As you can find in all the literature, the age of the universe can be computed easily $$t_0= (H_0)^{-1} = 14Gy$$

For the accepted values of Hubble constant $H_0$, but recently I read that this result takes the assumption of "constant expansion" (the source is in fact this one this), that is related to the density parameter $\Omega$,

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

And this constant expansion we imposed to find the age of the universe happens when $\Omega = 0$, i.e. the universe is empty.

  1. Why is this taken as an approximation, if we know part of the universe is not empty for sure, approximately,

$$\Omega_m = 0.05 \quad \Omega_{DM}=0.25$$

$\qquad $ For matter and dark matter respectively, the 70% left is sufficient to neglect this density?

Also I found that for a universe with the exact critical density $\rho=\rho_c$, i.e. $\Omega =1$. A universe that have the exact necessary density to maintain the expansion and not collapse, the age of the universe will be,


That is less than the accepted hypothesis ($14Gy$).

  1. How can this expression $\frac{2}{3H_0}$ be found? And how is related to the expansion rate $H$?

I don't know if I mixed some concept wrong, and I will be thankful for any help.

  • 2
    $\begingroup$ As you can find in all the literature… No. As you can find in Wikipedia, “the inverse of the Hubble constant … is slightly different from the age of the universe. The Hubble time is the age it would have had if the expansion had been linear, and it is different from the real age of the universe because the expansion is not linear; they are related by a dimensionless factor which depends on the mass-energy content of the universe, which is around 0.96 in the standard ΛCDM model.” $\endgroup$
    – Ghoster
    Nov 9, 2022 at 18:16
  • $\begingroup$ Hubble parameter is not constant, but depends on time, like : $$ H(t) = \frac {1}{t_0 + (q + 1) t} ,$$ where $q$ is Deceleration_parameter. If $q \gt -1$, then Hubble constant is decreasing with time. If $q \lt -1$ (universe has phantom energy), then Hubble constant must increase with time. If $q = -1$, then Hubble constant stays the same. Accordingly $H_0 = H(t=\text{now})$, so it's oversimplified for trying to extract current universe age from it. $\endgroup$ Nov 9, 2022 at 21:24
  • $\begingroup$ Duplicate of physics.stackexchange.com/questions/328409/… ? $\endgroup$
    – ProfRob
    Nov 9, 2022 at 22:36

1 Answer 1


In ΛCDM, $H(t)^{-1}$ is larger than the age of the universe at early times and smaller than the age of the universe at late times. We just happen to live in the crossover era where they are close in value. $H(t)^{-1}$ isn't, or at least shouldn't be, taken as an approximation of the age of the universe.

If $a \propto t^k$ for some constant $k$ (which is not the case in ΛCDM), then $H_0 \triangleq a'(t_0)/a(t_0) = k/t_0$, and so $t_0 = k/H_0$. If the expansion speed is constant then $a\propto t$, $k=1$, and $t_0 = H_0^{-1}$. In a matter-dominated ($Λ=0$) critical-density cosmology, $a\propto t^{2/3}$, and $t_0 = \frac{2}{3H_0}$. But neither of those expressions for the age of the universe is correct in ΛCDM.


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