Gauge Transformations and Lorenz Gauge

Take a proton, sitting still. Its potential is the Coulomb one — $\varphi = kq/r$ , $\vec A = \vec 0$ — and it happens to satisfy two famous gauge conditions at once. The Lorenz condition $\tfrac{1}{c^2}\partial_t\varphi + \nabla\!\cdot\!\vec A = 0$ : yes, both terms vanish. The Coulomb condition $\nabla\!\cdot\!\vec A = 0$ : yes, trivially, since $\vec A$ is zero. Now walk past at $0.6c$ and look again. Same physical object, boosted. And a nanometre from the proton the two conditions come apart:

\nabla'\!\cdot\!\vec A' \;=\; -2.882\ \mathrm{T}, \qquad \frac{1}{c^2}\frac{\partial\varphi'}{\partial t'} \;=\; +2.882\ \mathrm{T}, \qquad \partial_\mu A'^\mu \;=\; \mathbf{0} .

The Coulomb gauge is broken — by nearly three tesla, which is not a rounding error. The Lorenz gauge survives exactly, and it survives because the two terms are equal and opposite to the last digit. Not a coincidence and not a lucky example: $\partial_\mu A^\mu$ is a Lorentz scalar — every index contracted — so if it vanishes in one frame it vanishes in all of them. $\nabla\!\cdot\!\vec A$ is not a scalar; it is three-quarters of one, and the missing quarter is exactly the piece a boost mixes back in. That is the article. Gauge freedom itself — that you may replace $A^\mu$ by $A^\mu + \partial^\mu\chi$ anywhere, everywhere, at any time, and not one measurable quantity moves — looks at first like an embarrassment: a theory with more variables than physics. It is instead the deepest structure in the subject. Demand that the redundancy be local, and electromagnetism is not merely permitted but forced into existence. And a warning that belongs in the first paragraph, because the literature gets it wrong constantly: the Lorenz gauge is named for Ludvig Lorenz, a Dane, who published it in 1867. It is not named for Hendrik Lorentz, the Dutchman of the Lorentz transformation — who turned fourteen that year.

Key takeaways

Signature $(+,-,-,-)$ ; $\;\partial_\mu = \big(\tfrac1c\partial_t,\ +\nabla\big)$ , $\;\partial^\mu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ ; $\;p\cdot p = +m^2c^2$ .
The freedom: $\boxed{A^\mu \to A^\mu + \partial^\mu\chi}$ for any smooth $\chi$ , and $F^{\mu\nu}$ is identically unchanged — because $\partial^\mu\partial^\nu\chi = \partial^\nu\partial^\mu\chi$ . Not approximately; identically. No experiment at any precision can see $\chi$ .
In three-vector language: $\varphi \to \varphi + \partial_t\chi$ and $\vec A \to \vec A - \nabla\chi$ — note the minus, forced by $\partial^i = -\partial_i$ . (The familiar $\varphi \to \varphi - \partial_t\lambda$ , $\vec A \to \vec A + \nabla\lambda$ is the same map with $\lambda = -\chi$ . The base text mixes the two and is internally inconsistent.)
Owned result — the Lorenz gauge is the only standard gauge every inertial frame agrees on. $\partial_\mu A^\mu$ is a scalar, so $\partial_\mu A^\mu = 0$ holds in every frame at once. $\nabla\!\cdot\!\vec A = 0$ is not: boost a Coulomb-gauge potential and it leaves the Coulomb gauge, by $-2.882\ \mathrm{T}$ .
You can always get there: given $\partial_\mu A^\mu = f$ , solve $\Box\chi = -f$ — always soluble. And a residual freedom survives: any $\chi$ with $\boxed{\Box\chi = 0}$ preserves the Lorenz condition. Hence $4 - 1 - 1 = 2$ — the two transverse photon polarisations.
Key idea: gauge freedom is not slack we tolerate. It is the reason for the theory. Insist that a convention — the phase of the electron's wavefunction — be choosable independently at every point, and you are forced to invent a compensating field transforming as $A^\mu \to A^\mu + \partial^\mu\chi$ . That field is the photon. Electromagnetism is the price of a local convention.

The core insight

$F^{\mu\nu}$ is the physics; $A^\mu$ is a coordinate on the space of potentials that produce it. All the $A^\mu$ related by $A^\mu \to A^\mu + \partial^\mu\chi$ form one gauge orbit, and every point on it describes the identical physical world. A gauge condition is a rule for picking one representative per orbit — a slice through the orbits. The question this article answers is which slice a boost respects. Lorenz's slice, $\partial_\mu A^\mu = 0$ , is cut by a scalar equation, so a boost maps it to itself. Coulomb's slice, $\nabla\!\cdot\!\vec A = 0$ , is cut by an equation that names a frame, so a boost drags a potential straight off it.

