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&SMANIA UNIVERSITY LIBRARY
53. O - Accession No. / 6 S 9
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A-S-KOMPANEYETS
TH
RETICAL
PHYSICS
FOREIGN LANGUAGES PUBLISHING HOUSE Moscow 1961
TRANSLATED FROM THE RUSSIAN EDITED BY GEORGE YANKOVSKY
This translation has been read and approved by the author, Professor A. S. Kompaneyets
Printed in the Union of Soviet Socialist Republics
CONTENTS
Page
From the Preface to the First Edition 7
Preface to the Second Edition 9
Part I. Mechanics 11
Sec. 1. Generalized Coordinates 11
See. 2. Lagrange's Equation 13
Sec. 3. Examples of Lagrange's Equations 24
Sec. 4. Conservation Laws 30
Sec. 5. Motion in a Central Field 41
Sec. 6. Collision of Particles 48
Sec. 7. Small Oscillations 57
Sec. 8. Rotating Coordinate Systems. Tnertial Forces 66
Sec. 9. The Dynamics of a Rigid Body 73
Sec. 10. General Principles of Mechanics 81
Part II. Electrodynamics 92
Sec. 11. Vector Analysis 92
Sec. 12. The Electromagnetic Field. Maxwell's Equations 104
Sec. 13. The Action Principle for the Electromagnetic Field 117
Sec. 14. The Electrostatics of Point Charges. Slowly Varying Fields . . 124
Sec. 15. The Magnetostatics of Point Charges 135
Sec. 16. Electrodynamics of Material Media 144
Sec. 17; Plane Electromagnetic Waves 162
Sec. 18. Transmission of Signals. Almost Plane Waves 173
Sec. 19. The Emission of Electromagnetic Waves 181
Sec. 20. The Theory of Relativity 190
Sec. 21. Relativistic Dynamics 211
Part III. Quantum Mechanics 229
Sec. 22. The Inadequacy of Classical Mechanics.
The Analogy Between Mechanics and Geometrical Optics 229
Sec. 23. Electron Diffraction 238
Sec. 24. The Wave Equation 244
6 CONTENTS
Page
Sec. 25. Certain Problems of Quantum Mechanics 252
Sec. 26. Harmonic Oscillatory Motion in Quantum Mechanics
(Linear Harmonic Oscillator) 265
Sec. 27. Quantization of the Electromagnetic Field 271
Sec. 28. Quasi-Classical Approximation 280
Sec. 29. Operators in Quantum Mechanics 291
Sec. 30. Expansions into Wave Functions 301
Sec. 31. Motion in a Central Field 312
Sec. 32. Electron Spin 323
Sec. 33. Many-Electron Systems 334
Sec. 34. The Quantum Theory of Radiation 353
Sec. 35. The Atom in a Constant External Field 368
Sec. 36. Quantum Theory of Dispersion 379
Sec. 37. Quantum Theory of Scattering 385
Sec. 38. The Relativistic Wave Equation for an Electron 394
Part IV. Statistical Physics 413
Sec. 39. The Equilibrium Distribution of Molecules in an Ideal Gas . . 413 Sec. 40. Boltzmann Statistics (Translational Motion of a Molecule. Gas
in an External Field) 430
Sec. 41. Boltzmann Statistics (Vibrational and Rotational Molecular
Motion) 447
Sec. 42. The Application of Statistics to the Electromagnetic Field and
to Crystalline Bodies 457
Sec. 43. Bose Distribution 474
Sec. 44. Fermi Distribution 477
Sec. 45. Gibbs Statistics 498
Sec. 46. Thermodynamic Quantities 512
Sec. 47. The Thermodynamic Properties of Ideal Gases in Boltzmann
Statistics 535
Sec. 48. Fluctuations 546
Sec. 49. Phase Equilibrium 557
Sec. 50. Weak Solutions 568
Sec. 51. Chemical Equilibria 576
Sec. 52. Surface Phenomena 582
Appendix . 586
Bibliography 588
Subject Index 589
FROM THE PREFACE TO THE FIRST EDITION
This book is intended for readers who are acquainted with the course of general physics and analysis of nonspecializing institutions of higher education. It is meant chiefly for engineer -physicists, though it may also be useful to specialists working in fields associated with physics — chemists, physical chemists, biophysicists, geophysicists, and astronomers.
Like the natural sciences in general, physics is based primarily on experiment, and, what is more, on quantitative experiment. However, no series of experiments can constitute a theory until a rigorous logical relationship is established between them. Theory not only allows us to systematize the available experimental material, but also makes it possible to predict new facts which can be experimentally verified.
All physical laws are expressed in the form of quantitative relation- ships. In order to interrelate quantitative laws, theoretical physics appeals to mathematics. The methods of theoretical physics, which are based on mathematics, can be fully mastered only by those who have acquired a very considerable volume of mathematical knowledge. Nevertheless, the basic ideas and results of theoretical physics are readily comprehensible to any reader who has an understanding of differential and integral calculus, and is acquainted with vector algebra. This is the minimum of mathematical knowledge required for an understanding of the text that followrs.
At the same time, the aim of this book is not only to give the reader an idea about what theoretical physics is, but also to furnish him with a working knowledge of the basic methods of theoretical physics. For this reason it has been necessary to adhere, as far as possible, to a rigorous exposition. The reader will more readily agree with the conclusions reached if their inevitability has been made obvious to him. In order to activize the work of the student, some of the applications of the theory have been shifted into the exercises, in which the line of reasoning is not so detailed as in the basic text.
In compiling such a relatively small book as this one it has been necessary to cut down on the space devoted to certain important
8 FROM THE PREFACE TO THE FIRST EDITION
sections of theoretical physics, and omit other branches entirely. For instance, the mechanics of solid media is not included at all since to set out this branch, even in the same detail as the rest of the text, would mean doubling the size of the book. A few results from the mechanics of continuous media are included in the exercises as illustrations in thermodynamics. At the same time, the mechanics and electrodynamics of solid media are less related to the fundamental, gnosiological problems of physics than microscopic electrodynamics, quantum theory, and statistical physics. For this reason, very little space is devoted to macroscopic electrodynamics: the material has been selected in such a way as to show the reader how the transition is made from microscopic electrodynamics to the theory of quasi- stationary fields and the laws of the propagation of light in media. It is assumed that the reader is familiar with these problems from courses of physics and electricity.
On the whole, the book is mainly intended for the reader who is interested in the physics of elementary processes. These considerations have also dictated the choice of material ; as in all nonencyclopaedic manuals, this choice is inevitably somewhat subjective.
In compiling this book, I have made considerable use of the excellent course of theoretical physics of L. D. Landau and E. M. Lifshits. This comprehensive course can be recommended to all those who wish to obtain a profound understanding of theoretical physics.
I should like to express my deep gratitude to my friends who have made important observations: Ya. B. Zeldovich, V. G. Levich, E. L. Feinberg, V. I. Kogan and V. I. Goldansky.
A. Kompaneyets
PREFACE TO THE SECOND EDITION
In this second edition I have attempted to make the presentation more systematic and rigorous without adding any difficulties. In order to do this it has been especially necessary to revise Part III, to which I have added a special section (Sec. 30) setting out the general principles of quantum mechanics; radiation is now considered only with the aid of the quantum theory of the electromagnetic field, since the results obtained from the correspondence principle do not appear sufficiently justified.
Gibbs' statistics are included in this edition, which has made it necessary to divide Part IV into something in the nature of two cycles: Sec. 39-44, where only the results of combinatorial analysis are set out, and Sec. 45-52, an introduction to the Gibbs' method, which is used as background material for a discussion of thermo- dynamics. A phenomenological approach to thermodynamics would nowadays appear an anachronism in a course of theoretical physics.
In order not to increase the size of the book overmuch, it has been necessary to omit the theory of beta decay, the variational properties of eigenvalues, and certain other problems included in the first edition.
I am greatly indebted to A. F. Nikiforov and V. B. Uvarov for pointing out several inaccuracies in the first edition of the book.
A. Kompaneyets
PART I MECHANICS
See. 1. Generalized Coordinates
Frames of reference. In order to describe the motion of a mechanical system, it is necessary to specify its position in space as a function of time. Obviously, it is only meaningful to speak of the relative position of any point. For instance, the position of a flying aircraft is given relative to some coordinate system fixed with respect to the earth; the motion of a charged particle in an accelerator is given relative to the accelerator, etc. The system, relative to which the motion is described, is called a frame of reference.
Specification of time. As will be shown later (Sec. 20), specification of time in the general case is also connected with defining the frame of reference in which it is given. The intuitive conception of a uni- versal, unique time, to which we are accustomed in everyday life, is, to a certain extent, an approximation that is only true when the relative speeds of all material particles are small in comparison with the velocity of light. The mechanics of such slow movements is termed Newtonian, since Isaac Newton was the first to formulate its laws.
Newton's laws permit a determination of the position of a mechanical system at an arbitrary instant of time, if the positions and velocities of all points of the system are known at some initial instant, and also if the forces acting in the system are known.
Degrees of freedom of a mechanical system. The number of inde-\ pendent parameters defining the position of a mechanical system in space is termed the number of its degrees of freedom.
The position of a particle in space relative to other bodies is defined with the aid of three independent parameters, for example, its Cartesian coordinates. The position of a system consisting of N particles is determined, in general, by 3JV independent parameters.
However, if the distribution of points is fixed in any way, then the number of degrees of freedom may be less than 3^. For example,
12 MECHANICS [Part I
if two points are constrained by some form of rigid nondeformable coupling, then, upon the six Cartesian coordinates of these points, xi> Vi> zi> xz> y%> Z2> *s imposed the condition
(x2-xl^ + (y2~yl^ + (z2 — z1)^Rl29 (1.1)
where J?12 is the given distance between the points. It follows that the Cartesian coordinates are no longer independent parameters: a relationship exists between them. Only five of the six values xl9 . . . , Z2 are now independent. In other words, a system of two particles, separated by a fixed distance, has five degrees of freedom. If we consider three particles which are rigidly fixed in a triangle, then the coordinates of the third particle must satisfy the two equations :
y^ + (z,-z1)^Rll, (1.2)
y2)* + (Zs-z2)*=:Rl,. (1.3)
Thus, the nine coordinates of the vertices of the rigid triangle are defined by the three equations (1.1), (1.2) and (1.3), and hence only six of the nine quantities are independent. The triangle has six degrees of freedom.
The position of a rigid body in space is defined by three points which do not lie on the same straight line. These three points, as we have just seen, have six degrees of freedom. It follows that any rigid body has six degrees of freedom. It should be noted that only such motions of the rigid body are considered as, for example, the rotation of a top, where no noticeable deformation occurs that can affect its motion.
Generalized coordinates. It is not always convenient to describe the position of a system in Cartesian coordinates. As we have already seen, when rigid constraints exist, Cartesian coordinates must satisfy supplementary equations. In addition, the choice of coordinate system is arbitrary and should be determined primarily on the basis of expediency. For instance, if the forces depend only on the distances between particles, it is reasonable to introduce these distances into dynamical equations explicitly and not by means of Cartesian coordinates.
In other words, a mechanical system can be described by coordinates whose number is equal to the number of degrees of freedom of the system. These coordinates may sometimes coincide with the Cartesian coordinates of some of the particles. For example, in a system of two rigidly connected points, these coordinates can be chosen in the following way: the position of one of the points is given in Cartesian coordinates, after which the other point will always be situated on a sphere whose centre is the first point. The position of the second point on the sphere may be given by its longitude and latitude.
Sec. 2] LAGRANGE'S EQUATION 13
Together with the three Cartesian coordinates of the first point, the latitude and longitude of the second point completely define the position of such a system in space.
For three rigidly bound points, it is necessary, in accordance with the method just described, to specify the position of one side of the triangle and the angle of rotation of the third vertex about that side.
The independent parameters which define the position of a mechanical system in space are called its generalized coordinates. We will represent them by the symbols g^, where the subscript a signifies the number of the degree of freedom.
As in the case of Cartesian coordinates, the choice of generalized coordinates is to a considerable extent arbitrary. It must be chosen so that the dynamical laws of motion of the system can be formulated as conveniently as possible.
Sec. 2. Lagrange's Equation
In this section, equations of motion will be obtained in terms of arbitrary generalized coordinates. In such form they are especially convenient in theoretical physics.
Newton's Second Law. Motion in mechanics consists in changes in the mutual configuration of bodies in time. In other words, it is described in terms of the mutual distances, or lengths, and intervals of time. As was shown in the preceding section, all motion is relative; it can be specified only in relation to some definite frame of refer- ence.
In accordance with the level of knowledge of his time, Newton regarded the concepts of length and time interval as absolute, which is to say that these quantities are the same in all frames of reference. As will be shown later, Newton's assumption was an approximation (see Sec. 20). It holds when the relative speeds of all the particles are small compared with the velocity of light; here Newtonian mechanics is based on a vast quantity of experimental facts.
In formulating the laws of motion a very convenient concept is the material particle, that is, a body whose position is completely defined by three Cartesian coordinates. Strictly speaking, this idealization is not applicable to any body. Nevertheless, it is in every way reasonable when the motion of a body is sufficiently well defined by the displacement in space of any of its particles (for example, the centre of gravity of the body) and is independent of rotations or deformations of the body.
If we start with the concept of a particle as the fundamental entity of mechanics, then the law of motion (Newton's Second Law) is formulated thus:
14 MECHANICS [Part I
~-
(2.1)
Here, F is the resultant of all the forces applied to the particle
i2 f
(the vector sum of the forces) —r^ is the vector acceleration, the Cartesian components of which are
d2z
The quantity m involved in equation (2.1) characterizes the particle and is called its mass.
Force and mass. Equality (2.1) is the definition of force. However, it should not be regarded as a simple identity or designation, be- cause (2.1) establishes the form of the interaction between bodies in mechanics and thereby actually describes a certain law of nature. The interaction is expressed in the form of a differential equation that includes only the second derivatives of the coordinates with respect to time (and not derivatives, say, of the fourth order).
In addition, certain limiting assumptions are usually made in relation to the force. In Newtonian mechanics it is assumed that forces depend only on the mutual arrangement of the bodies at the instant to which the equality refers and do not depend on the con- figuration of the bodies at previous times. As we shall see later (see Part II), this supposition about the character of interaction forces is valid only when the speeds of* the bodies are small compared with the velocity of light.
The quantity m in equality (2.1) is a characteristic of the body, its mass. Mass may be determined by comparing the accelerations which the same force imparts to different bodies; the greater the acceleration, the less the mass. In order to measure mass, some body must be regarded as a standard. The choice of a standard body is completely independent of the choice of standards of length and time. This is what makes the dimension (or unit of measurement) of mass a special dimension, not related to the dimensions of length and time.
The properties of mass arc established experimentally. Firstly, it can be shown that the mass of two equal quantities of the same substance is equal to twice the mass of each quantity. For example, one can take two identical scale weights and note that a stretched ' spring gives them equal accelerations. If we join two such weights and subject them to the action of the same spring, which has been stretched by the same amount as for each weight separately, the acceleration will be found to be one half what it was. It follows that the overall mass of the weights is twice as great, since the force depends only on the tension of the spring and could not have changed.
Sec. 2] LAGRANGE'S EQUATION 15
Thus, mass is an additive quantity, that is, one in which the whole is equal to the sum of the quantities of each part taken separately. Experiment shows that the principle of additivity of mass also applies to bodies consisting of different substances.
In addition, in Newtonian mechanics, the mass of a body is a constant quantity which does not change with motion.
It must not be forgotten that the additivity and constancy of masses are properties that follow only from experimental facts which relate to very specific forms of motion. For example, a very important law, that of the conservation of mass in chemical transformations involving rearrangement of the molecules and atoms of a body, was established by M. V. Lomonosov experimentally.
Like all laws deduced from experiment, the principle of additivity of mass has a definite degree of precision. For such strong interactions as take place in the atomic nucleus, the breakdown of the additivity of mass is apparent (for more detail see Sec. 21).
We may note that if instead of subjecting a body to the force of a stretched spring it were subjected to the action of gravity, then the acceleration of a body of double mass would be equal to the acceleration of each body separately. From this we conclude that the force of gravity is itself proportional to the mass of a body. Hence, in a vacuum, in the absence of air resistance, all bodies fall with the same acceleration.
Inertial frames of reference. In equation (2.1) we have to do with the acceleration of a particle. There is no sense in talking about acceleration without stating to which frame of reference it is referred. For this reason there arises a difficulty in stating the cause of the acceleration. This cause may be either interaction between bodies or it may be due to some distinctive properties of the reference frame itself. For example, the jolt which a passenger experiences when a carriage suddenly stops is evidence that the carriage is in nonuniform motion relative to the earth.
Let us consider a set of bodies not affected by any other bodies, that is, one that is sufficiently far away from them. We can suppose that a frame of reference exists such that all accelerations of the set of bodies considered arise only as a result of the interaction between the bodies. This can be verified if the forces satisfy Newton's Third Law, i.e., if they are equal and opposite in sign for any pair of particles (it is assumed that the forces occur instantaneously, and this is true only when the speeds of the particles are small compared with the speed of transmission of the interaction).
A frame of reference for which the acceleration of a certain set of particles depends only on the interaction between these particles is called an inertial frame (or inertial coordinate system). A free particle, not subject to the action of any other body, moves, relative to such a reference frame, uniformly in a straight line or, in everyday
16 MECHANICS [Part I
language, by its own momentum. If in a given frame of reference Newton's Third Law is not satisfied we can conclude that this is not an inertial system.
Thus, a stone thrown directly downwards from a tall tower is deflected towards the east from the direction of the force of gravity. This direction can be independently established with the aid of a suspended weight. It follows that the stone has a component of acceleration which is not caused by the force of the earth's attraction. From this we conclude that the frame of reference fixed in the earth is noninertial. The noninertiality is, in this case, due to the diurnal rotation of the earth.
On the forces ot friction. In everyday life we constantly observe the action of forces that arise from direct contacts between bodies. The sliding and rolling of rigid bodies give rise to forces of friction. The action of these forces causes a transition of the macroscopic motion of the body as a whole into the microscopic motion of the constituent atoms and molecules. This is perceived as the generation of heat. Actually, when a body slides an extraordinarily complex process of interaction occurs between the atoms in the surface layer. A description of this interaction in the simple terms of frictional forces is a very convenient idealization for the mechanics of macro- scopic motion, but, naturally, does not give us a full picture of the process. The concept of frictional force arises as a result of a certain averaging of all the elementary interactions which occur between bodies in contact.
In this part, which is concerned only with elementary laws, we shall not consider averaged interactions where motion is transferred to the internal, microscopic, degrees of freedom of atoms and molecules. Here, we will study only those interactions which can be completely expressed with the aid of elementary laws of mechanics and which do not require an appeal to any statistical concepts connected with internal, thermal, motion.
Ideal rigid constraints. Bodies in contact also give rise to forces of interaction which can be reduced to the kinematic properties of rigid constraints. If rigid constraints act in a system they force the particles to move on definite surfaces. Thus, in Sec. 1 we con- sidered the motion of a single particle on a sphere, at the centre of which was another particle.
This kind of interaction between particles does not cause a transition of the motion to the internal, microscopic, degrees of freedom of bodies. In other words, motion which is limited by rigid constraints is completely described by its own macroscopic generalized co- ordinates qa.
If the limitations imposed by the constraints distort the motion, they thereby cause accelerations (curvilinear motion is always accelerated motion since velocity is a vector quantity). This ac-
Sec. 2] LAGRANGE'S EQUATION 17
celeration can be formally attributed to forces which are called reaction forces of rigid constraints.
Reaction forces change only the direction of velocity of a particle but not its magnitude. If they were to alter the magnitude of the velocity, this would produce a change also in the kinetic energy of the particle. According to the law of conservation of energy, heat would then be generated. But this was excluded from consideration from the very start.
To summarize, the reaction forces of ideally rigid constraints do not change the kinetic energy of a system. In other words, they do not perform any work on it, since work performed on a system is equivalent to changing its kinetic energy (if heat is not gener- ated).
In order that a force should not perform work, it must be perpen- dicular to the displacement. For this reason the reaction forces of constraints are perpendicular to the direction of particle velocity at each given instant of time.
However, in problems of mechanics, the reaction forces are not initially given, as are the functions of particle position. They are determined by integrating equations (2.1), with account taken of constraint conditions. Therefore, it is best to formulate the equations of mechanics so as to exclude constraint reactions entirely. It turns out that if we go over to generalized coordinates, the number of which is equal to the number of degrees of freedom of the system, then the constraint reactions disappear from the equations. In this section we shall make such a transition and will obtain the equations of mechanics in terms of the generalized coordinates of the system.
The transformation from rectangular to generalized coordinates. We take a system with a total of 3N==n Cartesian coordinates of which v are independent. We will always denote Cartesian coordinates by the same letter xi, understanding by this symbol all the co- ordinates x, y, z ; this means that i varies from 1 to 3jY, that is, from 1 to n. The generalized coordinates we denote by #a (l^Ca^v). Since the generalized coordinates completely specify the position of their s\'stem, Xi are their unique functions:
xl^xi(ql, ^2, ... ?a, ...,£v). (2.2)
From this it is easy to obtain an expression for the Cartesian com- ponents of velocity. Differentiating the function of many variables Xi ( . . . g«) with respect to time, we have
dt
In the subsequent derivation we shall often have to perform summations with respect to all the generalized coordinates qat
2-0060
18 MECHANICS [Part I
and double and triple sums will be encountered. In order to save space we introduce the following summation convention.
