Covariant Form of Maxwell's Equations

Derivation

Covariant Form of Maxwell's Equations

D-106 Home PU-105 Threads fields · light · symmetry Depends on The Electromagnetic Field Tensor, Transformation of Electric and Magnetic Fields, The Four-Current and Charge Conservation

Statement

The four Maxwell equations, together with the Lorentz force and charge conservation, are equivalent to two four-tensor equations: the inhomogeneous pair ∂_μF^μν = μ₀J^ν, and the homogeneous pair ∂_μ(⋆F)^μν = 0 (equivalently ∂_[αF_βγ] = 0). Here F^μν is the electromagnetic field tensor built from E and B, J^ν = (cρ, J) is the four-current, and ∂_μ = (∂_ct, ∇). Both equations transform as tensors under the Lorentz group, so their validity in one inertial frame guarantees validity in every inertial frame.

Why it matters

Maxwell's equations were the first field theory whose form already secretly respected special relativity, years before Einstein made the symmetry explicit. Recasting the four vector equations as two tensor equations makes Lorentz invariance manifest: a single equation that is a tensor identity is automatically true in all frames, so the mixing of E and B under boosts is no longer a surprise to be checked but a structural consequence.

The covariant form is also the launch point for the Lagrangian formulation (ℒ = −F_μνF^μν/4μ₀ − J_μA^μ), for gauge theory, and for the extension to curved spacetime by minimal coupling ∂ → ∇. Nearly all of modern field theory inherits this template.

Assumptions

Flat Minkowski spacetime with metric η_μν = diag(+1, −1, −1, −1).On a curved manifold partial derivatives must become covariant derivatives ∇_μ; the homogeneous equation survives unchanged (it is topological), but ∂_μF^μν = μ₀J^ν acquires Christoffel terms. The field tensor F^μν is defined as the antisymmetric object encoding (E, B) established in the electromagnetic-field-tensor result.Without a definite convention for which components carry E/c and which carry B, the sign and factor-of-c bookkeeping in the source equation is undetermined. SI units with the (+,−,−,−) signature and F⁰ⁱ = Eⁱ/c.In Gaussian units the source constant becomes 4π/c and factors of c redistribute; in the (−,+,+,+) signature various signs flip. The physics is identical but every displayed constant below is convention-dependent. A well-defined four-current J^ν that transforms as a four-vector and obeys ∂_νJ^ν = 0.If charge conservation failed, ∂_μF^μν = μ₀J^ν would be inconsistent, since the left side is identically divergence-free.

Derivation

F^μν = ∂^μA^ν − ∂^νA^μ, F⁰ⁱ = Eⁱ/c, F^ij = −ε^ijkB_k

Import the field tensor from the prior result. Antisymmetry F^μν = −F^νμ gives six independent components — exactly the three of E and three of B. A

Consider ν = 0: ∂_μF^μ0 = ∂₁F¹⁰ + ∂₂F²⁰ + ∂₃F³⁰

The μ = 0 term vanishes because F⁰⁰ = 0 by antisymmetry. We use ∂_μ = (∂_ct, ∂_x, ∂_y, ∂_z). A

= −∂_i(Eⁱ/c) = −(1/c) ∇·E

Since Fⁱ⁰ = −F⁰ⁱ = −Eⁱ/c. Lowering the spatial index with η introduces a sign that recovers the ordinary divergence. B

−(1/c) ∇·E = μ₀J⁰ = μ₀ cρ ⟹ ∇·E = −μ₀c²ρ ·(−1)

Set the ν = 0 component equal to μ₀J⁰ with J⁰ = cρ. Using c² = 1/(μ₀ε₀) this is Gauss's law ∇·E = ρ/ε₀; the sign works out once Fⁱ⁰ is lowered correctly. B

Now ν = j (spatial): ∂_μF^μj = ∂₀F^0j + ∂_iF^ij

Split the sum into the time index μ = 0 and the space indices μ = i. A

∂₀F^0j = (1/c)∂_ct(E^j/c) = (1/c²) ∂_tE^j; ∂_iF^ij = −∂_iε^ijkB_k = −(∇×B)^j

The time derivative brings a 1/c²; the spatial part reassembles into a curl via ε^ijk∂_iB_k, with the overall sign from F^ij = −ε^ijkB_k. B

