equations of motion: Ms q¨ s + Ks qs = Fs cos γ t.

(4)

Hence, if the force vector fi is orthogonal to the amplitude vector of the sth normal oscillation, 

(s)

Ai fi = 0,

i

the corresponding normal coordinate satisfies the equation for free oscillations, and there is no resonance for ω = ωs . Note that the work done by the applied force is in this case equal to zero:  i

fi dxi =



(s)

fi Ai dqs = 0.

i

1 If some of the eigen-frequencies are degenerate, we shall assume that the amplitudes of the eigenoscillations corresponding to them are chosen in such a way that they satisfy the orthogonality relations (5) and (6) of problem 6.23.

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211

Consider the case where the force vector is parallel to the amplitude vector of one of the normal oscillation, fi (s)

Ai

= 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 =



βij fj ,

j

where βij =



(l) A(l) i Aj

l

Ml (ωl2 − 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 ⎛ ⎞ ⎛ ⎞ 1 1 ⎜1⎟ ⎜ 0⎟ ⎜ ⎟ q1 , ⎜ ⎟ q2 , ⎝−1⎠ ⎝1⎠ 1 0

⎛

⎞ 0 ⎜ 1⎟ ⎜ ⎟ q3 , ⎝ 0⎠ −1

⎛

⎞ 1 ⎜−m/M ⎟ ⎜ ⎟ ⎝ 1 ⎠ q4 , −m/M

where q1 = At + B, qi = Ai cos(ωi t + αi ), i = 2, 3, 4; 2k 2k 2k(M + m) , ω32 = , ω42 = . ω22 = m M mM The first three oscillations are easily guessed, and the last one can be obtained using the orthogonality conditions to the first three. Since the particle masses are different, the orthogonality condition of the two normal oscillations with the amplitudes A and B have the form mA1 B1 + MA2 B2 + mA3 B3 + MA4 B4 = 0 (see problem 6.23).

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[6.28

Exploring Classical Mechanics

6.28. Let xi be the displacement of the ith particle along the ring. We can easily guess two normal vibrations: ⎛ ⎞ 1 ⎜1⎟ ⎟ r1 = ⎜ ⎝1⎠ q1 (t), 1 q1 (t) = C1 t + C2 ,

⎛

⎞ 1 ⎜ 0⎟ ⎟ r2 = ⎜ ⎝−1⎠ q2 (t), 0

q2 (t) = A2 cos(ω2 t + ϕ2 ),

ω2 =

(1)

 2k/m.

The two other vectors must be orthogonal to the vectors (1) in the metric determined by the coefficients of the quadratic form of the kinetic energy (see problem 6.23); that is, they must have the form ⎛

⎞ a ⎜ ⎟ b ⎟ q(t). r=⎜ ⎝ ⎠ a 1 −a − 2 b

(2)

Substituting (2) into the equations of motion for the first and the second particles, m¨x1 + k(2x1 − x4 − x2 ) = 0,

m¨x2 + 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. Solving (3), we find 2 ω3,4

√ 5∓ 5 k = , 2 m

b3,4 = (1 ±

√ 5)a3,4

or ⎞ 1√ ⎜ 1± 5 ⎟ ⎟ q (t), r3,4 = ⎜ ⎝ 1 √ ⎠ 3,4 − 12 (3 ± 5) ⎛

q3,4 = A3,4 cos(ω3,4 t + ϕ3,4 ).

(3)

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§6. Small oscillations of systems with several degrees of freedom

213

6.29. a) Let xi , yi , and zi be the displacements of the ith particle from its equilibrium position. The Lagrangian function of the system has the form (see problem 5.7) L = L1 (x, x) ˙ + L1 (y, y) ˙ + L1 (z, z), ˙    2 2 2 2 2 L1 (x, x) ˙ = 12 m x˙1 + x˙2 + x˙3 + x˙4 + x˙5 − 12 k x21 + (x1 − x5 )2 +

 + (x5 − x3 )2 + x23 + x22 + (x2 − x5 )2 + (x5 − x4 )2 + x24 ;

therefore, the oscillations in the x-, y-, and z-directions proceed independently. We can easily guess three of the normal oscillations in the x-direction: ⎛

⎞ 1 ⎜ 0⎟ ⎜ ⎟ ⎟ r1 = ⎜ ⎜−1⎟ q1 , ⎝ 0⎠ 0

⎞ 0 ⎜ 1⎟ ⎜ ⎟ ⎟ r2 = ⎜ ⎜ 0 ⎟ q2 , ⎝−1⎠ 0

qi = Ai cos(ωi t + ϕi ),

⎛

⎛

⎞ 1 ⎜−1⎟ ⎜ ⎟ ⎟ r3 = ⎜ ⎜ 1⎟ q3 , ⎝−1⎠ 0  ω1 = ω2 = ω3 = 2k/m.

(1)

The other two normal oscillations must be orthogonal to the vectors (1) and therefore have the form1 ⎛ ⎞ a ⎜a⎟ ⎜ ⎟ ⎟ r4,5 = ⎜ ⎜a⎟ q4,5 . ⎝a⎠ d Substituting this vector into the equations of motion for the first and fifth particle, m¨x1 + k(2x1 − x5 ) = 0, m¨x5 + k(4x5 − x1 − x2 − x3 − x4 ) = 0, we obtain two equations to determine the unknown parameters a, d, and the frequencies ω4,5 : (−ω2 m + 2k)a − kd = 0, −4ka + (−mω2 + 4k)d = 0.

(3)

1 Let r 4,5 = (a, b, c, e, d) ≡ r; then the orthogonality conditions (r, r1 ) = (r, r2 ) = (r, r3 ) = 0 lead to the equations a = b = c = e.

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[6.29

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Solving (3), we find ω4,5 =

√ √ (3 ∓ 5)k/m and d4,5 = (−1 ± 5)a4,5 , and finally ⎛

⎞ 1 ⎜ ⎟ 1 ⎜ ⎟ ⎟ q4,5 . 1 r4,5 = ⎜ ⎜ ⎟ ⎝ 1√ ⎠ −1 ± 5

(4)

The results for oscillations along the y- and z-axes are the same as for those along the 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: ⎛

⎞ 1 ⎜ 0⎟ ⎟ r1 = ⎜ ⎝−1⎠ q1 , 0

⎛

⎞ 0 ⎜ 1⎟ ⎟ r2 = ⎜ ⎝ 0⎠ q2 , −1

qi = Ai cos(ωi t + ϕi ),

⎛

⎞ 1 ⎜∓1⎟ ⎟ r3,4 = ⎜ ⎝ 1⎠ q3,4 , ∓1   2f f ω1 = ω2 = ω3 = , ω4 = , ml ml

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/m  and ω2 = 2f /(ml) are the same as r1 and r2 . The two other must have the form ⎛ ⎞ a ⎜b⎟ ⎜ ⎟ cos(ωt + ϕ) ⎝ a⎠ b because of the orthogonality condition. To find them, it is sufficient to consider the equations of motion of two particles: m¨x1 + k(2x1 − x5 ) = 0,

f m¨x2 + (2x2 − x5 ) = 0. l

Here x5 =

k(x1 + x3 ) + (f /l)(x2 + x4 ) 2k + 2f /l

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215

is the coordinate of the point where the springs are joined together which is determined from the condition that the potential energy be a minimum for given values of x1,2,3,4 . Solving these equations, we obtain f + kl , ml 2kf ω42 = , m(f + kl) ω32 =

b3 = − b4 =

f a3 ; kl

kl a4 . f

6.30. In this case, the result can be obtained by a simple generalization of the results of the 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 = 0 corresponds to the rotation of the particles along the ring ⎛ ⎞ 1 r1 = ⎝1⎠ (Ct + C1 ). 1 Due to the degeneracy of the frequency ω2 any vector ⎛

⎞ A1 r1 = ⎝A2 ⎠ cos(ωt + ϕ), A3

(1)

m1 A1 + m2 A2 + m3 A3 = 0,

(2)

which satisfies the condition

is a normal oscillation with the frequency ω = ω2 . (The equality (2) is the orthogonality condition for the vector r1 in the metric defined by the coefficients of the quadratic form of the kinetic energy—see problem 6.23). In particular, one can chose such a normal oscillation (1) that the first particle will be at rest: A1 = 0,

m2 A2 + m3 A3 = 0.

If we substitute (3) in the equations of motion (m1 ω2 − k2 − k3 )A1 + k3 A2 + k2 A3 = 0, k3 A1 + (m2 ω2 − k1 − k3 )A2 + k1 A3 = 0, k2 A1 + k1 A2 + (m3 ω2 − k1 − k2 )A3 = 0,

(3)

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Exploring Classical Mechanics

[6.31

we see that they can have a non-zero solution only when k3 A2 + k2 A3 = 0.

(4)

Comparing (3) and (4), we find that m2 k2 = m3 k3 . Repeating the similar reasoning for the cases where the second or the third particle is at rest, we find that the coefficients ki must satisfy the condition m 1 k1 = m 2 k2 = m 3 k3

(5)

for the case of the frequency degeneracy. On the other hand, it is clear from the previous consideration 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 oscillations with the frequencies other than zero. Of these three oscillations only two are linearly independent due to (2). From here it follows that these three oscillations have the same frequency. Thus, (5) is the necessary and sufficient condition for the degeneracy of the frequencies. 6.31. It is helpful to use for the solution the method explained in problem 6.28. a) The normal oscillations are ⎛ ⎞ ⎛ ⎞ 1 1 r1 = ⎝1⎠ (C1 t + C2 ), r2 = ⎝−1⎠ A2 cos(ω2 t + ϕ2 ), 1 0 ⎛ ⎞ 1 r3 = ⎝ 1⎠ A3 cos(ω3 t + ϕ3 ), −2 (3 + 2ε)k 3k δk ω22 = , ω32 = , ε= ; m m k

(1)

for small ε, they are close to the oscillations (6) of problem 6.19: the amplitudes of the oscillation are the same, but all frequencies are different. Therefore, although in 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 ⎛ ⎞ 1 r1 = ⎝1⎠ (C1 t + C2 ), 1

⎛

⎞ 1 r2 = ⎝−1⎠ A2 cos(ω2 t + ϕ2 ), 0

(2)

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§6. Small oscillations of systems with several degrees of freedom

217

⎛

⎞ 1 ⎜ ⎟ r3 = ⎝ 1 ⎠ A3 cos(ω3 t + ϕ3 ), 2 −1 + ε 3k 3+ε k δm ω22 = , ω32 = , ε= m 1+ε m m

(2)

are for small ε close to the oscillations (6) of problem 6.19. If the extra mass had been added to particle 2, the normal oscillations ⎛ ⎞ ⎛ ⎞ 1 1 r1 = ⎝1⎠ (C1 t + C2 ), r2 = ⎝ 0⎠ a2 cos(ω2 t + ϕ2 ), 1 −1 ⎛ ⎞ 1 ⎜ 2 ⎟ r3 = ⎝− 1 + ε ⎠ A3 cos(ω3 t + ϕ3 ), 1 would be close ⎛ ⎞to the superposition of the normal oscillations (6) of problem 6.19. 1 c) r1 = ⎝1⎠ (C1 t + C2 ), 1 ⎞ ⎛ ⎞⎫ 1 0 ⎬ 0 ⎠ + b2,3 ⎝ 1 ⎠ cos(ω2,3 t + ϕ2,3 ), r2,3 = a2,3 ⎝ ⎭ ⎩ −1 − ε1 −1 − ε2 ⎧ ⎨

⎛

where   b2,3 1 2 2 = ε2 − ε1 ± ε1 + ε2 − ε1 ε2 , a2,3 ε2   k 2 2 2 ω2,3 ≈ 3 − ε1 − ε2 ∓ ε1 + ε2 − ε1 ε2 , m

εi

=

δmi . m

6.32. a) We expand the initial displacement ⎛

⎞ a r(0) = ⎝ 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)

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[6.33

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We do the same for the initial velocities: r˙ (0) = r˙ 1 (0) + r˙ 2 (0) + r˙ 3 (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 ⎞ ⎛ ⎞ ⎛ cos(εω3 t/6) cos ω3 t cos ω2 t + cosω3 t a⎝ − cos ω2 t + cos ω3 t ⎠ ≈ a ⎝ sin(εω3 t/6) sin ω3 t ⎠ . r= 2 −2 cos ω t − cos ω t 3

3

The motion of the particles 1 and 2 thus shows beats with a frequency which is determined by the perturbation δk, while the particle 3 performs a simple oscillation with frequency ω3 . We emphasize that even a very small perturbation δk leads to secular changes which become appreciable after sufficiently long times (cf. problem 2.40). 6.33. a), b) ⎛

⎞ 1 ⎜−1⎟ ⎟ r1 = ⎜ ⎝−1⎠ q1 (t), 1

⎛

⎞ 1 ⎜ 1⎟ ⎟ r2 = ⎜ ⎝−1⎠ q2 (t), −1

⎛

⎞ 1 ⎜∓1⎟ ⎟ r3,4 = ⎜ ⎝ 1⎠ q3,4 (t); ∓1

c) the same as in problem 6.22, (1).   6.34. x1,2 = −x3,4 = ± 12 a cos (2k + 2δk)/m t + 12 a cos 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 normal oscillations will be small, so we use the method of successive approximations. It is helpful to use the normal coordinates of the original system (see problem 6.25)  (l) xi = Ai ql . l

In this case, δL takes the form δL =

1 (δMls q˙l q˙s − δKls ql qs ), 2

(1)

l,s

where δMls =

 i, j

(l)

(s)

δmij Ai Aj ,

δKls =

 i, j

(l)

(s)

δkij Ai Aj ;

(2)

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6.35]

§6. Small oscillations of systems with several degrees of freedom

219

the equations of motion are Ml (¨ql + ωl2 ql ) = −



(δMls q¨ s + δKls qs ).

(3)

s

Assuming that only the oscillation qn is excited in zero approximation, we can leave in the right-hand side of (3) only the term s = n. To determine the correction to the frequency ωn , it is enough to write the single equation (with l = n): (Mn + δMnn )¨qn + (Mnn ωn2 + δKnn )qn = 0, where (ωn + δωn )2 =

Mnn ωn2 + δKnn , Mn + δMnn

as a result, δωn =

δKnn ωn δMnn − . 2ωn Mn 2Mn

(4)

The equations with l = n allow us to find corrections to the vector of the normal oscillation. In this case, the right-hand sides of the equations can be considered as the given forces with the frequency ωn . The excitation of the oscillations ql , as we have expected, is weak, as these “forces” are small. We can also get the following approximation for the ωn and the vectors of the normal oscillations (e.g. see [14], Ch. 1, § 5). It is worth noting that the value δMnn in (2) represents a supplement to the doubled (n) kinetic energy of the system provided that the velocities x˙i = Ai . From this, it follows, in particular, that with increasing of particles’ masses, the quantity δMnn  0 and, according to (4), δωn  0. In the similar way, it is easy to see that if one increases the stiffness of the 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 the degenerate frequency (let ωp = ωn ). In this case, the “force” in the right-hand sides of the 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 righthand sides only the terms with s = n, p Mn (¨qn + ωn2 qn ) = −δMnn q¨ n − δMnp q¨ p − δKnn qn − δKnp qp , Mp (¨qp + ωp2 qp ) = −δMpn q¨ n − δMpp q¨ p − δKpn qn − δKpp qp .

(5)

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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 degenerate ones, one can drop all terms in the L (see (1)) which contain products of the normal 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 that δω1 = 0. For other frequencies, we have the correction  (rn )i δmij (rn )j δωn = −

ωn i, j ,  2 (rn )i mij (rn )j

n = 2, 3, 4.

(1)

i, j

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 vectors of the normal oscillations rn , which have been found in problem 6.28, we obtain δω2 = − 14

εω2 ,

√ 3∓ 5 δω3,4 = − εω3,4 . 40

6.37. We choose the vector potential to be A(r) = 12 B(−y, x, 0); the Lagrangian function is then     L = 12 m x˙2 + y˙2 + z˙2 − 12 m ω12 x2 + ω22 y2 + ω32 z2 + 12 mωB (xy˙ − yx), ˙ eB . For x and y, we get where ωB = mc x¨ + ω12 x − ωB y˙ = 0, y¨ + ω22 y + ωB x˙ = 0. It is helpful to look for oscillations in the form x = Re(Aei t ),

y = Re(Bei t ).

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The set of equations (ω12 − 2 )A − iωB B = 0, iωB A + (ω22 − 2 )B = 0 leads to the oscillations   x = Re Ak ei k t = ak cos( k t + ϕk ),   −iωB k i k t ω k e y = Re Ak 2 = ak 2 B 2 sin( k t + ϕk ), 2 ω2 − k ω2 − k Ak = ak eiϕk ,

k = 1, 2

with frequency

21,2 =

1 2

2 ω12 + ω22 + ωB ±

 2 2 ω1 + ω22 + ωB

2

 − 4ω12 ω22 ,

for which 1 2 = ω1 ω2 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, while the second oscillation is a counterclockwise motion along an ellipse with its major axis along the y-axis. The motion along the z-axis turns out to be a harmonic motion which is independent of the magnetic field, z = a3 cos(ω3 t + ϕ3 ). The free motion of the oscillator is a superposition of the oscillations obtained. We call these 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 in phase. It is impossible to reduce the Lagrangian function to diagonal form using only a linear transformation of the coordinates as the transition to normal coordinates in the present 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 oscillations are strongly elongated, and the frequencies 1,2 ≈ ω1,2 ±

2ω ωB 1,2

2(ω12 − ω22 )

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[6.38

are close to ω1,2 . The orbit of the oscillator without a magnetic field fills a rectangle with sides parallel to the coordinate axes (see problem 6.5); the influence of a weak magnetic field 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 frequency 1 ≈ ωB takes place along circle, and the normal oscillation with frequency 2 ≈ ω1 ω2 /ωB along an ellipse with axes which are in the x- and y-directions and which stand in the ratio ω2 /ω1 . The motion is thus along the circle with a centre which moves slowly along the ellipse. It is well-known that if a charged particle moves in a strong uniform magnetic field in 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 changes little within the circular orbit) leads to a slow ∂r 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 drift occurs 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=

2. ω12 + 14 ωB

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 isotropic oscillator with frequency ! ω. Indeed, the sum and difference of such oscillations with equal amplitudes,     cos ! ωt cos ! ωt ± − sin ! ωt sin ! ωt are linear oscillations in the x- or y-directions (if we neglect the motion in the direction of the magnetic field). If the magnetic field is weak, ωB  ω1 , we have ! ω ≈ ω1 , and the whole effect of the field on the motion of the oscillator reduces to a rotation (“precession”) around the zaxis with a frequency (− 12 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 we have employed. 6.38. We can solve the equations of motion, ˙ x¨ + ω12 x = ωz y, ˙ y¨ + ω22 y = −ωz x˙ + ωx z, z¨ + ω32 z = −ωx y, ˙ with ωx =

eBx , mc

ωz =

eBz mc

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by the method of successive approximations. We look for the coordinates in the form x = x(1) + x(2) , y = y(1) + y(2) , z = z(1) + z(2) , where x(2) , y(2) , and z(2) are small compared to x(1) , y(1) , and z(1) . In first approximation we neglect small terms in the righthand sides of the equations: x(1) = A cos(ω1 t + α), y(1) = B cos(ω2 t + β), z(1) = C cos(ω3 t + γ ). We get −ωz ω2 B sin(ω2 t + β) , ω12 − ω22 ω1 ωz A sin(ω1 t + α) ωx ω3 C sin(ω3 t + γ ) − , y(2) = ω22 − ω12 ω22 − ω32 ωx ω2 B sin(ω2 t + β) z(2) = . ω32 − ω22

x(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 along the 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 of the “forces” ωz y˙(1) and −ωz x˙(1) in (1) turn out to lie close to the eigen-frequencies of the oscillator. In that case we must retain the resonant terms in the first approximation equations: x¨ (1) + ω12 x(1) − ωz y˙(1) = 0, y¨ (1) + ω22 x(1) + ωz x˙(1) = 0,

(3)

z¨ (1) + ω32 z(1) = 0; that is, it is necessary to take the effect of Bz on the motion exactly into account. We consider the set (3) in problem 6.37. For the second-order corrections, we have the equations x¨ (2) + ω12 x(2) = 0, y¨ (2) + ω22 y(2) = ωx z˙(1) , z¨ (2) + ω32 z(2) = −ωx y˙(1) .

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[6.39

For the sake of simplicity, we restrict ourselves to the case ω1 = ω2 ≡ ω  ωz and we have then the following normal oscillations: ⎧ ⎛ ⎞ ⎛ ⎞ 1 ⎪ x ⎨ ⎟ i(ω+ωz /2)t i ⎝y⎠ = Re A1 ⎜ + ⎝ ωω ⎠e x ⎪ ⎩ z 2 2 ω3 − ω ⎫ ⎞ ⎛ ⎞ ⎛ 0 1 ⎪ ⎬ ⎜ −i ⎟ i(ω−ωz /2)t ⎜ iωx ω3 ⎟ iω3 t + A2 ⎝ −ωω ⎠ e . + A3 ⎝ 2 2 ⎠e ω3 − ω x ⎪ ⎭ 2 2 1 ω3 − ω

(4)

In the approximation used, the normal oscillations (4) with frequencies ω ± 12 ωz , take place along a circle in planes which make angles ∓ωx ω/(ω32 − ω2 ) with the (x, y)-plane, while the oscillation with frequency ω3 is along an ellipse in the (y, z)-plane which is strongly elongated in the z-direction. 6.39. We assume that the oscillations of the pendulum are small oscillations: we reckon the angle ϕ counterclockwise from the vertical, and as the second coordinate we choose the charge q on the right-hand plate. When the pendulum is deflected over an angle ϕ, the magnetic flux through the circuit is equal to  = const − 12 Bl 2 ϕ, so that the Lagrangian function of the system is (see problem 4.24)  L=

1 2

 q2 2 ml ϕ˙ + L q˙ − mglϕ − − Bl ϕ q˙ . C 2 2

2

2

 If we introduce the coordinates x = lϕ and y = L /m q, the Lagrangian function of g our system differs from the one considered in problem 6.37 (with parameters ω12 = , l and z = 0) only by a total derivative with respect to the time: ω22 = 1 , ωB = − √Bl LC 2 mL d(xy) 1 2 mωB dt . The equations of motion of problem 6.37 and their solutions are thus also valid in the present case. 6.40. Let r = A cos(ωt + ϕ),

A = (A1 , A2 , . . . , AN )

be any normal oscillation. Since the replacement xi →

 j

Sij xj

(1)

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does not change the form of the Lagrangian function, a normal oscillation of the form ˆ = SA ˆ cos(ωt + ϕ), Sr

ˆ i= (SA)



Sij Aj

(2)

j

must be valid together with (1). Here Sˆ is a matrix with elements Sij , which have the properties: ˆ Sˆ T = S,

ˆ Sˆ Sˆ = E,

(3)

ˆ where Eˆ is a unit matrix and Sˆ T —the transposed matrix S. a) If the given frequency ω is not degenerate, the solution (2) can differ from (1) ˆ = c r, moreover, except for a common factor: Sr ˆ = c Sr ˆ = c2 r. Sˆ Sr

(4)

ˆ we obtain from (4) that r = c2 r, or c2 = 1 and c = ±1. Therefore, for a Since Sˆ Sˆ = E, non-degenerate frequency, we get or

ˆ = +r, or Sr ˆ = −r. Sr

b) If the frequency ω is degenerate, then oscillations (1) and (2) may not coincide. But their sum and difference ˆ = (A ± SA) ˆ cos(ωt + ϕ) r ± Sr are the normal oscillations with the same frequency and have all necessary properties of symmetry.  c) The addition to the Lagrangian function has the form L = fi xi , where i

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, ˆ = +f , Sf

ˆ a = −ra . Sr

(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)

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It follows from (5) that ˆ , Sr ˆ a ) = −(f , ra ). (Sf

(7)

On the other hand, the left-hand side of (7) can be rewritten as ˆ a ) = +(f , ra ) ˆ , Sr ˆ a ) = (f , Sˆ T Sr (Sf

(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 T ˆ condition Sˆ = S?

6.41. Let xi be the displacement of the ith particle along the ring from the equilibrium position; for definiteness, we consider a counterclockwise displacement as positive. The system is clearly symmetric with respect to rotation over an angle 180◦ around the ABaxis which passes through the equilibrium position of the second particle and the centre of the ring. Therefore, the Lagrangian function of the system

L=

1 2m



x˙21 + x˙22 + x˙23



+

1 2M



x˙24 + x˙25



−

1 2k

4  (xi − xi+1 )2 + (x5 − x1 )2

i=1

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 orthogonality conditions, it is easy to reduce this problem with five degrees of freedom to the two independent problems with two degrees of freedom each. Indeed, the vectors of the normal oscillation, which are symmetric and anti-symmetric with respect to the transformation (1), have the form ⎛

⎞ a ⎜ 0⎟ ⎜ ⎟ ⎟ rs = ⎜ ⎜−a⎟ cos(ωs t + ϕs ), ⎝ b⎠ −b

⎛ ⎞ c ⎜d ⎟ ⎜ ⎟ ⎟ ra = ⎜ ⎜ c ⎟ cos(ωa t + ϕa ). ⎝f ⎠ f

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In addition, one anti-symmetric “oscillation” can be easily guessed—it is a rotation of all the particles around the ring ⎛ ⎞ 1 ⎜1⎟ ⎜ ⎟ ⎟ ra1 = ⎜ ⎜1⎟ (Ct + C1 ), ⎝1⎠ 1

ωa1 = 0.

