Potentials and Gauge Freedom

Derivation

Potentials and Gauge Freedom

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Statement

Because the two homogeneous Maxwell equations, ∇·B = 0 and ∇×E = −∂B/∂t, are constraints of pure structure, the fields can be written as B = ∇×A and E = −∇φ − ∂A/∂t in terms of a vector potential A(r,t) and a scalar potential φ(r,t). These potentials are not unique: the simultaneous replacement A → A + ∇χ, φ → φ − ∂χ/∂t, for any smooth scalar field χ(r,t), leaves both E and B unchanged. This is the gauge freedom of electromagnetism.

Why it matters

Two of Maxwell's four equations carry no source and no dynamics of their own — they are identities that any physical field must satisfy. Trading the six components of (E,B) for the four components of (φ,A) makes those two equations automatic: they hold by the vector-calculus identities ∇·(∇×A) = 0 and ∇×(∇φ) = 0, whatever the potentials do. Only the two inhomogeneous equations remain to be solved, and they become second-order equations for (φ,A).

The gauge freedom that comes with this trade is not a nuisance to be tolerated but the organising principle of the whole subject. It is redundancy we can spend: choosing χ to impose a convenient condition (Coulomb, Lorenz, temporal) decouples the equations. Promoted to a local symmetry, this same freedom is the template on which the entire Standard Model of particle physics is built.

Assumptions

The field domain is topologically simple (star-shaped / simply connected).On a domain with holes, a divergence-free B need not be a global curl and a curl-free field need not be a global gradient; potentials then exist only locally, and quantities like magnetic flux through a non-contractible loop become gauge-invariant physical data (Aharonov–Bohm).

The fields are sufficiently smooth (C¹, and the potentials C²).Poincaré's lemma and the interchange of mixed partials ∂_t∇ = ∇∂_t both require continuous derivatives; at a genuine field discontinuity (a surface current or charge sheet) the potentials must instead be matched by boundary conditions.

We work in a single inertial frame with a fixed choice of coordinates and a flat metric.The split E = −∇φ − ∂A/∂t mixes under Lorentz boosts; the frame-independent object is the four-potential A^μ = (φ/c, A). Dropping this makes the 3-vector decomposition frame-dependent, though the gauge structure survives as A_μ → A_μ + ∂_μχ.

Derivation

∇·B = 0

Homogeneous Maxwell equation (no magnetic monopoles); one of the two source-free constraints inherited from the assembly of Maxwell's equations. A

B = ∇×A

On a simply connected domain a field with zero divergence everywhere is the curl of some vector field (Poincaré lemma / Helmholtz). This defines A; the identity ∇·(∇×A) ≡ 0 then makes Step 1 hold automatically. B

∇×E = −∂B/∂t = −∂(∇×A)/∂t

Substitute Step 2 into Faraday's law, the second homogeneous equation. A

∂(∇×A)/∂t = ∇×(∂A/∂t)

Space and time derivatives commute for a C² field (Clairaut/Schwarz); the curl is a spatial operator that passes through ∂/∂t. B

∇×(E + ∂A/∂t) = 0

Combine Steps 3–4 and move both curls to one side; the curl is linear, so ∇×E + ∇×(∂A/∂t) = 0 collects into a single curl. A

E + ∂A/∂t = −∇φ

A field whose curl vanishes on a simply connected domain is a gradient (Poincaré lemma again). The minus sign and the symbol φ are conventions chosen so that in statics E = −∇φ recovers the electrostatic potential. This defines the scalar potential φ. B

E = −∇φ − ∂A/∂t

Rearrange Step 6 for E. Both homogeneous equations are now built into the form of the fields. A

A′ = A + ∇χ

Now probe uniqueness. Add the gradient of an arbitrary scalar field χ(r,t) to A. The identity ∇×(∇χ) ≡ 0 guarantees B′ = ∇×A′ = ∇×A = B: the magnetic field is untouched for any χ. B

E′ = −∇φ − ∂A′/∂t = −∇φ − ∂A/∂t − ∇(∂χ/∂t)

Compute E with the shifted potential, keeping φ fixed for the moment, and use ∂(∇χ)/∂t = ∇(∂χ/∂t) (Step 4's commutation). The change in A has leaked an unwanted term −∇(∂χ/∂t) into E. B

φ′ = φ − ∂χ/∂t

To cancel the leaked term, shift the scalar potential too. Then −∇φ′ = −∇φ + ∇(∂χ/∂t), which exactly annihilates the extra piece in Step 9. This fixes the required companion transformation of φ. B

