## Lorentz Invariance of the Four-Potential Lorentz invariance is at the core of any (special) relativistic theory. Sometimes, however, we encounter expressions that seem to transform in a wrong way. This is especially true for the four-potential calculated from the retarded four-current. Find out why the four-potential is indeed Lorentz invariant!

## Problem Statement

The well-known retarded integrations over the charge density $$\rho\left(x^{\nu}\right)$$ and current density $$\mathbf{j}\left(x^{\nu}\right)$$ give rise to the scalar and vecor potential:$\begin{eqnarray*} \phi\left(x^{\nu}\right)&=&\frac{1}{4\pi\varepsilon_{0}}\int\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\rho\left(\mathbf{r}^{\prime},t-\left|\mathbf{r}-\mathbf{r}^{\prime}\right|/c\right)dV^{\prime}\ \text{and}\\\mathbf{A}\left(x^{\nu}\right)&=&\frac{\mu_{0}}{4\pi}\int\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\mathbf{j}\left(\mathbf{r}^{\prime},t-\left|\mathbf{r}-\mathbf{r}^{\prime}\right|/c\right)dV^{\prime}\ . \end{eqnarray*}$At a first glance, these equations seem to be not Lorentz invariant. Show that the four-potential $$A^{\mu}\left(x^{\nu}\right)=\left(\phi\left(x^{\nu}\right)/c,\,\mathbf{A}\left(x^{\nu}\right)\right)$$ obeys indeed Lorentz invariance:

• express $$A^{\mu}$$ in terms of the four-current $$j^{\mu}$$,
• re-forumlate the integral to go over space and time, $$dV^{\prime}\rightarrow dV^{\prime}dt^{\prime}$$ using the $$\delta$$-distribution and finally,
• $\text{use}\quad\delta\left[f\left(x\right)\right] = \sum_{\text{roots }x_{i}}\frac{\delta\left(x-x_{i}\right)}{\left|f^{\prime}\left(x_{i}\right)\right|}$ to find a forumulation of the integral without the $$1/\left|\mathbf{r}-\mathbf{r}^{\prime}\right|$$.

## Hints

Remember that we use SI units in which $$j^{\mu}\left(x^{\nu}\right)=\left(c\rho\left(x^{\nu}\right),\mathbf{j}\left(x^{\nu}\right)\right)$$.

For the last part of the problem: Think of the variables of the argument of the $$\delta$$-distribution (a function!) such that the derivative of it is $$\propto\left|\mathbf{r}-\mathbf{r}^{\prime}\right|$$.

## Four-Potential and Four-Current

Let us first write down how we calculate the four-potential for a given coordinate system $$x^{\mu}=\left(ct,\mathbf{r}\right)$$ with respect to some charges and currents. In the necessary retarded formulation we have $A^{\mu}\left(x^{\nu}\right)=\left(\begin{array}{c}\phi\left(x^{\nu}\right)/c\\\mathbf{A}\left(x^{\nu}\right) \end{array}\right) = \frac{1}{4\pi}\int\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\left(\begin{array}{c}\rho\left(\mathbf{r}^{\prime},t-\left|\mathbf{r}-\mathbf{r}^{\prime}\right|/c\right)/\varepsilon_{0}c\\\mu_{0}\mathbf{j}\left(\mathbf{r}^{\prime},t-\left|\mathbf{r}-\mathbf{r}^{\prime}\right|/c\right)\end{array}\right)dV^{\prime}\ .$Now we can remember that $$j^{\mu}\left(x^{\nu}\right)=\left(c\rho\left(x^{\nu}\right),\mathbf{j}\left(x^{\nu}\right)\right)$$ and conclude: $A^{\mu}\left(x^{\nu}\right) = \frac{\mu_{0}}{4\pi}\int\frac{j^{\mu}\left(\mathbf{r}^{\prime},t-\left|\mathbf{r}-\mathbf{r}^{\prime}\right|/c\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}dV^{\prime}\ ,$since $$\varepsilon_{0}\mu_{0}\equiv1/c^{2}$$ and so $$1/\varepsilon_{0}c=\mu_{0}c$$. This result is of course nice but we probably knew it already as solution to $$\Box A^{\mu}=j^{\mu}$$. Nevertheless, the integral does not seem to be Lorentz invariant at first sight. So, what can we do to show it?

## Reformulation of the Integral

We can try to bring the integral into a form such that it's an integral with invariant quantities each, i.e. no retardation. Ok, we may simply write$A^{\mu}\left(x^{\nu}\right) = \frac{\mu_{0}}{4\pi}\int\frac{j^{\mu}\left(\mathbf{r}^{\prime},t^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\delta\left(c\left(t-t^{\prime}\right)-\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)\theta\left(t-t^{\prime}\right)dV^{\prime}dt^{\prime}$and keep in mind that $$t>t^{\prime}$$ since we only care about the retarted interaction, not the advanced one. This fact is reflected with the Heaviside-theta and will be important later on.

Now we want to get rid of anything looking non-Lorentz-invariant. The only chance we have is try something with the $$\delta$$-distribution - maybe we can get rid of the distance $$\left|\mathbf{r}-\mathbf{r}^{\prime}\right|$$!

## $$\delta$$-Gymnastics and Lorentz-Invariance

Ok, let us use$\delta\left[f\left(x\right)\right] = \sum_{\text{roots }x_{i}}\frac{\delta\left(x-x_{i}\right)}{\left|f^{\prime}\left(x_{i}\right)\right|}$to re-formulate the expression inside the $$\delta$$-distribution regarding it at a fixed spacial distance $$\left|\mathbf{r}-\mathbf{r}^{\prime}\right|$$ and with variable $$\tau\equiv c\left(t-t^{\prime}\right)$$:$\begin{eqnarray*} \delta\left[\left(x^{\nu}-x^{\nu\prime}\right)^{2}\right]&=&\delta\left[-\tau^{2}+\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right]\\&=&\delta\left[\left(-\tau+\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)\left(\tau+\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)\right]\\&=&\frac{1}{2\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\left[\delta\left(-\tau+\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)+\delta\left(\tau+\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)\right]\ . \end{eqnarray*}$Then, we end up with$A^{\mu}\left(x^{\nu}\right) = \frac{\mu_{0}}{2\pi}\int j^{\mu}\left(x^{\nu\prime}\right)\delta\left(\left(x^{\nu}-x^{\nu\prime}\right)^{2}\right)\theta\left(t-t^{\prime}\right)dV^{\prime}dt^{\prime}\ .$
Note that the decomposition of the $$\delta$$-distribution resulted in two contributions- retarded, with "$$-\tau$$" and advanced, with "$$+\tau$$". Luckily, the $$\theta$$-distribution fixes the issue. The result is astonishing since we end up with just an integration over the four-current which looks somehow unfamiliar. The given integral is naturally invariant since any quantity involved is Lorentz-invariant: the integral measure, the current and $$\left(x^{\nu}-x^{\nu\prime}\right)^{2}=\eta_{\mu\nu}x^{\mu}x^{\nu\prime}$$.