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Closed Universe - FRW Equation

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Show that the equations satisfy FRW equations.
    (b) Show the metric when ##\eta## is taken as time

    frw1.png
    2. Relevant equations


    3. The attempt at a solution


    The FRW equation is:
    [tex] 3 \left( \frac{\dot a}{a} \right)^2 = 8\pi G \rho [/tex]

    Using ##\frac{da}{dt} = \frac{da}{d\eta} \frac{d\eta}{dt}##:

    [tex] \dot a = \frac{\frac{1}{\eta_*} sin \left( \frac{\eta}{\eta_*}\right)}{1 - cos \left( \frac{\eta}{\eta_*} \right)} [/tex]
    [tex]\dot a = \frac{C sin \left( \frac{\eta}{\eta_*} \right)}{\eta_* a} [/tex]

    The LHS is then

    [tex] 3 \left( \frac{\dot a}{a} \right)^2 = 3 \left[ \frac{C^2 sin^2 (\frac{\eta}{\eta_*})}{\eta_*^2 a^4} \right] [/tex]


    Not sure how to show this equals RHS..
     
  2. jcsd
  3. Mar 1, 2015 #2

    Dick

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    In a closed universe the FRW equation has a spatial curvature term with a 'k' in it. You omitted that.
     
    Last edited: Mar 1, 2015
  4. Mar 2, 2015 #3
    FRW equation is given by:
    [tex]3 \left( \frac{\dot a}{a} \right)^2 + \frac{kc^2}{a^2} = 8\pi G \rho [/tex]

    Curvature parameter is given by ##\Omega = \frac{8\pi \rho G}{3H_0^2}##.

    [tex] \frac{C^2 sin^2 (\frac{\eta}{\eta_*})}{\eta_*^2 a^4} + \frac{kc^2}{a^2} = \frac{\Omega}{H_0^2} [/tex]

    At ##t=0##, ##\frac{\eta_0}{\eta_*} = sin (\frac{\eta_0}{\eta_*})##, so ##a_0 = C\left[ 1 - \sqrt{1 - (\frac{\eta_0}{\eta_*})^2} \right]##.

    Rearranging, ##(\frac{\eta_0}{\eta_*})^2 = (\frac{a_0}{c})(1 - \frac{a_0}{c})##.

    How am I to relate ##\eta## to ##\eta_0##?
     
    Last edited: Mar 2, 2015
  5. Mar 2, 2015 #4
    I'm just confused at this stage, shouldn't ##a_0 =1##? I have a feeling this problem is much simpler than it seems..Would appreciate some help
     
    Last edited: Mar 2, 2015
  6. Mar 2, 2015 #5
    5th Attempt

    [tex] \left( \frac{\dot a}{a}\right)^2 = \frac{8\pi G \rho}{3} - \frac{kc^2}{a^2} [/tex]

    Using ##\rho = \Omega \rho_c = \Omega \cdot \frac{3H^2}{8\pi G}##:

    [tex] \left( \frac{\dot a}{a}\right)^2 = \Omega H^2 - \frac{kc^2}{a^2} [/tex]

    [tex] \dot {a}^2 = \Omega \dot {a}^2 - kc^2 [/tex]

    [tex] \frac{da}{dt} = c \sqrt{\frac{k}{\Omega -1}} [/tex]

    But there is no factor of ##C## on the LHS, as ##\frac{da}{dt} = \frac{sin(\frac{\eta}{\eta_*})}{\eta_* \left[1-cos(\frac{\eta}{\eta_*})\right]}##
     
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