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rta~~~@[F Mrn©OO~~rl©~Translated from the Russianby YURI ATANOV

fDMir Publishers Moscow

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CBS PUBLISHERS & DISTRIBUTORS4596/1A1JJ PClr'J~ Gan1 New Delhi - 110 002 (India)--, , -

First published t9RORevised from tho 1n7R Russian e,lition

© H3AaTOJlhCTOO «BhlCIlIRll mKOJlU, 1!)7R© English translat.ion. Mir Publishers, HI;;!

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CONTENTS

PrefaceNotatiouIntroduction

PART ONECLASSICAL MECHANICS

.,,\l

tt

First Indian eJitic'n : 1QQ4

Reprint : 2001Reprint:2002

Thl" edilion has t-..:..:n puhli,h<:.! ::~ !ndia "~. :IlT:,ngcli1C!l1 ..nthMir Publishers. ~h,,;..:,lW.

© English tran'iali,'n. Mir Put-li;.hcr,. 1980

For sal..: in India only.

All ri~:ht~ reserved. No part nl this publil'ation may ~' reproduced.sh'reJ In a retricval svstem. ,'r tran:.milled in any tom' -)1" by any~W·IO~. without the' p;i,'r 1'"'fmissi,1O ,ll th.· publishl:!

Published by S.K. Jain for CBS Publishers &: Distributor,.459611 A. II Darya Ganj, New Delhi - 110 002 (India)

Printed <1l

;";:tzia Printers. Delhi· 110006

ISBN: 81-239-0304-9

Chapter 1. Essentials of Kinematics . . . . . . 14§ 1. 1. Kinematics of a Point . . . . . . . . 14§ 1.2. Kinematics of a Solid . . . . . . . . . . . 21§ 1.3. Transformation of Velocity and Acceleration on

Transition to Another Reference Frame 30Problems .to Chapter 1 . . . . . . . . . 34

~:

Chapter 2. The Basic Equation of Dynamics . . 41

§ 2.1. Inertial Reference Frames . . . . . . . . . . .• 41§ 2.2. The Fundamelltal Laws of Newtonian Dynamics 44§ 2.3. Laws of Forces . . . . . . . . . . . . . 50§ 2.4. The Fundamental Equation of Dynamic8 52§ 2.5. Non-inertial Reference Frames. Inertial Fore.cs 56Problew.s to Chapter 2 . . '" . 61

Chapter 3. Energy Conservation Law 72

§ 3.1. On Conservation Laws . . 72§ 3.2. Work and Power . . . . . 74§ 3.3. Potential Field of Forces . . . .. ... 79§ 3.4. Mechanical Energy of a Particle in a Field 90§ 3.5. The Energy Conservation Law for a System 94Problems to Chapter 3 . . . . . . . . . . . . . 104

Chapter 4. The Law of Conservation of Momentum 114§ 4.1. Momentum. The Law of Its Conservation 114§ 4.2. Centre of Inertia. The C Frame . . . . 120§ 4.3. Collision of Two Particles . . . . . . 126§ 4.4. Motion of a Body wit.h Variable Mass 136Problems to Chapter 4 . . . . . . . . . . • 13!)

Chapter 5. The Law of Conservat.ion or Anltular Momentum 147

§ 5.1. Angular Momentum of a Particle. Moment of Force 147§ 5.2. The Law of Conservation or Angular Momentum 154§ 5.3. Internal Angular Momentum HiO§ 5.4. Dynamics of a Solid . . . . . . . . . . . . .. 164Problems to Chapter 5 • . . . . . . . . . . . . " t 78

6

PART TWORELATIVISTIC MECHANICS

Chapter 6. Kinematics in the Special Theory of Relativity 189

§ 6.1. Introduction . . . . . . . . . . . . . . . " 189§ 6.2. Einstein's Postulates . . 194§ 6.3. Dilation of Time and' Co~t;acii~n'o'f 'L~ngtl~ 198§ 6.4. Lorentz Transformation . . . . . . . . . . . " 207§ 6.5. Consequences of Lorentz Transformation . . " 210§ 6.6. Geometric Description of Lorentz Transformation 217Problems to Chapter 6 . . . . . . .. 221

Chapter 7. Relativistic Dynamics . 227

§ 7.1. Relativistic Momentum . . . . . . . . . . . .. 227§ 7.2. Fundamental Equation of Relativistic Dynamics 231§ 7.3. Mass-Energy Relation . . . . . . . . . . . . .. 233§ 7.4. Relation Between Energy and Momentum of a Par-

ticle . . . . . . . . . . . . . . . . . . . . .. 238§ 7.5. System of Relativistic Particles 242Problems to Chapter 7 . . 249

Appendices . . . . . . . . . . . . . . 257

1. Motion of a Point in Polar Coordinates 2572. On Keplerian Motion . . . . . . . . . 2583. Demonstration of Steiner's Theorem . . . 2604. Greek Alphabet . . . . . . . . . . . . . 2625. Some Formulae of Algebra and Trigonometry 2626. Table of Derivatives and Integrals '. . 2637. Some Facts About Vectors . . . . . . . . . . .. 2648. Units of Mechanical Quantities in the SI and CGS

Systems . . . . . . . . . . . . • . . 2669. Decimal Prefixes for the Names of Units 266

to. Some Extrasystem Units 26711. Astronomic Quantities 26712. Fundamental Constants 267

Index -.. . . . . . • . . • 269

PREFACE

The objective of the book is to draw the readers' atten­tion to the basic laws of mechanics, that is, to the laws ofmotion and to the laws of conservation of energy, momen­tum and angular momentum, as well as to show how theselaws are to he applied in solving various specific prohlems.At the same time the author has excluded all things ofminor i.mportance in order to concentrate on the questionswhich'lire the hardest to comprehend.

The hook consists of two parts: (1) classical mechanicsand (2)relativist1c mechanics. In the first part the laws ofmechanics are treated in the Newtonian approximation, Le.when mo~ion velocities are much less than the velocity oflight, while in the second part of the book velocities com­parable to that of ligh~ are considered.

Each chapter opens with a theoretical essay followed bya number of the most instructive and interesting examplesand problems, with' solutions provided. There are about80 problems altogether; being closely associated with theintroductory text, they develop and supplement it andtherefore their examination is of equal importance.

A few corrections and refinements have been made in thepresent edition to stress the physical essence of the prob­lems studied. This holds true primarily for Newton's secondlaw and the conservation laws. Some new examples andproblems have been provided.

The book is intended for first-year students of physicsbut can' also be useful to senior students and lecturers.

I. E. Irodov

'+

I

NOTATION

Vectors are designated by roman bold-fac.e type (e.g. r, F);the same italicized letter~ (r. F) designate the norm of avector.

Mean values are indicated by crotchets ), e.g. (v),(N).

The symbols 11, d. 6 (when put in front of a quantity)signify:11, a finite increment of a quantity, i.e. a difference betweenitsflllal and initial values. e.g. I1r=rz- r l • I1U= U 2 - U1;

d. a differential (an inftnitesimal increment), e.g. dr. dU;6, an elp,mental'y value of a quantity, e.g. 6A is an elemen­tary work.

If!'lUnit vectors:i, j, k are unit vecJors of tbe Cartesian coordinates x. y. z;ep , e.,. e; are unit, vectors of the cylindrical coordinates

p. <p, z;n, 't are unit vectors of a normal and a tangent to a path.

Reference frames are denoted by the italic letters K, K'and C.

The C frame is a reference frame fixed to the centre ofinertia and translating relative to inertial frames. All quan-tities in the C frame are marked with a tilde, e.g. p, E.

A, work,c, velocity of light in vacuo.E, total mechanical energy. the total energy,E, electric field strength.e. elementary electric eharge.F. force. .H. fteld strength.~. free fall aceeleration.f. moment of inertia.

L. angular momentum with respect to a point,L Zl angnlar momentum with respect to an axis,

10 NoltJlton

•

1, arc coordinate, the arm of a vectorM, moment of a force w~th respect to ~ point,

M z, moment of a force wIth respect to an axism, mass, relativistic mass, mo re~t mass,N, power,p, momentum,g, electric charge,r, radius vector,s, path, interval,t, time,

T, kinetic energy,U, potential energy,v, velocity of a point or a particle,

w, acceleration of a point or a particle,~, angular acceleration,~, velocity expressed in units of the velocity of light,y, gravitatjonal constant, the Lorentz 'factor,e, energy of a photon,

x, elastic (quasi-elastic) force constant,~, reduced mass,p, curvature radius. radius vector of the shortest distance

to an axis, density,lp, azimuth angle, potential,ro, angular velocity,Q, solid angle.

INTRODUCTION

Mechanics is a branch of physics treating the simplestform of motion of matter, mechanical motion, that is, themotion of bodies in space and time. The occurrence of mechan­ical plM!nomena in space and time can be seen in any mechan­ical law involving, explicitly or implicitly, space-timerelations, Le. di&f.ances and time intervals.

The position 01 a body in space. can be determined onlywith respect to other bodies. The same is true for the motionof a body, Le. for the change in its position over time. Thebody (or the system of mutually immobile bodies) servingto define the position of a particular body is identified asthe reference body.

For practical purposes', a certain coordinate system, e.g.the Cartesian system, is fixed to the reference body when­ever motion is describett. The coordinates of a body permitits' position in space to be established. Next, motion occursnot only in space but also in time, and therefore the descrip­tion of the motion presupposes time measurements as well.This is done by means of a clock of one or another type.

A referen.ce body to which coordinates are fixed and mu­tually synchronized clocks form the so-called reference frame.The notion of a reference frame is fundamental in physics.A space-time description of motion based on distances andtime intervals is possible only when a definite referenceframe is chosen.

Space and time by themselves' are also physical objects,just as any others, even though immeasurably more impor-

tanto The properties of space and time can be investigatedby observing bodies moving in them. By studying the char­acter of the motion of bodies we determine the propertiesof space and time. .

Experienc.e shows that as long as the velocities of bodiesare small in comparison with the velocity of light, linearscales and time intervals remain invariable on transitionfrom one reference frame to another, Le. they do not dependon the choice of a reference frame. This fact finds expres­sion in the Newtonian concepts of absolute space and time.Mechanics treating the motion of bodies in such cases isreferred to as classical.

When we pass to velocities comparable to that of light, itbecomes obvious that the character of the motion of bodieschanges radically. Linear scales and time intervals becomedependent 011 the choice of a reference frame and are differ­ent in different reference frames. Mechanics based on theseconcepts is referred to as relativistic. Naturally, relativisticmechanics is more general and becomes classical in the caseof small velocities.

The actual motion characteristics of bodies are so complexthat to investigate them we have to neglect all inSignificantfactors, otherwise the problem would get so complicated asto render it praetically insoluble. For this purpose notions(or abstractions) are employed whose application dependson the specific nature of the problem in question and on theaccuracy of the result that we expect to get. A particularlyimportant role is played by the notions of a mass point andof a perfectly rigid body.

A mass point, or, briefly. a particle, is a body whose dimen­sions can be neglected under the eonditions of a given prob­lem. It is clenr that the same body can be treated as a masspoint in some eases and as an extended object in others.

A perfectly rigid body, or, briefly, a solid, is a system ofmass points separated by distances which do not vary dur­ing its motion. A real body can be treated as a perfectlyrigid one provided its, deformations are negligible underthe conditions of the problem considered.

Mechanics tackles two fundamental problems;f. The investigation of various motions and the general­

ization of the results obtained in thp form of laws of mo-

tion, Le. laws that. can be e~ployed in predicting thecharacter of motion lD each speCific case. .

2 The s~arch for general properties that are typical ofany' system regardless of the specific interactions betweenthe bodies of the system. .

The solution of the first problem ended up With .the ~o­

called dynamic laws established by Newton and ElI~stelD,

while the solution of the second problem resulted In thediscovery of the laws of conservation for such fundamental

uantities as energy, momentum and angula~ momentum.q The dynamic laws and the laws of conservatIOn of ~nergy,momentum and angular momentum represent the baSIC. lawsof mechanics. The investigation of these laws constitutesthe subject matter of this book.

12I,.'roduction

•

Introduction 13

Fig. 1

- r l · The ratio I1r/l1tl is called the mean velocity vector (v)during the time interval M. The direction of- the vector(v) coincides with that of I1r. Now let us define the veloc-'ity vector v of the point at a given moment of time as thelimit of the ratio I1r/l1t as I1t __ 0, I.e.

v=lim A!..=..!k. (1.1)l\t..o L1t dt

This means that the velocity vector v of the point at a givenmoment of time is equal to the derivative of the radiusvector r with respect to time, and its directlon, like that of

PART ONE

CLASSICAL MECHANICS

CHAPTER 1

ESSENTIALS OF KINEMATICS

Kinematics is the subdivision of mechanics treating waysof describing motion regardless of the causes inducing it.Three probl~ms will be considered in this chapter: kinematicsof a point, kinematics of a solid, and the transformation ofvelocity and acceleration on transition from one referenceframe to another.

i

Essenttals 0/ Kinematics is

<tf>

§ 1.1. Kinematics of a Point

There are three ways to describe the motion of a point:the first employs vectors, the second coordinates, and t.hethird is referred to as natural. Let us examine them inorder.

The vector method. With this method the location of agiven point A is defined by a radius vector r drawn froma certain stationary point 0 of a chosen reference frame tothat point A. The motion of the point A makes its radiusvector vary in the general case both in magnitude and indirection, I.e. the radius vector r depends on time t. The locusof the end points of the radius vector r is referred to as thepath of the point A.

Let us introduce the notion of the velocity of a point.Suppose the point A travels from point 1 to point 2 in thetime interval I1t (Fig. 1). It is seen from the figqre thatthe displacement vector I1r of the point A represents theincrement of the radius vector r in the time f!t: I1r = r 2 _

the vector dr, along the tan~ent to the path at a given pointcoincides with the direction of motion of the point A. Themodulus of the vector v is equal to*

v= I v I = I dr/dt I.The moHon of a point is also characterized by acceleration.

The acceleration vector w defines the rate at which the ve-H

locity vector of a point varies with time:

w = dV/dt, (1.2)

I.e. it is equal to the derivative of the velocity vector withrespect to time. The direction of the vector w coincides withthe direction of the vector dv which is the increment of thevector v during the time interva~ dt. The modulus of the

• Note that in the general case I dr I =1= dr, where r is the modulusof the radius vector r, and 'v =1= dr/dt. For example, when r changesonly in direction, that is the point moves in a circle. then r = const,'dr = O. but I dr I =1= O. .

l·'.····....~..·i...••' ..•.\;"

t6 Essentials of Kinematics f7

--9tl2

(

• The motion of a point in polar coordinates is' consideredin Appendix 1.

2-0539

t

M = ) v (t) dt::>'" vot +wt2/2.o

To find the radius vector r (t), the location ro of t4e pointat the initial moment of time must be known. Then r == ro + ilr, or

r = ro +vot +wt2J2.Let us consider, for example, the motion of a stone thrown

with the initial velocity Vo at an angle to the horizontal.Assuming the stone to move with the constant accelerationw = g, its location relativeto the pQint fo = 0 from whichthe stone was thrown is de­fined by the radius. vector

r = vot +gt¥2,

Integrating this relation with respect to t9e' function v (t),we obtain the increment of the radius vector during theinterval from t ~ 0 to t:

I.e. in this case r representsthe sum of two vectors as O~~~~77i7,77i7,77i7,'l'mWJ').;wtshown in Fig. 2.

Thus, the tcomplete solu- Fig. 2tion of the problem of a movingpoint, that is, the determination of its vcleoity v and itslocation r as functions of time, requires knowing not onlythe dependence w (t), but also the initia.l conditions, I.e. thevelocity Vo and the location fo of the point at the initialmoment of time.

The method of coordinates. In this method a certain coor­dinate system (Cartesian, oblique-angled or curvilinear)is fixed to a chosen reference body. The choice of a coordinatesystem is stipulated by various considerations: the char­acter or) the symmetry of the problem, the formulation ofthe problem, the quest for a simpler solution. Wf3 shallconfine ourselves hare· to Cartesian coordinates i, y, z.

...

t

L\v- Jwde=we.o

However, the quantity L\v is not the required velocity v.To find v, we must know th~ velocity Vo atthe initial mo­ment of time. Then v = "10 + L\v, or

v " >(I!- wt.

The radius vector r (t) of the point is found. in a silBilar.manner. According to Eq. (1.t) the elementary '.nerement oft.he radius vector during the time interval dl IS dr = v dt.

vector w is defined in much the same way as that Qf thevector v.

Example. The radius vector of a point depends on time t asr = at + btl /2,

where a and b are constaQ&, vcc~. Let us lind the velocity v of thepoint and its accelera\io1t w:

v = dr/dt = a + bt, W = dv/dt = b =;= const.

The modulus of the velocity vector

IJ = Y Vi = ya..,1....+~2a-:b~t-:+....,b:"::l~tl.

Thus, knowing the funcLion r (t), one can find the velocityv of a point and its acceleration w at any moment of time.

Here t.he reverse problem arises: can we find v (t) andr (t) if the time dependence of the acceleration w (t) isknown?

It. turns out that the dependence w (t) is not sufficient toget a single-valued solution of this problem; one needs alsoto know the so-called initial condilions, namely, the veloc­ity Vo of the point and its radius vector r o at 8 certaininitial moment t = O. To make sure, let us examine thesimple case when the acceleration of the point remains con­stant in the course of time.

First, let us determine the velocity v (I) of the point. Inaccordance with Eq. (1.2) the elementary velocity incre­ment during the time interval dl is equal to dv = w dt.Integrating this relation with respect to time betweent = 0 and I, we obtain the velocity vector increment dur­ing this interval:

•

18 Cltultcal Mechanics Essentials of Kinematic, i9

where v~ = dUdt is the projection of the vector v on thedirection of the vector 't, with V~ being an algebraic quan-tity. Besides, it is obvious that .

Iv~1 = Ivl =v.Acceleration ofa point.. Let us differentiate Eq. (1.5)

with tespect to time:dv dv~ d"f

W = dt = dt 't +v~ "'i1t (1.6)

(1.5)

Fig. 3

covered on time, the dependence of the velocity on theposition of a point etc.

The reverse problem, that is, the determination of thevelocity and the law of motion of a point from a given accel­eration, is solved, as in the vector method, by integration(in this case, integr~tion of acceleration projections withrespect to time); this problem also has a single-valued solu­tion provided that in addition to the acceleration the initialconditions are also available, I.e. velocity projections andthe coordinates of a point at the initial moment.

The "natural" method. This method is employed whenthe path of a point is known in advance. The location ofa point A is defined by the arc coordi­nate l, that is, the distance from thechosen origin 0 measured along thepath (Fig. 3). In so doing, the pos­itive~ direQ:tion of the coordinatel is adopted at will (e.g. as shownby an arrow in the pgure).

The motion of a point is deter­mined provided we know its path,the origin 0, the positive direction of the arc coordinate landthe law of motion of the point, I.e. the function 1 (t).

Velocity of a point. Let us introduce the unit vector'tfixed to the moving point A and oriented along a tangent tothe path in the direction of growing values of the arc coordi­nate l (Fig. 3). It is obvious that 't is a variable vector sinceit depends on l. The velocity vector v of the point A isoriented along a tangent to the path and therefore can berepresented as follows:

I-v-=~v~'t-,I.. .

Let us write the projections of the radius vector r (t)on the axes x, y, z to characterize the position of the pointin question relative to the origin 0 at the moment t:

x = x (t); y = y (t); z = z (t)...Knowing the dependence of these coordinates on time, thatis, the law of motion of. the point, we can find the positionof the point at any moment of time, as well as its velocityand acceleration. Indeed, from Eqs. (1.1) and (1.2) we caneasily obtain the formulae defining the projections of thevelocity vector and the acceleration vector on the x axis:

v,; = dx/dt, (1.3)

where dx is the projection of the displacement vector dr onthe x axis;

w= = dv=/dt = d2x/dt2, (1.4)

where dv= is the projection of the velocity increment vectordv on the x axis. Similar relations are obtained for y and zprojections of the respective vectors. It is seen from theseformulae that the velocity and acceleration vector projec­tionsare equal respectively to the first and second time de­rivatives of the coordinates.

Thus, the functions x (t), y (t), z (t), in essence, complete­ly define . the motion of a point. Knowing them, one canfind not only the position of a point, but also the projectionsof its velocity and acceleration, and, consequently, themagnitude and direction of vectors v and w at any momentof time. For example, the modulus of the velocity vector

v= vv:+v:+v:;

the direction oithe vector v is defined by the directional co­sines as follows:

cos a = vJv; cos ~ = v,/v; cos 'V = viv,

where a, ~, 'V are the angles formed by the vector v with theaxes x, y, z respectively. Similar formulae define the mag­nitude and direction of the acceleration vector.

Besides, some more questions can be solved: one can de­termine the path of a point, the dependence of the distance

2*

o

(

21

urn

Essentials 0/ Kinematics

Here the first term is called the tangential acceleration w'\"and the second one, the normal (centripetal) acceleration Wn :

W'\"= (dv'tldt)t; wn=(v2/p)n. (1.10)

Thus, the total acceleration W of a point can be represent­ed as the sum of the tangen­tial w'\" and normal W n accele­rations.

The magnitude of the totalacceleration of a point is

w= -Vw~+w~=" ' .....

" f ....., I"J

Example. Point A travels along aan arc of a circle of radius p .(Fig. 5). Its velocity depends on FIg. 5the arc cO<J1dinate l as v = a VIwhere a is1l constant. Let us calculate the angle ex between the vec­tors of the total acceleration and of the velocity of the point as a func­tion of the x coordinaw l.

It is seen from Fig. 5 that the ang.le ex can be found by means ofthe formula tan ex = wnlw'\". Let us find wn and W't:

wn=~=~; W't= dv'\" = dv'\" ~=_a_a Vl=~.p p dt dl dt 2 VI 2

Whence tan ex = 2lfp.

§ 1.2. Kinematics of a Solid

Being important by itself, the theory of motion of a solidis also essential in another respect. I t is well known thata reference frame used for describing various kinds of. mo­tion in space and time can be fixed to a solid. Therefore,the study of motion of solids is actually equivalent to thestudy of motion of corresponding reference frames. Theresults to be obtained in this section will be repeatedly usedhereafter.

Five kinds of motion of a solid are identified: (1) trans­lation, (2) rotation about a stationary axis, (3) plane motion,(4) motion about a stationary point, and (5) free motion.The first two kinds of motion, that is, translation and rota­tion about a stationary axis, are the basic kinds of motion

(1.9)

Classical Mechanics

dv-c Viw=-'f+-n.dt . P

Fig. 4

point O. The point 0 is referred to as the centre of curvatureof the path at the given point, and the radius p of the cor­responding circle as the radius of curvature of the path atthe same point.

It is seen from Fig. 4 that the angle ~a = I dl lip == I d'f \11, whence .

I dtldl I = tip;

at the same time, if dl -+ 0, then dt -L 'f. Introducing a unitvecto~ n of the normal to the path at point 1 directed towardthe centre of curvature, we write the last equality in a vec­tor form:

d'fldl ...:.. nIp. (1.8)

Now let us substitute Eq. (1.8) into Eq. (1.7) and thenthe expression obtained into Eq. (1.6). Finally we get

Then transform the last term of this expression:d" d't dl 2 d't 2 d" 1

v't"iit=v't(jf"iit=vc(jf=v (jf' ( .7)

Let us determine the increment of the vector t in the inter­val dl (Fig. 4). It can be strictly shown that when point 2approaches point 1, the segment of the path between themtends to turn into an arc of~e with centre at some

20

;-

22 Claulcal Mechanics Essentials of Klnemattcs 23

• In the case of a finite rotation through the angle]~q> the lineardisplacement of the point A can be found from Fig. 6:

~ M I = r sin 9·2 sin (~q>/2).

Whence it is immediately seen that the displacement M cannotbe represented as a vector cross product of ~q> and r. It is only pos­sible in the case of an infinitesimal rotation dcp when the radius vec­tor r can be regarded invariable.

dqJ -- dfJJtr+ dqJ2' (1.12)

i.e. the two given rotations, dfJJt and dqJ2' are equivalent toone rotation through the angle dqJ = dqJI + dqJ2 about theaxis coinciding with the vector dqJ and passing throughthe point O.

Note that in treating such quantities as radius vector r,velocity v, acceleration w we did not hesitate over the choiceof their direction: it naturally followed from the propertiesof the quantities themselves. Such vectors are referred toas polar. As distinct from them, such vectors as dqJ whose

dr = [dqJ, rl. (1.11)

Note that this equality holds only for an infinitesimalrotation dqJ. In other words, only infinitesimal rotationscan be treated as vectors.·

Moreover, the vector introduced (dqJ) can be shown tosatisfy the basic property of vectors, that is, vector addition.Indeed, imagine"a solid performing two elementary rota­tions, dfJJt and dqJ2' about different axes crossing at a sta­tionary point O. Then the resultant displacement dr ofan arbij-rary point A of the body, whose radius vector withrespect to the point 0 is equal to r, can be represented asfollows:,

dr = drl +~dr2 = [dfJJt, r] + [dqJ2' r] = [dqJ, rJwhere

or in a vector form

a certain point 0 on the rotation axis. Then the linear dis­placement.of the end point of the radius vector r is associ­ated with the rotation angle dq> by the relation (Fig. 6)

I dr I = r sin e dq>,

Fig. 6

o

of a s~lid. All the other kinds of motion of a solid prove to ber~ducIbl~ to .one of the basic motions or to their combina­tIOn. T~IS wII.I be shown by the example of plane motion.

I~ thIS sectI?n we shall deal with the first three kinds ofn:t0tIOn and wIth the problem of summing angular veloci­ties.. Translation. ~n this k.ind of motion of a solid any straight

hne fixed to It remams parallel to its initial orientation0' all-th~time. Examples: a car

travelling along a straight sec­tion of a road, a Ferris wheelcage, etc.

When moving translationa­ry,. all points of a solid tra­verse equal distances in thesame time interval. Thereforevelocities, as well as aocelera­tions, are of the same valueat all points of the body at thegiven moment of time. Thisfact allows the study of trans­lation of a solid to be reducedto the study of motion of anindividual point belonging to

that solid, Le. to the problem of kinematics of a point.Th?-s, the tr~nslation of a solid can be comprehensively

descrIbed provIded the dependence of the radius vector ontime r (t) for any point of that body.is available as well asthe position of that body at the initial moment.

Rotation about a stationary axis. Suppose a solid, whilerotating about an axis 00' which is stationary in a givenreference frame, accomplishes an· infinitesimal rotationduring the time interval dt. We shall describe Ithe corre­sponding rotation angle by the vector dqJ whose modulus isequal to the rotation angle and whose direction coincides~ith the axis 00', with. the rotation direction obeying therIght-hand screw rule With respect to the direction of thevector dqJ (Fig. 6).

Now let us find the elementary displacement of any pointA of the solid resulting from such a rotation. The locationof the point A is specified by the radius vector r drawn from

24 Classical Mechanics Essentials of Kinematics 25

(1.15)

(1.16)

(1.17)

Fig. 8

w

o

Of

Thus, knowing the function <p (t), the law of rotation ofa body, we can find the angular velocity and angular accel­eration at each moment of time by means of Eqs. (1.15) and(1.16) On the other hand, knowing the time dependence ofangular acceleration and the initial conditions, Le. theangular velocity 0>0 and the angle <Po at the initial moment oftime, we can find 0> (t) and <p (t).

Exam,le. A solid rotates about a stationary axis in accordancewith the law cp = at - bt"/2 where a and b are positive constants.Let us determine the motion characteristics of this body.

In accordance with Eqs. (U5) and (U6)00% = a - bt; p% = -b = const.

Whence it is seen that the body performs a uniformly deceleratedrotation (P% < 0), comes to a standstill at the moment to = alband then reverses its rotation direction (due to 00% ehanging its signto the opposite).

Note that the solution of all problems on rotation of a solidabout a stationary axis is similar in form to that of prob­lems on rectilinear motion of apoint. It is sufficient to replace thelinear quantities x, V:c and W x by the

- corresponding angular quantities <p,ro z and ~ % in order to obtain allcharacteristics and relationships forthe case of a rotating body.

Relationship between linear andangular quantities. Let us find thevelocity v of an arbitrary point A ofa solid rotating about a stationaryaxis 00' at an angular velocity ro.Let the location of the point A rel­ative to some point 0 of the rota­tion axis be defined by the radiusvector r (Fig. 8). Dividing bothsides ofEq. (1.11) by the correspond­ing time interval dt and taking into account thatdr/dt = v and dfPldt = 0>, we obtain

Iv = [o>r], ILe. the velocity v of any point A of a solid rotating aboutsome axis at an angular velocity 0> is equal to the cross

The direction of the vector p coincides with the direction ofdo>, the increment of the vector 0>. Both vectors, p and 0>,

are axial.The representation of angular velocity:and angular accel­

eration in a vector form proves to be very beneficial, espe­cially in the study of more complicated kindsof motion'of a solid. In many cases this makesa problem more explicit, drastically simplifiesthe analysis of motion and the correspondingcalculations.

Let us write the expressions for angular ve­locity and angular acceleration via projec-

Fig. 7 tions on the rotation axis z whose positive. direction is associated with the positive di-

re~tIon of ~he coordinate <p, the rotation angle, in accordancewith the right-hand screw rule (Fig. 7). Then the projectionsro % and ~ % of the vectors 0> and pon the z axis are defined bythe following formulae:

0>% = d<p/dt,

~% = dro z/dt.

direction is specified by the rotation direction are calledaxial.

Now let us introduce the vectors of angular velocity andangular acceleration. The angular velocity vector 0> isdefined as

0> = dfP/dt, (1.13)

where dt is the time interval during which a body performsthe rotation dfP. The vector 0> is axial and its direction coin-cides with that of the vector dfP. .

The time variation of the vector 0> is defined by the angu­lar acceleration vector p~

p = aro/dt. (1.14)

Here 0) z and ~% are algebraic quantities. Their sign specifiesthe direction of the corresponding vector. For example, ifro z > 0, then the direction of the veetor 0> coincides withthe positive direction of the z axis; and if ro z < 0, then thevector 0> has the opposite direction. The same is true forangular acceleration.

/

26 Cla,t.cal JlIIChtUdC' ~senttals of Kinemattcs 27

or

time, we find the acceleration w of the point A:

_w = [dco/dt, r] + [co, dr/dt]

the time during the motion. Example: a cylinder rollingalong a plane without slipping (in a similar case a cone per­forms a much more complicated motion).

It is easy to infer that the position of a solid in plane mo­tion is unambiguously determined by the position of theplane figure <D within the stationary plane P. The study ofthe plane motion of a solid thus reduces to the study ofmotion of a plane figure within its plane.

Let the plane figure <D move within its plane P, whichis stationary in the K reference frame (Fig. 10). The posi­tion of the figure <D in the plane can be defined by specify­ing the radius vector r o of an arbitrary point 0' of the figureand the angle <p between the radius vector r' rigidly fixedto the figure and a certain selected direction in the K refer­ence frame. The plane motion of the solid is then describedby the two equations .

<it ro = ro (t); <p = <p (t).

It is cl~ar that if the radius vector r' of the point A(Fig. 10) ~turns through the angle dq> during the time inter­val dt, then any segment fixed to the figure will turn throughthe same angle. In other words, the rotation of the figurethrough the angle dqJ does not depend on the choice of thepoint 0'. This means that the angular velocity co of thefigure does not depend on the choice of the point 0', andwe have the right to call ro the angular velocity of the solidper se.

Now let us find the velocity v of an arbitrary point Aof a solid in plane motion. Let us introduce the auxiliaryreference frame K' which is rigidly fixed to the point 0'of the solid and translates relative to the K frame (Fig. 10).Then the elementary displacement dr of the point A in the Kframe can be written in the following form: .

dr=dro+dr't

where dro is the displacement of the K' frame, or the point0', and dr' is the displacement of the point A relative tothe K' frame. The translation dr' is caused by the rotationof the solid about the axis which is at rest in the K' frameand passes through the point 0'; according to Eq. (1.11)dr' = [dept r'J. Substituting this relation into the previous

/('

!/

Fig. 10Fig. 9

product of]ro and the'radius vector r of the point A relative toan arbitrary point 0 of the rotation axis (Fig, 8),

The modulus of the vector (1.17) is v = cor sin at or

v = cop,

where p is the radius of the circle which the point A circum­scribes. Having differentiated Eq. (1.17) with respect to

Iw = U~r] + [ro [ror)). I (1.18)

In this case (when the rotation axis is stationary) "" ro,and therefore the vector [Ilr] represents the'tangential accel­eration W't. The vector [ro [ror)) is the normal accelerationwn • The moduli of these vectors are

whence the modulus of the total acceleration w is equal to

w= y w~+w~=p Y~2+C04.

Plane motion of a solid. In this kind of motion each pointof a solid moves in a plane which is parallel to a certainstationary (in a given reference frame) plane. In this casethe plane figure <D formed as "a~result of cutting the solid bythat stationary plane P (Fig. 9) remains in that plane all

28 Cltu,fcal MechanicsEssentials of Kinematics 29

Fig. 11

OL_--=:===::::::::

(1.20)0) = 0)0+ 0)'.

Thus, thll resultant motion of the solid in the K frame isa pure rotation with the angular velocity 0) about an axiscoinciding at each moment with the vector 0) and passingthrough the point 0 (Fig. 12). This axis is displaced relativeto the K frame: it rotates together with the OA axis aboutthe axis OB at the angular velocity 0)0'

It is not difficult to infer that e.ven when the angularvelocities 0)' and 0)0 do not change their magnitudes, thebody in the K frame will possess the angular acceleration~ directed, according to Eq. (1.14), beyond the plane(Fig. 12). The angular acceleration of a solid is analysed indetail in Problem 1.10.

Generally speaking, the position of the instantaneous axisyaries with time. For exa~ple, in the case of a cylinder roll­mg over a plane surface the instantaneous axis coincides atany moment with the line of contact between the cylinderand the plane.

Angular velocity summation. Let us analyse the motionof a solid rotating simultaneously about two intersectingaxes. We shall set into rotation a cer-tain solid at the angular velocity 0)' Babout the axis OA (Fig. 12), and thenwe shall set this axis into rotationwith the angular velocity 0)0 about "'0the axis OB which is stationary in theK reference frame. Let us find the re­sultant motion in the Kframe.

We shall introduce an auxiliary ref-erence ff3.me K' fixM rtgidly to the Fig.!12axes OA and OB. It is clear that thisframe rot(tes with ~ the angular velocity 0)0' while thesolid rotates relative to this frame with the angularvelocity 0)'.

During the time .Interval dt the solid will turn throughan angle d«p' about the axis OA in the K' frame and simul­taneously through d«po about the axis OB together with theK' frame. The cumulative rotation followsJrom Eq. (1.12):dq> d«po + d«p'. Dividing both sides of this equality bydt,we obtain

~ Note that Eq. (1.19) also holds for any complex motion ofa sohd.

.. The point M may turn out to be outside the solid.

one and dividing both sides of the expression obtained bydt, we get

v=vo+[O)r'J, (1.19)

i.e. the velocity of any point A of a solid in plane motion·comprises the velocity Vo of an arbitrary point 0' of thatsolid and the velocity v' = [O)r'J caused by rotation of thesolid about the axis passing through the point 0'. Onceagain we would like to stress that v' is the velocity of thepoint A relative to the translating reference frame K' which

K is rigidly fixed to the point 0'.g In other words, plane motion

Of-a-solid can be represented asa combination of two basic kindsof motion: translation (togetherwith an arbitrary point 0' of thesolid) and rotation· (around anaxis passing through the point0').

Now we shall demonstratethat plane motion can be re­ducedto a purely rotational mo-tion. Indeed, in plane motion

!he velocity: Vo of the arbitrary point 0 of the solidIS normal to the vector 0) which means that we can alwaysfind a certain point M which is rigidly fixed to the solid"and whose velocity v = 0 at a given moment. The locationof the point M, i.e. its radius vector rM relative to thepoint 0' (Fig. 11), can be found from the condition 0 == Vo + [O)rM], The vector rM is] perpendicular to 0) andvo, its direction corresponding to the vector cross productVo = -[O)rMJ and its magnitude rM = vo/ro.

The point M defines the position of ·another importantaxi~ (coinciding with the direction of the vector 0). At~ gIven moment of time the motion of a solid representsa pure rotation about this axis. Such an axis is referred to asan instantaneous rotation axis.

30 Cla"lcal Mechanics Essentials 01 Kinematic, 31

And here is one more remark. Since the angular velocityvector 0) satisfies the basic property of vectors, vector sum­mation, 0) can be expanded into a sum of vector componentsprojected on definite directions, I.e. 0) = 0)1+ 0)2 + ... ,where all vectors belong to the same reference frame. Thisconvenient and beneficial routine is frequently employedto analyse complex motions of a solid.

m~nt dr' rel~tive to the K' fra~e: dr = dro + dr'. DividingthIS expreSSIOn by dt, we obtam the following formula forthe velocity transformation:

Iv =vo+v'. (1.21)

Differentiating Eq. (1.21) with respect to time, we immedia­tely get the acceleration transformation formula:

(6)

Iw=wo+w'·1 (1.22)

~hence it is seen; specifically, that if Wo - 0 and w = w',I.e. when the K frame moves without acceleration theacceleration .of the point A relative to the K frame will bethe same in both frames.

2. The ~' frame 'rotates a.t ,the constant angu­lar velo.~ty (J) about an ax'l,s. 'Which is station­ary in the K frame.

Let us as~ume the cnigins of the reference frames K and K'to be located at an,arbitrary point 0 on the rotation axis

Fig. 14

(Fig. 144). Then the radius vector of the point A will bethe same in both reference frames: r ==r'.. If ~he point A is at rest in the K' frame, this means thatIts ~lsplacement dr in the K frame during the time intervalcit IS caused only by the rotation of the radius vector rthrough the angle cUp (together with the K' frame) and in

A

Fig. 13

/(

§ 1.3. Transformation of Velocity and Accelerationon Transition to Another Reference Frame

Prior to entering upon the studY of this problem we shouldrecall that within the bounds OfCrassical mechanics thelength of scales and time are considered absolute. A scaleis' the same in different reference frames, I.e. it does notchange during motion. This is also true of time runninguniformly throughout all frames. ,

Formulation of the problem. There are two. arbitraryreference frames K and K' moving relative to each otherin a definite manner. The velocity v and the acceleration w

. of a point A in the K frame areknown. What are the correspond­ing values v' and w' of thispoint in the K' frame?

We shall examine the threemost significant cases of relativemotion of two reference framesin succession.

1. The K' fra me trans,latesrelqtive to the K f'rame•

Suppose the origin of theX' frame is determined by. theradius vector ro in the K frame,and its velocity and accele­

ration by the vectors Vo and woo If the location of the pointA in the K frame is determined by the radius vector randin the K' frame by the radius vector r', then apparentlyr = ro + r' (Fig. 13). Next, let during the t!me interval dtthe point A accomplish the elementary dlsplace~ent drin the K frame. This displacement,is made up ofthe d~splace­ment dro (together with the K frame) and the dlsplace-

l.~.

accordance with Eq. (1.11) is [equal to ~the ivector crossproduct [dq>, rl.

If the point A moves at the velocity v' relative to the K'frame, it will cover an additional distance v'dt during thetime interval dt (Fig. 14a), so that

dr = v'dt + [dq>, rl. (1.23)

Dividing this expression by dt, we obtain the velocitytransformation formula as follows: ..

Iv=v'+[ror], I (1.24)

where v and v' are the velocity values which characterizethe motion of the point A in the ~' frames respec­tively.

Now let us pass over to acceleration. In accordance withEq. (1.24) the increment dv of the vector v during the timeinterval dt in the K frame must comprise the sum of theincrements of the vectors v' and [rorl, Le.

dv = dv' + [ro, drl. . (1.25)

Let us find dv'. If the point A moves in the K' frame witha constant velocity (v' = const), the increment of this vec­tor in the K fJ:'ame is caused only by this vector turningthrough the angle dq> (together with the K' frame) and isequal, as in the case of r, to the vector cross product [dq>, v'l.To make sure of this, let us position the beginning of thevector v' on the rotation axis (Fig. 14b). But if the point Amoves with the acceleration w' in the K' frame, the vectorv' will get an additional increment w'dt during the timeinterval dt, and consequently

dv' = w'dt + [dq>, v'l. (1.26)

Now let us substitute Eqs. (1.26) and (1.23) into Eq. (1.25)and then divide the expression obtained by dt. Thus weshall get the acceleration transformation formula:

w = w' + 2 [roy'] + [ro [rorll, (1.27)

where wand w' are the acceleration values of the point Aobserved in the K and K' frames. The second term on theright-hand side of this formula is referred to as the Coriolis

Essentials of Kinematics

acceleration wCor and the third term is the axipetal accel­eration wop directed toward the axis.

(1.28)

(1.29)

(1.30)

woP = [ro [rorj].

Iv = v' +vo+ [ror].1

Iw == w' +2 [rov']- ro2p.1

WCor = 2 [rov'],

3. The K' frame 'rotates with the constant an­gU;lar velocit~ ro about the axis tJ'anslatingwdh the veloc~ty Vo ana acceleration Wo rela­tive to the K frame.

This case combines the two previous ones. Let us intro­d.uce an. auxiliary S, frame which is rigidly fixed to the rota­tIOn 'aXIS of the K frame and translates in the K frame.Suppose v and v s are the velocity values of the point A inthe K and S frames;. then in accordance with Eq. (1.21)v = Vo + ~ s· Replacmg v,s in accordance with Eq. (1.24)by. v s = v t [rorl, where r is the radius vector of thepomt A relatlV~ to the arbitrary point on the rotation axisof t~e K' frame, we obtain the following velocity transfor-matIOn formula: '

Thus, the acceleration w of the point relative to the Kframe is equal to the sum of three accelerations: the accel­eration w' relative.to the K' frame, the Coriolis acceleratio~WCar ane}.. the axipetal a~celeration wop. .

The aXIpetal acceleratIOn can be represented in the formwop = _co 2p where p is the radius vector which is normalto th? rotatioJ?- aXi~ and describes the position of the point Arelative to thIS aXIS. Then Eq. (1.27) can be written as fol­lows:

. • This ax!petal acceleration should not be confused with conven­honal (centrIpetal) acceleration.3-0539

Classical Mechanics32

Classtcal MechanicsEssentials of Kinematics 35

Fig. 16

y

Fig. 15

Solution. Differentiating r with respect to time twice, we obtainw = -U>s (a sin u>t + b cos mt) = -U>2r ,

i.~. the .vector w i.s always oriented toward the point 0 while its mag­m~ude IS proportIOnal to the distance between the particle and thepomt O. .

Now let us determine the trajectorY equation. Projecting r on thex and y axes, we obtain

x = a sin U>t, y =b cos U>t.

Eliminating u>t from these two equations, we getx2/a2 + y2/bs = 1.

T~is is the equation of an ellipse, and a and b are its semi-axes (seeFig. 15; the arrow shows the direction of motion of the particle A).

. et..2·ID.isplact:ment andl distance. At the moment t = 0 a.par­ticle l~ set l!l mo.tlon .at the velocity Vo whereupon its velocity beginschangmg With time m accordance with the law

... = Vo (1 - tIT),

where '1' is a positive constant. Find:(1) the displacement vector t:.r of the particle and(2) the distance s covered by it in the first t se~onds of motion

.Solution. 1. In accordance with Eq. (1.1) dr = v dt = v (1"':- tIT) dt. Integ~ating this equation with respect to time bet~een 0and t, we obtam

t:.r = vot (1 - t/2'1')•

. 2. The distance s covered by the particle in the time t is deter­mmed by

t

s=JVdt,o

v' = v - [mp).

The acceleration w' can be found from Eq. (1.29), taking into accountthat in this case w = 0 since v = const. Then w' = -2 [mv' ) ++ mSp. Substituting the expression for v' into this formula we obtain

w' = 2 [vm) - U>sp.

In a similar fashion, using Eqs. (1.22) and (1.29), we obtainthe acceleration transformation formula·:

Iw=w' +wo+2[IDV'l-ro2p·1 (1.31)

Recali that inthe last two formulae v, v' and w, w' are thevelocities and acceleratiol!s of_the point A in the K and K 'frames respectively, Vo and Wo are the velocity and accele­ration of the rotation axis of the K ' frame in the K frame, ris the radius vector of the point A relative to an arbitrarY.point on the rotation axis of the K ' frame, and p is theradius vector perpendicular to the rota~onaxis and describ­ing the location of the point A relative 0 this axis.

In conclusion, let us examine the f lowing example.

Problems to Chapter 1

eL1. The radius vector describing the position of the particle Arelative to the stationary point 0 changes with time according to thefollowing law:

r = a sin u>t + b cos mt,

where a and b are constant vectors, with a J.b; U> is a positive constant.Find the acceleration w of the particle and the equation of its pathy (x), assuming the x and y axes to coincide with the directions ofthe vectors a and b respectively and to have the origin at the point O.

• Note that in the most general case when 0) =i= const, the ri!ht­hand side of Eq. (1.31) will feature one more term, namely [ rl,where II is the angular acceleration of the K' frame, r is the ra iusvector describing the position of the point located on the rotationaxis and taken for the origin in the· K frame.

Example. A discrrotates with a""constant angular~velocity m aboutan axis fixed to the table. Point A moves along the disc withthe constant velocity v relative to the table. Find the velocity v'and acceleration w' of the point A relative to the disc at the momentwhen the radius vector describing its position relative to the rota-tion axis is equal to p. .

In accordance with Eq. (1.24) the velocity v' of the point A isequal to

3*

II dll:= (a - bx) dx.

,;2/2 = ax - bxl/2 or II = V (2a - bx) z.

Integrating this equation (the left-hand side between 0 and II andthe right-hand side between 0 and x), we get

37Essentials 0/ Kinematics

Integrating this expression, we obtain the following equation:

x

y = J(bla) x dx = (bl2a) Xl,

oi.e. the path of the point is a parabola.

.1.5. The motion law for the point A of the rim of a wheel rollinguniformly along a horizontal path (the x axis) has the form

a: = a. (rot - sin rot); y = ~ (1 - cos rot),

where a and ro are po~itive constants. Find the velocity v of the 'pointA, the distance 8 which it traverses between two successive contactswith the roadbed, as well as the magnitude and the direction of theacceleration w of the point A. .

Solution. The velocity II of the point A and the distance s it coversare determined by the formulae

11.= V '1+'1=aro V2 (1-cos rot) =2aro sin (rotI2),~(

tt

s= Jv(t) dt=4a [1-cos(rotl/2)j,o

where tl is the time interval bet~een two successive contacts. Fromy (t) we find that y (~) = 0 at rotl = 2n. Therefore, s = 8a.

The acceleration of the point A

w= Vu{+w~-arol.

Let us show that the vector w, constant in its magnitude, is alwaysdirected toward the centre of the wheel, the point C. In fact, in theK' frame fixed to the point C and translating uniformly relativeto the roadbed the point A moves uniformly along a circle about thepoint C. Consequently, its acceleration in the K' frame is directedtoward the centre of the wheel. And since the K' frame moves uniform­ly, the vector w is the same relative to the roadbed.

.1.6. A point moves along a circle of radius r with deceleration;at any moment the magnitudes of its tangential and normal accelera­tions are equal. The point was set in motion with the velocity 110'Find the velocity v and the magnitude of the total acceleration wof the point as a function of the distance s covered by it.

Solution. By the hypothesis, dv/dt = -vl/r. Replacing dt by dS/II,we reduce the initial equation to the form

dv/v = -ds/r.

The integration' of this expression with regard to the initial velocityyields the following result:

v=voe-S/ T •

1:'.':'"

;j(.'t

~Cla.ltcal Mechanics

From this equation it, can be immediately seen that .the distancebetween the stations, that is, the value Xii correspondmg to II = 0is equal to z = 2alb. The maximum velOCity ~n be found fro~ thecondition dv?dx = 0, or, simply, from the condltl.on for the ~lmumvalue of the radicand. The value Zm correspondmg to vmax IS equalto Zm = alb and llmaz = alyb: . ~

.1.4. A panicle moves in the x, !I plane from the pomt z = !I = 0with the velocity v = ai + bZj, where a and b ~re constants .and iand j. are the unit vectors of tlie x and !I axes. Fmd the equatiOn ofits path y (z). ., d'

Solution. Let us write the increments of the z and y coor matesof the particle in the time interval dt: dy = v1/ dt, dx = v:a: dt, whereV1/ = bz, IIX = a. Taking their ratio, we get

dy = (bla) z dx.

where II is the modulus of the vector v. In this case

{110 (1 - t/1;) , if t ~ 't,

1I=1I011-tl'tl = 110 (tl't-1), if t> 'to

From this it follows that if t > 't~the integral for calculating thedistance should be subdivided into two parts: between 0 and 't andbetween 't and t. Integrating in the two cases (t < 't and t > 't),we obtain

{1I0t (1 - tI2't), if t~'t,

s= 1I0't [1+(1.,..-tl't)I)/2, if t>'t.

Fig. 16 illustrates the plots II (t) and s (t)~he dotted lines showthe time dependences of the projections IIx a !J.x of the vectors vand !J.r on the x axis oriented along the vector .

.1.3. A street car moves rectilinearly from sta ion A ,to the nextstop B with an acceleration varying accordin~ tl? the .law w = a - bxwhere a and b are positive constants and x IS ItS distance fro~ sta­tion A. Find the distance between these stations and the maximumvelocity of the street car. . . .

Solution. First we shall find how the velOCity depends on x . .ourmgthe time interval dt the velocity increment dll = W ~t. Makmg useof the equation dt = dxlll, we reduce the last expressiOn to the formwhich is convenient to integrate:

36

In this case I w't I == wn' 'and therefore the total accelerationW = Vi'" wn = y2vSlr, or

w= Y2 v5/re2Blr,

el.7. A point moves along a plane path so that its tangentialacceleration W't = a and the normal acceleration wn = bt·, wherea and b are positive constants and t is time. The point started movingat the moment t = O. Find the curvature radius p of its path and itstotal acceleration W as a function of the distance s covered by the point.

Solution. The elementary velocity increment of the point dv == w't dt. Integrating this equation, we get v = at. The distancecovered s = at2/2. .

In accordance with Eq. (1.10) the curvature radius of the pathcan be represented as p = ,v2lwn = a2lbt2, \

p = a3/2bs.The total acceleration

W= Vw~+w~=a y 1+(4bs2/a3 )2.

e1.8. A particle moves uniformly with the velocity v along a para­bolic path y = ax2, where a is a positive constant. Find the accel­eration W of the particle at the point x = O.

Solution. Let us differentiate twice the path equation with respectto time:

The integration of this expression with regard to the initial condition(OOz = 0 at q> = 0) yields ooy2 = ~o sin q>. From this it follows that

OOz = ± y2~0 sin q>.

The }llot OOn (q» is shown in Fig. 17. It can be seen that as the angle q>grows, the vector m first increases, coinciding with the direction ofthe vector 110 (m > 0), reaches the maximum at q> = n12, then startsdecreasiIlg and !nallY turns into zero at q> = n. After that the bodystarts rofle.ting in the opposite direction in a similar fashion (m~ < 0).Aa a result, thelbody will oscillate about' the;position q> = 1t/2 withan amplitude equal 10 n12.

el.l0. A round cone having the height h....and the base radius rrolls without,slipping"'along""the table surface' as shown in Fig. 18.

39Essentials of KinematicsClassical Mechanlcs,38

Since the particle moves uniformly, its acceleration at all pointsof the path is purely normal and at the point x = 0 it coincides withthe derivative dlVldt2 at that P,oint. Keeping in mind that at thepoint x = 0 I dxldt I 7.= v, we get

Note that in this solution method we have avoided calculating thecurvature radius of the path at the point x = 0, which is usuallyneeded to determine the normal acceleration (wn = v2/p).

el.9. Rotation of a solid. A solid starts rotating about a station­ary axis with the angular acceleration II = 110 cos q>, where 110 isa constant v~ctor and q> is the angle of rotation of the solid from theinitial position. Find the angular velocity OOz of the solid as a func­tion of q>.

Solution. Let us choose the positive direction of the z axis alongthe vector 110' In accordance with Eq. (1.16) dooz = ~z dt. UsingEq. (1.15) to repldce dt by dq>/ooz' we reduce the previous equationto the following form:

OOz dooz = ~o cos q> dq>.

The cone apex is hinged at the point 0 which is exactly level withthe point C, the cone base centre. The point C moves at the constantvelocity v. Find:

(1) the angular velocity m and '(2) the angular acceleration p. of the cone relative to the table.Solution. 1. In accordance WIth Eq. (1.20) m = mo + m' where

mo and m' are the angular velocities of rotation about the a~es 00'and DC respectively. The magnitudes of the vectors mo and m' can beeasily found from Fig. 18:

Wo=vlh, m'=v/r.

Their ratio molm' = rlh. It follows that thel vector m coincides atany moment with the cone generatrix which passes through the con­tact point A.

The magnitude of the vector m is equal to

00 = y 003+m'2 = (vIr) Y 1+(rlh)2.

2. In accordance with Eq. (1.14) the angular acceleration II of thecone is represented by the derivative of the vector m with respectto time. Since mo = const, then '

p = dmldt = dm'ldt.

Fig. 18Fig. 17d2y r ( dx ) 2 d2x ]dt2 =2a L lit +x dtl •

dy dxlit=2ax(jt i

ClaBBtcal Mechanics

The vector 0' rotating about the 00' axis with the an¥Ular velocityCOo retains its magnitude. Its increment in the time mterval dt isequal to I d0' (=00' '000 dt ,or in vector form to d0'=[ 000') dt. Thus,

II = [(00)').

.The magnitude of this vector ~ is equal to ~ = rf'/rh.• t.tt. Velocity and acceleration transformatioD. A horizontal

bar rotates with the constant angular velocity 0 about a verticalaxis which is fixed to a table and passes through one of the ends ofthat bar. A small coupling moves alon.g the bar. Its velocity relativeto the bar obeys the law v' = ar where a is a constant and r is theradius vector determining the distance between the coupling and therotation axis. Find:

(1) the velocity v and the acceleration w of the coupling relativeto the table and depending on r;

(2) the angle between the vectors v and w in the process of motion.Solution. 1. In accordance with ~24)

v = ar + [0r).

T4e magnitude of this vector v = r y at + oot., The acceleration w is found from Eq. (1.29) where in this case

" = dv'/dt = atr. Then" = (at - ( 1) r + 2a [0r).

