# In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential: A^alpha = nu_0 j^alpha, to get A in terms of J, however, we have to use a considerably uglier formula

In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential:
$◻{A}^{\alpha }={\mu }_{0}{J}^{\alpha }$
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:
${A}^{\alpha }\left(t,\mathbf{r}\right)=\frac{{\mu }_{0}}{4\pi }\int \frac{{J}^{\alpha }\left(t-\frac{1}{c}‖\mathbf{r}-{\mathbf{r}}^{\prime }‖,{\mathbf{r}}^{\prime }\right)}{‖\mathbf{r}-{\mathbf{r}}^{\prime }‖}{d}^{3}{\mathbf{r}}^{\prime }$
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 $\left(t,\mathbf{r}\right)$ and the integration goes over only the spatial dimensions.
Can we rewrite the second equation in a covariant form? If not, then why not?
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ululatonh
The integral expression is Lorentz-covariant, too, and it may be made manifestly Lorentz-covariant, too.
The integral measure $\int {d}^{3}{r}^{\prime }/||\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }||$ is equal to and may be rewritten as the four-dimensional integral with a delta-function (and step function) added:
$2\int {d}^{4}{x}^{\prime }\cdot \delta \left[\left(x-{x}^{\prime }{\right)}^{2}\right]\cdot \theta \left(t-{t}^{\prime }\right)$
It's understood that $J$ is substituted at the point $J\left({x}^{\prime }\right)$.
Note that the step function $\theta$ (equal to one for positive arguments and zero otherwise) is Lorentz-covariant assuming that the points $x,{x}^{\prime }$ 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 $x,{x}^{\prime }$
The step function guarantees that the cause precedes its effect.
The argument of the delta-function is a Lorentz invariant, $\left(x-{x}^{\prime }{\right)}^{2}$ which means $\left(x-{x}^{\prime }{\right)}^{\mu }\left(x-{x}^{\prime }{\right)}_{\mu }$. 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 ${t}^{\prime }$. The theta-function implies that we only integrate over the semi-infinite line ${t}^{\prime }. The delta-function implies that the integral is only sensitive on the value of $J\left({t}^{\prime }\right)$ where $c|t-{t}^{\prime }|=|r-{r}^{\prime }|$ where the delta-function vanishes.
Finally, the delta-function also automatically generates the $1/|\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }|$ factor because
$\delta \left({y}^{2}\right)=\delta \left({y}_{0}^{2}-|\stackrel{\to }{y}{|}^{2}\right)=\delta \left[\left({y}_{0}+|\stackrel{\to }{y}|\right)\left({y}_{0}-|\stackrel{\to }{y}|\right)\right]=\dots$
which is equal, because $\delta \left(kX\right)=\delta \left(X\right)/|k|$, to
$\cdots =\frac{\delta \left({y}_{0}-|\stackrel{\to }{y}|\right)}{{y}_{0}+|\stackrel{\to }{y}|}=\frac{\delta \left({y}_{0}-|\stackrel{\to }{y}|\right)}{2|{\stackrel{\to }{y}}^{\prime }|}$
You see that the factor of two was needed, too.