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

[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:
\begin{equation}
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) \,
\end{equation}
We consider an ansatz for a spherically symmetric metric:
\begin{equation}
ds^2 = - e^{2g(r)} dt^2 + e^{2f(r)} dr^2 + r^2 d\Omega^2
\end{equation}
It says the unique spherically symmetric solution to this problem is the Reissner-Nordström solution:

\begin{equation}
\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}
\end{equation}

The question:
My primary question is: how is \begin{equation} F_{\theta \phi} = p \sin \theta \end{equation} 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:
\begin{equation}
q = \frac{1}{4 \pi} \oint {}^{\star} F \;, \;\;\,
p = \frac{1}{4 \pi} \oint F \;,
\end{equation}
and if I calculate the magnetic field:
\begin{equation}
B^r = \frac{\epsilon^{0 r j k}}{\sqrt{-g}} F_{jk} = \frac{p}{r^2}
\end{equation}
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?

Thanks for your help!

 

[George Jones]

Are you familiar with Lie derivatives?

 

[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.

 

[Jakob_L]

Alright, thanks for the answers.

So, I've read up on Lie derivatives today.
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.

I still find this a bit strange, though :-) Is there a more qualitative argument?