- What is the smallest mass of pure hydrogen that can ignite fusion?
- That is can population III stars have tiny masses?
- How would such stars develop?
- How long would such a star last?
[Physics] Pure hydrogen star
astrophysicsfusionhydrogenstarsstellar-evolution
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I think there are really three questions that need to be answered for this to make sense:
- is there a "normal" limit to how large a star can be?
- how can population III stars form with such large masses?
- how can population III stars retain their large masses?
An answer to the first question is tricky. We expect large stars to be rare, and the largest stars to be the rarest. On top of this, they'll lead the shortest lives. Getting observational constraints has thus been tricky. There might be a limit to the amount of mass that is available to turn into stars when they form. As for the "normal" limits on the masses of stars, most (as far as I know) involve around pulsational instability. But the recent discovery of massive stars in and near the cluster R136a suggests that stars with masses over 150 solar can form even in material that has a non-negligible metal content. So whether there is a "normal" limit is open question.
The second question is much better understood, thanks to a lot of numerical work. Tom Abel recently wrote an article for Physics Today that summarizes current understanding of pop III star formation. Basically, the smallest amount of gas unstable to collapse under its own gravity, the Jeans Mass, increases with temperature (like T3/2). So the cooler the gas can become, the smaller the fragments we expect to see. What determines how cool the gas can become? The atoms and molecules that radiate within it, and whether this radiation can escape. In metal-polluted gas, various molecular and atomic lines allow the gas to cool to tens of K. In metal-free material, the most effective coolant (in terms of the low temperatures it can achieve) is molecular hydrogen, which will only cool to around 200 K. This is a higher temperature, so we expect more massive fragments. This is a gross simplification! The situation really involves complex dynamics, shock formation, and all sorts of other stuff. Even the question of whether or not molecular hydrogen can form is contested.
Finally, if a massive pop III star formed, would it keep its mass? We know that the some massive stars in the local universe, like Eta Carinae, are violent beasts. This kind of episodic, pulsational mass loss could be present in Pop III stars, but since such mass loss is so poorly understood, this is often ignored. More generally, we expect that the metals in the atmospheres of massive stars absorb enough of the radiation created inside the star to be driven away in a wind. Again, there aren't any metals in metal-free gas, so we expect this effect to be much smaller in Pop III stars.
So, we expect Pop III stars to be larger because there is more gas available, because the gas fragments less owing to its higher temperature, and because we don't think the stars lose as much mass as modern stars do. And, we aren't even sure that there's a limit on how massive stars can be in the first place!
Short answer: gravitational potential energy is converted into heat.
Let's look at the Sun as an example. Its mass is $M_\odot = 2.0\times10^{30}\ \mathrm{kg}$ and its radius is $R_\odot = 7.0\times10^8\ \mathrm{m}$. If its density were uniform, its gravitational binding energy would be $$ U_{\odot,\,\text{uniform}} = -\frac{3GM_\odot^2}{5R_\odot} = -2.3\times10^{41}\ \mathrm{J}. $$ In fact the Sun's mass is centrally concentrated, so $U_{\odot,\,\text{actual}} < U_{\odot,\,\text{uniform}}$.
Where did the Sun come from? Something like a giant molecular cloud with a density of $2\times10^{-15}\ \mathrm{kg/m^3}$. The mass of the Sun would thus have been extended over something like a sphere of radius $6\times10^{14}\ \mathrm{m}$, for a gravitational binding energy of $$ U_\text{cloud} = -3\times10^{35}\ \mathrm{J}, $$ which is negligible in comparison with $U_\odot$.
All of the $2.3\times10^{41}\ \mathrm{J}$ had to go somewhere, and the only place to dump energy is into heat. The gas particles gain velocity as they fall into the potential well, but they don't lose that velocity because they never climb back out of the well.
Not worrying about whether the heating is isobaric or isochoric or somewhere in between, the heat capacity of monatomic gas is about twice the ideal gas constant, or $8.3\times10^3\ \mathrm{J\,K^{-1}\,kg^{-1}}$. At this amount, in order to heat all of $M_\odot$ by the average temperature of the Sun (say $10^7\ \mathrm{K}$, somewhere between the core and surface temperatures), you would need about $1.7\times10^{41}\ \mathrm{J}$ of energy. There is enough energy released by gravitational collapse to heat the Sun to its current temperature. You can do a more detailed analysis taking into account how much cooling occurs during the collapse, but the steep temperature dependence of the Stephan-Boltzmann law makes it difficult to lose heat to space until the object is already hot. I'm also neglecting a factor of $2$ that comes from splitting the energy between heating the gas and compressing it.
Once the material is this hot, it simply glows like any blackbody emitter. The energy lost to space is replenished by nuclear fusion in the core. In fact, fusion acts as a regulator: too much of it and the star expands and cools, slowing down fusion; too little and the star collapses further, heating up more and increasing the fusion rate.
In summary, gravitational collapse provides the initial energy to heat a star. As it uses up this energy source, it begins to tap into fusion. Ultimately it reaches an equilibrium where the energy produced by fusion is balanced by the energy radiated into space.
Best Answer
As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium.
Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by electron degeneracy pressure. They will never become hot enough to fuse anything beyond deuterium, hydrogen and lithium. Instead they will slowly fuse all their hydrogen, but because they are fully convective they will be able to resupply the core with fresh fuel until it is all gone. This scenario is discussed by Laughlin, Bodenheimer & Adams (1997) in the context of a solar composition - they suggest this phase takes around $10^{13}$ years for a star just above the minimum mass for hydrogen burning. Once the fuel is exhausted, the star will continue to cool and will contract a little more until its electrons become completely degenerate. At which point it can continue to cool, but now at constant radius since the pressure becomes independent of temperature. Such a star (a helium white dwarf) will just sit and cool forever, until their protons decay, or something else externally happens to them or the universe.
The answer to the first part must come from theoretical models. The canonical work by Chabrier & Baraffe (1997) find a "minimum mass" for hydrogen burning (which they define as that mass where hydrogen burning can supply the stars luminosity after 1 billion years) is about $0.072M_{\odot}$ for a star with solar metallicity, but rises to about $0.083M_{\odot}$ for stars which have less than a tenth of the solar metallicity. Earlier work by Baraffe et al. (1995) determined a limit of $0.09M_{\odot}$ for a metallicity 30 times less than the Sun. The review of Burrows et al. (2001) says that the minimum mass for a "zero metallicity" population III star is $0.092M_{\odot}$. This limit originates from modelling work performed by Saumon et al. (1994).
I am not aware of anything more recent, nor of anything more recent that tackles the entirely hypothetical question of a pure hydrogen star. However, one could wave ones hands and say that the virial theorem tells us that the central temperature is proportional to $\mu M/R$, where $\mu$ is the mean atomic mass in the gas. If hydrogen burning starts at a fixed central temperature, then the mass at which that occurs will be proportional to $\mu^{-1}$. If we remove the helium from the mixture, then $\mu$ goes down from around 0.6 to 0.5. So this very crude argument (it implicitly assumes a perfect gas and that the envelope opacity and hence radius is not changed significantly) might suggest the minimum mass for hydrogen burning rises to about $0.6\times 0.092/0.5 = 0.11M_{\odot}$.
Further edit: Your revised question now asks about fusion, rather than hydrogen fusion. Well, deuterium burning can occur at much lower masses - probably about 15 Jupiter masses.