# Wald Problem 6.3: Reissner-Nordstrom Metric

1. Apr 13, 2013

### WannabeNewton

Hi guys. This question is related to Problem 6.3 in Wald which involves deriving the Reissner-Nordstrom (RN) metric. We start with the source free Maxwell's equations $\nabla^{a}F_{ab} = 0,\nabla_{[a}F_{bc]} = 0$ in a static spherically symmetric space-time which, in the coordinates adapted to the hypersurface orthogonal time-like killing vector field and the spherical symmetry, takes the form $ds^2 = -f(r)dt^{2} + h(r)dr^2 + r^2d\theta^{2} + r^2\sin^{2}\theta d\phi^{2}$. The solution to Maxwell's equations, in these coordinates, takes the form $F_{ab} = 2A(r)(e_{0})_{[a}(e_{1})_{b]} + 2B(r)(e_{2})_{[a}(e_{3})_{b]}$ which one can justify on physical grounds; in addition we are only concerned with the case $B(r) = 0$. The only relevant orthonormal basis fields from the tetrad are given by $(e_{0})_{a} = f^{1/2}(dt)_{a}, (e_{1})_{a} = h^{1/2}(dr)_{a}$.

Hence, in the tetrad basis, the solution takes the form $F_{\mu\nu} = A(r)f^{1/2}h^{1/2}(\delta^{t}_{\mu}\delta^{r}_{\nu} - \delta^{t}_{\nu}\delta^{r}_{\mu})$. It's pretty clear from this that the only non-vanishing components will be $F_{rt} = -F_{tr}$ hence we can just look at $F_{rt} =-A(r)f^{1/2}h^{1/2}$. Using Maxwell's equations, $\nabla^{\mu}F_{\mu t} = g^{\mu t}\nabla_{\mu}F_{tt} + g^{\mu r}\nabla_{\mu}F_{rt} + g^{\mu \theta}\nabla_{\mu}F_{\theta t} + g^{\mu \phi}\nabla_{\mu}F_{\phi t} = g^{tt}\nabla_{t}F_{tt} + g^{rr}\nabla_{r}F_{rt} + g^{\theta \theta}\nabla_{\theta}F_{\theta t} + g^{\phi \phi}\nabla_{\phi}F_{\phi t}$. Now, $\nabla_{t}F_{tt} = -\Gamma ^{\alpha}_{tt}F_{\alpha t} -\Gamma ^{\alpha}_{tt}F_{t\alpha} = 0$, $g^{\theta \theta}\nabla_{\theta}F_{\theta t} = -\frac{1}{r^{2}}\Gamma^{r}_{\theta \theta}F_{r t} = \frac{1}{r}g^{rr}F_{rt} = g^{\phi \phi}\nabla_{\phi}F_{\phi t}$, and $\nabla_{r}F_{rt} = \partial _{r}F_{rt} - \Gamma ^{r}_{rr}F_{r t} - \Gamma ^{t}_{tr}F_{rt} = \partial _{r}F_{rt} - \frac{1}{2}F_{rt}h^{-1}\partial _{r}h - \frac{1}{2}F_{rt}f^{-1}\partial _{r}f$.

Putting all this together gives us $\nabla^{\mu}F_{\mu t} = -\partial _{r}(Af^{1/2}h^{1/2}) + \frac{1}{2}Af^{1/2}h^{-1/2}\partial _{r}h + \frac{1}{2}Af^{-1/2}h^{1/2}\partial _{r}f -\frac{2}{r}Af^{1/2}h^{1/2} = 0$ which, after performing the derivative, yields $-\partial _{r}A -\frac{2}{r}A = 0$. Since $A = A(r)$ the solution to this is just $A = \frac{C}{r^{2}}$ for some undetermined constant $C$. Now Wald says the solution is supposed to be $A = -\frac{q}{r^{2}}$ where $q$ "may" be interpreted as the total charge. How exactly would I get this from my solution for $A$? I figured since we are dealing with a static spherically symmetric field from a source of compact support in an asymptotically flat space-time, if I go far away from the source i.e. $r\rightarrow \infty$ I should recover the classical coloumb field $A(r) = -\frac{q}{r^{2}}$ where $q$ is the total charge as measured far away from the source (so measured at infinity). I am not sure about this argument however. Could someone comment on all this? Thanks in advance.

2. Apr 14, 2013

### micromass

Have you tried logarithms?

3. Apr 15, 2013

### Mentz114

I'm not sure what the question is so I'll assume it's about interpreting the constant C that comes from the integration. The clincher is that the SET got from the Einstein tensor is that of a point charge with q = C.

4. Apr 15, 2013

### WannabeNewton

How are we supposed to know this before solving Einstein's equations? That's why I used my limiting case argument above; I just wasn't content with the non-rigorous nature of the argument.

5. Apr 15, 2013

### Bill_K

Sure, that's valid. You've found the static spherically symmetric solution of Maxwell's source-free equations, and all that remains is to fix the constant in front, as you've done.

6. Apr 15, 2013

### WannabeNewton

Cool, thanks Bill!