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Say we have got a system in GR that is described by the Schwazschild metric. Then we perform a coordinate transform that gives the metric in a rotating system.

Why is the transformed metric not the Kerr metric in some form?

My suspicion is that this is due to the requirement that both Kerr and Schwarzschild metric tend to flat space far from the central mass. This assumption is used in the derivation of the two metrics. But why is this physical? And if what i have said so far is correct are there experimental tests of the "flatness" far from bodies that are assumed to be Schwarzschild/Kerr in our universe? (i.e. to test how useful these solutions to Einsteins equations are in modelling real objects).

Wolpertinger
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  • Is this any different from classical mechanics? In a rotating coordinate system gravity would seemingly become repulsive (because of fictional forces) at some distance, would it not? – CuriousOne Mar 01 '16 at 08:28
  • @CuriousOne I feel like your question my be related to Mach's principle, although I don't know enough about GR nor the principle (which I think isn't even necessarily part of GR see http://physics.stackexchange.com/q/5483/) to say anything about it. – Wolpertinger Mar 01 '16 at 08:46
  • A rotating coordinate system is clearly locally detectable, I am merely pointing out that the situation doesn't require general relativity to occur. Maybe I misunderstand your question? – CuriousOne Mar 01 '16 at 09:10
  • @CuriousOne I want to know why you can't apply a rotation transform to the Schwarzschild metric to get the Kerr metric, i.e. in what way they are 2 distinct metrics rather than the same metric in the a different coordinate system. Everything about my question seems GR to me, what do you mean by we can do it in classical mechanics? About the locally detectable: another suspicion i had was that rotation transforms would not be "allowed" in some sense. I don't see why though. Is that what you're saying? – Wolpertinger Mar 01 '16 at 09:34
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    I got that part. My concern is that a rotating coordinate system is not the same thing as e.g. a rotating planet. The Newtonian gravity of a rotating planet is the same as that of a non-rotating one, while the fictional force field that results from a rotating coordinate system overlaid on the Newtonian gravitational acceleration of a point mass shows a repulsive term that grows with distance. This is not a relativistic phenomenon, at all, so I would not expect GR "to make this right" (since it turns into Newtonian physics in the low speed, weak field limit). – CuriousOne Mar 01 '16 at 09:56
  • @CuriousOne Ah now I see your point :) I agree that you get this phenomenon in the classical system you described, but shouldnt GR still yield this result naturally? As in when you have the Minkowski metric and do a rotation transform the geodesics of the new metric should be exactly what is described by the fictional forces in the Newtonian framework (I hope... if anything of what I said in this comment is wrong then I clearly don't understand GR at all). Would you agree on that?

    If so then my question is just why this doesn't work for rotating the Schwarzschild metric.

     – Wolpertinger Mar 01 '16 at 10:27
  • I am not a relativist and I never went trough a formal derivation of the Kerr-metric, I have, however, seen Kerr's comments on the significant difficulties that he and others had to overcome to derive it. It seems to look trivial when going backwards from the final, less than overwhelming result... but that's not how it seems to have played out in the discovery process, which indicates that there is something non-trivial going on there. We have a question about it here: http://physics.stackexchange.com/questions/150446/derivation-of-kerr-metric-is-there-any-reference. – CuriousOne Mar 01 '16 at 10:32
  • @CuriousOne thanks for the link, the question is related I think. I think the non-triviality is that we impose that the metric ought to tend to flat space far from the object. My question is mainly if that is correct (because it would explain why you can NOT obtain the Kerr metric from a rotation of Schwarzschild. The systems would simply be different, since Kerr is flat at infinity while rotating Schwarzschild is not, since there are still fictitious forces there). Then as a follow-up from this I asked why we impose this in the first place and if there is any validation for it. – Wolpertinger Mar 01 '16 at 10:44
  • I'll move this to chat since it is not really relevant to the question, but rather additional discussion. – Wolpertinger Mar 01 '16 at 11:02
  • Let us continue this discussion in chat. – Wolpertinger Mar 01 '16 at 11:02
  • The underlying point here is that a coordinate transformation makes no difference to the physics at all: it's just a bookkeeping exercise. You can never get from one physical system to another by bookkeeping, unless those systems are, in fact, the same system, which Kerr and Schwarzschild are not. –  Mar 01 '16 at 22:08

2 Answers2

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One of the key features of the Kerr metric is that the black hole horizon is rotating with respect to the space at infinity -- you get frame dragging effects that cause the notion of "rest" to change as you get closer to the black hole. In fact, energy must be exerted if you want to stay stationary with respect to infinity, until you finally reach a surface, called the ergosphere (which is actually outside the event horizon), where it is actually impossible to be at rest relative to infinity.

