Typical rotation speed of a black hole?

[Warp]

General relativity predicts that when a large-enough rotating star collapses at the end of its life, it will collapse into a ring singularity, and this ring conserves angular momentum.

When a typically-sized star (large enough to collapse into a black hole) rotating at a typical speed collapses into a ring singularity, how fast does the ring rotate (in terms of revolutions per second)? Also, what's the diameter of the ring?

What would the rotational speed and diameter of the ring be for a humongous star, such as VY Canis Majoris?

Also, in both cases: What would the dimensions and shape of the event horizon be?

 

[phyzguy]

Rotating black holes are described by the Kerr metric. Here's a good link to get started, and here's a nice paper with more detail. The event horizon of a Kerr black hole is still spherical, but there is an oblate spheroidally shaped surface called the ergosphere which is outside the event horizon. Between the ergosphere and the event horizon, everything must co-rotate along with the black hole.

The event horizon is rotating as though it were a solid body, so can be characterized by an angular velocity or rotation rate in RPM. For a 10 solar mass black hole rotating at the maximal rate, it is rotating at about 10^4 radians/second or about 10^5 RPM. Larger black holes will rotate at lower RPM rates, but at the maximal rotation rate, a point on the equator of the event horizon is always rotating at 1/2 the speed of light.

 

[Warp]

The event horizon is rotating as though it were a solid body

I don't really understand this. The event horizon does not consist of matter (or even any kind of energy, if I understand correctly). It's simply a zone in spacetime which geometry has certain characteristics. There's "nothing" there to rotate. (Rotation would imply that something physically moves around the central axis.)

Perhaps what you meant is that spacetime geometry at the event horizon is the same as if the event horizon were a physical massive surface that rotates? (The difference between a non-rotating and a rotating black hole can be determined by this difference in spacetime geometry, and said difference is caused by so-called frame dragging.)

In this sense, does the event horizon "rotate" at a different speed than the ring singularity?

 

[phyzguy]

You're right of course - hence the phrase "as though" I didn't mean to imply there was anything there to rotate, I was just trying to answer your question about rotation speed. As far as any reference to the "ring singularity", we don't really have a working theory for what is inside the event horizon, so I don't have any comment.

 

[timmdeeg]

The event horizon does not consist of matter (or even any kind of energy, if I understand correctly). It's simply a zone in spacetime which geometry has certain characteristics. There's "nothing" there to rotate. (Rotation would imply that something physically moves around the central axis.)

As there is "nothing there to rotate", one can add testparticles. Besides beeing accelerated towards the singularity they experience frame dragging in case the black hole is spinning and thus show this very special behaviour of that spacetime.

 

[stevebd1]

General relativity predicts that when a large-enough rotating star collapses at the end of its life, it will collapse into a ring singularity, and this ring conserves angular momentum.

When a typically-sized star (large enough to collapse into a black hole) rotating at a typical speed collapses into a ring singularity, how fast does the ring rotate (in terms of revolutions per second)? Also, what's the diameter of the ring?

What would the rotational speed and diameter of the ring be for a humongous star, such as VY Canis Majoris?

Also, in both cases: What would the dimensions and shape of the event horizon be?

Technically, the quantity $a$ in Kerr metric is the coordinate radius for the ring singularity. This may seem initially odd as it is greater than the coordinate radius of the inner Cauchy horizon $(r_-)$ though when using elliptical coordinates, $a$ would reside within the inner horizon as shown on page 35 of http://casa.colorado.edu/~ajsh/phys5770_08/bh.pdf and the bottom two images on this web page. $a$ is established by $a=J/mc$ where J is angular momentum, m is mass and c the speed of light. In most cases, it is predicted that the ring singularity is spinning at c. It's worth noting though that due to frame dragging, space itself is already rotating at c at the ergosphere relative to infinity with tangential velocity increasing to infinite at the event horizon (this can be calculated using equations derived from the Kerr metric). If you want to calculate $a$ for VY Canis Majoris then you would need to know its angular momentum.

 

[bcrowell]

Possibly relevant: https://www.physicsforums.com/showthread.php?t=520173

 

[Nabeshin]

There are suggestions that the majority of black holes (>80%, perhaps) are near the extremal regime with a>0.9 or so. Depending on the modeling done, this can even be pushed up past 0.95. But other studies suggest that accretion might halt at a much lower spin, so it's somewhat up in the air at the moment.