# Deceleration parameter and critical densities

1. May 24, 2012

### dingo_d

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

Using Friedmann equations find the parameter of deceleration $q_0$ using relative densities $\Omega_M,\ \Omega_R,\ \Omega_\Lambda$.

2. Relevant equations

Friedmann equations, deceleration parameter:

$q_0=-\frac{\ddot{a_0}a_0}{\dot{a_0}^2}$

3. The attempt at a solution

So from second Friedmann equation (the one with k parameter which tells us whether the Universe is open, flat or close) and got to this point:

$$\left(\frac{\dot{a}}{\dot{a_0}}\right)^2 = \left(\frac{a_0}{a}\right)^2\Omega_R+\left(\frac{a_0}{a}\right)\Omega_M+\Omega_K+\left(\frac{a}{a_0}\right)^2\Omega_\Lambda$$

a is the scale factor (sometimes denoted as R in the books).

And I'm stuck. The book here

equation (11.55)

Says that I should get

$q=\frac{\Omega_M}{2}+\Omega_R-\Omega_\Lambda$

I tried using first Friedmann equation, but then my equation depends on parameter w, which connects pressure and density: $p=w\rho$, and I don't get correct answer :\

I found one presentation which says that I should somehow (no explanation, of course -.-") connect the equation I got with the deceleration parameter using: "a bit of math".

I tried connecting the second derivative of the scale factor in the definition of deceleration parameter, so that I don't need to use the equation which has w in it, but I really have no idea what to do.

Any help would be appreciated...

Last edited by a moderator: May 24, 2012
2. May 24, 2012

### George Jones

Staff Emeritus
From this, I get the correct result.

What do you get when you differentiate with respect to time the equation in "The Attempt at a solution"?

3. May 25, 2012

### dingo_d

You mean I should differentiate the expression with omegas? But that would give me really complicated expression on the right, wouldn't it?

$\frac{2 \dot{a} \ddot{a}}{\dot{a_0}^2}-\frac {2 \dot{a}^2 \ddot{a_0}}{\dot{a_0}^3}$

Hmmm...

If the a_0 is scale factor at the present time, than it's constant, no? That means that it doesn't depend on time, and I just derive the numerator, right? And if, after that I put that I am looking at t=t_0 than I get deceleration parameter.

Am I going in the right direction?

Last edited: May 25, 2012
4. May 25, 2012

### George Jones

Staff Emeritus
When differentiating, $\dot{a}_0 = \dot{a} \left( t_0 \right)$ and $a_0 = a \left( t_0 \right)$ are constants.

Last edited: May 25, 2012
5. May 25, 2012

### dingo_d

Oh that simplifies things :D Thanks, I dk why I thought it depended on time :\

6. May 26, 2012

### dingo_d

So when I do the derivation and when I simplify things a bit, I need to presume that I'm looking at the t=t_0, so that my scale factors would all cancel out, right?

7. May 26, 2012

### George Jones

Staff Emeritus
Yes.

8. May 27, 2012

### dingo_d

I got the correct answer :) Thanks for all the help ^^ thumbs up :D