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Robertson-Walker Models Conformally Flat

  1. Oct 3, 2012 #1
    As an understanding exercise in my general relativity course my professor recommended proving to ourselves that all Robertson-Walker models are conformally flat. However, I am unsure of how to approach such a proof. Thanks in advance for any help.
     
  2. jcsd
  3. Oct 3, 2012 #2

    bcrowell

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    By homogeneity, tidal forces vanish, i.e., the Weyl tensor is zero. Since the Weyl tensor is zero, the spacetime is conformally flat.
     
  4. Oct 5, 2012 #3

    Bill_K

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    Is it homogeneity or isotropy? The Robertson-Walker metric is both. But you can have a cosmology with different expansion rates in different directions:

    ds2 = dt2 - a(t)dx2 - b(t)dy2 - c(t)dz2

    which is homogeneous but not isotropic. I'd guess the Weyl tensor for this type of cosmology is nonzero.
     
  5. Oct 5, 2012 #4

    bcrowell

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    Yep, thanks for the correction.
     
  6. Oct 5, 2012 #5
    I don't get this. If the expansion rates are anisotropic then it would seem that the space could only be homogeneous at a particular moment in time. From that point becoming inhomogeneous from then on.
    ???
     
  7. Oct 6, 2012 #6

    Bill_K

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    Each hypersurface t = const in the spacetime is a flat 3-space, hence it is always homogeneous. But the expansion rate is different in different directions. This is an example of a Bianchi cosmology, of which there are nine types, each with a different symmetry. See for example Ken Jacobs' thesis.
     
  8. Oct 7, 2012 #7
    I'm doubtful that a non-isotropic cosmology can be made flat thru a conformal transformation. How do you preserve the angles when going from anisotropy to
    isotropy?

    Edit:eek:ops, I realize this seems to be precisely your point.
     
    Last edited: Oct 7, 2012
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