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Let's say I have the ability to crunch a given quantity of particles into a black hole.

I now use a sizeable quantity of electrons to do this, with the restriction being that all of them have $+\frac{1}{2}$ spin.

Apart from the charge, does this intrinsic angular momentum exhibit itself in the black hole?

Would the resulting black hole resemble a Reissner–Nordström black hole or a Kerr-Newman black hole?

Would it be different from a black hole that I make out of equal amounts of $+\frac{1}{2}$ and $-\frac{1}{2}$ electrons? It should be, shouldn't it? (because the information about those spins should be conserved)

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  • $\begingroup$ I have to ask: why would intrinsic spin not contribute to the total a.m. of the black hole? Sorry, I am not well up enough in the field to see an obvious reason it would not, so if you judge it's too complex to explain to a newbie, that's OK. It's pure curiosity, is all. Thanks $\endgroup$ – user154420 Jun 25 '17 at 21:25
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    $\begingroup$ I've seen people mention that this "spin" has more to do with magnetic properties than anything else , and not to do with actual spinning, so that raises suspicion for me. (regarding whether it'd count) $\endgroup$ – Hritik Narayan Jun 25 '17 at 21:28
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    $\begingroup$ Well, was it not originally postulated to explain energy difference in the spectrum (in other words, a.m. seemed (and still seems) to explain it best, to Uhlenbeckand & Goldschmit even though I guess we both would agree it's not a settled issue, 100 years later. Also, an atom incorporates intrinsic spin in calculations of its total spin a.m. energy, (hope I'm right here:) Anyway, thanks for a fast reply and for giving me another way to think about it, in terms of magnetism. I do take your point that it could equally be thought of in the way you say. best of luck with it. $\endgroup$ – user154420 Jun 25 '17 at 21:51
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    $\begingroup$ @HritikNarayan: I've seen people mention that this "spin" has more to do with magnetic properties than anything else , and not to do with actual spinning It's a form of angular momentum that counts just the same as any other form of angular momentum. It's true that you can't generate spin 1/2 solely from orbital motion of particles, but that doesn't mean that it's not a form of angular momentum. $\endgroup$ – user4552 Jun 25 '17 at 23:51
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Spin angular momentum plays the same role in standard general relativity as any other form of angular momentum. (Nonstandard models of gravity can have torsion, in which case spin can interact differently with gravity. Experimental searches for gravitational torsion have given null results.) Angular momentum is conserved in GR in an asymptotically flat spacetime, so if a black hole forms by gravitational collapse, and the infalling matter has angular momentum, it makes absolutely no difference whether the angular momentum is spin or orbital angular momentum.

Actually there is no way in general to take a chunk of matter and determine how much of its spin is due to the intrinsic spins of its constituent fermions. For example, you can think of the spin 1/2 of a neutron as being partly due to the spins of its quarks and its virtual gluons, and partly due to the orbital angular momentum of the quarks and gluons.

But in any case, you can't form a black hole out of a polarized set of electrons. WP's article on the Kerr-Newman metric has a brief discussion of this:

An electron's a and Q (suitably specified in geometrized units) both exceed its mass M, in which case the metric has no event horizon and thus there can be no such thing as a black hole electron — only a naked spinning ring singularity.

This is referring to a single electron, but when you use multiple electrons, a stays the same while Q gets bigger, so the problem gets even worse.

But if a beam of polarized electrons makes up part of what went into a black hole, then it certainly does give a different black hole than if the beam had been unpolarized. Angular momentum is one of the characteristics of a black hole.

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  • $\begingroup$ "you can't form a black hole out of a polarized set of electrons" You're supposing that they are all moving in the same direction, right? $\endgroup$ – Mitchell Porter Jun 26 '17 at 2:42
  • $\begingroup$ @MitchellPorter: Good point. If they're in motion relative to one another at relativistic speeds (e.g., you're colliding them), then it requires a different analysis. $\endgroup$ – user4552 Jun 27 '17 at 1:22

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