I have read this:


In General Relativity the black hole solutions which have so far been found form a four parameter family called the generalized Kerr-Newman family of black holes. The four parameters are mass M, angular momentum J, charge Q, and the cosmological constant Λ [6].

A black hole is a region of spacetime exhibiting gravitational acceleration so strong, that nothing, not even light can escape from it.


There is no consensus whether the mass of the black hole lies in the singularity or the event horizon. For an outside observer, the mass of the black hole may lie on the event horizon.

How can a singularity in a black hole rotate if it's just a point?

The singultarity of a rotating Kerr black hole is not a point, but a ring in the plane of rotation, though still zero volume.

Some physicists even believe that singularities are not real.

Though, none of the answers on this site explain what is exactly rotating in the black hole. Is it the singularity, like a point particle (with the mass of a black hole), then it is like the spin of an elementary particle.

Or is it the event horizon, that is rotating, since to an outside observer, all the infalling mass is frozen on the horizon.

Some answers say that the gravitational field (potential) of the black hole (its stress energy) is determined by the (though the stress energy is determined by mass, angular momentum, charge, cosmological constant) event horizon (its size).

Some answers say that a singularity of a black hole is not made of any matter, since it is not a place in space, rather a place in time.

Like Luboš Motl's comment to this question:

What happens when a black hole and an "anti-black-hole" collide?

So there is a contradiction, because none of the answers specifically state what is actually rotating in a rotating Kerr black hole.


  1. What is actually rotating in a rotating Kerr black hole, is it the singularity or the event horizon?
  • 1
    $\begingroup$ Who says that there needs to be a physical "thing" that "rotates" in order for a system to have angular momentum? It isn't the case in electromagnetism - why would it need to happen in general relativity? $\endgroup$ Commented Sep 21, 2019 at 11:08
  • 3
    $\begingroup$ @EmilioPisanty Say, I take a loop wire with a current producing a magnetic field. Do electrons rotate there in a circle? Now I take the same electrons and make them rotate around a positive charge on the same radius producing the same magnetic field. Do they still rotate? But now I can use the Schrodinger equation to solve the electrons rotation around the charge and then... they no longer rotate? Rotation is a classical concept. Its quantum equivalent is spin. So spin is the quantum representation of rotation. Besides, this question is purely classical, so your comment is moot either way :) $\endgroup$
    – safesphere
    Commented Sep 27, 2019 at 7:41

1 Answer 1


So let us first ask ourselves: What makes us believe that anything is rotating in the Kerr space-time?

The answer is that we go very far away from the black hole and look at the asymptotics of the gravitational field as it becomes weaker and weaker. From there we can read off in a well defined way a set of mass and current multipoles "contained in the space-time". The lowest-order multipoles can be defined almost unambiguously and they would be interpreted as "total linear momentum in the space-time" and "total angular momentum in the space-time". Then what you read of for the Kerr metric is an angular momentum of the magnitude $J =Ma$.

What makes us assign such meaning to these quantities is matching in the limit when there are only small amounts of dilute matter (which can be treated by linearized gravity). In that limit these asymptotic quantities exactly match the meaning of total linear and angular momentum of the dilute matter as defined in your classical-mechanics course. So a similar asymptotic field as for the Kerr metric would be generated by a cloud of dilute matter with a total mass $M$ and rotating with a total angular momentum $Ma$. So that is what makes you believe that there is something rotating in Kerr space-time.

But as larger amounts of dense matter are added, you cannot just sum all the momenta of the matter particles in a space-time to get the total linear and/or angular momentum. The problem is that the space-time is curved and you stop being able to easily do things such as sum two vectors at different points of the space-time. But another issue is that no matter how hard you try to sum the contributions of momentum in the matter, you keep missing something, it just does not add up to the asymptotic momenta. It turns out that somehow the gravitational field itself carries momentum. But pinning down how and where this energy and momentum of the gravitational field is notoriously difficult. There simply is not a universally valid formula for that, the energy and angular momentum of the gravitational field is stored non-locally.

So what is, really, rotating in Kerr space-time? The curvature singularity in the center of the space-time is a ring, and it would be very easy to say it is a rotating ring carrying the entire angular momentum $Ma$. But consider a very compact object very close to a black hole such as a neutron star. There you cannot assign the angular momentum only to the matter, an increasingly larger portion is stored nonlocally in the gravitational field, the space-time is rotating as well. On the other hand, the gravitational field (the space-time) would never rotate on its own, it will only rotate if the neutron star is rotating as well. This is true for any situation with matter - the field and its source rotate in tandem, each having a significant contribution to the angular momentum.

So what about black holes? There is really no rigorous argument to extend the observations of the last paragraph to the black hole, though. But consider this: in a rather precise sense, the nonsingular part of the black hole field "causes" the existence of the curvature singularity inside. You can slice space-time in a particular way, put one of the slices not including the black hole singularity in the computer and let it evolve the slice into the future according to GR. To your surprise, it will spontaneously evolve to the singularity (in the case of the Kerr space-time it would actually evolve a crumpled singularity called the Cauchy horizon). So in this sense it is the field that causes the singularity to exist rather than the other way around. So I am inclined to say that black holes can be seen as the limit where the fraction of mass and angular momentum in the gravitational field has actually converged to the entire mass and angular content of the space-time.

  • $\begingroup$ If I understand it correctly, your conclusion is that what rotates is the spacetime around the horizon. This is also evident by the frame dragging effect +1. $\endgroup$
    – safesphere
    Commented Sep 27, 2019 at 7:26

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