Where this lives today

The gauge principle is the organising idea of modern physics, and it runs in the opposite direction to the one this article took. We started with $A^\mu$ and discovered a redundancy. The modern construction starts with the redundancy and derives $A^\mu$ : postulate that the phase of the electron's wavefunction, $\psi \to e^{i\theta}\psi$ , is a convention choosable independently at every point of spacetime, and the free Dirac equation immediately fails to be invariant. Repairing it requires a compensating field transforming as $A^\mu \to A^\mu + \partial^\mu\chi$ — and there is only one way to do it. Electromagnetism is not discovered; it is deduced, and the photon's masslessness, the universality of its coupling, and charge conservation all fall out of the same demand. Yang and Mills generalised the argument in 1954 by replacing the phase $e^{i\theta}$ (the group $U(1)$ ) with a non-commuting group, and the Standard Model is the result: $SU(3) \times SU(2) \times U(1)$ , whose gauge fields are the eight gluons, the $W$ and $Z$ , and the photon. And the Lorenz gauge remains the working choice wherever covariance matters. Every calculation of radiation uses it, because it turns Maxwell's equations into $\Box A^\mu = \mu_0 J^\mu$ with its retarded solution; quantum electrodynamics is quantised in it — the Gupta–Bleuler and covariant-gauge formalisms — because the alternatives force you to choose a frame and then defend the choice. Ludvig Lorenz's 1867 condition, written to make retarded potentials work, turned out to be the only one Einstein's 1905 postulates would have permitted anyway.

Origins · 2 minHermann Weyl and how this was discovered

Hermann Weyl

1885–1955 · Eichinvarianz 1918 · the phase reinterpretation · the gauge principle

The word "gauge" is a fossil of a beautiful mistake. In 1918, in "Gravitation und Elektrizität," Hermann Weyl tried to unify gravity and electromagnetism by proposing that the scale of a measuring rod — its Eichung, its calibration or gauge — might be a local convention, choosable independently at each point of spacetime, with electromagnetism as the compensating field keeping the geometry consistent. He called the symmetry Eichinvarianz, and the name has stuck to every theory built this way since. As physics it was wrong, and Einstein said so almost immediately. If lengths really depended on the path taken through an electromagnetic field, two identical atoms with different histories would have different sizes — and therefore different spectral lines. Atomic spectra are sharp. Weyl's theory predicted they should not be, and he eventually conceded the point. The repair came from quantum mechanics, and it was not Weyl's. In 1926 Vladimir Fock noticed that the Schrödinger equation for a charged particle is invariant under a local phase change of the wavefunction provided the potential shifts by a gradient; in 1927 Fritz London saw that this was Weyl's idea exactly, with the scale factor made imaginary — a rotation of phase rather than a stretching of length. Nothing measurable stretches, so Einstein's objection evaporates: a phase is not a size. Only in 1929, in "Elektron und Gravitation," did Weyl adopt the quantum reading, recasting his symmetry as local $U(1)$ phase invariance and giving the gauge principle its definitive form — which Pauli's 1941 review then made standard. The credit therefore divides, and it is worth dividing honestly. The idea of a local, arbitrarily choosable convention is Weyl's, from 1918, and it was wrong in its first incarnation. The correct reading — that the convention is a phase, not a scale — is Fock's and London's. The synthesis Yang and Mills would generalise in 1954 is Weyl's again, from 1929. The misleading name he coined in the wrong theory is the one we still use.

History

1867

— Ludvig Lorenz (Danish), in "On the Identity of the Vibrations of Light with Electrical Currents," introduces the retarded potentials and the condition now called the Lorenz gauge. Hendrik Lorentz, of the Lorentz transformation, turns fourteen that year.

1918

— Hermann Weyl, in "Gravitation und Elektrizität," proposes that the scale of lengths is a local convention and coins Eichinvarianz — gauge invariance. Einstein objects at once: path-dependent lengths would smear atomic spectra, which are sharp.

1926

— Vladimir Fock shows that the Schrödinger equation for a charged particle is invariant under a local change of the wavefunction's phase, provided the potential shifts by a gradient.

1927

— Fritz London identifies Fock's phase symmetry as Weyl's idea with the scale factor made imaginary: a rotation of phase, not a stretching of length. Einstein's objection dissolves.

1929

— Weyl, in "Elektron und Gravitation," adopts the quantum reading and recasts his symmetry as local U(1) phase invariance, giving the gauge principle its modern form.

1941

— Pauli's review article establishes the modern terminology and makes gauge invariance standard.

1954

— Yang and Mills generalise the gauge principle to non-commuting groups; the Standard Model is built on the result.

