<p>Black holes are solutions of the Einstein equations, the fundamental equations of Einstein's relativity, that link together space-curvature to density of matter and energy. The black holes solutions are found in the void, when you let a single point of the space-time be singular (the 'center of the black hole'). If you further assume that the black hole is static, you find three solutions:</p>

<ol>
<li>the Schwarzschild solution, this is a non-rotating non-charged black hole, which is entirely described by its mass M ;</li>
<li>the Kerr solution, this is a rotating non-charged black hole, entirely described by its mass and spin (how fast it rotates) ;</li>
<li>the Kerr-Newmann solution, describing a rotating charged black hole, entirely described by its mass, spin and charge.</li>
</ol>

<p>It is a reasonable interpretation to think about a black hole as being a void with an infinitely small singularity in which lays all the mass, momentum and charge. In this sense, a black hole is empty, except for the region not described by the space-time framework (and that is not subject to the Einstein equations). So it is correct to think that black holes are only distorted region of space-time. The space-time is distorted so that going from the infinity (where there is no mass and no spin) you can reach the singularity continuously. If you fall onto a rotating black-hole, you will end up with the same spin as the black hole once you hit the singularity, (almost) regardless of how much spin you had initially.</p>

<p>Schwarzschild black holes force you to fall inward once you cross the 'horizon' of the black hole, regardless of how much acceleration you can produce. Kerr black holes have the same horizon, as well as a so-called 'ergosphere'. This 'sphere' is a region of space where everything including light is forced to rotate in the direction of the spin of the black-hole.</p>

<p>In the case of the Earth, you cannot directly use the Kerr solution to approximate the rotating Earth (as stated previously, this is a solution of the equations <em>in void</em>) how much drag you will have, but this gives you the correct idea, as stated by @James K.</p>

CC BY-SA 3.0
2016-09-12T08:19:49.380Z
2016-09-12T08:19:49.380Z
cphyc
<p>What we call a black hole is a solution of Einstein's equations of gravity.</p>

<p>We know several exact solutions: Empty and flat space corresponds to special relativity. Other known solutions include that of a point mass, a charged point mass, and a solution, known as the Kerr solution in which spacetime around a point mass is being "dragged" </p>

<p>This last solution can be interpreted as corresponding to a black hole that would form if a rotating mass were to collapse. As you note, a black hole contains no matter: it is a vacuum solution of Einstein's equations. However a black hole does have properties like mass and angular momentum.</p>

<p>The shape of space around the Earth can be approximated the Kerr solution of a spinning black hole, with a radius of about 1cm. We would therefore expect some dragging of spacetime around the Earth. In 2008 the <a href="https://en.wikipedia.org/wiki/Gravity_Probe_B">Gravity B probe</a> had measured the gravity around the Earth and confirmed that the Frame Dragging effect predicted was present and its results were consistent with  General relativity. Frame dragging around the Earth is fantastically small and the Gravity B probe was working right at its limit of sensitivity. It causes a gyroscope to deviate from its spin by 0.00001 degrees per year.</p>

CC BY-SA 3.0
2016-09-13T00:09:00.770Z
So, when considering conservation of angular momentum (and energy for that matter) do we have to consider all of space time and not just matter? (I suppose that would make sense since matter is part of space-time). In other words, can there be transfer of energy or angular momentum between what we know as matter and energy and the "fabric" of space-time.
Jack R. Woods
CC BY-SA 3.0
2016-09-15T04:19:53.903Z
Also, I calculated that if a 10M(sun) black hole started out as 30M(sun) star with 30R(sun) radius and 80km/s rotational velocity (the values given for the star Alpha Camelopardalis) then conservation of momentum figured classically would say that the minimum radius would be about 1680km since then the rotational velocity would be the speed of light. I'm not sure how frame dragging and relativistic equations would change this?????? (Even with smaller rotational velocities, etc. there would still be an R(min) classically.
Jack R. Woods
CC BY-SA 3.0
2016-09-11T12:46:16.053Z
2016-09-11T12:46:16.053Z
James K
<p>This seems like a pretty basic question, but I couldn't find a duplicate upon quick inspection. From what I understand about black holes they actually contain no matter. All of the mass that maintains the black hole (ie. keeps space warped) is found in the energy bound in the space itself. I may not be stating this quite right, but this is basically what Kip Thorne states in "The Science of Interstellar". Therefore, it seems that to maintain angular momentum during black hole formation that space itself must "spin". This would imply that space is actually "spinning" even in the vicinity of the Earth (to a much lesser degree of course). Is this true?</p>

<p>As an aside, I don't take everything that Kip says as gospel truth. He states several times that Io is the innermost moon of Saturn in Chapter 7.</p>

<black-hole>
How does a black hole spin?
Jack R. Woods
# How does a black hole spin?

