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

  1. Nov 28, 2011 #1
    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!
     
  2. jcsd
  3. Nov 29, 2011 #2

    George Jones

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    Gold Member

    Are you familiar with Lie derivatives?
     
  4. Nov 29, 2011 #3

    fzero

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

    [itex]\omega_2 \sim \sin\theta ~d\theta\wedge d\phi.[/itex]

    Similarly, the metric in angular variables must have the correct factor of [itex]\sin\theta[/itex] 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 [itex]x_i[/itex] to see the symmetry. For the 2-form, we have

    [itex] \omega_2 \sim \sum_{i,j,k} \epsilon_{ijk} x^i dx^j\wedge dx^k,[/itex]

    while the metric can be written in the form

    [itex] ds^2 = A \sum_i (dx^i)^2 + B \left( \sum_ix^i dx^i \right)^2.[/itex]

    From these expressions we can see the spherical symmetry.
     
  5. Nov 30, 2011 #4
    Alright, thanks for the answers.

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

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