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Deriving a scale factor expression for FLRW model

  1. Dec 1, 2011 #1
    Hello, I am reading this paper about quantum gravity, trying and failing to follow along in the derivation of this, eq (204) p. 72, expression for the scale factor:
    [itex]
    a(t) = a_1 \left( 1 + \frac 3 2 ( 1 + \omega ) H_1 (t - t_1) \right)^{\frac 2 {3(1+\omega)}}
    [/itex]

    (I'm not sure what the subscript 1 means, it's not clear from the text)

    This we apparently can get from the equation

    [itex]
    3 H^2 - c^2 \Lambda = \frac{8 \pi G} {c^2} \rho
    [/itex]

    and

    [itex]
    \rho (t) = \rho_1 (\frac {a(t)} {a_1} )^{-3(1+\omega)}.
    [/itex]

    Could anyone help me with this?

    /Rettaw
     
  2. jcsd
  3. Dec 1, 2011 #2

    marcus

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    This is Woodard's equation (205)
    [itex]
    a(t) = a_1 \left( 1 + \frac 3 2 ( 1 + \omega ) H_1 (t - t_1) \right)^{\frac 2 {3(1+\omega)}}
    [/itex]

    "(I'm not sure what the subscript 1 means, it's not clear from the text)"

    I think he means pick any arbitrary time t1 as a base time. Then a1 is just the scalefactor at that time. Shorthand for a(t1).

    Likewise H1.

    That's the only thing substantive i have to say right now. I don't want to plow through the elementary algebra steps. So I'm not helping you in the way you asked. Basically just commenting. But maybe this will help a little bit and someone else will provide additional explanation as needed.
    ===================
    This is his equation (199)

    [itex]
    3 H^2 - c^2 \Lambda = \frac{8 \pi G} {c^2} \rho
    [/itex]

    and this is his equation (204)

    [itex]
    \rho (t) = \rho_1 (\frac {a(t)} {a_1} )^{-3(1+\omega)}.
    [/itex]

    ===================
    Here are a few loosey goosey intuitive remarks FWIW
    Equation (199) looks like it's just a form of the familiar Friedmann equation, It often appears with (200) which is one version of the Friedmann "acceleration" equation. Note that rho is an ENERGY density in his treatment, not a mass density as it sometimes is in other books.

    Notice also that (204) is only true under the very restrictive assumption that the EOS remains constant. w is constant! He has already absorbed Lambda into the energy density rho. So this assumption can only be approximately right over a limited timespan. It is a drastic simplification, but still useful.

    The intuitive content of (204) is straightforward. I think of dividing both sides by rho1 so you have a ratio of densities equal to a ratio of scalefactors to the -3 (....) power.


    Suppose momentarily that w = 0 (pressureless dust) so that the exponent really is -3.

    then it is exactly what you expect. PICK ANY BASE TIME WHATEVER as your t1
    and then look at some other time t, the ratio of densities is going to be equal to the ratio of linear scale raised to the -3 power.

    Double the size and you divide the density by 8.
    ==========================

    You still have to crunch the equations :smile:, but maybe this will help you make interpretive sense.
     
    Last edited: Dec 1, 2011
  4. Dec 4, 2011 #3
    Well, actually "elementary algebra" wasn't that helpful. But thinking to myself about how to explain why it wasn't helpful WAS helpful, so thanks for the help :)


    Anyway, for completeness and posterity this is (my guess at) the solution:
    Starting with

    [itex]
    3 H^2 - c^2 \Lambda = \frac{8 \pi G} {c^2} \rho
    [/itex]

    we move the [itex]\Lambda[/itex] term over to the rhs

    [itex]
    3 H^2 = \frac{8 \pi G} {c^2} \left( \rho + \frac {c^4}{8 \pi G} \Lambda \right)
    [/itex]

    and "renormalize" our density as

    [itex]
    \rho_{ren} = \rho + \frac {c^4}{8 \pi G} \Lambda
    [/itex]

    thus giving

    [itex]
    3 H^2 = \frac{8 \pi G} {c^2} \rho_{ren}. \qquad (*)
    [/itex]

    Now, we have the following expression for the density:

    [itex]
    \rho (t) = \rho_1 \left( \frac {a(t)} {a_1} \right)^{-3(1+\omega)}. \qquad (**)
    [/itex]

    and we wish to express [itex]\rho_1[/itex] in terms of the Hubble constant. So, understanding that [itex]\rho_1[/itex] means [itex]\rho_{ren}(t=t_1)[/itex], we solve for it in eq (*) with [itex]t=t_1[/itex] giving:

    [itex]
    \rho_1 = \frac{3 c^2} {8 \pi G} H_1^2
    [/itex]


    Putting this into (**) gives

    [itex]
    \rho(t) = \frac{3 c^2} {8 \pi G} H_1^2 \left( \frac {a(t)} {a_1} \right)^{-3(1+\omega)}
    [/itex]

    and putting that into (*) we get

    [itex]
    H(t) = H_1 \left( \frac {a(t)} {a_1} \right)^{- \frac 3 2 (1+\omega)}
    [/itex]

    where factors have been eliminated, and a square root has been taken. Substituring in [itex] H = \frac { \dot{a(t)}} {a(t)} [/itex] and moving all factors of [itex]a(t)[/itex] to the lhs, we get

    [itex]
    a^{\frac{1+ 3 \omega} 2 } \frac{da}{dt} = H_1 a_1^{\frac{3} 2 ( 1+ 3 \omega)}.
    [/itex]

    Here I use the physics trick of just using differentials like I want and move [itex]dt[/itex] to the rhs, and integrate between [itex]t_1[/itex] and t

    [itex]
    \int_{t_1}^t a^{\frac{1+ 3 \omega} 2 }da = \int_{t_1}^t H_1 a_1^{\frac{3} 2 ( 1+ 3 \omega)} dt
    [/itex]

    then I pretend I named things properly and evaluate this to

    [itex]
    \frac 2 {3(1+\omega)} \left( a(t)^{\frac{3} 2 ( 1+ 3 \omega)} - a_1^{\frac{3} 2 ( 1+ 3 \omega)} \right) = (t-t_1) H_1 a_1^{\frac{3} 2 ( 1+ 3 \omega)}
    [/itex]

    and this then is rewritten as

    [itex]
    a(t) = a_1 \left( 1 + \frac 3 2 ( 1 + \omega ) H_1 (t - t_1) \right)^{\frac 2 {3(1+\omega)}}.
    [/itex]

    *phew*

    So yes, not entirely elemental algebra, but not much interesting going on besides defining [itex]\rho_1[/itex] in terms of the hubble constant
     
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