ESSENTIAL PHYSICS Part 1 RELATIVITY, PARTICLE DYNAMICS, GRAVITATION, AND WAVE MOTION FRANK W. K. FIRK Professor Emefitus of Physics Yale University   3 CONTENTS PREFACE 7 1 MATHEMATICAL PRELIMINARIES 1.1 lnvariants 11 12 Some geometical invarants 12 1.3 Elements of differential geometry 15 1.4 Gaussian coordinates and the invarant line dement 17 1.5 Geometry and groups 20 1.6 Vectors 23 1.7 Quatemions 24 1.8 3-vector analysis 26 1.9 Linear algebra and n-vectors 28 1.10 The geometry of vectors 31 1.11 Linear operators and matrices 34 1.12 Rotation operators 36 1.13 Components of a vector under coordinate rotations 38 2 KINEMATICS: THE GEOMETRY OF MOTION 2.1 Velodty and acceleraton 42 22 Differental equations of kinematics 45 2.3 Velodty in Cartesian and polar coordinates 49 2.4 Acceleraton in Cartesian and polar coordinates 50 3 CLASSICAL AND SPECIAL RELATMTY 3.1 The Galilean transformaton 56 32 Einstein's space-tme symmetry: the Lorentz transformation 58 3.3 The invarant interval: contravarant and covarant vectors 61 3.4 The group structure of Loreneztransformadons 63 3.5 The rotaton group 66 3.6 The relatvity of simultaneity: lime dilalion and length contraclion 68 3.7 The 4-velocity 71 4 NEWTONIAN DYNAMICS 4.1 The law of inerta 75  4 42 Newton's laws of motion 77 4.3 Many interacting particles: conservaton of linear and angular momentum 77 4.4 Work and energy in Newtonian dynamics 84 4.5 Potental energy 86 4.6 Partide interaclions 89 4.7 The moton of rigid bodies 94 4.8 Angular velocity and the instantaneous center of rotaton 97 4.9 An applicaton of the Newtonian method 98 5 INVARIANCE PRINCIPLES AND CONSERVATION LAWS 5.1 Invariance of the potental under translatons: conservaton of linear momentum 105 52 Invariance of the potental under rotations: conservation of angular momentum 105 6 EINSTEINIAN DYNAMICS 6.1 4momentum and the energy-momentum invariant 108 62 The relatvistc Doppler shift 109 6.3 Relatvistc collisions and the conservaton of 4-momentum 110 6.4 Relatvistc inelastc collisions 113 6.5 The Mandelstam variables 115 6.6 Positron-eledron annihiladon-in-flight 117 7 NEWTONIAN GRAVITATION 7.1 Properties of moton along curved paths in the plane 122 72 An overview of Newtonian gravitaton 124 7.3 Gravitaton: an example of a central force 129 7.4 Moton under a central force: conservaton of angular momentum 131 7.5 Kepler's 2nd law explained 131 7.6 Central orbits 133 7.7 Bound and unbound orbits 137 7.8 The concept of the gravitatonal field 139 7.9 The gravitadonal potendal 143 8 EINSTEINIAN GRAVITATION: AN INTRODUCTION TO GENERAL RELATMTY 8.1 The principle of equivalence 147 8.2 Time and length changes in a gravitadonal field 149 8.3 The Schwarzschild line element 149  5 8.4 The metric in the presence of matter 153 8.5 The weak field approximaton 154 8.6 The refracve index of space-time in the presence of mass 155 8.7 The deflection of light grazing the sun 155 9 AN INTRODUCTION TO THE CALCULUS OF VARIATIONS 9.1 The Eulerequadon 160 92 The Lagrange equatons 162 9.3 The Hamilton equatons 165 10 CONSERVATION LAWS, AGAIN 10.1 The conservaton of mechanical energy 170 102 The conservaton of linear and angular momentum 170 11 CHAOS 11.1 The general moton of a damped, drven pendulum 173 112 The numerical solution of differenlial equations 175 12 WAVE MOTION 12.1 The basicform of awave 179 122 The general wave equation 182 12.3 Lorenlz invarant phase ofa wave and the relativisic Doppler shift 183 12.4 Plane harmonic waves 186 12.5 Spherical waves 187 12.6 The superposition of harmonic waves 188 12.7 Standing waves 189 13 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 13.1 Definitions 192 13.2 Some trigonometric identities and their Fourier series 193 13.3 Determination of the Fourier coefficients of a funclion 195 13.4 The Fourier series of a periodic saw-tooth waveform 195 APPENDIX A: SOLVING ORDINARY DIFFERENTIAL EQUATIONS 200 BIBLIOGRAPHY 211  6  7 PREFACE Throughout the decade of the 1990's, I taught a one-year course of a specialized nature to students who entered Yale College with excellent preparaton in Mathematcs and the Physical Sciences, and who expressed an interest in Physics or a dosely related field. The level of the course was that typified by the Feynman Lectures on Physics. My one-year course was necessarly more restricted in content than the two-year Feynman Lectures. The depth of treatment of each topic was limited by the fact that the course consisted of a total of fifty-two lectures, each lastng one-and-a-quarter hours. The key role played by invariants in the Physical Universe was constanty emphasized. The material that I covered each Fall is presented, almost verbatm, in this book. The first chapter contains key mathematcal ideas, induding some invariants of geometry and algebra, generalized coordinates, and the algebra and geometry of vectors. The importance of linear operators and their matiix representatons is stressed in the eary lectures. These mathematcal concepts are required in the presentaton of a unified treatment of both Classical and Spedal Relatvity. Students are encouraged to develop a "relatvistc outlook"at an eary stage. The fundamental Lorentz transformaton is developed using arguments based on symmetizing the dassical Galilean transformaton. Key 4-vectors, such as the 4-velodty and 4-momentum, and their invariant norms, are shown to evolve in a natural way from their dassical forms. A basic change in the subject matter occurs at this point in the book. It is necessary to introduce the Newtonian concepts of mass, momentum, and energy, and to discuss the conservaton laws of linear and angular momentum, and mechanical energy, and their associated invariants. The discovery of these laws, and their applications  8 to everyday problems, represents the high point in the scientific endeavor of the 17th and 18th centuries. An introduction to the general dynamical methods of Lagrange and Hamilton is cblayed until Chapter 9, where they are induded in a discussion of the Calculus of Variatons. The key subject of Einsteinian dynamics is treated at a level not usually met in at the introductory level. The 4momentum invariant and its uses in relatvisic collisions, both elastc and inelastic, is discussed in detail in Chapter6. Further developments in the use of relatvistc invariants are given in the discussion of the Mandelstam variables, and their application to the study of high-energy collisions. FolloMng an overview of Newtonian Gravitaton, the general problem of central orbits is discussed using the powerful method of [p, r] coordinates. Einstein's General Theory of Relatvity is introduced using the Principle of Equivalence and the noton of "extended inerdal frames" that indude those frames in free fall in a gravitational field of small size in which there is no measurable field gradient. A heuristc argument is given to deduce the Schwarzschild line element in the 'weak field approximaton"; it is used as a basis for a discussion of the refractive index of space-tme in the presence of matter. Einstein's famous predicted value for the bending of a beam of light grazing the surface of the Sun is calculated. The Calculus of Variatons is an important topic in Physics and Mathematcs; it is introduced in Chapter 9, where it is shown to lead to the ideas of the Lagrange and Hamilton functions. These functions are used to illustrate in a general way the conservaton laws of momentum and angular mromentum, and the relaton of these laws to the homogeneity and isotropy of space. The subject of chaos is introduced by considering the modon of a damped, driven pendulum. A method for solving the non-linear equadon of modon of the pendulum is outined. Wave modon is treated from the point-of-view of  9 invariance prndples. The form of the general wave equaton is derived, and the Lorentz invariance of the phase of a wave is discussed in Chapter 12. The final chapter deals with the problem of orthogonal functions in general, and Fourier series, in particular. At this stage in their training, students are often under-prepared in the subject of Differental Equations. Some useful methods of solving ordinary differental equations are therefore given in an appendix. The students taking my course were generally required to take a parallel one-year course in the Mathematcs Department that covered Vector and Matrix Algebra and Analysis at a level suitable for potental majors in Mathematics. Here, I have presented my version of a first-semester course in Physics - a version that deals with the essentials in a no-frills way. Over the years, I demonstrated that the contents of this compact book could be successfully taught in one semester. Textbooks are concerned with taking many known facts and presenting them in dear and concise ways; my understanding of the facts is largely based on the wriings of a relatvely small number of celebrated authors whose work I am pleased to acknowledge in the bibliography. Guilord, Connecticut February, 2000 I am grateful to several readers for poin ng out errors and undear statements in my first version of this book. The comments of Dr Andre Mirabelli were particularly useful, and were taken to heaL March, 2003  10  11 1 MATHEMATICAL PRELIMINARIES 1.1 Invariants It is a remarkable fact that very few fundamental laws are required to descilbe the enormous range of physical phenomena that take place throughout the universe. The study of these fundamental laws is at the heart of Physics. The laws are found to have a mathematical stidure; the interplay between Physics and Mathematcs is therefore emphasized throughout this book. For example, Galileo found by observaton, and Newton developed within a mathematcal framework, the Principle of Relatvity: the laws governing the motions of objects have the same mathematical form in all inertial frames of reference. Inertial frames move at constant speed in straight lines with respect to each other- they are mutually non-acceleradng. We say that Newton's laws of moton are invariant under the Galilean transformation (see later discussion). The discovery of key invariants of Nature has been essental for the development of the subject. Einstein extended the Newtonian Prndple of Relatvity to indude the motons of bearrs of light and of objects that move at speeds dose to the speed of light This extended principle forms the basis of Special Relatvity. Later, Einstein generalized the principle to include acceleradng frames of reference. The general principle is known as the Principle of Covarianoe; it forms the basis of the General Theory of Relatvity (a theory of Gravitaton).  12 A review of the elementary properties of geometrical invariants, generalized coordinates, linear vector spaces, and matiix operators, is given at a level suitable for a sound treatment of Classical and Special Relatvity. Other mathematcal methods, induding contra- and covariant 4-vectors, variatonal prndples, orthogonal funcions, and ordinary differential equatons are introduced, as required. 12 Some geometrical invariants In his book The Ascent of Man, Bronowski discusses the lasting importance of the discoveries of the Greek geometers. He gives a proof of the most famous theorem of Euclidean Geometry, namely Pythagoras' theorem, that is based on the invariance of length and angle (and therefore of area) under translatons and rotatons in space. Let a right-angled tiangle with sides a, b, and c, be translated and rotated into the following four positons to form a square of sicbac: 24 c c a <- (b-a) --I The total area of the square = c= area of four tiangles +area of shaded square. If the right-angled tiangle is translated and rotated to form the rectangle:  13 a a 1 4 b b 2 3 then the area of four tilangles =2ab. The area of the shaded square area is (b-a)2= b2-2ab + a2 We have postulated the invariance of length and angle under translatons and rotatons and therefore c2=2ab+(b-a) = a2+ b2. (1.1) We shall see that this key result characterizes the locally flat space in which we live. It is the only forn that is consistent with the invariance of lengths and angles undertranslafions and rotations. The scalar product is an important invariant in Mathematics and Physics. Its invariance properties can best be seen by developing Pythagoras' theorem in a three-dimensional coordinate form. Consider the square of the distance between the points P[x1, yi, zi] and Q [x2, y2, z2] in Cartesian coordinates:  14 Z 0 X7 X2X We have (PQ)2 = (x2-X1)2 + (y2-yl)2 + (Z2-Z1)2 = X2- 2X1X2 + X12 + Y22- 2y1Y2 + y,2 + Z22-2Z1z2 + Z12 =(Xi2 + yi2+ Z2) + (x2+ y2 + zz ) -2(XIX2 + y1y2 + ZZ) _ (OP)2 + (OQ)2 - 2(xlx2 + y1y2 + ZIZ2) (1.2) The lengths PQ, OP, OQ, and their squares, are invariants under rotations and therefore the entire ~ght-hand side of this equation is an invariant The admixture of the coordinates (XIX2 + y1y2 + ZIZ2) is therefore an invariant under rotations. This term has a geometiic interpretation: in the triangle OPQ we have the generalized Pythagorean theorem (PQ)2 = (OP)2 + (OQ)2 - 20P.OQ cosu, OP.OQ CoscL = xlx2 +y1y2 + zlz2 =the scalar product 13 (1.3)  15 Invariants in space-tme with scalar-product-like forms, such as the interval between events (see 3.3), are of fundamental importance in the Theory of Relatvity. Although rotatons in space are part of our everyday experience, the idea of rotatons in space-tme is counter-intuitve. In Chapter 3, this idea is discussed in terms of the relative motion of inertialobservers. 1.3 Elements of differential geometry Nature does not presdbe a particular coordinate system or mesh. We are free to select the system that is most appropriate for the problem at hand. In the familiar Cartesian system in which the mesh lines are orthogonal, equidistant, straight lines in the plane, the key advantage stems from our ability to calculate distances given the coordina1as - we can apply Pythagoras' theorem, directly. Consider an arbitrary mesh: v-direction iP[f3..4 Origin 0 1a 2g 3i0u-direction  16 Given the point P[3,, 4], we cannot use Pythagoras' theorem to calculate the distance OP. In the infinitesimal parallelogram shown, we might think it appropriate to write ds2= du2+ dv2 + 2dudvcosa. (ds2 = (ds)2, a squared "length") This we cannot do! The differentals du and dv are not lengths - they are simply differences between two numbers that label the mesh. We must therefore multply each differental by a quantity that converts each one into a length. Introdudng dimensioned coefficients, we have ds2= giidu2 + 2gl2dudv + g2dv2 (1.4) where Vg11du andVg22 dv are now lengths. The problem is therefore one of finding general expressions for the coefficients; it was solved by Gauss, the pre-eminent mathematcian of his age. We shall restrict our discussion to the case of two variables. Before treatng this problem, it will be useful to recall the idea of a total differenfla/assodated with a funcion of more than one variable. Let u = f(x, y) be a function of two variables, x and y. As x and y vary, the corresponding values of u descilbe a surface. For example, if u = x2 + y2, the surface is a paraboloid of revolution. The partial dervadves of u are defined by af(x, y)/ax = limit as h ->0 {(f(x + h, y)-f(x, y))Ih} (treaty as a constant), (1.5) and af(x, y)/ay =limit as k ->0 {(f(x, y + k) -f(x, y))Ik} (treat x as a constant). (1.6) For example, if u =f(x, y) =3x2 +2y3 then afa= 6x, A2/ax2 =6, E3f/ax3 =0  17 and af/By = 6y2, a2f/ay2 =1 2y, a3fay3=12, and E4f/ay4 =0. If u = f(x, y) then the total differentalof the funcion is du = (aflax)dx + (af/By)dy corresponding to the changes: x--> x + dx and y -- y + dy. (Note that du is a function of x, y, dx, and dy of the independent varables x and y) 1.4 Gaussian coordinates and the invariant line element Consider the infinitesimal separaton between two points P and Q that are described in either Cartesian or Gaussian coordinates: y+dy Q v+dv ds ds y P vP x x+dx u u+du Cartesian Gaussian In the Gaussian system, du and dv do not represent distances. Let x=f(u, v) and y= F(u, v) (1.7 a,b) then, in the infinitesimal limit dx = (ax/au)du +(axlav)dv and dy = (ay/au)du + (aylv)dv.  18 In the Cartesian system, there is a direct correspondence between the mesh-numbers and distances: ds2=dx2 + dy2. (1.8) But dx2 = (8x/au)2du2 + 2(ax/au)(8x/av)dudv + (ax/av)2dv2 and dy2 = (ay/au)2du2 + 2(aylau)(aylav)dudv + (aylav)2dv2. We therefore obtain ds2 = {(8xlau)2 + (ay/au)2}du2 + 2{(ax/u)(8x/8v) + (aylau)(aylav)}dudv +{(ax/av)2 + (ayav)2}dv2 = g11du2+2g12dudv+g2dv2. (1.9) If we put u = ui and v = u2, then ds2 = g iduiduj where i,j = 1,2, (a general form in n-dimensional space: i, j =1,2,3, ...n) (1.10) iij Two important points connected with this invariant differenflal line element are: 1. Inteipretaflon of the coefficients gi;: consider a Eudidean mesh of equispaced parallelograms: v R ds dv P du Q u In PQR ds2= 1.du2+ 1.dv2+2cosadudv  19 = g11du2+ g2dv2 +2g12dudv (1.11) therefore, g11= g22=1 (the mesh-lines are equispaced) and g12 = cosa where ais the angle between the u-v axes. We see that if the mesh-lines are locally orthogonal then g12=0. 2. Dependence ofthe gs on the coordinate system and the local values of u, v. A specific example will illustrate the main points of this topic: consider a point P described in three coordinate systems - Cartesian P [x, y], Polar P [r, $], and Gaussian P [u, v] - and the square ds2 of the line element in each system. The transformaton [x, y] -- [r,$4] is x= rcos$ and y= rsin$. (1.12 a,b) The 1ransformadon [r,$)] -- [u, v] is direct, namely r=uand$=v. Now, ax/ar= cos, ay/ar= sin$, ax/a$=-rsin, ayla$)= rcos$ therefore, ax/au = cosv, ay/au = sinv, ax/av = - usinv, ay/av = ucosv. The coefficients are therefore gii= cos2v+sin2v= 1, (1.13 a-c) g2=(-usinv)2 +(ucosv)2=u2  20 and g12 = cos(-usinv) + sinv(ucosv) =0 (an orthogonal mesh). We therefore have ds2 = dx2 + dy2 (1.14a-c) = du2+u2dv2 = dr2 + r2d4p. In this example, the coefficient g22= f(u). The essental point of Gaussian coordinate systems is that the coefficients g . completely characterize the surface - they are intiinsic features. We can, in principle, determine the nature of a surface by measuring the local values of the coefficients as we move over the surface. We do not need to leave a surface to study its form. 1.5 Geometry and groups Felix Klein (1849-1925), introduced his influential Erlanger Program in 1872. In this program, Geometry is developed from the viewpoint of the invariants associated with groups of transfonnaflons. In Eudidean Geometry, the fundamental objects are taken to be rigid bodies that remain fixed in size and shape as they are moved from place to place. The noton of a rigid body is an idealizaton. Kein considered transformadons of the entre plane - mappings of the set of all points in the plane onto itself. The proper set of rigid modons in the plane consists of translatons and rotatons. A refleclion is an improper rigid motion in the plane; it is a physical impossibility in the plane itself. The set of all rigid modons - both proper and improper - forms a group that has the proper rigid motions as a  21 subgroup. A group G is a set of distnct elements {g} for which a law of composition 0"is given such that the compositon of any two elements of the set satsfies: Closure: if g , g jbelong to G then g k = g.io g jbelongs to G for all elements gi, gj, and Associativity for all g , g , g kin G, g i(g jog k)=(g iog j) g k.. Furthermore, the set contains A unique identity, e, such that g ioe = eg i= g ifor all glin G, and A unique inverse, g r1, for every element gi in G, such that gio gr1=gFr1ogi=e. A group that contains a finite number n of distnct elements gon is said to be a finite group of order n. The set of integers Z is a subset of the reals R; both sets form infinite groups under the composition of additon. Z is a "subgroup"of R. Permutatons of a set X form a group Sx under compositon of functions; if a: X -- X and b: X -- X are permutations, the composite funcion ab: X -- X given by ab(x) = a(b(x)) is a permutation. If the set X contains the first n positve numbers, the n! permutatons form a group, the symmetric group, Sn. For example, the arrangements of the three ntxnbers 123 form the group S3={123, 312, 231, 132, 321, 213}.  22 If the vertices of an equilateral triangle are labelled 123, the six possible symmetry arrangements of the tilangle are obtained by three successive rotatons through 120>about its center of gravity, and by the three reflecdions in the planes I, II, III: 2 3 II III This group of "isometries" of the equilateral triangle (called the dihedral group, D3) has the same structure as the group of permutations of three objects. The groups S3 and D3 are said to be isomorphic. According to Klein, plane Eudidean Geometry is the study of those properties of plane rigid figures that are unchanged by the group of isometiles. (The basic invariants are length and angle). In his development of the subject, Klein considered Simladity Geometry that involves isometries with a change of scale, (the basic invariant is angle), Afne Geometry, in which figures can be distorted under transformadons of the form x'= ax+ by+ c (1.15 a,b) y'=dx+ey+f,  23 where [x, y] are Cartesian coordinates, and a, b, c, d, e, f, are real coefficents, and Projecflve Geomety, in which all conic sections: drdes, ellipses, parabolas, and hyperbolas can be transformed into one another by a projeclive transformadon. It will be shown that the Lorenlz transformadons - the fundamental transformadons of events in space and time, as described by different inertal observers- form a group. 1.6 Vectors The idea that a line with a definie length and a definite direclion - a vector- can be used to represent a physical quandty that possesses magnitude and direction is an ancient one. The combined aclion of two vectors A and B is obtained by means of the parallelogram law, illustrated h the following diagram B A The diagonal of the parallelogram formed by A and B gives the magnitude and direction of the resultant vector C. Symbollically, we write C= A+ B (1.16) in which the ""sign has a meaning that is dearly different from its meaning in ordinary arithmetic. Galileo used this empirically-based law to obtain the resultant force acling on a body. Although a geometric approach to the study of vectors has an intuitve appeal, it wMill often be advantageous b use  24 the algebraic method - partcularly in the study of Einstein's Special Relatvity and Maxwell's Electromagnetism. 