Covariant Maxwell Equations

Four equations. Two with sources, two without; one scalar, one vector, one scalar, one vector; space and time entering on visibly different footings. That is how Maxwell's theory is first met, and written that way its most important property — that it is already relativistic, and was so forty years before relativity — is completely invisible. Package $\vec E$ and $\vec B$ into the antisymmetric field tensor $F^{\mu\nu}$ , package $\rho$ and $\vec J$ into the four-current $J^\nu = (c\rho, \vec J)$ , and the four equations collapse into two:

\partial_\mu F^{\mu\nu} = \mu_0 J^\nu ,\qquad \partial_\mu G^{\mu\nu} = 0 .

Nothing is lost in the collapse — the count is exact. Four vector equations carry $1 + 3 + 1 + 3 = 8$ scalar equations; the two tensor equations carry $4 + 4 = 8$ components. The $\nu = 0$ slot of the first is Gauss's law, its $\nu = i$ slots are Ampère–Maxwell, and the second does the same for $\nabla\cdot\vec B = 0$ and Faraday. What is gained is that both sides of each equation are now tensors of the same rank, so an equation true in one inertial frame is true in every inertial frame, automatically. Two things then fall out for free, and neither was put in. The antisymmetry of $F^{\mu\nu}$ forces $\partial_\nu J^\nu = 0$ — charge conservation is a theorem, not a postulate. And writing $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ makes the second equation an identity: not a law about fields at all, but the shadow of the potential's existence.

Key takeaways

Signature $(+,-,-,-)$ , $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. Dictionary: $A^\mu = (\phi/c, \vec A)$ , $J^\mu = (c\rho, \vec J)$ , $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ .
The whole of classical electromagnetism: $\boxed{\partial_\mu F^{\mu\nu} = \mu_0 J^\nu}$ (Gauss at $\nu=0$ , Ampère–Maxwell at $\nu=i$ ) and $\boxed{\partial_\mu G^{\mu\nu} = 0}$ (no monopoles at $\nu=0$ , Faraday at $\nu=i$ ). Eight scalar equations in, eight components out — the collapse loses nothing.
Covariance is now free. Both sides of each equation are tensors of the same rank, so $\partial_\mu F^{\mu\nu} - \mu_0 J^\nu = 0$ in one frame means zero in all of them — which is why Maxwell's equations were Lorentz-covariant before anyone knew what a Lorentz transformation was.
Charge conservation is a theorem. Hit the sourced equation with another $\partial_\nu$ : $\mu_0\partial_\nu J^\nu = \partial_\nu\partial_\mu F^{\mu\nu} = 0$ , because a symmetric double derivative contracted with an antisymmetric $F$ vanishes. It is forced — and Maxwell's displacement current is what makes it true.
Owned result: the sign of $F^{0i}$ is not a convention. Given $(+,-,-,-)$ , $F^{ij} = -\varepsilon^{ijk}B_k$ and $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ , demanding $\nabla\cdot\vec E = +\rho/\varepsilon_0$ forces $\boxed{F^{0i} = -E_i/c}$ . Flip it and you get $\nabla\cdot\vec E = -\rho/\varepsilon_0$ and a displacement current of the wrong sign. Gauss's law fixes the dictionary; you do not choose it.
Key idea: a charging capacitor is the whole argument in one object. With $R = 5.00\ \mathrm{cm}$ and $I = 1.00\ \mathrm{A}$ , the rim field is $B = 4.00\ \mathrm{\mu T}$ whether you count conduction current through a flat surface or displacement current through a bulging one — because $\varepsilon_0\,\partial_t\vec E$ carries exactly $1.00\ \mathrm{A}$ across the gap. Delete that term and $\partial_\nu J^\nu \ne 0$ .

The core insight

The four-divergence $\partial_\mu(\cdot)^{\mu\nu}$ eats one index of a rank-2 tensor and leaves a four-vector, so each free value of $\nu$ hands you exactly one of Maxwell's laws — the time slot gives the scalar equations, the space slots the vector ones. Apply it to $F$ and you get the sourced pair; apply it to the dual $G$ and you get the source-free pair. Everything else in this article follows from one algebraic fact: $F$ is antisymmetric and $\partial_\mu\partial_\nu$ is symmetric. That single mismatch delivers charge conservation, makes the homogeneous equations an identity, and is why the theory needed a displacement current in the first place.

Where this lives today

Every antenna, every optical fibre, every MRI scanner and every radar is a boundary-value problem for $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ , and the codes that solve them — finite-difference time-domain, method of moments, finite-element solvers — are discretisations of exactly this pair. Structure-preserving schemes are built to keep $\partial_\mu G^{\mu\nu} = 0$ satisfied to machine precision, because a discretisation that lets $\nabla\cdot\vec B$ drift from zero develops spurious charges and blows up. The identity has to stay an identity, even on a grid. Beyond engineering, this is the template for every gauge theory written since. $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ is the abelian case of Yang–Mills, $D_\mu F^{\mu\nu a} = g J^{\nu a}$ , and $\partial_\mu G^{\mu\nu} = 0$ is the abelian Bianchi identity; the strong and weak interactions differ principally by making the derivative covariant and letting the field strength carry charge. The 1865 structure survived being rewritten in tensors, then in fibre bundles, then in the Standard Model — because the thing that makes it work, the antisymmetry of $F$ , is the same in every one of them.

