What I know about black holes (correct me if I'm wrong) is that they are the most compact objects in the universe that have been discovered. Due to all that gravity, wouldn't black holes be a perfect spheroid, sort of like planets are spheroids (due to centrifugal forces)? Can you measure the geometry of a black hole due to its power of warping space-time itself?


The shape of a black hole's event horizon depends on who is asking. Observers who are moving quickly towards a hole, for example, will see a different shape compared to those who are not.

Per @benrg the event horizon of a static black hole is not observer dependent, similarly to how the shape of an expanding flash of light is not.

In the coordinates appropriate to very distant "inertial" observers, the event horizon of a nonspinning uncharged black hole in equilibrium is spherical. If the hole is rotating, the event horizon will bulge out along the equator like any other rotating object.

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    $\begingroup$ I guess with the coordinates of an observer at infinity you mean cartesian Kerr-Schild coordinates? $\endgroup$ – Yukterez Jul 25 '16 at 1:43
  • $\begingroup$ Good point -- that the shape of an object depends on the observer's inertial system is nothing special to black holes (to light, the universe is flat). $\endgroup$ – Peter A. Schneider Jul 25 '16 at 16:54
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    $\begingroup$ The Wikipedia article about ergosphere seems to indicate that the event horizon (which I would say "is" the black hole for our purposes) stays spherical for rotating black holes, while the ergosphere bulges. I swear I didn't just write that ;-). $\endgroup$ – Peter A. Schneider Jul 25 '16 at 17:04
  • $\begingroup$ @PeterA.Schneider While the caption to that diagram is certainly misleading, I think it means to indicate that the event horizon is (inaccurately) spherical in the diagram, to distinguish it from the ergosphere. See the image on page 24 of arxiv.org/abs/0706.0622 (or the Wikipedia article on Kerr spacetime). $\endgroup$ – AGML Jul 25 '16 at 18:18
  • $\begingroup$ The first paragraph is wrong. Because there's a null Killing vector field on the Kerr-Newman horizon, the geometry of any spacelike slice through it is the same. In other words, the event horizon doesn't Lorentz-contract. Much like an expanding sphere of light in special relativity, it has the same shape with respect to any "observer". That shape isn't a sphere, though, or even an oblate spheroid. See page 27 of arXiv:0706.0622. $\endgroup$ – benrg Jul 26 '16 at 7:08

Yes, they are perfect spheres. But let's understand what is the sphere and why.

The spherical surfaces are the horizons. They are surfaces in space that have perfect spherical geometry. Now, that only holds true for various conditions:

1) they are static black holes, and if they arose from matter/energy collapsing they are in their equilibrium states. There are black holes which are not spherical, which are rotating, and which are called stationary. The spherically symmetric ones are called Schwarzchild black holes, the others Kerr (and Kerr Newman if charged).

2) they must be exactly spherically symmetric or they are not static. A little bump that forms in it (say due to in falling matter) quickly disappears as gravitational (and other) radiation is produced outside the horizon and settles the whole black hole to perfectly symmetrical. The same is true if they are Kerr black hole, any disturbances will settle down back to Kerr (or Kerr Newman if it acquired some charge from the in falling matter). For Kerr or Kerr Newman black holes the rotation causes a non spherical geometry, more ellipsoidal

3) the above is true for so called perennial a black holes. This is a little of a fine point that really doesn't need to be thought of much in this view. Astrophysical black holes approach the perfect black hole geometries exponentially fast, but there is still some controversy whether it ever finally finishes. For all physical and astrophysical purposes except for quantum gravity considerations, that is all irrelevant. People will say sometimes that it is an apparent horizon, not a final one. You can still think of it as an horizon.

4) why does that happen, that they get rid of any protuberances and small deviations from spherical, or from Kerr? Because GR requires black holes that do not have those symmetries to radiate away the perturbations. There is a famous theorem that says 'Black Holes have no hair'. The hair is meant to be all those perturbations. In essence anything that can be radiated away is, and the only properties that stay with the black hole, and define the geometry and metric, are its mass (how large), its angular momentum (rotation) and charge. Those are preserved because conservation laws require it. The conservation laws are due to deeper symmetries.

