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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?

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More on retarded $4$-potential: physics.stackexchange.com/q/67533/2451 –  Qmechanic Jun 20 '13 at 6:47
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Lorentz covariance of the solution is provided with the Lorentz covariance of the four-current and the Lorentz invariance of the retarded Green's function given in Lubosh's answer. –  Vladimir Kalitvianski Jun 20 '13 at 11:19
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up vote 5 down vote accepted

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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I think we also need an extra factor of $c$ in order to make the units work out. –  Brian Bi Jun 20 '13 at 7:20
    
Right. I just wasn't sure which powers of $c$ I wanted to add to the definition of the 4-vector because this depends on some conventions, too. Adult physicists work in the $c=1$ units, anyway. –  Luboš Motl Jun 20 '13 at 8:33
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