The Four-Potential

Put a proton at rest and stand next to it. All you have is a scalar potential: $\varphi = 1.440\ \mathrm{V}$ at a nanometre out, and $\vec A = \vec 0$ . No magnetic field, nothing to see. Now walk past it at $0.6c$ . Of course there is a magnetic field now — but look where it came from. Your scalar potential has grown to $\varphi' = \gamma\varphi = 1.800\ \mathrm{V}$ , and a vector potential has appeared from nothing: $A'_x = 3.602\times10^{-9}\ \mathrm{V\,s\,m^{-1}}$ . Nobody switched on a current. It condensed out of $\varphi$ the way $x$ condenses out of $t$ when you boost. And the numbers are not merely suggestive — they are exact: $|\vec A'\,|\big/(\varphi'/c) = 0.6000 = \beta$ , precisely the ratio in which the same boost trades $ct$ for $x$ . The quantity $\varphi/c$ — not $\varphi$ — sits in the time slot, carrying the same units as $\vec A$ , and the four components move as one object, $A^\mu = (\varphi/c,\ \vec A\,)$ . For a century the potentials were mathematical scaffolding — convenient, unphysical, defined only up to a gauge, with $\vec E$ and $\vec B$ as the real things. Relativity inverts the hierarchy: $A^\mu$ is a genuine four-vector, while $\vec E$ and $\vec B$ are derived from it, one derivative away, and are not four-vectors at all. The scaffolding was the building. Quantum mechanics finishes the job. In the Aharonov–Bohm effect an electron is steered around a solenoid through a region where $\vec E = \vec B = \vec 0$ exactly — and it still picks up a phase, set by $\oint A_\mu\,\mathrm dx^\mu$ . Send it around one superconducting flux quantum $h/2e$ and the shift is exactly $\pi$ . The fields are zero everywhere the electron goes. The four-potential is not.

Key takeaways

Signature $(+,-,-,-)$ ; $\;\partial_\mu = \big(\tfrac1c\partial_t,\ +\nabla\big)$ but $\;\partial^\mu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ — the minus matters. $\;p\cdot p = +m^2c^2$ .
The potentials are a four-vector: $\boxed{A^\mu = (\varphi/c,\ \vec A\,)}$ . A boost mixes $\varphi$ into $\vec A$ exactly as it mixes $ct$ into $x$ — which is where the magnetic field of a moving charge comes from. The $c$ is not decoration: $\varphi/c$ carries the units of $\vec A$ .
Owned result — that $A^\mu$ is a four-vector must be PROVED, not asserted. Two routes: $\Box A^\mu = \mu_0 J^\mu$ with $\Box$ a scalar operator and $J^\mu$ a four-vector leaves $A^\mu$ no choice; and boosting the Coulomb potential reproduces $\vec A = \varphi\vec v/c^2$ with nothing put in by hand.
The fields are its derivatives: $\vec E = -\nabla\varphi - \partial_t\vec A$ , $\;\vec B = \nabla\times\vec A$ , both at once as $\boxed{F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu}$ — so the two homogeneous Maxwell equations become identities, true for free.
The dynamics: in the Lorenz gauge, $\boxed{\Box A^\mu = \mu_0 J^\mu}$ — four decoupled wave equations, each slot sourced by the matching component of the current. This is what you actually solve. Minimal coupling $q\!\int A_\mu\,\mathrm dx^\mu$ is a Lorentz scalar with no derivatives — which is why $A^\mu$ , not $\vec E$ and $\vec B$ , appears in the quantum Hamiltonian.
Key idea: the potentials looked like bookkeeping because in three-vector language they are a scalar and a vector with no visible relationship. In four-vector language they are one object — and it is $\vec E$ and $\vec B$ , six components that refuse to be four-vectors, that need explaining. Relativity did not make the potentials respectable; it revealed they were fundamental all along.

The core insight

Ask why a moving charge has a magnetic field and the three-vector answer is a shrug: "because Ampère." The four-vector answer is a theorem. A static charge has $A^\mu = (\varphi/c, \vec 0\,)$ — pure time component. Boost it, and a four-vector with only a time component cannot stay that way: the Lorentz transformation drags part of $A^0$ into $A^1$ , so $\vec A \neq \vec 0$ , and $\vec B = \nabla\times\vec A$ cannot vanish. Magnetism is not an extra force bolted onto electricity — it is what the same four-vector looks like from a moving chair.

