If Hubble's time is the inverse of a changing Hubble's "constant" and it estimates the age of the universe up to now, since Hubble's parameter is converging to an actual constant value, how can the inverse of a constant continue to measure the time since the Big Bang towards the future?
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1$\begingroup$ The Hubble time is not equal to the age of the universe, and it certainly isn't its definition: it's just a rough estimate. $\endgroup$– JavierCommented Oct 26, 2018 at 14:29
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$\begingroup$ Why does it give a rough estimate now and will apparently cease to do so when the Hubble's parameter becomes actually constant in the future? $\endgroup$– Giordano MottaCommented Oct 26, 2018 at 14:37
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2 Answers
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I think this plot answers your question. It is taken from Frieman (2008) and shows the evolution of the scale factor for four cosmological models, each of which has the same value of the Hubble parameter (and scale factor) measured at the current epoch.
The approximation that the age of the universe is $H_0^{-1}$ is exactly true for an "empty universe" (dotted line), because the current expansion rate can be just extrapolated backwards to when the scale factor was zero.
We think that we live in a universe that isn't empty(!). On its own (dashed lines) this would lead to a decelerating expansion and an age younger than $H_0^{-1}$. However, the addition of dark energy (solid line) means the expansion decelerates, then accelerates. The net result is that the age of our universe is very similar to that of an empty universe with the same $H_0$ (solid and dotted lines are almost coincident when the scale factor is zero).
The bottom line is that we live now in a universe where $\Omega_M$ and $\Omega_{\Lambda}$ are similar. This will not be the case in the future and the age will not be given by $H_0^{-1}$.
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Hubble's constant is not actually a constant. It is true the Hubble parameter is constant in regards to position and velocity of objects in the Universe at a certain time, meaning that for any measurement you get that $v \propto d $, where v is the velocity of the object away from us and d is its distance, Hubble's constant being exactly this proportionality factor. But this only means that for any given time the paremeter is the same for all objects, meaning that if you "look' at the sky at any given time you will find such a paremeter such that $v \propto d $ for all objects. This does not mean that the paremeter is constant over time, meaning that for two different times you still get that $v \propto d $, but with a different proportionality factor, so the rate of expansion actually changes.
The parameter is an estimate of the age of the universe, because the actual calculation of the age of the universe is: $$t_0=\frac {F( \Omega_m, \Omega_{\Lambda},...)}{H_0}$$
Where $F$ is a function of the relative densities of matter ($\Omega_m$), of dark energy ($\Omega_{\Lambda}$)... For a proper calculation this needs to be taken into account, but for an approximate calculation it is fine to take $F \approx 1$, because that is actually close to the current accepted value of $F$, according to the accepted model, the $\Lambda $CDM. The issue with the parameter becoming constant, it won't, although it will assymptotically approach a fixed value. In spite of that, accounting for the changes on $F$, it would still be possible to calculate the age of the universe that way.
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$\begingroup$ It is not constant today, but is it not converging to a constant value due to dark energy dominance? $\endgroup$ Commented Oct 26, 2018 at 14:39
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$\begingroup$ Is it then a coincidence that F≈1? It puzzles me that the current radius of the Hubble sphere is about 14 billion light years, which is about the same as the age of the universe in years. Since the Hubble sphere determines the frontier between superluminal and subluminal recession, when the universe was younger, say, 5 billion years old, the frontier at which objects were receding faster than the speed of light was also 5 billion light years from the center of Hubble Sphere? This apparently does not make sense if the expansion is accelerated. But I'm sure there is some fault in my reasoning. $\endgroup$ Commented Oct 26, 2018 at 19:49