1 Summary of Electrostatics

Transcription

1 1 Summary of Eletrostatis Classial eletrodynamis is a theory of eletri and magneti fields aused by marosopi distributions of eletri harges and urrents. In these letures, we reapitulate the basi onepts of lassial eletrodynamis. 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. Then we derive Maxwell s equations and study some of their solutions. We end up with the disussion of two lassial radiation problems. 1.1 Laws of Eletrostatis Eletrostatis is the study of eletri fields produed by stati harges. It is based entirely on Coulomb s law. This law defines the fore that two eletrially harged bodies point harges exert on eah other F x = k q 1 q 2 x 1 x 2 x 1 x 2 3, 1 where k is Coulomb s onstant depends on the system of units used 1, q 1 and q 2 are the magnitudes of the two harges, and x 1 and x 2 are their position vetors as presented in Figure 1. One an introdue the onept of an eletri field E as the fore per unit harge 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 Coulomb s law, we obtain an expression for the eletri 1 In SI 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 2 to one oulomb per volt. One an also use the Gauss system of units CGS. 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. 2

2 q 1 x 1 x 2 q 2 Figure 1: Two harges q 1 and q 2 and their respetive position vetors x 1 and x 2. The harges exert an eletri fore on one another. 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. 2 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 N ρ x = q i δ x x i. 4 i=1 Substituting this formula into eq.3, one reovers eq.2. 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 1.2, dn = E q n ds = r n ds r3 = q r os r, n ds = q 2 r 2 ds, 3

3 E n q Figure 2: The eletri flux through a surfae, whih is proportional to the harge within the surfae. where on the first step we have used that E = q r r 3. By the definition of ds, we observe that it is positive for an angle θ between E and n less than π 2 and negative otherwise. We introdue the solid angle dω dω = ds r 2. 5 Plugging this relation into eq.5 leaves us with the following expression for the flux dn = q dω. 6 By integrating eq.6, we obtain the following equation for the flux N { E 4πq if q is inside the surfae n ds = 0 otherwise S 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 ρx 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 4 V

4 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. 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 = x ρ x x x d3 x = grad ρ x 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. 5

5 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 ρx ϕx = x x d3 x, 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 ϕx does not hange the orresponding eletri field E. What is the physial interpretation of ϕx? 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. 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 l = q A A = q B A ϕ x ϕ ϕ dx + dy + y z dz dϕ = q ϕ B ϕ A. 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. 6

6 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. 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, 9 E x = rot E x = Charged Surfaes The eletri field E x of harged surfaes an be omputed by using the Gauss theorem. Let us define the surfae harge density q σ x = lim S 0 S = dq ds. One onsiders the flux N through suh an arbitrary surfae, and the harge enlosed in this surfae S is q = S σ: the harge density times the area of the surfae. As shown in Figure 1.4, we onsider a prism through whih the eletri field passes. The height of the prism is denoted by the paramater 7

7 n 2 S q n 1 Figure 4: The flux through a small surfae element S with harge q. dl. From this, the flux an be omputed. First, onsider the Gauss theorem whih allows one to write N = E n ds = 4πq = 4πS σ x. As before, we an express the flux in terms of its omponents as follows N = E 1 os E 1, n 1 S + E 2 os E 2, n 2 S + N. Here N is the ontribution from the sides horizontal flux. Note that the vetors n 1 = n and n 2 = n whih an be observed in the figure. Now by letting the height dl of the prism approah zero, we maintain that the prism strethes above and below the surfae, yet the horizontal ontributions N beome negligible. This leaves the following simplified relation N = E 1n + E 2n S = 4πσ S, where E 1n and E 2n are projetions of E on N. The flux is then just a measure of the jump in the eletri fields through the harged surfae. By removing ommon fators on both sides, we arrive at the following expression E n2 E n1 = 4πσ x. 8

8 Thus, normal omponents of E at two lose points separated by a harged surfae differ from eah other by 4πσ. 1.3 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 9, we obtain div ϕ x = 4πρ x. This gives 2 ϕ x ϕ x = 4πρ x. 11 Equation 11 is known as the Poisson equation. In ase ρ x = 0, i.e. in a region of no harge, the left hand side of 11 is zero, whih is known as the Laplae equation. Substituting into 11 the form salar potential ϕ, given by 8, we get 2 ϕ x = 2 ρ x x x d3 x = 1 d 3 x ρ x 2. 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 2 1 x x = 2 1 r = 1 r d 2 r 1 = 0. dr 2 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 and using the Gauss law, one obtains V 2 1 r d 3 x = = V S 1 div r 1 n r r d 3 x = nds = 9 S S n 1 r ds r 1 r 2 dω = 4π. r

9 Therefore, or 2 x 1 2 = 4πδx, r 1 x x = 4πδ x x. Thus, we find 2 ϕ = ρx 4πδx x d 3 x = 4πρx. Hene, we have proved that 1 solves the Poisson equation with the point r harge soure. In general, the funtions satisfying 2 ϕ = 0 are alled harmoni funtions. 1.4 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 ϕx = ρx d 3 x x x 12 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 12 an not 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. 10