The physical picture

Start by counting, because the counting is the argument. Maxwell's four vector equations are really eight scalar statements: Gauss ( $1$ ), Ampère–Maxwell ( $3$ ), no-monopoles ( $1$ ), Faraday ( $3$ ). Any repackaging that claims to be equivalent must carry eight numbers. The covariant pair does exactly that: $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ has one free index, so four components; $\partial_\mu G^{\mu\nu} = 0$ another four. Eight in, eight out. The collapse is not compression — it is re-filing. What the re-filing buys is the thing three-vector notation destroys. $\vec E$ and $\vec B$ are not vectors under boosts — they mix into each other — so a Maxwell equation written in $\vec E$ and $\vec B$ has no transformation law. But $F^{\mu\nu}$ is a tensor, $J^\nu$ is a four-vector, and $\partial_\mu F^{\mu\nu} - \mu_0 J^\nu$ is therefore a four-vector too — and a four-vector that vanishes in one frame vanishes in all of them, because $\Lambda^\mu{}_\nu 0 = 0$ . Lorentz covariance stops being a property to verify and becomes a property of the notation. And then the antisymmetry pays dividends nobody asked for. $F^{\mu\nu} = -F^{\nu\mu}$ contracted with the symmetric $\partial_\mu\partial_\nu$ gives zero, twice over: once to make $\partial_\nu J^\nu = 0$ compulsory, once to make $\partial_\mu G^{\mu\nu} = 0$ an identity. The first is charge conservation — a physical law extracted from an index symmetry. The second says the two source-free equations were never dynamics at all.

Figure

FOUR EQUATIONS BECOME TWO — and nothing is lost. The four vector equations carry 1 + 3 + 1 + 3 = 8 scalar statements; the two tensor equations carry 4 + 4 = 8 components. The ν = 0 slot of ∂_μF^{μν} = μ₀J^ν is Gauss's law; its ν = i slots are Ampère–Maxwell. The ν = 0 slot of ∂_μG^{μν} = 0 is "no magnetic monopoles"; its ν = i slots are Faraday. The collapse is a re-filing, not a compression — and the payoff is that both sides of each equation are now tensors of the same rank, so covariance is automatic.

The core equations

The sourced equation — Gauss and Ampère–Maxwell in one line:

\boxed{\;\partial_\mu F^{\mu\nu} = \mu_0 J^\nu\;}\ ,\qquad J^\nu = (c\rho,\ \vec J)\ ,\qquad \partial_\mu = \Big(\tfrac1c\partial_t,\ \nabla\Big) \qquad (1)

where: $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ — the antisymmetric field tensor; $\mu_0$ — the vacuum permeability, with $\mu_0\varepsilon_0 c^2 = 1$ . Physical reading: one free index, four components, two of Maxwell's laws. $\nu = 0$ gives Gauss's law; $\nu = i$ gives the three components of Ampère–Maxwell. The displacement current is not an add-on here — it is the $\partial_0 F^{0i}$ term, and it appears whether you want it or not. The component dictionary — derived, not chosen (see the first derivation):

F^{0i} = -\frac{E_i}{c}\ ,\qquad F^{ij} = -\varepsilon^{ijk}B_k \qquad\Longleftrightarrow\qquad F_{0i} = +\frac{E_i}{c}\ ,\qquad F_{ij} = -\varepsilon_{ijk}B_k \qquad (2)

where: indices are lowered with $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , so one lowered spatial index flips a sign and two do not. Physical reading: this is the only sign assignment compatible with equation (1). Flip $F^{0i}$ to $+E_i/c$ and (1) returns $\nabla\cdot\vec E = -\rho/\varepsilon_0$ together with $\nabla\times\vec B = \mu_0\vec J - \mu_0\varepsilon_0\,\partial_t\vec E$ — a negative displacement current. Both are wrong, and both are wrong for the same reason. Gauss's law fixes the dictionary. The source-free equation, and its equivalent cyclic form:

\boxed{\;\partial_\mu G^{\mu\nu} = 0\;}\ ,\quad G^{\mu\nu} = \tfrac12\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta} \qquad\Longleftrightarrow\qquad \boxed{\;\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0\;} \qquad (3)

where: $\varepsilon^{\mu\nu\alpha\beta}$ — the Levi-Civita symbol with $\varepsilon^{0123} = +1$ ; $G^{\mu\nu}$ — the dual tensor. Physical reading: $\nu = 0$ is $\nabla\cdot\vec B = 0$ , $\nu = i$ is Faraday. But substitute $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ and every term of the cyclic sum cancels against another, because $\partial_\mu\partial_\nu = \partial_\nu\partial_\mu$ . (3) is an identity, not a law. It says only that a four-potential exists. Charge conservation, which nobody put in:

