- #1
Jakob_L
- 3
- 0
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!
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!