# Spin (or not) of a Kerr Black Hole

• I
• gerald V

#### gerald V

The Kerr solution describes the gravitational field of a rotating black hole. Oftenly, the hole is said to be „spinning“, what appears as misleading to me. My questions:

1.) Is it correct to say that angular momentum in this way is treated like orbital angular momentum, not like spin?

2.) Can a black hole have spin like an elementary particle (say, a massive gauge Boson or a Fermion, or even an entire atomic nucleus)? By which equations would this have to be described?

3.) If an elementary particle with spin falls radially into a Schwarzschild black hole, what is the result? Will a Kerr hole be generated? Or will this hole have spin in the sense the infalling elementary particle had?

Thank you very much in advance

The Kerr solution describes the gravitational field of a rotating black hole. Oftenly, the hole is said to be „spinning“, what appears as misleading to me. My questions:

1.) Is it correct to say that angular momentum in this way is treated like orbital angular momentum, not like spin?
GR is a classical field theory, so it's definitely classical angular momentum.
2.) Can a black hole have spin like an elementary particle (say, a massive gauge Boson or a Fermion, or even an entire atomic nucleus)? By which equations would this have to be described?
It's not quantum spin. A black hole is not a (quantised) elementary particle.
3.) If an elementary particle with spin falls radially into a Schwarzschild black hole, what is the result? Will a Kerr hole be generated? Or will this hole have spin in the sense the infalling elementary particle had?

Thank you very much in advance
All black holes must have a small angular momentum - but in some cases it's negligible. An elementary particle falling into a black hole would have negligible angular momentum. And, not just spin, but from its trajectory into the black hole, which is unlikely to be perfectly radial.

vanhees71
1.) Is it correct to say that angular momentum in this way is treated like orbital angular momentum, not like spin?
Strictly speaking, it's neither. It's a global property of the spacetime geometry. See further comments below.

If you are viewing the hole from far away, and don't have to worry about its internal structure or things falling into it, its angular momentum will act like a classical spin (the same as the spin of the Earth or any other astronomical body). But of course not all scenarios can be handled using that approximation.

2.) Can a black hole have spin like an elementary particle (say, a massive gauge Boson or a Fermion, or even an entire atomic nucleus)?
Black holes are classical objects, not quantum objects.

3.) If an elementary particle with spin falls radially into a Schwarzschild black hole, what is the result?
Black holes are classical objects, not quantum objects. So your thought experiment here would be better formulated as a classical object with spin (such as a spinning ball) falling into a black hole.

Assuming that re-formulation has been done, angular momentum in GR has to be viewed, as I noted above, as a global property of the spacetime geometry. So it's not really possible to have a spacetime with nonzero angular momentum in it that contains a Schwarzschild black hole. At the very least, one has to be very careful about how the scenario is constructed: you would need a spacetime with multiple regions having different geometries and you would need to make sure things match at the boundaries between the different regions. It's not going to be anything simple like "Schwarzschild hole turns into Kerr hole"--there is no way to connect those two geometries and have conditions match at a boundary.

vanhees71
3.) If an elementary particle with spin falls radially into a Schwarzschild black hole, what is the result? Will a Kerr hole be generated? Or will this hole have spin in the sense the infalling elementary particle had?
Remember that both Kerr and Schwarzschild black holes are eternal, unchanging solutions. So if a non-rotating hole gains angular momentum by some process it was never a Schwarzschild nor Kerr black hole, strictly speaking. That said, we would expect the hole to look (externally) very like a Schwarzschild black hole in the past and very like a Kerr black hole in the future, with a messy bit in the middle with gravitational waves and whatnot as the mass and angular momentum is absorbed. Not sure what the interior would look like.

We don't have a quantum theory of gravity, so you have to hedge a bit if you're talking about absorbing quantum objects. For stellar mass black holes a single electron will make no measurable difference to anything. Presumably the hole would spin up to conserve angular momentum, but it's such a tiny change how would you ever know? For holes that are small enough that a single electron would matter you'd need a quantum theory of gravity - but we'd expect it to evaporate almost instantly, so I don't think there'd be much opportunity to study it.

vanhees71
It's not going to be anything simple like "Schwarzschild hole turns into Kerr hole"--there is no way to connect those two geometries and have conditions match at a boundary.
Perhaps a related question would be, suppose that we started with a central spherical ball of dust, sufficiently large to collapse, and an outer spherical shell of dust, with a large spherical shell of vacuum between them. Suppose further that the outer shell is initially rotating uniformly about some axis with some given total angular momentum. How does the spacetime evolve?

Even though the central ball has a well known analytical solution, and even though there are rotating black hole solutions, I suspect that this combination would require a numerical solution.

vanhees71
Suppose further that the outer shell is initially rotating uniformly about some axis with some given total angular momentum.
If it is, then I don't think it's possible for the inner shell to be spherically symmetric. Indeed, I don't think it's possible for the spacetime geometry in general inside a rotating shell to be spherically symmetric. The shell theorem only applies to a non-rotating, spherically symmetric shell.

Dale and vanhees71
If it is, then I don't think it's possible for the inner shell to be spherically symmetric. Indeed, I don't think it's possible for the spacetime geometry in general inside a rotating shell to be spherically symmetric.
You can, for example, match an exterior Kerr solution onto an interior (spherically symmetric) Minkowski solution, with the constraint equations at the boundary being enforced by a thin, slowly-rotating shell (see e.g. E. Poisson's treatment of the thin-shell formalism). But I don't know about Schwarzschild?

Dale and vanhees71
You can, for example, match an exterior Kerr solution onto an interior (spherically symmetric) Minkowski solution, with the constraint equations at the boundary being enforced by a thin, slowly-rotating shell (see e.g. E. Poisson's treatment of the thin-shell formalism).
Do you have a reference for the Poisson treatment? I dimly remember seeing something like this, but it's been quite some time.

It's chapter 3.10 in his textbook "A relativist's toolkit"/"An advanced course in GR". The stuff on junction conditions is the most difficult section of the whole book, IMO.

vanhees71
It's chapter 3.10 in his textbook "A relativist's toolkit"/"An advanced course in GR".
Ok, got it. But, as you noted, this solution is only valid for a slowly rotating shell, where "slowly rotating" ends up meaning "rotating slowly enough that we can ignore the centrifugal force on the shell due to the rotation, which would make the shell non-spherical". In other words, the matching to the Minkowski geometry inside only works because the shell itself is being approximated as spherical, which means the shell theorem is taken to be approximately valid for this case.

Also note that the "Minkowski" geometry inside the shell is "rotating" with respect to infinity, just as the shell itself is. In other words, observers inside the shell who are inertial will find that they are not at rest with respect to observers at infinity; they are slowly rotating with the shell (because of the frame dragging of the shell). This effect I would expect to still be there for a shell that was not slowly rotating, but I would not necessarily expect the geometry inside it to be Minkowski.

ergospherical