<p>There are multiple solutions to general relativity which allow for multiple different types of black holes. The "normal" black hole you see most people talk about, with a zero-volume, point singularity, is known as the <a href="https://en.wikipedia.org/wiki/Schwarzschild_metric#Singularities_and_black_holes" rel="noreferrer">Schwarzschild black hole</a>. If the black hole is spinning, the Schwarzschild solution no longer applies and you're talking about a different type of black hole. This new type is referred to as a <a href="https://en.wikipedia.org/wiki/Rotating_black_hole" rel="noreferrer">Kerr black hole</a> (other types include the <a href="https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_black_hole" rel="noreferrer">Reissner-Nordstr&ouml;m</a> and the <a href="https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric" rel="noreferrer">Kerr-Newman</a>).</p>

<p>In a Kerr black hole, the singularity at the center is still zero-volume, but it is no longer a point. Instead it is a disk of zero height, often referred to as a <a href="https://en.wikipedia.org/wiki/Ring_singularity" rel="noreferrer">ring singularity</a> or "Ringularity". The angular momentum of the spinning black hole is then the angular momentum around the axis of rotation passing through this ring.</p>

CC BY-SA 3.0
2017-12-17T22:12:34.413Z
Please excuse my ignorance, but there is something that has always bothered me. I think it's been explained to me, but I can't remember.
Jack R. Woods
CC BY-SA 3.0
2017-12-17T22:22:22.087Z
Above comment continued: If you use the spinning ice skater analogy and the conservation of angular momentum, it seems to me that a spinning object getting smaller and smaller would reach a point to where tangential surface spin velocity would approach c. I would think that this would limit the size of the object before it becomes a singularity.
Jack R. Woods
CC BY-SA 4.0
2020-11-03T10:28:00.570Z
@JackR.Woods old comment but fun question.  Because I like spheres, consider the Earth not an ice skater.   If you shrink the Earth to it's Schwarzschild radius, it becomes a black hole.  But if you conserve angular momentum, every time you shrink the Earth's radius by 1/2 you've increased it's rotational velocity by 4 to maintain angular momentum.   If you shrink Earth enough - to about 10 meters, according to Newton, Earth is now rotating at the speed of light at it's equator.   That's ofcourse, ridiculous, but that's the idea.  The angular momentum either goes somewhere (into a ring)
userLTK
CC BY-SA 4.0
2020-11-03T10:32:11.100Z
@JackR.Woods or it discipates, somehow, and the black hole has no angular momentum.   Nobody knows what really happens, but there are models where the black hole retains angular momentum and the singularity is a ring, just as there are models for ringed shaped planets that are gravitationally stable (but they may not exist in reality) because mass concentrations are likely to destabilize a ring planet into a binary planet.  https://io9.gizmodo.com/what-would-the-earth-be-like-if-it-was-the-shape-of-a-d-1515700296  That doesn't mean ring singularities exist, but mathematically they work.
userLTK
CC BY-SA 4.0
2017-12-11T15:08:14.147Z
2019-07-29T14:45:39.527Z
2019-07-29T14:45:39.527Z
zephyr
<p>Don't think of the singularity as being an object made of matter. A black hole is a vacuum solution to the relativity equations. That means there is <em>nothing</em>  inside the black hole. </p>

<p>A black hole doesn't contain matter, but it still has mass. The mass of the black hole can be observed in the curvature of space-time around the black hole. It also has angular momentum, this causes space-time around a spinning black hole to be dragged. This is weird. But this is what relativity predicts, and it's predictions are well supported by observations.</p>

<p>In a black hole, there is no object in the middle, it is the space-time itself that is rotating.</p>

