ELECTRODYNAMICS: PHYS 30441

Transcription

1 . Relativisti Eletromagnetism. Eletromagneti Field Tensor How do E and B fields transform under a LT? They annot be 4-vetors, but what are they? We again re-write the fields in terms of the salar and vetor potentials, A E and B= A = φ. The six equations are written in terms of = and A. t x Firstly, the E-field is given by: Ei A A A A = = = = x x i i φ = ; sine A,A and x i ( x, x ), x ( x, x i E A A i i = = i The B-field is obtained F i ) A A A A B = = = F Similarly, B A A A A = = = F and A A A A B = = = F Collating terms we have: E / E / E / A A E/ B B F = = A A = = F x E / B B E / B B E i and B i are the six elements of the antisymmetri 4x4 matrix F. This matrix is referred to as the eletromagneti field tensor. As it transforms a tensor of rank under a LT: A A F =ΛΛ F, with F = = A A Exerises:. Show that F is a rank Lorentz tensor.. Show that Maxwells nd (M) and rd (M) equations may be expressed as: λ λ λ F + F + F =, and Maxwells st (M) and 4th (M4) by: F = ρ (M).E = ε (M).B = (M) xe = B t (M4) xh = J +ε E t J Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester --

2 In Q we onsider F : A A F = A A = in frame O. Also we know that ontravariant vetors transform as: Firstly we know, A =Λ A and A =Λ A Hene: F A A A A x = =Λ Λ Also, as x =Λ x and we also found earlier x = ( Λ ) x x = Λ x. Thus it is lear that = Λ and = Λ F A A A A =ΛΛ ΛΛ =ΛΛ =ΛΛF x x Thus, F is a Lorentz rank tensor.. In Q. taking F λ + λ F + F λ = with =, =, λ = : B E E B F + F + F = + = + E t, N.b. =, and =,. t t t z with =, =, λ = : x = ( t,x,x,x ) = ( t,r ) B E E B and x = ( t, x, x, x) = ( t, r) F + F + F = + = + E = t, t with =, =, λ = : B E E B F + F + F = E t + = + = x x t x B Collating all three omponents gives Faraday s law or (M): E = t Now for =, =, λ = : B B B F + F + F = = This we reognize as B= or (M). y. Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester --

3 In the seond part of Q F = J stands for 4 equations one for eah value of. For example, for = the first equation reads: F F F F F = E E E = J + + = = ρ E E E ρ ρ + + = ρ = = or E ε ε -.ie. Gauss s law or (M). For = we obtain: F F F F F = E B B E + = + = = x z y B J J x t y z t x gives Ampere s law or (M4): E B= J+ t making a similar expansion for =, Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester --

4 . Lorentz Transform of E and B Fields: Consider the speifi LT orresponding to a boost along the x = x axis. In this ase F transforms as: F =Λ Λ F or in matrix form F = ΛFΛ Thus, evaluating these equations expliitly will give the requisite transformation of E and B fields: F γ γ E / E / E / γ γ E/ B B γ γ γ γ E / B B E / B B = γe γe E E γe γe B B γe γb γ E+γB B γ E +γ B E B B γ γ E γ E B γ E +B E γ B E γ B + E F = γ E B γ B E B γ E + B γ B + E B Also: F E / E / E / = E/ B B E / B B E / B B Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester -4-

5 Thus omparing terms: E = E B = B E =γ( E B ) B =γ B + E E =γ ( E+ B ) B =γb E E =γ E+ γ E x+γ B z B y ( ) ˆ ( ˆ ) ˆ ( ) ( v.e ) v vxb γ v ( γ) ( ) E =γ E+ v.e v+γvxb v ( γ) B B ( v.b ) γ =γ + v vxe v Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester -5-

6 Lienard-Wiehert Eletri Field From A Diret Lorentz Transform We investigate a harge q, initially at rest in the moving frame O. We seek the fields in frame O. Eletri field from a harge q in uniform motion along ˆx. At t = t = both the stationary frame O and the moving frame O overlap. Within O the eletri field follows from Coulombs law: q E = r and B=. 4πε r Also, at time t = from the perspetive of O the fields are given by: E = E B = B = E = E =γ( ) =γ + = =γ E E B B B E E E E = γe E =γ ( E + B ) B =γb E= Thus, Charge q situated at the origin of the moving frame. Initially, at t= both frames oinide. q E = E, γe, γ E = x, γy, γz, and using x= γ x-ut = γx, at t= and y=y, z=z t ( ) ( ) ( ) 4πε r q qγ = γ ( x,y,z) = r. 4πε r 4πε r Note, the eletri field lines point to the present position. Finally, let us write the position vetor in O in terms of variables in O: r = x + y + z =γ x + y + z =γ θ+ θ=γ θ+ θ r os r sin r [os ( )sin ] =γ θ r[ sin ]. γ Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester -6-

7 Collating all terms we obtain: qγr E = = 4 πε γ r sin θ 4 πε r sin θ ( ) q ( ) r ( ) / / This is of ourse no more than the Lienard-Wiehert formula for the E-field of a uniformly moving point harge! We note that the field lines emanate from the present position of the harge and not the retarded position. For a partile at rest the field is isotropi. However, for relativisti veloities along the diretion of motion (θ =, π) the field is redued by a fator of γ - relative to the isotropi ase. Transverse to the diretion of motion (θ = π/) the field is enhaned by a fator of γ. This ompression of the lines of fore in the transverse diretion an be viewed as a onsequene of the FitzGerald- Lorentz ontration. E-field of a harge at rest (γ = ) E-field of a harge in uniform motion (γ = ) Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester -7-

8 Minkowski Fore and Lorentz Fore Covariane How does this relate to the Lorentz fore? We seek a ovariant representation. E dt Reall: U = ( U, U) = ( γ, γ ), p =, pand =γ N.b. Here, E is the partile energy. The Lorentz Fore: = q ( E + v B ), N.b. Here, E is the eletri field. We will adhere to this representation. dt in terms of the momentum 4-vetor beomes the Minkowski fore: = qγ E+ B = q γ E+γ B = U E+ U B q ( ) ( ) ( ) The spatial part of is The time omponent is the rate of hange of energy: =qu.e If the fore and energy hange equations are Lorentz ovariant, the right hand side of the rate of hange of 4-momentum equation must also form the omponents of a 4-vetor They involve harge (learly invariant from frame to frame), 4-veloity and e.m. fields. Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester -8-

9 Let us propose the following ovariant representation: du = m = qf U Writing this out in matrix form, eah omponent is readily identified. E / E / E / γ E/ B B γv γ γ = q E / B B v E / B B v Thus, the zero omponent: E E E q = qγ v + v + v = γe.v And as p denotes the energy of the partile, this learly represents a statement that the rate of hange of energy (or power) is equal to the fore multiplied by partile veloity. We now turn our attention to the spatial omponents. The first omponent (x) is: E = qγ + Bγv Bγv ( ) = qγ E + v B B v Dividing both sides by γ, and with = dt/γ, we reognise this as the first omponent (x) of the Lorentz fore equation. Similarly, for the other omponents (as an exerise, prove the remaining omponents yourself!). Eletrodynamis PHYS 44, Relativisti Eletromagnetism Part, R.M. Jones, University of Manhester -9-