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 ( c t , x , y , z ) (ct, x, y, z) ( c t , x , y , z ) in a four-dimensional Spacetime. The spacetime interval between two events is:
d s 2 = − c 2 d t 2 + d x 2 + d y 2 + d z 2 ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2 d s 2 = − c 2 d t 2 + d x 2 + d y 2 + d z 2
This interval is invariant under Lorentz transformations --- all inertial observers agree on its Value. We use the metric signature η μ ν = d i a g ( − 1 , + 1 , + 1 , + 1 ) \eta_{\mu\nu} = \mathrm{diag}(-1, +1, +1, +1) η μν = diag ( − 1 , + 1 , + 1 , + 1 ) .
Lorentz transformations. For a boost with velocity v v v along the x x x -axis, define β = v / c \beta = v/c β = v / c and γ = 1 / 1 − β 2 \gamma = 1/\sqrt{1-\beta^2} γ = 1/ 1 − β 2 :
Λ ν μ = ( γ − γ β 0 0 − γ β γ 0 0 0 0 1 0 0 0 0 1 ) \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} Λ ν μ = γ − γ β 0 0 − γ β γ 0 0 0 0 1 0 0 0 0 1
Coordinates transform as x ′ μ = Λ ν μ x ν x'^\mu = \Lambda^\mu_{\ \nu}\,x^\nu x ′ μ = Λ ν μ x ν (Einstein summation convention Implied).
7.2 Four-Vectors A four-vector A μ = ( A 0 , A 1 , A 2 , A 3 ) A^\mu = (A^0, A^1, A^2, A^3) A μ = ( A 0 , A 1 , A 2 , A 3 ) transforms as A ′ μ = Λ ν μ A ν A'^\mu = \Lambda^\mu_{\ \nu}\,A^\nu A ′ μ = Λ ν μ A ν Under Lorentz transformations. The inner product A μ B μ = η μ ν A μ B ν A_\mu B^\mu = \eta_{\mu\nu}A^\mu B^\nu A μ B μ = η μν A μ B ν is a Lorentz scalar (invariant).
Key four-vectors in electromagnetism:
Position: x μ = ( c t , x , y , z ) x^\mu = (ct, x, y, z) x μ = ( c t , x , y , z )
Four-velocity: U μ = d x μ d τ = γ ( c , v x , v y , v z ) U^\mu = \frac{dx^\mu}{d\tau} = \gamma(c, v_x, v_y, v_z) U μ = d τ d x μ = γ ( c , v x , v y , v z ) where τ \tau τ is proper time.
Four-momentum: p μ = m U μ = ( E / c , p x , p y , p z ) p^\mu = mU^\mu = (E/c, p_x, p_y, p_z) p μ = m U μ = ( E / c , p x , p y , p z ) With E = γ m c 2 E = \gamma mc^2 E = γ m c 2 .
Four-current density:
J μ = ( c ρ , J x , J y , J z ) J^\mu = (c\rho, J_x, J_y, J_z) J μ = ( c ρ , J x , J y , J z )
The continuity equation ∇ ⋅ J + ∂ ρ / ∂ t = 0 \nabla \cdot \mathbf{J} + \partial\rho/\partial t = 0 ∇ ⋅ J + ∂ ρ / ∂ t = 0 becomes the Manifestly covariant:
∂ μ J μ = 0 \partial_\mu J^\mu = 0 ∂ μ J μ = 0
Four-potential:
A μ = ( V / c , A x , A y , A z ) A^\mu = (V/c, A_x, A_y, A_z) A μ = ( V / c , A x , A y , A z )
The Lorenz gauge condition ∇ ⋅ A + 1 c 2 ∂ V ∂ t = 0 \nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial V}{\partial t} = 0 ∇ ⋅ A + c 2 1 ∂ t ∂ V = 0 Becomes:
∂ μ A μ = 0 \partial_\mu A^\mu = 0 ∂ μ A μ = 0
7.3 The Electromagnetic Field Tensor The six components of E \mathbf{E} E and B \mathbf{B} B are unified in the antisymmetric field Tensor F μ ν F^{\mu\nu} F μν Defined by:
F μ ν = ∂ μ A ν − ∂ ν A μ F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu F μν = ∂ μ A ν − ∂ ν A μ
In matrix form:
F μ ν = ( 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 ) 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} F μν = 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
The dual field tensor is:
F ~ μ ν = 1 2 ε μ ν ρ σ F ρ σ \tilde{F}^{\mu\nu} = \frac{1}{2}\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma} F ~ μν = 2 1 ε μν ρ σ F ρ σ
Where ε μ ν ρ σ \varepsilon^{\mu\nu\rho\sigma} ε μν ρ σ is the totally antisymmetric Levi-Civita symbol with ε 0123 = + 1 \varepsilon^{0123} = +1 ε 0123 = + 1 . In matrix form:
F ~ μ ν = ( 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 ) \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} F ~ μν = 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
The dual tensor is obtained from F μ ν F^{\mu\nu} F μν by the replacement E / c → − B \mathbf{E}/c \to -\mathbf{B} E / c → − B , B → E / c \mathbf{B} \to \mathbf{E}/c B → E / c .
