Black hole and angular momentum

Main Question or Discussion Point

It's theorized that most black holes have rotational speed. Also, I'm guessing, event horizons are always spherical or close to spherical because they are a function of the gravity well extending from center mass of the black hole. My question is this, could a black hole ever rotate with such high speed that it flattens out to the point that some of the material of the black hole travels outside the event horizon? I assume not but don't know how to do the math to check.

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mfb
Mentor
Rotating black holes are not spherical, see Kerr metric for details.

could a black hole ever rotate with such high speed that it flattens out to the point that some of the material of the black hole travels outside the event horizon?
There are solutions of General Relativity without an event horizon, but it is very questionable if they exist in the universe (probably not).

K^2
Event horizon of the Kerr black hole is still spherical. There are two singularities in the Kerr metric. One is spherical, the other oblong. The area in between is the ergosphere, and you can still accelerate things out of the ergosphere by sacrificing some mass to the black hole. So OP's question holds.

The bigger problem is that OP seems to picture mass in black hole as some finite volume object. It is not. For a non-rotating black hole, it's a point. A single point in space with finite mass and therefore infinite density. I don't think it is known yet exactly how the mass in a rotating black hole is distributed. Ring shape is a popular hypothesis, which apparently agrees with equations, but not proven to be the lowest energy state. At any rate, it's not something that has volume. So it doesn't flatten under rotation exactly as you picture it.

Still, if it is a ring shape, the radius of the ring probably depends on angular momentum. Personally, I wouldn't know where to begin verifying that. That leaves us with two possibilities. Either the rotational energy provides enough mass for the black hole to ensure that event horizon is always further out, or there is an actual maximum angular momentum a black hole can have. I do not know which it is.

stevebd1
Gold Member
My question is this, could a black hole ever rotate with such high speed that it flattens out to the point that some of the material of the black hole travels outside the event horizon? I assume not but don't know how to do the math to check.
From what I could gather, as gravity increases relative to the metric the closer you get to the black hole, the centripetal acceleration decreases relative to the metric. For instance, the Newtonian equation for overall gravitational acceleration taking into account centripetal acceleration is-

$$a_t=a_g-a_c$$

where at is total gravitational acceleration, ag is gravitational acceleration and ac is the centripetal acceleration of the orbiting object. This can be rewritten as-

$$a_t=\frac{Gm}{r^2}-\frac{v_t^2}{r}$$

where G is the gravitational constant, m is the mass of the central object, r is the radius between the two objects and vt is the tangential velocity of the orbiting object.

The relativistic version of this (i.e. for a static black hole) is-

$$a_t=\frac{Gm}{r^2}\frac{1}{\sqrt{1-2M/r}}-\frac{v_t^2}{r}\sqrt{1-2M/r}$$

Where M=Gm/c2 This ties with the statement that once inside the event horizon, all world lines lead to the singularity. Centripetal acceleration becomes zero at the EH and negative once inside the EH. (Note: If you set at as zero, vt as c and solve for r, you will get the photon sphere radius).

This gets a little more complex for rotating black holes as frame dragging is taken into account but the principle still applies. It also explains why objects that approach the ergosphere (which is spinning at c relative to infinity) are still pulled into the BH. The whole idea of centripetal acceleration reducing in extreme gravity is demonstrated to some extent in the following paper-

Geometric transport along circular orbits in stationary axisymmetric spacetimes
http://arxiv.org/abs/gr-qc/0407004

Hrmm, I understand the notion of something being a point in space but can't really visualize it. If the black hole is rotating, is it still a point? It being ring shaped was mentioned. Is it possible to have a ring shaped distribution of some description while still being a point?

stevebd1
Gold Member
Hrmm, I understand the notion of something being a point in space but can't really visualize it. If the black hole is rotating, is it still a point? It being ring shaped was mentioned. Is it possible to have a ring shaped distribution of some description while still being a point?
Technically the quantity $a$ in the Kerr metric is the radius of the ring singularity. If we consider the the equation for $a$ which is a=J/mc, we can see that it is a rearrangement of the general equation to calculate the angular momentum of a spinning ring which is J=vmr where v becomes c and r becomes $a$. If we use general spherical coordinates with Kerr metric then $a$ doesn't have much significance (see top diagram in the picture* attached to this post) but if we use elliptical coordinates then $a$ has significance and the spheres do become oblate, this oblateness reduces and becomes more spherical the larger r is (see the bottom diagram in the attached picture). It's worth noting that within the inner (Cauchy) horizon timelike geodesics are reinstated so the ring 'singlarity' may not be a singularity in the strictest sense as there are no spacelike geodesics pushing the singularity to a point as is the case in a static black hole.

*The picture attached is a Kerr BH with a spin parameter of a/M=0.95, the maximum (maximal) being a/M=1.

Elliptical coordinates are also used on the following web page (scroll down)-

K^2
I understand that metric singularities depend on choice of coordinate system, but shouldn't event horizon be defined unambiguously? I thought that the inner metric singularity of the Kerr metric in spherical coordinates was the event horizon, in which case, it would still be spherical even if you chose a coordinate system where the singularity itself has a different shape. (E.g., there are coordinate systems where Schwarzschild metric has no metric singularities, but the event horizon is still there.)

Or did I misunderstand something about the above post?

stevebd1
Gold Member
I understand that metric singularities depend on choice of coordinate system, but shouldn't event horizon be defined unambiguously? I thought that the inner metric singularity of the Kerr metric in spherical coordinates was the event horizon, in which case, it would still be spherical even if you chose a coordinate system where the singularity itself has a different shape. (E.g., there are coordinate systems where Schwarzschild metric has no metric singularities, but the event horizon is still there.)

Or did I misunderstand something about the above post?
I think I understand what you're saying. If a ring singularity was to form within a Cauchy horizon, there is no way the workings of the inner horizon could have any causal effect on the shape of the exterior horizon so while there might be an oblate inner (Cauchy) horizon with a ring singularity within, the outer horizon would still appear spherical (in some papers the Cauchy horizon is a weak singularity itself and considered the boundary of predictability which means the shape of the ring (if there was any) may have no impact on the shape of that either). Elliptical coordinates simply show the Kerr BH EH as oblate as this is a tidy way of incorporating the ring with a radius of $a$ (I've seen some papers where spherical coordinates are used and the ring singularity is shown at the coordinate radius of $a$ in the spacelike area between the outer (r+) and inner (Cauchy or r-) horizon which doesn't seem right as there is no stable r in this zone).