If a Greek symbol is met twice on one side of an equation, it will be understood as denoting a summation from unity to v, that is, over all the generalized coordinates. (It is not convenient to use this convention for the Latin characters which denote the Cartesian coordinates.)
Then the velocity --.— can be rewritten thus:
_^a'»_ __ djri ^?« /o Q\
dt ~~ #7a dt ' ' *~" '
Here the summation sign is omitted.
The total derivative with respect to time is usually denoted by a dot over the corresponding variable:
d* __ . dq* .
~dt *" "dT^V"- (~' '
In this notation, (2.3) is written in an even more abbreviated form:
• f) .?*|* . i c\ f\
«=-t~q*. (2.5)
Differentiating (2.5) with respect to time again, we obtain an expression for the Cartesian components of acceleration:
The total derivative in the first term is written as usual: A
dt
The Greek symbol over which the summation is performed is denoted by the letter (3 to avoid confusion with the symbol a, which denotes the summation in the expression for velocity (2.5). Thus, we obtain the desired expression for £, :
82xi . . , dxi .. /rt 0.
Xi = -q*q* + q'" (2'6)
The first term on the right-hand side contains a double summation with respect to a and p.
Potential of a force. We now consider components of force. In many cases, the three components of the vector of a force acting on a particle can be expressed in terms of one scalar function U according to the formula:
Sec. 2j LAGRANGE'S EQUATION 19
Such a function can always be chosen for the force of Newtonian attraction, and for electrostatic and elastic forces. The function U is called the potential of the force.
It is clear that by far not every system of forces can, in the general case, be represented by a set of partial derivatives (2.7), since if
then we must have the equality
< F\ =_ ?Fk _ <*V
c^k Pxi c'.r, cVfc
for all i, k, which is not, in advance, obvious for the arbitrary functions F, , Fk . The definite form of the potential, in those cases where it exists, will be given below for various forces.
Expression (-.7) defines the potential function U to the accuracy of an arbitrary constant term. U is also called the potential energy of the system. For example, the gravitational force F = — mg, while the potential energy of an elevated body is equal to mgz, where {7^980 cm/sec2 is the acceleration of a freely falling body and z is the height to which it has been raised. It can be calculated from any level, which in the given case corresponds to a determi- nation of U to the accuracy of a constant term. A more precise ex- pression for the force of gravity than F — — mg (with allowance made for its dependence on height also admits of a potential, which we shall derive a little later [see (3.4)].
We denote the component reaction forces of rigid constraints by Fi. We now note that
n
£f"*dxi=09 (2.8)
i -i
if the displacements are compatible with the constraints. Indeed, (2.8) expresses precisely the work performed by the reaction forces for a certain possible displacement of the system; but this work has been shown to be equal to zero.
Lagrange's equations.* We will now write down the equations of motion with the aid of (2.7) and (2.8) as
fcfc +-£»•)— J^- <"»
Here, of course, m1 = m2 = ms is equal to the mass of the first particle, w4 = ms=m6 equals the mass of the second particle, etc.
* In the first reading, the subsequent derivation up to equation (2.18) need not be studied in detail.
20 MECHANICS [Part I
Let us multiply both sides of this equation by -~- and sum from 1
to n over i.
Let us first consider the right-hand side. We obviously have
in accordance with the law for differentiating composite functions. For the forces of reaction we obtain
since this equality is a special case of (2.8), in which the displace- ments dxi are taken for all constants q except for #Y; that is why we
retain the designation of the partial derivative -~- . It is clear that
in such a special displacement the work done by the reaction forces of the constraints is equal to zero, as in the case of a general displace- ment.
After multiplication by --— and summation, the left-hand side
of equality (2.9) can be written in a more compact form, without resorting to explicit Cartesian coordinates. It is precisely the purpose of this section to give such an improved notation. To do this we ex- press the kinetic energy in terms of generalized coordinates:
1-1 1-1
Substituting the generalized velocities by using (2.5), we obtain
The summation indices for q must, of course, be denoted by different letters, since they independently take all values from 1 to v inclusive. Changing the order of summation for Cartesian and generalized co- ordinates we have
n
Tl • . v~T fix* dxt ,~ , ~N
~T«-ftZ>15;ri£- <2-13>
Henceforth, T will have to be differentiated both with respect to generalized coordinates qa and generalized velocities qa. The co- ordinate qa and its corresponding generalized velocity #a are in-
Sec. 2] LAGKANGE'S EQUATION 21
dependent of each other since, in the given position in which the coordinate has a given value #« , it is possible to impart to the system an arbitrary velocity </a permitted by the constraints. Naturally, qa
<) T and </p^ « are also independent. It follows that in calculating
all the remaining velocities g^-^a. and all the coordinates, including <7a, should be regarded as constant.
d T
Let us calculate the derivative -x-— . In the double summation
3qr
(2.13), the quantity y can be taken as the index a and also the index (3, so that we obtain
8 T 1 . 1 e-r<)xi l • 1 d Sj
Both these sums are the same except that in the first the index is denoted by [3 and in the second by a. They can be combined, replacing (3 by a in the first summation ; naturally the value of the sum does not change due to renaming of the summation sign. Then we obtain
n
^ /rt - ..
(2.14) v x
Let us calculate the total derivative of this quantity with respect to time:
d ^T » »
Here we have had to write down the derivatives of each of the three factors of all the terms in the summation (2.14) separately.
8 T
Now let us calculate the partial derivative - — . As has been
d<?Y
o rri
shown, ga, q& are regarded as constants. Like -^r~, the derivative
COf
dT -Q — consists of two terms which may be amalgamated into one.
Differentiating (2.13), we obtain
3T . . V-T 82xi dxi /0 ,-.
•fl - ==9ra 3& \ mi^ - -5 - 7T-- (2.16)
d- ww ^ v ;
22 MECHANICS [Part 1
Subtracting (2.16) from (2.15), we see that (2.16) and the last term of (2.15) cancel. As a result we obtain
%i /0 im
(2J7)
However, the expression on the right-hand side of (2.17) can
dX'
also be obtained from (2.9) if we multiply its left-hand side by -~- and sum over i. For this reason, (2.17), in accordance with (2.10),
Q J-J
is equal to — « — . Thus we find
^ dgY
d dT _ W __3U^
dt dq-f c)</Y dq-f ' \-1* )
In mechanics it is usual to consider interaction forces that are independent of particle velocities. In this case U does not involve #«, so that (2.18) may be rewritten in the following form:
The difference between the kinetic and potential energy is called the Lagrangian function (or, simply, Lagrangian) and is denoted by the letter L:
L^T—U. (2.20)
Thus we have arrived at a system of v equations with v independent quantities ga , the number of which is equal to the number of degrees of freedom of the system:
d d L d L A ^ / o o i \
j, a" -- a -~-=v, l^a<v. (2.21)
dt dqa. dqoi '
These equations are called Lagrange's equations. Naturally, in (2.21) L is considered to be expressed solely in terms of </a and </a, the Cartesian coordinates being excluded. It turns out that this type of equation holds also in cases when the forces depend on the velocities (see Sec. 21).*
The rules for forming Lagrangc's equations. Since the derivation of equations (2.21) from Newton's Second Law is not readily evident we will give the order of operations which, for this given system, lead to the Lagrange equations.
* In this case, the Lagrangian function does not have the form of (2.20), where U is a function of generalized coordinates only. However, the form of equations (2.21) is still valid.
Sec. 2] LAGRANGE'S EQUATION 23
1) The Cartesian coordinates are expressed in terms of generalized coordinates :
2) The Cartesian velocity components are expressed in terms of generalized velocities:
3) The coordinates are substituted in the expression for potential energy so that it is defined in relation to generalized coordinates:
U=U (ql9 . . ., g«, • • ., #v).
4) The velocities are substituted in the expression for kinetic energy
- — 2 ^'".~, ,
which is now a function of #a and ga . It is essential that in generalized coordinates, T is a function both of #« and </a.
5) The partial derivatives -TTT- and -^- are found.
dq& dq<x.
6) Lagrange's equations (2.21) are formed according to the number of degrees of freedom.
In the next section we will consider some examples in forming Lagrange's equations.
Exercises
1) Write down Lagrange's equation, where the Lagrangian function has the form:
2) A point moves in a vortical plane along a given curve in a gravitational field. The equation of the curve in parametric form is x = x (s), z = z (a). Write down Lagrange's equations.
The velocities are
dx . , . . dz . , .
x = _ 8 ^ x 8 z — _ 8 _= z 8
ds ds
The Lagrangian has the form:
i = -y las'1 + *'1)i1— %»*(«)• Lagrango's equation is
~ m [(x'* + 2/2) «] — m s* (x'x" + z'z") + mgz' = 0 .
24 MECHANICS [Part I
Sec. 3. Examples ol Lagrange's Equations
Central forces. Central forces is the name given to those whose directions are along the lines joining the particles and which depend only on the distances between them. Corresponding to such forces, there is always a potential energy, U, dependent on these distances. As an example, we consider the motion of a particle relative to a fixed centre and attracting it according to Newton's law. We shall show how to find the potential energy in this case by proceeding from the expression for a gravitational force.
Gravitational force is known to be inversely proportional to the square of the distance between the particles and is directed along the line joining them:
F-— a -2 -r . (3.1)
r2 r x '
Here a is the factor of proportionality which we will not define more precisely at this point, r is the distance between the particles, and
— is a unit vector. The minus sign signifies that the particles attract
each other, so that the force is in the opposite direction to the radius vector r. According to (3.1), the attractive-force component along x is equal to
since x is a component of r. But r=V#2 + ?/2 + 22 > so
and similarly for the two other component forces. Comparing (3.3) and (2.7), we see that in the given case
I/ = — y. (3.4)
WTe note that the potential energy U is chosen here in such a way that U (oo) = 0 when the particles are separated by an infinite distance. The choice of the arbitrary constant in the potential energy is called its gauge. In this case it is convenient to choose this constant so that the potential energy tends to zero at infinity.
It is obvious that an expression similar to (3.4) is obtained for two electrically charged particles interacting in accordance with Coulomb's law.
Spherical coordinates. Formula (3.4) suggests that in this instance it is best to choose precisely r as a generalized coordinate. In other words, we must transform from Cartesian to spherical coordinates. The relationship between Cartesian and spherical coordinates is
Sec. 3]
EXAMPLES OF LAGRANGE'S EQUATIONS
shown in Fig. 1. The z-axis is called the polar axis of the spherical coordinate system. The angle & between the radius vector and the polar axis is called the polar angle; it is complementary (to 90°) to the "latitude." Finally, the angle 9 is analogous to the "longitude" and is called the azimuth. It measures the dihedral angle between the plane zOx and the plane passing through the polar axis and the given point.
Let us find the formulae for the transformation from Cartesian to spherical coordinates. From Fig. 1 it is clear that
z = rcos&. (3.5)
The projection p of the radius vector onto the plane xOy is
Whence,
p=r sin <0.
x = p cos 9 = r sin & cos 9 , y = p sin 9 =r sin £ sin 9 .
(3.6)
(3.7) (3.8)
We will now find an expression for the kinetic energy in spherical coordinates. This can be done either by a simple geometrical con- struction or by calculation according to the method of Sec. 2.
Fig. 1
Although the construction is simpler, let us first follow the compu- tation procedure in order to illustrate the general method. We have :
z = cos
x = r sin
= r sin
— 'r sn ft, cos 9 + r cos sin + ^ cos
cos 9 sin
r sn & sn 99, sin & cos 99.
Squaring these equations and adding, we obtain, after very simple manipulations, the following:
26 MECHANICS [Part I
T= vm(i2 + y2 + z2) = -T- (r* + r2&2 + r2sin2£<p2) . (3.9)
2t &
The same is clear from the construction shown in Fig. 2. An arbitrary displacement of the point can be resolved into three mutually per- pendicular displacements: dr, rd$ and pe&p— r sin ^9. Whence
sin2 «• dy*. (3.10)
Since the square of the velocity va=h , (3.9) is obtained from
(3.10) simply by dividing by (dt )2 and multiplying by ~ .
Hence, in spherical coordinates, the Lagrangian function is expressed as
L = 1" (r 2 + r2sin2&cp2 + r* &2) —U(r). (3.11)
z
Now in order to write down Lagrange's equations it is sufficient to calculate the partial derivatives. We have:
OL . dL on dL 2 • 2 a -
- — _ = mr , — - = mrz $ , — = mrz sin2 -9- cp ;
'dr V* d? T
ft L • «> rv • 9 , no & U & ] L o • t\ r\ • 9 & Lt ~
^— •=•- mr sin2 v ®2 4- mr t>2 — - ., — , -^— = mr2 sin fr cos 9- cp2 , - -— — 0 .
dr T c)r #9- T c?9
These derivatives must be substituted into (2.21), which, however, we will not now do since the motion we are considering actually reduces to the plane case (see beginning of Sec. 5).
Two-particle system. So far we have considered the centre of at- traction as stationary, which corresponds to the assumption of an infinitely large mass. In the motion of the earth around the sun, or of an electron in a nuclear field, the mass of the centre of attraction is indeed large compared with the mass of the attracted particle. But it may happen that both masses are similar or equal to each other (a binary star, a neutron-proton system, and the like). We shall show that the problem of the motion of two masses interacting only with one another can always be easily reduced to a problem of the motion of a single mass.
Let the mass of the first particle be ml and of the second ra2. We call the radius vectors of these particles, drawn from an arbitrary origin, rx and r2, respectively. The components of rx are xl9 yl9 21; the components of r2 are #2, ?/2, z2. We now define the radius vector of the centre of mass of these particles B by the following formula:
j 2 v '
Synonymous terms for the "centre of mass" are the "centre of gravity" and the "centre of inertia."
Sec. 3] EXAMPLES OF LAGRANGE's EQUATIONS 27
In addition, let us introduce the radius vector of the relative position of the particles
r = r1— r2 (3.13)
Let us now express the kinetic energy in terms of B and r. From (3.12) and (3.13) we 'have
(3.14)
v '
r2=R -- -. (3.15)
2 -- v '
The kinetic energy is equal to
Differentiating (3.14) and (3.15) with respect to time and substi- tuting in (3.16), we obtain, after a simple rearrangement,
y = l 2 2 t z _ .2 (3 17)
2 ^ 2 (ml + m2) \ • /
Tf we introduce the Cartesian components of the vectors R (X, Y, Z) and r (x, y, z), then we obtain an expression for the kinetic energy in terms of Cartesian components of velocity.
Since no external forces act on the particles, the potential energy can be a function only of their relative positions: U=U(x,y,z). Thus, the Lagrangian is
Transition to the centre-of-mass system. Let us write down La- grange's equations for the coordinates of the centre of mass. We have
Hence, in accordance with (2.21)
X=Y=Z = 0. These equations can be easily integrated:
X = XQt + X0, Y=Y0t+Y0, Z=Z0t+ZQ, (3.18)
where the letters with the index 0 signify the corresponding values at the initial time.
28 MECHANICS [Part I
Combining the coordinate equations into one vector equation, we obtain
Thus, the centre of mass moves uniformly in a straight line quite independently of the relative motion of the particles.
Reduced mass. If wo now write down Lagrange's equations for relative motion in accordance with (2.21) the coordinates of the centre of mass do not appear. It follows that the relative motion occurs as if it were in accordance with the Lagrangian
(where r2 = i2 + ?/2 + z2), formed in exactly the same way as the Lagrangian for a single mass m equal to
(3.20)
. rn1 -f- w2
This mass is called the reduced mass.
The motion of the centre of mass does not affect the relative motion of the masses. In particular, we can consider, simply, that the centre of mass lies at the coordinate origin R = 0.
In the case of central forces (for example, Newtonian forces of attrac- tion) acting between the particles, the potential energy is simply equal to U (r) [this is taken into consideration in (3.10)], where
f z2. Then, if we describe the relative motion in spherical coordinates, the equations of motion will have the same form as for a single particle moving relative to a fixed centre of attraction.
The centre of mass can now be considered as fixed, assuming R = 0. From this, in accordance with (3.14) and (3.15), we obtain the distance of both masses from the centre of mass:
__ B t _
1 t -- ' 2
x -f-
We see that if one mass is much smaller than the other, w2 <^ ml9 then rl <^r2, i.e., the centre of mass is close to the larger mass. This is the case for a sun-planet system. At the same time the reduced mass can also be written thus:
From here it can be seen that it is close to the smaller mass. That is why the motion of the earth around the sun can be approximately described as if the sun were stationary and the earth revolved about it with its own value of mass, independent of the mass of the sun.
Sec. 3]
EXAMPLES OF LAGRANGE'S EQUATIONS
29
Simple and compound pendulums. In concluding this section we shall derive the Lagrangian for simple and compound pendulums.
The simple plane pendulum is a mass suspended on a flat hinge at a certain point of a weightless rod of length Z. The hinge restricts the swing of the pendulum. Let us assume that swinging occurs in the plane of the paper (Fig. 3). It is clear that such a pend- ulum has only one degree of freedom. We can take the angle of deflection of the pendulum from the ver- tical 9 as a generalized coordinate. Obviously the veloc- ity of particle m is equal to /<p, so that the kinetic energy is
m m j2 . 2
77?*r-
The potential energy is determined by the height of the mass above the mean position z-— / (1 — cos 9). -pi<r 3 Whence, the Lagrangian for the pendulum is
L = ~-l*y* — mgl(l — coa<p). (3.22)
A double pendulum can be described in the following way: in mass m there is another hinge from which another pendulum, which is forced to oscillate in the same plane (Fig. 4), is suspended. Let the mass and length of the second pendulum be ml and Zx, respectively, and its angle of deflection from the verti- cal, i. The coordinates of the second particle are
Xi = I sin 9 + Zt sin <J> , zl = l (1 — cos 9) + ^ (1 — cos <L) . Whence we obtain its velocity components:
Zj = / sin 9 9 + Zj_ sin Ad> .
Squaring and adding them we express the kinetic energy
of the second particle in terms of the generalized coordinates 9, <|; and
the generalize^ velocities 9, <L :
Tl =
2 Ul cos (9 —
The potential energy of the second particle is determined in terms of zt. Finally, we get an expression of the Lagrangian for a double pen- dulum in the following form:
T L =
m -\-
m,
i cos (9 — <]>) 9 fy —
— (m
— 0039) —
(3.23)
30
MECHANICS
[Part I
All the formulae for the Lagrangian functions (3.11), (3.22), and (3.23) will be required in the sections that follow.
Exercises
1) Writo down Lagrango's equation for an elastically suspended pendulum.
J/t
For such a pendulum, the potential energy of an elastic force is U — — (Z — Z0)2,
2*
whore J0 is the equivalent length of the unstretched rod and k is a constant, characteristic of its elasticity.
2) Writo down the kinetic; energy for a system of three particles with masses mt, w2, and m3 in the form of a sum of the kinetic energy of the centre of mass and the kinetic energy of relative motion, using the following relative coordi- nates :
Sec. 4. Conservation Laws
The problem of mechanics. If a mechanical system has v degrees of freedom, then its motion is described by v Lagrangian equations. Each of these equations is of the second order with respect to the time derivatives q [see (2.17)]. From general theorems of analysis we conclude that after integration of this system we obtain 2 v arbitrary constants. The solution can be represented in the following form:
(4.1)
Differentiating these equations with respect to time, we obtain expressions for the velocities:
(4.2)
Let us assume that equations (4.1) and (4.2) are solved with respect to the constants C^, . . . , C^v, so that these values are expressed in terms of t and qlf . . ., gv, qv . . .,qv.
Then
Sec. 4] CONSERVATION LAWS 31
From the equations (4.3) we see that in any mechanical system described by 2v second-order equations there must be 2 v functions of generalized coordinates, velocities, and time, which remain constant in the motion. These functions are called integrals of motion.
It is the main aim of mechanics to determine the integrals of motion.
If the form of the function (4.3) is known for a given mechanical system, then its numerical value can be determined from the initial conditions, that is, according to the given values of generalized coordinates and velocities at the initial instant.
In the preceding section we obtained the so-called centre -of -mass integrals R0 and B0 (3.18).
Naturally, Lagrange's equations cannot be integrated in general form for an arbitrary mechanical system. Therefore the problem of determining the integrals of motion is usually very complicated. But there are certain important integrals of motion which are given directly by the form of the Lagrangian. We shall consider these integrals in the present section.
Energy. The quantity
is called the total energy of a system. Let us calculate its total deriv- ative with respect to time. We have
___ ___ ___
~dt~ - a ~0$7 « ~dt "0 "fcT ~J~q^"~ "FfcT « ~ ~di" '
The last three terms on the right-hand side are the derivatives of the Lagrangian L, which, in the general form, depend on g, q and t. In determining $ and its derivative we have made use of the sum-
7 f\ T
mation convention. The quantity ^- - .— in Lagrange's equations
*\ -r
can be replaced by -= — . The result is, therefore,
uC[y.
dff 8L
-dr=—w <4-5>
Consequently, if the Lagrangian does not depend explicitly on time
(O T \
~7j— = 01 , the energy is an integral of motion. Let us find the con-
ditions for which time does not appear explicitly in the Lagrangian.
