where $\partial_\mu\partial^\mu\!=\!-\Box^2$ is the wave operator and $\delta^{(4)}(x\!-\!x')$ is the four-dimensional delta function. Integration of this wave equation yields the explicit form: $G\!=\!\delta\{t'\!-t+R/c\}/(4\pi R)$. The function $G$ satisfies the property $\partial^\mu G\!=\!-\partial'^\mu G.$ We now evaluate (20) at the source point $x'$ and multiply the resulting equation by $\mu_0G$ and integrate over all spacetime, obtaining $$\begin{aligned}
\int\! \mu_0G\partial'_\nu J^\nu d^4x'=0.\end{aligned}$$ After an integration by parts in (21), in which we use the relation $G\partial'_\nu J^\nu\!=\! \partial_\nu (G J^\nu) + \partial'_\nu (G J^\nu)$ and the fact that the surface integral originated by the term $\partial'_\nu (G J^\nu)$ vanishes at spatial infinity, we take out the operator $\partial_\nu$ from the integral in (21) and obtain $$\begin{aligned}
\partial_\nu\!\! \int\!\mu_0 G J^\nu d^4x'=0.\end{aligned}$$ The integral in (22) must have some significant interpretation, we call it the four-potential $$\begin{aligned}
A^\nu\!= \!\mu_0\!\int\!\!G J^\nu d^4x',\end{aligned}$$ in terms of which (22) becomes elegantly simple compact $$\begin{aligned}
\partial_\nu A^\nu=0.\end{aligned}$$ In the next step we take the wave operator $\partial_\mu\partial^\mu$ to (23), use the result $\partial_\mu\partial^\mu G\!=\!\delta^{(4)}(x-x')$ and integrate over all spacetime to obtain the wave equation $$\begin{aligned}
\partial_\mu\partial^\mu A^\nu=\mu_0J^\nu.\end{aligned}$$ This is the covariant equation we were looking for. It clearly provides an expression for the four-current $J^\nu=\partial_\mu\partial^\mu A^\nu/\mu_0$ that 