10 Consider an arbitrary vetor field 2 A. We have div A d 3 x = A n ds. 13 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 + ϕ 2 ψ. ϕ ψ x i Substituting this bak into eq.13, we get ϕ ψ + ϕ ψ 2 d 3 x = ϕ ψ n ds = V S = ϕ ψ x i x i 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 V ϕ 2 ψ ψ 2 ϕ d 3 x = S ϕ dψ dn ψ dϕ ds. 14 dn 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 2 ψ = 4πδ x. 15 Substituting it into the seond Green formula 14 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 2 Now introdued for mathematial onveniene, but it will later prove to be of greater importane. 11 S

11 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 ϕx as ϕ x = V ρ x R d3 x + 1 4π S [ 1 ϕ R n ϕ n 1 R ] ds, 16 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 R, then the surfae integral turns to zero and ϕx turns into our general solution given by eq.12. 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 n Cauhy boundary onditions we overdetermined the problem. 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 3. One an prove, with the help of the first Green formula, that the Poisson equation 2 ϕ = 4πρ, 3 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. 12

12 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 2 whih both have the same boundary onditions. Consider U = ϕ 2 ϕ 1. It solves 2 U = 0 inside V and has either U = 0 on S Dirihlet or U = 0 n on S Neumann. In the first Green formula one plugs ϕ = ψ = U, so that V U 2 + U 2 U d 3 x = S U U n ds. 17 Here the seond term in the integral vanishes as 2 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 4 V U 2 = 0 U = 0 U = 0 18 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 2 everywhere, i.e. there is only one solution. Similarly, the solution under Neumann boundary onditions is also unique up to unessential boundary terms. 1.5 Method of Green Funtions This method is used to find solutions for 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 4 If and only if. 2 1 x x = 4πδ x x

13 However, the funtion 1 x x is just one of many funtions whih obeys 2 ψ = 4πδ x x. The funtions that are solutions of this seond order differential equation are known as Green s funtions. In general, 2 G x, x = 4πδ x x, 20 where G x, x = 1 + F x, x x x, so that 2 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 16 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 4π S [ G x, x ϕ x ϕ x G x, x 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 4π S ϕ x G x, x n ds, where G x, x is referred to as the bulk-to-bulk propagator and G x, x is n the bulk-to-boundary propagator. For the Neumann ase we ould try to hoose G x, x = 0 when x S. n However, one has G x, x ds = G n ds = div G n d 3 x = 2 G d 3 x S = 4π δx x d 3 x = 4π. 21 For this reason we an not demand G x, x = 0. Instead, one hooses another n simple ondition G x, x = 4π, where S is the total surfae area, and the n S left hand side of 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 22 S S 14 ] ds.

14 S 2 S 1 Figure 5: For an arbitrary hoie of surfaes S 1 and S 2, where S is the area between them, then when we let them expand then the last term in equation 22 would vanish. The last term represents ϕ, the averaged value of the potential on S. If one takes the limit S = S 1 +S 2, where S 1 and S 2 are two surfaes enlosing the volume V and suh that S 2 tends to infinity, this average disappears. 1.6 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. 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 2 + y 2 + z 2, z θ = aros, 23 x2 + y 2 + z 2 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, 24 z = r os θ. 15

15 z θ P r, θ, φ φ r y x Figure 6: Spherial oordinate system. In terms of spherial oordinates our differential operators look different. The one we will be most interested in, the Laplae operator, beomes 2 = 1 r 2 r r2 + 1 r r 2 sin θ θ sin θ θ r 2 sin 2 θ φ. 2 Hene, in these oordinates the Laplae equation reads as 2 ϕ = 1 2 r r rϕ + 1 sin θ ϕ 1 2 ϕ + 2 r 2 sin θ θ θ r 2 sin 2 θ φ = 0. 2 We use the ansatz that ϕ r, θ, φ = Ur P θ Q φ. Upon substituting this r r into the Laplae equation and multiplying both sides by 2 sin 2 θ, one UrP θqφ obtains 2 φ = r 2 sin 2 θ 1 U 2 U + r 2 1 r 2 sin θp P sin θ Q θ θ Q φ. 2 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 2 Q Q φ = 2 m2. 16

16 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 1 d sin θ dp ] + [ll + 1 m2 sin θ dθ dθ sin 2 P = 0, θ and for U r one obtains d 2 U l l + 1 U = 0, dr2 r 2 where we have just again onveniently piked ll + 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 Ur 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 2 dp dx ] + [ll + 1 m2 1 x 2 ] P = 0, whih one an reognize as the so-alled generalized Legendre equation. Its solutions are the assoiated Legendre funtions. For m 2 = 0, we obtain the Legendre equation [ d 1 x 2 dp ] + ll + 1P = 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 P x = x α j=0 a j x j, 26 17

17 where α is unknown. Substituting our trial solution 26 into the Legendre equation 25, we obtain α + j α + j 1 a j x α+j 2 j=0 [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 2 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+2 = α + j α + j + 1 l l + 1 a j α + j + 1 α + j + 2 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 2 < 1 for any l. 2. 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 suh a way that they are all equal to identity at x = 1. These are the Legendre polynomials P l x: P 0 x = 1 ; P 1 x = x ; P 2 x = 1 3x 2 1 ; 2 P 3 x = 1 5x 3 2x ; 3 P l x = 1 d l x 2 1 l. 2 l l! dx l 18