If the formulae expressing the generalized coordinates q in terms of
Cartesian coordinates x do not contain time explicitly (which corre-
* a is summed from 1 to v (see Sen. 2).
32 MECHANICS [Part 1
spends to constant, time-independent constraints) then the transfor- mation from x to q cannot introduce time into the Lagrangian.
O TT
Besides, in order that -7-- — 0 , the external forces must also be
G t
independent of time. When these two conditions — constant constraints and constant external forces — are fulfilled, the energy is an integral of motion. To take a particular case, when no external forces act on the system its energy is conserved. Such a system is called closed.
When frictioiial forces act inside a closed system, the energy of macroscopic motion is transformed into the energy of molecular microscopic motion. The total energy is conserved in this case, too, though the Lagrangian, which involves only the generalized coordi- nates of macroscopic motion of the system, no longer gives a complete description of the motion of the system. The mechanical energy of only macroscopic motion, determined by means of such a Lagrangian, is not an integral. We will not consider such a system in this section.
Let us now consider the case when our definition of energy (4.4) coincides with another definition, $=T+U. Let the kinetic energy be a homogeneous quadratic function of generalized velocities, as expressed in equation (2.13). For this it is necessary that the con- straints should not involve time explicitly, otherwise equation (2.5) would have the form
where the partial derivative of the function (2.2) with respect to time is taken for all constants #«. But then terms containing qa in the first degree would appear in the expression for T.
Since we assume that the potential energy U does not depend on velocity [see (2.18) and (2.19)], then
8L BT and the energy is
JP 2* T I 4 £±\
6—<ta-Q-q ^> • (4.6)
But according to Euler's theorem, the sum of partial derivatives of a homogeneous quadratic function, multiplied by the corresponding variables, is equal to twice the value of the function (this can easily be verified from the function of two variables ax2 + 2 bxy + cy2). Thus,
that is, the total energy is equal to the sum of the potential energy
and the kinetic energy, in agreement with the elementary definition.
We note that the definition (4.4) is more useful and general also
in the case when the Lagrangian is not represented as the difference
Sec. 4] CONSERVATION LAWS 33
L—T — U. Thus, in electrodynamics (Sec. 15) L contains a linear term in velocity. For the energy integral to exist, only one condition
is necessary and sufficient : -^~- = 0 (if, of course, there are no friction-
al forces).
The application of the energy integral to systems with one degree of freedom. The energy integral allows us, straightway, to reduce problems of the motion of systems with one degree of freedom to those of quadrature. Thus, in the pendulum problem considered in the previous section we can, with the aid of (4.7), write down the energy integral directly:
_ cos 9) m (4 S)
The value £ is determinable from the initial conditions. For example, let the pendulum initially be deflected at an angle cp0 and released without any initial speed. It follows that (p0 — 0. Whence
1— cos 90). (4.0)
Substituting this in (4.8), we have
From this, the relationship between the deflection angle and time is determined by the quadrature
—— — — 9 — . (4.11)
\/cos <p — cos 90
Tt is essential that the law governing the oscillation of a pendulum depend only on the value of the ratio l/g and is independent of the mass. The integral in (4.11) cannot be evaluated in terms of elementary functions.
A system in which mechanical energy is conserved is sometimes called a conservative system. Thus, the energy integral permits reducing to quadrature the problem of the motion of a conservative system with one degree of freedom.
Tn a system with several degrees of freedom the energy integral allows us to reduce the order of the system of differential equations and, in this way, to simplify the problem of integration.
Generalized momentum. We shall now consider other integrals of motion which can be found directly with the aid of the Lagrangian. To do this we shall take advantage of the following, quite obvious, consequence of Lagrange's equations. If some coordinate </a does not
appear explicitly in the Lagrangian (-^- =0), then in accordance
\ 09* I with Lagrange's equations
34 MECHANICS [Part I
44^- = 0- (4.12)
dt dqa v '
But then
r\ 7"
pa S3 — - - = const, (4-13)
i.e., it is an integral of motion. The quantity pa is called the generalized momentum corresponding to a generalized coordinate with index a. This definition includes the momentum in Cartesian coordinates:
pL "I-
px=zmvx = ~ — . Summarizing, if a certain generalized coordinate does
vVx
not appear explicitly in the Lagrangian, the generalized momentum corresponding to it is an integral of motion, i.e., it remains constant for the motion.
In the preceding section we saw that the coordinates X, F, Z of the centre of gravity of a system of two particles, not subject to the action of external forces, do not appear in the Lagrangian. From this it is evidentthat
Z = Pz (4.14)
are constants of motion.
The momentum of a system of particles. The same thing is readily shown also for a system of N particles. Indeed, for N particles we can introduce the concept of the centre of mass and the velocity of the centre of mass by means of the equations:
(4.15a)
(4.15b)
JTm;
i The velocity of the ith mass relative to the centre of mass is
r/ = rr-R (4.16)
(by the theorem of the addition of velocities). The kinetic energy of the system of particles is
N N
T = ~ JTwn if = |2>, ( V + R)2 =
f-i ;~i
N N N
= y^wi*/1 + B 2>;r;' + y2><*a- (4.17)
l-l i-l .'-1
Sec. 4] CONSERVATION LAWS 35
However, from (4.15b) and (4.16) it can immediately bo seen that
N
JTraif/^O, by the definition of r,' and R. Therefore, the kinetic
» = i
energy of a system of particles can be divided into a sum of two terms :
the kinetic energy of motion of the centre of mass
and the kinetic energy of motion of the mass relative to the centre of mass
N
The vectors f,' are not independent; as has been shown, they are
N
governed by one vector equation J^ w,-r,' = 0. Consequently, they can
be expressed in terms of an N- 1 independent quantity by determining the relative positions of the ith and first masses. For this reason the kinetic energy of N particles relative to the centre of mass is, in general, the kinetic energy of their relative motion, and is expressed only in terms of the relative velocities i^ — r,. By definition, no external forces can act on the masses in a closed system, and the interaction forces inside the system can be determined only by the relative positions rt— r,.
Thus, only R appears in the Lagrangian, and R does not. Therefore, the overall momentum is conserved:
ar ( N \ P - ~ = \£nii tt = const . (4.18)
Equality (4.18), which contains a derivative with respect to a vector, should be understood as an abbreviated form of three equations:
T^ dL -n 8L -o dlj
For more detail about vector derivatives see Sec. 11.
We have seen that the overall momentum of a mechanical system not subject to any external force is an integral of motion. It is impor- tant that it is what is known as an additive integral of motion, i.e., it is obtained by adding the momenta of separate particles.
36 MECHANICS [Part I
It may be noted that the momentum integral exists for any system in which only internal forces are operative, even though they may be frictional forces causing a conversion of mechanical energy into heat.
If we integrate (4.18) with respect to time once again, the result will be the centre-of-mass integral similar to (3.18). This will be the so- called second integral (for it contains two constants); it contains only coordinates but not velocities. (3.18) and (4.11) are also second integrals.
Properties of the vector product. The angular momentum of a particle is defined as
M-Frp]. (4.19)
Here the brackets denote the vector product of the radius vector of the particle and its momentum. We know that (4.19) takes the place of three equations,
Mx = ypz—zpy, MY -= zpx- ~xpz, Mz — xpy — ypx,
for the Cartesian components of the vector M.
Recall the geometrical definition for a vector product. We construct a parallelogram on the vectors r and p. Then [rpj denotes a vector numerically equal to the area of the parallelogram with direction perpendicular to its plane. In order to specify the direction of [rp] uniquely, we must agree on the way of tracing the parallelogram con- tour. We shall agree always to traverse the contour beginning with the first factor (in this case beginning with r). Then that side of the plane will be considered positive for which the direction is anticlockwise. The vector [rp] is along the normal to the positive side of the plane. In still another way, if we rotate a corkscrew from r to p, then it will be displaced in the direction of [rp]. The direction of traverse changes if we interchange the positions of r and p. Therefore, unlike a conven- tional product, the sign of the vector product changes if we inter- change the factors. This can also be seen from the definition of Carte- sian components of angular momentum.
The area of the parallelogram is rp sin a, where a is the angle between r and p. The product r sin a is the length of a perpendicular drawn from the origin of the coordinate system to the tangent to the trajec- tory whose direction is the same as p. This length is sometimes called the "arm" of the moment.
The vector product possesses a distributive property, i.e.,
Hence, a binomial product is calculated in the usual way, but the order of the factors is taken into account.
[a + d, b + c] = [ab] + [ac] + [db] + [dc] .
Sec. 4] CONSERVATION LAWS 37
The angular momentum of a particle system. The angular momentum of a system of particles is defined as the sum of the angular momenta of all the particles taken separately. In doing so we must, of course, take the radius vectors related to a coordinate origin common to all the particles:
N
We shall show that the angular momentum of a system can be separated into the angular momentum relative to the motion of the particles and the angular momentum of the system as a whole, similar to the way that it was done for the kinetic energy. To do tin's Ave must represent the radius vector of each particle as the sum of the radius vector of its position relative to the centre of mass and the radius vector of the centre of mass ; we must expand the expression for the particle velocities in the same way. Thou, the angular momentum can he written in the form
N
N N N N
1-1 1=1 »«l J-l
111 the second and third sums, wo can make use of the distributive property of a vector product and introduce the summation sign inside the product sign. However, both these sums are equal to zero, by definition of the centre of mass. This was used in (4.17) for veloc- ities. Thus, the angular momentum is indeed equal to the sum of the angular momenta of the centre of mass (M0) and the relative motion (M'):
M = [RP] + JTfr' Pi'J - M0 + M'. (4.21)
Let us perform these transformations for the special case of a system of two particles. We substitute rx and r2 expressed [from (3.14) and (3.15)] in terms of r and R. This gives
M = friPil + [r2P2] = [tt,Pi + P2] + (
Further, we replace ^ by ml r1? p2 by mz r2 and P!+P2 by P, after which the angular momentum reduces to the required form:
38 MECHANICS [Part I
M = [RP] + -w^- [r r] = [RPJ + [rp] . (4.22)
Here, "m ^^ r = wr = p is the momentum of relative motion.
We shall now show that the determination of angular momentum of relative motion does not depend on the choice of the origin. Indeed, if we displace the origin, then all the quantities r/ change by the same amount ri'=r,-" + a.
Accordingly, angular momentum for relative motion will be
N N N
N
because
N N
Thus, the determination of angular momentum for relative motion does not depend on the choice of the origin of the coordinate system.
Conservation ol angular momentum. We shall now show that the angular momentum of a system of particles not acted upon by any external forces is an integral of motion.
Let us begin with the angular momentum of the system as a whole. Its time derivative is equal to zero:
because P = 0 for any system not acted upon by external forces, and R is in the direction of P, so that the vector product [RP] = 0.
We shall now prove that angular momentum is conserved for relative motion. The total derivative with respect to time is
i - 1
Here, the first term on the right-hand side is equal to zero since r/ is in the direction of p,'. We consider the second term. Let us choose the origin to be coincident with any particle, for example, the first. As a result, M', as we have already seen, does not change. The potential energy can only depend on the differences rx — r2, rx — ra, . . . , ^ — T&, . . .
Sec. 4] CONSERVATION LAWS 39
The other differences are expressed in terms of these, for example, Yk — ?I = (TI — r/) — (rx — r^). We introduce the abbreviations.
p -rj t\ -J-T n j-j
Then the derivatives-^ — , -^ — -, . . . , -« — . . . will be expressed in terms of the variables px, . . . , pfe _ i , . . . as follows :
fe-1
su su
Substituting this in (4.23) we obtain
dt '
k~l /?=»!
N-1 n N-l N-l
In this expression, only the relative coordinates px, ...,PN-I remain. We shall now show that, for a closed system, the right-hand side is identically equal to zero. The potential energy is a scalar function of coordinates. Hence, it can depend only on the scalar arguments pfe2> P/2> (p/ Pfe)> totally irrespective of whether the initial expression was a function only of the absolute values | r; — r* | , or whether it also involved scalar products of the form (r» — rk, r/ — r«). An essential point is that the system is closed (in accordance with the definition, see page 32), and the forces in it are completely defined by the relative positions of the points and by nothing else. Therefore, the potential energy depends only on the quantities r,- — r*, and only in scalar combinations (r/ — rfe, r/ — rrt) (in particular, the subscripts i, I; k, n can even be the same; then the scalar product becomes the square of the distance between the particles i, k).
To summarize, the potential energy U depends on the following arguments:
u==u Oi> P'> • • •> 9k, • • -, PN-I; (Pi P2)> - - •, (pfep/)] -
40 MECHANICS [Part I
In order to save space we will, in future, perform the operation for two vectors, though this operation can be directly generalized to any arbitrary number. We obtain
CJU __ dU j)(pf) dU fr(plPa)
~" "~ ~r ~ '
_£ [V = dU _
The partial derivatives of the scalar quantities pj, (px p2) with respect to the vector arguments are in the given case easily evaluated. Thus.
Each of these equations is a shortened form of three equations referring to the components (the components of p, are &, TJ/, &):
Henco,
Substituting (4.25) into (4.24) for the case of the two variables, we obtain
r/M' or . dU or -, dU /r lir -^ 8U
-ir = - 2 [PlpJ ^.y - 2 fp2 p2l. --(W _ ([Pl pj + [p2 p J) ^_p-- .
But the sign of a vector product depends on the order of the factors [Pi p2.1 ^ — T?2 Pi]- Hence it can also be seen that [Pl Pl] = — [Pl Pl] = 0 and [pspo]^^- Tlicrefore, , =0, as stated.
The integral, like the angular momentum, can also be formed when the forces are determined not only by the relative position of the par- ticles but also by their relative velocities. This is the case, for example, in a system of elementary currents interacting in accordance with the Biot-Savart law.
Additive integrals of motion for closed systems. We have thus shown that a closed system has the following first integrals of motion: energy, three components of the momentum vector and three compo- nents of the angular-momentum vector. Momentum and angular momentum are always additive, while energy is additive only for the non interacting parts of a system.
All the other integrals of motion are found in a much more compli- cated fashion and depend on the specific form of the system (in the sense that one cannot give a general rule for their definition).
SeC. 5] MOTION IN A CENTIiAL FIELD 41
Exorcises
Describe the motion of a point moving along a cycloid in a gravitational field.
The equation of the cycloid in parametric form is
z = — R cos s, x= The kinetic energy of the point is
T=~ (x2 + z2) - 2
^
The potential energy is U = — rtujR cos s. The total-energy integral is
<? = 2 7ft ft- cos2 — - • $2 — mgR cos 5 = const .
2i
The value $ can be determined on the condition that the velocity £ is equal to zero when the deflection is maximum s — s0; the particle moves along the cycloid from that position. Hence,
<f = tug R COS «S'0.
After separating the variables and integrating, wo obtain
— cos s0
Calling sin -y = u, wo rewrite the integral in the form :
arc sin -
In order to find the period of the motion, we must take the integral between the limits — - MO and -\- UQ and double the result. This corresponds to the oscilla- tion of the particle within the limits 8= — s0 arid s = «v
Thus the total period of oscillation is equal to 4:c I/ —
Hence, as long as the particle moves on the cycloid, the period of its oscillation does not depend on the oscillation amplitude (Huygens' cycloidal pendulum). The period of oscillation of an ordinary pendulufn, describing an arc of a circle, is known, in the general case, to depend on the amplitude [see (4.11)].
Sec. 5. Motion in a Central Field
The angular-momentum integral. We shall now consider the motion of two bodies in a frame of reference fixed in the centre of mass. If the origin coincides with the centre of mass, then R = 0. As was shown in the preceding section, the angular momentum of relative motion is conserved in any closed system; specifically, it is also conserved in
42 MECHANICS [Part I
a system of two particles. If the radius vector of the relative position of the particles r— FJ — r2, and the momentum of relative motion is
(
' V '
then the angular-momentum integral is reduced to the simple form :
M=[rp]= const. (5.2)
It follows that the velocity vector and the relative position vector all the time remain perpendicular to the constant vector 31 ; in other words, the motion takes place in a plane perpendicular to M (Fig. 5). When transforming to a spherical coordinate system, it is advisable to choose the polar axis along M. Then the motion will take place in the
plane xy or -9- = -"- , sin £ = 1, <i> = 0.
The potential energy can depend only on the absolute value r, because this is the only scalar quantity which can be derived solely from the vector r. In accordance with (3.9), the Lagran- gian for plane motion, with fr = 0, sin o> = l, is
L=-TT(r* + r*<?*)—U(r)9 (5.3)
Fig. 5
where m is the reduced mass. Angular momentum as a generalized momentum. We shall now show
Q •*•
that M z — M is nothing other than ^p- , i.e., the component of angular
momentum along the polar axis is a generalized momentum, provided the angle of rotation cp around that axis is a generalized coordinate. Indeed, in accordance with (5.2), the angular momentum M is
M = Mz — xpy — ypx ~ mr cos 9 (r sin 9+7* cos 99 ) —
— mr sin 9 (r cos 9 — r sin <py)=mr2 9 (cos2 9 + sin29)=wr2 9.
On the other hand, differentiating L with respect to 9 we see that
o y P* = ~0T = ^^ = MZ * (5<4)
The expression for angular momentum in polar coordinates can also be derived geometrically (Fig. 6). In unit time, the radius vector r moves to the position shown in Fig. 6 by the dashed line. Twice the area of the sector OAB, multiplied by the mass w, is by definition equal to the angular momentum [cf. (5.2)]. But, to a first approxi- mation the area of the sector is equal to the product of the modulus
r and -$- . The height h is proportional to the angle of rotation in unit
Sec. 5] MOTION IN A CENTRAL FIELD 43
time and to the radius itself so that the area of the sector is 1/2 r2 9. Thus, a doubled area multiplied by the mass m is indeed equal to the angular momentum.
The quantity 1/2 r2 9 is the so-called areal velocity, or the area described by the radius vector in unit time. The law of conservation of angular momentum, if interpreted geometrically, expresses constancy of areal velocity (Kepler's Second Law).
The central field. If one of two masses is very much greater than the other, the centre of mass coincides with the larger mass (see Sec. 3). In this case, the particle with the smaller mass moves in the given central "field of the heavy particle. The potential energy depends only on the distance between the particles and does not depend on the angle 9. Then, in accordance with (4.12), p^ — Mz is the integral of motion. However, since one particle is considered at rest, the origin should be chosen coincident with that particle and not with some arbitrary point, as in the case of the relative motion of two particles. In the case of motion in a central field, angular momentum is conserved only relative to the centre.
Elimination ol the azimuthal velocity component. The angular- momentum integral permits us to reduce the problem of two-particle motion, or the problem of motion of a single particle in a central field, to quadrature. To do this we must express 9 in terms of angular mo- mentum and thus get rid of the superfluous variable, in as much the angle 9 itself does not appear in the Lagrangian. Such variables, which do not appear in L, are termed cyclic.
In accordance with (4.7), we first of all have the energy integral
f=~(r* + r*i*)+U(r). (5.5)
Eliminating 9 with the aid of (5.4) we obtain
U(r). (5-6)
This first-order differential equation (in r) is later on reduced to quadrature. Before writing down the quadrature, let us examine it graphically.
The dependence of the form of the path on the sign of the energy. For such an examination, we must make certain assumptions about the variation of potential energy.
From (2.7), force is connected with potential energy by the relation
44 MECHANICS [Part I
The upper limit in the integral can be chosen arbitrarily. If F (r)
00
tends to zero at infinity faster than — , then the integral I F dr is
00 f
convergent. Then we can put U (r)^( F dr, or £/(oo)— 0. In other
words, the potential energy is considered zero at infinity. The choice of an arbitrary constant in the expression for potential energy is called its yawje.
In addition we shall consider that at r~0, U (r) does not tend to
infinity more rapidly than - , as, for example, for Newtonian attrac-
oo
, . fT I* (t , tt
turn U --- „ dr- --- .
J rz r
r
Let us now write (5.6) as
The left-hand side of this equation is essentially positive. For /-=oo the last two terms in (5.7) tend to zero. Thus, for the particles to be able to recede, from each other an infinite distance, the total energy must be positive when the gauge of the potential energy satisfies f7(oo)=0. Given a definite form of U, we can now7 plot the curve of the function
The index M in U denotes that the potential energy includes the
I/2 "centrifugal" energy ./ ^r. The derivative of this quantity with
Af2 respect to r, taken with the opposite sign, is equal to ""—£-. If we put
M ~ wra 9, the result will be the usual expression for "centrifugal force." However, henceforth, we will call a mechanical quantity of different origin the "centrifugal force" (see Sec. 8). Let U <() and monotonic. Since U (oo) = 0, we see that U (r) is an increasing function
of r. It follows that the force has a negative sign (since F — — 4—1 , i.e., it is an attractive force. Let us assume, in addition, that at infinity I W ('*) ! > V M,2~* Let us summarize the assumptions that we have
made concerning UM (r):
1) U\t(r) is positively infinite at zero, where the centrifugal term is predominant.
