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Friedmann Equation

  1. Jan 23, 2007 #1
    1. The problem statement, all variables and given/known data
    By substituting in

    [tex]\left( \frac{\dot{a}}{a_0} \right)^2 = H^2_0 \left(\Omega_0 \frac{a_0}{a} + 1 - \Omega_0 \right)[/tex]

    show that the parametric open solution given by

    [tex]a(\psi)=a_0 \frac{\Omega_0}{2(1-\Omega_0)}(\cosh{\psi} - 1)[/tex]

    and

    [tex]t(\psi)=\frac{1}{2H_0} \frac{\Omega_0}{(1 - \Omega_0)^{3/2}}(\sinh{\psi} - \psi)[/tex]

    solve the Friedmann equation.


    2. The attempt at a solution
    I get

    [tex]\dot{a} = a_0 \frac{\Omega_0}{2(1 - \Omega_0)}(\dot{\psi}\sinh{\psi})[/tex]

    and

    [tex]\dot{\psi}=\frac{2H_0(1-\Omega_0)^{3/2}}{\Omega_0(\cosh{\psi}-1)}[/tex]

    but I can't get to the first equality. Is this the correct approach?
     
  2. jcsd
  3. Jan 23, 2007 #2

    Physics Monkey

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    You may find it easier to write [tex] \dot{a} = \frac{\frac{da}{d\psi}}{\frac{dt}{d\psi}} [/tex]. Dividing by [tex] a_0 [/tex] and squaring should leave you with some expression involving only [tex] \psi [/tex]. A little algebra is then all you need to re-express things in terms of [tex] a [/tex].
     
  4. Jan 23, 2007 #3
    I get the right hand side to equal

    [tex]H^2_0 \frac{2(1-\Omega_0}{\cosh{\psi}-1}[/tex]

    and the LHS

    [tex]H^2_0 \frac{(1-\Omega_0)\sinh^2{\psi}}{(\cosh{\psi}-1)^2}[/tex]

    but I can't get them equal...
     
  5. Jan 23, 2007 #4

    Physics Monkey

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    I agree with your second expression from computing da/dt, but I think you've made a mistake with your first expression.
     
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