18 ϕ = 0 S Figure 7: The field ϕ x, whih obeys the Laplae equation, has no maximum or minimum inside a region S. The general expression given in the last line is also known as the Rodriges 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 P l, multiplies it by P l, and then integrates or 1 P l [ d dx 1 x2 dp l dx + ll + 1P l ] dx = 0, [ x 2 1 dp ] l dp l dx dx + ll + 1P l P l dx = 0. Now subtrat the same equation, but with the interhange of l and l, suh that the following expression is left 1 [l l + 1 ll + 1] P l P l = 0. 1 The equation above shows that for l l the polynomials are orthogonal 1 1 P l P l = 0. By using the Rodriges formula, one an get an identity 1 1 P l xp l xdx = 2 2l + 1 δ l,l. 19

19 For any funtion defined on 1 x 1 fx = A l P l x, l=0 A l = 2l fxp l xdx. Note that this expansion and its oeffiients are not different to any other set of orthogonal funtions in the funtion spae. In situations where there is azimuthal symmetry, one an take m = 0. Thus, ϕ r, θ = Al r l + B l r l+1 P l os θ. l=0 If harge is absent anywhere in the viinity of the oordinate system, one should take B l = 0. Take a sphere of radius a with the potential V θ. Then V θ = A l a l P l os θ l=0 so that A l = 2l + 1 2a l π 0 V θp l os θ sin θdθ. Example: find the potential of an empty sphere of radius r = a whih has two semi-spheres with separate potentials V θ, suh that the potential is equal to V for 0 θ < π and equal to V for π < θ π. For suh a 2 2 system, the salar potential is given by ϕr, θ = V 1 j 1 2j 1Γj a 2j P2j 1 os θ π j! r j=1 [ 3 r = V P 1 os θ 7 r 3 P3 os θ + 11 r 5 P5 os θ...]. 2 a 8 a 16 a Here Γ z for R z > 0 is defined as Γ z = 0 20 t z 1 e t dt.

20 Finally, we would like to omment on the solutions of the Laplae equation ϕ = 0. It is not diffiult to show that one annot have an absolute minimum or maximum in the region in both diretions, x and y beause for an < 0 implying that in the other diretion the seond derivative must have an opposite sign. extremum to exist one requires ϕ x i = 0 whih results in 2 ϕ x 2 i > 0 or 2 ϕ x 2 i 2 Eletrodynamis Here we will treat eletrodynamis as a lassial relativisti field theory. We will rewrite the basi equations of eletrodynamis in the manifestly Lorentzovariant form. We will also study the orresponding solutions. 2.1 Relativisti Partile in Eletormagneti Field Let us first revisit some of the basis of speial relativity written using tensor notation. The Minkowski metri η µν that we will use has the signature +,,, and we will use the onvention that the Latin indies run only over the spae oordinates i.e. i, j, k... = 1, 2, 3, whereas the Greek indies will inlude both time and spae oordinates i.e. µ, ν, σ, ρ... = 0, 1, 2, 3. Additionally, in speial relativity we will have to distinguish between 3-vetors those with only spae omponents and 4-vetors having both spae and time omponents. The onvention that we will use is that A will denote a 3-vetor, whereas A µ will denote a 4-vetor. Using these definitions, we an define the Lorentz invariant relativisti interval given by the expression ds 2 = x µ x µ = 2 dt 2 dx i The ation for a relativisti partile has the following form Rewriting 27, we get S = α b a b ds2 = α ds. a ds = dx µ dt dx µ dt dt2 = dx µ dt dx µ dt dt

21 A Figure 8: The simplest form of ation is given by the length of the spae-time interval between points A and B. B Here we have used the onvention V µ V µ = η µν V µ V ν, where η µν is the Minkowski metri. dx µ =, v, ds = dt v 2 = 1 v2. 2 Therefore, t1 S = α 1 v2 t 0 dt, 2 where in non-relativisti physis we assume v2 1. In general, S = 2 t1 t 0 L dt where L is the so-alled Lagrangian of the system, whih in the non-relativisti limit is given by: L = α 1 1 v2 α v α + α v If we want to reover the usual form of the Lagrangian L = Kin Energy V ext for a free partile V ext = 0 hene L = 1 2 m v2, we need to set α = m. When we do so, equation 29 turns into Thus, one an rewrite L as L = m m v2. L = m ẋ µ ẋ µ. When we use the anonial momentum p µ defined as the derivative of L with respet to ẋ µ, we get p µ = L ẋ µ = m ẋ µ ẋν ẋ ν. 22

