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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:

    [itex]ds^{2}[/itex]=[itex]a^{2}(t)\left(\frac{dr^{2}}{1-Kr^{2}}+r^{2}(d\theta^{2}+sin(\theta)^{2}d\phi^{2})\right)-dt^{2}[/itex]

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

    [itex]ds^{2}[/itex]=[itex]f(x^{\gamma})\eta_{\alpha\beta}dx^{\alpha}dx^{β}[/itex]

    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

    Bill_K

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