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If light orbits a black hole without ever getting closer or further from it, and the black hole pulls on the light from all directions nearly simultaneously, does the black hole extend at its equator? Would more light make this change more extreme?

nelomad
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  • Light can't orbit a black hole, it will either be pulled in or escape, the limit between the two cases is not stable. In any case, adding any form of mass-energy in a bound configuration will only add to the total mass of the black hole. Can one, theoretically, tether to a ring around a black hole? Yes, there are many theoretical scenarios which use that to attempt to calculate what's happening at the event horizon. – CuriousOne Jun 21 '16 at 20:53
  • doesn't it gain stability as it moves out of range of the event horizon? – nelomad Jun 21 '16 at 20:56
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    Have you seen http://physics.stackexchange.com/questions/25657/black-hole-photon-sphere? It's not a duplicate of your question, but it may answer some aspects. There is a good question here, by the way, I just don't think that one should ask it with light but with matter. A neutron star should, theoretically, get quite close to a black hole before being pulled apart. The distortion of the black hole should be measurable, and you are basically asking for the proper physical interpretation of the effective event horizon, I believe. Does that make sense? – CuriousOne Jun 21 '16 at 21:11
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    @Adamawesome no: there's a single radius (which is $3R_S/2$, where $R_S$ is the Schwarzschild radius) where photons could orbit, but these orbits are not stable: any deviation inwards results in them falling in, and any deviation outwards leads to an escape to infinity. This radius is, however, the lower (exclusive) limit for stable orbits for massive bodies. It's usually called the photon sphere, and it's the reason for the strange halo you see around simulated black holes. –  Jun 21 '16 at 21:48
  • I think your question is fanciful and speculative rather than serious and well motivated. – sammy gerbil Jun 22 '16 at 12:36
  • @sammygerbil I am just curious, and what you said is true, though I disagree with it being "fanciful" – nelomad Jun 22 '16 at 16:14

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When we talk about a (non-spinning) black hole we normally mean the spacetime geometry called the Schwarzschild metric. This is one of the simpler solutions to the Einstein equations, but it's important to understand that the solution is based on a number of assumptions and as a result it is only an approximation to a real black hole. The assumption that is important here is that the black hole is the only thing that exists i.e. the Schwarzschild metric describes a universe containing just the black hole and nothing else.

In the real universe black holes aren't isolated. Many of them will have accretion disks, and even those black holes that don't have (significant) accretion disks are likely to form in proximity to other matter. For example the supermassive black hole in our galaxy probably doesn't have much of an accretion disk because it seems pretty quiet. However it has lots of stars that orbit near it, and of course it's surrounded by the whole of the Milky Way.

The presence of other matter perturbs the spacetime geometry so it is no longer exactly described by the Schwarzschild metric. Where this is a small effect, e.g. where the black hole is much more massive than the matter around it, we call the perturbation back-reaction.

In your question you describe a black hole surrounded by (I assume) a ring of light orbiting at the photon sphere. As the comments have mentioned this woudn't happen in practice because the orbit is unstable, but there's no reason we can't ask in principle what would happen. And the answer is that the presence of the light would indeed change the shape of the event horizon so it was no longer perfectly spherical.

However I have to confess I don't know how the shape would change. There is no analytic solution to the field equations for a black hole surrounded by a ring, and I haven't seen any attempts at a perturbative calculation (though they may well exist). What we can say is that the effect is likely to be vanishingly small. The spacetime curvature is related to the energy density, where we consider matter to be equivalent to an amount of energy given by Einstein's famous equation $E=mc^2$. Because $c^2$ is so large this means that unless the light is ridiculously intense its gravitational field is negligible when compared to even small amounts of matter.

John Rennie
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  • Is there any reason why light would orbit in one particular plane, in preference to any other? If all orbital planes are possible, I think there will be no effect on shape, only (possibly) on size. – sammy gerbil Jun 22 '16 at 16:29
  • @sammygerbil: Well if you're considering a single light ray then it would remain in the one plane. I'm guessing that's what the OP is thinking of. – John Rennie Jun 22 '16 at 16:43
  • But there is no preferential plane for the orbit? – sammy gerbil Jun 22 '16 at 16:47
  • @sammygerbil: no. The Schwarzschild BH is spherically symmetric. The obvious question is what the photon sphere looks like for a Kerr BH, and actually I can't remember - I'd have to go and look it up. – John Rennie Jun 22 '16 at 16:55
  • @sammygerbil: if you're interested you should take this to the chat room though we're currently discussing 70s games consoles :-) – John Rennie Jun 22 '16 at 16:56