\partial_\nu\big(\partial_\mu F^{\mu\nu}\big) = \mu_0\,\partial_\nu J^\nu \quad\text{and}\quad \partial_\nu\partial_\mu F^{\mu\nu} \equiv 0 \qquad\Longrightarrow\qquad \boxed{\;\partial_\nu J^\nu = \frac{\partial\rho}{\partial t} + \nabla\cdot\vec J = 0\;} \qquad (4)

where: the identity holds because $\partial_\nu\partial_\mu$ is symmetric in $(\mu,\nu)$ while $F^{\mu\nu}$ is antisymmetric, and a symmetric object fully contracted with an antisymmetric one vanishes. Physical reading: charge conservation is a consistency condition of the field equations, not a postulate. Turn it around and it becomes a constraint on Maxwell: any candidate for the right-hand side of (1) must be divergence-free. Ampère's law without the displacement current fails that test — which is precisely why the term had to exist.

Reading Maxwell out of the tensor equations

Derivation AThe sourced equation gives Gauss and Ampère–Maxwell — and in doing so fixes the sign of F⁰ⁱ

The goal is to extract two of Maxwell's laws from $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ — and, along the way, to settle something usually waved through as "a matter of convention". The component dictionary $F^{0i} = \pm E_i/c$ is not free: once the signature, the equation, and the sign of $F^{ij}$ are fixed, Gauss's law picks the sign for you. Carry it as an unknown $s = \pm 1$ and watch it get determined.

Step 1. Write $F^{0i} = s\,E_i/c$ with $s$ unknown, and $F^{ij} = -\varepsilon^{ijk}B_k$ (fixed by putting $\vec B$ in the space-space block with the standard orientation). Antisymmetry gives $F^{i0} = -s\,E_i/c$ . Why: the only thing genuinely undetermined here is one overall sign in the time-space block. Everything else is definitions.

Step 2. Take $\nu = 0$ . Since $F^{00} = 0$ , only spatial derivatives survive: \partial_\mu F^{\mu 0} = \partial_i F^{i0} = -\frac{s}{c}\,\nabla\cdot\vec E \;=\; \mu_0 J^0 = \mu_0 c\rho . Using $\mu_0 c^2 = 1/\varepsilon_0$ , this rearranges to \nabla\cdot\vec E = -s\,\frac{\rho}{\varepsilon_0} . Why: there it is. Gauss's law is an experimental fact with a $+$ sign, so $s = -1$ and $\boxed{F^{0i} = -E_i/c}$ . The sign was never a convention; it was a measurement in disguise.

Step 3. Now $\nu = i$ , with $s = -1$ in hand: \partial_\mu F^{\mu i} = \underbrace{\tfrac1c\partial_t\Big(-\tfrac{E_i}{c}\Big)}_{\text{from }F^{0i}} + \underbrace{\partial_j\big(-\varepsilon^{jik}B_k\big)}_{\text{from }F^{ji}} = -\frac{1}{c^2}\frac{\partial E_i}{\partial t} + (\nabla\times\vec B)_i \;=\; \mu_0 J^i , using $-\varepsilon^{jik} = +\varepsilon^{ijk}$ . Rearranged, and with $1/c^2 = \mu_0\varepsilon_0$ : \boxed{\;\nabla\times\vec B = \mu_0\vec J + \mu_0\varepsilon_0\,\frac{\partial\vec E}{\partial t}\;} Why: the Ampère–Maxwell law, displacement current included, with no separate step. The $\partial_t\vec E$ term is simply the $\mu = 0$ contribution to the divergence. In the 3-D telling it had to be invented; here it would take effort to leave out.

Step 4. Check what the other sign would have done. Setting $s = +1$ gives $\nabla\cdot\vec E = -\rho/\varepsilon_0$ from Step 2, and reversing the $F^{0i}$ term in Step 3 gives $\nabla\times\vec B = \mu_0\vec J - \mu_0\varepsilon_0\,\partial_t\vec E$ . Why: it fails twice — like charges would attract, and the displacement current would run backwards, destroying the wave equation. A sign that breaks two independent laws is not a convention. It is an error. *Verified in verify_aq-sr-covariant-maxwell-equations.py: $F^{\mu\nu}$ is rebuilt from $A^\mu = (\phi/c, \vec A)$ via $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ with $\vec E = -\nabla\phi - \partial_t\vec A$ , confirming $F^{0i} = -E_i/c$ independently; then $\partial_\mu F^{\mu\nu}$ is contracted symbolically for arbitrary $\vec E(t,\vec x)$ , $\vec B(t,\vec x)$ and matched against Gauss and Ampère–Maxwell — and the $s = +1$ alternative is shown to produce $\nabla\cdot\vec E = -\rho/\varepsilon_0$ and a sign-flipped displacement current.*

Go deeper · Derivation B · 2 minFaraday's law from the cyclic identity — and the sign the base article gets backwards

Derivation BFaraday's law from the cyclic identity — and the sign the base article gets backwards

The homogeneous pair hides inside the cyclic sum $\partial_{[\lambda}F_{\mu\nu]} = 0$ . Pulling Faraday out is three lines of index bookkeeping, and it is exactly where the AllQFT base article goes wrong: its own inserted components give $+\partial_t B_z$ , while its stated conclusion says $-\partial_t B_z$ . The error is worth reproducing, because it propagates from a single wrong sign in the lowered dictionary.