22 A ϕ, A Figure 9: In the presene of the vetor potential A µ = ϕ, A the ation of a harged partile ontains an additional term desribing an interation with the vetor potential. B Now when we take p 2 p µ p µ = m 2 2 ẋ µ ẋ µ ẋν ẋ ν 2 = m2 2. Hene, the partile trajetories whih minimize the ation must satisfy the onstraint p 2 m 2 2 = 0, whih is referred to as the mass-shell ondition. Let us now define the vetor potential, whih is an underlying field a Lorentz invariant 4-vetor in eletrodynamis that we will base our further derivations on. It reads A µ = ϕ x, A x. Notie that A µ A µ = η µν A ν = ϕ x, A x. The properties of a harged partile with respet to its interation with eletromagneti field are haraterized by a single parameter: the eletri harge e. The properties of the eletromagneti field itself are determined by the vetor A µ, the eletromagneti potential introdued above. Using these quantities, one an introdue the ation of a harged partile in eletromagneti field, whih has the form S = m b a ds e A µ dx µ. Using Hamilton s priniple, stating that partiles follow paths that minimize their ation δs = 0, we an derive the equations of motion in whih we neglet the bak reation of the harge on the eletromagneti field dxµ 0 = δs = m ds dδxµ e [δa µ dx µ + A µ dδx µ ]

23 Using 28, the term δs in the first integral beomes δds = dx µdδx µ, dxµ dx µ whereas in the seond integral we have simply used the produt rule of differentiation. Let us onsider for a moment the U µ = dxµ term, whih we will ds refer to as 4-veloity. The expliit form of U µ is U µ = dxµ ds = dx µ = 1 v, v2 dt 1 v2 1 v and it has an interesting property that U µ U µ = dx µ ds dx µ ds = 1. Note that this result is only valid for the signature of the metri that we hose. If we were to invert the signature, the result would be 1 instead. Using the fat that δa µ = A µ x ν + δx ν A µ x ν = ν A µ δx ν +, we an rewrite equation 30 as follows δs = m du µ δx µ + e ν A µ dx ν δx µ ν A µ δx ν dx µ = 0. This imposes the following ondition for the extremum m du µ ds + e νa µ µ A ν U ν = 0. Identifying the tensor F νµ of the eletromagneti field ν A µ µ A ν = F νµ = F µν, we an write the equation of motion of the harge in the eletromagneti field as follows m du µ ds = e F µν U ν. 32 This expression an also be written in a more suggestive form if we define the momentum p µ = mu µ whih is onsistent with the requirement p 2 = m 2 2 sine U 2 = 1, so that one an express the aeleration term du µ as = d2 x µ ds ds 2 dp µ ds = dpµ dt dt ds = e F µν U ν, 33 24

24 where the right hand side of the equation is referred to as the Lorentz fore, whereas the left hand side is simply the rate of hange of momentum with respet to the relativisti interval. This equation is omparable with the Newtonian statement: fore is the rate of hange of momentum. Note that this derivation has assumed that the eletromagneti field is given fixed and that we vary the trajetory of the partile only the endpoints remain fixed. 2.2 Gauge Invariane and Maxwell s Equations All the physial properties of the eletromagneti field as well as the properties of harge in the eletromagneti field are determined not by A µ, but rather by F µν. The underlying reason for this is that eletrodynamis exhibits an important new type of symmetry 5. To understand this issue, we may deide to hange the vetor potential in the following way A µ A µ + µ χ, 34 whih an be rewritten in a less abstrat form of spae and time omponents separately: A A + χ and ϕ ϕ 1 χ t. 35 These transformations are referred to as the gauge transformations. Let us see what effet they have on the tensor of the eletromagneti field: δf µν = µ A ν + ν χ ν A µ + µ χ F µν = µ ν χ ν µ χ = Thus, the transformation 34 does not hange the form of the eletromagneti field tensor. For this reason eletromagnetism is a gauge invariant theory! The tensor of the eletromagneti field an be then written as 0 E x E y E z F µν = E x 0 H z H y E y H z 0 H x 37 E z H y H x 0 5 This symmetry extends to many other physial theories besides eletrodynamis. 25

25 and, therefore, F µν = η µσ η νρ F σρ 0 E x E y E z E x 0 H z H y E y H z 0 H x E z H y H x 0, 38 where we have defined the F 0i omponents to be the eletri fields and the F ij omponents to the magneti fields. From the eletri and magneti fields one an make invariants, i.e. objets that remain unhanged under Lorentz transformations. In terms of the tensor of th eletromagneti field two suh invariants are F µν F µν = inv ; 39 ε µνρσ F µν F ρσ = inv. 40 Let us inspet the gauge invariane of the eletri and magneti fields E and H, whih from the form and their in terms of the eletromagneti field tensor omponents an be expressed in terms of the vetor potential as E = ϕ 1 A t and H = rot A. 41 One an easily see that in the first ase an extra ϕ term anels with an extra A term and in the seond ase we have the gauge transformation ontribution vanishing due to the fat that rot gradχ = 0. We look bak at the expression for the Lorentz fore and try to write it in terms of eletri and magneti fields. Rearranging 33, we get dp i dt = = e F i0 U 0 + e ds F ij U j dt = e 1 e Ei + 1 F ij v 1 v2 v v2 2 2 We an thus rewrite the expression for the Lorentz fore as dp i dt = eei + e [ v, H ]

