The Hubble time is about 14 billion years. The estimated current age of the Universe is about 13.7 billion years. Is the reason these two time are so close (a) a coincidence, or (b) a reflection that for much of its history the Universe has been expanding at a constant rate?


2 Answers 2


It is in fact a reflection of the fact that the rate of expansion has been nearly constant for a long time.

Mathematically, the expansion of the universe is described by a scale factor $a(t)$, which can be interpreted as the size of the universe at a time $t$, but relative to some reference size (typically chosen to be the current size). The Hubble parameter is defined as

$$H = \frac{\dot{a}}{a}$$

and the Hubble time is the reciprocal of the Hubble parameter,

$$t_H = \frac{a}{\dot{a}}$$

Now suppose the universe has been expanding at a constant rate for its entire history. That means $a(t) = ct$. If you calculate the Hubble time in this model, you get

$$t_H = \frac{ct}{c} = t$$

which means that in a linear expansion model, the Hubble time is nothing but the current age of the universe.

In reality, the best cosmological theories suggest that the universe has not been expanding linearly since the beginning. So we would expect that the age of the universe is not exactly equal to the Hubble time. But hopefully it makes sense that if any nonlinear expansion lasted for only a short period, then the Hubble time should still be close to the age of the universe. That is the situation we see today.

For more information on this, I'd suggest you check out these additional questions

and others like them.

  • 1
    $\begingroup$ Thanks-- I've been randomly thinking about the meaning of the Hubble Time and although I could readily understand that it is the reciprocal of Hubble's Constant, I didn't see why this should be synonymous with the age of the Universe. In thinking about your explanation I still don't have a deep physical intuition about it (I am not sure why I am slightly embarrassed by that), although I am satisfied with the more abstract mathematical explanation. $\endgroup$ Commented Aug 3, 2012 at 5:46
  • $\begingroup$ I asked a question here physics.stackexchange.com/questions/402706/… What is the meaning of $H^{-1}$? @DavidZ $\endgroup$
    – SRS
    Commented Apr 28, 2018 at 17:49
  • $\begingroup$ This answer is wrong, and Camiel Wijffels's answer is correct. 1/H significantly overestimated the age in the past (by a factor of 3/2) and will underestimate it in the future. The present era happens to be close to the crossover point. $\endgroup$
    – benrg
    Commented Aug 5, 2020 at 3:30

It is a coincidence.

The reason is that the Hubble constant H is not constant, and varies over time. For example 6 billion years ago, when the universe was 7.5 billion years old, the Hubble constant was about 100 (km/s)/Mpc, what means the Hubble time was 9.78 billion years. When the universe is 24 billion years of age, H will be 60 (km/s)/Mpc, and the Hubble time will be 16.3 billion years.

Not even close to the age of the universe.

  • $\begingroup$ You can find a detailed calculation of the Hubble time in this answer of mine. The figures I get broadly agree with Camiel's. $\endgroup$ Commented Jan 22, 2015 at 18:34
  • $\begingroup$ What is said is factually true but I don't think it is necessarily coincidence that we observe the universe when $\Lambda$ and $\Omega$ are similar, galaxies and planets have had time to form etc. The coincidence could be an example of the weak anthropic principle. $\endgroup$
    – ProfRob
    Commented Apr 7, 2018 at 13:55
  • $\begingroup$ The conclusion is consistent with the answer above. > When the universe is 24 billion years of age, H will be 60 km/s/Mpc, > and the Hubble Time will be 16.3 billion years. > Not even close to the age of the universe. 16.3E9 years is VERY close to 13.8E9 years. Much less than a factor of 2. Given the approximations involved, that is good agreement. So the inverse of H is approximately the age of the universe, no coincidence. But of course, the agreement isn't perfect because we're using a highly approximate model. $\endgroup$
    – Gerry Harp
    Commented Jul 20, 2018 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.