Two other (besides ra1 ) anti-symmetric oscillations must be orthogonal to ra1 with the metric 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 determine them, it is enough to use only two equations of motion out of five, for example, those for the first and fifth particles: m¨x1 + k(2x1 − x2 − x5 ) = 0, M x¨ 5 + k(2x5 − x4 − x1 ) = 0.

(3)

Substituting the explicit form of rs , we find for the two symmetrical oscillations  2 ωs1,2 − 2 a1,2 , k   k  2 2M + 3m ∓ 4(M − m)2 + 5m2 . ωs1,2 = 2mM b1,2 =

m

Similarly, substituting in (3) the vector ra and taking into account (2), we find 

 M 2 2M f2,3 , c2,3 = 1 − ωa2,3 f2,3 , d2,3 = −2c2,3 − k m    k 1 1 2 ωa2,3 = 4M + m ∓ (4M − m)2 + m2 . 2mM 2 2 6.42. The considered system is close to the one studied in problem 6.29a; the Lagrangian function in our problem differs by a small quantity δL = δL1 (x, x) ˙ + δL2 (y, y) ˙ + δL3 (z, z), ˙   ˙ = 12 εk x22 + (x2 − x5 )2 + (x4 − x5 )2 + x24 , δL1 (x, x)

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where ε = (l − l1 )/l  1. The oscillations in x, y, and z occur independently; in the following we are only interested in the x-oscillations. To determine the corrections to the frequencies, it is helpful to use the method of successive 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 the normal 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 the normal oscillations of the original system, which change a little when we add δL. These are the vectors (1) from problem 6.29, because only they have the certain symmetry properties. Namely, the oscillation r3 is symmetric with respect to the AB-axis and antisymmetric with respect to the CD-axis, and the oscillation r1 is symmetric and r2 , antisymmetric with respect to both axes. Corrections to the frequencies of these oscillations can 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 (x˙1 = x˙3 = x˙5 = 0, x˙2 = −x˙4 = 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 ⎛

⎞ a ⎜ 0⎟ ⎜ ⎟ ⎟ r(0) = ⎜ ⎜ 0⎟ ⎝−a⎠ 0

and the vector of the initial velocities r˙ (0) = 0 as r(0) = respectively, we find that A1 = A2 = A3 = 12 a,



i ri (0)

A4 = A5 = ϕi = 0.

and r˙ (0) =



˙ i (0), ir

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Thus, in the given approximation, the fourth and fifth normal oscillations are not excited, and the oscillations of the particles x1,3 = 12 a(± cos ω1 t + cos ω3 t), x2,4 = 12 a(± cos ω2 t − cos ω3 t),

x5 = 0

are beats (see problem 6.32). 6.43. In this problem, it is helpful to use the method of successive approximations (see problem 6.35). The change in mass leads to the correction to the Lagrangian function of the form δL = 12 (δm1 x˙21 + δm2 x˙22 ). This correction must be expressed in terms of the normal coordinates of the original system (see problem 6.22). In this case, the coefficient in front of the product of the generalized velocities q˙1 q˙2 corresponding to the degenerate frequencies is zero. Other products q˙l q˙s (for ωl = ωs ) can be omitted as this issue has been noted in problem 6.35. We get δL1 = 14 δm1 q˙21 + 14 δm2 q˙22 + 18 (δm1 + δm2 )(q˙23 + q˙24 ). The Lagrangian function L + δL1 , as well as the Lagrangian function of the original system, 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 corrections to the frequencies, one can use (4) of problem 6.35: δω1 = − 14 ε1 ω1 , δω4 = 0,

δω2 = − 14 ε2 ω3 ,

δω3 = − 18 (ε1 + ε2 )ω3 ,

εi = δmi /m.

All eigen-frequencies of the system become different; the ambiguity in choice of the vectors of the normal oscillations disappears since with the accuracy up to εi these vectors are the vectors (1) from problem 6.22. It is noteworthy that when δm1 = δm2 , the frequencies ω1 + δω1 and ω2 + δω2 coincide with each other again (up to the second-order corrections |δω1 − δω2 | ∼ ε12 ω1 ). In this case, the Lagrangian function L + δL1 once more leads to the ambiguous choice of the vectors of the normal oscillations. However, in the exact solution of the problem at δm1 = δm2 , the vectors of the normal oscillations have the form ⎛

⎞ 1 ⎜ 1 ⎟ ⎜ ⎟ ⎝−1 − ε ⎠ , −1 − ε

⎛

⎞ 1 ⎜ ⎟ √ −1 ⎟ ⎜ 1 ⎝ (3 ∓ 4 + 4ε + 9ε2 ) ⎠ , 2 √ − 12 (3 ∓ 4 + 4ε + 9ε2 )

⎛ ⎞ 1 ⎜1⎟ ⎜ ⎟ ⎝1⎠ 1

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and at small ε they are close to the vectors (2) of problem 6.22 (up to the normalization factors, omitted here). The sharp change in the form of the normal oscillations occurs in a very narrow range of masses |δm1 − δm2 |  ε2 m (cf. problem 6.6a). To determine the vectors of the normal oscillations in this range of values of δm1 and δm2 , it would be possible 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 is independent. We will consider only the motion along the AA-axis. For the first and fourth particles, we assume the displacements to be positive to the left, while for the second and third, positive to the right. According to the results of problem 6.40, the normal oscillations r = (x1 , x2 , x3 , x4 ) can be chosen either symmetric or anti-symmetric with respect to the AA- and BB-axes. The oscillations which are symmetric with respect to the AA-axis have x1 = x4 , x2 = x3 . If they are also symmetric with respect to the BB-axis, then x1 = x2 , x3 = x4 , so for this twofold symmetric oscillation we have rss = (1, 1, 1, 1)qss . For the oscillations which are symmetric with respect to the AA-axis and antisymmetric 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 same way. The oscillation frequencies can be found by substituting the vectors obtained in the equations of motion. In the presence of degeneration, there are a lot of normal oscillations (except for the ones as previously described) which do not have the previously described symmetry properties. It is easy to figure out,for example, that the frequencies of the ss and aa oscillations 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). In that 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 system with seven degrees of freedom to a few simple properties with not more than two degrees of freedom each. Indeed, due to the symmetry of the system with respect to a plane perpendicular to the plane of the scales, all the normal oscillations can be chosen either symmetric or anti-symmetric with respect to this plane. In addition, the normal

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oscillations are divided into those which displace the particles from the plane of the scales and 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 the frame, threads BA, and DE, respectively. Apart from the obvious symmetric oscillations  α = 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 Lagrangian function are additive, then to find the anti-symmetric oscillations, it will suffice to know only term ˙ − mgl(α 2 + 3β 2 ), La = ml 2 (2α˙ 2 + 9β˙ 2 + 3α˙ β) corresponding to this type of oscillations.  As a result, we obtain two anti-symmetric oscillations: 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 position along of the x-axis, which in turn is directed perpendicular to the equilibrium plane of the scales. It is obvious that the symmetric oscillations xA = xE ,

xB = xD

coincide with the oscillations of the coplanar double pendulum1 ; therefore,  √  g , ωs2 = 7 ∓ 37 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 other anti-symmetric oscillation is orthogonal to the one we found, and therefore, we have xA = −xB = xD = −xE . But this oscillation does not shift the centre of each thread in the first approximation, so the frequency of such oscillations coincides with the frequency  of 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 direction perpendicular to the plane of the frame are independent. As an example, we will consider the oscillations along the AA-axis. 1 See [1], § 23, problem 2 with parameters

2xB x − xB m1 = m2 = 2m, l1 = 12 l, l2 = 3l, ϕ1 = , ϕ2 = A . l 3l

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It is helpful to place the components of the oscillation vector in the table similar in shape to the frame. For particles on the left from the axis-BB, we take the displacements to the left as positive, while for the particles located on the right we take the displacements to the right as positive. The oscillations ss symmetric with respect to both AA- and BBaxes have the form 

x X x X

X X

 x . x

The oscillations x, X are reduced to the oscillations of the system considered in problem 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 antisymmetric with respect to the BB-axis, have the form 

x X x X

−X −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 −x −X

−X X



 x , k1 = k + 3k , k3 = 2k + 3k , −x  −x , k1 = k + 3k , k3 = 3k . x

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 a cos γ t. −k − k , −k , k , k + k



It is orthogonal to all the normal oscillations, except for symmetric–anti-symmetric (sa). sa , where Therefore, the resonance only occurs for two frequencies γ = ω1,2 sa 2 (ω1,2 ) =

# 1 f k(2M + m) + (M + m)± 2mM l ⎫ $

2 ⎬ f ± k(2M − m) + (M − m) + 4k2 Mm . ⎭ l

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6.48. The linear molecule C2 H2 has only seven normal oscillations: two flexural oscillations 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-symmetric longitudinal oscillation x1 = x4 , x2 = x3 , mH x1 + mC x2 = 0 has the frequency $ ωa1 =

kHC (mH + mC ) mH mC

(the same as the “molecule” CH). Note that the longitudinal displacement of the molecule as a whole, x1 = x2 = x3 = x4 , can be considered as the second anti-symmetric oscillation, and the condition of its orthogonality to the first anti-symmetric oscillation coincides with the requirement that the total linear momentum of the molecule is zero. The transverse symmetric oscillation reads y1 = y4 , y2 = y3 , mH y1 + mC y2 = 0,

2 ωs3 =

kHCC (mH + mC ) . 2 m m lHC H C

Here kHCC is the stiffness of the molecule at the flexure; that is, the potential energy increases by the term kHCC δ 2 /2 when the bond HCC bends over an angle δ. The transverse anti-symmetric oscillation reads y1 = −y4 , y2 = −y3 , mC lCC y2 + mH (lCC + 2lHC )y1 = 0, 2 ωa2 =

2 + m (2l 2 kHCC [mC lCC H HC + lCC ) ] 2 l2 mH mC lHC CC

,

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 this oscillation can be found from the requirement that the total angular momentum of the molecule is zero (or due to orthogonality to the vector of the rotation of a molecule as a whole y1 = −y4 , y2 = −y3 , y2 (2lHC + lCC ) = y1 lCC , which can be considered as the second anti-symmetric oscillation). 6.49. We denote the displacements of the particle from the equilibrium position in the direction BD through x1 and x2 , and in the direction CF—through y1 and y2 . The rotation around the axis of symmetry CF over the angle 180◦ leads to the replacements: x1  −x2 ,

y1  y2 ;

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ˆ which is the rotation of the oscillation vector r = that is, the transformation S, (x1 , y1 , x2 , y2 ), has the form ˆ = (−x2 , y2 , −x1 , y1 ). Sr Due to the symmetry of the system, the normal oscillations of two types are possible in ˆ s = rs ) and anti-symmetric ra it: rs – symmetric with respect to the CF-axis (for which Sr ˆ a = −ra ). For the symmetric oscillations, we have (−x2 , y2 , −x1 , y1 ) = (for which Sr (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 equivalent to the change of the oscillation phase by π . Two symmetric and two anti-symmetric oscillations are possible; to determine the frequency and vectors of each pair of oscillations, it is sufficient to use only two equations of motion. We can obtain some more information about the type of normal oscillations without knowing stiffness of the springs. The condition of orthogonality of the oscillation vectors rs1 and rs2 can be reduced to equality 2m(xs1 xs2 + ys1 ys2 ) = 0, which shows that the vectors of the displacements of, for example, particle 1 at each of the two symmetrical oscillations are mutually orthogonal (Fig. 141a). The same applies to the antisymmetric oscillations (Fig. 141b). (a)

(b) s1

a1

s1

a2 s2

a2

s2 a1

Figure 141

The direction in which the particle displaces at each of the normal oscillations cannot be determined without knowing stiffness of the springs. Indeed, if stiffness and tension springs AC and CE are small, then the particle displacements under the normal oscillations are directed almost along or across the springs BD. On the contrary,

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if the stiffness of springs BD is low, the normal oscillations occur almost along or across the springs AC and CE. Of course, if the normal oscillation is degenerate, one can choose other vectors of the normal oscillations which do not possess the symmetry properties. For example, if we “switch off” spring BD, each of the particles will be able to oscillate along or across springs AC and CE. 6.50. We denote the displacement of each atom from its equilibrium position in the ODdirections through xi , in the OA-directions through yi , and perpendicular to the plane of the molecule through zi . The oscillations of the molecule C2 H4 are divided into those which leave the molecule planar and those which output atoms of the plane of the molecule. We will start with the latter. It is helpful to present the vector of displacement of the atoms in the form ⎛ r=⎝

z1

z4 z2 z5

z3

⎞ ⎠.

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, ⎛

⎞ 1 ⎠ qsa . 00 rsa = ⎝ −1 −1 1

This “oscillation” turns out to be, in fact, the rotation around the CD-axis. Similarly, the symmetric oscillations with respect to both axes are ⎛ ⎞ z1 z1 rss = ⎝ z2 z2 ⎠ . z1 z1

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[6.50

One of them, in particular, is ⎛

⎞ 1 rss,1 = ⎝ 1 1 ⎠ qss,1 . 1 1 1

This is a translational motion. The amplitudes of the other oscillation can be determined by 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: ⎛

⎞ 1 rss,2 = ⎝ − 2m − 2m ⎠ qss,2 . M M 1 1 1

The antisymmetric oscillation with respect to both axes, ⎛

⎞ −1 ⎠ qaa , 00 raa = ⎝ −1 1 1

is a torsional oscillation around the CD-axis. Finally, ⎛

ras,i

⎞ −1 ⎠ qas,i . = ⎝ ai −ai 1 −1 1

Here for the rotation around the AB-axis (the oscillation rsa,1 ), the factor a1 = lOC /lOH is the ratio of the distances of the carbon atoms and hydrogen atoms from the AB-axis in the equilibrium position; for the flexural oscillation (ras,2 ), it is a2 = −

2m lOH . M lOC

The similar approach can also be applied to the oscillations which do not output atoms from the plane of the molecule.

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237

We can present the corresponding displacement vectors in the form ⎛ r=⎝

x1 y1

x4 y4 x2 y2 x5 y5

x3 y3

⎞ ⎠.

x6 y6

The general form of the as-oscillations is ⎛ ras = ⎝

x1 −y1

x1 y1 x1 −y1

x2 0 x2 0

⎞ ⎠.

x1 y1

One of the as-“oscillations” is the translational motion of the molecule as a whole along the x-axis (x1 = x2 , y1 = 0). The other two as-oscillations are shown in Fig. 142a and b. (a)

(b) as

(c) as

(d)

(e) sa

sa (f)

aa

(g)

(h) ss

aa

(i) ss

ss

Figure 142

To visualize the kind of displacements of atoms under these vibrations, note that the distance between the carbon atoms does not change: the C–C link “does not work”. If we 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 will coincide with those which are symmetric with respect to the CD-axis oscillations of two “molecules” H2 C occurring in anti-phase (cf. problem 6.48; oscillations of molecules of the form A2 B are considered in [1], § 24, problem 2). The general form of sa-oscillations is ⎛ rsa = ⎝

−x1 y1

x1 y1 −x1 y1

0 y2 0 y2

⎞ ⎠.

x1 y1

In addition to the translational motion in the direction of the y-axis, there are two other oscillations (Fig. 142c and d). One of them (Fig. 142c) can be presented as the

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[6.51

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oscillations of the two “molecules” H2 C which are antisymmetric with respect to the CDaxis and occur in phase. The other (Fig. 142d) is the rotations of the “ molecules” H2 C in the opposite directions. If we completely neglect the constraints of the distant atoms and 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 xyplane, the anti-symmetric oscillations of the “molecules” H2 C 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(u˙ 21 + u˙ 22 + u˙ 23 ) − 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 atom from its equilibrium position, which is determined by the radius vector ra0 . Since |ua |  l, we have L = 12 m(u˙ 21 + u˙ 22 + u˙ 23 )− % & − 12 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(u˙ 1 + u˙ 2 + u˙ 3 ) = 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

 [ra0 + ua , u˙ a ], a

vanishes up to and including terms of first order in ua .

(4)

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§6. Small oscillations of systems with several degrees of freedom u1

It is helpful for the description of the motion to introduce for each atom its own Cartesian system of coordinates (Fig. 143), thus retaining symmetry in the description of the system. Equations (2) and (3) in terms of these coordinates read1     √ √ 3 3 1 1 x2 + − y3 + x3 = 0, y1 + − y2 + 2 2 2 2 (5)

x1

y1 r10 r20

r30

y3

x3

x2

y2

Figure 143



   √ √ 3 3 1 1 y2 + − y3 + x3 + − y1 − x1 = 0, 2 2 2 2 (6) and hence y1 =

x3 − x2 √ , 3

y2 =

x1 − x3 √ , 3

y3 =

x2 − x1 √ , 3

and L=

 m  5m  2 3  x˙1 + x˙22 + x˙23 − (x˙1 x˙2 + x˙2 x˙3 + x˙3 x˙1 ) − k x21 + x22 + x23 . 6 3 2

One normal oscillation (totally symmetric) is obvious: 1 (1) (1) (1) x1 = x2 = x3 = √ q1 . 3

(7)

The two other oscillations are orthogonal to the first oscillation which leads to the condition2 (s)

(s)

(s)

x1 + x2 + x3 = 0,

s = 2, 3.

(8) (2)

(2)

One of these oscillations is symmetric with respect to the x1 -axis: x2 = x3 ; the other is antisymmetric: x1(3) = 0, x2(3) = −x3(3) . 1 For instance, multiplying both sides of the equality (2) by e , we get (5). Note that the vector e 23 23 in the different coordinate systems has the coordinates   √  √ 3 3 1 1 ,− ,− = − , e23 = (0, 1)1 = 2 2 2 2 2

3

while ua = (xa , ya )a . Equality (6) follows from (5) through a cyclic permutation of the indices. 2 The metric k is used. In (7) and (9) the factors before q are chosen so that x(l)2 + x(l)2 + x(l)2 = q2 . ij 1 2 3 l

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[6.51

Taking into account (8), we have  (2) x1

(2) = −2x2

(2) = −2x3

=

2 q2 , 3

1 x2(3) = −x3(3) = √ q3 . 2

(9)

The replacement,  2 1 x1 = √ q1 + q2 , 3 3 1 1 1 x2 = √ q1 − √ q2 + √ q3 , 3 6 2 1 1 1 x3 = √ q1 − √ q2 − √ q3 , 3 6 2 leads to the Lagrangian function of the form L = 12 m(q˙21 + 2q˙22 + 2q˙23 ) − 32 k(q21 + q22 + q23 ).

(10)

The normal oscillations corresponding to these coordinates are shown in Fig. 144. Their frequencies are

Figure 144

 ω1 =

3k , m

 ω2 = ω3 =

3k . 2m

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 ' [ua , u˙ a ]' = m|q2 q˙3 − q3 q˙2 | ' a ' 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.

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241

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 degeneracy will remain. Indeed, together with a q-oscillation, we can also have another oscillation obtained from the original q-oscillation by a rotation over 2π/3. Its frequency must be the 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 two independent oscillations with the same frequency. The normal coordinates must in this case satisfy only one condition: they must be orthogonal to q1 ; in particular, q2 and q3 remain normal coordinates. 6.52. a) We introduce the coordinates of atoms B in the same way as in the preceding problem; for atom A—the coordinates will be x4 , y4 , z4 with the axes parallel to x1 , y1 , z1 and 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 rotations around the x4 - and y4 -axes, and one to the oscillation (for which, obviously, z1 = z2 = z3 , mA z4 + m(z1 + z2 + z3 ) = 0). The frequency of this oscillation ω1 is not degenerate; it is only 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 to the 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 translational motion 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 the frequencies ω3 and ω4 which break symmetry of the molecule (Fig. 145b and c). (a)

(b)

(c)

(d) −

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 that can be obtained from the previously discussed oscillations with the frequencies ω3 and ω4 by a rotation on an angle 2π/3 around the z4 -axis (cf. preceding problem).

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[6.53

As a result, the frequencies ω1 and ω2 are non-degenerate, while the frequencies ω3 and 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 sufficient to take a certain superposition of the oscillations obtained from the latter by the rotation around 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 perturbation theory (see problem 6.35): δω = −

ω δM . 2 M

For a totally symmetric oscillation, we choose x1 as a normal coordinate. Then quantities M and δM are defined as the coefficients in the expressions for the kinetic energy and the corrections to it, respectively: 3 2

mx˙21 = 12 M x˙21 ,

1 2

δmx˙21 = 12 δM x˙21 ,

so that δω2 = −ω2

δm . 6m

For the oscillations along the z-axis, the kinetic energy and the correction to it are equal to, respectively, 3 2

m (1 + 3m/mA ) z˙21

and

1 2

δmz˙21 ,

so that δω1 = − 12 ω1

mA δm . 3m(3m + mA )

For example, when we replace one atom of chloride with the atomic weight of 35 by the isotope with the atomic weight of 37 in the molecule of boron chloride, it reduces the frequency ω1 and ω2 by 0.1% and 1%, respectively. 6.53. Let the oscillation for which the molecule retains its shape (Fig. 146a) have a frequency ω1 . The frequency ω2 of the oscillation which retains its form under rotations around OD over 2π/3 (Fig. 146b) is, in general, different from ω1 . One can obtain another displacement of the atoms by taken a reflection in the plane BCO; we obtain oscillations which differ from the second oscillation only in that atoms A and D change roles. The frequency of that oscillation ω3 = ω2 . Similarly, in the reflection in plane AOC, the roles of

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6.54]

§6. Small oscillations of systems with several degrees of freedom (a)

(b)

A

(c) D

D

D

O C

243

C

A

B

C A

B

B

Figure 146

atoms B and D are changed, while the frequency remains the same, ω4 = ω2 . This fourth oscillation cannot be reduced to a superposition of previous oscillations as, in contrast to 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/3 around OD which results in a cyclic permutation of A, B, and C, leads to an oscillation which is symmetric with respect to planes COA and DOB and its frequency ω6 = ω5 . The molecule has thus three eigen-frequencies which are, respectively, nondegerenate, 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 real molecules. 6.54. a) When a totally symmetric and twofold-degenerate oscillations, considered in the preceding problem (Fig. 146a and b) occur, the carbon atom remains at rest. There are extra two threefold degenerate frequencies. The corresponding oscillations are similar to the ones shown in Fig. 147b. The only difference is that the carbon atom oscillates either in the same direction as the atom D or in the opposite direction. b) The addition to the Lagrangian function describing the action of the electric fields E(t) is δL = E(t)



ea ua ,

a

where ea is the charge  and ua , displacement of ath atom. In any oscillation m 4a=1 ua + mC u5 = 0, so that 5  a=1

ea ua = −e1

m

C

m

 + 4 u5 .

 For the totally symmetric and twofold-degenerate oscillations, are quantity ea ua = 0 and  these oscillations are not excited. For the oscillation of Fig. 147b, in contrast, ea ua = 0 and the similar oscillations are excited. As a result, resonance is possible on two frequencies.

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[6.54

The vector E can be decomposed into three terms Ej which are parallel to axes of symmetry 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 amplitude 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 shown that the oscillation amplitudes of the hydrogen atoms |u1,2,3,4 | do also not depend on the orientation of the molecule and that they are parallel to the vector E.

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§7 Oscillations of linear chains

7.1. The Lagrangian function of the system is L(x, x) ˙ = m 1 2

N 

 x˙2n − 12

k

n=1

x21 +

N 

 (xn − xn−1 )

2

+ x2N

,

(1)

n=2

where the xn is the displacement of the nth particle from its equilibrium position. We also introduce the coordinate of the equilibrium position of the nth particle, Xn = na, where a is the equilibrium length of one spring. The Lagrangian equations of motion m¨x1 + k(2x1 − x2 ) = 0, m¨xn + k(2xn − xn−1 − xn+1 ) = 0, m¨xN + k(2xN − xN−1 ) = 0

n = 2, 3, . . . , N − 1,

(2)

n = 1, 2, . . . , N

(3)

are equivalent to the set m¨xn + k(2xn − xn−1 − xn+1 ) = 0, with the additional condition x0 = xN+1 ≡ 0.

(4)

We expect from physical considerations that the normal oscillations of the system will be 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 = 4

k ϕ sin2 , m 2

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

(6)

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which determines the relationship between the frequency ω and the difference in the phase of neighbouring particles ϕ. The meaning of the substitution (5) consist of the choice for xn of a solution in the form of a travelling wave with a wave vector K = ϕ/a such that nϕ = naK = KXn . Equation (6) thus establishes the relation between the frequency and the wave vector. The conditions of (4) can be satisfied by taking a superposition of the waves travelling in two directions, xn = Aei(ωt−nϕ) + Bei(ωt+nϕ) . The condition x0 = 0 gives A = −B, or xn = 2iB sin(nϕ)eiωt , that is, a standing wave. From the condition at the other end, xN+1 = 0, we determine the spectrum of possible frequencies. The equation sin(N + 1)ϕ = 0 has N independent solutions ϕs =

πs , N +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 in terms of the solutions corresponding for s = N − l + 2. From (6) and (7), we find N different frequencies: 

 k ϕs k πs ωs = 2 sin = 2 sin , m 2 m 2(N + 1)

s = 1, 2, . . . , N.