26 Conerning this result,it is interesting to point out that de kin dt = d dt m 2 dp i = v 1 dt = e E v. v2 2 This is the work of the eletromagneti field on the harge. Hene, the magneti field does not play any role is kineti energy hanges, but rather only affets the diretion of the movement of the partile! Using basi vetor alulus and the definitions of the eletri and magneti fields 41, the first two Maxwell s equations are attained div H = div rot A = 0 div H = 0 ; 44 rot E = 1 rot grad ϕ 1 t rot A rot E = 1 H t. 45 Equation 44 is known as the no magneti monopole rule and 45 is referred to as Faraday s law, whih we have already enountered in the previous setion, but then the right hand side was suppressed due to time independene requirement. Together these two equations onstitute the first pair of Maxwell s equations. Notie that these are 4 equations in total, as Faraday s law represents three equations - one for every spae diretion. Additionally, notie that Faraday s law is onsistent with eletrostatis; if the magneti field is time independent then the right hand side of the equation is equal 0, whih is exatly equation 10. These equations also have an integral form. Integrating 45 over a surfae S with the boundary S and using Stokes theorem, we arrive at rot E ds = E d l = 1 Hd S t. 46 S S S For eq.44 one integrates both sides over the volume and uses the Gauss- Ostrogradsky theorem to arrive at div HdV = H ds = V V 2.3 Fields Produed by Moving Charges Let us now onsider the ase where the moving partiles produe the fields themselves. The new ation will be then S = S partiles + S int + S field, 27

27 where we have added a new term S field, whih represents the interation between the partiles and the field that they have produed themselves. We will write is as S field F µν F µν d 4 x = F µν F µν dt d 3 x. Then adding the proportionality onstants the total ation is written as S = m ds e A µ dx µ 1 F µν F µν dt d 3 x, 16π where we have adopted the Gauss system of units, i.e. µ 0 = 4π and ε 0 = 1. 4π Note that we an rewrite the seond term as e A µ dx µ = 1 = 1 ρa µ dx µ dv = 1 j µ A µ dv dt = 1 2 dx µ ρa µ dv dt dt j µ A µ d 4 x, 48 where in the seond line we have introdued, the urrent j i = ρ dxi = ρ, ρ v. dt Inluding this, we an now write the ation of the moving test harge as S = m ds 1 j ρ A 2 ρ d 4 x 1 F µν F µν dtd 3 x. 16π Keeping soures onstant and the path unhanged i.e. δj µ = 0 and δs = 0, we an write the deviation from the ation as follows δs = 1 j ρ δa 2 ρ d 4 x 1 F µν δf µν dtd 3 x 8π = 1 [ 1 j ρ δa ρ d 4 x + 1 ] F µν 4π x δa µdtd 3 x, 49 ν where in the last term in the first line, we have used that δf µν = µ δa ν ν δa µ. To find the extremum, we need to satisfy δs = 0, whih due to eq.49, is equivalent 1 2 jµ 1 4π µ F µν = 0. 28

28 Upon rearrangement, this gives us the seond pair of Maxwell s equations F µν x ν = 4π jµ. Notie that for vanishing urrents, these equation resemble the first pair of Maxwell s equations given by 54, when urrents are to vanish i.e. j µ = 0. Identifying the respetive omponents of the eletromagneti tensor we an rewrite the seond pair of Maxwell s equations in a more familiar form rot H = 4π j + 1 E t and div E = 4πρ, 50 where 4π j and 4πρ are the soures and 1 E is the so-alled displaement t urrent. The first expression is Ampére s law also known as the Biot-Savart law, whereas the seond one is Coulomb s law, whih we have already found before, but using a different priniple. Finally, we notie that the ovariant onservation of the urrent jµ = 0 is equivalent to the ontinuity equation x µ ρ t + divj = 0. Below we inlude here a short digression on the tensor of the eletromagneti field. It is easy to hek that, using the definition of the tensor, the following is true: df = F µν x + F νσ σ x + F σµ µ x ν = With a hange of indies, this takes the form ε µνσρ F νσ x ρ = 0, 52 whih are four equations in disguise, sine we are free to pik any value of the index µ. Let us introdue the so-alled dual tensor Then we an rewrite equation 52 as F µν = 1 2 εµνρσ F ρσ. 53 F µν x ν =

29 Omitting the urrents in the seond pair, the first and seond pair of Maxwell s equations are similar. Indeed, we have F µν x µ = 0, F µν x µ = 0. The main differene between them is that the first pair it never involves any urrents. This has a deeper meaning. The magneti field, as opposed to the eletri field, is an axial vetor, i.e. one that does not hange sign under refletion of all oordinate axes. Thus, if there would be soures for the first pair of Maxwell equations, they should be an axial vetor and a pseudosalar 6. The lassial desription of partiles does not allow to onstrut suh quantities from dynamial variables assoiated to partile. 2.4 Eletromagneti Waves When the eletri harge soure and urrent terms are absent, we obtain the eletromagneti wave solutions. In this ase the Maxwell equations redue to rote = 1 H t, div E = 0, roth = 1 E t, div H = 0. These equations an have non-zero solutions meaning that the eletromagneti fields an exist without any harges or urrents. Eletromagneti fields, whih exist in the absene of any harges, are alled eletromagneti waves. Starting with the definitions of the eletri and magneti fields given in terms of the vetor potential in equation 41, one an hoose a gauge, i.e. fix A µ, whih will simplify the mathematial expressions as well as the alulations, we will be dealing with. The reason why we are allowed to make this hoie is that gauge symmetry transforms one solution into another, both solutions being physially equivalent 7. By making a gauge hoie one 6 A physial quantity that behaves like a salar, only it hanges sign under parity inversion e.g. an improper rotation. 7 Both solutions belong the same gauge orbit. 30

