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Electric and magnetic field transformations Consider inertial frames F’ and F. Suppose we know the electric and magnetic fields, E’(x’,t’) and B’(x’,t’), as observed in frame F’. Picture: Then what are the electric and magnetic fields, E(x,t) and B(x,t), as observed in frame F? Suppose F’ moves with velocity v relative to F. WLOG, take v to be in the x direction; v = v i. Fields are present, which are measured, by the forces that they produce, with respect to the two different frames of reference, F and F’. Electric and magnetic field transformations To figure out E(x,t) and B(x,t): (1) We already know the field tensor F’μν for the inertial frame F’. (2) We know how to transform the field tensor, Fμν = Λμρ Λνσ F’ρσ where Λμρ is the Lorentz transformation matrix. (3) Finally deconstruct Fμν into E(x,t) and B(x,t). (1) 0 E’x/c F’μν = −E’x/c 0 −E’y/c −B’z −E’z/c B’y (2) xμ = Λμρ x’ρ γ γv/c Λμρ = γv/c γ 0 0 0 0 ; E’y/c E’z/c B’z −B’y 0 B’x −B’x 0 Fμν = Λμρ Λνσ F’ρσ 0 0 0 0 1 0 0 1 (3) {Ex, Ey, Ez } = { c F01 , c F02 , c F03} {Bx, By, Bz } = { F23, F31, F12} μν = 01 μν = 12 F01 = Λ0ρ Λ1σ F’ρσ = Λ00 Λ11 F’01 + Λ01Λ10 F’10 Ex = γ2 E’x + γ2 (v2/c2) (-E’x ) Ex = E’x μν = 02 F02 = Λ0ρ Λ2σ F’ρσ = Λ00 Λ22 F’02 + Λ01 Λ22 F’12 Ey = γ E’y + c γ(v/c) B’z F12 = Λ1ρ Λ2σ F’ρσ = Λ10 Λ22 F’02 + Λ11 Λ22 F’12 Bz = γ (v/c) (Ey /c) + γ B’z Bz = γ (B’z + (v/c2) Ey ) μν = 31 similarly By = γ (B’y − (v/c2) Ez ) μν = 23 F23 = Λ2ρ Λ3σ F’ρσ = Λ22 Λ33 F’23 Bx = B’x Ey = γ (E’y + v B’z ) μν = 03 similarly Ez = γ (E’z − v B’y ) Tables of coordinate transformations and field transformations Suppose reference frame F’ moves with velocity v = v i with respect to frame F. Picture: Coordinate transformations (x μ ← x’ μ ) xμ = Λμρ x’ρ Λμρ = γ γv/c γv/c γ 0 0 0 0 0 0 0 0 1 0 0 1 t = γ (t’ + (v/c2) x’) x = γ (x’ + v t’) y = y’ z = z’ Field transformations (F μν← F’ μν) Ex = E’x Ey = γ (E’y + v B’z ) Ez = γ (E’z − v B’y ) Bx = B’x By = γ (B’y − (v/c2) Ez ) Bz = γ (B’z + (v/c2) Ey ) Example. Consider a line of charge that is at rest in reference frame F’, lying along the x’ axis; charge per unit length = λ. Use the field transformations to calculate the electric and magnetic fields in the frame F . NOTA BENE: the line of charge moves in the x direction relative to frame F. Picture: E’ = λ/(2πε0 r’) { (y’/r’) j + (z’/r’) k } On the other hand B’ = 0 because there is no current. The transformed fields (relative velocity = v i): Ex = E’x = 0 Ey = γ E’y Ez = γ E’z E = γλ/(2πε0 r’) { (y’/r’) j + (z’/r’) k } E = γλ/(2πε0 r) { (y/r) j + (z/r) k } ; this is the same as a line of charge with density γλ, larger than λ because of Lorentz contraction. Solution: First, what are the fields E’ and B’? By Gauss’s law, the electric field around a charged line points radially away from the line, with magnitude λ/(2πε0 r’) where r’ is the perpendicular distance from the line. Bx = B’x = 0 By = − γ (v/c2) Ez Bz = γ (v/c2) Ey B = γ(v/c2) λ/(2πε0 r’) { (−z’/r’) j + (y’/r’) k } B = μ0 I /(2πr) { (−z/r) j + (y/r) k } ; this is the same as a line of current with I = γ λ v, by Ampere’s law. Next lecture -calculate the electric and magnetic fields of a single charge that moves with constant velocity v.