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Lifetime of the universe - what are the limits?

  1. Apr 25, 2015 #1
    I am studying general relativity from Hobson and came across the term 'lifetime' of a closed (k>0) universe, ##t_{lifetime}##.

    I suppose at late times the curvature dominates and universe starts contracting? Are they simply referring to ##\int_0^{\infty} dt##? If so, would the bottom expression be right?

    [tex]\int_0^{\infty}dt = \int_1^{0} \frac{1}{a(t) H(t)} dt[/tex]

    Do you think this makes sense since ##a(t)## is decreasing and will eventually reach ##0##?

    Then for an open universe (k<0), wont the universe simply keep expanding? If I read right, our universe is mostly flat, right? Then what is driving the expansion? Dust?
  2. jcsd
  3. Apr 25, 2015 #2


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    In the closed universe (without dark energy), there will be a past singularity as well as a future singularity, i.e., there will be a maximal value for the cosmological time t.

    This is purely academic as we seem to be living in a universe with a dark energy component that has already started to dominate, meaning the universe will forever undergo accelerated expansion.
  4. Apr 25, 2015 #3
    Ok, so it would be
    [tex]\int_{t_0}^{t_{life}}dt = \int_1^{0} \frac{1}{a(t) H(t)} dt[/tex]

    So there is a 'cosmological constant' after all? Meaning since ##\rho_{\Lambda} = constant##, then solving the friedmann equation gives ##a(t) \propto e^{mt}##.
  5. Apr 30, 2015 #4
  6. Apr 30, 2015 #5


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    No. What happens is that the density of ordinary matter in the universe is sufficient to cause it to recollapse. The second Friedmann equation, for ##\ddot{a} / a##, is the key to the dynamics; note that there is no curvature term in this equation.

    No. This just gives ##\infty - 0 = \infty##.

    No. I don't understand what you think this integral represents.

    There is a cosmological constant according to our best current model, yes. But I don't see what the integral you wrote down has to do with it.
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