30 breaks the gauge symmetry. This removes the exessive, unphysial degrees of freedom, whih make two physially equivalent solutions to the equations of motion appear different. Obviously the simpliity of these equations and their solutions drastially depends on the gauge hoie. One of the onvenient gauge hoies involves setting µ A µ = 0, whih is the ovariant gauge hoie known as the Lorenz gauge 8. This however is not a omplete gauge hoie, beause, as will be shown later, there are still the gauge transformations that leave the eletromagneti field tensor unhanged. A further speifiation of the Lorenz gauge known as the Coulomb gauge sets the divergene of the vetor or the salar potential equal to zero, i.e. div A = 0 and ϕ = 0. We will return bak to the omparison of these gauge hoies later. To see the proess of gauge fixing and how we an use it to simplify the equations of motion, onsider the gauge transformations A A + f, ϕ ϕ 1 f t. If f does not depend on t, ϕ will not hange, however A will. On the other hand, div A does not depend on t by the Maxwell equations 9. Thus, in this gauge, equations 41 beome E = ϕ 1 H = rot A. A t = 1 A t, Plugging this into 50, our Maxwell s equation desribing the url of the 8 Often erroneously referred to as the Lorentz gauge, due to the similarity with the name Lorentz as in Lorentz transformations, developed by Duth physiist Hendrik Lorentz. However it was a Danish physiist, Ludvig Lorenz, who atually introdued the Lorenz gauge. 9 Under the gauge transformation with the time-independent funtion f we have div A div A + div f, therefore, the funtion f should be determined from the Poisson equation f = div A. 31

31 magneti field, we obtain rot H = rot rot A = 1 t 1 A = 1 2 A t t, 2 A + grad div A = A t 2. In this gauge we an hoose f, suh that the term involving the divergene of A disappears. The equation that remains is known as d Alembert s equation or the wave equation A A t 2 = 0. When we only onsider the plane-wave solutions i.e. only x-dependene, then the equation redues to 2 f x 1 2 f 2 2 t = 0. 2 It an be further written in the fatorized form t x t + f = 0. x With a hange of variables ξ = t x and η = t + x 2 f solution to the equation is ξ η = 0. Hene, the f = f ξ + f η. Changing our variables bak to x and t, we find that the general solution for f is given by f = f 1 t x + f 2 t + x. Notie that this solution simply represents the sum of right- and left-moving plane waves of any arbitrary profile, respetively. Let us return to the issue of the Coulomb versus the Lorenz gauge hoie, and first onsider the later. The Lorentz gauge ondition reads as follows 0 = Aµ x µ = div A + 1 φ t. 32

32 We see that under gauge transformations the Lorenz gauge ondition transforms as µ A µ + µ χ = Aµ x µ + µ µ χ and it remains unhanged provided µ µ χ = 0. Thus, the Lorenz gauge does not kill the gauge freedom ompletely. We still have a possibility to perform gauge transformations of the speial type µ µ χ = 0. Hene there will be still an exessive number of solutions that are physially equivalent and transform into eah other under gauge transformations involving harmoni funtions. This problem is fixed with the introdution of the omplete gauge hoie. Starting over, one an always fix ϕ = 0 by hoosing a suitable funtion χ x, t, i.e. a funtion suh that ϕ = 1 χ. Under the gauge transformations t we have ϕ ϕ 1 χ ϕ = 0. t Transforming the new ϕ = 0 with a new, only spae-dependent funtion χ x, y, z, we obtain 10 Sine E = ϕ 1 0 = ϕ ϕ 1 A t div E = 1 χ t = 0 and A A + χ. and ϕ = 0, we find t div A and div E = 0, where the right hand side has to be equal to zero from our original assumption - lak of soures of eletromagneti fields. From the above equation we an infer that t div A = 0. We an use yet another gauge freedom to set the spae-dependent and time-independent χ, suh that div A = div χ, whih means that we have reahed the Coulomb gauge div A div A + div χ = 0. Having fixed the gauge, let us now onsider plane wave solution to the d Alambert equation. In this ase the derivatives of the y and z omponent of the vetor potential with respet to y and z omponents respetively 10 Note that χ t = 0. 33

