# How can a black hole rotate if time dilation stops time at the event horizon?

How does a black hole rotate if time is dilated to infinity (e.g. stopped) at the event horizon? Note: this is relevant to this question, but different: How can a singularity in a black hole rotate if it's just a point?

Edit: I am considering this from an external reference frame (e.g. like what we would see from Earth, or perhaps even an object somewhat close to the black hole, but not very close to its event horizon).

• Time isn't 'dilated to infinity' at the event horizon. For example, a clock free-falling towards the horizon doesn't stop at the horizon but 'keeps on ticking' as it crosses the horizon and stops only when (if) its world line terminates on the singularity. Dec 22 '17 at 1:48
• @AlfredCentauri: the statement that time is dilated to infinity depends on the reference frame of the observer. The OP probably considers an external observer, while your statement refers to an infalling reference frame. From the perspective of the external observer the clock would indeed stop ticking when it reaches the event horizon. Dec 22 '17 at 4:15
• @AlfredCentauri: actually it does. One can use GR to compute that is happening to the clock from the perspective of the external observer. From a scientific perspective one could argue that the external observer should be regarded as the objective' observer. In that sense the clock stops. Dec 22 '17 at 4:40
• @flippiefanus, would you make the same argument that a Rindler observer is the objective observer and that clocks objectively stop at the Rindler horizon? Dec 22 '17 at 4:47
• @flippiefanus, it is relevant to this piece of, what I regard as, nonsense: "From a scientific perspective one could argue that the external observer should be regarded as the objective' observer." Dec 22 '17 at 4:55

The statement "time is dilated to infinity at the horizon" is a (very imprecise) way of saying that the event horizon is a null/lightlike surface. However, as is clear from light-rays, being null/lightlike is no impediment to moving. In particular it is possible for a null/lightlike surface to rotate. (Just as photons move `despite' that "time is dilated to infinity on a light-ray".

A spinning black hole is characterized by a mass M and an angular momentum L. No further parameters, hence the question on the angular speed is not applicable.

• Just a notation amendment, the angular momentum of a spinning black hole is usually indicated as J. Jan 10 '18 at 17:42
• Charge is another parameter, just for completeness.
– Jens
Jun 14 '21 at 11:02

Michele Grosso and mmeent have relevant points. The thing that must be remembered is the event horizon does not in any way causally influence the outside world. It really is the other way around. There is then angular momentum associated with a Killing vector $K_\phi$ for the entire spacetime. This carries the angular momentum information with it, or in a Noether theorem sense defines an isometry that defines angular momentum as an invariant. This is what is relevant.

If we think of there being a Gaussian surface around the black hole that also acts as a sort of cloak, then it matters not whether this is a black hole or a compact object of the same mass and angular momentum. The source of the external gravity field is irrelevant. So it matters not whether there is a compact star or a black hole behind the Gaussian surface/cloak. If there is a black hole the we can think of the horizon as a congruence of null rays that spiral as a sort of barber pole. However, the causal influence is from material that went into the black hole, not the horizon influencing the outside world.

Ultimately a black hole is a quantum object, and its angular momentum is similar to the intrinsic spin of an elementary particle. As a result the angular momentum is ultimately not derived from anything, which Michele makes a correct point on. A complete understanding of a black hole according to quantum theory would most likely have the angular momentum as an eigenstate similar to intrinsic spin.