<p>Your intuition is correct, as far as classical dynamics is concerned. The main problem, I think, arises in that you are referring to a strictly quantum mechanical phenomenon - what is the physical singularity? In General Relativity (GR), we know these singularities exist, but we cannot really know more than that (aside from events that occur not too close to the physical singularity). But, it's deeper than that, because in GR we cannot get information out of a black hole, so we'd never really know. That's why people in the comments are referring to a "theory of quantum gravity" which would take over where GR breaks down (at the physical singularity). </p>

<p>For a non-rotating (Schwarzschild) BH, the physical singularity is at <span class="math-container">$r = 0$</span>, but the coordinate singularity is at the event horizon, <span class="math-container">$r = 2M$</span>. </p>

<p>For a rotating (Kerr) BH, the physical singularity is now a ring with no thickness but non-zero radius - this is <strong>the</strong> ringularity. We literally don't know what should happen here (need "quantum gravity"), but in GR we can't even know in principle, because if someone were to try to fall into the ringularity we would never receive any signals back from them (as far as GR tells us).</p>

<p>OKay, with all that out of the way, we can still make a similar comment like what you want to <strong>within</strong> the framework of GR:</p>

<p>Consider a Kerr BH (in Boyer-Lindquist coordinates) with dimensionless spin <span class="math-container">$a$</span> whose symmetry axis is along the <span class="math-container">$z$</span>, so that the ringularity lies in the <span class="math-container">$x$</span>-<span class="math-container">$y$</span> plane. One can show that at <span class="math-container">$z = 0$</span> and at <span class="math-container">$r = 0$</span> (where <span class="math-container">$r$</span> is not the physical radial distance), the equation of the ringularity is</p>

<p><span class="math-container">$$ x^2 + y^2 = a^2$$</span></p>

<p>and so the radius of the ringularity is the spin. Therefore, as the spin of the Kerr BH increases, the radius of the ringularity also increases. This is the opposite of what your classical intuition leads to, but then again, this is a very nontrivial spacetime metric, and it exists on the fringe between classical and quantum physics.</p>

<blockquote>
  <p>Why doesn't a singularity spin infinitely fast?</p>
</blockquote>

<p>the ringularity is not spinning itself, rather it is defined by the spin of the BH. Mathematically, we can construct some kind of "spin" of the ringularity, but it wouldn't make much physical sense. </p>

<p>Now, if the BH was spinning infinitely fast, i.e. <span class="math-container">$a \rightarrow \infty$</span>, then the ringularity would have an infinite radius!!! And THAT's certainly problematic (but interesting!). Fortunately, as far as we know, BH's cannot have a spin greater than 1 - the maximum spin of a Kerr BH is capped by the <a href="https://en.wikipedia.org/wiki/Naked_singularity" rel="nofollow noreferrer">naked singularity</a> <a href="https://en.wikipedia.org/wiki/Cosmic_censorship_hypothesis" rel="nofollow noreferrer">hypothesis</a>. You see, if BH's could have a spin greater than 1, then the ringularity would extend "outside" the event horizon, which is problematic because it would mean that we could observe the collapse of an object into infinite density. ;)</p>

CC BY-SA 4.0
2019-01-21T15:08:58.103Z
2019-01-21T16:28:07.000Z
2019-01-21T16:28:07.000Z
Daddy Kropotkin
<p>A question for a random reader passing through: because of the nature of things spinning faster when it gets smaller, wouldn't a singularity spin infinitely fast as it's infinitely small, if it has a spinn from the beginning that is? Btw I know what a "ringularity" but does it automaticly become one when rotational velocity is added? Is it just "fake news"?</p>

CC BY-SA 4.0
2019-01-16T20:57:58.393Z
The singularity is not a thing. A mathematical singularity is just a place where math breaks down. That's always a sign that our understanding is incomplete. Currently we only have general relativity to use when thinking about black holes. But we would also need quantum mechanics for the treatment to be complete. The problem is, we don't know how to make GR and QM play together. Using GR alone, math explodes at the center - you get divisions by zero and other absurdities. This means we basically don't know what happens at the center. It's all speculation for now.
Florin Andrei
CC BY-SA 4.0
2019-01-16T21:53:05.623Z
I agree that we don't know much, but if spacetime contracts indefinitely around these singularities, going the opposite direction once your inside of the event horizon is the same as going back in time, making so that everything around stuff in there is the "singularity". We don't know how math behave at that point, but we know that it must be a "point" if time only lets things pass in one direction. And a point can't spinn.
Das Stein
CC BY-SA 4.0
2019-01-17T17:51:20.967Z
It is also worth noting that in the Kerr metric the singularity is not a point, but a circle-like region.
Anders Sandberg
CC BY-SA 4.0
2019-01-20T11:59:05.047Z
Not only the angular will be infinite in the singularity, also curvature and energy density. Math can describe such object, but the actual physics is probably highly different.
peterh
CC BY-SA 4.0
2019-01-21T11:05:53.493Z
I strongly suspect that a quantum gravity theory will eliminate the singularity. However, in standard GR,  as [Ben Crowell says](https://physics.stackexchange.com/a/144458/123208), a singularity in GR is like a piece that has been cut out of the manifold. It's not a point or point-set at all. Also note that a BH singularity is never in the past of any observer.
PM 2Ring
<singularity><infinite>
Why doesn't a singularity spin infinitely fast?
Das Stein
# Why doesn't a singularity spin infinitely fast?

