1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
    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:

    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} \\

    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:
    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?
    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook