Classical Field Theory

Transcription

1 Preprint typeset in JHEP style - HYPER VERSION Classial Field Theory Gleb Arutyunov a a Institute for Theoretial Physis and Spinoza Institute, Utreht University, 3508 TD Utreht, The Netherlands Abstrat: The aim of the ourse is to introdue the basi methods of lassial field theory and to apply them in a variety of physial models ranging from lassial eletrodynamis to marosopi theory of ferromagnetism. In partiular, the ourse will over the Lorentz-ovariant formulation of Maxwell s eletromagneti theory, advaned radiation problems, elements of soliton theory. The students will get aquainted with the Lagrangian and Hamiltonian desription of infinite-dimensional dynamial systems, the onept of global and loal symmetries, onservation laws. A speial attention will be paid to mastering the basi omputation tools whih inlude the Green funtion method, residue theory, Laplae transform, elements of group theory, orthogonal polynomials and speial funtions. Last Update G.Arutyunov@phys.uu.nl Correspondent fellow at Steklov Mathematial Institute, Mosow.

2 Contents 1. Classial Fields: General Priniples 1.1 Lagrangian and Hamiltonian formalisms 3 1. Noether s theorem in lassial mehanis Lagrangians for ontinuous systems Noether s theorem in field theory Hamiltonian formalism in field theory 0. Eletrostatis 1.1 Laws of eletrostatis 1. Laplae and Poisson equations 6.3 The Green theorems 7.4 Method of Green s funtions 9.5 Eletrostati problems with spherial symmetry 31.6 Multipole expansion for salar potential Magnetostatis Laws of magnetostatis Magneti (dipole moment Gyromagneti ratio. Magneti moment of eletron Relativisti Mehanis Newton s relativity priniple Einstein s relativity priniple Defining Lorentz transformations Lorentz group and its onneted omponents Struture of Lorentz transformations Addition of veloities Lie algebra of the Lorentz group Relativisti partile Classial Eletrodynamis Relativisti partile in eletromagneti field 6 5. Lorentz transformations of the eletromagneti field Momentum and energy of a partile in a stati gauge Maxwell s equations and gauge invariane Fields produed by moving harges Eletromagneti waves 7 1

3 5.7 Hamiltonian formulation of eletrodynamis Solving Maxwell s equations with soures Causality priniple Radiation Fields of a uniformly moving harge Fields of an arbitrary moving harge Dipole Radiation Appliability of Classial Eletrodynamis Darvin s Lagrangian Advaned magneti phenomena Exhange interations One-dimensional Heisenberg model of ferromagnetism Non-linear phenomena in media Solitons Appendies Appendix 1: Trigonometri formulae Appendix : Tensors Appendix 3: Funtional derivative Appendix 4: Introdution to Lie groups and Lie algebras Problem Set Problems to setion Problems to setion Problems to setion Problems to setion Problems to setion Problems to setion Problems to setion Reommended literature Classial Fields: General Priniples Classial field theory is a very vast subjet whih traditionally inludes the Maxwell theory of eletromagnetism desribing eletromagneti properties of matter and the Einstein theory of General Relativity. The main sope of lassial field theory is

4 to onstrut the mathematial desription of dynamial systems with an infinite number of degrees of freedom. As suh, this disipline also naturally inorporates the lassis aspets of fluid dynamis. The basi mathematial tools involved are partial differential equations with given initial and boundary onditions, theory of speial funtions, elements of group and representation theory. 1.1 Lagrangian and Hamiltonian formalisms We start with realling the two ways the physial systems are desribed in lassial mehanis. The first desription is known as the Lagrangian formalism whih is equivalent to the priniple of least ation 1 (Maupertuis s priniple. Consider a point partile whih moves in a n-dimensional spae with oordinates (q 1,..., q n and in the potential U(q. The Newtons equations desribing the orresponding motion (trajetory are m q i = U q i. (1.1 These equations an be obtained by extremizing the following funtional S = t t 1 dt L(q, q, t = t t 1 ( m q dt U(q. (1. Here S is the funtional on the spae of partile trajetories: to any trajetory whih satisfies given initial q i (t 1 = qin i and final q i (t = qf i onditions it puts in orrespondene a number. This funtional is alled the ation. The speifi funtion L depending on partile oordinates and momenta is alled Lagrangian. Aording to the priniple of stationary ation, the atual trajetories of a dynamial system (partile are the ones whih deliver the extremum of S. Compute the variation of the ation δs = t t 1 [ d dt dt (m qi + U ] δq i + total derivative, q i where we have integrated by parts. The total derivative term vanishes provided the end points of a trajetory are kept fixed under the variation. The quantity δs vanishes for any δq i provided eq.(1.1 is satisfied. Note that in our partiular example, the Lagrangian oinides with the differene of the kineti and the potential energy L = T U and it does not expliitly depend on time. In general, we simply regard L as an arbitrary funtion of q, q and time. The equations of motion are obtained by extremizing the orresponding ation δs δq = d ( L L i dt q i q = 0 i 1 More aurately, the priniple of stationary ation. 3

5 and they are alled the Euler-Lagrange equations. We assume that L does not depend on higher derivatives q,... q and so on, whih reflets the fat that the orresponding dynamial system is fully determined by speifying oordinates and veloities. Indeed, for a system with n degrees of freedom there are n Euler-Lagrange equations of the seond order. Thus, an arbitrary solution will depend on n integration onstants, whih are determined by speifying, e.g. the initial oordinates and veloities. Suppose L does not expliitly depend on t, then Substituting here L q i Therefore, we find that dl dt = L q i qi + L q i qi. from the Euler-Lagrange equations, we get dl dt = L q i qi + d ( L q i = d ( L dt q i dt q i qi. d ( L dt q i qi L = 0 (1.3 as the onsequene of the equations of motion. Thus, the quantity H = L q i qi L, (1.4 is onserved under the time evolution of our dynamial system. For our partiular example, H = m q L = m q + U(q = T + U E. Thus, H is nothing else but the energy of our system; energy is onserved due to equations of motion. Dynamial quantities whih are onserved during the time evolution of a dynamial system are alled onservation laws or integrals of motion. Energy is our first non-trivial example of a onservation law. Introdue a quantity alled the (anonial momentum p i = L q i, p = (p 1,..., p n. For a point partile p i = m q i. Suppose that U = 0. Then ṗ i = d ( L = 0 dt q i by the Euler-Lagrange equations. Thus, in the absene of the external potential, the momentum p is an integral of motion. This is our seond example of a onservation law. This is homogenuity of time. 4