Step 1. Lower both indices of (2) with $\eta$ . One spatial index flips a sign, two flip twice: F_{0i} = \eta_{00}\eta_{ij}F^{0j} = (+1)(-\delta_{ij})\Big(-\frac{E_j}{c}\Big) = +\frac{E_i}{c},\qquad F_{ij} = -\varepsilon_{ijk}B_k . Why: this is the line that decides everything. $F^{0i}$ is negative, $F_{0i}$ is positive. Carry the wrong one into the cyclic sum and Faraday comes out with its sign reversed — which is precisely the failure below.

Step 2. Take the index triple $(\lambda,\mu,\nu) = (0,1,2)$ in the cyclic identity: \partial_0 F_{12} + \partial_1 F_{20} + \partial_2 F_{01} = 0 . With $F_{12} = -B_z$ , $F_{20} = -F_{02} = -E_y/c$ , $F_{01} = +E_x/c$ , and $\partial_0 = \tfrac1c\partial_t$ : -\frac1c\frac{\partial B_z}{\partial t} - \frac1c\frac{\partial E_y}{\partial x} + \frac1c\frac{\partial E_x}{\partial y} = 0 . Why: every term carries the same $1/c$ , so the physics is entirely in the bracket. Multiply through by $-c$ and read.

Step 3. Rearrange: \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} = -\frac{\partial B_z}{\partial t} \qquad\Longleftrightarrow\qquad \boxed{\;(\nabla\times\vec E)_z = -\frac{\partial B_z}{\partial t}\;} and the triples $(0,2,3)$ and $(0,3,1)$ give the $x$ and $y$ components. The triple $(1,2,3)$ , which contains no time index, gives $\nabla\cdot\vec B = 0$ . Why: Faraday's law of induction, with the minus sign that makes Lenz's law work and transformers not explode. All four homogeneous components, from four index triples.

Step 4. Now the base's version. Its derivation declares $F_{0i} = -E_i/c$ — the lowering of the wrong $F^{0i} = +E_i/c$ — and so inserts $F_{20} = +E_y/c$ , $F_{01} = -E_x/c$ . Its cyclic sum reads $-\tfrac1c\partial_t B_z + \tfrac1c\partial_x E_y - \tfrac1c\partial_y E_x = 0$ , which rearranges to $(\nabla\times\vec E)_z = +\partial_t B_z$ , the opposite of what its own next line asserts. Why: the algebra is faithful to the dictionary; the dictionary is wrong; and the article papers over the gap by writing down the right answer at the end. A derivation that does not track its own signs is decoration. Here they are tracked, and Step 1 is where they are earned. *Verified in verify_aq-sr-covariant-maxwell-equations.py: the cyclic sum $\partial_0F_{12} + \partial_1F_{20} + \partial_2F_{01}$ is evaluated symbolically with $\vec E$ and $\vec B$ arbitrary functions of $(t,\vec x)$ and shown to equal $-\tfrac1c[(\nabla\times\vec E)_z + \partial_tB_z]$ ; the base's dictionary is then substituted and shown to yield $+\partial_tB_z$ , contradicting the base's printed conclusion.*

Go deeper · Derivation C · 1 minCharge conservation is a theorem — and it is what forces the displacement current to exist

Derivation CCharge conservation is a theorem — and it is what forces the displacement current to exist

Two claims, and the second is the interesting one. That antisymmetry implies $\partial_\nu J^\nu = 0$ is three lines. That the implication runs backwards — that demanding a conserved current forces the displacement term into Ampère's law — is the historical heart of the theory, and the covariant form makes it obvious.

Step 1. Apply $\partial_\nu$ to both sides of (1): \partial_\nu\partial_\mu F^{\mu\nu} = \mu_0\,\partial_\nu J^\nu . Why: the left side is a full contraction of the double derivative against $F$ on both of its indices. Nothing is left free.

Step 2. Kill the left side. $\partial_\nu\partial_\mu$ is symmetric under $\mu\leftrightarrow\nu$ (partial derivatives commute), $F^{\mu\nu}$ is antisymmetric, and relabelling gives \partial_\nu\partial_\mu F^{\mu\nu} = \partial_\mu\partial_\nu F^{\nu\mu} = -\,\partial_\nu\partial_\mu F^{\mu\nu} \;\Longrightarrow\; \partial_\nu\partial_\mu F^{\mu\nu} = 0 . Hence \boxed{\;\partial_\nu J^\nu = \frac{\partial\rho}{\partial t} + \nabla\cdot\vec J = 0\;} Why: a quantity equal to minus itself is zero. Charge conservation drops out of an index symmetry — no physics was assumed beyond "the field equation holds".