(Asked by: username_0 on 2019-01-16)

A question for a random reader passing through: because of the nature of things spinning faster when it gets smaller, wouldn't a singularity spin infinitely fast as it's infinitely small, if it has a spinn from the beginning that is? Btw I know what a "ringularity" but does it automaticly become one when rotational velocity is added? Is it just "fake news"?

**Tags:** singularity, infinite

> **Comments:**
> **username_1**: The singularity is not a thing. A mathematical singularity is just a place where math breaks down. That's always a sign that our understanding is incomplete. Currently we only have general relativity to use when thinking about black holes. But we would also need quantum mechanics for the treatment to be complete. The problem is, we don't know how to make GR and QM play together. Using GR alone, math explodes at the center - you get divisions by zero and other absurdities. This means we basically don't know what happens at the center. It's all speculation for now.  
> **username_0**: I agree that we don't know much, but if spacetime contracts indefinitely around these singularities, going the opposite direction once your inside of the event horizon is the same as going back in time, making so that everything around stuff in there is the "singularity". We don't know how math behave at that point, but we know that it must be a "point" if time only lets things pass in one direction. And a point can't spinn.  
> **username_2**: It is also worth noting that in the Kerr metric the singularity is not a point, but a circle-like region.  
> **username_3**: Not only the angular will be infinite in the singularity, also curvature and energy density. Math can describe such object, but the actual physics is probably highly different.  
> **username_4**: I strongly suspect that a quantum gravity theory will eliminate the singularity. However, in standard GR,  as [Ben Crowell says](https://physics.stackexchange.com/a/144458/123208), a singularity in GR is like a piece that has been cut out of the manifold. It's not a point or point-set at all. Also note that a BH singularity is never in the past of any observer.  

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## 1 answers

### Answer by username_5 on 2019-01-21. Score: 1

Your intuition is correct, as far as classical dynamics is concerned. The main problem, I think, arises in that you are referring to a strictly quantum mechanical phenomenon - what is the physical singularity? In General Relativity (GR), we know these singularities exist, but we cannot really know more than that (aside from events that occur not too close to the physical singularity). But, it's deeper than that, because in GR we cannot get information out of a black hole, so we'd never really know. That's why people in the comments are referring to a "theory of quantum gravity" which would take over where GR breaks down (at the physical singularity).

For a non-rotating (Schwarzschild) BH, the physical singularity is at $r = 0$, but the coordinate singularity is at the event horizon, $r = 2M$.

For a rotating (Kerr) BH, the physical singularity is now a ring with no thickness but non-zero radius - this is **the** ringularity. We literally don't know what should happen here (need "quantum gravity"), but in GR we can't even know in principle, because if someone were to try to fall into the ringularity we would never receive any signals back from them (as far as GR tells us).

OKay, with all that out of the way, we can still make a similar comment like what you want to **within** the framework of GR:

Consider a Kerr BH (in Boyer-Lindquist coordinates) with dimensionless spin $a$ whose symmetry axis is along the $z$, so that the ringularity lies in the $x$-$y$ plane. One can show that at $z = 0$ and at $r = 0$ (where $r$ is not the physical radial distance), the equation of the ringularity is

$$ x^2 + y^2 = a^2$$

and so the radius of the ringularity is the spin. Therefore, as the spin of the Kerr BH increases, the radius of the ringularity also increases. This is the opposite of what your classical intuition leads to, but then again, this is a very nontrivial spacetime metric, and it exists on the fringe between classical and quantum physics.

> Why doesn't a singularity spin infinitely fast?

the ringularity is not spinning itself, rather it is defined by the spin of the BH. Mathematically, we can construct some kind of "spin" of the ringularity, but it wouldn't make much physical sense.

Now, if the BH was spinning infinitely fast, i.e. $a \rightarrow \infty$, then the ringularity would have an infinite radius!!! And THAT's certainly problematic (but interesting!). Fortunately, as far as we know, BH's cannot have a spin greater than 1 - the maximum spin of a Kerr BH is capped by the [naked singularity](https://en.wikipedia.org/wiki/Naked_singularity) [hypothesis](https://en.wikipedia.org/wiki/Cosmic_censorship_hypothesis). You see, if BH's could have a spin greater than 1, then the ringularity would extend "outside" the event horizon, which is problematic because it would mean that we could observe the collapse of an object into infinite density. ;)


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