The magnitude of this vector ID = (at + 001) r.2. To calcul8;te the angle a betweerithe vectors v and w we shall

make use of theIr scalar product, from which it follows thai cos a == vw/uw. After the requisite transformations we obtain

cos a = tIl' t +(oola)t.

It is seen fr~m this formula that in this case the angle a remainsconstant durmg the motion.

CHAPTER 2

THE BASIC EQUATION OF DYNAMICS

§ 2.1. Inertial Reference Frames

The law of inertia.'Kinematics, being concerned with de­scribing motion irrespective of its causes, makes no essentialdifference between various l'l'ference frames and regardsthem as equivalent. It is quite different with dynamics,which deals with laws of motion. Here we detect the intrin­sic difference between various reference frames and identifythe advantages of one class of frames over others.

Basically, we can use anyone of the infinite numberof reference frames. But the laws of mechanics have, gener­ally speaking; a different form in different reference frames;it may tnen happen that in an arbitrary reference frame thelaws goveJ;Iling simple phenomena prove to be very complicat­ed. Thus, we face the problem of choosing a reference frame.in which the laws Of mechanics take the simplest form. Sucha reference frame is obviously most suitable for describingmechanical phenomena.

With. this aim in view let us consider acceleration of amass point relative to an arbitrary reference ftame. Whatcauses the acceleration? Experience shows that it can bedue to some definite bodies acting on this point, as well asto the properties of the reference frame itsels (in fact, in thegeneral case the acceleration is different relative to differ­ent reference frames).

We can, however, assume that there is a reference framein which acceleration of a mass point arises solely due to itsinteraction with other bodies. Then a free mass point exper­iencing no action from any other' bodies moves rectili­nearly and uniformly, relative to.such a frame, or, in otherwords, due to inertia. Such a reference frame is called inertial.

The statement of the existence of inertial referenceframes formulates the content of the first law of claSsical me­chanics, the law of inertia of Galileo and Newton.

The existence of inertial frames is corroborated by exper­iments. By early tests it was established that the Earthrepresents such a frame. Subsequently, the more accurate

•

• It should be pointed out that in many cases the reference framefixed to the Earth can be regarded practically inertial.

experi~ents (Foucault's experiment and the like) arguedt~at thIS reference. frame is not totally inertial*, viz., somekmds of acceleration were detected whose occurrence can­not be explain.ed by any definite bodi~ acting in this frame.At the same.t1m~ the observation of acceleration of planetsproved the mertlal character of the heliocentric referenceframe fixed to the centre of the sun and "stationary" stars.At the present time the inertial character of the heliocentric~eference frame is confirmed by the whole totality of exper­Imental facts.

Any other. reference frame moving rectilinearly and uni­formly r~latlVe to the heliocentric~ame is also inertial.In fact~ If th~ acceleration of a bodY'Is equal to zero in~he helIocentrIc reference frame, it will be equal to zerom any other of these reference frames. .

T~us, there is a vast number of inertial reference framesmovmg relative to one a!10ther rectilinearly and uniformly.~efe~ence frames executmg accelerated motion relative tomertIal ones are called non-inertial.

On sym~etry. properties of t!me and space. An importantfeature of mertlal fr8;mes conSIsts in the fact that time and.space posse~s definite symmetry properties with respect tothem~ SpeCIfically, experience shows that in such framestime is uniform while space is both uniform and isotropic.

The uniformity of time signifies that physical phenomenaproceed identically at different moments when observedunder the same conditions. In other words different momentsof time are equivalent in terms of their physical properties.

!he uniformity and isotropy of space mean that the prop­ertIe~ of ~pac~ are ident!cal at all points (uniformity)and m all dIrectlOll8 at each point (isotropy).

Note that space is non-uniform and anisotropic with re­spect to non-inertial reference frames. This means that ifa certain body does not interact with any other bodies itsdifferent orientations are still not equivalent in mecha~icalterms. In the general case this is also true for time which isnon-uniform, i.e. different moments of time are not equiva-

43

A

Fig. 19

/ii'/,

The Basic Equation of Dynamics

lent. It is clear that such properties of space and time wouldcomplicate the description of mechanical phenomena verymuch. For example, a body experiencing no action from·other bodies could not be at rest: even though its velocityis equal to zero at an initial moment of time, the next mo­ment the body would start moving in a definite direction.

Galilean relatiVity. In inertial reference frames the fol­lowing principle of relativity is valid: all inertial framesare equivalent in their mechanical properties. This meansthat no mechanical tests performed "inside" a given inertialframe can detect whether that frame moves or not. Through-out all inertial reference ,frames the properties of 9 K !J,l/{space and time, as well as Iall laws of mechanics, are Iidentical. I r

This $tatement formu­lates .the content of theGalilean principle oj-relativ­ity, one ·of the most impor­tant principles of Classicalmechanics. This principle zis a generalization of prac­tice and is confirmed byall multiform applicationsof classical mechanics to motion of bodies whose veloc­ity is considerably less than that of light.

Everything that was said above clearly demonstrates theexceptional nature of inertial reference frames, which as arule' makes them indispensable in studies of mechanicalphenomena.

,the Galilean transformation. Let us find the coordinatetransformation formulae describing a transition from oneinertial frame to another. Suppose the inertial frame K'moves relative to the inertial frame K with the velocity V.Let us take the x', y', z' coordinate axes of the K' frameparallel to the respective x, y, z axes of the K frame, sothat the axes x and x' coincide and are directed along thevector V (Fig. 19). The moment when the origins 0' and 0coincide is to be taken for the initial reading of time. Letus write the relation between the radius vectors r' and r

Classical Mechanics42

Classical Mechanics

I?iffer~ntiating this expression with respect to time andtakmg mto acocunt that V = const we obtain w' = wi.e. the point accelerates equally in'all inertial referenc~frames.

45The Basic Equ.ation 0/ Dynamics

The influence of another bod) (or bodies) causing the accel­eration of the body A is referred to as a force. Therefore,a body accelerates due to a force acting on it.. One of, the most significant features of a force is its mate­

rial origin. When speaking' of a force, we always implicitlyassume that in the absence of extraneous bodies the forceacting on the body in question is equal to zero. If it be­comes evident that a force is present, we try to identify itsorigin as one or another concrete body or bodies.

All the forces which are treated in mechanics are usuallysubdivided into the forces emerging due to the direct con­tact between bodies (forces of pressure, friction) and theforces arising due to the fields generated by interacting bodies(gravitational and electromagnetic forces). We should pointout, however, that such a classification of forces is condition­al: the interacting forces in a direct contact are essentiallyproduced 9Y some kind of field generated by molecules andatoms of ,J)odies. Consequently, in the final analysis all for­ces of interaction between bodies are caused by fields. Theanalysis ofthe nature of interaction forces, lies'~"utside thescope of mechanics /and is considered in other .visions ofphysics.

Mass. Experience shows that every body "resists", anyeffort to change its velocity, both in magnitude and direc­tion. This property expressing the degree of unsusceptibili­ty of a body to any change in its velocity is called inertness.Different bodies reveal this property in different degrees.A measure of inertness is provided by the quantity calledmass. A body possessing a greater mass is more inert, andvice versa.

Let us introduce the notion of mass m by defining theratio of masses of two different bodies via the inverse ratioof accelerations imparted to them by equal forces:

ml/m, = W,/WI' (2.5)

Note that this definition does not require any preliminarymeasurements of the forces. It is sufficient to meet the crite­rion of equality of forces. For example, if two different bodieslying on a smooth horizontal surface are pulled in succes­sion by the same spring oriented horizontally and stretchedto the same length, the influence of the spring on the bodies

(2.4)

t' = t.

Iv'=v-v·1

and, besides,

of the same point A in the K' and K frames:

r' = r - Vt (2.1)

(2.2)

The length ~f rods and time rate are assumed to be indepen­dent of motion here and, consequently, are identical in thetwo reference frames. The assumption that space and timeare absolute underlies the concepts of classical mechanicswhich are based on extensive experimental data pertainingto the study of motion whose velocity is substantially lessthan that of light.

The relations (2.1) and (2.2) are referred to as the Galileantransformations. These transformations can be written ina coordinate form as follows:

l-;':x-vt, y'=y, z'=z, t'=t. I (2.3)

Di~erentiating Eq: (2.1) with respect to time, we get theclassIcal law of veloCIty transformation for a point on tran­sition from one inertial reference frame to another:

§ 2.2. The Fundamental Laws of NewtonianDynamics

Investigating various kinds of motion in practice wediscover that in inertial reference frames any acceler~tionof a body is caused by some other bodies acting on it. Thedegree of influence (action) of each of the surrounding bodieson the state of motion of the body A in question is a prob­lem whose solution in a concrete case can be obtained throughexperiment.

is equal in both cases, Le. the force is identical in bothcases.

Consequently, a comparison of the masses of two bodiesexperiencing the action of the same force reduces to thecomparison of accelerations of these bodies. Having adopteda certain body for a mass standard, we may compare themass of any body against the standard.

Experience shows that in terms of Newtonian mechanicsa mass determined that way possesses the following twoimportant properties:. (1) ma~ is an additive quantity, i.e. the mass of a compo­

sIte body IS equal to the sum of the masses of its constituents'(~). the. mass. of a ~ody proper is a constant quantity, re:

maIllIllg Illvariable III the process of motion. 'Force. Let us get back to the experiment in which

we compared the accelerations of two different bodies sub­jected to the action of an equally stretched spring. The factthat the spring was stretched equally in both cases permit­ted. us to claim an identical force exerted by thesprIllg.

On the other hand, a force makes a body accelerate. Theaccelerations of different bodies under the action of the sameequally stretched spring are different. Our task is to definea force in such a way as to make it the same despite the differ­ence in accelerations of different bodies in the case consid­ered.

To do this, we have to clear up the following thing first:what quantity is the same in this experiment? The answeris obvious: it is the product mw. It is then natural to adoptthis quantity for a definition of force. Besides, taking intoaccount that acceleration is a vectorial quantity, we shallalso assume a force t~ be a vector coinciding in its direction­with the acceleration vectorw.

Thus, in Newtonian mechanics a force acting on a body ofmass m is defined as a product mw. Apart from the maximumsimplicity and convenience, this definition of a force isof course justified only by the subsequent analysis of allconsequences following from it.

Newton's second law. Examining in practice the interac­tion of various mass points with surrounding bodies, weobserve that mw depends on the quantities characterizing

both the state of the mass point itself and the state of sur­rounding bodies.

This significant physical fact underlies one of the mostfundamental generalizations of Newtonian mechanics, New­ton's second law:

the product of the mass of a masS point by its accelerationis a function of the position of this point relative to surround­ing bodies, and sometimes a function of its velocity as well.This function is denoted by F and is called a force.

This is exactly what constitutes the actual content ofNewton's second law, which is usually formulated in a briefform as follows:

the product of the mass of a mass point by its accelerationisequal to the force acting on it, i.e.

This eq'\Iation is referred to as the motion equation of amass point.

It shou~ be in1Inediately emphasized that Newton'ssecond law and Eq"'2.6) acquire specific meaning only afterthe function F is established, that is, its dependence on thequantities involved, or the law of force, is known. Determin­ing the law of force in each specific case is one of the basicproblems of physical mechanics.

The definition of 'a force as mw (Eq. (2.6») has the remar­kable merit of presenting the laws of force in a very simpleform. The study of motions at relativistic veloc­ities, however, showed that the laws of force should bemodified to make the forces dependent on the velocity of amass point in an intricate way. The theory would thus turnout to be cumbersome and confusing.

However, there is an easy way to dispose of the problem;the definition of a force should be slightly modified as fol­lows: a force is a derivative of the momentum p of a mass pointwith respect to time, that is, dp/dt; Eq. (2.6) should then berewritten as dp/dt = F.

In Newtonian mechanics this definition of a force is iden­tical to mw since p = mv, m = const and dp{dt '..:- mw,while in relativistic mechanics, as we shall see, momentumdepends on the velocity of a mass point in a more complicat-

\

4'1

(2.6)jmw=F·1

The Basic Equatton of DynamicsCl4uicaZ Mechanics46

{.

This implies that interaction forces always appear in pairs.The two forces are applied to different mass points; besides,they are the forces of the same nature. .

The law (2.7) holds true for systems comprising any num­ber of mass points. We proceed from the assumption that

ed way. But something different is important here. Whenforce is defined as dp/dt, the laws of forces prove to remainthe sa~e in the r~lativistic ca~ as well. Thus, the simpleexpressIOn of a given force via the physical surroundingshould not be changed on transition to relativistic mechan­ics. This fact will be employed later.. On summation 01 forces. Under the given specific condi­

tIOns any mass point experiences, strictly speaking, onlyone force F whose magnitude and direction are specifiedby the position of that point relative to -all surroundingbodies, and sometimes by its velocity as well. And stillvery often it is convenient to depict this force F as a cumu­lative action of individual bodies, or a sum of the forces F

l,

F lI' •• '.' Experience shows that if the bodies acting assources of force exert no influence on each other and so donot change their state in the presence of other bodies, then

F = F1 + Fli + ...,where F i is the force which the ith body exerts on the givenmass point in the absence of other bodies.

If that is the case, the forces Fl , F lI , ••• are said to obeythe principle of superposition.This statement should be regard­ed as a generalization of experimental data.

Newton's third law. In all experiments involving onlytwo bodies A and B, body A imparting acceleration to B, itturns out that B imparts acceleration to A. Hence, we cometo the conclusion that the action of bodies on one anotheris of an interactive nature.

Newton postulated the following general property of allinteraction forces, Newton's third law:

two mass points act on each other with forces which are alwaysequal in magnitude and oppositely directed along a straightline connecting these points, i.e.

49The Balic Equation of Dynamic.

in this case as well the interaction reduces to the forcesof paired interaction between 'mass points.

In Newton's third law both forces are assumed to be equalin magnitude at any moment of time regardless of the motionof the points. This statement correspon~s to th~ Newto.nianidea about the instantaneous propagatIOn of mteractIOns,an assumption which b; identified in classical mech~nics ~s

the principle of long-range action. In accordance w~th thiSprinciple the interaction between bodies propag~tes m spaceat an infinite velocity. In other words, havmg changedthe position (state) of one body, we can immediately det~et

at least a slight variation in the other bodies interactmgwith it, however far they may be located.

Now we know that this is actually not the case: there doesexist a finite maximum velocity of interaction propag~tion,

being equjil to the velocity of light in vacuo. :'-ccor~mgly,Newton'lfihird law (as well as the second one) IS vahd0!1lywithin certain bounds. However, in classical mechamcs,treating bQdies mofing with velocities substantially lowerthan the velocity of light, both laws hold true with a veryhigh accuracy. This is evidenced, for example, by orbits ofplanets and artificial satellites computed with an "llstro­nomical" accuracy by the use of Newton's laws.

Newton's laws are the fundamental laws of claSSical me­chanics. They make it possible, at least in principle, tosolve any meehani~al problem. Besides, all the other lawsof classical mechanics can be derived from Newton'slaws.

In accordance with the.Galilean principle of relativitythe laws of mechanics are identical throughout all inertial ,reference frames. This means, specifically, that Eq. (2.6)will have the same form in any inertial reference frame. Infact, the mass m of a mass point per se does not depend onvelocity, i.e. is the same in all reference frames. Moreover,in all inertial reference frames the acceleration w of a pointis also identical. The force F is also independent of thechoice of a reference frame since it is determined only bythe position and velocity of a mass po~nt relative t~ .su~­rounding bodies, and in accordance WIth non-relatlvls~IC

kinematics these quantities are equal in different inertialreference frames.4-05311

(2.7)

Classical Mechanics48

5f

(2.9)

The Basic Equation of Dynamics

The Coulomb force acting between two point chargesql and q2t

where r is the distance between the charges and k is a pro­portionality constant dependent on the choice of a systemof units. As distinct from the gravitational force Coulomb'sforce can be both attractive and repulsive.

It should be pointed out that Coulomb's law (2.9) does nothold precisely when the charges move. The electrical inter­action of moving charges turns out to be dependent ontheir motion in a complicated way. One part of that inter­action which is caused by motion is referred to as magnetic

. force (hence, another name of this interaction: the electro­magnetic one).At low (non-relativistic) velocities the magnet­ic force constitutes a negligible part of an electric interac­tion, wl}!ch is described by the law (2.9) with a high degreeof accuracy.

In spit~~ of the raet that gravitational and electrical inter­actions underlie .11 innumerable mechanical phenomena,the analysis of these phenomena, especial~y macroscopicones, would prove to be very complicated if we proceededin all cases from these fund·amental interactions. Therefore,it is convenient to introduce some other, approximate, lawsof forces which can in principle be obtained from the funda­mental forces. This waywecan simplify the problem in mathe­matical terms and to turn it into a practically soluble one.

With this in mind, the following forces can be, for example,introduced.

The uniform force of gravity

F = mg, (2.10)

where m is the mass of a body and g is gravity acceleration. *The elastic force is proportional to a displacement of a

mass point from the equilibrium position and directed to-

• Note t~at in contrast to the force of gravity the weight P isthe force WhIC~ a body exerts on a support or a suspension which ismotionless relatIve to this body. For example, if a body with its sup­po~t ~suspe~sion) is at .rest with respect to the Earth, the weight P~omcides wIth .the gravIty force. Otherwise, P = m (g - w), where w18 the acceleratIon of the body (with the support) relative to the Earth.4*

(2.8)

Cla,,'cal Mechanics

In accordance with Eq. (2.6) the motion laws of a particlecan be determined in strictly mathematical terms providedwe know the laws of forces acting on this particle, that is,the dependence of the force on the quantities determining it.In the final analysis, each law of this kind is obtained fromthe processing of experimental data, and, basically, alwaysrests on Eq. (2.6) as a definition of force.

Gravitational and electrical forces are the most fundamen­tal forces underlying all mechanical phenomena. Let usdescribe briefly these forces in the simplestform when inter­acting masses (charges) are at rest or move with a low(non-relativistic); velocity.

The gravitational force acting between two mass points.In accordance with the law of universal gravikttion this forceis proportional to the product of -the masses of points m1 andm l , inversely proportional to the square of the distance rbetween them and directed along the straight line connect­ing these points:

Thus, the three quantities m, wand F appearing inEq. (2.6) do not change on.transition from one inertial refer­ence frame to another, and therefore Eq. (2.6) does notchange either. In other words, the equation mw = F isinvar~ant with respect to the Galilean transformation.

where,\, is the gravitation constant.The masses involved in this law are called gravitational

in distinction to inert masses entering Newton's second law.It was established from experience, however, that a gravi­tational mass and an inert mass of any body are strictlyproportional to each other. Consequently, we can regardthem equal (Le. to take the same standard for measuringthe two masses) and speak just of mass, whether it appearsas a measure of inertness of a body or as a measure of gravi­tational attraction.

§ 2.3. Laws of Forces

__ 50-----------------~=:::::.:::::....:::.::..:..::::::..:.:.::..:

where k is a positive coefficient intrinsic to a giv.en bodyand a given medium. Generally speaking, this coefficientdepends on the velocity v, but in many cases at low veloci­ties it can be regarded practically constant.

/

53

(2.15)

The Basic Equation of Dynamics

where F x, F y' F z are the projections of the vector F on thE'x, y, z axes. It should be borne in mind that these projec­tions are algebraic quantities: depending on the orientationof the vector F they may be both positive and negative. Thesign of the projection of the resultant force F also definesthe sign of the projection of the acceleration vector.

Let us show a concrete example of the standard methodof solving problems through the use of Eq. (2.15).

Example. A small bar of mass m slides down an inclined planeforming an anglo a \vith the horizontal. The friction coefficient isequal to k. Find tho acceleration of the bar rotative to the plane. (Thisreference frame is assumed to be inertial.)

First of all we should depict all the forces acting on the bar: theforce of gravity mg, the normal force of reaction R of the plane and

Basically, Eq. (2.14) is a differential equation of motionof a point in vector form: Its solution constitutes the basicproblem of dynamics of a mass point. Two antithetic formu­lations of the problem are possible here:

(1) to find the force F acting on a point if the mass m ofthe point and the time dependence of its radius vectorr (t) are known, and

(2) to find the motion law of a point, Le. the time depend­ence of its radius vector r (t), if the mass m of the pointand the force F (or the forces F i ) are known together withthe initial conditions, the velocity voand the position .roof the point at the initial moment of time.

In the first case the problem reduces to differentictingr (t) with respect to time a.nQ in the second to integrctingEq. (2.14). The mathematicaf aspects of this problem werediscussed at length when we treated kinematics of a point.

Depen"lling on the nature and formulation of a specificproblem Eq. (2.14) is solved either in vector form or incoordinat~ or in ,Projections on the tangent and on thenormal to the trajectory at a given point. Let 'Us see howEq. (2.14) is written in the last two cases.

In projections on the Cartesian coordinate axes. Projectingboth sides of Eq. (2.14) on the x, y, z axes, we get three dif­ferential equations

(2.13)

(2.14)

(2.11)

~ta"tcal Mechanics

F= -kv,

§ 2.4. The Fundamental Equation of Dynamics

The fundamental equation of dynamics of a mass pointis nothing but a mathematical expression of Newton'ssecond law:'

where k is the sliding friction coefficient depending on thenature and condition of the contacting surfaces (specifically,their roughness), and Rn is the force of the normal pressuresqueezing the rubbing surfaces together. The force F isdirected oppositely to the motion of a given body relativeto another body. .

The resistance force acting on a body during its transla­tion through fluid. This force depends on the velocity v ofa body relative to a medium and is directed oppositelyto the v vector:

where r is the radius vector describing the displacement ofa particle from the equilibrium position; and x is a positiveconstant characterizing the "elastic" properties of a particu­lar force. An example of such a force is that of elastic defor­mation arising from an extension (constriction) of a springor a bar. In accordance with Hooke's law this force is definedas F - xAl, where Al is the magnitude of elastic deforma­tion.

The sliding' friction force, emerging when a given bodyslides over the surface of another body

F = kRn , (2.12)

ward the equilibrium position:

F = -xr,

55

mg

Fig. 22Fig. 21

The Basic Equation of Dynamics

where F 1: and Fn are the projections of the vector F on theunit vectors 'f and n. In Fig. 21 both projections are posi­tive. The vectors F1: and Fn are referred to as the tangentialand normal components of the force F.

Recall that the unit vector 'f is oriented in the directionof growing arc coordinate l while the unit vector n is directedto the centre of curvature of the trajectory at a given point.

Eqs. (2.16) are convenient to use provided the trajectoryof a mass point is known.

Example. A small body A slides off the top of a smooth sphereof radius r. Find the velocity of the body at the momeD~ it loses con­tact with the surface of the sphere if its initial velocity is negligible./,

Ft''I: .............\ R

\\\,,,

" "''''-Ffn, ;......-\ .....--,,"--

Let us depi,ct the forces acting on the body A (which are the forceof gravity ml( and the normal force of reaction R) and write Eqs. (2.16)via projections on the unit vectors l' and n (Fig. 22):

m dv/dt = mg sin e,mv2/r = mg cos a - R;

since the subindex 'f is inessential here, it has been omitted.The first equation should be transformed to make it more conven­

ient to integrate. Taking into consideration that dt = ~l/v = r ~a/vwhere dl is an elementary path the body A covers durmg the tImeinterval dt, we shall write the first equation in the following form:

v dv = gr sin e de.I"ntegrating the left-hand side of this expression between the limits 0and v and the right-hand side between 0 and a, we find

VS = 2gr (1 - cos e).Next, at the moment the body loses contact with the surface R = 0,and therefore the second initial equation takes the form

u2 = /{r cos e.(2.16)

VS

mp=Fn ,

Fig. 20

mg

And finally,

Classical Mechanics

R=mg cosct and Ffr=kR=kmg cos ct.

the friction force Ffr (Fig. 20) directed oppositely to the motion ofthe, bar.

After that let us fix the coordinate system x, y, z to the "inclined­plane" reference frame. Generally speaking, a coordinate system canbe oriented at will, but in many eases (and in this one, in particular)the direction of th~ axes is specified by the character of motion. In thisease, for example, the direction in which the bar moves is knownin advance, and therefore the coordinate axes should be so laid outthat one of them coincides with the motion "direction. Then the prob­lem reduces to the solution of only one of the equations (2.15). Thus,

_ let us choose the x axis as shownR U, in Fig. 20, and indicate its po-

I . sitive direction by an arrow.And only now we can set about!. working out Eq. (2.15): the left-

..... hand side contains the product of..... ..... the mass m of the bar by the pro-

'" jection of its acceleration W:e' andx the right-hand side the projections

of all forces on the x axis:

mwx=mgx+Rx+Ffrx·

In this ease g:e = g sin ct, R ==0 and Fir :e=- Ffr , and therelore

mW:e = mg sin ct - Ffro

Since the bar moves only along the.. x axis, the sum of projections of

~ll forces on any.dIrectlOn p~rpendicular to the.x axis is equal to zerom accordance wIth Newton s second law. Takmg the y axis as sucha direction (Fig. 20), we obtain

In projections on the tangent and the normal to the trajec­tory at a given point. Projecting both sides of Eq. (2.14)on the travelling unit vectors 'f and n (Fig. 21) and makinguse of the tangential and normal accelerations appearingin Eq. (1.10), we can write

mW:e = mg sin ct - kmg cos ct.

If the right-hand side of this eq~ati~n is positive, then also W:e > 0,and consequently the vector w IS dIrected down along the inclinedplane, and vice versa.

w~ere v and 9 correspond to the moment when the body loses contactWith the surface. Eliminating cos 9 from the last two equalities weobtain v = y 2gr/3. '

§ 2.5. Non-inertial Reference Frames.Inertial Forces

The fundamental equation of dynamics in a non-inertialtrame..As mentioned above, the fundamental equation ofdynamIcs holds true only in inertial reference frames. StilltherE> are many eases when a specific problem needs to bes?lved in a non-inertial reference frame (e.g. motion of as~mple p~ndulum in ~ carriage moving with an accelera­tIon, motIOn of a s~telhte relative to the Earth surface etc.).Hence, the followmg question arises: how to modify the!und~mental equation of dynamics to make it valid in non­mertIal reference frames?W~th t~is in mind let us consider two reference frames:

the mertzal frame K and non-inertial frame K'. Supposethat ~e kno~ the mass m of a particle, the force F exertedon thI~ partIclebr surrounding bodies and the characterof motIon of the K frame relative to the K frame.

Let us examine a sufficiently general case when the K'fr8;me r~tates with a constant angular velocity ro about anaXIs. whIch translates relative to the K frame with the accel­e~atIOn woo We shall employ the acceleration' transforma­tion formula (1.31), from which it follows that the accelera­tion of the particle in the K' frame is

w' = w- wo+ ro2p+2 [v'roj, (2.17)where v' is the velocity of the particle relative to the H'f~ame a!1d p is the radius vector perpendicular to the rota­tion aXIS and describing the position ofithis particle withrespect to this axis.

Multiplying both sides of Eq. (2.17) by the mass m ofthe particle and taking into account that in an inertialreference frame mw = F, we obtain

fmw' =F-mwo+mro2p+2m [V'ro).' (2.18)

This is the fundamental equation of dynamics in a non-inertialreference frame rotating with a constant angular velocity (0

about an axis translating with the acceleration Wo' It indi­cates that even if F = 0, the particle will move in thisframe with an acceleration (which in 'tile general casediffers from zero), as if under the influence of certain forcescorresponding to the last three terms of Eq. (2.18). Theseforces are referred to as inertial.

Eq. (2.18) shows that the introduction of inertial forcesmakes it possible to keep tlm format of the fundamentalequation of dynamics in non-inertial reference frames aswell: the left-hand side is the product of the mass of theparticle by its acceleration (but this time relative to thenon-inertial reference frame), and the right-hand side con­tains the forces. However, apart from the force F caused bythe influence of surrounding bodies (interaction forcf1s), itis necessary to take into account inertial forces (the remain­ing terms ,on the right-hand side of Eq. (2.18»).

Inertiakforces. Let us write Eq. (2.18) in the followingform: '

\'

57

(2.19)mw'·= F+Fin +Fet +F Cor,

The Basic Equation of DynamIcs

where

IFin=-mWol (2.20)

is the inertial force caused by the translation of the non­inertial reference frame;

IFet = mro2pI (2.21)

is the. centrifugal force of inertia;

IFcor =2m[v'ro)I (2.22)

is the CorioUs force. The last two forces emerge due to rota­tion of the reference frame.

Thus, we see that the inertial forces depend on the charac­teristics of the non-inertial reference frame (wo, ro) as wellas on the distance p and the velocity v' of a particle in thatreference frame.

For example, if a non-inertial reference frame translatesrelative to an inertial one, a free particle in that frameexperiences only the force (2.20) whose direction is opposite

Classical MechanIcs56

58 Classical Mechanics The Basic Equation of Dynamics

•

59

to the acceleration roo of the given reference frame. Recallhow a sudden braking of the carriage we travel in makes usswing forward, that is, in the direction opposite to woo

Here is another example: a reference frame rotates abouta stationary axis with the angular velocity ro, and the bodyA is at rest in that frame (e.g. you are on a rotating plat­form in an amusement park). Apart from the forces of inter­action with surrounding bodies, the body A experiencesthe centrifugal force of irtertia (2.21) directed along theradius vector p from the rotation axis. As long as the body Ais at rest relative to the rotating platform (v' = 0), thisforce makes up for the interaction force. But as soon as thebody begins to move, Le. the velocity v' appears, thereoriginates the Coriolis force (2.22) whose direction is deter­mined by the vector cross product [v'ro). Note that theCoriolis force crops up to supplement the centrifugal force ofinertia appearing irrespective of whether the body is at restor moves with respect to the rotating reference frame.

It was pointed out that the reference frame fixed to theEarth's surface can be regarded in many cases as practicallyinertial. However, there are some phenomena whose inter­pretation in this reference frame is impossible unless itsnon-inertial nature is taken into account.

For instance, free-fall acceleration is the greatestat the Earth's poles. Approaching the equator. one observesa decrease in this acceleration caused not only by the devia­tions of the Earth from a spherical shape, but also by thegrowing action of the centrifugal force of inertia. There arealso such phenomena as a deviation of free-falling bodies tothe East, a wash-out of right banks of rivers in the NorthernHemisphere and left banks in the Southern Hemisphere,a rotation of the Foucault pendulum oscillation plane, etc.Phenomena of this kind are associated with the motion ofbodies relative to the Earth's surface and can be explainedby the Coriolis force.

Example. A train of mass m moves along a meridial!- at the la~i­tude q> with the velocity v'. Fi.nd the lateral force whIch the tramexerts on the rails.

In the reference frame fixed to the Earth (rotating at the angularvelocity (I) the train's acceleration component normal to the. me~id­ian plane is equal to zero. Therefore, the sum of the prOJectiOnsof forces acting on the train in this direction is also equal to zero. And

I;I'

1I

this means that the Coriolis force FCar (Fig. 23) must be counterbal­~nced by the lateral force R !lxert~d by the r!ght rail on the train,I.e. FCor = -R. In accordance wIth Newton s third law the trainacts on that rail in the horizontal direction with the force R' = -R.

Fig. 23

Consequently, R' = Fear = 2 m [v' (I)). The magnitude of the vectorR' is equal to R' = 2mv' w sin'q>.

The following simple example illustrates how the inertialforces "appear" on transition from an inertial referenceframe to a non-inertial one.

Example. A horizontal disc D freely rotates over the surface ofa table about a. vertical axi~ with a constant angu~ar velocity (I).

A sphere possessmg mass m IS suspended over the dISC as shown inFig. 24a. Let us consider the behaviour of that sphere in the K framefixed to the table and assumed inertial, and in the K' frame fixed tothe rotating disc.. .. In the inertial K frame the sphere is subjected to two forces thegravity force and the stretching force of the thread. These f~rcesequalize each other so that the sphere is at rest in the K frame.

!n the .non-inertial K' frame t~e sphere moves uniformly alonga Circle wIth the normal acceleratiOn w2p, where p is the distancebetween the sphere and the rotation axis. One can easily see that thisacceleration is due to inertia~ forces. Indeed, in the K' frame, apartfrom the two counterbalancmg forces mentioned above there arealso the centripetal force of inertia and the Coriolis force'(Fig. 24b).Taking the projections of these forces on the normal n to the path at

60 Classical MechanicsThe Basic Equation 01 Lynamtcs

the point where the sphere is located, we write:

mw~= FCor - Fe! = 2mv'00 - moollp = moo·p,

where it is taken into consideration that in this case v' = oop. Hence,w~ = oolp. .

Properties of inertial forces. To summarize, we shalllist the most significant properties of these forces in order todiscriminate them from interaction forces:

1. Inertial forces are caused not by the interaction ofbodies, but by the properties of non-ine:dial reference framesthemselves. Therefore inertial forces do not obey Newton'sthird law.

2. To avoid misunderstandings is should be firmly bornein mind that these forces exist only in non-inertial referenceframes. In inertial reference frames there are no inertialforces at all, and the notion of force is employed in theseframes only in the Newtonian sense, that is, as a measure ofinteraction of bodies.

3. Just as gravitational forces, all inertial forces areproportional to the mass of a body. Consequently, in a uni­form field of inertial forces, as in the field of gravitationalforces, all bodies move with the same acceleration regardlessof their masses. This highly important fact has far-reachingconsequences.

The principle of equivalence. Since inertial forces, justas gravitational ones, are proportional to the masses ofbodies, the following important conclusion can be made.Suppose we are in a certain closed laboratory and are deprivedof observing the external world. Moreover, let us assumethat we are not aware of the whereabouts of our laboratory:outer space or, e.g. the Earth. Observing the bodies fallingwith an equal acceleration regardless of their masses, wecannot determine the cause of this acceleration from onlythis fact. The acceleration can be brought about by a gravi­tational field, by an accelerated translation of the laboratoryitself, or by both causes. In such a laboratory no experimentwhatsoever on free fall of bodies can distinguish the uniformfield of gravitation from the uniform field of inertial forces.

Einstein argued that no physical experiments of any kindcan be of use to distinguish the uniform field of gravitationfrom. the uniform field of inertial forces. This suggestion;

ra~se~ to a postulate, provides a basis for the~;;nc;;l~ o{ e'huivalence of gravitational and inerti:tf:~~l:s~t r P ~slca p enomena proceed in the uniform field of gr ..fi~l~not i~~~~:~17:":':s~e way as in the corresponding unif~~'::ttiaThr far-reaching analogy between gravitational and iner

orces was used by Einstein as a startin oint' .-

td.e,:etl?pmhent of the general theory of relativit/:r. theI~er~~IVIS IC t eory of gravitation. ' .

In conclusion it should be pointed out that .problefm can bUe solved in both inertial and nO:~~::i~~~~~~~1ence rames. sually the choice of h -~~am~ is. determined ~y ~he formulat~~eofrt~:o;r:~l~~e~~n~ye

e esue to solve It 1Il as straightforwardPfssible. ~n so doing, we quite often find thatan:::'~~::ti:~r:;~en~~9~~~~)~ are most convenient to apply (see Prob-

.Problems to Chapter 2

.2. t. A bar of mass ml is placed on a lank of .~sts on a smooth horizontal plane (Fig 25f Th ffis~ ml

, whIChtlOn between the surfaces of the bar and 'th . I ke?oe clent of fric­e p an IS equal to k. The

---------rFig. 25

plank is subjected to the horizontal force F d d' .F = at (a is a constant) F' d' epen mg on time t as

(1) the moment of tim' t IUt . h' h hunder the bar; e 0 a w IC t e plank starts sliding from

(2) the accelerations of the b d f hess of their motion. ar WI an 0 t e plank wbn the proc-Solution. 1 Let us write th f d .

for the plank a'nd the bar havine t::k a"hntal .e9uati~n of dynamicsaxis as shown in theftgure:" g en t e posltlVe direction of the x

mlwl =F!r, mswl=F-F!r' (t)

As the force F grows so does the f . t' fmOlllcul it represC'uts the friction of :lCt)lOlnI orce F!r (at. t~e initiales . owever, the frictIOn force

(3)

The Basic Equation oj Dynamics

the motion direction, but only decrease the acceleration. Thus, thedirection of the friction force acting on the body mz is determined ifwe find the acceleration direction of this body in the absence of fric­tion (k = 0). Accordingly. we shall begin with that.

Let us write the fundamental equation of dynamics for both bo­dies in terms of projections, having taken the positive directions ofthe .xl and .xz axes as shown in Fig. 27:

mlwX = mig - T. mzwx = T - mIg sin a.

where T is the tensile force of the thread. Summing up termwise theleft- and right-hand sides of these equations, we obtain

TJ -sin aW x = TJ+1 g.

After the substitution TJ = 2/3 and a = 300 this expression yIeldsWx > 0, Le. the body mz moves up the inclined plane. Consequently.tlie friction force acting on this body is. directed oppositely. Takingthis into account, we again write the equations of motion:

mlw~ =:? mig - T', mzw; = T' - mzg.sin a - kmzg cos a.Hence,

W • 1t-sin a-k cos a _ 005x TJ+1 g-. g.

.2.3. A non-stretchablu thread with masses m). and mz attachedto its ends (ml > mz) is thrown over a pulley block (Fig. 2lt). Webegin to lift the pulley block willi the acceleration Wo relative to theEarth. Assuming the thread to slide over the pulley block withoutfriction, find the acceleration WI of the mass ml relative to the Earth.

Solution. Let us designate the positive direction of the .x axisas shown in Fig. 28 and write the fundamental equation of dynamicsfor the two masses m terms of projections on this axis:

mlwlX=T-mlg. (1)

mawzx= T-1lI.aK· (2)

These two equations contain three unknown quantities: wlX ' wax,and T. The third equation is provided by the kinematic relationsnipbetween the accelerations:

Wl=WO+W'. wz=wo-w'.

where w' is the acceleration of the mass ml with respect to the/ulleyblock. Summing up termwise the left-hand and the right-han sidesof these equations, we get

WI + Wz = 2wo,

or in terms of projections on the .x axis

wu:+ wu :=2wn.

TII

m,ItXt

Fig. 27

Clallsic.al Mechanics

t

Fig. 26

. - 0 10 The masSes of the pulley block and thethe body ma Ii. k'hl 'F" d the magnitude'and direction of the .a~~l­~ds afrethenegblgdl e. iffue system is set'into motion from an InItialeratIOn 0 0 Y mlstate of rest. . _l. Id t ckle the problem associated with the

Solution. Fust we !Wou a. h b d m Otherwise. wedirection ?f ththe frfictdion fOfi :~~~lio~0\ dyn~Jics aior the body ~Icannot write e. u~ amen a h roblem roves to be uncertam.in terms of proJectloril an~ t ;pEse that i~ the absence of friction

We shall argue as 0 ts°WSl"ds~ng Gay. upward along the inclinedf thebodym star SII ...... . I t-...orcas ., I " th· f . t'on forcas we obVIOUS y canno reve.....plane. "SWltchmg on e ric I •

2. If t';;;;; to. then

WI = WI = at/(~ + ma);

and if t~;;;;' to. thenWI = kg = const, Wa = (at - kmlg)/mZ'

The 1 ts (t) and w (t) are shown in Fig. 26. h1.2. I:the arrang:ment of Fig. 27 the. inclined plane forms t .e, = 300 with the horizontal. The ratIO of the masses shown IS

an!e a/ _ 2/3 The coefficient of friction between the plane andf)-mlml- .

Wz;;;;' WI'

Substituting here the values of WI and Wz ta~en from Eq. (1) andtaking into account that FIr = kmlg. we obtam

(at - kmlg)/mZ ;;;;. kg,.

where the sign "=" corresponds to the moment t = to. Hence.to = (ml + ml ) kg/a.

62

F has the ultimate value Fjr. maX = kml~' Unless this vall;le isIr h d both bodies move as a smgle whole with equal accelerat.t3!ls.

B~~ a~ ~on as the force FIr reaches the limit. the plank starts s11 mgfrom un~er the bar, Le.

65

wt

Fig.3i

o

T3m

Fig. 30

(aJ

(6)

T.~_T-=:::r2~ -e'~~

;jtf

to a force which is a geometric sum of two;vectors each of which hasthe magnitude of the tension sought T (Fig. 30b). Consequently,in accordanoe with Newton's second law

lim ·oo'r = T ·lIa. (1)

Since lim = (m/21t) lIa. and r = l/21t (where l is the length of the cordin the state of rotation), Eq. (1) takes the form

moo.·l!4n· = T. (2)

On the other hand, in accordance with Hooke's lawT =x (1- 'o). (3)

Eliminating l from Eqs. (2) and (3), we obtain

T "'04n·x/moo·-1 .

Note that in the caseofa non-stretchable cord (x = 00) T = moo·,o/4n'.• 2.6. Integration of motion equations. A particle of mass m moves

due to the action of the force F. The initial conditions, that is, itsradius vector r (0) and velocity v (0) at the moment t = 0, are known.Find the position of the particle as a function of time if

(1) F = Fosin oot, r (0) = O,v (0) = 0;(2) F= -kv, r (0) =0, ~ (0) = vo;(3) F=-xr, r(O)=ro, v(O)=vo, with voUro.

Here F. is a constant vector, and 00, k, x are positive c0D8tants.

Cl-Gli31l

The Basic Equation of Dynamics

• 2.5. A thin uniform elastic cord of mass m and length lo (ina non-stretched state) ha.s a c?efficient of elasticity x. After havingthe ends of the cord splIced, It was placed on a smooth horizontalplane, shaped as a circle and set into rotation with the angular veloc­ity 00 about the vertica!) axis passing through the centre of the circle.Find the tension of tlie cord in this state.

Solution. Let us single out a small element of the cord of maaelim as shown in Fig-. 30a. This element moves along the circle due

Fig. 29Fig. 28

ItIIIIIIIIIIIIII

I!lfluicfll Atechantc.

with the horizontal. Find how the velocity v of the disc depends onthe angle q> between the veetor v and the z axis (Fig. 29) if at theinitial moment v = liD and q> = n/2.

Solutton. The acceleration of the disc along the plane is determinedby the pro~ection of the force of gravity on this plane F% = mg sin aand the friction force Ffr = kmg cos a. In our case k = tan a andtherefore

Ffr= Fs=mg sin a;

Let us find the projections of the acceleration on the d~ction of thetangent to the trajectory and on the z axis:

mWt=Fs cos q>-Ffr=mg sin a (cosq>-i),mws=Fx-Ffr cos q>=mg sin a (i-cos q».

It is .seen fr?m .this tha~ W t = -ws , which means that the velocity vand Its proJectlOn Vs differ only by a constant value C which doesnot change with time, Le.

v= -vs + C,

v.:h~re v% = v cos q>. The constant C is found from the initial con­dltlOn v = Vo, whence C = Vo' Finally we obtain

v = vo/(i + cos q».

In the course of time q> -+- 0 and v-+- v0/2.

The simultaneous solution of Eqs. (i), (2) and (3) yields

WIX= (2m,wo+(m.-ml) g)/(ml+",.).Whence it is seen that for a given Wo the sign of WIX depends on theratio of the masses ml and mi'

.2.4. A small disc moves along an inclined plane whose frictioncoefficient k = tan a, where Q is the angle which the plane form!

66TAe B.-Ie Efutlon of Dynamic, 87

Fig. 33

if~t__---

y

O'---------O::x

Fig. 32

Fig. 32 shows the plots of the velocity v and the' path covered,as fwictions of time t (in our case , = r).

3. In this case the panicle moves along the straight line coincidingwith the radius vector r. Choosing the x axis in this direction, we canimmediately write the fundamental equation of dynamics in termaof the projection on this axis:

-; +ooIX=O, (1)

where -; is the second derivative of the coordinate with respect totime, i.e. the projection of the acceleration vector, ro! = x/m. Eq. (1)is referred to as the equation of harmonic IJfbratlolll.

It can he shown mathematically that the general solution of thisequation takes the form

x (t) = A cos oot + B sin oot, (2)

where A and B are arbitrary constants. The restrictions imposedOn these constants are usually determined from the initial conditions.For instance, in our case at the moment t = 0

x.(O)=xo and ,Vx (0) = VOx, (3)

where x, and Vf)X are the projections of the ro and Vo vectors on the xaxis. Alter substituting Eq. (2) into Eq. (3) we get: A = zo, B == vttx!oo. All the rest is obvious.

• 2.7. A panicle of mass m moves in a certain plane due to theforce F whose magnitude is constant and whose direction rotates withthe constant angular velocity 00 in that plane. At the moment t = 0the velocity of the particle is equal to zero. Find the magnitude of thevelocity of the particle as a function of time and the distance that thepanicle covers between two consecutive stops.

Solutton. Let us fix the z, 11 coordinate system to the given plane(Fig. 33), taking the x axis in the direction along which the forcevector was oriented at the moment t = O. Then the fundamentalequation of dynamics expressed via the projections on the z and 11.

..-

. t

r(t)-r(O)={Fo/moo) J(1-cosoot)dt.o .

Integrating this expression and taking into account that r (0) = 0,we get

r (t) = (Fo/moot) (oot-sin oot).

Fig. 31 illustrates the plots Vx (t) and x (t), t~e time d~pendencesof projections of, the vec.tors.v and r .on t~e x aXIS chosen m the par­ticle motion direction, I.e. m t~e d!rection of the Fo vector.

2. In this case the acceleratiOn 1,Sdv/dt;::: - (kim) v.

To integrate this equation we must pass to the scalar form, that is,to the modulus of the vector v:

dvlv = - (kim) dt.

Integration of this equation with allowance ma~e for .the i~itialConditions yields: In (vlvl)) = - (kim) t. After takmg antl1oganthr..:.swe return to the vector form:

v= voe- ltt1m•

Integr~t~ng the l~s.t equation on~ more (and again taking into accountthe imtlal condItIons), we obtam

t

r= ~ vdt=(mvolk)(1-e-It'Jm).o

t

V(t)-v (O)= (Fo/m) ~ sin oot dt.o

Taking into account that v (0) = 0, we oMain after integrationv (t) =(F 01 moo) (1 - cos oot).

Now let us find the elementary d~splace~ent dr, .or the inc~eme~of the radius vector r of the particle durmg th~ mterva~ dt. dr ­= v (t) dt. The increment of the radius vector durmg the tIme from 0to t is equal to

Solution. 1. In accordance with the fund~mental equation ofdynamics the acceleration is

dv/dt = (FoIm) sin oot.

We obtain the elementary increment of the velocity. vector d.v duringthe time dt and then the increment of this vector durmg the tIme fromo to t:

69

(2)

FeormII

I

I

n

r

F

Fig. 34

v' = v - oor,

Th. Ballc Equation of DynamIcs

where v is the velocity of the satellite in the K frame (Fig. 34), andof the equation of motioD of the satellite in the K frame

mvA!r = ymMlr~, (3)

from which v is found. Solving simultaneously Eqs. (1), (2) and (3),we obtain

w' =(1-oor V r/yM)2 '(Mlr2•

~r--Specifically, w' = 0 when r = 'J' '(Mlm2 = 4.2.103 km. Sucha satellite is called stattonary: it is motionless relative to the Earth'ssurface.

• 2.10. A small sleeve of mass m slides freely along a smoothhori­zontal shaft which rotates with the constant angular velocity 00about a fixed vertical axis passing through one of the shaft's ends.Find the horizontal component of the force which the shaft exertson the sleeve when it is at the distance r from the axis. At the initialmoment the sleeve was next to the axis and possessed a negligiblevelocity.

Solutton. Let us examine the motion of the sleeve in a rotatingreference frame fixed to the shaft. In this reference frame the sleevemoves rectilinearly. This means that the sought force is balanced

around the Sun, find the acceleration w' of the satellite in the ref­erence frame fixed to the Earth.

Solutton. Suppose K is an inertial reference frame in which theEarth's rotation axis is motionless, and K' is a non-inertial referenceframe fixed to the Earth and rotating with the angular velocity 00with respect to the - K frame.

To derive the acceleration w' of the satellite in the K' frame wemust first of all depict all the forces acting on the satellite in thatreference frame: the gravity F, theCoriolis force Feor and the centri- ~ wrfugal force' Fe! (Fig. 34, the view \from the Earth's North Pole). u. u'

Now let us make use of Eq.(2.18), assuming Wo = 0 (in accor­dance with the conditions of theproblem). Since in the K' framethe satellite travels along a circle,Eq. (2.18) can be immediately writ­ten via projections on the tra­jectory's NOrmal n:

mw' = F - 2mv'00 - m002r, (1)

where F =' ymMIr2, ~and m andM are the" masses of the satelliteand the Earth respectively. Now we have only to find the velocity v'of the satellite in the K' frame. To do this, we shall make use of thekinematic relation (1.24) in scalar form

'j~\

il:l:r~;/ 'i!,;:~·i,' 0(

r' .~ ,*

;.

f,)

(3)

(1)

ClassIcal Mechanics

Wmax=Vw~+(vtIR)2, (2)

where v is the velocity of the automobile at the moment its accelera­tion reaches the maximum value. This velocity and the sought dis­tance s are interrelated by the following, formula:

Eliminating v and Wmax from Eqs. (1), (2) and (3), we obtain

s= (RI2) V (kg IW,;)2 -1.

It is not difficult to see that the solution is meaningful only if theradicand is positive, Le. if W,; < kg.

• 2.9. Non-inertial reference frames. A satellite m~V8ll in theEarth's equatorial plane along a circular orbit of radius r m the ~t­east direction. Disregarding the acceleration due to the Earth's motIOn

On the other hand,

ax811 takes the formm dvJdt = F cos oot, m d~,/dt = F sin oot.

Integrating these equations with respect to time with allowancemade for the initial condition v (0) = 0, we obtain

vx=(Flmoo) sin oot, vII = (Flmoo) (1 - cos oot).

The magnitude of the velocity vector is equal to

V=V~+v~=(2F1moo) sin (ootI2).

It is seen from this that the velocity v turns into zero after the timeinterval /}"t, which can be found from the relation 00 /),t12 = n. Conse­quently, the sought distance is

At

s= ) v dt=8Flm002•o

• 2.8. An automobile moves with the constant tangential acceler­ation W,; along the horizontal plane circumscribing a circle of ra~ius R.The coefficient of friction between the wheels of the automobile andthe surface fs equal to k. What distance .. will be covered by the auto­mobile without slipping in the case of zero initial velocity?

Solutton. As the velocity increases, so do both the normal andthe total acceleration of the automobile. There is no slip,ping as longas the total acceleration required is provided by the friction force.The maximum possible value of that force Fmax = kmg, where mis the mass of the automobile. Therefore, in accordance with thefundamental equation of dynamics, mow = F. the maximum valueof the total acceleration is '

WJII(I% = kg.

68

'10 Clcu'cal Mechardc,The Ballic Equation of Dynam'c, 7t

The right-hand side of this equation contains the projections of thecentrifugal force and gravity. Taking: into account' that Fe! == mro2r sin 8, we rewrite the foregoing expression as follows:

F", - sin 8 (cos 8- glro2r). (i)

From the equilibrium condition (F", = 0) we can find the two valuesof the angle 80 ensuring that equilibrium: sin 80 = 0 and cos 80 == glro2r. The first condition can be satisfied for any value of ro, Whilethe second one only if glro2r < 1. Thus, in the case of low ro valuesthere is only one equilibrium position, at the bottom point (80 = 0);but in the case of large ro values (ro > Vglr) another equilibriumposition, defined by the second condition, is possible.

A certain equilibrium position is stable provided the force F'"appearing on withdrawal of the sleeve from that position (in anydirection) is directed back, to the equilibrium position, that is. thesign of F", must be opposite to(that of the deflection ,i8 from theequilibrium angle 80,

At low deflections d8 from the 80 angle the appearing force {jF'"may be found as a differential of expression (i):

{jF", - [cos 80 (cos 80 -g/ro2r)-sin2 80] d8 .

At the bottnl equilibrium position (80 = 0){jF", ,.., (i-glro2r) d8. (2)

This equilibri~m positi~n is stable provided the expression put inparentheses is negative; i.e. when ro < V glr.

At the other equilibrium position (cos 80 = glrolr){jF", _-sin2 80 d8.

It is seen that this equilibrium position (if it exists) is always stable.Thus, as long as there is only the bottom equilibrium position

(with ro < V glr) , it is always stable. However, on the appearanceof the other equilibrium position (when ro > Vglr) the bottom. posi­tion becomes unstable (see Eq. (2», and the sleeve immediately passesfrom the lower to the upper position, which is always stable.Fer

(2)

mg~ 0

rig. 36

v' = ror.

Fig. 35

Substituting Eq. (2) into Eq. (i), we get

R = 2mro [mr].

• 2.11. 'l:he stability of motion. A wire ring of radius r rotateswith the constant angular velocity m about the vertical axis 00'

0'

out by the Coriolis force (Fig. 35):R= -Fcor=2m [cov']. (t)

Thus, the problem reduces to determining the velocity v' of the sleeverelative to the shaft. In accordance with Eq. (2.i9)

dv'ldt = Fe!lm = rotr.

Taking into account that dt = drll!, the last equation can be trans­formed to

I! dl! = rolr dr.

Integrating this equation with allowance made for the initial condi­tions (I! = 0, r = 0), we find I! = ror, or in a vector form

passin~ through its diameter. A small sleeve A can slide .along thering WIthout friction. Find the angle 8 (Fig. 36) correspondmg to thestable position of the sleeve.

Solutton. Let us examine the behaviour of the sleeve in a referenceframe fixed to the rotating ring. Its motion along the ring is char­acterized by the resultant force projection F", on the unit vector 'I'at the point A. It is seen from Fig. 36 that

F",=Fef cos 8-mg sin 8.

the sum of the corresponding values for the individual con­stituent parts (incidentally, in the case of momentum andangular momentum additivity holds true even in the pres­ence of interaction)..It is additivity that makes these threequantities extremely important.. Later on it became known that the laws of conservation

of energy, momentum and angular momentum intrinsicallyoriginate from the fundamental properties of time and space,uniformity and isotropy. By way of explanation, theenergy conservation law is associated with uniformity oftime, while the laws of conservation of momentum andangular momentum with uniformity and isotropy of spacerespectively. This implies that the conservation laws listedabove can be derived from Newton's second law supplement­ed with' the corresponding properties of time and space sym­metry. We shall not, however, discuss this problem inmore detail.

