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Is there something like ds'^2=ds^2 in GR?

  1. Dec 19, 2013 #1
    So in special relativity we have [itex]ds'^{2}=ds^2[/itex], which is another way of saying [itex]\Lambda^{T}\eta\Lambda=\eta[/itex]. Where [itex]\eta=diag(-1,1,1,1)[/itex].

    It seems in GR the symmetry group for transformations is GL(R,4) or Diff(M) depending on who you ask: http://physics.stackexchange.com/questions/65688/group-theory-in-general-relativity

    However (I might have missed it) the answers in the above link don't mention if there is an explicit formula akin to [itex]\Lambda^{T}\eta\Lambda=\eta[/itex] for allowed transformations, or a formula for a conserved quantity like [itex]ds'^{2}=ds^2[/itex]?
     
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  3. Dec 19, 2013 #2

    WannabeNewton

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    ##ds'^2 = g_{\mu'\nu'}dx^{\mu'}dx^{\nu'} \\= g_{\alpha\beta}\frac{\partial x^{\alpha}}{\partial x^{\mu'}}\frac{\partial x^{\beta}}{\partial x^{\nu'}}\frac{\partial x^{\mu'}}{\partial x^{\gamma}}\frac{\partial x^{\nu'}}{\partial x^{\delta}}dx^{\gamma}dx^{\delta} \\= g_{\alpha\beta}\delta^{\alpha}_{\gamma}\delta^{\beta}_{\delta}dx^{ \gamma } dx^{\delta} \\= g_{\mu\nu}dx^{\mu}dx^{\nu} = ds^2##.

    In SR the symmetry group is the Poincare group (with the subgroup of Lorentz transformations usually restricted to the proper, orthochronous Lorentz transformations) and the matrix representations of the proper, orthochronous Lorentz group have an explicit formula akin to what you wrote down. However in GR we are no longer dealing with special covariance but rather general covariance; there is no explicit formula for allowed transformations in the context of GR.
     
  4. Dec 19, 2013 #3

    bcrowell

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    GR is diffeomorphism-invariant. The only transformations not allowed would be those that were not diffeomorphisms (e.g., were not differentiable). This is actually the same as SR, at least from the modern point of view. The way people today define the distinction between SR and GR is that SR applies when there's no curvature. By that definition, there is nothing wrong with using any coordinate system you like for SR. For example, you can use Rindler coordinates, which can be interpreted as an accelerated frame of reference. It is true in both SR and GR that [itex]ds'^2=ds^2[/itex], and in both cases this invariance is ensured in the same way: when you change coordinates, in general the metric will also change.

    In old-fashioned presentations of special relativity, the form of the metric as diag(-1,1,1,1) was taken to be holy. This is where we got the idea of allowed and forbidden sets of coordinates.
     
  5. Dec 19, 2013 #4

    tom.stoer

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    I think [itex]ds'^{2}=ds^2[/itex] is trivial b/c it is a coordinate-free statement for a manifold.

    Is the distance between New York and Berlin the same as the distance between New York and Berlin? Yes!
     
  6. Dec 19, 2013 #5

    bcrowell

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    Maybe it's trivial for a manifold with metric, but the assumption that spacetime has a metric is highly nontrivial. In Galilean spacetime, the distance between where New York is at a certain time and where Berlin is at some other time is frame-dependent.
     
  7. Dec 19, 2013 #6

    tom.stoer

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    The thread is about GR, so we have a manifold and a metric.
     
  8. Dec 19, 2013 #7

    bcrowell

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    The title of the thread is this: "Is there something like ds'^2=ds^2 in GR?" It looks like we both agree that it's trivial that a "yes" answer is equivalent to the existence of a metric.

    If someone asks why and whether proposition P is true, and everyone agrees that P is trivially logically equivalent to proposition Q, then that's all good, but it doesn't constitute an answer to the question unless someone also gives an explanation of why Q is true.
     
  9. Dec 19, 2013 #8

    tom.stoer

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    How would a spacetime manifold w/o metric look like and what has all this to do with GR?
     
  10. Dec 19, 2013 #9

    ChrisVer

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    I guess a manifold without metric would be a manifold where you cannot define distances angles etc. Non riemannian manifolds.

    Now for the ds invariance, I would suggest also the mathematics equivalence that is the 1st Fundamental form....
     
  11. Dec 19, 2013 #10

    tom.stoer

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    If you cannot define a distance then ds is not defined
     
  12. Dec 19, 2013 #11

    K^2

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    You can have distances defined, but still not have a metric. You have to have distances follow certain axioms, such as the triangle inequality, in order for it to be a metric space. So yes, you can have a manifold with ds² defined, but which is not a metric manifold, and where ds'² ≠ ds².
     
  13. Dec 19, 2013 #12

    tom.stoer

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    Can you give an example for manifold with definition of a distance but w/o metric? What is the mathematical structure? And in which sense is this relevant for GR?
     
  14. Dec 19, 2013 #13

    WannabeNewton

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    It won't be relevant to GR at all but a topological manifold can always be endowed with a metric (and by metric I mean an actual metric, the central tool in the theory of metric spaces in real analysis); a metric is a very different object from a metric tensor but it defines a notion of distance nonetheless.
     
  15. Dec 20, 2013 #14

    ShayanJ

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    What about the speed of light?Is it still frame independent and the maximum speed?
    I remember reading somewhere that it is locally like SR but globally can be different but didn't understand it clearly!
     
  16. Dec 20, 2013 #15

    WannabeNewton

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  17. Dec 20, 2013 #16

    K^2

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    Sure. x,y in ℝ. d²(x,y) = Sin²((x-y)(x+y)). Works basically like a distance locally; lets you have an element ds² = 4x² dx². Fails some of the metric axioms, and is not invariant under coordinate transformation x' = x+c.

    Absolutely irrelevant to GR, because GR manifolds are always pseudo-Riemannian.
     
  18. Dec 20, 2013 #17

    ShayanJ

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  19. Dec 20, 2013 #18

    tom.stoer

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    This is not a distance in the sense of a metric space. The fact that it works locally is not sufficient.
     
  20. Dec 20, 2013 #19

    K^2

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    Having a distance that is one in the sense of a metric space is the definition of having a metric space.

    Not every distance is a metric. That's kind of what the whole discussion so far has been about. You asked for an example of a distance that's not a metric, and I gave you one. And now you're not happy that it's not a metric?
     
  21. Dec 20, 2013 #20

    tom.stoer

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    A distance in the sense of a metric space must fulfill the following axioms:

    A) d(x, y) ≥ 0
    B) d(x, y) = 0 if and only if x = y
    C) d(x, y) = d(y, x)
    D) d(x, z) ≤ d(x, y) + d(y, z)

    Your example definitly violates (B), and I bet it violates (D)

    So it's not a distance in that sense. But what else shall a distance be?
     
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