Wave Equation for the Four-Potential

Maxwell wrote down his field equations in 1865, and buried in them was a number he had not put there. Combining the laws of electricity and magnetism produced a wave, and the wave's speed was $1/\sqrt{\mu_0\varepsilon_0}$ — two constants measured with coils and capacitors, no optics anywhere in sight. It came out equal to the speed of light. Written in spacetime, that whole story compresses to one line:

\Box A^\mu = \mu_0 J^\mu .

Four coupled vector equations for $\vec E$ and $\vec B$ become one wave equation for one four-vector, driven by one four-current. And the operator on the left is no new invention: $\Box = \partial_\mu\partial^\mu = \frac{1}{c^2}\partial_t^2 - \nabla^2$ is the Minkowski metric written as a differential operator. The relative minus sign between the time and space terms — the only thing separating a wave equation from Laplace's — is the signature $(+,-,-,-)$ itself. So when $\Box A^\mu = 0$ forces its plane waves onto the light cone, $k_\mu k^\mu = 0$ and hence $\omega = c|\vec k|$ , that speed is not an input. It is the metric, read back out. A null four-vector stays null under every Lorentz transformation, so every inertial observer measures the same wave speed — Einstein's second postulate, sitting inside Maxwell's equations forty years before anyone thought to postulate it.

Key takeaways

Signature $(+,-,-,-)$ , $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. Dictionary: $A^\mu = (\phi/c,\vec A)$ , $J^\mu = (c\rho,\vec J)$ , $\partial_\mu = (\tfrac1c\partial_t, \nabla)$ .
The equation: $\boxed{\Box A^\mu = \mu_0 J^\mu}$ , in the Lorenz gauge $\partial_\mu A^\mu = 0$ . Four uncoupled wave equations, each driven by the matching component of the four-current.
The d'Alembertian is the metric. $\Box \equiv \partial_\mu\partial^\mu = \eta^{\mu\nu}\partial_\mu\partial_\nu = \frac{1}{c^2}\partial_t^2 - \nabla^2$ . Flip to a Euclidean signature and the minus becomes a plus: the wave equation collapses into Laplace's — no light, no retardation, no causality. Waves exist because the metric is not positive-definite.
The gauge is a scalar condition, and that is why this works. $\partial_\mu A^\mu$ is a four-divergence: vanish in one frame, vanish in all.
Counting the photon. $A^\mu$ has 4 components; the Lorenz condition removes 1; the residual gauge freedom ( $\Box\lambda = 0$ ) removes 1 more. $4 - 1 - 1 = \mathbf{2}$ — the two polarisations of light, obtained by bookkeeping.
Key idea: $c$ is derived, not assumed. $\Box A^\mu = 0 \Rightarrow k_\mu k^\mu = 0 \Rightarrow \omega = c|\vec k|$ , and $k_\mu k^\mu$ is a Lorentz scalar, so the same $c$ emerges in every frame. Numerically $1/\sqrt{\mu_0\varepsilon_0} = \mathbf{2.99792458\times10^{8}\ m\,s^{-1}}$ — a capacitor constant and a coil constant, with no optics anywhere in it.

The core insight

The wave operator and the metric are the same object. $\Box = \eta^{\mu\nu}\partial_\mu\partial_\nu$ — feed two derivatives to the metric and you get the d'Alembertian, exactly as feeding it two displacements gives the interval. Everything follows: the minus sign that makes waves instead of harmonic functions, the light cone that makes causality, the null condition that makes $c$ frame-independent. Maxwell's equations were never "made" relativistic. They were relativistic — and it took forty years to notice that the wave operator physics had used since 1747 already had Minkowski spacetime inside it.

Where this lives today

Every antenna, fibre, radar and radio link is a boundary-value problem for $\Box A^\mu = \mu_0 J^\mu$ , and the engineer's version — retarded potentials, near and far zone, the $1/R$ radiation term — is the three-dimensional shadow of this one equation. Finite-difference time-domain solvers step the hyperbolic operator forward on a grid, and the reason they can step forward at all, with a finite domain of dependence and a stability condition tied to $c\,\Delta t/\Delta x$ , is the light cone that the minus sign in $\Box$ puts there. The equation's deeper career is as a template. Replace $A^\mu$ by the metric perturbation $\bar h^{\mu\nu}$ and the harmonic gauge plays exactly the role of the Lorenz gauge: the linearised Einstein equations become $\Box \bar h^{\mu\nu} \propto T^{\mu\nu}$ — same operator, same gauge logic, two propagating polarisations after the same subtraction, and gravitational waves travelling at $c$ for the same reason light does. When LIGO timed gravitational waves against a gamma-ray burst in 2017 and found their speed equal to $c$ to a part in $10^{15}$ , what was tested was that the same $\Box$ — the same signature — governs both fields. In quantum field theory this page's $4-1-1=2$ becomes the photon's two helicities.