6 Now we remind the seond desription of dynamial systems whih exploits the notion of the Hamiltonian. The onserved energy of a system expressed via anonial oordinates and momenta is alled the Hamiltonian H H(p, q = 1 m p + U(q. Let us again verify by diret alulation that it does not depend on time, dh dt = 1 m p iṗ i + q i U q i = 1 m m q i q i + q i U q i = 0 due to the Newton equations of motion. Having the Hamiltonian, the Newton equations an be rewritten in the form q j = H p j, ṗ j = H q j. These are the fundamental Hamiltonian equations of motion. Their importane lies in the fat that they are valid for arbitrary dependene of H H(p, q on the dynamial variables p and q. In the general setting the Hamiltonian equations are obtained as follows. We take the full differential of the Lagrangian dl = L q i dqi + L q i d qi = ṗ i dq i + p i d q i = ṗ i dq i + d(p i q i q i dp i, where we have used the definition of the anonial momentum and the Euler-Lagrange equations. From here we find d(p i q i L = q i dp i ṗ i dq i. }{{} H From the differential equality the Hamiltonian equations follow. Transformation H(p, q = p i q i L(q, q q i p i is the Legendre transform. The last two equations an be rewritten in terms of the single equation. Introdue two n-dimensional vetors ( ( H p x =, H = q p j H q j and n n matrix J: J = ( Then the Hamiltonian equations an be written in the form ẋ = J H, or J ẋ = H. 5

7 In this form the Hamiltonian equations were written for the first time by Lagrange in A point x = (x 1,..., x n defines a state of a system in lassial mehanis. The set of all these points form a phase spae P = {x} of the system whih in the present ase is just the n-dimensional Eulidean spae with the metri (x, y = n i=1 xi y i. To get more familiar with the onept of a phase spae, onsider a one-dimensional example: the harmoni osillator. The potential is U(q = q. The Hamiltonian H = p + q, where we hoose m = 1. The Hamiltonian equations of motion are given by ordinary differential equations: q = p, ṗ = q = q = q. Solving these equations with given initial onditions (p 0, q 0 representing a point in the phase spae 3, we obtain a phase spae urve p p(t; p 0, q 0, q q(t; p 0, q 0. Through every phase spae point there is one and only one phase spae urve (uniqueness theorem for ordinary differential equations. The tangent vetor to the phase spae urve is alled the phase veloity vetor or the Hamiltonian vetor field. By onstrution, it is determined by the Hamiltonian equations. The phase urve an onsist of only one point. Suh a point is alled an equilibrium position. The Hamiltonian vetor field at an equilibrium position vanishes. The law of onservation of energy allows one to find the phase urves easily. On eah phase urve the value of the total energy E = H is onstant. Therefore, eah phase urve lies entirely in one energy level set H(p, q = h. For harmoni osillator p + q = h and the phase spae urves are onentri irles and the origin. The matrix J serves to define the so-alled Poisson brakets on the spae F(P of differentiable funtions on P: {F, G}(x = (J F, G = J ij i F j G = The Poisson braket satisfies the following onditions {F, G} = {G, F }, n j=1 ( F p j G q j F q j G p j. {F, {G, H}} + {G, {H, F }} + {H, {F, G}} = 0 3 The two-dimensional plane in the present ase. 6

8 for arbitrary funtions F, G, H. Thus, the Poisson braket introdues on F(P the struture of an infinitedimensional Lie algebra. The braket also satisfies the Leibnitz rule {F, GH} = {F, G}H + G{F, H} and, therefore, it is ompletely determined by its values on the basis elements x i : whih an be written as follows {x j, x k } = J jk {q i, q j } = 0, {p i, p j } = 0, {p i, q j } = δ j i. The Hamiltonian equations an be now rephrased in the form ẋ j = {H, x j } ẋ = {H, x} = X H. It follows from Jaobi identity that the Poisson braket of two integrals of motion is again an integral of motion. The Leibnitz rule implies that a produt of two integrals of motion is also an integral of motion. The algebra of integrals of motion represents an important harateristi of a Hamiltonian system and it is losely related to the existene of a symmetry group. In the ase under onsideration the matrix J is non-degenerate so that there exists the inverse J 1 = J whih defines a skew-symmetri bilinear form ω on phase spae ω(x, y = (x, J 1 y. In the oordinates we onsider it an be written in the form ω = j dp j dq j. This form is losed, i.e. dω = 0. A non-degenerate losed two-form is alled sympleti and a manifold endowed with suh a form is alled a sympleti manifold. Thus, the phase spae we onsider is the sympleti manifold. Imagine we make a hange of variables y j = f j (x k. Then ẏ j = yj ẋ k = A j }{{} x k k J km x mh = A j km yp kj x m y H p A j k 7

9 or in the matrix form ẏ = AJA t y H. The new equations for y are Hamiltonian with the new Hamiltonian is H(y = H(f 1 (y = H(x if and only if AJA t = J. Hene, this onstrution motivates the following definition. Transformations of the phase spae whih satisfy the ondition AJA t = J are alled anonial 4. Canonial transformations 5 do not hange the sympleti form ω: ω(ax, Ay = (Ax, JAy = (x, A t JAy = (x, Jy = ω(x, y. In the ase we onsidered the phase spae was Eulidean: P = R n. This is not always so. The generi situation is that the phase spae is a manifold. Consideration of systems with general phase spaes is very important for understanding the struture of the Hamiltonian dynamis. Short summary A Hamiltonian system is haraterized by a triple (P, {, }, H: a phase spae P, a Poisson struture {, } and by a Hamiltonian funtion H. The vetor field X H is alled the Hamiltonian vetor field orresponding to the Hamiltonian H. For any funtion F = F (p, q on phase spae, the evolution equations take the form df dt = {H, F } = X H F. Again we onlude from here that the Hamiltonian H is a time-onserved quantity dh dt = {H, H} = 0. Thus, the motion of the system takes plae on the subvariety of phase spae defined by H = E onstant. 4 In the ase when A does not depend on x, the set of all suh matries form a Lie group known as the real sympleti group Sp(n, R. The term sympleti group was introdued by Herman Weyl. The geometry of the phase spae whih is invariant under the ation of the sympleti group is alled sympleti geometry. 5 Notie that AJA t = J implies that A t JA = J. Indeed, multiplying by J both sides of the first equality from the right, we get AJA t J = J = 1, whih further implies A t J = J 1 A 1 = JA 1. Finally, multiplying both sides of the last expression from the right by A, we obtain the desired formula. 8