Lorentz force. The four-force on a charge q q q is:
K μ = d p μ d τ = q F μ ν U ν K^\mu = \frac{dp^\mu}{d\tau} = qF^{\mu\nu}U_\nu K μ = d τ d p μ = q F μν U ν
The spatial components reduce to F = q ( E + v × B ) \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) F = q ( E + v × B ) And the time component gives the power equation d E / d t = q E ⋅ v dE/dt = q\mathbf{E} \cdot \mathbf{v} d E / d t = q E ⋅ 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):
∂ μ F μ ν = μ 0 J ν \boxed{\partial_\mu F^{\mu\nu} = \mu_0 J^\nu} ∂ μ F μν = μ 0 J ν
Homogeneous equations (Gauss’s law for magnetism + Faraday’s law):
∂ μ F ~ μ ν = 0 \boxed{\partial_\mu \tilde{F}^{\mu\nu} = 0} ∂ μ F ~ μν = 0
Verification: $\nu = 0$ gives Gauss's law For ν = 0 \nu = 0 ν = 0 :
∂ μ F μ 0 = μ 0 J 0 = μ 0 c ρ \partial_\mu F^{\mu 0} = \mu_0 J^0 = \mu_0 c\rho ∂ μ F μ 0 = μ 0 J 0 = μ 0 c ρ
Since F μ 0 = ( 0 , − E x / c , − E y / c , − E z / c ) F^{\mu 0} = (0, -E_x/c, -E_y/c, -E_z/c) F μ 0 = ( 0 , − E x / c , − E y / c , − E z / c ) :
∂ 0 F 00 + ∂ 1 F 10 + ∂ 2 F 20 + ∂ 3 F 30 = 0 + ∂ ∂ x ( − E x c ) + ∂ ∂ y ( − E y c ) + ∂ ∂ z ( − E z c ) \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) ∂ 0 F 00 + ∂ 1 F 10 + ∂ 2 F 20 + ∂ 3 F 30 = 0 + ∂ x ∂ ( − c E x ) + ∂ y ∂ ( − c E y ) + ∂ z ∂ ( − c E z )
− 1 c ∇ ⋅ E = μ 0 c ρ -\frac{1}{c}\nabla \cdot \mathbf{E} = \mu_0 c\rho − c 1 ∇ ⋅ E = μ 0 c ρ
∇ ⋅ E = − μ 0 c 2 ρ = ρ ε 0 \nabla \cdot \mathbf{E} = -\mu_0 c^2 \rho = \frac{\rho}{\varepsilon_0} ∇ ⋅ E = − μ 0 c 2 ρ = ε 0 ρ
This is Gauss’s law, using c 2 = 1 / ( μ 0 ε 0 ) c^2 = 1/(\mu_0\varepsilon_0) c 2 = 1/ ( μ 0 ε 0 ) . ■ \blacksquare ■
Field transformations. Under a Lorentz boost with velocity v = v x ^ \mathbf{v} = v\,\hat{\mathbf{x}} v = v x ^ The fields transform as:
E x ′ = E x , E y ′ = γ ( E y − v B z ) , E z ′ = γ ( E z + v B y ) E'_x = E_x, \quad E'_y = \gamma(E_y - vB_z), \quad E'_z = \gamma(E_z + vB_y) E x ′ = E x , E y ′ = γ ( E y − v B z ) , E z ′ = γ ( E z + v B y )
B x ′ = B x , B y ′ = γ ( B y + v c 2 E z ) , B z ′ = γ ( B z − v c 2 E y ) 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) B x ′ = B x , B y ′ = γ ( B y + c 2 v E z ) , B z ′ = γ ( B z − c 2 v E y )
Components parallel to the boost are unchanged; perpendicular components mix E \mathbf{E} E and B \mathbf{B} B .
Lorentz invariants. The following quantities are the same in all frames:
F μ ν F μ ν = 2 ( B 2 − E 2 c 2 ) , F μ ν F ~ μ ν = − 4 c E ⋅ B F_{\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} F μν F μν = 2 ( B 2 − c 2 E 2 ) , F μν F ~ μν = − c 4 E ⋅ B
These invariants classify electromagnetic fields:
If E 2 > c 2 B 2 E^2 \gt c^2 B^2 E 2 > c 2 B 2 in some frame, there exists a frame where B = 0 \mathbf{B} = \mathbf{0} B = 0 (purely electric). If c 2 B 2 > E 2 c^2 B^2 \gt E^2 c 2 B 2 > E 2 There exists a frame where E = 0 \mathbf{E} = \mathbf{0} E = 0 (purely magnetic). If E ⋅ B = 0 \mathbf{E} \cdot \mathbf{B} = 0 E ⋅ B = 0 and E = c B E = cB E = c B The field is a null field (electromagnetic wave).