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Cosmology - Analytic solution to Friedmann Equation in a matter-radiation Universe

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Rewrite Friedmann equation using conformal time and density parameters [itex]\Omega_m[/itex] and [itex]\Omega_r[/itex]. Is there a relation between the two? How many parameters define the problem?


    2. Relevant equations
    Friedmann equation
    [tex]\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3c^2}\left(\frac{\epsilon_{0m}}{a^3}+\frac{ε_{0r}}{a^4}\right)[/tex]

    Conformal time definition
    [tex]dt=a(\eta) d\eta[/tex]

    Density parameter:
    [tex]\Omega\equiv \frac{ε(t_0)}{ε_c(t_0)}=\frac{3c^2}{8\pi G}H_0^2[/tex]

    3. The attempt at a solution
    First part is rather simple: just a matter of changing the variable in Friedmann Equation, noting that:

    [tex]\frac{d}{dt}=\frac{d\eta}{dt}\frac{d}{d\eta}=\frac{1}{a}\frac{d}{d\eta}[/tex]
    so Friedmann Equations turn out to be:
    [tex]\left(\frac{da}{d\eta}\right)^2=\frac{8\pi G}{3c^2}(aε_{m0}+ε_{r0})[/tex]
    or in terms of the density parameter:
    [tex]\left(\frac{da}{d\eta}\right)^2=H_0^2(a\Omega_m+Ω_r).[/tex]

    My trouble starts now. So, normally the densities parameters are constrained due to scale factor normalization, that is to say, they must obey the constrain equation [itex]1=\Omega_m+\Omega_r[/itex]. That relation should hold regardless of the coordinate system I choose to write Friedmann equation, but I cannot see the connection, unless I postulate that there must exist a [itex]\eta_0[/itex] such that [itex]a(\eta_0)=1[/itex], and later find its relation to physical time (and I can only find that once I know [itex]a(\eta)[/itex], i.e, when I solve Friedmann equation). But that does not seem to me as a correct assumption, for the next exercise on my list ask me to make precisely this assumption, indicating that there must exist another way of constraining the parameters. I think, thus, that there must exist a constrain between the density parameters that does not involve normalization of scale factor, but I can't think about anything to solve that. Can anyone help me out?

    Thank you!
     
  2. jcsd
  3. Feb 3, 2013 #2

    fzero

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    Re: Cosmology - Analytic solution to Friedmann Equation in a matter-radiation Univers

    You're already assuming that [itex]a_0=1[/itex] when you write the Friedmann equation this way. The equation is really

    [tex]\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3c^2}\epsilon,[/tex]

    where [itex]\epsilon[/itex] is the energy density (usually people use [itex]\rho[/itex], but I will keep your notation). For a specific equation of state [itex]p=w \epsilon[/itex], we'll find

    [tex] \frac{\epsilon}{\epsilon_0} = \left( \frac{a}{a_0} \right)^{-3(1+w)}.[/tex]

    Obviously the factors of [itex]a_0[/itex] have been set to one in your version of the formula.

    You can solve the problem by defining a density parameter [itex]\Omega(t) = \epsilon(t)/\epsilon_c(t)[/itex] instead of the way that you've defined it (only at [itex]t=t_0[/itex]). Rewrite the Friedmann equation in terms of these and you'll get what you were expecting without extra scale factors.
     
  4. Feb 3, 2013 #3
    Re: Cosmology - Analytic solution to Friedmann Equation in a matter-radiation Univers

    First, thank you by your time!

    Well, I used the equation that the problem statement have gave me. I agree with you that this assumption is implicit in the equation, but then can I take for granted that the densities will add up to one due to this assumption?
     
  5. Feb 3, 2013 #4

    fzero

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    Re: Cosmology - Analytic solution to Friedmann Equation in a matter-radiation Univers

    The density parameters will add up to one because the Friedmann equation you've written is for a flat universe. This is really the definition of "critical" in the critical density. It is the value of the total density such that the universe is flat.
     
  6. Feb 4, 2013 #5
    Re: Cosmology - Analytic solution to Friedmann Equation in a matter-radiation Univers

    Ok, thank you very much. I think I've done it :)
     
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