Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
A surface on which light can orbit a black hole is called a photon sphere. The Kerr solution has infinitely many photon spheres, lying between an inner one and an outer one. In the nonrotating, Schwarzschild solution, with a=0, the inner and outer photon spheres degenerate, so that all the photons sphere occur at the same radius. The greater the spin of the black hole is, the farther from each other the inner and outer photon spheres move. A beam of light travelling in a direction opposite to the spin of the black hole will circularly orbit the hole at the outer photon sphere. A beam of light travelling in the same direction as the black hole's spin will circularly orbit at the inner photon sphere. Orbiting geodesics with some angular momentum perpendicular to the axis of rotation of the black hole will orbit on photon spheres between these two extremes. Because the space-time is rotating, such orbits exhibit a precession, since there is a shift in the φ variable after completing one period in the θ variable.
This may or may not help, but as far as I understand it, the stuff with horizons is all geometry: the Schwarzschild spacetime is, at its most complicated, topologically $\mathbb{R}^4 - \{0\}$ (as far as I understand it - I’ve not studied topology in GR properly) (Edit: this is patently absurd, as Tim vB points out in #17 below. I meant $(\mathbb{R}^3 - \{0\})\times \mathbb{R}$, of course). The interesting stuff depends on the manifold/metric structure. A photon sphere is defined in terms of lightlike paths, but this is metric dependent, even though it defines a submanifold with mildly interesting topology (it is $S^2\times \mathbb{R}$ in simplest form). Interestingly, this is not too dissimilar to how Poincare discovered/introduced the idea of topology, by looking at stability of planetary orbits and if they were closed or not asymptotically in time. The size/shape of the orbit was not as important as the topological properties of the worldlines.
Please also consider the difference between saying spacetime is rotating and saying space is rotating. Spacetime is (for argument’s sake) the whole 4-dimensional Einstein manifold. It doesn’t rotate, it just is. If however one thinks of what happens to test objects (which are only used to see what is happening in the metric) in a time-evolving spacelike slice, then one can observe rotational behaviour. Compare the statement: ’the worldline of the particle is orbiting the sun’. I’m sure you’d agree it is meaningless. The worldline of the particle is a static object, because there is not another ’external time’ by which to say it is doing anything dynamic. I know you clarify that your impression is perhaps not correct, so this is just to give you the explanation from my point of view of the ’static topological’ description.
Anyway, I’ll let the real experts give you a proper answer.
And how would a mathematical physicist react if GP-B data can’t improve the accuracy of the frame-dragging data?
Easy - get the experimentalists to make better instruments :P The ’worry’ would be if the accuracy was say an order of magnitude higher than the predicted size of the effects, and no frame-dragging was observed.
Urs mentioned that nothing actually rotates in the Kerr metric.
No.
Here is the exchange we had in the spacetime thread, the first quote from you, the second from me:
The example I gave there is the Kerr metric that explicitly has rotation built into the metric. Rotation does not necessarily have anything to do with gravity.
You realize that there is no thing that rotates in a Kerr spacetime except the gravitational field itself? It’s a vacuum solution to Einstein’s equations.
the embedded massive object had a measurable angular momentum.
No. There is no massive embedded object in a Kerr spacetime. It is a vacuum solution. It still has non-vanishing angular momentum. This is entirely carried by the gravitational field itself.
No. There is no massive embedded object in a Kerr spacetime. It is a vacuum solution. It still has non-vanishing angular momentum. This is entirely carried by the gravitational field itself.
So what causes the spacetime to curve in a Kerr metric?
Let’s say we have a rotating massive star, then the Kerr metric is supposed to be a good approximation if you keep a certain distance from the star. The Kerr spacetime is the mathematically idealization of the situation that you let the radius of the star go to zero. If you yourself can move along timelike curves in the Kerr spacetime without disturbing it (neglecting your effect on the spacetime, that is), you won’t ever experience a collision of any sort.
I don’t intend to put words in Urs’ mouth, but that would be the way I would explain “there is no massive object in Kerr spacetime”.
So, if the angular momentum is carried by the gravitational field, does that mean it is separate from spacetime itself and thus while the field rotates, the spacetime manifold on which it is defined does not?
