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Density in the friedmann equation

  1. Jun 1, 2010 #1
    I'm a little confused about the density [tex]\rho [/tex] in the equation:

    [tex] H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho - \frac{kc^2}{a^2} [/tex]

    Measuring [tex]\rho [/tex] at a single instant in time seems easy. But [tex]\rho [/tex] changes with time. The time dependence of [tex]\rho [/tex] is given as [tex]\rho=\frac{M}{a(t)^3} [/tex] where M is a constant. But to determine M from a measurment of [tex]\rho [/tex], doesn't one have to know a(t), which is what the equation is trying to find?
     
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  3. Jun 1, 2010 #2

    bcrowell

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    No, it's a second-order differential equation, so the general solution is going to have two adjustable constants in it that have to be adjusted to match the boundary conditions. Fo comparison, the differential equation [itex]\ddot{y}=-y[/itex] has solutions of the form [itex]y=A\cos(t+b)[/itex]. You don't need to know A and b in order to determine that that's the general form of the solution.
     
  4. Jun 1, 2010 #3

    George Jones

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    Suppose [itex]H_0[/itex] and [itex]\rho_0[/itex], the values of [itex]H[/itex] and [itex]\rho[/itex] at the present time, are measured. Evaluating the Friedmann equation at the present time gives [itex]a_0[/itex], the present value of the scale factor.

    [tex]\rho = \frac{M}{a^3},[/tex]

    then gives

    [tex]\rho = \rho_0 \frac{a_0^3}{a^3}.[/tex]
     
    Last edited: Jun 1, 2010
  5. Jun 1, 2010 #4
    Assume k=0. Then if you measure [tex]\rho_0[/tex], then don't you have [itex]H_0[/itex]? Or vice versa: if you measure [itex]H_0[/itex] don't you have [tex]\rho_0[/tex]?

    That's what seems to be implied by the equation:[tex]
    H^2 = \frac{8 \pi G}{3}\rho
    [/tex], so I'm still not sure how to get [tex]a_0[/tex]
     
  6. Jun 1, 2010 #5

    George Jones

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    This isn't the equation that you gave in your original post.
     
  7. Jun 1, 2010 #6
    I think he just said he assumes k to be zero.

    You have to be careful here: As bcrowell said, [tex]a_0[/tex] and [tex]\rho_0[/tex] are adjustable according to the conditions of material distribution. If you don't have a given [tex]a_0[/tex], then you cannot predict what probably the value of [tex]\rho_0[/tex] would be even when the universe is isotropically flat i.e. when k=0. This is simply because you still have those adjustable constants involved within the Friedmann equation so assuming here we must have a known [tex]H_0[/tex] due to reading it off the equation [tex] H^2 = \frac{8 \pi G}{3}\rho [/tex] immediately is correct but yet again the equation [tex] H = \left(\frac{\dot{a}}{a}\right)[/tex] is an ODE involving a constant of integration to be set by hand.

    AB
     
  8. Jun 1, 2010 #7

    George Jones

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    Yikes, I should be banned from making posts from home when my daughter is awake (like now). Sorry RedX, I neither read nor thought k = 0. I'll get back to this tomorrow.
     
  9. Jun 1, 2010 #8

    George Jones

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    Okay, assume matter only and suppose measurements of [itex]H_0[/itex] and [itex]\rho_0[/itex] satisfy

    [tex]H_0^2 = \frac{8 \pi G}{3} \rho_0,[/tex]

    which means that [itex]k = 0[/itex]. This, in turn, means that the scale factor is not pinned down. In spite of this, one can solve for the time evolution of [itex]\rho[/itex] and [itex]H[/itex]. Daughter is asleep, but wife wants to watch a mystery movie, so I'll only start the exercise.

    Write

    [tex]\rho = \rho_0 \left( \frac{a_0}{a} \right)^3.[/tex]

    Use this in the Friedmann equation, integrate the Friedmann equation, and find [itex]\rho \left( t \right)[/itex] and [itex]H \left( t \right)[/itex]. The value of [itex]a_0[/itex] is not needed to do this.
     
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