Cosmological constant deceleration parameter

[ChrisVer]

Homework Statement

Give q(t) the deceleration parameter, as a function of:
\Omega_{\Lambda},
the cosmological constant density,
and
\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}
where a's the scale factors

have already defined τ = H_{0}t 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.

Homework Equations

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

\frac{1}{\bar{a}} \frac{d \bar{a}}{dτ}= \frac{H}{H_{0}}

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

The Attempt at a Solution

I made the observation that q(t) is given by:

q(t) \propto \frac{1}{H^{2}} \frac{dH}{dt}

proof:
\frac{dH}{dt} = \frac{\ddot{a}}{a} - \frac{\dot{a}^{2}}{a^2}
so
\frac{1}{H^{2}} \frac{dH}{dt} = \frac{1}{H^{2}} (\frac{\ddot{a}}{a} - H^{2})
the first term is -q(t). The second term is 1...
\frac{1}{H^{2}} \frac{dH}{dt} = -q(t) -1

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

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

\frac{H^{2}}{H_{0}^{2}} = \Omega_{\Lambda} - \frac{k}{H_{0}^{2}a^{2}}

or:

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

I take its derivative wrt to t:

\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}}

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

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

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

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

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

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

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

\frac{k}{H_{0}^{2} a^{2}} = \frac{\Omega_{\Lambda} - 1}{\bar{a}^{2}}q(t)= -1 - \frac{\Omega_{\Lambda} - 1}{\bar{a}^{2}} \frac{1}{\Omega_{\Lambda} - \frac{\Omega_{\Lambda} - 1}{\bar{a}^{2}}}

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

Do you see any flaw?
Am I always possible to define a_{0}=1 in order to make it disappear?
PLEASE DON'T GIVE SOLUTION

 

[ChrisVer]

Corrected some mistakes^