2) at infinity, where U (r) predominates, UM (r) tends to zero from a negative direction.
Sec. 5] MOTION IN A CENTRAL FIELD 45
Consequently, the curve UM (r) has the form shown in Fig. 7, since we must go through a minimum in order to pass from a decrease for small values of r to an increase at large values of r.*
In this figure we can also plot the total energy 6\ But since the total energy is conserved, the curve of $ must have the form of a horizontal straight line lying above or below the abscissa, depending on the sign of 6.
For positive values of energy, the line <? = const lies above the curve UM (r) everywhere to the right of point A . Hence the difference $ — UM(T) to the right of A is positive. The particles can approach each other from infinity and recede from each other to infinity. Such motion is termed infinite. As we will see later in this section, in the case of Newtonian attraction, we obtain hyperbolic orbits.
For <5"<0, but higher than the mini- mum of the curve UM (r), the difference
&* — UM (r), i.e., ^— , remains positive
only between the points B and B' (finite
motion). Thus, between these values of
the radius there is included a physi- Fig. 7
cally possible region of motion, to which
there correspond elliptical orbits in the case of Newtonian
attraction. In the case of planetary motion around the sun, point B
is called the perihelion and point B1 the aphelion.
For «? = () the motion is infinite (parabolic motion).
If U (/•)>(>, which corresponds to repulsion, the curve UM(T) does not possess a minimum. Then finite motion is clearly impossible.
Falling towards the centre. For Newtonian attraction, U (0) tends to infinity like — 1/r. If we suppose that U (0) tends to — oo more rapidly than — 1/r2, then the curve UM(^) is negative for all r close to zero. Then, from (5.7), f 2 is positive for infinitesimal values of r and tends to infinity when r tends to zero. If r < 0 initially, then r does not change sign and the particles now begin to move towards collision. In Newtonian attraction this is possible only when the particles are directed towards each other; then "the arm" of the angular momentum is equal to zero and, hence, the angular momentum itself is obviously equal to zero, too, so that UM (r) = U (r). If an initial "arm" exists within the distance of minimum approach, then
M2 * Ff I U (r) | < — 2" at infinity, then the curve approaches zero on the
positive side, and there can bo a further small maximum after the minimum. This form of UM (r) applies to the atoms of elements with medium and large atomic weights.
46 MECHANICS [Part I
the angular momentum M = mvp ^ 0 (p is the "arm") and the motion can in no way become radial.
In the case of Newtonian or Coulomb attraction for a particle with angular momentum not equal to zero, there always exists a distance
M2 r0'for which- — r becomes greater than $ — U (r0). This distance
determines the perihelion for the approaching particles.
However, if U (r) tends to infinity more rapidly than -- ^ then,
as r->0, there will be no point at which UM (r) becomes zero. In place of a hyperbolic orbit, as in the case of Newtonian attraction, a spiral curve leading to one particle falling on the other results. The turns of the spiral diminish, but the speed of rotation increases so that the angular momentum is conserved, as it should be in any central field. But the "centrifugal" repulsive force turns out to be less than the forces of attraction, and the particles approach each other indef- initely.
Of course, the result is the same if the energy is negative (for example, part of the energy is transferred to some third particle, which then recedes). In the case of attractive forces increasing more rapidly than 1/r3, no counterpart to elliptical orbits exists.
If three bodies in motion are subject to Newtonian attraction, two of them may collide even if, initially, the motion of these particles was not purely radial. Indeed, in the case of three bodies, only the total angular momentum is conserved, and this does not exclude the collision of two particles.
Reducing to quadrature. Let us now find the equation of the tra- jectory in general form. To do this we must, in (5.6), change from differentiation with respect to time to differentiation with respect to 9. Using (5.4) we have
dt = lj£-d<p. (5.9)
Separating the variables and passing to 9 in (5.6) gives
M dr
9 =
Here, the lower limit of the integral corresponds to 9 = 0. If we cal- culate 9 with respect to the perihelion, then the corresponding value r = r0 can be easily found by noting that the radial component of velocity r changes sign at perihelion (r has a minimum, and so dr = 0). From this we find the equation for the particle distance at perihelion :
*=•?£* +UM- <6-n>
Sec. 5] MOTION IN A CENTRAL FIELD 47
Kepler's problem. Thus, the problem of motion in a central field is reduced to quadrature. The fact that the integral sometimes cannot be solved in terms of elementary functions is no longer so essential. Indeed, the solution of the problem in terms of definite integrals contains all the initial data explicitly; if these data are known, the integration can be performed in some way or other.
But, naturally, if the integral is expressed in the form of a well-known function, the solution can be more easily examined in the general form. In this sense an explicit solution is of particular interest.
Such a solution can be found in only a few cases. One of these is the case of a central force diminishing inversely as the square of distance. The forces of Newtonian attraction between point masses (or bodies possessing spherical symmetry) are subject to this law.
It will be recalled that the laws of motion in this case were found empirically by Kepler before Newton deduced them from the equations of mechanics and the law of gravitation. It was the agreement of Newton's results with Kepler's laws that was the first verification of the truth of Newtonian mechanics. The problem of the motion of a particle in a field of force diminishing inversely with the square of the distance from some fixed point, is called Kepler's problem. The prob- lem of the motion of two bodies with arbitrary masses always reduces to the problem of a single body when passing to a frame of reference fixed in the centre of mass.
The expression "Kepler's problem" can also be applied to Coulomb forces acting between point charges. These can either be forces of
attraction or repulsion. In all cases we shall write U = — , where a < 0 for attractive forces and a>0 for repulsive forces.
If we replace in (5.10) by a new variable x, the integral in the
Kepler problem is reduced to the form
rr4- — dx + M
. 2a
___ ffi ._ _ /J» „ I
X +
— = arc cos
„ _ ,
M m M*
-— — ] a*
If ,
V M*
M
x ~ - — mr
M x =
At the lower limit, the expression inside the arc cos sign is equal to unity [as will be seen from (5.11)], since the lower limit was chosen on the condition that dr = 0. But arc cos 1—0. Rearranging the result of integration and reverting to r, we obtain, after simple manipu- lations,
M*
r = «*» . (6J2)
. , M ] a* , 2^ V '
— 1 H tf -5TrT H COS <p
a V M2 m
48 MECHANICS [Part I
(5.12) represents the standard equation for a conic section, the
eccentricity being equal to I/ 1 4 \2 . As long as this expression
is less than unity, the denominator in (5.12) cannot become zero, because cos cp< 1. But this is true for " <$<(). Thus, when
<^<0, the result is elliptical orbits. For this it is necessary that a<0, i.e., that there be an attraction, otherwise (5.12) would lead to r<0, which is senseless.
For ff > 0 the eccentricity is greater than unity and the denominator in (5.12) becomes zero for a certain 9 = 900. Thus, the orbit goes to infinity (a hyperbola). The direction of the asymptote is obtained by
putting r =00 in (5. 12). This requires that cos 900 — -»*- • — .
•M- I/ a2 2<^
V ~M* + ~w~
The angle between the asymptotes is equal to 2 900, when the particles repel each other, and to 2 (7^—900), when the particles attract. An example of a trajectory, when the forces are repulsive, is shown in Fig. 8, Sec. 6.
Exercise
ar2
Obtain iho equation, of the trajectory when U = — — $ > 0.
2
See. 6. Collision of Particles
The significance of collision problems. In order to determine the forces acting between particles, it is necessary to study the motion of particles caused by these forces. Thus, Newton's gravitational law was established with the aid of Kepler's laws. Here, the forces were determined from iinite motion. However, infinite motion can also be used if one particle can, in some way, be accelerated to a definite velocity and then made to pass close to another particle. Such a process is termed 4 'collision" of particles. It is not at all assumed, however, that the particles actually come into contact in the sense of "collision" in everyday life.
And neither is it necessary that the incident particle should be artificially accelerated in a machine: it may be obtained in ejection from a radioactive nucleus, or as the result of a nuclear reaction, or it may be a fast particle in cosmic radiation.
Two approaches are possible to problems on particle collisions. Firstly, it may be only the velocities of the particles long before the collision (before they begin to interact) that are given, and the problem is to determine only their velocities (magnitude and direction) after they have ceased to interact. In other words, only the result of the collision is obtained without a detailed examination of the process. In this case, some knowledge of the final state must be available (or
Sec. 6] COLLISION OF P ARTICLES 49
specified) beforehand: it is not possible to determine, from the initial velocities alone, all the integrals of motion which characterize the collision, and, hence, it is likewise impossible to predict the final state. With this approach to collision problems, only the momentum and energy integrals are known.
However, another approach is possible : it is required to precalculate the final state where the precise initial state is given.
Let us first consider collisions by the first method. It is clear that if only the initial velocities of the particles are known, the collision is not completely determined: it is not known at what distance the par- ticles were when they passed each other. This is why some quantity relating to the final state of the system must be given. Usually the problem is stated as follows : the initial velocities of the colliding par- ticles and also the direction of velocity of one of them after the colli- sion are specified. It is required to determine all the remaining quanti- ties after the collision. In such a form the problem is solved uniquely. Six quantities are unknown, namely the six momentum components of both particles after the collision. The conservation laws provide four equalities: conservation of the scalar quantity (energy) and conservation of the three components of the vector quantity (momen- tum). Therefore, with six unknowns, it is necessary to specify two quantities which refer to the final state. They are contained in the determination of the unit vector which specifies the direction of the velocity of one of the particles ; an arbitrary vector is defined by three quantities, but a unit vector, obviously, only by two. Actually, only the angle of deflection of the particles after the collision need be given, i.e., the angle which the velocity of the particle makes with the initial direction of the incident particle. The orientation (in space) of the plane passing through both velocity vectors is im- material.
Elastic and inelastic collisions. A collision is termed elastic, if the initial kinetic energy is conserved when the particles separate after the collision at infinity, and inelastic, if, as a result of the collision, the kinetic energy changes at infinity. In nuclear physics, studies are very often made of collisions of a more general character, in which the nature of the colliding particles changes. These collisions are also inelastic. They are called nuclear reactions.
The laboratory and centre-o!-mass frames of reference. When colli- sions are studied in the laboratory, one of the particles is usually at rest prior to collision. The frame of reference fixed in this particle (and in the laboratory) is termed the laboratory frame. However, it is more convenient to perform calculations in a frame of reference, relative to which the centre of mass of both particles is at rest. In accordance with the law of conservation of the centre of mass (3.18), it will also be at rest in its own frame after the collision. The velocity of the centre of mass, relative to the laboratory frame of reference, is
60 MECHANICS [Part I
(6.1) v '
Here V0 is the velocity of the first particle (of mass m^ relative to the second (with mass m2). In so far as the second particle is at rest in the laboratory system, v0 is also the velocity of the first particle relative to this system.
The general case of an inelastic collision. The velocity of the first particle relative to the centre of mass is, according to the law of addition of velocities, equal to
(6.2) v
and in the same system, the velocity of the second particle is
(6.3)
Thus, mx V10 + m2 V2o = 0, as it should be in the centre-of-mass system.
In accordance with (3.17), the energy in the centre-of-mass system is
(6-4)
Here, the reduced mass is indicated by a zero subscript, since in nuclear reactions it may change.
Let the masses of the particles obtained as a result of the reaction be ws and ra4, and the energy absorbed or emitted Q (the so-called "heat" of the reaction). If Q is the energy released in radiation, then, strictly speaking, one should take into account the radiated momen- tum (see Sec. 13). But it is negligibly small in comparison with the momenta of nuclear particles.
Thus, the law of conservation of energy must be written in the follow- ing form:
-=£+«--=£-. (6.5)
Here, m = - ~-^ — is the reduced mass of the particles produced in
^3 *T~ ^4
the nuclear reaction, and v is their relative velocity.
In order to specify the collision completely, we will consider that the direction of v is known, since the value of v is determined from (6.5). Then the velocity of each particle separately will be
They satisfy the requirement mz V30+w4 v40 = 0, i.e., the law of conservation of momentum in the centre-of-mass system, and give the necessary value for the kinetic energy
Sec. 6] COLLISION OF PABTICLES 51
Now, it is not difficult to revert to the laboratory frame of reference. The velocities of the particles in this system will be
V - ___ Y
Equations (6.7) give a complete solution to the problem provided the direction of v is given.
Elastic collisions. The computations are simplified if the collision is elastic, for then mz = mi, w4 = w2> Q = 0. It follows from (6.5) that the relative velocity changes only in direction and not in magnitude. Let us suppose that its angle of deflection x is given. We take the axis Ox along v0, and let the axis Oy lie in the plane of the vectors v and v0 (which are equal in magnitude in the case of elastic collisions). Then
VX = VQ COS X, Vy = VQ SHI X •
From (6.6), the components of particle velocity in the centre-of-mass system will, after collision, be correspondingly equal to
.. _
i 20y
Since the velocity of the centre of mass is in the direction of the axis Ox, we obtain, from (6.7), the equations for the velocities in the laboratory frame of reference:
xjtjo m2v0sinx
Viy ~ V"v ~ -m~+m~ '
_ _ m,v0sinx
"" - ~
- - 114 + tn, J 2y
By means of these equations, the deflection angle 6 of the first particle in the case of collision in the laboratory system can be related to the angle x» (i- e., its deflection angle in a centre-of-mass system):
(6.8)
The "recoil" angle of the second particle 0' is defined as
v0y sin x 4. X
-~^-T-COSX~CO y (69)
6' — -_-* 2 2 '
52 MECHANICS [Part I
The minus sign in the definition of tg 0' is chosen because the signs of vlY and v2y are opposite.
The case of equal masses. Equation (6.8) becomes still simpler if the masses of the colliding particles are equal. This is approximately true in the case of a collision between a neutron and proton. Then, from (6.8).
tan 6--- tan --, 0--->
i.e., the particles fly off at right angles and the deflection angle of the neutron in the laboratory system is equal to half the deflection angle in the ceritre-of-mass system. Since the latter varies from 0 to 180°, 0 cannot exceed 90°. And, in addition, the velocity of the incident particle is plotted as the "resultant" velocity of the diverging particles. The collision of billiard balls resembles the collision considered here of particles of equal mass, provided that the rotation of the balls about their axes is neglected.
The energy transferred in an elastic collision. The energy received by the second particle in a collision is
JP __ mzml (l---cosx) vj
(mi + "!,)•
Its portion, relative to the initial energy of the first particle, is <?2 __ 2mtw2 (1 — cos/)
\2
(6.10)
& 7
From this we obtain ~~ = sin2 —• =sin20 for particles of equal
0 .
mass. Accordingly, the portion of the energy retained by the first
particle is - ==cos2 0. In a "head-on" collision x^18()0, 0-90°.
0Q
The first particle comes to rest and the second continues to move forward with the same velocity. This can easily be seen when billiard balls collide.
The problem of scattering. Let us now examine the problem of colli- sion in more detail. We shall confine ourselves to the case of elastic collision and perform the calculations in the centre-of-mass system. The transformation to the laboratory system by equations (6.7) is elementary.
It is obvious that for a complete solution of the collision problem, one must know the potential energy of interaction between the par- ticles U (r) and specify the initial conditions, so that all the integrals of motion may be determined. The angular-momentum integral is found in the following way. Fig. 8. which refers to repulsive forces,
Sec. 6] COLLISION OF PARTICLES 53
represents the motion of the first particle relative to the second. The path at an infinite distance is linear, because no forces act between the particles at infinity. Since the path is linear at infinity, it possesses asymptotes. The asymptote for the part of the trajectory over which the particles are approaching is represented by the straight line Al\ and FB is the asymptote for the part where they recede. The collision parameter. The distance of an asymptote from the straight line OG, drawn through the second point and parallel to the relative velocity of the particles at infinity, is called the collision parameter f'aimmg distance"). It has been denoted by p, since, as can be seen from Fig. 8, p Ls also the "arm" of the angular momentum. If there were 110 interaction between the particles, they would pass each other at a distance p ; this is why p is called the collision parameter. But we know that the angular momentum is very simply expressed in terms of p. In the preceding section it was shown that it is equal to ravp. Let us draw the radius vector OA to some very distant point A. Then the angular momentum is
M = mvr sin a (the angle a is shown on the diagram). But rsina = p, so that
(6.11)
Recall that here ra is the reduced mass of the particles and v is their relative velocity at infinity.
The energy integral is expressed in terms of the velocity at infinity thus :
since U (oo) = 0.
The deflection angle. The deflection angle x is equal to | re — 29 ,, | , where 9CJ is half the angle between the asymptotes. The angle 9 0 corresponds to a rotation of the radius vector from the position OA, where it is infinite, to a position OF, where it is a minimum. Hence, from equation (5.10) the angle 900 is expressed as
oo
•/•
M dr ,„ 1Qv
*n7aT "Tr^f-T7" --TrrT7o =-"--- v > (Q.Lt)
r0 is determined from (5.11). In place of M and $ we must substitute into (6.13) the expressions (6.11) and (6.12).
54
MECHANICS
[Part I
The differential effective scattering cross-section. Let us suppose that the integral (6.13) lias been calculated. Then 9^, and therefore x> are known as functions of the collision parameter p. Let this relation- ship be inverted, i.e., p is determined as a function of the deflection angle :
P = P(X)« (6-14)
In collision experiments, the collision parameter is never defined in practice; a parallel beam of scattering particles is directed with identical velocity at some kind of substance, the atoms or nuclei of which are scatterers. The distribution of particles as to deflection angles x (or> more exactly, as to angles 0 in the laboratory system) is observed. Thus a scattering experiment is, as it were, performed very many times one after the other with the widest range of aiming distances.
Let one particle pass through a square centimetre of surface of the scattering substance. Then, in an anuulus contained between p and p-Mp, there pass 2?rp dp particles. We classify the collisions according to the aiming distances, similar to the way that it is done on a shooting target with the aid of a concentric system of rings. If p is known in relation to x> then it may be stated that dcr= 2;rp dp =
= 27rp~— dx particles will be deflected at the angle between x
and
Let us suppose that the scattered particles are in some way detected at a large distance from the scattering medium. Then the whole scatterer can be considered as a point and we can say that after scattering the particles move in straight lines from a common centre. Let us consider those particles which occupy the space between two cones that have the same apex and a common axis ; the half -angle of the inner cone is equal to x> and the external cone x + ^X- The space between the two cones is called a solid angle, similar to the way that the plane contained between two straight lines is called a plane angle. The measure of a plane angle is the arc of a circle of unit radius drawn about the vertex of the angle, while the measure
of a solid angle is the area of a sphere of unit radius drawn about the centre of the cone. An elementary solid angle is shown, in Fig. 9 as that part of the surface of a sphere covered by an element of arc r/x when it is rotated about the radius OC. Since OC=\y the radius of rotation of the element dx is equal to sin x- Therefore, the surface of the sphere which it covers is equal to 2rc sin xdx* Thus, the elemen- tary solid angle is
Fig. 9
Sec. 6] COLLISION OF PARTICLES 55
(6.15)
The number of particles scattered in the element of solid angle is, thus,
da==p*£.J*<L. (6.16)
^ dx sinx V '
The quantity da has the dimensions of area. It is the area in which a particle must fall in order to be scattered within the solid-angle element d£l. It is called the effective differential scattering cross-section in the element of solid angle dQ.
Experimentally we determine just this value, because it is the angular distribution of the scattered particles that is dealt with [in (6.16) we consider that p is given in relation to xl- If there are n scatterers in unit volume of the scattering substance, then the attenu- ation of the primary beam J in passing through unit thickness of the substance, due to scattering in an elementary solid angle dii, is
d Jn = — Jndc= — Jnp ~ —. — particles/cm.
If we examine da as a function of x, we find a relationship between the collision parameter and the deflection angle. And this allows us to draw certain conclusions about the nature of the forces acting between the particle and the scattering centre.
Rutherford's- formula. A marvellous example of the determination of forces from the scattering law is given by the classical experiments of Rutherford with alpha particles. As was pointed out in Sec. 3, the Coulomb potential acting on particles decreases with distance
according to a — law, in the same way as the Newtonian potential.
Consequently, the deflection angle can be calculated from the equations of Sec. 5. Let us first of all find the angle <pco. It can be determined from equation (5.12) by putting r = oo, a > 0 (the charges on the nucleus and alpha particle are like charges). Hence,
The integrals of motion S and M are determined with the aid of (6.11) and (6.12). We therefore have
(6.18)
I since 9 = -*- — -f I . We now form the equation for the effective di
ential scattering cross-section in the centre-of-mass svstear wt the aid of (6.16):
66 MECHANICS [Part I
«*£.. (6.19)
If the scattering nucleus is not too light, this equation, to a good approximation, also holds in the laboratory system.
Thus, the number of particles scattered in the elementary solid angle d£l = 2n sin x ^X> *s inversely proportional to the fourth power of the sine of the deflection half-angle. This law is uniquely related to the Coulomb nature of the forces between the particles.
Studying the scattering of alpha particles by atoms, Rutherford showed that the law (6.19) is true for angles up to y, corresponding to collision parameters less than 10~12 cm. It was thus experimentally proved that the whole mass of the atom is concentrated in an ex- ceedingly small region (recall that the sixe of an atom is ^lO"8 cm.). Thus, experiments on the scattering of alpha particles led to the discovery of the atomic nucleus and to an estimation of the order of magnitude of its dimensions.
