astrophysics - Minimum size of a "water star"

So I had this weird idea the other day. We know that stars form from simple matter (dust, gasses, etc) forming a gravity well, attracting to each other, until the amount of mass and density in the gravity well creates enough heat and pressure sufficient to spontaneously start a fusion reaction (of course I am greatly oversimplifying).

What if we started with something quite dense and resistant to compression, like, say, water? Naturally this would be close to impossible to occur in nature, but it might theoretically be possible to do on purpose.

I'm trying to figure out how much pure water would need to gather in one spot to generate a fusion reaction... and would it be star-ish, if we started with just hydrogen and oxygen in a 2:1 ratio, or would that be inherently unstable and pretty much explode right away... or the opposite, immediately start collapsing into a black hole?

Answer

I think a "water-star" is possible at lower masses than for a star with a more normal composition - as I show below. There is no issue with beginning hydrogen fusion in a similar way to a normal star but the details are changed by the odd mixture. You probably need a full-blown stellar evolution model to answer this precisely and I am unaware of any such models for an oxygen-dominated star.

A first guess would be similar to a metal-rich star - i.e. about 0.075 times the mass of the Sun. Any less than this and the brown dwarf (for that is what we call a star that never gets hot enough at its centre to initiate significant fusion) can be supported by electron degeneracy pressure.

A star/brown dwarf with a "water" composition would be a different. The hydrogen and oxygen ions would be thoroughly and homogeneously mixed by convection. Note that other than a thin layer near the surface, the water would be completely dissociated and the hydrogen and oxygen atoms completely or mostly ionised respectively. Hence the density of protons in the core would be lower for the same mass density than in a "normal star". However, the temperature dependence is so steep I think this would be a minor factor and nuclear fusion would be significant at a similar temperature.

Of much greater importance is that there would be fewer electrons and fewer particles at the same density. This decreases both the electron degeneracy pressure and normal gas pressure at a given mass density. The star is therefore able to contract to much smaller radii before degeneracy pressure becomes important and can thus reach higher temperatures for the same mass as a result.

For that reason I think that the minimum mass for hydrogen fusion of a "water star" would be smaller than for a star made mainly of hydrogen.

A back of the envelope calculation could use the virial theorem to get a relationship between perfect gas pressure and the temperature, mass and radius of a star. Let gravitational potential energy be $\Omega$, then the virial theorem says $$ \Omega = -3 \int P \ dV$$ If we only have a perfect gas then $P = \rho kT/\mu m_u$, where $T$ is the temperature, $\rho$ the mass density, $m_u$ an atomic mass unit and $\mu$ the average number of mass units per particle in the gas.

Assuming a constant density star (back of the envelope!) then $dV = dM/\rho$, where $dM$ is a mass shell and $\Omega = -3GM^2/5R$, where $R$ is the "stellar" radius. Thus $$\frac{GM^2}{5R} = \frac{kT}{\mu m_u} \int dM$$ $$ T = \frac{GM \mu m_u}{5k R}$$ and so the central temperature $T \propto \mu MR^{-1}$.

Now what we do is say that the star contracts until at this temperature, the phase space occupied by its electrons is $\sim h^3$ and electron degeneracy becomes important.

A standard treatment of this is to say that the physical volume occupied by an electron is $1/n_e$, where $n_e$ is the electron number density and that the momentum volume occupied is $\sim (6m_e kT)^{3/2}$. The electron number density is related to the mass density by $n_e = \rho /\mu_e m_u$, where $\mu_e$ is the number of mass units per electron. For ionised hydrogen $\mu_e=1$, but for oxygen $\mu_e=2$ (all the gas would be ionised near the temperatures for nuclear fusion). The average density $\rho = 3M/4\pi R^3$.

Putting these things together we get $$h^3 = \frac{ (6m_e kT)^{3/2}}{n_e} = \frac{4\pi \mu_e}{3}\left(\frac{6 \mu}{5}\right)^{3/2} (Gm_e R)^{3/2} m_u^{5/2} M^{1/2}.$$
Thus the radius to which the star contracts in order for degeneracy pressure to be important is $$ R \propto \mu_e^{-2/3} \mu^{-1} M^{-1/3}.$$ If we now substitute this into the expression for central temperature, we find $$ T \propto \mu M \mu_e^{2/3} \mu M^{1/3} \propto \mu^2 \mu_e^{2/3} M^{4/3}.$$

Finally, if we argue that the temperature for fusion is the same in a "normal" star and our "water star" (after all, it is still just hydrogen that is fusing via the pp chain), then the mass at which fusion will occur is given by the proportionality $$ M \propto \mu^{-3/2} \mu_e^{-1/2}.$$

For a normal star with a hydrogen/helium mass ratio of 75:25, then $\mu \simeq 16/27$ and $\mu_e \simeq 8/7$. For a "water star", $\mu = 18/11$ and $\mu_e= 9/5$. Thus if the former set of parameters leads to a minimum mass for fusion of $0.075 M_{\odot}$, then by increasing $\mu$ and $\mu_e$ this becomes smaller by the appropriate factor $(18\times 27/11\times 16)^{-3/2} (9\times 7/5\times 8)^{-1/2} = 0.173$.

Thus a water star would undergo H fusion at $0.013 M_{\odot}$ or about 13 times the mass of Jupiter!

NB This only deals with hydrogen fusion. The small amount of deuterium would fuse at lower temperatures. A similar analysis would give a minimum mass for this to occur of about 3 Jupiter masses.