10 1. Noether s theorem in lassial mehanis Noether s theorem is one of the most fundamental and general statements onerning the behavior of dynamial systems. It relates symmetries of a theory with its onservation laws. It is lear that equations of motion are unhanged if we add to a Lagrangian a total time derivative of a funtion whih depends on the oordinates and time only: L L + d G(q, t. Indeed, the hange of the ation under the variation will be dt δs δs = δs + t t 1 dt d G δg(q, t = δs + dt q i δqi t=t t=t 1. Sine in deriving the equations of motion the variation is assumed to vanish at the initial and final moments of time, we see that δs = δs and the equations of motion are unhanged. Let now an infinitezimal transformation q q + δq be suh that the variation of the Lagrangian takes the form (without usage of equations of motion! 6 of a total time derivative of some funtion F : δl = df dt. Transformation δq is alled a symmetry of the ation. Now we are ready to disuss Noether s theorem. Suppose that q = q + δq is a symmetry of the ation. Then δl = L q i δqi + L q i δ qi = L q i δqi + L q i d dt δqi = df dt. By the Euler-Lagrange equations, we get δl = d ( L δq i + L d dt q i q i dt δqi = df dt. This gives δl = d ( L dt q i δqi = df dt. As the result, we find the quantity whih is onserved in time dj dt d ( L dt q i δqi F = 0. This quantity J = L q i δqi F = p i δq i F is alled Noether s urrent. Now we onsider some important appliations. 6 As we have already seen, a variation of the Lagrangian omputed on the equations of motion is always a total derivative! 9

11 Momentum onservation. Momentum onservation is related to the freedom of arbitrary hoosing the origin of the oordinate system. Consider a Lagrangian Consider a displaement L = m q i. q i = q i + a i δq i = a i, q i = q i δ q i = 0. Obviously, under this transformation the Lagrangian remains invariant and we an take F = 0 or F = any onstant. Thus, J = p i δq i = p i a i, Sine a i arbitrary, all the omponents p i are onserved. Angular momentum onservation. Consider again and make a transformation Then, L = m q i q i = q i + ɛ ij q j δq i = ɛ ij q j. δl = m q i ɛ ij q j. Thus, if ɛ ij is anti-symmetri, the variation of the Lagrangian vanishes. Again, we an take F = 0 or F = any onstant and obtain J = p i δq i = p i ɛ ij q j, Sine ɛ ij is arbitrary, we find the onservation of angular momentum omponents J j i = p i q j p j q i. Partile in a onstant gravitational field. The Lagrangian L = m ż mgz. Shift z z + a, i.e. δz = a. We get δl = mga = d ( mgat. Thus, the dt quantity J = mżδz F = mża + mgat is onserved. This is a onservation law of the initial veloity ż + gt = onst. 10

12 Conservation of energy. Energy onservation is related to the freedom of arbitrary hoosing the origin of time (you an perform you experiment today or after a several years but the result will be the same provided you use the same initial onditions. We derive now the onservation law of energy in the framework of Noether s theorem. Suppose we make an infinitesimal time displaement δt = ɛ. The Lagrangian response on it is δl = dl dt ɛ. On the other hand, δl = L q i δqi + L q δ i qi + L t δt = d ( L δq i + L dt q i q δ i qi, where we have used the Euler-Lagrange equations and assumed that L does not expliitly depends on time. Obviously, δq i = q i ɛ and δ q i = q i ɛ, so that δl = d dt ( L q i q i ɛ + L q i qi ɛ = dl dt ɛ. Canelling ɛ, we reover the onservation law for the energy dh dt = 0, H = p i q i L. Finally, it remains to note that in all the symmetry transformations we have onsidered so far the integration measure dt in the ation did not transform (even for in the last example dt d(t + ɛ = dt. 1.3 Lagrangians for ontinuous systems So far our disussion onerned a dynamial system with a finite number of degrees of freedom. To desribe ontinuous systems, suh as vibrating solid, a transition to an infinite number of degrees of freedom is neessary. Indeed, one has to speify the position oordinates of all the points whih are infinite in number. The ontinuum ase an be reahed by taking the appropriate limit of a system with a finite number of disrete oordinates. Our first example is an elasti rod of fixed length l whih undergoes small longitudinal vibrations. We approximate the rod by a system of equal mass m partiles spaed a distane a apart and onneted by uniform massless springs having the fore onstant k. The total length of the system is l = (n + 1 a. We desribe the displaement of the ith partile from its equilibrium position by the oordinate φ i. Then the kineti energy of the partiles is T = n i=1 m φ i. 11

13 The potential energy is stored into springs and it is given by the sum U = 1 n k (φ i+1 φ i. i=0 Here we assoiate φ 0 = 0 = φ n+1 with the end points of the interval whih do not move. The fore ating on ith partile is F i = U φ i : F i = k(φ i+1 + φ i 1 φ i. This formula shows that the fore exerted by the spring on the right of the ith partile equals to k(φ i+1 φ i, while the fore exerted from the left is k(φ i φ i 1. The Lagrangian is L = T U = n m φ i 1 n k (φ i+1 φ i. i=1 i=0 At this stage we an take a ontinuum limit by sending n and a 0 so that l = (n + 1 a is kept fixed. Inreasing the number of partiles we will be inreasing the total mass of a system. To keep the total mass finite, we assume that the ratio m/ a µ, where µ is a finite mass density. To keep the fore between the partiles finite, we assume that in the large partile limit k a Y, where Y is a finite quantity. Thus, we have L = T U = 1 n i=1 ( m a φ i 1 a n ( φi+1 φ i a(k a. a i=0 Taking the limit, we replae the disrete index i by a ontinuum variable x. As a result, φ i φ(x. Also φ i+1 φ i a Thus, taking the limit we find L = 1 φ(x + a φ(x a l 0 dx [µ φ ] Y ( x φ. x φ(x. Also equations of motion an be obtained by the limiting proedure. Starting from m a φ i k a φ i+1 + φ i 1 φ i a = 0, and using φ i+1 + φ i 1 φ i lim a 0 a = φ x xxφ 1

14 we obtain the equation of motion µ φ Y xx φ = 0. Just as there is a generalized oordinate φ i for eah i, there is a generalized oordinate φ(x for eah x. Thus, the finite number of oordinates φ i has been replaed by a funtion of x. Sine φ depends also on time, we are dealing with the funtion of two variables φ(x, t whih is alled the displaement field. The Lagrangian is an integral over x of the Lagrangian density The ation is a funtional of φ(x, t: S[φ] = t t 1 L = 1 µ φ 1 Y ( xφ. dt l 0 dx L (φ(x, t, φ(x, t, x φ(x, t. It is possible to obtain the equations of motion for the field φ(x, t diretly from the ontinuum Lagrangian. One has to understand how the ation hanges under an infinitesimal hange of the field The derivatives hange aordingly, This gives δs[φ] = S[φ + δφ] S[φ] = Integrating by parts, we find δs[φ] = φ(x, t φ(x, t + δφ(x, t. (1.5 t φ(x, t t φ(x, t + δφ(x, t, t (1.6 φ(x, t φ(x, t + δφ(x, t. x x x (1.7 + t t 1 l 0 dt t t 1 l 0 dt l 0 [ L L dx δφ + φ φ tδφ + L ] ( x φ xδφ. [ L dx φ L t φ L ] x δφ ( x φ dx L ( t φ δφ t=t t=t 1 + t t 1 L dt ( x φ δφ x=l x=0. (1.8 The ation priniple requires that the ation priniple be stationary w.r.t. infinitezimal variations of the fields that leave the field values at the initial and finite time unaffeted, i.e. δφ(x, t 1 = δφ(x, t = 0. 13

