<p>I'm sure I worked this out in detail in another answer, but I can't find it now. Just to add a point addressing some comments above. The limiting angular momentum of a black hole is (in suitable units) the square of its mass, while the Schwarzschild radius grows as the mass. So consider a large (near) maximally spinning black holes of mass <span class="math-container">$M$</span> which will have Schwarzchild radius <span class="math-container">$2M$</span>.</p>

<p>The maximum orbital angular momentum you can add to it by firing a particle of mass <span class="math-container">$m$</span>  just inside the event horizon and velocity almost <span class="math-container">$c$</span> (which is 1 in these units) is therefore <span class="math-container">$2Mm$</span>. If that particle is a smaller maximally spinning black hole of mass <span class="math-container">$m$</span> and angular momentum <span class="math-container">$m^2$</span> then the total angular momentum of the coalesced holes is <span class="math-container">$M^2+2Mm+m^2$</span> which is exactly <span class="math-container">$(M+m)^2$</span>, so the new black hole is still just maximally spinning.</p>

CC BY-SA 4.0
2024-02-14T19:19:08.773Z
The horizon radius of a (nearly) maximally spinning black hole is $M$ not $2M$. See https://doi.org/10.1016/0003-4916(74)90125-0 for the correct version of this argument.
TimRias
CC BY-SA 4.0
2019-06-24T13:10:15.027Z
2019-06-24T13:10:15.027Z
Steve Linton
<p>From a quick drive around the InformationSuperHighway, I'd say the answer will remain a complicated mess :-) .  I did find a reasonably non-mathematical discussion at <a href="http://www.universetoday.com/109308/how-fast-do-black-holes-spin/" rel="nofollow noreferrer">universetoday</a></p>

<blockquote>
  <p>The speed limit is set by the event horizon, eventually, at a high
  enough spin, reaches the singularity. You can’t have what’s called a
  naked singularity. You can’t have a singularity exposed to the rest of
  the Universe. That would mean that the singularity itself could emit
  energy or light and somebody outside could actually see it. And that
  can’t happen. That’s the physical limitation of how fast it can spin.
  Physicists use units for angular momentum that are cast in terms of
  mass, which is a curious thing, and the speed limit can be described
  as the angular momentum equals the mass of the black hole, and that
  sets the speed limit.”</p>
  
  <p>Just imagine. The black hole spins up to the point that it’s just
  about to reveal itself. But that’s impossible. The laws of physics
  won’t let it spin any faster. And here’s the amazing part. Astronomers
  have actually detected supermassive black holes spinning at the limits
  predicted by these theories.</p>
  
  <p>One black hole, at the heart of galaxy NGC 1365 is turning at 84% the
  speed of light. It has reached the cosmic speed limit, and can’t spin
  any faster without revealing its singularity.</p>
</blockquote>

