Deceleration parameter and critical densities

In summary, the conversation discusses using the Friedmann equations to find the parameter of deceleration, q_0, by using relative densities \Omega_M, \Omega_R, and \Omega_\Lambda. The first Friedmann equation with k parameter and the second Friedmann equation were used to reach a certain point in the solution. It was then suggested to differentiate the equation in "The Attempt at a Solution" with respect to time to find the correct result. It was clarified that a_0 is the scale factor at the present time and is therefore a constant. By simplifying the derivative and presuming that t=t_0, the scale factors cancel out and the correct answer for q_0 is obtained.
  • #1
dingo_d
211
0

Homework Statement



Using Friedmann equations find the parameter of deceleration [itex]q_0[/itex] using relative densities [itex]\Omega_M,\ \Omega_R,\ \Omega_\Lambda[/itex].

Homework Equations



Friedmann equations, deceleration parameter:

[itex]q_0=-\frac{\ddot{a_0}a_0}{\dot{a_0}^2}[/itex]

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:

[tex]\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[/tex]

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

[itex]q=\frac{\Omega_M}{2}+\Omega_R-\Omega_\Lambda[/itex]

I tried using first Friedmann equation, but then my equation depends on parameter w, which connects pressure and density: [itex]p=w\rho[/itex], 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...
 
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  • #2
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
You mean I should differentiate the expression with omegas? But that would give me really complicated expression on the right, wouldn't it?

[itex]\frac{2 \dot{a}
\ddot{a}}{\dot{a_0}^2}-\frac
{2 \dot{a}^2
\ddot{a_0}}{\dot{a_0}^3}[/itex]

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?
 
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  • #4
When differentiating, [itex]\dot{a}_0 = \dot{a} \left( t_0 \right)[/itex] and [itex]a_0 = a \left( t_0 \right)[/itex] are constants.
 
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  • #5
Oh that simplifies things :D Thanks, I dk why I thought it depended on time :\
 
  • #6
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
dingo_d said:
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?

Yes.
 
  • #8
I got the correct answer :) Thanks for all the help ^^ thumbs up :D
 

1. What is the deceleration parameter?

The deceleration parameter, denoted as q, is a measure of the rate at which the expansion of the universe is slowing down. It is defined as the negative of the ratio of the acceleration of the universe to the square of its current expansion rate.

2. How is the deceleration parameter related to the critical density?

The deceleration parameter is directly related to the critical density of the universe. If the deceleration parameter is less than 1, the universe's density is below the critical density, indicating that the universe's expansion is accelerating. If the deceleration parameter is greater than 1, the universe's density is above the critical density, indicating that the universe's expansion is decelerating. If the deceleration parameter is equal to 1, the universe's density is equal to the critical density, indicating that the universe's expansion is neither accelerating nor decelerating.

3. What is the critical density?

The critical density, denoted as ρc, is the density at which the universe is considered flat. It is the minimum amount of matter and energy required for the universe to eventually stop expanding and collapse in on itself. It is calculated by dividing the square of the speed of light by the gravitational constant.

4. How is the deceleration parameter used to study the expansion of the universe?

The deceleration parameter is a key parameter in cosmology that is used to understand the evolution and fate of the universe. By measuring the deceleration parameter, scientists can determine whether the universe is expanding at an accelerating or decelerating rate, which provides insights into the amount and nature of matter and energy in the universe.

5. What is the current value of the deceleration parameter and critical density?

The current value of the deceleration parameter is estimated to be around -0.55, indicating that the expansion of the universe is accelerating. The current value of the critical density is estimated to be around 9.47 x 10^-27 kg/m^3, which is equivalent to about 5.9 protons per cubic meter. However, these values are subject to ongoing research and refinement.

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