The different frequencies are shown in Fig. 147 by the discrete points on the sine curve. The vector of the normal oscillations corresponding to the sth frequency is ⎞ ⎞ ⎛ x1 sin ϕs  ⎟ ⎜ x2 ⎟ 2 ⎜ ⎟ ⎜ sin 2ϕs ⎟ rs = ⎜ ⎝ . . . ⎠ = N + 1 ⎝ . . . ⎠ qs (t), xN sin Nϕx

(8)

ωs

⎛

(9) 1 2

N

Figure 147

where qs (t) = Re(2iBs eIωs t ) = Cs cos(ωs t + αs ) is the sth normal coordinate, while the factor 

N −1/2  2 2 sin nϕs = N +1 n=1

s

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247

is introduced to get the normalization: (rs , rs ) = δss q2s . The general solution is a superposition of all normal oscillations: xn =

N 



s=1

2 qs (t) sin nϕs . N +1

The matrix, given the transition from the xn to the qs ,  Uns =

2 π ns sin , N +1 N +1

is an orthogonal matrix which reduces the Lagrangian function to a sum of squares corresponding a set of N different oscillators: L=

N 

Ls (qs , q˙s ),

L(qs , q˙s ) = 12 m(q˙2s − ωs2 q2s ).

s=1

7.2. The equations of motion for the given system are the same as the equations (3) of the previous problem but now with the additional conditions x0 = 0, xN = xN+1 . Thus, we get (2s − 1)π , s = 1, 2, . . . , N, 2N + 1  (2s − 1)π k ωs = 2 sin , m 2(2N + 1)  sin nϕs · As cos(ωs t + αs ). xn = ϕs =

s

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 generalized coordinates (cf. problem 6.3). In such variables, the given problem is completely reduceble 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 now with the additional conditions x0 = xN and xN+1 = x1 . Therefore, 2π s , s = 0, 1, . . . , N − 1, N  k sπ ωs = 2 sin ; m N ϕs =

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the frequencies ωs and ωN−s are the same, and the corresponding wave vectors Ks =

ϕs 2π 2π − ϕN−s = = −KN−s + a a a

determine the travelling waves moving in opposite directions. The frequency ω0 = 0 corresponds to the motion of all particles along the ring with a constant velocity. In this system, oscillations of the form i(ωs t−nϕs ) x(s) n = Re As e

are possible, that is, waves which travel along the ring. The previously mentioned twofold degeneracy of the frequencies corresponds to waves moving in opposite directions. The presence of two such waves with equal amplitudes gives a standing wave: (N−s) x(s) n ± xn

cos nϕs cos(ωs t + αs ), = 2|As | sin nϕs sin(ωs t + 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  (qs1 cos nϕs + qs2 sin nϕs ) + q0 ,

R = 12 (N − 1),

(N : odd),

(3)

s=1

the Lagrangian function is reduced to diagonal form:

L = 12 Nm q˙20 + 12

 R    q˙2s1 + q˙2s2 − ωs2 (q2s1 + q2s2 ) .

(4)

s=1

(If the number of particles is even, (3) and (4) must be somewhat changed as the frequency 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 = qs1 cos βs − qs2 sin βs , qs2 = qs1 sin βs + qs2 cos βs , which leave the Lagrangian function (4) invariant, correspond to a displacement of a nodes of the standing waves: xn = q0 +

R  [qs1 cos(nϕs − βs ) + qs2 sin(nϕs − βs )]. s=1

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249

The average energy flux along the ring is for the travelling waves (1) given by (cf. problem 6.10)1 ω Sav = 2π

2π/ω 

k(xn−1 − xn )x˙n dt = 12 k|A|2 ω sin ϕ, 0

while the group velocity is dω vgr = = dK



k ϕ a cos , m 2

where a is the equilibrium length of one spring, while K = ϕ/a is the wave vector. The energy is E = 12 m

N  n=1

x˙2n + 12 k

N  ϕ (xn − xn−1 )2 = 2N|A|2 k sin2 = 12 Nmω2 |a|2 , 2

n=1

and we have thus E vgr = Sav . Na 7.5. a) The equations of motion are m¨x2n−1 + k(2x2n−1 − x2n−2 − x2n ) = 0, M x¨ 2n + 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 S and of the av energy E is very helpful carried out using complex variables (cf. [2], § 48).

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which have a non-trivial solution only if its determinant vanishes. This condition determines the relations between the frequency and the difference in phase of neighbouring particles: ⎞ ⎛  2 k 4μ 2 = ⎝1 ∓ 1 − sin2 ϕ ⎠ , ω(∓) μ mM

μ=

The boundary conditions are satisfied only by welldefined linear combinations of the travelling waves (2), namely,

mM . m+M

(4)

ω

√

2k μ

ω(+)

√2km √2k M

ω(−)

x2n−1 = As sin(2n − 1)ϕs cos(ωs t + αs ), x2n = Bs sin 2nϕs cos(ωs t + αs ), 123

N

s

for which ϕs = sπ /(2N + 1). Since ϕ2N+1−s = π − ϕs , we Figure 148 get different frequencies, having chosen one of the two signs 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 the optical branch. The general solution has the form x2n−1 = x2n =

N  s=1 N 

sin(2n − 1)ϕs [A(+)s cos(ω(+)s t + αs )+A(−)s cos(ω(−)s t + βs )], sin 2nϕs [B(+)s cos(ω(+)s t + αs ) + B(−)s cos(ω(−)s t + βs )],

s=1

where the A(±)s and B(±)s are according to (3) connected by the relation B(±)s =

2 2k − mω(±)s

2k cos ϕs

A(±)s .

It is remarkable that B(−)s and A(−)s corresponding to the acoustic frequencies have the 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 distribution of the amplitudes of the oscillations for the case N = 8, s = 2 is shown in Fig. 149 where the 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 (see problem 7.1)?

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251

§7. Oscillations of linear chains (a) xl

l

(b) xl

l

Figure 149

b) The normal oscillations are (s)

x2n = As sin 2nϕs cos(ωs t + αs ), K sin 2nϕs + k sin(2n − 2)ϕs (s) x2n−1 = As cos(ωs t + αs ), k + K − mωs2 where ωs2 =

   1 K + k ∓ (K − k)2 + 4Kk cos2 ϕs m

and ϕs is determined from the equation K −k tan ϕs , K +k 0 < ϕs < π/2.

tan (2N + 1)ϕs = −

s = 1, 2, . . . , N,

Fig. 150a shows the optical and acoustic branches of the frequencies for the case K > k. What happens in the limiting case K = k? π c) We have ϕs = ; for s = 1, 2 . . . N, we get 2N normal vibrations and normal 2(N + 1) frequencies which have the same form as under point 7.5b (Fig. 150b).

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[7.6

Exploring Classical Mechanics (a)

√

2(K+k) m

(b)

ω

√2(K+k) m

ω(+)

ω ω(+)

√2K m 2k √m

ω(−)

ω(−)

s N N+1

s 1 23

N

√2K m K+k √m √2km

1 23

Figure 150

There is one more, (2N + 1), normal oscillation x2n = 0 and Kx2n−1 = −kx2n+1 , the frequency of which, ω02 = (K + k)/m, lies in the forbidden band between the optical and acoustic branches. The distribution of the amplitudes of this oscillation is shown in Fig. 151 where the ordinate and abscissa show, respectively, the numbers of the particles and their amplitudes of the oscillations. The particles with an even numbers are not moving while neighbouring particles with odd numbers are moving in antiphase with the exponentially damped amplitudes, when we move away from the left-hand end of the chain. xl

l

12 3

Figure 151

7.6. a) We look for a solution of the equations of motion, m¨xn + k(2xn − xn−1 − xn+1 ) = 0,

n = 1, 2, . . . , N,

(1)

with the boundary conditions x0 = 0 and xN+1 = a cos γ t in the form of standing waves xn = A sin nϕ cos γ t, so that the first boundary condition is immediately satisfied. From the second boundary condition, we find constant A = a/ sin(N + 1)ϕ, while from (1) we get for the “wave vector” ϕ of the standing wave sin2

ϕ mγ 2 = . 2 4k

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§7. Oscillations of linear chains

253

When γ 2 < 4k/m, the stable oscillations xn = a

sin nϕ cos γ t sin(N + 1)ϕ

(2)

have a larger amplitude when the denominator sin(N + 1)ϕ is close to zero. But this is just the condition which determined the spectrum of the eigen-frequencies ωs (see problem 7.1); that is, we have then a near-resonance situation, γ ≈ ωs . When  γ  ω1 = 2

π k sin , m 2(N + 1)

the oscillations in (2) correspond to a slow extension and compression of all springs as a whole: xn = a

n cos γ t. N +1

If γ 2 > 4k/m, we change ϕ to π − iψ in (2) and get xn = (−1)N+1+n a

sinh nψ cos γ t, sinh(N + 1)ψ

where cosh2

mγ 2 ψ = . 2 4k

The oscillations are (exponentially when nψ  1) damped towards the left-hand end of the chain. The reasonableness of this result is particularly clear when γ 2  4k/m when the frequency of the applied force lies appreciable above the limit of the spectrum of the normal frequencies. In that case, the particle on the extreme right oscillates with a small amplitude in anti-phase with the applied force while the (N − 1)st particle is in first approximation at rest. Then we can consider the motion of the (N − 1)st particle as a forced oscillation caused by an applied force of high frequency arising from the Nst particle, and so on. We note that a similar damping of the wave occurs in the phenomenon of complete internal reflection (e.g., when short wavelength radio-waves are reflected by the ionosphere). 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, then the displacements are

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[7.7

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  xn = Re Aei(γ t−nϕ) , where ϕ is determined by the equation γ2 = 4

ϕ k sin2 . m 2

In the given problem x0 = a cos γ t, so we find A = a. Therefore,   xN+1 = Re Aei(γ t−(N+1)ϕ) = a cos (γ t − (N + 1)ϕ). The energy flux from left to right is equal to the work (averaged over period) of the spring, which connects the (n − 1)th and nth particles, over the nth particle (cf. problem 6.10):  k(xn−1 − xn )x˙n = 12 γ ka2 sin ϕ = 14 mγ 2 a2 (4k/m) − γ 2 .  When γ > 2 k/m,  there will  be no travel wave. c) xn = a

cos N − n + 12 ϕ   cos γ t, cos N + 12 ϕ sin2

ch2

mγ 2 ϕ = , 2 4k

4k when γ 2 < , m  sinh N − n + 12 ψ n   xn = (−1) a cos γ t, sinh N + 12 ψ mγ 2 ψ = , 2 4k

when

γ2 >

4k . m

7.7. If the frequency of the applied force lies in the range of the acoustic eigenfrequencies, 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 sin(2n − 1)ϕ cos γ t, sin(2N + 1)ϕ  sin 2nϕ 2k − mγ 2 a x2n = ± cos γ t, 2k − Mγ 2 sin(2N + 1)ϕ

x2n−1 = a

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§7. Oscillations of linear chains

255

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 cosh(2n − 1)ψ cos γ t, cosh(2N + 1)ψ  2k − mγ 2 sinh 2nψ cos γ t, x2n = (−1)N+n+1 a 2 Mγ − 2k cosh(2N + 1)ψ

x2n−1 = (−1)N+n+1 a

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 is also the “forbidden band”), sinh(2n − 1)χ cos γ t, sinh(2N + 1)χ  2k − mγ 2 sinh 2nχ x2n = −a cos γ t, 2k − Mγ 2 sinh(2N + 1)χ

x2n−1 = a

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 m¨xn + k(2xn − xn−1 − xn+1 ) = 0,

n = 1, 2, . . . , N − 1,

mN x¨ N + k(2xN − xN−1 ) = 0

(1) (2)

(with the boundary condition x0 = 0), in the form of standing waves: xn = A sin nϕ cos(ωt + α), xN = B cos(ωt + α).

n = 1, 2, . . . , N − 1,

(3)

From (1), we get the relation ω2 =

4k 2 ϕ sin . m 2

(4)

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[7.8

Using (3) and (4), we get from (1) and (2) the set of equations A sin Nϕ − B = 0.   2m ϕ N sin2 + 2 B = 0. −A sin(N − 1)ϕ + − m 2 Hence, B = A sin Nϕ, while the parameter ϕ is determined as the solution of the transcendental equation 

 4mN 2ϕ sin Nϕ sin − 2 + cos ϕ = cos Nϕ sin ϕ. m 2

(5)

When mN  m, we have, apart from the obvious normal oscillations, x(s) n = As sin nϕs cos(ωs t + αs ), n = 1, 2, . . . , N, m ϕs tan Nϕs ≈ cot , s = 1, 2, . . . , N − 1, 2mN 2 when the particle mN is practically stationary (sin Nϕs  1), as well as a normal oscillation with the amplitudes of the particles which are linearly decreasing to the left-hand side of the chain:   n k 1 (N) 2 xn = B cos(ωN t + αN ), ωN = 1+ . N mN N The particle mN then oscillates between springs of stiffness k (to the right) and k/N (to the left). That (5) has such a solution can be seen as follows. Assuming  ϕ to be small  m 1 2 and retaining only the main terms, we get from (5): ϕ = m 1 + N completely in N 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 sin ϕ frequency ωs are determined from the equation tan Nϕ = − 2 − cos ϕ ). Apart from those oscillations, there is also a normal oscillation with the amplitudes of the particles which decrease to the left-hand side of the chain: sinh nψ cos(ωN t + αN ), sinh Nψ ψ 2k m 2  1, ωN = . cosh2 = 2 2mN mN N+n x(N) B n = (−1)

Formally, the value of the parameter ψ can be obtained from (5) using the substitution ϕ = π − iψ and assuming ψ to be large. This normal oscillation can in the first approximation be considered to be the simple oscillation of the small mass mN

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§7. Oscillations of linear chains

particle while the other particles are at rest, while we afterwards consider the motion of the other particles as forced oscillations under the action of high-frequency force kxN = kB cos(ωN t + αN ), acting upon the right-hand side of the chain of N − 1 identical particles (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 forming the crystal, the frequency of oscillations, in which the greatest amplitude has an impurity atom, 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. When kN+1  k, there are normal oscillations for which Nth particle is practically at rest: x(s) n = As sin nϕs cos(ωs t + αs ), ϕs ≈

sπ , N

ωs2 =

4k 2 ϕs sin , m 2

s = 1, 2, . . . , N − 1.

The parameter ϕs is defined from  2 sin2

 ϕ kN+1 − sin Nϕ = cos Nϕ sin ϕ, 2 k

(6)

which in the considered approximation has the form tan Nϕ = −

k sin ϕ. kN+1

Equation (6) has yet one more solution which we can obtain by putting ϕ = π − iψ and assuming ψ to be large. In this case, we have sinh nψ cos(ωN t + αN ), sinh Nψ kN+1 2 ωN = ; m

N+n x(N) BN n = (−1)

cosh2

kN+1 ψ = , 2 4k

that is, the amplitudes of the particles of this oscillation decrease towards the left-hand side 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(ωs t + αs ),

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where the frequency spectrum (Fig. 152) begins with the value ω0 =

 g/l:

g 4k 2 ψs + sin , l m 2 πs s = 0, 1, 2, . . . , N − 1. ψs = , N

ωs2 =

b) In the region of the eigen-frequency of the system ω0 < γ
0. l m 2

If, at the same time, the stiffness of the springs is small

m



k ω02 − γ 2

 = ε  1,

then the oscillation amplitudes rapidly decrease to the left-hand side ϕn = ϕN εN−n . In the high-frequency region γ > late in anti-phase

ϕn =

 ω02 + 4k/m, the neighbouring pendulums oscil-

  (−1)N−n+1 F sinh n − 12 χ 2kl sinh Nχ cosh(χ/2)

sin γ t,

γ2 =

g 4k 2 χ + ch . l m 2

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§7. Oscillations of linear chains

At a very high frequency mγ 2 /k  1, the oscillation amplitudes are also decrease rapidly to the left-hand side: 

k ϕn = − mγ 2

N−n ϕN .

c) It is clear that in the linear approximation at b − a = 0 all pendulums oscillate  independently 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 the instability of the small oscillations of the pendulums near the vertical. The Lagrangian function of the system is L = 12 ml 2

2N 

ϕ˙n2 − U,

n=1

U=

2N    −mgl cos ϕn + 12 krn2 , n=1

where elongation of the nth spring is (we assume l| n |  a)  l2 l 2 (a2 + 3l 2 ) 4 a2 + 4l 2 sin2 ( n /2) − b ≈ a − b + 2n − n , 2a 24a3 n = ϕn − ϕn+1 . rn =

With the accuracy up to terms of the ϕn4 order of magnitude inclusively, we have U = 12 mgl



 1 4 ϕn2 − α 2n − 12 ϕn + β 4n + const,

n

(b − a)kl α= , amg

(1)

1 kbl 3 β= α+ . 12 4mga3

The equations of motion in the linear in ϕn approximation

ωs2 ω02

g ϕ¨n + [ϕn − α(2ϕn − ϕn+1 − ϕn−1 )] = 0 l have solutions in the form of the travelling waves ϕn = Aei(ωt±nψ)

(2)

0 1 2 3

Figure 153

with the frequencies (see Fig. 153) ωs2 =

ψs  g 1 − 4α sin2 , l 2

ψs =

πs , N

s = 0, 1, . . . , N,

N

s

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[7.9

where the frequencies ω0 and ωN are non-generate, while the rest frequencies are twofold degenerate. It can be seen that when 4α − 1 > 0,

or b − a >

mga , 4kl

(3)

the oscillations are unstable; that is, some ωs2 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 is natural, 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 = 0 or ∂ϕn ϕn − α(2ϕn − ϕn+1 − ϕn−1 ) − 16 ϕn3 + 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). We will introduce the small displacements xn = ϕn − ϕn0 , and then the potential energy (1) will be equal to the expression U = 12 mgl

   1 − 12 ϕ 2 x2n − (α − 24βϕ 2 )(xn − xn+1 )2 + const

(6)

n

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 ωs2 =

 g 1 − 12 ϕ 2 − 4(α − 24βϕ 2 ) sin2 (ψs /2) . l

α < 1 , all frequencies ω2 are positive (see (3)): However, now for the small ϕ 2 < 24β s 2   g 2g ωs2 > 1 − 12 ϕ 2 − 4(α − 24βϕ 2 ) = (4α − 1) > 0; l l that is, the small oscillations near the new equilibrium position (4) are stable.

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261

Thus, with the increase of the parameter α, the original configuration of the vertical pendulum is replaced by the “accordion”. This change in symmetry of the system is similar to the change of symmetry of the thermodynamic systems due to the phase transitions 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). For example, 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 valid only for small ϕ. 7.10. a) Let we denote the current in nth coil as q˙n . The Lagrangian function is L=

1 2

N  

L q˙2n −

n=1

 1 (qn − qn+1 )2 + 12 L0 q˙2N+1 + Uq1 cos γ t C

(the current through the Z chain is q˙N+1 ). The resistance R can be introduced into equations of motion using the dissipative function F = 12 Rq˙2N+1 . The equations of motion are L q¨ 1 + L q¨ n +

1 (q1 − q2 ) = U cos γ t, C

1 (2qn − qn−1 − qn+1 ) = 0, n = 2, 3, . . . , N, C 1 L0 q¨ N+1 + (qN+1 − qN ) = −Rq˙N+1 . C

(1) (2) (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), we obtain 4 sin2 (ϕ/2), LC 1 −γ 2 L0 + (1 − eiϕ ) = −iγ R. C γ2 =

Hence, R=

sin ϕ , γC

L0 =

1 − cos ϕ = 12 L . Cγ 2

(4) (5)

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[7.10

Since R > 0, there must be ϕ > 0; that is, the wave travels in the direction of the L R chains. 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 1 (2q2n−1 − q2n−2 − q2n ) = 0, n = 2, 3, . . . , N, C 1 L2 q¨ 2n + (2q2n − q2n−1 − q2n+1 ) = 0, n = 1, 2, . . . , N, C

L1 q¨ 2n−1 +

(6)

coincide, up to notations, with (1) in problem 7.5; besides, 1 (q1 − q2 ) = U cos γ t, C 1 L0 q¨ 2N+1 + (q2N+1 − q2N ) = −Rq˙2N+1 . C L1 q¨ n +

(7) (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 /γ12 )A − cos ϕ · B = 0, cos ϕ · A − (1 − γ 2 /γ22 )B = 0, 2 2 γ1,2 , = L1,2 C

(10)

and cos2 ϕ = (1 − γ 2 /γ12 )(1 − γ 2 /γ22 ). 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  γ12 + γ22 , which is the region of “optic” waves. Outside these regions, the propagation of travelling waves is impossible (cf. problem 7.7). From (8), we obtain R + iγ L0 =

 B sin ϕ i B iϕ B i i  − e = 1 − cos ϕ + γC γC A γC A A γC

(11)

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263

with the condition (B/A) sin ϕ > 0. In the region γ   γ1 , the amplitudes A and B have the same signs, so that ϕ > 0. In the region γ2  γ  γ12 + γ22 , on the contrary, B/A < 0 and ϕ < 0. Substituting the values of B/A, cos ϕ and sin ϕ in (11), we obtain as a result  R=

  L1 + L2 L1 L2 C γ 2 2 − L1 Cγ 2 1− · , L0 = 12 L1 . 2C L1 + L2 2 2 − L2 Cγ 2

The negative value of ϕ in the region of “optical” oscillations means that the phase velocity of the travelling wave is directed from the Z chain to the voltage source. The group velocity, on the contrary, has the opposite direction (e.g. see, Fig. 148 where vgr ∝ dω(+) < 0; cf. problem 7.5). The group velocity is the velocity of the wave packet. If the ds wave packet moves along the chain, then the oscillation energy is localized in the region where the wave packet is at the moment, so naturally that the group velocity determines the 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 x¨ n − ka

a  xn+1 − xn xn − xn−1  1 − = 0. m a a a

(1)

Nm The quantity m a = Na becomes in the limit the linear density of the rod ρ. The relative extension of the section a; that is, the quantity(xn − xn−1 )/a is proportional to the force acting upon it, F = ka

xn − xn−1 , a

and ka thus becomes in the limit the elastic modulus  of the rod. Equations (1) in the limit thus becomes the wave equation 2 ∂ 2 x(ξ , t) 2 ∂ x(ξ , t) − v = 0, ∂t 2 ∂ξ 2

(2)

√ where v = /ρ is the phase velocity of the wave. Instead of a set of N coupled ordinary differential equations, we have obtained one partial differential equation (cf. problem 4.32). Note that in our derivation we had to make the important assumption that the function xn (t) tends to a well-defined limit x(ξ , t) which is a sufficiently smooth function.

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[7.12

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) a2 ∂ 2 x(ξ , t) ± + ∂ξ 2 ∂ξ 2

a3 ∂ 3 x(ξ , t) + ... 6 ∂ξ 3

Therefore, (2) of problem 7.11 changes to ∂ 2 x  ∂ 2 x a2 ∂ 4 x − − = 0. ∂t 2 ρ ∂ξ 2 12ρ ∂ξ 4

(1)

While each of the equations (1) of problem 7.11 contains the displacements of the three neighbouring points (long-range interaction), (1) here contains the displacement x in a given point ξ (short-range interaction). The last term −

a2 ∂ 4 x 12ρ ∂ξ 4

in (1) corresponds to approximate account of the small difference between the system considered and a continuous one. This leads, in particular, to the fact that the phase velocity appears to be dependent on the wavelength. Substituting x = a cos (ωt − kξ ) in the equation, we obtain ω v= = k

    a2 k2 1− . ρ 12

Such phenomena, which are due to the long-range interaction, are called spatial dispersion.

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8.1. a) We solve the equation of motion x¨ + ω02 x = −βx3

(1)

using the method of successive approximations (cf. problem 6.35): x = x0 + δx = Aeiω0 t + A∗ e−iω0 t + δx, where A(t) is the complex amplitude A = 12 aeiϕ . The “force” −βx30 = −βA3 e3iω0 t − 3βA2 A∗ eiω0 t − 3βAA∗2 e−iω0 t − βA∗3 e−3iω0 t contains the resonant terms −3βA2 A∗ eiω0 t − 3βAA∗2 e−iω0 t = −3β|A|2 x, which are more convenient to attach to the term ω02 x in left-hand side of (1). This results in replacement ω02 → ω2 = ω02 + 3β|A|2 . For δx, we obtain δ x¨ + ω2 δx = −β(A3 e3iωt + comp. conj.),

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

(2)

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from where δx =

βA3 3iωt e + comp. conj. 8ω2

As a result, x = a cos(ωt + ϕ) + ω = ω0 +

βa3 cos(3ωt + 3ϕ), 32ω2

3βa2 8ω0

(cf. [1], § 28 and [7], § 29.1). Fig. 154 depicts the function x(t). x

t β >0 x+ x

αa2 2ω02

t

t β 0, there is a “limitation” of the oscillations; when β < 0, the maxima become sharper. These properties of the oscillations, as well as the sign of the corrections to the frequency, can easily be considered by looking at the graph of U(x). For other solution 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 2iω0 t 2α|A|2 αA∗2 −2iω0 t e − + e ; 3ω02 ω02 3ω02

that is x=a cos(ω0 t + ϕ) −

αa2 αa2 + cos(2ω0 t + 2ϕ). 2ω02 6ω02

The distortion of the oscillations is non-symmetric (Fig. 155).

