As is stated or illustrated in many different articles, including this one, rotating black holes are oblate spheroids. It is likewise stated in some sites that this fact should be intuitive, as all rotating bodies are oblate spheroids. However, this is 1) not true, as objects which are perfectly inelastic will retain their original shape when spun, and 2) as far as I understand, planets and most other bodies gain this shape because of hydrostatic pressure, which obviously cannot apply to light and, again in my understanding, does not even exist in a black hole.

My question, then, is: why are rotating black holes not perfect spheres? Isn't gravity uniform in all directions? Or does the rotation of the black hole disrupt this uniformity?

Also, as a little bonus question: the first-ever direct image of a black hole, published today, appears to be oval-shaped. Is this because the black hole is rotating or some other effect?

  • $\begingroup$ "objects which are perfectly non-elastic will retain their original shape when spun" - but black holes are not objects, black holes are regions of spacetime from which nothing, not even light, can escape. $\endgroup$ Apr 10, 2019 at 23:42
  • $\begingroup$ @AlfredCentauri exactly, which is why I don't think comparisons to other spinning objects are of any help. $\endgroup$
    – Max
    Apr 10, 2019 at 23:46
  • $\begingroup$ Eeeh, the dimensions of the region we see are directly linked to the matter and energy within, so I don't think it's an entirely unreasonable comparison. The light that gets trapped might not have mass, but it would affect the energy level of the mass that is there. I think acceleration/compression toward the center would dwarf any perpendicular velocity by enough to accomplish something very close to a sphere though. Wikipedia compares "Approximately" to "Precisely" spherical. If it's rotating, it has momentum. $\endgroup$
    – K H
    Apr 11, 2019 at 0:21
  • $\begingroup$ @Max The question can be turned around: Why can't a spherically-symmetric black hole have a non-zero angular momentum? A spherically-symmetric black hole is uniquely characterized by its mass. This is Birkhoff's theorem. So if a spherically-symmetric black hole can have zero angular momentum, then it can't have non-zero angular momentum. An explicit answer to your question would therefore involve understanding why Birkhoff's theorem is true, and understanding how "angular momentum" is defined for a black hole. $\endgroup$ Apr 11, 2019 at 1:07
  • $\begingroup$ @Max Here's another approach: A spinning black hole causes "frame dragging." Light (or an object) falling radially inward from far away won't continue falling radially inward. Instead, it will be dragged in the direction of the hole's rotation as it approaches the black hole. Since the event horizon is defined to be the border of the region from which light cannot escape, there is no reason to expect the event horizon of a spinning black hole to be spherically symmetric. $\endgroup$ Apr 11, 2019 at 1:19

1 Answer 1


It is pretty difficult to answer "why" questions of this type. But to gain an intuition consider the following points

  • The rotating black hole is created by the collapse of a rotating cloud of gas or other matter. The particles on the overall axis of rotation carry no angular momentum with respect to the center, particles away from the axis do carry angular momentum on average. This means that the collapse is not spherical, the cloud is an oblate object at every step of the collapse. Why exactly should an object that is never spherically symmetric collapse to a single point? Instead, it first collapses to a disk-like structure with an oblate gravitational field. This geometry also survives in the curvature singularity of the black hole.
  • The rotation of the black hole drags frames in its vicinity with a finite speed - that is basically how we came to the conclusion that the gravitational field represents a rotating black hole! However, the rotation speed has to disappear at some rotational axis and the gravitational field thus cannot be spherically symmetric.
  • In a coordinate sense of the word, the horizon actually does appear at a single Boyer-Lindquist radius $r = r_H$. But the properties of the gravitational field at every point of this topological sphere are different when the black hole rotates (see previous points). So when we visualize or embed the surface $r=r_H$ in any reasonable way based on the actual physical properties of the gravitational field (space-time geometry), we end up showing it as an oblate surface as well.

As for your last note, the EHT image of a black hole shadow comes from a black hole immersed in a glowing plasma. The plasma is probably behind, next to, and even in front of the black hole. The shape of the silhouette you see cannot be quite understood as the shape of the black hole itself, but it is true that the spin plays a major role in the resulting image.

Consider the following images, it is obvious from the sequence that an oblate shape of the shadow in the bottom right region is determined more by the geometry of the plasma than the oblateness of the BH field itself. The bottom right image actually corresponds, up to a rotation, quite closely to what EHT people believe is happening in the image of M87. (the image is from the simulations of Mościbrodzka et al. (2014)) enter image description here

  • $\begingroup$ I see, +1-ed and will probably accept your answer soon, but I've still got a few questions that arise from your answer: the first point actually makes sense and I didn't think about that; however, I still don't see how your second point impacts the actual event horizon. The gravitational field may be asymmetric due to frame-dragging, but won't that only affect the rotation speed of objects in it? For example, if light were to be emitted from a heavily frame-dragged region, wouldn't it still travel away from the black hole at the speed of light, but also with some apparent rotational speed? $\endgroup$
    – Max
    Apr 11, 2019 at 10:49
  • $\begingroup$ If my understanding is correct, the shape and size of the event horizon shouldn't be affected. Finally, does your last sentence imply that the silhouette observed is also affected by the shape and structure of the (plasma) accretion disk? If so, can you give me an example of how an accretion disk feature may cause such an oval-ish shape? $\endgroup$
    – Max
    Apr 11, 2019 at 10:49
  • $\begingroup$ @Max The curvature singularity in the center of a rotating black hole is disk-like and the gravitational field must show effects that are not spherically symmetric from quite general arguments. That is as far as intuition takes you and the rest is in the actual equations - and those speak very clearly as well, you cannot just "understand them away". Note that the black hole horizon is not a "surface" or a "membrane" with physical properties you can easily understand by any other intuition. It is "nothing" locally. Instead, it is a global feature of the gravitational field. $\endgroup$
    – Void
    Apr 11, 2019 at 12:07
  • $\begingroup$ Yeah, the first bit makes sense. Of course, I know that the event horizon isn't a surface - for this reason, I have used the word "region" instead of "surface". I understand that the event horizon is just the region of space in which light cannot escape upon emission, which is what appears confusing to me, as frame-dragging shouldn't affect this region (for reasons described in earlier comment). I just thought, perhaps, there was a way to interpret those equations in a relatively intuitive manner, but if there isn't, as you seem to suggest, the first point is enough for my intuition. $\endgroup$
    – Max
    Apr 11, 2019 at 13:04

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