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Solving expansion rate for a variant of the Friedmann equation

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data
    For the equation [tex] H^2 = \frac{8 \pi G \rho_m}{3} + \frac{H}{r_c} [/tex] how do I find the value of H for scale factor [itex] a \rightarrow \infty [/itex], and show that H acts as though dominated by [itex] \Lambda [/itex] (cosmological constant) ?

    2. Relevant equations
    [tex] \rho_m \propto \frac{1}{a^3} [/tex]
    [tex] H > 0 [/tex]

    3. The attempt at a solution
    I'm not sure how to show that H is driven by [itex] \Lambda [/itex], but have tried to sub in the scale factor in place of matter density and make the scale factor go to infinity.
    As in,
    [tex] H^2 = \frac{8 \pi G }{3 a^3} + \frac{H}{r_c} [/tex]
    This gets rid of the [itex] \frac{8 \pi G \rho_m}{3} [/itex] leaving [itex] H = \frac{1}{r_c} [/itex]
     
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  3. Oct 12, 2014 #2

    George Jones

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    What is ##r_c##? Is it constant?
     
  4. Oct 12, 2014 #3
    Yes [itex] r_c [/itex] is a constant which is called the cross over scale. I don't think we need to know the value of it
     
  5. Oct 12, 2014 #4

    George Jones

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    Okay. Now, solve

    $$H = \frac{1}{r_c}$$

    for the scale factor ##a##.
     
  6. Oct 12, 2014 #5

    George Jones

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    Alternatively, what is ##H## for a universe that has a non-zero cosmological constant ##\Lambda##, and that is otherwise empty, i.e., that has no matter or radiation content?
     
  7. Oct 12, 2014 #6
    ok, so [itex] H = \frac{1}{r_c} [/itex]
    [tex] H = \frac{1}{a}\frac{da}{dt} = \frac{1}{r_c} [/tex]
    [tex] \int_{0}^{a} \frac{1}{a} da = \int_{0}^{t} \frac{1}{r_c} dt [/tex]
    [tex] ln(a) = \frac{1}{r_c} t [/tex]
    [tex] a = \exp(\frac{t}{r_c}) [/tex]

    Which is akin to dark energy domination [itex] a \propto \exp(Hr_c) [/itex] ?
     
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