# Homework Help: Cosmological constant deceleration parameter

1. Jul 7, 2014

### ChrisVer

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

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.

2. Relevant equations

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

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

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

3. 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?