15 On the other hand, sine the rod is lamped, the displaement at the end points must be zero, i.e. δφ(0, t = δφ(l, t = 0. Under these irumstanes we derive the Euler-Lagrange equations for our ontinuum system ( L + ( L L t ( t φ x ( x φ φ = 0. Let us now disuss the solution of the field equation φ Y xx φ = 0, = µ, where is the propagation veloity of vibrations through the rod. This equation is linear and, for this reason, its solutions satisfy the superposition priniple. Take an ansatz φ(x, t = e ikx a k (t + e ikx b k (t. If we impose φ(0, t = 0, then b k (t = a k (t and we an refine the ansatz as φ(x, t = a k (t sin kx. Requiring that φ(l, t = 0 we get sin kl = 0, i.e. k k n = πn l. Coeffiients a k(t then obey ä k + k a k (t = 0 a k (t = e iω kt a k, where ω k = ±k is the dispersion relation. Thus, the general solution is φ(x, t = ( sin k n x A n os ω n t + B n sin ω n t, ω n = k n, n and the onstants A n, B n are fixed by the initial onditions, whih is an initial profile φ(x, 0 and an initial veloity φ(x, 0. Salar and Vetor Fields The generalization to ontinuous systems in more spae dimensions is now straightforward. In two-dimensions one an start with two-dimensional lattie of springs. The displaement of a partile at the site (i, j is measured by the quantity φ ij, whih is a two-dimensional vetor. In the limit when we go to a ontinuum, this beomes a displaement field φ(x, y, t of a membrane subjeted to small vibrations in the (x, y-plane. In three dimensions we get a vetor φ ijk. The ontinuous limit yields a three-dimensional displaement field φ(x, y, z, t of a ontinuous solid vibrating in the x, y, z diretions with eoms of a partial differential equation type: φ 1 xx φ yy φ 3 zz φ 4 xy φ 5 yz φ 6 xz φ = 0, 14

16 the oeffiients i enode the properties of the solid. In general, fields depending on the spae-time variables are tensors, i.e. they transforms under general oordinate transformations in a definite way. Namely, a tensor field φ i 1...i p j 1...j q of rank (p, q under general oordinate transformations of the oordinates x i : x i x i (x j transforms as follows 7 φ k 1...k p l 1...l q (x = x k 1 x x kp i 1 x ip x j1 x l 1 xjq x lq φi 1...i p j 1...j q (x. Here tensor indies are ated with the matries x i whih form a group GL(d, R. x j This is a group of all invertible real d d matries. A simplest example is a salar field that does not arry any indies. Its transformation law under oordinate transformations is φ (x = φ(x. We stress that a point with oordinates x in the original frame and a point with oordinates x in the transformed frame is the one and the same geometri point. 1.4 Noether s theorem in field theory In order to fully desribe a dynamial system, it is not enough to only know the equations of motion. It is also important to be able to express the basi physial harateristis, in partiular, the dynamial invariants, of the systems via solutions of these equations. Noether s theorem: To any finite-parametri, i.e. dependent on s onstant parameters, ontinuous transformation of the fields and the spae-time oordinates whih leaves the ation invariant orresponds s dynamial invariants, i.e. the onserved funtions of the fields and their derivatives. To prove the theorem, onsider an infinitezimal transformation x i x i = x i + δx i, i = 1,..., d, φ I (x φ I(x = φ I (x + δφ I (x. As in the finite-dimensional ase, the variations δx i and δφ I are expressed via infinitezimal linearly independent parameters δω n : δx i = Xnδω i n, δφ I (x = Φ I,n δω n. (1.9 1 n s 1 n s Here all δω n are independent of the oordinates x. Suh transformations are alled global. The oeffiients X i n and Φ I,n may depend on x and the fields, and they 7 There is a simple rule to remember the appearane of primed and unprimed indies in the tensor transformation rule. Assuming that all indies on the left hand side of the tensor transformation formula are primed, then they must label primed oordinates in the right hand side of the formula. 15

17 desribe a response of oordinates and fields on the infinitezimal transformation with a parameter δω n. Obviously, partiular ases of the transformations above arise, when X k n = 0 or Φ I,n = 0. In the first ase the oordinates x i do not hange under symmetry transformations at all, while the fields are transformed aording to φ I (x φ I(x = φ I (x + δφ I (x. In the seond ase the symmetry ats on the spae-time oordinates only and the ondition Φ I,n = 0 implies that φ I (x = φ I (x, i.e. the fields under onsiderations are salars. We point out that in the ase when φ I is not a salar but rather a tensor, Φ I,n is not zero even if the symmetry ats on the spae-time oordinates only! To illustrate this point, onsider a vetor field φ i (x. Under oordinate transformation x i x i = x i + δx i one gets φ i (x = x i x j φj (x = (xi + δx i φ j (x = φ i (x + δxi x j x j φj (x }{{} δφ i, whih implies that the orresponding quantity Φ I is non-trivial; the trivial ase ours only when δx i does not depend on oordinates, i.e. it is a onstant. In the general ase symmetry transformations at on both the spae-time oordinates and the fields, f. eq.(1.9. Consider φ I(x = φ I(x + δx = φ I(x + k φ I(xδx k +... = φ I(x + k φ I (x X k nδω n +... It is important to realize that the operations δ and / x do not ommute. This is beause δ is the variation of the fields due to both the hange of their form and their arguments x i. We therefore introdue the notion of the variation of the form of the field funtion δφ I (x = φ I(x φ I (x = (Φ I,n k φ I X k nδω n. Variation of the form does ommute with the derivative / x. For the variation of the Lagrangian density we, therefore, have L (x = L (x + dl dx k δxk = L (x + L (x L (x + dl }{{} dx k δxk. δl (x The hange of the ation is 8 δs = dx L (x dx L (x = dx [L (x + δl (x + dl dx k δxk ] dx L (x. 8 We onsider a field theory in d-dimensions, so that the integration measure dx must be understood as dx = dx 1 dx... dx d d d x. 16