These are all physical, frame-independent effects. In particular, it is possible to transfer energy from the black hole to infinity by clever explosions within the ergosphere, through soemthing called the Penrose process. A mere coordinate transformation would not be able to replicate effects like this. And a coordinate change describing a simple rigid rotation around the Schwarzschild black hole would leave "the sphere at infinity" rotating at the same rate as the hole.

Zo the Relativist
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  • thank you for your answer. So by "rotating with respect to the space at infinity" you mean that for large distances the metric approaches Minkowski spacetime as I said in the question? I understand how this causes all the effects you mentioned, but could you say the reason why we impose this for the Kerr and Schwarzschild metric? As in why do we expect the metric for rotating star or black hole in our universe to be flat spacetime at infinity? – Wolpertinger Mar 02 '16 at 09:57
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    @Numrok: yes. In fact, these spacetimes are called "asymptotically flat", which has a technical meaning in the GR literature. And the reason why we expect this to be the case is that we don't seem to feel the effects of black holes in Andromeda on Earth. – Zo the Relativist Mar 02 '16 at 18:02
  • since your argument is now using that we do not see the effects of that would imply I'd like to ask if you know about any experimental verofication for that? – Wolpertinger Mar 03 '16 at 08:54
  • @Numrok: other than the classical tests of general relativity and local flatness of the solar system, etc.? – Zo the Relativist Mar 03 '16 at 14:59
  • I don't see how the classical tests of GR would show this since the metric at infinity being non-flat wouldnt render GR invalid. what do you mean by local solar system flatness? is that a particular set of experiments/data you are referring to? – Wolpertinger Mar 04 '16 at 00:48
  • also i'll try to reformulate again what i'm actually asking in light of what we are discussing now. You pointed out in your answer that the "black hole horizon is rotating wrt to the space at infinity" which i already new (its even in the question). Then you listed some physical effects which are all correct but really not much to do with the question. At the end you compare thid to Schwarzschild, which is answering part of the question in that it confirms the suspicion i stated. – Wolpertinger Mar 04 '16 at 00:54
  • i also asked and still find it non-trivial why flat space at infinity is a good physical assumption though. as in do we have evidence that rotating objects in our universe have a Kerr metric rather than a rotated Schwarzschild metric? – Wolpertinger Mar 04 '16 at 00:56
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    @Numrok: flat space at infinity is a reasonable approximation because space is locally Minkowski, minus cosmological effects. We don't observe random background effects, locally, from distant black holes. All of the solar system gravity tests have this result -- there's no bulk anisotropy in gravity depending on the phase of plentary orbits, for example. – Zo the Relativist Mar 26 '16 at 22:21
  • ty, that was what i was looking for! – Wolpertinger Mar 27 '16 at 07:56
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I think the real question is, why would it be? A rotating coordinate frame is not the same as physically rotating object. This is easier to see in Galilean relativity, where we know perfectly well that only uniform motion is relative: a rotating star is not the same as an observer rotating around a static star, because the latter experiences a centrifugal force.

Suppose we take the Kerr and the rotating Schwarzschild metrics, which according to you should be the same, and let the black hole's mass go to zero. The Kerr metric goes to the Minkowski metric, which is reasonable, since you're standing still in empty spacetime. But the rotating Schwarzschild metric goes to a rotating Minkowski metric, which is different from regular Minkowski! You have centrifugal forces and so on. Therefore, the original two metrics are not the same.

Moving to a rotating coordinate system is not a symmetry of nature. That's all there is to it, really.

Javier
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  • thank you for your answer. I think your are pointing out what I mean by that we impose that the metric goes to Minkowski metric at infinity. but why is that? what makes us believe this is true for modelling e.g. black holes in our universe? Is there a theoretical reason? Or experimental evidence? – Wolpertinger Mar 02 '16 at 15:57
  • @Numrok: I never said that we impose that the metric goes to Minkowski at infinity; for the black holes we're talking about here, it goes to flat spacetime at spatial infinity but not at temporal infinity. The second paragraph is just elaboration, the first and last are more important. – Javier Mar 02 '16 at 19:02