The physical picture

Start with the covariant field equation, $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ — owned by `aq-sr-covariant-maxwell-equations`. Substituting $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ turns it into an equation for the potential, and something awkward appears:

\Box A^\nu - \partial^\nu\big(\partial_\mu A^\mu\big) = \mu_0 J^\nu .

The first term is what we want: a clean wave operator acting on one component at a time. The second is a menace — it couples all four components of $A^\mu$ through their common divergence. But $A^\mu$ is not an observable. The fields $F^{\mu\nu}$ are unchanged by $A^\mu \to A^\mu - \partial^\mu\lambda$ for any scalar $\lambda$ , so we may choose $A^\mu$ to make our lives easy. Choose $\partial_\mu A^\mu = 0$ — the Lorenz gauge, owned by `aq-sr-gauge-transformations-lorenz-gauge`. The menacing term vanishes identically, and four uncoupled wave equations remain. Two things make this more than a trick. First, $\partial_\mu A^\mu = 0$ says a Lorentz scalar vanishes — zero in one frame, zero in every frame — so the gauge can be imposed globally and covariantly. (Contrast the Coulomb gauge $\nabla\cdot\vec A = 0$ : enormously useful, but it picks out a time axis and is not preserved by boosts.) Second, once the equation reads $\Box A^\mu = \mu_0 J^\mu$ , every symbol is a tensor — $\Box$ a scalar operator, $A^\mu$ and $J^\mu$ four-vectors. It has the same form in every inertial frame by inspection, which is what "manifestly covariant" means, and it is why this equation, not the four three-dimensional ones, is the honest statement of electromagnetism.

Figure

THE d'ALEMBERTIAN IS THE METRIC WEARING DERIVATIVES. Feed the metric two displacements and you get the interval, ds² = c²dt² − dx² − dy² − dz². Feed it two derivatives and you get the wave operator, □ = (1/c²)∂ₜ² − ∇². It is the SAME contraction with the SAME η^{μν} — the minus signs come from exactly one place, the signature. And the consequence is total: with the relative minus sign the equation is HYPERBOLIC, its solutions are waves travelling at c, and it has a light cone. Make the signature Euclidean (+,+,+,+) and the same operator becomes the four-dimensional LAPLACIAN — elliptic, no waves, no propagation speed, no causality, no light. The single minus sign in the metric is the entire difference between a universe with light in it and a universe without.

The core equations

The equation this page owns:

\boxed{\;\Box A^\mu \;=\; \mu_0 J^\mu\;}\qquad\text{in the Lorenz gauge }\ \partial_\mu A^\mu = 0 \qquad (1)

where: $A^\mu = (\phi/c, \vec A)$ is the four-potential; $J^\mu = (c\rho, \vec J)$ the four-current. Physical reading: four uncoupled wave equations. The $\mu = 0$ component reads $\Box\phi = \rho/\varepsilon_0$ (using $\mu_0 c^2 = 1/\varepsilon_0$ ), the spatial ones $\Box\vec A = \mu_0\vec J$ . Charge density sources the scalar potential; current density sources the vector potential; nothing else couples. The operator, spelled out:

\Box \;\equiv\; \partial_\mu\partial^\mu \;=\; \eta^{\mu\nu}\partial_\mu\partial_\nu \;=\; \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \;-\; \nabla^2 \qquad (2)

where: $\partial_\mu = \big(\tfrac1c\partial_t,\ \nabla\big)$ and $\partial^\mu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ . Physical reading: raising the index is where the minus sign is born, and there is no other source of it. Contract two four-gradients with $\mathrm{diag}(+1,-1,-1,-1)$ and you get one plus and three minuses — hyperbolic. Contract with $\mathrm{diag}(+1,+1,+1,+1)$ and you get the four-dimensional Laplacian: elliptic, with no waves in its solution set at all. The vacuum limit, and the dispersion relation it forces:

\Box A^\mu = 0,\quad A^\mu = \varepsilon^\mu e^{-i k_\nu x^\nu} \qquad\Longrightarrow\qquad \boxed{\;k_\mu k^\mu = \frac{\omega^2}{c^2} - |\vec k|^2 = 0\;}\qquad\Longrightarrow\qquad \boxed{\;\omega = c\,|\vec k|\;} \qquad (3)

where: $k^\mu = (\omega/c,\ \vec k)$ is the wave four-vector and $\varepsilon^\mu$ the polarisation four-vector. Physical reading: the wave four-vector is null — light lives on the light cone, not inside it. And because $k_\mu k^\mu$ is a Lorentz scalar, a null $k$ is null in every frame: every inertial observer finds $\omega' = c|\vec k'|$ with the same $c$ . The frame-independence of the speed of light is not an assumption bolted onto Maxwell; it is a corollary of the operator in (2). The gauge condition, and what it does not do:

\partial_\mu A^\mu = \frac{1}{c^2}\frac{\partial\phi}{\partial t} + \nabla\cdot\vec A = 0, \qquad\text{with residual freedom}\qquad A^\mu \to A^\mu - \partial^\mu\lambda\ \ \text{for any}\ \ \Box\lambda = 0 \qquad (4)

where: $\lambda$ is any scalar field obeying the homogeneous wave equation. Physical reading: the Lorenz gauge does not fix the potential — only its four-divergence. A residual gauge function survives, and it is exactly enough to remove one more component of $A^\mu$ . That is where the photon's polarisation count comes from.