There are some mistakes I repeat a hundred time before I get it :-) While I think I can guess what you mean by “the spacetime manifold rotates”, you will see, that the participants of this thread already agreed that this statement does not make any sense :-)
Same goes for “nothing is really moving”…You can only move relative to the gravitational field, the gravitational field itself defines what is movement and what is “stationary”, ergo there cannot be a movement of the gravitational field. It’s like: define space to be $\mathcal{R}^3$, choose an orthonormal basis $\{e_1, e_2, e_3\}$, question: is $e_2$ moving?
The Kerr spacetime is asymtotically Minkowskian, that is if we are far away we could play a game and say:
Kerr spacetime, in the area where we are, = Minkowski spacetime + some small, strange effects.
Kerr spacetime is a family of vacuum solutions with two parameters M and a, if you set M = a = 0 you get the Minkowski spacetime, so question is: can we use these small, strange effects to measure M and a? (I guess you already know the answer, what I intend to achieve is to provide context that serves as a common ground where we can meet :-)
…it acts like an angular momentum, but nothing is really moving, per se…
Well, maybe the electron does spin, who knows? :-) You won’t see it easily unless you paint a dot on it somewhere… But seriously: The spin of an electron is an intrinsic property of the electron that cannot be changed by any process (well, at least that’s one of the assumptions of QFT). While you cannot change spacetime (see above), you certainly can influence and change the gravitational field in your spatial environment…
I may have mentioned it before: When I learned a bit about GR the book “Quantum Gravity” by Carlo Rovelli was very helpful. You don’t have to be interested in quantum gravity, the chapter 2 about GR will be very worthwhile to read by all means.
Ian, you write:
So what causes the spacetime to curve in a Kerr metric? […] So, if the angular momentum is carried by the gravitational field, does that mean it is separate from spacetime itself and thus while the field rotates, the spacetime manifold on which it is defined does not? Or are you saying that angular momentum here is a little like “spin” in QM - it acts like an angular momentum, but nothing is really moving, per se? I think I have spent waaaaay too much time studying Eddington’s Fundamental Theory…
This is good: you are now asking questions and no longer making statements! This is the way to progress. Very good.
I think I’m more confused than ever so please pardon my apparent lug-headedness.
All lug-headedness is very much pardoned here, if coming with the willingness to learn.
Tim gave some good answers already. If you feel like understanding the answers to these questions more formally, you might start looking into what is called “ADM formalism”. See the index of your copy of Misner-Thorne-Wheeler for that. There is an “ADM mass” and also an “ADM angular momentum” defined for asymptotically flat spacetimes. I think MTW supply lots of heuristic discussion for this in their chapter.
(And do not look at the Wikipedia entries on this. Like almost all Wikipedia entries on physics – and in stark contrast to those on math – they are no good.)
Kerr spacetime is a family of vacuum solutions with two parameters M and a, if you set M = a = 0 you get the Minkowski spacetime, so question is: can we use these small, strange effects to measure M and a? (I guess you already know the answer, what I intend to achieve is to provide context that serves as a common ground where we can meet :-)
you certainly can influence and change the gravitational field in your spatial environment...
…I’m still working under what apparently is the old adage, “space tells matter how to move, matter tells space how to curve,” or whatever the quote was (was that Wheeler?).
That’s still true! (I, too, think that it was Wheeler). In GR, everything that is not the gravitational field, is matter. And you can get a non-trivial gravitational field without matter, like the Kerr spacetime. This is not a contradiction to Wheeler’s statement.
…do you view that as saying that the field and the spacetime are one and the same thing or that they are separate but related by Einstein’s equations?
If you state the field equations of GR as
$G _{ab}= 8 \pi T_{ab}$(using the notation of the book by Wald), then the (gravitational) field and the spacetime are both defined by the left side, while all matter is on the right side (your question seems to imply that you see the field or spacetime on the right side). The field equations define the influence of gravity on matter and of matter on gravity, just like Wheeler said :-)
The second is that this idea that spacetime itself can impart an angular momentum seems to suggest it possesses some itself which seems odd since it’s static.
