17
$\begingroup$

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?)?

$\endgroup$
21
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Where does $\vec J_{\rm max}$ come from? Are we assuming that Naked Singularities are disallowed (ie $GM^{2} > J$)? $\endgroup$ – Aron Jun 16 '16 at 5:18
  • 2
    $\begingroup$ Yes, exactly, the statement that $GM^2/c\lt J$ is forbidden is exactly equivalent to the statement that the angular momentum $a=1$ i.e. $J=GM^2/c$ is the maximal one. ;-) The extrapolation of the solution to $a\gt 1$ has the property that there are no horizons, see en.wikipedia.org/wiki/Kerr_metric#Overextreme_Kerr_solutions for details $\endgroup$ – Luboš Motl Jun 16 '16 at 5:21
10
$\begingroup$

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?)?

Yes, exactly. For a spinning black hole there are two event horizons, an inner and an outer horizon. The positions of the horizons are given by:

$$ r = \tfrac{1}{2}\left(r_s \pm \sqrt{r_s^2 - 4\left(\frac{J}{Mc}\right)^2}\right) \tag{1} $$

where $J$ is the angular momentum and $r_s = 2GM/c^2$ is the Schwarzschild radius. If the black hole isn't spinning then $J=0$ and equation (1) gives the horizon positions as $r=0$ and $r=r_s$ i.e. just a Schwarzschild black hole. As we increase the angular momentum the inner horizon moves outwards and the outer horizon moves inwards, and when:

$$ r_s^2 = 4\left(\frac{J}{Mc}\right)^2 \tag{2} $$

the two horizons meet and disappear to leave a naked singularity. This is then called an extremal black hole, and a simple rearrangement of (2) gives the extremal angular momentum as:

$$ J_\text{ex} = \frac{Mc}{2}r_s = \frac{GM^2}{c} $$

The value of $0.2$ means $0.2J_\text{ex}$.

I was going to ramble on about the physical significance of this, but Luboš has beaten me to it. See his answer for the details.

$\endgroup$
  • $\begingroup$ Please see my comment on the OP. From the paper "We find that at least one of the component black holes has spin greater than 0.2." This is not the final value. $\endgroup$ – m4r35n357 Jun 16 '16 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.