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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).

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  • $\begingroup$ 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? $\endgroup$
    – CuriousOne
    Commented Mar 1, 2016 at 8:28
  • $\begingroup$ @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 physics.stackexchange.com/q/5483) to say anything about it. $\endgroup$
    – Wolpertinger
    Commented Mar 1, 2016 at 8:46
  • $\begingroup$ 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? $\endgroup$
    – CuriousOne
    Commented Mar 1, 2016 at 9:10
  • $\begingroup$ @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? $\endgroup$
    – Wolpertinger
    Commented Mar 1, 2016 at 9:34
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    $\begingroup$ 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). $\endgroup$
    – CuriousOne
    Commented Mar 1, 2016 at 9:56

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Wolpertinger
    Commented Mar 2, 2016 at 9:57
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    $\begingroup$ @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. $\endgroup$
    – Zo the Relativist
    Commented Mar 2, 2016 at 18:02
  • $\begingroup$ 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? $\endgroup$
    – Wolpertinger
    Commented Mar 3, 2016 at 8:54
  • $\begingroup$ @Numrok: other than the classical tests of general relativity and local flatness of the solar system, etc.? $\endgroup$
    – Zo the Relativist
    Commented Mar 3, 2016 at 14:59
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    $\begingroup$ @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. $\endgroup$
    – Zo the Relativist
    Commented Mar 26, 2016 at 22:21
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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.

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  • $\begingroup$ 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? $\endgroup$
    – Wolpertinger
    Commented Mar 2, 2016 at 15:57
  • $\begingroup$ @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. $\endgroup$
    – Javier
    Commented Mar 2, 2016 at 19:02

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