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Flatness Problem

  1. Feb 18, 2007 #1
    Reading about the flatness problem in the standard cosmological model I have came to this relation

    \frac{1-\Omega(t)}{\Omega(t)}=\frac{1-\Omega_0}{\Omega_0} \frac{1}{1+z}


    \Omega = \frac{\rho}{\rho_c}

    and [itex]z[/itex] is the redshift.

    I would like to know where this relation cames from. I supose it cames from the Friedmann equations but I am not getting there...
  2. jcsd
  3. Feb 18, 2007 #2


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    Start with the first Friedmann equation. Set [itex]\Lambda = 0[/itex], multiply both sides by [itex]3/8 \pi G \rhp[/itex] and rearrange terms to get:

    [tex]\left( \frac{1 - \Omega}{\Omega} \right) \rho a^2 = constant[/tex]

    Evaluate this equation for today with [itex]\Omega_0[/itex], [itex]\rho_0[/itex] and [itex]a = 1[/tex], and for another cosmological epoch with [itex]\Omega[/itex], [itex]\rho[/itex] and [itex]a[/tex].

    Then, consider that the energy density is matter-dominated and you will get the result.
  4. Feb 18, 2007 #3
    Ok. I get it! Thanks.
    But this leads me to another question.
    I supposed that the flatness problem was that if [itex]\Omega_0[/itex] was a little different from 1, then the universe at an early stage would have had a very different [itex]\Omega[/itex] and so the curvature would have been very different from what it is today. By other words, flat at the beginning flat for ever.
    But I can see from your reply that the curvature is suppose to be constant (no matter if [itex]\Omega[/itex] changes or not) in time and so it seems that I didnt understand the flatness problem at all...
    So my next question is - what is the flatness problem? I would be pleased if someone could give me a explanation or a good reference where I could learn more about it.
    Last edited: Feb 18, 2007
  5. Feb 19, 2007 #4


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    The universe must have always been very close, if not exactly 'flat', for the reasons you have already noted. Any deviations pile up exponentially when you run the clock forward [or backwards] from t=0 to the present epoch.
  6. Feb 19, 2007 #5


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    In a matter or radiation dominated universe the curvature is only constant if it is exactly zero. Otherwise, as Chronos points out, you can see in your formula that curvature strongly deviates from flatness in a matter (or radiation) dominated universe and that [itex]\Omega[/itex] must be extremely close to one at [itex]z \rightarrow \infty[/itex] for [itex]\Omega_0[/itex] to be of order unity today.

    You could try solve this problem in two ways. Either assume that the initial conditions were such that [itex]\Omega[/itex] was (extremely close to) one, or assume that a mechanism exists that leads the curvature to zero regardless of the initial conditions. The first option is actually no solution to the problem because it merely shifts it. The second option is inflation.
    Last edited: Feb 19, 2007
  7. Feb 19, 2007 #6
    But in the derivation of that relation I got
    \left( \frac{\Omega - 1}{\Omega} \right) \rho a^2 = \frac{3k}{8 \pi}

    then you say that in every other epoch the [itex]3k/8 \pi[/itex] has the same value so that the left side of the relation can be evaluated in any instant [itex]t[/itex] . But that seems to go against the Flatness problem because [itex]k[/itex] changes with [itex]\Omega[/itex] and hence it should change with time. What am I missing?
  8. Feb 19, 2007 #7


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    Note that [itex]k[/itex] is an integer with three possible values: -1, 0 or 1, and it cannot change from one value to another. The value [itex]3k/8 \pi[/itex] is therefore constant in time.
  9. Feb 19, 2007 #8
    Ok. Now I understand. I got confuse with the fact that the universe is almost flat but not flat and because of this I forgot that k was an integer.
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