CC BY-SA 3.0
2017-03-06T13:33:04.473Z
2017-03-06T13:33:04.473Z
Carl Witthoft
<p>Since I like math, let's throw some math into this. I'll try to keep it as simple as possible though.</p>
<h3>Kerr Black Holes</h3>
<p>A rotating black hole is known as a <a href="https://en.wikipedia.org/wiki/Kerr_metric" rel="nofollow noreferrer">Kerr Black Hole</a> (named after <a href="https://en.wikipedia.org/wiki/Roy_Kerr" rel="nofollow noreferrer">Roy Kerr</a> who found the numerical solution to GR equations for rotating black holes). In the case of a rotating black hole, there are two important parameters used to describe the black hole. The first is of course the mass of the black hole <span class="math-container">$M$</span>. The second is the spin <span class="math-container">$a$</span>. Really <span class="math-container">$a$</span> isn't the spin itself <span class="math-container">$-$</span> it's defined by <span class="math-container">$a=J/M$</span> <sub>(see footnote)</sub> where <span class="math-container">$J$</span> is the angular momentum of the black hole <span class="math-container">$-$</span> but it's a good proxy for spin so often you'll see scientists get lazy and just call it the spin of the black hole. The mathematics will tell you that Kerr black holes have the limitation that</p>
<p><span class="math-container">$$0 \le a/M \le 1$$</span></p>
<p><strong>Black Hole Event Horizon</strong></p>
<p>The important parameter we want to calculate is the black hole's radius. If you run through the math, you find that this radius is given by</p>
<p><span class="math-container">$$r_{\text{e}} = M+(M^2-a^2)^{1/2}$$</span></p>
<p>In the case when <span class="math-container">$a/M=0$</span> (and thus <span class="math-container">$a=0$</span>), this reduces to just <span class="math-container">$r_{\text{e}}=2M$</span>, or in regular units (instead of geometrized units) <span class="math-container">$r_{\text{e}}=2GM/c^2$</span>. Hopefully you can see that this just reduces to the normal <a href="https://en.wikipedia.org/wiki/Schwarzschild_radius" rel="nofollow noreferrer">Schwarzschild radius</a> for a non-rotating black hole and thus the equation above is a generalization to account for spin. Let's look at the other limit when <span class="math-container">$a/M=1$</span> (and thus <span class="math-container">$a=M$</span>). In this case, you find that the radius is <span class="math-container">$r_{\text{e}}=M$</span>. When <span class="math-container">$a/M=1$</span>, you have a <em>maximally rotating</em> black hole, and your radius is half the normal Schwarzschild radius of a non-rotating black hole. This equation defines the radius of the Event Horizon, the point after which there is no returning from the black hole.</p>
<p><strong>Ergosphere</strong></p>
<p>As it turns out, when you define your equation to calculate the radius of the black hole, there are actually multiple solutions! The section above shows one such solution, but there's another important solution as well. This radius, sometimes called the <em>static limit</em> is given by the equation</p>
<p><span class="math-container">$$r_{\text{s}} = M+\left(M-a^2\cos^2(\theta)\right)^{1/2}$$</span></p>
<p>Notice that this is almost exactly the same as above, except for that extra <span class="math-container">$\cos^2(\theta)$</span>. This defines a different, slightly larger and somewhat &quot;pumpkin-shaped&quot; horizon which encompasses the inner Event Horizon defined above. The region between this outer horizon and the inner horizon is known as the <a href="https://en.wikipedia.org/wiki/Ergosphere" rel="nofollow noreferrer">Ergosphere</a>. Without getting into the nitty gritty details, I'll just say that one important point about the Ergosphere is that anything inside it (that is, <span class="math-container">$r_{\text{e}}&lt;r&lt;r_{\text{s}}$</span>) must rotate exactly with the black hole - it is physically impossible to stay still here!</p>
<h3>Answers</h3>
<blockquote>
<p>They stopped short of saying that the tangential velocity of this spin rate is &quot;c&quot; (and how can a singularity have a &quot;tangential velocity&quot;?)</p>
</blockquote>
<p>When you talk about the tangential velocity, there are multiple components of this black hole you/they may be talking about. One such tangential velocity is the tangential velocity of the event horizon (defined by <span class="math-container">$r_{\text{e}}$</span> above). We can take a look at the case of a maximally rotating black hole and say that the angular momentum, based on the equations above, of such a black hole is given by</p>
<p><span class="math-container">$$J_{\text{max}} = a_{\text{max}}Mc = M^2c$$</span></p>
<p>Note that I've dropped the geometrized units just to be completely explicit. This has introduced an extra <span class="math-container">$c$</span> now. Remember that <span class="math-container">$a_{\text{max}}$</span> is achieved when <span class="math-container">$a/M=1$</span>.</p>