1.7 Quatemions In the decade 1830-1840, the renowned Hamilton introduced new kinds of numbers that contain fourcomponents, and that do not obey the commutatve property of multplicaton. He called the new numbers quatemions. A quatemion has the form u+dxi+yj+zk (1.17) in which the quantities i, j, k are akin to the quantity i = -i in complex numbers, x + iy. The component u forms the scalar part, and the three components xi+ yj + zk form the vector part of the quatemion. The coefficients x, y, z can be considered to be the Cartesian components of a point P in space. The quantities i, j, k are qualitative units that are directed along the coordinate axes. Two quatemions are equal if their scalar parts are equal, and if their coefficients x, y, z of i, j, k are respectively equal. The sum of two quatemions is a quatemion. In operations that involve quatemions, the usual rules of multiplication hold except in those terms in which products of i, j, k occur - in these terms, the commutative law does not hold. For example jk=i, kj=-i, ki=j, ik=-j, ij=k, ji=-k, (1.18) (these products obey aright-hand rule), and j2=j2= k2=-1. (Note the relation to|2=_1). (1.19) The product of two quatemions does not commute. For example, if  25 p=1+2i+3j+4k, and q=2+3i+4j+5k then pq=-36+6i+12j+12k whereas qp=-36+23i-2j +9k. Multplicaton is assodative. Quatemions can be used as operators to rotas and scale a given vector into a new vector: (a+ bi+ cj+ dk)(xi+yj+ zk)= (xi+y'j+ z'k) If the law of composition is quatemionic multplicaton then the set Q={±1,±i,±j,±k} is found to be a group of order 8. It is a non-commutatve group. Hamilton developed the Calculus of Quatemions. He considered, for example, the properties of the differential operator: V = i(8/3x) +j(8/y) + k(313z). (120) (He called this operator "nabla"). If f(x, y, z) is a scalar point function (single-valued) then Vf = i(Df/3x) +j(Df/Jy) + k(DfI3z) , a vector. If V=vi+v2j+v3k  26 is a contnuous vector point function, where the v's are functions of x, y, and z, Hamilton introduced the operaton Vv = (i/3x + jIy + k3/3zvii + v2j +v3k) (121) = - (JviI/x + Jv2Iy + Jv3/3z) + (JvJ3y-3v2/3z)i + (Jvi/3z-Jv3x)j + (3v213x-3v1/y)k = a quatemion. The scalar part is the negatve of the "divergence of V"(a term due to Clifford), and the vector part is the "cud of V" (a term due to Maxwell). Maxwell used the repeated operator V2, which he called the Lapladan. 1.8 3-vector analysis Gibbs, in his notes for Yale students, written in the period 1881 - 1884, and Heaviside, in artides published in the Electician in the 1880's, independency developed 3-dimensional Vector Analysis as a subject in its own right- detached from quatemions. In the Sciences, and in parts of Mathematcs (most notably in Analytical and Differential Geometry), their methods are widely used. Two kinds of vector multplicaton were introduced: scalar multplicaton and vector multplicaton. Consider two vecbrs v and v' where v=vie1 +v~e+v3e3 and v'=vi'ei+v2"e2+v3'e3.  27 The quantites e1, e2, and e3 are vectors of unit length pointing along mutually orthogonal axes, labeled 1,2, and 3. i) The scalar multplicaton of v and v'is defined as V - V =V1v1'+V2V2'+V3V3', (1.22) where the unit vectors have the properties e1- e1=e2 e2=e3 e3=1, (1.23) and e1- e2=e2 e1=e1 e3=e3 - 1=e2 - e3=e3 e2=0. (124) The most important property of the scalar product of two vectors is its invariance under rotatons and translatons of the coordinates. (See Chapter1). ii) The vector product of two vectorsv and v'is defined as e1 e2 e3 v x v'= v1 v2 v3 (whereI.. .lis the determinant) (125) v1 v2 v3' = (Vv3'-v3v2')e1+ (v3v1' -v1v3')e2 + (v1v2'-v2v1')e3. The unit vectors have the properties e1 x e1 =e2 x e2 =e3 x e3 =0 (1.26 a,b) (note that these properties differ from the quatemionic products of the i, j, k's), and ei xe2=e3,e2xe1=-e3,e2xe3=ei,e3xe2=-ei,e3xel1=e2,el xe3=-e2  28 These non-commuting vectors, or "cross products" obey the standard rghthand-rule. The vector product of two parallel vectors is zero even when neither vector is zero. The non-associatve property of a vector product is illustrated in the folloMng example ei xe2xe2 = (e1 x e2) xe02=e3 x e2=-e1 =e1 x(e2xe2)=0. Important operations in Vector Analysis that follow directly from those introduced in the theory of quatemions are: 1) the gradient of a scalarfunction f(xi, x2, x3) Vf = (3f/3xi)ei + (/3x2)e2+ (3f/3x3)e3, (127) 2) the divergence of a vector funcdion v V - v=v1/3x1i+Jv2/3x2+v3/x3 (1.28) where v has components vi, v2, v3 that are functions of xi, x2, x3 , and 3) the culof a vector functionv e1 e2 e3 Vx v= 3/3x1 3/3x2 3/x3 . (1.29) v1 v2 v3 The physical significance of these operatons is discussed later. 1.9 Linear algebra and n-vectors A major part of Linear Algebra is concerned with the extension of the algebraic properties of vectors in the plane (2-vectors), and in space (3-vectors), to vectors in higher dimensions (n-vectors).  29 This area of study has its origin in the work of Grassmann (1809 -77), who generalized the quatemions (4-component hyper-complex numbers), introduced by Hamilton. An n-dimensional vector is defined as an ordered column of numbers x1 xn= . (1.30) It will be convenient to write this as an ordered row in square brackets xn,=[x1 , x2, ... xn,]. (1.31) The transpose of the column vector is the row vector xnT = (x1,x2, ...xn,). (1.32) The numbers x1, x2, ...xn are called the components of x, and the integer n is the dimension of x. The order of the components is important, for example [1,2, 3] # [2, 3, 1]. The two vectors x= [x1, x,...xn] andy= [y1, y2, ...yn] are equal if xi=yi (i=lton). The laws of Vector Algebra are 1. x+y =y+ x. (1.33 a-e) 2. [x+y]+ z= x+[y+ z] . 3. a[x+y]= ax+ay where ais ascalar . 4. (a+ b)x= ax+ by where a,b are scalars .  30 5. (ab)x = a(bx) where a,b are scalars . If a=1 and b =-1 then x + [-x]=0, where 0= [0,0, ...0] is the zero vector. The vectors x = [x1, x,...xn]and y = [y1, y2 ...yn] can be added to give their sum or resultant x+y = [x1+Y1, x2+y2,...,xn+yn]. (1.34) The set of vectors that obeys the above rules is called the space of all n-vectors or the vector space of dimension n. In general, a vector v = ax + by lies in the plane of x and y. The vector v is said to depend linearly on x and y - it is a linear combinaton of x and y. A k-vectorv is said to depend linearly on the vectors u1, u2, ...uk if there are scalars a such that v=au1+a2u2+...akuk. (1.35) For example [3,5,7] = [3,6,6] + [0,-1, 1] = 3[1, 2,2] +1[0,-1, 1], a linearcombinaton of the vectors [1,2,2] and [0,-1, 1]. A set of vectors u1, u2, ...uk is called linearly dependent if one of these vectors depends lineady on the rest For example, if u1=a2u2+a3u3+...+akuk., (1.36) the set u1, ...uk is linearly dependent.  31 If none of the vectors u1, u2, ...uk can be written linearly in terms of the remaining ones we say that the vectors are lineady independent. Altematvely, the vectors u1, U2, ...uk are linearly dependent if and only if there is an equation of the form cIu1+ c2u2+...Qkuk=0 , (1.37) in which the scalars o are not all zero. Consider the vectors e obtained by putting the it-component equal to 1, and all the other components equal to zero: e1= [1, 0, 0,...0] e2= [0,1,0,...0] then every vector of dimension n depends lineady on ei, e2, ...en, thus x = [x11,)x2,...xn] = xie1+ x2e2+...xnen. (1.38) The el's are said to span the space of all n-vectors; they form a basis. Every basis of an n-space has exactly n elements. The connection between a vectorx and a definite coordinate system is made by choosing a set of basis vectors el. 1.10 The geometry of vectors The laws of vector algebra can be interpreted geometically for vectors of dimension 2 and 3. Let the zero vector represent the origin of a coordinate system, and let the 2-vectors, x and y,  32 correspond to points in the plane: P [x1, x2] and Q [yi, y2]. The vector sum x + y is represented by the point R, as shown R [xi+y1, x2+y2] 2nd component x2- P [x1, x2 01[0, 0]1 x1 yi 1 component R is in the plane OPQ, even if x and y are 3-vectors. Every vector point on the line OR represents the sum of the two corresponding vector points on the lines OP and OQ. We therefore introduce the concept of the directed vector lines OP, OQ, and OR, related by the vector equation OP+OQ=OR. (1.39) AvectorV can be represented as a line of length OP pointng in the direction of the unit vectorv, thus P =v.OP 0 A vectorV Vis unchanged by a pure displacement: V1= V2  33 where the "="sign means equality in magnitude and diredion. Two dasses of vectors will be met in future discussions; they are 1. Polar vectors: the vector is drawn in the direction of the physical quantity being represented, for example a velocity, and 2. Axial vectors: the vector is drawn parallel to the aids about which the physical quantty acts, for example an angular velocity. The associatve property of the sum of vectors can be readily demonstrated, geometically C V B A We see that V=A+B+C=(A+B)+C=A+(B+C)=(A+C)+B. (1.40) The process of vector additon can be reversed; a vector V can be decomposed into the sum of n vectors of which (n -I) are arbitrary, and the ne vector closes the polygon. The vectors need not be in  34 the same plane. A special case of this process is the decompositon of a 3-vector into its Cartesian components. A general case A special case V V5 V Vz V4 ViV3 Vx Vy V2 V1, V2, V3, V4: arbitrary Vz doses the polygon V5 doses the polygon The vector product of A and B is an axial vector, perpendicular to the plane containing A and B. z B y AxB aunit vector,+n A perpendicular to the A, B plane x Ax B= AB sinan=-B x A (1.41) 1.11 Linear Operators and Matrices Transformadons from a coordinate system [x, y] to another system [x', y'], without shift of the origin, or from a point P [x, y] to another point P' [x', y'], in the same system, that have the form  35 x= ax+ by y'=cx+dy where a, b, c, d are real coefficients, can be written in matrix notaton, as follows x' a b x (1.41) y c d y Symbolically, x'= Mx, (1.42) where x = [x, y], and x'= [x', y'], both column 2-vectors, and a b M= c d a 2 x 2 matrix operator that "changes" [x, y] into [x', y']. In general, M transforms a unit square into a parallelogram: y/\ y'[a+b,c+d] [b~d [0,0] [1,0] x This transformadon plays a key rdle in Einstein's Special Theory of Relatvity (see later discussion).  36 1.12 Rotation operators Consider the rotation of an x, y coordinate system about the origin through an angle$: y y P[x,y]orP'[x',y' y y' x 0,0' x From the diagram, we see that x'= xcos$+ysin$ x and y'=-xsin$+ycos$ or x' cos$ sin$ x y' -sin$ cos$ y Symbolically, (1.43) where  37 cos sin$ = is the rotation operator. -sin$ cos$ The subscript c denotes a rotaton of the coordinates through an angle+$3 The inverse operator,Mc-1($) is obtained by reversing the angle of rotaton:+$ ---$. We see that matrix product 9C-1($)1() = 9cT()lc($)M= 1 (1.44) where the superscript T indicates the transpose (rows <-> columns), and 1 0 I = is the identfy operator. (1.45) 0 1 Eq.(1.44) is the defining property of an orthogonalmatix. If we leave the axes fixed and rotate the point P[x, y] to P'[x', y'], then we have y, y'P'[x',y'] y P[x, y] 0 x' x x From the diagram, we see that x' = xcos$-ysin$, and y' = xsin$ +ycos$)  38 or P'=ty($)P(1.46) where cos$ -sin$ 91y($)= ,the operator that rotates a vector through +$. sin$ cos$ 1.13 Components of a vector under coordinate rotations Consider a vectorV [v, vy], and the same vectorV' with components [v,vy], in a coordinate system (primed), rotated through an angle +$. y' y^ vy V = V' 0,0' vx x We have met the transformaton [x, y] -- [x', y'] under the operation c1($4); here, we have the same transformaton but now it operates on the components of the vector,vx and vy, [v', vy]=9tc($)[vx, v]. (1.47) PROBLEMS 1-1li) If u =3 x4show that 3u/3x = (3 xVn3)Iy and Bu/Jy =(-3 xldn3)/y2. ii)lIf u=lIn{(x3 +y)/x2} show that 3u/3x = (x3 -2y)I(x(x3 +y)) and Bu/Jy =1I(x3 +y).  39 1-2 Calculate the second partial dervadves of f(x, y) = (1/ry)exp{-(x-a)24y}, a = constant 1-3 Check the answers obtained in problem 1-2 by showing that the funcdion f(x, y) in 1-2 is a solution of the partial differential equaton 3f/3x2 -8Df/Jy =0. 1-4 If f(x, y, z) =1/(x2 + y2+ z2)12=1/r, show that f(x, y, z) =1/r is a solution of Laplace's equation 41/3x2+323/y2+2f/8z2=0. This important equation occurs in many branches of Physics. 1-5 At a given instant, the radius ofa cylinder is r(t) = 4cm and its height is h(t) =10cm. If r(t) and h(t) are both changing at arate of 2 cm.s-1, show that the instantaneous increase in the volume of the cylinder is 192T cm3.s-1. 1-6 The transformaton between Cartesian coordinates [x, y, z] and spherical polar coordinates [r, 0,4] is x = rsin6cos$, y = rsin6sin, z = rcos0. Show, by calculating all necessary partial derivatives, that the square of the line element is ds2 = dr2 + r2sin20d$2 + r2d62. Obtain this result using geometrical arguments. This form of the square of the line elemmt wMill be used on several occasions in the future. 1-7 Prove that the inverse of each element of a group is unique.  40 1-8 Prove that the set of positve ratonal numbers does not form a group under division. 1-9 A finite group of order n has n2 products that may be written in an nxn array, called the group multplicaton table. For example, the 4th-roots of unity {e, a, b, c} = {±1, ±i}, where i = 4-1, forms a group under multplicaton (1i = i, i(-i)=1, i2=_1, (_i)= -1, etc. ) with a multiplicaton table e=1 a=i b=-1 c=- e 1 i -1 - a i -1 i 1 b -1 -i 1 i c - 1 i -1 In this case, the table is symmetiic about the main diagonal; this is a characteristc feature of a group in which all products commute (ab = ba) - it is an Abelian group. If G is the dihedral group D3, discussed in the text, where G = {e, a, a2, b, c, d}, where e is the identty, obtain the group multiplicaton table. Is it an Abelian group?. Notce that the three elements {e, a, a2} form a subgroup of G, whereas the three elements {b, c, d} do not; there is no identty in this subset The group D3 has the same multplicaton table as the group of permutatons of three objects. This is the conditon that signifies group isomorphism. 1-10 Are the sets and  41 ii) {[1, 3, 5,7], [4,-3, 2, 1], [2, 1, 4, 5]} lineady dependent? Explain. 1-11 i) Prove that the vectors [0,1,1], [1,0,1], [1, 1, 0]form a basis for Eudidean space R3. ii) Do the vectors [1, i] and [i,-1], (i = J-1), form a basis for the complex space (2? 1-12 Interpret the linear independence of two 3-vectors geometrically. 1-13 i) If X = [1,2,3] and Y = [3,2, 1], prove that their cross product is orthogonal to theX-Y plane. ii) If X and Y are 3-vectors, prove thatXxY =0 iffX and Y are linearly dependent 1-14 If aii a12 a13 T = a21 a22 a23 0 0 1 represents a linear transformaton of the plane under which distance is an invaiant, show that the following relatons must hold: a112+a212=a12+a222=1, and a11a12+a21a22=0.  42 2 KINEMATICS: THE GEOMETRY OF MOTION 2.1 Velocity and acceleration The most important concepts in Knematics - a subject in which the properties of the forces responsible for the motion are ignored - can be introduced by studying the simplest of all motions, namely that of a point P moving in a straight line. Let a point P [t, x] be at a distance x from a fixed point 0 at a time t, and let it be at a point P' [t', x']= P'[ t + At, x + Ax] at a time At later. The average speed of P in the interval At is = AX/At. (2.1) If the ratio Ax/At is not constant in time, we define the instantaneous speed of P at lime t as the limiting value of the ratio as At -->0: vp=vp(t)=limit as At ->0 of Ax/At= dx/dt= x=vx. The instantaneous speed is the magnitude of a vectorcalled the instantaneous ve/ocityof P: v = dx/dt, a quantity that has both magnitude and direction. (2.2) A space-time cuve is obtained by plotting the positions of P as a function of t x* vp P O  43 The tangent of the angle made by the tangent to the curve at any point gives the value of the instantaneous speed at the point The instantaneous acceleration, a, of the point P is given by the lime rate-of-change of the veloity a = dv/dt = d(dx/dt)/dt = d2x/dt2 = x. (2.3) A change of variable from t to x gives a = dv/dt = dv(dxldt)/dx = v(dv/dx). (2.4) This is a useful relation when dealing with problems in vwhich the velodty is given as a function of the posilion. For example v VP P v 0 N Q x The gradient is dv/dx and tana = dv/dx, therefore NQ, the subnonal, = v(dv/dx) = ap, the acceleration of P. (2.5) The area under a curve of the speed as a function of time between the limes ti and t2 is [A][ rg = J~t1e]v(t)dt = J~t1e v(dx/dt)dt = JtxlQ]dx =(x2- x1) = distance traveled in the lime t2-t1. (2.6)  44 The solution of a kinematical problem is sometimes simplified by using a graphical method, for example: A point A moves along an x-axis with a constant speed vA. Let it be at the origin 0 (x =0) at lime t = 0. It continues for a distance xA, at which point it decelerates at a constant rate, finally stopping at a distance XfromOat lime T. A second point B moves away from 0 in the +x-direction with constant acceleration. Let it begin its motion at t = 0. It continues to accelerate until it reaches a maximum speed vBnex at a tine tBx when at xBnex from 0. At xnx, it begins to decelerate at a constant rate, finally stopping at X at time T: To prove that the maximum speed of B during its motion is vBx = vA{l - (xA2X)}-1, a value that is independent of the lime at which the maximum speed is reached. The velocity-time curves of the points are v , A possible path for B vA B A 0 t=0 t ~ e T t x=0 xAXx X The areas under the curves give X =vALA + vA(TI- tA)/2 = vBeT2, so that  45 VBmax = vA(1 + (tvT)), but vAT =2X- x, therefore vBmx = vA{1 - (xV2X)}-1: f(tm). 22 Differential equations of kinematics If the acceleration is a known function of lime then the differential equation a(t) = dv/dt (2.7) can be solved by performing the integrations (either analytcally or numerically) fa(t)dt= idv (2.8) If a(t) is constant then the result is simply at + C = v, where C is a constant that is given by the initial conditions. Let v=uwhen t=0 then C= uand we have at + u = v. (2.9) This is the standard result for motion under constant acceleration. We can continue this approach by wriing: v=dx/dt=u+at Separating the variables, dx = udt + atdt. Integrating gives x = ut + (112)at2 + C' (for constant a). If x=0 when t= 0 then C' =0, and x(t) = ut+ (1/2)at2. (2.10) Mulliplying this equalion throughout by 2a gives  46 2ax = 2aut + (at)2 =2aut+ (v-u)2 and therefore, rearranging, we obtain v2=2ax-2aut+2vu -u2 =2ax+2u(v-at)-u2 =2ax+u2. (2.11) In general, the acceleration is a given funcdion of lime or distance or velodty: 1) Ifa=f(t) then a = dv/dt =f(t), (2.12) dv = f(t)dt, therefore v = (f(t)dt + C(a constant). This equation can be written v = dx/dt = F(t) + C, therefore dx = F(t)dt + Cdt Integrating gives x(t) =JF(t)dt+ Ct+ C'. (2.13) The constants of integralion can be determined if the velocity and the posilon are known at a given lime.  47 2) If a = g(x) = v(dv/dx) then (2.14) vdv= g(x)dx. Integratng gives V2= 2Jg(x)dx+D, therefore v= G(x)+D so that v = (dx/dt)= ±(G(x)+D). (2.15) Integratng this equaton leads to fdx/{\(G(x)+D)}= t+D'. (2.16) Altemadvely, if a = d2x/dt2= g(x) then, muliplying throughout by 2(dx/dt)gives 2(dxdt)(d2x/dt2) = 2(dxldt)g(x). Integratng then gives (dxldt)2 = 2Jg(x)dx + D etc. As an example of this method, consider the equadon of simple harmonic modon (see later discussion) d2x/dt2 =-o. (2.17) Muldply throughout by 2(dx/dt), then 2(dx/dt)d2x/dt2 =-2wdx(dx/dt).  48 This can be integrated to give (dx/dt)2=-x2 +D. If dx/dt =0 when x = A then D = w2A2, therefore (dx/dt)2 =o(A2- x2)=y2, so that dx/dt = ±o(A2- x2). Separatng the variables, we obtain - dx/{/(A2 -x2)}=wdt. (The minus sign is chosen because dx and dt have opposite signs). Integratng, gives cos-1(x/A) =ot+D'. But x = A when t=0, therefore D'=0, so that x(t) = Acos(wt), where A is the amplitude. (2.18) 3) Ifa= h(v), then (2.19) dv/dt = h(v) therefore dv/h(v)= dt, and Jdv/h(v) =t+ B. (2.20) Some of the techniques used to solve ordinary differental equadons are discussed in Appendix A.  49 2.3 Velocity in Cartesian and polar coordinates The transformadon from Cartesian to Polar Coordinates is represented by the linear equations x= rcos$ and y= rsin$, (221 a,b) or x=f(r,$) and y= g(r,$) The differentals are dx = (D/Ir)dr + (3/)d$ and dy = (3g13r)dr + (3I)d$. We are interested in the transformaton of the components of the velodty vector under [x, y] -- [r, 4]. The velodty components involve the rates of change of dx and dy Mth respect to time: dx/dt= (D/Ir)dr/dt+(DI38)d$/dt and dy/dt=(3g/Br)dr/dt +(3g/8)d$/dt or x= (//r)r+ (8f/8)$ and y = (3g/r)r + (1g/$)$. (2.22) But, D//r = cos$,3f/&$ = -rsin$, 3g/3r = sin$, and og/&$ = rcos$, therefore, the velodty transformadons are x=cos$r-sin$(r$)=vx (223) and y = sin$i r + cos$(r$4)= vy. (2.24) These equadons can be written  50 vx cos$ -sin$ dr/dt v sin$ cos$ rd$/dt Changing $ ---$, gives the inverse equations dr/dt cos$ sin$ vx rd$/dt -sin$ cos$ v or vr Vx = 9($) .(2.25) The velocty components in [r,$] coordinates are therefore V IvIl= r$= rd$/dt |r= r =dr/dt r P [r,] +$, anticlockwise o x The quandty d$/dt is called the angular velocityof P about the origin 0. 2.4 Acceleration in Cartesian and polar coordinates We have found that the velocity components transform from [x, y] to [r, 4)] coordinates as follows vx= cos$)r-sin(r$)) =x  51 and vy= sin$ r+cos$(r$)=y. The acceleration components are given by ax= dvxdt and vy = dv/dt We therefore have ax=(d/dt)cos r-sin$(r$)} (226) = cos$(r-r4V)-sin$(2r$+ r$) and ay= (dldtXsin$ r + cos$(r$)} (2.27) = cos$(2r$+ r$)+sin$(r-r4g). These equations can be written ar cos$ sin$ ax (2.28) a, -sin$ cos$ ay The acceleration components in [r,$] coordinates are therefore A A la,| = 2r$ + r$ 0 x  52 These expressions for the components of acceleration will be of key importance in discussions of Newton's Theory of Gravitation. We note that, if r is constant, and the angular velocityo is constant then a, =r$=rw=0, (229) ar =-r$2=-ro2=-r(v/r)2=_ y 2/r, (2.30) and v,= r4= rw. (2.31) These equations are true for circularmotion. PROBLEMS 2-1 A point moves with constant acceleration, a, along the x-axis. If it moves distances Axi and Ax2in successive intervals of lime Ati and A, prove that the acceleration is a = 2(v2-vi)/T where vi= Axi/Ati, v2 = Ax2/At, and T = Ati1+ At. 2-2 A point moves along the x-axis with an instantaneous deceleration (negative acceleration): a(t) o -vn*1(t) where v(t) is the instantaneous speed at lime t, and n is a positive integer. If the inilial speed of the point is u (at t =0), show that ko={(un -vn)I(uv)"}In, where ko is a constant of proportionality,  53 and that the distance travelled, x(t), by the point from its initial positon is kox(t)= {(un-1-v-1)/(uv)-1}/(n - 1). 