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Is hyperbolic space consistent with homogeneity?

  1. Sep 29, 2014 #1
    The FRW metric is usually expressed as
    $$ds^2 = -dt^2 + a(t)^2 ( \frac{dr^2}{1-kr} + r^2 d\Omega^2))$$
    where ##k=-1,0,+1## respectively for a hyperbolic, flat or spherical space. The spatial part of this metric can be derived by considering a 3-sphere embedded in a four-dimensional flat space -- any sphere obviously being homogeneous.

    Similarly, the k=-1 case can be derived by considering a hyperboloid embedded in a flat four dimensional space. Now, the hyperboloid is only a homogeneous space when embedded in a flat minkowskian space -- in Euclidean space the point at the tip of the hyperboloid (corresponding to r=0) is certainly special.

    Since it is the euclidean distance we measure when measuring distances to other galaxies, it seems like the k=-1 case is not consistent with homogeneity. Is this correct, or is my thinking wrong?
     
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  3. Sep 29, 2014 #2

    PeterDonis

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    The distance we measure is not the Euclidean distance; it's the distance in a spacelike slice of constant cosmological time (i.e., ##dt = 0## in the metric you wrote down), given the metric of the spacelike slice. (Note that we don't directly "measure" this distance either; we calculate it based on other "distance" observations.) None of this has anything to do with how, or whether, you can embed that spacelike slice in a higher-dimensional Euclidean or Minkowskian space. The only requirement for homogeneity is that no point is "special" with reference to the intrinsic metric of the spacelike slice. The ##k = -1## hyperbolic space satisfies that requirement.
     
  4. Sep 30, 2014 #3
    I agree that the spatial distance measure is not euclidean; its Minkowskian (see Weinberg Gravitation p. 391) -- so the space is homogeneous only with respect to this metric. However, I would argue that we measure distances in the universe by the Euclidean distance measure, and thus, to us, I would not think that it appears homogeneous regarding the distribution of galaxies etc.
     
  5. Sep 30, 2014 #4
    So I agree that the space is homogeneous with regards to the intrinsic metric. However, as we do not use this metric (which has signature -1), my argument is that we do not __see it__ as being homogeneous?
     
  6. Sep 30, 2014 #5

    PeterDonis

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    Why do you think that? The "Euclidean distance measure" is a measure in a higher-dimensional embedding space that doesn't even exist (it's just a mathematical fiction used by some people for modeling).

    I'm confused. What "intrinsic" metric do you think we are not using? (If you think we are using the Euclidean metric, see above.)
     
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