E′ = −∇φ′ − ∂A′/∂t = −∇φ − ∂A/∂t = E

Substitute Step 10 into Step 9; the ∇(∂χ/∂t) terms cancel identically. Both E and B are invariant under the joint transformation. A

Result

B = ∇×A, E = −∇φ − ∂A/∂t

A → A + ∇χ, φ → φ − ∂χ/∂t ⟹ (E,B) unchanged

Reading. Every electromagnetic field can be generated from a single scalar and a single vector potential, and the homogeneous Maxwell equations are then satisfied identically. The potentials carry one extra function's worth of redundancy: any χ(r,t) reshuffles φ and A without moving a single measurable field line. Physics lives in the equivalence classes of potentials modulo gauge, not in the potentials themselves.

Units check. With [B] = T and [∇] = m⁻¹, B = ∇×A gives [A] = T·m = Wb/m = V·s/m. Then [∂A/∂t] = V·s·m⁻¹·s⁻¹ = V/m = [E] ✓, and [∇φ] = V/m ✓. For the gauge scalar, [∇χ] = [A] = V·s/m ⟹ [χ] = V·s = Wb, and consistently [∂χ/∂t] = V = [φ] ✓.

Limiting cases

Statics (∂/∂t = 0): the two potentials decouple, E = −∇φ and B = ∇×A, recovering electrostatics and magnetostatics; the residual gauge freedom is A → A + ∇χ with time-independent χ.
Pure radiation (no sources, temporal/radiation gauge φ = 0): the field is carried entirely by A, with E = −∂A/∂t and B = ∇×A.
Coulomb gauge (∇·A = 0): φ obeys the instantaneous Poisson equation ∇²φ = −ρ/ε₀, so the scalar potential tracks the charge distribution with no retardation while all propagation sits in A.
Lorenz gauge (∇·A + (1/c²)∂φ/∂t = 0): φ and each component of A satisfy the same wave equation □φ = ρ/ε₀, □A = μ₀J, a manifestly Lorentz-covariant form.

Breaks when

The domain is not simply connected. Around a hole (an excluded solenoid, a wire threading a loop) a locally curl-free or divergence-free field can fail to admit a single-valued global potential. The line integral ∮A·dl = Φ then becomes a gauge-invariant physical quantity, observable as the Aharonov–Bohm phase shift even where B = 0.
Magnetic monopoles exist. If ∇·B = μ₀ρ_m ≠ 0, then B is no longer a global curl and a smooth A cannot cover all of space; one is forced into patched potentials (a Dirac string / fibre-bundle description), and charge quantisation emerges as the consistency condition.
Fields are discontinuous or singular. At a surface current or a point source the derivatives in Steps 4 and 6 are ill-defined; the potentials must be constructed piecewise and matched by jump conditions, not by the smooth Poincaré-lemma construction used here.
Strong-field / quantum electrodynamics. When photon self-interactions matter (vacuum birefringence, light-by-light scattering) the effective field equations become nonlinear in A; the linear superposition and the clean split above are only leading-order approximations.

Failure modes

Transforming A but forgetting φ. Shifting A → A + ∇χ alone leaves B right but corrupts E by −∇(∂χ/∂t). The two shifts are a package deal whenever χ depends on time.
Sign errors in the companion shift. Writing φ → φ + ∂χ/∂t (wrong sign) doubles the spurious term instead of cancelling it. The signs are fixed by the −∇φ and −∂A/∂t conventions in the definition of E.
Believing A is "the real field." Treating A or φ as physically measurable pointwise; only gauge-invariant combinations (E, B, and holonomies ∮A·dl) are observable.
Confusing a gauge choice with a physical assumption. Reading ∇·A = 0 or φ = 0 as a restriction on the physics rather than a free bookkeeping choice available for any field.
Applying Poincaré's lemma globally on a multiply connected region. Assuming "curl-free ⟹ gradient" or "div-free ⟹ curl" everywhere, ignoring the topological caveat that makes Aharonov–Bohm possible.
Dropping the vector identity justification. Claiming B = ∇×A "obviously" satisfies ∇·B = 0 without invoking ∇·(∇×A) ≡ 0 — the whole point is that it is an identity, not a dynamical statement.

