Exercises in field theory

Transcription

1 Exercises in field theory Wolfgang Kastaun April 30, 2008

2 Faraday s law for a moving circuit Faradays law: S E d l = k d B d a dt S If St) is moving with constant velocity v, it can be written as St) E k v B) d B l = k St) d a

3 Faraday s law for a moving circuit We have d B d a = ) B + v dt St) St) ) B d a Applying the rule ) ) ) ) ) w 1 w 2 = w 2 w 1 w 1 w 2 + w 1 w 2 w 2 w 1 with w 1 = B, w 2 = v, and using B = 0, we obtain v ) ) B = B v

4 Faraday s law for a moving circuit d B d a = dt St) = St) St) B ) + v ) B d a B B v) ) d a B = St) d a = 1 E d k l St) St) B v) d l Thus, St) E k v B) d B l = k St) d a

5 The constant in Faraday s law Consider circuit moving in a constant magnetic field. Compute force F acting onto the charges inside. Comoving System: Faradays law F = qe = qk F v B Laboratory system: Lorentz-force F = qk L v B Thus, k L = k F Actually, k = 1 c

6 Differential form of Faraday s law 1 c S B d a = = S S E d l E ) d a Thus, Since S is arbitrary, 0 = S E + 1 c E + 1 c B B ) d a = 0

7 Maxwell displacement current Amperes law for stationary currents: H = 4π c J From charge continuity equation, ρ = J = c 4π ) H = 0 For nonstationary currents, using Coulomb s law, 0 = ρ + J = 1 4π D + J = J + 1 4π D )

8 Maxwell displacement current Replacing by H = 4π c H = 4π c J J + 1 4π D ) makes the equation compatible with charge conservation. It also turned out compatible with reality.

9 Gauge transformations Replacing the vector and scalar potentials like A = A + Λ, Φ = Φ 1 c Λ does not change the observable fields E, B, since B = A = ) A + Λ = A = B E = Φ 1 A = c Φ 1 ) Λ 1 ) A + Λ c c = Φ 1 A c = E A condition to choose only one or a subset) of the equivalent representations of a physical state is called gauge.

10 Lorentz gauge In the Lorentz gauge, we require that A + 1 c Φ = 0 This can always be achieved after a gauge transformation 0 =! A + 1 Φ = c ) A + Λ + 1 Φ 1 ) Λ c c 1 c t Λ Λ = A + 1 c Φ This inhomogenous wave equation in Λ can always be solved, e.g. using Green functions.

11 Longitudinal and transverse parts of vector field Every vector field can be split into longitudinal and transverse parts, i.e. w = w L + w T, with w T = 0, w L = 0 If w has compact support, we can write w T x) = 1 4π w x ) x x d 3 x w L x) = 1 4π w x ) x x d 3 x

12 Longitudinal and transverse parts of vector field Using ) w = ) w w, we have w T x) = 1 4π w x ) x x d 3 x = 1 4π w x ) x x d 3 x 1 w x ) 1 4π x x d 3 x = 1 4π w x ) 1 x x d 3 x + w x )δ x x ) d 3 x After partial integration of the first term we get w T x) = + 1 4π w x )) 1 x x d 3 x + w x) = w L x) + w x)

13 Coulomb gauge Is defined by A = 0 The Maxwell equation for the scalar potential then reduces to the Poisson equation Φ = 4πρ which is solved by Φ x) = ρ x ) x x d 3 x

14 Coulomb gauge Together with the continuity equation we get ρ = J and hence Φ x) = ρ x ) J x x d 3 x ) x = x x d 3 x Φ = 4π J L The Maxwell equation for A in Coulomb gauge becomes A 1 c A t = 4π c J + 1 c Φ = 4π c J T

15 Conservation Laws Many physical laws can be written as a conservation law w x, t) = f w, x, t) + sw, x, t) where f is the flux and s the source. It follows W = f d a + S where W = V S w d 3 x, S = V s d 3 x If the system is completely contained inside V and there are no sources, the quantity W is conserved, i.e. W = 0.

16 Poynting s theorem The Maxwell equations lead to a conservation law for the energy. The conserved density is the energy density of the field w = 1 ) E D + B H 8π The flux is given by the Poynting-vector S f = S c 4π E H The source is the mechanical work done by the charges s = J E

17 Poynting s theorem Linear momentum The Maxwell equations also lead to a conservation law for the linear momentum. Since the conserved quantity is a vector now, we write the conservation law in index notation t w a = b f ab + s a In this notation we implicitly sum over indices occuring twice in a term, in this case over b. The flux corresponding to a vector-valued conserved density w is a tensor f ab, since we need a flux vector for every component w a of w

18 Poynting s theorem Linear momentum The momentum density of the EM field is given by w a = 1 ) E H 4πc = 1 a c 2 S a Thus the momentum density of the EM field is proportional to the flux of it s energy density, i.e. the Poynting-vector. The flux of the momentum is given by the Maxwell stress tensor T ab f ab = T ab 1 E a E b + B a B b 1 ) 4π 2 δ ab E c E c + B c B c ) The source of momentum equals minus) the electromagnetic force acting on the charges s a = ρe + 1 ) c J B a

19 Poynting s theorem Linear momentum Force density) acting on the charges F = ρ E + 1 c J B Using the Maxwell equations ρ = 1 4π E, J = c 4π B 1 c E ) we get 4π F = E ) E + B 1 c E ) B

20 Poynting s theorem Linear momentum Use to get 4π F = E 4π F = E E ) E + B 1 c B = E B E ) B E B ) ) ) E + B B + 1 E c B 1 ) E B c

21 Poynting s theorem Linear momentum 4π F = E ) ) E + B B + 1 E c B 1 ) E B c Together with the Maxwell equations we obtain 4π F + 1 c 1 B c ) E B = E, B = 0 = E ) E + B E ) E ) B B ) B

22 Poynting s theorem Linear momentum To go on, we switch to index notation, where E b E b, )) E E a b E b a E) a ɛ abc b E c Where ɛ abc is the totally antisymmetric symbol. Using ɛ cab ɛ cde = δ ad δ be δ ae δ bd we can write E E )) a = ɛ abc E b ɛ cde d E e = δ ad δ be δ ae δ bd ) E b d E e = E b a E b E b b E a

23 Poynting s theorem Linear momentum Further ) E E E )) E a = E a b E b E b a E b + E b b E a = b E a E b 1 ) 2 E ce c δ ab The same holds with B instead of E, yielding ) E E E ) E + B ) B B )) B = 4π b T ab Therefore, we can finally write F a + 1 ) E B 4πc = bt ab a a

24 Poynting s theorem Linear momentum We have thus proven the conservation law for the linear momentum S a t c 2 = F a + b T ab or, in integral form t P a = Fa T + T ab n b da S with the total electromagnetic force acting on the charges Fa T = F a d 3 x and the total momentum of the field S a P a = c 2 d 3 x V V