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Maxwell's Equations in Covariant Form

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The four Maxwell equations are two Lorentz-covariant statements about one antisymmetric tensor. This packaging is more than notation: it fixes how electric and magnetic fields mix between observers, separates evolution from constraints, and makes charge conservation a consistency condition rather than an extra postulate.

Required background. Special Relativity supplies the Minkowski metric, Lorentz covariance, four-vector contractions, and index raising and lowering; Calculus and Linear Algebra supplies partial derivatives, integration by parts, and vector identities. Helpful background. Ordinary Differential Equations supplies initial-value reasoning; Differential Forms and Integration supplies exterior differentiation and the local distinction between closed and exact forms.

Use rationalized Heaviside–Lorentz units and the inherited spacetime conventions. In a contractible patch, choose a potential

Aμ=(−Φ,A),Fμν=∂μAν−∂νAμ.A_\mu=(-\Phi,\mathbf A), \qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.

Then

E=−∇Φ−∂tA,B=∇×A,\mathbf E=-\boldsymbol\nabla\Phi-\partial_t\mathbf A, \qquad \mathbf B=\boldsymbol\nabla\times\mathbf A,

and therefore

F0i=−Ei,Fij=ϵijkBk.F_{0i}=-E_i, \qquad F_{ij}=\epsilon_{ijk}B^k.

All six independent components are visible in the matrix

(Fμν)=(0−Ex−Ey−EzEx0Bz−ByEy−Bz0BxEzBy−Bx0).(F_{\mu\nu})= \begin{pmatrix} 0&-E_x&-E_y&-E_z\\ E_x&0&B_z&-B_y\\ E_y&-B_z&0&B_x\\ E_z&B_y&-B_x&0 \end{pmatrix}.

The current is Jμ=(ρ,J)J^\mu=(\rho,\mathbf J), so the source coupling has the familiar component form

JμAμ=−ρΦ+J⋅A.J^\mu A_\mu=-\rho\Phi+\mathbf J\mathbin{\cdot}\mathbf A.

Writing F=dAF=dA is locally justified by the Poincaré lemma when dF=0dF=0. A single global potential exists only when the cohomology class [F][F] vanishes. In particular, a nonzero period ∫CF\int_C F over a closed two-cycle CC obstructs global exactness. Such a cycle need not bound a three-chain inside the physical domain, so this is compatible with dF=0dF=0. See Guillemin and Haine 2019, §2.6, printed pp. 64–65; §4.6, pp. 147–151; and §5.1, pp. 171–173, Open PDF for local exactness, Stokes’ theorem, and de Rham cohomology.

With the conventions above, take

S[A]=∫Ωd4x(−14FμνFμν+JμAμ).S[A]=\int_\Omega d^4x\left( -\frac14F_{\mu\nu}F^{\mu\nu}+J^\mu A_\mu \right).

Antisymmetry turns the field variation into

−12FμνδFμν=−Fμν∂μδAν.-\frac12F^{\mu\nu}\delta F_{\mu\nu} =-F^{\mu\nu}\partial_\mu\delta A_\nu.

After integration by parts,

δS=∫Ωd4x (∂μFμν+Jν)δAν−∫∂ΩdΣμ FμνδAν.\delta S =\int_\Omega d^4x\, (\partial_\mu F^{\mu\nu}+J^\nu)\delta A_\nu -\int_{\partial\Omega}d\Sigma_\mu\, F^{\mu\nu}\delta A_\nu.

Fixed boundary potential, or another boundary condition that removes the surface term, gives

∂μFμν=−Jν.\boxed{\partial_\mu F^{\mu\nu}=-J^\nu.}

This action calculation is the site-convention translation of Tong 2015, §5.5.3, printed pp. 125–126, PDF. Tong uses a different simultaneous choice of metric, tensor, and unit conventions, so individual signs should not be imported in isolation.

The definition F=dAF=dA also gives

∂[λFμν]=0,\boxed{\partial_{[\lambda}F_{\mu\nu]}=0,}

because commuting partial derivatives makes the cyclic sum vanish. The two boxed equations are the inhomogeneous Maxwell equation and the homogeneous Bianchi identity.

Recovering the four three-vector equations

Section titled “Recovering the four three-vector equations”

Raising indices gives F0i=EiF^{0i}=E^i and Fi0=−EiF^{i0}=-E^i. The tensor components then reproduce the familiar equations without guesswork:

Tensor componentCalculationThree-vector equation
ν=0\nu=0 in ∂μFμν=−Jν\partial_\mu F^{\mu\nu}=-J^\nu∂iFi0=−∇⋅E=−ρ\partial_iF^{i0}=-\boldsymbol\nabla\mathbin{\cdot}\mathbf E=-\rho∇⋅E=ρ\boldsymbol\nabla\mathbin{\cdot}\mathbf E=\rho
ν=i\nu=i in ∂μFμν=−Jν\partial_\mu F^{\mu\nu}=-J^\nu∂tEi−(∇×B)i=−Ji\partial_tE^i-(\boldsymbol\nabla\times\mathbf B)^i=-J^i∇×B−∂tE=J\boldsymbol\nabla\times\mathbf B-\partial_t\mathbf E=\mathbf J
Three spatial Bianchi indicesCyclic derivatives of FijF_{ij}∇⋅B=0\boldsymbol\nabla\mathbin{\cdot}\mathbf B=0
One temporal Bianchi indexCyclic derivatives of F0iF_{0i} and FijF_{ij}∇×E+∂tB=0\boldsymbol\nabla\times\mathbf E+\partial_t\mathbf B=0

The tensor construction and covariant component equations are developed in Tong 2015, §§5.3.2 and 5.4, printed pp. 106–108 and 113–115, PDF, translated consistently to the conventions used here.

