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I Area of the event horizon of a rotating black hole

  1. Jun 9, 2017 #1
    The Kerr metric for a black hole of mass ##M## and angular momentum ##J = aM## is

    $$ds^{2} = - \frac{\Delta(r)}{\rho^{2}}(dt-a\sin^{2}\theta d\phi)^{2} + \frac{\rho^{2}}{\Delta(r)}dr^{2} + \rho^{2} d\theta^{2} + \frac{1}{\rho^{2}}\sin^{2}\theta (adt - (r^{2}+a^{2}) d\phi)^{2},$$

    where ##\Delta(r) = r^{2} + a^{2} - 2Mr##, ##\rho^{2} = r^{2} + a^{2} \cos^{2}\theta## and ##- M < a < M##.

    The event horizon is at ##r_{+} = M + \sqrt{M^{2} - a^{2}}##. This is obtained by solving the equation ##\Delta(r) = 0##.


    How do you use this to compute the area of the horizon?

    My idea is to simplify the metric to obtain

    $$ds^{2} = - \left( \frac{r^{2} + a^{2} - 2Mr - a^{2} \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dt^{2} + \left( \frac{r^{2} + a^{2} \cos^{2}\theta}{r^{2} + a^{2} - 2Mr} \right) dr^{2} - \left( \frac{4aMr \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dtd\phi + \left( r^{2} + a^{2} \cos^{2}\theta \right) d\theta^{2} + \sin^{2}\theta \left( \frac{(a^{2} + r^{2})^{2} - a^{2} \sin^{2}\theta (a^{2}-2Mr+r^{2}) }{r^{2} + a^{2} \cos^{2}\theta} \right) d\phi^{2}.$$

    Then, I think that the area of the horizon is given by

    $$A = \int d\theta\ d\phi\ g_{\phi\phi}g_{\theta\theta}|_{r=r_{+}}.$$

    Am I wrong?
     
    Last edited: Jun 9, 2017
  2. jcsd
  3. Jun 9, 2017 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    You don't need to go through that step. Just evaluate the original metric for the case ##dt = dr = 0## and ##r = r_+##, and integrate over ##\theta## and ##\phi##.
     
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