Where this lives today

Every calculation of radiation begins with $\Box A^\mu = \mu_0 J^\mu$ . You do not solve for $\vec E$ and $\vec B$ directly if you can help it: you solve four decoupled scalar wave equations for $A^\mu$ , take the retarded solution, and then differentiate. That is how the Liénard–Wiechert potentials are obtained, and hence the Larmor formula, synchrotron radiation, and the pattern of every antenna ever built. The reason is Derivation 3: in the Lorenz gauge the four components do not talk to each other. In quantum theory $A^\mu$ is the photon: the quantised field has four components, of which gauge invariance removes two, leaving the two transverse polarisations we observe. Minimal coupling, $\partial_\mu \to \partial_\mu + iqA_\mu$ , is how every charged particle in the Standard Model talks to light — and it is not an ansatz: demanding invariance under local phase rotations forces a compensating four-vector field with exactly this coupling. Electromagnetism is what you get when you insist the phase of the electron field be a local convention. And the Aharonov–Bohm effect settled the ontological question in the laboratory. The same phase is now a workhorse of condensed-matter physics: the SQUID magnetometer, interferometry in mesoscopic rings, the flux-threading arguments behind the quantum Hall effect. Heaviside wanted the potentials murdered. They outlived the fields.

Origins · 2 minFour-potential and how this was discovered

Four-potential

1857–present · Maxwell's electrotonic state · Minkowski's four-vector · Aharonov–Bohm

The potentials are older than the fields they generate. James Clerk Maxwell introduced the vector potential in the 1850s and 1860s under a name borrowed from Faraday — the electrotonic state — and in his formulation it, not $\vec B$ , was primary: the thing whose rate of change drives an electromotive force. He took it as the mathematical expression of Faraday's intuition that the space around a magnet is genuinely altered, whether or not anything is there to feel it. The next generation disagreed, vehemently. Oliver Heaviside, who recast Maxwell's system into the four vector equations we now teach, thought the potentials metaphysical baggage and wanted them "murdered"; Hertz agreed. Their objection was sharp and, on its own terms, correct: the potentials are not unique. Add the gradient of any function to $\vec A$ , adjust $\varphi$ to match, and nothing measurable changes. A quantity you cannot measure, they argued, is not a quantity. Relativity answered by changing the question. In 1908 Hermann Minkowski, in the Grundgleichungen, showed that $(\varphi/c, \vec A\,)$ transforms as a four-vector, whose components mix under boosts exactly as $(ct,\vec x\,)$ do: the scalar and vector potentials were not two things but one, and their apparent arbitrariness sat alongside a rigid geometric structure. The gauge freedom that so offended Heaviside turned out, in Hermann Weyl's hands and then in Yang and Mills', to be the organising principle of all fundamental interactions. The last word was experimental: in 1959 Yakir Aharonov and David Bohm showed that an electron travelling through a region of strictly zero field can still be measurably deflected — an effect anticipated by Werner Ehrenberg and Raymond Siday (1949) and overlooked — and Akira Tonomura's group confirmed it in the 1980s with a toroidal magnet wrapped in a superconductor. The potentials are physical. Heaviside lost.

History

1857

— Maxwell, following Faraday's notion of an "electrotonic state," introduces the vector potential; in his formulation it, rather than the magnetic field, is primary.

1865

— Maxwell's "A Dynamical Theory of the Electromagnetic Field" gives the field equations in terms of the potentials.

1867

— Ludvig Lorenz publishes the retarded-potential solution and the condition now called the Lorenz gauge, in "On the Identity of the Vibrations of Light with Electrical Currents."

1908

— Minkowski's "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern" shows that the scalar and vector potentials are components of a single four-vector.

1949

— Ehrenberg and Siday note that the vector potential can shift an electron's interference pattern in a field-free region; the result is overlooked.

1959

— Aharonov and Bohm publish "Significance of Electromagnetic Potentials in the Quantum Theory" (Phys. Rev. 115, 485), making the case decisively.

1986

— Tonomura and collaborators confirm the effect with a toroidal magnet shielded by a superconductor, excluding any leakage-field explanation.

The physical picture

The potentials entered physics as a labour-saving device. $\nabla\!\cdot\!\vec B = 0$ means $\vec B$ is a curl, $\vec B = \nabla\times\vec A$ ; substituting into Faraday's law makes $\vec E + \partial_t\vec A$ a gradient, $\vec E = -\nabla\varphi - \partial_t\vec A$ . Two of Maxwell's four equations are solved identically by introducing $\varphi$ and $\vec A$ . It was a trick, and everyone treated it as one — Heaviside wanted the potentials "murdered." Three things reversed the verdict. They are a four-vector: $\varphi$ and $\vec A$ are the time and space parts of one object, rotated into each other by a boost with the same $\gamma$ and $\gamma\beta$ that mix $ct$ and $x$ — and the magnetic field of a moving charge is nothing but that rotation, seen through $\vec B = \nabla\times\vec A$ . They make the fields derived objects: once $A^\mu$ exists, $\vec E$ and $\vec B$ are its derivatives, packaged as $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , whose very existence makes the homogeneous Maxwell equations identities. You do not impose $\nabla\!\cdot\!\vec B = 0$ ; you cannot violate it. And they are what matter couples to: the interaction is $q\!\int A_\mu\,\mathrm dx^\mu$ , a Lorentz scalar with no derivatives. That is minimal coupling, it is how electromagnetism enters the quantum Hamiltonian ( $\vec p \to \vec p - q\vec A$ ), and it is why the Aharonov–Bohm effect exists at all. The price is redundancy: $A^\mu$ has four components, the physical field two propagating degrees of freedom. The excess is gauge freedom — add $\partial^\mu\chi$ and not a single field changes — and Gauge Transformations and the Lorenz Gauge owns it entirely. Here we need one consequence: the Lorenz condition $\partial_\mu A^\mu = 0$ turns the field equations into four decoupled wave equations, and it is the only common choice every frame agrees on.