<p>We can also define the angular momentum using the standard equation from physics 101, <span class="math-container">$J=rMv_\perp$</span>, where of course <span class="math-container">$r$</span> is the radius of your object, and <span class="math-container">$v_\perp$</span> is the perpendicular, or else tangential, velocity of your spinning object. Recall from above that for a maximally rotating black hole, <span class="math-container">$r_{\text{e}}=M$</span> so we also have that</p>
<p><span class="math-container">$$J_{\text{max}} = r_{\text{e}}Mv_{\perp} = M^2v_\perp$$</span></p>
<p>You can see that these two equations for <span class="math-container">$J_{\text{max}}$</span> only equal each other if the tangential velocity <span class="math-container">$v_\perp$</span> is equal to the speed of light <span class="math-container">$c$</span>. So yes, you are correct to presume that at the fastest possible rotations, the black hole's event horizon is rotating at the speed of light!</p>
<p>I said though that there are multiple components you could talk about when discussing rotating black holes. The other, as you allude to, is the rotating singularity. You correctly point out - &quot;how can a singularity have a tangential velocity&quot;? As it turns out, Kerr black holes don't have point singularities, they have <a href="https://en.wikipedia.org/wiki/Ring_singularity" rel="nofollow noreferrer">ring singularities</a>. These are &quot;rings&quot; of mass with zero width but some finite radius. Almost like a disk of no height. These rings of course can then have a tangential velocity. You were correct to be suspicious of a point singularity having tangential velocity though. That is not possible.</p>
<blockquote>
<p>They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a &quot;naked black hole&quot; that would defy the laws of physics (GR).</p>
</blockquote>
<p>We know the equation exactly, since I defined it above. The radius of a stellar black hole (that is a black hole with mass exactly equal to the mass of the Sun, <span class="math-container">$M_\odot$</span>) is given by</p>
<p><span class="math-container">$$r=\frac{GM_{\odot}}{c^2}=1.48\:\text{km}$$</span></p>
<p>So yes, they were correct on their radius. They also state that spinning any faster results in a naked singularity. This is entirely true. To see this, go back to the equation for the event horizon. Remember that our upper spin limit is that <span class="math-container">$a=M$</span>. What happens to our event horizon radius when <span class="math-container">$a&gt;M$</span> (and thus <span class="math-container">$a/M&gt;1$</span>)? For arguments say, lets say <span class="math-container">$a=2M$</span>. Then our event horizon radius becomes</p>
<p><span class="math-container">$$r_{\text{e}}=M-(M^2-a^2)^{1/2}=M-(M^2-4M^2)^{1/2}=M-(-3M^2)^{1/2}=M-i\sqrt{3}M$$</span></p>
<p>Suddenly our radius is complex and has an imaginary component! That means it is not physical and thus <em>cannot exist</em>. Now that we don't have an event horizon, our singularity can't hide behind it and is &quot;naked&quot;, exposed to the universe for anyone to see. GR tells us such an event shouldn't be allowed to happen because it results in all sorts of violations of physics. So somehow, something has to prevent black holes from spinning faster than a maximal black hole.</p>
<blockquote>
<p>Shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down.</p>
</blockquote>
<p>Yes, that is true in general. All black holes should spin extremely fast, simply because of conservation of angular momentum. In fact, I don't think I can come up with a case where a black hole was found to be not spinning. Shown below is a plot from <a href="http://www.nature.com/news/spin-rate-of-black-holes-pinned-down-1.13512" rel="nofollow noreferrer">this Nature paper</a> which shows the measured spin of 19 supermassive black holes. They're all spinning pretty fast with some of them almost at the speed of light. None of them are even close to not spinning.</p>
<p><a href="https://i.sstatic.net/4JYOz.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4JYOz.jpg" alt="enter image description here" /></a></p>
<hr />
<p><sub>Footnote: In GR, to make the math easier, often scientists adopt special units known as <a href="https://en.wikipedia.org/wiki/Geometrized_unit_system" rel="nofollow noreferrer">geometrized units</a>. These are units chosen in just such a way that the gravitational constant, <span class="math-container">$G$</span>, and the speed of light, <span class="math-container">$c$</span>, are equal to one. There's infinitely many units which allow this. Essentially this means no GR equations have <span class="math-container">$G$</span> or <span class="math-container">$c$</span> in them, but they're implicitly there, they're just equal to one and so not shown.</sub></p>