Reducing Maxwell to a wave equation

Derivation AFrom ∂_μF^{μν} = μ₀J^ν to □A^ν = μ₀J^ν — and why the gauge choice must be a scalar one

The goal is to take the covariant inhomogeneous Maxwell equation, substitute the potential, and watch four coupled equations fall apart into four independent ones. The manoeuvre turns on a single term, $\partial^\nu(\partial_\mu A^\mu)$ , and on our being allowed to make it vanish. What deserves attention is not that we can set $\partial_\mu A^\mu = 0$ , but that this condition is a Lorentz scalar equation — and that a non-scalar gauge, however convenient, would wreck the covariance we are trying to exhibit.

Step 1. Insert the potential. With $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , \partial_\mu F^{\mu\nu} = \partial_\mu\partial^\mu A^\nu - \partial_\mu\partial^\nu A^\mu = \mu_0 J^\nu . Why: nothing is assumed yet — an identity, true in any gauge. The construction of $F^{\mu\nu}$ from $A^\mu$ belongs to `aq-sr-electromagnetic-field-tensor`; we use it, we do not rebuild it.

Step 2. Factor the second term. Partials commute, so $\partial_\mu\partial^\nu A^\mu = \partial^\nu(\partial_\mu A^\mu)$ : \Box A^\nu - \partial^\nu\big(\partial_\mu A^\mu\big) = \mu_0 J^\nu . Why: this is the coupling term, and the only obstacle. It ties the evolution of $A^0$ to the divergence of $\vec A$ and back. Every component of $A^\mu$ sits inside it, which is exactly what stops the four equations separating.

Step 3. Impose the Lorenz gauge. Gauge freedom shifts $A^\mu \to A^\mu - \partial^\mu\lambda$ without touching $F^{\mu\nu}$ , and shifts the divergence by $\partial_\mu A^\mu \to \partial_\mu A^\mu - \Box\lambda$ . To reach $\partial_\mu A^\mu = 0$ from any starting potential we need only solve $\Box\lambda = \partial_\mu A^\mu$ — an inhomogeneous wave equation, which always has a solution. The gauge is reachable from anywhere. With it imposed the coupling term dies: \boxed{\;\Box A^\nu = \mu_0 J^\nu\;} Why: reachability is not a detail. If $\Box\lambda = \partial_\mu A^\mu$ had no solutions the Lorenz gauge would be an empty wish. It has them because $\Box$ is invertible on suitable sources — the very fact that makes retarded potentials exist.

Step 4. Check the choice was covariant. $\partial_\mu A^\mu$ contracts a lower index with an upper one: a Lorentz scalar. So $\partial'_\mu A'^\mu = \partial_\mu A^\mu = 0$ in every frame at once, from one imposition. Why: the Lorenz gauge is the common choice that is a scalar condition, which is exactly why it is the one that yields (1). The Coulomb gauge $\nabla\cdot\vec A = 0$ constrains the spatial part alone: it singles out a time axis, it is not a scalar, and a boost destroys it.

Step 5. Read off the three-vector content. With $\mu_0 = 1/(\varepsilon_0 c^2)$ and $J^\mu = (c\rho, \vec J)$ , the $\nu = 0$ component of (1) is $\Box(\phi/c) = \mu_0 c\rho$ , i.e. \Box\phi = \frac{\rho}{\varepsilon_0},\qquad\text{and}\qquad \Box\vec A = \mu_0\vec J . Why: the two equations of ordinary electrodynamics — but now inside one four-vector equation, and their relative normalisation, which looks arbitrary in three-dimensional language, is fixed by nothing more than $J^\mu$ being a four-vector. *Verified in verify_aq-sr-wave-equation-four-potential.py: $F^{\mu\nu}$ is rebuilt symbolically from $A^\mu = (\phi/c,\vec A)$ with arbitrary functions of $(t,x,y,z)$ ; the identity $\partial_\mu F^{\mu\nu} = \Box A^\nu - \partial^\nu(\partial_\mu A^\mu)$ is confirmed component by component; and the reduction to $\Box\phi = \rho/\varepsilon_0$ , $\Box\vec A = \mu_0\vec J$ is checked.*

Go deeper · Derivation B · 2 minOwned result: the speed of light is an output, and the null condition makes it frame-independent

Derivation BOwned result: the speed of light is an output, and the null condition makes it frame-independent

Set $J^\mu = 0$ and ask what the equation permits. The answer is the most consequential deduction in classical physics, and it splits into two claims usually run together. Claim A: the vacuum equation has wave solutions of speed $c$ . Claim B: every inertial observer measures that same speed. Claim A is nineteenth-century physics and needs only the operator. Claim B needs the operator and the fact that $k_\mu k^\mu$ is a scalar — and it is the whole of Einstein's second postulate, arrived at without postulating anything at all.