18 Transformation of the integration measure is x 1 x d x dx x 1 = J dx det.. dx = det } x 1 x d x d x {{ d } Jaobian Thus, at leading order in δω n we have dx = dx(1 + k δx k Plugging this into the variation of the ation, we find dl ] δs = dx [ δl (x + dx k δxk + k δx k L = We further note that δl (x = L φ I δφi δx1 δx d x 1 x 1.. δx 1 x d dx [ δl (x + L ( L ( k φ I k δφ I = k δφi + ( k φ I 1 + δxd x d dx. d ] dx (L k δxk. L ( k φ I k δφ I = = k ( L ( k φ I δφ I, where we have used the Euler-Lagrange equations. Thus, we arrive at the following formula for the variation of the ation δs = dx d [ L dx k ( k φ I δφ ] I +L δx k = dx d [ L ] dx k ( k φ I (Φ I,n m φ I Xn m +L Xn k δω n. Sine the integration volume is arbitrary we onlude that where dj k n dx k = 0 divj n = 0, Jn k = L ( k φ I (Φ I,n m φ I Xn m L Xn k and n = 1,... s. Thus, we have shown that the invariane of the ation under the s-parametri symmetry transformations implies the existene of s onserved urrents. An important remark is in order. The quantities J k n are not uniquely defined. One an add J k n J k n + m χ km n, where χ km n = χ mk n. Adding suh anti-symmetri funtions does not influene the onservation law k J k n = 0. Now we are ready to investigate onrete examples of symmetry transformations and derive the orresponding onserved urrents. 17

19 Energy-momentum tensor. Consider the infinitezimal spae-time translations x k = x k + δx k = x k + δ k nδω n = X k n = δ k n and Φ I,n = 0. Thus, the onserved urrent Jn k beomes in this ase a seond rank tensor Tn k Tn k = L ( k φ I nφ I δnl k. Here, as usual, the sum over the index I is assumed. The quantity Tn k is the so-alled stress-energy or energy-momentum tensor. If all the fields vanish at spaial infinity then the integral 9 P n = d n 1 xt 0 n is a onserved quantity. Here 0 signifies the time diretion and the integral is taken over the whole (n 1-dimensional spae. Indeed, dp n dt = dx dt 0 n dt = d n 1 x dt i n dx i = dω ( T n n, Ω where Ω is a (n -dimensional sphere whih surrounds a n 1-dimensional volume; its radius tends to infinity. The vetor n is a unit vetor orthogonal to Ω. Angular momentum. Consider infinitezimal rotations x n x n + x m δω nm, where δω nm = δω mn. Beause of anti-symmetry, we an hoose δω nm = δω nm with n < m as linearly independent transformation parameters. We find δx k = X k j δω j = n<m = m<l x l δ k mδω ml + m>l X k nmδω nm = x l δω kl = x l δ k mδω ml x l δ k mδω ml = m<l(x l δ k m x m δ k l δω ml. (1.10 From here we dedue that X k nm = x m δ k n x n δ k m, n < m. If we onsider a salar field then φ (x = φ(x and δφ = 0. Φ I,n = 0. Using the general formula As a result, Jn k = L ( k φ I (Φ I,n m φ I Xn m L Xn k, 9 Here we expliitly distinguished a time diretion t and write the integration measure in the ation as dx = dt d n 1 x. 18

20 we therefore find the following angular momentum tensor M k lm = L ( k φ ( lφ x m m φ x l + L (x l δ k m x m δ k l. Notie that the last formula an be written in the form ( L ( L Mlm k = x m ( k φ lφ L δl k x l ( k φ mφ L δm k where T k l is the stress-energy tensor. = x m T k l x l T k m, If we onsider now a vetor field φ i, then aording to the disussion above, we will have δφ i = Φ i mlδw ml = δxi x j φj (x = ( (x x j l δm i x m δlδω i ml m<l m<l so that Φ i ml = (g jl δ i m g jm δ i lφ j = φ l δ i m φ m δ i l, where g ij is a spae-time metri. Aording to our general formula, the set of orresponding Noether urrents will have the form Jmn k = L ( k φ i (Φi mn l φ i Xmn l L Xmn k. Substitution of all the quantities gives Jmn k = L [ φn δ ( k φ i m i φ m δn i l φ i (x n δm l x m δ l n] L (x n δm k x m δn k. We, therefore, see that for the vetor field, the angular-momentum tensor takes the form ( L Jmn k = x n Tm k x m Tn k ( k φ n φ m L ( k φ m φ n. The first piee here, whih depends on the stress-energy tensor is alled the orbital momentum and the seond piee haraterizes polarization properties of the field and is related with a notion of spin. The final remark onern ontinuous s-parametri transformations whih leave the ation invariant up to a total derivative term (in the original formulation of the Noether s an exat invariane of the ation was assumed! δs = δω n dx k Fn k. These transformations also lead to onservation laws. It obtain them, it is enough to subtrat from the anonial urrent J k n the term F k n : J k n = J k n F k n. One an verify that this new urrent is onserved k Jn k equations of motion. as the onsequene of the 19

21 1.5 Hamiltonian formalism in field theory As was disussed above, in the Lagrangian formalism the dynamis of lassial fields φ i is desribed by the ation funtional S = Ldt = dtd x L (φ i, µ φ i, where L is the Lagrangian density being a funtion of φ i and µ φ i taken at the same point x. The transition to the Hamiltonian formalism is performed by introduing the anonial momenta onjugate to the oordinates φ i : p i (x = δl δ φ i (x = L φ i (x. The Hamiltonian has the form H = d x H, H = L φ φ i (x L, i (x where in the right hand side of the last formula one has to substitute the expression for φ i (x via p i (x and φ i (x. The definition of the Poisson brakets is also generalized to the field-theoreti ase. For any two loal in time funtionals F and G of fields and their momenta we define their Poisson braket as the following funtional [ δf δg {F, G} = d x δp i (x δφ i (x δg δf ], δp i (x δφ i (x where F and G are taken at the same moment of time. The Hamiltonian equations are then φ i = {H, φ i }, ṗ i = {H, p i }. The anonial Poisson brakets are {φ i (t, x, φ j (t, y} = 0, {p i (t, x, p j (t, y} = 0, {p i (t, x, φ j (t, y} = δ j i δ( x y. Note that all the fields for whih the brakets are omputed are taken at the one and the same moment of time! Consider the simplest example of a real massive salar field φ desribed by the Lagrangian density L = 1 ( µφ µ φ m φ. The momentum is p(x = L = φ(x φ(x and, therefore, the Hamiltonian density is H = 1 ( p i φ i φ + m φ. 0

