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Homework Help: Deceleration parameter and critical densities

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    Friedmann equations, deceleration parameter:


    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:

    [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


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

    George Jones

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    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"?
  4. May 25, 2012 #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}
    {2 \dot{a}^2


    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
  5. May 25, 2012 #4

    George Jones

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    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.
    Last edited: May 25, 2012
  6. May 25, 2012 #5
    Oh that simplifies things :D Thanks, I dk why I thought it depended on time :\
  7. May 26, 2012 #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?
  8. May 26, 2012 #7

    George Jones

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  9. May 27, 2012 #8
    I got the correct answer :) Thanks for all the help ^^ thumbs up :D
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