# Deceleration Parameter Computation

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~t1/2. 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/ρcrit in this case? I know that ρcrit=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.

Last edited:

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.

1 person
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~t1/2, 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.

1 person
George Jones
Staff Emeritus
Gold Member
A radiation-dominated universe has positive deceleration (negative acceleration).

Since for a radiation dominated universe we had the scale factor a~t1/2, 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.

1 person
Thank you very much. :)