Can the angular velocity of a black hole be measured? I found this article explaining how to measure the spin of a black hole can be measured, though I'm unclear whether 'spin' means angular velocity or momentum. In the case of a charged black hole, could you derive the angular velocity from the electromagnetic field?

If angular velocity can be measured, assuming angular momentum can be measured (via the radius of the ergosphere?), couldn't you derive an effective radius of the singularity from that?

  • $\begingroup$ I'm curious, for no practical reason. Everything I've read talks about singularities having zero dimensions, even if its a ring (what is a zero dimensional ring?), but if we can reasonably come up with numbers for angular velocity and angular momentum, than we should be able to calculate an effective radius, yes? $\endgroup$ Commented Nov 26, 2020 at 3:40
  • $\begingroup$ " I've read talks about singularities having zero dimensions" the singularities in GR are little more pathological then infinities in other theories. The singularity cannot be made part of the spacetime at all. It is not ring, it has no dimensions, it is simply place where spacetime ends and we have no idea what (or if anything) is there. $\endgroup$
    – Umaxo
    Commented Nov 26, 2020 at 9:41

1 Answer 1


The angular velocity of a black hole is not a very well defined concept.

The best we can do is define an angular velocity to the horizon. This essential the rate at which objects get frame dragged along the horizon, and an outside observer should be able to measure it by observing objects falling into the black hole.

However, there is no compelling reason to assign this angular velocity to other parts of the black hole (such as the singularity).

That being said, we can ignore all this and see what happens.

The horizon angular velocity of a Kerr black hole is given by $$ \omega_H = \frac{a}{2Mr_{+}} $$ where $a = L/M$ the "spin" or more precisely the the angular momentum per unit mass of the black hole, and $r_{+}$ is the radius of the outer horizon.

Naively, the moment of inertia $I$ of a ring with mass $M$ is

$$ I = M R_{ring}^2$$

Combining these through $L = \omega I$ gives

$$R_{ring} = \sqrt{2Mr_+}$$

which leads to the dubious conclusion that $R_{ring} > r_{+}$.

Note that from the analytic extension of the Kerr metric, one can calculate the radius of the ring singularity, which is simply given by $a/c$.

  • $\begingroup$ how is the angular momentum of a black hole defined in your equation for $a$? $\endgroup$
    – Umaxo
    Commented Nov 26, 2020 at 9:30
  • $\begingroup$ @Umaxo, through either 1) the ADM angular momentum, 2) the Komar integral associated with the axial killing vector, or 3) the Bondi angular momentum. (For a Kerr black hole these all give the same answer.) The Bondi angular momentum is the one most relevant to a distant observer. $\endgroup$
    – TimRias
    Commented Nov 26, 2020 at 9:42

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