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Struggling with Friedmann equations

  1. Feb 19, 2015 #1

    I'm struggling to derive the accelerated Friedmann equation (shame on me!)...

    I'll tell you what I'm doing and maybe you can find where the mistake is.

    First of all, we know that:

    [tex] H^2 = \frac{\rho}{2M^2_p}+\frac{\Lambda}{3}[/tex]

    Now, using the Einstein Equations and doing the trace of these, we get:
    [tex]-R-4\Lambda = \frac{T}{M^2_p} = \frac{-\rho+3p}{M^2_p}[/tex]
    Where [itex]R=6[ \frac{\ddot{a}}{a}+H^2][/itex].
    Using this, I get:
    [tex] \frac{\ddot{a}}{a} = -\Lambda - \frac{1}{6}\frac{\rho+3p}{M^2_p}[/tex]

    The problem is that I shouldn't get a minus sing in the [itex] \Lambda [/itex] and divided by 3, but I just CAN'T see what I'm doing wrong, as you can see here:
    I've tried to look for a book that does this derivation but I haven't found any...

    If you could help me I would really appreciate it!
  2. jcsd
  3. Feb 19, 2015 #2


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    Why do you have the trace of the [itex]\Lambda[/itex] as [itex]-4\Lambda[/itex]? I don't think that's right.
  4. Feb 19, 2015 #3

    George Jones

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    The trace of the ##\Lambda g_{\mu \nu}## term is ##+4 \Lambda##, which then gives a ##+ \Lambda/3## in the expression for ##\ddot{a}/a##.
  5. Feb 20, 2015 #4
    Oh my god, thank you, I was using
    [tex] R_{\mu \nu} -\frac{1}{2}R g_{\mu \nu} -\Lambda g_{\mu \nu} = 8\pi G T_{\mu \nu}[/tex]
    Because that's what the problem said, but I see now that it is a typo in the sign... I'm going to kill my professor he has done 3 typos in this problem already T_T

    Thanks a lot people!
  6. Feb 20, 2015 #5


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    Please don't kill your professor
  7. Feb 20, 2015 #6


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    Are you sure about your result? It doesn't seem like only a matter of sign, but also a matter of your factors (like 1/6 in front of the density/momentum)

    I find it easier to take the FE and take the time derivative of it. Doing so (and using the fluid continuity equation) the answer would be like:

    [itex] \frac{\ddot{a}}{a} = \frac{\Lambda}{3} - \frac{1}{2} \frac{\rho +3p}{2M_{Pl}^2}[/itex]

    This coincides also with the bibliography of [itex] 8 \pi G /3 [/itex] since in your case you have written that as [itex] \frac{1}{2 M_{Pl}^2}[/itex]. If you prefer ##G##, it would be:

    [itex] \frac{\ddot{a}}{a} = \frac{\Lambda}{3} - \frac{4 \pi G }{3} (\rho +3p)[/itex]
    Last edited: Feb 20, 2015
  8. Feb 21, 2015 #7
    Just kidding ;)

    As I said, it was a problem of the sign and also the factor 3.
    I've already solved my problem, thanks a lot!! ^^
  9. Feb 21, 2015 #8


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    There is an ancient tradition for murder that inflicts all the victims responsibilities - including servicing their spouse.
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