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A non-rotating black hole has a Schwarzchild radius of $2GM/c^2$. A rotating black hole of the same mass has a smaller outer horizon radius, down to a limit of $GM/c^2$ at the fastest possible rotation.

Why does faster rotation shrink the outer event horizon? I would expect rotation to be a form of energy, which would increase the black hole's effective mass, and I would expect that larger mass to result in a larger effect horizon. Clearly this intuition is wrong, so what's a more accurate intuition?

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2 Answers 2

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First, let us note that “horizon radius” (at least when talking about black holes without spherical symmetry) is a coordinate dependent term, so it is better to use coordinate independent measure of size such as “event horizon area”. Then we can say that

… A rotating black hole of the same mass has a smaller area of event horizon.

Next,

Why does faster rotation shrink the outer event horizon?

It is more correct to say that spinning up the black hole increases its mass but does not lead to corresponding increase in horizon area.

I would expect rotation to be a form of energy, which would increase the black hole's effective mass, …

This is indeed correct.

… and I would expect that larger mass to result in a larger effect horizon.

And this is wrong, because this additional rotational energy is located outside the event horizon, and so does not lead to increase in horizon area.

A useful notion when discussing rotational energy of a black hole is irreducible mass, which can be seen as the mass of nonrotating (and uncharged) black hole with the same area of event horizon. For more discussion of irreducible mass see this answer, but here let us mention that in an idealized situation of reversible changes in black hole states it is possible to take a nonrotating black hole with mass $M_\text{irr} $ and “spin it up” by transferring to it angular momentum $J$ without increasing its irreducible mass, so that the full mass of this rotating black hole $$ M^{2}=M_\text{irr}^{2}+{\frac {J^{2}c^{2}}{4M_\text{irr}^{2}G^{2}}}.$$ Subsequently (again in an idealized situation) it would be possible to extract all the rotational energy and arrive back at a nonrotating black hole with the same $M_\text{irr} $.

Of course, losses would make perfectly reversible processes impossible, so at each stage of (classical) evolution the area of event horizon (and thus irreducible mass $M_\text{irr} $) would be increasing ($\delta M_\text{irr} > 0$) but this increase (at least in principle) could be made quite small.

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    $\begingroup$ can you suggest where can I read more about the comment "because this additional rotational energy is located outside the event horizon" please? $\endgroup$
    – S.G
    Commented Sep 10, 2023 at 12:50
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    $\begingroup$ @S_G As you may know, in GR there is no way to unambiguosly define gravitational energy density at a point. However, we can define quasilocal energy (QLE) to evaluate the energy contained within specific surface. For Kerr black hole this QLE evaluated just outside the event horizon does not depend on angular momentum and is simply $2M_\text{irr} $ which means that all the rotational energy is outside horizon. See for example here. $\endgroup$
    – A.V.S.
    Commented Sep 10, 2023 at 14:37
  • $\begingroup$ @A.V.S. You state that QLE evaluated just outside the horizon suggests that the rotational energy is “outside”, meaning farther away from the horizon. However, the paper you link presents the opposite trend: “The energy is a positive, monotonically decreasing function of the boundary surface radius.” - If I read this right, QLE evaluated farther away from the horizon is not larger, but smaller. Furthermore, the paper states that QLE matches ADM at infinity, but isn’t it true that ADM doesn’t depend on the angular momentum either? If so, then where is the rotational energy? $\endgroup$
    – safesphere
    Commented Sep 11, 2023 at 4:57
  • $\begingroup$ @A.V.S. Also, this statement appears confusing: "additional rotational energy is located outside the event horizon, and so does not lead to increase in horizon area." - When we observe a massive object $m$ falling to a (non-rotating) black hole $M$, in our coordinates this object always remains outside the horizon, but the area of the horizon increases practically to $4\pi \left( 2(M+m) \right)^2$ in a second. So energy being "outside" is not a justification for not affecting the area of the horizon. To have no contribution, this energy also must be far away. Is this logic correct? $\endgroup$
    – safesphere
    Commented Sep 11, 2023 at 5:23
  • $\begingroup$ @safesphere Re: monotonically decreasing, you must compare the QL energy for different values of angular momentum. And if we hold fixed $M_\text{irr} $ and vary angular momentum then ADM mass is a function of both those quantities. $\endgroup$
    – A.V.S.
    Commented Sep 11, 2023 at 6:29
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isaacg asked: "Why does faster rotation shrink the outer event horizon?"

