Deceleration Parameter Computation

[roam]

Hi everyone,

Can anyone point me into the right direction on how to compute the deceleration parameter ($q_0$) for a radiation dominated universe?

I know that q₀ is given by

$q_0=-\frac{\ddot{a}(t_0)}{a(t_0) H^2 (t_0)} = - \frac{\ddot{a}(t_0) a(t_0)}{\dot{a}^2(t_0)}=\frac{\Omega_0}{2}$​

So for a radiation dominated universe, we have $\omega = + \frac{1}{3}$. And ρ~1/a⁴, and a~t^1/2. I'm not sure how to relate this to that equation.

My notes say $\Omega _{rad} = 8.2 \times 10^-5$ but I'm not sure how this was calculated. How do I compute Ω₀=ρ/ρ_crit in this case? I know that ρ_crit=3H²/8πG, but what is the density ρ?

Using that value I get the following but I'm not sure if it's correct:

$q_0 = \frac{\Omega_0}{2} = \frac{8.2 \times 10^-5}{2} = 4.1 \times 10^{-5}$.

I couldn't find any info on this computation online. So any help is greatly appreciated.

 

[bapowell]

You have a(t), and since H(t) = \dot{a}(t)/a(t) and q(t) is fully determined by a(t), you should be good to go...

Your value for \Omega_{rad} is the current radiation density in the actual universe, which isn't radiation dominated and not appropriate for use with w=1/3, etc.

 

[roam]

Thank you for the clarification regarding $\Omega$, it makes perfect sense now.

But I'm still not sure how the calculation goes (my book lacks any worked problems)

Since for a radiation dominated universe we had the scale factor a~t^1/2, with its time derivatives $\dot{a}=\frac{1}{2 t^{1/2}}$, and $\ddot{a}=\frac{-1}{4t^{3/2}}$, then

$q_0 = \frac{\ddot{a}}{a \left( \dot{a}/a \right)} = \frac{-1}{16t^{6/2}}$

Is this correct?

 

[Chalnoth]

A radiation-dominated universe is one where \Omega_{rad} is very close to one. I haven't checked your math at all, but hopefully that helps you get the right answer.

 

[George Jones]

A radiation-dominated universe has positive deceleration (negative acceleration).

Since for a radiation dominated universe we had the scale factor a~t^1/2, with its time derivatives $\dot{a}=\frac{1}{2 t^{1/2}}$, and $\ddot{a}=\frac{-1}{4t^{3/2}}$, then

$q_0 = \frac{\ddot{a}}{a \left( \dot{a}/a \right)} = \frac{-1}{16t^{6/2}}$

Is this correct?

You forgot the initial minus sign.

 

[roam]

Thank you very much. :)