Figure

PUNCHLINE — a boost turns φ into A, and that IS magnetism. A proton at rest has a pure time component: Aᵐ = (φ/c, 0). Boost to 0.6c and the four-vector tilts exactly as (ct, x) does — φ grows by γ and a vector potential condenses out of nothing. The ratio |A|/(φ/c) comes out at 0.6000, which is β to four figures. B = curl A is then forced to be non-zero: the magnetic field of a moving charge is a Lorentz rotation of the electric potential

The core equations

The object, and the reason for the $c$ :

\boxed{\;A^\mu \;=\; \left(\frac{\varphi}{c},\ A_x,\ A_y,\ A_z\right)\;} \qquad (1)

where: $\varphi$ — scalar potential (volts); $\vec A$ — vector potential ( $\mathrm{V\,s\,m^{-1}}$ ). Physical reading: the $c$ is forced by units, exactly as in $x^\mu = (ct,\vec x\,)$ — $\varphi/c$ and $\vec A$ share dimensions, and a boost forms linear combinations of the slots, so slots carrying different units are impossible. Writing $A^0 = \varphi$ is a dimensional error, catchable before any physics. The fields, as derivatives of it — and both at once:

\vec E = -\nabla\varphi - \frac{\partial\vec A}{\partial t}, \qquad \vec B = \nabla\times\vec A \qquad (2) \qquad\qquad \boxed{\;F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu\;} \qquad (3)

where: $\partial^\mu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ — note the minus, whence every sign in the field tensor. Physical reading: (2) are not two unrelated formulae but the components of the one tensor equation (3), as Derivation 2 shows. And (2) guarantees $\nabla\!\cdot\!\vec B = \nabla\!\cdot\!(\nabla\times\vec A) \equiv 0$ and $\nabla\times\vec E = -\partial_t\vec B$ : the two homogeneous Maxwell equations are identities the moment a potential exists. The matrix of $F^{\mu\nu}$ belongs to The Electromagnetic Field Tensor; here it is enough that $\vec E$ and $\vec B$ are both derivatives of $A^\mu$ , and neither is fundamental. The dynamics, in the Lorenz gauge:

\boxed{\;\Box A^\mu = \mu_0 J^\mu\;}, \qquad \Box \equiv \partial_\nu\partial^\nu = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2, \qquad\text{given}\quad \partial_\mu A^\mu = 0 \qquad (4)

where: $J^\mu = (c\rho, \vec J\,)$ — the four-current; $\partial_\mu A^\mu = \tfrac{1}{c^2}\partial_t\varphi + \nabla\!\cdot\!\vec A$ is the Lorenz condition. Physical reading: four decoupled wave equations — $\Box\varphi = \rho/\varepsilon_0$ and $\Box\vec A = \mu_0\vec J$ , each slot hearing only its own source. All of radiation theory starts here. And this is the fastest proof that $A^\mu$ is a four-vector: $\Box$ is a Lorentz scalar and $J^\mu$ is a four-vector, so $A^\mu$ has no choice. And the coupling to matter:

S_{\text{int}} = q\!\int A_\mu\,\mathrm dx^\mu = q\!\int A_\mu U^\mu\,\mathrm d\tau \qquad (5)

where: $U^\mu = \gamma(c,\vec v\,)$ — the four-velocity. Physical reading: minimal coupling — a Lorentz scalar with no derivatives, contracting the potential with the particle's worldline. It is why the quantum Hamiltonian contains $\vec p - q\vec A$ rather than $\vec E$ or $\vec B$ , and why the Aharonov–Bohm phase survives where the fields are exactly zero.

Derivation AWhy the potentials form a four-vector — proved twice, because asserting it is not enough

Every textbook writes $A^\mu = (\varphi/c, \vec A\,)$ and moves on, as though stacking four numbers in a column made them a four-vector. It does not. A four-vector is defined by how it transforms, $A'^\mu = \Lambda^\mu{}_\nu A^\nu$ — a claim about nature, which has to be earned. Here are two independent proofs, one structural and one by brute boost, and they had better agree.

