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I know all about how black holes form and why their gravity is so strong. However, is the gravity equally powerful in all directions? Will the event horizon appear as a perfect sphere?

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  • $\begingroup$ One can not directly see the event horizon, but one can see radiation from the infall of material, which might give visual cues to the event horizon. The event horizon would be a perfect sphere, but it would not appear as a perfect sphere because of gravitational lensing. If you were stationary or moving directly towards or away from the black hole, it would appear as a perfect circle that was apparently larger than the actual black hole, also because of lensing. $\endgroup$ – Aabaakawad Nov 6 '15 at 1:28
  • $\begingroup$ @Aabaakawad What about a spinning black hole? Surely that would have some effect on the event horizon, maybe making it more oval-shaped? $\endgroup$ – Sir Cumference Nov 6 '15 at 2:35
  • $\begingroup$ you might think that because you are imagining the mass of the black hole to be distributed throughout the volume within the event horizon. It is not. All the mass is either concentrated into a point in the center of the black (a singularity), or an ultracompact object being held up from being a singularity by forces not yet known, (not likely, but who knows). Yes, a singularity can have spin, and I do not know how that works. It can also have an electric charge and/or a magnetic moment. $\endgroup$ – Aabaakawad Nov 6 '15 at 3:33
  • $\begingroup$ See this Physic.SE Q&A, or even the wikipedia page. $\endgroup$ – zibadawa timmy Nov 6 '15 at 7:46
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You cannot see the event horizon. That being said:

A non-rotating black hole, free of external influences, has perfect spherical symmetry. All its properties are exactly the same in any direction, period. This is the Schwarzschild metric.

https://en.wikipedia.org/wiki/Schwarzschild_metric

Even if it is electrically charged, if it's non-rotating, and free of external influences, it is still perfectly spherically symmetrical - this is the Reissner–Nordström metric.

https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric

As soon as it begins rotating, the black hole is no longer in perfect spherical symmetry. It acquires an ergosphere, which is shaped like a flattened ball. This is the Kerr metric.

https://en.wikipedia.org/wiki/Kerr_metric

Kerr metric with ergosphere

Significant external factors can distort the shape of the event horizon. Two black holes merging will go through a process where spherical symmetry is lost temporarily:

https://www.youtube.com/watch?v=JOQMvyLYmd4

Technically, anything near a BH would slightly distort the metric of spacetime, but in practice this would not be easily measurable unless the external factor is very large (another very massive object). So for most practical purposes, the metrics described above would apply.

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The Schwarzschild metric by definition can be written in a spherically symmetric form, the Kerr metric is not spherically symmetric, but in Boyer–Lindquist coordinates the event horizon has a constant radial coordinate.

For most people this will be good enough to say the event horizon of Kerr black holes (including Schwarzschild black holes) are perfect spheres. A physical black hole has a quasi-neutral charge, but the more general case of a charged Kerr-Newman black hole still has a perfectly spherical event horizon.

Now of course no physical situation actually possess the exact symmetries of Kerr black holes. However the outer Kerr metric is stable, which means that the event horizon mantains a near-perfect spherical shape, except in extreme situations such as when two black holes are undergoing a merger.

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As I understand it, general relativity says that there's gravitation everywhere in the universe, and this gravitation creates dips in space, so to speak, often represented in 2D as a weight on a rubber sheet, like the picture below.

enter image description here

Source,

so, a black hole might generate a perfectly spherical event horizon, but it generates it on a not perfectly flat 3 dimensional surface, and I think the gravitation from other objects makes it not quite a perfect sphere. For example, a star or planet that orbits a black hole would drag around a ripple on the event horizon as it orbits the black hole.

Precisely what shape that ripple would be . . . I'm not sure. That said, if you had a black hole as the only object in a universe, then I think the event horizon might be a perfect sphere, or as perfect as possible, given quantum fluctuation, hawking radiation and the impossibility of precise observation and all that good stuff.

Non black holes tend to have lumpy/inconsistent gravity - see here. Even Neutron stars have some inconsistency, but Black holes probably avoid that because the matter is condensed to a point, so there's equal gravitational pull from all directions from the singularity.

Now a Kerr black hole, that's a whole different question. I'm not smart enough to try to answer that one. That might not be a sphere at all.

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    $\begingroup$ I would avoid using the rubber sheet analogy here. It's misleading in a number of aspects, and doesn't help your points too much. $\endgroup$ – HDE 226868 Nov 6 '15 at 13:32
  • $\begingroup$ @HDE226868 are you saying the gravitational field doesn't effect the perfect sphere of the event horizon? $\endgroup$ – userLTK Nov 6 '15 at 21:37
  • $\begingroup$ So, the rubber sheet analogy is not ideal, but is it wrong? Can someone say why it's wrong rather than just voting down? It's not bad logic. The Kerr metric is obviously bigger and I said that. Why is this wrong for a non rotating black hole? Inquiring minds want to know. It might be wrong, but what's the explanation? $\endgroup$ – userLTK Nov 7 '15 at 6:27
  • $\begingroup$ The rubber sheet analogy isn't wrong, per se, but it can be misleading. Eg, what causes bodies to move down into the depression, some kind of meta-gravity? $\endgroup$ – PM 2Ring Jan 15 at 17:22
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    $\begingroup$ "there's equal gravitational pull from all directions from the singularity". It's a bit more subtle than that. We can't get any information from inside the EH (event horizon), and that includes mapping the shape of the black hole's core; gravity cannot escape from the EH anymore than light can. The gravitational response of bodies outside the EH is completely determined by the spacetime curvature that was created as the BH was formed. For further details, please see math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… $\endgroup$ – PM 2Ring Jan 15 at 17:29
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Because the gravitational pull of a non-rotating black hole in a void space is equal at all sides (as a result of it being a point) the event horizon should in theory be a perfect sphere. Because of the nature of a blackhole not letting light escape you would need a background for it to be seen. Also, in real world physics a blackhole would need to interact with light to be "seen" which "correct me if I'm wrong" has it's own very small gravitational pull, it would slightly bend the sphere a bit, the observer will give the same effect.

If rotating it would (as pointed out before) form an ergosphere around the event horizon connecting to it at the poles an being the farthest over the equator, thus forming a more oval shape.

Also, you know who to subscribe to right?

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