CC BY-SA 3.0
2017-12-11T14:45:58.970Z
I'm not sure I agree with this at all. Do you have sources to back this up?
zephyr
CC BY-SA 3.0
2017-12-11T20:19:40.863Z
I suppose the contentious point is that "there is nothing inside a black hole". This is a consequence of the Kerr metric being a [vacuum solution](https://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)). The Kerr metric is discussed in [Kerr](https://arxiv.org/pdf/0706.0622.pdf) where it notes " this manifold
is Ricci flat, Rab = 0, and so satisfies the vacuum Einstein field
equations".  The metric has several singularities (ring singularities). These are not matter, they are not objects, there are singularities in the gravitational field. There is no rotating object.
James K
CC BY-SA 3.0
2017-12-11T20:42:33.297Z
I've never heard this interpretation before. My understanding is that the singularity is the point where GR breaks down and also where all the matter physically exists. It's not incorrect to say having this matter in the singularity contradicts the fact that it is a vacuum solution simply because the singularity is not part of the solution, it cannot be due to its infinite nature. Hence why people say all the mass is concentrated in the singularity. I'd really like a source saying the mass is within the space-time curvature if you have one.
zephyr
CC BY-SA 3.0
2017-12-11T21:12:25.540Z
I think the notion that the singularity is "where the matter is" leads to misconceptions implicit in questions like https://astronomy.stackexchange.com/questions/19734/how-can-gravitational-waves-escape-from-the-gravity-of-black-hole After all, if the mass in the singularity is generating the gravity, how does that gravity escape from black hole? The source I linked notes that the pictures of black holes assume a fictional Minkowski space inside the black hole. This is a useful fiction, but of course the actual shape of spacetime is curved, and so can't be easily drawn.
James K
CC BY-SA 3.0
2017-12-11T21:14:24.053Z
The singularity is not part of spacetime, it is not a "place" or a "time", so matter can't be "there" because the singularity isn't a "there".
James K
CC BY-SA 3.0
2017-12-12T04:54:53.973Z
Hmmm. True, a  Schwarzchild black hole is a pure vacuum solution of the GR field equations, but it's a rather artificial entity. It's eternal, so it wasn't formed by the gravitational collapse of matter, and it's totally isolated: there's nothing else in its universe. (And of course it has no spin). So it's a very simple model of a physical black hole. OTOH, we don't yet have a fully-working theory of quantum gravity, so we can't completely answer the question of the final fate of matter that falls into a BH, either during its formation, or later.
PM 2Ring
CC BY-SA 3.0
2017-12-12T04:55:04.403Z
(cont) We _can_ say that such matter rapidly falls to the centre, but we don't know what happens when it gets there. We can be fairly confident that the core of a BH is tiny, smaller than an atom, and that quantum effects _probably_ prevent the formation of a mathematical singularity. And that whatever _does_ happen down there has to be pretty weird since Pauli exclusion makes it hard to pack a lot of fermionic matter into such a small space. ;)
PM 2Ring
CC BY-SA 3.0
2017-12-10T07:48:44.220Z
2017-12-10T07:54:14.153Z
2017-12-10T07:54:14.153Z
James K
<p>One way to think of a black hole is that it is what is left behind when some matter (or energy) collapses so far that an event horizon forms. After that, no information of any kind can get out past the event horizon, so what happens inside has no effect on the rest of the universe. The externally visible properties of the black hole (basically the gravitational field outside the event horizon) sustains itself from then on (that's what we mean by saying it's a vacuum solution to the field equations)</p>

<p>If the matter that collapsed had angular momentum, that fact is reflected in the gravitational field, and (among other things) will drag nearby matter around the event horizon. That is what we talk about when we say the black hole is spinning.  More precise is to say that its gravitational field is spinning, and still more precise to simply say that the gravitional field matches the Kerr solution to Einstein's equations.</p>

<p>What is actually happening inside the event horizon is (in straight GR, not quantum gravity) not something we can ever know directly. GR lets us predict some of it, but we can never check those predictions and there are things GR is silent about.</p>

CC BY-SA 4.0
2019-07-29T18:15:03.980Z
2019-07-29T18:15:03.980Z
Steve Linton
<p>So I've heard that all black holes that have been observed <em>rotate</em> to some degree. But if it has zero radius, how is it even possible for the black hole itself to be spinning? Or is it just all the junk inside the event horizon that's rotating? 
Maybe it spins with infinite angular velocity and a finite angular momentum? Please explain how this works.</p>

CC BY-SA 3.0
2017-12-12T05:00:25.623Z
What junk? Anything that crosses the event horizon _has_ to fall to the centre of the BH _very_ quickly. We don't know what the core of a BH is like, we need a QG (quantum gravity) theory to talk about that, and of course, even when (if) we do have a QG theory we'll never be able to observe what goes on inside the event horizon directly, we'll need to use other evidence to validate QG.
PM 2Ring
CC BY-SA 3.0
2017-12-13T00:46:11.987Z
Everything we have observed in the Universe has been observed to rotate.  A collapsing star rotates, and angular momentum must be conserved, so the black hole must also rotate (which we have also observed). The scientific community would probably be quite astonished to find something that doesn't rotate, and the first response would be to check the instruments and data.
Mick
CC BY-SA 3.0
2017-12-13T02:29:49.357Z
What I was wondering is, if it is rotating, mustn't the angular velocity be essentially infinite, since the moment of inertia must be 0? It does make sense that it would have angular momentum because that needs to be conserved, but that 0 moment of inertia is bothering me.
Nathanael Vetters
CC BY-SA 4.0
2019-07-30T18:29:46.800Z
@Mick "angular momentum must be conserved" - When a black hole evaporates angular momentum is NOT conserved. The hawking radiation comes out in straight lines.
SurpriseDog
CC BY-SA 4.0
2019-07-31T06:13:51.693Z
@Benjamin You are aware, of course, that there are three fundamental quantities which are conserved in Newtonian mechanics, and of the energy–momentum relation in relativistic physics? Whether Hawking radiation is emitted or not, and whether it travels in straight lines or not, has no bearing on the fundamental assumption of conservation.
Mick
<black-hole><high-energy-astrophysics>
How can a singularity rotate?
Nathanael Vetters
# How can a singularity rotate?