33 H E diretion of propagation should vanish as we will only look at osillations in the x diretion. This implies that div A = 0 = A x x + A y y + A z z A x x = 0. If A x x form = 0 everywhere, then 2 A x x 2 = 0, whih leaves the wave equation in the 2 A x x A x t 2 = A x = 0 2 A x 2 t 2 t 2 = 0 A x t = onst. Sine we are not interested in a onstant eletri field E x, we need to fix A x = 0. Sine E = 1 A and H = rot A, then t [ H = t x, A ] = 1 [ n, ] A t = [ n, E ], where [ ] A, B denotes the ross-produt of two vetors. From the definition of the ross produt one an see that the eletri field E and the magneti field H are perpendiular to eah other. Waves with this property are referred to as transversal waves. Eletromagneti waves are known to arry energy; we an define the energy flux to be S = [ ] E, H [ [ ]] = E, n, E. 4π 4π 34

34 Sine [ a, [ b, ]] = b a, a, b, where a, b denotes the salar produt between vetors a and b, we find the following result S = 4π n E 2, where due to orthogonality of n and E the ontribution of the seond term vanishes. The energy density is given by W = 1 E 2 + H 8π 2. For eletromagneti waves E = H, so that W = 1 4π E 2. Hene, there exists a simple relationship S = W n. We define the momentum assoiated to the eletromagneti wave to be p = S 2 = W n. For a partile moving along n, we have p = W. Consider a partile moving with veloity v. We then have p = ve whih for v beomes 2 p = E ; the dispersion relation for a relativisti partile moving at the speed of light photon. Continuing, we are now interested in the ase of fields reated by moving harges. So far we have disussed 1. Time-independent fields reated by harges at rest 2. Time-dependent fields but without harges We will now study time-dependent fields in the presene of arbitrary moving harges 11. Consider x ν µ A ν ν A µ = F µν x ν = 4π jµ, 2 A ν 2 A µ = 4π x ν x µ x ν x ν jµ. 11 The motion of the harges has to be stritly defined, i.e. even though the harges produe an eletromagneti field, their motion will not be influened by the presene of external eletromagneti fields. 35

35 Imposing the Lorenz ondition we obtain from the previous equation A ν x ν = 0, 2 x ν x ν A µ = 4π jµ. The last equation an be split into two A A t 2 = 4π j, ϕ ϕ t 2 = 4π ρ. These wave equations represent a struture, whih is already familiar to us, namely ψ 1 2 ψ = 4πf x, t t2 To solve this problem, as in eletrostatis, it is useful to first find the Green s funtion G x, t; x, t, defined as a solution of the following equation x 1 2 G x, t; x, t = 4πδ x x δ t t t 2 Note that G x, t; x, t is not unique and it has to be speified in a number of ways. Additionally, it is referred to as the propagator espeially in the field of quantum eletrodynamis. The solution to equation 55 reads ψ x, t = G x, t; x, t f x, t d 3 x dt. To hek that this is atually the solution, one an apply the operator x 1 2 and move it into the integral - two delta funtions will emerge by virtue 2 t 2 of 56, whih upon integration will turn f x, t into f x, t. In what follows we will need the Fourier transforms of all the elements of equation 56 δ x x δ t t = 1 2π 4 G x, t; x, t = d 3 k d 3 k 36 dω g dω e i k x x e iωt t, k, ω e i k x x iωt t.

36 Plugging these into the equation, we obtain g k, ω k 2 + ω2 = 4π 1 2 2π = 1 4 4π, 3 whih amounts to g k, ω = 1 4π 3 1 k2 ω2 2. From this one an find an integral expression for G x, t; x, t G x, t; x, t = 1 d 3 k 4π 3 dω ei k x x iωt t. k2 ω2 2 The omplex funtion inside the integral is singular at k 2 = ω2 2 and thus has two first order poles at ω = ± k. We have to find the proper way to treat this singularity. This is done by using the following physial reasoning. The Green funtion is a wave-type perturbation produed by a point soure sitting at x and emanating during an infinitesimal time at t = t. We an expet that this wave propagates with the speed of light as a spherial wave. Thus, we should require that a G = 0 in the whole spae for t < t b G is a diverging wave for t > t We shall see that the above only represents one of the possible Green s funtions, sine a different treatment of the poles produes different Green s funtions - an advaned or a retarded one: Retarded Green funtion states G = 0 if t < t Advaned Green funtion states G = 0 if t > t Notie that the differene of the two G adv G ret, alled the Pauli Green s funtion G P auli, satisfies the homogenous equation. Consider the retarded Green s funtion. For t > t, it should give a wave propagating from a point-like soure. Let us define τ = t t, R = x x and R = R. Then we have e iωt t e Iωτ, 37

37 sine τ > 0. Thus we need to require that Iω < 0 in order to have a deaying funtion at large ω, hene we have to integrate over the lower omplex plane. In other words, for t < t, the ontour over whih we integrate in the upper half of the omplex plane should give zero ontribution due to the aforementioned physial reasons. As a result, one ould infinitesimally shift the poles into the lower half plane when performing the analyti ontinuation. Aording to this presription, the Green s funtion is speified as follows G x, t; x, t = 1 4π 3 d 3 k e i kr iωτ dω k 2 1 ω + iε. 2 2 We an onveniently rewrite the previous statement, by making use of partial frations G x, t; x, t = 57 = 1 d 3 k dωe i kr [ ] 1 4π 3 2k k iε ω 1 e iωτ. k iε ω In the limit ε 0, using Cauhy s theorem, we find G x, t; x, t = 1 4π 3 = 2π 2 = 2 πr = 1 πr = 1 = 0 4πR 1 2πR d 3 ke i k R 2πi [ e ikτ e ikτ] 58 2k d 3 k ei k R k sin kτ dk sinkr sinkτ 59 k R d k sin sin k τ 60 dx e ix R e ix e R ixτ e ixτ 61 = 1 R δ τ R = 1 R δ τ R dx e ix τ R e ixτ+ R 1R δ τ + R Note that in the meantime we have used: partial frations 57, the Cauhy theorem in 57-58, swithed to spherial oordinates and integrated over the 38

