6.3 Coulomb gauge

Definition of Coulomb gauge

The Coulomb gauge is a choice of gauge that constrains the vector potential to be divergence-free:

$\nabla \cdot \mathbf{A} = 0$

This single condition is what defines the gauge, and it has a direct physical consequence: the vector potential becomes purely transverse. That is, $\mathbf{A}$ has no longitudinal component. This transversality is why the Coulomb gauge is sometimes called the transverse gauge or radiation gauge.

By imposing this constraint, you reduce the gauge freedom in the system. The vector potential is then uniquely determined (up to a constant) once you also specify appropriate boundary conditions.

Coulomb gauge vs Lorenz gauge

These are the two most common gauge choices, and they impose different conditions on the potentials.

Divergence conditions

• Coulomb gauge: $\nabla \cdot \mathbf{A} = 0$
• Lorenz gauge: $\nabla \cdot \mathbf{A} = -\frac{1}{c^2} \frac{\partial \phi}{\partial t}$

The Lorenz condition is Lorentz covariant, meaning it takes the same form in every inertial frame. The Coulomb condition is not covariant; it picks out a preferred frame, which is why it's less natural for relativistic problems.

Implications for the potential equations

In the Coulomb gauge, the scalar potential satisfies Poisson's equation:

$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$

This is an instantaneous equation with no time derivatives, so $\phi$ responds immediately to changes in $\rho$. The vector potential satisfies a more complicated equation:

$\nabla^2 \mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J} + \frac{1}{c^2} \frac{\partial}{\partial t}(\nabla \phi)$

Notice the coupling to $\phi$ on the right-hand side. The scalar and vector potential equations are not fully decoupled in the time-dependent case.

In the Lorenz gauge, both potentials satisfy symmetric wave equations:

$\Box^2 \phi = -\frac{\rho}{\epsilon_0}, \qquad \Box^2 \mathbf{A} = -\mu_0 \mathbf{J}$

where $\Box^2 = \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}$ is the d'Alembertian. The equations are decoupled and manifestly covariant.

When to use each

• Coulomb gauge: Best for electrostatics, magnetostatics, and non-relativistic problems. Also the standard choice for quantizing the EM field in QED (since it isolates the physical transverse degrees of freedom).
• Lorenz gauge: Best for radiation, relativistic electrodynamics, and any problem where covariance matters.

Poisson's equation in Coulomb gauge

Because $\nabla \cdot \mathbf{A} = 0$, the equation for the scalar potential reduces to Poisson's equation:

$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$

This holds even in the time-dependent case. The scalar potential is determined entirely by the instantaneous charge distribution, with no retardation effects. The solution is:

$\phi(\mathbf{r}, t) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{r}', t)}{|\mathbf{r} - \mathbf{r}'|} \, d^3r'$

Note the key feature: the integrand uses $\rho$ evaluated at time $t$, not the retarded time. This looks like an instantaneous action-at-a-distance, but it's not a physical problem. The observable fields $\mathbf{E}$ and $\mathbf{B}$ still propagate at $c$. The apparent instantaneity is an artifact of the gauge choice; it gets cancelled by corresponding terms in $\mathbf{A}$.

Solving Poisson's equation

Standard techniques apply:

1. Green's function method: The free-space Green's function $G(\mathbf{r}, \mathbf{r}') = -\frac{1}{4\pi|\mathbf{r} - \mathbf{r}'|}$ gives the integral solution above directly.

2. Separation of variables: Useful when the geometry has a definite symmetry (spherical, cylindrical, Cartesian).

3. Numerical methods: Finite difference or finite element approaches for complex geometries.

The choice depends on the symmetry of the charge distribution and the boundary conditions.

Ampère's law in Coulomb gauge

The vector potential satisfies:

$\nabla^2 \mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J} + \frac{1}{c^2} \frac{\partial}{\partial t}(\nabla \phi)$

The right-hand side contains both the current density and a term involving the time derivative of $\nabla\phi$. You can interpret this more cleanly by decomposing the current density into transverse and longitudinal parts: $\mathbf{J} = \mathbf{J}_T + \mathbf{J}_L$. The $\nabla\phi$ term exactly cancels the longitudinal current contribution, so the equation effectively becomes:

$\nabla^2 \mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}_T$

This makes physical sense: a transverse (divergence-free) vector potential is sourced only by the transverse part of the current.

Solving for the vector potential

The same techniques used for Poisson's equation apply here, though the full time-dependent case requires solving a wave equation with a source. In the static limit (next two sections), the time derivatives drop out and the problem reduces to a vector Poisson equation.

Coulomb gauge in electrostatics

When all fields are time-independent, the Coulomb gauge gives the cleanest possible equations. The time derivatives vanish, and the potential equations fully decouple:

• Scalar potential: $\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$ (Poisson's equation)
• Vector potential: $\nabla^2 \mathbf{A} = 0$ (Laplace's equation, since $\mathbf{J} = 0$ in pure electrostatics)

With appropriate boundary conditions (e.g., $\phi \to 0$ and $\mathbf{A} \to 0$ at infinity), Laplace's equation for $\mathbf{A}$ gives $\mathbf{A} = 0$. The entire problem reduces to solving Poisson's equation for $\phi$, and the electric field is simply $\mathbf{E} = -\nabla\phi$.