2-3 A point moves along the x-axis with an instantaneous deceleraton kv3(t), where v(t) is the speed and k is a constant. Show that v(t) = ul(1 + kux(t)) where x(t) is the distance travelled, and u is the inital speed of the point 2-4 A point moves along the x-axis with an instantaneous acceleration d2x/dt2 = - 2/x2 where o is a constant If the point starts from rest at x= a, show that the speed of the partide is dxldt = - o{2(a - x)/(ax)}12. Why is the negatve square root chosen? 2-5 A point P moves with constant speed v along the x-axis of a Cartesian system, and a point Q moves with constant speed u along the y-ads. At time t =0, P is at x =0, and Q, moving towards the origin, is at y = D. Show that the minimum distance, dn, between P and Q during their moton is dmin= D{1/(1 + (ulv)2)}1'2. Solve this problem in two ways:1) by direct minimizadon of a funclion, and 2) by a geometical method that depends on the choice of a more suitable frame of reference (for example, the rest frame of P).  54 2-6 Two ships are sailing with constant velocities u and von straight courses that are inclined at an angle 0. If, at a given instant, their distances from the point of intersection of their courses are a and b, find their minimum distance apart. 2-7 A point moves along the x-axis with an acceleration a(t)= kt2, where t is the lime the point has been in motion, and k is a constant.If the initial speed of the point is u, show that the distance travelled in lime t is x(t) = ut+ (1/12)kt4. 2-8 A point, moving along the x-axis, travels a distance x(t) given by the equation x(t) = aexp{kt} + bexp{-kt} where a, b, and k are constants. Prove that the acceleration of the point is proportional to the distance travelled. 2-9 A point moves in the plane with the equations of motion d2x/dt2 -2 1 x d2y/dt2 1 -2 y Let the following coordinate transformalion be made u = (x+y)/2andv= (x-y)12. Show that in the u-v frame, the equalions of molion have a simple form, and that the tine-dependence of the coordinates is given by u = Acost + Bsint, and  55 v = Ccos<3 t + Dsinl3 t, where A, B, C, D are constants. This coordinate transformadon has "diagonalized" the original matiix: -2 1 -1 0 1-2 0-3 The matrix with zeros everywhere, except along the main diagonal, has the interestng property that it simplyscales the vectors on which it acts- it does not rotate them. The scaling values are given by the diagonal elements, called the eigenvalues of the diagonal matrix. The scaled vectors are called eigenvectors. A small industry exists that is devoted to finding optmum ways of diagonalizing large matrices. Illustrate the motion of the system in the x-y frame and in the u-vframe.  56 3 CLASSICAL AND SPECIAL RELATMTY 3.1 The Galilean transformation Events belong to the physical world - they are not abstractions. We shall, nonetheless, introduce the idea of an ideal event that has neither extension nor duration. Ideal events may be represented as points in a space-time geometry. An event is descilbed by a four-vector E[t, x, y, z] where t is the lime, and x, y, z are the spatial coordinates, referred to arbitrarlychosen origins. Let an event E[t, x], recorded by an observer 0 at the origin of an x-axis, be recorded as the event E'[t', x'] by a second observer 0', moving at constant speed V along the x-axis. We suppose that their docks are synchronized at t = t'=0 when they coincide at a common origin, x = x'=0. At lime t, we write the plausible equations t'= t and x'= x-Vt, where Vt is the distance travelled by 0' in a lime t. These equalions can be written E' =GE (3.1) where 1 0 -VI1 G is the operator of the Galilean transformalion.  57 The inverse equatons are t =t' and x = x'+Vt' or E = G-1E' (3.2) where G-1 is the inverse Galilean operator. (It undoes the effect of G). If we muliply t and t' by the constants k and k', respecively, where k and k'have dimensions of velocity then all terms have dimensions of length. In space-space, we have the Pythagorean form x2+ y2 = r2 (an invariant under rotatons). We are therefore led to ask the question: is (kt2+ x2 an invariant under G in space-tme? Direct calculation gives (kt)2+ x2 = (k't')2 + x'2+2Vx't' + \2t'2 =(k't')2+x'2 onlyifV=O! We see, therefore, that Galilean space-tme does not leave the sum of squares invariant. We note, however, the key role played by acceleraflon in Galilean-Newtonian physics: The velocdes of the events according to 0 and 0' are obtained by diferentatng x'=-Vt+ xwith respect to Sme, givng v'=-V+v, (3.3)  58 a result that agrees with everyday observations. Differentiating v' with respect to lime gives dv'Idt'= a'= dv/dt = a (3.4) where a and a'are the accelerations in the two frames of reference. The classical acceleration is an invariant under the Galilean transformalion. If the relationship v'= v- V is used to describe the motion of a pulse of light, moving in empty space at v = c 3 x 108 m/s, it does not fit the facts. For example, if V is 0.5c, we expect to obtain v'= 0.5c, whereas, it is found that v'= c. Indeed, in all cases studied, v'= cfor all values of V. 32 Einstein's space-time symmetry: the Lorenlz transformation It was Einstein, above all others , who advanced our understanding of the nature of space- lime and relative motion. He made use of a symmetry argument to find the changes that must be made to the Galilean transformalion if it is to account for the relative motion of rapidly moving objects and of beams of light. Einstein recognized an inconsistency in the Galilean-Newtonian equations, based as they are, on everyday experience. The discussion will be limited to non-acceleraing, or so called inertal, frames We have seen that the dassical equations relating the events E and E'are E'= GE, and the inverse E = G-1E' where 1 0 1 0 G = andG-1 =- -VI1 VI1  59 These equations are connected by the substitution V < -V; this is an algebraic statement of the Newtonian prndple of relativity. Einstein incorporated this principle in his theory. He also retained the linearty of the dassical equations in the absence of any evidence to the contrary. (Equispaced intervals of lime and distance in one inerdal frame remain equispaced in any other inertal frame). He symmetizedthe space-time equations as follows: t' 1-V t (3.5) x -V 1 x Note, however, the inconsistency in the dimensions of the time-equation that has now been introduced: t'= t-Vx. The term Vx has dimensions of [L]2/[T], and not [T]. This can be corrected by introdudng the invariant speed of light, c - a postulate in Einstein's theory that is consistent with the result of the Michelson- Morey experiment ct'= ct-Vx/c so that all terms now have dimensions of length. Einstein went further, and introduced a dimensionless quantityy instead of the scaling factor of unity that appears in the Galilean equations of space-time. This factor must be consistent with all observations. The equalions then become ct'= yct- pyx x'= -pyct + yx , where p3=V/c. These can be written  60 E' =LE, (3.6) where Y -p3Y L =, -3Y Y and E=[ct,x]. L is the operator of the Loreneztransformaton. The inverse equaton is E = L-1E' (3.7) where Y ' This is the inverse Loreneztransformadon, obtained from L by changing 3 - -p (V -- -V); it has the effect of undoing the transformadon L. We can therefore write LL-1 = 1 (3.8) Carrying out the matiix multiplications, and equatng elements gives therefore, y =1WI(1-$12) (taking the positve root). (3.9) As V -- 0, p3 -- 0 and therefore y -- 1; this represents the dassical limit in which the Galilean transformation is, for all pracical purposes, valid. In particular, lime and space intervals have the same  61 measured values in all Galilean frames of reference, and acceleration is the single fundamental invariant 3.3 The invariant interval: contravariant and covariant vectors Previously, it was shown that the space-tme of Galileo and Newton is not Pythagorean under G. We now ask the question: is Einsteinian space-time Pythagorean under L ? Direct calculaton leads to (ct)2+ x2 = y2(l+ p2)(ct'y + 42x'ct' +y2(1 + p2)x2 (ct')2+ x2ifP>0. Note, however, that the difference of squares is an invariant (ct)2-x2 = (ct')2-x2 (3.10) because y2(1_32) =1. Space-tme is said to be pseudo-Eudidean. The negatve sign that characterizes Lorenlz invariance can be induded in the theory in a general way as follows. We introduce two kinds of 4vectors x= [xO, x1, x2, x3], a contra variant vecto, (3.11) and xp= [xo, x1, x2, x3], a covariant vector, where x= [xO, -x1, -x2, _x3]. (3.12)  62 The scalar (or inner) product of the vectors is defined as xpTxp=(xO, x1, x2, x3)[xO, x1, _x2, _x3], to conform to matrix multplication T T row column =(x0)2_-(x1)2 + (x2)2 + (x3)2). (3.13) The superscript T is usually omitted in wriing the invariant; it is implied in the form . The event 4-vector is E = [ct, x, y, z] and the covariant form is Ep= [ct,-x,-y, -z] so that the invariant scalar product is EMEp= (ct)2_-(x2+y2+ z2). (3.14) A general Lorentz 4-vector xP transforms as follows: xU = Lxu (3.15) where y-3y 0 0 L= -py y 0 0 0 0 1 0 0 0 0 1 This is the operator of the Lorentz transformadon if the modon of 0' is along the xaxis of 0's frame of reference, and the inital dmes are synchronized (t = t' =0 at x = x' =0). Two important consequences of the Lorenlz transformation, discussed in 3.5, are that intervals of lime measured in two different inerdal frames are not the same; they are rebted by the equalion At'=yAt (3.16)  63 where At is an interval measured on adock at rest in 0's frame, and distances are given by Al'= AVy (3.17) where Al is a length measured on a ruler at rest in 0's frame. 3.4 The group structure of Lorenlz transformations The square of the invariant interval s, between the origin [0,0,0,0] and an arbitrary event x = [xO, x1, x2, x3] is, in index notaton s2= xfy,= x'x',,(sum over[L=0, 1, 2, 3). (3.18) The lower indices can be raised using the metric tensoriv = diag(1 , -1, -1, -1), so that 2= xxv = 'x'yv, (sum over [Land v). (3.19) The vectors now have contravariant forms. In matrix notation, the invariant is S2= XT1 X= x-1 x'. (320) (The transpose must be written explicity). The primed and unprimed column matrices (contravarant vectors) are related by the Lorentzmatrix operator, L x'=Lx. We therefore have xT1x = (L)T (L) = xTLT1 x. The x's are arbitrary, therefore  64 LT1L =1. (3.21) This is the defining property of the Lorentz transformatons. The set of all Lorenlz transformadons is the set L of all 4 x 4 matrices that satsfies the defining property L={L: LTL=1; Lall4x4 realmatices;i = diag(1,-1,-1,-1}. (Note that each L has 16 (independent) real matrix elements, and therefore belongs to the 16- dimensional space, R16). Consider the result of two successive Lorentz transformadons L1 and L2 that transform a 4- vectorx as follows x-->X - x where x'= Lix, and x" = L2x'. The resultant vector x" is given by x"= L2(Lix) = L2Lix where Lc= L2Li(Li followed by L2). (3.22)  65 If the combined operation L, is always a Loreneztransformadon then it must satsfy We must therefore have (L2L1)T1 (LL1) = 1 or L1T(L2T1L2)L1 =1 so that L1T1L1=q, (Li, L2E L) therefore Lc=L2L1EL. (3.23) Any number of successive Lorenz transformatons may be caried out to give a resultant that is itself a Lorentz transformaton. If we take the determinant of the defining equaton of L, det(LT1 L) = detri we obtain (detL)2=1 (detL = detLT) so that detL =±1. (3.24) Since the determinant of L is not zero, an inverse transformadon L-1exists, and the equadon L-1L =1, the identity, is always valid.  66 Consider the inverse of the defining equation (LT1L)-1=.nq-1, or L-1i-1(LT)-1=-1. Using =1-1, and rearranging, gives L-11(L-1)T =1 . (3.25) This result shows that the inverse L-1 is always a member of the set L. The Lorenlz transformadons L are matrices, and therefore they obey the associatve properly under matrix muliplicaton. We therefore see that 1. If L, and L2E& L , then L2 L1E& L 2.If L E L , then L-1E L 3. The identityl= diag(1, 1,1, 1) E L and 4. The matrix operators L obey associaivity. The set of all Lorentz transformatons therefore forms a group. 3.5 The rotation group Spatal rotatons in two and three dimensions are Lorenlz transformations in which the lime- component remains unchanged. In Chapter 1, the geometrical properties of the rotalion operators are discussed. In this section, we shall consider the algebraic structure of the operators.  67 Let 9t be a real 3x3 matrix that is part of a Loreneztransformation with a constant lime-component, 1000 0 (326) L= 0 91 0 In this case, the defining property of the Loreneztransformations leads to 1000 1000 1000 1000 0 0-100 0 0-100 0 9T 00-1 0 0 91 = 00-10 (327) 0 000-1 0 000-1 so that 91TR = 1, the identity matrix, diag(1,1,1). This is the defining property of a three-dimensional orthogonal matrix. (The related two -dimensional case is treated in Chapter 1). If x=[x1, x2, x3] is a three-vector that is transformed under91 to give x' then x'Tx'= xT 9x= xTx= x12+ x22+ x32=invarant under91. (328) The action of 91 on any three-vector preserves length. The set of all 3x3 orthogonal matrices is denoted by0(3), O(3)={J: 91R9 =IrIEReals}. The elements of this set satisfy the four group axioms.  68 3.6 The relativity of simultaneity: time dilation and length contraction In order to record the lime and place of a sequence of events in a particular inertal reference frame, it is necessary to introduce an infinite set of adjacent "observers", located throughout the entire space. Each observer, at a known, fixed posilion in the reference frame, caries a dock to record the lime and the characteristic property of every event in his immediate neighborhood. The observers are not concerned with non-local events. The docks caried by the observers are synchronized - they all read the same time throughout the reference frame. The process of synchronization is discussed later. It is the job of the chief observer to collect the information concerning the lime, place, and charateristic feature of the events recorded by all observers, and to construct the world line (a path in space-ime), associated with a partcular characteristicfeature (the type of pardde, for example). Consider two sources of light, I and 2, and a point M midway between them. Let E1 denote the event lash of light leaves 1", and E2 denote the event lash of light leaves 2". The events E1 and E2 are simutaneous if the flashes of light from 1 and 2 reach M at the same lime. The fact that the speed of light in free space is independent of the speed of the source means that simultaneity is relative. The docks of all the observers in a reference frame are synchronized by correcting them for the speed of light as follows: Consider a set of docks located at xo, x1, x2, x3, ... along the x-axis of a reference frame. Let xo be the chiefs dock, and let a flash of light be sent from the dock at x when it is reading to~ (12 noon, say). At the instant that the light signal reaches the dock at xi, it is set to read to + (x1/c), at the instant  69 that the light signal reaches the dock at x2, it is set to read to + (x2/c) , and so on for every dock along the x-axis. Al docks in the reference frame then "read the same lime"- they are synchronized. From the viewpoint of all other inertial observers, in their own reference frames, the set of docks, sychronized using the above procedure, appears to be unsychronized. It is the lack of symmetry in the sychronization of docks in different reference frames that leads to two non-intuitive results namely, length contraclion and time dilation. Length contraction: an application of the Lorentz transfomnafion. Consider a rigid rod at rest on the x-axis of an inerdal reference frame S'. Because it is at rest, it does not matter when its end-points xi' and x2'are measured to give the rest-, or proper-length of the rod, Lo' =x2-X1. Consider the same rod observed in an inerdal reference frame S that is moving with constant velocity- V with its x-axis parallel to the x'-axis. We wish to determine the length of the moving rod; we require the length L = x - x1 according to the observers in S. This means that the observers in S must measure x1 and x2 at the same time in their reference frame. The events in the two reference frames S, and S' are related by the spatial part of the Lorentz transformalion: x=-pyct + yx and therefore x2'- x1'-yc(t2-t)+y(x2- x1). where p= V/cand y =1/N(1- 2)  70 Since we require the length (x2 - xi) in S to be measured at the same tme in S, we must have t2- t1= 0, and therefore Lo'= X2- X1' =y(x2-x1), or Lo'(at rest) =yL (moving). (329) The length of a moving rod, L, is therefore less than the length of the same rod measured at rest, Lo because y>1. Time dilafion Consider a dock at rest at the origin of an inertal frame S', and a set of synchronized docks at xo, x1, x2, ... on the x-axis of another inertal frame S. Let S' move at constant speed V relatve to S, along the common x -, x'- axis. Let the docks at xo, and xo' be sychronized to read to , and to' at the instant that they coincide in space. Aproperfime interval is defined to be the time between two events measured in an inertial frame in which the two events occur at the same place. The tme part of the Lorenlz transformaton can be used to relate an interval of lime measured on the single dock in the S' frame, and the same interval of time measured on the set of synchronized docks at rest in the S frame. We have ct =yct'+ px or c(t2-t1)=yc(t2'-t1') + py(x2'- x1'). There is no separaion between a single dock and itself, therefore xe' - x1' =0, so that  71 c(t2-t1)(moving) =yc(t'-t1')(at rest) (y> 1). (3.30) A moving dock runs more slowly than adock at rest. In Chapter 1, it was shown that the general 2 x2 matiix operator transforms rectangular coordinates into oblique coordinates. The Lorenlz transformadon is a special case of the 2 x 2 matilces, and therefore its effect is to transform rectangular space-tme coordinates into oblique space-tme coordinates: x x tanr-S E [ct, x]or E'[ct', x] ct' ct The geometrical form of the Lorentz transformation The symmetry of space-time means that the transformed axes rotate through equal angles, tan-13. The relativity of simultaneity is clearly exhibited on this diagram: two events that occur at the same tine in the ct, x -frame necessarily occur at different ines in the oblique ct', x'-frame. 3.7 The 4-velocity A differenlial lime interval, dt, cannot be used in a Lorenaz-invariant way in kinematics. We must use the proper lime differenlial interval, dr, defined by  72 (cdt)2-dx2 = (cdt')2 - dx'2 = (ct)2. (3.31) The Newtonian 3-velodty is VN = [dx/dt, d yldt, dz/dt], and this must be replaced by the 4velodty V\ = [d(ct)/d, dx/dt, dy/dt, dz/dt] = [d(ct)Idt, dx/dt, dy/dt, dz/dt](dtldt) = [yC, yvN]. (3.32) The scalar product is then \1Vp = (yc)2- (yvN)2 (the transpose is understood) =(yc)2(l - (vN c)2) =c2. (3.33) The magnitude of the 4velodty is therefore|Vul = c, the invariant speed of light.  73 PROBLEMS 3-1 Two points A and B move in the plane with constant velocities|VAl = 2 m.s1 and IvBI = 212 m.s1. They move from their inital (t =0) positons, A(0)[1,1] and B(0)[6, 2] as shown: y,m 6 5 4 3 2 VA ---"" 0 0 1 2 3 4 5 6 7 8x,m Show that the closest distance between the points is IRimin = 2.529882..meters, and that it occursI .40...seconds after they leave their inital positions. (Remember that all inertal frames are equivalent, therefore choose the most appropriate for dealing with this problem). 3-2 Show that the set of all standard (moton along the common x-axis) Galilean transformadons forms a group. 3-3 A flash of light is sent out from a point xi on the x-axis of an inerdal frame S, and it is received at a point x2 = x1 +1. Consider another inerdal frame, S', moving with constant speed V = p3c along the x-axis; show that, in S':  74 i) the separation between the point of emission and the point of reception of the light is I'=I{(1-)(1+j3)}12 ii) the lime interval between the emission and reception of the light is At'= (VcX(1-3)1(1+ p3)}I 3-4 The distance between two photons of light that travel along the x-axis of an inertal frame, S, is alwaysI. Show that, in a second inertalframe, S', moving at constant speed V= Pc along the x-axis, the separation between the two phot ons is Ax'=l{(1+1p)/(1- p)}1. 3-5 An event [ct, x] in an inertial frame, S, is transformed under a standard Lorentz transformalon to [ct', x]in a standard primed frame, S', that has a constant speed V along the x-axis, show that the velodty components of the point x, x'are related by the equation vx= (vx' + V)/(1+(vxV/c2)). 3-6 An object called a KO-meson decays when at rest into two objects called--mesons (n±), each with a speed of 0.8c. If the K-meson has a measured speed ofO.9c when it decays, show that the greatest speed of one of the-mesons is (85/86)c and that its least speed is (5/14)c.  75 4 NEWTONIAN DYNAMICS Although our discussion of the geometry of motion has led to major advances in our understanding of measurements of space and lime in different inertial systems, we have yet to come to the crux of the matter, namely - a discussion of the effects of forces on the motion of two or more interacting partides. This key branch of Physics is called Dynamics. It was founded by Galileo and Newton and perfected by their followers, most notably Lagrange and Hamilton. We shall see that the Newtonian concepts of momentum and kinetic energy require fundamental revisions in the light of the Einstein's Special Theory of Relativity. The revised concepts come about as a result of Einstein's recognition of the crucial rle of the Prndple of Relativity in unifying the dynamics of allmechanical and optical phenomena. In spite of the conceptual dificulties inherent in the classical concepts, (dificulties that will be discussed later), the subject of Newtonian dynamics represents one of the great triumphs of Natural Philosophy. The successes of the dassical theory range from accurate descriptions of the dynamics of everyday objects to a detailed understanding of the motions of galaxies. 