Static means that the whole future and past is described by spacetime. But that travelling on a timelike curve changes your state by interaction with gravitation is not a surprise at all, you experience it when you fall down :-) A more dramatic example is a spacecraft that is torn apart by tidal forces, but I’ll set that aside for the next Bruckheimer blockbuster…
The first is that, since photons travel on geodesics, and photons orbiting in the opposite direction of the “rotation” (or whatever) behave differently, geodesics going one way around a Kerr solution must be fundamentally different in some way from those going around in the other direction. How is this explained geometrically…?
That’s tricky, finding some exact but easy to understand explanation of this situation. If I come up with something useful I’ll mention it, but please don’t be disappointed if I don’t :-)
And you can get a non-trivial gravitational field without matter, like the Kerr spacetime.
your question seems to imply that you see the field or spacetime on the right side
I'll set that aside for the next Bruckheimer blockbuster...
That's tricky, finding some exact but easy to understand explanation of this situation. If I come up with something useful I'll mention it, but please don't be disappointed if I don't :-)
Ian wrote:
The second is that this idea that spacetime itself can impart an angular momentum seems to suggest it possesses some itself which seems odd since it’s static. Now, if you think in terms of the marbles, the naive way to explain it is to say that the spacetime is this manifold on which the marbles (matter) move
if you are visualising marbles, then you are looking at point-in-time snapshots, and so it is not appropriate to talk about spacetime. Spacetime contains geodesics which represent the entire history of the marbles, and the interpretation that marble ’move along’ the geodesics requires a second ’time’ dimension to allow this process to occur. If you are in the reference frame of the marble, then the spacelike slice (just a local bit of it) is not static: it changes dynamically. This should then account for the fact the marble changes it angular momentum.
Time vB wrote:
you certainly can influence and change the gravitational field in your spatial environment…
And Ian replied
Right, and I think that nails my misunderstanding right on the head. So, would you say spacetime is defined by the associated metric and all the associated curvature stuff (Reimann tensor, Ricci scalar, etc.) and the field is defined by the stress-energy tensor and then Einstein’s equations link them?
This is where I repeat my warning: this whole discussion should take place with no reference to the gravitational field. I think Tim was being slightly colloquial: really he was saying (if I may take the liberty of putting words in his mouth) that you can curve spacetime in your vicinity, i.e. you create a disturbance in the Force (whoops, wrong movie ;-)
OK, so the empiricist in me then asks, what causes the curvature in this situation? And what happens to the mass of a rotating star when it collapses to a Kerr black hole if the Kerr solution is for a vacuum?
Note that the Kerr spacetime is ’just’ a manifold with a metric that satisfies some equations, with a ’hole’ where the singularity should be. From a more physical pov, technically the only place in the Kerr spacetime which is not vacuum is the worldline of the singularity, and then it is very much a philosophical question whether the singularity is part of spacetime or not. The metric is not defined at the singularity, and so (at least mathematically) that bit is removed. Really we know that a quantum theory of gravity is needed to deal with what happens at the singularity, and so this is not a question we are allowed to ask under strictly classical GR.
Ian asked:
OK, so the empiricist in me then asks, what causes the curvature in this situation? And what happens to the mass of a rotating star when it collapses to a Kerr black hole if the Kerr solution is for a vacuum?
David answered:
…Really we know that a quantum theory of gravity is needed to deal with what happens at the singularity, and so this is not a question we are allowed to ask under strictly classical GR.
Indeed, as an empiricist you should view the Kerr spacetime as a mathematical idealization, much like in classical mechanics where you would model the sun and the planets in the solar system as points to calculate their trajectories. And although it is common sense that GR predicts the existence of black holes, and astronomors are sure that such objects exist, we should keep in mind that we do not really know what an observer would observe after falling through the event horizon, because we do not really know what “physically existing black holes” look like (BTW: the movie was about black holes, wasn’t it? I mean the “Event Horizon” had a wormhole drive).
David wrote some time ago:
the stuff with horizons is all geometry: the Schwarzschild spacetime is, at its most complicated, topologically $\mathcal{R}^4$−{0} (as far as I understand it - I’ve not studied topology in GR properly). The interesting stuff depends on the manifold/metric structure.
Interesting question, but I think you mean something like $\mathcal{R}^4$ with a one dimensional submanifold removed, or else the singularity would exist for an infinitesimal time only :-) Note that both the Schwartzschild and the Kerr metric can be used to describe the (vacuum) spacetime outside of a spherically symmetric mass distribution of radius $\gt 0$, so there does not need to be a singularity.
