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Surface gravity (Killing vecotrs) of Kerr black hole

  1. Apr 14, 2014 #1
    Hello,
    Im interested in surface gravity of Kerr black hole it means, I need to find killing vector for Kerr which is null on outer horizont. Is it true? How should I do that? I guess it could be some linear combination of vecotrs ∂/∂t and ∂/∂[itex]\phi[/itex]. So can I looking for that this way?[itex][/itex]
    [itex]\chi=(a,0,0,b)[/itex] and than from limit [itex]g_{\mu\nu}\chi^{\mu}\chi^{\nu}=0[/itex], [itex]r→r_{+}[/itex], where [itex]r_{+}[/itex] is outer horizon?
    I find out [itex]b=a\frac{-g_{\phi t}\pm \sqrt{g_{\phi t}^2-g_{tt}g_{\phi \phi}}}{g_{\phi \phi}}[/itex] for [itex]r→r_{+}[/itex].
    Is it right? What is it telling me?
     
  2. jcsd
  3. Apr 14, 2014 #2

    WannabeNewton

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    Hi Vrbic! You're almost there actually but let me just quickly walk you through the overview of the calculation. First of all you should realize that surface gravity does not have a straightforward interpretation for rotating black holes. For non-rotating black holes we can define the surface gravity as the force that must be exerted by an observer at infinity in order to hold in place a test mass in the limit as one approaches the event horizon. However for rotating black holes an observer at infinity cannot even in principle hold a test mass in place in the limit as one approaches the event horizon so the surface gravity's usual physical interpretation breaks down. However the surface gravity can still be computed using the same formula as for a non-rotating black hole.

    To do this we do as you said and find the time-like Killing field which becomes null on the event horizon as well as normal to it, and therefore acts as the null generator of the horizon. So how do we know what Killing field to pick? Well in the non-rotating case we chose the stationary Killing field ##\xi^{\mu}## generating time translations, the orbits of which correspond to observers at rest in the gravitational field. In Schwarzschild space-time we know that ##\xi^{\mu} \propto \nabla^{\mu} t## i.e. the observers at rest in the gravitational field have a well-defined one-parameter family of space-like hypersurfaces orthogonal to their 4-velocities.

    In rotating black hole space-times this is no longer true i.e. ##\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} \neq 0## so the observers following orbits of ##\nabla^{\mu} t## differ from those following orbits of ##\xi^{\mu}##. Naturally we want to work with the former since they have well-defined global rest spaces so we consider the vector field ##\nabla^{\mu}t## or equivalently the vector field ##\chi^{\mu} = \xi^{\mu} + \omega \psi^{\mu}## where ##\omega = -\frac{g_{t\phi}}{g_{\phi\phi}}## is the angular velocity of the observers following orbits of ##\nabla^{\mu} t## and ##\psi^{\mu}## is the axial Killing field; physically these are the stationary observers with zero angular momentum along ##\psi^{\mu}##, and they circle around the black hole with the above angular velocity (which varies with radius and polar angle!). You can check that in the limit as ##r \rightarrow r_+##, ##\chi^{\mu}## becomes a null Killing field normal to the event horizon, which therefore itself rotates with angular velocity ##\lim_{r\rightarrow r_+}\omega := \omega_H= \frac{a}{r_+^2 + a^2}##. In other words you can just take the null generator to be ##\chi^{\mu} = \xi^{\mu} + \omega_H \psi^{\mu}##.

    See section 5.3 of the following notes for more details: http://www.physics.uoguelph.ca/poisson/research/agr.pdf
     
  4. Apr 15, 2014 #3
    Ou very nice. Thank you very much.
     
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