The physical picture

The potentials are not unique, and it is easy to say exactly how. Take any smooth $\chi(t,\vec x\,)$ — literally any — and shift the four-potential by its four-gradient. The field tensor is a four-curl, $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , and a curl annihilates a gradient: the extra piece is $\partial^\mu\partial^\nu\chi - \partial^\nu\partial^\mu\chi$ , which vanishes identically because partial derivatives commute. So $\vec E$ , $\vec B$ , every force, every energy: unchanged. Heaviside thought this made the potentials fraudulent. The modern reading is the reverse. The redundancy has a shape, and the shape is a symmetry — one with an arbitrary function's worth of parameters at every point of spacetime, a vastly bigger object than the handful of parameters in a rotation. Symmetries that large are not decorations; they are constraints. The price of redundancy is that you must eventually choose. Gauge fixing is picking one $A^\mu$ per physical situation, and the standard choices differ enormously in convenience — see the table below. None of them changes the physics. All of them change how much work you do. And exactly one of them is a statement every observer in the universe can agree to without re-deriving it in their own frame.

Figure

PUNCHLINE — the same potential, boosted. LEFT: a proton at rest, whose Coulomb potential satisfies BOTH gauge conditions at once. RIGHT: the same potential seen at 0.6c, one nanometre out. The Coulomb condition is broken by −2.882 T. The Lorenz condition survives EXACTLY, because its two terms come out at ∓2.882 T and cancel to the last digit — which they must, since ∂ₘAᵐ is a Lorentz SCALAR while ∇·A is not

The core equations

\boxed{\;A^\mu \;\longrightarrow\; A'^\mu \;=\; A^\mu + \partial^\mu\chi\;} \qquad\Longrightarrow\qquad \boxed{\;F'^{\mu\nu} = F^{\mu\nu}\;} \qquad (1)

where: $\chi(t,\vec x\,)$ — any smooth scalar function. Physical reading: substitute into $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ and the extra term is $\partial^\mu\partial^\nu\chi - \partial^\nu\partial^\mu\chi = 0$ , because mixed partials commute. An identity, not an approximation. Nothing measurable can depend on $\chi$ . In three-vector components — and mind the sign:

\varphi \;\to\; \varphi + \frac{\partial\chi}{\partial t}, \qquad\qquad \vec A \;\to\; \vec A - \nabla\chi \qquad (2)

where: the minus on $\nabla\chi$ is forced by $\partial^i = -\partial_i$ . Physical reading: the textbook form $\varphi \to \varphi - \partial_t\lambda$ , $\vec A \to \vec A + \nabla\lambda$ is the same transformation with $\lambda = -\chi$ ; both are correct, and mixing them is not.

\boxed{\;\partial_\mu A^\mu \;=\; \frac{1}{c^2}\frac{\partial\varphi}{\partial t} + \nabla\!\cdot\!\vec A \;=\; 0\;} \qquad\text{(the Lorenz condition)} \qquad (3)

where: $\partial_\mu A^\mu$ is fully contracted — one upper index, one lower, summed. Physical reading: it is therefore a Lorentz scalar, and a scalar that vanishes in one frame vanishes in every frame. That one line is the entire argument for the Lorenz gauge's privileged status. The Coulomb condition $\nabla\!\cdot\!\vec A = 0$ contracts only the spatial indices, which is not a Lorentz-covariant operation — it names a frame's notion of "space" — so it does not survive a boost.

\Box\chi = -\partial_\mu A^\mu \quad\text{(always soluble)} \qquad (4) \qquad\qquad \boxed{\;\Box\chi = 0\;} \quad\text{(residual freedom)} \qquad (5)

where: $\Box = \partial_\mu\partial^\mu = \tfrac{1}{c^2}\partial_t^2 - \nabla^2$ . Physical reading of (4): start from any potential; if $\partial_\mu A^\mu = f \ne 0$ , solve the inhomogeneous wave equation $\Box\chi = -f$ (it always has a retarded solution) and the shifted potential satisfies (3). The Lorenz gauge is reachable from anywhere. Reading of (5): the Lorenz condition does not fix $A^\mu$ completely — any further $\chi$ with $\Box\chi = 0$ preserves it. That surviving freedom removes the second unphysical component, and it is why the photon has two polarisations, not three.

The gauge freedom, and the one gauge every frame agrees on

Derivation AThe field tensor cannot see the gauge — and the three-vector sign that trips everyone

Two things to establish, and the second is where the errors live: that $F^{\mu\nu}$ is identically invariant under $A^\mu \to A^\mu + \partial^\mu\chi$ — a two-line proof that nonetheless carries the whole meaning of the word "gauge" — and then the three-vector form of that transformation, whose sign is not a matter of taste and which the base text for this article gets internally inconsistent.

Step 1. Substitute the shifted potential into the definition: $\;F'^{\mu\nu} = \partial^\mu(A^\nu + \partial^\nu\chi) - \partial^\nu(A^\mu + \partial^\mu\chi) = F^{\mu\nu} + \big(\partial^\mu\partial^\nu - \partial^\nu\partial^\mu\big)\chi$ . Why: nothing but the definition and linearity of the derivative.