22 . Eletrostatis Classial eletrodynamis is a theory of eletri and magneti fields aused by marosopi distributions of eletri harges and urrents. Within the field of eletrodynamis, one an study eletromagneti fields under ertain stati onditions leading to eletrostatis (eletri fields independent of time and magnetostatis (magneti fields independent of time. First, we fous on the laws of eletrostatis..1 Laws of eletrostatis Eletrostatis is the study of eletri fields produed by stati harges. It is based entirely on Coulomb s law (1785. This law defines the fore that two eletrially harged bodies (point harges exert on eah other F ( x = k q 1 q x 1 x x 1 x 3, (.1 where k is Coulomb s onstant (depends on the system of units used 10, q 1 and q are the magnitudes of the two harges, and x 1 and x are their position vetors (as presented in Figure 1. q 1 x 1 x q Figure 1: Two harges q 1 and q and their respetive position vetors x 1 and x. The harges exert an eletri fore on one another. One an introdue the onept of an eletri field E as the fore experiened by a point-like harge q in the limit of vanishing q F ( x E ( x = lim. q 0 q We have used the limiting proedure to introdue a test harge suh that it will only measure the eletri field at a ertain point and not reate its own field. Hene, using 10 In SI units (SI the international system of units, the Coulomb s onstant is k = 1 4πɛ 0, while fore is measured in newtons, harge in oulombs, length in meters, and the vauum permittivity ɛ 0 is given by ɛ 0 = 107 4π = F/m. Here, F indiates farad, a unit of apaitane being equal to one oulomb per volt. One an also use the Gauss system of units (CGS entimetre-gramseond. In CGS units, fore is expressed in dynes, harge in statoulombs, length in entimeters, and the vauum permittivity then redues to ɛ 0 = 1 4π. 1

23 Coulomb s law, we obtain an expression for the eletri field of a point harge x x E ( x = kq x x. 3 Sine E is a vetor quantity, for multiple harges we an apply the priniple of linear superposition. Consequently, the field strength will simply be a sum of all of the ontributions, whih we an write as E ( x = k i q i x x i x x i. (. 3 Introduing the eletri harge density ρ ( x, the eletri field for a ontinuous distribution of harge is given by E ( x = k ρ ( x x x x x 3 d3 x. (.3 The Dira delta-funtion (distribution allows one to write down the eletri harge density whih orresponds to loal harges ρ ( x = N q i δ ( x x i. (.4 i=1 Substituting this formula into eq.(.3, one reovers eq.(.. However, eq.(.3 is not very onvenient for finding the eletri field. For this purpose, one typially turns to another integral relation known as the Gauss theorem, whih states that the flux through an arbitrary surfae is proportional to the harge ontained inside it. Let us onsider the flux of E through a small region of surfae ds, represented graphially in Figure, dn = ( E q n ds = ( r n ds r3 = q r os ( r, n ds = q r ds, where on the first step we have used that E = q r. By the definition of ds, we r 3 observe that it is positive for an angle θ between E and n less than π, and negative otherwise. We introdue the solid angle dω dω = ds r. (.5 Plugging this relation into eq.(.5 leaves us with the following expression for the flux dn = q dω. (.6

24 q E n Figure : The eletri flux through a surfae, whih is proportional to the harge within the surfae. By integrating eq.(.6, we obtain the following equation for the flux N S ( E n ds = { 4πq if q is inside the surfae 0 otherwise Equivalently, using the fat that the integral of the harge distribution over volume V is equal to the total harge enlosed in the volume, i.e. q = ρ (x V d3 x, one finds a similar expression ( N = E n ds = 4π ρ(x d 3 x. S By making use of the Gauss-Ostrogradsky theorem, one may rewrite the above integral in terms of the volume integral of the divergene of the vetor field E ( E n ds = div E ( x d 3 x. S Realling that the left hand side is equal to 4πq, a relation between the divergene of the eletri field and the harge density arises [ 0 = div E ] ( x 4πρ ( x d 3 x. V Sine the relation holds for any hosen volume, then the expression inside the integral must equal to zero. The resulting equation is then div E ( x = 4πρ ( x. This is known as the differential form of the Gauss (law theorem for eletrostatis. This is the first equation from the set of four Maxwell s equations, the latter being the essene of eletrodynamis. V 3

25 The Gauss theorem is not enough, however, to determine all the omponents of E. A vetor field A is known if its divergene and its url, denoted as div A and rot A respetively, are known. Hene, some information is neessary about the url of eletri field. This is in fat given by the seond equation of eletrostatis rot E = 0. (.7 The seond equation of eletrostatis is known as Faraday s law in the absene of time-varying magneti fields, whih are obviously not present in eletrostatis (sine we required all fields to be time independent. We will derive this equation in the following way. Starting from the definition of the eletri field (Coulomb s law given by equation (.3, we rewrite it in terms of a gradient and pull the differential operator outside of the integral E ( x = ρ ( x x x x x 3 d3 x = ρ ( x 1 x x x d3 x = ρ ( x x x x d3 x ρ( x = grad x x d3 x. (.8 From vetor alulus we know that the url of gradient is always equal to zero, suh that rot (grad f = 0 rot E = 0. This derivation shows that the vanishing of rot E is not related to the inverse square law. It also shows that the eletri field is the minus gradient of some salar potential E( x = grad ϕ. From the above, it then follows that this salar potential is given by ϕ(x = ρ(x x x d3 x, (.9 where the integration is arried out over the entire spae. Obviously, the salar potential is defined up to an additive onstant; adding any onstant to a given ϕ(x does not hange the orresponding eletri field E. What is the physial interpretation of ϕ(x? Consider the work whih has to be done to move a test harge along a path from point A to B through an eletri field E B W = A B F d l = q A E d l. 4

26 B d l E A Figure 3: The work that has to be done over a harged partile to move it along the path from A to B through an eletri field E. The minus sign represents the fat that the test harge does work against the eletri fores. By assoiating the eletri field as the gradient of a salar potential, one obtains B B W = q gradϕ d ϕ ϕ ϕ l = q dx + dy + A A x y z dz tb ( ϕ dx = x dt + ϕ dy y dt + ϕ dz dt = q (ϕ B ϕ A, z dt t A where we have parametrized the path as (x(t, y(t, z(t. The result is just a differene between the potentials at the end points of the path. This implies that the potential energy of a test harge is given by V = q ϕ. In other words, the potential energy does not depend on the hoie of path (hene, the eletri fore is a onservative fore. If a path is hosen suh that it is losed, i.e. A = B, the integral redues to zero E d l = 0. This result an also be obtained from Stokes theorem ( E d l = rot E ds = 0, where we have used the fat that rot E = 0. S To summarize, we have derived two laws of eletrostatis in the differential form E ( x = div E ( x = 4πρ ( x, (.10 E ( x = rot E ( x = 0. (.11 5

