Without degeneracy, when would Solar cores collapse?

[PeterDonis]

if you are considering an adiabatic process, why bring in the Fermi energy at all?

There might be some confusion about the term "Fermi energy". I am using it in the sense that S&T use it; but the way S&T calculate it has implications that seem to me to make it relevant in any process.

S&T calculate what they call the Fermi energy as follows: first they calculate the "Fermi momentum" $p_F$ by using the uncertainty principle: for an object containing $N$ fermions in a star of radius $R$, the fermion number density is $n = N / R^3$ and the uncertainty in position of a given fermion (i.e., the size of the "cell" in space that it occupies) is $\approx 1 / n$. The uncertainty in momentum is then $p_F \approx \hbar n^{1/3}$.

The Fermi energy is then obtained using the obvious formula for whichever regime is under consideration: in the non-relativistic regime, $E_F = p_F^2 / 2m$, and in the relativistic regime, $E_F = p_F c$.

This calculation seems to me to be relevant regardless of the particular value of $R$ or the particular process under consideration; it is, so to speak, giving at least a minimum value of kinetic energy per electron. The electron would have exactly this kinetic energy at zero temperature in an object containing $N$ electrons with that radius $R$; it could have more at finite temperature; but it can't have less.

 

[Ken G]

For a given mass of electron degenerate matter, the white dwarf equilibrium solution for that mass has the property of being the state of minimum energy.

Agreed.

So of course that is the state we would expect an isolated white dwarf of that mass to equilibrate to, by losing heat if necessary. If we then add some mass to this isolated white dwarf, we would expect it to re-equilibrate to the new white dwarf equilibrium configuration for its new mass.

Agreed.

For the same mass of electron degenerate matter inside a surrounding fusion shell, obviously we can't assume that that degenerate matter will be able to get to the same equilibrium state as an isolated white dwarf.

It can be close enough for our purposes though, so far I'm not seeing any problem with that until the mass gets added. At that point one has to decide if it's adiabatic or going degenerate. In the former case, you have to watch the energy that you are giving it, in the latter case, you don't care about that.

If nothing else, the white dwarf equilibrium state assumes zero temperature, but a degenerate core inside a surrounding fusion shell would not be expected to be at zero temperature. And similar remarks would apply if we add mass to the degenerate core in the form of fusion ash from the surrounding shell. But it still seems like the degenerate core would be colder than the surrounding shell, because of the effect mentioned earlier, that degenerate electrons steal kinetic energy from the ions, and at least a large piece of that kinetic energy is independent of temperature.

Degenerate electrons are very good conductors, the core generally is maintained at the T of the shell. In the case of stellar cores at earlier stages, like the core of the Sun when it becomes a red giant, the usual assumption is the core is isothermal, at the T of the fusion shell. (It's also highly degenerate, so the electron kinetic energies are way higher than that, it is a very tight ball that is virialized all on its own and is like a white dwarf that doesn't much care, for its hydrostatic equilibrium, about its T > 0.)

So what I'm trying to understand is, for the case of the core with the surrounding shell, if it isn't the zero temperature white dwarf equilibrium state that governs what happens, what does govern what happens? Or is there no particular state that the system necessarily gets driven towards, it depends on the details of how fast the mass gets added vs. other processes that are going on?

It will depend on the various timescales, like the timescale for adding mass, and the timescale for heat loss. We can generally assume force balance is very fast, so it will stay virialized at whatever its M, but if it gains mass faster than it can lose heat, we will have the adiabatic case that requires we know how much kinetic energy the mass comes in with. The Fermi energy will never matter there, the core won't stay degenerate. A more standard assumption is that the heat loss is fast enough to maintain degeneracy, or at least that the mass is not dropped in from infinity, but perhaps introduced at the surface with very little kinetic energy (that's what I was picturing). The latter assumption is probably not that much different from letting it stay degenerate, as it seems like a pretty "minimum energy" kind of configuration.

