"Expressed in inverse years, the value 70 km/s/Mpc comes to about one divided by 14 billion years - the approximate age of the universe."

Could the Hubble Constant be an artifact of the structure, and thus a way to roughly directly measure its age? The fact expansion is, or seems to be, accelerating (Reiss et al) would strengthen this hypothesis.

Could perhaps an advanced civilization could measure it with far more precision than we could, and would find the Hubble Constant a direct "age tag" marker? Or is this merely a remarkable coincidence?

Would a hypothetical observer in the distant future be able to use it to directly measure the universes`s age then? (Current theory says "no, it just seems coincidental today").

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from A buyer’s guide to the Hubble Constant (Paul Shah et al., September 2021) https://arxiv.org/pdf/2109.01161.pdf with thanks to Henry Norman.

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    $\begingroup$ Hi, what do you mean by "artifact of the structure"? What structure? $\endgroup$
    – Prallax
    Commented Sep 9, 2021 at 9:39
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    $\begingroup$ Are you asking whether or not it's a coincidence that $1/H_0$ is roughly equal to the age of the Universe? $\endgroup$
    – pela
    Commented Sep 9, 2021 at 21:33
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    $\begingroup$ @User123 not really, I`m asking if a hypothetical observer in the distant future will be able to use it to directly measure the universes's age then. Current theory says "no, it just seems coincidental today". $\endgroup$ Commented Sep 12, 2021 at 11:53
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    $\begingroup$ @pela It sure looks like it $\endgroup$
    – uhoh
    Commented Sep 13, 2021 at 0:33
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    $\begingroup$ @uhoh Seems so, but I realized I didn't actually answer the question, so I added a paragraph on whether or not $1/H_0$ can be used to calculate the age of the Universe. $\endgroup$
    – pela
    Commented Sep 13, 2021 at 14:01

2 Answers 2


I'm a bit uncertain if I understand your question correctly, but if I do, you're asking whether or not it's a coincidence that $1/H_0$ is roughly equal to the age of the Universe.

If so, the answer is "Somewhat, but not really".

Not a complete coincidence

The reason it is not exactly a coincidence is that, if the Universe had expanded linearly with time since the Big Bang, its age would indeed be exactly equal to $1/H_0$. To see this, use the definition of the Hubble constant: $$ H_0 \equiv \frac{\dot{a}}{a}, $$ where $a$ is the scale factor of the Universe, i.e. the factor by which (cosmological) distances scale with time. If the Universe expands linearly with time, then $\dot{a}=const.$, so $a \propto 1/H_0$. This factor — the inverse of $H_0$ — is called the Hubble time, and is hence the age of a linearly expanding universe.

The fact that the age of our Universe is close to $1/H_0$ tells you that it has been expanding roughly, but not exactly, linearly since the Big Bang. In fact it slowed down for the first some 10 billion years (due to the mutual attraction of all matter), and then accelerated lately (due to dark energy, we think).

A bit of a coincidence

When I nevertheless say that it is somewhat of a coincidence, it is because in principle our Universe might have decelerated or accelerated much more that it did, so that its age would not at all be close to $1/H_0$.

Really a coincidence?

Whether or not this is really a coincidence is beyond the scope of this answer — there could be a good anthropic explanation for this, since a too decelerating universe might collapse before it creates life, while a too accelerating universe might not have the time to create structure before everything is too apart.

So, can $1/H_0$ be used to measure the age of the Universe?

Coincidentally or not, the inverse Hubble constant is only an approximation of the age of the Universe. The exact age requires exact knowledge of the evolution of the Universe which, in turn, requires a cosmological model as well as observationally constrained values of the parameters entering the model. The accepted model is the Friedmann model, in which the age of the Universe can be calculated as $$ t = \frac{1}{H_0} \int_{z=0}^{z=\infty} \frac{dz'}{(1+z')\sqrt{\Omega_m (1+z')^3 + \Omega_k (1+z')^2 + \Omega_\Lambda}}. $$ Here $\Omega_{m,k,\Lambda} \simeq \{0.3,0,0.7\}$ are the density parameters of matter, curvature, and dark energy, respectively.

If the expansion continues as it is now, dark energy will dominate more and more, i.e. $\Omega_m\rightarrow0$ and $\Omega_\Lambda\rightarrow1$. For a completely dark energy-dominated Universe, the scale factor takes the simple from $a(t) \propto e^{H_0 t}$, i.e. exponentially accelerating expansion.

As can be seen from the equation above, if the $\Omega_{m,k}$ terms vanish, the integral diverges. A future cosmologist will hence reach the conclusion that the Universe is infinitely old or, more likely, that the age cannot be calculated from available data.

So the answer to your question is "no"…


I'll include an answer that I considered satisfactory from another forum. To the point, succinct, yet comprehensive.

"The Hubble Constant is not simply the inverse of the “Age of the Universe”, but an approximation. This is due to the fact that the Hubble Constant (otherwise named the “Hubble Parameter” because it is not really a constant) is a function of the various energy densities that make up the universe at various stages in its evolution, all of these densities varying from each other. For example, in the very early stages of the universe, radiation (photons and neutrinos) “dominate” as their combined density is greater than matter. Then there is a transition to matter domination, which includes dark matter and baryonic matter. Finally, there is the transition to dark energy driven expansion, which has a constant density throughout evolution. If the universe were empty, that is only consisting in space and entirely driven by dark energy, then the Hubble parameter would be directly related to the age of the universe." -Shawn Fitzgibbons


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