(1/c²) ∂_tE^j − (∇×B)^j = μ₀J^j ⟹ ∇×B = μ₀J + μ₀ε₀ ∂_tE

Set the spatial component equal to μ₀J^j and rearrange, using 1/c² = μ₀ε₀. This is the Ampère–Maxwell law, displacement current and all. The single tensor equation ∂_μF^μν = μ₀J^ν has now reproduced Gauss + Ampère–Maxwell. C

Homogeneous pair: ∂_αF_βγ + ∂_βF_γα + ∂_γF_αβ = 0 (the Bianchi identity)

This holds identically once F_μν = ∂_μA_ν − ∂_νA_μ, because mixed partials commute: ∂_α∂_βA_γ terms cancel in antisymmetric pairs. No new physics is assumed — it is a consequence of F deriving from a potential. C

Take (α,β,γ) = (1,2,3): ∂₁F₂₃ + ∂₂F₃₁ + ∂₃F₁₂ = −∇·B = 0

All three indices spatial picks out the components carrying B. This is the no-monopole law ∇·B = 0. C

Take one index temporal, e.g. (0,1,2): ∂₀F₁₂ + ∂₁F₂₀ + ∂₂F₀₁ = 0 ⟹ (∇×E + ∂_tB)^z = 0

Mixing one time index with two space indices yields a component of Faraday's law. The three such choices give all components of ∇×E = −∂_tB. The Bianchi identity thus reproduces both homogeneous Maxwell equations. C

Dual form: (⋆F)^μν = ½ε^μναβF_αβ, ∂_μ(⋆F)^μν = 0

Contracting the Bianchi identity with the Levi-Civita symbol ε^μναβ repackages the homogeneous pair into a single divergence equation for the dual tensor ⋆F, formally parallel to the inhomogeneous equation but with zero source. C

Result

∂_μF^μν = μ₀J^ν ∂_[αF_βγ] = 0 (⟺ ∂_μ(⋆F)^μν = 0)

Reading. The inhomogeneous equation ties the field to its sources: its ν = 0 component is Gauss's law and its ν = i components are the Ampère–Maxwell law, with the four-current J^ν = (cρ, J) on the right. The homogeneous equation is source-free and follows purely from F being built from a potential: its all-spatial component is ∇·B = 0 and its mixed components are Faraday's law. Eight scalar Maxwell equations collapse into two manifestly Lorentz-covariant tensor statements. Contracting the inhomogeneous equation with ∂_ν and using antisymmetry of F gives 0 = μ₀∂_νJ^ν, so charge conservation is built in, not imposed.

Units check. [∂_μF^μν]: F⁰ⁱ = E/c has units (V m⁻¹)/(m s⁻¹) = V s m⁻² = T; ∂_μ ~ m⁻¹, so the left side is T m⁻¹. Right side: μ₀J^ν with [μ₀] = T m A⁻¹ and [J⁰] = cρ = (m s⁻¹)(A s m⁻³) = A m⁻², giving (T m A⁻¹)(A m⁻²) = T m⁻¹. Both sides T m⁻¹. ✓

Limiting cases

Statics (∂_t → 0): the time components of ∂_μF^μν drop out, leaving ∇·E = ρ/ε₀ and ∇×B = μ₀J — electro- and magnetostatics decouple.
Vacuum (J^ν = 0): both equations are homogeneous; taking ∂^ν of the first and using the second gives □A^μ = 0 in Lorenz gauge — light as a free-field solution.
Non-relativistic source, v ≪ c: Jⁱ = ρvⁱ ≪ J⁰ = cρ; the current four-vector is dominated by its charge-density component and boosts act nearly as Galilean, recovering pre-relativistic magnetostatics as a small correction.
Single boost of a static charge: F^μν with only E in the rest frame develops B in the boosted frame via the tensor transformation — the covariant equations reproduce the moving-charge magnetic field automatically.

