Conservation Laws and Poynting

Transcription

1 Chpter 11 Conservtion Lws n Poynting Vector In electrosttics n mgnetosttics one ssocites n energy ensity to the presence of the fiels U = 1 2 E B2 = (electric n mgnetic energy)/volume (11.1) In the sttic sitution this is constnt. However for ynmic time epenent sitution we expect U = U( r, t) Let us look t the time vrition U t = E. E t + B. B t (11.2) n inserting in Mxwell s eqution U t = j. E + c( E. B B. E) (11.3) U t + j. E +.(c E B) = 0 (11.4) 78

2 Let us integrte over some volume contining chrge t 3 ru = S.(cE B) + 3 r j. E (11.5) The Lorentz force F = e E + e v /c B (11.6) F.v = e E.v + e v B.v = e v. E (r, t) (11.7) rte of the work one F. v = e v. E ( r, t) (11.8) 3 rj. E = 3 r e v (t)δ 3 (r r (t)) E(r, t). v = e v. E ( r, t) = F v (11.9) The sum is over ll the prticles in volume V t 3 ru = S.(c E B) + F v (11.10) t 3 ru is the time rte of Loss of energy (E.M.) in V n F v is the rte of work one on chrges in V. Conservtiuon of energy les to the interprettion S.(c E B) = rte of energy flowing out of V cross the surfce S (11.11) we re le to the following interprettion P = (c E B) = flux of energy = energy/(time re) flowingcross the surfce P (11.12) P x = energy/(time re) flowingcross the surfce x xis (11.13) P = Poynting Vector (11.14) 79

3 Now let the Volume becomes ll spce n the surfce S be the surfce t r. If our E n B fiels fll off fst enough, i.e. E 1 r, B 1, for r (11.15) 3/2+ɛ r3/2+ɛ Then the energy integrl converges, i.e., E 3 ru = totl e.m. energy (11.16) The surfce integrl then vnishes S = ˆnr 2 Ω, Ω = sinθθφ (11.17) S( E B) r 2 1 Ω( r 3/2+ɛ )2 (11.18) Also by Newton s Lw F v = 1 p m t. p = t 1 p 2 = 2 m t T (11.19) T=totl prticle kinetic energy Hence we hve (E + T ) = 0 (11.20) t which is conservtion of totl energy Thus the fiel hs energy which when e to prticle energy is conserve. Tht is the fiel cts just like nother mechnicl system. If the fiel is like mechnicl system it shouls lso crry momonetum like prticle oes so tht the totl momentum (prticle plus fiel) is conserve. To see if this is true, let us look t Newton s lw F = t F = t p (11.21) p = p; p = totl momentum (11.22) t 80

4 Now from the lorentz force t p = Hence F = = V = 3 r[ρ(r, t) E(r, t) + 1 c j(r, t) B(r, t)] V e [ E(r (t), t) + v c B (r (t), t) (11.23) 3 r[ e δ 3 (r r (t) E(r, t) + e v c B (r (t), t)] 3 r[(.e)e + ( B) B 1 c E t B] E ( E) 1 c E t B (11.24) t p = t 3 r 1 c (E B) + 3 r[ E ( E) + (.E)E B ( B)] (11.25) Let us look t the E pieces first Q = E ( E)+(.E)E = j E j E j +(E. )E+(.E)E = 1 2 E2 +(E. )E+(.E)E (11.26) In component form Q i = i ( 1 2 E2 ) + j j (E j E i ) = j j [ 1 2 δ ije 2 + E i E j ] (11.27) We cn o the sme thing with the B term R = B ( B) + (.B)B (11.28) B i = j j [ 1 2 δ ijb 2 + B i B j ] (11.29) Thus t p i + t 3 r 1 c (E B) i = 3 r j j T ji (11.30) 81

5 where T ji = T ij = 1 2 δ ij(e 2 + B 2 ) (E i E j + B i B j ) (11.31) The RHS is ivergence. Hence t ( p i + 3 r 1 c (E B) i) = S j T ji (11.32) S j If our fiels fll off s before, we cn exten the volume to ll spce n the surfce integrl vnishes t ( p i + 3 r 1 c (E B) i) = 0 (11.33) Thus we hve conservtion lw of monentum, i.e., the prticles momentum is not conserve but the sum of prticle + e.m. momentum is conserve. We cn therefore ssocite momentum to the em fiel P totl em = 3 r 1 (E B) (11.34) c n P totl em + P totl prticle = 0 (11.35) Further, for n rbitrry volume, we cn interpret 3 r 1 (E B) = e.m momentum in volume V (11.36) c Hence the right hn sie of Eq is the flux of the momentum out of V cross surfce S. T ji = flux of ith component of electromgnetic momentum cross surfce jth irection=(em momentum)/(time re) crossing surfce jth irection Since V is rbitrry we cn interpret G = 1 (E B) = e.m. momentum ensity (11.37) c 82

6 P = c 2 G (11.38) The em. fiels= poesses energy, momentum just like mechnicl system. Now it is time to look for the ngulr momentum of the e.m. fiel. For prticle pf ngulr momentum is: L = r p (11.39) n the ngulr momentum of set of prticles is L prticle = r p (11.40) If G = momentum ensity, we woul expect ( r G) i = ɛ ijk x j G k = ith component of e.m. momentum ensity (11.41) Returning to our energy eqution let us consier sitution with just em fiel n no prticles with the fiels flling off to zero t infinity suffiecient rpily to neglect ll surfce integrl. We efine R(t) ru(r, t) 3 r U(r, t)3 r (11.42) = center of energy coorinte of EM fiels energy. For this sitution E em = U 3 r = const (11.43) Differentiting E R em t = 3 r r t U = 3 r r.(ce B) = c 2 r. G (11.44) We cn integrte by prts x i.g = x i j G j = j (x i G j ) ( j x i )G j (11.45) 83

7 Hence x i (, G) = j (x i G i ) G j (11.46) Hence E R em t = c2 G 3 r c 2 3 r j ( rg j ) (11.47) Hence (E em /c 2 ) V = 3 rg = P em (11.48) In prticle ynmics we write p = m v (11.49) So we see tht the istribution of EM energy s if it hs n effective mss M em = E EM /c 2 (11.50) 84