TITLE: Can we write the electromagnetic potential covariantly in terms of the four-current?
QUESTION [7 upvotes]: In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential:
$$\Box A^{\alpha} = \mu_0 J^{\alpha}$$
To get $A$ in terms of $J$, however, we have to use a considerably uglier formula; or, at least, this is the formula presented in textbooks:
$$A^{\alpha}(t, \mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{J^{\alpha}\left(t - \frac{1}{c}\|\mathbf{r} - \mathbf{r}'\|, \mathbf{r}' \right)}{\|\mathbf{r} - \mathbf{r}'\|} d^3\mathbf{r}'$$
The first equation is evidently Lorentz covariant. The second one, on the other hand, doesn't look covariant at all. The integrand has some messy dependence on $(t, \mathbf{r})$ and the integration goes over only the spatial dimensions.
Can we rewrite the second equation in a covariant form? If not, then why not?

REPLY [7 votes]: The integral expression is Lorentz-covariant, too, and it may be made manifestly Lorentz-covariant, too.
The integral measure $\int d^3 r' / ||\vec r - \vec r'||$ is equal to and may be rewritten as the four-dimensional integral with a delta-function (and step function) added:
$$2\int {d^4 x'} \cdot \delta[(x-x')^2]\cdot \theta(t-t') $$
It's understood that $J$ is substituted at the point $J(x')$.
Note that the step function $\theta$ (equal to one for positive arguments and zero otherwise) is Lorentz-covariant assuming that the points $x,x'$ aren't spacelike-separated (because the ordering of a cause and its effect is frame-independent), and they're not spacelike-separated as guaranteed by the delta-function that is only non-vanishing near/at the null separation of $x,x'$
The step function guarantees that the cause precedes its effect.
The argument of the delta-function is a Lorentz invariant, $(x-x')^2$ which means $(x-x')^\mu(x-x')_\mu$. The sign convention for the metric doesn't matter becase this invariant is the argument of an even function (delta-function). 
Finally, the equivalence of the two integrals may be shown by performing the integral over $t'$. The theta-function implies that we only integrate over the semi-infinite line $t'\lt t$. The delta-function implies that the integral is only sensitive on the value of $J(t')$ where $c|t-t'| = |r-r'|$ where the delta-function vanishes. 
Finally, the delta-function also automatically generates the $1/|\vec r - \vec r'|$ factor because
$$\delta(y^2) = \delta (y_0^2-|\vec y|^2) = \delta[(y_0+|\vec y|)(y_0-|\vec y|)]=\dots  $$
which is equal, because $\delta(kX) = \delta (X)/ |k|$, to
$$\dots = \frac{\delta(y_0-|\vec y|)}{y_0+|\vec y|}=\frac{\delta(y_0-|\vec y|)}{2|\vec y'|}$$
You see that the factor of two was needed, too.
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