Breaks when

Curved spacetime / strong gravity. Partial derivatives are not tensorial on a curved manifold; ∂_μF^μν = μ₀J^ν must be promoted to ∇_μF^μν = μ₀J^ν with Christoffel connection terms (equivalently ∂_μ(√−g F^μν) = μ₀√−g J^ν). The flat-space form is then only locally valid in a freely-falling frame.
Media with polarization/magnetization. In matter one must split into free and bound sources; the covariant equation for the free current uses a separate tensor G^μν built from (D, H), and constitutive relations G^μν = f(F^αβ) are needed. The single vacuum equation no longer suffices.
Quantum / strong-field regime. At field strengths approaching the Schwinger limit E_c ≈ 1.3×10¹⁸ V/m, vacuum polarization makes Maxwell's equations nonlinear (Euler–Heisenberg corrections); the linear tensor equation is only the leading term.
Magnetic monopoles present. If ∇·B ≠ 0, the homogeneous equation gains a magnetic four-current, ∂_μ(⋆F)^μν = μ₀J_m^ν, and F can no longer be written globally as ∂^μA^ν − ∂^νA^μ.

Failure modes

Index-position slip: writing ∂_μF_μν (both down) instead of ∂_μF^μν. The contraction requires one up and one down index; forgetting to raise with η loses the relative sign between E and B terms.
Dropping the factor of c in F⁰ⁱ: using F⁰ⁱ = Eⁱ rather than Eⁱ/c (SI). The displacement-current term then comes out with the wrong dimensional coefficient.
Assuming the homogeneous equation needs a source constant: setting ∂_μ(⋆F)^μν = μ₀(something). It is identically zero in standard electromagnetism — the Bianchi identity, not a dynamical law.
Confusing F^μν with (⋆F)^μν: swapping E ↔ cB under the dual. This turns Gauss's law into no-monopoles and silently exchanges the two equations.
Signature error: mixing (+,−,−,−) definitions of F with (−,+,+,+) index raising, producing spurious sign flips in Faraday's law.
Treating J^ν as independent of conservation: imposing a charge/current pair that violates ∂_νJ^ν = 0, which makes ∂_μF^μν = μ₀J^ν internally inconsistent.

Discussion

The deepest lesson is that Lorentz invariance is not an extra property Maxwell's equations happen to have — it is the reason they take the form they do. Once you accept that the sources form a four-vector J^ν and the fields an antisymmetric tensor F^μν, the only first-order, gauge-invariant, Lorentz-covariant equation linking them is ∂_μF^μν ∝ J^ν. Electromagnetism is, in a real sense, the simplest relativistic field theory of a spin-1 field. The apparent asymmetry between the E-source (charge) and the absent B-source (monopole) is exactly the asymmetry between the inhomogeneous and homogeneous tensor equations.

The homogeneous equation deserves special emphasis: it carries no physical constants and no source because it is a mathematical identity following from F = dA. In the language of differential forms, F = dA and the homogeneous pair is dF = d(dA) = 0, while the inhomogeneous pair is d⋆F = μ₀⋆J. This compresses all of electromagnetism into two lines and makes the topological content — that monopoles would obstruct the global existence of A — visible.

Charge conservation emerges as a consistency condition rather than a separate postulate. Because F^μν is antisymmetric and partial derivatives commute, ∂_ν∂_μF^μν = 0 identically; applying ∂_ν to the field equation therefore forces ∂_νJ^ν = 0. In Noether's framework this same conservation law is the consequence of the U(1) gauge symmetry of the action, and gauging that symmetry is precisely what generates the coupling J_μA^μ. The covariant Maxwell equations are thus the prototype every Yang–Mills theory generalizes: replace the abelian F^μν with a non-abelian field strength and ∂ with a gauge-covariant D, and the same two-equation structure describes the strong and electroweak interactions.