Step 1 — the structural route. Take $\Box A^\mu = \mu_0 J^\mu$ as given (derived in the third card). Inspect its ingredients: $\Box = \partial_\nu\partial^\nu$ is a Lorentz scalar operator — every index contracted, so it is the same operator in every frame — and $J^\mu$ is a four-vector, earned in The Four-Current Density as $\rho_0 U^\mu$ . Why: an equation of the form (scalar operator)(X) = (four-vector) can only hold in every frame if X is itself a four-vector. There is nothing else it could be.

Step 2 — read off the conclusion. If $A^\mu$ transformed as anything else, the two sides of $\Box A^\mu = \mu_0 J^\mu$ would transform differently, and the equation would hold in one frame and fail in the next. Since Maxwell's equations hold in every inertial frame, $A^\mu$ must be a four-vector. Why: we computed nothing — covariance did the work. The caveat is real, though: this holds within the Lorenz gauge class, because that is where (4) holds. See the second warning.

Step 3 — the brute-force route. Put a charge $q$ at rest in $S$ : $\vec A = \vec 0$ , $\varphi = q/4\pi\varepsilon_0 r$ , so $A^\mu = (\varphi/c,\ 0,0,0)$ , purely timelike. Boost to $S'$ moving at $-v\hat x$ , so the charge moves at $+v\hat x$ in $S'$ . With $\Lambda^0{}_0 = \Lambda^1{}_1 = \gamma$ and $\Lambda^0{}_1 = \Lambda^1{}_0 = +\gamma\beta$ :

A'^0 = \gamma\frac{\varphi}{c} + \gamma\beta\cdot 0 = \frac{\gamma\varphi}{c}, \qquad A'^1 = \gamma\beta\frac{\varphi}{c} + \gamma\cdot 0 = \frac{\gamma v\varphi}{c^2} .

So $\;\boxed{\varphi' = \gamma\varphi}$ and $\;\boxed{\vec A' = \dfrac{\gamma v\varphi}{c^2}\hat x = \dfrac{\varphi'\vec v}{c^2}}$ . Why: a vector potential has appeared, and nobody put it there. It is $\varphi$ , tilted. And $\vec A = \varphi\vec v/c^2$ is exactly the standard result for a uniformly moving charge — here derived, not looked up.

Step 4 — collect the magnetic field, and the punchline. Since $\vec A' \ne \vec 0$ , $\vec B' = \nabla\times\vec A' \ne \vec 0$ : the moving charge has a magnetic field, and it came from the scalar potential of a static charge. Why: this is the thesis of the whole chapter, in four lines. The magnetic field is not a separate entity accompanying moving charges — it is what the electric potential of a stationary charge looks like from a moving chair. The ratio $|\vec A'|/(\varphi'/c) = \beta$ (Worked Example 1 checks it to four figures) is precisely the ratio in which a boost trades $ct$ for $x$ . The two routes agree. *Verified in verify_aq-sr-four-potential.py: the boost of $A^\mu = (\varphi/c, \vec 0\,)$ with an explicit $\Lambda$ , asserted equal to $(\gamma\varphi/c,\ \gamma v\varphi/c^2, 0, 0)$ ; the identity $\vec A' = \varphi'\vec v/c^2$ ; and $\Box = \partial_\mu\partial^\mu$ , so the structural argument's premise is checked, not assumed.*

Go deeper · Derivation B · 2 minThe potentials in, the fields out — and two of Maxwell's equations for free

Derivation BThe potentials in, the fields out — and two of Maxwell's equations for free

The goal is to show that $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is not a definition imposed from outside but the inevitable four-dimensional packaging of $\vec E = -\nabla\varphi - \partial_t\vec A$ and $\vec B = \nabla\times\vec A$ — and then to collect the bonus: half of Maxwell's equations stop being laws and become identities. The whole calculation turns on one fact people forget: $\partial^\mu$ has a minus sign on its spatial part.

Step 1 — get the index position right. $\partial_\mu = \partial/\partial x^\mu = \big(\tfrac1c\partial_t,\ +\nabla\big)$ has a lower index and a plus. Raising costs the metric: $\partial^\mu = \eta^{\mu\nu}\partial_\nu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ . Why: every sign in the field tensor descends from this one minus, and there is no freedom in it. Get it wrong and every entry of $F^{\mu\nu}$ is wrong.

Step 2 — evaluate one electric component. With $A^0 = \varphi/c$ , $A^1 = A_x$ :

F^{01} = \partial^0 A^1 - \partial^1 A^0 = \frac1c\frac{\partial A_x}{\partial t} - \left(-\frac{\partial}{\partial x}\right)\frac{\varphi}{c} = \frac1c\left(\frac{\partial A_x}{\partial t} + \frac{\partial\varphi}{\partial x}\right) = -\frac1c\left(-\frac{\partial\varphi}{\partial x} - \frac{\partial A_x}{\partial t}\right) = -\frac{E_x}{c} .