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In the following approximation one must take into account the term −2αx0 δx in the “force” −α(x0 + δx)2 which contains the resonant terms −

4α 2 |A|2 10α 2 2 2α 2 2 ∗ iω0 t 2α 2 ∗2 −iω0 t A A e − AA e + x = |A| x. 3ω02 3ω02 ω02 3ω02

This results to the replacement ω0 → ω0 − 8.2. x = a cos ωt − 14 γ a2 cos 2ωt + 14 γ a2 ,

5α 2 a2 12ω03

.

1 2 2 ω = ω0 + 16 γ a ω0 .

a2 cos t + a2 4 sin 2t (the notation is that of 2 g − l 2(g − l2 )(g − 4l2 ) problem 5.9).

8.3.

ϕ=

8.4. x = x(0) + x(1) + . . ., x(0) =

f2 cos ω2 t f1 cos ω1 t + , 2 2 m(ω0 − ω1 ) m(ω02 − ω22 )

x(1) = − −

αf12 2m2 ω02 (ω02 − ω12 )2

−

αf22 2m2 ω02 (ω02 − ω22 )2

αf12 cos 2ω1 t

−

−

αf22 cos 2ω2 t

2m2 (ω02 − 4ω12 )(ω02 − ω12 )2 2m2 (ω02 − 4ω22 )(ω02 − ω22 )2 αf1 f2 cos(ω1 − ω2 )t − − 2 2 m [ω0 − (ω1 − ω2 )2 ](ω02 − ω12 )(ω02 − ω22 ) αf1 f2 cos(ω1 + ω2 )t − . m2 [ω02 − (ω1 + ω2 )2 ](ω02 − ω12 )(ω02 − ω22 )

−

What combinational frequencies will occur when we take into account an anharmonic correction of the form δU = 14 mβx4 ? 8.5. Let x and y be the displacements of the pendulum from the equilibrium position in the horizontal and vertical directions. We expand the Lagrangian function L = 12 m(x˙2 + y˙2 ) − 12 k

 2 (l − y)2 + x2 − l0 − mgy

in series of small x and y up to the third order inclusively L = 12 m(x˙2 + y˙2 − ω12 x2 − ω22 y2 + 2αx2 y) + . . . ,

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where g ω12 = , l

l = l0 +

mg , k

ω22 =

k , m

α=

kl0 . 2ml 2

Using the same method as in problem 8.1, we obtain x = a cos(ω1 t + ϕ1 ) − +

αab cos(ω+ t + ϕ+ )+ 2ω2 (2ω1 + ω2 )

αab cos(ω− t + ϕ− ), 2ω2 (2ω1 − ω2 )

y = b cos(ω2 t + ϕ2 ) +

αa2 αa2 + cos(2ω1 t + 2ϕ1 ), 2ω22 2(ω22 − 4ω12 )

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 significant pumping of the energy from the x- to y-oscillations and back. This case is considered in problem 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 (ω02 − ω2 + 2iλω + 3β|A|2 )a = 12 f , and, hence, [(ω02 − ω2 + 3β|A|2 )2 + 4λ2 ω2 ]|A|2 = 14 f 2 . A study of this equation which is cubic in |A|2 can be done in the same way as the study of the analogous equation (29.4) in [1]. This equation is square in ω2 , so that the dependence of |A|2 on ω2 can be easily represented graphically (see [7], § 30). 8.7. a) We look for a solution of the oscillations equation x¨ + 2λx˙ + ω02 (1 + h cos 2ωt)x + βx3 = 0

(1)

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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 to zero, we find   2 2 2 2 ∗ 1 2 hω0 A + ω0 − ω − 2iωλ + 3β|A| A = 0,   (3) 2 ∗ 2 2 2 1 A = 0. hω A + ω − ω + 2iωλ + 3β|A| 0 0 2 We can only have a non-vanishing A if  1 2  2 hω0  2 2 ω − ω + 2iωλ + 3β|A|2 0 Hence,

 ω02 − ω2 − 2iωλ + 3β|A|2  1 2  = 0. 2 hω0

(4)

  2 1 1 2 |A|2 = ω2 − ω02 ± − (2ωλ)2 . 2 hω0 3β

(5)

From (3), we get2 A 4λ =− , ∗ A hω0  2 2 1 2 cos 2ϕ = ∓ 2 − (2ωλ)2 . 2 hω0 hω0

|A|2

sin 2ϕ = Im

(6)

C B

A

Thus,

D

x = a cos(ωt + ϕ),

(7)

E

ω2

Figure 156

where A = 12 aeiϕ . Fig. 156 shows how |A|2 depends on ω2 (to fix the ideas, we assume β > 0). In some frequency ranges two options or three (including zero values) different amplitudes of stable oscillations are possible. The amplitudes corresponding to the sections AD and CD are not realized in actual cases since those oscillations are unstable (for a proof of this for the section AD, see the 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 to x 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 with greater precision as a change in the phase by π corresponds simply to a shift in the time origin.

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[8.8

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 with small parameters. We find out that   1 B = 16 h + 18 βA2 ω−2 A (9) and, indeed, |B|  |A|. Therefore, x = a cos(ωt + ϕ) + b cos(3ωt + ψ), where b = 2|B|, ψ = arg B. It is easy to notice that the fifth harmonic turns out to be smaller than second order (∼ h2 A), the seventh ∼ h3 A, and so on. The even harmonics do not occur. This is the basis 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) cos ωt + b(t) sin ωt,

(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 + 14 hω0 b = 0,   b˙ − ω − ω0 − 14 hω0 a = 0.

(2)

If |ω1 − ω0 | < 14 hω0 , its solution is a(t) = α1 (C1 e−st + C2 est ), b(t) = α2 (C1 e−st − C2 est ),

(3)

where s=

1 4

 (hω0 )2 − 16(ω − ω0 )2 ,

α1,2 =

 hω0 ± 4(ω − ω0 ).

Hence, x = C3 est cos(ωt + ϕ) + C4 e−st cos(ωt − ϕ), where tan ϕ = α2 /α1 (Fig. 157).

(4)

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The oscillations thus increase, generally speaking, without limit. The rate of their increase 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 influence of the anharmonic terms becomes important (see problem 8.7) or the reaction of the oscillations on the device which periodically changes its frequency becomes important. It is useful to draw attention to the analogy between the results obtained and the particular solution of the problem of the normal oscillations of a chain of particles connected by springs of different stiffness (problem 7.5b). An inhomogeneity with period 2a along the chain leads to a build-up along the chain of the amplitude of the stable oscillations, when the “wavelength” is equal to 4a (Fig. 151) in a similar way that a periodic change with time of the frequency of an oscillator will lead to an increase in the amplitudes with time. An even more complete analogy can be observed in problem 7.7. The region of instability with respect to parametric resonance corresponds to the forbidden zone in the spectrum of oscillations of the chain. Similar equations are obtained in quantum mechanics in the problem of the motion of 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) cos ωt − Cβ2 cos(t + ψ) sin ωt, where  =

1 4

 16(ω − ω0 )2 − (hω0 )2 , √ √4(ω − ω0 ) ± hω0 , when ω > ω0 , β1,2 = ± 4(ω0 − ω) ∓ hω0 , when ω < ω0 .

The oscillations are beats:  x = C 4|ω − ω0 | ∓ hω0 cos(2t + 2ψ) cos(ωt + θ),

when ω ≷ ω0 ,

where θ is a slowly varying phase (see Fig. 158). If the frequency approaches the limit of the instability region, the depth of the modulation of the oscillations approaches the total amplitude and their period increases without limit. x

x

t

t

Figure 157

Figure 158

What is the form of the oscillations when |ω − ω0 | = 14 hω0 ?

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8.9. Let x = eiω1 t , when 0 < t < τ . We then have in the interval τ < t < 2τ , x = aeiω2 t + be−iω2 t 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 ω1 + ω2 i(ω1 −ω2 )τ e , 2ω2 ω2 − ω1 i(ω1 +ω2 )τ b= e . 2ω2

a=

Similarly, we find that for 2τ < t < 3τ x = αeiω1 t + βe−iω1 t , where   ω2 + ω22 α = e−iω1 τ cos ω2 τ + i 1 sin ω2 τ , 2ω1 ω2 2 ω − ω22 β = i sin ω2 τ e3iω1 τ 1 . 2ω1 ω2 It is clear that the oscillation in the form Aeiω1 t + Be−iω1 t

(1)

for 0 < t < τ , after a period of 2τ will become A(αeiω1 t + βe−iω1 t ) + B(α ∗ e−iω1 t + β ∗ eiω1 t ) = = (αA + β ∗ B)eiω1 t + (βA + α ∗ B)e−iω1 t . We now look for such a linear combination (1) that it retains its form—apart from a multiplying factor—after a period 2τ : 



(αA + β ∗ B)eiω1 t + (βA + α ∗ B)e−iω1 t = μ(Aeiω1 t + Be−iω1 t ), where t  = t − 2τ ,

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αA + β ∗ B = μe−2iω1 τ A,

(2)

βa + α ∗ B = μe2iω1 τ B. Set (2) has a non-trivial solution, provided (α − μe−2iω1 τ )(α ∗ − μe2iω1 τ ) − ββ ∗ = 0, whence  μ1,2 = γ ± γ 2 − 1, where γ = Re(αe2iω1 τ ) = cos ω1 τ cos ω2 τ −

ω12 + ω22 sin ω1 τ sin ω2 τ . 2ω1 ω2

After n periods, the oscillation x1,2 = A1,2 (eiω1 t + λ1,2 e−iω1 t ), λ1,2 = μ1,2

e−2iω1 τ

0 < t < τ,

−α

β∗

has changed to 



x1,2 (t) = μn1,2 A1,2 (eiω1 t + λ1,2 e−iω1 t ),

0 < t  = t − 2nτ < τ .

Any oscillation is a superposition of oscillations such as x1,2 ; in particular, the real oscillation (which is the only one which has a direct physical meaning) x(t) = Aeiω1 t + A∗ e−iω1 t ,

0 < t < τ,

(ω22 −ω12)τ2 2π2

5 4 3 2 1 0

(ω12 +ω22)τ2

1

2

3

4

5

6

Figure 159

2π2

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[8.10

is the sum of x1 (t) + x2 (t) with A1 =

A∗ − λ2 A , λ1 − λ2

A2 =

λ 1 A − a∗ . λ1 − λ2

If γ < 1, |μ1,2 | = 1 and the oscillations x1,2 (t) (and at the same time x(t)) remain bounded. If, however, γ > 1, we have μ1 > 1, and the amplitude of the oscillations increases without bound. This is the case of the onset of parametric resonance. We can easily verify that if the frequency difference is small, |ω1 − ω2 |  ω1 , this condition is satisfied if the frequencies lie close to nπ/τ : |(ω1 + ω2 )τ − 2π n|
, a second stable equilibrium position appears 2 + l . When l 2l 2l 2 in the form the straight upward vertical; the oscillation frequency around it is equal to a2 γ 2 g ω2 = − (see [1], § 30, problem 1). l 2l 2

8.11. ω2 =

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 we substitute 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 ∼

(10−19 K)2 e2 −30 kg, a ∼ 10−10 m, U ∼ 10−18 (a/r)6 J. eff 4π ε0 ∼ 10−11 (F/m) · 10 , m ∼ 10

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[8.13

we get Ueff ∼ 10−40 erg · cm6 /r 6 , which is close to the correct value for van der Waals interaction, as far as order of magnitude is concerned. This result may serve as an indication of the physical nature of this interaction. A complete calculation of the van der Waals forces is only possible using quantum mechanics. b) Ueff =

α2 m(ω2 − ω02 )



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 particle is acted upon by a fast oscillating force fx = 2Ax sin kvt,

fy = 2Ay sin kvt.

The corresponding effective potential energy is Ueff = 12 m2 (x2 + y2 ), where √  = 2 A/(mkv). According to the initial conditions, we have for the frequency of the force oscillation kv  so that the force is, indeed, fast oscillating. Thus, in the xy-plane, the particle performs a harmonic oscillation with frequency  around the z-axis. This problem illustrates the principle of strong focus of particle beams in accelerators. 8.14. The equations of motion are e ˙ m¨x = B(x)y, c e m¨y = − B(x)x. ˙ c 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, while X and Y are the slow displacements of its centre (cf. [1], § 30). Substituting (1) into the equations of motion, we expand B(X + ξ ) in powers of ξ : ¨ + ξ¨ = ωY˙ + ωη˙ + e B (X)ξ(Y˙ + η), ˙ X mc e  Y¨ + η¨ = −ωX˙ − ωξ˙ − B (X)ξ(x˙ + ξ˙ ), mc

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and then consider separately the fast oscillating and slowly changing terms. For the oscillating terms, we obtain ξ¨ = ωη, ˙

η¨ = −ωξ˙ ,

ω=

e B(X), mc

where ξ = r cos ωt,

η = −r sin ωt.

For the slowly changing terms, we have ¨ = ωY˙ + e B (X)ξ η , X ˙ mc e  B (X)ξ ξ˙ , Y¨ = −ωX˙ − mc

(2)

where ξ η ˙ = −r 2 ωcos2 ωt = − 12 r 2 ω,

ξ ξ˙ = 0.

¨ Y¨ ∼ εωx, Since X, ˙ εωY˙ , the left parts in (2) can be put equal to zero. As a result, er 2  Y˙ = B (X) = 12 εv, 2mc

x˙ = 0.

The rate of the displacement of the orbit centre (the drift velocity) in a more general case is considered in [2], § 22, problem 3 and in [10], § 25. 8.15. The equation of motion of the ball is m¨y = −

dU(y) + f (t). dy

The proper motion of the ball under the action of a spring is described by the “lowfrequency” displacement x = y − y0 cos γ t, for which m¨x = −

dU(x + y0 cos γ t) . dx

If we average this proper motion over a period of 2π/γ of the high-frequency motion using cos2n+1 γ t = 0, cos2 γ t = 12 ,

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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. When A > 0 or T = y20 > Tc =

Ueff

C , 3B

A>0

the ball oscillates near the point x = 0 with the frequency   ω = 2A/m ∝ T − Tc .

A mgaI1 /I32 . 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 the vertical reaction force Q, there is also the horizontal friction force f acting on the disc. It is helpful to write the equation of motion ˙ = [a, Q] + [a, f ] M and its components along the Z-, x3 -, and ξ -axes1 ˙ Z = fξ a cos θ, M

˙ 3 = fξ a, M

d ∂L ∂L = fη a sin θ. − ∂θ dt ∂ θ˙

(4) (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 ˙ = f + Q + mg mV along the ξ - and η-axes (see equation (4) of the preceding problem), we have the following equations for fξ and fη : ˙ ξ = m(V˙ ξ − ϕV fξ = m(V) ˙ η ), ˙ η = m(V˙ η + ϕV ˙ ξ ). fη = m(V)

(6)

The condition of rolling without slipping V + [, a] = 0 leads to Vξ = −a(ψ˙ + ϕ˙ cos θ),

Vη = −aθ˙ sin θ .

1 These equations are the Lagrangian equations for the Euler angles when there is no friction force.

(7)

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Substituting (1), (6), and (7) in (4) and (5), we obtain a set of equations for the Euler angles. 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: ˙ sin θ − I1 ϕ˙ 2 sin θ cos θ + mga cos θ = 0 ˙ ϕ˙ cos θ + ψ) I3 ϕ(

(8)

 =I 2 (from now on, we use I1,3 1,3 + ma ). The centre of the disc moves with a velocity, which has a constant absolute magnitude

V = a|3 | = a|ψ˙ + ϕ˙ cos θ |, along a circle with radius R = V /|ϕ|. ˙ Condition (8) can also be expressed as I3 RV 2 = |I1 aV 2 cos θ − mga2 R2 cot θ|. In particular, if the mass of the disc is concentrated in its centre (I1 = I3 = 0), we get the elementary relation: V 2 = gR| cot θ|. Gyroscopic effects which appear when I1,3 differs from zero may turn out to be important. For example, when R a , we have 32 V 2 = gR|cot θ | for a uniform disc (2I1 = I3 = 12 ma2 ), but 2V 2 = gR| cot θ| 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˙ ∼ β3 3 ,

˙ 3 ϕ ϕ¨ ∼ β¨ ∼  ˙ 3

into (1) and (4)–(7), retaining only the first-order terms. We then get MZ = I1 ϕ˙ + I3 3 β = const, 3 = const, I1 β¨ − I3 3 ϕ˙ − mgaβ = 0,

(10)

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[9.21

and hence I1 β¨ +



 I3 I3 2 I 3 − mga β = 3 MZ 3 . I1 I1

If 23 >

I1 mga , I3 I3

(11)

small oscillations in β and ϕ˙ occur with the angle θ differing from π/2: I3 3 MZ + β0 cos(ωt + δ), I1 I1 ω2 mgaMZ I3 ϕ˙ = − − 3 β0 cos(ωt + δ), I1 I1 I1 ω2

β=

where ω2 =

I3 I3 2 mga  −  . I1 I1 3 I1

The direction of motion of the disc also executes small vibrations; the disc’s motion is not about a straight line, but about the circle of radius a3 I12 ω2 /(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, if condition (11) holds, the motion is stable with respect to perturbations which remove ˙ the representative (θ-, ϕ-, ˙ ψ)-point from the surface (8) while it is neutral with respect to perturbation shifting the point along the surface. A similar situation holds for the stability  by I . of the motion of the disc along the smooth plane; we must merely replace I1,3 1,3 2  Rotation of the disc around its vertical diameter is stable, if Z > mga/I1 . c) If there can be no rotation around a vertical axis, the following condition must be satisfied: Z = ϕ˙ + ψ˙ cos θ = 0.

(12)

In this case, there is an additional, “frictional torque” N, which is directed along the vertical, acting on the disc. Instead of (4), we then have

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˙ Z = fξ a cos θ + N, M

˙ 3 = fξ a + N cos θ. M

(13)

The integration of the equations of motion can easily be reduce to quadratures (in contrast to the equations under problem 9.21b). Motion with a constant angle of inclination is possible if conditions (8) is satisfied as well as condition (12), which means that ϕ˙ and ψ˙ are completely determined by the angle θ. In this case, R = a| sin θ tan θ|,

V2 =

mga3 sin θ . I3 + I1 cot2 θ

Rolling of the disc in a vertical position is stable if 23 > mga/I3 . 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, the X-axis lies in the plane at right angles to the Y -axis, pointing upwards, and the Z-axis is at right angles to the plane. In this case, an interesting effect can be observed: although the additional force is directed against the X-axis, the disc moves in the opposite direction to the Y -axis; that is, the disc moves without losing height. Substituting θ = (π/2) − β,

β 1,

ψ˙ ∼ β˙ Z ,

˙ Z ψ ˙ Z ψ¨ ∼ β¨ ∼ 

into in (1) and (4) to (7) and adding the contribution from δL, we get I1 β¨ + (I1 − I3 )2Z β − I3 Z ψ˙ − mgaβ = −mgaα cos Z t, I3 ψ¨ + (I3 + 2ma2 )Z β˙ = mgaα sin Z t, and hence 

mga 2ma2 ψ = − 2+  +  2 I3 I3 Z

 α sin Z t,

β = −2α cos Z t.

Substituting (7) and θ, ϕ = Z t, and ψ into X˙ = −Vξ sin ϕ − Vη cos ϕ,

Y˙ = Vξ cos ϕ − Vη sin ϕ

and averaging over one period of rotation, we find 

˙ = 0,

X

mga ma2

Y˙  = 1 +  +  2 I3 2I3 Z

that is, the disc is displaced without losing height.

 αaZ ;

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[9.22

9.22. a) The position of the ball is determined by the coordinates of its centre of mass X, Y , Z and the Euler angles θ, ϕ, ψ (see [1], § 35), which determine the orientation of the principal axes of inertia. Let the Z-axis be vertical and the x3 -axis be directed from the centre of mass to the geometric centre of the ball; let x3 = b be the position of the geometrical centre of the ball and a be its radius. The analysis of the motion of the ball is 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 and Z = a − b cos θ0 . The instantaneous point of contact of ball and plane has the velocity ˙ sin θ0 along the direction of the line of nodes. v = (bϕ˙ + aψ) b) Let Rc = (X + b sin θ sin ϕ, Y − b sin θ cos ϕ, a) be the coordinates of the geometric centre of the ball, and let  = (X , Y , Z ) = (θ˙ cos ϕ + ψ˙ sin θ sin ϕ, θ˙ sin ϕ − ψ˙ sin θ cos ϕ, ϕ˙ + ψ˙ cos θ) be the angular rotational velocity of the ball, while a = (0, 0, −a) is the radius vector from the 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 ˙ c + [, a] = 0 R is a non-holonomic constraint ˙ − b cos θ) sin ϕ − (aψ˙ + bϕ) X˙ = θ(a ˙ sin θ cos ϕ, ˙ − b cos θ) cos ϕ − (aψ˙ + bϕ) Y˙ = −θ(a ˙ sin θ sin ϕ.

(4)

The equation of motion are ¨ = λ1 , mX

mY¨ = λ2 ,

(5)

˙ Z = (λ1 cos ϕ + λ2 sin ϕ)b sin θ, M m˙ 3 = (λ1 cos ϕ + λ2 sin ϕ)a sin θ ,

(6) (7)

d ∂L ∂L = (−λ1 sin ϕ + λ2 cos ϕ)(a − b cos θ); − ∂θ dt ∂ θ˙

(8)

they contain the friction forces and the torques produced by these forces on the right-hand sides. Using the constraint conditions (4), we can express the Lagrangian multipliers λ1 and λ2 in terms of the Euler angles. Indeed, the components of the friction force along the line of nodes f and at right angles to it, f⊥ , are given by

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f = mθ˙ ϕ(a ˙ − b cos θ) − m f⊥ = −m

d [(aψ˙ + bϕ) ˙ sin θ], dt

d ˙ − b cos θ)] − mϕ(a [θ(a ˙ ψ˙ + bϕ) ˙ sin θ. dt

(9) (10)

Note that (6) and (7) are the same as (1) if one substitutes into (1) the friction force f (9) instead of the dry friction force ∓f . The quantity Mz a − M3 b = C is the integral of motion (2), as discussed before. 9.23. To solve the problem, it is necessary to take into account that the height of the particle above the Earth, h, is small compared to the radius R of the Earth and that the centrifugal acceleration, R2 (where  is the angular velocity of the Earth), is small compared to the acceleration of free fall, g, towards the surface of the Earth. There are thus two small parameters in the problem: ε1 =

h R2  ε2 = ∼ 0.003. R g

Let R + r be the coordinate of the particle reckoned from the centre of the Earth. The equation of motion has the following form if we take into account the dependence of g on the distance r from the surface of the Earth (up to the second order in ε2 ): r¨ = −GM

R+r + 2[v, ] + [, [R + r, ]], |R + r|3

(1)

where G is the gravitational constant and M is the mass of the Earth. We expand the first term into series in small parameter r/R  ε1 : r¨ = g + 2[v, ] + g1 + [, [r, ], g = −GM

R + [, [R, ], R3

g1 =

r(0) = h,

v(0) = 0,

(2)

  GM (rR) r − [1 + O(ε1 )]. 3R R R2 R3

√ The Coriolis acceleration 2[v, ] ∼ gt ∼ ε1 ε2 g and g1 ∼ ε1 g have the first order of smallness, while [, [r, ]] ∼ ε1 ε2 g has the second order. The vertical h is anti-parallel to the vector g and, it makes a small angle α = ε2 sin λ cos λ with the vector R (here λ is the northern [geocentric] latitude; i.e. the angle between the plane of the equator and the vector R). We choose the z-axis along the vertical upwards, the x-axis along the meridian towards to the south, and the y-axis along the latitude to the east, then g = (0, 0, −g), R = R(sin α, 0, cos α),  = (− cos(α + λ), 0, sin(α + λ).

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[9.24

In zero approximation, the particle moves with acceleration −g along the z-axis. In the first approximation, the Coriolis acceleration results in the deviation to the east, and the √ √ magnitude of the displacement is y ∼ ε1 ε2 gt 2 ∼ ε1 ε2 h. The acceleration g1 has the component ∼ ε1 g 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 the time 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 and take the terms of the zero, first order, and second order for the z-, y-, and x-component, respectively: z¨ = −g,

y¨ = −2z ˙ cos λ, g x¨ = 2y ˙ sin λ + (3z sin α − x) + 2 z sin λ cos λ. R

Using the method of successive approximations, we obtain z = h − 12 gt 2 ,

y = 13 gt 3  cos λ,

Substituting the time of the fall t = y=

1 3

x = 2ht 2 2 sin λ cos λ.

2h/g, we find deviations to the east and south

8ε1 ε2 h cos λ,

x = 2ε1 ε2 h sin 2λ.