Isotropic scattering. As may be seen from equation (6.19), the scattering has a pronounced maximum for small deflection angles. This maximum relates to large aiming distances since particles passing each other at these distances are weakly deflected, while large distances predominate since they dciiiie a larger area. Thus, if the interaction force between the particles does not identically convert to zero at a finite distance, then, for small deflection angles, the expression for c/a will always have a maximum. This maximum is the more pronounced, the more rapidly the interaction force decreases with distance, for in the case of a rapidly diminishing force, large aiming distances correspond to very small deflection angles.
However, particles that are very little deflected can in no way be detected experimentally as deflected particles. Indeed, the initial beam cannot be made ideally parallel. For this reason, when investi- gating a scattered beam, one must always neglect those angles which are comparable with the angular deviation of the particles in the initial beam from ideal parallelism.
For a sufficiently rapid attenuation of force with distance, the region of the maximum of da in relation to the angle % can refer to such small angles that the particles travelling within these angles will not be distinguished as being scattered because of their small deflection angles. On the other hand, the remaining particles will be the more uniformly distributed as to scattering angle, the more rapidly the forces fall off with distance.
This can be seen in the example of particles scattered by an im- permeable sphere (Exercise 1). Such a sphere may be regarded as the limiting case of a force centre repulsing particles according to
the law UJir) = U0 1— J ; when n tends to infinity : if r < r0, then U (r)->oo,
Sec. 7] SMALL OSCILLATIONS 57
and if r >r0, then U (r)->0. When n== oo the scattering is completely isotropic. If n is large, the angular distribution of the particle is almost isotropic, and only for very small deflection angles has the distribution a sharp maximum. Hence, a scattering law that is almost isotropic indicates a rapid diminution of force with distance.
The scattering of neutrons by protons in the centre-of-mass system is isotropic up to energy values greater than 10 Mev (1 Mev equals 1.6 x 10~6 erg). An analysis of the effective cross -section shows that nuclear forces are short-range forces; they are very great at close distances and rapidly diminish to zero at distances larger than 2 x 10~13 cm. It must be mentioned, however, that a correct investi- gation of this case is only possible on the basis of the quantum theory of scattering (Sec. 37).
Exercises
1) Find tho differential effective scattering cross -section for particles by an impermeable sphere of radius r0.
The impermeable sphere can be represented by giving tho potential energy in tho form U (r) = 0 for r > r0 (outside the sphere) and U (r)^oo for r< r0 (inside tho sphere). Then, whatever tho kinetic energy of tho particle, penetra- tion into the region r <rQ is impossible.
In reflection from the sphere, the tangential component of momentum is conserved and the normal component changes sign. Tho absolute value of the momentum is conserved since tho scattering is elastic. A simple construction shows m that the collision parameter is related to the deflection angle by
p « r0 cos — , if p < rQ. Hence the general equation gives
so that the scattering occurs isotropically for all angles. Tho total effective scattering cross-section a is equal, in this case, to 7cr02, as expected. Note that if the interaction converted to zero not at a finite distance but at infinity, the total scattering cross-section would tend to infinity since to any arbitrarily large approach distance p there would correspond a certain deflection angle,
and the integral 2 :rp dp diverges. In the quantum theory of scattering,
a is also finite when the forces diminish fast enough with distance.
2) Tho collision of particles with masses ml and ra2 is considered (the mass of the incident particle is mj. As a result of the collision, particles with the same masses are obtained whose paths make certain angles 9 and >\> with the initial flight direction of the particle of mass m±. Determine the energy Q which is absorbed or emitted in the collision.
See. 7. Small Oscillations
In applications of mechanics, we very often meet a special form of motion known as small oscillations. We devote a separate section to the theory of small oscillations.
58
MECHANICS
[Part I
The definition of small oscillations of a pendulum. In the problem of pendulum oscillations in Sec. 4 it was shown that the equation relating the deflection angle 9 to time led, in the general case, to a nonelementary (elliptical) integral (4.11). A simple graphical in-
vestigation shows that the function 9 (t) is periodic. Fig. 10 shows the curve U (<p)=mgl (1 — cos 9), which gives the relationship be- tween potential energy and deflection angle. The horizontal straight line corresponds to a certain constant value of S. If $ < 2 mgl, the motion occurs periodically with time between the points — 90 and 90.
The problem is greatly simplified if 90^!* i-e-> ^e angle 90 is small in com- parison with a radian. Then cos 90 can
be replaced by the expansion 1 — -?"-. Since |9|<90» cos 9 can also
be replaced by 1 -- —-. After this the integral (4.11) can be easily evaluated :
lo
1/J f rf9 I
= — V— •—.—--- =-=11—
r 9 J > - 2 ' ^
9 arc cos — .
(7.1)
<Po
Inverting relation (7.1), we get the angle as a function of time
*. (7.2)
The result is a periodic function. As can be seen from (7.2), the period 9 is equal to 2n J/— . The quantity I/™ is called the frequency
of oscillation
(7.3)
This quantity gives the number of radians by which the argument of the cosine in (7.2) changes in one second. Sometimes the term frequency denotes a quantity that is 2 re times less and equal to the number of oscillations performed by the pendulum in one second.
The inverse value -^- , is the period of small oscillations of the pen- dulum. An important point is that the period and the frequency of small oscillations do not depend on the amplitude of oscillation 90. The general problem of small oscillations with one degree of freedom. In order to solve the small-oscillation problem, we need not, initially, reduce to quadrature the problem of arbitrary oscillations; we can first perform an appropriate simplification of the Lagrangian.
Sec. 7] SMALL OSCILLATIONS 59
First of all, we note that any oscillations, both large and small, always occur about a position of equilibrium. Thus, a pendulum oscillates about a vertical. On deflection from the position of stable equilibrium, a restoring force acts on the system in the opposite direction to the deflection. In the equilibrium position, this force obviously becomes zero (by definition of the "equilibrium" concept).
Force is equal to the derivative of potential energy with respect to the coordinate taken with opposite sign. The equilibrium condition written in terms of this derivative is
Let us denote the solution of this equation by q = q0- We assume, initially, that the system has only one degree of freedom and expand U in a Taylor series in the vicinity of the point qQ:
U(q)-U (q0) + ° (q-q0) + ±? (g-j.)' + . . . (7.5)
The linear term relative to q — q0 vanishes in accordance with (7.4). We denote (-0-5- ) ^J the letter (3. Then, confining ourselves to these terms of the series, we obtain
U(<D-^U(q») + -t(q-q0)*. (7.6)
The force near the equilibrium position is
F to) = ~ *dT = — P to — ?o) - (7.7)
For this force to be a restoring force (i.e., for it to act in the direction opposite to the deflection), the following inequality must hold:
This is the stability condition for the equilibrium ; the function U (q) must increase on both sides of the point q=qQ, It follows that the potential energy at that point must be a minimum. This is shown in Fig. 10 at 9 = 0.
Let us now examine the expression for kinetic energy. If, in the general formula for the kinetic energy of a particle,
we put x = x(q), y = y(q), z = z(q), then T reduces to the form
. +(dq
60 MECHANICS [Part I
The quantity in the brackets depends only on </; and so the kinetic energy of a particle can be represented in the form
T= J-a(?)j». (7.9)
Let us now expand the coefficient a (q) in a series, in terms of q~ qQf in the vicinity of the equilibrium position:
In order that the particle should not move far from the equilibrium position, its velocity must be small. In other words, the zero member
of the kinetic energy expansion -t>- a (ql}) q2 is already of the same order of smallness for small oscillations as tho second term in the expansion
Q
of U ((/), i.e., ^-(q — </0)2 . When q^q0 all the energy of oscillation
is kinetic, while for maximum deflection all the energy is potential.
i A
Therefore ^ a (q0) q* and (q — <?0)2 are of the same order of magnitude,
and the remaining terms in the series [including those containing (q — f/0)</2] can be neglected. We shall show that the mean values
of both the quantities .^a (</0) q2 and | (q — qQ)2 are the same after
we determine q as an explicit function of time.
In future, the coordinate q will be measured from the equilibrium position, i.e., we shall put </0=^0. Then [omitting U (0), which does riot affect the equations of motion] the Lagrangian can be written in the following form:
Thus, Lagraiige's equation will be written as
oc(0)§ + pgr = 0. (7.11)
Denoting
w - a(0) a(0) ' v>>i"'
we reduce (7.12) to the general form for the oscillation equation:
>2g = 0. (7.13)
Various forms for the solutions of small-oscillation problems. The
general solution of this equation, which contains two arbitrary constants, may be written in one of three forms:
Sec. 7] SMALL OSCILLATIONS 61
q = Cl cos co t + C» sin co£, (7.14a)
g = (7cos(co£ + y), (7.14b)
7 = Re{C"cfwl}. (7.14c)
The symbol Ke{} signifies the real part of the expression inside the braces. The constant C" inside the braces is complex: C' = Cl — iC2. The constants are chosen in accordance with the initial conditions. The constant y is called the initial phase, and C is the amplitude. If we are only interested in the frequency of small oscillations, and not the phase or amplitude, it is sufficient to use equation (7.12),
verifying that the second derivative I V-a-1 is positive.
A system which is described by equation (7.13) is called a linear harmonic oscillator.
It can be seen from equations (7.10), (7.12), and (7.14b) that the averages of the potential energy and kinetic energy of the oscillator during one period are the same because the averages of the squares of a sine or cosine are equal to one half:
sin* (<o* + Y) - sin2 (a* + y) dt = - ; cos* (at + y) =
Small oscillations with two degrees of freedom. We shall now con- sider oscillations with several degrees of freedom. As an example, Jet us first take the double pendulum of Sec. 3. If we confine ourselves to small oscillations, we must consider that the deflections 9 and <p are close to zero, (i.e., the pendulum is close to a vertical position). Then, by substituting the equilibrium values of the coordinates 9 and <];, cos (9 — ^) in the kinetic energy must be replaced by cos 0 = 1 as in the problem of oscillations with one degree of freedom where cos 9 and cos <L, in the expression for potential energy, must be re-
2 t 2
placed by 1 -- |- and 1 -- ^- . Then the Lagrangian will have the form
L = , V + L Hy + mMty _ _ i. lg(?2 _ ^
(7.15) Let us examine this in a somewhat more general form:
U (0) —
62 MECHANICS [Part I
Here, the coefficients an, a12, and a22 are assumed to be constant numbers expressed in terms of the equilibrium values gx and q2. Comparing (7.15) and (7.16), we find that in the problem of the double pendulum
an = (m + raj) I2 , a12 = mlll1, a22 = ml l\ ;
In the general case, the coefficients pn. p12 and (322 are expressed by the equations
where the derivatives are also taken in the equilibrium positions. For the equilibrium to be stable, we must demand that the following inequality be satisfied:
^(?)— ^(0) = Y(Pii«J + 2p12?lgt + paag;)>0. (7.17)
Under this condition, U has a minimum at the point ^ = 0, q^ = 0. Let us rewrite the left-hand side of (7.17) in identical form
This expression remains positive for all values of ql and q2, provided the coefficients of both quadratics in q are greater than zero:
Pn>0, (7.18)
(7.19)
In future, we shall consider that the conditions (7.18) and (7.19), together with analogous conditions for an, a12, and oc22, are satisfied. We shall now write down Lagrange's equations. We have
Whence
a ar = °» 1 (7
In order to satisfy these equations, we shall look for a solution in the form
As in (7.14c), the real part of the solution (7.21) must be taken.
Sec. 7] SMALL OSCILLATIONS 63
The equation for frequency. Substituting (7.21) in (7.20), we obtain equations relating A1 and Az:
" ' * \ (7.22)
/ Q *,*x2\/*l/n ..9.\yl /^l ^ '
(p12 — a12co } £
Transferring terms in A2 to the right-hand side of the equation and dividing one equation by the other, we eliminate Al and A%:
Reducing (7.23) to a common denominator, we arrive at the bi- quadratic equation
(<xna22 — afa) to4 — ( Pna22 + £22*11 — 2a12(i12) to2 +
+ PiiP22— Ma = 0. (7.24)
Substituting here the expressions for a,*, (3ifc from (7.15), we obtain an equation for the frequencies of a double pendulum
If we introduce still another contraction in notation (for the given problem) -~- — \ — - = fji, the expression for frequencies will be of the following form:
It is easy to see that this expression yields only the real values of the frequencies. However, we shall show this in more general form for equation (7.24). Let us assume that the following function is given:
F ( co2) = (aua22 — of,) co* — ( pna22 + p22an — 2p12a12) <»* + PiiP22 — Pfi >
which passes through zero for all values of G> that satisfy equation (7.24). F (co2) is positive for to2 = 0 and for to2 = 00, since Pnp22 — — Pi2 > ®> ana22 — ai2 > ^- Let us now substitute into this function
Q
the positive number co2=~1-. After a simple rearrangement we obtain
P22 - Pl2 «22)2 < 0 .
Thus, as co2 varies from 0 to oo, F (co2) is first positive, then negative, and then again positive. Hence, it changes sign twice, so that equation (7.24) has two positive roots coj, cojj and, as was asserted, all the values for frequency are real.
64 MECHANICS [Part I
The quantity co has four values, both pairs of which are equal in absolute value. If we represent the solution in the form (7.21), it is sufficient to take only positive G>.
Normal coordinates. Let us put these roots in (1.22). To each of
^
them there will correspond a definite ratio of the coefficients -~- . For
Ai
i — I, 2 we have
n
' «•
A "
Pl2
According to (7.23), the same ratio is also obtained from the second equation of (7.22). For example, for the double pendulum
6i?
£;=• //^A ? ; * i's equal to 1 or 2 depending on the sign in front
of the root in the solution for co2.
Each frequency co; defines one partial solution of the system (7.20). Since the system is linear, the general solution is the sum of these particular solutions. Let us write this as
*' . (7.26)
We must, of course, take only the real parts of the expressions on the right.
We now introduce the following notation:
4(JV«i' == Qi , A<? €*<**' = Qt . (7.27)
According to (7.27), the quantities Qi and Q2 satisfy the differential equations
Qj + co^-O; (32 + col92-0. (7.28)
Each of these equations can be obtained from the Lagrangiaii
Lt=*2Q1—l'<*1<K, (7.29)
which describes oscillation with one degree of freedom.
Thus, in terms of the variables Qi, the problem of two related oscillations with two degrees of freedom q^ q2 has been reduced to the problem of two independent harmonic oscillations with one degree of freedom Ql and Q2. The coordinates Ql and Q2 are termed normal.
In equations (7.20), we cannot arbitrarily put 31 = 0 or g2 = 0: if the quantity ql oscillates, then it must cause q2 to oscillate. In contrast, the oscillations of the quantities Ql and Q2 are in no way related [as long as we limit ourselves to the expansion (7.16) for L].
Sec. 7] SMALL OSCILLATIONS 65
From equations (7.26), we can express Qi and Q% i*1 terms of ql and q2 :
If, for example, we choose the initial values of q and q so that at this instant Q± = 0 and Q± = 0, then the oscillation with frequency G^ will riot occur at all. For this it is sufficient, at t = 0, to take the co- ordinates and velocities in a relationship such that £2 9i — #2 == 0 and £2£i — ?2==^- In °ther words, only the frequency co2 will occur, and the oscillations will be strictly periodic. When both frequencies Oj and co2 are excited they are generally speaking incommensurable, i.e., their ratio cannot be expressed as a rational fraction), the oscil- lation q is no longer periodic, since the sum of two periodic functions with incommensurable periods is not periodic.
Expressing energy in normal coordinates. From the form of the Lagrangian (7.29) it can be immediately concluded that the expression for energy in normal coordinates reduces to the form
-h <*>?#?), (7.31)
since L = T — U and (^ = T+U. This result is true for small oscil- lations with any number of degrees of freedom.
We must note that if the normal coordinates are expressed directly
by equations (7.30), then the separate energy terms -^ (Qf + ^fQi), will also be multiplied by certain numbers a,-. However, if we replace Qi by Qi Vo7, then these numbers are eliminated from the expression for energy, which is then reduced to the form (7.31). An example of this procedure is given in the exercises.
Thus, the energy of any system performing small oscillations is reduced to the sum of the energies of separate, independent linear harmonic oscillators. As a result of this, consideration of oscillation problems is greatly simplified since the linear harmonic oscillator is, in many respects, one of the most simple mechanical systems.
The reduction to normal coordinates turns out to be a very fruitful method in studies of the oscillations of polyatomic molecules, in the theory of crystals, and in electrodynamics. In addition, normal coordinates are useful in technical applications of oscillation theory.
The case ol equal frequencies. If the roots of equation (7.24) coincide, the general solution must not be written in the form (7.26), but somewhat differently, namely,
ql = A cos co£ 4-B sin co£, )
At TV - \ (7-32)
q2=A cos co£ + B sin co£ . J
5-0060
66 MECHANICS [Part I
Four arbitrary constants appear in this solution, and this is as it should be in a system with two degrees of freedom.
An example of such a system is a pendulum suspended by a string instead of a hinge. In the approximation (7.32), it turns out that the pendulum describes an ellipse centred about the equilibrium position. Account taken of the subsequent terms in the expansion of the potential energy in powers of deflection shows that the axes of the ellipse do not remain stationary, but rotate.
Exercise
Find the natural frequencies and normal oscillations of a double pendulum, taking the ratios of load masses y. = 3/4 and the rod lengths X = 5/7.
From the equation for the oscillation frequencies of a double pendulum,
we obtain <o* = ~ ~r » <°a = T^ ~r • Further, ^ - — 7/3, £2 = 7/5.
L I 1U I
Let us now write down the expression for kinetic energy. For simplicity, we write l = g~ m—l so that only ratios of X and JJL will appear in all the equa- tions. This gives ocn — 1 -f (Jt —7/4, a12 = (xX — 15/28, a22 = jiX2 = 75/196; pn — 1 +jjt = = 7/4, p12 = 0, p22 = (jiX— 15/28. Let us determine the coefficients a,-. To do this we must calculate the kinetic energy
7 • 2 15 •
4 l 2 14 l
V5"
Consequently, wo must put at = — -~ , a2 = 1/2 .
Denoting — -- - and —^- again by the letters <?i and Q2, we have the
A £i
expression for potential energy
as it should be according to (7.10). The generalized coordinates are related to the normal coordinates by
5\/3"/7
Thus, if 7 9 = — 3 <];, and 7 9 = — 3 \ initially, then we have Q2 = 0 for all time, so that both pendulums oscillate with one frequency colf with the constant relationship between the deflection angles 7 9 — — 3 ^ holding all the time. Both pendulums are deflected to opposite sides of the vertical. The other normal oscillation, with frequency w2, occurs for a constant angular relationship 7 9 = 5 4>.
Sec. 8. Rotating Coordinate Systems. Inertial Forces
The equivalence of inertia! coordinate systems. The particular significance of inertial coordinate systems in mechanics was pointed out in Sec. 2. In such systems, all accelerations are produced by
Sec. 8] ROTATING COORDINATE SYSTEMS INERTIAL FORCES 67
interaction between bodies. It is impossible to find a strictly inertial system in nature (any system is noninertial if the motions of bodies in it are observed over a sufficiently long period of time).
In the exercise at the end of this section we shall consider the Foucault pendulum, whose plane of oscillation rotates with a speed depending only on the geographical latitude of its location. This rotation cannot be explained by an interaction with the earth, because the gravitational force cannot make the pendulum rotate from east to west instead of from west to east. * However, if we consider several oscillations, then the rotation of the plane is still insignificant and can be ignored. Then it is sufficient to consider that gravity alone is acting on the pendulum and that the coordinate system fixed in the earth is approximately inertial over a period of several oscil- lations.
The concept of an inertial system is meaningful as an approximation and is a very convenient idealization in mechanics. In such a co- ordinate system, the interaction forces are measured by the acceler- ations of the bodies.
Let a coordinate system be defined for which it is known that, to the required degree of accuracy, it can be regarded as inertial. Then another coordinate system, moving uniformly relative to it, is also inertial within the same degree of accuracy. Indeed, if all the accelerations in the first system are due to interaction forces between bodies, then no additional accelerations can appear in the second system either. Therefore, both systems are inertial. Either of them may be considered at rest and the other moving, since motion is always relative.
The principle of relativity. One of the basic principles of mechanics is that all laws of motion have an identical form in all inertial co- ordinate systems, since these systems are, physically, completely equivalent. This principle of the equivalence of all inertial systems is known as the relativity principle, for it is connected with the relativity of motion.
It should be noted that this in no way signifies that inertial and noninertial coordinate systems are equivalent: in the latter, not all the accelerations can be reduced to interaction forces, so that there is no physical equivalence between two such systems.
Mathematically, the principle of relativity is expressed by the fact that equations of motion for one inertial system preserve their form after the variables have been transformed to another inertial system.
The equations for the transformation from one inertial system to another can be obtained only on the basis of certain physical
* In this case the plane of oscillation must pass through the vertical, since, otherwise, the pendulum would have an initial angular momentum relative to the vertical and would describe an ellipse whose semiaxes rotate (see end of Sec. 7).