The lawlf of conSQl'vation of energy, momentum and angu­lar momentum fall ipto the category of the most fundamentalprinciples of physics, whose significance cannot be overes­timated. These laws have become even more significant sinceit was discovered that they go beyond the scope (If mechanicsand represent universal laws of nature. In any case, nophenomena have been observed so far which do not obeythese laws. They "work" reliably in all quarters: in the fieldof elementary particles, in outer space, in atomic physicsand in solid state physics. They are among the few mostgeneral laws underlying contemporary physics.

Having made possible a new approach to treating variousmechanical phenomena, the conservation laws turned intoa powerful and efficient' instrument of research used byphysicists. The importance of the conservation principlesas a research instrument is due to several reasons.

1. The conservation laws do not depend on either the pathsof particles or the nature of acting forces. Consequently, theyallow us to draw some general and essential conclusionsabout the properties of various mechanical processes withoutresorting to their detailed analysis by means of motionequations. For example, as soon as it turns out that a cer­tain process is in conflict with the conservation laws, onecan be sure that such a process is impossible and it is nQ

CHAPTER 3

ENERGY CONSERVATION LAW

§ 3.1. On Conservation Laws

Any body (or an assembly of bodies) represents, in fact,a system of mass points, or particles. If a system changes inthe course of time, it is said that its state varies. The stateof a system; is defined by specifying the concurrent coordi­nates and velocities of all constituent particles.

Experience shows that if the laws of forces acting on asystem's particles and the state of the system at a certaininitial moment are known, the motion equations can helppredict the subsequent behaviour of the system, Le. find itsstate at any moment of time. That is how, for example, theproblem of planetary motion in the solar system has beensolved.

However, an_analysis of a system's behaviour by the use ofthe motion equations requires so much effort (e.g. due tothe complexity of the system itself), that a comprehensivesolution seems to be practically impossible. Moreover, suchan approach is absolutely out of the question if the laws ofacting forces are not known. Besides, there are some prob­lems in which the accurate consideration of motion of indi­vidual particles is meaningless (e.g. gas).

Under these circumstances the following question natural­ly comes up: are there any general principles following fromNewton's laws that would help avoid 'these difficulties byopening up some new approaches to the solution of theproblem.

It appears that such principles exist. They are calledconservation laws.

As it was mentioned, the state of a system varies in thecourse of time as that system moves. However, there aresome quantities, state functions, which possess the veryimportant and remarkable property of retaining their valuesconstant with time. Among these constant quantities,energy, momentum and angular momentum play the mostsignificant role. These three quantities have the importantgeneral property of additivity: their value for a system com­posed of parts whose interaction is negligible is equal to

Energy Con,ervatlon Law 73

. § 3.2. Work and Power

75

(3.2)2 2

A= JFdr= JF,ds.1 1

Energy Conllrr1tsfton LaID

where a. is the angle between the vectors F and dr, ds == I dr I is the elementary path, and Fa is the projection ofthe vector F on the vector dr (Fig. 37).

Thus, the elementary work of the force F over the dis­placement dr is

«SA = Fdr = Fads. (3.1)

The quantity ~A is algebraic: depending on the angle betweenthe vectors F and dr, or on the sign of the projection Fa ofthe vector F on the vector dr, it can be either positiveor negative, or, in particular, equal to zero (when F ...L dr,i.e. Fa = 0).

Summing up (integrating) the expression (3.1) over allelementary sections of the path from point 1 to point 2, wefind the work of the force' F over the given path:

The expr~sion' (3.2) can be graphically illustrated. Letus plot F as a function of the particle position along thepath. Suppose, for example, that this plot has the shapeshown in Fig. 38. From this figure the elementary work-6Ais seell to be numerically equal to the area of the shadedstrip, and the work A over the path from point 1 to poi~t 2is equal to the area of the figure enclosed by the curved hne,ordinates 1 and 2, and the s axis. Here the area of the figurelying over the s axis is taken with the plus sign (it co~re­

sponds to positive work) while the area of the figure lymgunder the s axis is taken with the minus sign (it correspondsto negative work).

Let us consider a few examples involving calculationsof work.

The work of the elastic force F = -xr, where r is the radiusvector of the particle A relative to the point 0 (Fig. 39).Let us displace the particle A experiencing the action ofthat force along an arbitrary path from point 1 to point 2.We shall first find the elementary work performed by theforce F over the elementary displacement dr:

«SA = Fdr = ..;.-. xrdr.

Classtcal Mechantcs

Fig. 37

F. ,..;'\\$/ \

\\

\\

\\

\\

74

Work. Let a particle travel along a path 1-2 (Fig. 37)under the action of the force F. In the general case the

force F may vary during the2 motion, both in magnitude and

direction. Let us consider theelementary displacement dr, dur-.ing which the force F can beassumed constant.

The action of the force Fover the displacement dr ischaracterized by a quantityequal to the scalar product Fdr

and called the elementary work of the force F over the dis­placement dr. It can also be presented in another form:

Fdr = F cos a.ds = F,ds,

use trying to accomplish it.2. Since .the conservation laws do not depend on acting

forces, they may be employed even when the forces are notknown. In these cases the conservation laws are the onlyand indispensable instrument of research. This is the pres­ent trend in the physics of elementary particles.

3. Even when the forces are known precisely, the con­servation laws can help substantially to solve many problemsof motion of particles. Although all these problems can besolved with the use of motion equations (and the conserva­tion laws provide no additional information in this case), theutilization of the conservation laws very often allows thesolution to be obtained in the most straightforward andelegant fashion, obviating cumbersome and tedious calcula­tions. Therefore, whenever new problems are ventured, thefollowing order of priorities should be established: first,one after another conservation laws are applied and onlyhaving made sure that they are inadequate, the solution issought through the use of motion equations.

We shall begin examining the conservation laws with theenergy conservation law, having introduced the notion ofenergy via the notion of work.

2

71

mg1

z

k

Energ1l C01Ueruatlon LaUJ

(3.5)

The work of the uniform forreof gravity F = mg. Letus write this force as F = - mgk, where k is the unit vector ofthe vertical z axis whose positive direction is chosen upward(Fig. 40). The elementary work of gravity over the dis­placement dr is

6A = Fdr = - mgkdr.

The scalar.froduct kdr= (dr)k, where (dr)k is the projectionof dr on the unit vector k and is equal to dz, the z coordinateincr~ment. Therefore,. k dr = dzand

6A = - mgdz = - d (mgz).

The total work of this force per­formed over the whole path frompoint 1 to point 2 is

2

A= - Jd(mgz)=mg(zt-z'l,)~1

The forces considered are Fig. 40interesting in that the workperformed by them between points Ijand 2 does notliependon the shape of the path and is determined only by their ;positions (see Eqs. (3.3)-(3.5». However, this very significantpeculiarity of the forces considered is by no means a prop­erty of all forces. For example, the friction force does notpossess this property: the work performed by this forcedepends not only on the positions of the initial and finalpoints but also on the shape of the path connecting them.

As in the previous case, the scalar product r dr = r dr,so that

6A· = adrlr2 = - d (air).

The total work performed by this force over the whole pathfrom point 1 to point 2 is

2 .

A = - Jd (aIr) =a.1rc-a.lr'l,' (3.4)1

2

Classical Mechanics

Fig. 39Fig. 38

76

6A = F dr = (air) rdr.

The scalar product r dr = r (dr)r, where (dr)r is the projec­tion of dr on the vector r. This projection is equal to dr, theincrement of the magnitude ofthe vector r. Therefore r dr == rdr and

6A = - xr dr = - d (xr/2).

Now, to calculate the work performed by the given force overthe whole path, we should integrate the last expression be­tween point 1 and point 2:

2

A = - Jd (xr2/2) = xr~/2 - xr:/2. (3.3)1

The work of a gravitational (or Coulomb's) force. Leta stationary point mass (charge) be positioned at the point 0

of the vector r (Fig. 39). We shall find the work of the gra­vitational (Coulomb's) force performed during the dis­placement of the particle A along an arbitrary path frompoint 1 to point 2. The force acting on the particle A may berepresented as follows:

F = (alr)r,

h . _ {~Y mlm2, the gravitational interaction,were a - kQ1Q2, the Coulomb interaction.

Let us first calculate the elementary work performed by. this force over the displacement dr:

Cl.,,'eal Meeha1dc, Energy Conservatton Law 79

So far we have discussed the work performed by a singleforce. But if during its motion a particle experiences severalforces whose resultant is F = F1 + F, + ..., it can beeasily shown that the work performed by the resultant forc,e Fover a certain displacement is equal to the algebraic sumof the works performed by all the forces over the same 'dis­placement. In fact,

A= J(F1+FlI+···)dr== JFtdr+ JF2 dr+ ... ====A t +A2+... • (3.6)

Power. To characterize intensity of the work performed,the quantity called power is introduced. Power is defin,ed asthe work performed by a force per unit time. If the-force Fperforms the work F dr during the time interval dt, the powerdeveloped by that force at a given moment of time is equalto N = F drldt; taking into account that drldt == v, we obtain

(3.7)'

Thus, the power developed by the force F is equal to thescalar product of the force vector by the vector of velocitywith which the point moves under the action of the givenforce. Just like work, power is an algebraic quantity.

Knowing the power of the force F, we can also find thework which that forcA performs during the time interval t.Indeed, expressing the integrand in formula (3.2) as F dr === Fvdt . N dt, we get

t

A= JN d.t.o

As an example, see Problem 3.1.Finally, one very essential circumstance should be pointed

out. When dealing with work (or power), in each specific caseone should indicate precisely what force (or forces) performsthat work. Otherwise, misinterpretations are, as a rule,inevitable.

§ 3.3. Potential Field of Forces

A field of force is a region of space at whose each pointa particle experiences a force varying regularly from pointto point, e.g. the Earth's gravitational field, or the field oft~e resistance forces in a fluid stream. If the force at eachpoint of a field of forces does :r1Ot vary in the course of time,such a field is referred to a,s stationary. Obviously, a station­ary field of forces may turn into a non-stationary field ontransition from one reference frame to another. In a station­ary field of forces the force is determined only by the posi­tion of a particle.

Generally speaking, the work performed by the forces ofthe field during the displacement of a particle from point1 to point 2 depends on the path. However, there are somestationalt fields of forces in which that work does notdependon the path between points 1 and 2. This class of fields pos­sesses a n1pnber of \he most important properties in physics.Now we shall pro~eed to these properties.

The definition: a stationary field of forces in which the workperformed by these forces between any two points does notdepend on the shape of the path but only on the positions ofthese points, is referred to as potential, while the forcesthemselves are called conservative.

If this 'condition is not satisfied, the field of forces is notpotential and the forces of the field are called non-conserva­tive. Among them are, for example, friction forces (the workperformed by these forces depends on the path in the generalease).

An example of two stationary fields of forces, one of whichis potential and the other is not, is examined in Problem 3.2.

Let lis dexnonstrate that in a potential field the work per­formed by the field forces over any closed path is equal to zero.I~ fact: any closed path (Fig. 41) ma~ be arbitrarily subdi­v.Ided mto two parts: la~ and 2bl. Smce the field is poten­tIal, then by the.h~othesisAW= A ~~). On the other handit is obvious that A~~) = - AW. Therefore, '

A(B)+A(b) A(a) A(b) 012 21 = 12 - 12 = ,

which was to be proved.

81Energy C01/.servation Law

about by ()n~ motionless particle B. The elementary workperformed by tne brce (3.8) over the displacement dr isequal to 6A = F dr = f (r) er dr. Since er dr is equal to dr,the projection of the vector dr on the vector er or on thecorresponding radius vector r (Fig. 42), then 6A = / (r) dr.The work performed by this force over an arbitrary pathfrom point 1 to point 2 is

2

A12 = J/(r) dr.1

This expres~ion obvi~us]y depends only on the appearanceof the function f (r), I.e. on the type of interaction, and onthe values of r1 and r 2 , theinitial andl final, distances be­tween the'particles A and B. Itdoes not.....epend on ,the shapeof the path in any way. Thismeans that: the give'n field offorces is potential. -

Let us generalize the re­sult obtained to a stationaryfield of forces induced by a setof motionless particles exert- 'ing on the particle A the forces Fig. 42F1, F2' ••• , each of whichis central. Ih this case the work performed by the resultantforce during the displacemont of the particle A from onepoint to another is equal to the al~f)braic sum of the worksperformed by individual forces. But since the work per­formed by each of these forces does not depend on the shapeof the path, the work performed by the resultant force doesnot depend on it either.

Thus, any stationary field of central forces is indeed poten-tial. "

Potential energy of a particle in a field. Due to the factthat the work performed by potential field forces dependsonly on the initial and final positions of a particle we canintroduce the extremely important concept of potentialenergy.6-05311

Cltulleal AI,ell.llfe,

Fig. 41

80

Conversely, if the w(lrk~pel'formedby the field forces overany closed path is equal to zero, the work of the same forcesperformed between arbitrary points 1 and 2 does not dependon the path, Le. the field is potential. To prove this, let-ustake two arbitrary paths: la2 and lb2 (Fig. 41). W.e--can"'eonnect them to make closed path 1a2bl. The work performedover this closed path is equal to zero by the hypothesis,i.e. AW -t-A~~) = O. Hence, AW = ~ AW. But A~~) =.= ~AW and therefore

A~ai=AW.

Thus, when the work of the field forces perfomned overany closed path is equal to zero, we obtain the necessary

and sufficient condition for thework to be independent of the shapeof the path, a fact that can beregarded as a distinctive attributeof any potential field of forces.

The field of central forces. Anyfield of forces is brought about bythe action of definite bodies. Inthat field the particle A experi­

ences a force arising due to the interaction of this par­ticle with the given boJiies. Forces depending only onthe distance between interacting particles and directedalong the straight line connecting them are referred- t~. ascentral. An example of this kind of forces is provided bygravitational, Coulomb's and elastic forces.

The central force which the particle B exerts on the par­ticle A can be presented In general form as follows:

F = / (r) er , (3.8)

where / (r) is a function depending for the given type ofinteraction only on r, the distance between the particles;and er is a unit vector defining the direction of the radiusvector of the particle A with respect to the particle B(Fig. 42).

Let us demonstrate that any stationary field 0/ centralforces is potential.

To do this, we shall first finLthe work performed by thecentral forces in the case when the field of forces is brought

82 Classical Mechanics Energy Conservation Law

The expression on the right-hand side of this equation isthe diminution* of the potential energy, Le. the difference

• The variation of some quantitY.: X can be characterized eitherby its 'ncrement, or its diminution. The increment of the X quantityis the difference between its final (XI) and initial (Xl) values:

the increment !lX = XI - Xl'The diminution of the quantity X is taken to ~ the difference betweenits initial (Xl) and final (XI) values:

the diminution Xl - XI = -!lX,i.e. the diminution of the quantity X is equal to its increment inmagnitude but is opposite in sign.

The diminution and increment are algebraic quantities: if XI >> X17 then the increment is positive while the diminution is negative,and vice versa.

Suppose we displace a particle from various points P,of a potential field of forces to a fixed point O. Since thework performed by the field forces does not depend on theshape of the path, it is only the position of the point Pthat determines this work, provided the point 0 is fixed.This means that a given work is a certain function of theradius vector r of the point P.

Designating this function as U (r), we writeo

Apo = 1Fdr=U(r). (3.9)p

The function U (r) is referred to as the potential energy ofthe particle in a given field.

Now let us· find the work performed by the field forcesduring the displacement of the particle from point 1 .topoint 2 (Fig. 43). The work being independent of the shapeof the path, we can choose one passing through the point O.Then the wotk performed over the path 102 can be repre­sented in the form

A 12 = AJO +A 02 = AlO - A20

or, taking into account Eq. (3.9),

h - {-,\,mlm2 , gravitational interaction,were a - kqtq2' Coulomb's interaction;

(3) in the uniform field of gravity

U (z) = mgz. (3.13)

It should be pointed out once again that the potentialenergy U is a function determined with an accuracy of a

2

(3.12)

(3.11)

o

between the valu~ of the potential energy at the initialand final points of the path.

Thus, the work of the fit;ld forces performed over the path 1-2is equal to the decrease in the potential energy of the particlein a given field. (

Obviously, any potential energy value can be chosen inadvance and assigned to a particle positioned at the point 0of the field. Therefore, measuring the work makes it possibleto determine only the difference of the potential energies attwo points but not the ~bsolute val- 1ue of the) potential energy. How­ever, as soon as the potential energyat some point is specified, its valuesare uniquely determined at all otherpoints of the field by means of Eq.(3.10).

Eq. (3.10~ makes it possible to findthe expresston U(r) for any potential Fig. 43field of forces. It is spfficient to cal-culate the work perforpled by the field forces over any pathbetween two points and to represent it as the diminution ofa certain function which is the potential energy U (r).

This is exactly how the work was calCUlated in the casesof the fields of elastic and gravitational (Coulomb's) forces,as well as in the uniform field of gravity (see Eqs. (3.3)­(3.5». It is immediately seen from these formulae that thepotential energy of a particle in such fields of forces takesthe following form:

(1) in the field of elastic forces

U (r) = 'XrI2;

(2) in the field of a point mass (charge)

U (r) = aIr,

(3.10)2

A 12 = ) Fdr=U t -U2•

1

6*

Classical MechanicsEnergy Conservation Law 85

F s ds = -dU,

Recalling (see Eq. (3.1» that F dr = Fs ds, where ds == I dr I is the elementary path and Fs is the projection ofthe vector F on the displacement dr, we shall rewriteEq. (3.14) as

certain arbitrary addendum. This vagueness, however, isquite immaterial as all equations deal only with the differ­ence of the values of U at two positions of a particle. There­fore, the arbitrary addendum, being equal at all points ofthe field, gets eliminated. Accordingly, it is usually omittedas it is done in the three previous expressions.

There is another important point. Potential energy shouldnot be assigned to a particle but to a system consisting ofthis particle interacting with the bodies generating afield of force. For a given type of interaction the potentialenergy of interaction of a particle with given bodiesdepends only on the position of the particle relative tothese bodies.

Potential energy and force of a field. The interaction ofa particle with surrounding bodies can be described in twoways: by means of forces or through the use of the notion ofpotential energy. In classical mechanics both ways are ex­tensively used. The first approach, however, is more generalbecause of its applicability to forces in the case of whichthe potential energy is impossible to introduce (e.g. fric­tion forces). As to the second method, it can be utilizedonly in the case of conservative forces.

Our objective is to establish the relationship betweenpotential energy and the force of the field, or putting itmore precisely, to define the field, of forces F (r) from a givenpotential energy U (r) as a function of a position of a par­ticle in the field.

We have learned by now that the work performed by fieldforces during the displacement of a particle from one pointof l potential field to another may be described as thediminution of the potential energy of the particle, that is,Au = U1 - U 2 = -I1U. The same can be said about theelementary displacement dr as well: c5A = -dU, or

where -dU is the diminution of the potential energy in thedr displacement direction. Hence,

(3.15)F s = -iJU/as,

Le. the projection of the field force, the vector F, at a givenpoint in the direction of the displacement dr equals thederivative of the potential energy U with respect toa givendirection, taken with the opposite sign. The designation ofa partial derivative a/as emphasizes the fact of deriving withrespect to a definite direction. .

The displacement dr can be accomplished along any direc­tion and, specifically, along the x, y, z coordinate axes. Forexample, if the displacement dr is parallel to the x axis, itmay be described as dr = i dx, where i is the unit vector ofthe x axis and dx is the x coordinate increment. Then thework perfo~ed by the force F over the displacement drparallel to the x axis is

F rk = Fi dx = F x dx,

where Fx is the projection of the vector F on the unit vector i(but not on the dr displacement as in the case of Fs).

Substituting the last expression into Eq. (3.14), we get

Fx = -au/ax,

-where the partial derivative symbol implies that in theprocess of differentiation U (x, y, z) should be considered asa function of only one variable, x, while all other variabJesare assumed constant. It is obvious that the equations forthe F y and F z projections are. similar to that for Fx'

So, having reversed the sign of the partial derivatives ofthe function U with respect to x, y, z, we obtain the projec­tions Fx' F y and F z of the vector F on the unit vectors i, jand k. Hence, one can readily find the vector itself: F == Fxi + F yj + Fzk, or

F= _ ( aU i + aU j + aU k).ox ay azThe quantity in parentheses is referred to as the scalargradient of the function U and is denoted by grad U or V U.We shall use the second, more convenient, designation where

(3.14)Fdr = -dU.

86 ClaSBlcal MechanIcs Energy Conservatton Law 87

where m is the mass (or the charge) of the particle A, and Gis a certain vector depending both on the position of theparticle A and on the properties of surrounding bodies, thatis, the system B.

This opens up the possibility for another physical inter­pretation of interaction which is based on the concept of afield. In this case the system B is assumed to induce a spatialfield ch~racterized by the vector G (r). Expressed otherwi.se:the system B, the origin of a field, is assumed to prOVidesuch conditions (the vector G) at all points of space thata particle positioned at any point experiences the force(3.17). Moreover, the field (the vector G) is supposed to exist

Eq. (iU5) Fa> 0, Le. the vector F is oriented in the direc­tion of decreasing U values. As F is directed oppositely tothe vector V U, we may conclude that the gradient of U isa vector ortented along a normal to an equipotential surfacein the direction of increasing values of potential energy U.

Fig. 44 illustrating a two-dimensional case clarifies- whatwas said above. It shows a system of equipotentials (U1 <­< Ua< U 3 < U..), a gradient of the potential energ~ V Uand the corresponding vector of the force F at the pomt .4of the field. It pays to consider how u..these two vectors a.re directed, for ex- UZ~u.., t~ample, at the point B of the ~given "vufield. ~

In conclusion it should be pointed Aout that a gradient can he calculatednot only from the function U but fromany othlr scalar function of coordi- 0 8.::::nates. The 0 concept of a gradient perme-ates various divisitms of physics. Fig. 44

Concept of a field. Experience showsthat in the case of gravitational andelectrostatic interactions the force F that surrounding bodies(the system B) exert on a particle A is proportional to themass (or the charge) of that particle. In other words, theforce F may be represented as the product of two quan-

o tities:

(3.17)F=mG,

(3.16)

Le. the field force F is equal to the potential p-nergy gradient,taken with the minus sign, for a particle at a given point ofthe field. Put simply, the field force F Is equal to the anti­gradient of potential energy. The last equation permits thefield of forces F (r) to be derived from the function U (r).

Example. The potential energy of a particle in a certain fieldhas the following form:

(a) U (3:, y) = -axy, where C% is a constant;(b) U (r) = ar, where a is a cOnstant vector and r is the radius

vector of a point of the field.Let us find the field of force corresponding to each of these ,cases:

(a) F= - ( ~~ i+ :~ j) =C% (yi+xj);

(b) first, let us transform the function U to the following form:U = axX + al/Y + azz; then

( au. au. au k) . (. + .+' k)F=- --I+-J+- =- axl ayJ .az =-a.ax ay azThe meaning of a gradient becomes more obvious and

descriptive as soon as we in.troduce the concept of an equipo­tential surface at all of whose points the potential energy Uhas the same magnitude. It is clear that each value of Uhas a corresponding equipotential surface.

It follows from Eq. (3.15) that th~ projection of the vec­tor F on any direction tangential to the equipotential surfaceat a given point is equal to zero. This means tha~ the ve~tor

F is normal to the equipotential ~llrface at a given pomt.Next, let us consider the displacemeut iJs in the directionof decreasing U values; then aU < 0 and in accordance with

v ("nabla") signifies the symbolic vector or ,operator

• a +. a +k aV = I ax J BY 8Z'

Therefore V U may be formally regarded as the product ofa symbolic vector V by a scalar U. . '

Consequently, the relationship between the force of afield and the potential energy, expressed as a function ofcoordinates, can be written in the following compact form:

88 Classtcal Mechanics Energy Conservation Law 89

2. Instead of the three components of the vectorial func­tion G (r) it is simpler to specify the scalar function <P (r).

3. When,# field is produced by many sources, the potential<P is more easily calculated than the vector G: potentials arescalars, so that they can be summed up disregarding theforce directions. In fact, in accordance with Eqs. (3.18)

and' (3.19) G dr-: ~ (G, dr) = - ~ d<Pi =-d ~ <Pi = -dIp·

i.e.

Note that the potential <P, as well as the potential energy,can be determined only with an accuracy of some adden­dum,. whose magnitude is of no importance. Accordingly,it is usually omitted. .

Thus, a field can be described either in vector form G (r)or in scalar form <P (r). Both methods are adequate. How­ever, for all practical purposes the second method of de­scribing,a field (by means of the potential <p) turns out to befar more convenient in most cases. Here is why this is so.

1. If <P (r) is known, the potential energy U and the workA performed by the field forces can be immediately obtained:

where <Pi is the potential produced by the ith particle at agiven point of the field.

4. And finally, when the function <P (r) is known, one canreadily obtain the field G (r) as the antigradient of thepotential. <p:'

G = - V <p. (3.24)

This formula follows directly from Eq. (3.16).In conclusion let us examine an example involving the

determination of the field potential of centrifugal forcesof inertia.

Example. Let us find the field stren~h G and the potential lp /of centrifugal forces of inertia in a reference frame rotating wit1{a constant angular velocity about a stationary axis.

The field strength G= Fc/lm = CJ)lp, where p is"the radius vectorof a point of the field relative to the rotation axis.

(3.23)

(3.22)

<p (r) = ~ <P, (r),

irrespective of whether the particle A is~ actually in thefield or not·.

The vector G is referred to as the field strength.One of the most important properties of a field is that

the field induced by several sources is equal to the sum ofthe fields induced by all of them. Putting it more precisely,the strength G of the resultant field at an arbitrary point

G = ~ G" (3.18)

where Gi is the field strength of the ith source at the samepoint. This formula expresses the so-called principle ofsuperposition of fields.

Now let us direct our attention to the potential energy ofa particle. In accordance with Eq. (3.17) we can rewriteEq. (3.14) as mG dr = -dUo Dividing both sides by mand denoting Ulm = <P, we get .

G dr = -dip (3.19)

2fG dr = <P1 - <Ps· (3.20)1

The function <P (r) is called the field potential at the pointwhose radius vector is equal to r.

Eq. (3.20) allows the potential of any gravitational orelectrostatic field to be found. To do this, it is sufficient to

calculate the integral f G dr over an arbitrary path between

points 1 and 2 and then to present the expression obtainedas a diminution of a certain function, the potential <i> (r).For instance, the potentials of the gravitational field of apoint mass m and of the Coulomb field of a point charge qare determined, in accordance with Eq. (3.12), as follows:

<Pgr = - ymlr, <i>coul = kqlr. (3.21)

or

• While we confine ourselves to statics, the concept of a fieldmay be treated as a mere formality simplifying the description ofphenomena. However, when we pass to variable fields, the conceptof a field acquires a profound physical meaning: a field Is a physicalrealUy.

90 Cla.,rcal JI,chanfc, Energy Conservation Law 9f

Thus, the increment of the kinetic energy of a particle duringits elementary displacement is equal to

dT = 6A, (3.27)

and during the finite displacement from point 1 to point 2

(3.26)

(3.28)

(3.29)

IT=mv2/2·1

in pare;ltheses) referred to as kinetic energy:

IT2-T. =A12 , -I

I.e. the increment of the kinetic energy gained by a particleover a certain displacement is efJual to the algebraic sum ofthe wo~ks performed by all forces acting on the particle overthe sam," displacement. If Ai2 > 0, then T 2 > T I , I.e. thekinetic energy of a particle grows; but if Au < 0, thekinetic e~ergy deoreases.

Eq. (3.27) may be written in another form, having dividedits both sides by the corresponding time interval dt:

[ dT/dt=FV=N·1

This means that the derivative of the kinetic energy of aparticle with respect to time is equal to the power N devel­oped by the resultant force F acting on the particle.

Eqs. (3.28) and (3.29) hold true both in inertial and innon-inertial reference frames. I~ the latter frames, apartfrom the forces exerted on the considered particle by otherbodies (interaction forces), we must also take into accountinertial forces. That is why in these equations work (power)should be conceived as an algebraic sum of the works (powers)performed both by interaction forces and by inertial forces.

Total mechanical energy of a particle. In accordancewith Eq. ,(3.27) the increment of the kinetic energy of a par­ticle is equal to the elementary work performed by the resul­tant F of all forces acting 011 the particle. What kind offorces are they? If the particle is located in a potential field,this fIeld exerts a conservati ve force Fe on it. Besides, the

§ 3.4. Mechanical Energyof a Particle in a Field

Kinetic energy. Suppose a par-ticle of mass m moves under the Fig. 45action of a eertain force F,which in the general case maybe a resultant of several forces. Let us determine theelementary work performed by this force over an elementarydisplacement dr. Taking into account that F = m dv/dtanddr = v dt, we may write

6A = F dr = mv dv.

The scalar product v dv = v (dvh" where (dv)v is the projec­tion of the vector dv on the direction of the vector v. Thisprojection is equal to dv, the increment of the velocityvector magn~tude. Consequently, v dv = v dv and the ele­mentary work

fPcf= -oolpl/2. (3.25)

Fig. 45 shows fPd vs the distance p to the rotation axis. For the sakeof comparison also shown is the po-tential 'Pgr (p) of the gravitational rpfield pro<tuced by a point mass ol-o;;;;;:::----~~placed at the point p = O.

Now, utilizing Eq. (3.20), we integrate G over a path from point 1to point 2:

2 2 2

JG dr=ool Jpdr=ool Jp dp = 001 (PI-,pf)/2.1 t t

It is evident that this integral does not depend on the shape of thepath between points 1 and 2 and is defined by the positions of thegiven points. This means that the considered field of forces is poten­tial.

Comparing the result obtained with Eq. (3.20), we get

6A = mv dv = d (mv2/2).

It is seen from this that the work of the resultant force Fcontributes to an increase in a certain quantity (enclosed

92 Classtcal Mechanics Energy Conservation taw 93

or

Examfle. From. a cliff rising to the height h over a lake surfacea stone 0 mass m IS thrown with the velocity Vo' Find the work per­formed by air resistance forces if the stone falls on the lake surfacewith the velocity v.

When the motion of the body is considered in the Earth's gravita­tional field, the air resistance forces should be treated as outsideones, and in accordance with Eq. (3.32) the sought work Ares == Em - E1 = mvl/2 - (mvB/2 + mgh), or

Are. = m (v2-v8)/2-mgh.

The obtain~d value may turn out to be not only negative but positiveas well (thIS depends, for example, on the character of wind duringthe fall of the body).

Eq. (3.31) may be presented in another form if its bothsidas are divided by the corresponding time interval dt.Then

~

This implies that the derivative. of the total mechanicalenergy of a, particle"with respece to time is equal to thepower develbped by the resultant of all outside forces actingon the particle.-

We ha~e thus established that the total mechanical ,~nergyof a particle can change only due to outside forces. Hence,the ~aw ~f conservation of the total mechanical energy ofapartIcle m an external field follows directly:

when outside forces ar:e absent or such that the algebraic sumof powers developed by them during the time int"!rval consideredis equal to zero, the total mechanical energy of a particleremains constant during that interval. In other words,

E = T + U = const,

(3.33)dE/dt = Foutv:

(3.30)

particle may experience other forces of different origin.We shall call them outside forces Fout.

Thus, the resultant F of all forces acting on the particlemay be presented in the form F = Fe + Fout. The workperformed by all these forces results in an increment of thekinetic energy of the particle:

dT = 6Ac+6Aout •

In accordance with Eq. (3.14) the work performed by thefield forces is equal to the decrease in the potential energyof the particle, Le. t'lA c = -dUo Substituting this expres­sion into the previous one and transposing the term dUto the left-hand side, we obtain

dT+dU =d(T+U)=t'lAout •

It is seen from this that the work performed by the outsideforces results in an increment of the quantity T + U. Thisquantity, the sum of the kinetic and potential energies, isreferred to as the total mechanical energy of a particle ina field:

Note that the total mechanical energy E, just as the poten­tial energy, is defined with an accuracy up to an arbitraryconstant. .

So, the increment of the total mechanical energy acquiredby a particle over an elementary displacement is equal to

dE = 6Aout (3.31)

and over the finite displacement from point 1 to point 2

(3.32)

Le. the increment of the total mechanical energy acquired bya particle over a certain path is equal to the algebraic sum ofworks pp.rformed by all outside forces acting on the particleover the same path. When A out > 0, the total mechanicalenergy of a particle increases, and when A out < 0, it de­creases.

Imv2/2 +U (r) = const. I (3.34)

Even when written in such a simple form this conservationla,,: per~its some sig?ificant solutions to be obtained quiteeasIly wIthout resortmg to equations of motion associatedwith cumber~ome and tedious calculations. That is whythe conservatIOn laws prove to be a very efficient instrumentof research.

Classical Mechanics Energy Con,eruaUon LarD 95

~Fig. 46

The following example illustrates the capabilities andadvantages provided by the application of the conservationlaw (3.34).

Example. Suppose a particle moves in the unidime~sional poten­tial field U (x) shown in Fig. 46. I,n th~ ab&:nce of o~tslde forces thetotal mechanical energy of a particle m ,this field, I.e. E, does notvary during the motion, and we can easily solve, for example, thefollowing prol>lems. , .

1. To find v (x), the velocity of the particle as .a functIOn of.Itscoordinate, without solving the fundamental equ~tIOn of dynamlc~.

In accordance With Eq. (3.34), thisIi I can be done provided the specific

appearance of the potential 'curve1------------------ U (x) and the total energy value

E (the right-hand, side of thatequation) are precisely known.

2. To .establish the variationrange of the x coordinate of the par­ticle within which it can be locat­ed when the total energy value E

L..-.L-..L.--:.::-----:=---"::x is fixed. It is clear that the parti-D xDxt Xz IS cle cannot get in the region where

U > E since the potential energyU of the particle cannot exceed itstotal energy. 'Ij'rom this it imme-diately follows that when E = E1

(Fig. 4.6), the particle can move either in th~ region confined by ~he Xl

and x. coordinates (oscillation) or to the rIght of t~e Xa coordmate.The article cannot however pass from the first, r~g~on to the ~ondone Cor vice versa) due to the potential barrier ,dlvldlllg thes,e regions.Note that the particle moving in a confmed regIOn of a field IS referredto as being locked in a potential well (m our case, be.tween .Zl and z.).

The particle behaves differently when E ~ E. (F~g. 46). the ,w~?leregion lying to the right of Zo becomes access~ble to 1~. If at an mltlalmoment tbe particle was located at the pomt ~o, ~t travels to theright. The reader is advised to find ~ow the kllletic energy of theparticle varies depending on its coordlllate z.

§ 3.5. The Energy Conservation Lawfor a System

Until now we confined ourselves to treating the behaviourof a single particle in t.erms of its energy. Now we ~hall passover to a system of particles. That may be any obJect, gas,any device, the solar system, etc., . .

In the general case particles of ,a syst~m mteract both WIthone anot},ler and with other bodIes WhICh do not belong to

that system. The system of particle:! experiencing negligible(or no) action from external bodies is called closed (01' isolat­ed). The concept of a closed system i::; the natural general­ization of the concept of an isolated mass point and playsa significant part in physics.

Potential energy of 8 system. Let us consider a closedsystem whose particles interact only through central forces,Le. the forces depending (for a given type of interaction)only on a distance between the particles and directed alongthe straight line that connects them.

We shall show that in any reference frame the work per­formed by all these forces during the transition of the systemof particles frOm one position to another may be representedas a decrease of a certain function depending (for a giventype of interaction) only on the configuration of the system,that is, on the relative positions of its particles. We shallcall this function an internal potential energy of the system(in distinctilD to a potential energy characterizing theinteraction of a given system with other bodies).

First we shal) examine' a system consisting of two particles.Let us calculate the elementary work performed by the forceof interaction betweea the particles. Suppose that in anarbitrary reference frame the positions of the particles ata certain moment of time are defined by th,e radius vectorsrl and r 2 • If during the time dt the particles shift throughdrl and dr2 respectively, the work performed by the interac­tion forces F12 and F21 is equal to

6At. 2= FS2 drs +F2t dr2'Taking into account that in accordance with Newton's third)aw F21 = -F12 , the previous expression may be rewrittenin the following form:

6A1 , 2= F12 (drt - drz).Let us introduce the vector r12 = r1 - r 2 describing theposition of the first particle with respect to the second par­ticle. Then dr12 = dr1 - dr2, substituted into the expres­sion for the work, it yields

6At • 2 = Ft2 drt2'.F12 is a central force, and therefore the work performed

by this force is equal to the decrease in the potential energy

96 Cla"ical Mechanics Energy Conservation Law 97

(3.38)

lJ=~Un+Ula, (3.37)

internal potential energy of the nth part

The internal potential energy U of the system is a non­additive quantity, that is, in the general case it is not equalto the sum ~f the internal potential energies of the con­stituent parts. The potential energy of interaction U,'between the individual parts of the system should be al~~taken into account:

Ea~h slim ~n parenthe~es represents the potential energy Uiof mteractlOn of the lth particle with the remaining two.7-0539

where Un is theof the system.

It should be also borne in mind that the internal potentialenergy of a system, just as the potential energy of interac­tion for each pair of particles, is defined with an accurl;lCYup to an arbItrary constant which is quite insignificant,though.

In conclusion, we shall quote some useful formulae forcalculation, of the internal potential energy of a system.First of aIr, we shall demonstrate that that energy may berepresented I;\s

whe~e Ui ~s the' poten~ia.l energy of interactioJ;l of the ithpartIcle wlt.h a~l remammg particles of the system. Herethe ~ummatIOn IS performed over all particles of the system.

Fnst, let us make sure that this formula is valid for asystem of. three particles. It was shown earlier that theinternal potential energy of the given system is U = U12 ++ U13 + U23' This sum can be transformed as follows. Letus depict each term U ik in a symmetrical form: U ill == (Vik + U lli )/2, for it is obvious that U ill = Ulil' Then

1U ="2 (U 12 +U21 +U13 +U31 +U23 +U32).

Grouping together the terms with identical first subindexwe get •

1U ="2 [(U I2 +U13 )+ (U21 +U23 )+ (U31 +U32)].

of interaction of a given pair of particles, Le.

6A1,2= -dU1S•

Since the function Uu depends only on the distance rubetween the particles, the work 6A 1,2 obviously is not affectedby the choice of a reference frame. . .

Now let us direct our attention to a system compnsmgthree particles (in this case the. re~u]ts obtain~d can readilybe generalized for a system conslstmg of an arbItrary numberof particles). The elementary work performed by all int~r­action forces during elementary displacements of all partic­les may be presented as the sum of the elementary worksperformed by all three pairs of interactions, Le. BA == BA 1 2 + BA1 ,3 + BA 2,3' It was shown,however, thatfor each pair of interactions SA I ,II = -dUIII' and there­fore

6A= -d(U12+U13 +US3) = -dU,

where the function U is the internal potential energy of thegiven system of particles:

U =U12+U13+U~,"

Since each term of this sum depends on the distance between, th~ corr?sponding partic~es, the internal potential e.nergy l!of the gIven systeIQ obvIOusly depends on ~he relatl~e pOSI­tions of particles (at the same moment of time), or motherwords, on the configuration of the system.

Naturally, similar arguments hold true for a. systemcomprising any number of particles. Consequently, It can beclaimed that every configuration of an arbitrary system ofparticles is characterized by its internal potential energy '!'and the work performed by all internal central force~, WhLlethat configuration varies, is equal to the decrease of the mternalpotential energy of the system,i.e. .

6A = -dU; (3.35)

during the finite displacement of all particles of the systemA = U 1 - U 2 , (3.36)

where Uland U2 are the values of the potential energy ofthe system in the initial and final states.

98 Classical Mechanics Energy Conservation LaID 99

6A = ~ 6A , = ~ dT I = d 2J T I = dT,

(3.41 )

(3.42)

dT = 6A,

and in the case of a finite displacement

where T = ~ T i is the total kinetic energy of the system.Note .that. t~e kinetic energy of a system is an additive~ua.n~Ity: It IS equal to the sum of the kinetic energies of~ndividual .parts of a system irrespective of whether theymteract WIth one another or not.

So, the .increment of the kinetic energy of a system is equalt~ the work performed by all the forces acting on all the par­hcles of .that system. During an elementary displacement ofall partIcles

. ~q. (~.~) may: be represented in another form, havingdIvI~ed Its both &Ides by the corresponding time interval dt.Makmg allowance fQl" 6A t = Fivt dt, we obtain

·f dT/dt= ~ F,v" I (3.43)

Le. the time de.rivative of the kinetic energy of a system is equalt~ the cumulatwe power of all the forces acting on all the ar-t~cles of the system. p. Eq.s. (3.41)-(3.43) hold true both in inertial and in non~mert.Ial r.eference frames. It should be recognized that in~on-mertial frames the work performed by interaction forcesIS to b~ su~plemented by that of inertial forces. . .

ClassIfication of forces. The particles of the consideredsyste~ are kn?wn to ~e able to interact both with one anotherand WIth .bodies ?utside the given system. Accordingly, theforces of mteractlOn between the particles of the systemreferred to as internal while the forces caused by th t ?reof Qther b?dies. outside the given system are called :x~~:d~In a. non~mertIal reference frame the latter forces includealso mertIal ones.

• Furthermore: all forces are subdivided into potential~nd non-po~ent~alones. The forces are referred to as potentialIf, for a gIven type of interaction, they depend only on7*

(3.39)

(3.40)

When the masses (charges) are continuously distributedthroughout the system, summation is reduced to integration:

Consequently, the last expression can be rewritten as

1 1 ~U =-2 (U. +U2 +U3) =2' k.! Uft ;>.

(-1

identically with Eq. (3.38).The result obtained can be evidently applied to an arbi­

trary system since these arguments by no means depend onthe number of particles constituting the system.

Making use of the concept of potential, Eq. (3.38) can betransformed in the case of a system with the gravitationalor Coulomb iI}teraction between particles. Replacing thepotential energy of the ith particle in Eq. (3.38) by Ui == miqJlt where mi is the mass (charge) of the ith particleand qJi is the potential produced by all remaining particlesof the system at the site of the ith particle, we obtain

where p is the volume density of the mass (charge), and dV isthe volume element. Here integration is performed over thewhole volume occupied by the masses (charges).

The application of the last formula may be illustrated byProblem 3.11 in which the internal potential energy of grav­itational 'interaction of masses distributed over the surfaceand volume of a sphere is calculated.

Kinetic energy of a system. Let us examine an arbitrarysystem of particles in a certain reference frame. Suppose theith particle of the system has the kinetic energy T i at agiven moment. In accordance with Eq. (3.26) the incrementof the kinetic energy of each particle is equal to the workperformed by all forces acting on this particle: dT i = 6A t •Let us find the elementary work performed by all forcesacting on all particles of the system:

100 Classical MechanicsEnergy Conservation Law 101

. Energy conserva~ion. law. It was shown above that theIncrement of the kInetIC energy of a system is equal to thework perform?d .by all forces acting on all particles of thesystem. ClassIfyIng these forces into external and internal'ones, ~n~ th.e internal forces, in their turn, into potentialand dISSIpatIve ones, we write the foregoing statement inthe following form:

dT = 6Aext +6A int = 6Aext +6Af~~ +6Af~:.

:rhen we take .into account that the work performed byInternal potentIal forces is equal to the decrease of theinternal potential energy of the system, Le. 6A:Po~ = -dUoThe foregoing expression then takes the form tn

dT +dU = 6Aext +4At~:. (3.46)

Let us iptroduce the concept of the total mechanical energyo~ a s!,steJ1f, or, briefly, mechanical energy, as the sum of thekInetIc and, potential energies of the system:

~

Obv~ously, the energy E depends on the velocities of thepartIcles of the. system, the type of interaction between them~nd the configuration of the system. Besides, the energy E,Just as the p?tentIal energy U, is defined with an accuracy~p to an arbItrary constant, and is a non-additive quantity,I.e. the energy E of.a syst~m .is ~ot equal in the general caseto. the sum of energIes of Its IndIvidual parts. In accordancewIth Eq. (3.37).

E = ~ En +Uia , (3.48)

wher.e En is the ~echanical e~ergy of the nth part of the~yst.em. a~d U ia IS the potentIal energy of interactions ofItS IndIVIdual parts.

Let us return to Eq. (3.46). We shall rewrite" it with anallowance made for Eq. (3.47) in the following form:

dE = 6Aext -+ 6At~:. (3.49)

The last expression is valid for an infInitesimal variationof the configuration of the system. In the case of a finite

the configuration of a mechanical system. The work per­formed by these forces was shown to be equal to the decreaseof the potential energy of the system.

To non-potential forces we refer the so-called dissipativeforces, the friction and resistance forces. The significantfeature of these forces is that the total work performed byinternal dissipative forces of the considered system is nega­tive in any reference frame. Let us demonstrate that.

Any dissipative force may be represented in the formF = -k (v) v, (3.44)

where v is the velocity of a given body relative to anotherbody (or medium) with which it interacts and k (v) is thepositive coefficient depending in the general case on thevelocity v. The force F is always directed oppositely to thevector v. Depending on the choice of a reference frame thework performed by that force can be either positive or nega­tive. However, the total work performed by all internaldissipative forces is always a negative quantity.

To prove this, we should point out first of all that theinternal dissipative forces in a given system appear inpairs. In accordance with Newton's third law the forcesof each pair are equal in magnitude and opposite in direc­tion. Let us find the elementary work performed by an arbit­rary pair of dissipative forces of interaction between bodies1 and 2 in the reference 'frame where the velocities of thesehodies at a given moment are equal to vl'and V2:

6Adis = Fl2v1 dt +F2lV2dt.

Making allowance for F 21 = -F12, Vl -V2 = v (the veloc­ity of bodYl 1 relative to body 2) and F12 = -k (v) v, wemay transform the expression for work as follows:

6Adis = Fl2 (vl-v2)dt = - k(v) vv dt = - k (v) v2 dt.

It is seen from this that the work performed by an arbitrarypair of internal dissipative forces of interaction is alwa.ysnegative, and hence the total work performed by all paIrsof internal dissipative forces is always negative. Consequently

I At~: < o. I (3.45)

IE=T+U./ (3.47)

102 Classical Mechanics Energy Conservation Law 103

• The solar system can be regarded as a closed conservative systemwith an adequate degree of accuracy.

Such a system is called conservative*. Note that during themotion of a closed conservative system it is the total me­chanical energy that remains constant while the kineticand potential energies vary in the general case. Thesevariations, however, always happen so that the increase ofone of them is exactly equal to the decrease in the other,Le. /iT = -/iU. It is obvious that this is valid only ininertial reference frames.

Furthermore, it follows from Eq. (3.50) that when a closedsystem is non-conservative, Le. when there are dissipativeforces in it, the mechanical energy of such a system decreases,in accordance with Eq. (3.45):

Ea- E1 = A1~i < O. (3.54)

It may be stated that the decrease-of the mechanical energyis caused by the work performed agai~st the dissipative forcesacting in.the system. Yet this explanation is formal since itdoes not reveal the physical nature of dissipative forces.

A more profound .examination of the problem- has led' tothe fundamental C()nclusion about the existence of theuniversal law of energy conservation in nature:

energy is never generated or eliminated, it may only passfrom one form into another or b" exchanged between individualparts of matter. The concept of energy had to be expandedby introducing some new forms (in addition to mechanical):energy of an electromagnetic field, chemical and nuclearenergies, etc.

The universal law of energy conservation thus encompas­ses even those phenomena for which Newton's laws are notvalid. Therefore, it cannot be derived from these laws, butshould be treated as an independent law, one of the mostextensive generalizations of experimental data.

Returning to Eq. (3.54), we can state the following: everydecrease in mechanical energy of a closed system alwaysimplies the appearance of the equivalent amount of energyof different kinds which are not associated with visiblemotion. In this respect Eqs. (3.49)-(3.51) can be consideredas a more general formulation of the energy conservation

(3.53)

(3.52)

(3.51)

(3.50)

dEldt' Ne:l:t+Nt~:,

IE=T+u=const.!

Thus, we have arrived at a significant conclusion: themechanical energy of a system may vary both due to externalforces and) due to internal dissipative forces (or, more pre­cisely, due to the algebraic sum of the works performed by allthese forces). From this another important conclusion followsdirectly, the law of conservation of mechanical energy:

in an inertial reference frame the mechanical energy of aclosed system of particles, in which there are no dissipativeforces, remains constant in the process of motion, Le.

Le. the time derivative of the mechanical energy of a systemis equal to the algebraic sum of powers developed by all externalforces and all internal dissipative forces.

Eqs. (3.49)-(3.51) are valid both in inertial and in nOn­inertial reference frames. It should be borne in mind that ina non-inertial reference frame one has also to take intoaccount the work (power) of inertial forces acting as externalforces, Le. A ext has to be treated as an algebraic sum of theworks performed by the external interaction forces A~~t

and the work performed by the inertial forces A tn • Toemphasize this fact, we rewrite Eq. (3.50) in the form

change

IE2-El=Ae:l:t+Af~:, ILe. the increment of the mechanical energy of a system is equalto the algebraic sum of the works performed by all externalforces and all internal dissipative forces.

Eq. (3.49) may be presented in another forin, havingdivided both its sides by the ~orrespondingtime interval dt.Then

....

1M Classical Mechanics Energy Conservation Law 105

Solution. Let us find the work performed by the force of each fieldover the path from a certain point 1 (Xl' YI) to a certain point 2 (x2, Y2):

In the first case the integral depends on the type of the y (x) func­tion, that is, on the shape of the path. c.nsequently, the first fieldof force is not potential. In the second clse both integrals do not

P{X,g)!J

Fig. 48

jo~----~---l • M/.x,Oj .x

Fig. 47

x.

(1) llA=Fdr=ayidr=aydx; A=a Jydx;

x,(2) llA =F dr= (axi+ byj) dr= ax dx+by dYi

I<' XI Y.

A=a Jxdx+b Jy dy.x, y,

depend on the shaye of the yath: they are defined only by the coordi­nates of the initia and fina points of the path; therefore, the secondfield of force is potential.

• 3.3. In a certain potential field a particle experiences the force

F = a (yi + xj),

where a is a constant, and i and j are the unit vectors of the X and yaxes respectively. Find the potential energy U (x, y) of the particlein this field.

Solution. Let us calculate the work performed by the force Fover the distance from the point 0 (Fig. 48) to an arbitrary pointP (x, y). Taking advantage of this work being independent of theshape of the path, we choose one passing through the points OMPand consisting of two rectilinear sections. Then

M P

Aop = JFdr+ JF dr.o M

The first integral is equal to zero since at all points of the OM sectiony == 0 and F .1 dr. Along the section MP x = const, F dr = Fj dy =

• 3.1. A stone of massmisthrown from the Earth's surface at thea~gle (X to the horizontal with the initial velocity Vo' Ignoring theaIr drag, find the power developed by the gravity force t seconds afterthe begi~ning of the motion, and also the work performed by thatforce durmg the first t seconds of the motion.

Solutio~. T~e velocity of the stone t seconds after the beginningof the motion IS v = Vo + gt. The power developed by the gravityforce mg at that moment is

N=mgv= m (gvO+g2t).

In our case gvo = gvo cos (rr./2 + (X) = -gvo sin (x, therefore,

N = mg (gt - vo sin (X).

It is seen from this that if t < to = vo sin (X/g, then N < 0 while ift > to, then N > O.

The work performed by the gravity force during the firilt tsecondsis

Problems to Chapter 3

t

A= JNdt=mg(gt2/2--vosin (X·t).

oThe N (t) and ,A (t) plots are shown in Fig. 47.

• 3.2. There are two stationary fields of force:(1) F = ayi;(2) F = axi + byj,

where i, j are the unit vectors of the x and y axes, and a and b areconstants. Are these fields of force 'potential?

law in which the cause of the mechanical energy variationfor a non-closed system is indicated.

Specifically, the mechanical energy of a non-closed sys­tem may remain constant, but this happens only in thosecases when, in accordance with Eq. (3.50), the decreasein the energy due to the work performed to overcome inter-

. nal dissipative forces is equalized by the inflow vf energyat the expense of the work performed by external forces.

Finally, we should point out that most problems usuallyrequire the energy conservation law to be applied togetherwith the momentum conservation law, or the angular mo­mentum conservation law; sometimes all three laws areused simultaneously. The way this is done will be illustratedin the next two chapters.

106 Classical Mechanics Energy Conservation Law 107

= F" dy = ax dy and therefore,p

Aop=O+ax) dy=axy.M

In accordance with Eq. (3.10) this work must be equal to the decreasein the p.otential energy, i.e. Aop = Uo - Up. Assuming Uo = 0,we obtam Up = -A OP, or

U (x, y) = -axy.

Another way of finding U (x, y) is to resort to the total differentialof that function:

dU = ('Ufo:.) dx + (au/ay) dy.

Taking into account that au/ax = -Fx = -ay and au/ay = -F == -ax, we get "

dU = -a (y dx + x dy) = d (-axy).

Whence U (x, y) = -axy.• 3.4. A ban of mass m is suspended on a weightless elastic thread

whose stiffness factor is equal to x. Then the ban is lifted so that thethread is not stretched, and let fan with the zero initial velocity.Find the maximum stretch Xm of the thread in the process of the ban'smotion.

Solution. Let us consider several solution methods based on theenergy conservation law.

1. According toEq. (3.28) the increment of the kinetic energy ofthe ban is equal to the algebraic sum of the works performed by allforces acting on it. In our case those are the gravity force mg andthe elastic force Fel = xx of the thread (Fig. 49a). In the initialand final positions of the ban its kinetic energy is equal to zero forwhen the maximum stretch occurs the ban stops moving. Therefore,in accordance with Eq. (3.28) the sum of the works A gr + Ael = 0,or

X m

mgxm +) (- xx) dx = mgxm - xx~/2= O., 0

Whence XnL = 2mg/x. .2. The ban may be considered in the Earth's gravity field. This

approach requires the total mechanical energy of the ban in the Earth'sgravity field to be analysed.. In accordance with Eq. (3.32) the incre­ment of this energy is equal to the work performed by the externalforces. In this case it is the elastic force that should be treated as anexternal force while the increment of the total mechanical energy ofthe- ban is equal to that of only its potential energy in the Earth's

gravity field. Therefore,Xm

O-mgxm = ) (-xx) dx= -xx~/2.o

Whence we get the same result for xm•N?te that we could proceed in the opposite way, examining the

ban In the field of the elastic force and treating the gravity forceas an external one. It pays to make sure that the result obtained inthat approach is the same.

