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A Global coordinate system

  1. Apr 25, 2017 #1
    Is there a universal criteria to determine if a coordinate system is global?

    I think that it is sufficient for the determinant of the metric to be non-zero in order for a coordinate system to be global. Is this so?

    For example, take the metric

    ##ds^{2} = \ell^{2}(-\cosh^{2}\rho\ dt^{2} + d\rho^{2} + \sinh^{2}\rho\ d\phi^{2}).##

    I'd say that this metric is not global because ##\sinh\rho = 0## for ##\rho = 0##. What do you think?
     
  2. jcsd
  3. Apr 25, 2017 #2

    Dale

    Staff: Mentor

    I think that you want to learn about geodesic completeness.
     
  4. Apr 25, 2017 #3
    I suppose you're talking about genuine singularities of the manifold. My question is more pedestrian. It's to check if the metric is global, or if it is degenerate at some point.
     
  5. Apr 25, 2017 #4

    Dale

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    Yes, I think that is geodesic completeness, but I could be wrong.
     
  6. Apr 25, 2017 #5

    PeterDonis

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    By "global" I assume you mean "covers the entire manifold", correct?

    In order to answer this question, you first have to know what "the entire manifold" means. The accepted meaning in GR is the maximal analytic extension.

    No. For a counterexample, consider Painleve coordinates on maximally extended Schwarzschild spacetime. Their determinant is nonzero everywhere on the patch that they cover, but the patch that they cover is not the entire manifold.

    I think that a nonzero determinant is a necessary condition for a coordinate chart to cover the entire maximal analytic extension (certainly it seems intuitively obvious that it ought to be), but as the above example shows, it is not sufficient. I am not sure if it is known what the sufficient conditions are.
     
  7. Apr 25, 2017 #6

    PeterDonis

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    Not quite, because the maximal analytic extension of a manifold can still be geodesically incomplete (Schwarzschild spacetime is the obvious example). What I think you might mean here is that the chart covers all geodesics in the manifold as completely as possible, i.e., to all values of their affine parameters that are included in the maximal analytic extension. I think that works as a heuristic, but I don't know how you would formulate it mathematically in terms of properties of the line element expressed in the chart.
     
  8. Apr 25, 2017 #7
    Ah! So does that mean that the metric

    ##ds^{2} = \ell^{2}(-\cosh^{2}\rho\ dt^{2} + d\rho^{2} + \sinh^{2}\rho\ d\phi^{2})##

    is not global because ##\sinh^{2}\rho = 0## at ##\rho = 0##?
     
  9. Apr 25, 2017 #8

    PeterDonis

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    Global on what manifold? That's the question. Just writing down a line element doesn't tell you what the manifold is.

    What ##\sinh^2 \rho = 0## at ##\rho = 0## tells you is that you have a coordinate singularity at ##\rho = 0##. But that's not enough information to know whether the manifold extends beyond that point or not; just the fact that ##\rho < 0## is mathematically possible doesn't necessarily mean that coordinate range maps to anything in the manifold (##\rho## could be like ##r## in Schwarzschild spacetime, where only the range ##r > 0## maps to points in the manifold). In other words, it's not enough information to tell whether, for example, ##\rho = 0## is like the singularity at ##r = 0## in Schwarzschild spacetime, which really does mark an "end" of the manifold, or like the singularity at ##r = 2M## in Schwarzschild spacetime, which doesn't.

    As I said before, I don't know exactly what the right conditions are, but computing geometric invariants at ##\rho = 0## (such as the Ricci scalar or the Kretschmann scalar) would help to determine what is going on there. Also, computing the Einstein tensor for your metric would be useful since it would tell you what kind of stress-energy would be needed to produce such a metric.
     
  10. Apr 25, 2017 #9
    Global on AdS##_{3}##. See what's written below equation (9.4) in page 97 of these lecture notes http://www.hartmanhep.net/topics2015/gravity-lectures.pdf.
     
  11. Apr 25, 2017 #10

    Dale

    Staff: Mentor

    Yes, this is what I was thinking about. I didn't mention the maximal analytic extension, but that is what I was thinking about.
     
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