CC BY-SA 4.0
2019-03-07T21:07:50.247Z
Great answer. So what happens if you try to feed more angular momentum to a near-maximally spinning hole? One possibility I can imagine is the black hole would asymptotically approaching max spin (this goes against my intuitions about angular momentum). Another is that spinning matter won’t be able to enter the black hole without exceeding the speed of light.
cobbal
CC BY-SA 4.0
2019-03-08T15:35:41.353Z
Good questions. Feel free to ask this as a new question on this site. Comments are not the best place for answering such questions.
zephyr
CC BY-SA 4.0
2020-07-30T08:17:30.997Z
It may be worth noting in this answer, that not all black holes are highly spinning. In particular, the merging black holes observed by LIGO and Virgo generally seem to have had very little or no spin (with a couple of recent exceptions).
TimRias
CC BY-SA 4.0
2022-10-29T08:39:39.447Z
@cobbal The maximum spin that a Kerr black hole can obtain from accretion was shown to be $\simeq 0.998$ by Thorne in 1974 https://ui.adsabs.harvard.edu/abs/1974ApJ...191..507T/abstract
Daddy Kropotkin
CC BY-SA 4.0
2022-10-29T08:45:44.843Z
@TimRias The spins of individual binary black holes (BBHs) observed by LIGO/Virgo are not well constrained, and instead constrain the aligned effective spin $\chi_{\rm eff}$. Since $\chi_{\rm eff}$ is a linear combination of the spin magnitudes and misalignments (w.r.t. the orb. ang. mom.), although most BBHs have $\chi_{\rm eff} \sim 0$ this does not imply that the spin magnitudes are small, since the misalignments are unconstrained. On the other hand, the population of BBHs do show evidence of non-negligible spin magnitudes, e.g., so it's not clear that LIGO/Virgo BHs have small spins...
Daddy Kropotkin
CC BY-SA 4.0
2022-10-29T08:45:51.953Z
... see e.g. this https://arxiv.org/abs/2205.08574
Daddy Kropotkin
CC BY-SA 4.0
2024-06-05T21:06:19.980Z
Why must one move faster than light to stand still in the ergosphere? If it is spinning at the same speed as the black hole, then it's moving slower than light.
Melab
CC BY-SA 4.0
2017-03-06T15:59:03.203Z
2023-12-31T23:29:57.033Z
2023-12-31T23:29:57.033Z
zephyr
<p>I have just been watching a podcast called &quot;Deep Astronomy&quot; and the discussion was about a super fast spinning black hole discovered with the NuSTAR space observatory. This black hole was modeled with high confidence to be spinning at about 99% of the maximum spin rate. They stopped short of saying that the tangential velocity of this spin rate is &quot;c&quot; (and how can a singularity have a &quot;tangential velocity&quot;?) They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a &quot;naked singularity&quot; that would defy the laws of physics (GR).</p>
<p>Also, shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down. Could somebody clarify this whole &quot;black hole spin thing&quot; without getting too complicated??</p>

