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What do black holes spin relative to?
In other words, what is black hole spin measured in relation to?

Spinning black holes are different from non-spinning black holes. For instance, they have smaller event horizons. But what are the spins of black holes measured relative to?

Let's first look at an example with common objects.

Example

Let's say there's a disc on a table that rotates at 60 rpm. When you are standing still it spins at 60 rpm. But if you start running around it, it will move faster or slower relative to you. In this case, the disc has a ground speed, 60 rpm, because it has something to spin in relation to, in this case, the table.

Black Holes

Now, let's say that there is a spinning black hole. Because there is no control for the black hole to spin relative to, its spin must be relative to an object, for example, you. If you stand still, it spins at a constant rate. But if you start moving around the black hole in the same direction as the rotation, according to Newtonian physics, the black hole would spin at a slower rate relative to you. Since a faster spinning black hole has a smaller event horizon, in the first case, there would be a smaller event horizon.

Then how do scientists say that there are spinning and non-spinning black holes? Is that just in relation to Earth?

Ideas

First Idea
My first idea is also one that is more intuitive. When I move around the black hole, the black hole spins slower relative to me and consequently has a larger event horizon. In this idea, the black hole would just behave like a normal object. This would mean that if you went really fast around a black hole, you could get a lot closer to the black hole that if you were standing still.

This is kind of like a satellite that orbits Earth. The slower it moves, the easier it is to fall to the Earth. (I know this is a horrible analogy.)
Nothing special happens here.

Second Idea
My second idea is that when you move faster around the black hole, the relative rotational speed of the black hole doesn't change. Because of how fast it is/how dense it is and special relativity, moving around the black hole doesn't affect its speed.

This is like trying to accelerate past the speed of light.
No matter how much energy you spend, your speed barely changes.

I don't understand how this one would work. Why won't the rotational speed of the black hole stay the same?

Conclusion

What do black holes spin relative to? And what happens if you move around it? There are lots of questions that ask how black holes spin, or how fast they spin, but as far as I know, none of them address this question.

Deschele Schilder
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Joshiepillow
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6 Answers6

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But if you start running around it, it will move faster or slower relative to you. In this case, the disc has a ground speed, 60 rpm, because it has something to spin in relation to, in this case, the table.

Actually, this is fundamentally incorrect. The spinning of the disk has nothing to do with the table in principle. Acceleration, including spinning, is not relative. It can be measured without reference to any external object. For example, using a ring interferometer, or a gyroscope.

It does not matter if the object is a disk or a black hole or anything else, spinning is not relative like inertial motion is.

When I move around the black hole, the black hole spins slower relative to me, and consequently has a larger event horizon.

The event horizon is a global and invariant feature of the spacetime. Your motion does not change it. Of course, you can use whatever coordinates you like and make the coordinate size change as you wish. However, which events are on the event horizon is unchanged by your motion.