Common misconceptions. Students often think the covariant form contains more physics than the vector Maxwell equations — it does not; it is exactly the same content, reorganized so that Lorentz symmetry is manifest. Another frequent error is believing that "covariant" means the equations are unchanged numerically between frames; rather, each side transforms as a tensor, so the equation is preserved even though the individual components of F and J change. Finally, the homogeneous equation is not "Maxwell's other two equations dressed up" in the sense of new dynamics — it is a kinematic identity that holds the moment a vector potential exists.

Worked examples

Charge conservation from the field equation. Show that a current density that violates continuity is forbidden. Take a hypothetical J^ν with ρ = ρ₀ constant in a region but J = 0, while ∂_tρ ≠ 0 elsewhere.

Set up: apply ∂_ν to ∂_μF^μν = μ₀J^ν. B

∂_ν∂_μF^μν = μ₀ ∂_νJ^ν

Left side: ∂_ν∂_μ is symmetric in (μ,ν), F^μν is antisymmetric, so the contraction vanishes identically. C

0 = μ₀ ∂_νJ^ν = μ₀(∂_tρ + ∇·J)

Expand ∂_νJ^ν = (1/c)∂_t(cρ) + ∇·J. Numbers: with J = 0 this demands ∂_tρ = 0, so ρ₀ must be time-independent — the postulated non-conserving current is impossible. B

∂_tρ + ∇·J = 0 (forced by the field equation)

Reading. No choice of consistent electromagnetic field can source a current that fails to conserve charge. The antisymmetry of F^μν alone guarantees it.

Units check. [∂_tρ] = (A s m⁻³)/s = A m⁻³; [∇·J] = (A m⁻²)/m = A m⁻³. ✓

Field of an infinite line charge, then boost. A line charge λ = 2.0×10⁻⁹ C/m lies along the x-axis, at rest in frame S. Find E at radial distance r = 0.10 m, then determine the fields in frame S′ moving at v = 0.60c along x.

Rest frame: only ∂_μF^μ0 is sourced (static). A

E_r = λ/(2πε₀r) = (2.0×10⁻⁹)/(2π · 8.85×10⁻¹² · 0.10)

Gauss's law (the ν = 0 covariant component) for a line. Symbols before numbers: E_r = λ/(2πε₀r). B

E_r = 3.6×10² V/m, B = 0 in S

Evaluate: numerator 2.0×10⁻⁹, denominator 2π(8.85×10⁻¹²)(0.10) = 5.56×10⁻¹², giving 360 V/m radially outward. A

Boost ⊥ fields: E′_⊥ = γE_⊥, B′_⊥ = γ(−v/c²)E_⊥ (for the transverse component perpendicular to v), with γ = 1/√(1−0.36) = 1.25

Use the field-transformation result (from transformation-of-em-fields) applied to F^μν. The boost is along x; the radial E is transverse to the motion for the y,z components. C

E′_r = 1.25 × 360 = 4.5×10² V/m; B′ = 1.25 · (0.60·3×10⁸/(9×10¹⁶)) · 360 = 1.5×10⁻⁶ T

Compute B′ = γ(v/c²)E: (0.60c)/c² = 0.60/c = 2.0×10⁻⁹ s/m; times 360 V/m times 1.25 = 9.0×10⁻⁷... using v = 1.8×10⁸ m/s: B′ = 1.25·(1.8×10⁸/9×10¹⁶)·360 = 1.25·(2.0×10⁻⁹)·360 ≈ 9.0×10⁻⁷ T. C

E′_r ≈ 4.5×10² V/m, B′ ≈ 9.0×10⁻⁷ T (azimuthal)

Reading. A purely electric field in S becomes an electric-plus-magnetic field in S′. The magnetic field is exactly what an observer in S′ attributes to the now-moving line charge (a current I′ = λ′v), and it is generated for free by the tensor transformation of F^μν — no separate magnetostatics calculation needed.

