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In a recent question, we learned why hydrogen is currently the most abundant nucleon or element in the universe.

Here I ask a follow-up: For how long will hydrogen be the most abundant nucleus? It is most abundant now, but is slowly being consumed by nuclear fusion in the interior of stars across the universe. Most currently existing hydrogen hasn't even made it into a star yet. Will it ever? If so, how long will it take?

Essentially, how many gigayears (or exayears or whatever) until hydrogen is no longer the most abundant nucleus anymore?

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    $\begingroup$ This is an interesting question, but it will be difficult to answer. I don't believe there is certainty in even knowing how much of all Baryonic matter is in star form currently (vs. hydrogen clouds - leftover stuff in the IGM). Even if that ratio were known, different stars burn Hydrogen differently and at (vastly) different rates. Add to that that ambiguity compact objects gives to this question (e.g. how do you count neutrons in neutron stars in your ratio? How do black holes factor into this question?) and I think you have a very difficult to answer question. $\endgroup$ – enumaris Jul 23 '18 at 20:21
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    $\begingroup$ Related (though the answers there aren't terribly conclusive, TBH): How long will the Universe's hydrogen reserves last for? The point made there concerning intergalactic hydrogen is worth considering, though; the answer may be "never". $\endgroup$ – Michael Seifert Jul 23 '18 at 20:41
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    $\begingroup$ I asked a question some time ago about how metal rich a gas could be to form a star, but so far hasn't attracted the answer I think it deserves. $\endgroup$ – Kyle Kanos Jul 23 '18 at 22:59
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One can only answer on the basis of what we currently know about cosmological parameters. If indeed these have been correctly estimated, and that the cosmological constant is constant, then the universe will continue to expand at an accelerating rate.

Given that more than half the baryons in the universe currently exist outside galaxies, then it seems to me quite likely that these will never form part of stars and that the increasingly rarefied universe will cease to form (many) stars in the future.

In that scenario, isolated protons will always be the most common nuclei.

One complication is that the answer may be different for that part of the universe that we can see.

Gravitationally bound systems like the Milky Way, the local group and cluster will likely not take part in the accelerating expansion. Thus the baryons within this region could continue to be processed in stars, even if they do not consist of the majority of baryons in the universe. How to estimate this timescale? A lower limit is probably the free-fall timescale of the local supercluster, which has a mass of about $10^{15}$ M$_{\odot}$ in a diameter of 33 Mpc. The free-fall timescale is $\sim (G \rho)^{-1/2} \simeq 100$ billion years. However, this time must be increased to take account that most stars formed are low-mass red dwarfs with timescales to turn hydrogen into helium of a further trillion years. So within our local group of galaxies, I would estimate 100 billion years to incorporate all the gas in our local cluster into stars and a further trillion years for them to fuse most of the hydrogen.

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  • $\begingroup$ Of course, star formation (and death) itself can affect the timescales (in a galaxy, not IGM). I have no idea how to estimate the effects, though - it's thought that some young or huge stars and whatevernovae can both disrupt and encourage the formation of other stars. It sounds plausible that at least for galaxies, there is a limit, though. For the currently observable universe at large, I'd go with "probably never", especially if the expansion rate keeps steady or increases. $\endgroup$ – Luaan Jul 24 '18 at 6:52
  • $\begingroup$ @Luaan Recycling of material will continue so long as stars more massive than about half a solar mass are being born. Whether the distribution of stellar birth masses will remain constant time as the ISM become extremely He and metal-rich is one of the big uncertainties in any detailed calculation. $\endgroup$ – Rob Jeffries Jul 24 '18 at 6:59

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