68 MECHANICS [Part I
assumptions. In Newtonian mechanics it is always taken that the interaction forces between bodies, in particular, gravitational forces, are transmitted instantaneously over any distance. Thus, the dis- placement of any body immediately transmits a certain momentum to any other body, no matter where it is located. As a result, a clock located in a certain inertial system can be instantaneously synchronized with a clock moving in another inertial system. Thus, in Newtonian mechanics, time is considered universal. In transforming from one inertial system to another (the latter with a velocity F relative to the former) it is taken that the time t is the same in both systems.
Later on we shall see that this assump- tion is approximate and holds only when the relative velocity of the systems is considerably less than the velocity of light.
*i The Galilean transformation. Let us
I , construct coordinate systems in two -j — x inertial frames of reference such that their abscissae are in the direction of Fig. 11 the relative velocity V and the other
coordinate axes are also mutually parallel.
Then from Fig. 11 it will be immediately seen that the abscissa of point x in the system which we shall call stationary is related to the abscissa in the moving system by the simple relation
x=x' + Vt, (8.1)
provided that the origins coincided at the instant t — Q. The co- ordinate construction does not impose any limitations on the generality of the transformation equations. The remaining transformations lead simply to the identities
y = y', z=z'. (8.2)
The relationship t — t' is a hypothesis which is correct only for values of V considerably less than the velocity of light (Sec. 20).
Condition (8.1) is absolutely symmetrical with respect to both inertial systems: if we consider that the one in which the variables are primed is stationary and the other in motion, (8.1) retains the same form; one should, of course, replace F by — F. In the given case, symmetry exists because t' —t. If t^£t', the transformation equations x — x' + Vt and x' ~x — Vt' would contradict each other. But it would seem that equation (8.1) is obtained, quite obviously, from Fig. 11. Thus, if we do not consider time as identical for all inertial systems, the mathematical formulation of the relativity principle should be more complicated than that obtained on the basis of equation (8.1); and, we must definitely give up this "obvious-
Sec. 8] ROTATING COORDINATE SYSTEMS INERTIAL FORCES 69
ness," which is so rooted in our everyday experience with velocities that are small compared with the velocity of light.
The equations of Newtonian mechanics involve, on the right-hand side, the forces of interaction between particles. These forces depend on the relative coordinates of the particles and, for this reason, they do not change 'with transformation (8.1), since Vt is cancelled in the formation of differences between the coordinates of any pair of particles. The left-hand sides of the equations contain accelerations, i.e., the second derivatives of the coordinates with respect to time. But since time enters linearly in (8.1) and is the same in both systems, x=x'. Thus, the equations of mechanics are of iden- tical form in any inertial frame of reference.
To summarize, the equations of mechanics do not change their form when the variables undergo trans- formations (8.1). In other words, it is common to say that the equations of mechanics are invariant to these transformations, which are usually called Galilean transformations .
The constancy (invariance) of mechanical laws under Galilean transformations is the essence of the relativity principle of Newtonian mechanics.
Here we must bear in mind that the relativity principle, which expresses the- equivalence of all iner- tial coordinate systems, expresses a far more general law of nature than the approximate equations of transformation (8.1), (8.2). The extension of the relativity principle to electromag- netic phenomena involves the replacement of these equations by more general ones, which reduce to the former equations only when all velocities are much smaller than that of light.
Rotating coordinate systems. Several new terms appear in the equations of mechanics when transforming to rotating coordinate systems. Let us first obtain the equations for this transformation.
In Fig. 12 the axis of rotation is represented by a vertical line. The origin 0 is on the axis of rotation. Let r be the radius vector of a point A rotating around the axis. Then, for a rotation angular velocity co (radians per second) the linear speed of the point will be
v = co • r sin a , (8.3)
since the radius of rotation is p —r sin a (see Fig. 12). Let the rotation be anticlockwise. If point A lies in the plane of the paper, then the velocity v is perpendicular to the plane of the paper and directed towards the back of the paper. This permits us to obtain a relationship between the linear and angular velocities in vector form. We represent the angular velocity by a vector directed along the axis of rotation and associated with the direction of rotation by the corkscrew rule.
70 MECHANICS [Part I
Then, if the rotation occurs in an anticlockwise direction, the vector to is directed upwards from the paper. From this it follows that
v=[cor]. (8.4)
This expression ensures a correct magnitude and direction for the linear velocity of the point.
Let us assume that point A, in addition to rotation, is somehow displaced relative to the origin 0 with velocity v' = r. The resultant velocity of the point relative to a noiirotating system will be represented as the sum v' + v. The kinetic energy of the point relative
to the nonrotating system is -«- (v + v')2, and the Lagrangian is £ = -£ (v + v')2-C7(r) = ^(v' + [cor])2-J7(r). (8.5)
Let us now write down Lagrange's equations for motion relative to a rotating system, i.e., considering r a generalized coordinate.
In order to do this we must calculate the derivatives -^- and-0— :
or or
let it be noted that differentiation with respect to a vector denotes a shortened way of writing down the differentiation with respect to all of its three components. The general rules for such differentiations will be given in Sec. 11; here we shall calculate the derivatives for each component separately.
Let <o be along the direction of the z-axis. Then, in vector com- ponents, L will be of the form
L= - [(i-cot,)2 + (y + coo:)2 + z2] — U(x,y9z) . (8.6)
Whence we obtain
dL .. v dL /. , v dL
dL /• , v 8U dL .. . dU
— = m<*(y+<*x) — -^-, — = — mco (x — coy) — -^-
dL __ dU
dz ~~ dz *
Lagrange's equations in component form appear thus:
f\ T^ 7/1(0; — <oi/) — mco (y + cox) — may + -g — = 0 ,
o 7"7
m (y + coi) + mo (x — co?/) + m&x + -j— = 0 ,
Sec. 8] KOTATING COORDINATE SYSTEMS INERTIAL FORCES 71
Let us leave on the left only the second derivatives and rewrite the last three equations as a single vector equation:
m[co[r<o]]— --. (8.7)
Expanding the double vector product on the right by means of the equation [A [BC]] = B (AC) — C (AB), and transforming to com- ponents, we can see that (8.7) is equivalent to the preceding system of three equations. A direct differentiation with respect to the vectors r and f would have led to (8.7), without the expression in terms of components.
Inertial forces. The first three terms on the right in (8.7) essentially distinguish the equations of motion, written relative to a rotating coordinate system, from the equations written relative to a non- rotating system.
The use of a noninertial system is determined by the nature of the problem. For example, if the motion of terrestrial bodies is being studied, it is natural to choose the earth as the coordinate system, and not some other system related to the Galaxy (the aggregate of stars in the Milky Way). If we consider the reaction of a passenger to a train that suddenly stops, we must take the train as frame of reference and not the station platform. When the train is braked sharply, the passenger continues to move forwards "inertially" or, as we have agreed to say, he continues to move uniformly relative to an inertial system attached to the earth. Thus, relative to the carriage, it is the familiar jerk forward. At the same time it is obvious that the noninertial system is the train and not the earth, since no one experiencies any jerk on the platform.
The additional terms on the right of equation (8.7) have the same origin as the jerk when the train stopped; they are produced by noninertiality (in the given case, rotation) of the coordinate system. Naturally, the acceleration of a point caused by noninertiality of the system is absolutely real, relative to that system, in spite of the fact that there are other, inertial, systems relative to which this acceleration does not exist. In equation (8.7) this acceleration is written as if it were due to some additional forces. These forces are usually called inertial forces. In so far as the acceleration associated with them is in every way real, the discussion (which sometimes arises) about the reality of inertial forces themselves must be con- sidered as aimless. It is only possible to talk about the difference between the forces of inertia and the forces of interaction between bodies.
But if we consider the force of Newtonian attraction, we cannot ignore the striking fact that, like the forces of inertia, it is proportional to the mass of the body. As a result of this, the equations of mechanics can be formulated in such a way that the difference between gravi-
72 MECHANICS [Part I
tational forces and incrtial forces does not at all appear in the equations ; all these forces turn out to be physically equivalent. However, this formulation is, of course, connected with a re-evaluation and a substantial revision of the basis of mechanics. It is the subject of Einstein's general theory of relativity, which is discussed in somewhat more detail at the end of Sec. 20.
Coriolis force. Let us now consider in more detail the inertial forces appearing in (8.7), which are due to a rotating coordinate system.
The first term in (8.7) occurs as a result of nonconstancy of angular velocity. It will not interest us. The second term is called the Coriolis force. For a Coriolis force to appear, the velocity of a point relative to a rotating coordinate system must have a projection, other than zero, on a plane perpendicular to the axis of rotation. This velocity projection can, in turn, be separated into two components: one, perpendicular to the radius drawn from the axis of rotation to the moving point, and the other, directed along the radius. The most interesting, as to its action, is the component of the Coriolis force due to the radial component of velocity. It is perpendicular both to the radius and to the axis of rotation. If a body moves perpen- dicularly to a radius, then its Coriolis acceleration is radial, and therefore analogous in its action to the centripetal acceleration which will be considered a little further on.
We note that the Coriolis force cannot be related, even formally, to the gradient of a potential function U.
There are many examples of the deflecting action of the Coriolis force in nature. The water of rivers in the Northern Hemisphere which flow in the direction of the meridian, i.e., from north to south, or from south to north, experience a deflection towards the right- hand bank (if we are looking in the direction of flow). This is why the right-hand bank of such rivers is steeper than the left. It is easy to form the corresponding component of the Coriolis force. The angular-velocity vector of the earth's rotation is directed along the earth's axis, "upwards" from the north pole. The waters of a river, which flows southwards at the mean latitudes of the Northern Hemi- sphere, have a velocity component perpendicular to the earth's axis and directed away from the axis. This means that the Coriolis acceleration of the water, relative to the earth, is in a westerly direction or, relative to a river flowing southwards, to the right. If the river flows in a northerly direction, the deflection will be towards the east, i.e., again to the right. In the southern hemisphere the deflection occurs leftwards.
The warm Gulf Stream which flows northwards is deflected towards the east, which is of tremendous importance for the climate of Europe. In general, the Coriolis force considerably affects the motion of air and water masses on the earth, though when compared in magnitude with the gravitational force it is very insignificant. Indeed, the angular
SeC. 9] THE DYNAMICS OF A RIGID BODY 73
velocity of the earth, as it completes one rotation about its axis in 24 hours, is a little less than 10~4 rad/sec, while the velocity of a particle of water or air can be taken as having an order of magnitude of 102 cm/sec. From this the Coriolis acceleration has an order of magnitude of 10~2 cm/sec2, which is one hundred thousand times less than the acceleration caused by the force of gravity.
The Coriolis force also causes the rotation of the plane of oscillation of a Foucault pendulum. With the aid of the Foucault pendulum, we can prove the rotation of the earth about its axis without astro- nomical observations. In a nonrotating system, the plane of oscillation must be invariable in accordance with the law of conservation of angular momentum.
Centrifugal force. The third vector term in equation (8.7) is the usual centrifugal force. Indeed, it is perpendicular to the axis of rotation and, in absolute value, is equal to
| m [d/[tor]] | = mto | [cor] | — mi* (cor sin a) = mco2r sin a . (8.8)
Here, the first equality takes account of the fact that the vectors to and [tor] are perpendicular to each other, so that the absolute value of the vector product is equal to the product of their absolute values. But r sin a is equal to the distance from the axis of rotation, so that this force satisfies the usual definition of a centrifugal force.
Exercise
Let us consider the rotation of the plane of oscillation of a Foucault pendulum under the action of the earth's rotation about its axis.
The axis Ox at a given point on the earth is drawn in a northerly direction and the axis Oy in an easterly direction. Then, if <*B= <*> sin 0, where 0 is the latitude of the locality, we have the equation of motion
x= — 6)2#_2£cDB, y = _ .coj?/ + 2£toB, co2 = J7
Multiplying the first equation by y and the second by x and then sub- tracting, we got
-*-(yx-xy) = - A(yi+ si),^ .
Integrating and transforming to polar coordinates (x = r cos 9, y ~ r sin 9) :
r*v=r*<*B. Whence, after cancelling the r2's, we have
9 = WB = <o sin 0 , which gives the angular velocity of rotation of the piano of oscillation.
Sec. 9. The Dynamics of a Rigid Body
The dynamics of a rigid body is a large independent chapter of mechanics and is very rich in technical applications. Our aim is to give only a brief account of the basic concepts of this branch
74 MECHANICS [Part I
of mechanics inasmuch as it contains instructive examples of general laws. In addition, certain mechanical quantities that characterize a rigid body are necessary for an understanding of molecular spectra.
The kinetic energy of a rigid body. As was shown in Sec. 1, a rigid body has six degrees of freedom. Three of them relate to the trans- lational motion of the centre of mass of a body in space. The re- maining three degrees of freedom correspond to rotation (relative to this centre of mass).
In Sec. 4, it was shown that the kinetic energy of a system consists of the kinetic energy of the motion of the whole mass of the body concentrated at the centre of mass, and the kinetic energy of the relative motion of the separate particles of the system. In the case of a rigid body, relative motion reduces to rotation with the value of angular velocity co the same for all particles. Naturally, both the magnitude and the direction of to may vary with time.
Let us calculate the kinetic energy of rotation of a rigid body. In the general case, the density p of the body may not be uniform over the whole volume of the body, and may depend on the co- ordinates: p — p (#, y, z) = p (r). The mass of an element of volume dV is equal to dm~p dV. The velocity of rotation v is, from (8.4), [cor], Therefore, the kinetic energy of the volume element is equal to
y p [cor]2dF. The kinetic energy of the whole body is represented by the integral of this quantity with respect to the volume
(9.1)
Expressing the square of the vector product in terms of the compo- nents co, we have
[cor]2 = co2r2 sin2 a = co2r2 — co2r2 cos2 a = co2r2 — (cor)2. Here a is the angle between co and r. But
(cor)2 = (cojctf + coy«/ + co*z)2 =
Since the body is rigid, the components co*, coy, co* can be taken out of the volume integral. Combining terms which are similar in the components co, we obtain for T:
9xzdV — — coy co* J 9yz dV . (9.2)
Sec. 9] THE DYNAMICS OF A RIGID BODY 75
Moments of inertia. All the integrals appearing in (9.2) depend only on the shape of the body and its density distribution, and do not de- pend on the motion of the body (in a coordinate system fixed in the body). We denote them as follows:
=/ p (yz + *2) dV, Jxr=-j pxydV,
(9.3)
The quantities with the same indexes are called moments of inertia, while those with different indexes are called products of inertia. In the notation of (9.3), the kinetic energy has the form
T = — (JxxCOx + «/yyCOy + JzztoZ + 2Jxy6>xCOy + 2Jxztoxtoz + 2jy^COyCO^) .
(9.4)
With the aid of the summation convention used in Sec. 2, when eval- uating Lagrange's equations the kinetic energy can be written in the following concise form :
T^-^J «£<**<*£.
Principal axes of inertia. Let us suppose that Oxyz is a coordinate system fixed in a body. In this system all the quantities Jxx , . . . , Jyz are constant. Let us take another coordinate system Ox'y'z' which is also fixed in the body. The old coordinates of any point are expressed in terms of its new coordinates by the well-known formulae of analyti- cal geometry:
x = x' cos /. (#', x) + y' cos /_ (yr, x) + z' cos /_ (z', x) , y = x' cos L (#', y) + y' cos /_ (yf, y) + z' cos L (*', y) , z = x' cos L (x', z) + y' cos /. (y1, z) + z' cos /_ (zf, z) ,
or, if we denote cos < (xa'x$) by the symbol A^*, then, with the aid of the summation convention
*^3 ==*^<x -"•oc3*
The same formulae are used to express also the components of any vector, and in particular 03, relative to the old axes, in terms of the components coa' relative to the new axes.
Let us substitute these expressions into the kinetic energy (9.4) and collect the terms containing the products eoj/fcV, co^'co/, toy'ci)/ and
76 MECHANICS [Part I
the squares to*'2, o>y'2, to/2. We shall now show that we can always rotate the coordinate axes so that the coefficients of the new products tox'coy', cox'co*', aVco/ become zero. Indeed, any rotation of the coordinate system can be described with the aid of three independent parameters, for a coordinate system is like an imaginary rigid body and its posi- tion in space is defined by the three angles of rotation (see Sec. 1). These three angles can be chosen so that the sums of the products of the cosines of the angles between the axes, for tox'coy', co^'ox*' and coy'co*' become zero. The remaining expressions for co*'2, coy'2, and to*'2 will be called Jl9 J2, «/3, so that
The kinetic energy is written in the following form in the new coordi- nate axes:
T - \ (J>? + J*<*\ + Js<»t). (0.5)
These axes are called the principal axes of inertia of the body, they can be defined relative to any point connected with the body. By defi- nition, the products of inertia convert to zero in the principal axes of inertia. The moments of inertia in the principal axes are called princi- pal moments of inertia. They are denoted by Jl9 J2, J3.
The angular momentum of a rigid body. Let us now calculate a pro- jection of the angular momentum of a rigid body. From the definition of angular momentum we obtain
M* = J p [rvMF = J p [r [tor]]xdF - J p (to.vr2 — x (cor)) dV = ^ cox Jp(2/2 + z2)dV — coy f pxydV — <o* f pxzdF =
= Jxx CO* + + Jxy C0y + Jxz 6)z (9.6)
or, in shortened form,
Comparing (9.6) and (9.4), we see that
^T1
*—•£;'
My and Mz appear analogous. In vector form, we may write
(9.8)
Equations (9.7) and (9.8) again express the fact that the angular momentum is a generalized momentum related to rotation. In this sense, (9.7) corresponds to (5.4). The only difference is that the com- ponents to are not total time derivatives of some quantities. This will
Sec. 9] THE DYNAMICS OF A RIGID BODY 77
be shown a little later in the present section. In that sense, to*, in (9.7), is not altogether similar to 9 in (5.4).
If the coordinate axes coincide with the principal axes of inertia, then the expression for angular momentum is even simpler than (9.6):
AT
*i = ls: = Ji«i (9.9)
and similarly for the other components.
Moment of forces. Let us now find equations which describe the variation of angular momentum with time. The derivative of angular momentum of a particle is
where the first term becomes zero since r and p are parallel. Integrating this equation over the volume of the rigid body and taking advantage of the additive property of angular momentum, we have
K. (9.10)
The right-hand side of (9.10), which we denote by K, is called the resultant moment of the forces applied to the body. If F is the gravi- tational force (which occurs in the majority of cases) then K can also be written as
= — Jp<7[rz0]dF,
where z0 is the unit vector in a vertical direction. But since the vector z0 is a constant, it should be put outside the integration sign:
K = [z0, JpgrrdFJ,
If the body is supported at its centre of mass, then, by the definition of centre of mass, the integral for all three projections pr will be zero. Then K = 0 and the total angular momentum will be conserved. This occurs in the case of a gyroscope.
For the conservation of angular momentum of a rigid body it is sufficient that K = 0; but for any arbitrary mechanical system, angular momentum is conserved only when there are no external forces.
Euler's equations. Equation (9.6) gives a relationship between M and co. The quantities «/**, . . ., Jyz are constant only in a coordinate system fixed in the rigid body itself. If we write equation (9.10) for a stationary coordinate system, then, differentiating M with respect to time, we must also find the derivatives of Jxx, . . . , JYZ with respect to time, which is very inconvenient. Therefore, it is preferable to
78 MECHANICS [Part I
transform the equation to a coordinate system fixed in the body, taking into account the accelerated motion of that system. The varia- tion of the vector M relative to the moving axes consists of two com- ponents : one is due to the variation of the vector itself, while the other is due to the motion of the axes onto which it is projected. For the vector M this variation is equal to [u>M], similar to the way that it was equal to [cor] for the radius vector r in Sec. 8. When the coordinate system is rotated, any vector varies like a radius vector.
Let the coordinate axes be taken in the direction of the principal axes of inertia. Obviously, the moments of inertia relative to these coordinates are constant. For this reason, the time derivative of Ml = Jl<f)1 is
Ml = J± c^ + [coM]! = J1vl + o>2 M3 — co3 M2 = Jjco,, + (J3 — J2) co3o>2 .
Equating this expression to the magnitude of the projection of the moment of force on the first axis of inertia, and doing the same for the other axes, we obtain the required system of equations
(9.11)
These equations were obtained by L. Euler and are named after him. They can be reduced to quadrature for any arbitrary values of integrals of motion in the following cases :
1) JK1 — K2~K3 = 0 (point of support at the centre of mass) for arbitrary values of the moments of inertia;
2) J2 = J3 ^ Jj and the point of support lies on the axis of symmetry, relative to which two moments of inertia are equal. This is the so- called symmetrical top.
For more than a hundred years, no other case of a solution of system (9.11) by quadratures was known. Only in 1887 did S. V. Kovalevskaya find another example (see G. K. Suslov, Theoretical Mechanics, Gostekhizdat, 1944). Kovalevskaya showed that the three listed cases exhaust all the possibilities of integrating the system (9.11) by quadra- tures for arbitrary constants (integrals) of motion.
A free symmetrical top. All three cases, and in particular the Kova- levskaya case, are very complicated to integrate. Therefore, we shall only consider the simplified first case, when J2 = J3 (a free symmetrical top).