4.1 The law of inertia Galileo (1544.1642) was the first to develop a quanitative approach to the study of motion. He addressed the question - what property of motion is related to force? Is it the position of the moving object? Is it the velocity of the moving object? Is it the rate of change of its velocity? ...The answer to the queslion can be obtained only from observations; this is a basic feature of Physics that sets it apart from Philosophy proper. Galileo observed that force influences the changes in velocity (acceleralions)  76 of an object and that, in the absence of external forces (e.g: friction), no force is needed to keep an object in motion that is travelling in a straight line with constant speed. This observationally based law is called the Law of Inertia. It is, perhaps, difficult for us to appreciate the impact of Galileo's new ideas concerning motion. The fact that an object resting on a horizontal surface remains at rest unless something we call force is applied to change its state of rest was, of course, wellknown before Galileo's lime. However, the fact that the object continues to move after the force ceases to be applied caused considerable conceptual difficultes for the early Philosophers (see Feynman The Character ofPhysical Law). The observation that, in practice, an object comes to rest due to fricional forces and air resistance was recognized by Galileo to be a side effect, and not germane to the fundamental question of motion. Aristote, for example, believed that the true or natural state of motion is one of rest. It is instructive to consider Atistode's conjecture from the viewpoint of the Principle of Relativity - is a natural state of rest consistent with this general Prindple? According to the general Prindple of Relativity, the laws of motion have the same form in all frames of reference that move with constant speed in straight lines with respect to each other. An observer in a reference frame moving with constant speed in a straight line with respect to the reference frame in which the object is at rest would condude that the natural state or motion of the object is one of constant speed in a straight line, and not one of rest. All inertial observers, in an infinite number of frames of reference, would come to the same condusion. We see, therefore, that Aristode's conjecture is not consistent with this fundamental Principle.  77 42 Newton's laws of motion During his early twentes, Newton postulated three Laws of Moton that form the basis of Classical Dynamics. He used them to solve a wide variety of problems induding the dynamics of the planets. The Laws of Moton, first published in the Pdnciia in 1687, play a fundamental role in Newton's Theory of Gravitation (Chapter7); they are: 1. In the absence of an applied force, an object will remain at rest or in its present state of constant speed in a straight line (Galileo's Law of Inertia) 2. In the presence of an applied force, an object will be accelerated in the direction of the applied force and the product of its mass multplied by its acceleraton is equal to the force. and, 3. If a body A exerts a force of magnitudeIF1BI on a body B, then B exerts a force of equal magnitude IFBA on A.. The forces act in opposite directions so that FA=-FmA. In law number 2, the acceleration lasts only while the applied force lasts. The applied force need not, however, be constant in tme - the law is tue at all tmes during the moton. Law number 3 applies to "contact' interactions. If the bodies are separated, and the interaction takes a finite time to propagate between the bodies, the law must be modified to include the properties of the "field" between the bodies. This important point is discussed in Chapter7. 4.3 Systems of many interacting particles: conservation of linear and angular momentum  78 Studies of the dynamics of two or more interacting partidcles form the basis of a key part of Physics. We shall deduce two fundamental principles from the Laws of Moton; they are: 1) The Consewation ofLinear Momentum which states that, if there is a direction in which the sum of the components of the external forces acling on a system is zero, then the linear momentum of the system in that direction is constant, and 2) The Consewation of Angular Momentum which states that, if the sum of the moments of the external forces about any fixed axis (or origin) is zero, then the angular momertum about that axis (or origin) is constant. The new terms that appear in these statements will be defined later. The first of these prndples will be deduced by considering the dynamics of two interacting partidcles of masses mi and m2 wiith instantaneous coordinates [x, yi ] and [x2, y2], respecively. In Chapter 12, these prndples will be deduced by considering the invariance of the Laws of Moton under translations and rotatons of the coordinate systems. Let the external forces acting on the particles be F1and F2 , and let the mutual interactions be F21' and F12'. The system is as shown yF F1 F2 m2 F12' mi1 F21' 0 Resolving the forces into their x- and y-components gives  79 yA Fy1 Fx12"- F2 Fy21 1 F y1'"m 0__ x a) The equafions of motion The equatons of moton for each partide are 1) Resolving in the x-direclion Fx1 + Fa1'= mi1(d2x1/dt2) (4.1) and F-Fx12'=m2(d2x2/dt2). (4.2) Adding these equations gives Fx1 + F + (Fx1'- Fx12') = m1(d2xi/dt2) + m2(d2x2/dt2). (4.3) 2) Resolving in the y-direcdion gives a similar equaton, namely Fy1+ Fy2+(Fy1'-Fy12') = m1(d2y1/dt2) + m2(d2y2Idt2). (4.4) b) The role of Newton's 3rdLaw For instantaneous mutual interactions, Newton's 3rd Law gives F21'Il=1F12'I so that the x- and y-components of the internal forces are themselves equal and opposite, therefore lie total equations of motion are Fx1 + F = m1(d2x1/dt2) +m2(d2x2/dt2), (4.5)  80 and Fy1+ Fy2=m1(d2y1/dt2)+ m2(d2y2/dt2). (4.6) c) The consevafion of inear momentum If the sum of the external forces acing on the masses in the x-direction is zero, then Fx1 + Fx2=0 , (4.7) inwhich case, 0= m1(d2x1/dt2) + m2(d2x2/dt2) or 0= (d/dt)(mivx1) + (dIdt)(m2vx), which, on integraton gives constant=m vx1+m2vx2. (4.8) The product (mass x velocity) is the linear momentum. We therefore see that if there is no resultant external force in the x-direclion, the linear momentum of the two partides in the x-direcdion is conserved. The above argument can be generalized so that we can state: the linear momentum of the two partides is constant in any direclion in which there is no resultant external force. 4.3.1 Interaction of n-particles The analysis given in 4.3 can be carried out for an arbitray number of partides, n, with masses mi, m2, ...mo and with instantaneous coordinates [x1, yi], [x2, y2] ..[xn, yn]. The mutual interactions cancel in pairs so that the equadons of modon of the n-partides are, in the x-direclion  81 Fx1 + Fx2+ ... Fxn=mixi+m2x2 + ... maxn = sum of the x-components of (4.9) the external forces acling on the masses, and, in the y-direcion Fy1+Fy2+... Fyn=my1 + my2 + ...mnyn = sum of the y-components of (4.10) the external forces acing on the masses. In this case, we see that if the sum of the components of the external forces acing on the system in a particular direcion is zero, then the linear momentum of the system in that direclion is constant. If, for example, the direction is the x-axis then mvx1+m2vx2+ ...movxn = constant (4.11) 4.3.2 Rotation of two interacting particles about a fixed point We begin the discussion of the second fundamental conservation law by considerng the motion of two interacling partdes that move under the influence of external forces F1 and F2, and mutual interactions (internal forces) F21' and F12'. We are interested in the motion of the two masses about a fixed point 0 that is chosen to be the origin of Cartesian coordinates. The perpendiculars drawn from the point 0 to the lines of acion of the forces are R1, R2, and R'. The system is illustrated in the following figure.  82 y -2 F1 F12' /m2 m F21' + Moment R' O a) The moment of forces aboutearfxed oegin The totalmoment 12of the forces about the origin O is defined as I'12 = R1F1 + R2F2+(R'F12'- R'F21') (4.12) T T moment of moment of extemnal forces intemnal foroes A positive moment acts in a counter-clockwise sense. Newton's 3rd Law gives |F21'l =|F12'l, therefore the moment of the intemnal foroes about 0 is zero. (Their lines of action are the same). The total effecive moment about 0 is therefore due to the external foros, alone. Writing the moment in terms of the x- and y-components of F1and F2, we obtain F12= x1Fy1 +xFy2-y1Fx1-y2F~a (4.13)  83 b) The conseivation of angular momentum If the moment of the external forces about the origin 0 is zero then, by integration, we have constant= xipy+ x2py2-ypx1-y2px2. where pxi is the x-component of the momentum of mass 1, etc.. Rearranging, gives constant= (Xipyi -ypx) + (x2py2y2px2). (4.14) The rght-hand side of this equaton is called the angularmomentum of the two partides about the fixed origin, 0. Altematvely, we can discuss the conservaton of angular momentum using vector analysis. Consider a non-relativistc partide of mass m and momentum p, moving in the plane under the influence of an external force F about a fixed origin, 0: y F p m r < 0 x The angular momentum, L, of m about 0 can be written in vector form L = r x p. (4.15) The torque, F, associated with the external force F acting aboutO0 is T = r x F. (4.16) The rate of change of the angular momentum with lime is  84 dL/dt = r x (dp/dt) + p x (dr/dt) (4.17) = r x m(dv/dt)+mv x v = r xF(becausev xv=0) =T. If there is no external torque, T =0. We have, therefore F = dldt=0, (4.18) so that L is a constant of the moton. 4.3.3 Rotation of n-interacting particles about a fixed point The analysis given in 4.3.2 can be extended to a system of n-interacling partidcles. The moments of the mutual interaclions about the origin 0 cancel in pairs (Newton's 3rd Law) so that we are left Mth the moment of the external forces about 0. The equation for the total moment is therefore F1,2,....n= EF1,n] (xd(mivy)Idt-yd(miv)Idt). If the moment of the external forces about the fixed origin is zero then the total angular momentum of the system about 0 is a constant. This result follows direcly by integrating the expression for F1,2,...n=0. (4.19) If the origin moves with constant velocity, the angular momeritum of the system, relative to the new coordinate system, is constant if the extemaltorque is zero. 4.4 Work and energy in Newtonian dynamics 4.4.1 The principle of work: kinetic energy and the work done by forces Consider a mass m moving along a path in the [x, y]-plane under the influence of a  85 resultant force F that is not necessarily constant. Let the components of the force be Fx and Fy when the mass is at the point P[x, y]. We wish to study the moton of m in moving from a point A[x4, yA] where the force is FA to apoint B[xB, YB] where the force is FB. The equatons of moton are m(d2x/dt2)= F(4.20) and m(d2yldt2)= Fy (4.21) Multplying these equatons by dx/dt and dy/dt, respectively, and adding, we obtain m(dx/dt)(d2x/dt2) + m(dy/dt)(d2y/dt2) = Fx(dxIdt) + Fy(dyIdt). This equaton now can be integrated with respect tot, so that m((dx/dt)2 +(dy/dt)2)/2 = J(Fxdx+ Fydy). or mv2/2=J(Fxdx+ Fydy), (422) where v = ((dx/dt)2+ (dy/dt)2)1v is the speed of the partde at the point [x, y]. The term mv9/2 is called the dassicalkinetic energyof the mass m. It is important to note that the kinetic energy is ascalar. If the resultant forces aing on m are FA at A[XA, yA] at ime tA, and FB at B[xB, yB] at ime tB, then we have mvB2/2 - mvA2?/2 = [xA xB]Fxdx + ryA\yB]Fydy. (4.23) The terms on the right-hand side of this equalion represent the work done by the resultant forces acing on the partide in moving it from A to B. The equation is the mathemaical form of the general Pdnch/e  86 of Work the change in the kinetic energy of a system in any interval of lime is equal to the work done by the resultant forces acting on the system during that interval. 4.5 Potential energy 4.5.1 General features Newtonian dynamics involves vector quantities - force, momentum, angular momentum, etc.. There is, however, another form of dynamics that involves scalarquantities; a form that originated in the works of Huygens and Leibniz, in the 17th century. The scalar form relies upon the concept of energy, in its broadest sense. We have met the concept of kinetic energy in the previous section. We now meet a more abstract quantity called potentialenergy. The work done, W, by a force, F, in moving a mass m from a position sAto a position sB along a path s is, from section 4.3, W = fA BFds = the change in the kinetic energy during the motion, = [J[sAsB] Fdscosa, where ais the angle between F and ds. (424) If the force is constant, we can write W = F(sB - sA), where sB- sA is the arc length. If the motion is along the x-axis, and F = Fxis constant then W = FXxB -XA) the force multiplied by the distance moved. (4.25) This equation can be rearranged, as follows mv22- FxxB = mva2/2 -FxxA . (4.25)  87 This is a surprising result; the kinetc energy of the mass is not conserved during the moton whereas the quantty (mvx2/2 - Fxx) is consewed during the motion. This means that the change in the kinetc energy is exactly balanced by the change in the quantty Fxx. Since the quantity mv2/2 has dimensions of energy, the quandty Fxx must have dimensions of energy if the equation is to be dimensionally correct. The quandty-Fx is called the potental energy of the mass m, when at the positon x, due to the influence of the force Fx. We shall denote the potental energy by V. The negative sign that appears in the definiton of the potential energy will be discussed later when explicit reference is made to the nature of the force (for example, gravitatonal or electromagnetic). The energy equation can therefore be written TB+VB= TA+VA. (4.27) This is found to be a general result that holds in all cases in which a potential energy function can be found that depends only on the positon of the object (or objects). 4.5.2 Conservative forces Let Fx and Fy be the Cartesian components of the forces acting on a moving partide with coordinates [x, y]. The work done W1_a by the forces while the partde moves from the positon P1 [x1, yi] to another positon P2[x2, y2] is W1_2 = ftx1,Q4Fxdx + fy1,y2 F~dy (4.28) =fP,P2](Fxdx+F~dy). If the quantity Fxdx +F~dy is a perfect differental then a funcdon U =f(x, y) exists such that F=83UI8xand Fy =8UI8y. (4.29)  88 Now, the tota/differental of the funcion U is dU = (3U/3x)dx + (3U/Jy)dy (4.30) = Fxdx + Fydy. In this case, we can write jdU =f(Fxdx+ Fydy) = U =f(x, y). The definite integral evaluated between P1[xi, yi] and P2[x2, y2] is fll,P2(Fxdx+ Fydy)=f(x2, y2)-f(xl, y1)=U2-U1. (4.31) We see that in evaluatng the work done by the forces during the moton, no menton is made of the actual path taken by the partide. If the forces are such that the function U(x, y) exists, then they are said to be conservative. The funcdion U(x, y) is called the force function. The above method of analysis can be applied to a system of many parades, n. The total work done by the resultant forces acting on the system in movhg the partides from their inital configuraton, i, to their final configuraton, f, is Wit=_-1,n]Jkl,P] (Fimdxk+ Fyyk), (4.32) =Uf-Ui, a scalar quantty that is independent of the paths taken by the individual partdes. Pkl[xk1, yk1] and PK[xQ, ye] are the initial and final coordinates of the kth-partide. The potendal energy, V, of the system moving under the influence of conservahve forces is defined in terms of thefumnclion U: Vh-U. Examples of interaclions that take place via conservadve forces are:  89 1) gravitatonal interactions 2) electromagnetic interaclions and 3) interaclions between partides of a system that, for every pair of partides, act along the line joining their centers, and that depend in some way on their distance apart. These are the so-called central interactions. Frctional forces are examples of non-conservadve forces. There are two other major methods of solving dynamical problems that differ in fundamental ways from the method of Newtonian dynamics; they are Lagrangian dynamics and Hamiltonian dynamics. We shall delay a discussion of these more general methods untl our study of the Calculus of Variations in Chapter 9. 4.6 Particle interactions 4.6.1 Elastic collisions Studies of the collisions among objects, first made in the 17th-century, led to the discovery of two basic laws of Nature: the conservaton of linear momentum, and the conservation of kinetc energy associated with a special class of collisions called elastic collisions. The conservaton of linear momentum in an isolated system forms the basis for a quanttative discussion of all problems that involve the interaclions between partides. The present discussion wMill be limited to an analysis of the elastc collision between two partides. A typical two-body collision, in which an object of mass mi and momentum p1 makes a grazing collision with another object of mass  90 m2 and momentum p2(p2 < pi), is shown in the following diagram. (The coordinates are chosen so that the vectors p1 and p2 have the same directions). After the collision, the two objects move in directions characterized by the angles 0 and 4 with momenta p1'and p2'. Before After m e p1 m2 2 22 If there are no external forces acbng on the pardes so that the changes in their states of moton come about as a result of their mutual interaclions alone, the total linear momentum of the system is conserved. We therefore have P1+ p2 P1'+ p2' (4.33) or, rearranging to give the momentum transfer, P1-p1p2'-p2. The kinetc energy of a parade, T is related to the square of its momentum (T = p2/2m); we therefore form the scalar product of the vector equaon for the momentum transfer, to obtain p12-2pr p1'+ p1'2=p2'2-2p2' p2+ p22. (4.34) Introducing the scattering angles 0 and$4, we have  91 pi2-2pipi'c050 + pi'2= p3-2p2p2'cos4+ p2'2 . This equation can be written pi'2 (x2 - 2xc050+1 ) =p2'2(y2 -2ycos4 + 1) (4.35) where x= pi/pi" and y = pIp'. If we choose aframe in which p2 = 0 then y =0 and we have x2 - 2xcosO+1 = (p2'/p1 ')2. (4.36) If the collision is elastic, the kinetic energy of the system is conserved, so that T, + 0= Ti' + T2" (T2 = 0 because p2 =0). (4.37) Substituting T = p2/2m; and rearranging, gives (p2'Ipi ')2 = (m2/ml)(x2 - 1). We therefore obtain a quadratic equation in x: x2 + 2x(ml/(m2 - mi))cosO - [(m2 + ml)I(m2 - m,)] =0 . The valid solution of this equation is X = (T1iT1 ')1/2 = - (ml/1(m72 - mi1))coOe + {(ml/(m2 -11m)""'s2 + [(m12 + ml1)/(mY2 -11m1)]}1/2. (4.38) If mi = rm, the solution is x = 1/cosh, in which case have  92 p1+ (-p1') = p2',(4.40) leading to p1' $2 1 If the masses are equal then pi'=picOsO. In this case, the two partides always emerge from the elastic colision at right angles to each other (O+$=900). In the eary 1930's, the measured angle between two outgoing high-speed nudear partides of equal mass was shown to differ from 9Q>. Such experiments deary demonstrated the breakdown of Newtonian dynamics in these interacdions. 4.6.2 Inelastic collisions Collisions between everyday objects are never perfectly elastic. An object that has an internal structure can undergo inelastc collisions involving changes in its structure. Inelasccollisions are found to obey two laws; they are 1) the conservaton of linear momentum and 2) an empirical law, due to Newton, that states that the relatve velocity of the colliding objects, measured along their line of centers immediately after impact, is -e limes their relative velocity before impact.  93 The quanty e is called the coefficient of resttution. Its value depends on the nature of the materials of the colliding objects. For very hard substances such as steel, e is close to unity, whereas for very soft materials such as putty, e approaches zero. Consider, in the simplest case, the impact of two deformable spheres with masses m1 and m2. Let their velocities be vi and v2, and vi' and v2'(along their line of centers) before and afer impact, respecively. The linear momentum is conserved, therefore mivi1+ m2v2 = mivi'+ m2v2' and, using Newton's empirical law, v1'-v2'=-e(v1-v2). (4.41) Rearranging these equatons, we can obtain the values vi' and v2'after impact, in terms of their values before impact: vi'= [mlvi + m2v2-em2(v1 -v2)y(m1 + m2), (4.42) and v2'= [mlvi + m2v2 + emi(vi -v2)]/(m1 + m2). (4.43) If the two spheres initally move in directions that are not colinear, the above method of analysis is stll valid because the momenta can be resolved into components along and perpendicular to a chosen axis. The perpendicular components remain unchanged by the impact. We shall find that the classical approach to a quantitadve study of inelastc collisions must be radically altered when we treat the subject within the framework of Special Relatvity. It wMill be shown that the combined mass (mi1+ m2) of the colliding objects isnot conseivedin an inelastc collision.  94 4.7 The motion of rigid bodies Newton's Laws of Moton apply to every pointlike mass in an object of finite size. The smallest objects of praclical size contain very large numbers of microscopic partides - Avogadro's number is about 6 x 10F atoms per gram-atom. The motons of the individual microscopic partidcles in an extended object can be analyzed in terms of the moton of their equivalent total mass, located at the center of mass of the object. 4.7.1 The center of mass For a system of discrete masses, m, located at the vector positions, ri, the positon rCM of the center of mass is defined as rCM -i mir i mi =i mirI/M, where M is the total mass. (4.44) The center of mass (CM) of an (idealized) contnuous distribution of mass with a density p (mass/volume), can be obtained by considering an element of volume dV with an elemental mass dm. We then have dm = pdV. (4.45) The positon of the CM is therefore rCM = (1/M)Jrdm = (1/M)JrpdV. (4.46) The Cartesian components of rCM are xiCM = (I/M)JxpdV. (4.47) In non-uniform materials, the density is afuncdion of r. 4.7.2 Kinetic energy of a rigid body in general motion  95 Consider a rigid body that has both translalional and rotational motion in a plane. Let the angular velocity, o, be constant At an arbitrary lime ,t, we have y'y v - ' , thevelodtyofm y y r lativeto G O',G $ x' x' o=constant Total mass, M= m 0 x x Let the coordinates of an element of mass m of the body be [x, y] in the fixed frame (origin 0) and [x', y'] in the frame moving with the center of mass, G (origin 0'), and let u and v be the components of velodty of G, in the fixed frame. For constant angular velodty o, the instantaneous velodty of the element of mass m, relative to G has a direction perpendicular to the radius vectorr', and a magnitude v'= r'w. (4.48) The components of the instantaneous velocity of G, relative to the fixed frame, are u in the x-direction, and v in the y-direction. The velocity components of m in the [x, y]-frame are therefore u -r'osin$' =u -y'o in the x-direclion, and v + r'wcos$' = v + x'w in the y-direction.  