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Assuming it is somehow possible to get a small piece of a white dwarf (maybe a dice) and this piece escapes into free space. Would that piece of white dwarf matter keep its density/state, or would it expand, explode, or transform into another state without the huge gravitational force of the whole white dwarf it was captured in before?

There is a similar question here for neutron stars: What would happen to a teaspoon of neutron star material if released on Earth?. Neutrons decay without the large pressure inside the neutron star, so the fate of a piece of neutron star material is clear. But what would happen to white dwarf material? I think the state of white dwarf matter is different from neutron stars (which mainly consists of neurons...)?

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    $\begingroup$ The decay of neutrons is a minor (and slow) contributor to the energy released from unconstrained neutron star matter. $\endgroup$
    – ProfRob
    Commented May 11, 2021 at 23:11

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The answer is similar to What would happen to a teaspoon of neutron star material if released on Earth? but the energetics are not so extreme. The material consists of fully ionised carbon and oxygen nuclei plus a gas of highly degenerate electrons.

At typical densities for white dwarfs of $10^{9} - 10^{11}$ kg/m$^3$, the internal kinetic energy density of the electrons in the degenerate gas is very high - of order $10^{22}$ to $10^{25}$ J/m$^3$. This energy would be "released"/explode if the white dwarf material is unconfined by gravity. Note that this would be true, even if the white dwarf material was cool, since the kinetic energy density of a degenerate gas is independent of temperature and the ions in the gas are minor contributors to the overall kinetic energy.

The result would be widely dispersed electrons plus carbon and oxygen ions.

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  • $\begingroup$ Many thanks, that's plausible! Is this high kinetic energy of electrons in the degenerated gas actually the counterforce compensating or avoiding the gravitational collapse of a white dwarf? $\endgroup$
    – Charles Tucker 3
    Commented May 11, 2021 at 22:46
  • $\begingroup$ @CharlesTucker3 Yes. Kinetic energy density is pressure. $\endgroup$
    – ProfRob
    Commented May 11, 2021 at 23:09
  • $\begingroup$ When it comes to degenerate matter, it is correct to equate kinetic energy density to P? Or is just the kinetic part a contribution? I mean, degeneracy P should be there even in absence of motion, like we can compress matter virtually so slowly that it remains in thermal equilibrium with the surrounding. In other words, the virial theorem don't apply in this extreme case. Please clarify otherwise I think I must ask it... $\endgroup$
    – Alchimista
    Commented May 12, 2021 at 13:26
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    $\begingroup$ @Alchimista it isn't equal to $P$ except to a small numerical factor. The pressure is 2/3 the kinetic energy density for non-relativistic degeneracy and 1/3 for ultra-relativistic degeneracy. See any statistical mechanics textbook. There is no pressure without kinetic energy. That is "kinetic theory". $\endgroup$
    – ProfRob
    Commented May 12, 2021 at 13:38
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    $\begingroup$ @Alchimista $T$ plays no role in the pressure of a degenerate gas. $\endgroup$
    – ProfRob
    Commented May 12, 2021 at 15:22

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