(Asked by: username_0 on 2017-12-10)

So I've heard that all black holes that have been observed *rotate* to some degree. But if it has zero radius, how is it even possible for the black hole itself to be spinning? Or is it just all the junk inside the event horizon that's rotating?
Maybe it spins with infinite angular velocity and a finite angular momentum? Please explain how this works.

**Tags:** black-hole, high-energy-astrophysics

> **Comments:**
> **username_1**: What junk? Anything that crosses the event horizon _has_ to fall to the centre of the BH _very_ quickly. We don't know what the core of a BH is like, we need a QG (quantum gravity) theory to talk about that, and of course, even when (if) we do have a QG theory we'll never be able to observe what goes on inside the event horizon directly, we'll need to use other evidence to validate QG.  
> **username_2**: Everything we have observed in the Universe has been observed to rotate.  A collapsing star rotates, and angular momentum must be conserved, so the black hole must also rotate (which we have also observed). The scientific community would probably be quite astonished to find something that doesn't rotate, and the first response would be to check the instruments and data.  
> **username_0**: What I was wondering is, if it is rotating, mustn't the angular velocity be essentially infinite, since the moment of inertia must be 0? It does make sense that it would have angular momentum because that needs to be conserved, but that 0 moment of inertia is bothering me.  
> **username_3**: @username_2 "angular momentum must be conserved" - When a black hole evaporates angular momentum is NOT conserved. The hawking radiation comes out in straight lines.  
> **username_2**: @Benjamin You are aware, of course, that there are three fundamental quantities which are conserved in Newtonian mechanics, and of the energy–momentum relation in relativistic physics? Whether Hawking radiation is emitted or not, and whether it travels in straight lines or not, has no bearing on the fundamental assumption of conservation.  

---

## 3 answers

### Answer by username_4 on 2017-12-11. Score: 6

There are multiple solutions to general relativity which allow for multiple different types of black holes. The "normal" black hole you see most people talk about, with a zero-volume, point singularity, is known as the [Schwarzschild black hole](https://en.wikipedia.org/wiki/Schwarzschild_metric#Singularities_and_black_holes). If the black hole is spinning, the Schwarzschild solution no longer applies and you're talking about a different type of black hole. This new type is referred to as a [Kerr black hole](https://en.wikipedia.org/wiki/Rotating_black_hole) (other types include the [Reissner-Nordström](https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_black_hole) and the [Kerr-Newman](https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric)).

In a Kerr black hole, the singularity at the center is still zero-volume, but it is no longer a point. Instead it is a disk of zero height, often referred to as a [ring singularity](https://en.wikipedia.org/wiki/Ring_singularity) or "Ringularity". The angular momentum of the spinning black hole is then the angular momentum around the axis of rotation passing through this ring.

> **Comments:**
> **username_5**: Please excuse my ignorance, but there is something that has always bothered me. I think it's been explained to me, but I can't remember.  
> **username_5**: Above comment continued: If you use the spinning ice skater analogy and the conservation of angular momentum, it seems to me that a spinning object getting smaller and smaller would reach a point to where tangential surface spin velocity would approach c. I would think that this would limit the size of the object before it becomes a singularity.  
> **username_6**: @JackR.Woods old comment but fun question.  Because I like spheres, consider the Earth not an ice skater.   If you shrink the Earth to it's Schwarzschild radius, it becomes a black hole.  But if you conserve angular momentum, every time you shrink the Earth's radius by 1/2 you've increased it's rotational velocity by 4 to maintain angular momentum.   If you shrink Earth enough - to about 10 meters, according to Newton, Earth is now rotating at the speed of light at it's equator.   That's ofcourse, ridiculous, but that's the idea.  The angular momentum either goes somewhere (into a ring)  
> **username_6**: @JackR.Woods or it discipates, somehow, and the black hole has no angular momentum.   Nobody knows what really happens, but there are models where the black hole retains angular momentum and the singularity is a ring, just as there are models for ringed shaped planets that are gravitationally stable (but they may not exist in reality) because mass concentrations are likely to destabilize a ring planet into a binary planet.  https://io9.gizmodo.com/what-would-the-earth-be-like-if-it-was-the-shape-of-a-d-1515700296  That doesn't mean ring singularities exist, but mathematically they work.  