Step 1. Substitute a plane wave. Put $A^\mu = \varepsilon^\mu e^{-ik_\nu x^\nu}$ with $k^\mu = (\omega/c, \vec k)$ , so $k_\nu x^\nu = \omega t - \vec k\cdot\vec x$ . Each $\partial_\mu$ brings down $-ik_\mu$ , so $\Box \to -k_\mu k^\mu$ : \Box A^\mu = 0 \quad\Longrightarrow\quad k_\mu k^\mu\,\varepsilon^\mu = 0 \quad\Longrightarrow\quad k_\mu k^\mu = 0 for any non-zero polarisation. Why: the wave four-vector is forced onto the light cone in momentum space — and note what did the forcing: the operator, nothing else.

Step 2. Unpack the null condition. $k_\mu k^\mu = \omega^2/c^2 - |\vec k|^2 = 0$ , hence \omega = c\,|\vec k| ,\qquad v_{\text{phase}} = \frac{\omega}{|\vec k|} = c ,\qquad v_{\text{group}} = \frac{\mathrm{d}\omega}{\mathrm{d}|\vec k|} = c . Why: phase and group velocity coincide, both equal to $c$ . The vacuum is non-dispersive — so familiar it is easy to forget it is a derived property of the operator, and one that fails the moment the wave enters glass.

Step 3. Now the genuinely relativistic part. $k_\mu k^\mu$ contracts a four-vector with itself: a Lorentz scalar. Zero in one frame, zero in all. Boost along $x$ : \omega' = \gamma(\omega - \beta c k_x),\qquad c k'_x = \gamma(ck_x - \beta\omega),\qquad k'_y = k_y,\ k'_z = k_z, and direct computation gives $\omega'^2/c^2 - |\vec k'|^2 = \omega^2/c^2 - |\vec k|^2$ . Why: $\omega$ changes. $|\vec k|$ changes. Their ratio does not. That is the relativistic Doppler effect and the invariance of $c$ — the same statement seen from two sides.

Step 4. State it sharply. A moving observer sees a different frequency and a different wavelength — that is Doppler — but finds $\omega' = c|\vec k'|$ with the same numerical $c$ . \boxed{\;k_\mu k^\mu = 0 \ \text{in one frame} \;\Longleftrightarrow\; k_\mu k^\mu = 0 \ \text{in all frames}\;} Why: this is Einstein's second postulate, and here it is a theorem. Einstein made the invariance of $c$ an axiom in 1905 because he was prepared to give up Newtonian kinematics to keep Maxwell. The logic here runs the other way: grant Maxwell and the Minkowski metric, and the invariance of $c$ is not an assumption — it is a line of algebra you cannot avoid. *Verified in verify_aq-sr-wave-equation-four-potential.py: the boost matrix is applied to $k^\mu$ and $k'_\mu k'^\mu - k_\mu k^\mu$ is shown to vanish identically in $\beta$ ; $\omega = c|\vec k|$ is confirmed preserved for boosts up to $\beta = 0.999999$ in 50-digit arithmetic, where float64 silently fails.*

Go deeper · Derivation C · 2 minOwned result: 4 − 1 − 1 = 2. Counting the polarisations of light out of the gauge freedom

Derivation COwned result: 4 − 1 − 1 = 2. Counting the polarisations of light out of the gauge freedom

The four-potential has four components. Light has two polarisations. The gap is not swept under a rug — it is closed exactly, by two subtractions, and the second is the residual gauge freedom the Lorenz condition famously fails to remove. This is the cleanest place in classical electromagnetism to see a degree-of-freedom count done honestly, and it is the classical shadow of the statement that the photon is massless and has two helicities.

Step 1. The Lorenz condition removes one. Applying $\partial_\mu A^\mu = 0$ to a plane wave $A^\mu = \varepsilon^\mu e^{-ik\cdot x}$ gives k_\mu \varepsilon^\mu = 0 , one linear constraint on the four-component polarisation vector, leaving three. Why: the polarisation must be orthogonal to the propagation direction — in the four-dimensional sense. For null $k$ this is weaker than it looks, and that is the crux of the next step.

Step 2. The residual gauge removes another. Inside the Lorenz gauge we may still shift $A^\mu \to A^\mu - \partial^\mu\lambda$ provided $\Box\lambda = 0$ . Taking $\lambda = i\alpha\, e^{-ik\cdot x}$ — which satisfies $\Box\lambda = 0$ precisely because $k_\mu k^\mu = 0$ — shifts the polarisation by \varepsilon^\mu \;\to\; \varepsilon^\mu + \alpha\, k^\mu . Why: the change is proportional to $k^\mu$ itself, so any part of $\varepsilon^\mu$ along $k^\mu$ is pure gauge — set it to zero by choosing $\alpha$ . Note the economy: the residual gauge exists only because $k$ is null, the same condition that gave us $c$ .