Why: the middle step is the whole game — $\partial^1 = -\partial_x$ , so subtracting it adds. The bracket that emerges is exactly the definition of $E_x$ , so the sign is forced: $F^{0i} = -E^i/c$ , with a minus. This is a notorious trap, and the AllQFT base for The Electromagnetic Field Tensor gets it wrong — it prints $F^{0i} = +E^i/c$ while its own matrices print $-E_x/c$ . The derivation leaves you no choice.

Step 3 — evaluate one magnetic component. $\;F^{12} = \partial^1 A^2 - \partial^2 A^1 = -\partial_x A_y + \partial_y A_x = -(\nabla\times\vec A)_z = -B_z$ . Why: both indices are spatial, so both pick up a minus and the two do not cancel — they factor out, giving $F^{ij} = -\varepsilon^{ijk}B_k$ . The magnetic block therefore behaves differently from the electric block under index-raising, which is the structural fact behind the next two articles.

Step 4 — collect the bonus. $\;\nabla\!\cdot\!\vec B = \nabla\!\cdot\!(\nabla\times\vec A) \equiv 0$ (divergence of a curl), and $\;\nabla\times\vec E = -\nabla\times(\nabla\varphi) - \partial_t(\nabla\times\vec A) = -\partial_t\vec B$ (curl of a gradient). In four dimensions the same content is the cyclic identity $\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0$ , whose six terms cancel in pairs because partials commute. Why: these are no longer laws of nature — they are theorems of vector calculus. The moment you agree a potential exists, you could not construct a magnetic monopole or violate Faraday's law if you tried. Two of Maxwell's four equations have been consumed, which is exactly why only two remain to be solved. *Verified in verify_aq-sr-four-potential.py: $F^{\mu\nu}$ built symbolically from $A^\mu = (\varphi/c, \vec A\,)$ with arbitrary potentials, every entry asserted against $\vec E$ and $\vec B$ — in particular $F^{01} = -E_x/c$ (minus) and $F^{12} = -B_z$ ; antisymmetry; $\nabla\!\cdot\!\vec B = 0$ and $\nabla\times\vec E + \partial_t\vec B = 0$ as identities; and the cyclic Bianchi identity for all four index triples.*

Go deeper · Derivation C · 2 minFour decoupled wave equations, and what the Lorenz condition is buying

Derivation CFour decoupled wave equations, and what the Lorenz condition is buying

Two of Maxwell's equations are now identities, so only the source equation is left: $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ . Feed the potential into it and what comes out is a mess — until one condition is imposed, at which point it collapses into the most useful equation in classical electrodynamics: four independent wave equations, each sourced by the matching component of the current.

Step 1 — substitute and split. $\;\partial_\mu(\partial^\mu A^\nu - \partial^\nu A^\mu) = \underbrace{\partial_\mu\partial^\mu}_{=\,\Box} A^\nu - \partial^\nu(\partial_\mu A^\mu) = \mu_0 J^\nu$ . Why: the derivatives commute, so the four-divergence $\partial_\mu A^\mu$ can be isolated inside the second bracket. The equation has separated into a wave operator and an obstruction.

Step 2 — kill the obstruction. The obstruction is a four-divergence of the potential — exactly the quantity gauge freedom lets us adjust. Impose the Lorenz condition $\;\partial_\mu A^\mu = \tfrac{1}{c^2}\partial_t\varphi + \nabla\!\cdot\!\vec A = 0$ , and $\;\boxed{\Box A^\nu = \mu_0 J^\nu}$ . Why: it is always reachable — given any $A^\mu$ with $\partial_\mu A^\mu = f \ne 0$ , gauge-transform with a $\chi$ solving $\Box\chi = -f$ , which always has a solution. That proof, the residual freedom that survives, and the reason it is the only standard gauge every frame agrees on, all belong to Gauge Transformations and the Lorenz Gauge. Here we simply spend it.

Step 3 — read the four equations. With $J^\mu = (c\rho, \vec J\,)$ and $\mu_0\varepsilon_0c^2 = 1$ :

\Box\left(\frac{\varphi}{c}\right) = \mu_0 c\rho \;\Longrightarrow\; \boxed{\Box\varphi = \frac{\rho}{\varepsilon_0}}, \qquad\qquad \boxed{\Box\vec A = \mu_0\vec J} .

Why: four decoupled wave equations. $\varphi$ hears only about charge, each component of $\vec A$ only about the matching component of current, and none of them knows the others exist — against the coupled first-order tangle of Maxwell's equations in $\vec E$ and $\vec B$ . This is the practical reason the potentials won the argument: they are what you actually solve.

