The components of $\frac{\rho \mathrm{v}}{c}$ and $i\rho$ must transform as components of a four-vector, so that if measured in one coordinate system they are known in all coordinate systems. On the other hand, any operational definition of $\rho (\mathbf{r},t)$ must take account of the positions of all particles at the same time $t$, that of the $s$-th particle being ${\mathbf{r}}_s\mathrm{}\left(t\right)\mathrm{}$. Upon performing the Lorentz transformation these will be ${{\mathbf{r}}_s}^{\prime }\left({t_s}^{\prime }\right)$, and the transformed time ${t_s}^{\prime }$ will be different for each particle. Another observer, in measuring ${\rho }^{\prime }$, would use ${{\mathbf{r}}_s}^{\prime }\left(t^{\prime }\right)$, $t^{\prime }$ being the same for all particles. As particles are in motion ${{\mathbf{r}}_s}^{\prime }\left({t_s}^{\prime }\right)\ne {{\mathbf{r}}_s}^{\prime }\left(t^{\prime }\right)$, and there appears to be no necessary relation between $\rho (\mathbf{r},t)$ and ${\rho }^{\prime }({\mathbf{r}}^{\prime },t^{\prime })$, operationally defined in each coordinate system. It turns out, however, that if in each coordinate system the charge density is defined by $\rho (\mathbf{r},t)=\Sigma s^{}{e}_s\delta (\mathbf{r}-{\mathbf{r}}_s\mathrm{}(t\left)\right)$, then relativistic equations of transformation hold.

• Received 16 June 1947

DOI:https://doi.org/10.1103/PhysRev.72.624