Dale
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  • They are relative right? For instance, someone on the Earth doesn’t feel that Earth is spinning. But someone outside of Earth does. – Joshiepillow Sep 02 '20 at 13:51
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    @Joshiepillow: On the contrary, everyone on Earth is fully aware that it is spinning. In fact, high-precision navigation systems rely on it to orient themselves to the spin axis. See What is MEMS Gyrocompassing? – Dave Tweed Sep 02 '20 at 15:04
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    @Joshiepillow the Earth's spins slowly enough that the force we feel from its rotation is pretty small for everyday purposes, but see hurricanes and Foucault's pendulum for examples where it becomes quite noticeable. – hobbs Sep 02 '20 at 15:27
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    @Joshiepillow no, this is not relative. Someone on earth does feel that they are spinning. Our inner ear is not sensitive enough to make us dizzy because of the earth’s spin, but the effects are present and measurable by sufficiently sensitive experiments. Even if the atmosphere were opaque so we couldn’t see the sun or the stars we would still be able to measure our rotation rate of one rotation per sidereal day. We could use ring interferometers, gyroscopes, Foucault pendulums, or other measurements. Only inertial motion is relative – Dale Sep 02 '20 at 16:25
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    Except that there is no way to tell what a nonrotating frame is without an experiment. Space does not come with a grid that says "this isn't rotating". And not all experiments agree (Lense–Thirring effect). But, on a large scale, Mach's principle holds: rotation is relative to distant matter. More local experiments then agree with that once a Lense–Thirring correction is applied. – John Doty Sep 02 '20 at 21:52
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    The relative nature of a black holes even horizon has nothing to do with your motion. It has to do with your position. The closer you are to the black hole the bigger it is. Why? Because we measure space with light and gravity red shifts light. Keep getting closer and eventually the black hole is to big for you to ever leave. It's really just another way to look at time dilation. – candied_orange Sep 02 '20 at 23:32
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This is just Newton's bucket in modern garb. The best explanation of this effect that I have seen is in Carlo Rovelli's book Quantum Gravity, which explains it as rotation with respect to the gravitational field. According to Einstein's Theory of General Relativity, the gravitational field is a real physical entity. And Rovelli says about Newton's bucket (on page 56 of the 2005 hardback edition):

Einstein's answer is simple and fulgurating:

The water rotates with respect to a local physical entity: the gravitational field.

Rovelli regards this as so important that he underlines it, as well as putting it in italics; but my formatting skills don't run to that. And yes, fulgurating is a real word.

TonyK
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  • Formatting to make the sentence underlined see: http://detexify.kirelabs.org/classify.html – Deschele Schilder Sep 02 '20 at 23:10
  • @descheleschilder: thank you, but I like my answer as it is. But just as a matter of interest: where in that link does it tell me how to underline text? – TonyK Sep 02 '20 at 23:26
  • Can't you see a square in which you can draw a symbol? if so, then the corresponding MathJax formatting should be visible on the right. Maybe I should check the link. It popped up at the right of this site yesterday. Very handy! – Deschele Schilder Sep 02 '20 at 23:47
  • @descheleschilder: But that doesn't help at all with underlining, does it? – TonyK Sep 03 '20 at 10:01
  • To make the answer mathematically more specific: the water rotates with respect to the metric field. (The metric field is non-zero everywhere, even in regions where "gravity" is absent. It defines which local frames are inertial. Presumably Rovelli is using "gravitational field" as a synonym for metric field.) – Chiral Anomaly Sep 03 '20 at 12:57
  • @ChiralAnomaly: a gravitational field has a metric (see e.g. here, here, or here), but they are not the same thing. However, the water does rotate with respect to the metric, as you say, as well as with respect to the gravitational field. – TonyK Sep 05 '20 at 10:19
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What do black holes spin relative to?

Relative to an inertial reference frame infinitely far from the hole, in which the hole has no translational motion.

And what happens if you move around it?

A spinning black hole is azimuthally symmetric. It “looks” the same from any azimuthal angle. Its spin parameter $a$ in the Kerr metric has nothing to do with how fast you move around it.

G. Smith
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Let's say there's a disc on a table that rotates at 60 rpm. When you are standing still it spins at 60 rpm. But if you start running around it, it will move faster or slower relative to you. In this case, the disc has a ground speed, 60 rpm, because it has something to spin in relation to, in this case, the table.

No, the table is not necessary to observe a difference between a spinning disc and a stationary disc. If you are rotating with a spinning disc, and are using a reference frame in which the disc is stationary, that reference frame is not inertial. There will appear to be a "centrifugal force" pushing you away from the spinning disc, and to keep next to it, you will have to have a force pushing you towards the disc. So you can tell the difference between a rotating disc and a stationary disc because you can be in a co-moving reference frame with a stationary disc without a centrifugal force appearing.

Now, there is a phenomenon called frame dragging in which a rotating black hole will distort space-time around it. Near the black hole, this will reduce the apparent rotation. But far away from the black hole, frame dragging become negligible, and the black hole's rotation can be measured with respect to inertial reference frames.