Step 2. Kill the extra term. For any $\chi$ with continuous second derivatives, $\partial^\mu\partial^\nu\chi = \partial^\nu\partial^\mu\chi$ , so the bracket vanishes and $\;\boxed{F'^{\mu\nu} = F^{\mu\nu}}$ . Why: the antisymmetrised derivative of a gradient is zero — the four-dimensional version of $\nabla\times\nabla\chi = \vec 0$ . This is why the freedom exists at all: $F$ is a curl, and curls are blind to gradients. Because it is an identity, no experiment at any precision can ever measure $\chi$ .

Step 3. Unpack into components, carefully. $A^0 = \varphi/c$ and $\partial^0 = \tfrac1c\partial_t$ , so the time slot gives $\varphi/c \to \varphi/c + \tfrac1c\partial_t\chi$ , i.e. $\;\boxed{\varphi \to \varphi + \partial_t\chi}$ . The spatial slots have $\partial^i = -\partial_i$ , so $A^i \to A^i - \partial_i\chi$ , i.e. $\;\boxed{\vec A \to \vec A - \nabla\chi}$ . Why: the minus is forced, not chosen. It comes from raising the index on the gradient — the same minus that gave $F^{0i} = -E^i/c$ . Anyone who writes $\vec A \to \vec A + \nabla\chi$ alongside $A^\mu \to A^\mu + \partial^\mu\chi$ has contradicted themselves.

Step 4. Check the fields directly, which is the reassurance worth having. $\;\vec E \to -\nabla(\varphi + \partial_t\chi) - \partial_t(\vec A - \nabla\chi) = \vec E$ , the two $\chi$ terms cancelling because $\nabla$ and $\partial_t$ commute; and $\;\vec B \to \nabla\times(\vec A - \nabla\chi) = \vec B$ , since the curl of a gradient vanishes. Why: the cancellation in $\vec E$ requires the relative sign of (2) to be right. Flip one of the two and the $\chi$ terms add, and you have manufactured a spurious electric field $-2\nabla\partial_t\chi$ out of a choice of notation. That is exactly the symptom of the base's inconsistency, and exactly how you catch it. *Verified in verify_aq-sr-gauge-transformations-lorenz-gauge.py: $F'^{\mu\nu} = F^{\mu\nu}$ asserted entry-by-entry, symbolically, for arbitrary $\chi(t,x,y,z)$ and arbitrary potentials; the component form derived from $A^\mu \to A^\mu + \partial^\mu\chi$ rather than assumed; $\vec E$ and $\vec B$ asserted unchanged; and the base's mixed pair shown to change $\vec E$ by $-2\nabla\partial_t\lambda \ne 0$ .*

Go deeper · Derivation B · 2 minThe Lorenz gauge is the only standard gauge every frame agrees on

Derivation BThe Lorenz gauge is the only standard gauge every frame agrees on

This is the article's owned result, and the reason a Danish physicist's name is attached to a condition rather than a transformation. There are many gauges; the question is which a Lorentz boost respects. The answer follows from a single observation about index structure — and then, to make it concrete rather than merely formal, we boost an actual potential and watch one condition break.

Step 1 — count the indices. $\partial_\mu A^\mu$ has one lower index and one upper index, summed: fully contracted, hence a Lorentz scalar. Under a boost, $\;\partial'_\mu A'^\mu = \big(\Lambda_\mu{}^\alpha\partial_\alpha\big)\big(\Lambda^\mu{}_\beta A^\beta\big) = \delta^\alpha_\beta\,\partial_\alpha A^\beta = \partial_\alpha A^\alpha$ — the two $\Lambda$ 's cancel. Why: this is the whole theorem. A scalar has the same value in every frame, so if $\partial_\mu A^\mu = 0$ for me it is zero for you, with no re-fixing. Contrast $\nabla\!\cdot\!\vec A = \partial_i A^i$ : the sum runs over $i = 1,2,3$ only — not a Lorentz contraction but three-quarters of one — and "which components are spatial" is exactly the question a boost refuses to answer frame-independently.

Step 2 — a potential in both gauges at once. Take a charge at rest: $\varphi = kq/r$ , $\vec A = \vec 0$ , $\partial_t\varphi = 0$ . Then $\nabla\!\cdot\!\vec A = 0$ (Coulomb ✓) and $\tfrac{1}{c^2}\partial_t\varphi + \nabla\!\cdot\!\vec A = 0$ (Lorenz ✓). Why: a potential satisfying both conditions is the perfect test specimen — whatever happens under the boost is a property of the conditions, not of the potential.