3: ;And fiJ.lany, we can consider the ban in the field generated bythe ]omt actIOn of both the gravity force and the elastic force. Thenexternal forces are absent, and the total mechanical energy of the

~ U Uti /1 g~

I I F

o~I II

FelI I,." I

:;~~mg' -- Iz,.,r mI

.... I.... I mg

~ Ugr~"""" I(a) fb) '"

0/,

Fig. 49 Fig. 50

ban in such a field remains constant in the process of motion, i.e.I1E = I1T + I1U = O. During the transition of the ball from theinitial to the final (lower) position I1T = 0, and therefore I1U == I1Ugr + I1Uel = 0, or

- mgxm+xx~/2 = O.Again we get the same result.

Fig. 49b shows the plots Uf!7'. (x) and Uel (x), whose origins areassumed to be located at the point x = 0 (which is not mandatoryof course). The same figure illustrates the plot of the total potentiaienergy U (x) = Ugr + Uel. For a given choice of the potential energyreference value tue total mechanical energy of the ball E = O.

• 3.5. A body of mass m ascends from the Earth's surface withzero initial v~locitr due to t~e action of the two forces (Fig. 50): theforce F. ,:arymg WIth the heIght y as F = -2mg (1- oy), where aIS a pOSItIve constant, and the gravity mg. Find the work performed~y the force F over the .first half of· the ascellt and the correspondingIncrement of the potentIal energy of the body ill the Earth's gravitylield, which is assumed to he uniform.

Solution. First, let us find the total height of ascent. At the bel"in­lling and the end of the path the velocity of the body Is equal to zero,

108 Classical MechanicsEnergy Conservation Law 109

u'

the disc has the velocity VI whose direction forms the angle al withthe x axis (see Fig. 51b, top view). Find the motion.direction of thedisc after it reaches the top, Le. lind the. angle a 2 •

Solution. First of all we shall note that this problem cannot besolved by the use of the fundamental equation of dYIl.amics sinc~ theforce F acting on the disc in the regi,on. Xl < X <; X 2 IS not specIfie.d.All we know about this force is that It IS perpendIcular to the !J aXIS.

We shall make use of the energy conservation law: mvf = mv~ ++ 2mgh, whence

vi = vf -2gh. (1)

This expression can be rewritten as follows:

vix+V~y = vfx+Vfy - 2gh.

Since the fOJ,1le of the field is perpendicular to the y axis, it does notaffect the vy projection of the velocity; hence, v2y = Vly' Therefore,the previous expression may be reduced to v~x = v~x - 2gh, or

Vs cos as = y vf cos' al - 2gh, (2)

where Vs is defined from Eq. (1). As a result(4'

'. cos a, = y (vf cosS al - 2gh)/(vf - 2gh) .

Note that .this expression holds if the radicand in Eq. (2) is notnegative, i.e.' when VI OOS al > y 2gh. Otherwise the d!sc can~otovercome the hill that is it is "reflected" from the potentIal barrIer.

• 3.7. A pla~e spirai made of stiff smooth wire is rotated witha constant angular velocity 00 in a horizontal plane about the lixedvertical axis 0 (Fig. 52). A small sleeve!vI slides along that spiral withoutfriction. Find its velocity V' relative tothe spiral as a function of the distancep from the rotation axis 0 if the initialvelocity of the sleeve is equal to vo.

Sollltion. It is advisabl~ to solve thisproblem in a reference frame lixed to thespiral. We know that the increment ofthe kinetic energy of the sleeve must be Fig 52equal to the algebraic sum of the works .performed by all the forces acting on it.It can easily be reasoned that of all forces work is performed onlyby the centrifugal force of [inertia. The remaining forces, gravity,the force of reaction of the spiral, the Corioiis force, are perpendic­ular to the v' velocity of the sleeve and do not perform any work.

In accordance with Eq. (3.28),

m(v'S-vo')/2= Jmoo'pdr,

where m is the mass of the sleeve and dr is its elementary displace­ment relative to the spiral. Since p dr = p (dr)p = pdp, the integral

:r

.x

(a)

l.

Fig. 51

and therefore the increment of the kinetic energy of the body is alsoequal to zero. On the other hand, in accordance with Eq. (3.28) flTis tJqual to the algebraic sum of the works A 'performed by all theforces, i.e. by the force F and gravity, over this]path. However, since

h h

A = J(F + mg) dr = J(Fy - mg) dy =o 0

h

=mg J(1-2ay)dy=mgh(1-ah)=0.

o

flT = 0, then A = O. Taking into account that the upward directionis assumed to coincide with the positive direction of the y axis, wecan write

Whence h = 1/a.The work performed by the force F over the first half of the ascent is

h/2 h/2

AF = JFy dy=2mg J (1-ay)dy=3mg/4a.

o 0

The corresponding increment of the potential energy is

flU = mgh/2 = mg/2a.

• 3.6. A disc slides without friction up a hill of height h whoseprofile depends only on the x coordinate (Fig. 51a). At the bottom

HO Classical Mechanics Energy Conservation Law Hf

(3)

(6)

and (6):

tion with respect to x is to be carried out from a - r to a + r. Asa result.

\Pinside= -'1m /a, (4)

that is, the potential inside the layer does not depend on the fositionof the point P, and consequently it will be the same at al pointsinside the layer.. In accordance with Eq. (3.24) the field strength at the point PIS equal to

Gr= _ alp = { - ym/rz outside the layer,

ar 0 inside the layer.

The plots !P (r) and G (r) for a thin spherical layer are illustratedin Fig. 53b.

Now let us generalize the results obtained to a uniform sphere ofmass M and radius R. If the point P lies outside the sphere (r > R),then from Eq. (3) it immediately follows that

!Potltside= -yM/r. (5)

But if the point P lies inside 'the sphere (r < 11), the potential !Pat that point pat be represented as a sum:

!P = !PI + !PI'where !PI is the ,potential tH a sphere having the radius r, and !pz isthe potential of the layer of radii rand R. In accordance with Eq. (5),

M (r/R)3 M!PI = -'I r = -Y/i3 r Z•

The potential !pz produced by the layer is the same at all points insideit. The potential !pz is easiest to calculate for the point positioned atthe layer's centre:

R

!pz= -y r~=_~ 1'M (RI-rl)J r 2 R3 'r

where dM = (3M/R3) r2 dr is the mass of a thin layer botween theradii rand r + dr. Thus,

!Pinside=!PI +!pz= -('1M/2R) (3-rl /RI).

The field strength at the point P follows from Eqs. (5)

Gr

= _~={ -'yM/rl with r > R,ar -1'Mr1R3 with r < R.

The plots !P (r) and G (r) for a uniform sphere of radiuIJ R areshown in Fig. 54.

• 3.9. Demonstrate that the kinetic energy Tz which a body re~quires to escape the Earth's gravitational pull is twice as high as theenergy T1 required to launch that body into a circular orbit around

L...I~+--+- "",p Ol----+=----::~

x dx = ar sin II dll. (2)

Using Eq. (2) we reduce Eq. (1) to the form dIp = - (1'm/2ar) dx;then we integrate this equation over the whole layer. Then

r+a

!Poutside= -(1'm/2ar) ~ dx= -ym/r.r-a

Fig. 53

mass of this band is equal to dm and all its points are l?cated at thedistance x from the point P, then d!p= -1' dm/x. Allowmg for dm == (m/2) sin II dll, we obtain

dIp = - (1'm/2x) sin II dll. (it

Next from the cosine theorem (for the triangle OAP) it follows thatx2 ='aZ + rZ - 2ar cos ll. Taking the differential of this expression,we get

proves to be equal to moozpz/2. Hence, the sought velocity

v' = V v,?+oozpz.

• 3.8. Find the strength and the potential of the gravitationalfield generated by a uniform sphere of mass M and radius R as a func­tion of the distance r from its centre.

Solution. First of all we shall demonstrats that the potentialproduced by a thin uniform spherical layer of substance outside thelayer is such as if the whole mass of the layer were concentrated atits centre, while the potential within the layer is the same at allits points. "

Suppose the thin spherical layer has the mass m and radius a.Let us start with calculating the potential dIp at the point P (r > a)forming the elementary band dS of the given layer (Fig. 53a). If the

Thus, the potential at the point P outside the thin uniform sphericallayer is indeed such as if the whole mass of that layer were concen-trated at its centre. , .

When the point P is located inside the layer (r < a), the foregomgcalculations remain valid till the integration. In this case the integra-

it2 Classical MechanicsEnergy Conservation Law it3

the Earth (close to its surface). The air drag and the Earth's rotationare to be ignored.

Solution. Let us find the velocity VI of a body travelling alonga circular orbit. In accordance with the fundamental equation of dy­namics

mvf/R=mg,

where m is the mass of the bQdy, and R is the orbit radius equal approx­imately to the Earth's radius.

ZJ/. Hence, _R vI = Y gR = 7.9 km/s.

This is the so-called first cosmicvelocity.

jC----!!.!.------~r.,.,. To overcome the Earth's gravi-tationalpull, the body has to reachthe second cosmic velocity v2. Itsmagnitude can be found from theenergy conservation law: the kinet­ic energy of the body close to theEarth's surface must be eqU:al~o

Fig. 54 the height of the potential barrierthat the body must overcome.

The height of the barrier is equal to the increment of the potentialenergy of the body between the points r = Rand r = 00. Thus,

mv:/2=ymM/R,

where M is the Earth's mass. Hence,

V'. = Y2yM/R = y2gR = 11 km/s.

Consequently, V2 = y'2 VI and T 2 = 2T1•• 3.10. Three identical charged particles, each possessing the

mass m and charge +q, are placed at thecorpers of an equilateraltriangle with the side roo Then the particles are simultaneously setfree and start fiying apart symmetrically due to Coulomb's repulsionforces. Find:

(1) the velocity of each particle as a function of the distance rbetween them;

(2) the work .AI performed by Coulomb's forces acting on eachparticle until the particles fiy from one another to a very large distance.

Solution. 1. Since the given system is closed, the increment ofits kinetic energy is equal to the decrease in the potential energy, I.e.

3mvl /2= 2kq2/ro -3kql/r.Hence,

V= y2kql (r-ro)/mrro'

It is seen that as r ->- 00 the velocity of each particle approaches thelimiting value vmaz = y 2kq2/mro'

. 2. The work p~rformed by .the .interaction forces during the varia­tIOn of the system s configuratIOn IS equal to the decrease in the poten­tial energy of the system:

A=UI-U2=3kq2/ro,

where allowance is made for the fact that at the final position U2 == O. Hence, the .sought work is

A1 =A/3=kq2/ro ' (1)

Notl'. Here attention should be drawn to a frequent mistake madewhen solving this kind of problem. In particular, one may oftenhear the following .ar~ment: at the initial. position the potentialenergy of each particle IS equal to 2kq2/ro while at the final positionto zero. Hence, the sought work Al = 2kq2/ro . This result is twiceas great as the one of Eq. (1). Why is tilis so?

. The mistake is made because the field in which each particletravels is non-stationary and, consequently, non-potential (sincethe two other particles also displace relative to each other), and there­fore the work in such a field cannot be represented as a decrease inthe potential energy of the particle.

• 3.11. ifiaking use ofthe results obtained in solving Problem 3.8,find the internal potential energy of gravitational interaction ofmasses distributed unifolmly: . .

(1) over the surface of a sphere;(2) throughout the volume of a sphere.

The mass of the sphere is equal to M and its radius to R.Solution. 1. Since the potential at each point of a spherical surface

is cp = -yM/R, we obtain, in accordance with Eq. (3.40):

U = (cp/2) S dm = -yMS/2R.

2. In this case the potential inside the sphere depends only on r(see Problem 3.8):

cp =- (3yM/2R) (1 - r2/3R2).

Substituting this expression into Eq. (3.40) and integrating, weget

R

u=J.. r cp dm= _1.. yM22 J 5 R '

r-O

where dm is the mass of an elementary spherical layer confined betweenthe radii rand r + dr; dm = (3M/R3) r2 dr.

8-0539

CHAPTER 4 The Law 01 Conservation 01 Momentum it5

THE LAW OF CONSERVATION OF MOMENTUM

§ 4.1. Momentum. The Law of Its Conservation

The momentum· of a particle. Practical t:nowledge andanalysi~ of mechanical phenomena indicate th.at apart fromthe kinetic energy T = mv2/2 one needs to mtr9duce onemore quantity, momentum (p = mv), in order to. describethe mechanical motion of bodies. These quantities providethe basic measures of mechanical motion of bodies, theformer being scalar and the latter vectorial. Both of themplay a most significant part in the construction of me­chanics.

Let us proceed to a more detailed analysis of mome.ntum.First of all we shall write the fundamental equatiOn ofNewtonian dynamics (2.6) in another form, by the use ofmomentum:

Idp/dt = F, I (4.1)

Le. the time derivative of the momentum of a mass point isequal to the force acting on that point. Specifically, if F == 0,then p = const.

Note that in a non-inertial reference frame the force Fcomprises Dot only the forces of interaction between a givenparticle .and other bodies, but also inertial forces.

Eq. (4.1) allows the increment of the momen'tum of aparticle to be found for' any time interval provided the timedependence of the force F is known. In fact, it follows fromEq. (4.1) that the elementary momentum increment thatthe particle acquires during the time interval dt is equalto dp = F dt. Integrating this expression with respect totime, we find the momentum increment of a particle duringthe finite time interval t:

t

P2-Pl= JFdt.o

• It is sometimes called the quantity of motion.

(4.2)

When the force F = const, the vector F can be removed fromthe integrand and then P2 -- PI = Ft. The quantity on theright-hand side of this equation is referred to as the powerimpulse. Thus, the momentum increment acquired by aparticle during any time interval is equal to the power im­pulse over the same time interval.

Example. A particle which at the initial moment t = 0 llossessesthe momentum yo is subjected to the force F = at (1 - th:) duringthe time interva 't, a being a constant vector. Find the momentum pof the particle at the moment when the action of the force comes toan end.

In accordance with Eq. (4.2) we get p = Po + JF dt = Po +o+ aT;2/6 (Fig. 55).

The momentum of a system. Let us consider an arbitrarysystem o~articles and introduce the concept of the momen­tum of a system as a vector sum ofthe momenta of its lionstituent par-ticles: "

p ~ ~Pi' (4.3)

where Pi is the momentum of theith particle. Note that the mo-mentum of a system is an additive Fig. 55quantity, that is, the momentum ofa system is equal to the sum of the momenta of its indi­vidual parts irrespective of whether or not they interact.

Let us fmd the physical quantity which defines the sys­tem's momentum change. For this purpose we shall differen­tiate Eq. (4.3) with respect to time:

dp/dt = ~ dp/dt.

In accordance with Eq. (4.1)

dpt/dt = L Fill +Fj,II

where F til are the forces which the other particles of thesystem exert on the ith particle, Le. internal forces; F i isthe force which other bodies outside the system under con-8*

116 Classical Mechanics The Law of Conservation of Momentum 117

(4.6)

sideration exert on the same particle (external forces). Sub­stituting the last expression into the previous one, we get

dp/dt= ~ ~ F ilt +~ F i •i It i

The double summation symbol on the right-hand sidedenotes the sum of all internal forces. In accordance withNewton's third law the interaction forces between the sys­t-em's particles are pairwise identical in magnitude andopposite in direction. Consequently, the resultant force ofeach interacting pair is equal to zero, and therefore thevector sum of all internal forces is also equal to zero. As aresult, the last equation takes the following form:

Idp/d!~ F, I (4.4)

where FOis the resultant of all external forces, F = ~ F"

Eq. (4.4) implies that the time derivative of the momentumof a system is equal to the vector sum of all external forcesacting on the particles of the system.

As in the case of a single particle, it follows from Eq. (4.4)that the increment of momentum which the system acquiresduring the finite time interval t is equal to

(4.5)

Le. the increment of momentum or the system is equal tothe momentum of the resultant of all external forces overthe corresponding time interval. Here F is the resultant ofall the external forces.

Eqs. (4.4) and (4.5) hold true both in inertial and innon-inertial reference frames. It should be borne in mind,however, that in non-inertial reference frames one needsto take into account the' inertial forces, which act as externalforces, Le. in these equations F should be regarded as thesum F ia + Fin, where ·F ia is the resultant of all externalinteraction forces, and Fin is the resultant of all inertialforces.

II

The law of momentum conservation. We have drawn animportant conclusion: in accordance with Eq. (4.4) themomentum of a system may vary only due to external forces.Internal forces cannot change the momentum of a system.Hence, another important conclusion immediately followsfrom this, the law of momentum conservation: in an inertialreference frame the momentum of a closed system of particlesremains constant, Le. does not change in the course of time:

Ip = ~ Pi (t) = const. ,

Here the momenta of individual particles or parts of aclosed system may change with time, a fact emphasized inthe last expression. These changes, however, always happenso that the momentum increment of one part of the systemis equal Ito the momentum decrease pf another part of thesystem. "In other words, the individual parts of a closedsystem can only ipterchange momenta. Having detecteda momentum increment in a certain system, we can statethat this increment originated at the expense of a momentumdecrease in surrounding bodies.

In this regard Eqs. (4.4) and (4.5) should be treated asa more general formulation of the momentum conservationlaw. This formulation indicates that the momentum changeof a non-closed system is caused by the action of other bodies(external forces). What was said is of course valid only inreference to inertial reference frames.

The momentum of a non-closed system can remain constantprovided the resultant of all external forces is equal to zero.This follows immediately from Eqs. (4.4) and (4.5). In thesecases the conservation of :Qlomentum is of practical interest,for it permits the system to be studied in a sufficientlysimple fashion without going into a detailed analysis ofthe process.

One more thing. Sometimes in a non-closed system it isnot the momentum p itself that remains constant, but itsPx projection on a certain x direction. This happens whenthe projection of the resultant F of the external forces onthe x direction is equal to zero, Le. the vector F is per­pendicular to that direction. In fact, projecting Eq. (4.4),

p - mv = (R) t + mgt,

where (R) is the vector R averaged over the time t. It is helpful todepict this relationship graphically (Fig. 56). It can be immediatelyseen from the figure tllat the sought value (R) is defined by the for­mula (R) t = P sin a + mgt cos a.

Example 2. A man of mass ml..islocated on a narrow raft of mass m2afloat on the surface of a lake. The man travels through the distance~r' with respect to the raft and then stops. The resistance of the wateris negligible. We shall find the corresponding displacement M

2of

the raft relative to the shore.In this case the resultant of all external forces acting on the "man­

raft" system is equal to zero, and, therefore, the momentum of thatsystem does not change, remaining equal to zero in the process ofmotion:

mIvI + m2v2 = 0,

where v). and V2 are the velocities of the man and the raft with respectto the shore. But the velocity of the man relative to the shore may berepresented in the form VI = V2 + v', where v' is the velocity of theman rell!tive to the raft. Eliminating VI from these two equations,

tt9The Law of Conservation of Momentum

we obtain

~ mivi = const.

~ m/vi = const'.

The result obtained is in complete agreement with the Gali­lean relativity principle, according to which the laws ofmechanics are identical in all inertial reference frames.

Multiplying both sides by dt, we find the relationship between theelementary displacements of the raft dr2 and the man dr' relativeto the raft. Obviously, the same relationship also holds in the caseof finite displacements:

~r2= ml Ar'.ml+m2

It is seen from this that the displacement M 2 of the raft does notdepend on the character of the man's motion, I.e. on the law v' (t).

We emphasize once again that the momentum conserva­tion law holds only in inertial frames. This statement, how­ever, does not rule out the momentum of a system remainingconstant in non-inertial reference frames as well. Thishappens when in Eq. (4.4), which is also valid in non-inertialreference.lframes, the external force F (including inertialforces) is equal to zero. Clearly this situation occurs onlyunder specral conditions. Such special cases are fairly rare.

Let us now demonstrate that if the momentum of a system.remains constant in one inertial reference frame K, it alsodoes so in any other inertial frame K'. Suppose in the Kframe

If the K' frame moves relative to the K frame with thevelocity V, the velocity of the ith particle in the K framemay be written as Vi = vi + V, where vi is the velocityof that particle in the K' frame. Then the expression forthe momentum of the system may be transformed as follows:

~ m/vi + ~ m/V = const. The second term here does notdepend on time. This implies that the first term, the momen­tum of the system in the K' reference frame, does not, dependon time either, i.e.

(4.7)

Classical Mechanics

we get

118

whence it follows that if F,x15 0, then Pox = const. Forexample, when a system moves in a uniform field of gravity,the projection of its momentum on any horizontal directionremains constant whatever happens inside the system.

Let us consider examples involving constant and varyingm~@b. '

Example t. A cannon of mass m slides down a smooth inclinedplane forming the angle a with the horizontal. At the moment when

the velocity of the cannon reaches v, itfires a shell in a horizontal directionwith the result that the cannon stopsand the shell "carries away" the momen­tum p. Suppose that the firing durationis equal to t. What is the reaction forceR of the inclined plane averaged overthe time t?

Here the system "cannon-shell" isnon-elosed. During the time interval tthis system acquires a momentum in-

Fig. 56 erement equal to p - mv. The changeof the system's momentum is caused

by two external forces: the reaction force R (which is perpendicu­lar to the inclined plane) and gravity mg. 'l:herefore, we can write:

120 Classical Mechanics The Law 0/ Conservation 0/ Momentum 121

Let us find the velocity of the centre of inertia in a gIVenreference frame. Differentiating Eq. (4.8) with respect totime, we get

If the velocity of the centre of inertia is equal to zero, thesystem is said to be at rest as a whole. This provides a natu­ral generalization of the concept of a motionless particle.Accordingly, the velocity Vc acquires the meaning of thevelocity of the system moving as a whole.

With allowance made for Eq. (4.3) we obtain from Eq. (4.9)

p = mV c, (4.10)Le. the momentum of a system is equal to the product of themass of the system by the velocity of its centre of inertia.

The equation of motion for the centre of inertia. Theconcept oWl centre of inertia allows Eq. (4.4) to be rewrittenin a more convenient form. To do this, we have to sub­stitute Eq. ,(4.10) in~o Eq. (4.4) and take i~to account thatthe mass of a system per se has a constant value. Then weobtain

(4.9)

(4.11)dVc Fm7"= ,

where F is the resultant of all external forces acting on thesystem. This is the equation of motion for the centre of inertiaof a system, one of the nwst important equations of mechan­ics. According to this equation, during the motion of anysystem of particles its centre of inertia moves as if all the massof the system were concentrated at that point, and all externalforces acting on the system were applied to it. In this casethe acceleration of the centre of inertia is quite independentof the points to which the external forces are applied.

Next, it follows from Eq. (4.11) that if F == 0, thendV cldt = 0, and therefore Vc = const. In particular, thiscase is realized in a closed system (in an inettial referenceframe). Furthermore, if Vc = const, then in accordancewith Eq. (4.10) the momentum of the system p = const.

Thus, if the centre of inertia of a system moves uniformlyand rectilinearly, the momentum of the system remains con-

rc = ...!.. ~. m;ril (4.8)m .-;..,I

where mi and ri are the mass and theo radius vector of the ith particle, and

m is the mass of the whole systemF' 57 (Fig. 57).

19. It should be pointed out that thecentre of inertia of a system coincides

with its centre of gravity. However, this statement isvalid only when the gravitati?nal field can be assumeduniform within the limits of a gIVen system.

The validity of Newton's laws underlies the reasoningthat led us to th~ momentum ~onservation law. Specifically,the mass points of a closed system were assumed to interactin pairs and to obey Newton's third law. Now, what happensin systems whicft..do not obey Newton's laws, e/g. in lIystemsinvolving electromagnetic radiation?

Experience shows convincingly enough that the momen­tum conservation law is valid for such systems as well.However, in these cases one has to take into account in thegeneral equilibrium of momenta not only the momenta ofparticles, but also the momentum which, as electrodynamicsconfirms, the radiation field itself possesses.

Thus, experience shows that the momentum conservationlaw, when appropriately correlated, constitutes a fundamentallaw of nature which is valid without exceptions. But in thisbroad sense, this law is no longer a consequence of Newton'slaws, and should be regarded as an independent generalprinciple inferred from experimental data.

,§ 4.2. Centre of Inertia.The C Frame.

The centre of inertia. Any system of particles possessesone remarkable point C, the centre of inertia, or ~he .centreof mass, displaying a number of interesti~g. and sIgmfi?antproperties. It~ position relative to th~ onglI~ 0 of ! gIven

reference frame IS descrIbed by.the ra­dius vector redefined by the followingformula:

122 Classical Mechanics Th~ Law of Conservation of Momentum 123

balances g~avity (Fig. ?8,. right). As for the horizontal componento~ the tensIle strength, It IS constant in magnitude and permanentlydI~cte~ toward the .rotation ~xis. It follows from this that the centreof mertia of the cham, the pomt C, travels along the horizontal circlewhose radius p is easy to find via .Eq. (4.11), writing it as

mwap = mg tan 0,

where m is tho mass of the chain.In this case the point C is perma­nently located between the rota­tion axis and the thread, as shownin Fig. 58.

The 0 frame. In many caseswhen we examine only therelative motion of particleswithin a system, but not themotion of this system as a F' 58whole, it ~ most advisable to Ig.~esort to tlie reference frame in which the centre of inertiaI~ at rest. Then we ClUl significantly simplify both the analy­SIS of phenomena anll the calculations.

The reference frame rigidly fixed to the centre of inertiaof a given system of particles and translating with respect~o in~rtial fra~es is referred to as the frame of the centre ofmertta, or, brIefly, the C frame. The distinctive feature ofthe ~ fra~e is that the total momentum of the system ofpartIcles IS equal to zero; this immediately follows fromEq. (4.10). In other words, any system of particles as awhole is at rest in its C frame.

The C frame of a closed system of particles is inertial, while'that of a non-closed system is non-inertialin the general case.

Let an find the relationship between the values of themechanical energy of a system in the K and C referMceframes. L~t us begin with the kinetic energy T of the system.The velOCIty of t~e ith particle in !.he K frame may be repre-sent~d a~ Vi = Vi + Vc, where Vi is the velocity of thatpartIcle III the C frame and Vc is the velocity of the Cfra.me with respect to the K reference frame. Now we canwrIte

stant in the process of motion. Obviously, the reverse state­ment is also true.

Eq. (4.11) coincides in its form with the fundamentalequation of dynamics of a mass point and is its naturalgeneralization to a system of particles: the acceleration ofa system as a whole is proportional to the resultant of allexternal forces and inversely proportional to the totalmass of the system. Recall that in non-inertial referenceframes the resultant of all external forces includes bothforces of interaction with surrounding bodies and inertialforces. •

Let us consider three examples associated with motionof a system's centre of inertia.

Example 1. We shall show how the problem of a man on a raft(see Example 2 on p. 118) can be solved by resorting to the notionof the centre of inertia.

Since the resistance of water is negligibly small, the resultantof all external forces acting on the system "a man and a raft" is equalto zero. This means that the position of the centre of inertia of thegiven system does not change in the process of motion of the man(and the raft), i.e.

mlrl + mara = const,

where rl and r a are the radius vectors describing the positions of thecentres of inertia of the man and the raft relative to a certain pointon the shore. From this equality we find the relationship betweenthe increments of the vectors rl and ra:

ml M l + mt I1r2 = o.Taking into account that the increments M l and M a represent thedisplacements of the man and the raft with respect to the shore andthat M l = M a + M ' , we find the displacement of the raft:

I1r.= _ ml 11~.ml+ma

Example 2. A man jumps down from a tower into water. In thegeneral case his motion is quite complicated. However, if the airdrag is negligible, it can be immediately stated that the centre ofinertia of the jumper moves along a parabola, just as a mass pointexperiencing the constant force mg, where m is the man's mass.

Example 3. A closed chain connected by a thread to a rotatingshaft revolves around a vertical axis with the uniform angular veloc­ity CJ) (Fig. 58), the thread forming the angle 0 with the vertical.How does the centre of inertia of the chain··'ove?

First of all, it is clear that it does not move in the vertical direc­tion during the uniform rotation of the chain. This means that thevertical component of the tensile strength T of the thread counter-

124 Clas'sical Mechanics The Law of Conservation of Momentum 125

Since in the C frame 11 mjvj = 0, the previous expressiontakes the form

(4.12)

Eq. (4.13), we obtain

Ii = E _. p2/(2·2m) = mv3/4.

It is easy to realize that the internal energy 1]' is associated with therotation and oscillation of the given system, while at the initialmoment E was equal only to the rotational motion energy.

where I-l. is the so-called reduced mass of the system

jl-l.=ml m2/(ml +m2)./ (4.14)

mFig. 59

m

If the processes associated with a change in the totalmechaniclll energy take place in a closed system of particles,from Eq. (4.13) it follows that AE = AE, Le. the incrementof the total mechanical energy relativeto an arbitrary inertial reference frameis equal to the increment of the internalmechanical energy. In this case the kineticenergy resulting from the motion of thesystem of particles as a whole does notchange bec.«use in a closed system p == const.

Specifically, if a cLosed system is con­servative, its total ~chanical energy re­mains constant in all inertial referenceframes. This conclusion completely agrees with the Galileanrelativity principle. . ,

A system of two particles. Suppose the masses of the par­ticles are equal to mi and m 2 and their velocities in the Kreference frame to VI and V2, respectively. Let us find theexpressions defining their momenta and the total kineticenergy in the C frame.

The momentum of the first particle in the C system is

PI = mIvl = ml (vl-Vd,

where Vc is the velocity of the centre of inertia (of the Csystem) in the K reference frame. Substituting in this for­mula expression (4.9) for Vc, we obtain

The energy E = T + U is referred to as the internal me­chanical energy of the system.

Example. Two small discs, each of mass m, lying on a smoothhorizontal plane, are interconnected by a weightless sp.ring: Oneof the discs is set in motion with the velocity vo, as shown III Fig. 59.What is the internal mechanical energy Ii of this system in the processof motion? .

Since the surface is smooth, the ~ystem in the pro~ess of motIOnbehaves as a closed one. Therefore, its total mechanIcal. ~n~rgy Eand total momentum p remain constant and equal to the InItial :val­ues, Le. E = mvB/2 and p = mvo• Substituting these values lllto

where f = ~ mj~/2 is the total kinetic energy of the par­ticles in the C frame, m is the mass of all the system, and pis its total momentum in the K frame.

Thus, the kinetic energy of a system of particles comprisesthe total kinetic energy T in the C frame and the kinetic energyassociated with the motion of the system of particles as a whole.This important conclusion will be repea.tedly utiliz~d here­after (specifically, in studies of dynamIcs o~ a solId).

It follows from Eq. (4.12) that the kinetic energy of. asystem of particles is minimal in the C frame, another dlS­Hnctive feature of that frame. Indeed, in the C frame Vc == 0, and Eq. (a.12) yields T = T.

Now let us pass over to the total mechanical energy E.Since the internal potential ,energy U of a s¥stem depen?sonly on its configuration, the magnitude U IS th~ same lD

all reference. frames. Adding U to the left- and rIght-h~nd

sides of Eq. (4.12), we obtain the formula f.o~ transformationof the total mechanical energy on transItion from the Kto the C frame:

I-E-=-E---+-!!!:--~b-=-E-+-f--:-·I (4.13)

126 Classical Mechanics The Law of Conservation of Momentu.m 127

1 the momentum of the second particle in theSimilar y,C frame is

E=T+U, (4.17)where U is the potential energy of interaction of the given

particles. . t' t d'The formulae obtained play an 1m portant par m su les

of particle collisions.

Although we shall discuss collisions of particles, it shouldbe mentioned at once that all subsequent arguments andconclusions relate to collisions of any bodies. One only hasto substitute the velocity of the centre of inertia of eachbody for the velocity of a particle, and to replace the kinetic'energy of a particle by that part of the kinetic energy ofeach body that characterizes its motion as a whole.

In what follows we shall assume that(1) the initial reference frame K is inertial,(2) the system of two particles is closed,(3) the momenta (and the velocities) of the particles be­

fore and after a collision correspond to sufficiently largedistances between them; at the same time the. potentialenergy of interaction can be neglected.

In addition, the quantities relating to the system aftera collision will be marked with a prime, while those in theC frame WRh a tilde.

Now let us pass to the essence of the problem. Particlecollisions are classiood into three types: completely in­elastic, perfectly elastic, and inelastic (the intermedia.tecase).

Completely inelastic collision results in two partieles"sticking together", after which they move as a single unit.Suppose two particles with masses ml and ml! move withthe velocities VI and VI! before collision (in the K frame).After the collision a particle with mass ml + ml! is formedbecause of additivity of mass in Newtonian mechanics. Thevelocity v' of the formed particle can be immediately foundfrom the momentum conservation law:

The velocity v' is obviously equal to that of the system'scentre of inertia.

In the C frame this process is the most simple: prior tothe collision both particles move toward each other carry-ing equal momenta;' while after the collision. the formedparticle turns out to be stationary. In this case the totalkinetic energy T of the particles completely turns into theinternal energy Qof the formed particle, I.e. T = Q. Whence,

(4.16)

(4.15)

energy

IT= ])2/2/1 = /1V~ez/2·1interact, their total mechanicalIf the particles

in the C frame is

§ 4.3. Collision of Two Particles

In this secti on we shall examine 'arious cases of colli­sions of two particles, using only the momert~t and en~~;?iconservation laws as an investigatory too. ere we s 1see that the conservation laws enable us to draw som~ gentraand essential conclusiOl'S concerning the prope~tlles,0t agiven process irrespective of a specific law of partlc e III er-

action. . d t of theAt the same time we shall Illustrate the ~va? ages .C frame, whose utilization cons.iderably sImphfies analysIsof a process and many calculatIOns.

P2=/1(V2- VI)'

Thus the momenta of the two particles i~ the C frameare equ~l in magnitude and opposi~e in.directIOn; the modu­lus of the momentum of each particle IS

I ]) 0= /1vre lt 'I1 - I v - v I is the velo~ity of one particleWlere Vrel - 1 I!

relative to another. T 1 k' t'Finally, let us consider kinetic energy. .he tota me IC

energy of the two particles in the C frame IS

l' =' 1\+ 1'2 = p2/2m l +p2/2m2•

Since in accordance with Eq. (4.14) 1lml + 1iml! = 11/1,then

Now let us find the velocity of each particle after the col­lision in the K reference frame. For this purpose we shallmake use of the velocity transformation formulae for thetransition from the C to the K frame and also the foregoing

and V2 in the K reference frame (the particles either movetoward each other or one particle overtakes another). Whatare the velocities of these particles after the collision?

Let us first consider this process in the C frame, wherethe particles before and after the collision possess momentaequal in magnitude and opposite in direction (Fig. 60).Moreover, since the total kinetic energy of the particles isthe same before and after the collision, as well as theirreduced mass, then in accordance with Eq. (4.16) the mo­mentum of each particle only reverses its direction as aresult of the collision, its magnitude remaining unchanged,i.e. pi = -)1;, where i = 1, 2. The same can be said aboutthe velocity of each particle in the C frame:

129

(4.19)vix = 2Vex - Vix'

The Law 0/ Conservation of Momentum

~,

p =p.

vi = 2Vc -v" (4.18)

where i = 1, 2. In terms of the projections on an arbitraryx axis the last equality takes the form

equality. Then

, V -, -Vi= C--I--Vi= VC-Vt = Vc-(v,- Vc)=2VC - V i'

where ~c is the velocity of the centre of inertia (of the Cframe) 10 the K frame; t?is velocity is determined byEq. (4.9). Hence, the velocIty of the itth particle in 'the Kframe after the collision is

Specifically, when the particle masses are identical it iseasy to s~e that the particles exchange their velociiies asa result :~f the collision, i.e. v~ = V 2 and v;, = Vl'

2. A nOfl-central collision. We shall limit ourselves tothe case when one ~of the particles is motionless before thecoll~sion. Suppose a particle possessing the mass m andmomentum PI exp,eriences in the K frame a non-c:ntralelastic collision with a motionless particle of mass m .What are the possible momenta of these particles after thecollision?

First, let us examine this process' in the C frame. Here?S in the. previous case, the particles possess momentaequaiI~ magmtude and opposite in direction at any moment oftIme before and after the collision. Besides the momentumof e?~h pa:ticle does not change in magnitu'de following thecollIsIOn, l.e.

Ho~ever, the particles' rebound direction is different in thiscase. It forms a certain angle 8' with the initial motiondirection (Fig. 61), depending on the particle interaction lawand the mutual positions of the particles in the process ofcollision.

Now let u~ ~alc',llate the momentum of each particleafter the collIsIOn III the K reference frame. Making useof the velocity transformation formulae for the transition9-0U9

Classical Mechanics

p, ~'

pz before.. •... • afterp: ji;

Fig. 60

with allowance made for Eq. (4.16), we obtain

Q _ !1v~el _ -.!. m1m. (v _ V )2--2-- 2 m1+ m• 1 2'

Thus the value of Q for a given pair of particles depends onlyon their, relative velocity.

Perfectly elastic collision does not lead to any change'in the internal energy of the particles, so that the kineticenergy of the system does not change either. We shall con­sider two particular cases: central (head-on) and non-centralelastic collisions.

1. A head-on collision. Both particles move along thesame straight line before and after collision. Suppose thatprior to collision the particles move with the velocities VI

128

*

tOO Cla,,'cal Mechanics The Law 01 Con,eruatlon 01 MomentlJm tat

Fig. 62

elastic coriision of two particles (one of which rests initially)it is neces~ary:

. (1) first to depict'the section AB equal to the momentumPI of the bombarding particle;

(2) then through. the point B, the end point of the vectorPI' to trace a circle of radius

P = IJ.Vrez ml+m. PI'whose centre (point 0) divides the section AB into two partsin the ratio AO : OB = ml : mI'

This circle is the locus of all possible locations of theapex C of the momenta triangle ABC whose sides AC andCB represent the possible momenta of the particles afterthe collision (in the K reference frame).Dep~nding on the particle mass ratio the point A, the

beginning of the vector PI' can be located inside the givencircle,_on it, or outside it (Fig. 63). In all three cases theangle e can assume all the values from 0 to n. The possiblevalues of the angle 01 of scattering of the bombarding particleand the angle e of rebounding particles are as follows:

(a) m!<~ O<el~n 8>n/2(b) ml=~ 0<et :::;;;nI2 8=n/2(c) mI >m2 o<eI~ Otmax 8<n/2

where VI is the velociLy of the bombarding particle. Butinasmuch as ill our case VI = Vrel.

OB=/.l.Vrez=Pin accordance with Eqs. (4.14) and (4.15). Thus, in orderto draw a vector diagram of momenta corresponding to an

Fig. 61

itm, ••----"-....-

/

P~ +p; = (ml + m.) Vc = Pit

just as it should be in accordance with the momentum con­servation law.

Now let us draw the so-called vector diagram of momenta.First we depict the vector PI as the section AB (Fig. 62),

from the C to the K frame, we obtain:

P~ =mlv~=ml (Vc+v~)=mIVc+p~,

P; = ~v; = m. (Vc+v;) = ~Vc+Ii;, (4.20)

where Vc is the velocity of the C frame relative to the Kreference frame. .

Summing up separately the left- and right-han..? sides 2fthese equalities and taking into account that p~ = - p;,we get

and then the vectors p~ and p~, each of which represents,according to Eq. (4.20), a sum of t'Y0 vectors. _

Note that this drawing is valid regardless of the angle O.The point C, therefore, can be located only on the circle ofradius phaving.its centre at the point 0, which divides thesection AB into two parts in the ratio AO : OB. = mi : m 2 ,

Moreover, in the considered case (when the partIcle of massml rests prior to the collision) this circle pa~ses through ~hepoint B, the end point of the vector Pt, SlDce the sectIOllOB = p. Indeed,

V mtVtOB=m2 c= m. .+m.'. mi.

11*

Cla,,'cal Mechanics The Law of Con,erl1atlon of Momentum

Here 91mn is the limiting angle. It is defined by the formula

sin 91 max = ~/ffl.t, (4.21)

which directly follows from Fig. 63c: sin 91mu = OC'IAO == OBIAO = m2lml'

In addition, here is another interesting fact. In the lastcase (ml > m 2) the particle ml can be scattered by the same

angle 9 whether it possesses the momentum AC ?r AD(Fig. 63~), Le. the solution is ambiguous. The same IS truefor the particle_ m2 • •

And finally, from the same vector dIagram of moment,!,we can determine the relation between the angles 91 and 9

sine (422)tan 9t = _. ' .cos e+ml!m.

With this we have exhaust?d any information on thegiven process which can be derIved t~ough the use of onlythe momentum and energy conservatIOn laws. .

Thus we see that the momentum and energy con~er~atlOn

I s by'themselves permit us to draw a number of sIgmficantaw I' about the propert,ies of a given process. Mostconc USIons J ., I

essential here is the fact that these propertIes are umversain their nature, that is; they do not depend on the type ofinteraction between particles. . d

One fundamental fact, however, should be pOlllte out.The vector diagram of momenta based on t.he momentumand energy conservation laws provides us wIth a. complete

attern of all possible cases of rebounding partIcles; butfhis very significant result cannot indicate the .concrete casethat is actually realized. To answer that questIOn, we must

• Th~ aiming parameter is the distance between the straight linealong whICh the momentum of a bombarding particle is directed, an4lhe particle exposed to a "collision".

(4.23)1"-1'=Q.

analyse the collision process in more detail by means of themotion equations. In the process it becomes clear that thescatte....ing angle 91 of a bombarding particle depends on thetype of the interaction of colliding particles and on theso-called aiming parameter*, while the ambiguity of asolution in the case m1 > m2 is due to the fact that the samescattering angle 91 can occur with two different values ofthe aiming parameter irrespective of the law of particleinteraction.

The circumstance discussed here represents an inherent,fundamental property of all conservation laws in general.The conservation laws can never provide an unambiguouspicture of whl't is actually going to happen. But il, on thebasis of some other considerations, it becomes possible toinfer what exactly is going to happen, the conservation lawscan contribute information on how it must happen.

.Inelastw collision. After this kind of collision the inter­nal ener~ of rebounding particles (or one of them) changes,and therefOl'e the to~al kinetic energy of the system changesas well. .It' is custOlpary to denote the corresponding incre­ment of the kinetic' energy of the system by Q. Dependingon the sign of Qan inelastic collision is referred to as exoergic(Q> 0) or endoergic (Q< 0). In the former case the kineticenergy of the system increases while in the latter it decreases.In an elastic collision Q = 0, of course.

Our task is to determine possible momenta of particlesafter a collision.

This problem is easiest when solved in terms of the Cframe. By the hypothesis, the increment of the total kineticenergy of the system in the given process is equal to

Since in this case T' =1= T, this means, in accordance withEq. (4.16), that the momenta of particles change theirmagnitude after the collision. The momentum of each par-'ticle p' after the collision can be easily found if we replace

Fig. 63

(~fLLi-j

A 0 8 Am,-mz

(a~

( &LJA 0 8

m,<m,

iSS

(4.25)

The Law of Conservation of Momentum

Under what condition does such a process becomepossible?

Again, the problem is easiest when solved !n the ~ frame,where it is obvious that the total kinetic energy T of theparticles before the collision must in any case be not lessthan I Q I, Le. T~ I Q I. Whence it follows that thereexists the minimal value Tmin = I Q I, such that the kineticenergy of the system entirely turns into an increment of theinternal energy of the particles, and so the particles cometo a standstill in the C frame.

Let us consider the sll.meproblem in the K reference frame,where a particle of mass m i collides with a stationary par-ticle of mass m 2 • Since at Tmin the particles come to a stand­still after the collision in the C frame, this signifies thatin the K frame, provided the kinetic energy of the bombard­ing particJe is equal to the requisite threshold value T Ithr.

both particles move after the collision as a single unit whosetotal mom~ntum is equal to the momentum PI of the bom-.barding particle add the kinetic energy p~/2 (m i + m2).

Therefore

Taking into account that TWIT = p~/2mI and eliminating p:from these two equations, we obtain

T t thr= IQI+p~/2(m1 +~).

This is the threshold kinetic energy of the bombarding par­ticle sufficient to make the given endoergic process possiblein terms of energy.

It should be pointed out that Eq. (4.25) plays an impor­tant part in atomic and nuclear physics. It is used to deter­mine both the thresholds of various endoergic processes andtheir corresponding energies I Q I.

In conclusion we shall consider an example which, inessence, provides a model of an endoergic collision (seealso Problems 4.5 and 4.8).

Example. A small disc of mass m and a smooth hillock of mass Mand height h are located on a smooth horizontal plane (Fig. 65). What

1r... 'I~.·.·..··i .•,';"~':;

ii

If

tr

(4.24)

ClassIcal MechanIcs134

T' in Eq. (4.23) by its expression 'i' = [/2/211:

p' = V211(T+Q).

Now let us consider the same problilm in the K referenceframe, where a particle of mass ml with the momentum PIcollides with a stationary particle of mass m 2 • To determinethe possible cases of. particle rebounding after the collision,it is helpful to "resort to the vector diagram of momenta. Itis drawn similarly to the case of an elastic collision. The

momentum of a bombardingparticle PI = A B (Fig. 64)is divided by the point 0 intotwo parts proportional to themasses of the particles

L.-::.:!-+-_"';':;:;'-L..._--oOlofj (AO: OB = m] : m 2). Thenfrom the point 0 ~a circle is

Fig. 64 drawn with radius P' specifiedby Eq. (4.24). This circle is the

locus of all possible positions of the vertex C of the triangleABC whose sides AC and CB are equal to the momenta ofthe corresponding particles after the collision.

Note that in contrast to the case of an elastic collisionthe point B, the end point of the vector PI' does not lie onthe circle any more; in fact, when Q>O, this point is locat­ed inside the circle, and when Q < 0 outside it. Fig. 64illustrates the latter case, an endoergic collision.

Threshold. There are ml'!ny inelastic collisions in whichthe internal energy of particles is capable of changing bya quite definite value, depending on the properties of the

. particles themselves (e.g. inelastic collisions of atoms andmolecules). Nevertheless, exoergic collisions (Q> 0) canoccur fot an arbitrarily low kinetic energy of a bombardingparticle. In similar cases endoergic processes (Q < 0)possess a threshold. A threshold is the minimal kinetic energyof a bombarding particle just sufficient to make a givenprocess pO'lsihle in terms of energy.

So, suppose we need to carry out an endoergic collisionin which thl' intl'rnal energy of tlle pnrtirJl's is cnpahle ofilcQuiring an increment not less than a certain value I 0 I,

whence

137The Law of Conservation of Momentum

written thatm dv = F dt ± fJm ·u,

where the plus sign denotes the addition of mass and theminus sign denotes the separation. These cases OIln be com­bined, designating ±fJm as the mass increment dm of the bodyA (in fact, in the case of mass addition dm = +fJm, and inthe case of mass separation dm = -fJm). Then the fore­going equation takes the form

m dv = F dt + dm ·u.

Dividing this expression by dt, we obtain

I dv F dm Im CIt = +CIt u, (~26)

where u is the velocity of the added (separated) matter ithrespect t.•' the considered body.

This is the fundamental equation of dynamics of a asspoint with .tJariable '(ttass. It is referred to as the Meshcherskyequation. Obtained, in one irertial reference frame, thisequation is also valid, due to the relativity principle, inany other inertial ftame. It should be pointed out that ina non-inertial reference frame the force F is interpreted asthe resultant of both inertial forces and the forces of inter­action of a' given body with surrounding bodies.

The last term in Eq. (4.26) is referred to as the reactiveforce: R = (dmldt) u. This force appears as a result of theaction that the added (separated) mass exerts on a givenbody. If mass is added, then dmldt> 0 and the vector Rcoincides in direction with the vector u; if mass is separated,dmldt < 0 and the vector R is directed oppositely to thevector u.

The Meshchersky equation coincides in form with thefundamental equation of dynamics for a permanent masspoint: ~he left-hand side contains the product of the massof a body by acceleration, and the right-hand side containsthe forces acting on it, including the reactive force. Inthe case of variable mass, however, we cannot include themass m under the differential sign and present the left­hand side of the equation as the time derivative of themomentum, since m dvldt =1= d (mv)ldt. .

Classical Mechanics136

Vthr= V2 (1+m/M) gh.

mvfhr/2 = mgh (m+M)/M ,

Fig. 65

an increment of potential energy I1U = mgh. We shall regard thisprocess as endoergic in which I Q I = I1U. Then in accordance withEq. (4.25).

minimal velocity should be imparted to the disc1to make it capableof overcoming the hillock? . .

It is clear that the velocity of the disc must be at least sufficientfor. it to climb the-hillock and then to move together with it as a singleUnit. In the process, part of the system's kinetic energy turns into

§ 4.4. Motion ofa Body with Variable Mass

There are many cases when the mass of a body varies inthe process of motion due to the continuous separation or

, addition of matter (a missile, a jet, a flatcar being loadedin motion, etc.).

Our task is to derive the equation of motion of such abody.

Let us consider the solution of this problem for a masspoint, calling it a body for the sake of brevity. Supposethat at a certain moment of time t the mass of a moving bodyA is equal to m and the mass being added (or separated) hasthe velocity u relative ,to the given body.

Let us introduce an auxiliary inertial reference frame Kwhose velocity is equal to that of the body A at a givenmoment t. This means that at the moment t the body Ais at rest in the K frame.

Now suppose that during the time interval from t tot + dt the body A acquires the momentum m dv in theK frame. The momentum is gathered due to (1) the addition(separation) of the mass fJm bringing (carrying away) themomentum fJm 'u, and (2) surrounding bodies exerting theforce F, or the action of a field of force. Thus. it can be

f38 Classtcal Mechanics The Law of Conservation of Momentllm 139

Let us discuss two special cases.1. When u = 0, Le. mass is added or separated with zero

velocity relative to the body, R = °and Eq. (4.26) takesthe form

dvm (t)"dt = F, (4.27)

where m (t) is the mass of the body at a given moment oftime. This equation describes, for example, the motion ofa flatcar with sand pouring out freely from it (see Prob­lem 4.10, Item 1).

2. If u = -v, Le. the added mass is stationary in thechosen reference frame, or the separated mass becomes sta­tionary in that frame, Eq. (4.26) takes another form,

m (dvldt) + (dmldt) v = F,

In other words, in this case (and only in this one) the actionof the force. F determines the change of momentum of a bodywith variable mass. This is" realized, for example, duringthe motion of a flatcar being loaded with sand from a sta-'tionary hopper (see Problem 4.10, Item 2).

Let us consider an example in which the Meshcherskyequation is utilized.

Example. A rocket moves in the inertial reference frame K in theabsence of an external field of force, the gaseous jet escaping withthe constant velocity u relative to the rocket. Find how the rocketvelocity depends on its mass m if at the moment of launching the massis equal to mo.

In this case F = 0 and Eq. (4.26) yieldsdv = u iJ,,,tfm.

Integrating this expression with allowance made for the initial con­ditions, we get

v = - u In (mo/m), (f)

where the" minus sign shows that the vector v (the rocket velocity)• is directed oppositely to the vector u. It is seen that in this case (u =

= const) the rocket velocity does not depend on the fuel combustiontime: v is determined only by the ratio of t.he initial rocket mass moto the remaining mass m.

Note that if the whole fuel mass were momentarily ejected withthe velocity u relative to the rocket, the rocket velocity would be!lifferent. In fact, if the rocket initially is at rest in the chosen inertial

ord (mv)/dt = F. (4.28)

reference frame and after. the fuel ejection gathers the velocity v,the momentum conservatIOn law for the system "rocket-and-fuel"yields

o = mv + (mo - m) (u + v),

where u + v is the velocity of the fuel relative to the given referenceframe. Hence

v = - u(f - mlmo)' (2)

Ii! this case the rocket velocity v turns out to be less than in the pre­VIOUS case (!or equal values of the molm ratio). This is easy to demon­strate, havmg compared the dependence of v on mo/m in both cases.In t~e first case .(wh~n mat~er ~parates continuously) the rocketvel~cltr v grows mfimtely WIth mcreasing molm as Eq. (1) showswhlle. In the second case (when matter separates momentarily) th~velOCIty v tends to the limiting value - u (see Eq. (2».

Problems to Chapter 4

• 4. I,. A, particle moves with the momentum p (t) due to the forceF (t). Let _.!nd b be constant vectors, ~ith a 1,~. Assuming that

h(1) P (t) - a + t (1 - at) b, where a IS a posItIve constant find

t e vector F, at the moments of time when F .L p' '(2). F (t) ""' a + 2t b*and p (0) = Po, where Po is ~ vector directed

?pl;l0sltely to the vector a, find the vector p at the moment t whenIt IS turned through 900 with respect to the vector p 0

. Solution. 1. Th.e force F = dp/dt = (f - 2at) b, i.g: the vector F!S always perpendicular to the vector a. Consequently, the vector FIS te~pendlcular to .the vector p at those moments when the coefficient~ I~ the expressl~n for p (t) turns into zero. Hence, tl=O and t =- 1Ia, the respectIve values of the vector F are equal to 2

FI = b, F2 = - b.

:: The incremen~ of th~ vector. p during the time interval dt is~~ --:- F dt. .I!1tegratmg thIS equatIOn with allowance made for theImtlal conditIOns, we obtain

t

Ip-po= JFdt=at+btl ,

o

T~ere bi the hypothesis Po is directed oppositely to the vector ae ve~ or Phturns out to be perpendicular to the vector p at th~

momen to w en ato = Po. At this moment p = (poIa)' b 0

• 4.2. A rope th~own over a pulley (Fig. 66) has a ladder witha man A on one of ItS ends and a counterbaJancing mass M on its~~h:{ er(~lhe mrlan

h, whol!o mass is '!" climhs upward hy Ar' relative

I Ie a er an t en stops. Ignoring the masses of the pulle andtIC rope, as weB as the friction in tho pulley axis fmd the d' 'YJment of the centre of inertia of this system:' ISP acn-

140 Cla,BlcaZ Mechanic'The Law 0/ Con,ervation 0/ Momentum t4t

~----m m .r

Fig. 68

rI-. )0m

, Fig. 67 •

(1) ~he velo~ity Ve (t) of the centre of inertia of this system asa functIOn of time;

(2) the internal mechanical energy of the system in the process ofmotion. .. Solution. 1. In accordance with Eq. (4.11) the velocity vectorlDcre~nt of the centre of inertia is d Ve = g dt. Integrating thisequat.lOn, we get Ve (t) - Ve (0) = gt, where Ve (0) is the initialvelOCity of the centre of inertia. Hence

Vc (t) = (~Vl + mlva>/(~ + ml) + gt.