38 angles59, substituted k = x 60, expanded the trigonometri funtions in terms of their omplex exponentials 61, and identified Fourier transforms of delta funtions 62. On the last step we have rejeted δ τ + R, beause for τ, R, > 0, the result will always be zero. Substituting bak our original variables, we get δ t + x x t G ret x, t; x, t =. x x The result an be understood as the signal propagating at the speed of light, whih was emitted at t and will travel for x x and will be observed at time t. Thus, this Green funtion reflets a natural ausal sequene of events. The time t is then expressed in terms of the retarded time t t = t + x x Substituting this solution and integrating over t, we obtain the retarded potentials δ t + x x t ϕ x, t = ρ x, t d 3 x dt x x ρ x, t x x = d 3 x + ϕ x x 0, 64 A x, t = 1 = 1 j δ t + x x x x x, t x x x x. t j x, t d 3 x dt d 3 x + A 0, 65 where ϕ 0 and A 0 are the solutions of the homogeneous d Alembert equations those orresponding to the free eletromagneti field. Note that for ϕ in the ase of time-independent ρ and j we have ϕ = ρ x x x d3 x. 39

39 This is just the eletrostati formula for the salar potential. Moreover, if the urrent j is time-independent, we obtain Ax = 1 j x x x d3 x. This potential defines the following magneti field [ ] H = rot xa 1 rot x j x 1 = + x x x x x j x Note the use above of the following identity rotϕ a = ϕ rot a + ϕ a. d 3 x. 66 The first term in 66 vanishes, beause url is taken with respet to oordinates x, while the urrent j depends on x. This leaves H = 1 R j x d 3 x = 1 [ ] j x, x x d 3 x. R 3 x x 3 This is the famous law of Biot-Savart, whih relates magneti fields to their soure urrents. Let us now show that G ret is Lorentz invariant. We write δ t + x x t G ret x, t; x, t = Θ t t. x x Here the extra term Θ t t ensures that G ret x, t; x, t = 0 for t < t, beause { 0, t < t Θ t t = 1, t t When we use δ f x = i δ x f x o. In the last formula the derivative is evaluated at the set of points x o, suh that f o = 0. Realising that for a wave propagating at the speed of light 40

40 ds 2 = 0 and using some algebrai trikery, we get G ret x, t; x, t = 2Θ t t δ x x t t 2 x x = 2Θ t t δ x x t t x x + t t = 2Θ t t δ x x 2 2 t t 2, where the argument of the delta funtion is the 4-interval between two events x, t and x, t, whih is a Lorentz invariant objet. From this we an onlude that the Green s funtion is invariant under proper orthohronial ones that maintain ausality Lorentz transformations. 2.5 Causality Priniple A quik word on intervals. A spaetime interval we have already defined as ds 2 = 2 dt 2 dx 2 i 67 We refer to them differently depending on the sign of ds 2 : time-like intervals if ds 2 > 0 spae-like intervals if ds 2 < 0 light-like intervals also alled null intervals if ds 2 = 0 Consider Figure 1.9 representing the light-one built over a point X. Signals in X an ome only from points X, whih are in the pastlight-one of X. We say X > X X is later than X. The influene of a urrent j in X on potential A at X is a signal from X to X. Thus, the ausality priniple is refleted in the fat that AX an depend on 4-urrents jx only for those X for whih X > X. Thus, δax δjx GX X = 0 68 for X < X or points X that are spae-like to X. priniple for the Green funtion is Hene, the ausality GX X = 0, 69 in terms of the onditions desribed above. The retarded Green s funtion is the only relativisti Green s funtion whih has this property. 41

41 light-like absolute future X past X spae-like time-like Figure 10: At every point in time every observer has his past light one, whih is a set of all events that ould have influened his presene, and a future light one, the set of events whih the observer an influene. The boundaries of the light ones also define the split between different kinds of spae-time intervals. On the light one itself the intervals are all light-like, time-like on the inside and spae-like on the outside. 2.6 Appliability of Classial Eletrodynamis We onlude this setion by pointing out the range of appliability of lassial eletrodynamis. The energy of the harge distribution in eletrodynamis is given by U = 1 dv ρxϕx. 2 Putting eletron at rest, one an assume that the entire energy of the eletron oinides with its eletromagneti energy eletri harge is assumed to be homogeneously distributed over a ball of the radius r e m 2 e2 r e, where m and e are the mass and the harge of eletron. Thus, we an define the lassial radius of eletron r e = e2 m m. 42