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Homework Help: Cosmological constant deceleration parameter

  1. Jul 7, 2014 #1


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

    Give [itex]q(t)[/itex] the deceleration parameter, as a function of:
    the cosmological constant density,
    [itex]\bar{a}(t) = \frac{a(t)}{a(t_{0})}= 1+ H_{0} (t-t_{0}) - \frac{1}{2} q_{0} H_{0}^{2} (t-t_{0})^{2} [/itex]
    where [itex]a[/itex]'s the scale factors

    have already defined [itex] τ = H_{0}t [/itex] time parameter, and showed:

    PLEASE DON'T GIVE SOLUTION (I will repeat it at the end)

    I just want to reconfirm my result at first stage.

    2. Relevant equations

    [itex]q(t)= -\frac{1}{H^{2}} \frac{\ddot{a}}{a} [/itex]

    [itex] \frac{1}{\bar{a}} \frac{d \bar{a}}{dτ}= \frac{H}{H_{0}}[/itex]

    [itex] H^{2} + \frac{k}{a^{2}} = \frac{ \rho_{\Lambda}}{3M_{Pl}^{2}} [/itex]

    3. The attempt at a solution

    I made the observation that [itex]q(t)[/itex] is given by:

    [itex] q(t) \propto \frac{1}{H^{2}} \frac{dH}{dt} [/itex]

    [itex] \frac{dH}{dt} = \frac{\ddot{a}}{a} - \frac{\dot{a}^{2}}{a^2} [/itex]
    [itex] \frac{1}{H^{2}} \frac{dH}{dt} = \frac{1}{H^{2}} (\frac{\ddot{a}}{a} - H^{2})[/itex]
    the first term is [itex]-q(t)[/itex]. The second term is 1...
    [itex] \frac{1}{H^{2}} \frac{dH}{dt} = -q(t) -1 [/itex]

    [itex] q(t)= -1 - \frac{1}{H^{2}} \frac{dH}{dt} [/itex]

    So far I think I didn't lose any step.... Then I take Friedman equations, and have:

    [itex] \frac{H^{2}}{H_{0}^{2}} = \Omega_{\Lambda} - \frac{k}{H_{0}^{2}a^{2}} [/itex]


    [itex] H=H_{0} \sqrt{ \Omega_{\Lambda} - \frac{k}{H_{0}^{2}a^{2}}} [/itex]

    I take its derivative wrt to t:

    [itex]\dot{H} = \frac{1}{2} \frac{H_{0}}{\sqrt{ \Omega_{\Lambda} - \frac{k}{H_{0}^{2}a^{2}}}} \frac{2k \dot{a}}{H_{0}^{2} a^{3}} [/itex]

    [itex]\dot{H} = \frac{ H k}{H_{0} a^{2} \sqrt{ \Omega_{\Lambda} - \frac{k}{H_{0}^{2}a^{2}}}}[/itex]

    I insert this to the equation I got for [itex]q(t)[/itex], one H is going to be canceled:

    [itex] q(t)= -1 - \frac{1}{H^{2}} \frac{dH}{dt} [/itex]

    [itex] q(t)= -1 - \frac{ k}{ H_{0} a^{2} H \sqrt{ \Omega_{\Lambda} - \frac{k}{H_{0}^{2}a^{2}}}}[/itex]

    and using again that [itex]H[/itex] is the same square root multiplied by [itex]H_{0}[/itex]:

    [itex] q(t)= -1 - \frac{k}{H_{0}^{2} a^{2} (\Omega_{\Lambda} - \frac{k}{H_{0}^{2}a^{2}})}[/itex]

    Now I can generally determine [itex]k[/itex] from taking the Friedman equation today.
    [itex] \frac{k}{H_{0}^{2} a_{0}^{2}} = \Omega_{\Lambda} - 1 [/itex]

    [itex] \frac{k}{H_{0}^{2} a^{2}} = \frac{\Omega_{\Lambda} - 1}{\bar{a}^{2}} [/itex]

    [itex] q(t)= -1 - \frac{\Omega_{\Lambda} - 1}{\bar{a}^{2}} \frac{1}{\Omega_{\Lambda} - \frac{\Omega_{\Lambda} - 1}{\bar{a}^{2}}} [/itex]

    [itex] q(t)= -1 + \frac{1- \Omega_{\Lambda}}{(\bar{a}^{2}-1) \Omega_{\Lambda} +1} [/itex]

    Do you see any flaw?
    Am I always possible to define [itex]a_{0}=1[/itex] in order to make it disappear?
    Last edited: Jul 7, 2014
  2. jcsd
  3. Jul 7, 2014 #2


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    Corrected some mistakes^
    Last edited: Jul 7, 2014
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