Using a bit of classical reasoning I'm imagining black hole formation to be much like an ice skater pulling in her arms:

skater pulls in her arms to increase angular velocity

Now, the size difference between a star and its black hole can't even be effectively captured in an image. The black hole for our sun would be much less than a pixel in this image:

size comparison of our sun and planets

That suggests to me that even a very slowly rotating stars would have much more angular momentum than could be supported by their resulting black holes. I haven't done the calculation because I don't really understand the Kerr metric but even with a bunch of classical hand-waving I'd think that just about every black hole formed in a stellar-collapse would be spinning maximally.

So my question is, do we expect nearly all black holes to be spinning maximally? If so, (roughly) how much angular momentum is lost because the star had much more than the black hole could support? And, how is all of this extra angular momentum shed during collapse? Is it just in the form of tons of matter being ejected until the the angular momentum is low enough to allow for the formation of a Kerr black hole?

  • $\begingroup$ Related: physics.stackexchange.com/questions/34945/… $\endgroup$ – Brandon Enright Dec 23 '13 at 22:51
  • $\begingroup$ Quite a bit of the star's mass does not end up in the black hole and a lot of angular momentum can be carried away that way. No idea if that matters or not. $\endgroup$ – dmckee --- ex-moderator kitten Dec 23 '13 at 22:58
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    $\begingroup$ For an order of magnitude, if the Earth had uniform density but the same mass, radius, and rotation rate, it would have I think $\sim900$ times too much angular momentum to collapse to a black hole. As for actual black holes, I've asked this informally of AGN specialists, and I've never gotten anything but noncommittal answers. $\endgroup$ – user10851 Dec 24 '13 at 2:56
  • $\begingroup$ Hi @bobie, thanks for the edits and the question is still open. Of course I welcome another answer. $\endgroup$ – Brandon Enright Oct 16 '14 at 16:02

High angular momentum presents a barrier preventing collapse to a black hole (at least until this angular momentum is radiated away).

The parameter on which the formation of black hole depends is the ratio $q$ of angular momentum ($J$) to the square of mass ($M$). If $q=J/M^2 < 1$ (in relativistic units with $G=1$, $c=1$), then the black hole (non-extremal Kerr black hole, to be precise) can be formed. If $q>1$ then the black hole cannot be formed from all the matter without some mechanism for losing angular momentum (this means of course that some of the mass also has to be lost in the process).

For example, currently for the Sun this parameter is slightly more than 1. (Of course solar mass is too small to ever form a black hole, angular momentum or not).

If we are talking stellar collapse (see for instance [1]), during the late time evolution of the star it loses considerable portion of its outer shell. Since the outer layers carry most of angular momentum, it is quite possible than as the result of this process the rotation of the actual collapsing object would be slow enough to form the black hole outright.

Alternatively, if the rotation speed of the collapsing star is high enough, during the collapse the part of an angular momentum is retained in the accretion disk which is formed around the newly created black hole. The matter from such disk could then fall inside, potentially increasing the ratio $q$, but still it will never reach the limiting value of 1.

Note, that accretion disk (depending on the actual configuration) also can produce the opposite effect Blandford–Znajek process can slow the rotation of a black hole by extracting rotational energy.

Another possibility is that if the rotation is fast enough the black hole is never formed: for instance numerical simulations in [2] found that for $q>1$ the collapsing neutron star forms a torus which then fragments into nonaxisymmetric clumps (without ever producing horizon).

[1] Heger, A., Fryer, C. L., Woosley, S. E., Langer, N., & Hartmann, D. H. (2003). How massive single stars end their life. The Astrophysical Journal, 591(1), 288. arXiv:astro-ph/0212469.

[2] Duez, M. D., Shapiro, S. L., & Yo, H. J. (2004). Relativistic hydrodynamic evolutions with black hole excision. Physical Review D, 69(10), 104016. arXiv:gr-qc/040107.

  • $\begingroup$ Nice comment, but for emphasis, let's notice that this is an argument in favor of the answer "yes, black holes will tend to have maximum angular momentum". Because that maximum angular momentum ("extremal Kerr black hole") is J=M^2, hence q=1. So under black hole formation a star may generally have more than this critical angular momentum and black hole formation sets in right at the critical value. $\endgroup$ – Urs Schreiber Dec 25 '13 at 11:43

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