2. The internal. Plechanical energy of a system is its energy E inthe C frame. In thiS case t~e C frame moves with the acceleration g,so that each sphere experiences two external forces in. that frame'gravity mig and the inertial force -mig. The total work performed

by the external !?rces is thus equal to zero (in the C frame), and there­fore the. energy ~ ~<!es not change. To find the energy, it is sufficientto conSider the lDltIal ~ment of time, when the spring is not yetdeformed and the energy E is equal to the kinetic energy T in the Cframe. Making use of Eq. (4.16), we get 0

- - II. m mE=To=-2 (v1 -VI )a 1 I (vf+vl ).. 2(ml +m l ) I.

.•. ~.4. A ba~l possessin!!! the kinetic energy Tcollides head-on withan lD~tlallr sta.tlOna~ elastiC. du~bbeU (Fig. 68) and rebounds in theOPPOSite duectlOn With the kmetlC energy T'. The masses of all threeballs a~ ~he same. Find the energy of the dumbbell oscillations afterthe collISIOn.

Solution. Suppose p and p' are the momenta of the striking ballbefore and after the collision, Pc and To are the momentum and the~inetic em:rgy .of the dumbbell as a whole after the collision, and Eis the oSClllatlO~ energy. In accordance with the momentum andenergy conservatIOn laws .

p=-p'+Pc, T=T'+To+E.

~aking into account that T = pi/2m, we obtain from these two equa-tIOns: .

E=(T-3T' -2 yTT')/2.

tM

Ve =Vl = (m/2M) v'.

And finally, the sought displacement ism

M-m ~re = ) Ve dt = (mj2M) ) v'dt = (m/2M) M'.

Fig. 66 Another method of solution is based on a prop-erty of the centre of inertia. In the reference

frame fixed to the pulley axis the location of the centre of inertia ofthe given system is described by the radius vector

re = [Mrl + (M - m) r a + mrs]/2M,

where rlo ra, and ra are the radius vectors of the centres of inertiaof the mass M, the ladder, and the man relative to some point 0of the given reference frame. Hence, the displacement of the centreof inertia ~re is equal to

Me = [M M1 + (M - m) M a + m M sl/2M,

where ~rl' ~re, and M a are the displacements of the mass M, theladder, and tlie man relative to the given reference frame. Since~rl = -~ra and ~rs = ~ra + ~r', we obtain

Me = (m/2M) M'.

• 4.3. A system comprises two small spheres with masses ml and mainterconnected by a weightless spring. The spheres are set in motionat the velocities VI and Va' as shown in Fig. 67, whereupon the systemstarts moving in the uniform gravitational field of the Earth. Ignoringthe air drag and assuming that the spring is non-deformed at thejnitial moment of time, find: .

Solution. All the bodies of the system are initially at rest, andtherefore the increments of momenta of the bodies in their motionare equal to the momenta themselves. The rope tension is the sameboth on the left- and on the right-hand side, and consequently themomenta of the counterbalancing mass (PI) and the ladder with theman (Pa) are equal at any moment of time, i.e. P1 = Pa, or

MVl = mv + (M - m) va,

where Vlo V, and Va are the velocities of the mass, the man, and theladder, respectively. Taking into account that V~ = -VI and V =

= Va + v', where v' is the man's velocity relativeto the ladder, we obtain

v1=(m/2M) v' . (1)

On the other hand, the momentum of the wholesystem

P = PI + Pa = 2Plo or 2M Ve = 2Mvl,

where Ve is the velocity of the centre of inertiaof the system. With allowance made for Eq. (1)we get

ClalBtcal MechanicsThe Law of Conservation of Momentum t43

.4.7. What.fraction'1') of its kinetic energy does a particle of massml lose wjen It scatters, after an elastic collision through thethreshold .'tIngle on a stationary particle of mass m' (m > m )?

Solution. Suppose Tl , Pl, Ti, and pi are the value"s of Itheki~eticenergy. and momell:tul!\: of the striking particle before and after thescatter1llg, respectIvely; then

B

Fig. 70

P,AP,

Fig. 69

tion the .radical is greater than cos 8 and that pi is the vector's modu­lus, whICh cannot be negative.. But.when ~l > ~I' both signs have physical meaning; the solutionIS ambIguous 111 thIS case: the momentum of the particle scatteredthro~gh the. ~ng~e 8 may have one of two values (depending on therelatIve pos~t](~n1llg of the particles at the moment of collision). Thelatter case IS Illustrated by the vector diagram shown in Fig. 63c.

L11') = (Tl - Ti)/Tl = i-Ti/Tl = i-(pi/Pl)I, (i)

Le. the problem reduces to the de~rmination of pi/pl'Let us make use of the vector dIagram of momenta corresponding

~o the threshold angle 8thr (Fig. 70). From the right triangle ACOIt follows that

'1 ~ -- -PI =(Pl-p)l_pl=pf-2PtP,whence

(pj/Pl)"= i-2p/Pl = i-2ml/(mt +ml). (2)

Substituting Eq. (2) into Eq. (i), we obtain

1') = 2mJ(ml + m,,).

I. 41.8. t n atom of mass mJ.. collides inelastically with a stationary

mo ecu e 0 mass mI' After the collision the particles rebound atthe angle e between them with kinetic energies T' and T' respecti­yely. In the process the molecule is excited, that is its inter~al ent'rlDcreases by the definite value Q. Find Q and th~ threshold kinere;energy of the atom enabling the molecule to pass into the given excitedstate.. S~lution. From the energy and momentum conservation laws1D thIS process we can write

Tt=Ti+Ti+Q,

pi = pi'+pi'+2pipi cos e,

.4.5. In the K frame particle 1 of mass ml strikes a stationaryparticle 2 of mass nil' The charge of each particle is equal to +q.Find the minimal distance separating the particles during a head-oncollision if the kinetic energy of particle 1 is equal to Tl when it isfar removed from particle 2. _

Solution.Let us consider this process,both in the K frame and in the Cframe.

1. In the K frame the particles move at the moment of the closestapproach as a single unit with the velocity v, which can be determinedfrom the momentum conservation law:

PI = (Tn] + nil) v,

where PI is the momentum of the striking particle, PI =V 2m].Tl'On the other hand, it follows from the energy conservation law

thatTI = (nil +ml) v"/2+!J.U,

where the increment of the system's pot!lntialenergy AU = kql/rmtn'Eliminating v from these two equations, we get

rmln=(kq"/Tl ) (i+ml/ml)'

2. The solution is simplest in the C frame: here the total kineticenergy of the particles turns entirely into an increment of the poten­tial energy of the system at the moment of the closest approach:

T=!J.U,

where in accordance with Eq. (4.i6) f = !1vf/2 = TlmJ(Tn] + ml)•!J.U = kq"/rmtn.From this it is easy to find rmtn'

• 4.6. A particle of mass Tn] and momentum PI collides elasticallywith a stationary particle of mass mI' Find the momentum pi of thefirst particle after its collision and scattering through the angle 8relative to the initial motion direction.

Solutton. From the momentum conservation law we find (Fig. 69)

p;l=pf+pi"-2PIPt cos 8, (1)

where PI is the momentum of the second particle after the collision.On the other hand, from the energy conservation law it follows

that Tl = Ti + Ti, where Ti and Ti are the kinetic energies of thefirst and second panicles after the collision. Using the relation T == pI/2m, we can reduce Eq. (i) to the following form:

pil=(pl-pi") ml/m!' (2)

Eliminating P2" from Eqs. (i) and (2), we obtain

, cos8± Ycos'8+(ml/mf- i )PI =PI i+ /. ml ml

When mt < ml, only the plus sign (in front of the radical sign) hasphysical meaning. This follows from the fact that under this comli-

tM Classical MechantcsThe Law 01 Conservation 01 Momentum 145

where the primes mark the after-collision values. The second relationfollows immediately from the momenta triangle according to thecosine theorem. Making use of the formula pZ = 2mT, we eliminate T1

from these equations and getQ=(m./ml-l) T'+2 y(m./ml) TiTI cos 8

andT1thr=1 Q I (ml+mZ)/m••

.4.9.A particle with the momentum Po (in the Kframe) disintegratesinto 'two particles with masses ml and mz. The disintegrationenergy Qturns into kinetic energy. Draw the vector diagram of momentafor this process to find all possible momenta PI and PI of the generatedparticles.

Solution. This process appears to be the simplest in the C frame:here the disintegrating particle is at rest while the generated particles

Fig. 71

move in opposite directions with momenta equal in magnitude PI == pz = p. Since the disintegration energy Q turns entirely into thetotal kinetic energy T of the generated particles,

p= V2t1T= y 211Q,

where 11 is the reduced mass of the generated particles.Now let us find the momenta of these particles in the K frame.

Making use of the velocity transformation formula for the transitionfrom the C to the K frame, we can write:

PI =mlvi =ml (Vc+V';.)= mlVc +PI'

P.= m.v.= mz(Vc+Ys)= mlVc + PI

with Pi + PI = Po in accordance with the momentum conservationlaw.

Using these f.ormulae, we can draw the vector diagram of momenta(Fig. 71). First, we draw the segment AB equal to the momentum Po·Then we draw a circle of radiusp from the point 0 to divide the seg­ment AB into two parts in the ratio ml : mz. This circle. is the locusof all possible positionsofthevertex C of the momenta trIangle ABC.

• 4.10. A flatcar starts rolling at the moment t = 0 due to thepermanent force F. Ig~oring friction in the axes, find the time de­

. pendence of the velOCIty of the flatcar if(1) it is loaded with sa~d which pours ·out through a hole in the

b!?ttom at the constant rate 11 kg/s, and the initial mass of the flatcarWith sand at th~ moment t ~ 0 is equal to mo;

(2) the sand IS loaded on It from a stationary hopper at the per-.mhanent rate 11 kg/s, starting from the moment t = 0, when it hadt e mass mo. .

Solution. 1. In t.his case the reactive force is equal to zero and theMeshchersky equation (4.26) takes the form (m - lit) dv/dt = Fwhence 0 r ,

dv = F dt/(mo - I1t).

I~tegrating this expression with allowance made for the initial condi­tIOns, we get

v = (F/I1) In [mol(mo - I1t».

(h.?·hI~ tthhiS, calse the ho~izontal comp!?nent of the reactive force

w IC IS. 1/1- on y one of Interest here) IS R = 11 (-v), where v isthhe vldelobcltYkOf t~e flatcar. That is why the Meshchersky equations ou e ta en III the fs.>rm (4.28), or

d (mv) = F dt.

I~tegrating thi~ equation with allowance made for the initial c d'tIOns, we obtam on 1-

mv = Ft,

where m = mo + I1t. Hence,

v = Ft/(mo + lit).

I Ne.edlehss to say, the expressions obtained in both cases are vaiidon y m t e process o! unloading (or loading) a flatcar.

._ 4.11. A spaceship of mass mo moves with the constant,,· l·"itvVo III t~e abse~ce ?f a!field of force. To change the direction of I~~tion~ reactlv~ engIne IS started whose jet moves with the constant veloc:~ty thU With h.s~ec~.to t~e spaceship and is directed perpendicularo e ~p~ces IP.S uectIOn of motion. The engine stops when thesp~ceshlp s mass I~ equal to m. Fi.nd how much the course of the s ace-ship cha~ges durmg the operatIOn of the engine. p

SolutlO~. Let us. find. the increment of the spaceshi '5 velocitvector durmg the .tlme mterval dt. Multiplying both ~des of th~Meshchersky equatIOn (4.26) by dt and taking into account th tF = 0, we get a

dv = u dm/m.

10-0539

Here dm < O. Since the vectqr u is always perpendicular to the vec­tor v (the spaceship's velocity), the modu~us of the vector v does notchange and retains its original magnItude: I v I = I Vo I = Vo·lt follows from this that the rotation angle d(1. of the vector v thatoccurs during the time interval dt is equal to

d(1. = I dv Ilvo = (ulvo) I dmlm J.Integrating this equation, we obtain

(1. = (ulvo) In (molm).

146 Classical MechanicsCHAPTER 5

THE LAW OF CONSERVATION OF ANGULAR MOMENTUM

§ 5.1. Angular Momentum of a Particle.Moment of Force

The analysis of the behaviour of systems indicates thatapart from energy and momentum there is still anothermechanical quantity also associated with a conservation law,the so-called angular momentum·. What is this quantityand what are its properties?

First, let us consider one particle. Suppose r is the radiusvector describing its position relative to some point 0 ofa chosen reference frame and p is its momentum in thatframe. The angular momentum of the particle A relativeto the point 0 (Fig. 72) is the vector L equal to the vectorproduct oj the vectors rand p:

/

IL=[rp)·1 (5.1) .

'~-'

It follows from thi~ definition that L is an axial vector;Its direction is chosen so that the rotation about the point 0toward the vector p and the vector L correspond to a right-

•handed screw. The modulus of the vector L IS equal to

L = rp sin ~ = lp, (5.2)

where ex is the angle between rand p, and 1 = r sin ex is thearm of the vector p relative to the point 0 (Fig. 72);

The equation of moments. Let ut} determine what mechan­ical quantity is responsible for the variation of the vectorL in a given reference frame. For this purp6Sl:l we differen­tiate Eq. (5.1) with respect to time:

dUdt = [dr/dt, pI + [r, dpldt].

Since the point 0 is stationary, the vector dr/dt is equal tothe velocity v of the particle, I.e. coincides in its directionwith the vector p; therefore

[dr/dt, p] = O.

• The following names are also used: moment of momentum, mo­ment ofquanmy of motion, rotational moment,· or simply moment.10·

148 Classical MechanicsThe Law of Conservation of Angular Momentum 149

Fig. 75

P A

Fig. 74

I

"0 S

if the moment of all the forces acting on a particle relativeto a certain point 0 of a chosen reference frame is equal tozero during the time interval of interest to us, the angularmomentum of the particle relative to this point remainsconstant during that time.

Example 1. A planet A moves in the gravitational field of theSun S (Fig. 74). Find the point in the heliocentric reference framerelative to which the angular momentum of that planet does notchange in the course of time.

First of all, let us define what forces act on the planet A. In thegiven case it is only the gravitational force F of the Sun. Since duringthe motion of the planet the direction of this force passes through

the centre of the Sun, that point is the one relative to which the mo­ment of the force is equal to zeroa nd the angular momentum of theplanet remains constant. The momentum' I' of the planet changesm the process.

Exampie 2. A disc A moving on a smooth horizontal plane reboundsela.stically !rom a sm~oth vertical wall (Fig. 75, top view). I<'ind thepomt relatIve to whlCh the angular momentum of the disc remainsconstant in this process.

The disc experiences gravity, the force of reaction of the horizontalsurface, and the force R of reaction of the wall at the moment of theimpact against it. The first two forces counterbalance each otherleaving only the force R. Its moment relative to any point of th~line along which the vector R acts is equal to zero, and thereforethe angularmomentllm of the disc relative to any of these pointsdoes not change in the given process.

Example 3. On a horizontal smooth plane there are a motionlessvertical cylinder and a disc A connected to the cylinder by a thread AB(Fig. 76, top view). The disc is set in 1II0tion with the initial velocity vas shown in the ligure. Is there any point here relative to which theangular momentum of the disc is invariable in the process of motion?

LI/"

(5.5)

Fig. 73

I dL/dt=M. I

Fig. 72

to the point 0 (Fig. 73). Denoting it hy the letter M, wewrite

IM = [rFl·1 (5.3)

The vector M, like the vector L, is axial. Similarly to (5.2)the modulus of this vector is eqUal to '

M = iF, (5.4)

where i is the arm of the vector F relative to the point 0(Fig. 73). "

Thus the time derivative of the angular momentum Lof the p'article relative to some point 0 of the chosen referenceframe is equal to the moment M of the resultant force Frelative to the same point 0:

This equation is referred to as the equation of moments.Note that in the case of a noninertial reference frame themoment of the force M includes both the moment of theinteraction forces and the moment of inertial forces (relativeto the same point 0).

Among other things, from the equation of moments (5.5)it follows that if M == 0, then L = const. In other words,

Next in accordance with Newton's second law dp/dt = F,wher~ F is the resultant of all the forces applied to theparticle. Consequently,

dL/dt = [rFl.

The quantity on the right-hand side of this equation. 'isreferred to as the moment of force, or torque, of F relatIve

150 Classical MechanicsThe Law 01 Conservation 01 Angular Momentum 151

The quantity on the right-hand side of this equation isreferred to as the momentum of the force moment, or thetorque momentum. Thus, the increment of the a~gular mo­mentum of a particle during any time interval IS equal tothe momentum of the force moment during the same time.

Let us consider two ex-amplos.

Fig. 78

bt Z

lig. 77

a

Example 1. The angular momentum of a particle relative to a cer­tain point varies in the course of time t as L (t) = a + btS, wherea and b are certain constant vectors, with a .1 b. Find the force mo­ment M acting on the particle when the angle between the vectors Mand L is equal to 45°.

In accordance with Eq. (5.5) M = dL/dt = 2bt, i.e. the vector Mcoincides in its direction with the vector b. Let us depict the vectors M

//a

and L at somo, m~ment t,,(Fig. 77). It is seen from the figure that theangle a = 45° at the moment to, when a = bt~. Hence, to = V alband M = 2 y(iJfj b.

Example 2. A stone A of mass m is thrown at an angle to thehorizontal with the initial velocity Vo' Ignoring the air drag, findthe time dependence L (t) of the angular momentum of the stone rela­tive to the point 0 from which the stone was thrown (Fig. 78).

During the time interval dt the angular momentum of the stonerelative to the point 0 increases by dL = Mdt = [r, mg) dt. Sincer = vot + gt2/2 (see p. 17), dL = (vo, mg) t dt. Integrating thisexpression with allowance made for the in~tial condition. JL (0) = 0at t = 0), we get L -(t) = [vo, mg} t2/2. It IS seen from thIS that thedirection of the vector L remains constant in the process of motion(the vector L is directed heyond the plane of Fig. 78).

The angular momentum and the force moment relativeto an axis. Let us choose an arbitrary motionless axis zin a given reference frame. Suppose the angular momentumof the particle A relative to a certain point 0 of the z axisis equal to L and the force moment acting on the particleis equal to M.

The angular momentum relative to the z axis is the pro­jection of the vector L, defined with respect to an arbitrarypoint 0 of the given axis, on that axis (Fig. 79). The con­cept of a force moment relative to an axis is introduced ina similar fashion. They are denoted by L z and M z re-

(5.6)

The only uncompensated force acting on the disc A iJ.l this ~seis the tension F of the thread. It is easy to see that t~er:e IS ~o POIJ.lthere relative to which the moment of the for~ F IS I!1variable. mthe process of motion. Therefore, there is no pomt relatIve to whIch

the angular momentum of thedisc would vary.

This example illustrates thatsometimes a point relative to whichthe angular momentum of a particleis constant cannot be found at all.

The equation of momentaFig. 76 (5.5) make~ it possible to solve

, the followmg two problems:(1) find the force moment M relative to a certain point 0

at any moment of time t if the time dependence of theangular momentum L (t) of a particle relative to the samepoint is known;

(2) determine the increment of the angular moment.um ~f

a particle relative to a point 0 at any moment of tIme Ifthe time dependence of the force mo~~mt ~ (t) acting onthis particle (relative to the same pomt) IS known.

The solution of the first problem reduces to the calcula­tion of the time derivative of the angular momentum, thatis, dLldt, which is equal, in accordance with Eq. (5.5),to the sought force moment M. . .

The second problem is solved by mtegratmg Eq. (5.~).

Multiplying both sides of this e~~ation b! dt, we obtamthe expression dL = M dt determl~mg t~e mcrement of. thevector L. Integrating this expressIOn With respect !o tI~e,

we get the increment of the vector L over the fimte tImeinterval t:

o8

152 Classical Mechanics The Law of Conservation of Angular Momentum 153

spectively. We shall see later that L z and.Mz do not dependon the choice uf the point 0 on the z aXIS. .

Let us (OXamine the properties of these quantities. Pro­jecting Eq. (5.5) on the z axis, we obtain

dLz/dt=M z, (5.7)

Le. the time derivative of the angular momentum of theparticle relative to the z axis is equal to the force moment

direction of increasing coordinates, to the particle A (Fig. 81).In this coordinate system the radius vector r and momentump of the particle are written in the form

r = pep+ze, p = ppep+pq;elp +pzez,

where PP' Plp' P z are the projections of the vector p on thecorresponding unit vectors. It is known from vector algebra

i'

that the vector product [rp] can be represented via thefollowing determinant:

(5.9)

Fig. 81

o

Fig. 80

ep elp ez IL= [rp] = p 0 z

PP ptp pz IFrom this it i.mmediately seen that the angular momentumof the particle relative to the z axis is .

L z = PPlp' (5.8)

where p is the distance of the particle from the z axis. Letus reduce this expression to a form more suitable for prac­tical applications. Taking into account that Plp = mv'l' == mp<U z, we get

where (t) z is the projection of the angular velocity (i) withwhich the radius "ector of the particle rotates.

Fig. 7~

relative to the same axis. In particular, if M z == 0, thenL z = const. In other words, when the force moment relativeto a fixed axis z is equal t.o zero, the angular momentumof the particle relative to that axis remains constant. Thevector L can, however, vary in the process.

Example. A small body of mass m suspended on a th~ad movesuniformlY along a horizontal circle (Fig. 80) due to graVIty mg and/the tension T of the thread. The angular momentum of the bodyrelative to the point 0, the vector L, is located in the same planeas the z axis and the thread. During the motion of the body the ve~torL rotates continuously under the action of the moment M of ~ravIty.Le. it varies. As for the projection Lz' it remains constant smce thevector M is perpendicular to the z axis and M z = o.

Now let liS find analytical expressions for L z and !If z.It is easy to see that this pl'Oblem reduces to the determina­tion of the projections of the vector pro(\lIcts [rpl and [rFIon the z axis.

We shall make use of the cylindrical coordinate systemp, <p, z, fIxing the unit vectors cp , elp' e z, oriented in the

: I

r

154 Classtcal Mechanics The Law of Conservation of Angular Momentum 155

The force moment relative to the z axis is written similarlyto Eq. (5.8): .

M z = pF'fl' (5.10)

where Fq, is the projection of the force vector F on the unitvector e'll'

Note that the projections L z and M z are indeed indepen­dent of the choice of the point 0 on the z axis, relative towhich the vectors Land M are defined. Besides, it is seenthat L z and M z are algebraic quantities, their signs corre­sponding to those of the projections P'll and F 'fl'

Let us demonstrate that the total moment of all internalforces relative to any point is equal to zero. Indeed, theinternal forces are the forces of interaction between theparticles of the given system. In accordance with Newton'sthird law these forces are pairwise equal in magnitude, areopposite in direction and lie on the same straight line, thatis, have the same arm. Consequently, the force moments ofeach pair of interaction are equal in magnitude and oppo!'>itein direction, Le. they counterbalance each other, and hencethe total moment of all internal forces always equals zero.

As a result, the last equation takes the form

(5.11)

(5.12)

(5.13)

where M is the total moment of all external forces, M == ~Mi' "

Eq. (5.12) thus asserts that the time derivative of the angularmomentum of a system is equal to the total moment of all exter­nal forces. It is undefstood that both quantities, Land M,are determine'd relati.e to the same point 0 of a given refer­ence frame.

As in the case of a single particle, from Eq. (5.12)' itfollows that the increment of' the angular momentum of asystem during the finite time interval t is

Le. the increment of the angular momentum of a system isequal to the momentum of the total moment of all externalforces during the corresponding time interval. Of course,the two quantities, Land M, are also determined here rela­tive to the same point 0 of a chosen reference frame.

Eqs. (5.12) and (5.13) are valid both in inertial and non­inertial reference frames. However, in a non-inertial refer­ence frame one has to take into account the inertial forcesacting as external forces, i.e. in these equations M shouldbe regarded as the sum Mia -+- Min, where Mia is the totalmoment of all external forces of interaction and Min is the

Let us choose an arbitrary system of particles and intro­duce the notion of the angular momentum of that systemas the vector sum of angular momenta of its individualparticles:

§ 5.2. The Law of Conservationof Angular Momentum

where all vectors are determined relative to the same point 0of a given reference frame. Note that the angular momentumof the system is an additive quantity: the angular momentumof a system is equal to the sum of the angular momenta ofits individual parts, irrespective of whether they interactor not.

Let us clarify what quantity defines the change of theangular momentum of the system. For this purpose wedifferentiate Eq. (5.11) with respect to ¥me: dUdt == ~ dLi/dt. In thepr~n it was shown that thederivative dLi/dt is equal to the moment of all forces actingon the ith particle. We represent this moment as the sum ofthe moments of internal and external forces, Le. Mi + Mi.Then

dL/dt= ~ Mi +~ Mi'

Here the first sum is the total moment of all internal forcesrelative to the point 0 and the second sum is the total momentof all external forces relative to the same point.

156 Classlcal Mechanics The Law of Conservation of Angular Momentum 157

total moment of inertial forces (relative to the same point 0of the reference frame).

Thus, we have reached the following significant con­clusion: in accordance with Eq. (5.12) the angular momentumof a system can change only due to the total moment of allexternal forces. From this immediately follows another im­portant conclusion, the law of conservation of angularmomentum:

in an inertial reference frame the angular momentum ofa closed system of particles remains cunstant, i.e. does notchange with time. This statement is valid for an angularmomentum determined relative to any point of the inertial-reference frame.

Thus, in an inertial reference frame the angular momen­tum of a closed system of particles is

At the same time the angular momenta of individual partsor particles of a closed system can vary with time, a factemphasized in the last expression. These variations, how­ever, occur in such a way that the increment of the angularmomentum in one part of the system is equal to the angularmomentum decrease in another part (of course, relative tothe same point of the reference frame).

In this respect Eqs. (5.12) and (5.13) c be regarded asa more general formulation of the angular mentum con­servation law, a formulation specifying the ca e of varia­tion of the angular momentum of a system, whIch is theinfluence of other bodies (via the moment of external forcesof interaction). All this, of course, is valid only in inertialreference frames.

Once again we shall point out the following: the law ofconservation of angular momentum is valid only in inertialreference frames. This, however, does not rule out caseswhen the angular momentum of a system remains constantin non-inertial reference frames as well. For this, it issufficient that, in accordance with Eq. (5.12), which holdstrue also in non-inertial reference frames, the total momentof all external forces (including inertial forces) be equal to

Ii

I

I..!I

tI

IL = ~ L i (t) = const·1 (5.14)

zero. Such.circumstances are realized very seldom, and thecorrespondmg cases are exceptional.

T.he law of conservation of angular momentum is justas Important as the energy and momentum conservation~aws. In ~any cases this law by itself enables us to drawImportant I.nference~ab?ut the ~ssential aspects of particularprocesses WIthout gomg mto theIr detailed analysis. We shallIllustrate this by the following example.

Example. Tw.o identical spheres are mounted on a smooth horizon­tal bar along WhICh they can slide (Fig. 82). Initially the spheres arebrought together and connected by a thread. Then the whole assembly

Thread

f Fig. 82

i~ set into rotation about a vertical axis. After a period of free rota­tion, the thread is burned up. Naturally, the spheres fly apart towardthe ends of the bar. At the same time, the angular velocity of theassembly drops drastically. _

The o~served phenomenon is a direct consequence of the law ofconservation of ,angular momentum, for this assembly behaves asa clos,ed. system (t~e external forces counterbalance one another andthe frictIOn f<?rces In the axis a.re small). To assess quantitatively theangular velOCIty chaI!'ge, let us assume the mass of the whole assemblyto be concentrated In the spheres and their size to be negligibleThen, fro~ the.equali~Y. o.f the angular momenta of the spheres relativ~to the POInt c: In the IDltial and final states of the system 2m [r v J == 2m [r.v.) It follows that - I J

rfOOI =rIOOI'

It is seen that as the distance r from the spheres to the rotation axisgrows,. the angul~r mom~ntum of the assembly decreases (as 1/r2).And VIC~. versa, If the distance between the spheres decreases (dueto ~ome Internal forces), the angular velocity of the assembly increasesThIS general phenomenon is widely used by, for example figur~skaters and gymnasts. '. Note that the final result is quite independent of the nature ofInternal forces (here the friction forces between the spheres and thebar).

158 Classical Mechanics The Law of Conservation oj Angular Momentum 159

(5.16)

(5.17)

At the saae time the vector L, defined relative to an arbit­rary point 0 on that axis, may vary. For example, whena system moves in .. uniform gravitational field the totalm~ment o~all gravi~tionalforces relative to any 'stationarypomt 0 IS perpendICular to the vertical, and thereforeM z = 0 and L z = const relative to any vertical axis. ';fhiscannot be said about the vector L itself.

The reasoning leading to the law of conservation of angu­lar momentum is based entirely on the validity of Newton'slaws. But what about systems that do not obey those laws,e.g. the systems with electromagnetic radiation atomsnuclei, etc.? . "

Because of the immense role that the law of conservationof angular momentum plays in mechanics;' the conceptof angular momentum is extended in physics to non-mechan­ical; systems (which do not obey Newton's law) and the lawof conservation of angular momentum is postulated for allphysical processes.

The law of conservation of angular momentum thus extendedis no longer a consequence of Newton's laws; it representsan tndependent general principle generalized from experimentalfacts. Together with the energy and momentum conserva­tion laws the law of conservation of angular momentum isone of the most important, fundamental laws of nature.

where L i z and M t z are the angular momentum and the mo­ment of external forces relative to the z axis for the ithparticle of the system.

It follows from Eq. (5.15) that if M z == 0 relative to somestationary axis z in a given referenc~ frame, the angularmomentum of the system. relative to that axis does notchange:

Here L z andMz are the angular momentum and the totalmoment of external forces relative to the z axis:

:1I

Of special interest are ~ases in which the ang~lar ~omen­tum L remains constant m non-closed systems m whIch themomentum p is known to change with time. If the totalmoment of external forces relative to a point 0 of the chosenreference frame M == 0 during the time interval considered,then in accordance with Eq. (5.12) the angular momentumof the system relative to·thepoint 0 remains constan~ duringthis time interval. Generally speaking, such a pomt maynot exist in non-closed systems, so that the fact of its exis­tence should be established in every concrete case.

Example 1. The Earth-Moon system moving in the gravitatio~alfield of the Sun is non-closed. Its momentum continuously variesdue to gravitational forces. However, there is one point here relativeto which the moment of the gravitational forces acting on this systemis always equal to zero. This point is the centre of the Sun. Therefore,

it can be immediately claimedthat the angular momentum of theEarth-Moon system relative to theSun's centre remains constant.

Example 2. A rod OB lying ona smooth horizontal plane canrotate freely about a stationary:ver­

o A/ tical axis passing through the rod'sE tIrr::d end 0 (Fig. 83). A disc A moving

along the plane hits the rod's end B===========::$ and gets stuck there, whereuponthe whole system starts rotating

Fig. 83 as a single unit about the point O.It is clear that the disc and the

rod compose a non:closed sys~m:

apart from the forces counterbalancing each ~t~er m the vertl~aldirection. a horizontal force exert.ed by the. axiS IS generat«:d d~ngthe impact, while during the rotation the axiS exerts a ~orce Impelbngthe centre of inertia of the system to move along the cllcle. But bothof these forces tass through the point 0, andthe!'8fore the m.omentof these externa forces is always equal to zero (relatlVe to the pomt 0).Hence. the following conclusion can be.drawn: the a~gular momentumof this system remains constant relative to the pomt O.

Infrequently in non-elosed systems it is not t~e ang~larmomentum L itself that remains constant, but Its pro~ec­

tion on a stationary axis z. This happens when the proJec­tion of the total 'RomentM. of all ex;terl!al forces on thataxis I is equal to o. In fact, proJectmg Eq. (5.12) onthe z axis, we obtain

dLz/dt = M z' (5.15)

i

II I ~"I'"III

1,:1

1

11

1

.

Ill i

Iii!

111'11

',II

1

'1

1'·.. ·1.'.,1[1, I

'.1.:.1,,1,.1'III!I!

"I'

1.".'1.

11

.'1.I'"I'ii' I

"I' !

1.'11

"

"I'

II',.",I:j'i

161

(5.19)

The Law of Conservation of Angular Momentum

or

Indeed, the inertial force acting on each particle of thesystem Fi = -miwo, where Wo is the acceleration of theC fr~me. Consequently, the total moment of all these forcesrelatIve to the centre .of inertia C is

M~=~ [rt, -miwO] == - [(L miri), wol.

In accordance with Eq. (4.8) ~ miri = mrc, and as inour case rc == 0, then M~n = 0.

Internal angul~r momentum. Generally speaking, angu­lar .momentum, J.ust as the .force moment" depends on thechOIce of the POlllt 0 relatIve to which it is determinedWhen that ~oint is transferred by the distance ro (Fig. 84):the new radIUS vectors ri of the particles are related to theold ones ri by means of the formula ri = ri + roo Con­sequently, the angular momentum of the system relativeto the point 0 can be written as follows:

L = ~ [rtp,) = 2J [ripiJ + 2J [rop,),

the_C frame, and therefor.e in. accordance with Eq. (4.14)F - Fia + Fin ~ O..Ta~lllg mto account Eq. (5.18), wereach the followlllg sIgnIficant conclusion: in the C framethe total moment of all external forces, including inertialforces, does ,!ot depend. on the choice of the point O.

And here IS anothe~ Iml?ortant conclusion: in the C framethe total moment of lnert~al forces relative to the centre ofinertia is always equal to zero:

L = L' +[ropJ, (5.20)

where L' is the angular momentum of the system relativeto the point 0', and p = 2.1 Pi is the total momentum ofthe system.

From Eq. (5.20) it follows that if the total momentum ofthe system p == 0: then its angular momentum does notdepend on. the ~holce of the point O. This distinguishes HieC frame, III whIch t.he system of particles as a whole is atrest. Hence, we reach the third important conclusion: in theC frame the angular momentum of a system of particles is11-0539

(5.18)

Classical Mechanics

M=M' +[roFJ,

0'Fig. 84

o

or

where F = 2J F i is the resultant of all external forces.Eq. (5.18) shows that when F = 0, the total moment of

external forces does not depend on the choice of the pointrelative to which it is determined. In particular, such isthe case when a couple acts on a system.

In this respect the C frame possesses one interesting andimportant characteristic (rec I that this reference frameis rigidly fixed to the centre of in tia of a system of particlesand translates with respect to i tial frames). Since inthe general case the C frame is non-in rtial, the resultant ofall external forces must include not only the external forcesof interaction Fia but also the inertial forces Fin' On theother hand, the system of particles as a whole is at rest in

160

§ 5.3. Internal Angular Momentum

It was established in the foregoing section that the angularmomentum L of a system changes only due to the totalmoment M of all external forces; it is the vector M thatdefines the behaviour of the vector L. Now we shall examinethe most essential properties of these quantities together

with the significant conclusions fol­lowing from those properties.

The total moment of· externalforces. Just as the moment of an in­dividual force, the total moment offorces depends, generally speaking,on the choice of a point relative towhich the moment is determined.Let M be the total moment offorces relative to the point 0 and M'relative to the point 0' whose ra-dius vector is equal to ro (Fig. 84).

Let us find the relationship between M and M'.The radius vectors ri and ri of the point at which the force

Fi is applied are related as ri = ri + ro (Fig. 84). Conse­quently, M may be written in the following form:

,II~I

!L=L+[rcp),\ (5.23)

Le. the angutar momentum L of a system of particles comprisesits internal angular momentum L and the momentum IreP),associated with the motiun of the system of articles as a single

unit. 11'Let us consider, for example, a uniform here r? mg

down an inclined plane. Its angular momentu relative tosome point of that plane is composed of the angula~ m0!Den­tum associated with the motion of the centre of mertIa ofthe sphere and the internal angular II!-0men.tum associatedwith the rotation of the sphere about Its aXIs.

Specifically, it follows from Eq. (5.23) that if the centreof inertia of a system is at rest (the momentum of the s~stem

p = 0). then its angular momeutum L represents the mter-

independent of the choice of the puint relati ve to which it isdetermined. We shall refer to this quantity as the internalangular momentum of a system and denote it by 1.

Relation between Land L. Suppose L is the angular mo­mentum of a system of particles relative to the point 0of the K reference frame. Since the internal angular momen­tum L in the C frame does not depend on the choice of thepoint 0', this point may be taken coincid~nt with the pointo of the K frame at a given moment of time. Then at thatmoment the radius vectors of all the particles in both refer­ence frames are equal (ri = rl) and the velocities are relatedby the formula

Vt = Vt + Vc, (5.21)where V is the velocity of the C frame relative to the K

c 'tframe. Consequently, we milY Wfl e

L= ~ mt [rtvtl = ~ mt [rtVt) +~ mt [rtVel. (5.22)The first sum on the right-hand side of this equality is theinternal angular momentum -L. The second sum may bewritten in accordance with Eq. (4.8) as m [reVc1, or [rcp],where m is the mass of the whole system, rc is the radiusvector of its centre of inertia in the K frame and P is thetotal momentum of the system. Finally, we obtain

toil

(5.24)fdLldt=Mc, I

the Law oj Conservation of .Angular Momentum

nal angulai' momentum. We are already familial' with ihatcasc.. In another extremc casc, when L = 0, the angularmomentum of a system relative to some point is determinedonly via the momentum associated with the motion of thesystem as a single unit, Le. by the second term of Eq. (5.23).This is how, for example, the angular momentum of anysolid behaves during its translation.

Equation of moments in the C frame. We pointed outin the previous section that Eq. (5.12) holds true in anyreference jrame. Consequently, it is valid in the C frame aswell, and we can immediately write: dI.!dt = M, where AIis the total moment of external forces in the C frame.

Since the C frame is non-inertial in the general case, )fincludes not only the moments of external forces of interac­tion, but also the moment of inertial forces. On the otherhand, at the beginning of this section (see p. 161) the forcemoment Ii in the C frame was shown to be independent ofthe choice pf the point relative to which it is determined.Usually, the point C, the centre of inertia of the system, istaken as such a point: The choice of this point is advantageousbecause the total moment of inertial forces relative to itis equal to zero, so that one must take into accountonlgthe total moment M c of external forces of interaction.Thus.

Le. the time derivative of the internal angular momentumof a system is equal to the total moment of all external forcesof interaction relative to the centre of inertia of that system.

In particular, when Me = 0, then L = const, Le. theinternal angular momentum of a system does not vary.

When written in projections on the z axis passing throughthe centre of inertia ofa system, Eq. (5.24) takes the form

dlzldt = M Cz (5.25)where M Cz is the total moment of external forces of inter­action relative to the z axis fixed in the C frame and passingthrough the centre of inertia. Herc again. if JIll Cz = 0,then L z = const.

I

I

Classical Mechanics162

11*

etassical Mechanics The Law of Conservation of Angular Momentum i65

F

/

Fig. 85

of the ~orce F, but a line along which it acts. Knowing themodulI M and F of the corresponding vectors, we can findthe arm l of the force F (Fig. 85): l = M/F.

,!hus, if M ~ F, then the system of forces acting on variouspomts of a solId may be replaced by one eqUivalent forcewhich is equal to the resultant force F and produces a momentequal to the total moment Mof all external forces. M

A uniform field of force, e.g.the gravitational field, canserve as an example. In such afield each particle experiencesthe force F i = m"g. In this O'-'::-.....---\-..:'----~..... .....,case the total moment of grav- z.....itational forces relative toany point 0 is equal to '

M=L [rt~'mij:d = [(2} miri) gl.In accordl!nce with tEq. (4.8)the sum in parenthe~esis equal .to mre, where m is the mass of a body and re is the radiusvector of its centre of inertia relative t{) the point O. There-fore, ,

M = [mrc, gl = [re, mgl.

This implies that the equivalent force mg of the gravitationalforces passes through the centre of inertia of the body. It iscustomary to say that the equivalent force of gravity is"applied" to the centre of inertia of a body, or to its centreof gravity. Clearly, the moment of this force relative to thecentre of inertia of it body is equal to zero.

Now we shall move on to an' examination of the fourspecific cases of motion of a solid: (1) rotation about a sta­t~onary axis, (2) plane (two-dimensional) motion, (3) rota­tion about free axes, (4) the special case of motion when abody has only one motionless point (gyroscopes).

1. Rotation about a stationary axis. First let us find theexpression for the angular momentum of a solid body relativeto the ;otation axis 00' (Fig. 86). Making use of Eq. (.5.9),we write

§ 5.4. Dynamics of a Solid

In the general case the motion of a solid is defined by twovector equations. One of them is the equation of rn.-otion ofthe eentre of inertia (4.11), and the othertheequatiOn relat­ing momenta and moments in the C frame, Eq. (5.24):

mdVcldt =F; dLldt=Mc' (5.26)

If the laws of acting external forces, the points of theirapplication and the initial,conditions are ~~own, theseequations provide the velocIty and, the posItion o~ eachpoint of a solid at any moment of time, I.e. make It pos­sible. to solve the problem of motion of a solid completely.However, despite the apparent simple form of Eqs. (5.26),their solution in the general case is far from easy. First ofall this is because the relationship between the internalangular momentum Land the velocities of ind~vidual pointsof a solid in the C frame turns out to be complIcated, exceptfor a few special cases. We shall not consider this problemin the general case (it is solved in the general the.ory) andshall limit ourselves hereafter to only several speCIal cases.

But first we shall quote some considerations which followdirectly from the very appeara~ce ~f Eqs. (5:26). ~learly,translation of forces along the dnectIOn of then actIOn doesnot affect either the resultant F or the total moment M c'Eqs. (5.26) do not vary either, and therefore the mo.tion ofa solid remains the same. Consequently, the pomts ofapplication of external forces ca.n be transferred alon.g thedirection of their action, a techlllque used very extensIVely.

Equivalent force. In those cases when the total moment ofall extern!!.l. forces turns out to be perpendiCUla~o the resul­tant force;,<.l,e. M ..L F, all externa.l force~ ma . be reducedto one force F acting along a certam straIght me. In fact,if the total moment relative to some point 0 ..L F, thenwe can always find a vector ro ..L M (Fig. 85 , such thatwith the given M and F

M = [roFl.

Here the choice of 1'0 is not unambiguous: adding at;ly vect~rl' parallel to F, we do not violate the ~ast e9,ualIt.y. ~hI~means that this equality defines not a pomt of applIcatiOn

Hi6 Classical Mechanics 1'he Law of Conservation of Angular Momentum 167

(5.30)

In terms of mathematics the.calculation of the moment ofinertia of a solid body of an arbitrary shape relative to oueor another axis presents, generally speaking, a painstakingtask. In certain cases, however, the moment of inertia canbe calculated much more easily if one reserts to the Steinertheorem: the moment of inertia I relative to an arbitrary zaxis is equal to the moment of inertia I c relative to the Zcaxis which is parallel to the z axis and passes through thecentre of inertia C of the body, plus the product of the massm of the body by the square of the distance a between theaxes:

/1=Ic +ma2 .1 (5.29)

The theorem is proved in Appendix 3.Thus, when the moment of inertia leis known, the

moment of,lfnertia I is calculated with no effort. For example,the moment of inertia of a thin rod (of mass m and length I)relative to an axis JlOrpendicular to the rod and passingthrough its end is equal to

1=1; m12 +m(fr= ~ m12 •

The fundamental equation of rotation dynamics of a solidbody (stationary axis of rotation). This equation can beeasily obtained as a consequence of Eq. (5.15) by differen­tiating Eq. (5.27) with respect to time. Then

where M z is the total moment of all external forces relativeto the rotation axis. Specifically, from this equation themoment of inertia 1 is seen to determine the inertial Vl'Op­erties of a solid body during its rotation: the same momentof forces M z inducefll a smaller angular acceleration ~ zin bodies possessing greater moments of inertia.

Recall that force moments relative to an axis are algebraicquantities: their sign~ depend on both the choice of thepositive direction of the z axis (coinciding with the rotationaxis) and the "rotation" direetion of the corresponding forcemoment. For example, choosing the positive direct-ion of thez,. axis 8S shown in Fig. 87, we thereby specify the positive

Solid body I Zc axis I'Moment of Inertl~\ I c

A thin rod of Is perpendicular to \ .!- IIIlZlength l the rod 12

A uniform cylinder Coincides with the axis !. mRSof radius R of the cylinder 2

A thin disc of Coinc.ides with the !. mRSradiusR diameter of thl' II ise 4

A sphere of radius R Passes throllgh the ~ mRS('en tre of the sphere 5

h . and p. are the mass of the ith particle o.f a solidwere m r I • d "t I rand its distance from the rotation aXIS, an (iI z IS I .s angu avelocity. Denoting the quantity in parentheses by I, we get

1L z = I (ilz, I (5.27)

where I is the so-called moment of inertia of a solid relativeto the 00' axis:

. 1= }J mlPl. (5.28)

It is easy to see that tIre moment o~ inertia of a.so.lid dep~ndson the distribution of masses relatIve to th~ ~XIS \D que~tlOn,

and is an additIVe quantIty.The ~moment of inertia of a

hody is calculated by means ofthe following formula:'

I = ~ r 2 dm = ~ pr2 dV,. "

where dm -and dV are the massand the volume of an element ofthe body located at the ~distancer from the z axis chosen, and pis the density of the body at a

D given point.The 'moments of inertia of

Fig. 86 some liniform solids relativeto the Zc axis passing thl'Ough their centres of inertia arelisted in the following table (here m is the mass of the body):

168 Classtcal Mechanics The Law of Conservation of Angular Momentum 169

(5.33)q>

A= j M: dq>.

o

whenceIDZ= (11001: +I sIDSZ)1(l1 +12), (1)

Note that OOlz' ID 2z and IDz are algebraic quantities. If OOz > 0, thecorresponding vector 00 coincides with the positive direction of the zaxis, and vice versa.

The increment of the kinetic energy of i!rotation for this system is

~T = (11+ 12) oo~/2-(11ooiz/2+ 12OO~z/2).

Replacing OOz by its expression (1), we get

1112 2~T= 2(11+ 12) (OOlZ-OOU>. (2)

The minus sign shows that the kinetic ener- Fig. 88gy of the system decreases.

Note that the results (1) and (2) are quite similar in form andmeaning to the case of absolutely inelastic collision (see p. 127).

8.'

The work performed by external forces during the rota­tion of a solid body pbout a stationary axis. In accordancewith the law of vari~tion of mechanical energy of a systemthe elementary work of all external forces acting on a solidbody is equal only to the increment of kinetic energy of thebody, as its internal potential energy remains constant, Le.llA = dT. Making use of Eq. (5.31), we may write 6A== d (/0)2/2). Since the z axis coincides with the rotationaxis, 0)2 = O)~ and

6A = 10): dO) z.

But in accordance with Eq. (5.30), IdO): = M Z dt. Sub­stituting this expression into the last equation for l)Aand taking into account that 0) Z dt = dlp, we obtain

l)A = M: dlp. (5.32)

The work llA is an algebraic quantity: if M z and dlp haveidentical signs, then l)A > 0; if the signs are opposite, thenl)A < O.

The work performed by extern$ll forces during the rotationof a solid through an angle lp is equal to

(5.31)

Fig. 87

direction of reading the angle «p (both of these directionsare associated with the right~hand screw rule). Next, if acertain moment M i : "rotates" a body in the positive direc­tion of the angle «p, that moment is regarded as positive,and vice versa; In its turn, the sign of the total moment M zdetermines the sign of ~ z' the projection of the angularacceleration i1Vector on the z axis.

By integrating Eq. (5.30) with allowance made for theinitial conditions, the values 0)0: and lpo at the initial mo­

ment of time, we can obtain a compre­hensive solution of the problem of a sol­id body rotating about a stationaryaxis, Le. obtain the time dependence ofthe angular velocity 0): (t) and the rota­tion angle lp (t).

Note that Eq. (5.30) is valid in anyreference frame fixed rigidly to the rota-tion axis. However, if the reference frame

is nQn-inertial, it shouldibe borp.e in mind that the forcemoment M: consists not only of moments of forces of inter­action with other bodies, but also moments of inertialforces.. Kinetic energy of a rotating solid body (stationary axifof rotation). Since the velocity of the ith particle of a rotating solid body is Vi = riO), we may write

T - lj m,v:12 = (~ m, pn 0)2/2

or briefly,

where I is the moment of inertia of the body relative to therotation axis and 0) is its an~ular velocity.

Example. Disc 1 (Fig. 88) rotates about a smooth vertical axiswith the angular velocitYIDl' Disc 2 rotating with the angular veloc­ity IDs. falls on disc 1. Due to friction between them the discs soonstart rotating as a single unit. Find the increment of the rotationalkinetic energy of that system provided that the moments of inertiaof the discs relative to the rotation axis are equal to II and 12 respec-tively. . I

First let us find the steady-state angular velocity of rotation.From the law of conservation of the angular momentum of a systemrelative to the z axis it follows t.hat l 100lZ + 1zOOsz = (/1 + Is) IDz'

171The 'Jaw of Conservation of An/!ular Momentum

The simultaneous solution of all three equations allows the accelera~tions We and"p,' as welras-the Ftr force,· to· be found.

to use still another, additional condition specifying- therestrictions that existing bonds impose on the motion. Thiscondition provides a kinematic relationship between thelinear and angular accelerations.

Example. A uniform cylinder of mass m and radius r rolls withoutslipping down an inclined plane forming the angle a with the hori­zontal (Fig. 89). Find the motionequation of the ~ylinder.

The standard approach to thesolution of problems of this typeconsists in:rthe following. First of 0'1all, one should-identify the forces '::Z...racting on a given body and thepoints of their application (in thiscase the acting forces include m/(,gravity, R, the normal componentof the force of reaction of the in-clined pla~and F.f.r' the static fric mgtion force) .. Then It is necessary tochoose the positive directions of .the x axis and of the rotation angle Fig. 89q> (these directions shoutd be consis-tent so that the accelerations Wcx and ~z have the identical signs),e.g. as shown in Fig. 89, right. And only after that can the equationsof motion (5.34) be written in terms of projections on the chosen posi­tive directions of x and q>:

mwcx=mgsina-Fj" le'Pz=rFjr.

In addition, the absence of slipping provides the kinematic relation­ship between the accelerations:

Kinetic energy in the plane motion of a solid body. Sup­pose a body performs a plane motion in a certain inertialreference frame K. To find its kinetic energy T in this frame,w~ shall r~sort to Eq. (4.~2). ,!,he quantity.f' entering into'thiS ~quatlOn represe~ts m thiS case the kmetic energy ofrotatIOn of the body m the C frame about an ilxis passingthrough the body's centre of -inertia. In~ withEq. (5.31) f = I cCiJ"/2, therefore we may immediately

fI

(5.34)

Classical Mechanics

mwc=F; IcPz=Mcz ,

170

where m is the mass of the body, F is the resultant of allexternal forces, Ie and M cz are the moment of inertia andthe total moment of all external forces, both moments beingdetermined relative to the axis passing through the centreof inertia of the body.

It should be borne in mind that the moment M Cz includesonly external forces of interaction in spite of the fact thatin the general case the C frame is non-inertial. This is becausethe total moment of inertial forces is equal to zero bothrelative to the centre of inertia and relative to any axispassing through that point. Therefore it can be disregardedaltogether (see p. 163).

Note also that the angular acceleration ~ z' as well as co zand qJ, are equal in both reference frames since the C frametranslates relative to the inertial reference frame K.

Integrating Eqs. (fl.34) with allowance made for tIle ini­tial conditions, we can find the relationships r c (t) andqJ (t) defining the position of a solid body at any moment t.

When finding the motion of a non-free solid body one has

When M z = const, the last expression takes a simple form:A = MzqJ.

Thus, the work performed by external forces during therotation of a solid around a stationary axis is determined bythe action of the moment M z of these forces relative to thataxis. If the forces are such that their moment M z 55 0,then they perform no work.. 2. P18n~ motion of 8 solid body (see p. 26). During planemotion'the centre of inertia C of a solid body moves in acertain plane stationary in a given reference frame K whilethe angular velocity vector ro of the body remains perma­nently perpendicular to that plane. This'signifies that thebody in the C frame performs a purely rotational motionabout the stationary (in that frame) axis passing throughthe centre of inertia of the body. At the same time, therotation of a solid is defined by Eq. (5.30), which was shownto be valid in any reference frame.

Thus, we have the following two equations describingplane motion of a solid body: .

h,e Law oj Conservation of Angular Momentum

Thus, to maintain the rotation axis in a fixed direction, itshould be s)ibjected to the moment M of some external forcesF (shown in Fig. 90). However, it is easy to see.that whene = '1/2, the vector L coincides in its direction with thevector ro, and in this case M = 0, Le. the direction of therotation axis remains invariable in the absence of externalinfluence.

A rotation axis whose direction in space remains invariablein the absence of any external forces is referred to as a freeaxis of a body.

It is proved in the general theory that for any solid bodythere are three mutually perpendicular axes which passthrough the centre of inertia of a body and can serve as freeaxes. They are called the principal axes of inertia of a body.

The determination of the principal axes of inertia of abody of an arbitrary form is a complex mathematical prob­lem. It is much simpler, however, for bod~ngsome kind of symmetry, since the position of the centre ofinertia and the direction of the principal axes of inertiapossess in this case the same kind of symmetry.

For example, a uniform rectangular parallelepiped hasprincipal axes passing through the centres of opposite faces.When a body possesses a symmetry axis (e.g. a uniformcylinder), one of its principal axes of inertia may be repre-

where v is the velocity of the element. It can be easily seenfrom Fig. 90 that th" vector 6L is perpendicular to the rodand its direction is independent of the choice of the element6m. Consequently, the total angular momentum L of therod coincides with the vector 6L in direction.

Note that in this case the vector L docs not coincide illits direction with the vector rol

During the rotation of the rod the vector L also rotateswith the angular velocity ro. In the time interval dt thevector L acquires the increment dL. whose magnitude isseen from Fig. 90 to be equal to

I dL I = L sin ('1/2 - e) ro dt,

or, in a vector form, dL = [roLl dt. Dividing both sides ofthe last expression by dt, we obtain

M = [roLl.

(5.35)

Classical 1lfechanics

Ilews mV~ I

T=-2-+-2-'

write

172

where leis the moment of inertia of the body relative tothe rotation axis passing through its centre of inertia, rois the angular velocity of the body, m is its mass, and Veis the velocity of the body's centre of inertia in the K refer­ence frame.

Thus, the kinetic energy of a solid body in a plane motioncomprises the rotation energy in the C frame and the energy

associated with the motion of(J1 M the centre of inertia. '

3. Free axes. Principal axesof inertia. When a solid bodyis set in rotation and thenleft free, the direction of therotation axis changes in space,generally speaking. To makethe body's arbitrary rotationaxis keep its direction con­stant, some forces need to beapplied to that axis.

