This is what we know as electromagnetic stress-energy tensor $T^{\mu\nu}$. Now I want to know what is its direct relation with $\rho$, charge density?


$\frac {T^{\mu\nu}}{\rho}=?$


I think the fastest way to get a mathematical connection between the stress energy tensor of electromagnetics $T^{\mu\nu}$ and a continous charge density $\rho\left(\boldsymbol{r}\right)$ is to remember the concrete form of $T^{\mu\nu}$ (see here: http://en.wikipedia.org/wiki/Electromagnetic_stress%E2%80%93energy_tensor) and to remember the form of Gauss law either in its differential form $\nabla\cdot\boldsymbol{E}=\frac{\rho}{\epsilon_{0}}$ or its integral form $\boldsymbol{E}\left(\boldsymbol{r}\right)=\frac{1}{4\pi\epsilon_{0}}\int_{V}\frac{\hat{\boldsymbol{r}}^{\prime}}{\left|\boldsymbol{r}-\boldsymbol{r}^{\prime}\right|^{2}}\rho\left(\boldsymbol{r^{\prime}}\right)dV$. Inserting this into the explicit form of the components of $T^{\mu\nu}$ gives you the required connection.

  • $\begingroup$ Of course, note that this is true only if there is no current present (which also means that $\rho$ is independent of time) $\endgroup$ – Jerry Schirmer Jul 15 '13 at 2:59

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