Step 3 — boost it and compute both conditions. From The Four-Potential, $\varphi' = \gamma\varphi$ and $\vec A' = \varphi'\vec v/c^2$ , so $A'_x = \gamma v\varphi/c^2$ . Working at $t' = 0$ , where $x = \gamma x'$ and $\partial x/\partial t' = -\gamma v$ :

\nabla'\!\cdot\!\vec A' = \frac{\partial A'_x}{\partial x'} = \frac{\gamma v}{c^2}\cdot\gamma\,\frac{\partial\varphi}{\partial x} = \frac{\gamma^2 v}{c^2}\,\partial_x\varphi , \qquad \frac{1}{c^2}\frac{\partial\varphi'}{\partial t'} = \frac{\gamma}{c^2}\,(-\gamma v)\,\partial_x\varphi = -\frac{\gamma^2 v}{c^2}\,\partial_x\varphi .

Why: both terms are the same quantity with opposite signs. The chain rule supplies $+\gamma$ to the space derivative and $-\gamma v$ to the time derivative, and that relative minus is the entire mechanism.

Step 4 — read off both verdicts. Adding them, $\;\boxed{\partial_\mu A'^\mu = \tfrac{\gamma^2 v}{c^2}\partial_x\varphi - \tfrac{\gamma^2 v}{c^2}\partial_x\varphi = 0}$ — the Lorenz condition survives, exactly, for any static $\varphi$ whatsoever. Taken alone, though, $\;\boxed{\nabla'\!\cdot\!\vec A' = \tfrac{\gamma^2 v}{c^2}\,\partial_x\varphi \ne 0}$ — the Coulomb condition is broken. Why: the boost did not destroy the Coulomb condition by accident or by a large velocity; it destroyed it structurally, by mixing $\varphi$ into $\vec A$ . To work in Coulomb gauge in the new frame you must perform a second, frame-specific gauge transformation — which is precisely why the Coulomb gauge is not manifestly covariant. Worked Example 1 puts $-2.882\ \mathrm{T}$ on the broken condition. *Verified in verify_aq-sr-gauge-transformations-lorenz-gauge.py: $\partial'_\mu A'^\mu = \partial_\mu A^\mu$ proved from $\Lambda_\mu{}^\alpha\Lambda^\mu{}_\beta = \delta^\alpha_\beta$ ; the boosted static potential's Lorenz condition simplified to exactly zero for arbitrary static $\varphi(x,y,z)$ ; its $\nabla'\!\cdot\!\vec A'$ shown to be $\gamma^2v\,\partial_x\varphi/c^2$ , non-zero; and the printed $\mp 2.882\ \mathrm{T}$ recomputed from $\varphi = ke/r$ at $x' = 1.00\ \mathrm{nm}$ , $\beta = 0.6$ .*

Go deeper · Derivation C · 2 minReaching the Lorenz gauge, the freedom that survives it, and why light has two polarisations

Derivation CReaching the Lorenz gauge, the freedom that survives it, and why light has two polarisations

The Lorenz condition would be a curiosity if it were hard to satisfy. It is not: it is reachable from any starting potential by solving one wave equation. And it does not finish the job — a residual freedom survives, and that leftover is exactly what reduces the photon's four components to the two we observe, which is the deepest payoff in this article.

Step 1 — reachability. Suppose $\partial_\mu A^\mu = f(x) \ne 0$ . Transform with some $\chi$ : the new divergence is $\;\partial_\mu A'^\mu = \partial_\mu(A^\mu + \partial^\mu\chi) = f + \Box\chi$ . Setting it to zero requires $\;\boxed{\Box\chi = -f}$ — an inhomogeneous wave equation, which always has a solution. Why: the Lorenz gauge is always available. There is no potential in electrodynamics from which you cannot reach it, so imposing it costs nothing in generality.

Step 2 — what survives. Having arrived, ask what further $\chi$ keeps us there. Repeat Step 1 with $f = 0$ : we need $\;\boxed{\Box\chi = 0}$ . Why: the Lorenz condition is a partial gauge fixing, not a complete one. Any solution of the homogeneous wave equation — and there are infinitely many — moves $A^\mu$ while keeping $\partial_\mu A^\mu = 0$ . This slack is the residual gauge freedom, and it is not a nuisance: it is the tool that finishes the counting.

Step 3 — count for a free wave. Take a source-free plane wave $A^\mu = \epsilon^\mu e^{ik_\nu x^\nu}$ . The field equation $\Box A^\mu = 0$ forces $k_\mu k^\mu = 0$ (null — the wave travels at $c$ ). The Lorenz condition forces $\;k_\mu\epsilon^\mu = 0$ : one constraint, killing one of the four components of $\epsilon^\mu$ . Why: the polarisation vector must be Minkowski-orthogonal to the wave vector. That is the first of the two deletions.

