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R=6k or R=3k confused, FRW metric derivation, Maximally symm

  1. Apr 5, 2015 #1
    I'm looking at Tod and Hughston Introduction to GR and writing the metric in the two forms;

    [1]##ds^{2}=dt^{2}-R^{2}(t)(\frac{dr^{2}}{1-kr^{2}}+r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}))##

    [2] ##ds^{2}=dt^{2}-R^{2}(t)g_{ij}dx^{i}dx^{j}##

    where

    ##g_{ij}dx^{i}dx^{j}=d\chi^{2}+\chi^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}) ## for ##k=0##
    ##=d\chi^{2}+sin^{2}\chi{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}) ## for ##k=1##
    ##=d\chi^{2}+sinh^{2}\chi{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}) ## for ##k=-1##

    Now in solving for the form [2] Tod computes the Ricci scalar of ##ds^{2}=d\chi^{2}+f^{2}(\chi)(d\theta^{2}+sin^{2}\theta d\phi^{2})## and finds ##R=-(2\frac{f''}{f}-\frac{1}{f^{2}}+\frac{(f')^{2}}{f^{2}}## then integrates, uses ##R=3k## and solves for all 3 cases ##k=0,\pm 1##.

    My question

    ##R=3k## doesn't seem right to me, since in 3-d space we can write ##R_{ab}=2kg_{ab}##. Of course you could just define a constant ##K=2k##, but it uses the constant ##k## in the FRW metric of the form [1] not ##k##, comparing to Introduction to GR lecture notes by sean M.Caroll,I thought that this should be ##R=6k##

    ...In Caroll's notes he uses ##R_{ab}=2kg_{ab}## in the derivation and gives the FRW metric the same as in form [1] with small ##k##. So it doesn't look as though Tod has used ##K=2k##.
    Can anyone help explain how Tod uses ##R=3k##?

    Thanks.
     
  2. jcsd
  3. Apr 5, 2015 #2

    PeterDonis

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    Staff: Mentor

    That's what I get when I compute the Ricci scalar for the spatial part of [1] (i.e., the part inside the parentheses after ##R^2(t)## ).

    When I compute the Ricci scalar for this, I get twice the answer you are giving from Tod. So it looks to me like Tod is dividing by 2 somewhere in order to equate his answer for [2] with R = 3k for [1].
     
  4. Apr 6, 2015 #3
    Thanks for your reply. I can't see at all where Tod is dividing by 2 though.
    Could the answer possibly lie in the fact that simulatenously rescaling:
    ##k \to k/| k/##
    ##r \to \sqrt{| k|} r ##
    ##a \to a/\sqrt{|k|} ##
    The FRW metric is unchanged?
    If it could however, I'm unsure how to justify explicitly.
    Thanks.
     
  5. Apr 6, 2015 #4

    PeterDonis

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    Staff: Mentor

    I don't see how that would account for the factor of 2. I don't have Tod and Hughston, so I can't say where the discrepancy arises.
     
  6. Apr 20, 2015 #5
    I don't suppose anyone else has any ideas on where this extra factor of 2 is coming from?
    Is it possible that ##\chi## could have been re-scaled such that solving the case ##R=3k## is the same as solving for ##R=6k##.
     
    Last edited: Apr 20, 2015
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