(Asked by: username_0 on 2016-09-11)

This seems like a pretty basic question, but I couldn't find a duplicate upon quick inspection. From what I understand about black holes they actually contain no matter. All of the mass that maintains the black hole (ie. keeps space warped) is found in the energy bound in the space itself. I may not be stating this quite right, but this is basically what Kip Thorne states in "The Science of Interstellar". Therefore, it seems that to maintain angular momentum during black hole formation that space itself must "spin". This would imply that space is actually "spinning" even in the vicinity of the Earth (to a much lesser degree of course). Is this true?

As an aside, I don't take everything that Kip says as gospel truth. He states several times that Io is the innermost moon of Saturn in Chapter 7.

**Tags:** black-hole


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## 2 answers

### Answer by username_2 on 2016-09-11. Score: 5 | Accepted answer

What we call a black hole is a solution of Einstein's equations of gravity.

We know several exact solutions: Empty and flat space corresponds to special relativity. Other known solutions include that of a point mass, a charged point mass, and a solution, known as the Kerr solution in which spacetime around a point mass is being "dragged"

This last solution can be interpreted as corresponding to a black hole that would form if a rotating mass were to collapse. As you note, a black hole contains no matter: it is a vacuum solution of Einstein's equations. However a black hole does have properties like mass and angular momentum.

The shape of space around the Earth can be approximated the Kerr solution of a spinning black hole, with a radius of about 1cm. We would therefore expect some dragging of spacetime around the Earth. In 2008 the [Gravity B probe](https://en.wikipedia.org/wiki/Gravity_Probe_B) had measured the gravity around the Earth and confirmed that the Frame Dragging effect predicted was present and its results were consistent with General relativity. Frame dragging around the Earth is fantastically small and the Gravity B probe was working right at its limit of sensitivity. It causes a gyroscope to deviate from its spin by 0.00001 degrees per year.

> **Comments:**
> **username_0**: So, when considering conservation of angular momentum (and energy for that matter) do we have to consider all of space time and not just matter? (I suppose that would make sense since matter is part of space-time). In other words, can there be transfer of energy or angular momentum between what we know as matter and energy and the "fabric" of space-time.  
> **username_0**: Also, I calculated that if a 10M(sun) black hole started out as 30M(sun) star with 30R(sun) radius and 80km/s rotational velocity (the values given for the star Alpha Camelopardalis) then conservation of momentum figured classically would say that the minimum radius would be about 1680km since then the rotational velocity would be the speed of light. I'm not sure how frame dragging and relativistic equations would change this?????? (Even with smaller rotational velocities, etc. there would still be an R(min) classically.  

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### Answer by username_1 on 2016-09-12. Score: 2

Black holes are solutions of the Einstein equations, the fundamental equations of Einstein's relativity, that link together space-curvature to density of matter and energy. The black holes solutions are found in the void, when you let a single point of the space-time be singular (the 'center of the black hole'). If you further assume that the black hole is static, you find three solutions:

1. the Schwarzschild solution, this is a non-rotating non-charged black hole, which is entirely described by its mass M ;
2. the Kerr solution, this is a rotating non-charged black hole, entirely described by its mass and spin (how fast it rotates) ;
3. the Kerr-Newmann solution, describing a rotating charged black hole, entirely described by its mass, spin and charge.

It is a reasonable interpretation to think about a black hole as being a void with an infinitely small singularity in which lays all the mass, momentum and charge. In this sense, a black hole is empty, except for the region not described by the space-time framework (and that is not subject to the Einstein equations). So it is correct to think that black holes are only distorted region of space-time. The space-time is distorted so that going from the infinity (where there is no mass and no spin) you can reach the singularity continuously. If you fall onto a rotating black-hole, you will end up with the same spin as the black hole once you hit the singularity, (almost) regardless of how much spin you had initially.

Schwarzschild black holes force you to fall inward once you cross the 'horizon' of the black hole, regardless of how much acceleration you can produce. Kerr black holes have the same horizon, as well as a so-called 'ergosphere'. This 'sphere' is a region of space where everything including light is forced to rotate in the direction of the spin of the black-hole.

In the case of the Earth, you cannot directly use the Kerr solution to approximate the rotating Earth (as stated previously, this is a solution of the equations *in void*) how much drag you will have, but this gives you the correct idea, as stated by @username_2.


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