Step 4 — the source's consistency condition. Take $\partial_\nu$ of $\Box A^\nu = \mu_0 J^\nu$ : since $\Box$ and $\partial_\nu$ commute, $\Box(\partial_\nu A^\nu) = \mu_0\,\partial_\nu J^\nu$ , and the left side is zero by the Lorenz condition. So $\;\partial_\nu J^\nu = 0$ — charge conservation is the compatibility condition for the potential formulation to work at all. Why: Charge Invariance and Continuity's theorem, arriving from a completely different direction. Feed the theory a non-conserved current and the Lorenz condition cannot be maintained in time; the formalism seizes up. The solutions of (4) are the retarded potentials, $A^\mu(\vec r, t) = \tfrac{\mu_0}{4\pi}\int J^\mu(\vec r\,', t_r)/|\vec r - \vec r\,'|\,\mathrm d^3r'$ with $t_r = t - |\vec r - \vec r\,'|/c$ — the route to all of radiation, owned by The Wave Equation for the Four-Potential. *Verified in verify_aq-sr-four-potential.py: $\partial_\mu F^{\mu\nu} = \Box A^\nu - \partial^\nu(\partial_\mu A^\mu)$ checked for all four $\nu$ with arbitrary potentials; $\Box = \tfrac{1}{c^2}\partial_t^2 - \nabla^2$ ; the Lorenz condition; the collapse to $\Box\varphi = \rho/\varepsilon_0$ and $\Box\vec A = \mu_0\vec J$ using $\mu_0\varepsilon_0c^2 = 1$ ; and $\partial_\nu J^\nu = 0$ .*

Figure

The four-potential's job, in three panels — and the count that gives the photon two polarisations. LEFT: four decoupled wave equations, one per slot, each sourced only by the matching component of the current. CENTRE: the degree-of-freedom ledger, 4 minus 1 (Lorenz) minus 1 (residual gauge) = 2 transverse photon polarisations. RIGHT: Aharonov-Bohm — an electron round a solenoid, E = B = 0 everywhere on its path, yet one superconducting flux quantum h/2e shifts its phase by exactly pi

Limiting cases

Case	$A^\mu$	Meaning
electrostatics	$(\varphi/c,\ \vec 0\,)$	purely timelike. $\vec B = \vec 0$ — and a boost is guaranteed to produce $\vec A \ne 0$
magnetostatics	$(0,\ \vec A\,)$ , $\partial_t\vec A = 0$	steady currents: $\vec B = \nabla\times\vec A$ , $\vec E = \vec 0$ in that frame
charge at rest, boosted	$\varphi' = \gamma\varphi$ , $\;\vec A' = \varphi'\vec v/c^2$	the magnetic field of a moving charge, derived
plane wave, source-free	$\epsilon^\mu e^{ik_\nu x^\nu}$ , $\;k_\mu k^\mu = 0$	$\Box A^\mu = 0$ : null and transverse; $\epsilon^\mu$ carries the two polarisations
Coulomb gauge	not a four-vector class	the condition picks a frame — see the warning

Common trapA^μ is a four-vector only within a Lorentz-covariant gauge class. The Coulomb gauge breaks it.

The structural proof in Derivation 1 has a premise easy to skate past: it uses $\Box A^\mu = \mu_0 J^\mu$ , which holds in the Lorenz gauge. Gauge conditions are not all created equal. The Lorenz condition $\partial_\mu A^\mu = 0$ is a Lorentz scalar equation: impose it in one frame and every frame inherits it, so the boosted components really are $\Lambda^\mu{}_\nu A^\nu$ . The Coulomb condition $\nabla\!\cdot\!\vec A = 0$ is not a scalar — it names one frame's spatial slice. Boost a Coulomb-gauge potential and it no longer satisfies $\nabla\!\cdot\!\vec A' = 0$ ; re-fixing it by hand means the components you end up with are not $\Lambda^\mu{}_\nu A^\nu$ . None of this touches the physics — $F^{\mu\nu}$ is gauge-invariant, so $\vec E$ and $\vec B$ come out the same, and the Coulomb gauge is genuinely convenient. But it is not manifestly covariant, and " $A^\mu$ is a four-vector" is a statement about the Lorenz class.

Worked examples

1 · Boost a proton and watch the magnetic field condense out of the electric potential

A proton is at rest. At a field point $1.00\ \mathrm{nm}$ away, find $\varphi$ and $\vec A$ . Now view the same proton from a frame in which it moves at $0.6c$ . Confirm that the four components trade exactly as $(ct, x)$ do.

Step 1 — identify. In the rest frame there is no current, so $\vec A = \vec 0$ and the four-potential is purely timelike: $A^\mu = (\varphi/c, 0,0,0)$ . That is the entire input; everything else is a Lorentz transformation.

