How can the age of the universe be calculated as $1/H$, with $H$ the Hubble constant, when its expansion has been far from linear?

universe expansion under lambda-CDM with inflation, exponential at first, then linear and slowing down, then accelerating

The $1/H$ comes from Hubble’s law, and implies that the universe expanded like a cone, as opposed to the above.

Also, wouldn’t the Hubble constant naturally change over time? Has the change since its discovery been negligible? I know it’s changed in terms of the recognised value due to improvements in the way we measure it—I mean the actual value.


3 Answers 3


Indeed, the age is not just $1/H$. (In this answer I'll use units with $c=1$.

The age $T$ can be calculated by making repeated use of the chain rule, as well as $da/dt=aH$ and $a=1/(1+z)$ \begin{equation} T=\int_0^T dt = \int_0^1 da \frac{dt}{da} = -\int_0^\infty dz \frac{da}{dz} \frac{a}{H} = \int_0^\infty dz \frac{1}{(1+z)^3 H(z)} \end{equation} where I've assumed a flat universe so I can normalize the scale factor by $a(today)=1$, and I set the origin of time at the Big Bang a(t=0)=0.

However, using the Friedman equation and assuming a universe with matter, radiation, and a cosmological constant we can write \begin{equation} T=\frac{1}{H_0} \int_0^\infty dz \frac{1}{(1+z)^3 \sqrt{\Omega_{R,0}(1+z)^4+\Omega_{M,0} (1+z)^3 + \Omega_\Lambda}} \end{equation} Ignoring inflation, and using the $\Lambda$CDM best fit values for $\Omega_{I,0}$ ($I=R,M,\Lambda$), the integral is $O(1)$ (as you can check), so $1/H_0$ is a good approximation to the age of the universe.

  • 1
    $\begingroup$ Wolfram Alpha Plugged In $\endgroup$
    – AHusain
    Commented Jan 6, 2017 at 3:25
  • $\begingroup$ hi, can i ask why do you need to assume a flat universe to normalize the constant? $\endgroup$
    – Alucard
    Commented Oct 10, 2022 at 1:10

Suppose at $ t=0$ universe came into existence and at time $t$ the parts of the universe reached a distance $d=vt$. Now we have the Hubble's Law which states that the expansion of our universe is linear, that is if a galaxy A is moving away from you at speed $v$ then another galaxy B which is twice as far away from you is moving at speed $2v$. In mathematical terms, $v=H_0 d$, where $d$ is comoving distance. Now if you put $H=1/t$ in $d=vt$ you get $v=H d$. So we can use it to say that universe started at $t=1/H$. Actually it's not correct to evaluate age of the universe because it assumes the same expansion since the beginning. But still it gives very accurate results as other methods.

Yes the number will change again and that's why $H_0$ is used which tells you the present day Hubble's "constant". As the universe expands, $v$, $H$ and $d$ will change.


If the expansion is geometric, it doesn't matter how you extrapolate backwards, you never get to zero. You only can get backwards to zero if the expansion rate asymptotes to infinity at $t=0$. So there is no reason to believe the $1/H$ estimate. Besides $13.5\ \mathrm B$ years is barely enough to explain The Earth. We can determine that the earth is $4.5\ \mathrm B$ years old. But the iron in it had to come from a burned out star. It takes $10\ \mathrm B$ years for that to happen. Then, it would have had to have been the first to form after the Big Bang. So I favor an older age. Maybe $20\ \mathrm B$.

  • $\begingroup$ Big stars burn fast, so there has been plenty of time to create heavy elements, and for them to get homogenized in the interstellar medium. $\endgroup$
    – PM 2Ring
    Commented Mar 31, 2018 at 20:40
  • $\begingroup$ In the 1990's there was a question about the age of the universe. Considerations like other answers here favored an age of about 10 B years. But there were objects in the sky that appeared to be 20 B years old. It took a few years to resolve this. The objects were not as old as was first thought. Better analysis gives the current age of 13.8 B years. $\endgroup$
    – mmesser314
    Commented Nov 26, 2020 at 18:09

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.