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Does Friedmann equation allow for complex scale factor?

  1. Apr 24, 2012 #1
    Looking at the Friedmann equation
    [tex]H^2=\left[\frac{\dot{a}}{a}\right]^2=\frac{8\pi G\rho}{3}-\frac{kc^2}{a^2}[/tex]

    and considering positive curvature, then for the limit where the second term dominates, we're left with
    [tex]\left[\frac{\dot{a}}{a}\right]^2=-\frac{kc^2}{a^2}[/tex]

    This implies a complex scale factor, does it not?
     
  2. jcsd
  3. Apr 24, 2012 #2

    Ich

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    Either that, or it implies that the curvature in this limit is negative. Which is the case.
    IMHO, the equation makes more sense if you rearrange it:
    [tex]\frac{kc^2}{a^2}=\left[\frac{\dot{a}}{a}\right]^2=\frac{8\pi G\rho}{3}-H^2[/tex]
    Read: Curvature = positive contribution from energy density - negative contribution from expansion.
     
  4. Apr 24, 2012 #3
    Okay, so the curvature has a time dependence k=k(t)?
     
  5. Apr 24, 2012 #4

    Ich

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    Right. For example, in a closed universe, it could be its radius.
     
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