Step 2 — the rest-frame numbers. $\;\varphi = \dfrac{e}{4\pi\varepsilon_0 r} = \dfrac{(8.9875\times10^{9})(1.6022\times10^{-19})}{1.00\times10^{-9}} = \mathbf{1.440\ \mathrm{V}}$ , and $\;A^0 = \varphi/c = \mathbf{4.803\times10^{-9}\ \mathrm{V\,s\,m^{-1}}}$ — note the units: the same as $\vec A$ 's, which is the whole reason for the $c$ .

Step 3 — boost. $\beta = 0.6 \Rightarrow \gamma = 1/\sqrt{1-0.36} = \mathbf{1.25}$ , $v = 1.7987\times10^{8}\ \mathrm{m\,s^{-1}}$ . Applying $\Lambda$ to a purely timelike four-vector:

\varphi' = \gamma\varphi = 1.25\times 1.440 = \mathbf{1.800\ \mathrm{V}}, \qquad A'_x = \frac{v\varphi'}{c^2} = \frac{(1.7987\times10^{8})(1.800)}{8.9875\times10^{16}} = \mathbf{3.602\times10^{-9}\ \mathrm{V\,s\,m^{-1}}} .

Step 4 — the check that matters. A boost trades the time and space slots of any four-vector in a fixed ratio. Test it: $\;\dfrac{|\vec A'|}{\varphi'/c} = \dfrac{3.602\times10^{-9}}{1.800/(2.9979\times10^{8})} = \dfrac{3.602\times10^{-9}}{6.004\times10^{-9}} = \mathbf{0.6000}$ — that is $\beta$ , to four figures, precisely the ratio in which the same boost trades $x$ for $ct$ . The potentials are not merely called a four-vector; they transform as one.

Step 5 — the point. In the boosted frame $\vec A' \ne \vec 0$ , so $\vec B' = \nabla\times\vec A' \ne \vec 0$ : there is now a magnetic field, and it came from the scalar potential of a static charge. No current was switched on; nothing magnetic was assumed. The general result, $\vec A = \varphi\vec v/c^2$ , yields the full field of a moving charge — the business of The Field of a Uniformly Moving Charge. What this page owns is the mechanism: magnetism is a Lorentz rotation of the electric potential.

2 · The Aharonov–Bohm phase: exactly π per superconducting flux quantum

An electron is sent around a long solenoid, through a region where $\vec E = \vec B = \vec 0$ exactly. The solenoid encloses one superconducting flux quantum $\Phi_0 = h/2e$ . Find the phase shift — and decide whether the four-potential is physical.

Step 1 — identify. Minimal coupling, equation (5): the phase accumulates as $\Delta\phi = \tfrac{q}{\hbar}\oint A_\mu\,\mathrm dx^\mu$ . Outside an ideal solenoid $\vec B = \vec 0$ but $\vec A \ne \vec 0$ — the vector potential circulates, and $\oint\vec A\cdot\mathrm d\vec\ell = \Phi$ by Stokes' theorem, even though the flux it encloses is entirely inaccessible to the electron.

Step 2 — the phase. For an electron (magnitudes), $\;|\Delta\phi| = \dfrac{e}{\hbar}\oint\vec A\cdot\mathrm d\vec\ell = \dfrac{e\Phi}{\hbar}$ — a pure geometric loop integral, gauge-invariant, and blind to where the flux actually sits.

Step 3 — the numbers. $\;\Phi_0 = h/2e = (6.6261\times10^{-34})/(2\times 1.6022\times10^{-19}) = \mathbf{2.068\times10^{-15}\ \mathrm{Wb}}$ , so $\;|\Delta\phi| = \dfrac{e}{\hbar}\cdot\dfrac{h}{2e} = \dfrac{h}{2\hbar} = \dfrac{2\pi\hbar}{2\hbar} = \boldsymbol{\pi}$ .

Step 4 — check against the arithmetic. $\;(1.6022\times10^{-19})(2.068\times10^{-15})/(1.0546\times10^{-34}) = \mathbf{3.1416\ \mathrm{rad}} = \pi$ . The charges cancel exactly and the answer is a pure number — no material constants, no geometry, no solenoid radius. One superconducting flux quantum flips the interference pattern from bright to dark. (A full $2\pi$ needs $h/e = 4.136\times10^{-15}\ \mathrm{Wb}$ — twice as much, because the superconducting quantum is defined with the Cooper pair's charge $2e$ .)

Step 5 — the verdict. Set $\vec E = \vec B = \vec 0$ along the entire electron path — which an ideal solenoid does exactly — and the shift is still $\pi$ . A theory formulated only in terms of the fields has, at every point the electron visits, literally nothing to say: all six components of $F^{\mu\nu}$ vanish there. Yet the interference pattern moves, and it has been measured. Whatever carries the information is $A^\mu$ , along the path — which is why $A^\mu$ , not $\vec E$ and $\vec B$ , is the field that gets quantised.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §12.3.5 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Jackson, Classical Electrodynamics, 3rd ed., §11.9 · Aharonov & Bohm, Phys. Rev. 115, 485 (1959) · Tonomura et al., Phys. Rev. Lett. 56, 792 (1986).

