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A rotating black hole is believed to contain a ring singularity rather than a point.

However, if an astronaut is orbiting the black hole at exactly the same angular velocity as the blackhole (in direction and magnitude), then it is no longer rotating to him.

Would it appear as a point singularity? (if he could somehow see it)

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3 Answers 3

The ring singularity (assuming traditional GR is the whole story of gravity) is a definitive fact. It will remain a ring no matter what its observer does. This is in an absolutely identical sense to the objective difference between an inertial frame of reference (no pseudo-forces whatsoever) and a non-inertial frame (some kind of pseudo-forces).

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I'm not quite sure how to go about answering your question, as it seems likely that the angular momentum of the black hole would be sufficient that a 'geosynchronous' orbit would require the satellite to exceed the speed of light. According to Wikipedia the black hole GRS 1915+105 rotates at 1,150 times per second. How large its event horizon is I'm not in a position to say. But if we pencil in a 100km radius, a geosynchronous orbit outside this would require an orbital velocity of 100,000 * 2 * 3.14 * 1,150 m/s (~ 600,000,000 m/s) or 2c.

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We can, at least in theory, have a black-hole where the angular velocity at a distance slightly outside the horizon is less than the speed of light. Or am I wrong? –  Jus12 Aug 23 '11 at 11:56
@Jus12, no, there is no reason that one couldn't have a very small black hole suitable for study in a laboratory (if you can keep it from falling through the floor), or one that doesn't happen to be rotating very fast. I hadn't thought of that. –  Brian Hooper Aug 25 '11 at 0:15
GRS 1915 has a mass of ~15 Msun, so with a near-maximal spin, the horizon has a radius of 15*1.5 km = 22.5 km, so the orbital velocity is actually "only" about 1/2 c. –  Jeremy Nov 7 '11 at 15:32

I'm having a hard time answering this. If the question is really asking whether there is an observer for whom the Kerr metric (rotating BH) reduces to the Schwarzschild metric (BH) in local co-ordinates, then I think the short answer is no. The frame in which the $t\phi$ component of the metric vanishes appears to be counter-rotating with respect to the black hole, so I think it would still see a ring. Basically, it's not the Schwarzschild metric. See this section on Wikipedia.

There are a lot of problems with trying to descibe a stationary/synchronous orbit around a black hole, though. To start with, what is the angular velocity of a black hole? Is it found by considering the angular momentum at the event horizon? Or somewhere else? It might be intuitive to regard the event horizon as the "surface", but it's really a co-ordinate boundary, rather than a real one. Objects that fall in don't see a "surface".

In order to see the ring singularity in a rotating (and electrically neutral) black hole, the observer must be inside the second horizon, where spacetime becomes time-like again. I'd guess a stationary observer in this region would see the ring singularity as some sort of distorted ring. Then again, physics is pretty much breaking down here, so who knows what he'd see.

I'm also not sure whether a synchronous orbit is stable, if it exists. The innermost stable circular orbit is outside the event horizon. I've never seen (nor done myself) a calculation of orbits inside the space connected to the singularity, so I don't know if there are more stable orbits further in. But my instinct is that no stable stationary orbit exists inside the usual innermost one, so circular orbits in the vicinity of the singularity would be unstable too.

Finally, I guess some of this could be circumvented by having a black hole with super-maximal rotation, and therefore a naked singularity.

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yes that is the question I was asking (I think). –  Jus12 Aug 23 '11 at 12:48

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