27 . Laplae and Poisson equations In the previous setion it was shown that the url of the eletri field is equal to zero, thus the field is simply the gradient of some salar funtion, whih an be written as rot E ( x = 0 E ( x = ϕ ( x. Substituting the right hand side of this expression into equation (.10, we obtain This gives div ϕ ( x = 4πρ ( x. ϕ ( x ϕ ( x = 4πρ ( x. (.1 Equation (.1 is known as the Poisson equation. In ase ρ ( x = 0, i.e. in a region of no harge, the left hand side of (.1 is zero, whih is known as the Laplae equation. Substituting into (.1 the form salar potential ϕ, given by (.9, we get ϕ ( x = ρ( x x x d3 x = ( 1 d 3 x ρ( x. x x Without loss of generality we an take x = 0, whih is equivalent to hoosing the origin of our oordinate system. By swithing to spherial oordinates, we an show that 1 x x = 1 r = 1 ( d r 1 = 0. r dr r This is true everywhere exept for r = 0, for whih the expression above is undetermined. To determine its value at r = 0 we an use the following trik. Integrating over volume V, using the Gauss law and the fat that r = n, one obtains ( ( 1 d 3 x = div r 1 d 3 x = n r 1 r ds V V S ( ( 1 1 = n n ds = r dω r r r r }{{} = 4π. ds Therefore, S ( 1 = 4πδ( x, r S or Thus, we find x 1 x x = 4πδ ( x x. ϕ = ρ(x ( 4πδ(x x d 3 x = 4πρ(x. Hene, we have proved that 1 solves the Poisson equation with the point harge r soure. In general, the funtions satisfying ϕ = 0 are alled harmoni funtions. 6

28 .3 The Green theorems If in eletrostatis we would always deal with disrete or ontinuous distributions of harges without any boundary surfaes, then the general expression (where one integrates over all of spae ϕ(x = ρ(x d 3 x (.13 x x would be the most onvenient and straightforward solution of the problem. In other words, given some distribution of harge, one an find the orresponding potential and, hene, the eletri field E = ϕ. In reality, most of the problems deals with finite regions of spae (ontaining or not ontaining the harges, on the boundaries of whih definite boundary onditions are assumed. These boundary onditions an be reated by a speially hosen distribution of harges outside the region in question. In this situation our general formula (.13 annot be applied with the exeption of some partiular ases (as in the method of images. To understand boundary problems, one has to invoke the Green theorems. Consider an arbitrary vetor field 11 A. We have div A ( d 3 x = A n ds. (.14 V S Let us assume that A has the following speifi form A = ϕ ψ, where ψ and ϕ are arbitrary funtions. Then div A ( = div ϕ ψ ( = div ϕ ψ x i V = ϕ ψ + ϕ ψ. S = ( ϕ ψ x i x i Substituting this bak into eq.(.14, we get ( ( ϕ ψ + ϕ ψ d 3 x = ϕ ψ n ds = S ϕ ( dψ ds. dn whih is known as the first Green formula. When we interhange ϕ for ψ in the above expression and take a differene of these two we obtain the seond Green formula ( ϕ ψ ψ ϕ ( d 3 x = ϕ dψ dn ψ dϕ ds. (.15 dn V 11 Now introdued for mathematial onveniene, but it will later prove to be of greater importane. S 7

29 By using this formula, the differential Poisson equation an be redued to an integral equation. Indeed, onsider a funtion ψ suh that ψ 1 R = 1 x x ψ = 4πδ ( x. (.16 Substituting it into the seond Green formula (.15 and assuming x is inside the spae V integrated over, one gets ( [ 4πϕ( x δ ( x x + 4πρ( x d 3 x = ϕ d ( 1 1 ] dϕ ds. x x dn R R dn V Here we have hosen ϕ ( x to satisfy the Poisson equation ϕ ( x = 4πρ ( x. By using the sampling property of the delta funtion, i.e. ϕ V ( x δ ( x x = ϕ ( x, the expression above allows one to express ϕ( x as ρ ( x ϕ ( x = R d3 x + 1 [ 1 ϕ 4π R n ϕ ( ] 1 ds, (.17 n R V S whih is the general solution for the salar potential. The terms inside the integrals are equal to zero if x lies outside of V. Consider the following two speial ases: If S goes to and the eletri field vanishes on it faster than 1, then the R surfae integral turns to zero and ϕ( x turns into our general solution given by eq.(.13. For a volume whih does not ontain harges, the potential at any point (whih gives a solution of the Laplae equation is expressed in terms of the potential and its normal derivative on the surfae enlosing the volume. This result, however, does not give a solution of the boundary problem, rather it represents an integral equation, beause given ϕ and ϕ (Cauhy boundary onditions n we overdetermined the problem. S Therefore, the question arises whih boundary onditions should be imposed to guarantee a unique solution to the Laplae and Poisson equations. Experiene shows that given a potential on a losed surfae uniquely defines the potential inside (e.g. a system of ondutors on whih one maintains different potentials. Giving the potential on a losed surfae orresponds to the Dirihlet boundary onditions. Analogously, given an eletri field (i.e. normal derivative of a potential or likewise the surfae harge distribution (E 4πσ also defines a unique solution. These are the Neumann boundary onditions 1. 1 Note that both Dirihlet as well as Neumann boundary onditions are not only limited to eletrodynamis, but are more general and appear throughout the field of ordinary or partial differential equations. 8

30 One an prove, with the help of the first Green formula, that the Poisson equation ϕ = 4πρ, in a volume V has a unique solution under the Dirihlet or the Neumann onditions given on a surfae S enlosing V. To do so, assume there exist two different solutions ϕ 1 and ϕ whih both have the same boundary onditions. Consider U = ϕ ϕ 1. It solves U = 0 inside V and has either U = 0 on S (Dirihlet or U n (Neumann. In the first Green formula one plugs ϕ = ψ = U, so that V ( U + U U d 3 x = S U = 0 on S ( U ds. (.18 n Here the seond term in the integral vanishes as U = 0 by virtue of being the solution to the Laplae equation and the right hand side is equal to zero, sine we have assumed that the value of the potential (Dirihlet or its derivative (Neumann vanish at the boundary. This equation is true iff 13 V U = 0 U = 0 U = 0 (.19 Thus, inside V the funtion U is onstant everywhere. For Dirihlet boundary onditions U = 0 on the boundary and so it is zero uniformly, suh that ϕ 1 = ϕ everywhere, i.e. there is only one solution. Similarly, the solution under Neumann boundary onditions is also unique up to unessential boundary terms..4 Method of Green s funtions This method is used to find solutions of many seond order differential equations and provides a formal solution to the boundary problems. The method is based on an impulse from a soure, whih is later integrated with the soure funtion over entire spae. Reall 1 x x = 4πδ ( x x. (.0 1 However, the funtion is just one of many funtions whih obeys ψ = x x 4πδ ( x x. The funtions that are solutions of this seond order differential equation are known as Green s funtions. In general, G ( x, x = 4πδ ( x x, (.1 9