Let us consider this problemin detail using the followingexample. Suppose the middle

Fig. 90 point C of a uniform rod isrigidly fixed to a rotation axisso that the angle between

the axis and the rod is equal to e (Fig. 90). Let us find themoment M of the external forces which should be appliedto the rotation axis to keep its direction constant duringthe rod's rotation with the angular velocity ro. In accordancewith Eq. (5.12) this moment M = dL/dt. Thus, to determineM, one should first find the angular momentum L of therod and then its time derivative.

The angular momentum L can be determined most easilyrelative to the point C. Let us isolate mentally the rod'selement of mass 6m located at the distance r from the pointC. Its angular momentum relative to that point lIL=[r, 6mv),

175The Law of Conservation of Angular Momentum

L = [ro.

tOl? as an example. It is known from experience that if theaXIS of a spinning top is tilted from the vertical, the top doesn?t fall, .b?t performs so-called precession, with its axiscIrcumscnbmg a cone about the vertical with a certainangular velocity ro'. It turns out that with an increase inthe a!1gula~ velocity ro of spinning of the top, the angularvelOCity ro of precesl'lion decreal'lel'l.~uch behavio~r of a gyroscope can be readily explained

USl!1g the. e.quatlOn of momenta (5.12), assuming W ~ w'.ThIS condItIOn, by the way, elu-cidates what is meant by the rap-id spinning of a gyroscope. In fIJ'

fact, the angular momentum Lof a precessing gyJ:oscope rela­tive to the supporting point 0(Fig. 91) may be represented asthe sum of the angular momen­tum LCI)•.~sociated with the gy_roscopei spinning about its axisand some aidditional*angular mo­mentum L caused by the gy_roscope precession about. the ver­tical axis, Le.

L=LCI)+L'.

Since the gyroscope laxis coin- mgcides with one of the· principalaxes of inertia, then in accor- '"dance with Eq. (5.36) LCI) = [ro, F' 91where [is the moment of iner- Ig.tia of a gyroscope with respect .to that. axis. Moreover, it is clear that as the precession get.s .slower, t~e corresponding angular momentum L' diminishes.If W ~ OJ , then in all practical cases LCI) ~ L', a thereforethe resultant angular momentum L essent.ially oincideswith LCI) both in magnitude and direction. We c thusassume that

Knowing the behaviour of the vect.or L, we can find the mo­tion characteristics of a gyroscope axil'l.

Classical Mechanicst74

sonted by that symmetry axis while any two mutuallyperpendicular axes lying in the plane perpendicular to thosymmetry axis and passing through the body's centre ofinertia can be chosen as the other axes. Thus, in a body pos­sessing axial symmetry only one of the principal axes ofinertia is fixed. In a body possessing central symmetry (e.g.in a uniform sphere) any three mutually perpendicular axe~

passing through the body's centre can be chosen as the prin­cipal axes of inertia, that. is, not a single principal axis ofinertia is fixed with respect to the body.

Tt-..e important characteristic property of the principalaxes of inertia of a body is the fact that during the rotationof tlHi body about any of them the angular momentum L ofthe body coincides in its direction with the angular velocityro of the body and is determined by the formula

L = [ro, (5.36)

where [ is the moment of inertia of the body relative to thegiven principal axis of inertia. Here L does not depend onthe choice of the point relative to which it is determined(assuming the rotation axis to be stationary).

The validity of Eq. (5.36) may be easily demonstrated forthe case of a uniform hody possessing axial symmetry.Indeed, in accordance with Eq. (5.27) the angular momen­tum of a solid body relative to the rotation axis L z = [w z

(recall that L z is the projection of the vector L determinedrelative to any point on that axis). But when a body issymmetric relative to the rotation axis, it immediatelyfollows from the condition of symmetry that the vector Lcoincides in its direction with the vector ro, and therefore,L = [ro.

Once again, it should be pointed out that in the generalcase (when the rotation axis does not coincide with any ofthe principal axes of inertia though it passes through thecentre of inertia of a body) the direction of the vector Ldoes not coincide with that of .the vector ro, and the relation­ship between these vectors is very complex. This fact ac­counts for the complicated behaviour of rotating solids.

4. Gyroscopes. A gyroscope is a massive symmetric bodyrotating about its symmetry axis with a high angular veloc­ity. Let. us examine the behaviour of a gyroscope using a

177

Fig. 93Fig. 92

The Law of Conservation of Angular Momentum

~L i~ in~resting to note that the quantity 00' is independent of themc I~atlOn angle e o~ the gyroscope axis: In addition, the resultobtamed shows th!lt (I) is inversely proportional to (I), i.e. the higherthfe. angular ~eloclty of a gyroscope, the lower the angular v.elocityo Its preceSSIon. ' ..Gyr~opic. couple. Now let us examine a phenomenon

appearmg WIth the forced rotation of a gyroscope axis.,Suppose, for example, a gyroscope axis is mounted in a U­shaped support, which we rotate about the 00' axis as

shown. in ~ig. 92. If the' angular momentum L of the gyro­scope IS dIrected to the right, then during the time intervaldt In the process of that rotation the vector L gets an incre­ment dL, a vec~or directed beyond the plane of the figure.In ac~ordanceWIth Eq. (5.37) thisimplies that the gyroscopeexperIe~ces. the. force moment M coinciding with the dLvector m dIrectIOn. The moment M .is due to the appearanCeof the couple F t~at the sup~ort exerts on the gyroscope axis.In accordance WIth Newton s third law on the other handthe support it.self e.xperiences the forces 'F' developed by th~gyrosco~e aXIS (FIg. 92). These forces are referred to asgyros~op£c;. they form the gyroscopic couple M' = -M. Notethat m .th~s case .the gyroscope is not capable of resistingthe varIatIon of Its rotation axis direction.12-0539

(5.37)

Classical Mechanics

dL = Mdt,

It is seen from this equation that the moment of force Mdefines the angular precession velocity ro' (but not accelera­tionl). Therefore, an instantaneous elimination of the mo­ment M entails an instantaneous disappearance of precession.In this respect, one may say that precession possesses noinertia.

Note that the force moment M acting on a gyroscope maybe very different in nature. Continuous precession, Le. theconstant angular velocity ro', is maintained provided thevector M remains constant in magnitude and spins togetherwith the gyroscope axis.

Example. Find the angular velocity of preceS8ion of a tilted gyro­scope of mass m spinning with a high angular velocity (I) about itssymmetry axil! with respect to which the moment of inertia of thegyroscope is equal to [. The gyroscope's centre of inertia is located ata distance l from the supporting point.

In accordance with Eq. (5.38) 00'[00 sin e = mgl sin e, where eis the angle between the vertical and the gyroscope axis (Fig. !H).Hence,'

(I)' = mgll[w.

coinciding in direction with the vector M, the moment ofexternal forces relative to the same point 0 (in this casethat is the moment of the gravitational force mg). FromFig. 91 it is seen that dL ...L L. As a result, the vector L(and, consequently, the gyroscope axis) spins together withthe vector M about .the vertical, circumscribing a circularcone with the half-aperture angle e. The gyroscope precesses'about the vertical axis with some angular velocity ro'.

Let us find how the vectors M, Land ro' are interrelated.From the figure the modulus increment of the vector L duringthe time interval dt is seen to be equal to I dL I == L sin ew'dt, or in vector form dL = [ro'Ll dt. Substi­tuting this expression into Eq. (5.37), we obtain

[ro'Ll = M. (5.38)

But the behaviour of the vector L is described by theequation of moments (5.12). In accordance with that equa­tion the angular momentum L relative to the point 0(Fig. 91) acquires during the time interval dt the increment

176

. 1711The Law of Conservation of Angular Momentum

Fig. 94 Fig. 95

8"that equality, while the right-hand side contains its angular momen-tum at the maximum (minimum) distance r when r.l v.

From the energy censervation law it follows that. mvlj2J-ymMjro=mv·j2-ymMjr,

where.~ is ~he mass of the Sun and y is the gravitational constaht.Eltmmatmg v from these two equations, we get

r= 2 ro~ (I ± VI-~(2-~)sin·cp),

where ~ = rov~jyM. The plus sign in front of the radical sign corres­ponds to rma:w: a~d the minus sign to rmtn' .

. e.5.2. Particle 1 located far from particle 2 and· possessing thekmetl.c ~nergy To and mass ml strikes particle 2 of mass m2 throughthe almmg parameter l, the arm of the momentum vector relativeto particle 2 (Fig. 95). Each particle carries a charge +q. Find thesmallest distance between the particles when .

,..('1) ml « m.;(2) ml is comparable to m2•

Solution. 1. The condition litl « m2 means that in the processof interaction particle 2 is practically motionless. The vector of theforce acting on particle 1 passes continuously through the point atwhich particle 2 i.s located..Consequently, the angular momentumof particle 1 relative to motIOnless particle 2 is conserved. Hence,

lpo=rminP,

~here the left-hand side r~presents the angular momentum of par­ticle 1 located far from particle 2 and the right-hand side is the angu­lar momentum of particle 1 at the moment of the closest approach,

12*

Solution. Let us make use of the laws of conservation of angularmome~turn and energy. The centre of the Sun is the point relativeto which the angular momentum of the planet remains constant.Therefore,

romvo sin cp = rmv,

where m is t~e mass of the ll~met. The angular momentum of' theplanet at a g1Vlln moment 0 time enters into the left-hand side of

I

I

I

CiaBBlcat Mechanics

Problems to Chapter 5• 5.1. Find the maximum and mil1imum distances of the planet

A from the Sun S if at a certain moment of time it was at the distancero and travelled with the velocity v~, with the angl~ between theradius vector ra and the vector Vo bemg eijual to cp (Fig. 94).

I .

178

The appearance of gyroscopic forces is referred to asgyrostatic action, or gyroeffect. A gyroeffect associated withthe appearance of gyroscopic stresses in bearings is observed,for example, in turbinerotors of ships at the time of turning,rolling or pitching, in propeller-driven planes during turn-ing, etc.

Let us examine gyrostatic action in a gyroscope whoseaxis together with a bearing frame (Fig. 93) can rotate freelyabout the horizontal axis 00' of a V-shaped mount. Whenthe mount is set into a forced rotation about the verticalaxis as shown in the figure by the ro' vector, the angularmomentum L of the gyroscope receives during the timeinterval dt an increment dL1 , a vector directed beyond theplane of the figure. That increment is induced by the momentM

1of a couple exerted on the gyroscope axis by the frame.

The corresponding gyroscopic forces exerted by the gyroscopeaxis on the frame make the frame turn about the horizontalaxis 00'. In the process the vector L acquires an additionalincrement dL

2which is in its turn caused by the moment M 2

.-of the couple which the frame exerts on the gyroscope axis.Consequently, the gyroscope axis turns so that the vector L .tends to coincide in its direction with the vector ro'.

Thus, during the time interval dt the angular momentum Lof the gyroscope acquires an increment dL = dLl + dL2 == (M

1+M

2) dt. As this takes place, the frame experiences

the gyroscopic coupleM' = -(M1 +M2).

The component of that couple M ~= -M1 makes the frameturn about the horizontal axis 00', while the other compo­nent M~ = -Mil opposes the turning of the whole systemabout the vertical axis (in contrast to the previous case).

The gyroeffect underlies various applications of gyro­scopes: the gyrocompass, gyroscopic stabilizers, the gyro­sextant, etc.

181

m

A

m

The Law of Conservation of Angular Momentum

where m is the mass of the sphere, and v is its velocity in the positionwhen the thread forms the angle 11/2 with the vertical.

The sphere moves in the Earth's ~ravitational field under theinfluence of an external force, the tenSiOn of the thread. That forceis always perpendicular to the velocity vector of the sphere and there­fore does not perform any work. It follows that in accordance withEq. (3.32) the mechanical energy of the sphere in the Earth's gravi­tational field remains constant:

mv8/2=mvs/2+mgl cose, (2)where the right-hand side of theequality corresponds to the hori­zontal position of the thread.

The simultaneous solution ofEqs. (1) and (2) yields

vo = y2gl/cos e.

e5.4. Two identical small cou­plings are. positioned on a rigidwire ring df radius TO which canfreely rotate about the vertical axisAB (Fig. 97). They are"joined to­gether by a thread and set in theposition m-m. Then the whole as-sembly isset into rotation with the Fig. 97angular velocity coo, whereupon thethread is burnt at the point A.ASSUming the mass of the assembly to be concentrated mainly in thecouplings, find its angular velocity at the moment when the couplingshave slid d"wn (without friction) to the extreme lower position m'-m'.

Solution. Suppose that in the lower position the couplings arelocated at the distance T from the rotation axis and the angular velo­city of the assembly is co. Then, from the laws of conservation ofenergy and angular momentum relative to the rotation axis, it fol­lows that

rlcol - rBcoB = 2gh; rico = rlcoo,

where h is the difference in the height of the couplings in the upperand lower positions. Besides, from Fig. 97 it is seen that

rB=r'+h' .

These three equations, when solved simultaneously, yield

co=(1+ Y1+(4g/rocoB)')0>0/2.

., 5.5. A smooth rod rotates freely in a horizontal plane with theangular velocity COo about a stationary vertical axis 0 (Fig. 98) rela­tive to which the rod's moment of inertia is equal to I. A small couplingof mass m is located on the rod close to the rotation axis anll. is tiedto it by a thread. When the thread is burned, the coupling starts

I.

t·i

·s,';

1;

iI

•

Classical Mechanics

lPto = rmlnPt' (2)

where account is~ taken of thefact that ra 1. Pt at themo­ment of the closest approach(se.e Fig. 95). Moreover, in ac­cordance with the energy con-

2 servation law

To =1:+kql/rmin, (3)

where To and T are the totalkinetic energies of the particles

Fig. 96 in the C frame, respectively, atthe moment when they are far

from each other and at the moment of their close~ appro~ch. From.E s. (2) and (3) we get expression (1), only: wit~ To substItuted forT;. What is more, in this case (partIcle 2 IS orlgmally at rest)

T ml To,0= mt+ml

in accordance with Eq. (4.16). Note that if. fflt ~ ml' then To ~ Toand the expression for rmin is completely ldentl~al to Eq. (1). f

~ 5 3 A small sphere is suspended at the pomt 0 by means 0a li ht ~~n-stretchable thread of length I. Then the spb:e~ is s~ngthro~gh an angle e from the vertical and imparted an mltll~l velOCItyv perpendicular to the vertical plane in which the thread lSI loca~~?It what velocity vo is the maximum s~inging angle equa to n .

Solution. The swinging sphere experIences t~o fo~s: the graVI­tational force and the tension of the thread. ,It IS not dlffie;ult

lto .see

that the moment of these forces M z == 0 relatIve to the vertlca axIS Z

passing through the point O. COll!!~quentl¥, ~he angular momentumL of the sphe~ relative to the gIven axIS IS constant, orz. 1sin e .mllo =, l.mv, (1)

when r 1. p (Fig. 95.) Next, from the energy conservation law it fol-lows that .

To=T+kql/rmin,

where T is the kinetic energy of particle 1, at the mome!!t o.f the closestapproach. Having solved these two equatiOns (a~d takIng mto accountthe relationship between Po and To), we obtam

rmln= :;: (1+ y 1+(2lTo/kql)S). (1)

2 In this case one cannot as/lume that particle 2 i~ at rest ~n thero~ss of interaction. It is advisable to seek th~ ~olutlOn ~ere .m the~ frame in which the "collision" occurs the way It IS shown m. FI~. 96.Th's sy~tem of two particles is assumed closed, and therefore l~ mter-

1 nal angular momentum IS con-served:

1SO

183

x

Fig. 101Fig. 100

lp = [ro.

The Law of Conservation of Angular Momentum

system does not· vary as long as the bullet keeps moving in the rod.This signifies that in this case the horizontal component of the reactionforce at the point 0 is absent.

2. In this case the angular momentum of the system with respectto the point 0 also remains constant as long as the bullet moves inthe rod, and consequently in accordance with Eq. (5.23)

L+[lp]=L.

Here the angular momentum of the bullet relative to the point 0is written on the left-hand side and on the right-hand side we wrotethe angular momentum of the rod (together with the bullet stuckin it) immediately after the bullet stopped (see Fig. 100 where allthree vectors are located in the horizontal plane).

Let us calculate the vector L when the rod (with the bullet) acquiresthe angular velocity ro. We shall consider a small element of the rodof mass dm located at the distance r from the point O. Its angularmomentum relative to the point 0 is equal to

dL= [r. dmv] =dm·r2ro= (mro/lo) r2 dr,

where [ is the moment of inertia of the rod relative to the axis thuschosen, and CJ) is the angular velocity of the rod immediately afterthe bullet has stopped in it.

With allowance made for Vc = ror, where r is the distance fromthe point 0 to the centre of inertia of the rod, these two equationsyield

!1p = (31/2lo - 1) p.

It is seen that the sign of the increment !1p depends on the ratio lIlo•Specifically, when l/lo = 2/3, !1p = 0, i.e. the momentum of tile

write

c

-pFig. 99

Classical Mechantcs

A

Fig. 98

point 0 (Fig. 99). The bullet has the momentum p and hits the rodat a point lying at the distance l from the point O. Disregarding themass of the bullet, find

(1) the momentum increment of the bullet-rod system during thetime of motion of the bullet in the rod;

(2) the angular velocity acquired by the rod with regard to theinternal angular momentum of the bullet, which is equal to Landcoincides in direction with the vector p (the bullet spins about itsmotion direction). .

Solution. 1. The bullet-rod system is non-elosed: apart from thecounterbalancing forces a horizontal component of the reaction forceappears in the process of motion of the bullet at the point 0 of therod. That component brings about the momentum increment of thesystem:

!1p = mvc - p,

where Vc is the velocity of the centre of inertia of the rod after thebullet has hit.

/Since\all external forces in this process pass through the point 0,the angullar momentum of the system remains constant relative toany axis passing through that point as long as the bullet moves in therod. Choosing the axis that is at right angles to the figure plane, we

-.----~o

sliding along the rod. Find the velocity v' of the coupling relativeto the rod as a function of its distance r from the rotation axis.

Solution. In the process of motion of the given system the kineticenergy and the angular momentum relative to the rotation axis donot vary. Hence, it follows that

!ro;j=!ro2+mv2; !roo=(! +mrS) ro,

where v2 = V'2 + ro2r2 (Fig. 98). From these equations we obtain

v' = roor/l/1+ mrs/!.

.5.6. A bullet A flying horizontally hits and remains in a verticaluniform rod of mass m and length lo hinged by its upper end at the

182

i84 Class~cal MechanicsThe Law of Conservation of Angular Momentum i85

Fig. f03Fig. f02

The moment of the friction forces acting on the given ring is equal todM = rk (mglnrB) 2nr dr= (2kmg/rB) r l dr,

where m is the mass of each disc. Integrating this expression withrespect to r between 0 and ro, we get

M = (2/3) kmgro'

In accordance with Eq. (5.30), the angular velocity of the lower discincreases by doo over the time interval

dt = (IIM) doo = (3rol4 kg) dro.

Integrating this equation with respect to ro between 0 and 000/2,we find the sought time

t = 3roooo/8kg.

• 5.9. A uniform cylinder is placed on a horizontal board (Fig. f02).The coefficient of friction between them is equal to k. The board is

imparted a constant acceleration w in a horizontal direction at rightangles to the cylinder's axis. Find

(1) the acceleratibn of the cylinder axis in the absence of slipping;(2) the limiting value wllm for which there is no slipping.Solution. 1. Choosing the positive directions for z and cp as shown

in Fig. f02, we write the equation of motion of the cylinder axis andthe equation of moments in the C frame relative to that axis:

mWe=Fjr; I~=rFjr,

where m and I are the mass and the moment of inertia of the cylinderrelative to its axis. In addition, the condition for the absence ofslipping of the cylinder yields the kinematic relation of the accelera­tions:

W - We = ~r.

From these three equations we obtain We = wIs.2. Let us find from the previous equations the magnitude of the

friction force F/r ensuring that the cylinder rolls without slipping:FIr = mwl3. The maximum possible value of that force is equal

Thus,

where v is the velocity of the given element. Integrating this expres­sion over all elements, we obtjlin

L=mllro/3.

L+ [lp] = mlBro/3.

Using this formula, we obtain in accordance with Fig. fOO

00=3 Vi:,+,1p'/mlB.

From the same figure one can find also the direction of the vector ro(the angle l%) •

• 5.7. A uniform solid cylinder of mass mo and radius R can rotatewit~out friction about a s.tationary horizontal axi.s 0 (Fig.. 10i).A thin non-stretchable cord. of length l and mass m IS wound In onelayer over the cylinder. Find the angular a~celer.ation of the cy:lin~eras a function of the length z of the overhangmg pIece of cord. Slippmgis'assumed absent and the centre of gravity of the wound portion ofcord is supposed to be located at the cylinder's axis.

Solution. Let us make use of the equation of moments (5.15) rela­tive to the 0 axis. For this purpose we find the angular momentumof the system Lz: relative to tliegiven axis and the corresponding forcemoment Mz:. The angular momentum is

L z=Iroz+Rmv=(mo/2+m) R1ron

where allowance is made for the moment of inertia of the cylinderI = moRI/2 and v = rozR (no cord slipping). The moment of theexternal gravitational forces relative to the.o axis is

Mz: = Rmgzll.

Differentiating Lz: with respect to time and substituting dL/dt. and Mz: into the eqqation of moments, we obtain

pz: = 2mgzllR (mo + 2m).

• 5.8. A uniform disc of radius ro lies on a smooth horizontal plane~A similar disc spinning with the' angular velocity roo is carefullylowered onto the first disc. How soon do both discs spin with thesame angular velocity if the friction coefficient between them is equalto k?

Solution. First, let us find the)teady-state rotation angular veloc­ity 00. From the law of conservation of angular momentum it followsthat 1000 = 2Iro, where I is the moment of inertia of each disc relativeto the common rotation axis. Hence,

00 =ro0/2.

Now let us examine the behaviour of one of the discs, for example,the lower one. Its angular velocity varies due to the moment Mof the friction forces. To ciliculate M, we single out on the uppersurface plane of the disc an elementaryj ring with radii rand r + dr.

187

oFig. 105

F

The Law of Conservation of Antfular Momentum

point 0 Lis equal to

dMe! = (moo2Il) sin 0 cos ex2 dx •

Integrating this expression over the whole length of the rod, we get

Me! = (moo2l213) sin acos e. (2)

It follows from Eqs. (1) and (2) thatcos a = 3g/2oo2l. -(3)

Now let us determine the magnitude and direction of the vector R.In the reference frame where the rod rotates with the angular velocity00 its centre of inertia, the point C, moves along a horizontal circle.Consequently, from the law of motion of a centre of inertia, that is,Eq. (4.11), it immediately follows thatthe vertical component of the vector Ris RII = mg, while the horizontal compo­nent R.l is determined from the equationmWn = R.l' where wn is the normal ac­celeration..,of the centre of inertia C.Hence

R .1.= (mw2 l/2) sin 8. (4J•

The magnitude of the vector R isequal to

R= Y(mg)2+Rl =

= (moo2lj2) y 1+7g2j(4oo4l 2),

and its direction, specified by the angle6' between the vector R and the ver-~ical, ~s determined fr,om th~ formula cos 6' = mg/R. It islD.terestmg t~ no~e t~at a =1= a, I.e. the vector R does not coincide\~Ith the r~d I~ directIOn. One can easily make sure of this by expres­smg cos a via cos a:

cos 6' =4 cos aj V9+7 cos2 (l.

It is se~n fr~m this that cos 6' > cos 0, Le. 6' < e as it is in factshown lD Fig. 104. '

Note that the equivalent of the forces of inertia Fe! passes notthrough the point C but below it. Indeed, Fe! = R.l and is determinedby Eq. (4), whereas the resultant moment Me! is determined by Eq. (2).It follows from these formulae that the arm of the vector F relativeto the point 0 i~ e~ual to (2l/3) cos a (Fig. 104). e!

, • 5.12. 1\, splJlmng top of mass m whose axis forms an angle eWith the vertIcal pre«esses about the vertical axis passing throughthe POlDt ~f support 0, :rhe ,an~ular momentum of the top is equalto ,L,. and,~ts centre of ~nertla IS located at the distance l from thePOlllt 0 .. Fwd the magmtude and direction of the vector F whichlsthe hOrIzontal component of the reaction force at the point O.

Classical Mechanics

to kmg. Hence,Wlim=3kg.

• 5.10. A uniform sphere of radius r starts rolling .down with?utslipping from the top of another sphere of radius R (FIg. 103). Fmdthe angular velocity of the sphere after it leaves the surface of theother sphere.

Solution. First of all note that the angular velocity of the sphere,after it leaves does not change. Therefore the problem redu~ to thedetermination of its magnitude at the moment of breakmg-off.

Let us write the equation of motion for the centre of the sphere atthe moment of breaking-off:

mvl/(R + r) = mg cos a,where v is the velocity of the centre of the sphere at that moment,and a is the corresponding angle (Fig. 103). The velocity ~ can be

found from the energy conservatIon law:mgh = mvl/2 + 100'/2,

where 1 is the moment of inertia of thesphere relative to the axis passing throughthe sphere's centre. In addition,

v = oor; h = (R + r) (1 - cos a).From these four equations we obtain

00= Yl0g(R,+r)/17rl •

• 5.11. A thin uniform rod of massm and length l rotates with the constantangular velocity 00 about the vertical axispassing through the rod's suspensionpoint 0 (Fig. 104).Hn so doing, the rod

mg describes a conical surface with a half­aperture angle 8. Find the angle a aswell as the magnitude and direction of

Fig. 104 the reaction force R at the point O.Solution. Let us consider the frame

rotating about the vertical axis togeth·er with the rod. In this reference frame the rod experience.Enot only the gravity mg and the reaction .force ~ but also the centri­fugal force of inertia Fet. As the rod rests lD the given reference frame,that is, stays in the equilibr.ium position, the resultant momentof all forces relative to any pomt and the resultant of all forces areequal to zero. '

It is only gravity and the. centrifugal forces of i~ertia that producea moment relative to the POlDt O. From the equahty of the momentsof these forces it follows that

(mglI2)sina=Met. (1)

Let us calculate Met. The elementary moment of the force of inertiathat act,s on the rod element dx located at the distance x from the_

186

PART TWO

RELATIVISTIC MECHANICS

CHAPTER 6

KINEMATICS IN THE SPECIAL THEORYOF RELATIVITY

§ 6.1. Introduction

The special theory of relativity proposed by Einsteinin 1905 called for a review of all concepts of classiCal physicsand primFily the concepts of time and space. Therefore,this theory, in accordance with its basic contents, can bereferred to as a physv;:al study of time and space. The studyis called physical b~cause the properties of space and timeare analysed in this theory in close connection with thelaws governing physical phenomena. The term "special"implies that this theory considers phenomena only in iner­tial reference frames.

We shall begin this section with a brief review of pre­relativistic physics, dwelling in particular on the problemsthat led to tb.ll appearance of the theory of relativity.

Basic notions of prerelativistic physics. First, we shallrecall those notions of space and time that are associatedwith Newton's laws, i.e. that underlie classical mechanics.

1. Space, which has three dimensions, obeys Euclideangeometry.

2. Together with three-dimensional space and indepen­dent ttf it, there· exists time. Time is independent in thesame sense that the three dimensions are independent ofeach other. But for all that, time relates to space thrOugh

. the laws of motion. Specifically, time is measured by a clock,which is basically an instrument utilizing one or anotherperiodic process providing a time scale. Therefore, it is

,impossible to determine time irrespective of some periodicprocess. Le. irrespective of motion.

ClassIcal Mechanics

And finally,F = (m3gl [3ILI) sin 8.

It should be pointed out that if the point of SUPPt:t of the tO~hw:h:

~~::e~n~::I1:~ ~~~~~lrysb~t~b~~~n~h~th:e;~~ai~~i8 ~~~i:'g ~hroughthe centre of the top, the POInt C In FIg. 105.

Solution. In accordance with Eq. (5.38) the angular velocity ofprecession is '_ IlL

co - mg •

. f' t' f the top moves along the circle, the vec-~~~c; 1~eo~i~~tt:d °asI?li~~:r~ted in Fig. 105 (this vector precesses to-

getw:o~itfheth~oati~~ o:q~~~i~~P)ior the centre of inertia we obtain

mco'll sin 8=F.

188

3. Dimensions of solid bodies (scales) and time intervalsbetween given events are identical in different referenceframes. This corresponds to the Newtonian concept of abso­lute space and time, according to which their properties areassumed to be independent of the reference frame, that is,space and time are the same for all reference frames.

4. The Galileo-Newton law of inertia is assumed to bevalid, according to which a body experiencing no influence

from other bodies moves rec-K 0 tilinearly and uniformly.

I This law maintains the ex-II -y1--------1A istence of inertial reference

I I frames in which'Newton's lawsI • 1 GI I hold true (as we I as the a-I I lilean principle of relativity).'a' lx' 5. From the above notions

o .I the Galilean transformationfollows, expressing the space-

Fig. 106 time relation of any event indifferent ,inertial reference

frames. If the reference frame K' moves relative to the Kframe with the velocity V (Fig. 106) and the zero timereadin'g corresponds to the moment when the origins 0' ando of the two frames coincide, then*

x' = x - Vt; y' = y; t' = t. (6.1)

From this it follows' that the coordinates of any event arerelative, I.e. have different' values in different referenceframes; the moment of time at which an event occurs ishowever the same in different frames. This testifies to thefact that time flows identically in different reference frames.That seemed to be so obvious that it was not even stated asa special postulate.

From Eq. (6.1) the classical law of velocity transforma­tion (composition) follows immediately:

v' = v - V I (6.2)

where v' and v are the velocities of a point (particle) inthe X' and K frames respectively.

• Hereafter we shall limit ourselves to only two spatial coordi­nates x and /I. The z coordinate behaves as y in all respects.

6. The Galilean principJe of relativity holds: all inertialreference frames are equivalent in terms of mechanicsall laws of .mechanics are identical in these referenc~frames, or, m other words, are invariant relative to the Gali­lean transformation.

7. The principle of long-range action is valid: interactions~etw~en bo.dies propagate instantaneously, I.e. with anmfimtely hIgh velocity.

These ~otions of classical mechanics were in completeaccord wIth the totality of experimental data available atthat time (it should be noted, though, that those data relat­ed to the study of bodies moving with velocities much lowerthan the velocity of. light). The validity of these notions~as ~onfirmed.by the very successful development of mechan­ICS Itself. Therefore, the notions of classical mechanicsabout the properties of space and time were thought to beso. funda~ntalas not to raise any douhts\about their truth.

The fi.rs~ to be put. to the test was the Galilean principleof re~atIvlty, WhICh}~ ~nown to b~ applicable only in me­chamcs, t.he only dIVIsIOn of phYSICS advanced sufficientlyby .that time. As other branches of physics in particularoptIc~ and electrodynamics, were developi~g, the naturaiquestion arose: does the principle of relativity cover otherphenomena as well? If not, then using these (non-mechani­cal) phenomena one can in principle distinguish inertialreference frames and try to find a primary, or absolutereference frame. . '

One ?f su~h phenomena that was expected to occur differ­e.ntly m dIfferent reference frames is the propagation oflIght.. In accordanc~ with the predominant wave theory of~hat time. waves of lIght must propagate with a certain veloc­Ity relative to a certain hypothetical medium ("luminifer­ous ether") whose nature, however was debated amongscientists. Still, whatever the natur~ of that medium itsurely cannot rest in all inertial reference frames at o~ce.Consequently,one can distinguish one inertial frame theabsolute frame, which is stationary with respect t~ the"lum~niferous ether". It was supposed that in that (andonly ~n tha9 refere~ce f~ame light propagates with the equalvelOCIty C ,m all dIrectIOns. If a certain inertial referenceframe moves with the velocity V relative to the ether, the

191Kinematics In the $peclal Theory ollMaftvltyRelativistic Mechanics190

f93Kinematics in the SpeCial Theory of Relativity

Comparing the expressions for til and t.l' we see that lightmust take different time periods to Cover these paths. Bymeasuring the time difference til - t.l the velocity of theset-up (the Earth) relative to the ether can be determined.

.~ Despite the fact that the time difference was expectedto be extremely small, the set-up was capable of observingthat difference using a very sensitive interferometer tech-nique. .

For all that, the result of the experiment proved to benegative: a time difference was not detected. Surely, bysheer accident, the Earth could happen to be motionlessrelative to the ether at the time when the experiment wasconducted. But then in half a year, for example, the Earth'svelocity would have reached 60 km/s. The repetition of theexperiment in half a year, however, did not bring the resultexpected.

More aqc:urate experiments of the same kind performedlater corroborated the original result.

The negl1'tive result of Michelson's experiment contra­dicted what 'was expe;cted from the Galilean transformation(velocity composition). It also showed that motion relativeto the ether cannot be detected and the velocity of lightis independent of the motion of a light Source since itsmotion with respect to the ether is different at differentseasons of the year. Jf \:

Some astronomical observlltions (e.g. of double stars)also point to the fact that the velocity of light does notdepend on the velocity of a source. A number of specialexperiments carried out later gave the same evidence.

By the beginning of the twentieth century theoreticaland experimental physics faced a serious challenge. Onthe one hand, the theory predicted various effects permit­ting the principal (absolute) reference frame to be distin­guished from the great number of inertial frames. Onthe other hand, persistent attempts to detect these effectsexperimentally inevitably terminated in failure. The ex­periment perfectly confirmed the validity of the principle ofrelativity for all phenomena, including those which werethought incompatible with that principle.

A few attempts were ventured to explain the negativeoutcome of the Michelson experiment and some other simi­13-0539

Relativistic Mechanics

A

c-v--v-­I

c+v

Fig. 107

2l 2l 1t.l = ycl-V'=c V1-(Vlc)"

velocity of light c' in th~t r~f~;nce.f!ame must .obey ,theconventional law of velocIty composItIon (6.~), I.e. c ­= c - V. This assumption was tested experImentally byMichelson (together with Morley). . .

Michelson's experiment. The purpose of thIS expe~Iment

was to detect the "true" motion ?f the Earth relatIve tothe ether. Michelson in his experImen~ to~k advantage. ofthe motion of the Earth along its orbIt wIth the velo~Ity

30 km/s. The idea of the experiment was the fo~lowIll~.

The light from the source S (Fig. 107) was emItted III

two mutually perpendicular directions, renecte~ from themirrors A and B llocated atthe same distance l from thesource S and finally returnedto the point S. In this experi­ment a comparison was madebetween the time taken bylight to cover the path BASand the time taken to coverthe path SBS.

Let us suppose that at themoment of the experiment theset-up moves together withthe Earth so that its velocityVrelative to the ether is direct-

. edalong SA. If the velocityof light obeys the conventional law of velo?ity composit~on(6 2) light moves along the path SA wIth the velOCItyc .:-. 'v relative to the set-up (the Earth) and i!l the reversedirection with the velocity c + V. Then the tIme spent bylight to traverse the path SAS is equal to

l l 2l 1tU= c-V + c+V =7 1-(Vlc)' •

Along the path SBS the velocity of light relative to .theset-up is equal to c' = Vc2 - VI (Fig. 107) and the tImetaken to cover that path is equal to

192

'. f 1 ssical mechanics. However.lar experiments mdter~s t~ ~ea unsatisfactory in the finalall of .them turn~. °lu . lution of this problem was pro­analysIs. ITh~ caEr .msatei:?s theory of relativity.vided on y m m

§ 6.2. Einstein's Postulates

. f 11 ex erimental and theoreticalA profound analysIs 0 a. i~ of the twentieth century

data accumulated ~y t~~ b?g.l:~l ;otions of classical physicsled Einstein to reVIew e ml 1 ce and time. As a result,and primarily the ~ol~~pts;fo;~:latiVity,which prove-d tohe created the specla. eorf classical physics. .

-be a logical completIOn ~t d such concepts of classicalThis theory adopts una ere d the Galileo-Newton law

mechanics as Euclidean space an . the constancy of. . At· the statement concernmg f

of mertla.. so.. of time intervals in different re -size of solId bodl.es a?d ticed that these representationserence frames, Emstem n~. f bodies moving with lowemerged from the observa l~ns °tra olation to higher velo­v~l.ocit.ies, and theredfore Jhei~r e:his ~eason, incorrect. Onlycitles IS unwarrante an, .. the true proper-. . evidence concernmg .experIment can g~ve can be said about the GalI-ties of space and .tIme. J~~:a~rnciple of long-range action.lean transformatIOn an t 1 tes or principles as the

Einstein proposed two pos u a , 1 "t which werefoundation of the sp~cial tt~eo:Ytaof(~~;tprli~~rilY by thebacked up by experlmen a aMichelson experiment): ..

(1) the principle of relatlv~ty, l' ht of the velocity(2) independence of the velocIty of 19

of a source. '. eneralization of the GalileanThe flrstp08tul.at~ l~ ac!ver all physical processes: all

principle of relatIVIty 0 d 'de tically in all inertial refer­physical phenomena prjee / re

nand the equations describing

ence frames; a.ll la~s ok na ~heir form on transition from onethem are invanant, l.e. eep I other words all iner-inertial reference frame to ano.the[~nt n(indisCernibl~) in theirtial reference fra;nes ~re. eqftwa no experiment whllteverphysic~l. pr~Phertlese; of ~~~afr;~es as preferable.can dlstmguis on

kinematics in the Special Theory of Relativity

The second postulate states that the velocity of lightin vacuo is independent of the motion of a light source andis the same ~n all directions. This means that the velocity oflight in vacuo is the same in all inertial reference frames.Thus, the velocity of light holds a most unique position.In contrast to all other velocities, which change on transi­tion from one reference frame to another, the velocity oflight in vacuo is an invariant quantity. As we shall seelater, the existence of such a velocity essentially modi­fies the notions of space and time.

It also follows ffom Einstein's postulates that light prop­agates in vacuo at the ultimate velocity: no other signal,no interaction between any two bodies can propagate witha velocity exceeding that of light in vacuo. It is preciselydue to its limiting nature that the velocity of light is thesame in all reference frames. Indeed, in accordance withthe princirle of relativity the laws of nature must be iden­tical in al inertial reference frames. The fact that the veloc­ity of any signal capnot exceed the ultimate value is alsoa law of nature. Consequently, the ultimate velocity value,the velocity of light'in vacuo, must be the same in aU iner­tial reference frames.

Specifically, the existence of an ultimate velocity pre­supposes the limiting of velocities of moving particles bythe value c. If otherwise, particles could transmit signals(or interactions between bodies) with a velocity exceedingthe ultimate one. Thus, in accordance with Einstein'spostulates all possible velocities of moving bodies and ofinteraction propagation are limited by the value c. Thus,the principle of long-range action of classical mechanicsdoes not hold any more.

The whole content of the special theory of relativityfollows from these two postulates. By the present time,both Einstein's postulates, as well as all of their consequen­ces, have been convincingly confirmed by the totality ofexperimental data accumulated so far. .

Clock synchronization. Prior to drawing any conclusionsfrom these postulates Einstein carefully analysed the meth­ods of measuring spaCfl and time. First of all, he noticedthat neither a point of space nor a time moment at whicha certain event occurs possesses a physical. reality; it is13*

Relativisttc Mechanics194

the event itself that does. To describe an event in a givenreference frame, one must indicate the point and the mo­ment of time at which it occurs.

The location of the point at which the event occurs canbe determined in terms of Euclidean g~ometry b~ means ofrigid scales and expressed in CartesIan coo.rdmates. Inthis case classical mechanics resorted to qUlte work~ble

methods of comparing quantities being measured agamstreference standards. .

The corresponding moment of time can be determmed bymeans of a clock placed at the point of the reference fra~e

where the given event occurs. This method, however, ISnot satisfactory any more when we have to compare e.ventsoccurring at different points, or, which is the s&.~e, to m~er­compare the moments of time of events happemng at pomtsremoved from the clock.

Indeed, to compare the moments of time (time readings)at different points of a r~ference fram~ one has to definefirst what is the universal time for all pomts of the referenceframe. In other words, w:e have to ensure a synchronousrate of all clocks of the given reference frame. ..

It is clear that the synchronization of clocks posItionedat different points of the reference frame can b? accom~­

lished only by means of some signals. The fastest SIgnals S.Ul­table for the purpose are light and radio signals. propa~atmgwith the known velocity c. The choice of these SIgnals IS alsostipulated by the fact that their velocit~ is in.depe~dent ofthe direction in space and is the same m all mertIal refer­ence frames.

Next, we can do as follows. An observer located, for ex­ample at the origin 0 of a given reference frame broadcastsa tim~signal at the moment to by his clock. At the momentwhe;lithis signal reaches the clock lo~ated at the ~n~wndistance r from"the point 0 the clock IS set so tha! It reg­isters t = to +'r/c, Le. the signal delay ~s ta.ken .mto ac­count. The repetition of signals after defimte time I.ntervalspermits all observers to synchronize the rate of t~ell' clo~ks

with that of the clock at the point O. This operatIOn havmgbeen performed, one can claim that all the ~locks of thegiven reference frame register the same time at eachmoment.

Fig. 108

t97Kinematics in the Special Theory of Relativity

It is essential to point out that time defined that wayrefers only to the reference frame relative to which thesynchronized clocks are motionless.

Relationships between events. Let us investigate the spa­tial and temporal relationships between given events indifferent inertial reference frames.

Even in classical mechanics the spatial relationshipsbetween different events depend on the reference frame towhich they belong. For example, two consecutive flashesin a moving train occur at the same point of a referenceframe fixed to the train but at different points of a reference

A 0 B Vi • •

W////m//7//mr//7////t/7/7/7//7//#///7/7/7///////7//7////I(

frame fixed to the railroad bed. The statement that alter­native events occur at the same point or at a certain distancebetween them makes sense only if the reference frame towhich that statement refers is indicated.

By contrast, in classical mechanics temporal relation­ships between events are assumed independent of the refer­ence frame. This means that if two events occur simulta­neously in one reference frame they are simultaneous in allother frames. Generally, the time interval between thegiven events is assumed to be the same in all frames.

However, it is easy to see that this is not actually thecase: simultaneity (and therefore the rate of time flow) isa relative notion that makes sense only when the referenceframe to which that notion relates is indicated. By meansof simplf' reasoning we shall illustrate how two events si­multaneous in one reference frame prove to occur atdifferent time moments in another reference frame.

Imagine a rod AB moving with a constant velocity Vrelative to the reference frame K. A flashbulb is locatedat the middle point 0 of the rod, and photodetectors at theends A and B (Fig. 108). Snppose that al a certain momentthe bulb 0 flaSHes. Since the velocity with which light pro­pagates in the reference frame fixed to the rod (as in any

RelatilJiltle Mechanic;too

§ 6.3. Dilation of Time' and Contractionof Length

In this section we shall examine three important conse­quences of Einstein's postulates: the equality of transvel'sedimensions of bodies in different reference frames, thedilation of the rate of moving clocks, and the contractionof longitudinal dimensions of moving bodies. In § 6.4 weshall generalize the results obtained in the form of the perti­nent. formulae for transformation of coordinates and time.

Prior to solving these problems we recall that a referenceframe is undel'stoou as a reference body to which a coordi­nate grid is fixed together with a numbel' of identical synchro-

inertial reference frame) is equal to c in both directions,the light pulses reach the equidistant photodetectors Aand B at the same moment of time (in the reference frame"rod") and both photodetectors respond simultaneously.

Things are different in the K frame. In that referenceframe the velocity of light pulses propagating in bothdirections is also equal to c, whereas the paths traversedby them are different. In fact, as the light signals propagatetoward the points A and B, the latter shift to the right,and therefore the photodetector A responds earlier than thephotodetector B.

Thus, events that are simultaneous in one reference frameare not l'imultaneous in another one, Le. in contrast toclassical mechanics simultaneity here is a relative notion.This, in turn, means that time flows differently in differentreference frames.

If we had at our disposal signals that propagate instan­t'aneously, events simultaneous in one reference frame wouldalso be simultaneous in any other reference frame. Thisdirectly follows from the example just considered. In thiscase the rate of time flow would be independent of the refer­ence frame, and we could talk about the existence of theabsolute time that appears in the Galilean transformation.Thus, the Galilean transformation is, in fact, based on theassumption that clock synchronization is accomplished bymeans of signals propagating instantaneously. However,such signals do not exist.

nized stationary clocks. Next, the' cbordinate grids andclocks are assumed to be calibrated in a like manner in allinertial reference frames. Clearly, this can ,be accomplishedonly by means of length and time standards also similarlyrealized in all reference frames.

For this purpose we can utilize some natural periodicprocess providing a natural scale for both length and timemeasurements, e.g. monochromatic waves emitted by in­dividual atoms resting in a: given reference frame. Then inthat reference frame the wavelength can serve as a lengthstandard and the corresponding period of oscillation as atime standard. Using these standards we can construct astandard representing one metre as a definite number of thegiven wavelengths and a standard representing one secondas a definite number of periods of the given oscillations(it should be pointed out that this method has beenrealized). <If

A similar technique can 'be utilized in every inertialreference frame, using the same monochromatic wave emit­ted by the same atoms resting in each of these referenceframes. The method is justified, for in accordance with the:principle of relativity physical properties of stationary atomsare independent of the inertial reference frame in whichthese atoms rest. '

Having effected length and time sta'ndards in each re­ference frame, we can move on to l'olving the fundamentalproblem of comparing these standards in di £ferent refer­ence frames, or, in other words, the comparison of Qimensions

, of bodies and rates of time flow in these frames.Equality of transverse dimensions of bodies. Let us begin

with the comparison of transverse dimensions of bodiesin different inel·tial reference frames. Imagine two inertialreference frames K and K' whose y and y' axes are parallelto each other and perpendicular to the direction of motionof one frame relative to the other (Fig. 109), with the ori­gin 0' of the K' frame moving along the straight line pass­ing through the origin 0 of the K frame. Let us positionthe rods OA and O'A ' , which are lhe metre standanls inench of these frames, along the y and y' axes. Next,imag­ine that at the moment when the y and y' axes coincide theupper end of the left rod makes a marking on the y axis of

199Kinematics in the Special Theory of RelativityRelativistic Mechanicst98

the K frame. Does that marking coincide with the pointA, the upper end of the. right rod? . .

The principle of relativIty makes It possIble to answerthat question: yes, it does. Were it not so, one of the rodswould be, for example, shorter than the other from theviewpoint of both reference frames, an~ theref?r~ th~rewould appear an experimental opportumty to dIstmgUlsh

201

(6.4)

Ktnematics in the Special Theory of Relativity

The task is easiest when solved by means of the followingthought (i.e. basically feasible) experiment. Let us makeuse of a so-called light clock, a rod with mirrors at the endsbetween which a short light pulse travels back and forth.The period of such a clock is equal to the time intervalbetween two consecutive arrivals of a light pulse at agiven end of the rod.

Next, imagine two inertial reference frames K' and Kmoving relative to each other with the velocity V. Let thelight clock AB be at rest in the K' frame and orientedperpendicular to the direction of its motion with respectto the K frame (Fig. 110). Let us see what the rate of theclock is in each respective reference frame K' and K.

In the K' frame the clock is at rest, and its period is

dto == 2l/c,

where . l iM the distance between the mirrors.In theK frame, relative to which the clock moves, the

distance between tbe mirrors is also equal to l, for thetransverse dimensions of bodies are the same in differentinertial reference frames. However, the path of the lightpulse in that reference frame is different (the zigzag ofFig. 110): as the light pulse travels fram the bottom mirrortoward the upper one, the latter shifts somewhat to theright, etc. Consequently, to get back to the bottom mifr,or,the light pulse has to cover a longer distance in the Kframe while travelling with the same velocity c. Therefore,the light pulse takes a longer time to cover that distance ascompared to the case when the clock is motionless. Accord­ingly, the period of the moving clock is longer, i.e. interms of the K reference frame the clock's rate is slower.

Let us \ designate the period of the moving clock in theK frame by dt. It follows from the right triangle AB'A'(Fig. 110) that l2 + (Vdt/2)2 == (cdt/2)2, whence

M = (2Ud)/V1- (V /C)2.

But since 2l/: == Mo, then,------.....

L\t == L1toV1-~1 '

B'

Relattvistrc Mechantcs

Fig. ito

1 \1 \

I \1 \

1 \

/ \,"vi \....'1 l \

til \/ \ 1

/ \ /I. \ 1

/_!~tjL_ \ IA Ai -------A';---

B V--

a

A

!J K

Fig. to9

a'

200

one inertial reference frame from another by shorter trans­verse dimensions. This, however, contradicts the principleof relativity.

From this it follows that the transverse dimensions ofbodies are the same in all inertial referenee frames. Thisalso means that if the origins of the K and K' frames arechosen as indicated, the y' and y coordinates of any pointor event coincide, i.e.

y' == y. (6.3)This relation represents one of the sought transformations

. of 'coordinates.Dilation of time. Our next task is to compare th~

rate of time flow in different inertial reference frames.As we already mentioned, time is measured by a clock,which may be any device in which one or another periodicprocess is used. Accordingly, ~he theory of rel?tivi~y custom­arily deals with the companson of rates of Identical clockRin different inertial reference frames.

203Kinematics in the Special Theory of RelatiVity

the given clock moves). The same is true for all processesproceeding in reference frames moving with respect to anobserver.

Naturally, one may ask whether an observer in the K'frame, which moves relative to the K frame, realizes thathis clock goes slower compared to the clock of the K frame.Obviously, he does not. This immediately follows from theprinciple of relativity. If the K' observer also noticed thedilation of time in his reference frame, that wouldmean that for both observers K' and K time flows moreslowly in one of the inertial reference frames. From thatfact the observers might have inferred that one of the iner­tial reference frames differed from another, which is incontradiction with the principle of relativity.

From this it follows that the dilation of time is re­ciprocal and symmetric relative to both inertial referenceframes K IIf\.d K'. In other words, if in terms of the K framethe clock of the K' frame goes slower, then in terms of theK' frame it. is the dock of the K frame that goes slower(and with the same deceleration factor). This circumstanceis evidence that the dilation of time is a purely kinemat­ic phenomenon. It is an obligatory consequence of the

" velocity of light being invariant and cannot be attributedto some variation of clock properties caused by motion.

Eq. (6.4) has been experimentally confirmed by explain­ing the seemingly mysterious behaviour of muons travers­ing the Earth's atmosphere. Muons are unstable elementaryparticles whose average lifetime is 2 ·10-6 s, this time beingmeasured when the particles are at rest or moving withlow velocities. Muons are generated in the upper layers ofthe atmosphere at a height of 20 to 30 km. Were the life­time of muons independent of velocity, they would travel,even moving with the velocity of light, only about cM == 3 ·1OS·2 .10-6 m = 600 m. Observations indicate, howe­Vel', that a substantial number of muons nevertheless reachthe Earth's surface.

- The explanation is that the interval 2.10-6 s is the prop-er lifetime Ato of muons, time measured by a clock movingtogether wit.h them. Since the velocit.y of these particlesapproaches that of light a time interval measured by a clock

Fig. Ht

Relativistic Mechanics

o 0.1 0.2 0.; OA 0.5 O.b 0.7 a8 0.9/.0j3 =v/c

0

0

0 /

..- ,.,..0

2.

40

J.

where ~ = Vic and V is the velocity of the clock in theK frame.

This fl)rmula shows that At > M o' Le. the same ~lock hasdifferent rates in different inertial frames: the rate IS slowerin a reference frame relative to which the clock move~, ascompared to the reference frame iI. which the clock IS atrest. In other words, a moving clock goes slower than a ~ta­

tionary one. This phenomenon is referred to as the dtla-tion of time.' .

The time measured by the clock moving to.gether withthe body in which a certain process takes place IS called theproper time of that body. Itis denoted by Mo. As it fol­lows from Eq. (6.4) the propertime interval is the shortest.The time duration .At of thesame process in another ref­erence frame depends on thevelocity V of that frame withrespect to the body in whichthe process takes place. Thisdependence is quite apprecia­ble at velocities V comparableto that of light (Fig. 111).

Thus, as distinct from clas­sical mechanics the rate oftime flow actually depends onthe state of motion. There is . " .no such thing as universal time, and the notiOn of the t~meinterval between two given events" proves to be relative.The statement that a certain number of seconds has passedhetween two given events is I?ean.ing.ful o~ly ,,:,he~ t~e ref­erence frame, relative to which It is .val.ld, IS Illdl~ated.

The absolute time of classical mechalllcs is an approXImatenotion in the theory of relativity holding only for low (com­pared to the velocity of light) relative velocities of re!erenceframes. This follows directly from Eq. (6.4) fmd IS seenfrom Fig. 111: At = M o at V ~ c.. .

So we reach a fundamental conclUSiOn: III a referencefr~m~ moving together with a cloc!". time flows slower(from the standpoint of the observer With respect to whom

202

RelattvisUc Mechanics 205

I: t J 4- 5m V, ,~

! ,0 2 Jm

kinematics in tke Special Theory of llelativity

Fig. 113

V.and there~re registers "someone else's" time. In accordancewIth Eq. (6:'4) the observer's "own" time of flight is longer:

10 = V!1t.

From these two equations and Eq. (6.4) we get

lIlo= !1tol!1t = V1-p2or

point M. Then we may state the sought length of the rodIII the K frame to be equal to

I = V!1to'

AI?- observer fixed to the rod registers a different timeof fhght. In fact, from his point of view the clock whichregistered the time of flight !1to moves with the velocity

1291 Il = 10V~, I (6.5)

where p = Vic. The .Iength 10 measured in the referencefram~ where the rod IS at rest is referred to as thelength. proper

Thus, the longit.udinal length of a moving rod turns outto be shorter than ItS proper length i e 1< 1 ThI's h. II d ' . . o' P enom-enon IS. ca e the Lorentz contraction. Note that this~ontractlO.n occurs only in longitudinal dimensions of bod­Ies, that IS, dimensions along the motion direction where­as the transverse dimensions do not vary, as it w~s shown[bove. As c0';llp.ared to the shape of a body in the referencerame where It IS at rest, its shape in the moving referencef~ame may be characterized as oblate in the motion direc­tIon.. T~e lengt~ contraction of a moving rod is illustratedIII FIg. 113. III which it is seen that in the reference frame

.'8M

Fig. 112

For this purpose let us conduct the following imaginaryexperiment. Let us mark a point M on the x axis of theK frame and place a clock at that point. Using that clockwe can measure the time of flight I1tt) of the rod past the

on Earth is much longer (see Eq. (6.4)) and turns out tobe sufficient for muons to reach the Earth's surface.