Evolution, constraints, and charge conservation

Section titled “Evolution, constraints, and charge conservation”

The curl equations give the six first-order time derivatives of the three components of E\mathbf E and the three components of B\mathbf B. The two divergence equations contain no such time derivatives, so they constrain initial data. Their persistence follows from the evolution equations:

∂t(∇⋅B)=0,\partial_t(\boldsymbol\nabla\mathbin{\cdot}\mathbf B)=0,

and, using ∂tρ+∇⋅J=0\partial_t\rho+\boldsymbol\nabla\mathbin{\cdot}\mathbf J=0,

∂t(∇⋅E−ρ)=0.\partial_t(\boldsymbol\nabla\mathbin{\cdot}\mathbf E-\rho)=0.

Thus data satisfying both constraints continue to satisfy them in the exact continuum theory. A numerical scheme can still create constraint error, so propagation in the equations is not a substitute for monitoring it in a computation.

Charge conservation itself follows from antisymmetry:

∂νJν=−∂ν∂μFμν=0.\partial_\nu J^\nu =-\partial_\nu\partial_\mu F^{\mu\nu}=0.

The same requirement appears from gauge invariance of the source term. Under Aμ↦Aμ+∂μχA_\mu\mapsto A_\mu+\partial_\mu\chi,

ΔSsource=∫∂ΩdΣμ Jμχ−∫Ωd4x χ ∂μJμ.\Delta S_{\mathrm{source}} =\int_{\partial\Omega}d\Sigma_\mu\,J^\mu\chi -\int_\Omega d^4x\,\chi\,\partial_\mu J^\mu.

For compactly supported χ\chi, or boundary conditions that remove the flux term, a nonconserved prescribed current is incompatible with gauge invariance and with Maxwell’s equation.

For an observer with four-velocity uμu^\mu, define the electric field four-vector

E(u)μ=Fμνuν,E(u)μuμ=0.E_{(u)}^\mu=F^\mu{}_{\nu}u^\nu, \qquad E_{(u)}^\mu u_\mu=0.

Its three spatial components in the observer’s rest frame are the force per unit charge on a momentarily comoving test charge. Different observers split the same FμνF_{\mu\nu} into different electric and magnetic fields; they do not transform E\mathbf E while leaving B\mathbf B untouched.

The covariant Lorentz force is

dpμdτ=qFμνuν,\frac{dp^\mu}{d\tau}=qF^\mu{}_{\nu}u^\nu,

whose spatial part is dp/dt=q(E+v×B)d\mathbf p/dt=q(\mathbf E+\mathbf v\times\mathbf B). See Tong 2015, §5.4.1, printed pp. 115–116, PDF for the covariant force and its component recovery.

Changing one sign convention in isolation. The signs of A0A_0, F0iF_{0i}, the source action, the inhomogeneous equation, and the Lorentz force form one linked system. Test the full set against Gauss’s law and q(E+v×B)q(\mathbf E+\mathbf v\times\mathbf B).

Treating Gauss’s laws as optional boundary conditions. They restrict initial data everywhere on a time slice. The curl equations propagate valid constraints but do not repair invalid initial data.

Assuming a global potential without checking topology. The Bianchi identity implies F=dAF=dA locally. Global exactness is a stronger statement.

In vacuum, derive a wave equation for E\mathbf E from the curl equations, assuming the constraints hold.

Solution

Take the curl of Faraday’s law:

∇×(∇×E)+∂t(∇×B)=0.\boldsymbol\nabla\times(\boldsymbol\nabla\times\mathbf E) +\partial_t(\boldsymbol\nabla\times\mathbf B)=0.

Use ∇×B=∂tE\boldsymbol\nabla\times\mathbf B=\partial_t\mathbf E and

∇×(∇×E)=∇(∇⋅E)−∇2E.\boldsymbol\nabla\times(\boldsymbol\nabla\times\mathbf E) =\boldsymbol\nabla(\boldsymbol\nabla\mathbin{\cdot}\mathbf E)-\nabla^2\mathbf E.

Since ∇⋅E=0\boldsymbol\nabla\mathbin{\cdot}\mathbf E=0 in vacuum,

(∂t2−∇2)E=0.\boxed{(\partial_t^2-\nabla^2)\mathbf E=0.}

The same argument gives (∂t2−∇2)B=0(\partial_t^2-\nabla^2)\mathbf B=0. Omitting the divergence constraint would leave an unwanted longitudinal-gradient term.

Let J=0\mathbf J=0, ∂tA=0\partial_t\mathbf A=0, and E=−∇Φ\mathbf E=-\boldsymbol\nabla\Phi. Show that the covariant sourced equation gives Poisson’s equation with the expected sign.

Solution

The ν=0\nu=0 component is ∇⋅E=ρ\boldsymbol\nabla\mathbin{\cdot}\mathbf E=\rho. Substituting E=−∇Φ\mathbf E=-\boldsymbol\nabla\Phi gives

∇2Φ=−ρ.\boxed{\nabla^2\Phi=-\rho.}

For a positive point charge, the solution away from the origin decreases as 1/r1/r, and its outward electric field is positive. This checks the linked signs of A0A_0, F0iF_{0i}, and the source equation.

Potentials, Gauge Freedom, and Constraints next explains which parts of AμA_\mu are redundant and how gauge choices expose the physical radiative degrees of freedom.

  • Victor Guillemin and Peter J. Haine, Differential Forms, World Scientific, 2019, ISBN 978-981-3272-77-4, DOI 10.1142/11058, Open PDF.
  • David Tong, Electromagnetism (PDF), §§5.3.2, 5.4, 5.4.1, and 5.5.3, printed pp. 106–108, 113–116, and 125–126, University of Cambridge Mathematical Tripos lecture notes, Lent Term 2015.