The potentials are uniquely determined by the charge distribution and boundary conditions, which is one reason the Coulomb gauge is the natural default for electrostatics.

Coulomb gauge in magnetostatics

In magnetostatics, the fields are time-independent but steady currents are present. The equations become:

• Scalar potential: $\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$ (or $\nabla^2 \phi = 0$ if there are no free charges)
• Vector potential: $\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}$ (vector Poisson's equation)

The magnetic field is then obtained from:

$\mathbf{B} = \nabla \times \mathbf{A}$

The Coulomb gauge condition $\nabla \cdot \mathbf{A} = 0$ is automatically consistent with $\nabla \cdot \mathbf{B} = 0$, since the divergence of a curl is always zero. The integral solution for the vector potential is:

$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} \, d^3r'$

You can verify that this solution satisfies $\nabla \cdot \mathbf{A} = 0$ provided $\nabla' \cdot \mathbf{J} = 0$ (the continuity equation for steady currents).

Coulomb gauge in time-dependent fields

The Coulomb gauge can be used for time-dependent problems, but the equations are less elegant than in the Lorenz gauge.

The instantaneous scalar potential

As discussed above, $\phi$ satisfies Poisson's equation even when fields vary in time. The solution uses the instantaneous charge density, not the retarded one. This is a peculiar but valid feature of the Coulomb gauge.

The vector potential and retardation

All the retardation physics (finite propagation speed) is carried by the vector potential. The equation for $\mathbf{A}$ is a wave equation sourced by the transverse current $\mathbf{J}_T$, and its solution does involve retarded-time integrals.

The full retarded solution for $\mathbf{A}$ in the Coulomb gauge is more involved than in the Lorenz gauge, because you must first decompose $\mathbf{J}$ into transverse and longitudinal parts, which requires a nonlocal integral over the current distribution.

Limitations

• The equations are not Lorentz covariant, making them awkward for relativistic problems.
• The nonlocal decomposition of $\mathbf{J}$ into transverse and longitudinal parts can be computationally expensive.
• Radiation problems are generally easier in the Lorenz gauge or using Jefimenko's equations directly.

Gauge transformations and Coulomb gauge

The electric and magnetic fields are the physical observables, and they are invariant under gauge transformations:

$\phi \rightarrow \phi - \frac{\partial \Lambda}{\partial t}, \qquad \mathbf{A} \rightarrow \mathbf{A} + \nabla \Lambda$

where $\Lambda(\mathbf{r}, t)$ is an arbitrary scalar function.

Reaching the Coulomb gauge

If you start with potentials $(\phi, \mathbf{A})$ that don't satisfy $\nabla \cdot \mathbf{A} = 0$, you can find a gauge function $\Lambda$ that transforms you into the Coulomb gauge. The new vector potential $\mathbf{A}' = \mathbf{A} + \nabla\Lambda$ must satisfy:

$\nabla \cdot \mathbf{A}' = \nabla \cdot \mathbf{A} + \nabla^2 \Lambda = 0$

So you need to solve:

$\nabla^2 \Lambda = -\nabla \cdot \mathbf{A}$

This is just Poisson's equation for $\Lambda$, which always has a solution (given suitable boundary conditions). That's the proof that you can always find a gauge transformation into the Coulomb gauge.

Preserving the Coulomb gauge

Once you're in the Coulomb gauge, the residual gauge freedom consists of transformations with $\nabla^2 \Lambda = 0$. These are harmonic functions, and with standard boundary conditions ($\Lambda \to 0$ at infinity), the only solution is $\Lambda = \text{const}$. This means the Coulomb gauge essentially fixes the potentials uniquely.

Applications of Coulomb gauge

Electromagnetic waves

The transverse nature of $\mathbf{A}$ in the Coulomb gauge maps directly onto the transverse polarization of electromagnetic waves. In free space ($\rho = 0$, $\mathbf{J} = 0$), the scalar potential vanishes and the vector potential satisfies the free wave equation:

$\nabla^2 \mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = 0, \qquad \nabla \cdot \mathbf{A} = 0$

The condition $\nabla \cdot \mathbf{A} = 0$ enforces that only the two transverse polarization modes propagate, which is the correct physical content.

Quantum electrodynamics

The Coulomb gauge is the standard gauge for canonical quantization of the electromagnetic field. The reason: $\nabla \cdot \mathbf{A} = 0$ eliminates the unphysical longitudinal and scalar photon degrees of freedom from the start. You promote only the two transverse components of $\mathbf{A}$ to quantum operators, and each mode corresponds to a physical photon polarization. The Coulomb interaction between charges appears as an instantaneous potential (the same Poisson solution for $\phi$), while radiation effects come from the quantized transverse field.

Plasma physics

In plasma physics, the Coulomb gauge is useful for separating electromagnetic fields into transverse (radiative) and longitudinal (electrostatic) components. This decomposition aligns naturally with the distinction between electromagnetic waves propagating through the plasma and electrostatic oscillations (like Langmuir waves). The quasi-neutrality approximation in many plasma models also pairs well with the Coulomb gauge's treatment of the scalar potential.