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Explanation for Robertson-Walker being Conformally Flat

  1. Oct 8, 2012 #1
    I understand that in order for a space to be conformally flat that the vanishing of the Weyl tensor is a necessary condition. I was wondering if the following arguments are also (if not the same) in arguing that Robertson-Walker spacetime is conformally flat.

    Starting with the Robertson-Walker metric:


    Could it be argued that there exists coordinates [itex]x^{\gamma}[/itex] such that:


    where [itex]\eta_{\alpha\beta}[/itex] is flat. As such, the 4 dimensional space would possess signature of (+ or -) 2 and be describable as locally Minkowskian.
  2. jcsd
  3. Oct 10, 2012 #2


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