Skip to content

Special Relativity and Electromagnetism

7.1 Covariant Formulation

The laws of electromagnetism are inherently relativistic. In fact, it was the inconsistency of Maxwell’s equations with Galilean relativity that motivated Einstein’s 1905 theory.

Minkowski spacetime. Events are labelled by coordinates (ct,x,y,z)(ct, x, y, z) in a four-dimensional Spacetime. The spacetime interval between two events is:

ds2=c2dt2+dx2+dy2+dz2ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2

This interval is invariant under Lorentz transformations --- all inertial observers agree on its Value. We use the metric signature ημν=diag(1,+1,+1,+1)\eta_{\mu\nu} = \mathrm{diag}(-1, +1, +1, +1).

Lorentz transformations. For a boost with velocity vv along the xx-axis, define β=v/c\beta = v/c and γ=1/1β2\gamma = 1/\sqrt{1-\beta^2}:

Λ νμ=(γγβ00γβγ0000100001)\Lambda^\mu_{\ \nu} = \begin{pmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Coordinates transform as xμ=Λ νμxνx'^\mu = \Lambda^\mu_{\ \nu}\,x^\nu (Einstein summation convention Implied).

7.2 Four-Vectors

A four-vector Aμ=(A0,A1,A2,A3)A^\mu = (A^0, A^1, A^2, A^3) transforms as Aμ=Λ νμAνA'^\mu = \Lambda^\mu_{\ \nu}\,A^\nu Under Lorentz transformations. The inner product AμBμ=ημνAμBνA_\mu B^\mu = \eta_{\mu\nu}A^\mu B^\nu is a Lorentz scalar (invariant).

Key four-vectors in electromagnetism:

Position: xμ=(ct,x,y,z)x^\mu = (ct, x, y, z)

Four-velocity: Uμ=dxμdτ=γ(c,vx,vy,vz)U^\mu = \frac{dx^\mu}{d\tau} = \gamma(c, v_x, v_y, v_z) where τ\tau is proper time.

Four-momentum: pμ=mUμ=(E/c,px,py,pz)p^\mu = mU^\mu = (E/c, p_x, p_y, p_z)With E=γmc2E = \gamma mc^2.

Four-current density:

Jμ=(cρ,Jx,Jy,Jz)J^\mu = (c\rho, J_x, J_y, J_z)

The continuity equation J+ρ/t=0\nabla \cdot \mathbf{J} + \partial\rho/\partial t = 0 becomes the Manifestly covariant:

μJμ=0\partial_\mu J^\mu = 0

Four-potential:

Aμ=(V/c,Ax,Ay,Az)A^\mu = (V/c, A_x, A_y, A_z)

The Lorenz gauge condition A+1c2Vt=0\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial V}{\partial t} = 0 Becomes:

μAμ=0\partial_\mu A^\mu = 0

7.3 The Electromagnetic Field Tensor

The six components of E\mathbf{E} and B\mathbf{B} are unified in the antisymmetric field Tensor FμνF^{\mu\nu}Defined by:

Fμν=μAννAμF^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu

In matrix form:

Fμν=(0Ex/cEy/cEz/cEx/c0BzByEy/cBz0BxEz/cByBx0)F^{\mu\nu} = \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix}

The dual field tensor is:

F~μν=12εμνρσFρσ\tilde{F}^{\mu\nu} = \frac{1}{2}\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}

Where εμνρσ\varepsilon^{\mu\nu\rho\sigma} is the totally antisymmetric Levi-Civita symbol with ε0123=+1\varepsilon^{0123} = +1. In matrix form:

F~μν=(0BxByBzBx0Ez/cEy/cByEz/c0Ex/cBzEy/cEx/c0)\tilde{F}^{\mu\nu} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z/c & -E_y/c \\ B_y & -E_z/c & 0 & E_x/c \\ B_z & E_y/c & -E_x/c & 0 \end{pmatrix}

The dual tensor is obtained from FμνF^{\mu\nu} by the replacement E/cB\mathbf{E}/c \to -\mathbf{B}, BE/c\mathbf{B} \to \mathbf{E}/c.