Added minor edit: there is a lot that can be discerned about the geometry and shape of the horizon from the gravitational waves created. Gravitational waves will have quadruple moments but also a specific and well known n-pole moments which are more difficult to measure. One can measure them and obtain comparisons of measured vs modeled values and thus obtain some information on deviations from the expected Kerr black holes while the black holes are in spiraling in very closely and plunging into each other, and finally settling down. Any deviations will say something about the dynamics, and thus whether it follows GR to finer levels of approximation than what's been done so far. To do that we will need higher SNR's and also a wider range of frequencies to also cover even bigger black holes, including supermassive ones and cosmological ones. Typically that will need bigger interferometers, and for really big ones space based. The specifics are somewhat detailed. In any case, this is way beyond the question asked, and if interested ask another question.

See the various Wikipedia articles on black holes, horizons, and the two named black hole types.

  • $\begingroup$ Static black hole is a mathematical concept and in reality all black holes are rotating? BH without magnetic dipole moment don't exist? $\endgroup$ – HolgerFiedler Jul 25 '16 at 5:29
  • $\begingroup$ Wouldn't there probably be a very tiny amount of spin due to quantum effects in any static black hole and therefore a very tiny distortion (unless its composed of an even number of particles?) Probably its not significant though. $\endgroup$ – Steven Stewart-Gallus Jul 25 '16 at 21:14
  • $\begingroup$ This is a non quantum description. Nobody knows what a quantum horizon would look like $\endgroup$ – Bob Bee Jul 25 '16 at 22:54
  • $\begingroup$ I don't understand why you lead off with "Yes, they are perfect spheres" and then go on to point out (correctly) that they aren't. There is only spherical symmetry when $a=0$, which one would expect to occur with probability $0$ in the real world. (Note that the horizons and ergosphere boundaries are not oblate spheroids either; @Qmechanic, I'm not sure that your edit of the question really makes sense.) $\endgroup$ – benrg Jul 26 '16 at 7:46
  • $\begingroup$ For spherical symmetry they are perfect spheres. The less symmetrical Kerr and related black holes are probably the ones around most, but that is only one part of the answer. The most important part was to explain why in that case the spherical symmetry is perfect. Similarly for the axial symmetry case. I explained all I answered. I know ergospheres and the exact shapes for Kerr like BHs are more complicated, and I was not precise on purpose. For the OP it was my opinion that this answered the question w/o complicating things unecessarily $\endgroup$ – Bob Bee Jul 27 '16 at 0:25

A theorem known as Israel's theorem states that, given an asymptotically Minkowski spacetime, the only valid static spacetime solutions (with a number of regularity assumptions) of the Einstein Field Equations (EFEs) are Schwarzschild ones. The technical definition of "static" is excludes rotation, like the Kerr solution. A similar theorem, for "stationary" instead of "static" spacetimes, shows that the only solution is the Kerr solution, again under reasonable assumptions.

Israel's theorem was the start of black hole uniqueness theorems, very related to black hole "no hair" theorems. It is very easy to create asymptotically flat spacetimes. For example, if you have a chair sitting in spacetime, this will can create a very complicated stationary spacetime metric which is asymptotically flat. Black hole uniqueness theorems show that you can't create a complicated "chair-like metric" if you only have vacuum. It has to be one of these simple spacetimes.

This does not rule out things like black hole ringing, where the horizon is time dependent and decays in amplitude over time, but the oscillation magnitudes never exactly decreases, mathematically, to zero. In physics however, if an amplitude goes like $e^{-t}$ and $t=100$ seconds... the amplitude is zero. So yes, black holes are essentially perfect sphereoids because they will decay to the unique (regular-enough) asymptotically flat spacetimes given by the black hole uniqueness theorems.

(I learned all of this, including the chair visualization, from Lee Lindblom in the UCSD physics department)


It is not possible to know. The reason they are called holes is that they are effectively holes in space/time so concepts like length, size, distance and time no longer have meaning in the conventional sense. Equally, while they may be the most compact objects known, there is no evidence to suggest they must be uniform. If you are talking about the event horizon as opposed to the object, then it must be spherical by its very definition, being that distance from the center of gravity where the gravitational escape velocity exceeds c.

  • $\begingroup$ Downvotes without comments are not helpful. $\endgroup$ – Paul Smith Jul 26 '16 at 8:46
  • $\begingroup$ prudent and meaningful interpretation $\endgroup$ – user46925 Jul 27 '16 at 16:14

protected by Qmechanic Jul 24 '16 at 21:45

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