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Determinant of the metric tensor

  1. Feb 5, 2013 #1
    We are stating with equivalence principle that passing locally to non inertial frame would be analogous to the presence of gravitational field at that point, so [itex] g^'_{ij}=A g_{nm} A^{-1} [/itex] where g' is the galilean metric and g is the metric in curved space, and A is the transformation which eliminates gravity at the point.

    How do you prove that det[g] <0?

    Doesn't it follow from the above realtion that det[g] is exactly -2 because det[g'] = -2, and not just simply negative? What did I miss?

    Thanks!
     
  2. jcsd
  3. Feb 5, 2013 #2

    Bill_K

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    Det(AB) = Det(A) Det(B)
     
  4. Feb 5, 2013 #3

    bcrowell

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    Curvature, unlike a gravitational field, is coordinate-independent, so you can't introduce or eliminate curvature by a change of coordinates. "Galilean metric" also doesn't make sense, because there is no spacetime metric in Galilean relativity. (If you try to define one, it ends up being degenerate.)

    That would depend on what initial assumptions you were making. It could be taken as a postulate. There is nothing in GR, the field equations, etc., that prohibits a Riemannian rather than semi-Riemannian metric.
     
  5. Feb 5, 2013 #4

    PAllen

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    I would say the field equations without the assumption of locally SR = pseudo-Riemannian metric is a different theory, not GR.
     
  6. Feb 5, 2013 #5

    bcrowell

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    Sure. However, it's not like anything in the formalism breaks down if you assume a different signature. I believe the Ashtekar formulation of GR can even handle geometries in which the signature differs from point to point.
     
  7. Feb 5, 2013 #6
    Then how should I understand this paragraph from Landau Lifgarbagez "Classical Theory of Fields" (chapter 82):

    "We note that, after reduction to diagonal form at a given point, the metric of the quantities g_ik has one positive and three negative principal values. From this it follows in particular, that the determinant g, formed from the quantitiers g_ik, is always negative for a real spacetime: g<0."
     
  8. Feb 5, 2013 #7

    bcrowell

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    Like I said, it depends on what initial assumptions you're making. The first sentence states an initial assumption or something that follows from some other initial assumptions.
     
  9. Feb 6, 2013 #8

    martinbn

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    But if signature changes from point to point there should be points where the metric is degenerate, no?
     
  10. Feb 6, 2013 #9
    At a given point you can diagonalize the metric in case of curved spacetime.Real spacetime refers to spacetime of general relativity which will of course have g<0.
     
  11. Feb 6, 2013 #10

    bcrowell

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    Right. This is why the standard formalism of GR can't handle a change of signature; it's expressed using tensors with indices raised and lowered, which you can't do if you have a degenerate metric. In practice, if you get a degenerate metric in GR, it usually means you chose bad coordinates: http://lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.4 [Broken]
     
    Last edited by a moderator: May 6, 2017
  12. Feb 6, 2013 #11

    martinbn

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    OK, then Ashtekar formulation of GR is actually a generalization, not just a different formulation, is that right? By the way, where can a look at it?
     
    Last edited by a moderator: May 6, 2017
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