the four-potential is invariant under Lorentz transformations.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 \(1/\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\).

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