Step 3. Now run it backwards, which is the real content. Suppose Ampère's law had no displacement term: $\nabla\times\vec B = \mu_0\vec J$ . Take the divergence. The left is $\nabla\cdot(\nabla\times\vec B) \equiv 0$ , so $\nabla\cdot\vec J = 0$ always. Why: this is false. A charging capacitor has current flowing into a plate and nowhere out, so $\nabla\cdot\vec J = -\partial_t\rho \ne 0$ . Uncorrected Ampère is not merely incomplete — it is inconsistent with charge conservation.

Step 4. Repair it the way the tensor equation demands. Continuity says $\nabla\cdot\vec J = -\partial_t\rho = -\partial_t(\varepsilon_0\nabla\cdot\vec E) = -\nabla\cdot(\varepsilon_0\partial_t\vec E)$ , so $\vec J + \varepsilon_0\,\partial_t\vec E$ is divergence-free — and it is the unique local repair. Putting it on the right of Ampère gives \nabla\times\vec B = \mu_0\Big(\vec J + \varepsilon_0\frac{\partial\vec E}{\partial t}\Big) , exactly the law Step 3 of the first derivation produced from $\partial_\mu F^{\mu i} = \mu_0 J^i$ . Why: the displacement current is the price of charge conservation, and the covariant form charges it automatically. In $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ you could not omit the term if you tried: it is the $\mu = 0$ slot of a sum you are not allowed to truncate. *Verified in verify_aq-sr-covariant-maxwell-equations.py: $\partial_\nu\partial_\mu F^{\mu\nu} = 0$ is checked by explicit summation over all $4^3$ index combinations for an arbitrary $A_\mu(t,\vec x)$ ; $\partial_\nu J^\nu$ is confirmed to equal $\partial_t\rho + \nabla\cdot\vec J$ ; and $\nabla\cdot(\vec J + \varepsilon_0\partial_t\vec E) = 0$ is verified symbolically given Gauss's law and continuity.*

Limiting cases

Case	Result	Meaning
$J^\nu = 0$ (vacuum)	$\Box F^{\mu\nu} = 0$	every field component obeys the wave equation; light, at speed $c$
$\partial_t = 0$ , $\vec B = 0$	$\nabla\cdot\vec E = \rho/\varepsilon_0$ , $\nabla\times\vec E = 0$	electrostatics: the $\nu=0$ slot plus the curl-free condition
$\partial_t = 0$ , steady $\vec J$	$\nabla\times\vec B = \mu_0\vec J$	magnetostatics: Ampère, displacement term switched off
Lorenz gauge $\partial_\mu A^\mu = 0$	$\Box A^\nu = \mu_0 J^\nu$	four decoupled wave equations, one per potential component
$F = \mathrm{d}A$	$\partial_\mu G^{\mu\nu} \equiv 0$	the source-free pair is an identity, not dynamics
curved spacetime	$\nabla_\mu F^{\mu\nu} = \mu_0 J^\nu$	minimal coupling; the homogeneous pair is unchanged, connections cancel by antisymmetry
$\omega\varepsilon_0 \ll \sigma$	$\vec J_{\text{disp}} \ll \vec J_{\text{cond}}$	in copper at $60\ \mathrm{Hz}$ the ratio is $5.60\times10^{-17}$

Common trapThe sign of F⁰ⁱ is not yours to choose — and getting it wrong breaks Gauss AND the displacement current.

Textbooks differ on $F^{0i} = +E_i/c$ versus $-E_i/c$ , and the difference is usually filed under "conventions vary". In this signature it is not a convention. Fix $(+,-,-,-)$ , fix $F^{ij} = -\varepsilon^{ijk}B_k$ , and write $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ ; then the $\nu = 0$ component computes $\nabla\cdot\vec E = -s\,\rho/\varepsilon_0$ , where $s$ is the sign of $F^{0i}$ . Gauss's law says $+$ , so $s = -1$ , so $F^{0i} = -E_i/c$ . Full stop. The failure is not cosmetic. With $s = +1$ you get like charges attracting and $\nabla\times\vec B = \mu_0\vec J - \mu_0\varepsilon_0\partial_t\vec E$ — a displacement current of the wrong sign, which destroys the wave solutions. Both failures are recomputed in this article's verify script. The three ingredients — signature, the sign of $F^{0i}$ , and the sign on the right of the field equation — are not independent. Choose any two and the third is determined. State all three before comparing formulae with anyone, including your past self.

Common trapThe base article's Faraday derivation contradicts its own algebra, and its Gauss step contradicts its own dictionary.

The AllQFT base for this page carries at least three defects, all traceable to one wrong sign in the component dictionary.

One. Its Faraday derivation declares $F_{0i} = -E^i/c$ , correctly inserts $F_{20} = +E_y/c$ and $F_{01} = -E_x/c$ , correctly obtains $-\tfrac1c\partial_tB_z + \tfrac1c\partial_xE_y - \tfrac1c\partial_yE_x = 0$ — and then states the conclusion as $(\nabla\times\vec E)_z = -\partial_tB_z$ , when its own line gives $+\partial_tB_z$ . The algebra and the answer disagree, and the answer was written in from memory.

