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 Rob Jeffries'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).

So now I would like to ask the exact same question about white dwarfs (white dwarves?): 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.)

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    $\begingroup$ How would you distinguish between a planet and a white dwarf? Or a rock? $\endgroup$ – Rob Jeffries May 31 '18 at 19:04
  • $\begingroup$ @RobJeffries I think that is part of the question. But I propose a line of answer: keep removing mass from the white dwarf until (a) it blows up, (b) some other kind of discontinuity happens, or (c) its radius simply reaches a maximum. I suggest that maybe it ceases to be a white dwarf at that point, and my question is what that point is, and which case is true. $\endgroup$ – Gro-Tsen May 31 '18 at 19:24

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.

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    $\begingroup$ Cold very low mass hydrogen white dwarfs (a.k.a. gas giant planets) are not governed by ideal degeneracy pressure. $\endgroup$ – Rob Jeffries Apr 26 '19 at 19:33
  • $\begingroup$ And they are not called white dwarfs, but gas giant planets. $\endgroup$ – Yuan-Chuan Zou May 28 '19 at 16:06

There is no obvious lower limit to the mass of an object that can be supported by a cold, electron-degenerate equation of state.

If you had a carbon white dwarf, which would have a radius about that of the Earth, and you removed mass, then it would become larger, but would still be stable.

Below a few hundredths of a solar mass, the object would reach a maximum size of about a Jupiter radius (or perhaps a bit less, since carbon has more mass units per electron) and would essentially be a giant carbon planet.

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

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

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  • $\begingroup$ But there is a mass for which the radius is maximal, right? (In the question, I argue that this is a reasonable candidate for a “minimal white dwarf mass” because this is where the variation of radius with mass changes sign.) What is its order of magnitude, and how does it vary with, in particular, temperature? $\endgroup$ – Gro-Tsen Nov 18 '19 at 12:54
  • $\begingroup$ @Gro-Tsen I've edited. A Jupiter radius is about the maximum radius and would be achieved at around ~10-30 Jupiter masses. If you allow the temperature to vary, you could essentially have any radius you like. $\endgroup$ – Rob Jeffries Nov 18 '19 at 21:07

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