Discussion

The deep content of this derivation is a division of labour among Maxwell's four equations. The two homogeneous equations are not laws of motion at all; they are integrability conditions — statements that B is a curl and that E + ∂A/∂t is a gradient. By solving them structurally, once and for all, we reduce the dynamical content of electromagnetism to the two inhomogeneous equations, now written for (φ,A). The potentials are the natural variables in which Maxwell theory becomes a well-posed field theory with a Lagrangian and a canonical structure.

Gauge freedom is best understood as a redundancy of description: the map from potentials to fields is many-to-one, its kernel being exactly the pure-gauge configurations (δφ, δA) = (−∂χ/∂t, ∇χ). This redundancy is a feature we exploit. Because χ is completely free, we may impose one scalar condition on the potentials at no physical cost; Coulomb gauge kills ∇·A and instantaneises φ, while Lorenz gauge symmetrises space and time so that potentials propagate as waves at speed c. The physics is identical in every gauge — only the arithmetic differs.

The modern reading turns the logic inside out. Rather than deriving gauge freedom from Maxwell's equations, one postulates local phase invariance of a charged matter field, ψ → e^iqχ/ℏψ, and discovers that consistency forces the existence of a connection A_μ transforming as A_μ → A_μ + ∂_μχ — the very gauge transformation derived above. Electromagnetism is then the unique minimal theory of a U(1) gauge connection, and its field strength F_μν = ∂_μA_ν − ∂_νA_μ repackages E and B as the manifestly gauge-invariant curvature. Replacing U(1) by a non-abelian group yields the weak and strong interactions; this humble χ is the ancestor of the entire Standard Model.

Common misconceptions. Gauge freedom is not "arbitrariness in the fields" — the fields are rigidly fixed; it is arbitrariness in the labels. And potentials are not merely computational fictions: in quantum mechanics the phase (q/ℏ)∮A·dl is measurable (Aharonov–Bohm) even in regions where E = B = 0, showing that the gauge-invariant content of A is more than B alone encodes.

Worked examples

Example 1 — A vector potential for a uniform field, and the gauge linking two choices.

B = B ẑ, B = 0.50 T (uniform)

Given field; seek A with ∇×A = B. A

A_sym = ½B×r = ½B(−y, x, 0)

Symmetric gauge. Check: (∇×A)_z = ∂_x(½Bx) − ∂_y(−½By) = ½B + ½B = B ✓. B

at r = (2.0, 0, 0) m: A_sym = ½(0.50)(0, 2.0, 0) = (0, 0.50, 0) T·m

Evaluate; magnitude |A| = 0.50 T·m = 0.50 V·s/m. A

A_Lan = B(−y, 0, 0)

Landau gauge — a different potential for the same B: (∇×A_Lan)_z = −∂_y(−By) = B ✓. B

A_Lan − A_sym = B(−½y, −½x, 0) = ∇χ ⟹ χ = −½Bxy

Solve ∂_xχ = −½By, ∂_yχ = −½Bx; both give χ = −½Bxy. Static χ, so φ is unchanged. C

A_sym(2,0,0) = 0.50 ŷ T·m, linked to Landau gauge by χ = −½Bxy = −0.25xy (V·s)

Reading. The two potentials look completely different yet describe one field; the gauge function χ = −½Bxy is the exact relabelling between them. Units check. [χ] = T·m² = V·s ✓; ∇χ has units T·m = [A] ✓.

Example 2 — Fields from a plane-wave vector potential.

φ = 0, A = A₀ cos(kz − ωt) x̂

Radiation (temporal) gauge; a travelling wave along ẑ, polarised along x̂. A

E = −∂A/∂t = −A₀ω sin(kz − ωt) x̂

Differentiate; with φ = 0 the electric field is carried entirely by A. A

B = ∇×A = ∂_zA_x ŷ = −A₀k sin(kz − ωt) ŷ

Only ∂_zA_x survives; E⊥B⊥ẑ, a transverse wave. B

A₀ = 1.0×10⁻⁶ T·m, f = 1.0 GHz ⟹ ω = 2πf = 6.28×10⁹ s⁻¹, k = ω/c = 20.9 m⁻¹

Insert numbers; k from the vacuum dispersion relation ω = ck. A

E₀ = A₀ω = 6.28×10³ V/m, B₀ = A₀k = 2.09×10⁻⁵ T

Amplitudes. Ratio E₀/B₀ = ω/k = c = 3.0×10⁸ m/s — the wave is self-consistent. B

E₀ = 6.28×10³ V/m, B₀ = 2.09×10⁻⁵ T, E₀/B₀ = c

Reading. A single vector potential, with φ set to zero by gauge choice, reproduces a complete transverse electromagnetic wave with the correct E/B = c. Units check. [A₀ω] = (T·m)(s⁻¹) = T·m/s = V/m ✓; [A₀k] = (T·m)(m⁻¹) = T ✓.

