# Homework Help: Reissner-Nordström black hole: Spherical symmetry of EM field stregth tensor

1. Nov 28, 2011

### Jakob_L

The setup:

I am reading the review: arXiv:hep-th/0004098 (page 9-10).
In Einstein-Maxwell theory, the gravitational field equations read:

R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = \kappa^2
\left( F_{\mu \rho} F^{\rho}_{\;\;\nu} - \frac{1}{4} g_{\mu \nu}
F_{\rho \sigma} F^{\rho \sigma} \right) \,

We consider an ansatz for a spherically symmetric metric:

ds^2 = - e^{2g(r)} dt^2 + e^{2f(r)} dr^2 + r^2 d\Omega^2

It says the unique spherically symmetric solution to this problem is the Reissner-Nordström solution:

\begin{array}
ds^2 = - e^{2f(r)} dt^2 + e^{-2f(r)} dr^2 + r^2 d \Omega^2 \\
F_{tr} = - \frac{q}{r^2} \;, \;\;\;
F_{\theta \phi} = p \sin \theta \\
e^{2f(r)} = 1 - \frac{2M}{r} + \frac{q^2 + p^2}{r^2} \\
\end{array}

The question:
My primary question is: how is $$F_{\theta \phi} = p \sin \theta$$ to be considered spherically symmetric? Normally, I would consider something to be spherically symmetric, if it only depends on the radial coordinate. This is the case for F_{tr}, but not F_{\theta \phi}. Does this have to do with the fact, that we are exactly looking at the angular part of a two-form?
Furthermore, while we are at it, the spherically-symmetric metric tensor also has \theta dependence. How is my perception of spherical symmetry wrong?

Further discussion:
I see that this field strength gives the nice charges:

q = \frac{1}{4 \pi} \oint {}^{\star} F \;, \;\;\,
p = \frac{1}{4 \pi} \oint F \;,

and if I calculate the magnetic field:

B^r = \frac{\epsilon^{0 r j k}}{\sqrt{-g}} F_{jk} = \frac{p}{r^2}

it looks nicely spherically symmetric.

Is the given field strength tensor:
(1) actually not spherically symmetric, but arises from spherically symmetric electric and magnetic fields?
or
(2)
itself spherically symmetric, but this is just not obvious due to the two-form structure?

2. Nov 29, 2011

### George Jones

Staff Emeritus
Are you familiar with Lie derivatives?

3. Nov 29, 2011

### fzero

The field strength tensor is spherically symmetric since it's proportional to the volume form on the 2-sphere,

$\omega_2 \sim \sin\theta ~d\theta\wedge d\phi.$

Similarly, the metric in angular variables must have the correct factor of $\sin\theta$ to be consistent with the volume of the sphere.

In order to check spherical symmetry here, we'd want to write things in terms of the $x_i$ to see the symmetry. For the 2-form, we have

$\omega_2 \sim \sum_{i,j,k} \epsilon_{ijk} x^i dx^j\wedge dx^k,$

while the metric can be written in the form

$ds^2 = A \sum_i (dx^i)^2 + B \left( \sum_ix^i dx^i \right)^2.$

From these expressions we can see the spherical symmetry.

4. Nov 30, 2011

### Jakob_L

I see that my previous misconception was, that $\partial_\theta$ is not a Killing vector for spherical symmetry, but e.g. $R=\partial_\phi$ and $S=\cos \phi \partial_\theta - \sin \phi \cot \theta \partial_\phi$ are. This implies/follows from $\mathcal{L}_R F =0$ and $\mathcal{L}_S F =0$, where F is the two-form from my original question.