In conclusion a few words about the so-called "clock para­dox", or "twin paradox". Suppose there are two identicalclocks A and B, with clock A resting in some inertial refer­ence frame and clock B first moving away from clock Aand then returning to it. The clocks are assumed to showthe same time at the original moment, when they are locat­ed side by side.

In terms of clock A clock B moves, and therefore itsrate is slower and during its travel it will lag behind clockA. But in terms of clock B it is clock A that moves andtherefore on its return it will turn out to be slow. Thisevident incongruity constitutes the substance of the "para­dox".

Actually, we made a fundamental mistake when we ar­gued in terms of clock B since the reference frame fixed tothat clock is non-inertial (it first moves away with accelera­tion and then comes back), and we cannot in this case useresults related only to inertial reference frames. A detailedcalculation lying outside the special theory of relativityindicates that the clock moving with acceleration (in ourcase clock B) goes slower, and therefore it is this clock thatis behind when it returns to its initial point.

Lorentz contraction. Suppose rod AB moves relative tothe K reference frame with the constant velqcity V (Fig. 112)and the length of the rod in the reference frame K' fi~dto the rod is equal to 10 , Our task is to determine the lengthI of that rod in the K frame.

20~Kinematics in the Special theory 0/ Retativity

of dimensions of a given body and of time intervals ob­tained in different refer£Hlce frames are equivalent, that is,all of them are "true". These statements are difficult tocomprehend only because of our habit, based on routineexperience, to r~gard length and time intervals as absolutenotions while actually this is not so. The notions of lengthand time interval are as relative as those of motion and rest.

§ 6.4. Lorentz Transformation

Now we have to solve the fundamental problem of trans­formation of coordinates and time. Here we mean the for­mulae relating the coordinates and time moments of thesame event in different reference frames.

Could the Galilean transformation possibly serve thepurpose? Recall that this transformation is based on theassumptio,p that the length of bodies is invariable and timeflows at the same rate in different reference frames. In theprovious section, ho~ever, we found out that in fact thisis not so: the rate of time flow and the length of bodiesdepend on the reference frame. These are inevitable conse­quences of Einstein's. postulates. Therefore, we are com­pelled to reject the Galilean transformation, or, more precise­ly, to recognize it as a special case of some more generaltransformation.. Thus, we have to obtain the transformation formulae

which (i) take into account the dilation of time andthe Lorentz contraction (Le. are, in the final analysis,consequences of Einstein's postulates), and (ii) reduce tothe Galilean transformation formu~ae in the limiting caseof low velocities. Let us proceed to the solution of thisproblem. •

We shall consider two inertial reference frames K andK'. Suppose the K' frame moves with the velocity V rela­tive to the K frame. Let us orient the coordinate axes ofthe two frames as shown in Fig. 115: the x and x' axes coin­cide and are directed in parallel with the vector V, andthe y and y' axes are parallel to eaeh other. Let us posit.ionat various points of the two reference frames identical clocksand synchronize them, separately the clocks of the K frameand the clocks of the K' frame. And finally, let lis adopt

Relativistic Mechanic;

0./ 0.2 OJ 0.4 05 0.6 0.7 OJ 0.9 lOj3= Vic

t--. I-......I--,-.

r-....--~..

i'."\

1\

4 1\\

21

1/1 0I.a0,90.80.7D.50.5O.o.JO.

D.

o

fixed to the rod its length 10 = 5 m.. whereas i~ the Kframe relative to which the rod moves wIth,the velocIty V == 4/5 c its length I = 3 m. .

It follows from Eq. (6.5) that the degree of contractIOndepends on the velocity V. This dependence becom~s espe­cially pronounced at velocities V comparable wIth thevelocity of light (Fig. 114).

So in different inertial reference frames the length ofthe ~ame rod turns out to be different..In other wor?s,.length is a relative notion which is meamngful only wIthresllect to one or another reference frame. A statement ab?ut

the length of a body bemgequal to some number of me­tres has no sense unless thereference frame relative towhich the length is measuredis indicated.

As it follows from Eq. (6.5)and is seen in Fig. 114 1~~ 10 at low velocities (V ~c), so that the length of abody acquires an almost abso­lute meaning.

It should be pointed outthat the Lorentz contraction,

Fig. 114 just as the dil~tion lof tIT'mh~'must be reciproca. IS

means that if we compare two rods moving relative to eachother and having equal proper lengths, in terms of each ofthe rods the length of the other one is shorter in the sameproportion. Were it not so, there would have appe~red ~nexperimental possibility to distinguisl~ between the mertIalreference frames fixed to the rods, WhICh, however, contra­dicts the principle of relativity.

This points to the fact that the Lorentz contraction i,<:also a purely kinematic phenomenon: no stresses causingdeformations appear in a body. .

It should be emphasized that the Lorentz contractIOnof bodies in the direction of their motion as well as thedilation of time are real and objective facts by no meansassociated with any illusions of an obser~er. All the values

20G

209

(6.8)

(6.9)t = t' +x'V/c2 .1Y1_~2 '

t' = _t-=-:x=V=/C=2Y1-P2 '

y=y';

, x-Vt.X = . y' =y;

Y1_P2 '

I x'+Vt'x= .Y1-P2 '

Kinematics in Ihe Special Theory of Relativity

where ~ :::= Vic and Vis the velocity of the K' frame relativeto the K frame.

It. s~ould be imI:l\(ldiately emphasized that the symmetry(a SImIlar form) of Eqs. (6.8) and (6.9) is the consequenceo~ the co~plete e~uivalence of both reference frames.· (thed.Ifferent .sIgn ~f V. ~n these formulae is only due to the oppo­sIte motIOn dIrectIOn of the K and K' frames relative toeach other).T~e Lorentz transformation differs drastically from the

GalIlean transformation (6.1), but the latter can ,be obtainedfrom ~qs. (6.8) and (6.9) if the formal substitution c = 00 ismade In them. What does this mean?

At the end of the foregoing section it was mentionedt~at the Galilean transformation is based' on the assump­~Ion of clocks synchronized by means of signals propagatingInstanta~eously. From this fact it follows that the quantity~ plays In t~e Loren~z .transformation the part of the veloc­It~ of the. sIg~a~S utilIzed for clock synchronization. WhenthIS ve.locity IS In~n~tely great, we get the Galilean trans­formatIOn; when It IS equal to the velocity of light the~orentz transformation. Thus, the Lorentz transform~tionIS based on. the a~sumption of clock synchronization bymeans of lIght SIgnals possessing ultimate velocity.

The remarkable feature of the Lorentz transformation14-053'9

and for the reverse transition from the K' frame to the Kframe- .

of relativity. Using these formulae, the coordinates and timeo~ any event can be transformed on transition from one iner­tial reference frame to another.

Thus, the Lorentz transformation on transition from the Kframe to the K' frame takes the form

Relativistic Mechanics

Fig. 115

o

K'.I

: AY ---yj------l

I v I

~ :I II IL '-_ __

0' P X x'

208

the moment when the origins 0 and 0' coincide as the zerotime reading in both frames (t = t' = 0).

Suppose now that at the moment t at the point with thecoordinates x, y in the K frame a certain event A takesplace, e.g. a bulb flashes. Our task is to determine the coor­dinates x', y' and the time moment t' of that event in theK' frame.

The problem concerning the y' coordinate was solved atthe beginning of this section, where it was shown (see Eq.(6.3)) that y' = y. Therefore, we immediately pass tosearching the x' coordinate of the event. The x' coordinatedescribes the proper length of the segment 0'P resting inthe K' frame (Fig. 115). The length of the same segment in

the K frame where the measure­ment is taken at the momentt is -equal to x - Vt. The rela­tionship between these lengthsis specified by Eq. (6.5), fromwhich it follows that x-- Vt = x'V 1 - ~2. Whence

x' = (x- Vt)/V 1-~2.

(6.6)On the other hand, the x co-

ordinate describes the properlength of the segment OP at rest in the K frame. The lengthof the same segment in the K' frame where the measurementis taken at the moment t' is equal to x' + Vt'. Taking intoaccount Eq. (6.5) once again, we obtain x' + Vt' == xV1 - ~2, whence

x=(x' +Vt')/V1-~2. (6.6')The obtained formulae make it possible to determine the

relationship between the time moments t and t' of the eventA in both reference frames. For this purpose it is sufficientto eliminate x' or x from Eqs. (6.6) and (6.6'), whereuponwe get:t'-(t-xV/c2)/V1-~2; t=(t'+X'VIC2VV1-~2. (6.7)Eqs. (6.3), (6.6), (6.6') and (6.7) are referred to as the

Lorentz transformation. They playa key part in the theory

* Strictly speaking, it is also required that x/c ~ t, Le. thetimes of propagation of light signals over the distances typical forthe problems considered (x/c) should be less than the time intervalswe discuss here. When this condition is satisfied, the signals may beregarded as propagating instantaneously. .

§ 6.5. Consequences of LorentzTransformation

Concept of simultaneity. Suppose two events At (Xl' Yl't

1) and A 2 (X 2, Y2' t 2) occur in the reference frame K. Let

us find the time interval separating these events in theK' frame moving with the velocity V along the x axis as

211Kinematics in the SpeCial Theory of Relativity

~~ow~ in Fig. 115. In accordance with the time transfofma­IOn ormu a (6.8) the sought time interval is equal to

t' -t' _ (ts-t1)-(·rZ- x l) V/czs t- Y1-~S (6.10)

From this it follows that events which are simultaneous int~e K !rame (t 2 = t1) are not simultaneous in the K' frame(ts - tt =1= 0) .. The only exception is the case when the twoe~ents occur. III the K frame simultaneously at points with

ht e same values of the X coordinate (the y coordinate may

ave any value).Si~ultaneity is. thus a relative concept; events which

are slI~lUltaneous III one reference frame are not simulta­~?OuS 1;'1 the. general case in another reference frame. When(ISCUSSIllg sImultaneity of events, one has to specify thereferen~e Jrame relative to which simultaneity occursOtherWIse, the concept ?f. simultaneity loses its meaning:

I~ !ollows from relatIVIty of simultaneity that clocksP?sIhoned alon.g the" x' axis in the K' frame and synchro~~llzed together III that reference frame show different timeIII t~e K frame. Indeed, for the sake of simplicity let usconSIder the moment when the origins 0 and 0' f th treference frames. coincide and the clocks at th~se ;oi~~show the sa~e h~e: t = t' = O. Then in the K frame theclock. at a POIllt wlt.h the coordinate X shows at that momentth~ time t = 0, wIllIe the clock of the K' frame at the samePloIllt .shows a different time, t'. Indeed, in accordance witht Ie tIme transformation formula (6.8)

t' = -xV/c2 V1-~2.

It is seen from this that at the moment t = 0 (in the Kframe) .the clock of the K' frame shows a different time~c;;.en.dIllg on ~he ~ coordinate (the so-called loc~l time).. IS I.S sh?wn I.n FIg. 116a. In tenils of the 1(' frame the

SItuatIOn IS reciprocal (Fig. 116b), as it in fact should bedue to t?e. eqUIvalence of both inertial reference frames.e Nex~, It IS seen from Eq. (6.10) that for events simulta­neous III the I( frame the sign of the difference t' -- t' .defined by the sign of the expression -(x _ ) Vi C 1. ISq 1 tl . d'ff 2 Xl • onse­

uen y, III 1 erent reference frnmes (with different valuof the velocity V) the difference t~ - t; is different not onf;14*

Relativistic Mechanlc$

is the fact that at V ~ c it reduces* to the Galilean trans­formation (6.1). Thus, in the limiting case V ~ c the trans­formation laws of the theory of relativity and classicalmechanics coincide. This means that the theory of relativ­ity does not reject the Galilean transformation as incorrectbut includes it into the true transformation laws as a spe­cial case that is valid at V ~ c. In what follows we shallsee that this reflects the general relationship between thetheory of relativity and classical mechanics: the laws andrelations of the theory of relativity turn into the laws ofclassical mechanics 1n the limiting case of low velocities.

Next, from the Lorentz transformation it is seen thatat V> c the radicands become negative and the formulaelose physical meaning. This corresponds to the fact thatbodies cannot move with a velocity exceeding that of lightin vacuo. It is even impossible to use a reference frame mov­ing with the velocity V = c; in this case the radicandsturn into zero and the formulae lose physical meaning. Thismeans that no reference frame can in principle be fixed,for example, to a photon moving with the velocity c. Ex­pressed otherwise, there is no reference frame in which aphoton is at rest.·

And finally, it should be noted that the time transforma­tion formulae contain a spatial coordinate. This significantcircumstance reveals the inseparable relationship betweenspace and time. In other words, we should not speak sepa­rately of space and time but of unified space-:time in whichall physical phenomena take p~ace.

210

213

(6.12)

Kinematics in the Special Theonl of Relativity

1=10V1-~2. (6.11)

:rhus, the length 1 of the moving rod proves to be less thanIts proper length 10 , and in each reference frame it has itsown value. This result is in a complete-agreement with theresult obtained in, Eq. (6.5) .. I.t follows from the definition of the length that the rela­

tIVlt~ ?f t~e le.ngth of the given rod is a consequence of the.relatIvlty<!l.()f sImultaneity. The same pertains to the formof any body: its dimensions in the motion direction arealso different in d4fferent inertial reference frames.

Duration of processes. Suppose that at the point withthe coordinate x' of the reference frame K' a certain processtakes place whose duration in that frame is equal to tit ==. t; -- t~ (the. proper time of the process). Let us deter­mIne the duration of the given process tit = t

2- t

1in the

K frame relative to which the K' frame moves.For this purpose we shall resort to the Lorentz transforma­

tion of time. As the process takes place at the point withthe fixed coordinate x' of the K' frame it is more conve-nient to use Eq.(6.9): '

t2 - t1 =, (t;-t;)11/ 1-~2

In the K frame, relative to which the rocl moves itslength is defined as the distance 1 between the coordi~atesX 2 and Xl of. its ends taken at the same moment of time (t~ == t1)· M~kIng use of the Lorentz transformation (6.8)for the X and X coordinates, we get

10 =x;-x; = (x2 -xl)!V1-~2.= l/ V1-~2,whence

or

!t is.seen th~t th~ duration of the same pl'Ocess is differentIn dIfferent mertIal reference frames. In the K frame itscluration is longer (lit> lito), and therefore in that ref­erence frame it proceecls slowel' than in Ihe K' frame. Thisfact is ill a complete agreement with the result concerningthe rate of the same clock in (Iifferent inertial r'eferoncpframes! that is, Eq. (6.4).

Relativistic Mechanics212

raj

for definiteness, X 2 > Xl' we may write X 2 - Xl = V (t2 ­- t1). Substituting this equality into Eq. (6.10), we get

t~-t; = (t2-t1) (1-vV/c2)/V 1-~2.

The quantity in the second parentheses of the numerator isalways positive as V < c (even at v = c when the cause­and-effect relationship is determined by signals or inter­actions propagating with the highest possible velocity).It follows that if t2 > t1 , then t~ > t;, Le. the sequence ofthe cause-and-effect events is the same in all inertial ref­erence frames.

Lorentz contraction. Let us orient a rod resting in theK' frame along the x'axis, Le. along the motion directionof this reference frame relative to the K frame. Supposethe length of the rod in the K' frame is equal to 10 = x~ ­- .:r~ (the proper length).

Fig. 116

in magnitude but also in sign. This fact signifIes that thesequence of the events A 1 and A 2 may be either direct orrevel'se.

This, however, does not apply to events obeying thecausality principle. The sequence of such events (cause __-- effect) is the same throughout all reference frames. Thiscan be easily demonstrated through the following reason­ing. Let us consider, for example, a firing, event Al (Xl>t1), and a hitting of a target with a bullet, event A 2 (X 2 ,

assuming that both even,ts occur on the X axis. In the Kframe t2 > t1 , the velocity of the bullet is v, and assuming,

Interval. The relativity of spatial and time intervalsby no' means implies that the theory of relativity deniesthe existence of any absolute quantities. In fact, the oppo­site is true. The theory of relativity tackles the problemof finding such quantities (and laws) which are indepen­dent of the choice of the inertial reference frame.

The first of these quantities is the universal velocity withwhich interactions propagate; it is equal to the velocityof light in vacuo. The second invariant value, just as impor­tant, is the so-called interval 812 between events 1 and 2,whose square is defined as

IS~2 =C2t~2-l~2 = inv, I (6.13)

where t12 is the time interval between the events and l12is the distance between the two points at which the givenevents occur(l~2 = x:2 + Y:2 + Z~2)'

We can easily see that the interval is invariant, calculat­ing it directly in the reference frames K and K'. Makinguse of the Lorentz transformation (6.8) and taking intoaccount that Y;2 = Y12 and Z;2 = Z12' we can write:

2t'2 '2 _ 2 (t12- X 12V/c2)2 (X12- Vt12)2 _ 2t2 _ 2C 12"'-X12 -C 1-~2 - 1-62 -·c 12 X 12 •

Thus, it is clear that the interval is really invariant. Inother words, the statement "two events are separated by acertain interval s" has an absolute meaning for it is validin all inertial reference frames. The invariant· intervalplays a fundamental role in the theory of relativity a;ndprovides an efficient instrument of analysis and solutionof many problems (see, e.g., Problem 6.4).

Types of intervals. Depending on what component, spa­tial or temporal, preyails, an interval is referred to as eithet'

space-like (l12 > ct12), ortime-like (Ct12 > l12)'

In addition to these two types of intervals there is anothertype, light-like (Ct12 = l12)' .

If the interval between two events is space-like, a refer­ence frame K' can always be fOlllld in whieh these eventsoecur simultaneollsly (t;2 = 0):

c2t~2 -l~2 = -l~~.

If the interval is time-like, one can always find a referenceframe K' in which both events occur at the same point(l;2 = 0):

215Kinematics in the Special Theory of Relativity

c2t~2 -1~2 = c2t~~.

In the case of space-like intervals l12 >ct12 , Le. theevents cannot influence each other in any reference frame,even if communication between the events is carried outat the ultimate velocity c. This is not the case with time­like and light-like intervals, for which l12 -< ct12. Conse­quently, events separated by such intervals may be in acause-and-effect relationship with each other.

Transformation of velocity. Suppose in the x, Y planeof the K frame a particle moves with the velocity v, whoseprojections are equal to V x and v y. Using the Lorentz trans­formation (6.8), we find the velocity projections v~ andv;' of th~ particle in the K' frame moving with the velocityV as shown in Fig. 115:

, d£' dx' 1 vx-VvX =7= Cit dt'/dt = 1-vxV/c2 ,

, dy' dy' 1 Vy Y 1- ~2 "

vy =c= at' = (it dt' /dt = 1-vxVc2 , (6.14)

where ~ =.V/c. Hence, the velocity of the particle in theK' frame is

v' -" "Vv' 2+ '2 _ l/(vx- V)2+V~ (1-~2) " «(j.15)-: x vy - 1-vxV/c2 •

These formulae yield the so-called relativistic law of trans­formation of velocity.· At low velocities (V ~ c and v ~ c)they reduce, as one can easily see, to the classical formulaeof transformation of velocity:

v~ £'= vx - V: v;/ = vy,

or in a vector formv' = v-V.

Note that the last formula is valid only in the Newtonianapproximation; in t.he relativistic case it has" no meaning,for'" the simple law of velocity composition cannot be ap­plied here. This can easily be demonstrated by the follow­ing example. Suppose the velocity vector v of a particle

Relativistic Mechanics214

in the K frame is perpendicular to the x axis, Le. has theprojections Vx = 0 and v y = v. Then in accordanc~ wi~hEq. (6.14) the velocity projections of the same partiCle 10the K' frame are

v~= -V; v~=VyV1-p2. (6.1&)

This means that in the given case (v -L x axis) the v~ pro­jection diminishes on transition to the K' frame and clearlyv' =1= v - V (Fig. 117). .

Let us now examine the case when two partIcles movetoward each other with the same velocity v in the K refer-. ence frame. What is they K Y'1K' 1/,' 11 velocity v' of one particle

I V x with respect to the other?~ In the non-relativistic ap-i proximation the obvious an-I swer is 2v. In the case ofi high velocities we have toI apply the first formula of: (6.14), assuming the parti-I cles to move along the x10' axis. Let us fix the K' ref-

a i..O ------- --- x--;:- erence frame to one of theparticles, e.g. to the one

Fig. 117 moving in the positive di-rection of the x axis. Then

the problem reduces to finding the vel.ocit.y of the other par­ticle in that reference frame. SubstItut10g V x = - v andV = v into Eq. (6.14), we obtain

v~ = - 2v/[1 + (VIC)2).

The minus sign means that in the given case the secondparticle mo-ves in the negative direction of the x' axis ofthe K' reference frame. It should be pointed out that evenwhen both particles move alm~st with the highest po~sib~evelocity v "-' c, the velocity v cannot exceed c (whIch ISimmediately seen from the last formula). . . .

And finally, let us check directly whether the rel~hvI~t~cformulae of velocity transformation correspond to EmsteI? ssecond postulate concerning the constancy of the velOCItyof light c in all inertial reference frames. Suppose vector

217

Fig. 118

Ktnematics tn the Special Theory of Relativity

§ 6.6. Geometric Descriptionof Lorentz Transformation

c has the projections Cx and cy in the K frame, I.e. c2 == c~ + c~. Now let us transform the radicand in Eq.(6.15) as follows:

c~-2cxV+V2+(C2_C~) (1- ~:) = (c- c~v r.After this it is not difficult to get v' = c. Of course, in thegeneral case vector c' is oriented differently in the K' frame.

Let us now consider the relativistic concepts of space­time using the geometric approach developed by Minkowski,which helps to describe the substance of the Lorentz trans-formation in a differentlight. sk' r..ct 11:1 ~

Minkow i diagrams. Sup- /1 //Cpose there are two ineFtial ref-erence frames K and K', the _/ //latter moving relative to the I"f h 1 I ,,/ormer wit the ve ocity V. 1 /

First, let us draw the so-called fJ / ,',/ , ..... ':!'Ispace-time diagram for the 1 / ..... , ......

K frame, confining ourselves 1 /' _,..... ' .....1/ ~to the more simple and I~......' 8

graphic unidimensional case O~---'-------~:c

(Fig. 118). In this diagram theordinate axis usually marksnot the time t itself but thequantity 't = ct (where c is thevelocity of light). We may thus calibrate both axes, Oxand OT, in metres umng the. same scale.

Each point of the diagram, referred to as a world point,describes a certain event A (x, T). Each particle (e'ven astationary one) has a corresponding world line in this dia­gram. For example, the OT axis is the world line of a parti­cle resting at the point x = O. The Ox axis depicts the to­talityof events simultaneous with the event 0 irrespectiveof the x coordinate.

The world line corresponding to light propagating fromthe point 0 in the positive direction of the x axis is the

Relativistic Mechanics216

219

Fig. 120

/7:'I

II A

o

Kinematics in the Special Theory 0/ llelativity

whose points conform to the invariant interval s = 1(since when x = 0, 't =1 and s = 1). Its asymptote is theworld line of light. The hyperbola crosses the 't' axis at thepoint corresponding to a time unit in the K' frame. Indeed't'2 - X'2 = 1 and if x' = 0, then 't' = 1. '

The x and x' axes are calibrated in much the same way:a hyperbola x 2 -'t2 =_ 1 is drawn through the point x == 1, 't = °of the I( frame; then the point at which thehyperbola crosses the x' axis, and where 't' = ° marks aunit of length (x' _= 1) in the '1(' frame (since X'2 - 't'2 = 1and if 't' = 0, then x' =1).

The M inkowski diagram thusplotted illustrates the transitionfrom the I( to 1(' frame and con­forms to the Lorentz transforma­tion (6.8). In accordance withthe principle o~ relativity thereverse transitionlrom the 1('to K frame is illustrated by adiagram of a quite symmetricform: the coordinate grid of the1(' frame is rectangular and theI( frame is oblique-angled. We suggest that the readersthemsel ves should demonstrate this.. N.ow we s~all ~how how the Minkowski diagram assistsIn InterpretIng SImply and graphically such relativistice~ec~s as, for. example, the relativity of simultaneity, the_dIlatIOn of tIme, and the Lorentz contraction.

Relativity of shnultaneity follows immediately fromFig. 120. In fact, events A and B simultaneolls in' the I(frame turn out to occur at different moments in the 1('frame. The event A occurs later than event B by the~e-A't'. -_

I?i1ation of time. Let us consider two clocks I( and 1(' ~whIch show the same time 't = 't' = °at the moment whenthey are. at the same point, in space (x = x' = 0). Theclock I( IS nssumn!l to he at l'est in the K frnme and theclock 1(' in the K' hame.

Suppose !hat according to the clock I( a time unit elapses('t = 1); thIS corresponds to the event A in the diagram

Relativisttc Mechanics

11:'I

II

I

"C-1

218

S2 = 't2 - x2 = 't'2 _ X'2.,

Let us mark on the 't axis of the I( frame the point COl'l'e­sponding to a time unit in the K frame ('t = 1, Fig. 119)and then draw through that point the hyperbola

t 2- x2 = 1,

bisector OC of the right angle (the dotted line in Fig. 1-18).Let us plot the 't' and x' axes of the 1(' frame in this dia­

gram. Assuming x' = ° in the Lorentz transformation(6.8), we obtain the world line of the origin of the 1(' frame.Then x = Vt = ~'t, where ~ = VIc. This is the equationof the straight line forming the angle e with the 't axis,which can be determined from the formula tan e = ~.

The straight line obtained is the world line depicting thetotality of events occurring at the origin of the 1(' frame,

Le. it is the 't' axis.The x' axis of the 1('

frame is a straight line de­picting all the events whichare simultaneous with the­event 0 in the 1(' frame.Assuming t' = ° in theLorentz transformation(6.8), we get ct = xV/c,or 't = ~x. From this it fol-­lows that the x' axis formsthe same angle e (tan e ==~) with the x axis.

X=i x Thus, the 't' and x' axesF' 119 of the 1(' frame are ar-

19. ranged symmetrically with_ respect to the world line

OC of light, and the coordinate grid ('t', x') of the K' frameproves to be oblique-angled. The higher the velocity V ofthe K' frame, the more "oblate" its coordinate grid is. AsV -+ c it degenerates into the world line of light.

The last thing left to be done in the diagram is to scalethe 't, x, 't', x' axes of both reference frames. The easiest wayto do this is to utilize the 1nvariance of the interval:

22f

xFig. 123

Kinematics in the Special Theory of Relativity

Problems to Chapter 6

But in the K' frame the event () (the determination of theleft end of the rod) is simultaneous with the event B' repre­sented by the point of intersection of the world line of the~ight end of the .rod with the simultaneity line Ox'. Itis seen from the diagram that in the K' frame OB' < OA'Le. the rod moving relative to the K' frame is shorter tha~one metre.

I t can be "demonstrated just as easily that the Lorentzcontraction is reversible as well. If a one-metre rod is atr~st in t~e K' frame (segment OA '), then, drawing the worldhnes of itS ends in that frame (O't' and A'B) we see thatin the K frame OB < OA pro- 'vided the coordinates of the !{,~K'rod's ends are determined si- Y YI Vmultaneously, Le. the K' rod /--,._experiences the Lorentz con- Itraction .~ith respect to the IK hame. I

I Xl~--=L7-=--=-:.:z:===..,-_o 0'

(' 6.1. A stationary rod oflength I = 1.00 m is oriented at theangle 6 = 45° to the x axis of the K frame (Fig. 123). Find its length I'and the corres~onding angl~ 6' in the K' frame moving relative tothe K frame wIth the velocIty V = c/2 along the x axis.

Solution. The rod's length in the K' frame is

I' = V (L\x')2+ (L\y')2 =Y(L\X)2 (1-~2)+ (L\y)2.

Taking into account that" L\x = I cos 6 and L\y = I sin 6, we get

1'=IV1-~2cos26=0.94 m.

The angle 6' in the K' frame is found from its tangent:

tan fl' = L\y' L\y tan e .L\x' L\x V1-~2 V1-~2'

It shoul~ be. pointed out that the results obtained are independentof the dIrectIOn of the velocity of the K' frame.~

• 6.2. A ~o~ moves along a ruler with a certain constaIitvelocity.When th!l pOSItIOns of both ends of the rod are determined simulta­neously m the reference frame fixed to the ruler, the length of therod 11 = 4:0 m..However, when the positions of the ends of the rodare determmed simultaneously in the reference frame fixed to the rod,

x

Relativistic Mechanics

B A(x=/)

Fig. 122

D /7:'I

1

11

1/

1/

// " I/ / B

/ /' ,/'/ / " //,/ / /

1/ // 1t-::/ 1

r

oFig. 121

20

elapsed according to the moving clock K' when the clockK registers the passing of a time unit. This fact signifiesthat the clock K' goes slower.

Let us utilize the diagram to make sure that the effect ofthe dilation of time is reversible. Draw a straight line BA'parallel to the x' axis which describes all the events whichare simultaneous with the event A' in the K' frame ('t' == 1). The point B at which this straight line intersects theworld line of the clock K, the 't axis, shows that 't < 1,Le. it is the clock K whose rate is now slower with respectto the K' frame.

Lorentz contraction. Suppose a one-metre rod is at restin the K frame (the segment OA in Fig. 122). The straightlines O't and AD are the world lines of its ends. To measurethe length of that rod in the K' frame, we have to determinethe coordinates of its ends simultaneously in that frame.

(Fig. 121). Let us draw a hyperbola '(2 - x 2 = 1 throughthe point A and also a straight line AB' describing all theevents which are simultaneous with the event A in the Kframe. The intersection of the't' axis, Le. the world lineof the clock K', with the hyperbola gives the point A' ('t' == 1) and with the straight line OB' the point B' ('t' < 1).This means that in the K' frame a time unit has not yet

223

(1)

p

v }/ 1+~o 1-~

Fig. 125

A=cT-VT=(c-V) T=

= (c-- V) To •Y1-1}2

Therefore, the frequency re­ceived by the detector is equal to

C c -V1-~2V = _=o-=-:-_~

A To (c- V)orf3=V/C

Fig. 126

Ii III~

V1/

1/. V

/'V

I/'-to -as-0.5-0.4 -0.2 0 0.2 04 Q,S

Receding Approach

2.5

2.0

1.5

1.0

kinematics in the Special Theory of Relativity

0.5

v = vol(1 - ~) ~ Vo (1 + VIc).

where 011 the left-hand side we write the squared interval in the f!'amelixed to the particle and on the right-h!Uld side the llquared intervalin the K frame. From thill we ob-tain the same value of I. I

.6.5. Doppler effect. A station- f( tKary detector P of light signals is. Ilocated in the K frame (Fig. 125). s:lc~A source S of light signals approa- I

ches the detector with the velocity LV. In the reference frame fixed to O~-O~/;:-'-=-'=-==-=-=-=-'=-=-~.x~-.?the source the signals are emittedwith the frequency Yo. With whatfrequency v does the detector re­ceive these signals?

Solution. The time interval between two consecutive signals(pulses) in the K' frame fixed to the source is equal to To = 1/vo' As thisframe moyes with the velocity V, thelcorresporrding time interval

in the K frame is, in accordancewith Eq. (6.12), longer:

T= To/Y 1-~2, I}= VIc

The distance between two con!!Ocu­tive .pulses in the K frame is equalto .

yI=j32v=vo 1-~

When the source approaches(as in our case), then v > Yo, but

wben it moves away, v < Vo (in this caSe the sign of hanges to theopposite, Fig. 126). The formula obtained for t frequency v ex­presses the so-called radial Doppler "effect.

Note that in classical physics T = To since time is absolute.Therefore, the classical formula for the Doppler effect does not containthe factor y 1 - ~2, which is replaced by unity:

At the same time let us consider a more general case: in the Kframe the velocity V of the source forllls the angle a with the IiIll~ ofohservation (Fig. 127). In this case it is sufftcient to replace V in

kelatiuistic Mechanic.~

V A 8A'C!======-==::£\8'

Fig. 124

tube, fits in it completely. However, "in terms of" the rod it is thetube that becomes reduced by half, and consequently the rod (2.0 m)does not fit in the tube (0.5 m). Is there a contradiction here?

Solution. "In terms of" the tube the ends of the Dying rod coincidewith the ends of the tube simultaneously. "In terms of" the rod theends do not coincide simultaneously: first, the ends Band B' coincide(Fig. 124), and after the time interval M, the ends A and A'. Thetime interval I1t may be calculated as follows:

I1t=(Lo-l)/V=6.10-9 s,

where Lo = 2.0 m is the prope~ length of the rod, I =: q.5 m is .thelength of the tube moving relatIve to the rod, ani V IS Its velOCIty.The latter is found fr9m Eq. (6.11): V = c y3/2.

• 6.4. Find the distance which an unstable particle traversesin the K frame from the moment of its generation till decay, providedits lifetime in that reference frame is M = 3.0.10-6 s and its properlifetime is I1to = 2.2.10-6 S. . .

Solution. Using Eq. (6.12), we determlIle the velOCIty V of theparticle and then the distance sought:

1= I1t· V = I1t·c Y 1-(tlto/l1t)2=0.6 kill.

Another method of solution is based on the invarianr.e of theinterval:

the difference of readings made by the ruler is equal to 12 = 9.0 m.Find:

(1) the proper lengtb of the rod;(2) tho velocity of the rod relative to the ruler.Solution. The proper length of the rod lo is related to 11 and !2

via the following formulae:

II = 10 Y1- ~2; lo = 12 Y1- ~2,

where ~ is the velocity of the rod expressed in units of the velocityof light. From these formulae we obtain:

(1) 10 = Y lJ;"= 6 m; (2) ~ = Y 1 - 11/l2 = 5/3 ~ 0.75 or v = 0.75 c.

• 6.3. Rod-and-tube "paradox". A tube AB of length 1.0 m isat rest in the K frame. Let us take a rod A'B' of length 2.0 m andaccelerate it to such a velocity that its length in the K frame becomesequal to 1.0 m. Then at a certain moment the rod, flying through the

222

225

--------~

x x'

v..IIIIIIL_

0'

/(',II

o

Kinematics in the Special Theory of Relativity

Pair or

/"". 0' mW..rProper tim" Proper IPossible cause-events c·Ato. m distance and-effect reia-

&xo. m tlon

AB Time-like 4 - A_BAC Space-likej - 4 NoneBC Light-like! 0 0 C-B

Hint: use invariance of the interval.W 6.7. Two particles move in the K frame at right angles to each

other: the first one with the velocity VI and the second one with va'Find the velocity of one particle with respect to the other.

Solution. Let us choose the coordinate axes of the K frame asillustrated in Fig. 129. When the K ' frame is fixed to particle 1,the velocity of particle 2 in that reference frame is the value sought.Introducing V = VI and Vx = 0in Eq. (6.15), we get

vi~Yv;~+V;~==Y Vf+ vi-(VIValc)2.

~

Note that in accordance with theclassical law of vector composition

vi= Yvf+vi.. Ii 6.8. Velocity direction trans- Fig. 130

formation. AJlarticle moves in the Kframe with the velocity v at the angle 0 to the x axis. Find the co~sponding angle 0' in the K ' frame, which moves with the velocity Vas illustrated in Fig. 130.

Solution. Suppose the projections of the vector v in the K frameare equal to Vx and v". Then the following relation is true:

tan 6' .= vylvz •

Taking into account Eqs. (6.14) we obtain in the K ' frame

tan 0' =vi/v~=vy Y 1-~I/(vx-V). .After the substitution Vz = v cos 6 and v" =:JUlirfO we fmd

~

tane' sin 6· V1-~acos6-Vlv

As the last equation shows, the angle transformation law for velocitydiffers from that for segments (see Problem 6.1).

Ii 6.9. A rod oriented parallel to the x axis of the K reference framemoves in that frame with the velocity V in the positive direction ofthe y axis. Find the angle 6' between the rod and the x' axis of theK' frame travelling with the velocity V relative to the K frame in1Ii-0539

Relativistic Mechanics

p

Fig. 129

OL..--------~::;

K K'. J..I__~ --~_,1 2 z

Fig. 127

. y

s~Vcosa

Fig. 128

2 3 4 5 6 7 x,m

8

C

1A

o

t=ct,m6

5

4

3

2

Eq. (1) by V cos a. Then

V~v=vo 1-~cosa .

224

v=vo V1-~2,

in which the observed frequency always proves to be lower than theproper one Vo' Incidentally, the last expression is just a consequence

In particular, at a = 11/2 the so-called transverse Doppler effect isobserved:

of the dilation of time in a moving reference frame; it may alsobe obtained directly from Eq. (6,12): .

v=-!..= 1 vo Y1-~2.T ToIY1-~2

• 6.6. Relationships between events. Fig. 128 illustrates a sp.ace­time diagram. Each point of that diagram (a world point) descrIbes

a certain event, that is, a coordinate and a time moment. at whichthat event happens. Let us examine three events corresp~mdlI~g to t.heworld points A, B, and C. Prove that the following relatIOnshIps eXIsthetween these events:

the positive direction of its x axis. The x and x' axes coincide, the yand y' axes are parallel to each other.

Solutio~. Suppose t~at. at a certain moment the ends of the ~odcoincide wIth the x aXIs III the K frame. :rhese t\~o event~, whIChare simultaneous in the K frame, are not so III the K frame; In accor­dance with Eq, (6,10) they are separated by the time interval

. tJ.t' =tJ.xVlc2Y 1-~2,

where tJ.x is the proper length of the r?d. In that t~me t~e r~ght endof the rod shifts "higher" than the left one by fiy = vy M , wherev~ = v VI - ~2 (see,Eq. (6.W». Thus, in t~e K~ frame the rod isturned counterclockwise through the angle e , which may be deter­mined using the formula

tan e' = tJ.y'I tJ.x' =~v/c y 1-~2,

where tJ.x' = AxV1 - ~2 is the projection of the rod on the x' axisof the K' frame, and ~ = vic... ,

• 6.10. Relativistic transformation of a?celera~ion. A partIclemoves with the velocity v and th.e ac?eleratlO~ w III the (( fraI!1e.Find the acceleration of that particle III the K frame, ~hICh shiftswith the velocity V in the positive dir~ction of the x aXIs of the. Kframe. Examine the cases when the particle moves along the followlllgaxes of the K frame: (1) x, (2) y. .

Solution. 1. Let us write each projection of the acceleratIOn ofthe particle in the J(" frame as follows:

dv~ dv~ 1w~= dt' =(jt' dt'/dt .

Making use of the first of formulae (6.14) and the last one of (6.8),we get after differentiation:

, (1_~2)3/2 .w;e= (1-~vxlc)3 WX, wY=O.

2. Similar calculations yield the following results:w;=O; w;=(1-~2)wy•

In these formulae ~ = vIc.

226 Relativistic Mechanlc8 CHAPTER 1

RELATIVISTIC DYNAMICS

§ 7.1. Relativistic Momentum

Let us first recall two basic assumptions of Newtonianmechanics concerning momentum:·

(1) the momentum of a particle is defined as p = mv,and the mass m of the particle is supposed to be indepen­dent of its velocity;

(2) the momentum of a c~osed system of particles doesnot vary with time in any inertial reference frame.

Now we shall turn to relativistic dynamics. It is foundhere that the conservation law for Newtonian mnmentumis not valid in the case of a closed system of relativisticparticles. (We shall illustrate this later by a simple exam­ple.) Thus«we face the following choice: either to reject theNewtonian definition of momentum, or to discard the lawof conservation of that quantity.

Considering the immense significance of the conservationlaws, in the theory of relativity the momentum conserva­tion law is regarded as fundamental and the momentumitself is expressed accordingly·.

First of all we shall demonstrate that the requirementfor the momentum conservation law to hold in any inertialreference frame together with the relativistic transforma­tion of velocities on transition from one inertial referenceframe to another leads to the conclusion that the particle'smass must depend on the velocity of the particle (in contrast

• The following question arises: how can the momentum conser­vation law be of any value if momentum is defined so as to keep itconstant? To answe.r the question, let us imagine a particle collidingwith other. particles in the process of its motion. Having consideredthe first collision, we define momentum so that· it obeys the conserva­tion law in that collision. In th llowing collisions, however, thesituation is different: we know the mo a of the particles involvedin those collisions, and the momentum con vation law (if it reallyexists) is valid not according to definition but due to underlyinglaws of nature.

. Experience shows tbat momentum thus defined really obeys theconservation law. At least, not a single phenomenon has been ob­served up to now in which that law fails.

1Ii*

to Newtonian mechanics). For this purpose let us examinea completely inelastic collision of two particles, where the

-system is assumed closed.Suppose two identical particles 1 and 2 move toward

each other in an inertial reference frame K with the samevelocity Vo at an angle a to the x axis (Fig. 131a). In thatreference frame the total momentum of both particles ap­parently remains constant: it equals zero both before and afterthe collision (and the formed particle turns out to be mo­tionless, as follows from symmetry considerations).

Now let us see what happens in another inertial referenceframe. First, we choose two reference frames: the K1 frame

229

c 11

Fig. f32

o

mol---.....:::=-------i

Relatlvi,tlc Dynamic,

Now the y components of momenta of both particles maybe written in the K1 frame as m1u and m 2u'. In accordancewith Eq. (7.1) u' < u, and therefore it is easy to see thatthe momentum conservation law does not hold true in itsconventional (Newtonian) statement. Indeed, in our caseml = m 2 (the particles being identical) and therefore they component of the total momentum of the particles beforethe collision differs from zero while after the collision it isequal to zero, since the par­ticle formed moves only malong the x axis.

The momentum conserva­tion law becomes valid inthe K 1 frame if we assumemlu = m 2u' . Then fromEq. (7.1) we get

m2 =mi1V1-(Vlc)2.

When a -- 0 (Fig." 1§1),u -- 0 andml is thE\ massof the motionless particle;it is denoted by mo and isreferred to as the restmass. In that case the velocity V proves to be equalto v, the velocity of particle 2 with respect to particle1. Consequently, the last formula can be rewritten as

m = molV1- (V/C)2, (7.2)where m is the mass of the moving particle (recall that thetwo particles are identical). The mass m is referred to asrelativistic. As it is seen from Eq. (7.2), the latter is greaterthan the rest mass and depends on the particle's velocity(Fig. 132). ~ .

Thus, we have reached an impMtant conclusion: therelativistic mass of a particle depends on its velocity. In otherwords, the mass of the same particle is different in differentinertial reference frames.

In contrast to the relativistic mass the particle's rest massm o is an invariant quantity, Le. is the saIIJ.e in all referenceframes. For this reason we can claim that rest mass is achar~

acteristic property of a particle. But later on we shall(7.1)

Relatlvi,tfc Mechanic'

u' =uv 1-(Vlc)2.

in K frame((1)

y

228

moving to the right with the velocity Vl x and the K 2 framemoving to the left with the velocity V 2X (Fig. 131a). Clearly,particle 1 in the K1 frame and particle 2 in the K 2 framemove only along the y axis with velocities whose equalmoduli we denote by u.

Let 'us consider the collision in the K1 frame (Fig. 13tb)in which particle 1 has the velocity u. We find the y compo­nent of the velocity of particle 2 in that reference frame,denoting it by u'. As we mentioned, that particle moveswith the velocity u along the y axis in the K 2 frame and atthe same time translates together with the K 2 frame to theleft with the velocity V relative to the K1 frame. Therefore,in accordance with Eq. (6.16), the y component of the ve­locity of particle 2 in the K1 frame is equal to

frequently use the relativistic mass m to simplify someconclusions, reasonings, and calculations.

Now we shall take the last step. Using Eq. (7.2) we writethe momentum of a relativistic particle in the followingform:

231

(7.5)If (V 1~~V/C)2) = F .\

This is the fundamental equation of relativistic dynamics.

Relativistic Dynamic.

From the condition '11 = (p - pc/)Ip = 1 - Vi - (vlc)2 we get

vIc = VTJ (2 - '11).Whence

...!..={ 0.14 at '11=0.01,c 0.45 at '11 = 0.10.

Thus, the classical formula for momentum provides an accuracy betterthan one per cent at vIc ~ O.H·and better than ten per cent at vIc ~<: 0.45.

§ 7.2. Fundamental Equation of RelativisticDynamics

According to Einstein's principle of relativity all lawsof nature must be invariant with respect to inertial refer­ence frames. In other words, the mathematical formulationsof laws must be identical in all these reference frames.In partj(t,ular, this is true for the laws of dynamics.

However, detailed analysis shows that the fundamentalequation of dynamics of Newton mw = F does not satisfyEinstein's principle of relativity. The Lorentz transforma­tion totally changes the form of the equation on transitionto another inertial frame.

To satisfy the requirements of the principle of relativity,the fundamental equation of dynamics must have anotherform and only in the case of v ~ c turn into the Newtonianequation. It is shown in the theory of relativity that theserequirements are met by the equation

dpldt = F, (7.4)

where F is the force acting on the particle. This equationcompletely coincides in jormwith the fundamental equationof Newtonian dynamics (4.1). But the physical meaning isdifferent here: the left-hand side of the equation containsthe time derivative of a relativistic momentum defined byformula (7.3). Substituting Eq. (7.3) into Eq. (7.4), weobtain

(7.3)

Relattvistic Mechanics

Ip=mv= mov I~1-(VIC)1 •

Fig. 133

f

o

2

Then1 - ~2 = (1 + ~) (1 - ~) ~ 2 (1 - ~).

mlmo ~ 1/y2 (1 - ~) ~ 70.

Example 2. At what velocity does a particle's Newtonian momen­lum differ from its relativistic one by one per cent? by ten. per cent?

Example 1. modern giant accelerators protons can be accele-rated up to a veloc Tdiffering from that of light by 0.01 per cent.How many times does the relativistic mass of such protons exceedtheir restlmass?

In accordance with Eq. (7.2) mlmo = 1ly1 - ~1I, where ~ = vIc.Since ~ slightly differs from unity, the radicand should be trans-formed as follows: .

230

This is the so-called relativistic momentum of a particle.Experience confirms that the momentum thus defined does

obey the conservation lawregardless of the inertialreference frame chosen.

Note that if v ~ cEq.(7.3) yields the Newtoniandefinition of momentum:p = mov, where mo is in­dependent of the velocityv. The velocity dependencesof the relativistic andthe Newtonian momentumof a particle are comparedin Fig. 133. The differencebetween the momenta is

seen to grow substantially as the velocity of a particleapproaches that of light.

Let us consider two examples illustrating how Eqs. (7.2)and (7.3) re applied.

233Relativistic Dynamics

dT = Fvdt.

§ 7.3. Mass-Energy Relation

Kinetic energy of a relativistic particle. We shall de­fine this quantity in the same fashion as we did in classicalmechanics, i.e. as a quantity whose increment is equal tothe work performed by the force acting on a particle. Firstwe find the increment of the particle's kinetic energy dTdue to the force F acting over the elementary path dr == vdt:

takes the form

(mo + movs

/cs

) ~- FY 1-(v/c)s [1- (v/c)IIJsfB dt - ,

whence the acceleration written in vector form is

w = (F/mo) [1- (v/c)2]3/2.

I t is not ¥fficult to see that if the force F and the velocityv have the same values in both cases, the transverse forceimparts to the particle a greater acceleration than the longi­tudinal force.

The fundamental equatilJ:l of relativistic dynamics makesit possible to find the law relating to the force F actingon a particle provided the time dependence of the rela­tivistic momentum pet) is known, and?n the ot~er hand~ tonnd the equation of motion of the partIcle r ~t) If the actmgforce and the initial conditions, the velocIty Vo and theposition ro at the initial moment of time, are known.

The application of Eq. (7.5) is illustrated by problems7.1-7.3.

lnow/V1- (VIC)2 = F,

whence the acceleration

w=(F/mo) V1-(vlc)2;

(2) F II v (longitudinal force). In ~his c.ase Eq: (?5) ID:aybe written in scalar form; performmg dIfferentIatiOn WIthrespect to time on the left-hand side of the equation, weobtain

Relativistic Mechanics232

It can be easily seen that the equation written in thisform ensures the invariance of momentum for a free particleand turns into the fundamental equation of Newtoniandynamics (mw = F) at low velocities (v ~ c). .

Moreover, the fundamental equation of dynamics, whenwritten in this form, proves to be invariant relative to theLorentz transformation and, consequently, satisfies Ein­stein's principle of relativity. We shall not prove this here,

but shall only note that on tran­sition from one inertial referenceframe to another the force F istransformed in accordance withdefinite rules. In other words, theforce F is not an invariant in the

dm'r! theory of relativity and its ~mag-

at nitude and direction vary·.A surprising conclusion follows

Fig. 134 from the fundamental equation ofrelativistic dynamics: the accelera­

tion vector w of a particle does not coin~ide in the gen­eral case with the direction of the force vector F. To demon­strate this, we write Eq. (7.5) in the following form:

~ d(mv)/dt=F,

where m is the rel~tic mass of the particle. Differen­tiating with respect to time, we obtain

(dm/dt) v + m (dv/dt) = F. (7.6)

This expression is graphically illustrated in Fig. 134. Thus,the acceleration vector w is indeed not collinear with the'force vector F in the general case.

The acceleration w coincides in direction with 'the vectorF only in two cases:

(1)F-Lv (transverse force); in this case the magnitude ofthe vector v does not vary, i.e. v = const, and Eq. (7.5)

• As distinct from Newtonian mechanics where forces are absolute.in the theory of relativity the force projections perpendicular to thedirection of the relative velocity vector of the reference frames aredifferent in these frames. The projections have maximum valuesin the reference frame'where the particle is at rest at a given moment:

F~=Fx, Fy=F!I Y1-(vlc)s. '

Eliminating ~2 from these two equations, we get

2or2 + (4 - n) or - 2 (n - 1) = O.

The root of this equation is

't = [n - 4 + Y n (n + 8»)/4.

The minus sign in front of the radicand has no physical meaning(or cannot be negative) and thus can be omitted.

Here are the four values of or calculated from the last formula forthe following n:

235

0.62

2.0

Fig. 135

0 T

r,..

0 I11

V Tcl/ l.--V l.--'

60=po

o 0.1 0.2 0.3 0.4 0.5 0.5 0-7 u.s 0-9 1.0j3=V/c

't = TImoc2: 0.0067 0.065 0.32

Relativistic Dynamics

At ~ ~ 1 we can confine ourselves to the first two terms ofthe series, and then

T = moc2~2/2 = mov2/2.

Thus, at high velocities the kinetic energy of a particleis given by the relativistic formula (7.9), which is differentfrom mov2/2. It should be poin-ted out here that expression(7.9) cannot be representedas mv2/2, where m is the rela­tivistic mass.

The relativistic T rel andclassical Tel kinetic energiesplotted as functions of ~ arecompared in Fig. 135. Theirdifference becomes very pro­nounced.-.t velocities compara­ble to that of light.

Example!. A partic(e of mass momoves with the velocity at whichits relativistic kinetic energy T ex­ceeds by n times the kinetic energycalculated by means of the classical formula. Find T.

For the sake of simplicity we mtroduce the designation 't = Tlmoc2 •

Then the given condition T= ri.mov2 /2 may be written as

't = n~2/2,

where ~ = vIc. From Eq. (7.9) ~ can be expressed as

~2 = 1 - 1/(1 + or)'.

(7.9)

Let us find the differential of this expression, bearing inmind that m o and c are constants:

2mc2dm = 2mv2dm + 2m2vdv.

The right-hand side of this equality, when divided by 2m,coincides with the expression for dT. Hence, it follows that

dT = c2dm. (7.7)

In accordance with the fundamental equation of relativisticdynamics (7.4) Fdt = d (mv) = dm·v + mdv, where m isthe relativistic' mass. Therefore,

dT = v (dm·v + mdv) = v2dm + mvdv,

where 'the relation vdv = vdv is taken into account (seep. 90). This expression can be simplified by allowing forthe dependence of mass on velocity (Eq. (7.2». Squaringthe equation, we get

m2c2= m2v2+ m~c2.

Relativistic Mechanics

Thus, the increment of the kinetic energy of a particleis proportional to the increment of its relativistic mass.The kinetic energy of a motionless particle is equal to zeroand its mass is equal to the rest mass mo. Consequently,integrating Eq. (7.7), we obtain

T = (m-mo) c2, (7.8)or

where ~ = vic. This is the expression for the relativistickinetic energy of a particle. It can be seen how conspicuouslyit' differs from the classical mov

2/2. Let us make sure, how­ever, that at low velocities (~ ~ 1) expression (7.9) turn!'into the classical one. For this purpose we employ the bino­mial theorem, according to which

1 =(1_~2)~t/2=1+J...~2+2-W'+'"y1-~2 2 8

. I

237RelatiVistic Dynamic,

Rel.ation (7:10) may be written in another form if Eq."(7.8) IS taken mto account. Then the total energy of a bodyis

E=moc2+T,

where mois the rest mass of a body and T is its kinetic ener­gy. From this it follows directly that a motionless body(T = 0) also possesses the energy

Eo=moe2• (7.11)

This energy is referred to as the rest energy or the properenergy.

We see that the mass of a body which in non-relativisticmechan~cs manifested itself as a measure of inertness (inNewton s second law), or as a measure of gravitationalaction (in the law of universal gravitation), emerges nowas a meaS1l1'e of the energy content of a body. In accordancewith the theory of relativity even a body at rest has a cer­

.tain amount of storid-up energy, the rest energy.A ch~nge in the t9tal energy of a body (a system) is ac­

compamed by an equivalent change in its mass I!m == I!E/c2 and vice versa. In conventional macroscopic pro­cesses the change of mass of bodies turns out to be extreme­ly ~mall, so that its experimental detection is impossible.ThIS can be demonstrated by the following examples.

Examples. A. A satelli~ of{mass m = too kg tis launched intothe Eartij's !,rbit by a~celerating it to the velocity 11 = 8 km/s. Thismeans that Its ~ner~ mcroa~s by /),.E ~ mv2/2 (allowing for IJ « c).The correspondmg mcrease In the satellIte's mass is equal to

/),.m=/),.E/cll =mv'/2cs =3.5·fO-s kg.B. Heating one litre of, water from 0 to too °C requires the energy

/),.E = m~p /),.t, where cp =[4.2.1/(g.K) is the specific heat of waterand /),.t IS the temperature dIfference. The corresponding increasein the mass of the water is

/),.m=/),.E/c'=0.47·fO- l o kg.C. A spring of stiffness factor x = fOlI N/cm is compressed by

&l = 1 cm. In the process the spring acquires the energy /),.E == x (&l)'/2. The equivalent increment in its mass is equal to

/),.m=/),.E/c'=0.5·fO-16 kg.