Step 3. Count. Four components, minus one from $k\cdot\varepsilon = 0$ , minus one from $\varepsilon \sim \varepsilon + \alpha k$ : 4 - 1 - 1 = \boxed{\;2\;} and the survivors are the two directions genuinely transverse to $\vec k$ . Why: the two transverse polarisations of light, obtained by subtraction. Use the residual freedom to set $\varepsilon^0 = 0$ ; then $k_\mu\varepsilon^\mu = 0$ collapses to $\vec k\cdot\vec\varepsilon = 0$ — ordinary transversality, now visible as the leftover of a four-dimensional condition.

Step 4. Ask what a photon mass would do. A massive vector field obeys $(\Box + \kappa^2)A^\mu = 0$ with $\kappa = mc/\hbar$ , so $k_\mu k^\mu = \kappa^2 \ne 0$ — and Step 2 fails, because it required $k$ to be null. The second subtraction is lost: $4 - 1 = 3$ . Why: a massive spin-1 particle has three polarisations; a massless one has two. Light's missing longitudinal mode is not missing by accident — it is missing because $k_\mu k^\mu = 0$ , i.e. because $\Box$ has no mass term, i.e. because $c$ is the invariant speed. The polarisation count, the photon's masslessness and the constancy of $c$ are three names for one fact. *Verified in verify_aq-sr-wave-equation-four-potential.py: the shift $\varepsilon^\mu\to\varepsilon^\mu + \alpha k^\mu$ is shown to preserve both $\Box A^\mu = 0$ and $k\cdot\varepsilon = 0$ ; and the rank of the surviving polarisation space is computed to be exactly 2 for null $k$ and exactly 3 for timelike $k$ (the massive case).*

Limiting cases

Case	Equation	What it becomes
Vacuum, $J^\mu = 0$	$\Box A^\mu = 0$	plane waves, $k_\mu k^\mu = 0$ , $\omega = c\lvert\vec k\rvert$ — light
Static, $\partial_t \to 0$	$-\nabla^2 A^\mu = \mu_0 J^\mu$	Poisson: $\nabla^2\phi = -\rho/\varepsilon_0$ ⇒ Coulomb; $\nabla^2\vec A = -\mu_0\vec J$ ⇒ Biot–Savart
Point source, retarded	$A^\mu = \frac{\mu_0 c}{4\pi}\frac{q\,u^\mu}{u_\nu(x-w)^\nu}\big\vert_{\text{ret}}$	the Liénard–Wiechert potential (`aq-sr-lienard-wiechert-potentials`)
$c\to\infty$ (formal Galilean)	$-\nabla^2 A^\mu = \mu_0 J^\mu$ everywhere	instantaneous action at a distance; no radiation, no causal cone
Euclidean signature	$\nabla_4^2 A^\mu = 0$	elliptic — harmonic functions, no waves at all
Massive vector field	$(\Box + \kappa^2)A^\mu = 0$	$k_\mu k^\mu = \kappa^2 \ne 0$ : dispersive, three polarisations

Common trapLorenz is not Lorentz. Two men, two nationalities, thirty years apart — and the coincidence is genuinely unfortunate.

The gauge condition $\partial_\mu A^\mu = 0$ is the Lorenz gauge, named for Ludvig Valentin Lorenz (1829–1891), the Danish physicist who introduced it in the 1860s while writing electromagnetic theory in terms of retarded potentials. It is not named for Hendrik Antoon Lorentz (1853–1928), the Dutchman of the Lorentz transformation, force and factor. The confusion is compounded by an accident that could hardly be more perverse: the Lorenz gauge happens to be Lorentz invariant. The literature is thoroughly polluted with "Lorentz gauge", distinguished textbooks included.

Keep them straight by content, not spelling. Lorenz gave a gauge choice — a convention with no physical consequence. Lorentz gave a coordinate transformation — a physical statement about spacetime.

Common trapThe Lorenz gauge does NOT fix the potential — and if you forget that, you lose the photon's polarisation count.

It is tempting to read $\partial_\mu A^\mu = 0$ as "the gauge has been fixed" and move on. It has not. The condition constrains one scalar combination of the four components, and leaves a residual gauge freedom: any $\lambda$ with $\Box\lambda = 0$ generates a further $A^\mu \to A^\mu - \partial^\mu\lambda$ that preserves the Lorenz condition, since $\partial_\mu A^\mu \to \partial_\mu A^\mu - \Box\lambda = \partial_\mu A^\mu$ . This is not a wart. It is load-bearing. The residual freedom is exactly the second subtraction in $4 - 1 - 1 = 2$ ; without it the theory predicts three polarisations instead of two. A reader who believes the gauge is fully fixed has no way to remove the longitudinal mode, and will conclude that free electromagnetic waves have a component along $\vec k$ . They do not. The practical consequence: you may impose one more condition — most usefully $\varepsilon^0 = 0$ , the radiation gauge — precisely because the Lorenz condition left you the room.

