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Posting this as I have so far not been able to find a straightforward answer to the following question. The formula for the outer event horizon of a kerr black hole is given by the following equation:

$$ r_+ = \frac{GM}{c^2}\left(1+\sqrt{1-\frac{J^2c^2}{M^4G^2}}\right) $$

Where $J$ is the angular momentum of the black hole. My question is this: does the mass $M$ in this equation correspond to the irreducible mass of the black hole? Or does it correspond to the total mass equivalent of the black hole when the rotational mass-energy is included?

Mathematically speaking, does $M = M_{irr}$? Or does $M = \sqrt{M_{irr}^2 + \frac{J^2c^2}{4M_{irr}^2G^2}}$?

Edit: I should also mention that, despite my best efforts, I've yet to come across a textbook that answers this explicitly, with the distinction seemingly being taken for granted. Even when talking to others informally about this, there does not seem to be a consensus (there are already answers/comments in this thread stating opposing viewpoints). So if anyone has an answer, a source would be greatly appreciated if possible, thanks!

Scott
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From ${M}=\sqrt{M_{i r r}^{2}+\frac{J^{2}c^{2}}{4M_{i r r}^{2}G^{2}}}$ you can see, that the mass of the Kerr black hole increases, the bigger $J$ is. If the Kerr black hole is spun down to $J=0$, it is left with its irreducible mass and corresponds to a Schwarzschild black hole. Now, since the curvature of the spacetime depends on the total mass of the Kerr black hole, it is mandatory to use ${M}=\sqrt{M_{i r r}^{2}+\frac{J^{2}c^{2}}{4M_{i r r}^{2}G^{2}}}$ in the formula of $r_{+}=\frac{G M}{c^{2}}\left(1+\sqrt{1-\frac{J^{2}c^{2}}{M^{4}G^{2}}}\right)$, since $M_{irr}$ is the mass of the Kerr black hole only in the case of $J=0$.

Daniel
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For the equation, you provided, it is in fact the real Mass of the Black Hole.

You can derive it yourself by setting the $g_{tt}$ component of the Kerr-Metric to 0 and solving for the radius.

Stealing from Wikipedia, you have $g_{tt} = 1 - \frac{\frac{2GM}{c^2} r}{r^2+\frac{J^2}{M^2c^2}}$. By setting this to zero, you can get your outer and inner horizon.

Ken
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This has bothered me for a while. I have to look at my various GR books but I looked it up first in Wald’s GR book and he does have it as Mirr. Meaning it’s rest mass. Another point that indicates that’s right is that he also shows the BH entropy which comes from the area, which comes from r with M specifically labeled the irreducible mass in the Penrose process analysis.

It is pseudo confirmed by his analysis of the Penrose process where energy is extracted from the BH, and specifically from J. We know that the entropy can’t be reduced (from thermodynamics which applies) and he shows the entropy after the process with the equation with Mirr. Actually if not entropy decreases which is not allowed. In that process the max extraction allowed is by making J = 0 (btw it takes multiple, infinite times of the Penrose process to get there, that’s irrelevant here).

I still will check my other GR books, as I’ve had the same concerns as the author of this question. I even argued with the AI copilot and convinced him it was not the Mirr, but then later saw Wald.

PS/sorry about the lack of explicit use of underscripts. Hate MathJax

Bob Bee
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