Step 4 — spend the residual freedom, and land on two. Residual transformations with $\Box\chi = 0$ are themselves plane waves, $\chi = ia\,e^{ik_\nu x^\nu}$ with the same null $k$ ; they shift $\;\epsilon^\mu \to \epsilon^\mu - a\,k^\mu$ . Choose $a$ to set $\epsilon^0 = 0$ . With $\vec k$ along $z$ , $k^\mu = (\omega/c)(1,0,0,1)$ , the Lorenz condition $k_\mu\epsilon^\mu = (\omega/c)(\epsilon^0 - \epsilon^3) = 0$ then forces $\epsilon^3 = \epsilon^0 = 0$ as well. Only $\epsilon^1$ and $\epsilon^2$ survive — transverse to the propagation direction. Why: $\;4 - 1 - 1 = 2$ . Four components; Lorenz removes one; the residual freedom removes a second. That is why light has exactly two polarisation states — not three, as a naive vector field would suggest. *Verified in verify_aq-sr-gauge-transformations-lorenz-gauge.py: $\partial_\mu A'^\mu = \partial_\mu A^\mu + \Box\chi$ symbolically; an explicit instance of reachability ( $A^\mu = (0, \kappa x, 0, 0)$ , a pure-gauge potential with $\partial_\mu A^\mu = \kappa$ , gauged to zero by $\chi = \kappa x^2/2$ satisfying $\Box\chi = -\kappa$ ); the plane-wave conditions $k_\mu k^\mu = 0$ and $k_\mu\epsilon^\mu = 0$ ; the residual shift $\epsilon^\mu \to \epsilon^\mu - ak^\mu$ shown to preserve both; and the count $4-1-1=2$ carried out on an explicit $\epsilon^\mu$ .*

Figure

THE DEGREE-OF-FREEDOM LEDGER for a free electromagnetic wave. The four-potential starts with 4 components. The Lorenz condition kᵐεₘ = 0 deletes one. The residual gauge freedom □χ = 0 — which survives Lorenz precisely because it is a homogeneous solution — deletes a second, by shifting εᵐ → εᵐ − a kᵐ. What is left is two TRANSVERSE polarisations: 4 − 1 − 1 = 2. This is why light has two polarisation states and not three, and why a massive vector boson has three

Limiting cases

Gauge condition	Lorentz-invariant?	What it buys
Lorenz, $\partial_\mu A^\mu = 0$	YES — a scalar equation	$\Box A^\mu = \mu_0 J^\mu$ : four decoupled wave equations. The covariant workhorse
Coulomb, $\nabla\!\cdot\!\vec A = 0$	NO — it names a frame's space	$\nabla^2\varphi = -\rho/\varepsilon_0$ : $\varphi$ is the instantaneous Coulomb potential. Good for bound states, bad for covariance
Radiation, $\varphi = 0$ and $\nabla\!\cdot\!\vec A = 0$	no	reachable in source-free regions using the residual freedom; displays the two transverse polarisations directly
residual, $\Box\chi = 0$	—	the freedom that survives Lorenz, and removes the second unphysical component

Common trapLUDVIG LORENZ, not Hendrik Lorentz. Two men, two countries, one missing t.

The most-botched attribution in relativistic electrodynamics, and worth getting right on purpose.

Ludvig Valentin Lorenz (1829–1891) — Danish. In 1867, in "On the Identity of the Vibrations of Light with Electrical Currents," he introduced the retarded potentials and the condition $\tfrac{1}{c^2}\partial_t\varphi + \nabla\!\cdot\!\vec A = 0$ . The Lorenz gauge is his. He is also the Lorenz of Lorenz–Mie scattering and of the Lorenz–Lorentz relation, which is gloriously named after both of them.
Hendrik Antoon Lorentz (1853–1928) — Dutch. The Lorentz transformation, the Lorentz force, the Lorentz factor $\gamma$ , the 1902 Nobel Prize. He turned fourteen the year Lorenz published the gauge condition.

"Lorentz gauge" is therefore not a variant spelling; it is an error, and it is everywhere. The confusion is compounded by an accident: the condition genuinely is Lorentz-invariant, so the wrong name looks defensible. The condition is Lorentz-invariant and it is named after Lorenz. Both are true and they concern different people.

Worked examples

1 · Boost a proton: the Coulomb gauge breaks by 2.882 tesla; the Lorenz gauge does not move

A proton sits at rest. Its Coulomb potential satisfies both gauge conditions. View it from a frame moving at $0.6c$ and evaluate both conditions a nanometre from the proton, on the axis of motion.

Step 1 — identify. In the rest frame $\varphi = ke/r$ , $\vec A = \vec 0$ , $\varphi$ static. So $\nabla\!\cdot\!\vec A = 0$ (Coulomb ✓) and $\tfrac{1}{c^2}\partial_t\varphi + \nabla\!\cdot\!\vec A = 0 + 0 = 0$ (Lorenz ✓). Both hold; nothing yet separates them.

Step 2 — the boosted potential. From The Four-Potential, $\varphi' = \gamma\varphi$ and $A'_x = \gamma v\varphi/c^2$ , with $\beta = 0.6$ , $\;\gamma = \mathbf{1.25}$ , $\;v = \mathbf{1.7987\times10^{8}\ \mathrm{m\,s^{-1}}}$ .

