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Critical Density

  1. May 29, 2010 #1
    Using the FRW:

    \left( \frac {\dot{a}} {a} \right)^2 = \frac {8 \pi G \rho} {3} - \frac {k c^2} {a^2}

    We define critical density by setting k = 0 and rearranging to get:

    \rho_c = \frac {3 H^2} {8 \pi G}


    H = \left( \frac {\dot{a}} {a} \right)

    My question is does [tex]\rho[/tex] include the density contribution for Cosmological Constant (dark energy) [tex] \Lambda [/tex] or is this derivation only for a Universe with no cosmological constant?

    How does one then actually measure the density of Universe?

    I know that the density has been measured to be slightly less than the critical density, but I thought we are meant to live in a flat Universe? Is this due to the cosmological constant and how is this reconciled with [tex]\rho[/tex] not being exactly [tex]\rho_c[/tex]?

    Finally, I am right in saying that a Universe with [tex]\rho_c[/tex] will stop expanding after infinite time, one with [tex]\rho > \rho_c[/tex] will collapse back on itself and one with [tex]\rho < \rho_c[/tex] will expand forever?

    Many thanks.
    Last edited: May 29, 2010
  2. jcsd
  3. May 29, 2010 #2


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    Yes, [itex]\rho[/itex] includes contribution from the cosmological constant. In other words, we can write the density as a function of scale factor as
    \rho = \rho_c\left(\Omega_Ma^{-3} + \Omega_Ra^{-4} + \Omega_{\Lambda}\right)

    This is basically correct, although I don't think, "stop expanding after infinite time" is a well-defined notion.

    Fit supernova data and/or CMB data to different models and see what works best.

    The measurement of [tex]\rho[/tex] is within error of being less than, equal to, or greater than the critical density. People say we live in a "flat universe", because the measured value is very close to the critical density.
    Last edited: May 29, 2010
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