Lecture 18 Relativity & Electromagnetism

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 − |x| = c t − |x | = s

The maximum possible speed is c where β → 1, γ → ∞.

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

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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 ∇.J + = 0 ∂ J µ = 0 ∂t µ This shows that charge conservation is Lorentz invariant! Lorentz gauge condition: 1 ∂V + ∇.A = 0 ∂ Aµ = 0 c ∂t µ

Poisson’s equations: 2 2 2 1 ∂ A 2 1 ∂ V ρ ∇ A − 2 2 = −µ0J ∇ 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)

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8 Lorentz transformations of E and B

The ﬁelds in terms of the potentials are: 1 ∂A E = − − ∇V B = ∇ × 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 ˆr direction transverse to v:

0 0 0 0 Ex = Ex |E | Ey = γEy Ez = γEz

The magnetic ﬁeld is in the φˆ 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 |B0| = |E0|/c

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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   µν −Ex 0 Bz −By F =    −E −B 0 B   y z x     −Ez By −Bx 0   

13 Maxwell’s Equations in terms of F µν

Source equations: ∂F µν = J µ ∂xν

M1: ∇.E = ρ/0 µ = 0, ν = (1, 2, 3)

M4: ∇ × 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: ∇.B = 0 (µ, ν, σ) = (1, 2, 3) M3: ∇ × 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 ∇.B = 0 ⇒ ∇ .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 γ∇.B + x − z + y = 0 c ∂t ∂y ∂z The second bracket is zero from M3: ∇ × E = −∂B/∂t

Hence ∇.B = 0 and ∇0.B0 = 0 as required for Lorentz invariance.

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