It doesn't. In terms of the irreducible mass $\rm \mathcal{M}$ and the cartesian radius $\rm R=\sqrt{x^2+y^2+z^2}$

$$\rm \mathcal{M}= \sqrt{\frac{\sqrt{M^4-a^2 \ M^2}+M^2}{2}}$$

the equatorial horizon radius is always at $\rm R_{+}= 2 \mathcal{M}$, even if in terms of $\rm M$ and $\rm r$ it can be as low as $\rm r_{+}= M+\sqrt{M^2-a^2}=1 M$.

The horizon is not really spherical but ellipsoid though, so the polar radius is indeed smaller than the Schwarzschildradius even in terms of $\rm \mathcal{M}$ and $\rm R$, but due to the noneuclidean geometry the area is not reduced by that.

The area of the black hole $\rm A=8 \pi M r_{+}=16 \pi \mathcal{M}^2$ is only dependend on $\rm \mathcal{M}$, if you invest energy to spin the mass up the area in terms of square meters stays constant, see the plot below:

Kerr surface area

You could still ask why the input of rotational energy doesn't make the black hole larger, but it doesn't make it smaller either, at least not in terms of the area.

What happens is a deformation from spherical to ellipsoid, which gets obfuscated in pseudospherical coordinates where the horizons have constant $\rm r$, but the upside of these coordinates is that they make the metric a lot more elegant. You always have to keep in mind what they mean though.

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  • $\begingroup$ Thanks for the input; I have removed my answer. I have a question though: why using irreducible mass to calculate the radius is justified when the total mass is the sum of reducible and irreducible? $\endgroup$
    – S.G
    Commented Sep 10, 2023 at 9:30
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    $\begingroup$ Because you compared the area before and after the mass spun up, and since the mass was the irreducible mass before you spun it up and your rulers used that in the length scale Gℳ/c² you have to use the same ℳ before and after, otherwise you are comparing apples to bananas. $\endgroup$
    – Yukterez
    Commented Sep 10, 2023 at 9:41
  • $\begingroup$ @safesphere - Maybe you're confusing the cartesian radius with the proper radius, with a=0 we get R=r=C/2/π. The proper radius for stationary local observers is imaginary and therefore unphysical, but for free falling observers we get |gᵣᵣ|=1 in raindrop coordinates, so for them the proper radius even equals the coordinate radius (speaking of the Schwarzschild limit, Kerr is a little bit more complicated). The cartesian projection gives proper poloidial distance in the ZAMO frame, but the equatorial distance isn't proper then (with Kerr you have to choose one due to lacking symmetry) $\endgroup$
    – Yukterez
    Commented Sep 11, 2023 at 18:51
  • $\begingroup$ @safesphere - If I remember correctly, you already claimed something like that a few years ago, did you read this explanation about Schwarzschild's original coordinates in the meantime? $\endgroup$
    – Yukterez
    Commented Sep 11, 2023 at 19:02
  • $\begingroup$ @safesphere wrote: "Consider a particle falling in from a momentary rest at the horizon" - I did so here at Input 35, that is the E→0 particle. If it starts its free fall with v=0 in the limit of r→rs it has the longest proper time possible to reach the singularity, which is τ→πGM/c³. The v=0 at r→rs gets v→∞ inside the horizon by the way, but the faster the speed (relative to positions at fixed r) the longer the proper distance, since at r<rs with v>c you get Lorentz expansion, not contraction. $\endgroup$
    – Yukterez
    Commented Sep 12, 2023 at 6:16

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