Home exercise 5.5.3: The Lorenz gauge and charge conservation

The divergence commutes with ${\nabla }^2$ and with ${\partial }^2/\partial t^2$:

Multiply ${\nabla }^{2}V-\mu \epsilon {\partial }^{2}V/\partial t^2=-\rho /\epsilon$ by $\mu \epsilon$ and differentiate in time:

Adding the two equations,

If the potentials satisfy the Lorenz gauge, the left-hand side is zero and the sources must conserve charge. Conversely, if charge is conserved, the gauge expression $\Lambda =\nabla \cdot A+\mu \epsilon \partial V/\partial t$ obeys the homogeneous wave equation. If $\Lambda$ is zero before the sources are switched on (all fields and potentials zero), it stays zero for ever: the retarded potentials of sources that conserve charge satisfy the Lorenz gauge automatically.

Units: $\mu \nabla \cdot J$ and $\mu \partial \rho /\partial t$ are both in $\mathrm{H}A/{\mathrm{m}}^4$, and ${\nabla }^2(\nabla \cdot A)$ is in $Wb/{\mathrm{m}}^4=\mathrm{H}A/{\mathrm{m}}^4$.