The other answers saying that there is no need for there to be anything else to measure it with respect to are somewhat wrong, as frame dragging is dampened by the mass of the rest of the universe. If everything in the universe other than the black hole were to disappear, it would be impossible to observe the black hole as rotating.

Acccumulation
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You may think of this as an application of Mach's Principle. This represents an observed fact of physics and cosmology, not derivable from some other principle. The local, non-rotating frame appears to be determined by matter, mostly distant matter. General Relativity, partly inspired by this idea, covers the (small) effect of local matter, but does not demand Mach's Principle for the whole Universe. It's a proposition that has been tested to high precision.

John Doty
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    Calling Mach's Principle "an observed fact" is a bit of a stretch! Please see https://physics.stackexchange.com/questions/5483/is-machs-principle-wrong – D. Halsey Sep 02 '20 at 20:11
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    @D.Halsey Foucault's pendulum is a fine test of Mach's Principle. – John Doty Sep 02 '20 at 20:23
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    One should be careful to define what one means by Mach's principle. If the whole universe were set into rotation then GR says we could tell; Mach's principle says we could not. They can't both be right. – Andrew Steane Sep 02 '20 at 23:06
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    @AndrewSteane But Mach's principle works. It captures a key property of the universe. GR allows absolute rotation, but does not demand it, so it is hardly incompatible. They can both be right. – John Doty Sep 02 '20 at 23:17
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    @JohnDoty There is no universal agreement on what "Mach's principle" means in GR. Depending on how you formulate it, it might be compatible or incompatible with GR. – user76284 Sep 02 '20 at 23:36
  • @user76284 But experimentally we see that the frame defined by distant matter matches local measurements of rotation once the Lense–Thirring effect is accounted for. What theorists think it means in GR doesn't impact that empirical observation. In physics, unlike mathematics, experiment and observation have the final say. – John Doty Sep 03 '20 at 00:04
  • @JohnDoty The Lense–Thirring effect does not mean Mach's principle is true. You are misunderstanding this topic. There is no experimental confirmation of Mach's principle in its strong form. Please read this answer and the comments underneath it. – user76284 Sep 03 '20 at 00:18
  • @user76284 I never said the Lense-Thirring effect shows that Mach's principle is true. The Lense-Thirring effect is a perturbation that must be accounted for in the most precise experimental tests of Mach's principle, that's all. But you cannot test Mach's principle against GR, which is what you seem to want to do. Mach's principle is an empirical fact, not a prediction of GR. – John Doty Sep 03 '20 at 00:31
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    @JohnDoty Some versions of Mach's principle make predictions that have not been empirically observed. Some of them also contradict GR. See sections 1.3 and 1.4 of this paper. See also this paper for a survey of different versions of Mach's principle, some of which are inconsistent with each other. – user76284 Sep 03 '20 at 04:44
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    @JohnDoty With that in mind, I recommend editing the answer to state clearly that this is one version of what people call “Mach’s principle”, and to change the wording to match exactly that of Mach0 in the paper I linked to (in particular, removing the phrase “determined by”, since that makes predictions about counterfactual situations that we haven’t observed). Then you’d be talking about the purely empirical observation. – user76284 Sep 03 '20 at 04:45
  • @user76284 I believe what I said is entirely measured. I call Mach's Principle an observed fact. I said "appears to be determined by". As far as I know, no observation has ever contradicted Mach's Principle, which is all that matters for physics. That theorists cannot agree on how to understand it is irrelevant without experimental evidence. You're talking metaphysics, not physics. – John Doty Sep 03 '20 at 15:25
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    @JohnDoty You seem to have trouble understanding that “Mach’s principle” does not have a single meaning. Are you being deliberately obtuse? You also seem to be confused about the meaning of “metaphysics”. – user76284 Sep 03 '20 at 18:42
  • @user76284 What experiment has been performed to discriminate between the different meanings of Mach's Principle? Without such, you're making a "distinction without a difference" as far as physics is concerned. – John Doty Sep 03 '20 at 19:45
  • @JohnDoty Last sentence on page 10 of the second paper I linked to. – user76284 Sep 03 '20 at 20:39
  • Let us continue this discussion in chat. – John Doty Sep 03 '20 at 20:53
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    It would help enormously if you could state clearly what principle it is that you are here naming "Mach's Principle". If it is just a single empirical observation (such as "my car is red" or "the universe is not rotating") then it is not really a principle but a datum. – Andrew Steane Sep 03 '20 at 22:58
  • @AndrewSteane It's not merely a datum, since those who need to precisely define a non-rotating frame employ it as a practical tool: https://www.cosmos.esa.int/web/gaia/gaiadr2_rf#:~:text=The%20frame%20is%20constructed%20from,extragalactic%20sources%20known%20to%20date.&text=The%20Gaia%20Celestial%20Reference%20Frame%20(Gaia%2DCRF2)%20is%20realised,accurate%20positions%20of%20556%2C869%20quasars. – John Doty Sep 04 '20 at 00:00
  • And the location of Polaris (the pole star) is a convenient way to estimate the direction of north. I acknowledge the rotation example is more significant but it is not, according to our best theory of spacetime, a required result. – Andrew Steane Sep 04 '20 at 08:26
  • @AndrewSteane This is physics, not mathematics. Experimental and observational evidence must prevail over theory or it ceases to be a science. – John Doty Sep 04 '20 at 11:09
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    No one is denying any experimental observation; it is the usage and meaning of the word "Principle". There is no "Car Principle" saying that my car is red, even though empirical observation reveals it to be the case. As I understand it, Mach wanted to argue that local inertial frames must of necessity have no relative acceleration with the distant universe, on the grounds that the term "inertial" could only be defined that way. In this he was mistaken. – Andrew Steane Sep 04 '20 at 11:17
  • Let us continue this discussion in chat. – John Doty Sep 04 '20 at 11:18
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Let's say there's a disc on a table that rotates at 60 rpm. When you are standing >still it spins at 60 rpm. But if you start running around it, it will move faster >or slower relative to you. In this case, the disc has a ground speed, 60 rpm, >because it has something to spin in relation to, in this case, the table.