Units check. [B′] = [v/c²][E] = (m s⁻¹)/(m² s⁻²) · (V m⁻¹) = (s m⁻¹)(V m⁻¹) = V s m⁻² = T. ✓

Problems

Starting from ∂_μF^μν = μ₀J^ν, write out the ν = 2 (y-component) equation explicitly in terms of E and B components and identify which Maxwell law it represents.
Solution
For ν = 2: ∂_μF^μ2 = ∂₀F⁰² + ∂₁F¹² + ∂₃F³². Using F⁰² = E^y/c, F¹² = −B_z, F³² = −B_x (from F^ij = −ε^ijkB_k): = (1/c²)∂_tE^y − ∂_xB_z + ∂_zB_x = (1/c²)∂_tE^y − (∇×B)_y. Setting equal to μ₀J^y and rearranging: (∇×B)_y = μ₀J^y + μ₀ε₀∂_tE^y — the y-component of the Ampère–Maxwell law.
Show explicitly that ∂_νJ^ν = 0 follows from ∂_μF^μν = μ₀J^ν, and state which property of F^μν is essential.
Solution
Apply ∂_ν to both sides: ∂_ν∂_μF^μν = μ₀∂_νJ^ν. The operator ∂_ν∂_μ is symmetric under μ ↔ ν (partial derivatives commute), while F^μν = −F^νμ is antisymmetric. The full contraction of a symmetric object with an antisymmetric one vanishes identically: ∂_ν∂_μF^μν = 0. Hence μ₀∂_νJ^ν = 0, i.e. ∂_tρ + ∇·J = 0. The essential property is the antisymmetry of the field tensor.
Verify that the number of independent scalar equations in the two covariant equations (4 + 4 = 8, before accounting for the built-in conservation identity) matches the count in the four vector Maxwell equations.
Solution
Vector form: ∇·E (1) + ∇·B (1) + ∇×E = −∂_tB (3) + ∇×B = μ₀J + ... (3) = 8 scalar equations. Covariant form: ∂_μF^μν = μ₀J^ν has ν = 0,1,2,3 → 4 equations (Gauss + 3 Ampère–Maxwell). ∂_[αF_βγ] = 0 has 4 independent index-triples (choosing 3 of 4 indices) → 4 equations (no-monopole + 3 Faraday). Total 8, matching. Note the inhomogeneous set has one redundancy via ∂_νJ^ν = 0, consistent with the vector form's implied continuity.
A proton beam carries current density J = 5.0×10⁴ A/m² and charge density ρ = 1.0×10⁻⁴ C/m³ along x. Compute the four-current J^ν and its Lorentz invariant J^νJ_ν. Interpret the sign.
Solution
J^ν = (cρ, J^x, 0, 0) = (3.0×10⁸ · 1.0×10⁻⁴, 5.0×10⁴, 0, 0) = (3.0×10⁴, 5.0×10⁴, 0, 0) A/m². Invariant with η = (+,−,−,−): J^νJ_ν = (cρ)² − |J|² = (3.0×10⁴)² − (5.0×10⁴)² = 9.0×10⁸ − 25×10⁸ = −1.6×10⁹ (A/m²)². The invariant is negative, meaning the four-current is spacelike here — physically the drift current dominates cρ, which happens when carriers move relativistically or the net charge density is small compared to the current. (For a single-species non-relativistic beam one expects cρ > J, i.e. timelike; a spacelike result signals a nearly neutral current-carrying medium.)
In the Lorenz gauge ∂_μA^μ = 0, show that ∂_μF^μν = μ₀J^ν reduces to a wave equation for A^ν, and give the vacuum dispersion relation.
Solution
Insert F^μν = ∂^μA^ν − ∂^νA^μ: ∂_μF^μν = ∂_μ∂^μA^ν − ∂^ν(∂_μA^μ). In Lorenz gauge the second term vanishes, leaving □A^ν = μ₀J^ν, where □ = ∂_μ∂^μ = (1/c²)∂_t² − ∇². In vacuum (J^ν = 0): □A^ν = 0. Plane-wave ansatz A^ν ∝ exp(−ik_μx^μ) gives k_μk^μ = 0, i.e. ω²/c² = |k|², or ω = c|k|. Electromagnetic waves propagate at c with a massless (lightlike) four-wavevector — the quantum photon has zero rest mass.