CC BY-SA 4.0
2020-06-10T18:01:10.277Z
No black hole would comprise a singularity, and astrophysical black holes may not necessarily contain any functional one, per papers written by Nikodem J. Poplawski between 2009 and 2020 that are available free on Cornell University's Arxiv site, as well as thru published journals.  (Poplawski uses Einstein-Cartan Theory, developed in conversations between Einstein and the mathematician Cartan 14 years more recently than General Relativity:  Its math is, reportedly, extremely complex, but Poplawski's English verbiage is quite readable.)
Edouard
CC BY-SA 4.0
2022-11-16T17:15:46.877Z
ALL black holes have tangential veloity greater than speed of light.
Anixx
<black-hole><general-relativity>
Maximum spin rate of a black hole?
Jack R. Woods
# Maximum spin rate of a black hole?

(Asked by: username_0 on 2017-03-05)

I have just been watching a podcast called "Deep Astronomy" and the discussion was about a super fast spinning black hole discovered with the NuSTAR space observatory. This black hole was modeled with high confidence to be spinning at about 99% of the maximum spin rate. They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?) They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked singularity" that would defy the laws of physics (GR).

Also, shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down. Could somebody clarify this whole "black hole spin thing" without getting too complicated??

**Tags:** black-hole, general-relativity

> **Comments:**
> **username_1**: No black hole would comprise a singularity, and astrophysical black holes may not necessarily contain any functional one, per papers written by Nikodem J. Poplawski between 2009 and 2020 that are available free on Cornell University's Arxiv site, as well as thru published journals.  (Poplawski uses Einstein-Cartan Theory, developed in conversations between Einstein and the mathematician Cartan 14 years more recently than General Relativity:  Its math is, reportedly, extremely complex, but Poplawski's English verbiage is quite readable.)  
> **username_2**: ALL black holes have tangential veloity greater than speed of light.  

---

## 3 answers

### Answer by username_6 on 2017-03-06. Score: 33 | Accepted answer

Since I like math, let's throw some math into this. I'll try to keep it as simple as possible though.

### Kerr Black Holes

A rotating black hole is known as a [Kerr Black Hole](https://en.wikipedia.org/wiki/Kerr_metric) (named after [Roy Kerr](https://en.wikipedia.org/wiki/Roy_Kerr) who found the numerical solution to GR equations for rotating black holes). In the case of a rotating black hole, there are two important parameters used to describe the black hole. The first is of course the mass of the black hole $M$. The second is the spin $a$. Really $a$ isn't the spin itself $-$ it's defined by $a=J/M$ (see footnote) where $J$ is the angular momentum of the black hole $-$ but it's a good proxy for spin so often you'll see scientists get lazy and just call it the spin of the black hole. The mathematics will tell you that Kerr black holes have the limitation that

$$0 \le a/M \le 1$$

**Black Hole Event Horizon**

The important parameter we want to calculate is the black hole's radius. If you run through the math, you find that this radius is given by

$$r\_{\text{e}} = M+(M^2-a^2)^{1/2}$$

In the case when $a/M=0$ (and thus $a=0$), this reduces to just $r\_{\text{e}}=2M$, or in regular units (instead of geometrized units) $r\_{\text{e}}=2GM/c^2$. Hopefully you can see that this just reduces to the normal [Schwarzschild radius](https://en.wikipedia.org/wiki/Schwarzschild_radius) for a non-rotating black hole and thus the equation above is a generalization to account for spin. Let's look at the other limit when $a/M=1$ (and thus $a=M$). In this case, you find that the radius is $r\_{\text{e}}=M$. When $a/M=1$, you have a *maximally rotating* black hole, and your radius is half the normal Schwarzschild radius of a non-rotating black hole. This equation defines the radius of the Event Horizon, the point after which there is no returning from the black hole.

**Ergosphere**

As it turns out, when you define your equation to calculate the radius of the black hole, there are actually multiple solutions! The section above shows one such solution, but there's another important solution as well. This radius, sometimes called the *static limit* is given by the equation

$$r\_{\text{s}} = M+\left(M-a^2\cos^2(\theta)\right)^{1/2}$$

Notice that this is almost exactly the same as above, except for that extra $\cos^2(\theta)$. This defines a different, slightly larger and somewhat "pumpkin-shaped" horizon which encompasses the inner Event Horizon defined above. The region between this outer horizon and the inner horizon is known as the [Ergosphere](https://en.wikipedia.org/wiki/Ergosphere). Without getting into the nitty gritty details, I'll just say that one important point about the Ergosphere is that anything inside it (that is, $r\_{\text{e}}<r<r\_{\text{s}}$) must rotate exactly with the black hole - it is physically impossible to stay still here!

### Answers

> They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?)

