Electromagnetism - Lecture 18 Relativity & Electromagnetism • Special Relativity • Current & Potential Four Vectors • Lorentz Transformations of E and B • Electromagnetic Field Tensor • Lorentz Invariance of Maxwell's Equations 1 Special Relativity in One Slide Space-time is a four-vector: xµ = [ct; x] Four-vectors have Lorentz transformations between two frames with uniform relative velocity v: 0 0 x = γ(x − βct) ct = γ(ct − βx) where β = v=c and γ = 1= 1 − β2 The factor γ leads to timepdilation and length contraction Products of four-vectors are Lorentz invariants: µ 2 2 2 2 02 0 2 2 x xµ = c t − jxj = c t − jx j = s The maximum possible speed is c where β ! 1, γ ! 1. Electromagnetism predicts waves that travel at c in a vacuum! The laws of Electromagnetism should be Lorentz invariant 2 Charge and Current Under a Lorentz transformation a static charge q at rest becomes a charge moving with velocity v. This is a current! A static charge density ρ becomes a current density J N.B. Charge is conserved by a Lorentz transformation The charge/current four-vector is: dxµ J µ = ρ = [cρ, J] dt The full Lorentz transformation is: 0 0 v J = γ(J − vρ) ρ = γ(ρ − J ) x x c2 x Note that the γ factor can be understood as a length contraction or time dilation affecting the charge and current densities 3 Notes: Diagrams: 4 Electrostatic & Vector Potentials A reminder that: 1 ρ µ0 J V = dτ A = dτ 4π0 ZV r 4π ZV r A static charge density ρ is a source of an electrostatic potential V A current density J is a source of a magnetic vector potential A Under a Lorentz transformation a V becomes an A: 0 v 0 A = γ(A − V ) V = γ(V − vA ) x x c2 x The potential four-vector is: V Aµ = ; A c 5 Relativistic Versions of Equations Continuity equation: @ρ r:J + = 0 @ J µ = 0 @t µ This shows that charge conservation is Lorentz invariant! Lorentz gauge condition: 1 @V + r:A = 0 @ Aµ = 0 c @t µ Poisson's equations: 2 2 2 1 @ A 2 1 @ V ρ r A − 2 2 = −µ0J r V − 2 2 = − c @t c @t 0 2 µ µ @µA = −µ0J 6 Electric and Magnetic Fields The Lorentz force on a moving charge is: F = q(E + v × B) A static point charge is a source of an E field A moving charge is a current source of a B field Whether a field is E or B depends on the observer's frame Going from the rest frame to a frame with velocity v: 0 1 B = v × E c2 Going from a moving frame to the rest frame: 0 E = −v × B This formula was already derived from Induction (Lecture 6) 7 Notes: Diagrams: 8 Lorentz transformations of E and B The fields in terms of the potentials are: 1 @A E = − − rV B = r × A c @t Lorentz transformation of potentials: 0 0 v V = γ(V − vA ) A = γ(A − V ) x x x c2 Using this transformation and the Lorentz gauge condition the transformations of the electric and magnetic fields are: 0 0 0 Ex = Ex Ey = γ(Ey − vBz) Ez = γ(Ez + vBy) 0 0 v 0 v B = B B = γ(B + E ) B = γ(B − E ) x x y y c2 z z z c2 y 9 Fields of Highly Relativistic Charge A charge at rest has B = 0 and a spherically symmetric E field A highly relativistic charge has β ! 1, γ 1 The electric field is in the ^r direction transverse to v: 0 0 0 0 Ex = Ex jE j Ey = γEy Ez = γEz The magnetic field is in the φ^ direction transverse to v: 0 0 v 0 v B = 0 B = γ E B = −γ E x y c2 z z c2 y The electric and magnetic fields are perpendicular to each other with an amplitude ratio jB0j = jE0j=c 10 Notes: Diagrams: 11 Notes: Diagrams: 12 Electromagnetic Field Tensor The electric and magnetic fields can be expressed as components of an electromagnetic field tensor: @A @A F µν = ν − µ @xµ @xν where A = [V=c; A] and x = [ct; x] 0 Ex Ey Ez 0 1 µν −Ex 0 Bz −By F = B C B −E −B 0 B C B y z x C B C B −Ez By −Bx 0 C @ A 13 Maxwell's Equations in terms of F µν Source equations: @F µν = J µ @xν M1: r:E = ρ/0 µ = 0, ν = (1; 2; 3) M4: r × B = µ0(J + 0@E=@t) µ = 1, ν = (2; 3; 0) (similarly for µ = 2; 3) No-source equations: @F µν @F σµ @F νσ + + = 0 @xσ @xν @xµ M2: r:B = 0 (µ, ν; σ) = (1; 2; 3) M3: r × E = −@B=@t (µ, ν; σ) = (0; 1; 2)(3; 0; 1)(2; 3; 0) 14 Lorentz Invariance of Maxwell's Equations As an example we consider the Lorentz transformation of M2: 0 0 r:B = 0 ) r :B = 0 @B @B β @E @B β @E x + γ y − z + z + y = 0 @x0 @y0 c @y0 @z0 c @z0 We note that: 0 0 0 0 @B @B β @B x = γ(x + βct ) ct = γ(ct + βx ) x = γ x + x @x0 @x c @t Substituting into the previous equation and simplifying: γβ @B @E @E γr:B + x − z + y = 0 c @t @y @z The second bracket is zero from M3: r × E = −@B=@t Hence r:B = 0 and r0:B0 = 0 as required for Lorentz invariance. 15 Notes: Diagrams: 16.