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I Tetrads in rotating spacetime

  1. Nov 21, 2016 #1
    I'm interested in tetrad formalism for describing phenomenons near Kerr black hole. I've read some papers and I have a question about Localy Non-Rotating Frame (LNRF). In all papers is mentioned that most of astrophysically important cases are in equatorial plane (EP) and deals with EP only. Such tetrade looks:
    ##\omega^{(t)}{\mu}=(A,0,0,0) ##
    ##\omega^{(r)}{\mu}=(0,B,0,0) ##
    ##\omega^{(\theta)}{\mu}=(0,0,C,0) ##
    ##\omega^{(\phi)}{\mu}=(-\Omega_{LNRF} D,0,0,D) ##, where ##A,B,C,D, \Omega_{LNRF}##
    we can find out from definition of tetrad and metric.
    But what about out of EP? Is LNRF tetrad still in same form? Or there arise some extra expressions? Or how can I prove it that it is same on all ##\theta##?
     
  2. jcsd
  3. Nov 21, 2016 #2

    PeterDonis

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    First, I assume you are using Boyer-Lindquist coordinates, since in other coordinates the tetrad in the equatorial plane would not have the form you give.

    Second, where does the extra term in the fourth tetrad (the ##t## component) come from? Answering that should help to answer your questions about the form of the tetrad out of the equatorial plane?
     
  4. Nov 21, 2016 #3
    Definitely, Boyer - Lindquist coordinates.
     
  5. Nov 21, 2016 #4
    I was thinking about situation out of EP. And I know other thing, for example test particle can't orbit in plane parallel to EP. For me it says, there is force which acts on it in latitude direction. So it seems to me similar as in EP with draging in direction of rotation. So if any I would expect extra term ##\omega^{(\theta)}_t##.
     
  6. Nov 21, 2016 #5

    PeterDonis

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    Yes, that's correct.

    Why? A test particle can't orbit in a plane parallel to the equatorial plane because such an orbit would not be centered on the hole. That would be just as true in Schwarzschild spacetime, where there is no frame dragging.
     
  7. Nov 21, 2016 #6
    Yes, I see. Now I have new question: If I let the test particle to orbits in Schwarzschild spacetime out of my chosen equatorial plane (almost circle orbit - rosette shape -I don't know if it is good name) and then let the hole rotate (such that my equatorial plane will be real equatorial plane of rotating hole). The change of shape of orbit will be affected only by frame dragging or is it more complex?

    So from what you said, the tetrad should be same for all ##\theta##. Do you agree?
     
  8. Nov 21, 2016 #7

    PeterDonis

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    I don't understand what kind of orbit you are referring to. There are no orbits that are not centered on the hole. Are you just describing a non-circular orbit, which because of perihelion precession will not be a closed ellipse (as it would be in Newtonian gravity)?

    AFAIK the only effect that comes into play when the hole is rotating vs. not rotating is frame dragging. The reason I asked about where the extra term in the equatorial plane came from was to focus on how frame dragging is reflected in the metric in your chosen coordinates, and in particular whether there is any difference in how it is reflected out of the equatorial plane vs. in the equatorial plane.
     
  9. Nov 21, 2016 #8
    Yes, it is exactly what I've meant.
     
  10. Nov 21, 2016 #9
    Oh, ok. So may I say because LNRF tetrad coming from kerr metric in Boyer - Lindquist coordinate which are same for all planes or there is no change in equatorial plane, LNRF tetrad has to have same shape for all latitude. Do you agree?
     
  11. Nov 21, 2016 #10

    PeterDonis

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    Meaning the same components are nonzero? I believe so, yes. But the relative values of those components vary with latitude (and also with radius).
     
  12. Nov 21, 2016 #11
    Definitely, I mean that there are still same functions (coming from metric).
    So thank you very much for advices.
     
  13. Nov 21, 2016 #12

    PeterDonis

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    As long as by "the same functions" you mean the most general ones derived from the metric, including the ##\theta## dependence as well as the ##r## dependence, yes, this is true. When restricting attention to the equatorial plane, people often eliminate the ##\theta## dependence by assuming ##\sin \theta = 1##. But the functions you get when you do that are not the most general ones.
     
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