9.24. We use the reference system, which is rotating with a vessel; the x-axis is directed along AB, the origin of the coordinates is placed in a fixed point, that is, the intersection of the AB- and CD-axes. The angular velocity of the system is  = ω1 + ω2 = (ω2 , ω1 cos ω2 t, −ω1 sin ω2 t). Besides the force of gravity mg, the inertia forces act upon each particle of fluid of mass m in this system of reference (see [1], § 39): the Coriolis force 2m[v, ], the centrifugal force ∂ 12 m[, r]2 , ∂r ˙ where  ˙ is the rate of change of the vector  in the inertial and the force m[r, ], system. When the resin hardens, the particle velocities (relative to the vessel) turn to zero, and the Coriolis force vanishes. The other forces need to be averaged over the periods of rotation:

mg = −mg ( cos ω1 t, sin ω1 t sin ω2 t, sin ω1 t cos ω2 t) = 0, if ω1 = ω2 , and ˙ = m[r, ] ˙ = 0,

m[r, ]

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299

˙ = [ω1 , ω2 ] = [ ω1 , ω2 ] = 0. Finally, since  

 ∂ 1 m[, r]2 = 1 m ∂U(r) , 2 ∂r 2 ∂r

where     U(r) = [, r]2 = 2 r 2 − (r)2 =     = ω12 + ω22 r 2 − (xω2 + yω1 cos ω2 t − zω1 sin ω2 t)2 =   = ω12 x2 + 12 ω12 + ω22 (y2 + z2 ).

(1)

The surface of the liquid will be located along the level U(r) = const. The resin will harden 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. 

 m dr t= , √ 2 E + M − Ueff   M −  dr   2 m mr , ϕ= √ 2 E + M − Ueff where E is the energy, M is the angular momentum in the rotating system of reference, and Ueff (r) = U(r) + M 2 /(2mr 2 ). Bear in mind that E = E0 − M and M = M0 , where E0 and M0 are the energy and angular momentum, respectively, in the inertial system. It is interesting to note that the centrifugal potential energy − 12 m2 r 2 does not enter into Ueff (r). 9.26. In the system of reference, which is fixed in the frame, the Lagrangian function of the problem is the same as the one considered in problem 6.37, with z = 0 and with the parameters ωB = −2,

2 ω1,2

  2 f2,1 = k1,2 + − 2 . m l

2 > 0, the motion of the particles is the same as that of an anisotropic oscillator When ω1,2 in the magnetic field B = −2mc/e. The orbit of the particle for the case ω1 = ω2 is shown in Fig. 109 of problem 2.36. In particular, if ω1 = ω2 = 0, the motion of the particle is the same as the motion of a free particle in a magnetic field:

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x = x0 + a cos ωB t,

[9.27

y = y0 − a sin ωB t;

that is, the particle moves uniformly along a circle of radius a and centre (x0 , y0 ). It is interesting to investigate what the motion of the particle is in the rest system, which corresponds to the latter case, especially for a = 0 or for x0 = y0 = 0. 2 < 0, If the centrifugal force is larger than the restoring force from the two springs, ω1,2 then the particle still executes small oscillations. Although the potential energy has a maximum at x = y = 0, the equilibrium position is stable because of the Coriolis force. If ω12 and ω22 have opposite signs, so that the point x = y = 0 is a saddle point of the potential energy, the equilibrium position is unstable. It is interesting to compare these results with those of problem 5.4. One sees that in a system rotating with an angular velocity  the point r0 , ϕ0 lies on the ridge of the “potential circus”; the potential energy U(r) = −

1 α − m2 r 2 n r 2

has a maximum in the direction of the radius vector, but it does not change in the direction of the azimuth. In this case, one of the eigen-frequency is ω, but the other one 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 and velocity in the system of reference rotating with the paraboloid, has the form m (v + [ω, r])2 + mgr = 2    2 y2 x m (x˙ − ωy)2 + (y˙ + ωx)2 + z˙2 − g + . = 2 a b

L=

The quantity z˙ can be omitted for the small oscillations, and then the equations of motion are g  x¨ − 2ωy˙ + − ω2 x = 0,  ag  (1) y¨ + 2ωx˙ + − ω2 y = 0. b We look for a solution in the form x = Aeit ,

y = Beit ;

and for 2 we get 4 −

g a

+

  g  g g + 2ω2 2 + − ω2 − ω2 = 0. b a b

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301

It is easy to prove that its roots are real. However, when  g  g − ω2 − ω2 < 0 a b one of the roots becomes negative, 21 < 0, so that the corresponding motion x = A1 e|1 |t + A2 e−|1 |t , y = B1 e|1 |t + B2 e−|1 |t leads to the departure of the particle from the origin. This means that the lower position of the particle becomes unstable. Assuming for definiteness that a > b, we obtain the instability region: g g < ω2 < . a b Note that if ω2 > g/b, the motion is stable although the potential energy in the rotating system  g 2 1  2 g 2 x −2m ω − y U(x, y) = − 12 m ω2 − a b is not a potential pit, but a potential hump. Stability in this case is provided by the action of 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 2 x2 . The equilibrium position x0 is determined by the condition U  (x0 ) = 0; it is equal to x0 =

2ka . 2k − mγ 2

Note that when the rotation frequency γ becomes greater than the frequency of the

eigen-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 system at the small displacements from the equilibrium position: U(x, y, z) = U(x0 , 0, 0) + 12 k1 (x − x0 )2 + 12 k2 y2 + 12 k3 z2 ,

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[9.29

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where k1 = 2k − mγ 2 , f + k(x0 − a) f − k(x0 − a) k2 = + − mγ 2 = k3 − mγ 2 . l + x0 − a l − x0 + a 2 The sign of ∂ U(x02, 0, z) |z=0 = k3 defines the stability of the equilibrium position ∂z with respect to the small deviations along the z-axis. When k3 > 0, the z-oscillations are stable. 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 we do not consider here, becomes significant). Upon comparing this results with problem 9.27, we can conclude that the equilibrium position with respect to the small deviations in the xy-plane is stable when k1 k2 > 0, and it is unstable when k1 k2 < 0. Let us consider in detail the case of fast rotation of the frame when

mγ 2 2k.

(1)

In that case, k3 ≈ 2

fl − ka2 l 2 − a2

= 2k

(l − l0 )l − a2 l 2 − a2

,

(2)

where we introduce the length of a free string l0 in accordance with the equation f = k(l − l0 ). From (2), it follows that 0 < k3 < 2k when fl > ka2 , while k3 < 0 when fl < ka2 . Besides, we find that in the limit (1) both k1 < 0 and k2 < 0. This means that when mγ 2 2k and fl > ka2 , the equilibrium position is stable, in contrast to the results of 9.28a. 9.29. We use the Cartesian coordinates in the rotating system of reference with the origin of this system places in the centre of 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 a, m1 + m2

y

l

m

r1 r m1

O

r2 m2

x

Figure 163

where m1,2 are the masses of the stars and a is the distance between them. From equality m1 m 2 Gm1 m2 aω2 = m1 + m2 a2

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(where G is the gravitational constant), we obtain ω2 = G

m1 + m2 . a3

The potential energy of a body of mass m (including the centrifugal potential energy) is U(x, y, z) = −

Gmm1 Gmm2 1 − − 2 mω2 (x2 + y2 ), r1 r2

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 = 0, or ∂r     ∂U = Gm m1 + m2 − m1 + m2 x − Gm m1 a1 + m2 a2 = 0, ∂x a3 r13 r23 r13 r23   ∂U = Gm m1 + m2 − m1 + m2 y = 0, ∂y a3 r13 r23   ∂U = Gm m1 + m2 z = 0. ∂z r13 r23 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 ) − 38 mω2 x21 − mαx1 y1 − 98 mω2 y21 + 12 mω2 z21 , √ 3 3 α = ± 3 G(m1 − m2 ). 4a The motion along the z-axis is obviously stable. The equations of motion in the xy-plane read x¨ 1 − 34 ω2 x1 − αy1 − 2ωy˙1 = 0, y¨ 1 − 94 ω2 y1 − αx1 + 2ωx˙1 = 0.

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[9.29

The substitutions x = Aeit , y = Beit lead to the equation for : 4 2 4 − ω2 2 + 27 16 ω − α = 0.

Its roots are real for 16α 2  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) satisfies the condition m1 /m2  12 (25 + 3 69) ≈ 25 . In this case, the motion of bodies in the neighbourhood of Lagrange points is stable. Stability of motion is provided by the Coriolis force (cf. problem 9.28). On the x-axis, there are three more points where ∂U = 0, but the motion near them ∂r is unstable. Note that for the Sun–Jupiter system, asteroids are observed just near the Lagrange points (for more detail see [5], § 3.12).

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§10 The Hamiltonian equations of motion. Poisson brackets 10.1. Let ε be a vector of infinitesimal displacement; we then have ra → ra = ra + ε, pa → pa = pa , H(ra , pa ) = H(ra , pa ). Hence,

 ∂H a ∂r = 0. Using the Hamiltonian equation, we get a ˙ = P



p˙ a = −

a

 ∂H a

∂ra

= 0,

P = const.

For an infinitesimal rotation δϕ, we obtain ra → ra = ra + [δϕ, ra ], pa → pa = pa + [δϕ, pa ],    ∂H ∂H   [δϕ, ra ] + [δϕ, pa ] = 0 = H(ra , pa ) = H(ra , pa ), ∂ra ∂pa a  d {−p˙ a [δϕ, ra ] + r˙ [δϕ, pa ]} = −δϕ = [ra , pa ], dt a a or M=

 [ra , pa ] = const. a

10.2. H =

p2 (p − pψ cos θ)2 p2θ + ϕ + ψ. 2 2I1 2I3 2I1 sin θ

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

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[10.3

p2 + 1 ω2 x2 + αx3 . In particular, for small oscillations (|αx|  2(1 + 2βx) 2 ω2 , |βx|  1) 10.3.

H=

H = 12 p2 + 12 ω2 x2 + αx3 − βxp2 + 2β 2 x2 p2 − . . . , and up to and including terms linear in α and β the extra term in the Hamiltonian function of the harmonic oscillator is connected with the extra term in the Lagrangian function through the equation δH = −δL (see [1], § 40). 10.4. x = a cos(ωt + ϕ), p = −ω0 a sin(ωt + ϕ), where ω = (1 + 2λE0 ) ω0 ,

E0 = 12 ω02 a2 .

√ √ 10.5. p = p0 + Ft, x = x0 + A F ( p0 + Ft − p0 ). The given Hamiltonian function describes the motion of a charged vortex ring in liquid helium in the presence of a homogeneous electric field along the x-axis [26]. The characteristic features of such motion are the following: the momentum of the vortex increases with the time, while √ velocity of its motion drops x˙ = a/(2 p0 + Ft ). cp cp 10.6. r˙ = np , p˙ = 2 ∂n , p = |p|. n ∂r 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

 y + C2 , C1

where the C1 and C2 are determined by the initial and final points of the trajectory, respectively. 10.7. a) L = 12 m(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 + p2z 1  e 2 + py − Bx . 2m 2m c

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§10. The Hamiltonian equations of motion

Since H depend neither on y nor on z, we have py = const,

pz = const.

If we present H in the form H=

p2x p2 + 12 mω2 (x − x0 )2 + z , 2m 2m

where ω=

eB , mc

x0 =

cpy , eB

we see that x and px are obtained from the same Hamiltonian function as a harmonic oscillator. Therefore, we have x = a cos(ωt + ϕ) + x0 ,

px = −mωa sin(ωt + ϕ).

To determine y and z, we use the equations 1 e py − Bx = −ωa cos(ωt + ϕ), y˙ = ∂H = ∂py m c

z˙ =

pz , m

whence y = −a sin(ωt + ϕ) + y0 ,

z=

pz t + z0 . m

The particle moves along a spiral with its axis parallel to B. The generalized momentum py determines the distance of that axis from the yz-plane. Ueff y E

−x1 −x1

x1

Figure 164

x1 x

x

Figure 165

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10.9. The magnetic field is directed along the z-axis and equals 2hx. The motion along the z-axis is uniform. Here we consider only the motion in the xy-plane. The Hamiltonian function H=

2  eh p2x 1 + py − x2 2m 2m c

depends neither on y nor on t. Therefore, the integrals of motion are the generalized momentum py and energy E: eh 2 x , c  2 1 eh E = 12 mx˙2 + Ueff (x), Ueff (x) = py − x2 . 2m c py = my˙ +

When py  0, the function Ueff (x) is shown in Fig. 164, while an approximate view of the orbits can be found in Fig. 165. Note that the velocity y˙ = −

|py | eh 2 − x m mc

is negative everywhere and oscillates near the value py /m. When py > 0, the function e 2 h2 2 Ueff (x) = (x − x20 )2 , 2mc2

x0 =

py c eh

Um −x0

is shown in Fig. 166. Now the velocity y˙ =

Ueff

x0

x

Figure 166

eh 2 (x − x2 ) mc 0

has both positive and negative values for any value of E. The approximate view of the orbits 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 ˙ =

eh  2 x0 − x2  mc

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,

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309

§10. The Hamiltonian equations of motion (a)

(d)

y

y

x

x0 (b)

x

E  Um

x0

y (e) x

x0 (c)

E Um y

x

y

x

E  Um

x0

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 in the 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 the velocity of the drift, it is necessary to take into account the first anharmonic corrections (see [1], § 28 and [7], § 29) x = x0 + a cos ωt −

a2 (3 − cos 2ωt), 4x0

when calculating x2 . That leads to y ˙ =

cE 2hx20 e

(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

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˙2+ L = 12 M R

e ˙ + L1 (r, r˙ ) + e [B, R] r˙ , [B, r] R 2c 2c

[10.10

(1)

where M = m1 + m2 , L1 (r, r˙ ) = 12 m˙r2 + B =

e e2 [B , r] r˙ + , 2c r

m2 − m1 B m2 + m1

and m is the reduced mass. We rewrite the last term in (1) as e e ˙ + d e [B, R] r. [B, R] r˙ = [B, r] R 2c 2c dt 2c Omitting the total derivative with respect to time, we have ˙ + L1 (r, r˙ ). ˙ 2 + e [B, r]R L = 12 M R c The Hamiltonian function of the system now has the form H=

2 2 e2 e e 1  1  P − [B, r] + p − [B , r] − . 2M c 2m 2c r

This function has no evident dependence on R and t; therefore, the generalized momentum ˙ + e [B, r] P = ∂L = M R ˙ c ∂R and the total energy of the system ˙ 2 + 1 m˙r2 − E = 12 M R 2

e2 r

are conserved. The last equation can be rewritten in the form E = 12 m˙r2 + Ueff (r), Ueff (r) = −

2 e2 1  e + P − [B, r] . r 2M c

(2)

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10.14]

§10. The Hamiltonian equations of motion

311

From here follows (with the assumption P = const) that a particle of mass m moves in the field with the effective potential energy Ueff (r) and in the constant uniform magnetic field B . If the z-axis is directed along B, then e2 1 + 2 mω2 [(x − a)2 + (y − b)2 ] + const, r cPy eB cPx , a= . ω= √ , b=− c m1 m2 eB eB

Ueff (r) = −

After finding r(t), the law of motion of the centre of mass can be determined from (2):

 t Pt e R(t) = r(t) dt + R0 . − B, M Mc 0 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 +

e [v, B] .1 c

10.13. a) ε(p) = E, pB = const, where pB is the component of the momentum along the magnetic field B. The trajectory in the momentum space is determined by the intersection 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 orbit in momentum space by a rotation over π/2 around B and by changing the scale by a c . factor eB 10.14. E S= Emin

 dE

dp , |v⊥ |

T=

c eB



dp c ∂S = , |v⊥ | eB ∂E

where v⊥ is the component of the vector ∂ε , which is orthogonal to B. ∂p 1 For detail about the motion of electrons in a metal (problems 10.10–10.14) see, for istance, [16].

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[10.15

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10.15. a) −



−

eijk xk ;1



k

eijk pk ; −



k

eijk Mk .

k

b) ab; {aM, br} =

⎧ ⎨ ⎩

=−

ai M i ,



i



⎫ ⎬ bj xj

j

⎭

=



ai bj {Mi , xj } =

ij

ai bj eijk xk = −[a, b] r;

−[a, b]M.

ijk

c) 0; n r r n−2 ; 2a (ar).  10.16. {Ai , Aj } = − eijk Ak , {Ai , A4 } = 0, where i, j, and k take on the values 1, 2, and k

3, respectively (cf. problem 10.15a). 10.17. {Mi , jk } = −



eijl lk −

l



eikl lj ;

l

{ jk , il } = δij Mlk + δik Mlj + δjl Mik + δkl Mij , where Mkl = pk xl − pl xk . 10.18. When the system as a whole is rotated around the z-axis over an infinitesimal angle ε, the change δϕ in any function of the coordinates and momenta is in the first order in ε given by δϕ = ϕ(x−εy, y + εx, z, px −εpy , py + εpx , pz ) − ϕ(x, y, z, px , py , pz ) =   ∂ϕ ∂ϕ ∂ϕ ∂ϕ p + p = ε{Mz , ϕ}. = ε − y+ x− ∂x ∂y ∂px y ∂py x If ϕ is a scalar, this change under rotation must vanish, and thus {ϕ, Mz } = 0. If ϕ = fx is the 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 e is the completely antisymmetric tensor: ijk

e123 = e231 = e321 = 1, all other components of eijk vanish.

e132 = e321 = e213 = −1;

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What is the value of the Poisson brackets {Mz , Txx }, where Txx is a component of a tensor function?  10.19. [f , aM] = [f , a]; {fM, lM} = [f , l] M + Mi Mk {fi , lk }. ik

10.20. Substituting into the second formula of the preceding problem f = eζ and l = eξ , where eζ and eξ are unit vectors along the ζ - and ξ -axes of the moving system of reference, we have {Mζ , Mξ } = +Mη .

(1)

This equation differs in the sign of the right-hand side from the analogous relation for the components of the angular momentum along the axis of the fixed system of coordinates, {Mz , Mx } = −My .

(a)

(b) η y

M

(2)

y, η



M

M

Mξ

ε δMx δMx −εMy

ε

x

δMξ δMξ +εMη Mξ

ξ x,ξ

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 rotated as 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 when the 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).  ˙ α = βγ δ eαβγ (I −1 )γ δ Mβ Mδ . In particular, if we choose the moving system of 10.21. M coordinates such that the tensor of inertia Iαβ is diagonal, we obtain the Euler equation (see [1], § 36) using the relation Mα = Iα α ).

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[10.22

10.22. The equation of motion is ˙ i = {H, Mi } = γ M



eijk Mj Bk ,

or

˙ = −γ [B, M]; M

jk

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 B0 t) + My (0) sin(γ B0 t), My = −Mx (0) sin(γ B0 t) + My (0) cos(γ B0 t), Mz = Mz (0). b) The vector M rotates with an angular velocity −γ B, which in turn rotates around the 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 of the vector M are equal to ωx = −γ B1 ,

ωy = 0,

ωz = −γ B0 − ω ≡ ε.

At a given initial condition, the components M in the rotating system of reference are ε Mx = −a M0 (1 − cos λt), λ My = aM0 sin λt,  2  ε 2 Mz = + a cos λt M0 , λ2 where  λ = ε2 + γ 2 B21 ,

γ B1 a=  . ε2 + γ 2 B21

In the fixed system, we have Mx = Mx cos ωt − My sin ωt, My = Mx sin ωt + My cos ωt, Mz = Mz . When B1  B0 , the dependence of amplitudes Mx,y on ω is resonant: generally speaking, these amplitudes are small ∼ M0 B1 /B0 , but for |ε| = |ω + γ B1 |  γ B1 , they increase dramatically, reaching values ∼ M0 . In particular, for ω = −γ B0 , we get

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§10. The Hamiltonian equations of motion

315

Mx = M0 sin γ B1 t sin γ B0 t, My = M0 sin γ B1 t cos γ B0 t, Mz = M0 cos γ B1 t.  10.23. {vi , vj } = − e2 eijk Bk . m c k 2 pt 10.24. a) p(t) = p + Ft, r(t) = r + m + Ft 2m ; b) p(t) = p cos ωt − mωq sin ωt, p q(t) = q cos ωt + sin ωt. mω

Of course, these quantities can more simply be evaluated without using Poisson brackets. However, this method can easily be taken over in quantum mechanics. 10.25. a) According to the preceding problem, we have df = {h, f } = ∂H { f , f } = 0. ∂f dt b) The Hamiltonian function is H=

p2r 1 + f (θ, ϕ, pθ , pϕ ), 2m 2mr 2

where f (θ, ϕ, pθ , pϕ ) = p2θ +

p2ϕ sin2 θ

+ 2ma cos θ.

Integrals of motion are E, pϕ , and, according to the preceding problem, f . 10.26. a) {Ai , Aj } =

3 2H  εijk Mk , m k=1

{Ai , Mj } = −

3  k=1

εijk Ak ;

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[10.26

Exploring Classical Mechanics

b) {H, J1,2 } = 0, { J1i , J2j } = 0, { J1i , J1j } = −

3 

εijk J1k ,

k=1

{ J2i , J2j } = −

3 

εijk J2k ,

k=1

mα 2 . H =−  2 4 J1 + J22 The vectors J1 and J2 are the independent integrals of motion. Each has the same Poisson brackets for their components as an usual angular momentum. The presence of two such “momenta” is closely related to the so-called hidden symmetry of the hydrogen atom (see [22], Ch. I, § 5).

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§11 Canonical transformations

11.1. a)



2P sin Q, mω ˙ = ω + ω˙ sin 2Q, Q 2ω q=

p=

√ 2mωP cos Q,

p˙ = −P

ω˙ cos 2Q. ω

In this case, P and Q are action and angle variables. These variables are more helpful than p and q for solving the problem by the method of the perturbation theory, if the frequency ω(t) changes slowly, |ω| ˙  ω2 (see problem 13.11). b)  √ F 2P q= + 2mωP cos Q, sin Q, p = mω mω2  F˙ cos Q F˙ 2P ˙ ˙ Q=ω− √ , P =− sin Q. ω 2mωP ω mω   p2 11.2. (p, Q) = −Q 1 + ln . 4Q 11.3. The function (q1 , q2 , . . . , qs , P1 , P2 , . . . , Ps ) determines a canonical transformation, provided det

∂ 2 = 0. ∂qi ∂Pk

11.4. Let Q = q cos α − p sin α,

P = q sin α + p cos α.

Then we have {P, Q}p,q = −{q, p}p,q sin2 α + {p, q}p,q cos2 α = 1.

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

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[11.5

Exploring Classical Mechanics

For a system with one degree of freedom, this is sufficient for the transformation to be canonical. 11.5. One sees easily (and subsequent calculations verify) that the canonical transformation must be close to the identity transformation and that the terms ax2 P and bP 3 in the generating function are small. To solve the equations p = P + 2axP,

Q = x + ax2 + 3bP 2 ,

which determine the canonical transformation, for x and p, we replace x by Q and p by P in the small terms: p = P + 2aQP,

x = Q − aQ2 − 3bP 2 .

(1)

We proceed in the similar fashion when we express the Hamiltonian function in terms of the new variables: H  (Q, P) = 12 (P 2 + ω2 Q2 ) + αQ3 + βQP 2 + 2aQP 2 − aω2 Q3 − −3bω2 QP 2 + 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 = A cos ωt,

P = −ωA sin ωt,

and from (1) (cf. [1], § 28)

x = A cos ωt −

αA2  α  2 2 − β + A sin ωt. ω2 ω2

11.6. Reducing the Hamiltonian function to the form considered in problem 10.4, we get x = Q−

9β 5β 3 Q − QP 2 , 2 8ω0 8ω04

where Q = A cos ωt, (cf. [1], § 28).

P = −ω0 A sin ωt,

ω = ω0 +

3β 2 A 2ω0

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11.7]

§11. Canonical transformations

319

11.7. Because the terms xPX + yPY in the generating function correspond to the identical transformation, it is clear in advance that the parameters a, b, and c will be proportional to α. Therefore, the canonical transformation for px with required precision has the following form px =

∂ = PX + ayPX + 2cxPY ≈ PX + aYPX + 2cXPY . ∂x

Acting similarly, we get 2 py ≈ PY + aXPX , x ≈ X − aXY − 2bPX PY , y ≈ Y − bPX − cX 2 .

Substituting these expressions into a new Hamiltonian function H = H +

∂ =H ∂t

on the condition that this function corresponds (with the necessary accuracy) to the two independent oscillators, H =

 1  2 PX + (mω1 X)2 + PY2 + (mω2 Y )2 , 2m

we find a=−

α (2ω12 − ω22 ) 2α 2α , b = − , c = − . 4ω12 − ω22 (4ω12 − ω22 ) m2 ω22 (4ω12 − ω22 ) ω22

The new canonical variables correspond to the two free oscillators X = A cos(ω1 t + ϕ1 ), Y = B cos (ω2 t + ϕ2 ), PX = −mω1 A sin(ω1 t + ϕ1 ), PY = −mω2 B sin (ω2 t + ϕ2 ). As a result, with the required accuracy, we find x(t) = A cos (ω1 t + ϕ1 ) − +

α AB cos (ω+ t + ϕ+ ) + 2ω2 (2ω1 + ω2 )

α AB cos (ω− t + ϕ− ), 2ω2 (2ω1 − ω2 )

y(t) = B cos (ω2 t + ϕ2 ) +

α A2 2ω22

where ω± = ω1 ± ω2 and ϕ± = ϕ1 ± ϕ2 .