96 The kinetc energy of the body, EK, of mass M is therefore EK = (1/2)Zm{(u -y'o)2 + (v + x'o)2} (4.49) =(1/2)M(u2 + v2) + (112)w2Zm(x'2 + y2) - uwmy' + vwmx'. Therefore EK = (1/2)MvG2 + (l12)IGW2, (4.50) where VG = (u2 + v2)1/2 the speed of G, relatve to the fixed frame, Amy'= Zmx' =0, by definiton of the center of mass, and IG = Zm(x'2 + y'2) = Ymr'2 ,is called the moment of imertia of M about an axis through G, perpendicular to the plane. We see that the total kinetc energy of the moving object of mass M is made up of two parts, 1) the kinetc energy of translafion of the whole mass moving with the velodty of the center of mass, and 2) the kinetc energy of rotation of the whole mass about its center of mass.  97 4.8 Angular velocity and the instantaneous center of rotation The angular velocity of a body is defined as the rate of increase of the angle between any line AB, fixed in the body, and any line fixed in the plane of the moton. If $ is the instantaneous angle between AB and an axis Oy, in the plane, then the angular velocity is d5/dt. Consider a drcular disc of radius a, that rolls Mthout sliding in contact with a line Ox, and let$ be the instantaneous angle that the fixed line AB in the disc makes with the yaxis. At t = 0, the rolling begins with the point B touching the origin, 0: y a vy VA vyY y-- B <-- 0 x P (corresponds to$=0) x At tme t, after the rolling begins, the coordinates of B[x, y] are x = OP - asin$ = BP - asin$ = a$-asin$ = a($-sin$), and y=AP-acos$=a(1 -cos$). The components of the veloidty of Bare therefore v= dx/dt = a(d$/dt)(1 -cos$), (4.51) and v= dy/dt = a(d$/dt)sin$. (4.52) The components of the acceleration of B are  98 ax= dvJdt = (d/dt)(a(dIdt)(1-cos$)) (4.53) = a(d4/dt)2sin$ + a(1-cos$)(d24/dt2), and ay= dv/dt = (d/dt)(a(dIdt)sin$) (4.54) = a(d4/dt)2cos$ + asin$(d2/dt2). If$=0, dxldt =0 and dy/dt =0, which means that the point P has no instantaneous velocity. The point B is therefore instantaneously rotatng about P with a velocity equal to 2asin(/2)(d$/dt); P is a "center of rotaton". Also, d2x/dt2 = 0 and d2y/dt2 = a(d$/dt)2, the point of contact only has an acceleration towards the center. 4.9 An application of the Newtonian method The folloMng example illustrates the use of some basic principles of classical dynamics, such as the conservaton of linear momentum, the conservaton of energy, and instantaneous rotaton about a moving point Consider a perfectly smooth, straight horizontal rod with a ring of mass M that can slide along the rod. Attached to the ring is a straight, hinged rod of length L and of negligible mass; it has a mass m at its end. At lime t =0, the system is held in a horizontal position in the constant gravitalional field of the Earth.  99 Att=o: m L[\ x=0att=0 At t =0, the mass m is released and falls under gravity. At time t, we have gvx x x=0 L vx Lsin$(d/dt) Lcos$(d$/dt) L(d$Idt)=instantaneous velodty ofm about M There are no external forces acing on the system in the x-direclion and therefore the horzontal momentum remains zero: M(dxldt) + m((dx/dt) - Lsin$(d$/dt)) =0. (4.56) Integratng, we have Mx+mx +mLcos$= constant. (4.57) If x=0 and$=0 at t=0, then mL = constant, (4.58) therefore (M +m)x +mL(cos$ -1) =0,  100 so that x= mL(1-cos$)/(M + m). (4.59) We see that the instantaneous positon x(t) is obtained by integratng the momentum equation. The equation of conservaton of energy can now be used; it is (MI2)v?2+(mI2)(vx- Lsin$(d$/dt))2 + (m/2)(Lcos$(d/dt))2 =mgLsin$. (The change in kinetc energy is equal to the change in the potental energy). Rearranging, gives (M+m)v? -2mLsinv(d4$dt)+(mL2(d$/dt)2-2mgLsin$)=0. (4.40) This is a quadratc in vx with a solution (M +m)vx = mLsin$(d$/dt)[1 ±{1-[(M + m)(mL2(d$dt)2 - 2mLgsin$)]/[m2L2(d$/dt)2sin24]}1]. The left-hand side of this equaton is also given by the momentum equation: (M + m)vx = mLsin$(d$/dt). We therefore obtain, after substtuton and rearrangement, d$/dt={[2(M+m)gsin$][L(M+mcos24)]}12, (4.41) the angular velodty of the rod of length L at lime t. PROBLEMS 4-1 A straight uniform rod of mass m and length 21lis held at an angle6c to the vertical. Its lower end rests on a perfecily smooth horizontal surface. The rod is released and falls under gravity. At lime t after the molion begins, we have  101 Inital positon Mass m, length 21 mg If the moment of inerta of the rod about an axis through its center of mass, perpendicular to the plane of the moton, is mU/3, prove that the angular velocity of the rod when it makes an angle 0with the vertcal, is d0ldt = {6g(cos0o - cos)/l(1 + 3sin20)}l2. 4-2 Show that the center of mass of a uniform solid hemisphere of radius R is 3R18 above the center of its plane surface. 4-3 Show that the moment of inertiaof a uniform solid sphere of radius R and mass M about a diameter is 2MR2/5. 4-4 A uniform solid sphere of mass m and radius r can roll, under gravity, on the inner surface of a perfectly rough spherical surface of radius R. The moton is in a vertical plane. At tme t during the motion, we have rolling sphere, mass m and radius r O - A R  102 Show that d2OIdt2 + [5g/(7(R- r))]sinO =0. As a preliminary result, show that rw= (R - r)(dO/dt) for rolling motion without slipping. 4-5 A partidcle of mass m hangs on an inextensible string of lengthI and negligible mass. The string is attached to a fixed point 0. The mass oscillates in a vertical plane under gravity. At lime t, we have 0 0I O= d8/dt Tension, T m mg Show that 1) d20/dt2+ (g/)sin0 =0. 2) o2 = (2gI)[coso - cos0o], where 6o is the initial angle of the string with respect to the vertical, so thato= 0 when 0 = 60. This equation gives the angular velocity in any position. 4-6 Letlo be the natural length of an elastic string fixed at the point0. The string has a negligible mass. Let a mass m be attached to the string, and let it stretch the string unlil the equilibrium position is reached. The tension in the string is given by Hooke's  103 law: Tension, T = X(extension)/orginal length, whereX is a constant for agiven material. The mass is displaced vertcally from its equilibrium positon, and oscillates under gravity. We have 0 Equilibrium General position g]J lo yE y(t) --------- i T E mg T 4fng Show that the mass oscillates about the equilibrium positon with simple harmonic moton, and that y(t) =lo + (mgloX){1- cos[t J\/Xmlo]} (starts with zero velocity at y(0) = b) 4-7 A dynamical system is in stable equilibrium if the system tends to return to its original state if slighty displaced. A system is in a posidon of equilibrium when the height of its center of gravity is a maximum ora aminimum. Consider a rod of mass m with one end resdng on a perfectly smooth verdcal wall OA and the other end on a perfecly smooth indined plane, OB. Show that, in the position of equilibrium  104 cotO = 2tan$, where the angles are given in the diagram: g]J4 A Center of gravity y())($(fixed angle) 0 Find y = f(6), and show, by considering derivadves, that this is a state of unstable equilibrium. 4-8 A partidcle A of mass ma =1 unit, scatters elastcally from a stalionary partidcle B of mass mB =2 units. If A scatters through an angle 0, show that the ratio of the kinetc energies of A, before (TA) and after (TA') scattering is (T/TA') = (-cosO + /3 + cos2O)2. Sketch the form of the variaton of this rato with angle in the range 0 :0 .m (This problem is met in practice in low-energy neutron-deuteron scattering).  105 5 INVARIANCE PRINCIPLES AND CONSERVATION LAWS 5.1 Invariance of the potential under translations and the conservation of linear momentum The equaton of moton of a Newtonian partde of mass m moving along the x-axis under the influence of a force Fx is mdxldt2 = Fx. (5.1) If Fx can be represented by a potentalV(x) then mdxldt2 = - dV(x)/dx. (5.2) In the special case in which the potential is not a funcion of x, the equation of moton becomes mdxldt2 =0, or md(vx)/dt =0. (5.3) Integratng this equaton gives mvx = constant (5.4) We see that the linear momentum of the partidcle is constant if the potental is independent of the positon of the partide. 5.2 Invariance of the potential under rotations and the conservation of angular momentum  106 Let a Newtonian partide of mass m move in the plane about a fixed origin, 0, under the influence of a force F. The equations of motion, in the x-and y-direcions, are md2x/dt2= Fx and md2yldt2= Fy. (5.5 a,b) If the force can be represented by a potenalV(x, y) then we can write md2x/dt2 =-3V/xand md2yldt2 =-3VIy. (5.6 a,b) The total differentalof the potentalis dV = (3V/3x)dx+ (3V/3y)dy. Let a transformaton from Cartesian to polar coordinates be made using the standard linear equatons x= rcos andy= rsin$. The partial derivatives are x/3$=--rsin$=-y, x/r= cos$, 3y/34= rcos$= x, and Jy/3r = sin$. We therefore have 8V/8$=(3V/3x)(3x/8$) + (3VI3y)(3y&$) (5.7) = (3V/x)(-y) + (3V/3y)(x) = yFx + x(-Fy) =m(yax- xay) (ax and ay are the components of acceleraton) = m(d/dt)(yvx-xvy) (vx and vy are the components of velocity). If the potendal is independent of the angle$4 then 3V/8$=0, (5.8) in which case  107 m(d/dt)(yvx- xvy) =0 and therefore m(yvx- xvy) = a constant (5.9) The quantty on the left-hand side of this equaton is the angular momentum (ypx - xpy) of the mass about the fixed origin. We therefore see that if the potentid is invariant under rotatons about the origin (independent of the angle4), the angular momentum of the mass about the origin is conserved. In Chapter 9, we shall treat the subject of invariance prndples and conservaton laws in a more general way, using arguments that involve the Lagrangians and Hamiltonians of dynamical systems.  108 6 EINSTEINIAN DYNAMICS 6.1 4-momentum and the energy-momentum invariant In Classical Mechanics, the concept of momentum is important because of its role as an invariant in an isolated system. We therefore introduce the concept of 4-momentum in Relatvistc Mechanics in order to find possible Lorenlz invariants invoMng this new quandty. The contravariant 4-momentum is defined as: P=mV (6.1) where m is the mass of the particle. (It is a Lorentz scalar- the mass measured in the rest frame of the partidcle). The scalar product is PuPp=(mc)2. (6.2) Now, P= [myc, myvN] (6.3) therefore, P Pp=(myc)2 - (yvN)2. Writing M =yin, the relatvistic mass, we obtain Pp= (Mc)2 -(MvN)2 = (mc)2. (6.4) Multiplying throughout by c2 gives  109 M2c4-M2vN2c2 = m2c4. (6.5) The quandty Mc2has dimensions of energy; we therefore write E = Mc2, (6.6) the total energy of a freely moving partide. This leads to the fundamental invariant of dynamics c2PuP = E2-(pc)2 = E°2(6.7) where Eo=mc2 is the rest energy of the partidcle, and p is its relativistic 3-momentum. The total energy can be written: E = yEo = Eo + T, (6.8) where T = Eo(y -1), (6.9) the relativistic kinetic energy. The magnitude of the 4-momentum is a Lorenlz invariant IPI=mc. (6.10) The 4- momentum transforms as follows: P'p= LPM. (6.11) 6.2 The relativistic Doppler shift For relative modon along the x-axis, the equation P'" = LPs is equivalent to the equations E'= yE -pycpx (6.12)  110 and, cp'x= -PyE+ ycpx. (6.13) Using the Planck-Einstein equations E = hv and E = pc for photons, the energy equation becomes v'=yv- pyv =yv(1- ) = v(1- )/(1 - 2)1/2 = v{(1- O)(1 + p)}1/2. (6.14) This is the relativistic Doppler shift for photons of the frequency v', measured in an inertial frame (primed) in terms of the frequencyv measured in another inertial frame. 6.3 Relativistic collisions and the conservation of 4-momentum Consider the interaction between two partides, 1 and 2, to form two partides, 3 and 4. (3 and 4 are not necessarily the same as 1 and 2). The contravariant 4-momenta are PP": Before After 31 PP Pl P2[t 0 2 4 P4[i 1+2 - 3+4 All expeiments are consistent with the fact that the 4momentum of the system is conserved. We have, for the contravariant 4-momentum vectors of the interacting partides,  111 P1 + P2" = P3"+P4"(6.15) T T inital"free"state final "free"state and a similar equaton for the covarant 4-momentum vectors, P1I+ P2, = P3 + P4. (6.16) If we are interested in the change PI--> P3, then we require PP- P3"= P4- P2 (6.17) and P - P31 = P4[ - P2 .(6.18) Forming the invariant scalar products, and using P PPt= (EPIc)2, we obtain (E1O/c)2- 2(E1E/c2- p1.p3) + (E30/c)2 = (E40Ic)2-2(E2E4dc2-p2-p4) + (E2OIc)2. (6.19) Introdudng the scattering angles, 0 and 4, this equaton becomes E12-2(EIE3-c2plp3cos0) + E302= E202-2(E2E4-c2p2p4cos4) + E402. If we choose a reference frame in which partide 2 is at rest (the LAB frame), then p2= 0 and E2= E20, so that E102-2(E1E3-c2p1p3cos0) + E302 = E202-2E20E4+ E02. (620) The total energy of the system is conserved, therefore E1 + E2 = E3 + E4 = E1 + E22 (6.21)  112 or E4= E1+ E20- E3 Eliminatng E4 from the above "scalar product'equation gives E102-2(EIE3-c2plp3cosO) + E302= E402- E202-2E20(E1- E3). (622) This is the basic equation for all interaclions in which two relatvistc entities in the initial state interact to give two relativistic entities in the final state. It applies equally well to interacdions that involve massive and massless entties. 6.3.1 The Compton effect The general method discussed in the previous secion can be used to provide an exact analysis of Compton's famous experiment in which the scattering of a photon by a stationary, free electron was studied. In this example, we have E1= Es (the inddent photon energy), E2= Ee0 (the rest energy of the stationary electron, the "target"), E3 = Eph' (the energy of the scattered photon), and E4 = Ee' (the energy of the recoiling electron). The "rest energy"of the photon is zero: E= pc Ee0 The general equation (6.22), is now 0- 2(Eo~Em' - Eo~Eo'cos0) = Eeo2 -2Ee0(Epti+ Ee0 - Eph') + Eeo2 (6.23)  113 or -2EEp'(1- cos) = -2Ee0(Ep - Ep') so that Ep- E,'= EpEp'(1- cos)/EeO. (6.24) Compton measured the energy-loss of the photon on scattering and its cos-dependence. 6.4 Relativistic inelastic collisions We shall consider an inelastic collision between a partide 1 and a partide 2 (initially at rest) to form a composite partide 3. In such a collision, the 4-momentum is conserved (as it is in an elastic collision) however, the kinetic energy is not conserved. Part of the kinetic energy of partide 1 is transformed into excitation energy of the composite partide 3. This excitation energy can take many forms - heat energy, rotational energy, and the excitation of quantum states at the microscopic level. The inelastic collision is as shown: Before After 1 2 3 0 pi p2=0 p3 Rest energy: E10 E2 E30 Total energy: E1 E2= E20 E3 3-momentum: pi p2=0 p3 Knetic energy: T1 T2 =0 T3 In this problem, we shall use the energy-momentum invariants associated with each partide, directly: i) E12-(pc)2= E102 (6.25)  114 ii) Ei =20 (6.26) iii) E32 -(p3c)2 =E302. (6.27) The total energy is conserved, therefore El + E2= E = El + E20. (6.28) Introdudng the kinetic energies of the parades, we have (Ti + E10) + E22= T3 + EF°. (6.29) The 3-momentum is conserved, therefore pi+0= p3. (6.30) Using E3 - (p3c)2, (6.31) we obtain E3 (i+ E20)Y - (p3c)2 = E12+2E1E22+E222(pc2 = E102+2E1E29+ E2o2 = E102 + E202 + 2(T1 + E1O)E22 (6.32) or E3 °+ E20)2 + 2T1 E20 (E3°> E,° + E20). (6.33)  115 E3I= 2(y1 + 1)E1°02. (6.35) 6.5 The Mandeistam variables In discussions of relativistc interacions it is often usefrul to introduce additional Lorentz invariants that are known as Mandeistam variables. They are, for the spedial case of two par~des in the initial and final states (1 + 2 --> 3 +4): s = (Plk + P2")[P1 I + P2]J, the total 4-momentum invariant =((El + E2)Ic, (pi + p2))[(EI + E2)IC, -(pi + p2)] _ (El + E2)21c2 -(pi + p2)2, a Lorenlz invariant, (6.36) t = (Pl'k - P?)[P1~t - P3]J, the 4-momentum transfer (1 -->3) invariant = (E, - E3)21c2 - (p, - p3), a Lorenlz invariant, (6.37) and u = (Pi" - P4)[P1, - P4J], the 4-momentum transfer (1 -->4) invariant = (El - E4)2c2 - (p, - p4)2, a Lorenlz invariant (6.38) Now, sc2 = E12 + 2E1E2 + E-(pi2 + 2prp2 + p22)c2 = E1°02+ E292+2E1E2-2p.p2c2  116 The Mandelstam variable sc2 has the same value in all inertal frames. We therefore evaluate it in the LAB frame, defined by the vectors [E1L, piLc] and [E2L = E20, -p2c = 0], (6.40) so that 2(E1LE2L - piL.p2Lc2) =2E1LE20, (6.41) and SC2= E102+ E202+2E1LE20. (6.42) We can evaluate sc2 in the center-of-mass (CM) frame, defined by the conditon p1iM+ pP =0 (the total 3-momentum is zero): sc2=(E1M+ E2). (6.43) This is the square of the total CM energy of the system. 6.5.1 The total CM energy and the production of new particles The quantity cols is the energy available for the producdion of new pardcles, or forexcidng the internal structure of partidcles. We can now obtain the relaton between the total CM energy and the LAB energy of the inddent partide (1) and the target (2), as follows: sc2= E102+ E202+ 2E1LE20= (E1{m+ E2 )2=W2, say. (6.44) Here, we have evaluated the left-hand side in the LAB frame, and the rght-hand side in the CM frame. At very high energies, cis>> E10 and E20, the rest energies of the partices in the inital state, in which case, W2= sc2~2E2LE20. (6.45)  117 The total CM energy, W, available for the produclion of new partides therefore depends on the square root of the inddent laboratory energy. This result led to the development of colliding, or intersecing, beams of partides (such as protons and anti-protons) in order to produce sufficient energy to generate partides with rest masses greater than 100 limes the rest mass of the proton (~109e\/. 6.6 Positron-electron annihilation-in-flight A discussion of the annihilation-in-flight of a relativistic positron and a stationary electron provides a topical example of the use of relativistic conservation laws. This process, in which two photons are spontaneously generated, has been used as a source of nearly monoenergetic high- energy photons for the study of nudear photo-disintegration since 1960. The general result for a1 + 2 - 3+ 4 interaclion, given in section 6.3, provides the basis for an exact calculation of this process; we have E1= Eg (the inddent positron energy), E2= Ee0 (the rest energy of the stationary electron), F = Ep1 (the energy of the forward-going photon), and E4 = E@2 (the energy of the backward-going photon). The rest energies of the positron and the electron are equal. The general equation (622), now reads Eeo-2{EgEp1 -cpgEo1(cos)} +0=0- Ee2-2Ee0(Es- EphI) (6.46) therefore, Eph1{Epcs+ Ee0-[Eg02 - Eeo2]12 cos0} = (Eg + Ee0)Ee0, givng E1= EeO/(1 -kcosO) (6.47)  118 where k = [(Epos - EeO)/(Epos + Eel)]1. The maximum energy of the photon, E1"x occurs when 0 = 0, corresponding to moton in the forward direction; its energy is Ep1x= Eoe(1 -k). (6.48) If, for example, the inddent total positron energy is 30 MeV, and E& = 0.511 MeV then Ep1x= 0.51141 -(29.489/30.511)1]MeV = 3025 MeV. The forward-going photon has an energy equal to the kinetic energy of the incident positron (Ti= 30 -0.511 MeV) plus approximately three-quarters of the total rest energy of the positron-electron pair (2Ee0 = 1.02 MeV). Using the conservaton of the total energy of the system, we see that the energy of the backward-going photon is approximately 0.25 MeV. The method of positron-electron annihiladon-in-flight provides one of the very few ways of generatng neady monoenergetc photons at high energies. PROBLEMS 6-1 A partidcle of rest energy E0 has a relatvistc3-momentum p and a relativistickinetc energy T. Show that 1)|p1l= (l/c)(2TE)1'2{1 +(TI2EO)}12, and 2)|lvil= c{1 +[Eo2/T(T +2EO)]}-12, where v is the 3-velocity.  119 6-2 Two similar relatvistc pardcles, A and B, each with rest energy E, move towards each other in a straight line. The constant speed of each parade, measured in the LAB frame is V= 3c. Show that their total energy, measured in the rest frame of A, is E'(1 + p2)/( _ P2). 6-3 An atom of rest energy EAR is initally at rest. It completely absorbs a photon of energy Eo, and the excited atom of rest energy EP* recoils freely. If the excitaton energy of the atom is given by Eex= Ee*- EAR, show that Eex = -EA0 + EA0{1 + (2E/EA0)}1, exacly. If, as is often the case, Ep « EAP, show that the recoil energy of the atom is Ee=E,2/2EA0. Explain how this approximaton can be deduced using a Newtonian-like analysis. 6-4 A completely inelastic collision occurs betieen partide 1 and particle 2 (initally at rest) to form a composite parade, 3. Show that the speed of 3 is v3=v1/{1 +(E2E1)}, where vi and E1are the speed and the total energy of 1, and E is the rest energy of 2. 6-5 Show that the minimum energy that a y-ray must have to just break up a deuteron into a neutron and a proton isymin2.23 MeV, given Eno=939.5656 MeV, EoP = 938.2723 MeV, and  120 EdeuP=1875.6134 MeV. 6-6 In a general relativistic collision: 1 +2 -- n-partides - (3 + 4+...m) +(m+1, m+2. +...n) where the partdes 3 -- m are "observed", and the partides m+ 1-- n are "unobserved". We have E1+E2=(E3+ E4+...Em)+(Em+ +Em+2+...En), the total energy, = Eos+Euns and P1+ P2= pobs + pub. If Wun*Ic2 is the unobserved (missing) mass of the partides m+1 to n, show that, in the LAB frame (WuncbS)2 = (E1L+ E20- Z=3,m] EiL)2 - (piLC( -C=3,m] piL)2. This is the missing (energy)2 in terms of the observed quantties. This is the prndple behind the so-called "missing-mass spectrometers" used in Nudear and Partide Physics. 6.7 If the confravarant 4-force is defined as F1 = dP1/dt = [fT, f] where t is the proper lime, and Pis the contravariant 4-momentum, show that FV=0, where V~ is the covariant 4-velocity. (The 4- force and the 4-velocity are orthogonal). Obtain dEldt in terms of y, v, and f.  121 7 NEWTONIAN GRAVITATION We come now to one of the highlights in the history of intellectual endeavor, namely Newton's Theory of Gravitaton. This spectacular work ranks with a handful of masterpieces in Natural Philosophy - the Galileo-Newton Theory of Moton, the Camot-Clausius-KeMn Theory of Heat and Thermodynamics, Maxwell's Theory of Electromagnedsm, the MaxwellBoltzmann-Gibbs Theory of Statstcal Mechanics, Einstein's Theories of Spedal and General Relativity, Planck's Quantum Theory of Radiation, and the Bohr-deBroglie-Schrtdinger-Heisenberg Quantum Theory of Matter. Newton's most significant ideas on Gravitaton were developed in his eary twendes at a time when the University of Cambridge closed down because of the Great Plague. He retumed to his home, a farm at Woolsthorpe-by-Colsterworth, in Lincolnshire. It is a part of England dominated by vast, changing skies; a region buffeted by the winds from the North Sea. The thoughts of the young Newton naturally turned skyward - there was lite on the ground to str his imagination except, perhaps, the proverbial apple tree and the falling apple. Newton's work set us on a new course. Before discussing the details of the theory, it will be useful to give an overview using the simplest model, consistent with logical accuracy. In this way, we can appreciate Newton's radical ideas, and his development of the now standard "Scientific Method "in which a crucial interplay exists between the results of observatons and mathemadcal models that best account for the observatons. The great theories are often based upon relatvely small numbers of observations. The uncovering of  122 the Laws of Nature requires deep and imaginatve thoughts that go far beyond the demonstraton of mathemaical prowess. Newton's development of Differental Calculus in the late 1660's was strongly influenced by his attempts to understand, analytcally, the empirical ideas concerning moton that had been put forward by Galileo. In partcular, he investigated the analytical properties of motion in curved paths. These properties are required in his Theory of Gravitaton. We shall consider motion in 2-dimensions. 