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FRW metric derivation Q, Ricci vector algebra

  1. Feb 18, 2015 #1
    I'm looking at: http://arxiv.org/pdf/gr-qc/9712019.pdf,

    deriving the FRW metric, and I don't fully understand how the Ricci Vectors eq 8.5 can be attained from 7.16, by setting ##\partial_{0} \beta ## and ##\alpha=0##

    I see that any christoffel symbol with a ##0## vanish and so so do any Riemann tensors with a ##0##, and so only Ricci vectors with ##1,2,3## indices will be non-zero

    However, I thought the metric used to compute the Ricci vectors in eq 7.16 - 7.13- would need to reduce to 8.4.
    So I see ##\beta(t,r) -> \beta(t) ##, but I thought also the ##dt^{2}## coefficient would also have to vanish?

    Thanks in advance.
     
  2. jcsd
  3. Feb 18, 2015 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    It does. But that portion of the metric is only the spatial portion, with the ##t## dependence already factored out. See below.

    I think you mean ##\beta(r)##, correct? In equation 8.4, ##\beta## is a function of ##r## only; the ##t## dependence was already factored out in equation 8.1. See below.

    Setting ##\alpha = 0## means the coefficient of ##dt^2##, which is ##- e^{2 \alpha}##, becomes ##-1##. That's what is shown in equation 8.1, which also factors out the ##t## dependence of the spatial part of the metric into the scale factor ##a(t)##. Equation 8.4 is then just an equation for what's left in the spatial part in 8.1, i.e., the function ##\gamma_{ij}##.
     
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