Two. Its sourced-equation card declares " $F^{0i} = E^i/c$ (signature $+---$ )" and then writes $\partial_\mu F^{\mu0} = \partial_i F^{i0} = \tfrac1c\nabla\cdot\vec E$ — a line that requires $F^{i0} = +E_i/c$ , i.e. $F^{0i} = -E_i/c$ , the opposite of what it just declared. The printed algebra is right; the declared dictionary is wrong.

Three. The same card's spatial step prints $\partial_\mu F^{\mu i} = \tfrac{1}{c^2}\partial_tE^i - (\nabla\times\vec B)^i = -\mu_0 J^i$ — note the minus on the source, in a card whose own headline reads $\partial_\mu F^{\mu\nu} = +\mu_0 J^\nu$ . The expression is exactly $(-1)\times$ the correct divergence, and the error is hidden by flipping the right-hand side too, so the final Ampère–Maxwell law comes out looking right. Two errors that cancel are still two errors — and the moment anyone reuses the intermediate line, they propagate.

Worked examples

1 · The charging capacitor: two surfaces, one answer, and why the displacement current cannot be dropped

A circular parallel-plate capacitor with plate radius $R = 5.00\ \mathrm{cm}$ is charged by a steady current $I = 1.00\ \mathrm{A}$ . Find the magnetic field at the rim of the gap two different ways, and show they agree.

Step 1 — the geometry. Ampère's law in integral form relates $\oint\vec B\cdot\mathrm{d}\vec\ell$ around a loop to the current through any surface bounded by that loop. Take the loop to be the circle of radius $R$ midway between the plates. A flat disc spanning it passes through the gap and catches no conduction current. A bulging surface that balloons out and back through the wire catches the full $1.00\ \mathrm{A}$ . Uncorrected Ampère therefore gives two different answers — it is not even well defined.

Step 2 — route A, the conduction current. Through the bulging surface, $I_{\text{enc}} = I$ , so B = \frac{\mu_0 I}{2\pi R} = \frac{(1.256637\times10^{-6})(1.00)}{2\pi(0.0500)} = \mathbf{4.00\times10^{-6}\ T} = 4.00\ \mathrm{\mu T}.

Step 3 — route B, the displacement current. Through the flat disc there is no conduction current, only a changing $\vec E$ . The plate area is $A = \pi R^2 = 7.85398\times10^{-3}\ \mathrm{m^2}$ ; the surface charge grows at $\mathrm{d}\sigma/\mathrm{d}t = I/A = 127.324\ \mathrm{C\,m^{-2}\,s^{-1}}$ ; and with $E = \sigma/\varepsilon_0$ inside the gap, \frac{\mathrm{d}E}{\mathrm{d}t} = \frac{1}{\varepsilon_0}\frac{\mathrm{d}\sigma}{\mathrm{d}t} = \frac{127.324}{8.8541878\times10^{-12}} = \mathbf{1.438008\times10^{13}\ V\,m^{-1}\,s^{-1}}. The displacement current through the disc is I_d = \varepsilon_0\,\frac{\mathrm{d}E}{\mathrm{d}t}\,A = (8.8541878\times10^{-12})(1.438008\times10^{13})(7.85398\times10^{-3}) = \mathbf{1.000\ A}, exactly the conduction current, and therefore $B = \mu_0 I_d/(2\pi R) = \mathbf{4.00\times10^{-6}\ T}$ again.

Step 4 — what actually made this work. The agreement is not a coincidence to be admired; it is $\partial_\nu J^\nu = 0$ in disguise. The tensor equation $\partial_\mu F^{\mu i} = \mu_0 J^i$ carries the $\varepsilon_0\partial_t\vec E$ term as the $\mu = 0$ slot of a four-index sum, and the antisymmetry of $F$ then guarantees the divergence-free combination. Without Maxwell's term the two surfaces would disagree by the full $1.00\ \mathrm{A}$ — a $100\%$ discrepancy in a first-year lab experiment. The displacement current is not a correction. It is the term that makes electromagnetism consistent.

2 · Where the displacement current stops mattering, and where it takes over

A copper wire of radius $a = 1.00\ \mathrm{mm}$ carries $I = 1.00\ \mathrm{A}$ at mains frequency $f = 60\ \mathrm{Hz}$ . Copper has $\sigma = 5.96\times10^{7}\ \mathrm{S\,m^{-1}}$ . Compare the two source terms in the $\nu = i$ component of the field equation, and find where they become equal.

Step 1 — the conduction term. J = \frac{I}{\pi a^2} = \frac{1.00}{\pi(1.00\times10^{-3})^2} = \mathbf{3.18310\times10^{5}\ A\,m^{-2}}, \qquad E = \frac{J}{\sigma} = \frac{3.18310\times10^{5}}{5.96\times10^{7}} = \mathbf{5.34077\times10^{-3}\ V\,m^{-1}}.