---

### Answer by username_7 on 2017-12-10. Score: 4

Don't think of the singularity as being an object made of matter. A black hole is a vacuum solution to the relativity equations. That means there is *nothing* inside the black hole.

A black hole doesn't contain matter, but it still has mass. The mass of the black hole can be observed in the curvature of space-time around the black hole. It also has angular momentum, this causes space-time around a spinning black hole to be dragged. This is weird. But this is what relativity predicts, and it's predictions are well supported by observations.

In a black hole, there is no object in the middle, it is the space-time itself that is rotating.

> **Comments:**
> **username_4**: I'm not sure I agree with this at all. Do you have sources to back this up?  
> **username_7**: I suppose the contentious point is that "there is nothing inside a black hole". This is a consequence of the Kerr metric being a [vacuum solution](https://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)). The Kerr metric is discussed in [Kerr](https://arxiv.org/pdf/0706.0622.pdf) where it notes " this manifold is Ricci flat, Rab = 0, and so satisfies the vacuum Einstein field equations".  The metric has several singularities (ring singularities). These are not matter, they are not objects, there are singularities in the gravitational field. There is no rotating object.  
> **username_4**: I've never heard this interpretation before. My understanding is that the singularity is the point where GR breaks down and also where all the matter physically exists. It's not incorrect to say having this matter in the singularity contradicts the fact that it is a vacuum solution simply because the singularity is not part of the solution, it cannot be due to its infinite nature. Hence why people say all the mass is concentrated in the singularity. I'd really like a source saying the mass is within the space-time curvature if you have one.  
> **username_7**: I think the notion that the singularity is "where the matter is" leads to misconceptions implicit in questions like https://astronomy.stackexchange.com/questions/19734/how-can-gravitational-waves-escape-from-the-gravity-of-black-hole After all, if the mass in the singularity is generating the gravity, how does that gravity escape from black hole? The source I linked notes that the pictures of black holes assume a fictional Minkowski space inside the black hole. This is a useful fiction, but of course the actual shape of spacetime is curved, and so can't be easily drawn.  
> **username_7**: The singularity is not part of spacetime, it is not a "place" or a "time", so matter can't be "there" because the singularity isn't a "there".  
> **username_1**: Hmmm. True, a  Schwarzchild black hole is a pure vacuum solution of the GR field equations, but it's a rather artificial entity. It's eternal, so it wasn't formed by the gravitational collapse of matter, and it's totally isolated: there's nothing else in its universe. (And of course it has no spin). So it's a very simple model of a physical black hole. OTOH, we don't yet have a fully-working theory of quantum gravity, so we can't completely answer the question of the final fate of matter that falls into a BH, either during its formation, or later.  
> **username_1**: (cont) We _can_ say that such matter rapidly falls to the centre, but we don't know what happens when it gets there. We can be fairly confident that the core of a BH is tiny, smaller than an atom, and that quantum effects _probably_ prevent the formation of a mathematical singularity. And that whatever _does_ happen down there has to be pretty weird since Pauli exclusion makes it hard to pack a lot of fermionic matter into such a small space. ;)  

---

### Answer by username_8 on 2019-07-29. Score: 3

One way to think of a black hole is that it is what is left behind when some matter (or energy) collapses so far that an event horizon forms. After that, no information of any kind can get out past the event horizon, so what happens inside has no effect on the rest of the universe. The externally visible properties of the black hole (basically the gravitational field outside the event horizon) sustains itself from then on (that's what we mean by saying it's a vacuum solution to the field equations)

If the matter that collapsed had angular momentum, that fact is reflected in the gravitational field, and (among other things) will drag nearby matter around the event horizon. That is what we talk about when we say the black hole is spinning. More precise is to say that its gravitational field is spinning, and still more precise to simply say that the gravitional field matches the Kerr solution to Einstein's equations.

What is actually happening inside the event horizon is (in straight GR, not quantum gravity) not something we can ever know directly. GR lets us predict some of it, but we can never check those predictions and there are things GR is silent about.


---

