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Fluid Equation (Cosmology)

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    The fluid equation in cosmology is given as:

    [tex]\dot{\epsilon}[/tex] + 3*([tex]\dot{a}[/tex]/a)*([tex]\epsilon[/tex]+P) = 0

    Where [tex]\epsilon[/tex] is the energy density and a(t) is a scale factor.

    Using the equation of state, P = w*[tex]\epsilon[/tex], show how [tex]\epsilon[/tex] change with a(t).


    2. Relevant equations

    [tex]\dot{\epsilon}[/tex] + 3*([tex]\dot{a}[/tex]/a)*([tex]\epsilon[/tex]+P) = 0
    P = w*[tex]\epsilon[/tex]


    3. The attempt at a solution

    I can solve for the equation to the point where I re-arrange it to look like this:

    [tex]\dot{\epsilon}[/tex]/[tex]\epsilon[/tex] = -3*(1+w)*([tex]\dot{a}[/tex]/a)

    I do not know how to proceed from here. I know that this equation is supposed to end up like this,

    [tex]\epsilonw(a)[/tex] = [tex]\epsilonw,0[/tex]*a-3*(1+w)

    but I do not know how to get to this point. Can someone assist me please?
     
  2. jcsd
  3. Feb 9, 2010 #2
    can someone take a look at this? i'm pretty sure its a simple operation that i'm failing to realize.
     
  4. Feb 9, 2010 #3
    bump?
     
  5. Feb 10, 2010 #4
    You have your equation
    [tex] \frac{\dot\epsilon}{\epsilon} = -3(w+1)\frac{\dot a}{a} [/tex]
    From here you can eliminate the time-dependence
    [tex] \frac{d\epsilon}{\epsilon} = -3(w+1)\frac{da}{a} [/tex]
    and this is a differential equation involving just [tex]\epsilon[/tex] and [tex] a[/tex] you can solve by integrating both sides
     
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