The spinning black hole is a solution of the vacuum Einstein equation that describes the spacetime interval $ds^{2}$ which is an intrinsic property of the spacetime itself on which all the observers agree. In other words, what we call a spinning black hole is the spacetime itself, so the example described by the rotating disk on the table is not a good analogy.

Now, let's say that there is a spinning black hole. Because there is no control >for the black hole to spin relative to, its spin must be relative to an object, >for example, you. If you stand still, it spins at a constant rate. But if you >start moving around the black hole in the same direction as the rotation, >according to Newtonian physics, the black hole would spin at a slower rate >relative to you. Since a faster spinning black hole has a smaller event horizon, >in the first case, there would be a smaller event horizon.

Then how do scientists say that there are spinning and non-spinning black holes? >Is that just in relation to Earth?

The meaning of the spin is that any observer outside the black hole and sufficiently close (i.e. in the ergosphere) can not stand still. This is called the frame-dragging. The "least rotating observer" is a locally non rotating observer whose angular velocity, defined by an inertial observer at infinity is $$\Omega = \frac{d\phi}{dt} = -\frac{g_{t\phi}}{g_{\phi\phi}}$$ When we take the limit for $r \rightarrow r_{+}$ we have $\Omega \rightarrow \Omega_{Horizon}$.

What do black holes spin relative to? And what happens if you move around it? >There are lots of questions that ask how black holes spin, or how fast they spin, >but as far as I know, none of them address this question.

So the angular velocity of the horizon is the angular velocity of a local non-rotating observer (at the horizon) measured by an inertial observer at infinity. For a detailed discussion, you can find more material on "General Relativity" by R.M. Wald.

Pipe
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