= Solution
The <Lorenz gauge condition> is $\nabla\cdot\mathbf A=-c^{-2}\partial_t\phi$. Substitute the <electric potential> and <magnetic vector potential> into <Gauss's law>:
$$
-\nabla^2\phi-\partial_t(\nabla\cdot\mathbf A)=\rho/\epsilon_0,
\qquad
\Box\phi=-\rho/\epsilon_0.
$$
Next the <Ampère-Maxwell equation>, the <curl of the curl identity>, and the <Lorenz gauge condition> give
$$
\begin{aligned}
\nabla(\nabla\cdot\mathbf A)-\nabla^2\mathbf A
&=\mu_0\mathbf J-c^{-2}\nabla\partial_t\phi-c^{-2}\partial_t^2\mathbf A,\\
\Box\mathbf A&=-\mu_0\mathbf J.
\end{aligned}
$$
The <gradient> terms cancel. Since $\mu_0\epsilon_0c^2=1$, the time component obeys $\Box(\phi/c)=-\mu_0c\rho$, so together
$$
\boxed{\partial^\nu\partial_\nu A^a=-\mu_0J^a,\qquad A^\mu=(\phi/c,\mathbf A).}
$$
The <four-potential> can be chosen to transform as a <four-vector>. To make the argument rigorous, choose the <retarded potential> with no freely added homogeneous solution. If $\Box G_{\mathrm{ret}}(x)=\delta^{(4)}(x)$, this choice is
$$
A^\mu(x)=-\mu_0\int G_{\mathrm{ret}}(x-y)J^\mu(y)\,d^4y.
$$
The <retarded Green function> is a <Lorentz scalar> under <proper orthochronous Lorentz transformations>, and $d^4y$ is invariant. The <four-vector> rule for $J^\mu$ therefore proves $A'^\mu(x')=\Lambda^\mu{}_{\nu}A^\nu(x)$. More generally, transform any homogeneous initial data by this same rule. Both the <wave equation> and the <Lorenz gauge condition> are then preserved.
\b[Qualification.] The <wave equation> by itself cannot prove the transformation rule for every independently chosen <gauge potential>: the <d'Alembert operator> has a nontrivial <kernel of a linear map>. For a concrete counterexample, zero sources and zero fields admit $A^\mu=0$ in one <inertial frame> and $A'^\mu=(0,C,0,0)$ in another. Both satisfy the <wave equation> and the <Lorenz gauge condition>, but the latter is not the <Lorentz transformation> of zero. It is a <residual Lorenz gauge symmetry>, generated by $\chi=Cx'$. Thus the usual assertion is a covariant choice of <four-potential>, with a consistent <gauge transformation> and homogeneous data, rather than a uniqueness result from the displayed equation.
Back to article page
