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Limits FRW universe , rate of expansion, k=-1,0.

  1. Dec 28, 2014 #1
    Hi,

    I'm looking at deriving the limits of ##\dot{a}## as ## a-> \infty ## , using the Friedmann equation and conservtion of the ##_{00}## component of the energy momentum tensor for a perfect fluid. Both of these equations respectively are:

    ## \dot{a^{2}}=\frac{8\pi G}{3} \rho a^{2} + | k |## (from Friedmann and using ## k \leq 0## [1]

    ## 0= -\partial_{0} \rho - 3\frac{\dot{a}}{a}(\rho + p)## [2]

    ##a## is the scale factor, ##\rho## is the density

    I've just seen a proof when modelling this perfect fluid as dust, where dust obeys ##\rho a^{3} ## is a constant. I'm looking at how you generalise this proof to, radiation say - which obeys ##\rho a^{4} ## is constant. I will now state the solution proof for dust, and my attempt for the radiation proof. My question is whether my radiation proof is valid - I doubt it because the last step looks a bit dodgy - can you simply divided by ## a ## and have the limits still hold - that is ## \dot{a^{2}} -> | k | ## ...

    Dust Proof:

    ## \partial_{0} \rho a^{3} = a^{3} (\dot{\rho} + 3\frac{\rho\dot{a}}{a})= -3pa^{2}\dot{a}##.
    RHS ## < \leq 0 ## => ## \rho a^{2} -> 0 ## as ## a -> \infty ##
    => ## \dot{a^{2}} -> | k | ## from eq [1].

    My Radiation Proof:
    - Simply multiply eq [1] by a, eq[2] by ##a^{4}## as a pose to ##a^{3}## as done in the worked dust proof. I then have similar conclusions , so ##\rho a^{3} -> 0## as ##a -> \infty ## and so by eq[1] multiplied by a I conclude that ##a\dot{a^{2}} -> a | k | ## .

    This is the line I'm concerned with, dividing by ##a## am I then ok to claim that ##\dot{a^{2}} -> | k | ##

    Thanks in advance.
     
    Last edited: Dec 28, 2014
  2. jcsd
  3. Dec 28, 2014 #2

    mfb

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    ##\rho a^4 = c## implies ##\dot {a^2} = \frac{c'}{a^2} + |k|## with constants c and c'. As ##a \to \infty##, ##\dot {a^2} \to |k|##?

    Pressure will be different for dust and photons, that could be relevant in the derivation of those equations.
     
  4. Dec 28, 2014 #3
    Thanks. That looks like a better way to do it. Looking att the full derivation, apologie I don't have time to include it all here, pressure varying will not affect the derivation. So I conlude that ##\dot {a^2} -> 1## for an open universe with k=-1 and ##\dot {a^2} -> 0## when k=0. Are these the known limiting expansion values for the dust and radiation case? That they have the same limit, does this sound correct?
     
  5. Dec 28, 2014 #4

    mfb

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    You are considering the evolution of a universe with negligible mass and radiation and without cosmological constant. There is nothing that would accelerate or slow expansion, the solution looks natural.
     
  6. Dec 30, 2014 #5
    Sorry to re-bump. So post 2 proves the limits without needing the energy-momentum tensor or the conservation of energy? I'm just wondering why the source I used , used them if it can be shown without.

    Is what I did generally mathematically incorrect?
     
  7. Dec 30, 2014 #6

    mfb

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    In an empty universe (##\rho \approx 0##) without cosmological constant, the constant ##\dot a## is a direct consequence of the first Friedmann equation.
    The evolution until the universe is nearly empty is different and can need a more detailed analysis. If the initial matter density is too high, you never reach that point.
     
  8. Dec 31, 2014 #7

    DEvens

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    Not really relevant to the discussion above but...

    FRW cosmology is "introductory physics homework?" Wow. I wish I had gone to a school where that was the case.:)
     
  9. Dec 31, 2014 #8
    ha. it says undergrad in the description. I'm a final year :)
     
  10. Dec 31, 2014 #9
    Is the last line of what I did mathematially incorrect?
     
  11. Dec 31, 2014 #10

    mfb

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    If you know ##a## has a lower bound (at least after some time T), it works.
     
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