Problems

(A) Show explicitly that B = ∇×A satisfies ∇·B = 0 for any smooth A, and that E = −∇φ − ∂A/∂t satisfies ∇×E = −∂B/∂t.
Solution
∇·(∇×A) = ε_ijk∂_i∂_jA_k. The Levi-Civita symbol ε_ijk is antisymmetric in i,j while ∂_i∂_j is symmetric (mixed partials commute for C² fields); a symmetric object contracted with an antisymmetric one vanishes, so ∇·B = 0. For Faraday: ∇×E = ∇×(−∇φ) − ∇×(∂A/∂t) = 0 − ∂_t(∇×A) = −∂B/∂t, using ∇×∇φ ≡ 0 and the commutation of ∇× with ∂_t.
(B) A uniform field B = 2.0 T ẑ fills a region. Write a vector potential in the symmetric gauge and evaluate it at r = (0, 3.0, 0) m. Give magnitude and direction.
Solution
A = ½B×r = ½(2.0)(−y, x, 0) = (−y, x, 0) T·m (since ½B = 1.0). At (0,3.0,0): A = (−3.0, 0, 0) T·m, i.e. magnitude 3.0 T·m pointing in the −x̂ direction. Check: (∇×A)_z = ∂_xx − ∂_y(−y) = 1 + 1 = 2.0 T ✓.
(B) The potentials φ = 0, A = (0, 0, αt) with α = 5.0 T·m/s (equivalently 5.0 V/m), so that A_z = αt, describe a spatially uniform, time-varying A. Find E and B. Then transform to the gauge χ = −αtz and verify the fields are unchanged while φ and A change.
Solution
Fields: B = ∇×A = 0 (A is spatially uniform); E = −∂A/∂t = (0,0,−α) = −5.0 ẑ V/m. Gauge transform with χ = −αtz: A′ = A + ∇χ = (0,0,αt) + (0,0,−αt) = 0, and φ′ = φ − ∂χ/∂t = 0 − (−αz) = αz = 5.0z (V). New fields: E′ = −∇φ′ − ∂A′/∂t = −(0,0,α) − 0 = −5.0 ẑ V/m ✓, B′ = 0 ✓. Same field, described once by a time-dependent A and once by a static φ.
(C) Given the field B = B₀ ẑ in the Landau gauge A = B₀ x ŷ, find the gauge function χ that transforms it to the symmetric gauge A′ = ½B₀(−y, x, 0).
Solution
Need ∇χ = A′ − A = ½B₀(−y, x, 0) − (0, B₀x, 0) = ½B₀(−y, −x, 0). Integrate: ∂_xχ = −½B₀y ⟹ χ = −½B₀xy + g(y); then ∂_yχ = −½B₀x + g′(y) = −½B₀x ⟹ g′ = 0. Hence χ = −½B₀xy (up to an additive constant, which shifts nothing). Both potentials are static, so φ is unaffected.
(C) Starting from arbitrary potentials (φ,A) with ∇·A ≠ 0, show what equation the gauge function must satisfy to reach the Coulomb gauge ∇·A′ = 0, and explain why a solution always exists on ℝ³. Then, for the Lorenz gauge, state the analogous equation.
Solution
Coulomb: A′ = A + ∇χ ⟹ ∇·A′ = ∇·A + ∇²χ. Setting this to zero requires ∇²χ = −∇·A — a Poisson equation with source −∇·A. On ℝ³ with decaying sources it always has the solution χ(r) = (1/4π)∫ [∇′·A(r′)/|r−r′|] d³r′, so Coulomb gauge is always reachable. Lorenz: demanding ∇·A′ + (1/c²)∂φ′/∂t = 0 with φ′ = φ − ∂χ/∂t gives the inhomogeneous wave equation □χ ≡ (∇² − (1/c²)∂_t²)χ = −[∇·A + (1/c²)∂φ/∂t], which likewise has retarded-potential solutions. A residual freedom remains in each: harmonic χ (∇²χ = 0) for Coulomb, source-free □χ = 0 for Lorenz.