Worked examples

1 · The speed of light, computed from a capacitor and a coil

The vacuum wave equation gives phase speed $v = 1/\sqrt{\mu_0\varepsilon_0}$ . Evaluate it. Neither constant was measured with light.

Step 1 — where the speed comes from. With $\Box = \frac{1}{c^2}\partial_t^2 - \nabla^2$ and $c^2 \equiv 1/(\mu_0\varepsilon_0)$ , the vacuum equation $\Box\vec A = 0$ reads \mu_0\varepsilon_0\,\frac{\partial^2\vec A}{\partial t^2} = \nabla^2\vec A , a wave equation of speed $v = 1/\sqrt{\mu_0\varepsilon_0}$ . $\mu_0$ is fixed by the force between current-carrying wires; $\varepsilon_0$ by the force between static charges. Neither has anything to do with optics.

Step 2 — the numbers. With $\mu_0 = 4\pi\times10^{-7}\ \mathrm{H\,m^{-1}} = 1.25663706\times10^{-6}$ and $\varepsilon_0 = 8.8541878128\times10^{-12}\ \mathrm{F\,m^{-1}}$ : \mu_0\varepsilon_0 = 1.11265006\times10^{-17}\ \mathrm{s^2\,m^{-2}},\qquad v = \frac{1}{\sqrt{1.11265006\times10^{-17}}} = \mathbf{2.99792458\times10^{8}\ m\,s^{-1}} .

Step 3 — compare with light. The measured speed of light is $c = 2.99792458\times10^{8}\ \mathrm{m\,s^{-1}}$ — agreement to all nine figures. That is not a coincidence being reported; it is the reason we believe light is an electromagnetic wave. Maxwell's own version of the comparison, in 1865, was good to about one percent, and he drew the conclusion immediately.

Step 4 — the modern twist. Since the 2019 SI redefinition the logic runs backwards: $c$ is defined as exactly $299\,792\,458\ \mathrm{m\,s^{-1}}$ , and $\mu_0 = 1.25663706212(19)\times10^{-6}\ \mathrm{H\,m^{-1}}$ is now measured. The identity $c = 1/\sqrt{\mu_0\varepsilon_0}$ still holds exactly; what changed is which two of the three are handed to you and which you compute. The wave equation fixes the relation; the SI decides which end to nail down.

2 · A green laser, boosted — the frequency changes, the null condition does not

A $532\ \mathrm{nm}$ laser propagates along $+x$ . Compute $k^\mu$ , verify $k_\mu k^\mu = 0$ , boost to $\beta = 0.600$ along $+x$ , and verify the null condition survives.

Step 1 — the lab wave four-vector. $|\vec k| = 2\pi/\lambda = 2\pi/(532\times10^{-9}) = \mathbf{1.181050\times10^{7}\ m^{-1}}$ , so $\omega = c|\vec k| = \mathbf{3.540698\times10^{15}\ rad\,s^{-1}}$ and $\nu = \omega/2\pi = \mathbf{5.635197\times10^{14}\ Hz}$ . Hence k^\mu = \left(\frac{\omega}{c},\ k_x,\ 0,\ 0\right) = \big(1.181050\times10^{7},\ 1.181050\times10^{7},\ 0,\ 0\big)\ \mathrm{m^{-1}} .

Step 2 — the null check. k_\mu k^\mu = \frac{\omega^2}{c^2} - k_x^2 = (1.181050\times10^{7})^2 - (1.181050\times10^{7})^2 = \mathbf{0} \quad\text{exactly.} The time and space components are numerically equal. That is what "null" means — the arithmetic signature of a massless field.

Step 3 — boost at $\beta = 0.600$ along $+x$ . $\gamma = 1.2500$ , and both components transform the same way: \frac{\omega'}{c} = \gamma\left(\frac{\omega}{c} - \beta k_x\right) = k'_x = \gamma\left(k_x - \beta\frac{\omega}{c}\right) = 1.25\,(1 - 0.600)\times1.181050\times10^{7} = \mathbf{5.905249\times10^{6}\ m^{-1}} .

Step 4 — read the physics off. Both were multiplied by the same factor, \gamma(1-\beta) = \sqrt{\frac{1-\beta}{1+\beta}} = \sqrt{\frac{0.400}{1.600}} = \mathbf{0.500} , the relativistic Doppler factor. The observer running away at $0.6c$ sees the light redshifted by exactly two: $\lambda' = 2\times532 = \mathbf{1064\ nm}$ , $\nu' = \mathbf{2.817598\times10^{14}\ Hz}$ . And $1064\ \mathrm{nm}$ is the Nd:YAG fundamental whose frequency-doubled output is the $532\ \mathrm{nm}$ line we started from — a coincidence of the numbers chosen, but a memorable one.