Step 2 — the displacement term. With $E$ oscillating at $\omega = 2\pi f = 376.991\ \mathrm{rad\,s^{-1}}$ , |J_d| = \varepsilon_0\,\omega\,E = (8.8541878\times10^{-12})(376.991)(5.34077\times10^{-3}) = \mathbf{1.78272\times10^{-11}\ A\,m^{-2}}.

Step 3 — the ratio, and the fact that the wire drops out. \frac{|J_d|}{J} = \frac{1.78272\times10^{-11}}{3.18310\times10^{5}} = \mathbf{5.60059\times10^{-17}} = \frac{\omega\varepsilon_0}{\sigma}. The current, the radius and the field all cancelled: the ratio depends only on the frequency and the material — a useful simplification, and a good check that the algebra is right.

Step 4 — the boundary number. The two terms are equal when $\omega\varepsilon_0 = \sigma$ , i.e. f_\times = \frac{\sigma}{2\pi\varepsilon_0} = \frac{5.96\times10^{7}}{2\pi(8.8541878\times10^{-12})} = \mathbf{1.0713\times10^{18}\ Hz}, a photon energy of $\mathbf{4.43\ keV}$ and a wavelength of $\mathbf{0.280\ nm}$ — soft X-rays. So the displacement current is negligible in a mains cable ( $\sim 10^{-17}$ of the conduction term) and dominant for X-rays — which is why metals are shiny conductors at low frequency and nearly transparent windows in the X-ray band. One dimensionless ratio, read straight off the $\nu = i$ component of $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ , separates those two worlds.

Figure

THE DISPLACEMENT CURRENT IS NOT OPTIONAL — it is what makes Ampère's law well defined. Same loop (the rim of a charging capacitor, R = 5.00 cm, I = 1.00 A), two surfaces. The BULGING surface catches the full 1.00 A of conduction current. The FLAT disc through the gap catches ZERO conduction current — but it catches ε₀ ∂E/∂t = 1.438 × 10¹³ V m⁻¹ s⁻¹ × ε₀ × A = exactly 1.00 A of displacement current. Both give B = 4.00 μT at the rim. Without Maxwell's term the two surfaces disagree by 100%, and ∇·J = 0 would be forced — which a charging capacitor flatly violates. In the covariant form you could not omit the term if you tried: it is the μ = 0 slot of ∂_μF^{μi}.

Physics of the portrait

The equations are the portrait. The covariant form reveals that the four laws were never four laws — and that the strangest ingredient, the one Maxwell added by hand against no experiment, was the only one that could have been added at all.

Origins · 2 minMaxwell's equations and how this was discovered

Maxwell's equations

1861–present · the displacement current · light as an electromagnetic wave · relativistic before relativity

Maxwell's equations are the four laws — Gauss, Ampère–Maxwell, no-monopoles, Faraday — that between them exhaust classical electromagnetism, and their history is a long argument about the fourth term in the second one. Three of the laws were experimental: Gauss's, Faraday's induction law, and Ampère's circuital law. The fourth ingredient, the displacement current $\varepsilon_0\,\partial_t\vec E$ , was not. James Clerk Maxwell introduced it in "On Physical Lines of Force" (1861–62) on essentially theoretical grounds, and its immediate consequence was spectacular: the corrected equations admit wave solutions travelling at $1/\sqrt{\mu_0\varepsilon_0}$ , a number that came out equal — within the experimental error of the day — to the measured speed of light. In "A Dynamical Theory of the Electromagnetic Field" (Phil. Trans. R. Soc. 155, 459, 1865) Maxwell drew the conclusion in print: light is an electromagnetic disturbance. Heinrich Hertz generated and detected those waves at Karlsruhe in 1888, twenty-seven years after they were predicted and nine after Maxwell's death. The modern four-equation form is not Maxwell's own — he wrote twenty equations in twenty variables, in components — but Oliver Heaviside's, who recast them in vector notation from 1884. And the deepest fact about them was not appreciated until Einstein and Minkowski: the equations are exactly Lorentz-covariant, and always were — the first relativistic theory, written forty years before relativity, by a man who believed in an aether. The covariant pair is simply Maxwell's system with its symmetry made visible.

History

1861

— Maxwell, in "On Physical Lines of Force", introduces the displacement current on theoretical grounds and finds that the corrected equations support waves travelling at the speed of light.

1865

— Maxwell's "A Dynamical Theory of the Electromagnetic Field" (Phil. Trans. R. Soc. 155, 459) presents the field equations as a system and concludes in print that light is an electromagnetic wave.

1867

— Ludvig Lorenz publishes "On the Identity of the Vibrations of Light with Electrical Currents", introducing retarded potentials and the gauge condition now named after him; it is persistently misattributed to Hendrik Lorentz.

1873

— Maxwell's Treatise on Electricity and Magnetism codifies the theory in twenty equations in twenty variables.

1884

— Heaviside recasts Maxwell's system into the four vector equations used ever since, in the vector calculus he was inventing to do it.

1888

— Hertz, at Karlsruhe, generates and detects free-space electromagnetic waves, confirming Maxwell's prediction experimentally.