7.1 Properties of motion along curved paths in the plane The velocity of a point in the plane is a vector, drawn at the point, such that its component in any diredion is given by the rate of change of the displacement, in that direcdion. Consider the following diagram y B y + Ay PQ y 3A A Ax 0 x x x+Ax t t+ At P and Q are the posidons of a point moving along the curved path AB. The coordinates are P [x, y] at Sme t and Q [x + Ax, y + Ay] at Sme t + At. The components of the velocity of the point are lim(At--0) Ax/At = dx/dt=vx and  123 Iim(At-O) Ay/At = dy/dt=vy. Ax and Ay are the components of the vector PQ. The velocity is therefore lim(At-O) chordPQlAt. We have lim(Q-P) chord PQ/As =1, where s is the length of the curve AP and As is the length of the arc PQ. The velocity can be written Iim(At-O) (chordPQ/As)(As/At) = ds/dt (7.1) The direction of the instantaneous velocity at P is along the tangent to the path at P. The x- and y-components of the acceleration of P are lim(At-O) AvlAt = dvldt = d2xldt2, and lim(At-O) Avy/At= dv/dt= d2yldt2. The resultant acceleration is not directed along the tangent at P. Consider the motion of P along the curve APQB: y B v+ Av Q A V A O0 x  124 The change Av in the vectorv is shown in the diagram: v The vector Av can be written in terms of two components, a, perpendicular to the direction of v, and b, along the direction of v + Av: The acceleration is lim(At-)0) v/At, The component along a is lim(At--O) AaIAt=lim(At--O) vAy/At = lim(At--O) (vAy/As)(As/At) = v2(dy/ds)=v2Ip (7.2) where p = ds/dy, is the radius of curvature at P. (7.3) The direction of this component of the acceleration is along the inward normal at P. If the partide moves in a rle of radius R then its acceleration towards the center is v2/R, a result first given by Newton. The component of acceleration along the tangent at P is dv/dt = v(dv/ds) = ds/dt2. 7.2 An overview of Newtonian gravitation Newton considered the fundamental properties of motion, embodied in his three Laws, to be universal in character - the natural laws apply to all motions of all pardles throughout all space, at all times. Such considerations form the basis of a Natural Philosophy. In the Pnnchia, Newton wrote ..."  125 began to think of gravity as extending to the orb of the Moon..." He reasoned that the Moon, in its steady orbit around the Earth, is always acceleratng towards the Earth. He estmated the acceleration as follows: If the orbit of the Moon is drcular (a reasonable assumpton), the dynamical problem is v Moon Earth R The acceleraton of the Moon towards the Earth is IaR|= v2/R Newton calculated v = 2TRFF, where R =240,000 miles, and T = 27.4 days, the period, so that aR=4R2P12 0.007 ft/se2. (7.4) He knew that all objects, close to the surface of the Earth, accelerate towards the Earth with a value determined by Galileo, namely g 32 ft/sec2. He was therefore faced with the problem of explaining the origin of the very large difference between the value of the acceleraton a , nearly a quarter of a million miles away from Earth, and the local value, g. He had previously formulated his 2nd Law that relates force to acceeration, and therefore he reasoned that the difference between the acceleratons, a~ and g, must be associated with a propeby  126 of the force acling between the Earth and the Moon - the force must decrease in some unknown way. Newton then introduced his conviction that the force of gravity between objects is a universal force; each planet in the solar system interacts with the Sun via the same basic force, and therefore undergoes a characteristc acceleraton towards the Sun. He conduded that the answer to the problem of the nature of the gravitatonal force must be contained in the three empirical Laws of Planetary Moton announced by Kepler, a few decades before. The three Laws are 1) The planets descibe ellipses about the Sun as focus, 2) The line joining the planet to the Sun sweeps out equal areas in equal intervals of time, and 3) The period of a planet is proportional to the length of the semi-major axis of the orbit, raised to the power of 3/2. These remarkable Laws were deduced after an exhaustve study of the moton of the planets, made over a period of about 50 years by Tycho Brahe and Kepler. The 3rd Law was of particular interest to Newton because it relates the square of the period to the cube of the radius for acircular orbit: T2 oc R3 (7.5) or T2= CR3,  127 where C is a constant. He replaced the specific value of (R/T2) that occurs in the expression for the acceleration of the Moon towards the Earth with the value obtained from Kepler's 3rd Law and obtained a value for the acceleraton aR: aR = v2/R = 42R/T2(Newton) (7.6) but R/T2 =1/CR2 (Kepler) (7.7) therefore aR = 42(R/J2) = (4n2/C)(1/R2) (Newton). (7.8) The acceleration of the Moon towards the Earth vanes as the inverse square of the distance between them. Newton was now prepared to develop a general theory of gravitaton. If the acceleraton of a planet towards the Sun depends on the inverse square of their separaton, then the force between them can be written, using the 2nd Law of Moton, as follows F = M t at= Mt(4T2IC)(1/R2). (7.9) At this point, Newton introduced the first symmetiy argument in Physics: if the planet experiences a force from the Sun then the Sun must experience the same force from the planet (the 3rd Law of Moton!). He therefore argued that the expression for the force between the planet and the Sun must contain, explicity, the masses of the planet and the Sun. The gravitadonal force FG between them therefore has the form  128 FG= GMsunMpbneR2, (7.10) where G is a constant. Newton saw no reason to limit this form to the Sun-planet system, and therefore he announced that for any two spherical masses, M1 and M2, the gravitatonal force between them is given by FG= GM1M2/R2, (7.11) where G is a universal constant of Nature. All evidence points to the fact that the gravitatonal force between two masses is always attractive. Returning to the Earth-Moon system, the force on the Moon (mass MM) in orbit is FR = GMEMM/R2 = MMaR (7.12) so that aR = GME/R2, which is independent of MM. (The cancellaton of the mass M in the expressions for FR involves an important point that is discussed later in the section 8.1). At the surface of the Earth, the acceleraton, g, of a mass M is essentially constant It does not depend on the value of the mass, M, thus g = GMEIRE2, where RE is the radius of the Earth. (7.13) (It took Newton many years to prove that the entre mass of the Earth, ME, is equivalent to a point mass, ME, located at the center of the Earth when calculadng the Earth's gravitadonal interaction with a mass on its surface. This result depends on the exact 1/R2-nature of the force).  129 The rato of the accelerations, adlg, is therefore adlg = (GME/R2)/(GME/RE2) = (RE/R)2. (7.14) Newton knew from observatons that the rato of the radius of the Earth to the radius of the Moon's orbit is about 1/60, and therefore he obtained adlg~(1/60)2 = 1/3600. so that aR = g/3600 = (3213600)fVsec2 = 0.007...f/sec2. In one of the great understatements of analysis, Newton said, in comparing this result with the value for aR that he had deduced using aR = v2IR ..."that it agreed pretty nearly" ...The discrepancy came largely from the errors in the observed ratio of the radii. 7.3 Gravitation: an example of a central force Central forces, in which a particle moves under the influence of a force that acts on the partide in such a way that it is always directed towards a single point - the center of force - form an important class of problems. Let the center of force be chosen as the origin of coordinates: V m P [r,] F r Center of Force$ 0 * x  130 The description of partide motion in terms of polar coordinates (Chapter2), is well-suited to the analysis of the central force problem. For general motion, the acceleration of a point P [r, ] moving in the plane has the folloMng components in the r- and "$"-direclions ar = udd2r/dt2- r(d$/dt)2), (7.15) and a,= u,(r(d2f/dt2) + 2(drdt)(dIdt)), (7.16) where Ur and u, are unit vectors in the r- and4-direcions. In the central force problem, the force F is always directed towards 0, and therefore the component a,, perpendicular to r, is always zero: a,= u,(r(d2f/dt2) + 2(drdt)(dIdt) =0, (7.17) and therefore r(d24/dt2) + 2(drdt)(dIdt) =0. (7.18) This is the equation of motion of a partde moving under the influence of a central force, centered at 0. If we take the Sun as the (fixed) center of force, the motion of a planet moving bout the Sun is given by this equation. The differential equation can be solved by making the substitution CO= d$/dt, (7.19) givng rdoldt+ 2o(dr/dt) =0, (7.20) or rdo = -2odr.  131 Separatng the variables, we obtain dw/w = -2dr/r. Integratng, gives lo9eo =-2loger+ C (constant), therefore Ioge(wr2) = C. Taking antlogs gives rko = r2(d$/dt) = ec= k, a constant (721) 7.4 Motion under a central force and the conservation of angular momentum The above solution of the equation of motion of a partidcle of mass m, moving under the influence of a central force at the origin, 0, can be multiplied throughout by the mass m to give mr2(d$/dt) = mk (7.22) or mr(r(d$/dt)) = K, a constant for a given mass, (723) We note that r(d$/dt) = v,, the component of velocity perpendicular to r, therefore angular momentum of m about 0= r(mv,)= K, a constant of the motion fora central force. 7.5 Kepler's 2nd law explained The equation r2(d$/dt) = constant, K, can be interpreted in terms of an element of area swept out by the radius vector r, as follows  132 0 v x From the diagram, we see that the following inequality holds r2A4/2 x p>x C>x/d The moment of the velocity about 0 is then r(r(d$/dt) = pv = x(dy/dt)- y(dx/dt) = a constant, h, say. (725) The result r2(d$/dt) = constant for a central force can be derved in the folloMng altemative way: The lime dervalve of r2(d$/dt) is (d/dt)(r2(d$/dt)) = r2(d2f/dt2) +(d$/dt)2r(dr/dt) (7.26) If this equation is dMded throughout by rthen (1Ir)(d/dt)(r2(dIdt))= r(d24/dt2)+2(dr/dt)(d/dt) (727) = the transverse acceleration =0 for a central force. (7.28) Integraling then gives r2(d$/'dt)= constant for a central force. (7.29)  134 7.6.1 The law of force in [p, r] coordinates There are advantages to be gained in using a new set of coordinates- [p, r] coordinates- in which a point P in the plane is defined in terms of the radial distance r from the origin, and the perpendicular distance p from the origin onto the tangent to the path at P. (See following diagram). Let a partide of unit mass move along a path under the influence of a central force directed towards a fixed point, 0. Let ac be the central acceleraton of the unit mass at P, let the perpendicular distance from 0 to the tangent at P be p, and let the instantaneous radius of curvature of the path at the point P be p: Central orbit v Component of acceleraton P [r, p] along inward normal at P, al r p Center of Force -- 0 The component of the central acceleraton along the inward normal at P is a1 =acsina= v2/p= ac(p/r). (7.30) The instantaneous radius of curvature is given by p = r(dr/dp). (7.31) For a//central forces, pv=constant= h, (7.32)  135 therefore a1 = v2/p = (h2/p2)(1/r)(dp/dr)= a(p/r), (7.33) so that ac = (h2/p3)(dp/dr). (7.34) This differential equaton is the law of force per unit mass given the orbit in [p, r] coordinates. (It is left as a problem to show that given the orbit in polar coordinates, the law of force perunit mass is a,= h2u2{u + d2uId4p}, where u =1/r). (7.35) In order to find the law of force per unit mass (acceleraton), given the [p, r] equation of the orbit, it is necessary to calculate dp/dr. For example, if the orbit is parabolic, the [p, r] equaton can be obtained as follows Tangent at P p r Apex , A >x F , the Focus The triangles FAQ and FQP are similar, therefore p/a = r/p, where AF = a, (7.36) givng 1/p2=1/iar, the p-r equadon of a parabola. (7.37) Differentatng this equadon, we obtain  136 (1/p3)dp/dr=1/2ar2. (7.38) The law of acceleration for the parabolic central orbit is therefore ac = (h2/p3)dp/dr = (h2/2a)(1/r2) = constant/r2. (7.39) The instantaneous speed of P is given by the equadon v = h/p; we therefore find v=h/ ar. (7.40) This approach can be taken in discussing central orbits with elliptc and hyperbolic forms. Consider the ellipse Q R b pr2 x F1 F2 a The foci are F1and F2, the semi-major axis is a, the semiminor axis is b, the radius vectors to the point P [r, ] are ri and r2 , and the perpendiculars from F1and F2 onto the tangent at P are p, and p2. Using standard results from analytic geometry, we have for the ellipse 1) r1+ r2=2a, (7.41 a-c) 2) plp2= b2, and 3) angle QPF1 = angle RPF2. The triangles F1QP and F2RP are similar, and therefore p1/r1= p2/r2 (7.42)  137 or (p1 p2/rIr2)112= b/{r1(2a - r1)}1/2= pun1 so that b2/p,2 = 2alri-1. (7.43) This is the [p, r] equaton of an ellipse. The [p, r] equation for the hyperbola can be obtained using a similar analysis. The standard results from analytcal geometry that apply in this case are 1) plp2= b2, (7.44 a-c) 2) r2-r1=2a, and 3) the tangent at P bisects the angle between the focal distances. (b2 = a2(e2-1) where e is the eccenticity (e2> 1), and 2b2/a is the latus rectum). We therefore obtain b12/p,2=2a/r1 + 1. (7.45) This is the [p, r] equaton of an hyperbola. 7.7 Bound and unbound orbits For a central force, we have the equaton for the acceleraton in [p, r] form (h2/p3)dp/dr = ac. (7.46) If the acceleraton varies as 1/r2, then the form of the orbit is given by separatng the variables, and integradng, thus h2jdp/p3 = kjdr/r2, (7.47)  138 so that -h2/2p2=-klr, where k is a constant, or h21p2 = 2k/r + C, where the value of C depends on the form of the orbit. Comparing this form with the general form of the [p, r] equatons of conic sections, we see that the orbit is an ellipse, parabola, or hyperbola depending on the value of C. If Cis negative, the orbit is an ellipse, Cis zero, the orbit is a parabola, and if Cis positive, the orbit is an hyperbola. The speed of the partidcle in a central orbit is given by v = h/p. If, therefore, the partidcle is projected from the origin, 0(corresponding to r = ro) with a speed vo, then h2/p2=yv2=2k/ro + C, (7.48) so that the orbit is 1) an ellipse ifvo2 < 2k/ro, (7.49 a-c) 2) a parabola ifvo2 = 2k/ro, or 3) an hyperbola if vo2 >2k/ro. The escape velocity, the initial velocity required for the partide to go into an unbound orbit is therefore given by  139 vesop= 2k/ro = 2GMEIRE, for apartide launched from the surface of the Earth. This condition is, in fact, an energy equation (1/2)(m =1)v2sp= GME(m 1)RE. (7.50) kinetic energy potential energy 7.8 The concept of the gravitational field Newton was well-aware of the great difficulties that arse in any theory of the gravitational interaction between two masses not in direct contact with each other. In the Paincia, he assumes, in the absence of any experimental knowledge of the speed of propagation of the gravitational interaclion, that the interaction takes place instantaneously. However, in letters to other luminaries of his day, he postulated an intervening agent beMeen two approaching masses - an agent that requires a finite time to react. In the eary 17th century, the problem of understanding the interaclion between spatially separated objects appeared in a new guise, this lime in discussions of the electromagnetic interaction between chargedobjects. Faraday introduced the idea ofa field of force with dynamicalproperties In the Faraday model, an accelerating electic charge acts as the source of a dynamical electromagnetic field that travels at a finite speed through space-time, and interacts with a distant charge. Energy and momentum are thereby transferred from one charged object to another distant charged object. Maxwell developed Faraday's idea into a mathemaical theory- the electromagnetc theory of light- in which the speed of propagation of light appears as a fundamental constant of Nature. His theory involves the differential equations of motion of the electric and magnetic field vectors; the equations are not invariant under the Galilean transformation but they are invariant under the Lorentz transformaion.  140 (The discovery of the transformalon that leaves Maxwell's equations invariant for all inertial observers was made by Lorenlz in 1897). We have previously discussed the development of the Special Theory of Relativity by Einstein, a theory in which there is but one universal constant, c, for the speed of propagation of a dynamical field in a vacuum. This means that c is not only the speed of light in free space but also the speed of the gravitationalfield in the void between interacling masses. We can gain some insight into the dynamical properties associated with the interaclion between distant masses by investigating the effect of afinite speed of propagation, c, of the gravitational interaction on Newton's Laws of Motion. Consider a non-orbiting mass M, at a distance R from a mass mass Ms, simply falling from rest with an acceleration a(R) towards Ms. According to Newton's Theory of Gravitation, the magnitude of the force on the mass M is IF(R)I|= GMsM/R2 = Ma(R), (7.51) We therefore have a(R) = GMs/R2. (7.52) Let At be the lime that it takes for the gravitational interaclion to travel the distance R at the universal speed c, so that At= R/c. (7.53) In the lime interval At, the mass M moves a distance, AR, towards the mass1%i; AR = aAT/2 = (GMs/R2)At2/2 = (GMs/R2)(R/c)2/2. (7.54)  141 Consider the situation in which the mass M is in a drcular orbit of radius R about the mass,1%V. Let v(t) be the velocity of the mass M at lime t, and v(t + At) its velocity at t + At, where At is chosen to be the interaction travel time. Let us consider the motion of M if there were no mass Ms present, and therefore no interaclion; the mass M then would continue its motion with constant velocity v(t) in a straight line. We are interested in the difference in the positions of M at lime t + At, with and without the mass Ms in place. We have, to a good approximation: M v(t) -- - "extrapolated position" (no mass Ms) F(R) AR v(t + At) R R O Ms The magnitude of the gravitational force, FEX, at the extrapolated position, with Ms in place, is FEX = GMsM/(R + AR)2 (7.55) =(GMsM/R2)(1+ ARR)-2 (GMsM/R2)(1 - 2AR/R), for AR «R. (7.56) Substituting the value of AR obtained above, we find Fa GMsM\/lR2 -(GMsM/Rc2)(GMs/R2). (7.57) Nerwton's 3rd Law states that FMS,M =- FM, MS (7.58)  142 This Law is frue, however, for contact interactions only. For all interactions that take place between separated objects, there is a mis-match between the action and the reaction. It takes lime for one partidcle to respond to the presence of the other! In the present example, we obtain a good estimate of the mismatch by taking the difference between FEX(R + AR) and F(R), namely FEX(R + AR) - F(R)~(GMsMIRC2)(GMs/R2). (7.59) On the right-hand side of this equation, we note that the term (GMs/R2) has dimensions of "acceleration", and therefore the term (GMsMIRc2) must have dimensions of "mass". We see that this term is an estimate of the "mass" associated with the interaction, itself. The space between the interacting masses must be endowed with this effective mass if Newton's 3rd Law is to include non- contact interactions. The appearance of the term c2 in the denominator of this effective mass term has a special significance. If we invoke Einstein's famous relation E = MG, then AE = AMc2 so that the effective mass of the gravitational interaction can be written as an effeclive energy: AEGRAV= GMsMIR. (7.60) This is the "energy stored in the gravitational field" between the two interacling masses. Note that it has a hR-dependence - the correct form for the potential energy associated with a 1R2 gravitational force. We see that the notion of a dynamical field of force is a necessary consequence of the finite propagalion lime of the interaclion.  143 7.9 The gravitational potential The concept of a gravitatonal potental has its origins in the work of Leibniz. The potential energy, V(x), asssodated with n interacting parades, of masses m1, m2, ...mn, situated at x1, x2, ...xn, is related to the gravitatonal force on a mass M atx, due to the n pardcles, by the equation F(x) = -VV(x). (7.61) The exact forms of F(x) and V(x) are F(x) =-GM =1,n] m(x- xi)/Ix- xi3, (7.62) and V(x) =--GM yr,= 1,n]m/|x - xil In upper-index notaton, the components of the force are Fk(x)=-3V/3x, k=1, 2, 3. (7.63) The gravitational field, g(x), is the force per unit mass: g(x) = F(x)/M, (7.64) and the gravitatonal potental is defined as C(x) = V(x)/M =->I=in] Gm/Ix-xIl. (7.65) The sign of the potential is chosen to be negative because the gravitational force is alays attracdive. (This conventon agrees with that used in Electrostatcs). If the mass consists of a condnuous distribuWon that can be descibed by a mass density p(x), then the potential is dr < R r O Let the disc be divided into concentiic drdes. The potental at P, on the aids, due to the elemental ring of radius r and width dr is 2rrdrGQ/PQ, where o is the mass per unit area of the disc. The potental at P of the entire disc is therefore V=fro~a] 2lGordr/PQ, (7.68) where a is the radius of the disc. Therefore, V= 21GQJ[O,a rdr/(r2 + p2)1'2  145 =21TGo[(r2+ p2)1'],a] = 21TGQ(R - p), (7.69) where R is the distance of P from any point on the drcumfrence. PROBLEMS 7-1 Show that the gravitatonal potental of a thin spherical shell of radius R and mass M at a point P is 1) GMldwhered is the distance from P to the center of the shell if d >R, and 2) GM/R if P is inside or on the shell. 7-2 If d is the distance from the center of a solid sphere (radius R and densityp) to a point P inside the sphere, show that the gravitatonal potential at P is (Dp = 2TGp(R2-d2/3). 7-3 Show that the gravitatonal atracdion of adrcular disc of radius R and mass per unit area a, at a point P distant p from the center of the disc, and on the axis, is 2TGQ{[p/(p2 + R2)]-1 . 7-4 A partidcle moves in an ellipse abouta center of force at a focus. Prove that the instantaneous velocity vof the partide at any point in its orbit can be resolved into two components, each of constant magnitude:1) of magnitude ah/I6, perpendicular to the radius vector r at the point, and 2) of magnitude aheb2 perpendicular to the major axis of the ellipse. Here, a and b are the semi-major and semi-minor axes, e is the eccentricity, and h = pv = constant for a central orbit.  