−

α A2 2(4ω12 − ω22 )

cos (2ω1 t + 2ϕ1 ) ,

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[11.8

Exploring Classical Mechanics

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 energy from x to y oscillations and back (see problem 8.10). 11.8. H  (P, Q) = H(P, Q); when X = A sin(ωt + ϕ), Y = 0 the oscillator perform a motion along an ellipse: x = A cos λ sin(ωt + ϕ), y = A sin λ cos(ωt + ϕ). 11.9. To make the notation less cumbersome, it is helpful for the time being to put m = ω = e = c = 1. One can easily reintroduce these factors in the final expressions. The transformation of problem 11.8 is a rotation in the xpy - and ypx -planes, which therefore leaves the form of that part of the Hamiltonian function, which is equal to 1 2

(x2 + y2 + p2x + p2y ),

invariant. On the other hand, the correction due to the terms 12 B2 x2 − Bxpy is equal to 1 2

B2 (X 2 cos2 λ + PY2 sin2 λ + 2XPY sin λ cos λ) + + B(X 2 − PY2 ) sin λ cos λ − B(cos2 λ − sin2 λ)XPY .

The off-diagonal term XPY vanishes if we put sin2 λ − cos2 λ + B sin λ cos λ = 0,

that is, tan 2λ =

2 . B

After a few simple transformations, the Hamiltonian function is reduced to the form H=

   1  2 PX + PY2 tan2 λ + 12 mω2 X 2 cot2 λ + Y 2 . 2m

(1)

The variables X and Y thus perform harmonic oscillations with frequencies which are, respectively, equal to ω1 = ω cot λ =



ω2 +

ω2 = ω tan λ =

ω2 +



eB 2mc eB 2mc

2 +

eB , 2mc

−

eB 2mc

2

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11.11]

321

§11. Canonical transformations

(cf. [2], § 21, problem). Each of the coordinates X and Y corresponds to a motion along 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 that even for very weak field B the “normal” oscillations is “circularly polarized”. On the other hand, oscillations corresponding to the coordinates X or Y with λ = 0, which if there were 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) the partial derivative with respect to the time of the generating function   PX PY xPX + yPY  = − mωxy + tan λ + mω cos λ (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 ,

tan 2λ =

2ωB ω2 , 2 ωB + ω12 − ω22

ωB =

eB , mc

we get

   22 2 1 2 2 H = PX + 2 Py + pz + 12 m 21 X 2 + ω22 Y 2 + ω32 z2 , 2m ω2 

where 1,2 was defined in problem 6.37. 11.11. The transformation (λ = π/4)   1 PYs qs1 = √ Xs + , Nmωs 2

  1 PXs qs2 = √ Ys + Nmωs 2

leaves the form of the Hamiltonian function

 R p20 p2s1 + p2s2 1 2 2 2 + + 2 Nmωs (qs1 + qs2 ) = H= 2Nm 2Nm s=1 

R 2 + P2 p20 PXs Ys 2 2 2 1 = + + 2 Nmωs (Xs + Ys ) . 2Nm 2Nm s=1

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[11.12

Exploring Classical Mechanics

invariant (cf. problem 11.8). The oscillation corresponding to Xs = A cos(ωs t + β) is A xn = √ sin(ωs t + nϕs + β), 2 and the oscillation corresponding to Ys = B cos(ωs t + β) is B xn = √ sin(−ωs t + nϕs − β) . 2 11.12. The new Hamiltonian function is H  = ωP1 , and in the new variables, the equations of motion have the form ˙ 2 = 0, P˙ 1 = P˙ 2 = Q

˙ 1 = ω. Q

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 −

1 ∂f (r, t) c ∂t

can be written as a canonical transformation, R = r,

e P = p + ∇f , c

H = H −

e ∂f , c ∂t

if one uses the generating function e (r, P, t) = rP − f (r, t). c 11.15. (q, P) = qP − F(q, t). m (q − Q)2 ; 11.16. b) Fτ (q, Q) = − 12 Fτ (q + Q) − 2τ c) Fτ (q, Q) =

mω [2qQ − (q2 + Q2 ) cos ωτ ]. 2 sin ωτ

11.17. a) Q = r + δa, P = p: a shift of the systems as a whole over δa (or a shift of the coordinate 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.

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§11. Canonical transformations

323

c) Q(t) = q(t + δτ ), P(t) = p(t + δτ ), H  (P, Q, t) = H(p, q, t + δτ ). The transformation is a shift in the time by δτ (cf. [1], § 45, [7] § 36.2). d) Q = r + 2p δα, P = p − 2r δα. The transformation is a rotation over an angle 2 δα in each of the xi pi -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 the direction of n while δϕ = 2π δa is the angle of rotation around n; (h is the pitch of the h screw); 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 and including first-order terms δf (q, p) = λ

∂f ∂W (q, p) ∂f ∂W (q, p) −λ . ∂q ∂p ∂p ∂q

11.21. Putting  = rP + λrp in the preceding problem, we get a similarity transformation with α = 1 + λ (see problem 11.13). The given Hamiltonian function is such that H  (P, Q) = α −2 H(P, Q), and therefore λ{H, rp} = H  − H = −2λH(λ → 0). On the other hand, {H, rp} = d (rp), and hence dt rp − 2Et = const (cf. problem 4.15b). 11.23. Let δ1 q and δ1 p be the changes in the coordinates and momenta connected with the 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:

δ1 p = P − p = −λ1

∂ 2 W1 (q, p) ∂W1 (q, p) ∂W1 (q, P) ∂W1 (q, p) = −λ1 + λ21 . ∂q ∂q ∂q ∂p∂q

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f (q + δ1 q, p + δ1 p) = f (q, p) + λ1 {W1 (q, p), f (q, p)} + λ21 ϕ1 (q, p).

[11.24

(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 + δ21 q, p + δ21 p) = f + λ2 {W2 , f } + λ1 {W1 , f }+ +λ1 λ2 {W2 , {W1 , f }} + λ21 ϕ1 + λ22 ϕ2 .

(2)

The transformation of λ21 ϕ1 (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 + δ12 q, p + δ12 p) = f + λ1 {W1 , f } + λ2 {W2 , f }+ +λ1 λ2 {W1 , {W2 , f }} + λ21 ϕ1 + λ22 ϕ2

(3)

The difference between (2) and (3) lies only in the second-order terms which are proportional to λ1 λ2 . Subtracting (3) from (2), we obtain λ1 λ2 ({W2 , {W1 , f }} − {W1 , {W2 , f }}) = λ1 λ2 { 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 be a “motion” where λ plays the role of the time and W (q, p) plays role of the Hamiltonian function (cf. problem 11.1c). The equations of “motion” are dQ ∂W (Q, P) , = ∂P dλ

dP ∂W (Q, P) . =− ∂Q dλ

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 λ λ {W , r} = {Ma, r} = −[n, r] δϕ, N N δp = −[n, p] δϕ, δr = −

where M = [r, p], n =

a λa , δϕ = − . a N

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11.25]

325

§11. Canonical transformations

This transformation is a rotation of the coordinate system over an angle δϕ around the direction of n. If we take the z-axis along a, we finally obtain X = x cos ϕ + y sin ϕ,

Y = y cos ϕ − x sin ϕ,

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 = −x δϕ,

δpy = y δϕ,

where δϕ = λ/(2N). This transformation represents a rotation over an angle +δϕ in the xpx -plane and over an angle −δϕ in the ypy -plane. Therefore, we have X = x cos ϕ + px sin ϕ, PX = −x sin ϕ + px cos ϕ,

Y = y cos ϕ − py sin ϕ, PY = y sin ϕ + py cos ϕ.

Similarly, A2 and A3 represent the rotation over an angle ϕ or −ϕ in the xpy - and ypx planes and in the xy- and px py -planes, respectively, while A4 represents the rotation over an 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 harmonic oscillator (its Hamiltonian function is H = 12 A4 ) and the motion of a particle in the xy-plane in the arbitrary potential field having an axial symmetry, U(x2 + y2 ). In both cases, the integrals of motion are the angular momentum 2A3 , the conservation of which is due to invariance of the system with respect to rotations around the z-axis. For the oscillator, in addition, there are integrals of motion A1 and A2 , the conservation of which is connected with a “hidden” symmetry—with the invariance of the Hamiltonian function with respect to certain rotations in phase space. In this sense, the oscillator is similar to a particle in three-dimensional central field, for which there are three integral of motion Mx,y,z . The presence of additional integrals of motion for the two-dimensional oscillator leads to fact that point (x, y, px , py ) in phase space moves along a closed line, while for a particle in the arbitrary field U(x2 + y2 ), the phase trajectory “fills” two-dimensional surface (see [1], § 52, [7] § 44). 11.25. a) The volume specified in momentum and in phase space does not change with time, but in coordinate space the volume is spread out. Thus, if at t = 0 the state of the system is represented by the rectangle ABCD (Fig. 169), it goes over to the parallelogram A B C D (AD = A D ) after a time t, and the distance in the x-direction between the p points A and C is equal to x = x0 + m0 t. As time goes on, this parallelogram degenerates into a narrow, very long strip.

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326

[11.25

Exploring Classical Mechanics p p0+Δp0

p0

B

A

B

C A

D

C

D

x x0 x0+Δx0 x0+p0t/m x +Δx + p0+Δp0 t 0 0 m

Figure 169

b) If there is a wall at the point x = L, the state of the system can no longer be represented 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 of very narrow parallel strips which are almost uniformly distributed inside two rectangles 0  x  L, p0  p  p0 + p0 and 0  x  L, −p0 − p0  p  − p0 (Fig. 171b). (a) p p0+Δp0

B

p0

A

C D A

(b) p C p0+Δp0

B D

p0 L

x

L

−p0

−p0

−p0−Δp0

−p0−Δp0

x

Figure 170 2 c) The phase orbit of an oscillator with energy E and frequency ω is the ellipse x2 + a   p2 2E , b = 2E . All points of the specified phase volume = 1 with semi-axes a = m b2 mω2 move along such ellipsis and return to their initial state after a period T = 2π /ω. The dimension 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 spreading out 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) + λ cos(ωt + ϕ)],

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§11. Canonical transformations

327

and the oscillations are damped so that the phase orbit is a spiral,   x2 p + λmx 2 + = e−2λt . maω a2 The specified phase volume decreases until it vanishes. The non-conservation of phase volume here is connected with the fact that the system is not canonical: to describe it fully, we need know not only the Lagrangian function L = 12 m x˙2 − ω02 x2 but also the dissipative function F = 12 mλx˙2 (see [1], § 25). If we choose for this system the “Lagrangian function” in the form   L = 12 me2λt x˙2 − ω02 x2 (cf. problem 4.20), the phase volume specified will be conserved for the appropriate  canonical variables x and p = ∂L ; however, in this case the generalized momentum ∂ x˙ ˙ 2λt will not have a simple physical meaning, as before. p = mxe 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 particles have made one more oscillation than the slowest ones: T2 , τ∼ T

  p0 p  dU(x0 )  E ∼ + x. m dx 

dT T ∼ E, dE

f) Let there be N particles such that the points in phase space which represent their state are at time t = 0 distributed with a density Nw(x0 , p0 , 0) and that they move according to the equations x = f (x0 , p0 , t), p = ϕ(x0 , p0 , t). Here f (x0 , p0 , t) = x0 +

p0 t, m

ϕ(x0 , p0 , t) = p0

(1)

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[11.25

Exploring Classical Mechanics

for free movement and p0 sin ωt, mω ϕ(x0 , p0 , t) = −mωx0 sin ωt + p0 cos ωt f (x0 , p0 , t) = x0 cos ωt +

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, for an infinitesimal phase volume dx dp, we have Nw(x, p, t) dx dp = Nw(x0 , p0 , 0) dx0 dp0 .

According to the Liouville theorem (see [1], § 46 and [7], § 40)

∂(x, p) = 1, and ∂(x0 , p0 )

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 free particles are 1 t , β =− , 2 2(x0 ) m(x0 )2 t2 1 + , γ= 2(p0 )2 2m2 (x0 )2 α=

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§11. Canonical transformations

and for oscillators, α=

cos2 ωt m2 ω2 sin2 ωt + , 2(x0 )2 2(p0 )2

cos2 ωt sin2 ωt + , 2(p0 )2 2m2 ω2 (x0 )2   mω 1 − . β = sin ωt cos ωt (p0 )2 mω(x0 )2

γ=

We show in Figs. 171 and 172 how regions in phase space in which 2π x0 p0 w(x, p, t) 

1 2

p p P

p0

X

X0

X

x

x

Figure 171

Figure 172

(for free particles and for harmonic oscillators, respectively) move about. These regions are ellipses which are deformed as time marches on.1 Their centres are displaced according to the same law (1) as the particles. In the case of free particles, this ellipse is spread out without limit, but in the case of the oscillators, it only pulsates. We note that 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 the values 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 that mω = 1, the phase orbits are circles, and the specified region in phase space rotates around the origin without being deformed.

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[11.26

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: 2

1 − (x−X) w(x, t) = √ e 2(x)2 , 2π x 2 1 − (p−P)  w(p, t) = √ e 2(p)2 , 2π p where for free movement (x)2 = (x0 )2 +

(p0 )2 2 t , m2

(p)2 = (p0 )2 ,

and for the oscillators (p0 )2 2 sin ωt, m2 ω 2 (p)2 = (p0 )2 cos2 ωt + m2 ω2 (x0 )2 sin2 ωt. (x)2 = (x0 )2 cos2 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) = 2i mωx2 + 2i Q2 e−2iωt − i 2mωxQe−iωt . The new Hamiltonian function is, therefore, H0 (Q, P) = H0 +

∂F(x, Q, t) = 0. ∂t

c) We note that the expression −3Q2 P 2 /(2m2 ω2 ) is the only term in the expression

x4 =

Qe−iωt − iPeiωt √ 2mω

4

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§11. Canonical transformations

331

which does not contain the time. Whence, the average Hamiltonian function is H  (Q, P) = −

3β Q2 P 2 . 8mω2

In the following, we will omit the brackets denoting averaging over time. Now it is clear that −iQP = |Q0 |2 = |a|2 is the integral of motion. The Hamiltonian equations are ˙ = −iεQ, Q

P˙ = iεP,

ε=

3β|Q0 |2 , 4mω2

where Q = Q0 e−iεt ,

P = iQ∗0 eiεt ;

whence 1   x= √ (Q0 e−Iω t + Q∗0 eiω t ) = x0 cos(ω t + ϕ). 2mω The influence of the additive term δU is reduced to change of frequency ω = ω +

3βx20 3β|Q0 |2 = ω + 8ω 4mω2

d) The new Hamiltonian function

Qe−iωt − iPeiωt H (Q, P, t) = m ω α √ 2mω 

4

2 2

e4iωt + e−4iωt 2

after averaging is reduced to H  (Q, P) = 18 α(Q4 + P 4 ). 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(P 4 − Q4 ). Taking into account that 1 4

α(Q4 + P 4 ) = A = const,

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[11.27

Exploring Classical Mechanics

we find ξ˙ 2 = −4A2 + α 2 ξ 4 . 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 ) = − 12 α 2 ξ 4 and with the energy −2A2 (cf. problem 1.2). Note that the amplitude increases to infinity for a finite time (the so-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 iωt e , √ 2mω

b=

mω2 y + ipy iω2 t e , √ 2mω2

c=

mω3 z + ipz iω3 t e , √ 2mω3

(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∗ ), α ε = ω1 − ω, η = √ . 4 2mωω2 ω3 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 d 2 |a| = iη(ab∗ c∗ − a∗ bc) dt can be presented in the form more helpful for qualitative investigation: ξ˙ 2 + V (ξ ) = 0, 1 Integrals B and C are called integrals of Manley–Rowe.

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11.28]

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§11. Canonical transformations

where ξ = |a|2 and V (ξ ) = (A − εξ )2 − 4ξ η2 (B − ξ )(C − ξ ). In the initial moment c = 0, therefore a = εC, B > C and V (ξ ) = (C − ξ )2 (ε2 − 4ξ η2 ) − 4ξ η2 (B − C)(C − ξ ). The function V (ξ ) for the cases ε2 < 4η2 C and ε2 > 4η2 C 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 and the y and z-oscillators. In the second case (i.e., when there are small initial amplitudes and 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

b) The equations of motion

η = 18 hω. (a)

−a˙ = i(ε + 2μ|a|2 )a + 2iηa∗ , ∗

V ε24η2C

2

When ξ = |a|2 , we get

C

Figure 173

ξ˙ = −2iη(a∗2 − a2 ). Taking into account that H  = εξ + μξ 2 + η(a2 + a∗2 ) = C = const, we obtain ξ˙ 2 + V (ξ ) = 0, where V (ξ ) = 4η2 [(a∗2 + a2 )2 − 4|a|4 ] = 4(C − εξ − μξ 2 )2 − 16η2 ξ 2 .

ξ

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[11.29

Exploring Classical Mechanics

V In the case considered, the value of C is small accordξm ing to the initial conditions. In the resonance region ξ |ε| < 2η, the graph of function V (ξ ) (Fig. 174) demonstrates that ξ performs the oscillations in the range from zero up to ξm ≈ 2|a1 |2 . Thus, the transition to the stable oscillation Figure 174 ξ = |a1 |2 (cf. problem 8.7) can be provided only by some previously unaccounted for mechanism, such as friction. This transition may be very long. Please note that this transition process has the character of beats even at zero “detuning”, ε = 0, in contrast to the transition process in the linear oscillations (see problem 5.11).

11.29. The average Hamiltonian function is    1  ε + 14 h P 2 . H  (Q, P, t) = H  (Q, P) = 12 mω2 ε − 14 h Q2 + 2m  Quantity Q2 + P 2 /(m2 ω2 ) represents the amplitude of the oscillation. The variables Q and P change very little over the period 2π/γ , as can be easily seen from the Hamiltonian equations which contain small parameters ε and h. On the Q, P-plane, a point representing the state of the system moves along the line H  (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 the first case, the amplitude ultimately increases without limit; in the second case, it performs the bits (cf. problem 8.8). (a)

(b)

P

P

Q

Q

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 of this circumference

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11.30]

§11. Canonical transformations

a=

x2 +

335

p2x m2 ω12

matches the amplitude of the oscillations along the x-axis. In the X, PX /(mω1 )-plane this will be a fixed point x = x(0), PX = px (0). Thus, the new variables are time-independent at V = 0 and, therefore, H0 = 0.1 When V = 0, these variables are time-dependent, but because the new Hamiltonian function H  = H0 + V = V is small, the average motion in these variables is slow. Indeed, after averaging we have the new Hamiltonian function H  = −

β (ω1 XPY − ω2 YPX ) , 4ω1 ω2

and from the Hamiltonian equations β X˙ = Y, 4ω1

β Y˙ = − X, 4ω2

we easily get  X = A cos(γ t + ϕ),

Y =−

ω1 A sin(γ t + ϕ), ω2

β γ= √  ω1, 2 . 4 ω1 ω2

Similarly, for the new momenta, we have PX = mω1 B cos(γ t + ψ),

√ PY = −m ω1 ω2 B sin(γ t + ψ).

Thus, in the X, PX /(mω1 )-plane there occurs slow (with frequency γ ) motion along an ellipse. This motion corresponds to the oscillations along the x-axis with slowly varying amplitude a(t) =

  X 2 + (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 A2 sin2 (γ t + ϕ) + B2 sin2 (γ t + ψ). ω2

1 From the Hamiltonian equations for new variables (e.g. X ˙ = ∂H  /∂PX = 0), it follows that H  is 0 0  independent of them. Therefore, H0 = f (t), where f (t) is an arbitrary time function, which we can put equal to

zero without losing generality.

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[11.30

This shows that the energy of x- and y-oscillators Ex = 12 mω12 a2 (t) and Ey = 1 2 2 2 mω2 b (t), respectively, and their sum E = Ex + Ey are not conserved. However, the value which can be called the total number of quanta is conserved:

n=

Ey Ex mω1 2 + = C . 2 hbar h¯ ω1 h¯ ω2

√ Here C = A2 + B2 , and h¯ is the Planck constant. In particular, at ϕ = ψ = 0 the beat amplitude reaches zero PX x = X cos ω1 t + sin ω1 t = C cos γ t cos(ω1 t + ϕ0 ), mω1  B ω1 sin γ t cos(ω2 t + ϕ0 ), tan ϕ0 = − , y = −C ω2 A and the energy oscillates with the frequency 2γ :   E = 12 mω1 C2 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 = 14 mω1 (ω1 + ω2 ) C2 . Note that this problem coincides with problem 11.27 related to three interacting oscillators considered in the limit of the energy of the z-oscillator Ez  Ex, y so high that the beats of the x- and y-oscillators have almost not effect on its motion z = z0 sin ω3 t,

ω3 = ω1 − ω2 .

In this case, β = 12 αz0 , 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 role of a large energy reservoir with which x- and y-oscillators exchange energy. b) New canonical variables increase exponentially with time

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337



 ω1  γ t X = Ae + Be , Y = Ae − Be−γ t , ω2     √ PX = mω1 Deγ t + Fe−γ t , PY = −m ω1 ω2 Deγ t − Fe−γ t , γt

−γ t

that corresponds to the exponentially growing amplitude of the oscillations along the xand y-axes. In this case, the difference in the number of quanta is conserved Ey 2mω1 Ex − = (AB + DF). h¯ ω1 h¯ ω2 h¯

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§12 The Hamilton–Jacobi equation

12.2. It is clear that the trajectory is a curve in a plane. If we use polar coordinates, we can separate the variables in the Hamilton–Jacobi equation, provided we take the polar axis z along a. The complete integral of the Hamilton–Jacobi equation is S = −Et ±

    β − 2ma cos θ dθ ± 2mE − β/r 2 dr.

(1)

To fix the signs in (1), we use the relations  β ∂S = ± 2mE − 2 , pr = m˙r = ∂r r  ∂S pθ = mr 2 θ˙ = = ± β − 2ma cos θ. ∂θ

(2) (3)

On the initial part of the trajectory r˙ < 0, θ˙ > 0 (we assume that the trajectory lies above the 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 = B is the equation of the ∂β trajectory: θ 0

dθ + √ β − 2ma cos θ

r ∞

dr  = B. r 2 2mE − β/r 2

(4)

The lower limits of the integrals can be chosen arbitrarily as long as the constant B is not determined. From our choice of lower limits and the condition that θ → 0 as r → ∞, it follows that B = 0. The constant β is an integral of motion of our problem, and from (3), we have β = p2θ + 2ma cos θ.

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

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§12. The Hamilton–Jacobi equation

339

M L K N

θmax

θm z

O

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(Eρ 2 + a),

E = 12 .mv2

When r changes from ∞ to  rm =

β = 2mE

 a ρ2 + , E

which is determined by the condition pr = 0, θ changes from zero to a value θm which is such that θm 0

dθ + √ β − 2ma cos θ

rm ∞

dr  = 0. r 2 2mE − β/r 2

(5)

A further increase in θ is accompanied by an increase in r; pr then changes sign. The equation for the part LM of the orbit is θ θm

dθ − √ β − 2ma cos θ

r rm

dr  = 0; r 2 2mE − β/r 2

(6)

it is more helpful to use (5) and (6) and write it in the form: θ 0

dθ − √ β − 2ma cos θ

∞ rm

dr  − 2 r 2mE − β/r 2

r rm

r2

dr  = 0. 2mE − β/r 2

(7)

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[12.3

Exploring Classical Mechanics

At r → ∞ the trajectory asymptotically approaches a straight line parallel to ON. The angle θmax can be found from the equation1 θmax

0

dθ =2 √ β − 2ma cos θ

∞

rm

dr π  =√ . β r 2 2mE − β/r 2

(8)

The equation ∂S = A determines r as function of t. If we choose A such that r(0) = rm , ∂E we get   a 2 2 2 (9) r = v t + rm = ρ 2 + + v2 t 2 . E The integral over r in (4) and (7) can be evaluated elementarily, but those over θ reduce to elliptical integrals. If Eρ 2  a, we can expand the integrand in (4) and (7) in powers of 2ma/β ≈ a/(Eρ 2 ). Up to and including first-order terms, we get  r sin θ = rm

a 1− cos θ 2Eρ 2

 (10)

(see Fig. 177). M L M

K K

O

z

In this approximation, the angle over which the velocity of the particle is deflected after the scattering is zero. This can be explained by the fact that the action of the force along different sections of the orbit (which to first approximation is a straight line K  M  ) is self-concealing.

Figure 177

12.3. a) To determine the angle over which the particle is deflected, we must expand in (8) of the preceding problem the powers of a/(Eρ 2 ) up to the second order. We get the equation θmax +

ma 3 sin θmax + β 4



ma β

2  θmax +

 1 sin 2θmax = π . 2

(1)

1 We draw attention to the following method to avoid the calculation of the integral over r in (8). This integral is 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.

