Maxwell's equations in Coulomb gauge

Units are SI throughout, and the one four-vector statement uses the metric signature (+, −, −, −). The fields E = −∇φ − ∂A/∂t and B = ∇ × A do not change under A → A + ∇χ, φ → φ − ∂χ/∂t, so four potential components describe fewer physical ones. Fourier transform in space and split every vector along and across its wavevector: V∥(k) = k̂(k̂⋅V) and V⊥ = V − V∥, so that ∇ × V∥ = 0 and ∇⋅V⊥ = 0. Coulomb gauge sets ∇⋅A = 0, that is A∥ = 0. It can always be reached, because the χ solving ∇²χ = −∇⋅A removes any divergence (derived), and nothing is left over: a χ harmonic everywhere and vanishing at infinity is zero (quoted from potential theory). The labels refer to k, not to position. On the equatorial plane of a static dipole the curl-free field E∥ points along −p̂, at right angles to r̂. What remains per wavevector is φ, which the next step shows is not dynamical, and the two components of A across k.

Put E = −∇φ − ∂A/∂t into ∇⋅E = ρ/ε₀. The A term is −∂(∇⋅A)/∂t, which the gauge removes, leaving ∇²φ = −ρ/ε₀ and φ(r, t) = (1/4πε₀)∫ρ(r′, t)/|r − r′| d³r′ (derived). The time is the same on both sides. φ is fixed at every instant by where the charges are and has no initial data of its own, so it is a constraint, not a field to be quantised. Because A is transverse, so is ∂A/∂t, and the curl-free part of E is all in E∥ = −∇φ, the Coulomb field of the instantaneous charge distribution, while E⊥ = −∂A/∂t. Dimensions: ρ/ε₀ is C m⁻³ divided by C V⁻¹ m⁻¹, or V m⁻², the units of ∇²φ. Limiting case: for static charges this φ is the electrostatic potential, and it agrees with the Lorenz-gauge φ, which is a retarded integral; the two gauges part company only when charges move.

Ampère-Maxwell, ∇ × B = μ₀j + (1/c²)∂E/∂t, with ∇ × (∇ × A) = −∇²A in this gauge, becomes ∇²A − (1/c²)∂²A/∂t² = −μ₀j + (1/c²)∇(∂φ/∂t). Differentiate Poisson's solution in time and use ∂ρ/∂t = −∇⋅j: ε₀∇(∂φ/∂t) is a curl-free field whose divergence is ∇⋅j, and the only such field that vanishes at infinity is j∥. The last term is therefore μ₀j∥, it cancels the longitudinal current exactly, and ∇²A − (1/c²)∂²A/∂t² = −μ₀j⊥ (derived); μ₀j⊥ is N A⁻² × A m⁻² = T m⁻¹, the units of ∇²A. The longitudinal current only moves charge, which φ already follows. The price is locality, because the projector is local in k but not in r. Away from a point current element Il at the origin, its transverse part is j⊥ = [3r̂(r̂⋅Il) − Il]/(4πr³) (quoted from the transverse δ function). For Il = 0.010 A m that is 1.6 × 10⁻³ A m⁻² at 1 m along the axis: a source of A spread through all of space.

Write U = (ε₀/2)∫(E² + c²B²) d³r with E = E∥ + E⊥. The cross term is ε₀∫E∥⋅E⊥ d³r = −ε₀∫∇φ⋅E⊥ d³r = ε₀∫φ ∇⋅E⊥ d³r = 0, integrating by parts with fields that fall off at infinity. A second integration by parts and Poisson's equation turn (ε₀/2)∫E∥² d³r into ½∫ρφ d³r, so U = ½∫ρφ d³r + (ε₀/2)∫(E⊥² + c²B²) d³r exactly (derived). Only the integrals split: where E∥ and E⊥ cancel, their local energy densities do not simply add. For point charges ½∫ρφ is the sum over pairs of qᵢqⱼ/(4πε₀rᵢⱼ), a potential energy of the instantaneous positions with no degrees of freedom of its own, plus self-energies. With no charges the first term vanishes and all the energy is transverse, the free-field limit. In hydrogen the pair term is the whole binding: with the reduced mass it gives a binding energy of 13.598287 eV against the measured ionisation energy 13.598434 eV. The 1.47 × 10⁻⁴ eV left over is Dirac's fine structure, +1.81 × 10⁻⁴ eV, less the 1S Lamb shift, 3.4 × 10⁻⁵ eV (quoted).

φ responds at once to a distant charge, yet nothing measurable does, and gauge invariance is the reason. E and B are the same in every gauge, and in Lorenz gauge the potentials are retarded integrals over ρ and j, so the true E is the retarded field Eᵣₑₜ (stated, quoting the Lorenz-gauge solution). The curl-free part of any E is fixed by ∇⋅E = ρ/ε₀ at the same instant, so E∥ is always the instantaneous Coulomb field and E⊥ = Eᵣₑₜ − E∥. Where light from a change has not arrived, Eᵣₑₜ is unchanged and E⊥ cancels the change in E∥ exactly; A does change there, driven by the j⊥ that fills space. Ramp a dipole from 0 to p₀ over a time τ and watch its equatorial plane at distance r. In units of p₀/(4πε₀r³), the transverse part's largest excursion is 1.35 at r = 0.5cτ but only 0.033 at r = 0.1cτ, and it falls as (r/cτ)² as r/cτ → 0. That quasi-static limit, where the Coulomb field is the whole field, is the atom's regime: for hydrogen's Lyman-α line the role of cτ is played by c/ω = 19 nm, against an atom 0.1 nm across, so r/cτ ≈ 0.005.

∇⋅A = 0 is not a Lorentz-invariant statement. The potentials form a four-vector A^μ = (φ/c, A), and the covariant gauge is Lorenz's, ∂_μA^μ = 0, under which all four components obey □A^μ = μ₀j^μ with j^μ = (cρ, j) and □ = (1/c²)∂²/∂t² − ∇². A boost mixes φ and A, so a field in Coulomb gauge in one frame is not in it in another, and what one observer files as the Coulomb field of a moving charge another counts as partly transverse. For an atom this costs nothing, because the nucleus supplies a rest frame in which the electrons move at about αc, or c/137, the Coulomb term is treated exactly, and the coupling to the transverse field is a small perturbation; retardation of the electron-electron interaction enters only through it, at relative order (v/c)² ≈ α² = 5.3 × 10⁻⁵ (quoted, the Breit interaction). Quantising in Lorenz gauge keeps covariance at the price of four photon polarisations, two removed by a subsidiary condition (Gupta-Bleuler, named only). This course quantises the two transverse ones.