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Friedmann equations, Einstein equation with cosmo constant?

  1. Mar 5, 2015 #1
    Hi,

    I'm looking at 'Lecture Notes on General Relativity' by Sean M.Carroll.
    I have a question about p. 227, solving for ##a(t)## in the dark energy arrow-10x10.png case.
    So for dust and radiation cases it was Friedmann equations you solve.
    But in the case of a non-zero cosmological constant Eienstien equation, and consequently the Friedmann equations which are derived from Einstein's equation must differ.

    Below I have on the top the steps to deriving Friedmann for Einstein equation without the cosmological constant, and on the steps I believe with a non-zero cosmological constant.
    Please could someone let me know if this is correct? and my modified Friedmann equations are correct and what Carroll is referring to when solving here is this ?

    Thanks very much in advance:

    Zero Cosmological Constant:

    Einsteins equation:

    ##8\pi G ( T_{ab}-\frac{Tg_{ab}}{2} ) = R_{ab}##

    ##R_{00}: \frac{-3\ddot{a}}{a}=4\pi G( \rho + 3p) ##

    ##R_{ij}: \frac{\ddot{a}}{a}+2\frac{\dot{a}}{a}^{2}+\frac{2k}{a^{2}}=4\pi G( \rho-p)##

    Making ##\frac{\ddot{a}}{a} ## the subject in the 1st equation and plugging into the 2nd equation yields:

    ##\frac{\dot{a}}{a}^{2}=\frac{8\pi G}{3}(\rho)-\frac{k}{a^{2}}##

    Non-zero cosmological constant:

    Einsteins equation:

    ##8\pi G ( T_{ab}-\frac{Tg_{ab}}{2}+\Lambda g_{ab} ) =R_{ab}##

    ##R_{00}: \frac{-3\ddot{a}}{a}=4\pi G( \rho + 3p)-\Lambda ##

    ##R_{ij}: \frac{\ddot{a}}{a}+2\frac{\dot{a}}{a}^{2}+\frac{2k}{a^{2}}=4\pi G( \rho-p)+\Lambda ##

    Making ##\frac{\ddot{a}}{a} ## the subject in the 1st equation and plugging into the 2nd equation yields:

    ##\frac{\dot{a}}{a}^{2}=\frac{8\pi G}{3}(\rho)-\frac{k}{a^{2}}+\Lambda a - \frac{\Lambda}{3} ##
     
    Last edited by a moderator: Mar 5, 2015
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  3. Mar 5, 2015 #2

    PeterDonis

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    Your zero cosmological constant equation matches Carroll's equation 8.36, so that looks ok.

    Your nonzero cosmological constant equations start with ##\Lambda## in the wrong place; the correct initial equation (Einstein Field Equation) is:

    $$
    8 \pi G ( T_{ab} - \frac{1}{2} g_{ab} T ) = R_{ab} + \Lambda g_{ab}
    $$

    You should end up with the equation given here:

    http://en.wikipedia.org/wiki/Friedmann_equations#Equations

    (You're using units where ##c = 1##, so just leave the ##c##'s out.)
     
  4. Mar 5, 2015 #3
    ahhh thanks.
    I am now getting the sign of ##\Lambda## in ##R_{00}## and## R_{ij} ##the other way around . But then when I solve for Freidmann equation, I am getting ##\frac{-1\Lambda}{3}## as a pose to positive on the wiki page.
     
  5. Mar 5, 2015 #4

    PeterDonis

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    You'll need to show the steps of how you're deriving it.
     
  6. Mar 5, 2015 #5

    PeterDonis

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    Looking at your derivation again, I'm not sure it's right, even though the answer you wrote down is right. The usual way of deriving the second Friedmann equation is to take the trace of the EFE, not the ##ij## component. That means you should be looking at contracting both sides with ##g^{ab}##.
     
  7. Mar 5, 2015 #6
    ##\frac{\ddot{a}}{a}=\frac{-1}{3}4\pi G ( \rho + 3p) -\frac{\Lambda}{3} ##
    into the second:
    ##\frac{-1}{3}4\pi G (\rho +3p) - \frac{\Lambda}{3} + 2\frac({\dot{a}}{a})^{2}+\frac{2k}{a^{2}}=4\pi G(\rho-p)-\Lambda##
    so looking at ##\Lambda..##
    ##\frac{\dot{a^{2}}}{a}+...=\frac{1}{2}(-\Lambda + \frac{\Lambda}{3})=\frac{-\Lambda}{3}##

    \
     
  8. Mar 5, 2015 #7

    PeterDonis

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    Which I do not think is correct after reading through your OP again. See post #5. You need to take the trace of the EFE, not the ##ij## component.
     
  9. Mar 5, 2015 #8

    PeterDonis

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    Briefly, the second Friedmann equation comes from taking the trace of

    $$
    8 \pi G ( T_{ab} - \frac{1}{2} g_{ab} T ) = R_{ab} + \Lambda g_{ab}
    $$

    Taking the trace just means contracting with ##g^{ab}##, which gives

    $$
    - 8 \pi G T = R + 4 \Lambda
    $$

    Then you can proceed to substitute for ##R## (which is derived from the FRW metric) and ##T##, and go from there.
     
  10. Mar 15, 2015 #9
    Okay thanks, I've done this and I get an equation that looks like ##(a,b)=(i,j)## einstein equation with different coefficients on the lhs (as in the op) terms, so no ##2's## but all multiplied by ##3##. I'm unsure on what to do next?

    (Also carroll's lecture notes , page 223, seems to use the eliminating the second derivative method).
     
  11. Mar 15, 2015 #10

    PeterDonis

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    Can you show the expressions you substituted for ##R## and ##T## in the equation I posted? And how you got from there to the equation you're stuck at?
     
  12. Mar 16, 2015 #11
    Okay so Einstein equation with the zero cosmological constant, taking the trace gave ##-8\pi GT=R##, ##T=-\rho +3p##, ##R=\frac{6}{a^{2}}(a\ddot{a}+\dot{a^{2}}+k)##,
    gives ## -4\pi G(-\rho+3p)=\frac{3}{a^{2}}(a\ddot{a}+\dot{a^{2}}+k)##.

    I'm unsure what to do now, I know I need to eliminate the double derivaitive on ##a## but I don't know which is the 'correct' equation to use.
     
  13. Mar 23, 2015 #12
    Sorry anyone help on this, I'm still strugging, thanks.
     
  14. Mar 23, 2015 #13

    PeterDonis

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    Why? You're trying to derive the second Friedmann equation, which is an equation for ##\ddot{a} / a##. You just need to substitute the first Friedmann equation (which you got from the 00 component of the EFE) in what you have; this gives you a term in ##\rho## on the RHS in place of the terms in ##\dot{a}## and ##k##. Then you just rearrange the terms.
     
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