Lorentz force. The four-force on a charge qq is:

Kμ=dpμdτ=qFμνUνK^\mu = \frac{dp^\mu}{d\tau} = qF^{\mu\nu}U_\nu

The spatial components reduce to F=q(E+v×B)\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) And the time component gives the power equation dE/dt=qEvdE/dt = q\mathbf{E} \cdot \mathbf{v}.

7.4 Invariance of Maxwell’s Equations

All four Maxwell equations are contained in two covariant equations:

Inhomogeneous equations (Gauss’s law + Ampere-Maxwell law):

μFμν=μ0Jν\boxed{\partial_\mu F^{\mu\nu} = \mu_0 J^\nu}

Homogeneous equations (Gauss’s law for magnetism + Faraday’s law):

μF~μν=0\boxed{\partial_\mu \tilde{F}^{\mu\nu} = 0}

Verification: $\nu = 0$ gives Gauss's law

For ν=0\nu = 0:

μFμ0=μ0J0=μ0cρ\partial_\mu F^{\mu 0} = \mu_0 J^0 = \mu_0 c\rho

Since Fμ0=(0,Ex/c,Ey/c,Ez/c)F^{\mu 0} = (0, -E_x/c, -E_y/c, -E_z/c):

0F00+1F10+2F20+3F30=0+x ⁣(Exc)+y ⁣(Eyc)+z ⁣(Ezc)\partial_0 F^{00} + \partial_1 F^{10} + \partial_2 F^{20} + \partial_3 F^{30} = 0 + \frac{\partial}{\partial x}\!\left(-\frac{E_x}{c}\right) + \frac{\partial}{\partial y}\!\left(-\frac{E_y}{c}\right) + \frac{\partial}{\partial z}\!\left(-\frac{E_z}{c}\right)

1cE=μ0cρ-\frac{1}{c}\nabla \cdot \mathbf{E} = \mu_0 c\rho

E=μ0c2ρ=ρε0\nabla \cdot \mathbf{E} = -\mu_0 c^2 \rho = \frac{\rho}{\varepsilon_0}

This is Gauss’s law, using c2=1/(μ0ε0)c^2 = 1/(\mu_0\varepsilon_0). \blacksquare

Field transformations. Under a Lorentz boost with velocity v=vx^\mathbf{v} = v\,\hat{\mathbf{x}} The fields transform as:

Ex=Ex,Ey=γ(EyvBz),Ez=γ(Ez+vBy)E'_x = E_x, \quad E'_y = \gamma(E_y - vB_z), \quad E'_z = \gamma(E_z + vB_y)

Bx=Bx,By=γ ⁣(By+vc2Ez),Bz=γ ⁣(Bzvc2Ey)B'_x = B_x, \quad B'_y = \gamma\!\left(B_y + \frac{v}{c^2}E_z\right), \quad B'_z = \gamma\!\left(B_z - \frac{v}{c^2}E_y\right)

Components parallel to the boost are unchanged; perpendicular components mix E\mathbf{E} and B\mathbf{B}.

Lorentz invariants. The following quantities are the same in all frames:

FμνFμν=2 ⁣(B2E2c2),FμνF~μν=4cEBF_{\mu\nu}F^{\mu\nu} = 2\!\left(B^2 - \frac{E^2}{c^2}\right), \quad F_{\mu\nu}\tilde{F}^{\mu\nu} = -\frac{4}{c}\,\mathbf{E} \cdot \mathbf{B}

These invariants classify electromagnetic fields:

  • If E2>c2B2E^2 \gt c^2 B^2 in some frame, there exists a frame where B=0\mathbf{B} = \mathbf{0} (purely electric).
  • If c2B2>E2c^2 B^2 \gt E^2There exists a frame where E=0\mathbf{E} = \mathbf{0} (purely magnetic).
  • If EB=0\mathbf{E} \cdot \mathbf{B} = 0 and E=cBE = cBThe field is a null field (electromagnetic wave).