So far we have derived two electrostatic equations E = 0 (6.2) B = 0 (6.3) which are to be modified due to Faraday s observation,

Transcription

1 Chapter 6 Maxwell Equations 6.1 Maxwell equations So far we have derived two electrostatic equations and two magnetostatics equations E = ρ ɛ 0 (6.1) E = 0 (6.2) B = 0 (6.3) B = µ 0 J (6.4) which are to be modified due to Faraday s observation, E = ρ ɛ 0 (6.5) E + B = 0 (6.6) B = 0 (6.7) B = µ 0 J. (6.8) However, this is not a self consistent set of equation for time varyingconfigurations. Clearly, ( B) =µ 0 J =0 (6.9) and the continuity equation (5.4) allowonlysteadystatesconfigurationsof charges ρ =0 (6.10) 75

2 CHAPTER 6. MAXWELL EQUATIONS 76 which cannot be true in general. The problem was fixed by Maxwell who found a small modification to (6.8) whichmakesthenewsetof(maxwell)equationsself-consistent. By starting from the continuity equation (5.4) we get ρ + J = 0 (6.11) (ɛ 0 E) 1 ( B) (ɛ 0 E) µ 0 B 1 c 2 E = J (6.12) = J (6.13) = µ 0 J + C (6.14) where c = (µ 0 ɛ 0 ) 1/2 is the speed of light and C is an arbitrary timeindependent integration constant. If we set C =0,thenweobtain aself-consistentsetofmicroscopicmaxwellequations E = ρ ɛ 0 (6.15) E + B = 0 (6.16) B = 0 (6.17) B 1 E c 2 = µ 0 J. (6.18) In a media the macroscopic Maxwell equations are rewritten as D = ρ (6.19) E + B = 0 (6.20) B = 0 (6.21) H D = J. (6.22) where the (partial) fields H and D have the same definition as before E = D ɛ 0 P ɛ 0 (6.23) B = µ 0 H + µ 0 M. (6.24) Since the divergence equations (6.19) and (6.21) are not unchanged the boundary conditions are still give by (4.33) and(5.101), ˆn (D 2 D 1 ) = σ (6.25) ˆn (B 2 B 1 ) = 0. (6.26)

3 CHAPTER 6. MAXWELL EQUATIONS 77 In contrast the curl equations (6.20) and(6.22) are modified,but it turns out that the corresponding boundary (4.38) and(5.96) conditionsareunchanged, ˆn (E 2 E 1 ) = 0 (6.27) ˆn (H 2 H 1 ) = K. (6.28) This is due to the fact the integrals of finite D and B over infinitesimal area at the boundary vanish. 6.2 Gauge transformations The divergence equation (6.17) suggests that we can still introduce a vector potential, B A (6.29) and from (6.16) thefaraday slawisgivenby ( E + A ) =0 (6.30) which can be rewritten by defining a scalar potential E + A Φ (6.31) E = Φ A. (6.32) While (6.16)and(6.17)where used todefine the(vector andscalar)potentials the remaining two equations (6.15) and(6.18) canbeusedtodeterminethe dynamics of this potentials, i.e. 2 Φ+ ( A) = ρ (6.33) ɛ 0 2 A 1 ( 2 c 2 A A + 1 ) 2 c 2 Φ = µ 0 J. (6.34) Note that the observable quantities E and B defined by (6.29) and(6.32) are unchanged if we replace (or gauge transform) A new = A old + Λ (6.35) Φ new = Φ old Λ. (6.36)

4 CHAPTER 6. MAXWELL EQUATIONS 78 Then it is often convenient to choose the potentials such that A + 1 Φ=0 (6.37) c 2 This choice of gauge is known as Lorentz gauge (to be compared with Coulomb gauge, i.e A =0)whichcanbeusedtosimplify(6.33) and(6.34), 2 Φ 1 2 c 2 Φ 2 = ρ ɛ 0 (6.38) 2 A 1 2 c 2 A 2 = J. ɛ 0 (6.39) Lorentz gauge can always be constructed with desired gauge transformation (6.35) and(6.36), 0= A new + 1 c 2 Φ new = A old + 1 c 2 Φ old + 2 Λ 1 c Λ (6.40) where Λ is to be chosen to satisfy 2 Λ 1 c Λ= ( A old + 1 ) c 2 Φ old. (6.41) However, even in the Lorentz gauge there is still a remaining freedom to choose potentials (called restricted gauge freedom) with Λ satisfying 2 Λ 1 c 2 2 We have already discussed the Coulomb gauge, i.e. Λ=0. (6.42) 2 A =0 (6.43) in the context of magnetostatics and in electrodynamics the above choice is also known as transverse or radiation gauge. In this gauge (6.33) takesthe form of the Poisson equation 2 Φ= ρ ɛ 0 (6.44) with solution corresponding to the instantaneous Coulomb potential Φ(x,t)= 1 ρ(x,t) 4πɛ 0 x x d3 x (6.45)