From the first equation of (9.11), it immediately follows that a>1== const. For brevity, we write the value
Q. (9.12)
The second two equations of (9.11) are written thus:
Sec. 9]
THE DYNAMICS OF A RIGID BODY
79
<J>2 4- £}<o3 = 0 , o>3 — £ico2 = 0. (9.13)
Equations (9.13) are easily integrated if we represent the components co2 and co3 in the following form:
= coj^ cos fit , co3 —
(9.14)
Here, o>| + co| = co^ is a constant quantity. Thus, the angular- momentum projection on the axis of symmetry and the sum of the squares of the angular-momentum projections on the other two axes are conserved. This means that the angular-momentum vector rotates about the axis of symmetry, i. e., the first axis of inertia, with angular velocity fl; the vector makes with it a constant angle, the tangent
of which is -^~ . This is the situation in a system of moving axes.
Of course, in a system of stationary axes, the total angular momen- tum is conserved in magnitude and direction, since the resultant moment of force is equal to zero. In this system, the axis of symmetry of the top rotates about the angular-momentum direction making a constant angle with it. Such motion is called precession. Pre- cessional motion is only stable for rela- tively small external perturbations. The stabilizing action of gyroscopes is based on this principle.
Eulerian angles. We shall now show how to describe the rotation of a rigid body with the aid of parameters which specify its position. Such parameters are the Eulerian angles shown in Fig. 13. The figure depicts two coordinate systems: a fixed system Oxyz and a system Ox'y'z' fixed in the rigid body. It is most convenient to take x', y', z' along the principal axes of inertia through the point of support. Then the Eulerian angles are:
& is the angle between the axes z and z',
9 is the angle between the line OK of intersection between the planes xOy and x'Oy' and the #'-axis,
fy is the angle between the line OK and the #-axis.
If the angle fy varies, then the angular- velocity vector <J/ is directed along the axis Oz since that vector is perpendicular to the plane of
angle of rotation ^. Thus 9 must be taken along the axis Oz' and & along the line OK.
Let us now express the angular-velocity projections (i. e., &19 co2, co3,) onto the principal axes of inertia in terms of the generalized ve- locities 4», <p, &•
80 MECHANICS [Part I
co3 is the projection of the angular velocity on the axis Ozf (zr is the third axis). As was shown, 9 is projected exclusively on this axis and the projection of fy is equal to ^ cos &, since # is the angle between the axes Oz and Ozr. Hence,
0)3—9 -f^cosfl- . (9.15)
In order to find the projections of the angular velocity on the other two axes, we draw a line OL which lies in the plane x'Oy' and is per- pendicular to OK.
From Fig. 13 it can be seen that
/_LOx' = -J— 9 and /_zOL = ^ + a,
since the straight line OL lies in the plane zz' ', as do all lines perpendic- ular to OK. The projection of fy on OL is equal to — A sin &, and the projection on Ox' is equal to — ^ sin ^ cos iy — 9)= — <j> sin &
sin 9. The projection of ^ on Oy' is ^ sin -9- cos 9. The projection of & on Ox' and Oy' can be directly found by means of the diagram ; they are £ cos 9 and 9- sin 9. The result is therefore
co! — & cos 9 — ^sin#sin9 , (9.16)
(9.17)
From equations (9.15), (9.16), and (9.17) it will be seen that co1? co2 and co3 are not total time derivatives of any quantities and, in that sense, do not exactly agree with the usual notion of generalized velocities (as do 9, ij;, $-).
If we substitute into (9.5) the expressions for co1, o>2, co3 in terms of the Eulerian angles, we obtain the kinetic energy of a rigid body as a function of the generalized coordinates 9, <p, &.
The symmetrical top in a gravitational field. We shall find the Lagran- gian for a symmetrical top whose point of support lies on the axis of symmetry at a distance I below the centre of mass. Then the height of the centre of mass above the point of support is z = l cos #. Hence, the potential energy of the top is
U = mgz = mgl cos 0- . (9.18)
The kinetic energy of the top, expressed in terms of the Eulerian angles, is
T = ~ Jt (co? + col) + 4-^3^1 =
M £
= y«M^2 + <Psin2£) + ^ J3(^ + <j,cos£)2. (9.19)
The difference between the quantities (9.19) and (9.18) gives the Lagrangian for a symmetrical top. The sum gives the total energy & .
Sec. 10] GENERAL PBINCIPUSS OF MECHANICS 81
Since L does not contain time explicit^, the energy is an integral of motion:
£ = T + U = const . (9.20)
We can find two more integrals of motion, noting that the angles 9 and <J> do not appear explicitly in L (9 is eliminated only in the case of a symmetrical top). These integrals of motion are
p, 7"
#p = -p- = ^3 (? + ^ cos *) = const , (9-21)
o i"
p^ = —•- = e/x sin2 ^^ -|- J3 cos ft (9 + <j> cos ft) = const . (9.22)
If we eliminate 9 and ^ from equations (9.21) and (9.22) and substi- tute them into the energy integral, the latter will contain only the variable #, which allows us to reduce the problem to quadrature.
Substituting (9.21) in (9.22), we obtain
p^ = Jl sin2& fy -f- fa cos & , whence
The energy integral, after substituting p9 and p^ is
**-»"»** + *- + mgl cos » . (9.23)
Thus, the problem is reduced to motion with one degree of freedom ft, as it were. The corresponding "kinetic energy" is — J^2, and the
' 'potential energy" is represented by those energy terms which depend on &. This potential energy becomes infinite for £ = 0, and & — n. Hence, for 0 <0- <TT it has at least one minimum. If this minimum
corresponds to 9- > —> then the rotation of the top, whose centre of
mass is above the point of support, is stable. Small oscillations are possible near the potential energy minimum. These oscillations are superimposed on the precessional motion of the top which we have already noted. They are called nutations.
Sec. 10. General Principles of Mechanics
In this part of the book, mechanics is explained mainly through the use of Newton's equations (2.1). Going over to generalized coordinates, we obtain from them Lagrange's equations and a series of further deductions. In this section it will be shown that the system of Lagran- ge's equations can be obtained not only from Newton's Second Law, but also from a very simple assertion about the value of the integral
6-0060
82 MECHANICS [Part I
of the Lagrangian taken with respect to time. The basic laws of mechan- ics thus formulated are usually called integral principles.
The particular importance of these principles is that they allow us to understand, in a unified manner, the laws relating to various areas of theoretical physics (mechanics and electrodynamics), thus opening up a field for broad generalizations.
Action. For a certain mechanical system, let it be possible to define the Lagrangian
L=L(q,qJk), (10.1)
as dependent on the generalized coordinates q, velocities q, and the time t. We shall consider that all the coordinates and all the velocities are independent. Let us choose some continuous, but otherwise arbi- trary, dependence of the coordinates upon the time q (t). The functions q (t) can be in complete disagreement with the actual law of motion. The only requirement imposed on q (t) is that the functions q (t) should be smooth, i. e., that they should provide for differentiation and should correspond to the rigid constraints present in the system.
The time integral of the Lagrangian is called the action of the sys- tem:
'i
S = JL(q, q, t) dt. (10.2)
'o
The magnitude of this integral depends upon the law chosen for q (t), and is, in that sense, arbitrary. In order to examine the relation- ship between the action and the function q (t), it is convenient to cal- culate the change of S for a transition from some arbitrary law q (t) to another, infinitely close but also arbitrary, law qf (t).
Variation. Fig. 14 shows two such conceivable paths. Time is taken along the abscissa, and one of the generalized coordinates q, represent- ing the totality of generalized coordinates, is plotted on the ordinate axis.
For the specification of future operations, we shall consider that both paths pass through the same points, g0 and ql9 at the initial and final instants of time, -p. 14 The vertical arrow shows the difference
between two conceivable, infinitely close paths at some instant of time other than initial or final.
This difference is usually called the variation of q and is denoted by 8q. The symbol S should emphasize the difference between variation and the differential d\ the differential is taken for the same path at various instants of time, while the variation is taken for the same instant of time between different paths.
Sec. 10] GENERAL PRINCIPLES OF MECHANICS 83
Since the neighbouring paths in Fig. 14 have different forms, the speed of motion along them will also differ. Together with the variation of the coordinate 8q between paths, we can also find the variation
in velocity 8q. We shall show that 8q = -jr8q. Indeed, 8q = q' (t) — q (t),
at
where q' and q are values of the coordinates for neighbouring paths.
But the derivative of the difference -=- §q is equal to the difference of
at * ^
the derivatives q' (t) — q(t)=8q.
Let us now find the variation of the Lagrangian, i. e., the difference of the function for two adjacent paths. Since L = L (q, q, t) and the variation is taken at the same instant of time, i. e., S£ = 0, we obtain
Let us rearrange the second term. Taking advantage of the fact that
—
q = — 8q , we can write it thus :
The last equation simply expresses a transformation by parts. Substituting it into (10.3), we find
d
The integral of the variation of L is equal to the variation of action 8$, since the difference between integrals taken between the same limits is equal to the difference between the integrands.
The first term in (10.4) can be integrated with respect to time, because it is a total derivative. The variation of action is then reduced to the form
'o 'o
We have agreed to consider only those paths which pass through the same points, qQ and ql9 at the initial and final instants of time. Hence, at these instants the variation $q becomes zero by convention, and the integrated term disappears. The expression 8$ is reduced to the following integral:
. (10.6)
The extremal property of action. If the chosen path coincides with the actual path of motion, the coordinates satisfy Lagrange's equation:
84 MECHANICS [Part I
Substituting this in (10.6), we see that the variation of action tends to zero close to the actual path. The change in magnitude is equal to zero either close to its extreme, or close to the "stationary point" (for example, the function ?/ — #3 has such a point at x = Q, where 2/' = 0, iz/" = 0). Three cases can, in general, be realized: a minimum, a maximum and a stationary point.
For example, let a point, not subject to the action of any forces other than constraint reactions, move freely on. a sphere. Then its path will be an arc of a great circle. But through any two points on the sphere there pass two arcs of a great circle representing the largest and smallest sections of the circumference. One corresponds to a maxi- mum, and the other, to a minimum, 8. If the beginning arid end of the path are diametrically opposite, the result is a stationary point.
The principle o! least action. We have proven, on the basis of Lag- range's equations, that 8S — 0. We can proceed in a different way: by asserting that close to the actual path passing between the given initial and final positions of the system the increment of action is equal to zero, we can derive Lagrange's equations. Ordinarily, the action on an actual path is minimal, and therefore the assertion we have made is called the principle of least action. Action was written in the form (10.2) by Hamilton. Much earlier, the principle of least action was mathematically formulated by Euler for the special case of paths corresponding to constant energy.
For us, it is not essential that the action should be a minimum, but that it should be steady, 88 = 0.
Lagrango's equations are derived from the principle of least action by means of proving the opposite. We assume the right-hand side of equation (10.6) to be zero, 8$ = 0, and the variation 8q to be arbitrary. Then, if the expression inside the parentheses is not equal to zero, the sign of the variation 8q can always be chosen to be the same as
7\ T A y\ T
for the quantity -~— — ^- ^-. , because the variation is arbitrary. If,
f\ T ft P 7"
for example, the sign of the quantity -^ --- -^ -^p- changes several
times along the path of integration, then the sign of 8q must also be changed accordingly at those points so that the integrand of (10.6) should everywhere be non-negative. But the integral of a non-nega- tive function cannot equal zero unless the function is equal to zero
everywhere. Therefore, 8$ = 0 only when-^— • — -rr-^- becomes zero
^ oq at oq
along the whole path of integration, for otherwise the variation 8q can be so chosen that 88 >0. We have shown that if we proceed from the principle of least action as a requirement for the motion along an actual path, then that path must satisfy Lagrange's equations.
Sec. 10] GENERAL PRINCIPLES OF MECHANICS 86
The advantages of using action. The principle of least action may at first sight appear artificial or, in any case, less obvious than Newton's laws, to whose form we are accustomed. For this reason we shall try to explain where its advantages lie.
First of all, let it be noted that Lagrange's or Newton's equations are always associated • with some coordinates whose choisc is, to a significant extent, arbitrary. In addition, the choice of coordinate system, relative to which the motion is described, is also arbitrary. Yet the motion of particles along actual paths in a mechanical sys- tem expresses a certain set of facts which cannot depend on the arbitrary manner of their description. For example, if the motion leads to a collision of particles, that fact must always be represented in any description of the system.
But it is precisely the integral principle that is especially useful in a formulation of laws of motion not related to any definite choice of coordinates, the value of the integral between the given limits being independent of the choice of integration variables. The extremal property of an integral cannot be changed by the way in which it is calculated.
The integral principle S$ = 0 is equivalent, purely mathematically, to Lagrange's equations (2.21). But in order to apply it to any actual system, we must have an explicitly expressed Lagrangian. It may be found from those physical requirements which should be imposed on an invariant law of motion that is independent of the choice of coordi- nate axes and the frame of reference.
As a result of the invariance of the principle of least action, we can consider the laws of mechanics in a very general form, and this, therefore, opens the way for further generalizations.
The determinacy of the Lagrangian. Before finding an explicit form for the Lagrangian, we must put the qiiestion: Is the determined function we are looking for single-valued ? We shall show that if we add the total time derivative of any function of coordinates and time,
-IT- / (q, t), then Lagrange's equations remain unchanged. This can be
verified either by simple substitution into (10.7), or directly from the integral principle. Writing
L=L'+±f(q,t), (10.8)
we see that
'i 'i 'i ti
(10.9)
The variations of / appear in the variation of S only at the limits of integration. But since we have arranged that / depends on the coordi-
86 MECHANICS [Part I
nates and time, but not on the velocities, the variation of / is expressed linearly in terms of the variations of the coordinates, and is zero at the limits of integration. Therefore,
(10.10)
Hence, the Lagrangian is determined only to the accuracy of the total time derivative of the function of coordinates and time.
Defining forms of the Lagrangian. We shall now formulate in more detail those requirements which the integral principle expressing laws of mechanics must satisfy.
First of all we note that the form of this principle must be the same for different inertia! systems, since all such systems are equivalent. This statement follows from the relatively principle (see Sec. 8). The essence of the relativity principle consists in the fact that the choice of an inertial coordinate system is arbitrary, while the physical consequences of the equations of motion cannot be arbitrary.
Similarly arbitrary is the choice of the origin and the initial instant of time and also orientation of the coordinate axes in space.
It must, of course, be borne in mind that the form of action is by no means determined by speculation ; this form represents no less a gener- alization of physical experience than the laws of Newton. However, the principle of least action, best expresses the invariance of physical laws to the method of their formulation. Quite naturally, the form of the invariance (in relation to rotations, translations, reflections, etc.) is itself a certain, very broad, generalization of experience, and must by no means be considered as a priori.
Considering now the problem of finding the form of the Lagrangian, let us first of all determine the action of a free particle in an inertial coordinate system.* In such a system, the particle moves uniformly in a straight line, i. e., with constant velocity. (This statement is based on the experimental fact that inertial systems exist in nature). Thus, the Lagrangian for a free particle in an inertial system cannot con- tain any coordinate derivatives other than velocity.
By definition, a free particle is very far away from any other bodies with which it could interact. Therefore, its Lagrangian must not change its form upon displacement of the origin to any arbitrary point fixed in the given inertial system. In other words, the Lagrangian of such a particle does not depend explicitly on the coordinates.
In this way, one can conclude that the Lagrangian does not depend explicitly on time.
* See L. D. Landau and E. M. Lifshits, Mechanics, Fizmatgiz, 1958.
Sec. 10] GENERAL PRINCIPLES OF MECHANICS 87
The orientation of the coordinate axes is arbitrary as well as the choice of the origin. For the Lagrangian to be independent of the orien- tation of coordinate axes, it must be scalar quantity.
To summarize, then, the Lagrangian is a scalar that depends only on the velocity of the free particle relative to the given inertial system. The only scalar quantity which can be formed from a vector is the absolute value of the vector. Therefore,
The form of this function can be found from the relativity principle, in accordance with which the Lagrangian must not change with the transformation from one inertial system to another. In Newtonian mechanics, this transformation is effected with the aid of equations (8.1), (8.2), i.e., Galilean transformations. The Galilean transforma- tions led to the law of addition of velocities :
where V is the relative velocity of the inertial systems. Therefore, the Lagrangian must remain invariant with respect to Galilean trans- formations.
Since the Lagrangian is determined to a total derivative, it is suffi- cient (for its invariance) for the following equality to be satisfied:
£ = L(^) = L[(v' + V)2] = L(i;'2) + g) (J0.ll)
where the functions L (v2) and L (v'2) have the same form in accordance with the principle of relativity.
Any transformation (8.1), (8.2), in which the relative velocity V is finite, can be obtained by a set of infinitely small transformations applied successively. It is, therefore, sufficient to consider a transforma- tion in which the relative velocity of the inertial systems V is very much smaller than the particle velocity v. Then, to a very good approx- imation, the quantity (v'+V2) is equal to
where the term of the second order of smallness is discarded.
Expanding L [(v' + V)2)] in a series, we obtain, to the same approx- imation,
Comparing this with (10.11), we find:
3L gy'Y- dL gy<*r/
0(v'a) "" d(v'*) V dt ~~ dt
88 MECHANICS [Part I
However, the expression on the left-hand side of the equation can
/•) 7
be a total derivative of the function of coordinates only if g /a is independent of velocity. Introducing the notation
BL m ,
-7jT^r2y = -y = const ,
we obtain
di~ ~~~dim ' for, otherwise, . ( ,2) could not be put inside the derivative sign.
In this way we have shown that the Lagrangian for a free particle is equal to
£(V2) = -^-ra. (10.12)
The Lagrangian for a system of noninteracting particles is equal to the sum of the Lagrangians of these particles taken independently, since it is the only sum of quadratic expressions of the type (10.12) that changes by a total derivative when Vi==v,/+V (where i is the particle number) is substituted.
In order to write down L for a system of interacting particles, we must, of course, make certain physical assumptions about the nature of the interaction.
1) The interaction does not depend on the particle velocities. This assumption is justified for gravitational and electrostatic forces, and is not justified for electromagnetic forces. It should, however, be noted that electromagnetic interactions involve ratios of particle velocities and the velocity of light c, and therefore, to the approximation of Newtonian mechanics, they must be considered as negligibly small. The Lagrangian of Newtonian mechanics is not universal and is appli- cable only to a limited group of phenomena, when all V{ <^ c.
2) The interaction docs not change the masses of the particles.
3) The interaction is invariant with respect to Galilean transforma- tions.
From these conditions it can be seen that the interaction appears in the Lagrangian in the form of a scalar function determined only by the relative distribution of the particles:
(10.13)
From this expression, we can find the conservation laws for energy, linear momentum, and angular momentum (see Sec. 4).
The Hamiltonian function. We shall now use the principle of least action in order to transform a system of equations of motion to other variables. Namely, in place of coordinates and velocities we shall
Sec. 10] GENERAL PBINCIPLES OF MECHANICS 89
employ coordinates and momenta. Let us assume that velocities are eliminated from the relations
P = -ff. (10.14)
Since the Lagraiigian depends quadratically on the velocities, equa- tions (10.14) are linear in the velocities and can always be solved. We shall obtain for coordinates and momenta a more symmetrical system of equations than Lagrange's equations.
The passing from velocities to momenta was performed to some ex- tent when we substituted the integrals of motion in the expression for energy, for example, in (5.4), (9.21), (0.22).
Now, in place of the velocities we shall introduce into the energy the momenta for all the degrees of freedom, (and not only for the cyclic ones, i. e., those, whose coordinates do not appear explicitly in L). Energy expressed in terms of coordinates and momenta only is called the Hamiltonian function of the system or, for short, the Hamiltonian :
<?[q,q(p)]^Jf(q,p)=qp-L. (10.15)
Thus, for example, if we replace & by ~ in (9.23), we obtain the Hamiltonian for a symmetrical top:
(10.16)
Hamilton's equations. In order to derive the required system of equations, we write the expression for the principle of least action, expressing L in terms of jtf* :
ti
(10.17)
Here it is assumed that q is expressed in terms of p and q. Let us calculate the variation 88:
The second term inside the parentheses can be integrated by parts, similar to the way that it was done in (10.5). This gives
The integrated part becomes zero when limits of integration have been substituted. The independent variables are now p and q. The variation
90 MECHANICS [Part I
of p, as well as the variation of q, is completely arbitrary in sign. For 88 to bo equal to zero, the following equations must be satisfied:
This system of equations is more symmetrical than Lagrange's equations. Instead of v second-order Lagrangian equations, we have 2 v first-order equations (10.18). They are called Hamilton's equa- tions.
Reducing the order with the aid ol the energy integral. If #F does not depend on time, we can exclude time completely from the equations by dividing all the equations (10.18), except one, by the said equation. Then we have
if-— w- (10-19>
dp
Here, for simplicity, this operation has been performed for a system with one degree of freedom. The integration of (10.19) yields one con- stant. The second constant will be determined by quadrature from the equation
(10.20)
where „ - is a certain function q which can be obtained by integrating (10.19). The constant of integration in (10.20) is the initial instant £0.