1908

— Minkowski, in "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern", casts electrodynamics in four-dimensional tensor form; the two-equation covariant structure used today descends directly from that paper.

Going further

Texts. Jackson, Classical Electrodynamics, 3rd ed., §§11.9–11.11 · Griffiths, Introduction to Electrodynamics, 4th ed., ch. 12 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Maxwell, Phil. Trans. R. Soc. 155, 459 (1865) · Minkowski, Nachr. Ges. Wiss. Göttingen, 53 (1908).

Provenance. Signature $(+,-,-,-)$ , $\varepsilon^{0123} = +1$ ; $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass, and no velocity-dependent mass appears anywhere. The dictionary $F^{0i} = -E_i/c$ , $F_{0i} = +E_i/c$ , $F^{ij} = -\varepsilon^{ijk}B_k$ is not assumed — it is derived here from $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ and independently forced by Gauss's law, both routes recomputed in verify_aq-sr-covariant-maxwell-equations.py, which also parses this article's own printed digits. Base-defect repairs (logged): (i) the base's Faraday derivation contradicts its own algebra by a sign; (ii) its sourced-equation card declares $F^{0i} = E^i/c$ while its own printed line requires $F^{0i} = -E_i/c$ ; (iii) that card's spatial step silently flips the sign of the source, writing $= -\mu_0 J^i$ under a headline reading $= +\mu_0 J^\nu$ .

Scope. The construction of $F^{\mu\nu}$ belongs to The Electromagnetic Field Tensor; the dual and the monopole question to The Dual Field Tensor; the classification of fields to Electromagnetic Field Invariants; the motion of charges to The Covariant Lorentz Force Law; and $\Box A^\nu = \mu_0 J^\nu$ to the Wave Equation for the Four-Potential. This page owns the collapse of four equations into two, the sign that Gauss's law forces, and charge conservation from antisymmetry.

Builds on. The Electromagnetic Field Tensor · The Four-Current Density · The Dual Field Tensor · The Four-Potential. Leads to. Wave Equation for the Four-Potential · Charge Invariance and Continuity · The Covariant Lorentz Force Law.

If the homogeneous equation is only an identity, is half of Maxwell's theory empty?

Not empty — relocated. The physical content moved into the assumption that a four-potential exists, and that assumption does real work: it makes gauge invariance possible, it is what quantum mechanics actually couples to (the Aharonov–Bohm effect measures $A^\mu$ , not $F^{\mu\nu}$ ), and it is what a magnetic monopole would break. The content is still there — it has simply been recognised as a statement about the potential rather than about the fields.

Why does the displacement current have to be there, in one sentence?

Because $\nabla\cdot(\nabla\times\vec B) \equiv 0$ , so uncorrected Ampère would force $\nabla\cdot\vec J = 0$ everywhere and always — which a charging capacitor plainly violates — whereas $\vec J + \varepsilon_0\partial_t\vec E$ is divergence-free by Gauss plus continuity. In the covariant form this is free: the antisymmetry of $F$ forces $\partial_\nu J^\nu = 0$ , and the only source consistent with that is the corrected one.

Were Maxwell's equations really Lorentz-covariant before Lorentz?

Yes, and that is the historically strange part. The equations were written in 1861–65 by a man who believed in a material aether and expected the theory to pick out a preferred frame. It does not. The mismatch between Maxwell's exact Lorentz covariance and Newton's exact Galilean covariance is the crisis Einstein resolved in 1905 by keeping Maxwell and rebuilding mechanics. The covariant notation makes the property obvious; it did not create it.

Is ∂_μF^{μν} = μ₀J^ν the same equation in Gaussian units?

The structure is identical; only the constant changes. In Gaussian units it reads $\partial_\mu F^{\mu\nu} = \tfrac{4\pi}{c}J^\nu$ , and the field tensor holds $E_i$ rather than $E_i/c$ . Nothing physical differs, but every printed coefficient does — which is why a formula copied across unit systems without checking is one of the most reliable ways to introduce a factor of $c$ that nobody can later find.

Why can't I just check covariance by transforming E and B directly?

You can, and it works — it is simply enormous. You must transform six field components and four source components, substitute into four vector equations, and check that the mess reassembles into equations of the same form. The tensor argument replaces all of that with one observation: $\partial_\mu F^{\mu\nu} - \mu_0 J^\nu$ is a four-vector, and $\Lambda^\mu{}_\nu\,0 = 0$ . Manifest covariance is not a shortcut through the calculation; it is a demonstration that the calculation could not have come out any other way.

How does this survive in curved spacetime?

By minimal coupling, $\partial_\mu \to \nabla_\mu$ : the sourced equation becomes $\nabla_\mu F^{\mu\nu} = \mu_0 J^\nu$ . The homogeneous equation, remarkably, does not change at all — written as $\partial_{[\lambda}F_{\mu\nu]} = 0$ , the Christoffel terms that $\nabla$ would introduce cancel in the antisymmetrisation, so the flat-space form is already the curved-space form. Antisymmetry pays out one last time, and it is the main reason electromagnetism couples to gravity without the ambiguities that plague, say, spinors.