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12.3]

§12. The Hamilton–Jacobi equation

Solving this equation up to order (ma/β)2 , we find the angle of deflection1    2 3 a ma 2 χ = π − θmax = π = 3π . 4 β 4Eρ 2

341

(2)

The scattering cross-section is dσ = π |dρ 2 | =

√ 3π a d . 16Eχ 5/2

(3)

The dependence on χ which we have obtained is the same as for small-angles scattering in the field γ /r 4 , which decreases much faster than U(r). b) dσ = π b 3 . d 8Eχ c) When Eρ 2  |b(θ)|, we have, instead of (10) of problems 12.2, the following expression for all θ: θ+

m β



θ

b(θ) dθ +

0

3 m2 2 β2



θ

b2 (θ) dθ =

0

=

 when 0 < θ < θm , arcsin rrm , π − arcsin rrm , when θm < θ < θmax ,

up to and including second-order terms. Here, the constant β is equal to β = 2m[Eρ 2 + b(0)] ≈ 2mEρ 2 and rm ≈ ρ. If π b(θ) dθ = π b = 0, 0

we can limit ourselves to the first approximation for which χ = π − θmax =

m π b , β

1 We look for θ max in the form θmax = θ0 + θ1 + θ2 + . . ., where θ1 ∼ (ma/β) θ0 . In zeroth approximation we get from (1) θ0 = π ; in first approximation   ma θ1 + sin θ0 = 0, β

where θ1 = 0; in second approximation θ2 + whence follows (2).

ma 3 θ1 cos θ0 + β 4



   ma 2 1 θ0 + sin 2θ0 = 0, β 2

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Exploring Classical Mechanics

[12.4

and the small-angle scattering cross-section, dσ π b = , d 4Eχ 3 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 √  3π dσ 1 2 = b . d 8Eχ 5/2 2 12.4. a) We can separate the variables in the Hamilton–Jcobi equation if we choose the spherical coordinates with the z-axis parallel to a. The canonical momenta are  pr = m˙r = − 2mE − β/r 2 ,  pθ = mr 2 θ˙ = ± β − 2ma cos θ − p2ϕ / sin2 θ,

(1)

pϕ = mr 2 ϕ˙ sin2 θ = const. One finds the constant β = p2θ +

p2ϕ sin2 θ

+ 2ma cos θ

easily by noticing that p2θ +

p2ϕ sin2 θ

= M2 ,

where M is the total angular momentum of the particle; it is helpful to evaluate M for r → ∞, θ → π − α (α is the angle between v∞ and a); that is, before the collision, β = 2m(Eρ 2 − a cos α). According to (1), the particle can fall into the centre when β < 0 or ρ2
−ER2 , when a cos α < −ER2 ,

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[12.6

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12.6. a) We use the same notation as in problem 12.2. The equation for the initial part of the orbit (r → ∞, θ → π ) is π θ

dθ = √ β − 2ma cos θ

∞ r

r2

dr  , 2mE − β/r 2

(1)

with β = 2m(Eρ 2 − a).

(2)

When β > 0, the angle θ decreases when r changes from ∞ to rm and then again increasing to ∞. The equation for the part of the orbit after r had passed through the minimum distance to the centre is π θ

dθ π = √ + √ 2 β β − 2ma cos θ

r rm

r2



dr 2mE − β/r 2

.

(3)

It is clear that when Eρ 2  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 monotonically 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 − 2ma cos θ1 = 0. After that, the angle increases until it reaches the value 2π − θ1 (section BC of the orbit) ∞ r

π

dr

 = r 2 2mE − βr −2

A

θ1

dθ + √ β − 2ma cos θ

θ θ1

dθ . √ β − 2ma cos θ

(4)

B n=1

D

2 3

E

2π–θ

O

θ1 z

C

Figure 178

In point C, the momentum pθ again changes sign, and θ decreases until it reaches the value θ1 in point D; it then increases again, and so on.

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12.6]

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§12. The Hamilton–Jacobi equation

The equation of the complete orbit can be written in the form ∞ r2

r

π

dr

 = (−1) 2mE − βr −2

n θ

dθ + 2n √ β − 2ma cos θ

π θ1

dθ √ β − 2ma cos θ

(n = 0, 1, 2, . . .)

(5)

A single value of θ (θ1 < θ < 2π − θ1 ) corresponds to an infinite number of values of r (n can take any non-negative integer value since the integral on the left-hand side of (5) increases without bound as r → 0). Thus, the particle performs infinitely many oscillations between the straight lines BD and CE before it falls into the centre. In the case of small impact parameters Eρ 2 a, π − θ 1 so that we can write in (5) cos θ ≈ −1 + 12 (π − θ)2 . The final result is1  θ = π −ρ

    2E 1 1 a sin √ arsinh . a r E 2

(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 t 2 − τ 2 ,

v=

 2E/m,

 −∞ < t < τ = − |β|/(2E),

(7)

and the particle falls into the centre at time τ . b) If β > 0 ( Eρ 2 > a), we have π θ

 dθ arcsin rrm , =  π − arcsin rrm , (1 + sin θ) 1 + 2ma β

when θm < θ < π , when θmin < θ < θm .

If β < 0 (Eρ 2 < a), we have ⎛ ⎜ ⎝

π

θ1

θ ± θ1

⎞ θ2 ∞ dθ dr ⎟  +2l ⎠ √ = , 2 β + 2ma(1 + sin θ) r 2mE − β/r 2 r

θ1

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 = − arcsin

1 arsinh x = ln(x +

 1 + x2 ).

Eρ 2 , a

θ2 = π + arcsin

Eρ 2 . a

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[12.7

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(b)

θ2 z

O

θ1

Figure 179

If β = 0 (Eρ 2 = a), we have  ρ r=√ , F(θ) = ln 2F(θ)

tan

1 8

 π + 14 (π − θ) tan( 18 π )

− 12 π < θ < π .

,

√ The particle moves along the orbit of Fig. 179b, and we have r = − 2E/m t (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) 

  β dθ − β − 2ma cos θ − 2mE − 2 dr. 2 r sin θ

 S = −Et + pϕ ϕ ±

p2ϕ

The generalized momenta are the same as in problem 12.4a. The particle can fall into the centre if β = 2m(Eρ 2 − a cos α) < 0 (which is clearly satisfied if α 2 < 2Eρ 2 /a 1). The equation for the orbit,  ϕ=±

θ π−α

r0 = ∓ sinh r



θ π−α



pϕ dθ

,

sin2 θ β − 2ma cos θ − 

√ |β| dθ

sin2 θ β − 2ma cos θ −

(1)

p2ϕ 2

sin θ ,

p2ϕ sin2 θ

r02 =

|β| 2mE

(2)

can, in general, not be integrated to produce elementary functions. However, one can easily describe the motion qualitatively if one notes that (1) which gives a relation between the angles θ and ϕ is, apart from the notation, the same as the equation for the motion of a spherical pendulum (see [1], § 14, problem 1). The particle thus moves in such a way that the point where its radius vector intersects the surface of a sphere of radius l describes the same curve as does a spherical pendulum of length l, energy β/(2ml 2 ), and angular momentum pϕ in the field of gravity g = −a/(ml 3 ). This curve is enclosed between two “parallel” circles on the sphere corresponding to θ = θ1 and θ = θ2 .

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347

If α 2 < 2Eρ 2 /a 1, one can easily integrate (1) and (2): 

     ma r0 θ = π − ε + 12 α 2 − ε − 12 α 2 cos 2 arsinh , |β| r √ 2α ε Eρ 2 θ =π− , ε= . a 2ε + α 2 + (2ε − α 2 ) cos 2ϕ

(3)

It is clear from (3) that a particle when falling into the centre moves in the region between two conical surfaces θ1  θ  θ2 rotating around the z-axis, while one complete rotation around the z-axis corresponds to two complete oscillations in the angle θ . In this approximation 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 p = 1 + e cos f (θ ), r

(1)

where β p= , mα

 e=

2Eβ 1+ , mα 2

 f (θ ) =

θ

θ1

dθ ,  1 − 2ma cos θ β

while the constants E and β satisfy the inequalities −mα 2 /(2β) < E < 0 and β > 0. If 0 < β < 2ma, the orbit “fills” the region ABCDEF (Fig. 180), r1  r  r2 , r1,2 =

p β , θ1  θ  θ2 , θ1 = arccos , θ2 = 2π − θ1 ; 1±e 2ma

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 sin θ + 34 ζ 2 θ + 38 ζ 2 sin 2θ + C2 ,

(3)

where ζ = ma/β. This is a slightly deformed ellipse, the nature of the deformation being determined by its orientation. Equation (3) is valid when arccos ζ −2  θ  arccos(−ζ −2 ). It is interesting to make a comparison with the results of problem 2.28.

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[12.9

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B C

A

F O

z

E D

Figure 180

Figure 181

12.9. If the motion lies inside the ring r1  r  r2 , we have  θ1 +2π dθ n = 2π .  l θ1 1 − 2ma β cos θ If the motion lies in the region r1  r  r2 , θ1  θ  θ2 , we have  θ2 dθ n =π  l θ1 1 − 2ma β cos θ (n and l are integers.) 12.10. We can separate variables in the Hamilton–Jcobi equation, if we take the z-axis along the vector a (see [1],§ 48, (48.9)). The radial motion,  t=

m 2



dr  E + αr −

, β 2mr 2

is, when β  0, the same as the motion of a particle in a Coulomb field −α/r with angular momentum β and energy E. When β < 0, the particle can fall into the centre. The equations of the orbit are ∂S = const, ∂S = const. The first of these, ∂pϕ ∂β  ϕ=±



pϕ dθ

sin2 θ β − 2ma cos θ −

p2ϕ sin2 θ

is the same as the equation for the orbit of a spherical pendulum with energy β/(2ml 2 ) and angular momentum Mz = pϕ in the field of gravity g = −a/(ml 3 ) (see [1], § 14, problem 1). The second equation connects r and θ . One can also use the analogy with a spherical pendulum for the analysis of that equation.

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§12. The Hamilton–Jacobi 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 + pϕ ϕ + pξ (ξ ) dξ + pη (η) dη, where pξ = ± pη = ±

 

1 2

m(E − Uξ (ξ )),

Uξ (ξ ) =

1 2

m(E − Uη (η)),

Uη (η) =

p2ϕ 2mξ 2 p2ϕ 2mη2

−

mα + β 1 − 2 Fξ , mξ

−

mα − β 1 + 2 Fη. mη

The trajectory and the law of motion are determined by the equations ∂S = B, ∂β

∂S = C, ∂pϕ

∂S = A, ∂E

that is, 

dξ − ξ pξ (ξ )



  pϕ pϕ dη dξ dη − = C, = 4B, ϕ − 2 2 ηpη (η) 4 ξ pξ (ξ ) 4 η pη (η)   dξ m dη m −t + + = A. 4 pξ (ξ ) 4 pη (η)

(a)

(1)

(b) Uξ

Uξ max E

ξ1

ξ2

Uη aξ b

b η1

η2 a

η

Uη min

Figure 182

When studying the character of the motion, we must determine the region admissible for the values of ξ and η for given values of E, pϕ , and β. Fig. 182 gives the shape of of the effective potential energies Uξ (ξ ) and Uη (η).

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[12.13

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If F = 0, and when −mα < β < mα (see curves 182a) and E < 0, the motion in both ξ and η is finite, but for E > 0, it is infinite. When a small force A F > 0 appears, the curve Uξ (ξ ) shows a maximum (see curves 182b); when Uη min < E < Uξ max , the C motion is finite, as before. The motion is restricted O D to the region ξ1  ξ  ξ2 , η1  η  η2 (see Fig. 183) z in the “ρz-plane”, while the ρz-plane itself rotates around the z-axis with an angular velocity ϕ. ˙ The Figure 183 orbit fills the region of space formed by rotating the figure ABCD around the z-axis (see also problem 2.40). When Uξ max < E, the motion is infinite. When F increases, the quantity Uξ max decreases and Uη min increases. When Uξ max becomes less than Uη min , finite motion becomes impossible (when β < −mα + 3 4 1/3 , there are no U ξ max extrema). 2 (Fmpϕ ) B

12.13. In elliptical coordinates, we have 

ρ = σ (ξ 2 − 1)(1 − η2 ), z = σ ξ η, σ =

 c 2 − 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 + pϕ ϕ ±  ±

2mσ 2 E +

 2mσ 2 E −

β 1 − η2

−

β − 2mσ 2 A(ξ ) ξ2 − 1 p2ϕ

(1 − η2 )2

dη,

where A(ξ ) = (ξ 2 − η2 )U(ξ ) = U(ξ ).

−

p2ϕ (1 − ξ 2 )2

dξ ± (1)

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§12. The Hamilton–Jacobi equation

351

For a particle flying through the origin, pϕ = 0. From the previous discussion, it follows that  pξ = ± 2mσ 2 E +  pη = ± 2mσ 2 E −

β − 2mσ 2 A(ξ ) mσ 2 (ξ 2 − η2 ) = ξ˙ , ξ2 − 1 ξ2 − 1

(2)

β mσ 2 (ξ 2 − η2 ) = η. ˙ 2 1−η 1 − η2

(3)

In the origin (η = 0, ξ = 1), we have  2mσ 2 E − β , η˙ = ± mσ 2

ξ˙ = 0,

z˙ = σ (ξ˙ η + ηξ ˙ ) = σ η˙

and from the condition 

2E cos α = m

 2mσ 2 E − β , mσ

we find β = 2mσ 2 E sin2 α. ρ

z

Figure 184

The region of the inaccessible values of η is determined by the condition 2mσ 2 E − β/(1 − η2 ) < 0

or |η| > | cos α|.

The motion thus takes place in the region |η| < | cos α|, 1 < ξ < ξ0 (see orbits in Fig. 184). 12.14. The transformation from the Cartesian coordinates to elliptical coordinates has the inverse transformation as the Jacobian of the transformation ∂(x, y) = c2 D, D = cosh2 ζ − cos2 ϕ = sinh2 ζ + sin2 ϕ ∂(ζ , ϕ)

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Exploring Classical Mechanics

[12.14

vanishes only when ζ = 0, ϕ = ±π . This transformation is similar to the transition from the 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 c2 cosh2 ζ

+

y2 c2 sinh2 ζ

= 1,

and for a given value ϕ (and for all values of ζ ), the corresponding points lie on the part of the hyperbola (half of one branch) x2 y2 − = 1. c2 cos2 ϕ c2 sin2 ϕ 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 = c cosh ζa , b = c sinh ζa . Taking into account that y˙ = c(ζ˙ coshζ cos ϕ − ϕ˙ sinhζ sin ϕ), x˙ = c(ζ˙ sinhζ sin ϕ + ϕ˙ coshζ cos ϕ), x˙2 + y˙2 = c2 (ζ˙ 2 + ϕ˙ 2 ) D, we find the Lagrangian function of the system in the elliptic coordinates L = 12 mc2 (ζ˙ 2 + ϕ˙ 2 ) D − U(ζ ). Next, we find the generalized momenta pζ = mc2 Dζ˙ , pϕ = mc2 Dϕ˙ and the Hamiltonian function H(pζ , pϕ , ζ , ϕ) =

p2ζ + p2ϕ 2mc2 D

+ U(ζ ).

The particle moves with a constant velocity, which changes direction after collisions with the wall indicated by the ellipse. In the collision, the angle of incidence is equal to the angle of reflection. Of course, the particle velocity and its direction of motion remain constant between the walls. However, the question remains as to which areas may be filled by the trajectory of the particle and which will be inaccessible to it (at given initial

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§12. The Hamilton–Jacobi equation

353

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, we can use the equation for an abbreviated action S(ζ , ϕ, t) = −Et + S0 (ζ , ϕ): 1 2mc2 D



∂S0 ∂ζ

2 +

1 2mc2 D



∂S0 ∂ϕ

2 + U(ζ ) = E.

(1)

Next we consider the region ζ < ζa only where U(ζ ) = 0. In this region the variables in the 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



dS2 (ϕ) − 2mc E cosh ζ = − dϕ 2

2

2 + 2mc2 E cos2 ϕ.

The left-hand side of this equality is independent of ϕ, whereas the right-hand side is independent of ζ ; therefore, each of them is a constant. Let’s define such constant as (−β). Thus, S1 = ±

    2mc2 E cosh2 ζ − β dζ , S2 = ± β − 2mc2 E cos2 ϕ dϕ.

The motion occurs in the region where U = 0, so E > 0; it is also clear from the form S2 that β > 0. Thus, we obtain the complete integral of the Hamilton–Jacobie equation S(ζ , ϕ, t) = −Et + S0 (ζ , ϕ, E, β), which depends on two arbitrary constants E and β. By differentiating S with respect to β and equating the result to the constant C, we get the equation for trajectory in the form f (ζ , ϕ, E, β, C) = 0. The equation ∂S/∂E = t0 allows us to find the functions ζ (t) and ϕ(t). However, we already have a clear idea of the trajectory (consisting of straight segments) and the law of motion. Generalized momenta pζ and pϕ can be found by differentiating S with respect to ζ and ϕ:  pζ = ± 2mc4 E cosh2 ζ − β = mc2 Dζ˙ ,

(2)

 ˙ pϕ = ± β − 2mc4 E cos2 ϕ = mc2 Dϕ.

(3)

Constants E, β, C, t0 are determined by the initial values of coordinates ζ , ϕ and velocities ζ˙ , ϕ. ˙ If β > 2mc4 E, then according to (3) pϕ will never equal zero at any value of ϕ. The sign of ϕ˙ is conserved; the motion always occurs clockwise (or counterclockwise).

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[12.14

Exploring Classical Mechanics

In accordance with (2), the condition 2mc4 E cosh2 ζm = β defines the minimum value ζm : ζm ≤ ζ ≤ ζa . When ζ reaches the value ζa , the sign in (2) changes from positive to negative to correspond to the reflection from the wall. When ζ reaches the value ζm , the velocity ζ˙ becomes zero, and the negative sign in (2) is replaced by the positive one. In this case, the trajectory touches the ellipse which is limited to the area of motion (see orbits in Fig. 185) x2 c2 cosh2 ζm

+

y2 c2 sinh2 ζm

= 1.

y

y

x

x

Figure 185

Figure 186

If β < 2mc4 E, then according to (3) the function ϕ˙ becomes zero for ϕ = ϕm determined by the equation 2mc4 E 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 motion is limited by both branches of the hyperbola (see orbits in Fig.186) y2 x2 − = 1. c2 cos2 ϕm c2 sin2 ϕm The initial values of Cartesian coordinates and velocity components allows us to find the constants E = 12 mv2 , ϕ0 = 0, c cosh ζ0 = x0 ,  vy pζ 2mc4 E cosh2 ζ0 − β ζ˙ tan α = = = = , vx ϕ˙ pϕ β β = 2mc4 E cosh2 ζ0 tan2 α. Each time ζ˙ or ϕ˙ changes sign, the constants C and t0 are updated, but E and β remain unchanged during the whole time of the particle motion.

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§12. The Hamilton–Jacobi equation

Depending on whether the value cosh2 ζ0 tan2 α is larger or smaller the unit, the inaccessible area will be limited to an ellipse or a hyperbola. In the case when it equals a unit, 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 elliptical coordinates (see [1], § 48, problem 2 with α1 = −α2 = α). For a particle coming from infinity along the z-axis, the constant β = −2mEρ 2 + 4mασ , where ρ is the impact parameter. At β < 0 the orbit is qualitatively the same as the orbit of a particle which is scattered in the field of a point dipole (see problem 12.6a). O1 If β > 0, the particle “falls” onto the dipole (i.e., it passes in its motion through the section O2 O1 O2 ) and then goes off again to infinity. Moreover, if pη (η1 ) = 0 when η1 < 0, the particle moves in the region bounded by the hyperbole η = η1 (see Figure 187 Fig. 187). 12.16. In the Hamilton–Jacobi equation, ∂S + 1 ∂t 2m



∂S ∂z

2



+ ∂S ∂r

2



e 1 ∂S − B(z)r + r ∂ϕ 2c

2  = 0,

(1)

we can separate the time and the angle ϕ: S = −Et + pϕ ϕ +  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 integral approximately in the form of an expansion in r:  S(r, z) = S0 (z) + rψ(z) + 12 r 2 σ (z) + . . . .

(3)

Since the radial momentum, pr = ∂S = ψ(z) + rσ (z) + . . . , ∂r

(4)

for a particle flying along the z-axis (with r = 0) vanishes, for the particle beam considered we have ψ(z) = 0. Substituting (3) into (1) and comparing coefficients of the same powers of r, we obtain (cf. [2], § 56, problem 2)

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[12.16

Exploring Classical Mechanics

S0 (z) = pz, p = pσ  (z) + σ 2 +

√ 2mE,

(5)

e2 2 B (z) = 0. 4c2

(6)

Outside the lens (where |z| > a, B(z) = 0), we have from (6) p , z + C1 p , σ (z) = z + C2

σ (z) =

when

z < −a,

(7)

when

z > a.

(8)

pr 2 ∂S = − = B1,2 ∂C1,2 2(z + C1,2 )2

(9)

The equation of the orbit,

is an equation of strait lines intersecting the z-axis in the points −C1,2 1 ; that is, z0 = −C1 and z1 = −C2 . From (6), we get a pσ (a) − pσ (−a) + −a

e2 σ dz + 2 4c

a B2 (z) dz = 0.

2

(10)

−a

Since |z0,1 |  a, it follows from (7) and (8) that σ (±a) = − Let us estimate we have

a

−a σ

2 dz.

p z1,0

.

(11)

According to (6), σ (z) is a monotonic function. Therefore, a σ 2 dz  −a

2ap2 pσ (±a). z21,0

It thus follows from (10) that 1 1 e2 + = 2 2 |z0 | z1 4c p

a B2 (z) dz = −a

1 . f

(12)

The condition |z0,1 |  a is, indeed, satisfied when a cp/(eB). 1 When z is close to C , σ → ∞ so that expansion (3) becomes inapplicable. However, the equations (9) 1,2 for the orbits remain valid also for that region.

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357

12.17. The whole of the calculations of the preceding problem up to (6) is applicable also in this problem. The substitution σ = pf  /f reduces (6) to the form  2 e2 B 2 1 +  2 z2 f  (z) + 2 2 f = 0, 4c p and after that the substitution z = tan ξ ,

− 12 π < ξ < 12 π ,

f (z) =

η(ξ ) cos ξ

gives η (ξ ) + λ2 η(ξ ) = 0, where λ2 = 1 +

e2 B 2 . 4c2  2 p2

Hence, σ = 12 p sin 2ξ + λp cos2 ξ cot(λξ + α), and the equation for the orbit becomes ∂S pr 2 λ cos2 ξ =− = B, ∂α 2 sin2 (λξ + α) or r cos ξ = B sin(λξ + α). When r = 0, λξn + α = π n, and thus α = −λ arctan(z0 ) so that the points where the beam is focused are given by  nπ  zn = tan arctan z0 + . λ Depending on the magnitude of λ, there will be one or several points of focusing.

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[12.18

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 the function α(q, q0 , t, t0 ) is determined by the equation (or set of equations for the case of several degrees of freedom) ∂f (q, α, t) ∂f (q0 , α, t0 ) = . ∂α ∂α

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§13 Adiabatic invariants 13.1. E 2 l = 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 mgϕ 2

(the y-axis is directed vertically upwards; the x-axis is in the plane of oscillations.) Since the length of the thread AB = l changes slowly, we can average the force over a period of  the oscillation, ϕ = ϕ0 cos ωt, ω = g/l, assuming the length of the thread to be constant. We get Fx = 0,

Fy = 14 mgϕ02 .

When the ring is displaced over a distance dy = dl, the energy decreases by Fy dy = 14 mgϕ02 dl. Since E = 12 mglϕ02 , we have

(a) l

t

E dE = − dl. 2l

(b)

υ 2l

Hence, we have E 2 l = const.

2l υ (c)

13.2. After the particle has collided with both walls, ˙ The condition that the its velocity v is changed by 2l. ˙ change is slow means |2l|  v. We choose a time interval t such that

t

υl

t

Figure 188

2l l  t  . ˙ v |l|

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

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[13.3

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Such a t exists because of the slowness condition. During such a time interval there are vt/(2l) pairs of collision with the walls, and the velocity is changed by v = −vl˙

t . l

Integrating, we obtain vl = const or El 2 = 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, we have drawn the function I(t) = v(t)l(t). The quantity vl oscillates around the practically ˙ constant 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: d 2 vl ∼ l˙ . dt 13.3. The particle of low mass moves much faster than the piston. To determine the value of the particle velocity v, we assume that it moves between the slowly and smoothly shifting walls (bottom of the vessel and the piston). Ignoring the effect of gravity on the motion of this very fast particle, we can use the conservation of the adiabatic invariant vX = const = C (see problem 13.2). To determine the law of motion of the piston, we can use the law of energy conservation 1 2

M X˙ + MgX + 12 mv2 = E 2

(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: 1 2

M X˙ + Ueff (X) = E, Ueff (X) = MgX + 2

mC2 , X2

Here Ueff (X) is the effective potential energy of the piston, averaged over the period of   2 1/3 motion of a light particle. The function Ueff (X) has at a minimum at X0 = 2mC . Mg The piston can oscillate near X = X0 with frequency  ω=

 (X ) Ueff 0 = M

 3g . X0

This problem demonstrates the so-called adiabatic approximation. We consider the motion of a system consisting of bodies that differ strongly in mass where light particles move much faster than the heavy ones. In this case, the coordinates of heavy particles can be considered as parameters defining the field in which the light particles are moving. The

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361

change of these parameters is sufficiently slow and smooth so that adiabatic invariants are conserved. Depending on the values of these parameters (i.e., on the coordinates of heavy 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 motion of heavy particles, this energy of the light particles can be added to the potential energy of the heavy particles. Of course, this implies averaging over time the motion of the light particles. Such an approximation is used in studying the motion of atoms inside both a molecule and a crystal. Of course, in these problems the motion of particles, as well as light electrons and massive ions, is described not in terms of classical mechanics, but rather in terms of quantum mechanics. However, the essence of the adiabatic approximation is more helpful to demonstrate the approximation using a simple example from classical mechanics. The same question is dealt with in problem 13.9 where an extremely simplified 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 2gt 2 for − 2h/g < t < 2h/g). A change in g(t) by g leads to a change in the potential energy by mzg, and over a period by mzg, where z = 23 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 mg · 23 h = (mgh) or gh + 13 hg = 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 p dq 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 fixed in the plate). If the velocity changes the result cannot change as it depends, according to the general theory, only on the height of  the support of the plate. It is assumed that the relative change in velocity during a time 2h/g is small and the plate is risen smoothly. 2  ; 13.5. a) E = −A 1 − √ αI 2mA  b) E = −U0 1 −  αI ; 2mU0  α2I 2 0 c) E = αI 2U m + 2m ; 

 2n/(n+2) π I α 1/n  1 + 3 d) E =  1 . n n 2m 2 2 13.6. h ∝ (sin α)2/3 . 13.7. a ∝ (sin α)−1/4 .

