Lorenz gauge condition

Remarks

It is a gauge condition: if $L = \nabla \cdot A + c^{-2} \partial_t \phi$ for given potentials, a gauge transformation with a function $\chi$ that solves $\frac{1}{c^2} \partial_t^2 \chi - \Delta \chi = L$ yields potentials that satisfy it. Solutions $\chi$ of the homogeneous wave equation leave the condition unchanged, so the gauge is not complete.

In the Lorenz gauge, Maxwell's equations in vacuum with charge density $\rho$ and current density $J$ become the decoupled wave equations $$\frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} - \Delta \phi = \frac{\rho}{\varepsilon_0}, \qquad \frac{1}{c^2} \frac{\partial^2 A}{\partial t^2} - \Delta A = \mu_0 J.$$ Their retarded solutions give the retarded potentials.

Example: the static Coulomb potentials $\phi = q / (4 \pi \varepsilon_0 \abs{x})$, $A = 0$ satisfy the condition, since $\nabla \cdot A = 0$ and $\partial_t \phi = 0$.

Non-example: the potentials $\phi = 0$, $A = (x, 0, 0)$ describe the zero field, but $\nabla \cdot A + c^{-2} \partial_t \phi = 1 \neq 0$, so they are not in the Lorenz gauge.