# EM Field tensor of a point charge [closed]

If I say the Reissner-Nordstrom metric $$ds^2=-\left(1-\frac{2m}{r}+\frac{e^2}{r^2}\right)\text d t^2 + \left( 1-\frac{2m}{r}+\frac{e^2}{r^2}\right)^{-1}\text d r^2 + r^2 \text d \Omega^2$$ is the solution of the Einstein equation $G_{\mu\nu}=8\pi T_{\mu\nu}$ of a point charge, where $$T_{\mu\nu} = \frac{1}{4\pi}\left( F_{\alpha\mu}F^\alpha_{\phantom \alpha \nu} - \frac 1 4 g_{\mu\nu} F^{\alpha\beta}F_{\alpha\beta } \right)\;,$$ what does $F$ look like? Of course I need to write down $F$ for a point charge with mass $m$ and charge $q$. In one source I found that for a point charge we have $$A = A_\mu \text d x^\mu = \frac e r \text d r\;,$$ where $$F_{\mu\nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu\;.$$ But when I calculate it, I find: $$F_{\mu\nu} = \partial_\mu \left( \delta_{\nu r}\frac{e}{r} \right) - \partial_\nu \left( \delta_{\mu r}\frac{e}{r} \right) = -\frac e r^2(\delta_{\nu r}\delta_{\mu r} - \delta_{\mu r}\delta_{\nu r}) = 0\;.$$ So $F$ of a point charge is zero?

## closed as off-topic by ACuriousMind♦, Kyle Kanos, JamalS, Qmechanic♦Apr 7 '15 at 17:38

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• I have no idea what you did in your last equation, but the field strength of a point charge is certainly not zero, since $E^i = F^{i0}$, and the electric field of a point charge is certainly not zero. – ACuriousMind Apr 6 '15 at 15:56
• Yes, that's why I wrote this question ;-) . In the last equation I used the fact that $\nabla_\mu A_\nu - \nabla_\nu A_\mu = \partial_\mu A_\nu - \partial_\nu A_\mu$ and $A_\mu = \frac e r \delta_{\mu r}$. – thyme Apr 6 '15 at 16:04
• Try $A=\frac{e}{r}dt$. – Holographer Apr 6 '15 at 16:05

## 1 Answer

If four vector notation is less intuitive then refer back to three vectors \begin{align*} \vec{E} &= - \vec{\nabla}\phi - \frac{\partial\vec{A}}{\partial t} \\ \vec{B} &= \vec{\nabla}\times\vec{A} \end{align*} For a static point particle \begin{align*} \vec{E} &= \frac{e}{r}\hat{r}\\ \vec{B} &= 0 \end{align*} The solution up to gauge transformation is what you already know \begin{align*} A_o &= \phi = \frac{e}{r}\\ \vec{A} &=0 \end{align*} Or $A = \frac{e}{r}dt$. There for as @Holographer suggested the appropriate potential should be $A_\mu = (\frac{e}{r}, 0 ,0 , 0)$