In differential geometry we usually think of a topological manifold that gets additional geometric structure like a metric, in GR the situation is somewhat reversed: You assume that spacetime is a four dimensional Lorentzian manifold (Lorentzian by the equivalence principle), then the field equations spit out the metric. You can then go ahead and try to deduce information about the topology. For the Kerr spacetime it is possible to classify the spacetime up to homotopy, see Barrett O’Neill: “The geometry of Kerr black holes” chapter 3.9 “Topology of Kerr Spacetime”. (It has pictures!)
Indeed, as an empiricist you should view the Kerr spacetime as a mathematical idealization...
the movie was about black holes, wasn't it? I mean the "Event Horizon" had a wormhole drive
then the spacelike slice (just a local bit of it) is not static: it changes dynamically. This should then account for the fact the marble changes it angular momentum.
This is where I repeat my warning: this whole discussion should take place with no reference to the gravitational field.
then it is very much a philosophical question whether the singularity is part of spacetime or not.
For the Kerr spacetime it is possible to classify the spacetime up to homotopy, see Barrett O’Neill: "The geometry of Kerr black holes" chapter 3.9 "Topology of Kerr Spacetime". (It has pictures!)
I said:
Indeed, as an empiricist you should view the Kerr spacetime as a mathematical idealization…
Ian answered:
:-) Indeed, I view all of mathematical physics that way, to some extent…
Okay :-) But I was referring to the more concrete question “what rotates in the Kerr spacetime?” and meant “nothing, because that was a abstracted away when we let the radius of our rotating star go to zero”.
Right, though what happens if gravitons are discovered?
That discovery would leave the theorists with the same open question that they face now: “How do we reconcile quantum theories with gravity?” The existence of “black holes” (if you accept this as a fact) already shows that there is a large gap in our understanding, the discovery of gravitons would only support this.
Most of the folks I know think of it as part of the spacetime (I would assume) since they deal in very practical (?) matters of the masses of such things (e.g. the mass of the “black hole” at the center of the galaxy).
There is no contradiction here, a mathematician would say “spacetime is a Lorentzian manifold, ergo points with divergent metric cannot be part of it”, but that is just a mathematical definition. Wether or not you exclude the singularity itself from spacetime, the mass and angular momentum are still well defined and measurable. I think that is a point that Urs had in mind when he mentioned Misner-Thorne-Wheeler: Chapter 19 is about how to measure both e.g. in the weak field approximation far away.
…things like event horizons can, of course, be “eliminated” by changing one’s coordinate system, though it presumably doesn’t eliminate the physical effect…
There are “phenomena” that depend on the chosen coordinate system and phenomena that don’t. A “singularity” of the metric may depend on the chosen coordinate system, then it is not a true singularity that has any physical effect, or it does not depend on the chosen coordinate system. The difference becomes manifest if one takes a look at measurable quantities like the Riemann curvature tensor (divergent for a true singularity, not divergent for a coordinate singularity). The existence of an event horizon does not depend on the chosen coordinate system (“the event horizon is the boundary of the region where the escape to infinity is possible”).
Understood in this sense, the statement “the event horizon can be eliminated by changing one’s coordinate system” is wrong, strictly speaking. (But I guess you had something in mind that is similar to what I say above?).
Indeed, as an empiricist you should view the Kerr spacetime as a mathematical idealization...I was referring to the more concrete question "what rotates in the Kerr spacetime?" and meant "nothing, because that was a abstracted away when we let the radius of our rotating star go to zero".
Understood in this sense, the statement "the event horizon can be eliminated by changing one's coordinate system" is wrong, strictly speaking. (But I guess you had something in mind that is similar to what I say above?).
Ian said:
Empirically, I’m still interested in what’s physically there and what it’s doing…
Yes of course! This is a very fascinating question in theoretical physics, I share that feeling. But Kerr spacetime and more generally general relativity won’t tell us. If you ask “what’s really behind the event horizon?” the answer won’t come from general relativity.
