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Shape of Kerr event horizon in terms of reduced circumference

  1. Nov 19, 2012 #1
    The event horizon of a Kerr black hole is often depicted as being spherical, but this seems to be a reference to the horizon as defined in Boyer-Lindquist coordinates, where horizons appear at a constant value of r.

    However, Thorne describes the "black hole's horizon bulg[ing] out at its equator in the manner depicted in Figure 7.9" (text and figure on page 293 of my edition of Black Holes And Time Warps). Similar diagrams appear in his 1974 article for Scientific American, "The Search For Black Holes". In that article he states that "A hole rotating at moderate speed ... with a rotation parameter of .866 will be perfectly flat at the poles but will still be be rounded at the equator. Rotation does not affect the size of the equatorial circumference."

    This all seems to hang together with Taylor & Wheeler's Exploring Black Holes (1st ed.), section F-5, in which a circular polar cross-section of an extremal black hole is shown in BL coordinates, but it is noted that "When ... plotted in terms of the reduced circumference R/M instead of [BL coordinate] r/M, then the radius of the horizon is greater in the equatorial plane than along the axis of rotation, giving the horizon the approximate shape of a hamburger bun."

    I'd quite like to be able to plot this oblate event horizon in reduced circumference coordinates (although I understand that this isn't strictly possible in Euclidean space beyond the threshold of a = √[3]/2). However, although I can find extensive discussions of the applicability of reduced circumference to the equatorial plane of the Kerr metric, I can't seem to find any discussion of the overall shape of the event horizon in these coordinates, beyond that throwaway line from Taylor & Wheeler.

    Can anyone help?
  2. jcsd
  3. Nov 23, 2012 #2
    Well, that went well ... :smile:
    In the meantime, I found a couple of references that got me where I wanted to be:

    Properties of the instantaneous Ergo Surface of a Kerr Black Hole (Pelavas, Neary & Lake, 2001)
    The Kerr spacetime: A brief introduction (Visser, 2008)

    I'm cherishing this statement from Visser: Because the inner and outer horizons are specified by the simple condition r = r±, you might be tempted to deduce that the horizons are "spherical". Disabuse yourself of this notion.
    Is it possible he's a little tetchy on this topic?

    So now I have the (to me, at least) pleasing result that the "reduced circumference" radius of a Kerr black hole is, at the equator, the same as that of a Schwarzschild black hole of the same mass, whereas the corresponding radius at the poles is equal to that of a Schwarzschild black hole with mass equal to its irreducible mass.
  4. Nov 26, 2012 #3
    It's very hard to ask meaningful questions about the shape of an event horizon in the absence of symmetries. It's fair to say that the Schwarzschild event horizon is spherical because the spacetime is spherically symmetric and the horizon respects that symmetry. But in Kerr, what would the claim "this surface is spherical" even mean? I think the best you can do is to say that the horizon is axisymmetric. Maybe for small spin parameter a you can say more, using the background symmetry of Schwarzschild as a reference point.

    Sometimes people embed the intrinsic 2-geometry of the horizon as a surface in Euclidean 3-space, in which case you can talk about the shape of that surface in the usual way. (Though, as you seem to point out, it is not always possible to perform such an embedding.)
  5. Nov 27, 2012 #4
    Thanks for the reply.
    Visser's main point seems to be that the Ricci scalar varies with latitude along the horizon of a Kerr black hole, implying that the horizon's curvature is not constant. (I believe Thorne is making reference to the same thing when he says that a black hole is flat at the poles when a = 0.866.)

    Of course my own little radial calculations aren't compatible (in Euclidean terms) with the corresponding curvature, which I think serves to highlight your point about the difficulty of asking meaningful questions.
  6. Nov 27, 2012 #5

    George Jones

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    Since Kerr is a vacuum solution, the Ricci scalar is zero everywhere. Visser's (71) an expression for the Kretschmann curvature scalar.
  7. Nov 27, 2012 #6
    I'm clearly misunderstanding something (which comes as no surprise to me).
    On page 27 of the text I linked to earlier, just below his "disabuse yourself" comment, Visser discusses the intrinsic curvature of the horizon and introduces something he calls the Ricci scalar, in order to contrast its value at the equator of the event horizon with its value at the poles (equations 119 to 121).
  8. Nov 27, 2012 #7

    George Jones

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    Sorry, I thought that you meant the spacetime Ricci (which is zero) and/or Kretschmann scalars evaluated on the horizon.

    On page 127, Visser refers to the Ricci scalar for a particular 2-dimensional surface. To find this, first find the the 2-dimensional metric that results when the full spacetime metric is restricted to the surface. Now, find the Ricci scalar for this 2-dimensional metric.

    This is something quite different from the spacetime Ricci scalar restricted to the surface.
  9. Nov 27, 2012 #8
    Ah, I understand. Thank you.
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