Saturday, 3 August 2019

general relativity - What is black hole spin?


First of, congrats to the people at LIGO.


In this article, the BBC notes that the latest LIGO results show that a new black hole was formed with a spin of $0.2$ (dimensionless number).


What exactly is this number?


Is this simply the ratio of the angular momentum that the blackhole is observed to have as a ratio of the maximal angular momentum as limited by some physics (Kerr Metric?)?



Answer



Yes, the dimensionless spin such as $0.2$ in this case is simply the ratio $$ a= 0.2 = \frac{|\vec J_{\rm measured}|}{|\vec J_{\rm max}(M)|}$$ where the denominator is the maximum angular momentum allowed for the same value of the mass (as the measured mass). For the $d=4$ Kerr black hole, the maximum (the angular momentum of the extremal Kerr black hole) is $$ |\vec J_{\rm max}(M)| = \frac{GM^2}{c} $$ Note that at least one of the initial December 26th black holes had $a\geq 0.2$ while the final black hole's spin is estimated to be $a\approx 0.7$. That large value is coming mostly from the orbital angular momentum, the relative motion of the initial black holes.


Black holes with $a\gt 1$ are "overextremal" and they are prohibited because for this excessive value of the angular momentum (it's similar for too high electric charges), the black hole solution has no horizon at all, so it is not really a black hole (because the presence of the event horizon is what defines the black hole). Moreover, such a "naked singularity" is almost certainly prohibited in any complete theory of (quantum) gravity.


The limiting case $a=1$ of the extremal black holes is possible and interesting. It has various special properties, e.g. it radiates no Hawking radiation.


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