So your argument is akin to the standard argument that is given for electron spin - it’s not really rotating (forget the bizarre discreteness issue for a moment) since it’s a point particle, i.e. its radius is zero (which it has to be in order to not violate relativity).
Yes, true, I say that the theory won’t answer this kind of question and that we have to take this for granted when working with it. But I am still fascinated by the question if there is some deeper truth to be discovered here, and I don’t consider this to be a contradiction (believe or don’t believe?).
BTW: If I have some good ideas and some time I will expand the Kerr spacetime entry and announce that here, right now it only has a table with the metric in canonical coordinates for the Minkowski, Schwartzschild and Kerr spacetimes (any help with formatting this table is welcome :-).
…he finds category theory too abstract for him.
That’s unfortunate, did you notice that John Baez is giving a talk in Oxford today that may be of interest to you? And that Bob Coecke explained some of his thoughts and results on the nCafé here?
Mike made a comment recently that I meant to respond to, but work’s been hectic lately and it slipped by. If I can find it, I’ll point to it. The subject was “passing to cohomology”. It seems like the way a cohomologist might look at the world depends on his choice for what cohomology means. There are various flavors of cohomology and although they all have common roots, they can give different impressions of the same observations. This reminded me of quantum mechanics and experimentation. A choice of a cohomology theory seems like a choice of an experimental set up (loosely and vaguely speaking).
As we know from wave-particle duality, the outcome of an experiment depends to a large degree on how the experiment is set up and what its looking for. If you’re looking for waves, you’ll see waves. If you’re looking for particles, you’ll see particles. It would be neat to relate wave-particle duality to choices of cohomology. The link is tenuous at best, but my gut says there might be something there.
Anyway, I think what matters to an empirical physicist is what can be measured. Category theory seems perfectly suitable as a theory of measurement. You have probes testing spaces, etc. Have a look at:
motivation for sheaves, cohomology and higher stacks
Have you been following John’s recent discussions about copying classical states?
@Ian, Eric
Have either of you read Chris Isham’s Topos Methods in the Foundations of Physics? I find it fascinating, and it’s certainly backed up with a bunch of ’real maths’ (the article linked to is described as a conceptual description). But this is off topic.
If you ask "what's really behind the event horizon?" the answer won't come from general relativity.
That's unfortunate, did you notice that John Baez is giving a talk in Oxford today that may be of interest to you? And that Bob Coecke explained some of his thoughts and results on the nCafé here?
The subject was "passing to cohomology".
It would be neat to relate wave-particle duality to choices of cohomology. The link is tenuous at best, but my gut says there might be something there.
Have you been following John's recent discussions about copying classical states?
Have either of you read Chris Isham's Topos Methods in the Foundations of Physics?
Kerr spacetime now has a definintion of the Boyer-Lindquist blocks, and I think I can already make some of my statements a little more precise with the little material that is there:
Spacetime itself cannot move, of course, that is true for all spacetimes - to have it move you would need an extra time dimension, and maybe some ambient space, that is something that I called “stationary” above, but
in GR stationary is usually defined to be something else. Using the Boyer-Lindquist coordinates we can see that $\partial_t$ is a Killing vector, so Kerr spacetime is invariant under “time translations”, but
the problem with the last statement is of course that t is close to a classical global time coordinate on the Boyer-Lindquist block I only. When I said that GR won’t tell us what is behind the event horizon I meant that, of course, we can calculate what is going on in the blocks II and III, but do the results tell us that GR breaks down there or that the universe is very crazy? :-) I don’t know, I think we should expect that a “physically existing Kerr black hole” behaves differently from what pure GR predicts about blocks II and III.
There is nothing rotating in Kerr spacetime means that you can take the coordinate r from infinity to zero without finding any matter at any point :-)
Next, if time permits, I will add some additional statements about the blocks and then maybe a little bit about different geodesics…
There is nothing rotating in Kerr spacetime means that you can take the coordinate r from infinity to zero without finding any matter at any point :-)
Well, the problem I guess is whether the value of $r=0$ is an actual point on the geodesic since an empiricist would argue that matter does exist at that point.
To take the discussion one step further we will need a better definition of singularity and some facts about geodesics in Kerr spacetime, it will take me a while to prepare those. So, if the thread falls asleep, that does not imply that I lost interest :-)
1 to 27 of 27