When you talk about the tangential velocity, there are multiple components of this black hole you/they may be talking about. One such tangential velocity is the tangential velocity of the event horizon (defined by $r\_{\text{e}}$ above). We can take a look at the case of a maximally rotating black hole and say that the angular momentum, based on the equations above, of such a black hole is given by

$$J\_{\text{max}} = a\_{\text{max}}Mc = M^2c$$

Note that I've dropped the geometrized units just to be completely explicit. This has introduced an extra $c$ now. Remember that $a\_{\text{max}}$ is achieved when $a/M=1$.

We can also define the angular momentum using the standard equation from physics 101, $J=rMv\_\perp$, where of course $r$ is the radius of your object, and $v\_\perp$ is the perpendicular, or else tangential, velocity of your spinning object. Recall from above that for a maximally rotating black hole, $r\_{\text{e}}=M$ so we also have that

$$J\_{\text{max}} = r\_{\text{e}}Mv\_{\perp} = M^2v\_\perp$$

You can see that these two equations for $J\_{\text{max}}$ only equal each other if the tangential velocity $v\_\perp$ is equal to the speed of light $c$. So yes, you are correct to presume that at the fastest possible rotations, the black hole's event horizon is rotating at the speed of light!

I said though that there are multiple components you could talk about when discussing rotating black holes. The other, as you allude to, is the rotating singularity. You correctly point out - "how can a singularity have a tangential velocity"? As it turns out, Kerr black holes don't have point singularities, they have [ring singularities](https://en.wikipedia.org/wiki/Ring_singularity). These are "rings" of mass with zero width but some finite radius. Almost like a disk of no height. These rings of course can then have a tangential velocity. You were correct to be suspicious of a point singularity having tangential velocity though. That is not possible.

> They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked black hole" that would defy the laws of physics (GR).

We know the equation exactly, since I defined it above. The radius of a stellar black hole (that is a black hole with mass exactly equal to the mass of the Sun, $M\_\odot$) is given by

$$r=\frac{GM\_{\odot}}{c^2}=1.48\:\text{km}$$

So yes, they were correct on their radius. They also state that spinning any faster results in a naked singularity. This is entirely true. To see this, go back to the equation for the event horizon. Remember that our upper spin limit is that $a=M$. What happens to our event horizon radius when $a>M$ (and thus $a/M>1$)? For arguments say, lets say $a=2M$. Then our event horizon radius becomes

$$r\_{\text{e}}=M-(M^2-a^2)^{1/2}=M-(M^2-4M^2)^{1/2}=M-(-3M^2)^{1/2}=M-i\sqrt{3}M$$

Suddenly our radius is complex and has an imaginary component! That means it is not physical and thus *cannot exist*. Now that we don't have an event horizon, our singularity can't hide behind it and is "naked", exposed to the universe for anyone to see. GR tells us such an event shouldn't be allowed to happen because it results in all sorts of violations of physics. So somehow, something has to prevent black holes from spinning faster than a maximal black hole.

> Shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down.

Yes, that is true in general. All black holes should spin extremely fast, simply because of conservation of angular momentum. In fact, I don't think I can come up with a case where a black hole was found to be not spinning. Shown below is a plot from [this Nature paper](http://www.nature.com/news/spin-rate-of-black-holes-pinned-down-1.13512) which shows the measured spin of 19 supermassive black holes. They're all spinning pretty fast with some of them almost at the speed of light. None of them are even close to not spinning.

[![enter image description here](https://i.sstatic.net/4JYOz.jpg)](https://i.sstatic.net/4JYOz.jpg)

---

Footnote: In GR, to make the math easier, often scientists adopt special units known as [geometrized units](https://en.wikipedia.org/wiki/Geometrized_unit_system). These are units chosen in just such a way that the gravitational constant, $G$, and the speed of light, $c$, are equal to one. There's infinitely many units which allow this. Essentially this means no GR equations have $G$ or $c$ in them, but they're implicitly there, they're just equal to one and so not shown.