Step 3 — locate the field point and get the gradient. Evaluate at $x' = 1.00\ \mathrm{nm}$ on axis at $t' = 0$ ; the corresponding rest-frame coordinate is $x = \gamma x' = \mathbf{1.25\ \mathrm{nm}}$ . With $ke = (8.9875\times10^{9})(1.6022\times10^{-19}) = \mathbf{1.4400\times10^{-9}\ \mathrm{V\,m}}$ ,

\frac{\partial\varphi}{\partial x} = -\frac{ke}{x^2} = -\frac{1.4400\times10^{-9}}{(1.25\times10^{-9})^2} = \mathbf{-9.2158\times10^{8}\ \mathrm{V\,m^{-1}}} .

Step 4 — evaluate the two conditions. The prefactor is $\;\gamma^2 v/c^2 = (1.5625)(1.7987\times10^{8})/(8.9875\times10^{16}) = \mathbf{3.1272\times10^{-9}\ \mathrm{s\,m^{-1}}}$ , so

\nabla'\!\cdot\!\vec A' = \frac{\gamma^2 v}{c^2}\,\partial_x\varphi = (3.1272\times10^{-9})(-9.2158\times10^{8}) = \mathbf{-2.882\ \mathrm{T}} , \qquad \frac{1}{c^2}\frac{\partial\varphi'}{\partial t'} = \mathbf{+2.882\ \mathrm{T}} .

Step 5 — the boundary number, and the verdict. The Coulomb condition now reads $-2.882\ \mathrm{T}$ instead of zero: broken, by nearly three tesla. The Lorenz condition is the sum of the two terms, $-2.882 + 2.882 = \mathbf{0}$ : intact, exactly. The cancellation is not luck — Derivation 2 shows it is identically zero for any static $\varphi$ , because $\partial_\mu A^\mu$ is a scalar and it was zero before the boost. To keep working in Coulomb gauge in the new frame you would have to perform a second, frame-specific gauge transformation and re-solve; in Lorenz gauge you do nothing at all. That is the practical meaning of "manifestly covariant," priced in tesla.

2 · Why light has exactly two polarisations: 4 − 1 − 1 = 2

Take a source-free plane wave of wavelength $500\ \mathrm{nm}$ travelling along $z$ . Count its physical degrees of freedom by spending the Lorenz condition and then the residual gauge freedom.

Step 1 — identify. Write $A^\mu = \epsilon^\mu e^{ik_\nu x^\nu}$ . Naively $\epsilon^\mu$ has four independent components. The physical answer is two, and the two deletions have different origins.

Step 2 — the wave vector. $\;k = 2\pi/\lambda = 2\pi/(500\times10^{-9}) = \mathbf{1.2566\times10^{7}\ \mathrm{m^{-1}}}$ and $\;\omega = ck = \mathbf{3.7674\times10^{15}\ \mathrm{rad\,s^{-1}}}$ . The four-wavevector is $k^\mu = (\omega/c,\,0,\,0,\,k)$ , and $\;k_\mu k^\mu = (\omega/c)^2 - k^2 = \mathbf{0}$ — null, as $\Box A^\mu = 0$ demands.

Step 3 — spend the Lorenz condition (deletion 1). $\partial_\mu A^\mu = 0$ gives $k_\mu\epsilon^\mu = 0$ , which with $k^\mu = (\omega/c)(1,0,0,1)$ reads $(\omega/c)(\epsilon^0 - \epsilon^3) = 0$ , so $\;\epsilon^0 = \epsilon^3$ . One constraint; three components left.

Step 4 — spend the residual freedom (deletion 2). Gauge transformations with $\Box\chi = 0$ survive the Lorenz condition; taking $\chi = ia\,e^{ik_\nu x^\nu}$ with the same null $k$ shifts $\epsilon^\mu \to \epsilon^\mu - a k^\mu$ . This changes $\epsilon^0$ and $\epsilon^3$ by the same amount $a\omega/c$ , so it preserves $\epsilon^0 = \epsilon^3$ — as it must. Choose $a = \epsilon^0 c/\omega$ to set $\epsilon^0 = 0$ ; then $\epsilon^3 = \epsilon^0 = \mathbf{0}$ as well. A second component gone.

Step 5 — the boundary number, and the physics. What is left is $\epsilon^\mu = (0,\ \epsilon^1,\ \epsilon^2,\ 0)$ : two components, both perpendicular to the direction of travel. The count is $\;\mathbf{4 - 1 - 1 = 2}$ , and each deletion has a name — the Lorenz condition and the residual gauge freedom. This is the classical shadow of a deep quantum fact: a massless spin-1 particle has two helicity states, not three, and the missing state is precisely the one gauge invariance forbids.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §10.1.2 · Jackson, Classical Electrodynamics, 3rd ed., §6.3 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Jackson & Okun, "Historical roots of gauge invariance," Rev. Mod. Phys. 73, 663 (2001) · Weyl, "Elektron und Gravitation," Z. Phys. 56, 330 (1929).