Step 5 — the invariant. k'_\mu k'^\mu = (5.905249\times10^{6})^2 - (5.905249\times10^{6})^2 = \mathbf{0} . Both components changed by 50%. Their difference of squares did not change at all — it was zero and it stayed zero. Since $\omega' = c|\vec k'|$ follows immediately, the boosted observer measures the same $c$ from a wave whose colour they disagree about. That disagreement is the Doppler effect; that agreement is the second postulate; equation (3) delivers both at once.

Figure

THE WAVE FOUR-VECTOR SLIDES ALONG THE LIGHT CONE AND NEVER LEAVES IT. A 532 nm laser has k^μ = (1.181050×10⁷, 1.181050×10⁷, 0, 0) m⁻¹ — the time and space components are EQUAL, which is exactly what k_μk^μ = ω²/c² − |k|² = 0 means. Boost to β = 0.600 and BOTH components shrink by the Doppler factor √((1−β)/(1+β)) = 0.500: the wavelength stretches 532 nm → 1064 nm, the frequency drops 5.635197×10¹⁴ Hz → 2.817598×10¹⁴ Hz. The observers disagree about the colour of the light. They do NOT disagree about c, because a null vector rescaled is still null — the arrow slides ALONG the 45° line and cannot come off it. Every massive particle's four-momentum sits strictly INSIDE the cone and can be boosted to rest; light sits ON it and cannot. That is the whole difference.

Physics of the portrait

The portrait is an operator, not a person — because here the operator is the discovery.

Origins · 2 minD'Alembertian and how this was discovered

D'Alembertian

1747–present · the wave operator that turned out to be the metric · why the minus sign is the whole of light

The d'Alembertian $\Box = \partial_\mu\partial^\mu = \frac{1}{c^2}\partial_t^2 - \nabla^2$ is the Lorentz-invariant wave operator, and it carries the name of Jean le Rond d'Alembert, who in 1747 wrote the one-dimensional wave equation for a vibrating string and solved it as a superposition of a left-moving and a right-moving disturbance. That was a hundred and fifty years before anyone suspected the operator had anything to do with the geometry of spacetime. His equation was about strings; the "speed" in it was a property of tension and mass per unit length; and there was nothing invariant about it whatsoever — a wave on a string looks slower if you run alongside it, and that is the end of the matter. What relativity revealed is that the same operator, applied to the electromagnetic potential, is not a statement about a medium at all. It is $\eta^{\mu\nu}\partial_\mu\partial_\nu$ — the Minkowski metric with derivatives in its slots — and its "wave speed" is a property of spacetime, not of any substance, which is why no amount of running alongside changes the answer. The relative minus sign, which for d'Alembert was just what falls out of separating variables, is in Minkowski's reading the signature itself, and it is the sole reason the equation is hyperbolic — the reason its solutions have a light cone, a finite propagation speed, and somewhere for radiation to go. Everything that makes electromagnetism a theory of light rather than a theory of statics is encoded in one minus sign, and d'Alembert put it there ninety years before Maxwell and a hundred and sixty before Minkowski explained what it meant.

History

1747

— d'Alembert publishes the wave equation for a vibrating string, solved as a sum of left- and right-travelling disturbances.

1865

— Maxwell, in "A Dynamical Theory of the Electromagnetic Field", derives waves travelling at 1/√(μ₀ε₀) and identifies light as an electromagnetic wave.

1867

— Ludvig Lorenz publishes the retarded-potential formulation, introducing the gauge condition that now bears his name.

1873

— Maxwell's "Treatise on Electricity and Magnetism" presents the theory in full.

1905

— Einstein takes the invariance of c as a postulate rather than a puzzle, and rebuilds kinematics around it.

1908

— Minkowski's "Raum und Zeit" identifies the wave operator as the metric contraction ∂_μ∂^μ, making the covariance of Maxwell's theory manifest.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §10.1 and §12.3 · Jackson, Classical Electrodynamics, 3rd ed., ch. 6 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Landau & Lifshitz, The Classical Theory of Fields, 4th ed., §46–47.

Provenance. Signature $(+,-,-,-)$ , hence $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass and no velocity-dependent mass appears anywhere. The dictionary $A^\mu = (\phi/c,\vec A)$ , $\partial_\mu = (\tfrac1c\partial_t,\nabla)$ , $\partial^\mu = (\tfrac1c\partial_t,-\nabla)$ is used throughout, and $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is rebuilt from it in the verification rather than assumed. This article's base declares the correct signature and its physics is sound — one of the two clean bases in this chapter — but its degree-of-freedom count is absent, and is supplied and machine-checked here. Every number printed on this page is recomputed in verify_aq-sr-wave-equation-four-potential.py in `mpmath` at 50 digits, against this article's own printed digits.