> **Comments:**
> **username_7**: Great answer. So what happens if you try to feed more angular momentum to a near-maximally spinning hole? One possibility I can imagine is the black hole would asymptotically approaching max spin (this goes against my intuitions about angular momentum). Another is that spinning matter won’t be able to enter the black hole without exceeding the speed of light.  
> **username_6**: Good questions. Feel free to ask this as a new question on this site. Comments are not the best place for answering such questions.  
> **username_4**: It may be worth noting in this answer, that not all black holes are highly spinning. In particular, the merging black holes observed by LIGO and Virgo generally seem to have had very little or no spin (with a couple of recent exceptions).  
> **username_8**: @username_7 The maximum spin that a Kerr black hole can obtain from accretion was shown to be $\simeq 0.998$ by Thorne in 1974 https://ui.adsabs.harvard.edu/abs/1974ApJ...191..507T/abstract  
> **username_8**: @username_4 The spins of individual binary black holes (BBHs) observed by LIGO/Virgo are not well constrained, and instead constrain the aligned effective spin $\chi_{\rm eff}$. Since $\chi_{\rm eff}$ is a linear combination of the spin magnitudes and misalignments (w.r.t. the orb. ang. mom.), although most BBHs have $\chi_{\rm eff} \sim 0$ this does not imply that the spin magnitudes are small, since the misalignments are unconstrained. On the other hand, the population of BBHs do show evidence of non-negligible spin magnitudes, e.g., so it's not clear that LIGO/Virgo BHs have small spins...  
> **username_8**: ... see e.g. this https://arxiv.org/abs/2205.08574  
> **username_9**: Why must one move faster than light to stand still in the ergosphere? If it is spinning at the same speed as the black hole, then it's moving slower than light.  

---

### Answer by username_3 on 2019-06-24. Score: 3

I'm sure I worked this out in detail in another answer, but I can't find it now. Just to add a point addressing some comments above. The limiting angular momentum of a black hole is (in suitable units) the square of its mass, while the Schwarzschild radius grows as the mass. So consider a large (near) maximally spinning black holes of mass $M$ which will have Schwarzchild radius $2M$.

The maximum orbital angular momentum you can add to it by firing a particle of mass $m$ just inside the event horizon and velocity almost $c$ (which is 1 in these units) is therefore $2Mm$. If that particle is a smaller maximally spinning black hole of mass $m$ and angular momentum $m^2$ then the total angular momentum of the coalesced holes is $M^2+2Mm+m^2$ which is exactly $(M+m)^2$, so the new black hole is still just maximally spinning.

> **Comments:**
> **username_4**: The horizon radius of a (nearly) maximally spinning black hole is $M$ not $2M$. See https://doi.org/10.1016/0003-4916(74)90125-0 for the correct version of this argument.  

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### Answer by username_5 on 2017-03-06. Score: 2

From a quick drive around the InformationSuperHighway, I'd say the answer will remain a complicated mess :-) . I did find a reasonably non-mathematical discussion at [universetoday](http://www.universetoday.com/109308/how-fast-do-black-holes-spin/)

> The speed limit is set by the event horizon, eventually, at a high
> enough spin, reaches the singularity. You can’t have what’s called a
> naked singularity. You can’t have a singularity exposed to the rest of
> the Universe. That would mean that the singularity itself could emit
> energy or light and somebody outside could actually see it. And that
> can’t happen. That’s the physical limitation of how fast it can spin.
> Physicists use units for angular momentum that are cast in terms of
> mass, which is a curious thing, and the speed limit can be described
> as the angular momentum equals the mass of the black hole, and that
> sets the speed limit.”
>
> Just imagine. The black hole spins up to the point that it’s just
> about to reveal itself. But that’s impossible. The laws of physics
> won’t let it spin any faster. And here’s the amazing part. Astronomers
> have actually detected supermassive black holes spinning at the limits
> predicted by these theories.
>
> One black hole, at the heart of galaxy NGC 1365 is turning at 84% the
> speed of light. It has reached the cosmic speed limit, and can’t spin
> any faster without revealing its singularity.


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