5 CHAPTER 6. MAXWELL EQUATIONS 79 which is the reason why the gauge has its name. The other dynamical equation (6.34) incoulombgaugetakesthefollowingform 2 A 1 2 c 2 A = µ 0J c Φ. (6.46) 2 It is now convenient to split the current into longitudinal (or scalar J l = S) andtransverse(orvectorj t )partswhichcanbeaccomplishedbyfirst solving the Poisson equation and then by setting such that 2 S = J (6.47) J l = S. J t = J S. Then we can also split (6.46) intolongitudinal and transverse part J l = 0 (6.48) J t = 0. (6.49) µ 0 J l = 1 c Φ. (6.50) 2 2 A 1 2 c 2 A = µ 0J 2 t (6.51) which is the reason why the Coulomb gauge is also known as the transverse gauge. In a special case when there are no charges (i.e. free electromagnetism) equations (6.45) and(6.51) imply Φ = 0 (6.52) 2 A 1 2 c 2 A 2 = 0 (6.53) and the fields are determined form the vector potential E = A (6.54) B = A. (6.55) We note that due to symmetry of Maxwell equation one can define alternative (Lorentz and Coulomb) gauges where the magnetic scalar potential Φ M and electric vector potential A E are used instead to more conventional electric scalar potential Φ E and magnetic vector potential A M.

6 CHAPTER 6. MAXWELL EQUATIONS Green Functions The dynamics of potential usually involves the so-called wave equations (for example (6.38), (6.39),(6.51)) of the basic form 2 Ψ(x,t) 1 c 2 2 Ψ(x,t)= 4πf(x,t) (6.56) 2 where f(x,t) is a known function. The first step in solving this equation is to expand both functions into Fourier integral over frequencies (conjugate variable to time) Ψ(x,t) = 1 Ψ(x,ω) e iωt dω (6.57) 2π f(x,t) = 1 f(x,ω) e iωt dω. (6.58) 2π and insert the expansion into (6.56), ( 2 1c ) 2 1 Ψ(x,ω) e iωt dω = 2 2 2π ) ( 2 + ω2 Ψ(x,ω) e iωt dω = 4π c 2 1 2π f(x,ω) e iωt (6.59) dω f(x,ω) e iωt dω. (6.60) By equating each of the Fourier coefficients we get the inhomogeneous Helmholtz wave equation ( 2 + k 2) Ψ(x,ω)= 4πf(x,ω) (6.61) where k = ω/c is the wave number (Note that in the limit k = 0 the Helmholtz wave equation reduces to the Poisson equation). Equation (6.61) can be solved using the method of Green functions that obey ( 2 + k 2) G k (x, x )= 4πδ (x x ) (6.62) and appropriate boundary conditions. For example, if there are no boundaries then the solution mustonlyde- pend on r = x x and (6.62) canberewritteninsphericalcoordinates as 1 d 2 r dr (rg k)+k 2 G 2 k = 4πδ (r) (6.63) or for r 0 d 2 dr 2 (rg k)+k 2 (rg k )=0 (6.64)

7 CHAPTER 6. MAXWELL EQUATIONS 81 and for r 0 r d2 dr (rg k) 4πδ (r) (6.65) 2 since in this limit the second term in (6.63) ismuchsmallerthanthefirst term, 1 d 2 r dr (rg k) G k 2 r 2 k2 G k. (6.66) It follows that (6.64) hasasolutionintermsofoutgoing exp(ikr) and incoming exp( ikr) plane waves, i.e. rg k = Ae ikr + Be ikr (6.67) and (6.65) istheequationthatwealreadysolvedinelectrostatics,i.e. lim G k = 1 kr 0 r. (6.68) These two conditions determine the most general solution where G k = AG (+) k (r)+bg ( ) k (r) (6.69) G (±) k (r) = e±ikr (6.70) r A + B = 1. (6.71) To obtain a general solution for the Green functions in space-time we expand the functions in (6.62) usinginversefourier transformation, ) ( 2 + ω2 G c 2 k (x, x )e iω(t t) dω = 4π δ(x x )e iω(t t) (6.72) dω ( 2 1c ) 2 G 2 2 k (x, x )e iω(t t) dω = 4π δ(x x )e iω(t t) (6.73) dω ( 2 1c ) 2 G(x, x ; t, t ) = 4πδ(x x )δ(t t ) (6.74) 2 2 where G(x, x ; t, t ) 1 G k (x, x )e iω(t t) dω. (6.75) 2π Since we already solved for G(x, x ) we can write the two wave solutions of (6.74) as G (±) (x, x ; t, t )= 1 e ±ik x x t) dω. (6.76) 2π x x eiω(t

8 CHAPTER 6. MAXWELL EQUATIONS 82 In the case of nondispersive media (k = ω/c) theintegralof(6.76) yields G (±) (x, x ; t, t 1 1 [ ( ( ) = exp iω t t x ))] x (6.77) dω x x 2π c [ ( ( 1 = x x δ iω t t x ))] x (6.78) c These are the retarded (or causal) G (+) and advanced G ( ) Green functions. Then the solution of (6.61) intermsofgreenfunctionsisgivenby Ψ (±) (x,t)= G (±) (x, x ; t, t )f(x,t)d 3 x dt (6.79) but arbitrary solutions of the homogeneous wave equations 2 Ψ in (x,t) 1 2 c 2 Ψ in(x,t) 2 = 0 (6.80) 2 Ψ out (x,t) 1 2 c 2 Ψ out(x,t) 2 = 0. (6.81) can also be added to satisfy the incoming wave condition or the outgoing wave conditions, Ψ = Ψ in (x, t)+ψ (+) (x, t) (6.82) Ψ = Ψ out (x, t)+ψ ( ) (x, t). (6.83) In other words if either the initial wave configuration or the final wave configurations is known before any sources are turned on, then the combined solution is given by either (6.82) or(6.83) respectively.