146 7.5 A partide moves in an orbit under a central acceleraton a= k/r2where k= constant. If the partide is projected with an inital velodtyvo in a diredion at right angles to the radius vecttor rwhen at a distance ro from the center of force (the origin), prove (dr/dt)2 ={(2klro) -vo2(1 + (ro/r))}{(r/r)- I}. This problem involves the energy and momentum equations in r, coordinates. 7-6 A partide moves in a cardioidal orbit, r= a(1 + cos$), under a central force V a P [r,$ p IF r 1) show that the p-r equation of the cardioid is p2= r3/2a, and 2) show that the central acceleration is 3ah/r4, where h = pv = constant. 7-7 A planet moves in adrcular orbit of radius r aboutthe Sun as focus at the center. If the gravitational "constant" G changes slowly with lime- G(t), then show that the angular velocity, o, of the planet and the radius of the orbit change in lime according to the equations (l/o)(doldt) = (21G)(dGldt) and (I/r)(dr/dt) =(-1/IG)(dGldt). 7-8 A partide moves under a central acceleralion a = k(1/r3e) where k is a constant. If k = h2, where h = r2(d$/dt) = pv, then show that the path is 1/r= A$ + B, a "reciprocal spiral", where A and B are constants.  147 8 EINSTEINIAN GRAVITATION: AN INTRODUCTION TO GENERAL RELATMTY 8.1 The principle of equivalence The term "mass" that appears in Newton's equaton for the gravitatonal force between two interacting masses refers to "gravitational mass" - that property of matter that responds to the gravitational force...Newton's Law should indicate this property of matter FG = GMGmG/r2, where MG and mG are the gravitatonal masses of the interacting objects, separated by a distance r. The term "mass" that appears in Newton's equaton of motion, F = ma, refers to the "inertal mass" - that property of matter that resists changes in its state of moton. Newton's equaton of moton should indicate this property of matter F(r) = m'a(r), where m' is the inertal mass of the partide moving with an acceleration a(r) in the gra'Atational field of the mass MG. Newton showed by experiment that the inertial mass of an object is equal to its gravitatonal mass, m'= mG to an accuracy of 1 part in 13. Recent experiments have shown this equality to be true to an accuracy of1 part in 1012. Newton therefore took the equadons F(r) = GMGmG/r2 = m'a(r) (8.1) and used the conditon mG =m' to obtain a(r) = GMG/r2, (8.2)  148 Galileo had, of course, previously shown that objects made from different materials fall with the same acceleraton in the gravitatonal field at the surface of the Earth, a result that implies rn oc m'. This is the Newtonian Prndple of Equivalence. Einstein used this Prindple as a basis for a new Theory of Gravitaton! He extended tie axioms of Special Relativity, that apply to field-free frames, to frames of reference in "free fall". A freely falling frame must be in a state of unpowered motion in a unifon gravitational field. The field region must be suffiendy small for there to be no measurable variaton in the field throughout the region. If a field gradient does exist in the region then so called "idal effects" are present, and these can, in prndple, be determined (by distorting a liquid drop, for example). The results of all eperiments caried out in ideal freely falling frames are therefore fully consistent with Spedal Relatvity. All freely-falling observers measure the speed of light to be c, its constant free-space value. It is not possible to carry out experiments in ideal freely-falling frames that permit a distinclion to be made between the acceleraton of local, freely-falling objects, and their moton in an equivalent external gravitatonal field. As an immediate consequence of the extended Prndple of Equivalence, Einstein showed that a beam of light would be deflected from its straight path in a dose encounter with a suficiendy massive object. The observers would, themselves, be far removed from the gravitatonal field of the massive object causing the deflection. Einstein's original calculaton of the deflection of light from a distant star, grazing the Sun, as observed here on the Earth, induded only those changes in time intelas that he had predicted would occur in the near field of the Sun. His result tumed out to be in error by exactly a factor of two. He later obtained the "correct' value for the defleclion by including in the calculation the changes  149 in spatialintelvals caused by the gravitational field. A plausible argument is given in the sectbn 8.6 for introdudng a non-intuitive concept, the refractive index of spacetime due to a gravitational field. This concept is, perhaps, the characteristic physical feature of Einstein's revolutionary General Theory of Relativity. 82 Time and length changes in a gravitational field We have previously discussed the changes that occur in the measurement of length and lime intervals in different inertialframes. These changes have their origin in the invariant speed of light and the necessary synchronization of docks in a given inertial frame. Einstein showed that measurements of length and lime intervals in a given gravitational potential are changed relative to the measurements made in a different gravitational potential. These field-dependent changes are not to be confused with the Spedal-Relativistic changes discussed in 3.5. Although an exact treatment of this topic requires the solution of the full Einstein gravitational field equations, we can obtain some of the key results of the theory by making approximations that are valid in the case of our solar system. These approximations are treated in the folloMng sections. 8.3 The Schwarzschild line element An observer in an ideal freely-falling frame measures an invariant infinitesimal interval of the standard Special Relatvisticform ds2 = (cdt)2 -(dx2 + dy2 + dz2). (8.3) It is advantageous to transform this form to spherical polar coordinates, using the linear equations x = rsin~cos$, y = rsin~sin4p, and z = rcosO.  150 We then have z dr dl, the diagonal of the cube d$ z rde rdy x The square of the length of the diagonal of the infinitesimal cube is seen to be dl2= dr2 + (rdO)2 + (rsin6d)2. (8.4) The invariant interval can therefore be written ds2 = (cdt)2 - dr2 - r2(d62+ sin2Od$2). (8.5) The key question that now faces us is this: how do we introduce gravitaton into the problem? We can solve the problem by introdudng an energy equation into the argument. Consider two observers 0 and 0', passing by one another in a state of free fall in a gravitational field due to a mass M, fixed at the origin of coordinates. Both observers measure a standard interval of spacedme, ds according to 0, and ds' according to 0', so that ds2 = ds'2 = (cdt')2 -dr'2 -r'2(dO'2 + sin2O'd$'2) (8.6) The situadon is as shown  151 Z f vo(r) Mass, M (the source of the field) x Let the observer 0' just begin free fall towards M at the radial distance r, and let the observer 0, dose to 0', be freely falling away from the mass M. The observer 0 is in a state of unpowered motion with just the right amount of kinetic energy to "escape to infinity". Since both observers are in states of free fall, we can, according to Einstein, treat them as i they were 'inertial observers". This means that they can relate their local space-time measurements by a Lorentz transformation. In particular, they can relate their measurements of the squared intervals, dc2 and ds'2, in the standard way. Since their relative motion is along the radial direction, r, time intervals and radial distances will be measured to be changed: At=yAt' and yAr =Ar', (8.7 a,b) where y =1/{1- (v/c)212 , in which v = vo(r) because v'(r) ~0.  152 If 0 has just enough kinetc energy to escape to infinity, then we can equate the kinetc energy to the potental energy, so that vo2(r)/2 =1-.C(r) if the observer0 has unit mass. (8.8) fP(r) is the gravitational potentalat r due to the presence of the mass, M, at the origin. This procedure enables us to introduce the gravitatonal potental into the value of y in the Lorentz transformadon. We have vo2 = 2P(r) = v2, and therefore At = At'{1- 2P(r)c2}12, (8.9) and Ar= Ar'{1-2'c(r)/c2}12. (8.10) Only lengths parallel to r change, therefore r2(d62+ sin2Od4?) = r'2(d6'2+ sinO'd4p2), (8.11) and therefore we obtain ds2 = ds'2 = c2(1 - 2'P(r)/c2)dt2-dr2/(1 - 2'P(r)/c2) - r2(d62+ sin2Od4?). (8.12) If the potental is due to a mass M at the origin then fi(r) = GM/r, (r> R, the radius of the mass, M) therefore, ds2= c2(1- 2GM/rc2)dt2 -(1 - 2GM/rc2)1dr2 - r2(d02 + sin2Od4f). (8.13) This is the famous Schwarzschild line element, originally obtained as an exact soluton of the Einstein field equations. The present approach fortuitously gives the exact result!  153 8.4 The metric in the presence of matter In the absence of matter, the invariant interval of space-time is ds2= ,,dxdxv ([t,v=0, 1,2,3), (8.14) where ,%= diag(1,'-1,-1,-1) (8.15) is the metric of Special Relatvity; it "lowers the indices" dx = qdx". (8.16) The form of the Schwarzschild line element, ds§a, shows that the metric g,,in the presence of matter differs from qv. We have ds2s = gdxdx", (8.17) where dx0 = cdt, dx1 = dr, dx2 = rdO, and dx3 = rsined, and g, = diag((1-X), -(1-X)-, -(1-X)-,-(1- X)1 in which x=2GM/rc2. The Schwarzschild metric lowers the indices dx, = g,,dxv, (8.18) so that ds2 = dxdx,. (8.19)  154 8.5 The weak field approximation If x = 2GMIrc2 «<1, the coefficient, (1 - x)-1, of dr2 in the Schwarzschild line element can be replaced by the leading term of its binomial expansion, (1 +x ...) to give the 'weak field" line element: ds2w= (1- x)(cdt)2 -(1+ x)dr2 - r2(d02 + sin2Od4?). (820) At the surface of the Sun, the value of x is 42 x 8-6, so that the weak field approximaton is valid in all gravitational phenomena in our solar system. Consider a beam of light traveling radially in the weak field of a mass M, then ds2w =0 (alight-like interval) , and dO2 + sin2Od4V =0, (821) giving 0= (1-x)(cdt)2-(1 +x)dr2. (822) The "velocty" of the lightvt = dr/dt, as determined by observers far from the gravitational influence of M, is therefore vL=c{(1 -x)/(1 + X)}12 c if 'x 0. (823) (Observers in free fall near M always measure the speed of light to be c). Expanding the term {(1- x)/(1+ x)}12 to first order in x, we obtain vL(r) c (1 - X/2 ... )(1 - X/2 ...) =(1-x..).(8.24) Therefore vL(r)=c(1 -2GM/rc2..), (8.25) so that vL(r) R We have shown that the speed of light (moving radially) in a gravitatonal field, measured by an observer far from the source of the field, depends on the distance, r, from the source v(r)= c(1 -2GM/rc2) (828) where c is the invariant speed of light as r-- oo. We wish to compare dx with dx', the distances travelled in the x-direction by the wavefront at y and y + dy, in the interval dt. We have r2= (y+ R)2+ x2 (8.29)  157 therefore v(r) -- v(x, y) so that 2r(BrI8y)=2(y+ R), and 3r/Jy = (y+ R )Ir. (8.30) Very dose to the surface of the mass M (radius R), the gradient is 3r/JyIlyo -- R/r. (8.31) Now, 3v(r)I8y = (313r)(c(1 -2GM/rc2))(BrI8y) = (2GMIr2c)(BrI8y). (8.32) We therefore obtain 3v(r)/JyIlyo= (2GM/r2c)(RIr) = 2GMRIr3c. (8.33) Let the speed of the wavefront be v' at y + dy and v at y. The distances moved in the interval dt are therefore dx' =v'dt and dx = vdt (8.34 a,b) The first-order Taylor expansion of v'is V'= v + (v/y)dy, and therefore dx'-dx= (v+ (JvI8y)dy)dt-vdt= (3v/8y)dydt. (8.35) Let the corresponding angle of deflection of the normal to the wavefront be dar, then da =(dx' -dx)/dy  158 = (3v/8y)dt = (3v1y)(dxlv). (8.36) The total deflection of the normal to the plane wavefront is therefore Aa = f[-00,0](v/1y)(dx/v) (8.37) (1 /c)f[-, oo](8v/8y)dx . (v c cover most of the range of the integral). The portion of the wavefront that grazes the surface of the mass M (y- 0) therefore undergoes a total deflection Aa=(1/c)JH,-](2GMRdrc)dx (8.38) =2GMRc2Jf, dx/(R2+x2)32 =2GMRIc2[x(R2(R2 + x2)1/2- =2(GMRc2)(2/R2). so that Aa= 4GMIRc2. This is Einstein's famous prediction; puttng in the known values for G, M, R, and c, gives Aa=1.75 arcseconds. (8.39) Measurements of this very small effect, made during total eclipses of the Sun at various limes and places since1919, are fully consistent with Einstein's prediction. PROBLEMS 8-1 If a pardle A is launched with a velodty VGA from a point P on the surface of the Earth at the same instant that a partide B is dropped from a point Q, use the Principle  159 of Equivalence to show that if A and B are to collide then va must be directed along PQ. Q g]J B :A VoA P 8-2 A satellite is in a circular orbit above the Earth. It carries a dock that is similar to a dock on the Earth. There are two effects that must be taken into account in comparing the rates of the two docks. 1) the time shift due to their relative speeds (Special Relativity), and 2) the time shift due to their different gravitational potentials (General Relativity). Calculate the SR shift to second-order in (v/c), where v is the orbital speed , and the GR shift to the same order. In calculating the difference in the potentials, integrate from the surface of the Earth to the orbit radius. The two effects differ in sign. Show that the total relative change in the frequency of the satellite dock compared with the Earth dock is (Av/vE) ~ (gRE/c2){1 - (3RE2rs)}, where rs is the radius of the satellite orbit (measured from the center of the Earth). We see that if the altitude of the satellite is> RE/2 (- 3200 km) Av is positive since the gravitational effect then predominates, whereas at altitudes less than 3200 km, the Special Relativity effect predominates. At an atiude 3200 km, the docks remain in synchronism.  160 9 AN INTRODUCTION TO THE CALCULUS OF VARIATIONS 9.1 The Euler equation A frequent problem in Differental Calculus is to find the statonary values (maxima and minima) of a function y(x). The necessary conditon for a statonary value at x = a is dY/dXIx=a =0. Fora minimum, d2y/dX2Ix=a>0, and fora maximum, d2y/dX2Ix=a<0. The Calculus of Variatons is concerned with a related problem, namely that of finding a function y(x) such that a definite integral taken over a function of this function shall be a maximum or a minimum. This is dearly a more complicated problem than that of simply finding the stationary values ofaffunction, y(x). Explicity, we wish to find that funcion y(x) that will cause the definite integral J[xl,X4 F(x, y, dy(x)Idx)dx (9.1) to have a statonary value. The integrand F is a function of y(x) as well as of x and dy(x)Idx. The limits x and x2 are assumed to be fixed , as are the values y(x1) and y(x2). The integral has different values along different "paths" that  161 connect (x1, yi) and (x2, y2). Let a path be Y(x), and let this be one of a set of paths that are adjacent to y(x). We take Y(x) - y(x) to be an infinitesimal for every value of x in the range of integraton. Let the difference be defined as Y(x) -y(x)=by(x) (a"first-order change"), (9.2) and F(x, Y(x), dY(x)Idx)- F(x, y(x), dy(x)Idx)=SF. (9.3) The symbol S is called a varation; it represents the change in the quantty to which it is applied as we go from y(x) to Y(x)at the same value ofx Note 8x =0, and b(dyldx) = dY(x)/dx-dy(x)/dx = (dldx)(Y(x) - y(x)) = (dldx)(Oy(x)). The symbols b and (dldx) commute: b(d/dx) - (d/dx)b =0. (9.4) Graphically, we have y ' Y(x), the varied path yi y(x), the "tue"path 0 x1 x Using the definition of bF, we find F= F(x, y+ by, dy/dx+ b(dy/dx)) -F(x, y, dy/dx) (9.5) I I Y(x) (d/dx)Y(x)  162 = (3FIy)by + (3FIy')by' for fixed x. (Here, dyldx = y'). The integral j[xl,Xg F(x, y, y')dx, (9.6) is statonary if its value along the path y is the same as its value along the varied path, y + Sy = Y. We therefore require j[xl,x2] F(x, y, y')dx =0. (9.7) This integral can be written j[xl,X2 {(3FI3y)by+(FIy')by'}dx=0. (9.8) The second term in this integral can be evaluated by parts, giving [(3FIy')by]i2 J mFuxl](dldx)(3F/Jy')oydx. (9.9) But Sy1 = by2 = 0 at the end-points x1 and x2, therefore the term [ ]x12 = 0, so that the statonary conditon becomes JrX,4 {3FI3y-(dldx)3F/Jy'}5ydx=0. (9.10) The infinitesimal quantty by is positive and arbitrary, therefore, the integrand is zero: 3FIy - (dldx)3FI3y' = 0. (9.11) This is known as Eules equaton. 92 The Lagrange equations Lagrange, one of the greatest mathematicans of the 18th century, developed Eulers equadon in order to treat the problem of partide dynamics within the framework of generalized coordinates. He made the transformation  163 F(x, y, dy/dx) -- L(t, u, duldt) (9.12) where u is a generalized coordinate and duldt is a generalized velocity. The Euler equaton then becomes the Lagrange equafion-of-mofion: 3LI3u - (dldt)(3LI3u) =0, where u is the generalized velocity. (9.13) The Lagrangian L(t; u, u) is defined in terms of the kinetc and potental energy of a particle, or system of partidcles: L T -V. (9.14) It is instructive to consider the Newtonian problem of the motion of a mass m, moving in the plane, under the influence of an inverse-square-law force of attraction using Lagrange's equadons-of- moton. Let the center of force be at the origin of polar coordinates. The kinetc energy of m at [r,$4] is T = m((dr/dt)2+ r2(d$/dt)2)/2, (9.15) and its potental energy is V =- k/r, where k is a constant. (9.16) The Lagrangian is therefore L = T-V = m((dr/dt)2+ r2(d$/dt)2)/2 + k/r. (9.17) Put r = u, and$=v, the generalized coordinates. We have, for the "u-equadon" (d/dt)(3LUu) = (d/dt)(3LUr) = (d/dt)(m(drdt)) =mclr/dt2, (9.18) and BUIu =ULIr =mr(d$/dt)2 -k/r2 (9.19) Using Lagrange's equadon-of-modon for the u-coordinate, we have  164 m(d2rldt2) -mr(d$/dt)2+k/r2=0 (920) or m(d2rldt2-r(d$/dt)2)=-k/r2. (921) This is, as it should be, the Newtonian equaton mass x acceleraton in the r-direction=force in the r-direclon. Introdudng a second generalized coordinate, we have, for the 'V-equadon" (d/dt)(Uv)= (d/dt)(3Uj)= (d/dt)(mr24) (9.22) =m(r2$+$2r r), and 8Jv=3L8U$=0, (923) therefore m(r2$+2rr$)=0 so that (d/dt)(mr2$) =0. (924) Integratng , we obtain mr24= constant, (9.25) showing, again, that the angular momentum is conserved. The advantages of using the Lagrangian method to solve dynamical problems stem from the fact that L is a scaarfumnclion of generalized coordinates.  165 9.3 The Hamilton equations The Lagrangian L is a function of the generalized coordinates and velocities, and the tme: L = L(u, v,...;u, v, ...;t). (926) If the discussion is limited to two coordinates, u and v, the total differentalof L is dL = (3LI3u)du + (3LIv)dv + (3LI3u)du + (3Lv)dv + (3LI3t)dt Consider the simplest case of a mass m moving along the x-axis in apotential, so that u = x and u = x = vx, then L = T -V = mv2 -V (927) and BLavX = mvx = px, the linear momentum. (928) In general, it is found that terms of the formU3L/3u and 3L/Uv are "momentum" terms; they are called generalized momenta, and are written 3LI3u= pu, 3LI3v= pv,..etc. (929) Such forms are not limited to "linear"momenta. The Lagrange equation (d/dt)(3LUu) -8ULIu=0 (9.30) can be transformed, therefore, into an equadon that involves the generalized momenta: (d/dt)(pu) -8ULu =0, or  166 3L/Bu= pu. (9.31) The total differentalof L is therefore dL = pudu+ podv+ pudu+ pdv+(LI8t)dt. (9.32) We now introduce an important funcion, the Hamiltonian funclion, H, defined by H -puu+ pvV-L, (9.33) therefore dH ={pudu +udpu+ pdv+vdp}-dL. (9.34) It is not by chance that H is defined in the way given above. The definition permits the cancellaton of the terms in dH that involve du and dv, so that dH depends only on du, dv, dpo, and dp (and perhaps, t). We can therefore write H = f(u, v, pa, pv; t) (limitng the discussion to the two coordinates uand v). (9.35) The total differental of H is therefore dH = (3H/3u)du+ (3H/v)dv+(3HI/8)dpu+(HIp)dpv+(3HIt)dt (9.36) Comparng the two equatonsfor dH, we obtain Hamiton's equations-of-motion: H/Bu= -pu, 3H/3v = -p, (9.37) 3H/3pu= u, 3H/3p= v, (9.38) and H8= -3L/8t. (9.39) We see that  167 H = puu+pvv-(T -V). (9.40) If we consider a mass m moving in the (x, y)-plane then H = (mv)vx+ (mvy)vy- T + V (9.41) = 2(mvx2 + mvy/2) - T + V = T + V, the total energy. (9.42) In advanced treatments of Analytcal Dynamics, this form of the Hamiltonian is shown to have general validity. PROBLEMS 9-1 Studies of geodesics - the shortest distance between two points on a surface- form a natural part of the Calculus of Variations. Show that the straight line between two points in a plane is the shortest distance between them. 9-2 The surface generated by revolving the y-coordinate about the x-axis has an area 2liTyds where ds = {dx2 + dy2}12 Use Eulers variatonal method to show that the surface of revoluton is aminimum if (dy/dx) = {(y21a2) -1}1/2where a= constant. Hence show that the equaton of the minimum surface is y= acosh{(xla) + b} where b = constant, and y ;0. 9-3 The Principle of Least Time pre-dates the Calculus of Variatons. The propagaton of a ray of light in adjoining media that have different indices of refraction is found to be governed by this principle. A ray of light moves at constant speed v~ in a medium  168 (1) from a point A to a point Boon the x-axis. At Bo, its speed changes to a new constant value v2 on entering medium (2). The ray contnues untl it reaches a point C in (2). If the true path A -- Bo -> C is such that the total travel time of the light in going from A to C is a minimtm, show that (v1/v2) = xo{[yc2 + (d - xo/[yA2 + }12/(d - xo), (Snell's law) where the symbols are defined in the following diagram: yA YA 0 BB x Medium 2, speed v2 The path A -- B-> Cis an arbitrarly varied path. 9-4 Hamilton's PrindIple states that when asystemis moving under conservatve forces the lime integral of the Lagrange funclion is stalionary. (It is possible to show that this Principle holds for non-conservative forces). Apply Hamilton's Principle to the case of aprojecile of mass m moving in a constant gravitalional field, in the plane. Let the projecile be launched from the origin of Cartesian coordinates at time t =0.  169 The Lagrangian is L = m((dxldt)2 +(dy/dt)2)/2-mgy Calculate bJ[otl]Ldt, and obtain Newton's equatons of moton d2yldt2+ g =0 and d2x/dt2=0. 9-5 Reconsider the example discussed in section 92 from the point of view of the Hamiltonian of the system. Obtain H(r, 4, pr, pt), and solve Hamilton's equations of moton to obtain the results given in Eqs.9.21 and 925.  170 10 CONSERVATION LAWS, AGAIN 10.1 The conservation of mechanical energy If the Hamiltonian of a system does not depend explicitly on the lime, we have H = H(u, v, ...;pa, p,...). (10.1) In this case, the total differential dH is (for two generalized coordinates, u and v) dH = (3H/3u)du + (3H/Jv)dv + (3H/8p)dpu + (3H/3p)dp. (102) If the positions and the momenta of the partidcles in the system change with lime under their mutual interactions, then H also changes with lime, so that dHldt = (3H/3u)duldt + (3HI3v)dvldt + (3H/I3)dpu/dt + (3H/3p)dp/dt = (IJuU) + (-pv)+(up) + (vp) (10.3) =0, using Hamilton's equaions-of-moion. (10.4) Integration then gives H = constant. (10.5) In any system moving under the influence of conservative forces, a potenial V exists. In such systems, the total mechanical energy is H = T + V, and we see that it is a constant of the motion. 