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Metric tensor Determinant

  1. Aug 17, 2012 #1
    Hello,

    So, given two points, [itex]x[/itex] and [itex]x'[/itex], in a Lorentzian manifold (although I think it's the same for a Riemannian one). If in [itex]x[/itex] the determinant of the metric is [itex]g[/itex] and in the point [itex]x'[/itex] is [itex]g'[/itex]. How are [itex]g[/itex] and [itex]g'[/itex] related?By any means can [itex]g=g'[/itex]? In what conditions?

    I'm sorry if this is a dumb question but when prooving an equation I found out that it holds only if and only if [itex]g=g'[/itex] and I don't think that this is always true.

    Thank you.
     
  2. jcsd
  3. Aug 17, 2012 #2

    George Jones

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    These are two different points, right? If so, there is, in general, no relation between [itex]g[/itex] and [itex]g'[/itex].

    If it is actually two different coordinate representations of the same point, then [itex]g' = J^2 g[/itex], where [itex]J[/itex] is the Jacobian of the coordinate transformation.
     
  4. Aug 18, 2012 #3
    Thank you for the prompt response.
    The notation might have not been the best, sorry; it were two distinct points in a manifold.

    What if the point [itex]x'[/itex] were in a sufficiently small neighborhood of [itex]x[/itex] and vice versa (the neighborhoods may not be the same). Aren't tensors defined in a small neighborhood of a point: since they represent a multilinear map?

    Thank you, and sorry if i'm really confused about the concepts.
     
  5. Aug 18, 2012 #4

    bcrowell

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    Given a manifold, you can put any metric on it that you like. Just choose any smooth function that has the same signature everywhere.

    It would be unusual if the signature of the metric varied from point to point. If this happens in actual applications, it's usually a sign that you chose a bad coordinate system, which can be related via a singular transformation to some other, better system of coordinates. (Differential geometry is agnostic about coordinates, but only up to a smooth, one-to-one mapping, not up to any mapping at all.)

    I think the metric does have to be free of intrinsic singularities (as opposed to mere coordinate singularities). When a singularity turns up in general relativity, for example, we define the singularity not to be a point in the manifold at all. So for this reason, I think the answer to your #3 is that yes, g and g' are related if x and x' are close. For x and x' sufficiently close, g and g' can be made as close as desired.
     
    Last edited: Aug 18, 2012
  6. Aug 26, 2012 #5

    haushofer

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    I don't get this. The tensor transo. law of the metric can be considered to be active. Then I can use your formula. And relate g(x) and g'(x'), where x and x' are different points (different coordinates in the same chart).
     
  7. Aug 26, 2012 #6
    How is this different from cheating (mathematically speaking)?
     
  8. Aug 30, 2012 #7

    HallsofIvy

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    It isn't. But, after all, "cheating" is a well-respected mathematical method!
     
  9. Aug 30, 2012 #8
    :rofl:

    Well, in this particular case I interpreted that "we" in "we define..." to be referring to physicists, so it would be more of a well-respected "mathematical physics" method. :tongue2:
     
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