What is the theoretical lower mass limit for a white dwarf?

Question

This question in inspired by this other one, which asks what is the theoretical lower mass limit for a gravitationally stable neutron star (not the Chandrasekhar limit which is the upper mass limit for a white dwarf, or the effective lower mass limit of real neutron stars that are formed in the universe, but how little mass one could form a gravitationally stable neutron star in theory). According to ProfRob's answer to it, it probably lies somewhere between $0.087$ and $0.19$ solar masses (computations differ, but this gives us an order of magnitude).

I would like to ask the exact same question about white dwarfs: is there a theoretical lower mass limit at which they are stable, and if so, what is it? Again, I don't mean the lower mass limit at which real white dwarfs are formed in the universe, I mean the lowest mass for which it could remain stable.

Or to put it differently, if we remove mass from a white dwarf, it is well known that its radius increases (roughly as the inverse cube root of the mass): how far down does this relation hold, and what happens if we keep removing mass? Does the star eventually break apart? Or do we encounter some kind of discontinuity as matter "de-degenerates"? Or does the star's matter simply continuously become less and less degenerate as we remove it? If the last is correct, what is the order of magnitude of the mass for which the radius would be a maximum (and which is arguably the point at which the star ceases to be a white dwarf)?

The answer might depend a lot on the star's composition and temperature, but I just want a ballpark figure, not a detailed analysis. (Say, maybe a cold/black dwarf made of carbon.)

Answer 1

There is no obvious lower limit to the mass of an object that can be supported by a cold, electron-degenerate equation of state. Note though, that this would not be governed by ideal electron degeneracy pressure.

A typical carbon white dwarf of half a solar mass, would have a radius about that of the Earth. If you removed mass, then it would become larger (roughly as $M^{-1/3}$), but would still be stable, because $dM/d\rho$ is positive (where $\rho$ is the average density).

At around a few thousandths of a solar mass (maybe half a Jupiter mass), the object would reach a maximum size of about 4 Jupiter radii and would essentially be a giant carbon planet Zapolsky & Salpeter 1969). This maximum (which would not occur for an ideal electron-degenerate equation of state) is associated with a range of unavoidable, non-ideal interactions in the gas (e.g. Thomas-Fermi corrections) that harden the equation of state - the pressure depends more strongly on density.

If you continued to remove mass, then somewhere below about half a Jupiter mass, the planet would rapidly start to become smaller again, and might be referred to as a carbon "terrestrial" planet, but would still be stable at a density that is almost independent of the mass.

Finally, if you remove more mass, you have a lump of coal!

i.e. There is no equivalent to the minimum mass of a neuton star because the relationship between mass and average density retains a positive gradient at lower masses - which leads to stability. There is however likely to be an astrophysical lower limit to the smallest white dwarf that can be produced during stellar evolution - these are likely to be the lowest mass helium white dwarfs. Helium white dwarfs can be produced by the evolution of single stars of low-mass; but such objects could not yet be produced during the lifetime of the universe. Instead, the low-mass helium white dwarfs, perhaps as low as $0.1M_\odot$ could be produced by stripping the envelope away from a helium-cored red-giant (e.g. Althaus & Benvenuto 1997).

In terms of measured masses, the lowest mass white dwarf stars (if you exclude planetary mass objects from the definition) are about $0.2 M_\odot$ (e.g. see Why is the white-dwarf mass distribution highly peaked?).

Answer 2

For simplicity, let's consider a Hydrogen white dwarf. With the decreasing mass of the white dwarf, its Fermi energy $E_F$ is decreasing. Once the Fermi energy is comparable to the typical energy of an ideal gas $E_{\rm gas}$, we should say this is not a degenerated state but ideal gas. (I.e. the Fermi temperature is comparable to the real temperature.) Then we should not call it a white dwarf.

$E_F \propto n^{2/3}$, where $n$ is the number density (ref see here: https://en.wikipedia.org/wiki/Fermi_energy ).

$E_{\rm gas} = \frac{1}{2} k_{\rm B} T.$

Notice $E_F$ does not depend on temperature, while ideal gas does. This indicates that there is no minimum mass for a white dwarf. Once if you keep decreasing its temperature, any tiny mass can become degenerated. On the other hand, for a given temperature, it is possible to find the minimum mass.

However, you won't find a very low mass white dwarf, as it cannot form. For those low mass hydrogen blocks, they are called brawn stars. The universe lifetime is not long enough to cool them into a degenerated state (which can be called a white dwarf? but they are not white.).

Things become more complicated if we also consider the fact: the degeneracy is the degeneracy of electrons. If the temperature is too low, the hydrogen won't become ionized. And if the mass is very small, the gravity is easily be recovered by the pressure of ideal gas of atomic or molecular hydrogen. In this sense, there might be a minimum mass, that requires $E_F>13.6$ eV to keep the ionization. However, this is a rough estimite, as partly ionization can also support the white dwarf from collapsing.