Scope. The four-potential belongs to `aq-sr-four-potential`; gauge freedom and the Lorenz condition to `aq-sr-gauge-transformations-lorenz-gauge`; $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ to `aq-sr-covariant-maxwell-equations`; the four-current to `aq-sr-four-current-density`. The retarded solution for a point charge is developed in `aq-sr-lienard-wiechert-potentials`, the energy those waves carry in `aq-sr-electromagnetic-stress-energy-tensor`. This page owns the reduction of Maxwell to $\Box A^\mu = \mu_0 J^\mu$ , the derivation of $c$ from the null condition, and the $4-1-1=2$ polarisation count. A parallel treatment from the electromagnetism track lives at `aq-em-covariant-wave-equation`.

Builds on. `aq-sr-four-potential` · `aq-sr-gauge-transformations-lorenz-gauge` · `aq-sr-covariant-maxwell-equations` · `aq-sr-four-current-density`. Leads to. `aq-sr-lienard-wiechert-potentials` · `aq-sr-electromagnetic-stress-energy-tensor` · `aq-sr-covariant-lorentz-force`.

Is the appearance of c in the wave equation really a "derivation" of the speed of light, or is it circular?

It is not circular. $\mu_0$ and $\varepsilon_0$ are fixed by static experiments — the force between current-carrying wires, the capacitance of a parallel-plate capacitor. No light is involved in measuring either. The wave equation then predicts, with no further input, waves of speed $1/\sqrt{\mu_0\varepsilon_0}$ — and that number equals the independently measured speed of light. A prediction confirmed, not a definition unpacked. What is conventional in the modern SI is which of $\{c, \mu_0, \varepsilon_0\}$ you declare exact; the relation among them is physics, and was physics in 1865.

Why is the Lorenz gauge "better" than the Coulomb gauge? Are the physical fields different?

The physical fields are identical — that is what gauge invariance means. The Lorenz gauge is better for this purpose because it is a scalar condition: impose it once, it holds in every frame, and $\Box A^\mu = \mu_0 J^\mu$ is manifestly covariant. In the Coulomb gauge $\nabla\cdot\vec A = 0$ the scalar potential obeys $\nabla^2\phi = -\rho/\varepsilon_0$ , which looks like instantaneous action at a distance. It is not — the apparent instantaneity cancels against a compensating term in $\partial_t\vec A$ — but the cancellation is invisible, whereas in the Lorenz gauge causality is manifest term by term. Use Coulomb gauge to count photons; use Lorenz gauge to see relativity.

If A^μ has four components and light has two polarisations, where do the other two go?

One is eaten by the Lorenz condition, $k_\mu\varepsilon^\mu = 0$ . The other by the residual gauge freedom: because $k$ is null, $\lambda \propto e^{-ik\cdot x}$ satisfies $\Box\lambda = 0$ , so $\varepsilon^\mu \to \varepsilon^\mu + \alpha k^\mu$ changes nothing physical — the part of $\varepsilon^\mu$ along $k^\mu$ is pure gauge. Two genuinely transverse directions remain. The count is exact and it is not a coincidence: it is the classical version of a massless spin-1 particle having two helicities, and it fails for a massive vector field precisely because $k_\mu k^\mu \ne 0$ breaks the residual-gauge construction.

Does the Lorenz gauge condition constrain the physics, or only the bookkeeping?

Only the bookkeeping — with one caveat. Imposing $\partial_\mu A^\mu = 0$ is always possible (solve $\Box\lambda = \partial_\mu A^\mu$ ), so it excludes no physical configuration. But it is not automatic: consistency requires $\partial_\mu J^\mu = 0$ . Take $\partial_\mu$ of equation (1): $\Box(\partial_\mu A^\mu) = \mu_0\,\partial_\mu J^\mu$ . If charge were not conserved the right-hand side would not vanish, the Lorenz condition would not propagate, and the equation would be inconsistent. The wave equation and charge conservation are not independent statements — see `aq-sr-charge-invariance-continuity`.

Why "null" rather than just "the wave travels at c"? What does the language buy?

It buys frame-independence, which "travels at $c$ " does not obviously have. "The wave travels at $c$ " is a statement about a speed, and speeds are frame-dependent in every other context — which is exactly why the constancy of $c$ was a crisis rather than a triviality. " $k_\mu k^\mu = 0$ " says a scalar vanishes, and scalars are the same for everyone by construction. The null language does not merely restate the fact; it explains why the fact is frame-independent — and it transfers immediately to any field whose quanta live on the light cone.

The base article says the retarded solution "expresses causality". Is causality an input or an output here?

Both, in different senses. The light cone is an output: forced by the hyperbolic character of $\Box$ , which is forced by the signature. But choosing the retarded Green function over the advanced one is an input — $\Box G = \delta^4$ has both solutions, and the equation prefers neither. We impose retardation because effects follow causes. The wave equation gives you a cone; it does not tell you which way to go round it. That choice is developed in `aq-sr-lienard-wiechert-potentials`, whose invariant Green function $G_{\text{ret}} \propto \theta(\Delta t)\,\delta\big((\Delta x)^2\big)$ carries the step function that does the choosing.