. It is easy to see that in all three cases the mass changeshe far outside the capabilities of experimental technique.

(7.10)

Relativtlttc Mechantc,

It is seen that, for example, at T/mocl ~ 0.0067 the application of thecIa cal formula permits the kinetic energy to be determined withan ccuracy better than one per cent.

Examp'le 2. What amount of work must be performed to increasethe velocIty of a particle of rest mass mo from 0.6 c to 0.8 c? Comparethe result obtained with that calculated from the classical formula.

In accordance with Eq. (7.9) the work sought is equal to

A=TlI-T1 =mocS ( V f f )=0.42niocS.f-~I Vf-~f

The classical f~rmula yields the following value:

A=mo (vl-uf)/2= 0.f4mocs•

The difference between the two results is seen to be substantial.

Relation between mass and energy. It follows from Eq.(7.7)that the increment of kinetic energy of a particle is accompa­nied by a proportional increment of its relativistic mass.It is known, however, that various processes taking placein nature are connected with the transformation of one kindof energy into another. For example, the kinetic energy ofcolliding particles can be transformed into the internalenergy of a new particle formed after the collision. Therefore;it is natural to expect that the mass of a body grows notonly due to additional kinetic energy but also due to anyincrease in the total energy stored in the body, irrespectiveof what specific kind of energy is responsible for that in­crease.

Owing to this Einstein reached the following fundamen­tal conclusion: the total energy of a body (or a system ofbodies), whatever kinds of energy it comprises (kinetic,electric, chemical, etc.), is related to the mass of that bodyby the equation

This formula expresses one of the most fundamental lawsof nature, the relationship (proportionality) of the mass mand the total energy E of a body. To avoid misunderstand­ing, we should point out that the total energy E does notinclude the potential energy of a body in an external field,provided such a field acts on the body.

IIII1I

II, ,I,IIII

I

Relativistic Mechanics

In astronomical phenomena, however, associated, forexample, with exploration of stars mass can change by anappreciable amount. One may ascertain this by the exampleof solar radiation.

Example. According to astronomical observations the amountof energy carried by solar radiation each second to an area of 1 m2

of the Earth's surface oriented at right angles to the solar rays comesto about 1.4.103 11(s ·m2). This makes it possible to calculate thetotal energy radiated by the Sun per second:

.1E=1.4·103 ·4nR2=4·1026 lis,

where R is the distance between the Earth and the Sun. Consequentlyevery second the Sun loses the mass

.1m = .1Elc2 = 4.4.109 kg/s!

This value is stupendous on the Earth's scale, but when comparedto the mass of the Sun this loss is negligible: b.mlm = 2.10-2 S-I.

Things are quite different in nuclear physics. Here itbecam~ possible for the first time to experimentally checkand confirm the law relating mass and energy. This is be­cause nuclear processes and transformations of elementaryparticles are associated with very large changes of energycomparable with the rest energy of the particles themselves.We shall return to this problem in § 7.5.

§ 7.4. Relation Between Energy and Mo~entum

of a Particle

It is clear that both the energy E and the tllomentum pof a particle have different values in different referenceframes. There is, however, a quantity, a certain combinationof E and p, that is invariant, Le. has the same value indifferent reference 'frames. Such a quantity is E2 - p2e2.Let us make sure that this is so.

Making use of the formulae E = me2 and p = mv, wemay write

E2_ p2e2= m2e4-m2v2c2 1_:t;C)2 [1..:.....(vle)2}

or after cancelling

(7.12)

Relativistic Dynamics

The cancellation of the velocity v on the right-hand sidemeans that the value of E2 - p2e2 is independent of thevelocity of the particle, and consequently, of the referencefran: e. I? othe~ words, the quantity E2 - p2e2 is indeedan III varIant WIth the value m~e4 in all inertial referenceframe~; .

E2 ;- p2e2 = inv. (7.13)

~his .conclusion is of great importance since it allows,as It wIll be shown later, the analysis and solution of vari­ousproblems to be drasticall,\' simplified in many cases.

Here are two more relations which are very often usefulThe first one is .

Ip=mv=Ev/e2 / (7.14)

and the se~nd one relates the momentum and the kineticenergx T df a particle; it can be easily obtained by substi­tuting E = mOe2 + T,: into Eq. (7.12):

Ipc'= V T (T +2moC2)./ (7.15)

At T ~ moc2 the last relation turns into the classical one,p Tlr2moT, and when T ~ moe2

, it takes the form p=

t OEx5a1mMPle. Assuming the rest energy of an electron to be equalo. eV, calculate:

(1) th~ momentum· of an electron possessing a kinetic energy'equal to Its rest energy;

(2) the kinetic energy of an electron possessing the momentum0.51 MeVlc, where c is the velocity of light.

= 6,9 I~e~/~ moc2, we obtain from Eq. (7.15) p = lfamoc =

? This problem may also be solved by resorting to Eq (7 15)A SImpler way, though, is to utilize Eq. (7.12): . . .

T=E-moc2= V p2c2+mBc4-moc2=0.21" MeV.----. • Not,e ~~at now momenta of relativistic particles are expressed~n the UDltS f!nergy/c", 'Yhere c is the velocity of light. E.g. jf energy~s ~xpressed In ~eV UDltS. (1 MeV = 1.6.10-6 erg), then ~omentumIS m MkeYlc. The mtroductlOn of such a unit for momentum simplifiesmany mds of calculations '1uite noticeably.

The Lorentz transformation of momentum and energy. Let a par­ticle move with the velocity v = dl/dt in the K reference frame. FromEq. (6.i3) it follows that the elementary interval is

u= Vel (dt)I-(dl)l=c dt V i-(U/C)I.

Bearing that expression in mind, we present the projections of themomentum and the energy of the particle in the following form:

P _ me ~=.m.c dz. py=moc ••.~:;:1:- Vi-(u/c)1 dt dB ' ~....

moel dt dt c. dtE= -=moc8-=moc --.

Vi-(u/c). dt ds ds

From the invariance of the interval dB it immediately follows thaton transition to another inertial reference frame P:I: and Py are trans­formed as dz and dy, i.e. as z and y, whereas the energy E is trans­formed as cI dt, i.e. as the time t. Thus, the following correlations

24i

(7.i7)

(7.22)

Ilelatiuistic Dynamic,

may be pointed out:

P:I: - z, Pf/ - y, E/cl - t.

Rep.lacing .the indica~d quantities in the Lorentz transformation (6.8)we Ihmmedlately obtam the transformation of momentum and energy'lIOug t:

P'-= ;:I:-EVlcl

• E'-_ E-p:l:V- i ( P~ =PII'- Vlc)1 V i-(Vlc)1 '

where V is the velocity of the K' frame relative to the K fraThese formu~ae .express the transformation law for the mome~~m

fand energy proJectIOns of a particle on transition from the K to K'rame.

. ~o~e compac~ notation. At present all formulae of rela­tiVistiC mech~D1cs are customarily written in a more com­pact form, USlD~ . the follOWing abbreviations:

(1) the <uantItIes mc2 and pc are denoted simply by m and~n~:S~; expressed accordingly in energy units (e.g., in MeV

(2) all velocities ire expressed in units of the velocityof lIght and denoted by f}:

f} = vIc; (7.18)

(3) the frequently occurring factor 11V 1 - ~2 is denotedby Y. the BO.-called Lorentz jactor: __ .

y=1/V1-pz. (7.19)

These designations dramat,ically si~plify not.only the appear­an~e of the formulae but all transformations and calcu­~ahons as well. The basic formulae of relativistic dynamicsIn the new notation are given below:

relativistic momentum (7.3)

m,jl

Relatluistic Mechanics2«)'

In passing, let us examine the interesting possibility ofthe existence of particles with zero rest mass (mo = 0).From the equations

E = moe2/V 1-(vlc)2 and p= moV/V 1- (VIC)2

it follows that a particle whose rest mass m o = 0 may pos­sess energy and momentum only when it moves with thevelocity of light c. Then the last two formulae turn into theindeterminate ratio 010. This fact does not signify, how­ever, the indeterminacy of energy and momentum of sucha particle. The point is that both these quantities prove tobe independent of velocity. Moreover, the relationshipbetween the momentum p arid the energy E is specifiedby Eq. (7.14), where v = c, i.e.

p = Elc. (7.16)

Thus, in accordance with the theory of relativity theexistence of particles with zero rest mass is possible, provid­ed they move with the velocity c. This motion is not aresult of preceding acceleration but this is the only statein which such particles can ever exist. The stoppage of sucha particle is equivalent to its absorption (disappearance).At the present time there are two such particles known:the photon and the neutrino.

243

(7.28)

Relativistic Dynamic,

with other particles), and W is the total energy of interac­tion of all particles of the system.

In classical mechanics W is the potential energy of inter­action of a system's particles, a quantity depending onlyon the configuration of the system (for a given characterof interaction). It turns out that in relativistic dynamicsthere is no such concept as the potential energy of interac­tion of particles. This is due to the fact that the very con­cept of potential energy is closely connected with the con­cept of long-range action (instantaneous interaction trans­mission). Being a function of the system's configuration,potential energy is defined at every moment of time by therelative disposition of the system's particles. A change inthe configuration of a system must immediately induce achange in potential energy. Since there is no such thing inreality (int~actions are transmitted with a finite velocity),the concept of potential energy of interaction cannot beintroduced for a system of relativistic particles.

An expression for the interaction energy W, and thereforethe total energy E, 01 a system of interacting relativisticparticles cannot be written in the general case. The sa.rnecan be said about the system's momentum since in relati­vistic dynamics momentum is' not a quantity independentof the energy E. Things are as complicated in the case ofthe rest mass M o of the system. In the general case it isknown to be the mass in a reference frame where the givenmechanical system is stationary as a whole (Le. in theC frame).

OWing to the complications mentioned above the devel~opment of the dynamics of a system of relativistic particlesis restricted to a few simple cases, two of which will beexamined here: a system of non-interacting relativisticparticles and the case of two colliding particles, which isimportant from a practical point of view.

System of non-interacting particles. In this case the totalenergy E and momentum p possess additive propertieswhich can be given as

16*

where mi and Pi are the relativistic mass and the momentum

(7.26)

(7.12)-(7.15):

(7.23)

(7.24)

(7.25)

E2-p2=m~=inv,

p = E~,

P= VT (T+ 2rno);

relations between energy and momentum

Relativistic Mechanics

§ 7.5. System of Relativistic Particles

d t of a system. Up to nowAbout the energy an momens~d:ration of the behaviour

we re~tricted o~;seliesto t~~~ to the dynamics of a single

of a smgle partlC e. n con . d ics of a system ofpa.rticle, the develoPbment of c~emo~:a:IilPlicat~d task inparticles proves to e a mu ber of impor-the theory of relativitybeNe~~:~W:~:~'i: ~:: case as well.tant gene~al laws caJ.! e the motion. of a system as a whole,

If we wIsh to examm . the system andI t' the internal processes In

:~:~;i::~t~Cs~~~al d(imens:~cl:).t~:c~;J~ri:iy~a~:;s~:~ar:fed as a mass p~mt a par d scribed by the total energyrelativistic particles can be e M and the relations de-E tum p and rest mass 0' •

ri~e~~:'1~er can 'be considered valid for the system of parti-

cles as a whole. hI" h now how to interpret the totalWe have to esta IS and the rest mass M o of a sys-

energy E, the momen~um p, I case .if the system consists

~~mi::e:a~f:~e·r~fat;v~sf:cne~~rticle~, its total energy is

E= L~ mic2+ W, (7.27)

, of the ith particle (recallwhere mi

c2IS ~he dtotal etn~r~~de the energy of interaction

that this quantity oes no m

the Lorentz transformation of momentum and energy(7.17):

245

vI' x'

1('

'~'.

Fig. 136

RelatiVistic Dynamics

velocity of the system's centre of inertia.Collision of two particles. We shall coIfsider. the colli­

sion process as proceeding in two stages: first, the formationof a compound particle A * and then its decay into two par­ticles that, in the general case, may differ from the initialones:

Al+~-A*~A3+A4+'" .In the process of the convergence of particles A 1 and A 2

the interaction between them may not remain weak, and.Eq. (7.28) becomes inapplicable. However, after the result­ing particles have separated far from each other, Eq. (7.28)becomes applicable again.

In the given case the sum of the total energies of the twoinitial particles (when they are so far from each other thattheir interfction is negligible) can be shOwn to equal thetotal energy of the compound particle. The same is truefor the second stage pf the process, that is, the decay. Inother words, it may be shownthat the total energy conser­vation law proves to holdtrue for this process in thefollOWing form:

E1+Ez=E*=E3 +E4 + ....(7.34)

We shall demonstrate thatthis is really so by the fol­lowing simple example.

Let us imagine a collisionof two identical particles 1and 2 that results in the format~on of a certain compoundparticle. Suppose the particles move toward each other inthe K frame before the collision with the same velocityv as shown in Fig. 136. Let us consider this process in theK' frame moving to the left with the velocity V relativeto the K frame. Since in the K frame the velocity of eachparticle is perpendicular to the vector V, the two particlesin the K' frame, in accordance with Eq. (6.14), have xcomponent equal to V. The compound particle formed,whose relativistic mass is denoted by M, has the same

(7.29)

Relativistic Mechanics

·tl article of the system. Since there is no interac­~:O:lh~1I1tl~i; case, the velocities of all particles arc cons~antand consequently the total energy .and .the momen umf the whole system do not change With time. f

o Let us introduce the rest energy Eo for a ~ste~ho tP~riticles as its total energy in the C frame, were e 0 ~

t .. - - '''p-. = 0 and the system as a whole ISmomen um IS p --!.J l

at rest. Thus,

244

,,­Eo = ..:.J Ej,

h jf.' the total energy of the ith particle in the C~a:: Thi;~eans that the rest energy inc~ude~ no~ only t.he

t · . ~ of all particles but also their kllletIc energIesres energle", _ _'T· in the C frame: E i = m otc2 + T i •

I Obviously, the same is true for the rest mass of the sys-

tem: M E /2 (7.30),0= 0 c .

In Particular, it follows from this that the rest mass ofh of the rest masses ofthe system is not equal to t e sum

its constituent particles:

AIo> ~mot.

1.'he introduction of the rest energy and the redst mas~ ofE and M makes it possible to regal' a sys em

~fs;~~~Tn't~r~cting rei~tivistic particles as one particle~ith1 ener E = 2} mic2, the momentum p = .::.JPt,

thl e totta ,~_ E fc2 and to claim .Eqs. (7.12) and

t Ie res mass 0 - 0 , ft' 1 well'(7.14) to be valid for the system 0 par IC es as .

E2 _ p2C2 = .M~c4 = inv, 0 (7.31)

P = EVfc2 , (7.32)

where V is the velocity of the system a~ a ~hole, Le. t~evelocit of the C frame. In accordance WIth ~q. (7 .32~ thiSvelocit~ may be represented in the followmg form.

V= (2~ Pi)/(2~ mi). (7.33)

1 . . t' s of the ith particle of thewhero mi is the 1'0 atlvls IC ma s.. . f 'th thesystem. Note that Eq. ~7:3~) comcl.des, III orm ~~r thecorresponding lion-relativIstic expressIOn (4.9)

I

I

'I

velocity in the K' frame. Applying the momentum conser­vation law before and after the collision (to the x componentof the momentum), we obtain 2m (v') V = M· V, wherev' is the velocity of each initial particle in the K' frame.Hence,

247Relativistic Dynamics

Recalling that the total energy of each particle may be given as E == mocll + T, where mo is the rest mass of a nucleus and T is its ki­netic energy, we rewrite the preceding equality as

(m i +m2) ct +T12 = (ma+m,) cll +Ta4 ,

where T12 and Ta, are the total kinetic energies of nuclei before andafter the reaction. Hence,

Ts,-T12 = (m1+m.) ct-(ms+m,) ct.

The left-hand side of this equality is the increment of the overallkinetic energy of the nuclei of the given system. It is referred to asthe energy yield of a nuclear reaction and is denoted by the letter Q.Thus, .

Q= [(mi+m2)-'-(ma+ m,)J cll •

This quantity may have either sign depending on the nature of thenuclear reaction. Thus, the energy yield of a nuclear reaction is deter­mined by the difference of the cumulative rest masses of nuclei beforeand after the reaction. All the quantities involved in this relationcan be experimentally measured with a sufficiently high accuracy,verifying thereby the equality itself.

Let us.o!Eonsider the specific nuclear reaction

f ?Li + I H -+- 2'He.

The rest masses of the~ nuclei measured in atomic mass units (amu)are equal to 7.0160, 1.0078, and 4.0024 amu respectively. From thisit is not difficult to calculate that the sum of rest masses of the nucleidecreases by 0.019 amu as a result of the reaction. Since one amu cor­responds to an energy of 931.4 MeV, we find Q = 0.019 ·931.4 MeV == 17.7 MeV. This value agrees very accurately with experimentaldata.

Example 2. Decay of a particle. Suppose a stationary particle Aispontaneously decays producing two particles A. and A a: A I -+- A 2 ++ A a' In accordance with the law of conservation of total energy,

EI = E 2 + Ea.

As the total energy of each particle is E = moc. + T, the precedingequality takes the form

mIc·= (m ll+ma) cLf T2S ,

where Tsa is the overall kinetic energy of the resulting particles.This energy is referred to as the decay energy Q. Thus,

Q= [mI-(m2+ma)] c·.Since Q is essentially a positive quantity, the spontaneous decayof a particle is possible only if

mi > m2 + ma,

that is, if the rest mass of the initial particle exceeds the sum of the

Let us consider

Relativistic Mechanics246

Example 1. Energ, yield of nuclear reactions.a nuclear reaction 0 the type

Al + A 2 -+- A a + A"

where the initial nuclei are on the left-hand side and the reactionproduct nuclei on the right-hand side. We apply the law of conserva­tion of total energy to this reaction:

EI + E 2 = E a+ E,.

2m (v') = ¥,Le. the sum of the relativistic masses of the initial particlesis equal to the relativistic mass of the formed particle.Thesituation is similar in the K frame. Indeed, if the value ofV is small, the velocity v' is practically equal to v, and themass M to the rest mass M 0 of the formed particle, so thatin the K frame ~

. 2m (v) = 'Mo.

lt is seen that the rest mass of the formed article is greaterthan the sum of the rest masses of th initial particles.The kinetic energy of the initial particles experiences atransformation which causes the rest mass of the formedparticle to exceed the sum of the rest masses of the initialparticles.

Thus, we have shown that due to the system's momentumconservation the sum of the relativistic masses of the initialparticles equals the relativistic mass of the formed particle.The same is obviously true for the total energy. Therefore,we can assert that the conservation of the total energy inthe form described by Eq. (7.34) indeed occurs in the con­sidered stages of that process.

As we already mentioned at the end of § 7.3, the energyconservation law, when applied to nuclear processes, madeit possible to experimentally verify the validity of one ofthe fundamental laws of the theory of relativity, the mass­energy relationship. Let us consider some examples.

Example. In. the. K reference f.rame a particle possessing a restmass mo and a kmetic energy T strikes a stationary particle with thesame rest mass.. Let us find the rest mass M 0 and the velocity V of thecompound PartICle formed as a result of the collision.

249Relativistic Dynamics

M o= V 2mo(T+2rnocS)/c.

The velocity of the formed particle is the velocity of the C frame.In accordance with Eq. (7.32),

V = pcl/E=c VT (T+2mocl )/(T+2moc2)=c/ V1+2",oC'/~.

Whence,

Making use of the invariance of the quantity E' - pIeS, we write

EL.... plCS = Mlc4 ,

where the left-hand side of the equality relates to the K frame (priorto the collision) and the right-hand side to the C frame (after thecollision). In this case, E = T + 2mocs; besides, in accordance withEq. (7.15), plr,S= T (T + 2mocl), and therefore,

(T+2mocl )I-T (T+2mocl ) = M8c4 •

Problems ~, Chapter 7

Attention! In problems 7.4 through 7.11 we employ the abbreviatednotation described at the~nd of § 7.4. (e.g., p and mo are the abbrevi­ated forms of the quantities pc and mocS).

e 7.1. Motion due to'a longitudinal force. Aparticle of.rest ma~ mobegins moving under the action of a constant force F. Fmd the tImedependence of the particle's velocity.

'lolution. Multiply both sides of- Eq. (7.5) by dt. Then

d (mov/ V 1- (V/C)I) = F dt.

Integrating this expression and taking into account that" = 0 at theinitial moment, we obtain mov/V1 - (V/C)I = Ft. Whence

v (t) = (Ft/mo)/V 1+(Ft/modl •

Let us compare the expression thus obtained with the clas~ical one.According to Newton's second law, II) = Flmo and the velOCity Vel == Ft/mo, and that is why the preceding expression for the velocityv (t) may be presented as

v (t) =, vellV1+ (Vel/c)l.

From this it is seen that v < Vel' I.e. the actual velocity v of theparticle grows more slowly witli time as compared to Vel' and thevelocity V -+ c as t -+ 00 (Fig. 137). .

It is interesting to note that the momentum of the partIcle growslinearly with time: from the equation dp/dt = F it follows thatp = Ft. This is a characteristic property of relativistic motion: whilethe velocity of a particle approaches a certain limit (i.e. hecomespractically constant), the momentum of that particle keeps growing.

• 7.2. Motion due to a transverse force. A relativistic particleof rest mass "'0 and charge q moves in a stationary uniform magnetic

(7.35)

Relativistic Mechantcs248

Since the collision of particles and the subsequent decayof the compound particle do not involve any change inthe total ener~of the system (and consequently, its momen­tum), another Important concl~be inferred' fora •system, 'the quantity E2 -'- ptes is invariant not 'onlyw~th respect to different inertial reference frames but alsowIth respect to the above-mentioned stages of a collisionprocess.. Imagine, f?~ exam~le, two relativistic particles to expe­

rIen~e a C?IlISIOn whIch leads to the generation of a newpartIcle WIth rest mass Mo. If in the K frame of referencethe total energies of the particles are equal to E and Ebefore the collision (and their momenta to p and ~ respec:tively), we may write immediately that on iransiti~n fromthe K frame (prior to the collision) to the C frame (afterthe collision) the following equality holds:

~s~ ma~s of the ~ormed par~icles. Otherwise, spontaneous decayIS Impossible. IpxperImental eVidence fully confirms this conclusion.

L~t us consider, for example, the decay of a pi-meson. It is anexperlment~lfact that charged pi-mesons disintegrate into a mu-mesonand a neutrIno 'Y: 1t -+ /l. + 'Y. The tabulated data give the restmasses of th~se particles (in electron rest mass units) as 273.2, 206.8and 0 respectively. It follows that the rest mass decreases by 664 emuas a result of the decay. Since one emu corresponds to an en~rgy of0.51 MeV, the energy of this decay Q = 66.4.0.51 MeV = 34 MeVwhich accurately agrees with experimental data. '

in which it is taken into account that the formed particleis at rest in the C frame.

.The invariance of the quantity Et - pSes provides usWIth a means to investigate the various processes of decayan.d collisio? of relativistic particles. Its application sim­plIfies drastIcally both the analysis of the processes them­selves and the appropriate calculations.

I I

field who~ indu~~ion is equal to B.. The particle circumscribes a cir­cle of radiUs p In a plane perpendicular to the vector B. Find themomentum and t~e angular rotation frequency of the particle.

Solution. In this c~se the particle moves due to the Lorentz forceF = q [v~], where v IS the velocity of the particle. Since F.L v,the magmtude of the velocity of the particle v = const and Eq. (7.5)takes the form

mw = q [vB],

~here m is the re.Ia~ivistic mass of. the particle. Recalling that wIS a norma~ accelerat.IOn whose magmtude is equal to v2/p, we rewritethe prec~dm~ equatIOn as mv2/p = qvB. Hence, the momentum ofthe partIcle IS or

(1)

251Relativistic Dynamics'

.7.3. At the moment t = 0 a relativistic proton with momentum Poflies into a region where there is a transverse uniform electric fieldof strength E, with Po .L E. Find the time derendence of the angle eat which the proton is deGected from the initia direction of its motion.

Solution. Taking the x coordinate along the vector Po and the ycoordinate along E, we write Eq. (7.4) in projections on these axes:

dpJdt = 0, dpyldt = eE,

where e is the proton charge. From these equations it follows that

P:x; = Po, PIJ = eEt,

(1)

Relativistic Mechanics

p = mv = qpB.

250

whence

ThU.S, ,the product pB may serve as a ~easureof the relativisticmomentum of the given particle.

'Yith .allowance made for Eq. (1) the angu ar frequency of thepartIcle IS

ro = vip = pimp = qBlm.

It follo~s that the angular frequency ro depends on the velocit ofthe particle: the greater the velocity of the particle, and therefore,

Fig. 137

the relativisti? ~ass m, the lower the angular frequency ro. Howeverat low velOCItIes (v ~ c) m -+ mo, and '

ro = qBlmo = const,

Le. in this ,:elocity range the frequency ro is practically independentof the veioci ty.

From the ratio of the last two equalities we get

tan 0 = vylv:x; = eEtlpo'

It is interesting to point out that in contrast to the non-relativisticcase V:x; decreases with an increase in time here. To make sure ofthis, let ~ square both equalities (1) and then add separately theirleft-hand and right-hand sides:

~~ (vi +v~)1- (V/C)2 pB+ (eEt)2.

Recalling that ~+ v~ = vi, we obtain

( ~ )2:= [1+ pa;~:~t)2 r1.

Substituting this expression into the first equality of (1), we get

Vx= cry1+ (moc/po)2+(eEt/po)2,

Le. V x really decreases in the course of time t .• 7.4. Symmetric elastic scattering. A relativistic proton possessing

kinetic energy T collides elastically with a stationary proton withthe result that both protons move apart symmetrically relative to theinitial motion direction. Find the angle between the motion direc­tions of the protons after the collision.

Solution. In symmetrk scattering of the protons their momentaand energies must be equal in magnitude. This is immediately seenfrom the triangle of momenta (Fig. 138), which expresses the momen­tum conservation law. From that triangle we can write, in accordancewith the cosine theorem,

p2 = 2p'2+2p'2 cos e,cos 6= p2/2p'2_1.

Using Eq. (7.25) and taking into account that T = 2T', where T'

Substituting here p = e, p' = e' and Pe = V Te

(Te+ 2m

e) =

V (ll - e') (e - e' + 2me) , where me is the rest mass of theelectron, we obtain after simple transformations

253

£raj £ (b)

...N

l~II-:"

I': '"I~

II mJ+m*m,+m2 ~t

mJ+m.. mt+mZ

Fig. 140

Relativistic Dynamics

and in the reference frame fixed to one of the protons

[2 (T+ mp»)I=(T' +2mJl)I-T' (T' +2m p),

where m is the rest mass of a proton. From this it fonows thatp

. T' = 2T (T+2mp )/mp •

For example in the case of protons (mIJ ~ 1 GeV), if T = 50 Goy),then T' = 5'.10a GeV. The possibility of such a large energy "gam"underlies the method of head-on collision beams.

Ii 7.7. Energy diagram of a nuclear reaction. A.particle Al withkinetic energy T1 strikes a stationary nucleus A I (m the K frame).As a result of the reaction the nuclei A a and A, are formed:

A1+AI-Aa+A,.

The rest masses of the particles are equal to mh ml' ma, and m, respec­tively. Illustrate the energy level diagram of the nuclear reactionfor two cases: (a) ml + ml > mll. + mf,.' am! (b~ ml + m l < ma + m,.For the second case find the threshold kmetlC energy Tlthr of theincoming particle in the K frame.

Solution, From the law of conservation of the total energy itfollows th4 in the C £rallie

f n -t-'ml+ms = fa' +/TIa+m"

where ~I and Ta, are' the overall kinetic energies of the particles

before and after the reaction. Denoting the increment of the kineticenergy Ta,,-TII by Q, we may write the preceding expression u

Q = (~+ ml ) - (rna + m,,),

where Q is the energy yield of the nuclear reaction. The energy diagramof the reaction is illustrated in Fig. 140 for hoth cases. In case (a) theellect is positive, Q > 0: the overall kinetic energy incre&leS at the

Relativistic Mechanics

Fig. 139

~.pp

Fig. 138

252

is the kinetic energy of each proton after the collision, we find

pI T (T+2mo) -4 T+2mop'l T' (T' +2mo) - T+4mo '

where mo is the rest mass of a proton. Substituting this eXl.ressioninto the formula for cos e, we obtain

cos e = TI(T + 4mo).

Note that as distinct from the non-relativistic case when e = 1(/2,h8lll e < n/2.

• 7.5. A photon of energy e is scattered by a stationary free elect­ron. Find the energy ll' of the scattered - n if the angle betweenthe motion directions of the incoming p oton nd the scattered oneis equal to O.

Solution. Let us apply the momentum and energy conservationlaws to the given process: .

Te=e-e', Pe=P-p',

where Te and Pe are the kinetic energy and the momentum of therecoiled electron, and p and p' are the momenta of the incoming

and scattered photons. According to the cosine theorem it followsfrom the triangle of momenta (Fig. 139) that

p~= pl+ p's_2pp' cos O.

8'= B1+ (2e/me) sinl (8/2) •

'J 7.6. Two protons move toward each other with equal kineticenergies T (in the J( reference frame). Find the kinetic energy T'of one proton with respect to the other.

Solution. Let us take advantage of the inv.ariance of the quantityEI - pI, writing it in the K frame (which is also the C frame here)

Hence

mB= (el +el)I_(Pl +P2)I,

where PI and P are the momenta of the gamma photons. We trans­form the right-fu;nd side of this equation taking into account thatPI = e,. and P2 = el ; then

m5=28l82-2PlPI' or m5=2e182 (I-cos 8).

255RelatIvistic DynamIcs

~ = piE = 8/(e + mol.2. From the Lorentz transformation for energy (7.26) it follows

that the photon energy in the C frame is

- 1-~ ,;- 1-~e=Y(8-~p)=Y(8-~8)=8Yl_~1=8V 1+~'

Substituting the expression for ~ from the previous part, we obtain

B=8V mo/(28+mo)'

It is seen that pair production requires the photon energy to exceed2mo.

• 7.9. Decay of a moving particle. A relativistic :itO meson ofrest mass mo disintegrates during its motion into two gamma photonswith energies el and e l (in the K reference frame). Find the angle 8of divergence of the gamma photons.

Solution. Using the invariance of the expression EB - p2 , wewrite it in the C frame before the decay and in the K frame after thedecay:

sin (8/2) = mo/2Y e181 •

• 7.10. The total momentum and energy of a system of two non­interacting I6rticles are P = PI + PI and E = E l + Ef.Demonstrateexplicitly tliat the Lorentz transformation for the tota momentum pand energy E is consistent with the invariance of the quantity E2 _ pifor the given system. ..

Solution. Using the' Lorentz transformation for momentum andenergy (7.26), we find the projections of the total momentum andenergy in another (primed) reference frame possessing the velocity ~and the corresponding Lorentz factor y:

p~= pix+ P2X=Y (Pu+ PIX)-1'~ (El +E2)= l' (Px-~E);

Py = Pi.1I+P211 = PlY+PlY = Py;

E' = Ei + E; = l' (El + EI)-1'~ (PIX + PIX) = Y(E - ~Px)·

HenceE'I- p'l= E'I_(p~B+pyl) = EI- pl.

• 7.11. In the laboratory reference frame a photon with energy 8strikes a stationary particle A .of rest mass mo. Find:

(1) the velocity of the C frame for these two particles;(2) the energies of the photon and the particle in the C frame.Solution. 1. In accordance with Eq. (7.32) the velocity of the C

frame is

or(Tthr+ mo+Mo)2-Tthr (Tthr+2mo)= (ml +Inl +... )2.

Here we have taken into account that in the C frame the kineticenergy of the formed particles is equal to zero at the threshold of thereaction, and therefore their total energy equals the sum of the restmasses of the individual particles. From the latter equation we get

Tthr = [(ml+m2 + ... )2_ (mo+ M 0)2J/2Mo.

2. Let us write EI - p2 before the interaction in the referenceframe where the proton is at rest and after the interaction in the Cframe. At the threshold value of the energy e of the incoming photon

(Ethr+ M o)B-efhr = (Mo+2mo)l,

where M 0 is the rest. mass of a proton, and mg is the rest mass of anelectron (positron). Hence, '

ethr = 2mo (1 + 111 0/ IIIu)'

expense of a decrease in the sum of the rest masses of the'-system'sparticles; in case (b) the opposite is true. ..

In the latter case, as it is seen from Fig. 140b, the nuclear reactionis possible only if T12 ~ I Q I. H!re the equality sign correspondsto the thresho.l~ value of the energy T12• In accordance with Eq. (4.16),at low velocIties,

Itv~el m. mlvf milTu thr=-2-= ml+m. -2-= 111

l+m

lTllhr = I Q /,

whence

Relativistic Mechanics

Tlthr = I Q I (ml +m2)/ml •

!t 7.8. Threshold energy (the minimum energy required to activatea given process).. 1. A relativistic particle of rest mass mo strikes a stationary par­

tICle of rest mass Mo. As a result of the impact, particles of restmasses mit m2' . . . are generated according to the scheme

~ mo + M o --+- ml + m2+ ....Find t~e threshold kinetic energy T lhr of the incoming particle.

2. Fmd the threshold energy of a photon for electron-positron pairproduction in the field of a stationary proton.

Solution. 1. First of all, it is clear that threshold energy is meaning­ful only when the sum of the rest masses of the initial particles is lessthan. that. of the particles pr?duced. To find Tlhr, let us make use ofthe mvarlance of the quantity E2 - pl. Let us write this quantityprior to the collision for T = T Ihr in the reference frame where theparticle M 0 is at rest, and after the collision in the C frame:

EB_ p2=EI,

I

II

The energy of the particle A in the C frame is

EA = molV1-~I= (e-t lit) V-m-u/"'(2;::""e-+:--m-:"o).

The correctness of the formulae obtained may be checked by makinguse of the invariance of the expression EI - p2 on transition fromthe laboratory mference frame to the C frame:

(e+ InO)2-el = <E-t- RA)2.

A

rhl

O' .....--.......-L----O'

APPENDICES

In polar coordinates p, cp the position of a point A on a planeis defined if we know its distance p from the origin 0 (Fig. 14ia)and the a~gle .cp between the radius ,:ector p of the point anda chosen directIOn 00', the zero readmg of the angular coordi­nate cp.

Let us introduce unit vectors ell and e'll associated with the mov­ing point A and oriented in the direction of the increasingcoordinates p and cp as shown in Fig. i4ia. Unlike the unit vectors

1. Motion of a Point in Polar Coordinates

Relativistic Mechanics256

Fig. 141

of the Cartesian system of coordinates, ep and ecp are movable, thatis, they change their direction as the point A moves. Let us findtheir time derivatives, which will be required later. During themotion of the point A both unit vectors turn in the same directionthrough the same angle dcp in the time interval dt (Fig. f4ib) andacquire the increments:

clep= f.dcp.('cp; decp= i·dep·( -ep).

Dividing both expressions by 4t, we obtain

(i)

wheJe a dot over a letter signifies differentiation with respect totime.

Let us now determine the velocity and acceleration of the point A,writing its radiusveetor p in the form

p = pep. (2)

258 App.endlceBAppendices

(2)

m apu

Fig. 143

~.

o

VI == Va./mpo.

Fig. 142

0"'---.........--'"-- 0'

m· • a •T(pl+plcpl)- p=E; mplcp=L,

where E and L are the total mechanical energy and the angular mo­mentum of the particle relative to the point 0, the field centre. Bothof these quantities are easy to find from the initial conditions.

These equations are solved as follows. Initially, in the first equa­tion differentiation with respect to time is replaced by differentiation

does the trajectory of the particle have in polar coordinates p (cp)if P (0) = Po at cp = 0 and its velocity is perpendicular to the radiusvector and is equal to Vo (Fig. 143)?

To solve this problem, the laws of conservation of energy andangular momentum are usually utilized. In polar coordinates theselaws yield

259

with respect to cp; this can be done by Using the second equation:dt = (mpIIL) dcp. Then the variables p, cp are separated, i.e.. the ob­tained expression is reduced to the form dcp = / (P) dp. And finally,that equation is integrated, with account taken of the initial condi­tions. The result of integration yields the sought solution p (!p).

We shall not describe the rather cumbersome procedure for solVingthese equations here. If necessary, it may be found in almost anytextbook on theoretical physics or mechanics. We shall restrict our­selves to an analysis of the solution obtained, which has the form

p (cp) = po/[a + (1 - a) cos cpl, (1)where a = a.lmpovl.

It is known from mathematics that Eq. (1) describes a curve ofthe second order. Depending on the value of the parameter a thismay be an ellipse (circle), a parabola, or a hyperbola.

1. It is immediately seen that at 1.1= 1 P does not depend on cp,i.e. the trajectory is a circle. A particle has such a trajectory at a veloc­ity Vo equal to

17*

(7)

(6)

vec-

The velocity v 0/ a point. L~t us differentiate Eq. (2) with respectto time, allowing for Eq. (1).

• • (3)v=pep+pcpeq>'

i.e. the projections of the vector v on the movable unit vectors epand eq> aree~o •• (4)

v = p; vlJ' = pcp,d the magnitude of the velocity vector is v = Vpi 7- p2~2.

an The acceleration w 0/ a point. Differentiating Eq. (3) wIth respectto time once again, we get .

dv·· . d ,,.;..), +pq,~w=T=pep+pep + (if \1"1' ell' lJ"

d k' • t account Eq. (1):After simple transformations we fin , ta mg m 0

w={p_p~2) ep+(2i>~+pq;) elJ" (5)

i.e. the projections of the vector w on the unit vectors ep and elJ' areequal to

Wp=·p_p~I,

.. •• 1 d I')wq>=2pcp+pcp =p Cit (p cp •

f d . s In polar coordinates. TheThe fundamenta! equation 0mic~n~,::c= F in projections on the'fundamenta!' equatiOn of 1yna

is easy to obtain at once, making usemovable umt vectors ep an e'llof Eqs. (6):

2. On Keplerian Motion .., 1fi ld offorces that are mverse-

The mo~ion of a pahrtlcle m a cer~h~ d~stance from the field centrely proportiOnal to t e square o. . f s between massis called. Kep1!lrian. Th,: Newh~~i~:l~~~~~~~) ::i Coulomb forces~oints (or ~odles Possess~gf:rces of this kind. . .

eti:~lhl:tfi~~drfh: ~oten.tia1:di~n: :r!:r:b~efi~rd ~e~tre. rt~iwhere ex is a constant and p IS to. e the force acting on a particleus examin,: th~ cased when exd ~e fleid 'centre (attraction). What shapeof IIiass m IS duecte towar .

260 AppendicesAppendicel 261

(3)

Fig. 144

Proof. Let us draw through the ith element of the body a planeperpendicular to the I axis, and in this plane, three vectors Pi" p.and a (Fig. 145). The first two vectors describe the position of the nhelement of the body relative to the I and Ie axes while the vector aspecifies the position of the Ie axis relative to the z axis. Taking

Zc

Fig. 145

advantage of the relationship between these vectors (Pi = pi + a).we transform the .expression for the moment of inertia of the body ,relative to the I axis:

I = ~ m,pf = ~ m, {pi+8)2 = I mipiS+28 ~miPi + li mia2•

The first term on the right-hand side of that equality is the momentof inertia Ie of the body relative to the Ie ax~, al).d the last term isequal to maS, What is left is to show that the middle term equals zero.

Supposeri is the radius vector of the ith element of the bodyrelative to the centre of inertia; then the vector 2J mir, = 0 relativeto that centre. But pi is the vector projection of the vector r~ on the. \plane perpendicular to the I axis. Hence it is clear that if the compositevector is equal to zero, the sum of its vector projections on any planeis also equal to zero, i.e. I miPf = O. The theorem is thus proved.

p (n = pJ(2a - 1).

It is seen from this that p (n) is fim only when 2a > 1, i.e. whenthe velocity Vo < VII, where

VII = 11 2a/mpo'

2. For all values of the parameter a at which p is finite up to cp == n, the trajectory has the form of an ellipse. At cp = n, as it followsfrom Eq. (1),

3. If 2a = 1, i.e. Vo = VII, the ellipse degenerates into a parabola,which means that the particle does not come back again. '

4. At V > VII the trajectory has the shape of a hyperbola.All of ttese cases are illustrated in Fig. 144. It should be pointed

out that in elliptical orbits the field centre coincides with one of the

ellipse's focal points: namely, with the back foc,us if Vo < III. andwith the front focus if Vo > VI. '

Note that Eq.. (1) describes, for example, the traje.ctories of theplanets of the solar system, with a. = ymM, whereM IS the mass ofthe Sun. As applied to the motion of space vehicles, III and VII arethe orbital and escape velocities respectively. Obviously, their magni­tudes depend on the mass of the body that is the source of the lield.

3. Demonstration of Steiner's Theorem

Theorem: the moment of inertia I of a solid body relative to anarbitrary axis :& equals the moment of inertia Ie of that body relativeto the axis Ie parallel to the given one and passing through the body'.centre of inertia, plus the product of the mass m of the body and the~are of the distance a between the axes:

I =Ic +ma2•

282 Appendices Appendice, 263

... Greek Alphabet Concluded

5. Some Formulae of Algebra and Trigonometry

The basic formulae of trigonometry

sinl a.+cosl a.= 1 sin (a.±~)=

= sin a. cos ~±cos a. sin ~

cos a+cos ~=et+f' 'a-~

=2 cos -2- cos -2-

cos a.-cos ~=

2 . a.+~ . a.-f'=- sm-2-sm-2-

sin (a.±~)tan a.±tan f' cos a..cos ~

sin (a.±~)cot a.+cot II. = +. . A.

- t-' sma.·sm I'

2 sin a.·sin ~=cos (a.-~)­-cos(a.+~)

2 cos a.. cos ~=cos (a.-~)+

+cos (a+~)2 sin a.·cos f'=sin (a.- ~)+

+sin (a.+~)

2tan a..tan 2a. = 1-tanl a.

cot'a.-1cot 2a.= 2 tco a.

sin 2a.=2 sin a'cos acos 2a. = cos' a - sinl a.

. a. .. / i-cos a.smT=V· 2

a .. / 1+cosa.cosT= V .2

6. Table of Derivatives and Integrals

Function I DerivativeII

Function I Derivative

xn nxn- l sin x cos x

1 1 cos x -sinx- -7x 11 tan x coslxn

xn --;;m-1

1 cot x - sinlxV;" 2y-;" 1arcsin x y i-xlex ex

enx nenxarccos x 1

aX aXlna V~-x'

P, p-Rho~, a-SigmaT, 'I-Taur, u-Upsilon

cD, lp-PhiX, X-Chi'1', ¢-PsiQ, ro-Omega

sin a.=a.cos a.=1-a.1 /2

tan a.=a.

cos (a.±~)== cos a. cos ~+sin a. sin ~

t + tan a.+tan f'an (a_f') = 1+tan a..tan ~

cot (a.+~) cot a.·cot ~+1- cot ~+cot a.

sina+sin~=

=2sin a.t~ cos a.2~

sin a.-sin f'=

=2 cos a.t~ sin a.;-~

I, L-IotaK,x-KappaA, I.-LambdaM, ~-MuN, 'V-Nu3, ;-Xi0, a-OmicronII, 1t-Pi

cos a.·csc a. = 1

sin a.·sec a. = 1

secl a.-tanl a.= 1cscl a. - cotl a. = 1

tan ~·cot a.= 1

sin a. 1y 1+cotl a

A, a.-AlphaB, ~-Betar, v-Gamma.:1, !'I-DeltaE, e - EpsilonZ, b-Zeta

H, TJ-Eta8, a,it-Theta

The roots of the quadratic equation axl+bx+c=O

-b± Y.bl -4acXl,l = 2a •

Some approximate formulae. If a. ~ 1, then(1±a.)n= 1±na.

ea =1+a.In (1 +a.) =a

ab=ha=ab cos a;a (b+c)=ab+ae.

Function I Derivative ~] Function I Derivative!

Inz] 1 arctan x 1- 1+xlx

Vii"u'

arccot x 12yu - 1+xl

u' sinh x cosh xlin u - cosh x -sinh xu

u vu'-v'utanh x

1- viv coshl x

cothx 1sinhl x

Mixed, or vector-scalar, product of three vectors is a scalarequal numerically to the volume of the parallelepiped construct­ed from these vectors:

a [be] = b [cal = c lab];a [bel = -b [ae] = -a [eb].

265

Concluded

Double vector product:[a [be)) = b (ae) -e (ab).

Products of vectors in coordinate form. If

a=a1el+a~.+aaea,

b= b1e1 + ble2 + baea,where e1' e., ea are coordinate unit vectors which are~mutuallyperpendicular and form a right triad, then

~' ab=a1b1+a.b2+aaba;

e1 8 2 8 alab] = a1a2 aa = (a;ba-aab.) e1+ (aa b1- alba) 8 2+ (a1b.-a.b1) ea'

b1 b2 ba

The rules for differentiating vectors depending on a certainscalar variable t:

d da dbdt(a+b)=dt+dt ;

II da cIadt (aa) = (It a+a dt;

d da dbdt (ab)=dt b+a CIt;

:e [abl=[ :: b]+[a ~~ J.Gradient of the scalar function rp:

V = a<p 1+ arp j+ a(p k,<jJ ax iiy az

where I, i, k are the coordinate unit vectors of the x, y, z axes.

Appendlce,Appendlcll

Conclwhtl

i ~=tanx.I cos z

r~=-cotxJ smlx

) e%dx=e%

) 1~:. =arc tan x

J dx arcsin xy i-x·

Jy~=(n(x+Yx.-1)x·-1

JX 11+1

x1l dx=-'- n=l=-1n+1 '

Jd: =Inx

Jsin x dx= -cos x

rcos x dx=sin x

Jtan x dx = -In cos x

~ cot x dx = In sin oX

Scalar product:

264

7. Some Facts About Vectors

Cross product:[ab] = - [hal; I [ab) 1= ab sin a;

[a, b+e]=[ab]+[ac].

266 A"eJUJlcu Appendices 267

8. Units of Mechanical Quantities in the SI and eGS Systems 10. Some Extrasystem Units

11. Astronomic Quantities

1I1ass. kgMean radius, Mean orbit

m radius, m

Sun 1.97.1030 6.95·10sEarth 5.96 ·1OS4 6.37.106 1.50.1011

Moon 7.34.1022 1.74.106 3.84·10s

Unit ConversionQuantity

tactor, t 81

I unit/t CG881 CG8 unit

Length m Col 10STime s s 1Angle rad rad 1Area m2 em2 104

Volume m' em' 106

Velocity m/s emis 102Acceleration m/s2 cra/s2 102Frequencl Hz Hz 1Angular requency rad/s rad/s 1Angular velocity rad/s rad/s 1Angular acceleration rad/s2 radls2 1Mass kg g 103

pensity kg/m' g/cm3 10-3

Force N dyn 10:;Pressure Pa rlyn;'cm2 10Work, energy J erg 107Power W ergis 107Momentum kg·mis g.cm/s 10·Power impulse N·s dyn.s 10·Force moment (torque) N·m dyn.cm 107Angular momentum kfm2/s g.cm2/s 107Moment of inertia g.m2 g.cm2 107Torque momentum N·m·s dyn·cm·s 107Energy flux W erg/s 107Energy flux density W/m2 erg/(s. cm2) 103

9. Decimal Prefixes for the Names of Units

Length

Time

Mass

Force

Pressure

Energy",.

Power ..

1 A (angstrom) = 10-10 m1 AU (astronomical unit) = 1.496.1011 m1 light-year=0.946·1016 m1 parsec=3.086·1016 m1 in (inch)=0.0254 m;1 ft (foot)=0.3048 m; 1 yd (yard)=0.9144 m1 mile=1609 m1 day = 86400 s1 year=3.11·107 s1 amu = 1.66.10-27 kg; 1 oz (avdp)=0.028 kg1 ton=103 kg; 1 Ib (avdp)=0.454 kg1 kgf (kilogram-force) = 9.81 N1 tf (ton-foree)=9.81·103 N1 bar = 10· Pa (precisely)1 atm (atmosphere) = 1.01·10· Pa1 mm Hg (Torr) = 133 Pa1 in Hg=3386 Pa1 psi (pounds per square inch) = 6895 Pa1 eV=1.60·10-19 J1 Wh (Watt-hour)=3.6.103 J1 hp (horsepower) = 736 W

Examples: nm nanometer (10-9 m)kN kilonewton (103 N)

MeV megaelectronvolt (108 eV)I1W mierowatt (10-6 W)

12. Fundamental Constants

T tera (1012)

G giga (109)

1\1 mega (106 )

k kilo (103)

h hecto (102)da deca (101)

d deci (10-1)

c centi (10-2)m milli (10-3)

11 micro (10-8)

n nano (10-9)

p pico (10-12)

f femto (10-U )

a atto (10-18)

Velocity of light in vacuo

Gravitational constant

Standard freefall acceleration

c={1'= {

g={

2.998.108 m/s2.998.1010 cm/s6.67.10-11 m3J(kg.s2)6.67.10-8 cm3/(g.s2)

9.807 m/s2980.7 cm/s'

268

Avogadro constant

Elementary charge

Electron rest mass

Electron charge to mass ratio

Proton rest mass

Atomic mass unit . 1amu= {

Appendlcelt

Concludetl

6.025.1026 kmol-1

6.025.10" mol-1

1.602.10-19 C4.80.10-10 esu0.911.10-80 kg0.911.10-27 g0.511 Mev1.76 .1011 C/kg5.27.1017 esu/g1.672.10-27 kg1.672 .10-24 g1.660.10-27 kg1.660 .10-24 g931.4 MeV

INDEX

Acceleration 15angular 24centripetal (normal) 21CorioUs 32free fall 58of a point in polar coordi­

nates 298relativistic transformation of

226tangential 21transformation of 31

Aiming parameter 133Angular momentum 147

conservation of 156interna~61relative to an axis 15

Angular velocity 24summation of 29 •

Arc coordinate 19, 55Axes, free 173

of inertia, principal 173

Centre of mass (inertia) 120motion equation of 121

C frame 9, 123Clock synchronization 195Coefficient of friction 52Collision 126

completely inelastic 127endoergic 133exoergic 133head-on 128inelastic 133non-central 129perfectly elastic 128of two particles 126of two relativistic particles

245Conservative system 125Contraction of length 204Coordinates

arc 9, 55Cartesian 17polar 257

Coordinate system 5, 54Couple 160

Decay of a stationary particle247

of a moving particle 254Diagram of momenta, vector 130Dilation of time 200, 219Diminution 82Direction cosines 18Doppler effect 223

radial 223 .transverse 224

Einstein's postulates 194principle of relativity 194

Energyconservation of 93, 101decay 247internal 124internal potential, of a sys-

tem 96kinetic 90kinetic, of a solid body 171kinetic, of a system 98mechanical 90potential 81, 95potential,of a system 95of a relativistic particle 234of a rotating solid body 167threshold 254total mechanical 91

Energy conservation law 93, 101EneJ;gy content 237Energy diagram of nuclear reac­

tions 252Energy-momentum relationship

238Energy yield of nuclear reactions

246Equation

of dynamics, fundamental52

of harmonic vibrations 67of moments 147of moments in the C frame

163of relativistic dynamics, fun­

damental 231

I

270

Equipotential surface 86Equivalence principle 60Event 195

Fieldof central forces 80of elastic forces 83of force 79of gravity 83of a point mass (charge) 83potential 79stationary 79

Field potential 88Field strength 88Force 46

central 80centrifugal 57conservative 79Coriolis 57Coulomb 51dissipative 100elastic 51equivalent 164external 99of a field 84friction 52gravitational 50inertial 57, 60internal 99longitudinal 233Lorentz 249magnetic 51non-potential 99normal 55potential 99of reaction 48reactive 137resistance 52of sliding friction 52tangential 55transverse 232

Fundamental equation of dyna­mics 52

in polar coordinates 298of variable mass 136

Fundamental equation of rela­tivistic dynamics 231

Galilean relativity 43'Galilean transformation 34

Index

General theory of relativity 61Gradient 86Gyroeffect 178Gyroscope 174

precession of 175Gyroscopic couple (moment) 177Gyrostatic action 178

Hooke's law 52

Increment 82Inertness 45Initial conditions 16Instantaneous rotation axis 28Interval 214

light-like 214space-like 214time-like 214

Isotropy of space 42

Keplerian motion 298Kinematics of a point 14

of a solid body 21Kinetic energy 90

relativistic 234of a solid body 171

Lawof acceleration transforma­

tion, relativistic 226of conservation of angular

momentum 156of conservation of energy

93, 101of conservation of momentum

117of inertia. 41Newton's first 41Newton's second 46Newton's third 48of universal gravitation 50of velocity transformation,

relativistic 215Laws of ·forces 50Light clock 201Local time 211Long-range action 49L(lrentz contraction 205

IJUles

Lorentz factor 241Lorentz transformation 208

of momentum and energy240

Mass 45gravitational 50reduced 125relativistic 229rest 229

Mass-energy relation 236Meshchersky equation 137Michelson's experimlmt 192Minkowski diagram 217Moment of force 148

relative to an axis 151Moment of inertia 166Momentum 47, 114

conservaticln of 117relativistic 227, 230of a system 115total, of external fordbs 160

Path 14Perfectly rigid body 12Plane motion 26

of a solid body 170Potential 88Power 78Precession 175Principle of equivalence 60Proper length 205Proper time 202

Reference frame 11heliocentric 42inertial 41non-inertial 42

Relativistic mechanics 12Rod-and-tube "paradox" 222

271

Rotation 22of a solid body 38about a stationary axis 22,

165Rotation axis, instantaneous 28

Simultaneity 197, 210relativity of 219

Space-time diagram 224Steiner's theorem 301, 167Superposition of fields 88Symmetric elastic scattering 251Symmetry of space-time 42

Time dilation 200, 219Torque 148Transformation

of acceleration 31of velocity 31of velocity direction 225

Translation 22, 57

Uniformity of space 42of time 42

Vectoraxial 24displacement 14, 35mean velocity 15polar" 23

Vector diagram of momenta 130Velocity .

of a centre of mass (inertia)120

of light 49of a point 14

Weight 51Work 74

of Coulomb force 76of elastic force 75of gravitational force 77

World line 217World point 217

•

Ex Libris

Amit Dhakulkar

Klan Nosferatu

CBS OUTSTANDING PUBLICATIONS 1

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