The connection between momentum and action. We shall now show that if action is calculated for the actual paths of a system, then mo- mentum can be very simply expressed in terms of this action. For this we shall consider the change in action when the ends of the integra- tion interval are displaced along the actual paths. From (10.7), the expression under the integral sign in (10.5) is equal to zero on such paths. But the integrated part does not become zero ; only the varia- tions in it must be replaced by differentials, since we are considering the displacement of the ends of the integration interval along given paths. Therefore,
dS^^dq-^-dqQ=pdq-pQdq0 (10.21)
in agreement with the definition of momentum (4.13).
But action calculated along an actual path is uniquely determined by its initial and final points 8 = 8 (qQ, q). So
dS = --dq0+-8-dq. (10.22)
SeC. 10] GENERAL PRINCIPLES OF MECHANICS 91
Comparing (10.21) and (10.22), we obtain the very important rela- tionship between momentum and action
*-•£• *•---£• (10-23)
which is very essential for the formulation of quantum mechanics.
Exercise
Write down the Hamiltonian and Hamilton's equations for a particle in a central field.
PART II ELECTRODYNAMICS
Sec. 11. Vector Analysis
The equations of electrodynamics gain considerably in conciseness and vividness if they are written in vector notation. In vector notation, the arbitrariness associated with the choice of one or another coordi- nate system disappears, and the physical content of the equations becomes more apparent.
We have assumed that the reader is acquainted with the elements of vector algebra, such as the definition of a vector and the various forms of vector products. However, in electrodynamics, vector differential operations are also used. This section is devoted to a definition of vec- tor differential operations and to proofs of their fundamental proper- ties, which will be needed later.
The vector of an area. We first of all give a definition of the vector of an elementary area rfs. This is a vector in the direction of the normal to the area, numerically equal to its surface and related to the direction of traverse of the contour around the area by the corkscrew rule (Fig. 15).
d*
Fig. 15
Fig. 16
We shall make use of a right-handed coordinate system x, y, z, in which, if we look from the direction of the z-axis, the ar-axis is rotated towards the t/-axis in an anticlockwise sense (Fig. 16). In this system,
Sec. 11] VECTOR ANALYSIS 93
the vector area can be resolved into components which are expressed thus :
dsx—dydz, dsy = dzdx, dsx=dxdy.
Vector flux. Now suppose that a liquid of density 1 ("water") flows tlirough the area, the flow velocity being represented by the vector v. We shall call the angle between v and ds, a. Fig. 17 shows the flow lines of the liquid passing through ds. They are parallel to the velocity v. Let us calcu- late the amount of liquid that passes tlirough the ds^ area ds every second. Obviously, it is equal to the amount that passes through the area ds', placed ^ perpendicular to the flux and intersected by the same flow lines as pass through ds. This quantity is simply equal to v ds', because every second a liquid cylinder of base ds' and height v passes through the area ds'. But ds' — ds cos a, whence j?ig. 17 the quantity of liquid we are concerned with is
dJ=v ds' = v ds cos a^=vds. (11.1)
By analogy, the scalar product of any vector A (taken at the point of infinitesimal area) on ds is called the flux of the vector A across the area ds. Similar to the way that the flow of liquid across a finite area s is equal to the integral of dJ with respect to the surface,
J-Jvds, (11.2)
the integral
(11.3)
is called the flow (flux) of the vector A across any area.
The area vector is introduced so that we can make use of the noncoordinate and convenient notation of (11.3). The integrals appearing in (11.3) are double. In terms of the projections of (11.3) we can write
J=\ AdsE=J \Axdydz + (JAydzdx
where the limits of the double integrals are determined from the cor- responding projections, onto the coordinate planes, of the contour bounding the surface.
The Gauss-Ostrogradsky theorem. Let us now calculate the vector flux through a closed surface. For this we shall consider, first of all, the infinitesimal closed surface of a parallelepiped (Fig. 18). We shall make the convention that the normal to the closed surface will always be taken outwards from the volume.
94 ELECTRODYNAMICS [Part II
Let us calculate the flux of the vector A across the area A BCD (the direction of traverse being in agreement with the direction of the normal). Since the flux is equal to the scalar product of A by the vec- tor area A BCD, in the negative ^-direction (and hence equal to dydz), we obtain for this infinitely
small area
Fig. 18
dJABCD = — Ax (x) dy dz.
We get a similar expression for the area A'B'C'D', only in this case the projection dsx is equal to dy dz, and Ax is taken at the point x + dx instead of x. And so
dJA'ffc'D' — Ax (x + dx) dy dz.
Thus the resultant flux through both areas, perpendicular to the #~axis, is
+ dJADCD = [Ax(x + dx) — Ax (x)] dy dz = -—-dxdydz.
-—
(11.4)
We have utilized the fact that dx is an infinitely small quantity, and we have expanded Ax(x + dx) in a series. The resultant fluxes across the boundaries perpendicular to the y and z axes are formed similarly. The resultant flux across the whole parallelepiped is
A finite closed volume can be divided into small parallelepipeds, and the relationship (11.6) applied to each one of them separately. If we sum all the fluxes, the adjacent boundaries do not give any contribution, since the flux emerging from one parallelepiped enters the neighbouring one. Only the fluxes through the outer surface of the selected volume remain, since they are not cancelled by others. But the right-hand sides of ( 1 1.6) will be additive for all the elementary volumes dV — dx dy dz, yielding the very important integral theorem:
- <"-6>
It is called the Gauss-Ostrogradsky theorem.
The divergence of a vector. The expression appearing on the right- hand side under the integral sign can be written down in a much shorter form. We first of all notice that it is a scalar expression, since there is a scalar on the left-hand side in (11.6) and dV is also
Sec. 11] VECTOR ANALYSIS 95
a scalar. This expression is called the divergence of the vector A and is written thus:
The divergence can be defined independently of any coordinate system, if (11.5) is used. Indeed, from (11.5) the definition for diver- gence follows as
fAds divA=sUm^~ — . (11.8)
F->-0 V
The divergence of a vector at a given point is equal to the limit of the ratio of the vector flux through the surface surrounding the point to the volume enveloped by the surface, when the surface is contracted into the point.
Let us suppose that the vector A denotes the velocity field of some fluid. Then, from the definition (11.8), it can be seen that the divergence of the vector A is a measure of the density of the sources of the fluid, for it is obvious that the more sources there are in unit volume, the more fluid will flow out of the closed volume. If div A is negative, we can speak of the density of vents. But it is more convenient to define the source density with arbitrary sign. We note that from (11.7) there follows the quantity
since r has components x, y, z.
Contour integrals. We shall now consider the vector integral of a closed contour having the following form:
z). (11.10)
This single integral is called the circulation of the vector over the given contour. For example, if A is the force acting on any particle, then A dl = A dl cos a is the work done by the force on the contour element dl and C is the work performed in covering the whole contour.
Stokes5 theorem. We shall now prove that the circulation of the vector A around the contour can be replaced by the surface integral "pulled over" the y contour.
Let us consider the projection of an in- finitely small rectangular contour onto the plane yz. Let this projection also have the form of a rectangle shown in Fig. 19. We
shall calculate the circulation of A around Fig. 19
96 ELECTRODYNAMICS [Part II
this rectangle. The side A B contributes a component Ay (z) dy arid side CD the component — Ay(z + dz)dy, where the minus sign must be written because the direction of the vector CD is opposite to that of the vector AB. We obtain, for the sum due to the sides AB and CD,
(wo have expanded Ay (z + dz) in a series for dz), while for the sides BC and DA,
dA A; (y + dy) dz —Az (y) dz - - Q?- dydz .
The resultant value for circulation in the yz-plane is
The notation Bx is clear from the equation. Let us now find out what meaning this expression has. From the definition of (11.10), circulation is a scalar quantity and, hence, on the right side of equation (11.11) there must also be a scalar quantity. If the contour lies in the plane yz, this quantity is of the form dC~Bxdsx\ consequently, for an arbitrary orientation of the contour, the relationship (11.11) must have the form of the scalar product
where Bx, By, Bz must necessarily be the components of a vector, since, otherwise, dC could not be a scalar. From (11.11),
/? — 3AZ _ dAy .- ~.
*~~~~W. — aT- (ii.id)
In order to findrthe circulation for infinitely small contours in the xz, yz planes, it is sufficient to perform a cyclic permutation of the indices x, y, z. This permutation yields the components By, Bz :
The vector B has a special name: it is called the rotation or curl of the vector A and is denoted thus:
B=rot A.
rot A is expressed in terms of unit vectors i, j, k, directed along the coordinate axes:
C. 11] VECTOR ANALYSIS 97
Changing to the notation (11.16), we see that the component of rot A normal to the area appears in equation (11.11):
J
A dl = rot* A <fa , (11.17)
where the subscript n of rot A indicates that we must take the pro- jection of rot A normal to the area, i.e., coinciding with the vector ds. (11.17) permits us to define rot A in a noncoordinate manner, similar to the way that we defined div A in (11.8), namely:
(Adi rotn A --= lim - --- , (11.18)
5 -> 0 '9
or the projection of rot A, normal to the area at the given point, is the limit of the ratio of the circulation of A, over the contour of the area, to its value when the contour is contracted into the point.
So that the integral JA dl should not become zero, we must have closed vector lines, to some extent following the integration contour, which lines are simila.r to the closed lines of flow in a liquid during vortox motion. Hence the term curl, or rotation.
If the circulation is calculated from a finite contour then the contour can be broken up into infinitely small cells to form a grid. For the sides of adjacent cells, the circulations mutually cancel since each side is traversed twice in opposite directions; only the circulation along the external contour itself remains. The integral on the right- hand side of equation (11.17) gives the flux of rot A across the surface "pulled over" the contour. Thus, we obtain the desired integral theorem
[Adl-JrotAds, (11.19)
which is called Stokes' theorem.
Differentiation along a radius vector. The divergence and rotation of a vector are its derivatives with respect to the vector argument. They can be reduced to a unified notation by means of the following. We introduce the vector symbol V (nabla*) with components
Then, from (11.7), we obtain for the divergence of A:
* Nabla is an ancient musical instrument of triangular shape. This symbol is also called del.
7-0060
98 ELECTRODYNAMICS [Part II
(11.21)
x -
i.e., a scalar product of nabla and A. From (11.16), we have for the rotation
rot A ~- i (Vy Ax — Vz A y) -t- j ( V* Ax — V* Az) + k ( V* AY — Vy Ax
«LVA]. (11.22)
We use the identity symbol --. here in order to emphasize the fact that we are simply dealing with a new system of notation. We shall see, however, that this system is very convenient in vector analysis. We note, with reference to algebraic operations, that nabla is in all cases similar to a conventional vector. We shall use the expression "multiplication by nabla" if, when nabla operates on any expression, that expression is differentiated. Sometimes, nabla is multiplied by a vector without operating on it as a derivative, in that case it is applied to another vector [see (11.30), (11.32)].
Gradient. If we operate with V on a scalar 9, we obtain a vector which is called the gradient of the scalar 9:
«V9=i+J + k. (11.23)
Its components are:
V,9 = |J, V,9-|J, V.9-4J- (H-24)
From equations (11.24), it can be seen that the vector V9 is per- pendicular to the surface 9 = const. Indeed, if we take a vector dl lying on this surface, then, in a displacement dl, 9 does not change. This is written as
0> (11.25)
i.e., V9 is perpendicular to any vector which lies in the plane tangential to the surface 9 = const, at the given point, which accords with oiir assertion.
Differentiation of products. We now give the rules governing differential operations with V.
First of all, the gradient of the product of two scalars is calculated as the derivative of a product:
(11.26)
The divergence of a product of a scalar with a vector is calculated thus :
div 9 A = (Vep, 9 A) + (VA, 9 A) = (AV?) + 9 (VA) -
=^A grad 9 + 9 div A. (11.27)
Here the indices 9 and A attached to V show what V is applied to.
Sec. 11] VECTOR ANALYSIS 99
We find the rotation of 9 A in a similar manner:
rot 9 A= [V9, 9 A] + [V^, 9 A] = [grad 9, A] + 9 rot A . (11.28) Now we shall operate with V on the product of two vectors: div [AB]-(V [AB]) = (V^ [AB]) + (Vu [AB]).
We perform a cyclic permutation in both terms, since V can be treated in the same way as an ordinary vector. In addition, we put B after VB in the second term, and here, as usual, we must change the sign of the vector product. The result is
div [AB] = (B [VA A]) — (A [VB B])=B rot A — A rot B. (11.29)
Let us find the rotation of a vector product. Here we must use the relationship [A[BC]] -B (AC) — C (AB):
rot [AB] = [VA [AB]] + [VB [AB]] = (VA B) A — (VA A) B + ( VB B) A — — (AVB) B = (BV) A — B div A + A div B — (AV) B. (11.30)
Here we note the new symbols (BV) and (AV) operating on the vectors A and B. Obviously, (AV) and (BV) are symbolic scalars, equal, by definition of V, to
(AV) = A, Vx + AyVY + A,V, = Ax^ + Ar-^ + A, -J-- , (11.31)
and similarly for (BV). Then, (AV) B is a vector which is obtained
by application of the operation (11.31) to all the components of B.
Of the operations of this kind, we have yet to calculate grad AB :
grad (AB) =VA (AB) +VB (AB). We use the same transformation as in the preceding case:
grad (AB) = (BVA) A + [B [VAA]] + (AVB) B + [A [VBB]] =
= (BV) A + (AV) B + [B rot A] + [A rot B]. ( 1 1.32)
Certain special formulae. We note certain essential cases of operations involving V.
From the definition of divergence (11.7), we obtain from (11.27) and (11.9)
div^= -1 divr + r grad-» = ~ - -^ - 0 . (11.33) Further,
and in general
rotr = 0. (11.34)
100 KLECTBO DYNAMICS [Part II
We now take
and for all components of r at once
(AV)r = A. (11.35)
In addition, we apply V to a vector depending only on the absolute value of the radius vector. We note first of all that
fCf. (3.3.), where 1/r is differentiated], so that
Vr = y. (11.36)
Using the rule for differentiating a function of a function, we have div A (r) = -.- Vr = AL. (11.37)
Here A is a total derivative of A (r) with respect to the argument /*, i.e., a vector whose components are the derivatives of the three components of A (r) with respect to r: Ax, Av, As. Further,
rotA<r)=|ArJJ-] = J^-. (11.38)
Repeated differentiation. Let us investigate certain results con- cerning repeated operations with V.
The rotation of the gradient of any scalar is equal to zero:
rot grad 9 - | "V,V?I = [VVJ 9=0, (11.39)
since the vector product of any vector (including V) by itself is equal to zero. This can also be seen by expanding rot grad 9 in terms of its components. The divergence of a rotation is also equal to zero:
div rot A = (V [VA]) = ([VV] A)-=0 . (11.40)
Let us write down the divergence of the gradient of a scalar 9 in component form. From equations (11.7) and (11.24) we have
divgrad9HVV)<p=8-+ S- + -S--ACP. (11.41) Here A (delta) is the so-called Laplacian operator, or Laplaciaii:
A= *L+ a2_ + _^_
dx* ~r dy* "" dz* '
Sec. 11]
VECTOR ANALYSIS
101
Filially, the rotation of a rotation can be expanded as a double vector product:
rot rot A- [V [VA]]=V (VA)— (VV) A-grad div A— A A . (11.42)
The last eqiiation can be regarded as a definition of AA. In curvi- linear coordinates, A'cp and A A are expressed differently.
Curvilinear coordinates. We shall further show how the gradient, divergence, and rotation, as well as A of a scalar appear in curvilinear coordinates.
Curvilinear coordinates ql9 q2, q3 are termed orthogonal if only the quadratic terms dql, dql, dql appear in the expression for the element of length dl2, and not the products dq^ dq2, dql dq& dq% rf</3, similar to the way that dl2 = dx2 + dy2 + dz2 in rectangular coordinates. In orthogonal coordinates
dl2 = fil dql + hl dq\-\-hl dql . (11.43)
For example, in spherical coordinates qr = r, q2 = $,q3 = cp. The element of length is
so that
sn
Let us construct an elementary parallelepiped (Fig. 20). Then the components of the gradient will be
Cf
—
d q3
C
(11.44)
In order to find the divergence we repeat the proof of the Gauss-Ostrogradsky theorem for Fig. 20. The area ADCB is equal to h2 \ vector A through it is
Fig. 20 i dq2 dq%. The flux of
Here, A2 and A3 are also taken for a definite value of qv The sum of the fluxes through the areas ADCB and A'B'C'D' is
where we have used the expansion of the quantity h2 h% A± at the point ql + dqt in terms of dql9 jn a way similar to (11.4). The total flux across all the boundaries is
102 ELECTRODYNAMICS [Part II
dj ^ ?7 (h* h* AI) + r (AS hi AZ) + ~i (AI *• ^s) dqi dq* dq* *
Let us now take advantage of the definition of divergence (11.8):
dJ -div A- /4 A2 ^3 ^?i ^2 d#3 =div Adi V . Hence,
(11.45)
If, instead of Av A%, A3, we substitute the expressions (11.44), the result will be the Laplacian of a scalar in orthogonal curvilinear coor- dinates. Thus, in spherical coordinates it is
With the aid of Stokes' theorem, we can also calculate the rotation in curvilinear coordinates. We shall give it for reference without proof:
, . l / d A 7 8 f
rotl A = - A* 3 ~ '* 2
ro D
,
rot«
3
-A- ~~~" i " 7"*" l ~~/\ •**• i • ^i o -^*- 1 ' t'o l « Mi \ 0ft ^^i 3 3/
. I I d A i B 4 j \
A = ,—,- \ ~—A~h» — « - .4 T //! .
M| \a?i a«a J V
(11.47)
Exercises
Whore (from the requirements of the problem) expressing in terms of coordi- nates is not demanded, it is recommended that only the vector equations of the present section (11.2ti)-(11.42) be used.
1) Calculate the expressions: Answers:
a) A— (r^O). A — =divgrad— *=* — div — = 0.
b)div9(r)r, rot 9 (r) r . 89 4- 7-9; 0.
c) V (Ar), TAC A = const. At
d)V(A(r)r). A+l(rA).
e) div 9 (r) A (r), rot 9 (r) A (r) . 1 (rA) + 1 (pA) . _t [rAj + f [rA] t
f ) div [r [Ar]], A = const. — 2 (Ar) .
g) rot [r [Ar]], A = const. 3 [rA] .
h)AA(r) [CM. (11,42)]. i + -?- JL
r
i) V(A(r)B(r)). 1
r
j) rot[Ar], A = const. 2 A.
k) div [Ar], A - const. 0
DA!. ^a
r r>
See. 11] VECTOR ANALYSIS 103
2) Write down A^ in cylindrical coordinates.
3) Write down the three components of AA in spherical coordinates.
4) Two closed contours are given. The radius vector of points of the first contour is rx, of the second contour, r>. The elements of length along each contour are dl^ and d\2, respectively. Prove that the integral
is equal to zero, 4 TT, 8 w, and n-4 it, depending upon how many times the first contour is wound round the second, linking up with the latter. v\ denotes diffe- rentiation with respect to rx (Ampere's theorem).
Changing the order of integration and performing a cyclic permutation of the factors, we have
We apply Stokes' theorem (11.19) to the integral in d}l:
Wo use equation (11.30) ; rotj denotes differentiation with respect to tho compo- nents rx; and d\2 in such a differentiation may be regarded as a constant vec- tor:
|
In to |
accoi zero. rotx 1 ' |
LVll', 'dance Tliere |
with rerru |
| . -2J ' exercise ains, ther , di.]- |
— \ifi2 v la, the efore, (dlaVi |
if vi 1 r r 1 " last term containing At - \ * — /ji *-, |
. v |
1 l r r 1 |
equal |
|
1': i) |
i-M 1S i T7 |
||||||||
|
Vl|'i- |
-r2l |
)Vl|rx-ra| |
Vi I r r |
'2! ' |
since a function of the difference TI — r.2 is differentiated.
For short, we write r=rx — r2. Then the required integral will be
* = — J (d\2 V2) J <*Si Vi |ri_ra| = — J (dl* Vi) J ^»i V — •
We shall now explain tho geometrical sense of the second integrand, i.e.,
d^ v — = -- - — L-^- • -y . Tho scalar product - - - is the projection of
an element of the surface d^ pulled over the first contour, on the radius vector
r drawn from a point on the second contour. In other words, - - - is equal
to the projection of the area da ± on a plane perpendicular to r. This projection, divided by r2, is equal to the solid angle d& at a point r2 on the second contour
subtended by the area dsr The integral I - — 3 — - is, therefore, that solid angle
n which is obtained if a cone is drawn with vertex at the point r2, so that the generating line of the cone formed the contour lt.
The differential (d\2 V2) Q is the increment of solid angle O obtained in shifting along the contour 12 a distance d\2. Thus,
J (d\2 V2) Q
The integral of this quantity around a closed contour is equal to the total change in solid angle in traversing the contour 12. Let the initial point of cir- cumvention lie on the surface «x. Then the solid angle subtended by the surface
104 ELECTRODYNAMICS [Part TT
at the origin is — 2 TC. If the contours aro linked, then the solid angle will be 2 T: after the circumvention, since the ure,a is observed from a terminal point on the other side. If the contours