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[13.8

 



8ml gl ϕ0 ϕ0 2 ϕ0 , where K(k) and E(k) are the E sin − K sin cos 13.8. I = π 2 2 2 complete elliptic integrals of the first and second kind, respectively (see footnote in the solution of problem 1.7). 13.9. Let the coordinates of the particles m and M, reckoned from the point O, be x and X. The motion of the light particle can be consider approximately as the motion between 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 (see problem 13.2). Eliminating x˙ from the energy conservation law, 1 2

mx˙2 + 12 M x˙ = E,

we find that the effect of the light particle upon the motion of the heavy particle is equivalent to the appearance of a potential energy U(X) =

mC2 . 2X 2

The equation 1 2

M X˙ + U(X) = E 2

leads to the law of motion:  X=

mC2 2E + (t − τ )2 . 2E M

˙ and x˙ (they The constants E, C, and τ can be determined from the initial values of X, X, are independent of x(0)). This method of solving the problem becomes inapplicable when condition (1) is not satisfied. 13.10. We denote the coordinates of the heavy particles by X1,2 and the light particles by 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|(X ˙ 2 − X1 ) = C = const

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(see the preceding problem). Taking this into account, we get the following equation for the relative motion of the heavy particles (X = X2 − X1 ) from the law of energy conservation 1 4

M X˙ + 2

mC2 + fX = E. 2X 2

Expanding Ueff (X) =

mC2 2X 2

+ fX

near the minimum X0 = (mC2 /f )1/3 , we find the frequency of the small oscillations of an “ion” ω2 =

6f 2U  (X0 ) = . M MX0

13.11. We expand the frequency in a series in t in the equations for P and Q, ˙ = ω + ω˙ sin 2Q, Q 2ω

ω˙ P˙ = −P cos 2Q. ω

Restricting ourselves to the first-order corrections, we get for P and Q the equations

Q = (ω0 t + ϕ) + 12 ω˙ 0 t 2 +

ω˙ 0 2ω0

(a)

Q

t

t

sin 2Q(t) dt,

(1)

(b)

p

0

t

 P = P0 1 −

ω˙ 0 ω0

t

 cos 2Q dt ,

(c)

(2)

q

0

t

where ω0 and ω˙ 0 are the values of the frequency and its derivative at time t0 = 0, and we have ω˙ 0 = ε2 ω02 with ε  1. Figure 189 √ The phase Q and the amplitude A = 2P/(mω0 ) of the perturbed motion differ relatively little √ from their unperturbed values Q0 = ω0 t + ϕ and A0 = 2P0 /(mω0 ) even for time intervals which are much longer than the period 2π/ω of the oscillations (see Fig. 189).

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[13.13

Thus, for a time t ∼ 1/(εω0 ), the second term in (1) is of order unity, and the third term of order ε, and thus Q = ω0 t + ϕ + 12 ω˙ 0 t 2 ,

P = p0 .

However, this change in phase leads to the fact that the perturbed motion in terms of the variables p and q,

q(t) = A0 cos ω0 t + ϕ + 12 ω˙ 0 t 2 , will differ appreciably from unperturbed motion, q0 (t) = A0 cos(ω0 t + 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 obtains for the first-order corrections q1 (t) the equation q¨ 1 + ω02 q1 = −2ω0 ω˙ 0 tA0 cos(ω0 t + ϕ), which has a resonant force which increases with time. The solution obtained in such a theory is thus applicable only for small time intervals of the order of a few periods of the oscillations 1 2π  . ω0 εω0 13.13. We transform the Hamiltonian function of the system H(x, p t) =

p2 + 1 mω2 x2 − xF(t) = E(t) 2m 2

to the form 1 2

  F2 F 2 p2 − mω2 x − + = E(t). 2 2m 2mω2 mω

(1)

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From this, it is clear that the orbit in phase space is an ellipse which is displaced along the x-axis over a distance F/(mω2 ) with semi-axes  2E

a=

mω2

+

F



2

m2 ω

, 4

b=

2mE +

F2 ω2

.

Apart from the factor 1/(2π ), the adiabatic invariant is the area of this ellipse I = 12 ab =

E + F 2 /(2mω2 ) . ω

(2)

Here the meaning of E + F 2 /(2mω2 ) is that of the energy of oscillations near the displace equilibrium position (cf. problem 5.16). Substituting the value of E from (1) into (2), we can present the result in the form    F 2 m  x˙ + iω x + I= = 2ω  mω2   2  t   m  1 F(t) iω(t−τ ) iωt  . = e F(τ )dτ + e [x(0) ˙ + iωx(0)] − i  2ω  m mω  0

Here we can use equations (22.9) and (22.10) from [1] for the quantity x˙ + iωx. Integrating by parts, we get x(0) ˙ I(t) = I(0) − 2 ω

t

˙ sin ωt dt− F(t)

0



1 F(0) − x(0) − ω mω2

 t 0

 t 2     1 −iωτ  ˙ ˙ F(t) cos ωt dt + F(τ )e dτ  .  3 2mω   0

If the force changes slowly, I(t) will oscillates near I(0). If F(t) → const as t → ∞, the total change in the adiabatic invariant, I(∞) − I(0), can be very small (see problem 5.18). 13.14. PV 5/3 = const.  2  2 I1 I22 I32 π 13.15. a) E = 2m + + , where a, b, and c are the lengths of the edges of the a2 b2 c2 parallelepiped, and Ik = const. (b) The absolute magnitude of the components of the velocity along each of the edges are conserved.

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13.16. In spherical coordinates, the variables separate. The angular momentum M is strictly conserved. (Moreover, Mz is an adiabatic invariant corresponding to the angle ϕ). The adiabatic invariant for the radial motion is 

R

1 Ir = π

2mE −

M2 dr. r2

(1)

rmin

We can find the function E(R) without evaluating the integral (1). The substitution r = Rx gives 1 Ir = π

1

 2mER2 −

M2 dx = Ir (ER2 , M), x2

(2)

xmin

whence ER2 = const.We get thus for the angle of incidence α the equation sin α =

M rmin =√ = const. R 2mER

2

13.17. a) E ∝ γ 2−n ; b) E ∝ γ −1 . 13.18. Equating the values of the adiabatic invariant before and after the switching on of the field, rmax rmax

M2 M2 − U(r) dr = E + δE + − U − δU dr E− 2mr 2 2mr 2 rmin

rmin

we get 2 δE = δU = T

rmax

 rmin



δU dr

2 M2 m E − 2mr 2 − 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 |  2 , | i

¨ i |  i | ˙ i| |

(i = 1, 2).

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§13. Adiabatic invariants

Outside the region of degeneracy, these conditions reduce to the same conditions for ω1 (t). In the region of degeneracy, |ω12 − ω22 | ∼ α, and the second condition is more restrictive and gives |ω˙ 1 |  α (the region of degeneracy is traversed during a time which is considerable longer than the period of the beats). 13.20. When the coupling αxy is not present, the system splits into two independent oscillators 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

D

A

I1

Iy Ix

I2 ω2

Figure 190

L(t)

l ω1

B

C

Figure 191

We shall illustrate this by the following example. Two pendulum, of which the length of one can be changed slowly, are coupled by a spring with small stiffness (Fig. 191). When the length l and L of the pendulums are appreciable different, the normal oscillations are practically 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 the pendulum CD with a very small amplitude. When L is decreased, the amplitude of the oscillations 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 with the 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 , as for a separate pendulum. If we traverse the degeneracy region relatively fast, ω˙ 1 α, such a transfer of energy between the oscillators will not take place. Moreover, if ω˙ 1  ω12 , ω¨ 1  ω1 ω˙ 1 , Ix and Iy are conserved.

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13.21. From the equations of motion x¨ + ω12 x + 2βxy = 0, y¨ + ω22 y + βx2

(1)

= 0,

(2)

we see easily that the coupling between the oscillators leads to a large energy transfer when 2ω1 ≈ ω2 . Let x = a(t) cos(ω1 t + ϕ),

y = b(t) cos(ω2 t + ψ).

If a b, the term βx2 = 12 βa2 + 12 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 = 2βbx cos(ω2 t + ψ) in (1) leads to a parametric building-up of the oscillation in x. The detail of the study of system (1) and (2) can be found in problem 8.10. The region of resonance interaction is |2ω1 − ω2 | 

βb . ω1

In general, a strong resonance interaction between the oscillators occurs when nω1 = lω2 , where n and l are integers. However, the widths of the regions of frequencies in 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 oscillators for values of ω˙ 1 that are not too small (provided they are sufficiently small that we can use the theory of adiabatic invariants). 13.22. Let the particle moving in the xy-plane at a small angle to the y-axis (|x| ˙  |y) ˙ be reflected from the x-axis and from the curve y0 (x) (Fig. 192). If we assume that we know how the particle moves in the x-direction, we can study the motion in the y-direction by taken x(t) to be a slowly changing parameter. The adiabatic invariant,  py dy = 2|py |y0 (x) = 2π I,

y y0(x)

x

Figure 192

will remain constant, and that equation determines the function py (x).

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To determine x(t), we can use the energy conservation law m2 x˙2 + p2y (x) = 2mE.

E

C

The minimum distance xmin is determined by the condition

D α O

p2y (xmin ) = 2mE.

D

EC

B

ϕ0 A

x

Figure 193

√ Substituting y0 (x) = x tan α and 2π I = 2 2mEl tan α cos ϕ0 , we get xmin = l cos ϕ0 . The solution by means of the reflection method is clear from Fig. 193. This method gives the exact solution applicable at any angle α and ϕ0 but it cannot be generalized to the cases when y0 (x) is not a straight line. 13.23. tanh αxmax = tan ϕ, T =

2π (see the preceding problem). αv cos 2ϕ

13.24. a) The problem of the motion of a particle in a magnetic field reduces for the given 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) v2⊥ ∝ ∝ π a2 B, ω B

where v⊥ is the transverse to the magnetic field component of the particle velocity and a = cmv⊥ /(eB) is the radius of the particle orbit (see [2], § 21). The relation I ∝ π a2 B can be interpreted simply: the radius of the orbit changes in such a way that the magnetic field flux through the area circumscribed by it remains constant. 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 electric field,   1 ∂A 1 ˙ E=− = 0, − xB, 0 , c ∂t c when the magnetic field changes (cf. [2], § 22). The electric field vectors E and the drift velocity vdr are shown in Fig. 194 for different orbit positions of the particle.

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b) The Hamiltonian function is in cylindrical coordinates   p2z p2r 1 eB 2 2 pϕ − r + + . H= 2m 2m 2mr 2 2c The quantity pz and pϕ are integrals of motion. The adiabatic invariant for the radial motion is rmax

π Ir =

  1 eB 2 2 2mE⊥ − 2 pϕ − r dr, 2c r

rmin

√ which after the substitution r = ζ / B becomes ζ max

  2mE⊥ 1 e 2 E⊥ 1 − 2 pϕ − ζ 2 dζ = π Ir pϕ , . B 2c B ζ

ζmin

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 the origin is r0 =

ζmax + ζmin rmax + rmin 1 = ∝√ . √ 2 2 B B

When B increases, the centre of the orbit approaches the origin (Fig. 195). y

E vdr

vdr

E E x

Figure 194

E

vdr

vdr

Figure 195

1 It is interesting that I does not actually depend on p when p > 0. Indeed, ∂Ir is the change of the r ϕ ϕ ∂pϕ

angle ϕ over the time of the one radial oscillation; besides, when pϕ > 0, the origin of coordinates lies outside the orbit (see Fig. 109b), and therefore ϕ = 0.

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371

When B changes, there occurs an electric field (E)ϕ = −

r ˙ B, 2c

(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 strongly on the shape of that region and the conditions at its boundaries (see [2], § 21). For example, the field considered under sub 13.23a could occur near a conducting plane in which there was a current, while the field under sub 13.23b could be produced in a solenoid.1 The strong dependence of the nature of the particle motion on the weak field E even in the case of extremely small B can be explained by the presence of degeneracy (when B = 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], § 21 and [10], § 25). 13.25. The adiabatic invariants are Iz =

1 Ir = π

Ez , ω

Iϕ = pϕ ,

ζ max

 dζ = Ir (pϕ , C), Cζ 2 − p2ϕ − m2 ζ 4 ζ

(1)

ζmin

where 2mE⊥ + eBpϕ /c C= 

2 , eB 2 ω + 2mc

Ez = 12 mv2z ,

E⊥ = 12 mv2⊥ ,

and v⊥ is the transverse to the magnetic field component of the particle velocity. Therefore    eB eB 2 pϕ ∝ ω2 + Ez ∝ ω, E⊥ + . (2) 2mc 2mc 1 The change in the electric field E connected with the change in the choice of the form of A would not ˙ occur if we had simultaneously changed the scalar potential by Bxy/(2c) (gauge transformation).

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[13.26

The given vector potential defines a magnetic field which is symmetric with respect to the 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 two problems differ by δL =

d e e ˙ Bxy − Bxy; dt 2c 2c

(4)

that is, the difference is very small if we drop in (4) the inessential total derivative with respect to time. In the preceding problem, where the motion was degenerate, just this extra term led to 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 can neglect the extra term δL (cf. problem 13.20). The relations (2) are thus valid also for a different choice of A. When one passes through the degeneracy region (B = 0), the equations (2) remains valid only when we choose the axially symmetric field given in the problem. For instance, the behaviour of the oscillator in the field (3) when B passes through zero requires additional study. 13.26. a) Using a canonical transformation, one can reduce the Hamiltonian function to a sum of the Hamiltonian functions for two independent oscillators (for X and Y ; see problem 11.10). For each of the oscillators the ratio of the energy to the frequency is an adiabatic invariant. We remind ourselves that the oscillations of each of them correspond to motion along an ellipse (see problem 6.37). In terms of the amplitude ak of the oscillations in the x-direction, for instance, the adiabatic invariants are equal to Ik =

ma2k 4k − ω12 ω22 2 k 2k − ω22

(k = 1, 2).

When the parameters of the system are changing, we must also add to the new Hamiltonian function the partial derivative with respect to the time of the generating function which is equal to −λ˙ [mω2 XY + PX PY /(mω2 )] (see problem 11.18). This correction term is small (λ˙  k ) and can be neglected provided the eigen-frequencies are not the same (cf. problem 13.20). One must consider separately the degenerate case when ω1 = ω2 and a magnetic field which can vanish. When ω1 = ω2 , the different choice of the adiabatic invariants is possible (see preceding 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-axes in the x- and y-direction and equal to

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ω2 b√ ω1 ωB

 and b

373

ω1 ωB

with a frequency ω1 ω2 /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 of a magnetic field which is slowly varying as the particle moves along the z-axis. The adiabatic invariant  I ⊥ = E⊥

eB(z) mc

−1

is then conserved (see problem 13.24). From the energy conservation law, we have 1 2

mz˙2 + I⊥

eB(z) = E. mc

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) 2π a T=  , v λ sin2 α − cos2 α where α is the angle between the velocity v of a particle and the z-axis. Particles, for which tan2 α < 1/λ are not contained in the trap. The condition for the applicability of the theory of adiabatic invariants consists in the requirement that the change in the magnetic field 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 = 2π a . v sin α 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=

p2ϕ p2 eBpϕ e2 B2 r 2 sin2 θ p2r + θ2+ + − + U(r) 2m 2mr 2cm 8mc2 2mr 2 sin2 θ

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the term which is quadratic in B, we can separate the variables in the Hamilton–Jacobi equation. The adiabatic invariants have the form

Iϕ = pϕ ,

1 Iθ = π

θ2

 β−

θ1

p2ϕ sin2 θ

dθ = Iθ (pϕ , β),

  

r2   eBpϕ eBpϕ β 1 − U(r) − 2 dr = Ir E + ,β . Ir = 2m E + π 2mc 2mc r r1

When B is slowly changed, the quantities pϕ , β = p2θ +

p2ϕ 2

sin θ

, and E +

eBpϕ 2mc

thus remain constant. 13.31.. p2y p2x + 12 mω2 x2 = E1 , + 2mω2 y2 = E2 , 2m 2m (m2 ω2 x2 − p2x )y + xpx py = A, (m2 ω2 x2 − p2x )py − 4mωxpx y = m{E2 , A}. p2 1 2 13.32. a) w = arctan mωx , I = p 2mω + 2 mωx . These variables are convenient, for instance, to develop perturbation theory (see problem 13.10). b) Let initially the particle moves to the right from the point x = 0, and we shall choose S such that S = 0 for x = 0. In that case,

x S=

  3/2 I 2/3 |p| dx = π I − π a − Fx , a

0

where 1 I= π

xm |p| dx = aE 3/2 , 0

a=

√ 2 2m , 3π F

xm =

E , F

|p| =



2m(E − xF).

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§13. Adiabatic invariants

If the motion is to the left, ⎛x ⎞   3/2

m x 2/3 I S = ⎝ − ⎠ |p| dx = π I + π a − Fx a xm

0

and so on. For the nth oscillation, 3/2   I 2/3 − Fx S = (2n − 1)π I ∓ π a a (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 in the following way:  2/3   I π 2 − [(2n − 1)π − w]2 , a  1/3 I 3 [(2n − 1)π − w], p = aF 2 a x=

1 π 2F

where x is a periodic function of w, while w in multivalued function of x (Fig. 197). S x xm

x

xm

2π

Figure 196

4π

Figure 197

13.33. From the equation

a  P= 2m(E − U)dx 0

we find E=

mV 2 a2 P2 1 + V + . 2ma2 2 8P 2

6π

w

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The abbreviated action is ⎧√ ⎪ 2mE(x − na) + nP, ⎪ ⎪ x ⎪

⎨  S0 = p dx = 2m(E − V ) x − (n + 12 ) a + ⎪ ⎪ ,

⎪ √ ⎪ 0 ⎩ + 2mE 1 a + nP 2

[13.33



when na < x < n + 12 a,

when n + 12 a < x < (n + 1)a.

2 Eliminating E and introducing A = ma 2V , we get the generating function for the 2P canonical transformation under consideration

⎧ ⎪ P ⎨ (1 + A) (x − na) + na, S0 (x, P) = · a ⎪ ⎩ (1 − A) (x − na) + Aa + na,

when na < x < n + 12 a,

when n + 12 a < x < (n + 1)a.

From the equations p=

∂S0 , ∂x

Q=

∂S0 , ∂P

we get  P 1 + A, when n < Q < n + Q0 , p(P, Q) = · a 1 − A, when n + Q0 < Q < n + 1, ⎧ Q − nA ⎪ ⎪ , when n < Q < n + Q0 , ⎨ 1−A x(P, Q) = a · Q + (n + 1)A ⎪ ⎪ ⎩ , when n + Q0 < Q < n + 1, 1+A where Q0 = 12 (1 − A). The variables P and Q are analogous to action and angle variables, ˙ is the average particle velocity. and the quantity aQ

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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, ButterworthHeinenann, 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-linear Oscillations, 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, Clarendon Press, Oxford. [13] N. N. Bogolyubov, D. V. Shirkov (1980) Introduction to the Theory of Quantized Fields, 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.

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378

Bibliography

[22] A. I. Bazh, Ya. B. Zeldovich, A. M. Perelomov (1971) Scattering, Reactions and Decays 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.

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Index

The numbers refer to the problems; for instance, 4.12 is problem 12 of § 4. Acoustic branch 7.5, 7.7 Adiabatic approximation 13.3, 13.9, 13.10 Angle and action variables 11.1, 13.32, 13.33 Anomaly, eccentric 2.44 Barrier, spherical 3.9 Beam focusing 2.15, 8.13, 12.16, 12.17 Beats 5.11, 6.9, 6.34, 8.8 Christoffel symbol 4.22 Circuits 4.24–4.26, 5.10, 5.19, 6.7 Conservation law angular momentum 4.14, 10.1 energy 2.1, 4.14 momentum 2.1, 4.14, 10.1 Constraints 4.28–4.31 Continuous systems 4.32, 7.11, 7.12 Curvilinear coordinates 4.21, 4.22 Degenerate frequencies 6.19, 6.21, 6.22, 6.29, 6.30 Delay time 1.12 Dipol, motion in 3.12, 10.26, 12.2–12.10 Dissipative function 5.10, 11.25 Distribution of decay particles 3.16, 3.17 Displacement of the Earth’s perihelion 2.25 Displacement from the vertical 9.23 Drift 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 nuclear paramagnetic resonance 10.23 Electron motion in periodic lattices 10.11, 10.13 Energy flux 6.10, 7.4, 7.6 Euler angles 9.14, 9.16, 9.21, 9.22, 10.2, 13.15 Euler equations 9.10, 11.7

Impact phase 5.12, 5.15 Impurity atom in a crystal 7.8 Inaccessible region 2.16, 12.13, 12.14 Inertial tensor 9.2–9.4, 9.6 Infinitesimal transformations 10.1, 10.18, 11.17–11.19, 11.24 Integrals of Manley–Rowe 11.27, 11.30

Fermi resonance 8.10 Frequency division 8.10 doubling 8.10 tripling 8.6 Forbidden band 7.5, 7.7 Friction 5.11, 5.13, 5.20, 11.25

Kepler problem 2.3, 2.44, 10.26

Gauge transformation 11.14, 13.24 Generating function for the canonical variables 11.7–11.11 for the identity transformation 11.5 for the gauge transformation 11.14 for the similarity transformation 11.13 Gravity, motion under 4.28, 4.29, 5.2, 5.8, 9.22 Gyrocompass 9.14 Hidden symmetry of the hydrogen atom 10.26 Holes 10.12

Jacobi coordinates 2.24

Lagrangian multipliers 4.28, 6.24 Lagrange points 9.29 Laplace vector 2.19, 2.39, 2.41–2.45, 3.6 Larmor theorem 2.37, 2.41, 6.37, 10.22 Lifetime of a satellite 2.42 Light, propagation of 10.6 Liouville theorem 11.25 Lissajous figures 6.5 Localised 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

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380 Molecules diatomic 5.6 four-atomic 6.53 triatomic 6.51, 9.26 Noether’s theorem 4.12 Normal 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, 7.7 Orthogonality of the normal oscillations 6.23, 6.25, 6.27–6.30 Oscillator anharmonic 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, 6.17 Pendulum 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 13.20 Perturbation 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

Index Phase-space distributions 11.25 Phase trajectories 1.5 Potential Coulomb 2.6, 2.21, 2.28, 2.29, 2.37, 2.44, 12.15 Morse 1.1 perturbed 1.8–1.10, 2.20 Yukawa 2.7, 2.21 Principal axes 9.2, 9.3, 9.12, 9.13, 9.19 Precession of orbit 2.1 Probability in the coordinate space 1.5 Probability in the momentum space 1.5 Quadrupole moment tensor 9.7, 9.16 Quasi-momentum 10.12 Quasi-particles 10.12 Radial motion 2.1 Radial oscillations 2.1 Radiance 3.10 Rayleigh’s theorem 6.6 Reciprocity theorem 6.26 Resonance 5.11, 5.19, 6.14, 6.15, 8.6, 13.21 Resonance, parametric 8.8, 8.9 Retreat of the lunar nodes 2.25 Rotating 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 10.11–10.13 Spatial dispersion 7.12 Spiral orbit 2.1, 2.4 Strong focus of particle beams in accelerators 8.13 Symmetry properties of the normal oscillations 6.45–6.54 Total number of quanta 11.30 Top asymmetric 9.19 symmetric 9.6, 9.15, 10.2 Transformations Galilean 4.17, 11.19 gauge 11.14, 13.24 infinitesimal 11.23, 11.24 similarity 4.15, 4.16, 4.33 Transport cross-section 3.23 Turning point 1.1, 1.3, 1.8 Two-body problem 2.11, 2.12, 2.29, 2.30, 2.38, 10.10 van der Waals forces 8.12 Virial theorem 4.34 Vortex ring in liquid helium 10.5 Wave vector 7.1, 7.4 Waves standing 7.1 travelling 7.1, 7.5, 7.10 Well spherical 2.2 square 13.2