Provenance. Signature $(+,-,-,-)$ ; $\partial_\mu = (\tfrac1c\partial_t, +\nabla)$ , $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ ; SI units. No numeric defect was found in this article's base text — its derivation of $F^{i0} = +E_i/c$ is correct and agrees with the $F^{0i} = -E^i/c$ derived here. Every printed number is recomputed in verify_aq-sr-four-potential.py, which asserts against this article's own printed digits, including $|\vec A'|/(\varphi'/c) = 0.6000 = \beta$ and the exact $\pi$ of the Aharonov–Bohm example.

Scope. The gauge freedom $A^\mu \to A^\mu + \partial^\mu\chi$ — that $F^{\mu\nu}$ cannot see it, that the Lorenz gauge is always reachable and uniquely Lorentz-invariant, and the residual freedom $\Box\chi = 0$ — belongs entirely to Gauge Transformations and the Lorenz Gauge; it is stated here and spent once, not derived. The matrices and invariants of $F^{\mu\nu}$ belong to The Electromagnetic Field Tensor; the retarded solution of $\Box A^\mu = \mu_0 J^\mu$ to The Wave Equation for the Four-Potential.

Builds on. The Four-Current Density · Charge Invariance and Continuity · Four-Vectors · Index Notation and the Metric Tensor. Leads to. Gauge Transformations and the Lorenz Gauge · The Electromagnetic Field Tensor · The Covariant Lorentz Force Law.

If the potentials are not unique, how can they be physical?

Both halves are true and they do not conflict. $A^\mu$ is not unique — you may add $\partial^\mu\chi$ freely — and yet $\oint A_\mu\,\mathrm dx^\mu$ around a closed loop is gauge-invariant, because the added piece integrates to zero around any closed path. That loop integral is exactly what the Aharonov–Bohm phase measures. So the physical content of $A^\mu$ is real but non-local: it lives in loops, not in points. What is unphysical is the pointwise value, and nothing ever depends on it.

Why do the potentials make life easier than working with E and B directly?

Two reasons. First, half of Maxwell's equations disappear — $\nabla\!\cdot\!\vec B = 0$ and Faraday's law become identities, leaving two instead of four. Second, in the Lorenz gauge what remains is $\Box A^\mu = \mu_0 J^\mu$ : four decoupled second-order wave equations, each with a standard retarded solution, rather than the coupled first-order system in $\vec E$ and $\vec B$ . Every radiation calculation exploits this — find $A^\mu$ , then differentiate.

Is A^μ a four-vector in every gauge?

No. The claim is proved within the Lorenz gauge class, where $\partial_\mu A^\mu = 0$ is a scalar condition every frame inherits. The Coulomb condition $\nabla\!\cdot\!\vec A = 0$ is not a scalar — it singles out one frame's spatial slice — so a boosted Coulomb-gauge potential is no longer in Coulomb gauge, and re-fixing it needs an extra gauge transformation that spoils the simple transformation law. The physics is unaffected, since $F^{\mu\nu}$ is gauge-invariant, but manifest covariance is lost.

Where does the magnetic field of a moving charge actually come from?

From the scalar potential of the stationary one. A charge at rest has $A^\mu = (\varphi/c, \vec 0\,)$ — purely timelike. A boost cannot leave a four-vector purely timelike, any more than it can leave a purely temporal displacement purely temporal: it drags part of $A^0$ into $A^1$ , producing $\vec A = \varphi'\vec v/c^2$ , and then $\vec B = \nabla\times\vec A$ is forced to be non-zero. Nothing magnetic was added at any point. Worked Example 1 does it with numbers, and the ratio of the boosted components comes out at exactly $\beta$ .

What does "minimal coupling" actually mean?

The interaction between matter and the field is the simplest Lorentz scalar you can build from the potential and the particle's worldline: $S_{\rm int} = q\!\int A_\mu\,\mathrm dx^\mu$ — no derivatives, no extra constants, no freedom. In the classical Lagrangian it produces the Lorentz force; in quantum mechanics, $\vec p \to \vec p - q\vec A$ ; in field theory, $\partial_\mu \to \partial_\mu + iqA_\mu$ . "Minimal" signals that nothing has been added beyond what gauge invariance demands — and that is enough to reproduce all of electrodynamics.

Four components, but the photon has two polarisations. Where do the other two go?

Gauge freedom eats them. The Lorenz condition $\partial_\mu A^\mu = 0$ removes one component; but it does not fix $A^\mu$ completely — a residual freedom survives (any $\chi$ with $\Box\chi = 0$ ), and that removes a second. Four minus two leaves the two transverse physical polarisations we observe, matching the null plane-wave solutions with $k_\mu k^\mu = 0$ and $k_\mu\epsilon^\mu = 0$ .