Provenance. Signature $(+,-,-,-)$ ; $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ ; SI units. Base defect found and repaired: the AllQFT base's gauge-transformation card states the four-vector law as $A^\mu \to A^\mu + \partial^\mu\Lambda$ but prints a three-vector note in the opposite convention ( $\vec A \to \vec A + \nabla\lambda$ ) without recording that $\lambda = -\Lambda$ — as printed, the two lines are inconsistent and do not follow from one another. The consistent unpacking is $\varphi \to \varphi + \partial_t\chi$ , $\vec A \to \vec A - \nabla\chi$ (Derivation 1); the verify script shows the mismatched pair changes $\vec E$ by $-2\nabla\partial_t\lambda$ .

History correction (sourced). A common framing credits Weyl's 1929 paper with the phase reinterpretation of gauge invariance. That is not right: the phase reading is Fock's (1926) and London's (1927); Weyl's 1929 "Elektron und Gravitation" adopted it and gave it its definitive form. Weyl's own 1918 proposal was a scale invariance and was wrong as physics. See Jackson & Okun, Rev. Mod. Phys. 73, 663 (2001).

Scope. The four-potential itself, $\Box A^\mu = \mu_0 J^\mu$ , and minimal coupling belong to The Four-Potential; the field tensor's components and invariants to The Electromagnetic Field Tensor; the retarded solution in this gauge to The Wave Equation for the Four-Potential.

Is the Lorenz gauge named after Hendrik Lorentz?

No. It is named for Ludvig Lorenz (1829–1891), a Danish physicist who introduced the condition in 1867 along with the retarded potentials. Hendrik Antoon Lorentz (1853–1928) was Dutch and gave us the Lorentz transformation, the Lorentz force and the Lorentz factor — and he was fourteen when Lorenz published. The confusion is made worse by an accident: the condition genuinely is Lorentz-invariant, so the misspelling looks justified. It is not.

If the potentials are not unique, why bother with them at all?

Because they are the variables the equations are simple in. In the Lorenz gauge, Maxwell's coupled first-order system for $\vec E$ and $\vec B$ becomes four decoupled second-order wave equations $\Box A^\mu = \mu_0 J^\mu$ , each with a standard retarded solution. And quantum mechanically the potentials are not optional: the Aharonov–Bohm effect measures $\oint A_\mu\,\mathrm dx^\mu$ in a region where every field is exactly zero, so a theory of $\vec E$ and $\vec B$ alone is incomplete.

Does the Coulomb gauge give wrong answers, then?

No — it gives the same answers, because $F^{\mu\nu}$ is gauge-invariant and every measurable quantity is built from $F^{\mu\nu}$ . What it lacks is manifest covariance. It hides the relativity: $\varphi$ becomes the instantaneous Coulomb potential, which looks like action at a distance (it is not — the apparent instantaneity cancels against the vector potential in any physical field). And a Coulomb-gauge potential does not stay in Coulomb gauge under a boost, so you must re-fix by hand in each frame. For bound-state and quantum-optics problems, where you have a preferred frame anyway, that cost is worth paying.

What exactly is "residual" gauge freedom?

The Lorenz condition is one equation, so it removes one of the four components of $A^\mu$ — but not all of them. Any further $\chi$ satisfying the homogeneous wave equation $\Box\chi = 0$ shifts $A^\mu$ by $\partial^\mu\chi$ while leaving $\partial_\mu A^\mu$ still zero. That leftover slack is the residual freedom, and it is not a defect: it removes the second unphysical component, bringing the photon's count from four down to two.

Why does a massless photon have two polarisations rather than three?

Because gauge invariance deletes one. A massive spin-1 particle has three polarisation states — two transverse and one longitudinal. For the photon, the Lorenz condition kills one component and the residual gauge freedom kills a second, leaving $4 - 1 - 1 = 2$ , both transverse. The connection runs both ways: give the gauge boson a mass and you break the gauge symmetry, the residual freedom vanishes, and the longitudinal polarisation reappears — which is what the Higgs mechanism does to the $W$ and $Z$ .

Is gauge symmetry a real symmetry of nature, or just bookkeeping?

Both, and the tension is genuine. A gauge transformation relates two descriptions of the identical physical state, so unlike a rotation it generates no new physics or conserved charge on its own — in that sense it is redundancy, not symmetry. But the requirement of local gauge invariance is enormously restrictive, and that is where the physics is: it forces the existence of $A^\mu$ , forces the photon to be massless, forces the coupling to be universal, and ties electromagnetism to charge conservation. The redundancy is bookkeeping; the demand that the bookkeeping be locally arbitrary is a law of nature.