10.2 The conservation of linear and angular momentum If the Hamiltonian, H, does not depend explicity on a given generalized coordinate then the generalized momentum associated with that coordinate is conserved. For example, if H contains no  171 explicit reference to an angular coordinate then the angular momentum associated with that angle is conserved. Formally, we have dp/dt = -3H/8q , where p and q are the generalized momenta and coordinates. (10.6) Let an infinitesimal change in the jth-coordinate q be made, so that q -- q+ g, (10.7) then we have 8H =(3H/3q)6g. (10.8) If the Hamiltonian is invariant under the infinitesimal displacement og, then the generalized momentum p is a constant of the moton. The consewation of linear momentum is therefore a consequence of the homogeneity of space, and the consewation of angular momentum is a consequence of the isotropy of space. The observed conservaton laws therefore imply that the choice of a point in space for the origin of coordinates, and the choice of an axis of orientaton play no part in the formulaton of the physical laws; the Laws of Nature do not depend on an "absolute space".  172 11 CHAOS The behavior of many non-linear dynamical systems as a funcion of lime is found to be chaotic. The characteristic feature of chaos is that the system never repeats its past behavior. Chaotic systems nonetheless obey dassical laws of motion which means that the equations of motion are deterministic. Poincare was the first to study the effects of small changes in the initial conditions on the evolution of chaotic systems that obey non-linear equations of motion. In a chaotic system, the erratic behavior is due to the internal, or intrinsic, dynamics of the system. Let a dynamical system be described by a set of first-order differential equations: dxi/dt = f1(x1,x2,x3,...xn) (11.1) dx2/dt f2(xl,x2,x3,...xn) dxn/dt-fn(x1,x2,x3,...xn,) where the funcions fa are funcions of n-varables. The necessary conditions for chaolic molion of the system are 1) the equalions of molion must contain a non-linear term that couples several of the variables. A typical non-linear equalion, in which two of the variables are coupled, is therefore  173 dxildt= axi+ bx2 + CX1X2 + ... rxo, (a, d, c, ...r are constants) (112) and 2) the number of independent variables, n, must be at least three. The second conditon is discussed later. The non-linearity often makes the solution of the equations unstable for particular choices of the parameters. Numerical methods of solution must be adopted in all but a few standard cases. 11.1 The general motion of a damped, driven pendulum The equation of a damped, driven pendulum is ml(d26/dt2) + kml(dO/dt) + mgsinO = Acos(wDt) (11.3) or (d2OIdt2) + k(dO/dt) + (gAI)sinO= (A/ml)cos(wDt), (11.4) where 0 is the angular displacement of the pendulum, I is its length, m is its mass, the resistance is proportional to the velocity (constant of proportionality, k), A is the amplitude and OD is the angular frequency of the drMng force. Baker and Gollub in Chaotic Dynamics (Cambrdge,1990) write this equation in the form (d2OIdt2) + (Iq)(doldt) + sinO = Ccos(wDt), (11.5) where q is the damping factor. The low-amplitude natural angular frequency of the pendultn is unity, and tine is dimensionless. We can therefore write the equation in terms of three first-order differential equations doldt =-(1/q)o -sinO + Ccos($)) where 4)is the phase, (11.6)  174 dO/dt=w, (11.7) and d$/dt= W. (11.8) The three variables are (o, 0,$). The onset of chaotic motion of the pendulum depends on the choice of the parameters q, C, and oDm. The phase space of the oscillatons is three-dimensional: 0)0 A spiral with a pitch of 2r The - o trajectores are projections of the spiral onto thee -co plane. The motion is sensitve to WD since the non-linear terms generate many new resonances that occur when owor, is a ratonal number. (Here, ws is the angular frequency of the undamped linear oscllator). For pardcular values of q and oD, the forcing term produces a damped motion that is no longer periodic - the modon becomes chaotic. Periodic modon is characterzize by dosed orbits in the (0 - o) plane. If the damping is reduced considerably, the modon can become highly chaodc.  175 The system is sensitive to small changes in the inital conditions. The 1rajectores in phase space diverge from each other with exponential lime-dependence. For chaotic motion, the projection of the trajectory in (0, o, $ ) - space onto the (0 - o) plane generates trajectores that intersect. However, in the full 3 - space, a spiraling line along the )-axis never intersects itself. We therefore see that chaotic motion can exist only when the system has at least a 3- dimensional phase space. The path then converges towards the attractorwithout sel-crossing. Small changes in the initial conditions of a chaotic system may produce very different trajectores in phase space. These trajectores diverge, and their divergence increases exponentially with time. If the difference between trajectores as a funcion of time is d(t) then it is found that logd(t) Xt or d(t)~e4 (11.9) where X> 0- a positive quanlity called the Lyapunov exponent. In a weakly chaotic systemX « 0.1 whereas, in a strongly chaotic system, X>>0.1. 11.2 The numerical solution of differential equations A numerical method of solving linear differential equations that is suitable in the present case is known as the Runge-Kutta method. The algorithm for solving two equalions that are funcions of several variables is: Let dy/dx =f(x, y, z) and dz/dx = g(x, y, z). (11.10) If y= yoand z= zowhen x= xothen, for increments h in xo, k in yo, and lin zo  176 the Runge-Kutta equations are k1= hf(xo, yo, zo) k2= hf(xo+ h/2, yo+ ki/2, zo+li/2) k3= hf(xo + h12, yo + k2/2, zo +12/2) k4= hf(xo+ h, yo + k3, zo +13) k = (k1 +2k2+2k3+k4)/6 Ii = hg(xo, yo, zo) 12= hg(xo+ h12, yo+ k1/2, Zo+i1/2) b= hg(xo + h12, yo + k2/2, zo+12/2) = hg(xo + h, yo + k3, zo +13) and I =(11+212+213+k)/6. (11.11) The inital values are incremented, and successive values of the x, y, and z are generated by iterations. It is often advantageous to use varying values of h to optimize the procedure. In the present case, f(x, y, z) -- f(t, 0, o) and g(x, y, z) -- g(w). As a problem, develop an algorithm to solve the non-linear equation 11.5 using the Runge- Kutta method for three equations (11.6, 11.7, and 11.8). Write a program to calculate the necessary iterations. Choose increments in lime that are small enough to reveal the details in the 6-w plane. Examples of non-chaolic and chaotic behavior are shown in the following two diagrams.  177 -400 - -200 -1 I 1100 0 -2 The parameters used to obtain this plot in the 0o plane are: damping factor (1/q) =1/5, amplitude (C) =2, drive frequency (coD) = 0.7, and time increment, At = 0.05. All the initial values are zero.  178 1 150 -44 0 -100 -150 Points in the 8-w plane for a chaotic system The parameters used to obtain this plot in the O-w plane are: damping factor (lIq) = 1/2, amplitude (C) =1.15, drive frequency (WD) = 0.597, and tme increment, At=1. The intial value of the tme is 100.  179 12 WAVE MOTION 12.1 The basic form of a wave Wave moton in a medium is a collective phenomenon that involves local interaclions among the partides of the medium. Waves are characterized by: 1) a disturbance in space and tme. 2) a transfer of energy from one place to another, and 3) a non-transfer of materialof the medium. (In a water wave, for example, the molecules move perpendicularly to the velodty vector of the wave). Consider a kink in a rope that propagates with a velociyV along the +x-axis, as shown y Displacement - , the velocityof the waveform x xat Smet  180 Assume that the shape of the kink does not change in moving a small distance Ax in a short interval of tme At. The speed of the kink is defined to be V = Ax/At. The displacement in the y-diredion is a function of xand t, y=f(x, t). We wish to answer the queston: what basic prndples determine the form of the argument of the function, f ? For water waves, acoustical waves, waves along flexible strings, etc. the wave velocides are much less than c. Since y is a funcion of x and t, we see that all points on the wavebrm move in such a way that the Galilean transformaton holds for all inerdal observers of the waveform. Consider two inertal observers, observer #1 at rest on the x-axis, watching the wave move along the x-axis with constant speed, V, and a second observer #2, moving with the wave. If the observers synchronize theirdockssothat tl=t2=lt=Oatxl=x2=O,then x2 = X1 -Vt. We therefore see that the functional form of the wave is determined by the form of the Galilean transformadon, so that y(x, t) = f(x-Vt), (12.1) where V is the wave velodty in the particular medium. No other functional form is possible! For example, y(x, t) = Asink(x-Vt) is permitted, whereas y(x, t) = A(x2+V\Pt) is not. If the wave moves to the left (in the-x direction) then  181 y(x, t) = f(x+Vt). (122) We shall consider waves that superimpose ineatly. If, for example, two waves move along a rope in opposite directions, we observe that they "pass through each other". If the wave is harmonic, the displacement measured as a funcdion of lime at the origin, x =0, is also harmonic: yo(0, t) = Acos(wt) where A is the maximum amplitude, andco= 2-nv is the angular frequency. The general form of y(x, t), consistent with the Galilean transformation, is y(x, t) = Acos{k(x- Vt)} where k is introduced to make the argument dimensionless (k has dimensions of 1Iength]). We then have yo(0, t) = Acos(kVt) = Acos(wt). Therefore, CO= kV, the angular frequency, (12.3) or 2-nv = kV, so that, k =2nrvN= 2nNVTwhere T =1/v, is the period. (12.4) The general form is then y(x, t) = Acos{(2nNTf)(x- Vt)}  182 = Acos{(2irIA)(x-Vt)}, whereX= VT is the wavelength, = Acos{(2iix/X-2tT)}, = Acos(kx-2nt/T), where k = 21rIA, the "wavenumber', = Acos(kx-cot), = A cos(wt- kx), because cos(-O)= cos(h). (12.5) For a wave moving in three dimensions, the diplacement at apoint x, y, z at lime t has the form 1p(x, y, z, t) = Acos(wt- kr), (12.6) where Iki = 2r/X and r = [x, y, z]. 12.2 The general wave equation An arbitrary waveform in one space dimension can be written as the superposition of two waves, one travelling to the right (+x) and the other to the left (-x) of the origin. The displacement is then y(x, t) = f(x-Vt) + g(x+Vt). (12.7) Put u = f(x-Vt) = f(p), and v = g(x + Vt) = g(q), then y=u+v. Now, By/3x =8u/8x +8v/3x =(duldp)(8pI8x) +(dvldq)(3ql3x) = f'(p)(8pI8x) +g'(q)(3ql3x).  183 Also, 82y/ax2= (3/3x){(duldp)(p/3x) + (dv/dq)(3ql13x)} = f'(p)(&p/x2) + f"(p)(p/xy+ g'(q)(a/x2) + g"x(q)(q/y. We can obtain the second deiivahve of y with respect to time using a similar method: 32yI3t2 = f-( 2p/8t2)+f"(p)ap/t)+ g'(q)(32a/3t2) + g" (q)(q/t Now, 3p/3x = 1, 3q/3x = 1, 3p13t= -V, and 3q/3t = V, and all second derivatves are zero (V is a constant). We therefore obtain 3y/&x2=f"(P)+g"(q)) and 32y8t2=f(p\2+g"(q\/2. Therefore, 32y/8t2 = \P(32yI3x2). or 3y/x2-(1AP)(3zy/3t2) = 0. (12.8) This is the wave equation in one-dimensional space. For a wave propagatng in three-dimensional space, we have V2y-(1AP)(3jI/tg2)=0, (12.9) the general form of the wave equadon, in which l(x, y, z, t) is the general amplitude fundion. 12.3 The Lorentz invariant phase of a wave and the relativistic Doppler shift A wave propagadng through space and tine has a 'wave funclion"  184 y(x, y, z, t) = Acos(wt - kr) where the symbols have the meanings given in 122. The argument of this function can be written as follows V= Acos{(w/c)(ct) - kr). (12.10) It was not untl deBroglie developed his revolutionary idea of particle-wave duality in 1923-24 that the Lorenlz invariance of the argument of this function was fully appreciated! We have V= Acos{[/c, k]T[ct, -r]} = Acos{KEE>}= Acos$, where 4)is the "phase". (12.11) deBroglie recognized that the phase 4)is a Lorentz invariant formed from the 4-vectors K"= [olc, k], the "frequency-wavelength"4-vector, (12.12) and E = [ct, -r], the covariant "event"4-vector. deBroglie's discovery turned out to be of great importance in the development of Quantum Physics. It also provides us with the basic equations for an exact derivation of the relativistic Doppler shit. The frequency-wavelength vector is a Lorenlz 4-vector, which means that it transforms between inerdal observers in the standard way: K'= LK, (12.13) or  185 w'Ic y-3y 0 0 o/c kx' -3y y 0 0 kx k' = 0 0 1 0 k kz' 0 0 0 1 kz The transformadon of the first element therefore gives w'Ic=y(lc) - 3ykx, (12.14) so that 2nv'=y2nv - 3yc(2nT/X) or v'=yv -Vy(v/c), (where o = 2nmv, V/c = p(, and c = vX) therefore v'yv(1- p) or v= (v/(1- p2)11)(1 - P) giving v'=v{(1- p +)/(1+ )}12. (12.15) This is the relatvistc Doppler shift for the special case of photons - we have Lorenlz invariance in acdion. This result was derived in section 62 using the Lorentz invariance of the energy-momentum 4- vector, and the Planck-Einstein result E = hv for the relation between the energy E and the frequencyv of a photon. The present derivaton of the relativistic Doppler shift is independent of the Planck-Einstein  186 result, and therefore provides an independent verification of their fundamental equation E = hv for a photon. 12.4 Plane harmonic waves The one-dimensional wave equaton (12.8) has the solution y(t, x) = Acos(kx-wt), where co= kV and A is independent of both x and t. This form is readily shown to be a solution of (12.8) by direct calculation of its 2nd partial dervatves, and their substtuton in the wave equation. The three-dimensional wave equation (12.9) has the solution y(t, x, y, z) =y0cos{(kxx+kyy+ kzz)-wt}, where o = IkIV, and k = [kx, ky, kz], the wave vector. The solution y(t, x, y, z) is called a plane harmonic wave because constant values of the argument (kxx + kyy + kzz)-wt define a set of planes in space- surfaces of constant phase: z k, normal to the wave surface Equiphase surfaces of a plane wave y x  187 It is often useful to represent a plane harmonic wave as the real part of the remarkable Cotes-Euler equation e,= cosO+isinO, i= so that 9ocos((kr) - wt) = R.P.0er-A . The complex form is readily shown to be a solution of the three-dimensional wave equation. 12.5 Spherical waves For given values of the radial coordinate, r, and the time, t, the funcions cos(kr- wt) and e&) have constant values on a sphere of radius r. In order for the wave funcions to represent expanding spherical waves , we must modify their forms as follows: (lIr)cos(kr-wt) and (1/r)e-) (k along r). (12.16) These changes are needed to ensure that the wave functions are solutions of the wave equation. To demonstrate that the spherical wave (1/r)cos(kr- wt) is a solution of (12.9), we must transform the Lapladan operator from Cartesian to polar coordinates V2(x, y, z) -- V2(r, 0, (1/r2)[(8/Br)(r2(8/Br)) +(1/sin0)(8/80)(sin0(8/80)) + (1/sin2)(2/a94)]. (12.17) This transformaton is set as a problem.  188 If there is spherical symmetry, there is no angular-dependence, in which case, V2(r)= (11r2)(8/Ir)(r2(8/Ir)) =I3r2 + (2/r)(3/3r). (12.18) We can check that V= 4(1/r)cos(kr- ot) is a solution of the radial form of (12.9), Differentiating tMce, we find Yi/Blr2 = 0o[(-k2/r)cosu + (2k/r2)sinu + (21r3)cosu], where u = kr-wt, and j,/8t2= -9o(o2/r)cosu, o= kV, from which we obtain (1P)3 8I3t2 -[39BI3r2 + (2Ir)38I3r] =0. (12.19) 12.6 The superposition of harmonic waves Consider two harmonic waves with the same amplitudes, Vo, travelling in the same direction, the x-axis. Let their angular frequendes be slighty different - w ± 6±with corresponding wavenumbers k ± bk. Their resultant,'1, is given by = 0#-A+[lk-w)+ ei(-)] = 0e-2cos(bkx- bot)]  189 = Acos$, (1220) where A = 2oeA), the resultant amplitude, and $=bkx-bot, the phase of the modulaton envelope. The individual waves travel at a speed olk=v, the phase velodty, (1221) and the modulaton envelope travels at a speed 6olk= vG, the group velodty. (1222) In the limit of a very large number of waves, each differing slightly in frequency from that of a neghbor, dk -- 0, inwhich case doldk = vG. For electromagnetic waves travelling through a vacuum, vG= v, = c, the speed of light We shall not, at this stage, deal with the problem of the superpositon of an arbitrary number of harmonic waves. 12.7 Standing waves The superposidon of two waves of the same amplitudes and frequencies but travelling in opposite directions has the form 'P= ii192 = Acos(kx-owt) + Acos(kx+owt) = 2Acos(kx~cos~t). (12.23)  190 This form describes a standing wave that pulsates with angular frequency o, associated with the lime- dependent term cost. In a traveling wave, the amplitudes of the waves of all partides in the medium are the same and their phases depend on position. In a standing wave, the amplitudes depend on position and the phases are the same. For standing waves, the amplitudes are a maximum when kx =0, i, 2-u, 3n,... and they are a minimum when kx =1/2, 3n12, 5n12, ...(the nodes). PROBLEMS The main treatment of wave motion, induding interference and diffraction effects, takes place in the second semester (Part 2) in discussing Electromagnetism and Optics. 12-1 Ripples on the surface of water with wavelengths of about one centimeter are found to have a phase velodty v, = \(ak) where k is the wave number and ais a constant characteristicof water. Show that their group velodty is ;= (3/2)v,. 12-2 Show that y(x, t) = exp{x-vt} represents a travelling wave but not a periodic wave. 12-3 Two plane waves have the same frequency and they oscillate in the z-direcldon; they have the forms 9y(x, t) =4sin{20t + (lx/3) + 1}, and 9y(y, t) =2sin{20t + (ly/4) + 1}.  191 Show that their superposition at x =5 and y =2 is given by 1p(t) = 2.48sin{20t- (ir/5)}. 12-4 Express the standing wave y = Asin(ax)sin(bt), where a and b are constants as a combinaton of travelling waves. 12-5 Perhaps the most important application of the relatvistc Doppler shift has been, and contnues to be, the measurement of the velocides of recession of distant galaxies relatve to the Earth. The electromagnetc radiation associated with ionized calcium atoms that escape from a galaxy in Hydra has a measured wavlength of 4750 x 1010m, and this is to be compared with a wavelength of 3940 x1 0-10m for the same process measured fora statonary source on Earth. Show that the measured wavelengths indicate that the galaxy in Hydra is receding from the Earth with a speed v = 0.187c.  192 13 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 13.1 Definitions Two n-vectors An =[a1,a2,...an]and Bo,=[b1, b2,...b] are said to be orthogonalif 7_=1,n]ab = 0. (13.1) (Their scalar product is zero). Two functions A(x) and B(x) are orthogonalin the range x = a to x = b if fla,b] A(x)B(x)dx =0. (132) The limits must be given in order to specify the range in which the functions A(x) and B(x) are defined. The set of real, contnuous funcdions{$i(x),$2z(x), ...} is orthogonal in [a, b] if j[ab] 4(x)$n(x)dx =0 for m # n. (13.3) If, in additon, J[a,b]$n2(x)dx =Ifor all n, (13.4) the set is nomnal, and therefore it is said to be orthonomnal. The infinite set {cos0x, cosix, cos2x, ... sin~x, sinix, sin2x, ...} (13.5) in the range [-i-, -i] of x is an example of an orthogonal set. For example, ft-in cosx-cos2xdx =0 etc., (13.6)  193 and fr-in cos2xdx #0=1i, etc. This set, which is orthogonal in any interval of x of length 2ri, is of interest in Mathematcs because a large class of functions of x can be expressed as linear combinatons of the members of the set in the interval2n. For example we can often write $(x) = C1$1+ c$2+ where the c's are constants = aocos0x+alcos1x+a2cos2x+... + bosin0x + bisinix + b2sin2x +... (13.7) A large dass of periodic funcions ,of period 2n, can be expressed in this way. When a function can be expressed as a linear combinaton of the orthogonal set {1, cosix, cos2x, ...0, sin1x, sin2x,...}, it is said to be expanded in its Fourier series. 13.2 Some trigonometric identities and their Fourier series Some of the familiar trigonometric identties involve Fourier series. For example, cos2x=1-2sin2x (13.8) can be written sin2x=(1/2)- (1/2)cos2x and this can be written sin2x ={(1/2)cos0x +0cosix- (1/2)cos2x +0cos3x+.. +0{sin0x+ sinlx+ sin2x+ ...} (13.9)  194 the Fourier series of sin2x. The Fourier series of cos2x is cos2x = (1/2) + (1/2)cos2x. (13.10) More complicated tiigonometiic identes also can be expanded in their Fourier series. For example, the identty sin3x = 3sinx-4sin3x can be written sin3x = (3/4)sinx- (1/4)sin3x, (13.11) and this is the Fourier series of sin3x. The terms in the series represent the "harmonics"of the function sin3x. In a similar fashion, we find that the identity cos3x=4cos3x- 3cosx can be rearranged to give the Fourier series of cos3x cos3x = (3/4) + (1/4)cos3x. (13.12) In general, a combination of deMoivre's theorem and the binomial theorem can be used to write cos(nx) and sin(nx) (for n a positive integer) in terms of powersof sinx and cosx. We have cos(nx) + isin(nx) = (cosx + isinxr (i = J-1) (deMoivre) (13.13) and (a + by = an + nan-1b +(n(n-1)/2!) an-2b2 _.+b". (13.1l.) For example, if n =4, we obtain  195 cos4x = (1/8)cos4x+ (1/2)cos2x+ (3/8), (13.15) and sin4x =(1/8)cos4x-(1/2)cos2x+ (3/8). (13.16) 13.3 Determination of the Fourier coefficients of afunction If, in the interval [a, b], the function f(x) can be expanded in terms of the set {$1(X), $2(x), ...}, which means that f(x) =)[F1,0] C4k(x), (13.17) where {$(X), $(x), ...} is orthogonal in [a, b], then the coefficients can be evaluated as follows: to determine the kh-coefficient, ck, multply f(x) by4f(x), and integrate over the interval [a, b]: Jta,b]f(x)$k(x)dx=j[a,b]C1$1kdx+ ...ab]Qk2dx +... (13.18) = 0 + 0 + #0 + 0 ... The integrals of the products m n in the range [-in, n] are all zero except for the case that involves 42. We therefore obtain the kth-coefficient 0k=f[a,b]f(X)k(X)dX/Ja,b]k2(X)dX k=1, 2, 3,.. (13.19) 13.4 The Fourier series of a periodic saw-tooth waveform In standard works on Fourier analysis it is proved that every periodic contnuous funcion f(x) of period 21T can be expanded in terms of {1, cosx, cos2x, ...0, sinx, sin2x, ...}; this orthogonal set is said to be complete with respect to the set of periodic continuous funcdons f(x) in [a, b].  196 Let f(x) be a periodic saw-tooth waveform with an amplitude of± 1: f(x) +1 -2n -u ) n 2T x The funcdion has the following forms in the three intervals f(x)= (-2/T)(x + ) for-T< x:5 -Td2, =2x/i for-rI2 x TrI2, and = (-2/r)(x-iT) foruil2