The cosmology calculators I've been playing with have $\Omega_\Lambda$ and $\Omega_M$ as their main parameters, and calculate $\Omega_k$ with that.

Are there parameters for FLRW models that don't predict a finite age of the universe, and/or have lookback times that grow unbounded?


1 Answer 1


A universe with a pure positive cosmological constant, $\Omega_{\Lambda}=1$, $ \Omega_m=\Omega_k=0$, has an infinite age. The scale factor grows as $\sim e^{Ht}$ so is never zero at any finite $t$ in the past (or future).

However, if any matter or curvature is present, you would expect there to be a big bang. The reason is that at early times, the component with the largest equation of state parameter $w$ will dominate the Friedman equation, and matter ($w=0$) and curvature ($w=-\frac{1}{3}$) both have a larger $w$ than a cosmological constant ($w=-1$). The scale factor grows as a power law with a positive exponent for both matter and curvature, and a power law will always hit zero at a finite time in the past.

  • $\begingroup$ Thank you. Is that pretty much the only set of parameters to get an infinite age? Also, is this what's known as a de Sitter space? Or anti-de Sitter? $\endgroup$ Commented Mar 26, 2023 at 4:32
  • $\begingroup$ @MikeHelland It is pretty much the only physical parameters that will give you infinite age. (You should look at single component solutions of the Friedmann equation and see how the expansion rate and age depends on $w$). Also yes, a Universe with a positive cosmological constant and no matter is de Sitter space (a negative cosmological constant gives de Sitter space). However, FRW coordinates do not cover all of de Sitter space. $\endgroup$
    – Andrew
    Commented Mar 26, 2023 at 15:40

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