31 S S 1 Figure 4: Choosing arbitrarily the surfaes S 1 and S, where S is the area between them, we let them expand so that the average value of the salar potential tends to zero. where G ( x, x = 1 x x + F ( x, x, so that F ( x, x = 0, i.e. it obeys the Laplae equation inside V. The point is now to find suh F ( x, x, that gets rid of one of the terms in the integral equation (.17 we had for ϕ ( x. Letting ϕ = ϕ ( x and ψ = G ( x, x, we then get ϕ ( x = V ρ ( x G ( x, x d 3 x + 1 [ G ( x, x ϕ ( x ϕ ( x G ( x, ] x ds. 4π S n n By using the arbitrariness in the definition of the Green funtion we an leave in the surfae integral the desired boundary onditions. For the Dirihlet ase we an hoose G boundary ( x, x = 0 when x S, then ϕ( x simplifies to ϕ ( x = V ρ ( x G ( x, x d 3 x 1 ϕ ( x G ( x, x ds, 4π S n is the bulk- where G ( x, x is referred to as the bulk-to-bulk propagator and G( x, x n to-boundary propagator. For the Neumann ase we ould try to hoose G( x, x n one has G ( x, x ( ds = G n n S = 4π ds = = 0 when x S. However, div G d 3 x = G d 3 x δ( x x d 3 x = 4π. (. For this reason we an not demand G( x, x = 0. Instead, one hooses another simple n ondition G( x, x = 4π, where S is the total surfae area, and the left hand side of n S 13 If and only if. 30

32 the equation is referred to as the Neumann Green funtion. Using this ondition: ϕ ( x = ρ ( x G N (x, x d 3 x + 1 G N ( x, x ϕ ( x ds V 4π S n + 1 ϕ ( x ds (.3 S S The last term represents ϕ, the averaged value of the potential on S. If one takes the limit S = S 1 + S, where S 1 and S are two surfaes enlosing the volume V and suh that S tends to infinity, this average disappears. In any ase, the extra term 1 ϕ S S ( x ds is just a onstant (does not depend on x and, therefore, does not ontribute to the eletri field E = ϕ..5 Eletrostati problems with spherial symmetry Frequently, when dealing with eletrostatis, one enounters the problems exhibiting spherial symmetry. As an example, take the Coulomb law (.1, whih depends on the radial distane only and has no angular dependene. When enountering a symmetry of that sort, one often hooses a set of onvenient oordinates whih greatly simplifies the orresponding problem. z θ P( r, θ, φ φ r y x Figure 5: Spherial oordinate system. It is no surprise that in this ase, we will be making use of spherial oordinates, whih in terms of the Cartesian oordinates, are given by r = x + y + z, ( z θ = aros, (.4 x + y + z ( y φ = artan, x To obtain the Cartesian oordinates from the spherial ones, we use x = r sin θ os φ, y = r sin θ sin φ, (.5 z = r os θ. 31

33 In terms of spherial oordinates our differential operators look different. The one we will be most interested in, the Laplae operator, beomes = 1 r ( r r r + 1 ( r sin θ θ sin θ + θ Hene, in these oordinates the Laplae equation reads as ϕ = 1 r r (rϕ + 1 r sin θ ( sin θ ϕ + θ θ We use the ansatz that ϕ (r, θ, φ = U(r r Laplae equation and multiplying both sides by [( 1 r sin θ U U + r 1 r sin θp 1 r sin θ φ. 1 ϕ r sin θ φ = 0. P (θ Q (φ. Upon substituting this into the, one obtains r 3 sin θ U(rP (θq(φ ( ] P sin θ + 1 Q θ θ Q φ = 0. Sine we only have φ dependene in the last term we an state that, sine there are no other terms with φ, then this term has to be onstant (hosen here for onveniene with antiipation of the solution 1 Q Q φ = m. Hene the solution is Q = e ±imφ, where m is an integer suh that Q is single valued. This leaves us with two separated equations. For P (θ the equation simplifies to and for U (r one obtains ( 1 d sin θ dp ] + [l(l + 1 m sin θ dθ dθ sin P = 0, θ d U l (l + 1 U = 0, dr r where we have just again onveniently piked l(l + 1 to be the integration onstant suh that in our solution it will appear in a onvenient form. It is easy to verify that the solution to the equation for U(r is given by U (r = Ar l+1 + Br l, where l is assumed to be positive and A and B are arbitrary onstants. The seond equation, on the other hand, is a bit more ompliated and upon substitution os θ = x it transforms into d dx [ (1 ] ] x dp + [l(l + 1 m P = 0, dx 1 x 3

34 whih one an reognize as the so-alled generalized Legendre equation. Its solutions are the assoiated Legendre funtions. For m = 0, we obtain the Legendre equation [ d (1 x dp ] + l(l + 1P = 0. (.6 dx dx The solutions to this equation are referred to as the Legendre polynomials. In order for our solution to have physial meaning, it must be finite and ontinuous on the interval 1 x 1. We try as a solution the following power series (the Frobenius method P (x = x α j=0 a j x j, (.7 where α is unknown. Substituting our trial solution (.7 into the Legendre equation (.6, we obtain j=0 ( (α + j (α + j 1 a j x α+j [(α + j (α + j + 1 l (l + 1] a j x α+j = 0. For j = 0 and j = 1, the first term will have x α and x α 1 and the seond term will have x α and x α+1 respetively, whih will never make the equation equal to zero unless a 0 0, then α (α 1 = 0 so that (A α = 0 or α = 1 a 1 0, then α (α + 1 = 0 so that (B α = 0 or α = 1 For other j, one obtains a reurrene relation a j+ = (α + j (α + j + 1 l (l + 1 a j (α + j + 1 (α + j + Cases (A and (B are atually equivalent. We will onsider ase (A for whih α = 0 or 1. The expansion ontains only even powers of x for α = 0 and only odd powers of x for α = 1. We note two properties of this series: 1. The series is onvergent for x < 1 for any l.. The series is divergent at x = ±1 unless it is trunated. It is obvious from the reurrent formula that the series is trunated in the ase that l is a non-negative integer. The orresponding polynomials are normalized in 33

35 Figure 6: Profiles of a few Legendre polynomials. suh a way that they are all equal to identity at x = 1. These are the Legendre polynomials Pl (x: P0 (x = 1 ; P1 (x = x ; 1 3x 1 ; P (x = 1 P3 (x = 5x3 x ; 3 l 1 dl. Pl (x = l x 1 l! dxl The general expression given in the last line is also known as the Rodrigues formula. The Legendre polynomials form a omplete system of orthogonal funtions on 1 x 1. To hek whether they are indeed orthogonal, one takes the differential equation for Pl, multiplies it by Pl0, and then integrates Z 1 d dpl Pl 0 (1 x + l(l + 1Pl dx = 0, dx dx 1 or Z 1 1 dpl0 dpl + l(l + 1Pl0 Pl dx = 0. (x 1 dx dx 34