So it seems somewhat reasonable to imagine that the real electrons will maintain the minimum energy force balance as M increases, so will follow the white dwarf relation. But the weirdo electrons should lose more heat than that, so contract more, as you said. It would only be when the real electrons go relativistic that their degeneracy becomes an assist to contraction (that's the fascinating consequence of degeneracy no one talks about, especially if they think it is just some kind of extra outward force), since we know they must go to zero radius at 1.4 solar masses. The weirdos do not have that strict limit, they would sail past 1.4 without contracting to zero radius because of the nonrelativistic kinetic energy in their ions. But this all ignores the fact that endothermic runaway occurs before the electrons get completely relativistic, so it's a question for a more accurate simulation that considers all these competing timescales and energy scales.

That said, I do agree that the weirdos would likely have a tough time producing a larger radius than the real ones at any time before core collapse occurs, but it serves to demonstrate the more important fact that real electrons sow the seeds of their own demise by going degenerate, since that's what negates the potential saving graces of the nonrelativistic ions. That is what happens at the Chandra mass, it is the place where the electrons' capacity to steal kinetic energy from the ions makes them unable to revirialize. You can imagine the ions saying to the electrons, "we could have saved you from contracting to zero radius if you hadn't stolen all our kinetic energy." I think that's a more interesting way to explain what happens at that mass limit than it's where gravity overcomes the quantum mechanical forces of degeneracy pressure.

 

[Ken G]

There might be some confusion about the term "Fermi energy". I am using it in the sense that S&T use it; but the way S&T calculate it has implications that seem to me to make it relevant in any process.

I agree that at any M and R there is a Fermi energy, and it is meaningful as some kind of benchmark. The issue is, if the system is behaving adiabatically, it will generally not be at its minimum total energy, so the Fermi energy won't be realized.

S&T calculate what they call the Fermi energy as follows: first they calculate the "Fermi momentum" $p_F$ by using the uncertainty principle: for an object containing $N$ fermions in a star of radius $R$, the fermion number density is $n = N / R^3$ and the uncertainty in position of a given fermion (i.e., the size of the "cell" in space that it occupies) is $\approx 1 / n$. The uncertainty in momentum is then $p_F \approx \hbar n^{1/3}$.

Yes, I understand.

The Fermi energy is then obtained using the obvious formula for whichever regime is under consideration: in the non-relativistic regime, $E_F = p_F^2 / 2m$, and in the relativistic regime, $E_F = p_F c$.

Yes.

This calculation seems to me to be relevant regardless of the particular value of $R$ or the particular process under consideration; it is, so to speak, giving at least a minimum value of kinetic energy per electron. The electron would have exactly this kinetic energy at zero temperature in an object containing $N$ electrons with that radius $R$; it could have more at finite temperature; but it can't have less.

Yes, it's the minimum kinetic energy they can have at that radius, and the maximum kinetic energy the particles can actually have in a system with that M if the R is in force balance (the crossing of that minimum and maximum is the "go no further" signpost). We agree that is relevant as a signpost, I don't deny that, and it will be achieved if the system is actually degenerate, but I think it presented problems when it was being interpreted as the energy a system that is not losing heat would actually have. That's why I asked, why use it? You already know the energy if it's adiabatic, you don't need any signposts in the distance.

 

[PeterDonis]

Degenerate electrons are very good conductors, the core generally is maintained at the T of the shell. In the case of stellar cores at earlier stages, like the core of the Sun when it becomes a red giant, the usual assumption is the core is isothermal, at the T of the fusion shell.

Hm. I'll have to think about what that implies for the simple math I've been doing.

 

[Ken G]

Hm. I'll have to think about what that implies for the simple math I've been doing.

I think your simple math is fine, the kT of the real electrons is still very low compared to the Fermi energy, and for the weirdos we just have to treat them as adiabatic or else we have no idea what kT to give them. We wouldn't want to cool the weirdos down to the shell T, that would be a huge heat loss and they would definitely collapse. That might be what would really happen, but it wouldn't demonstrate the point. So we just say the mass is added faster than the heat can be lost, and do it adiabatically for both cases, but keep the real electrons close to the Fermi energy by bringing the mass in with the appropriate kinetic energy (which I suspect is a lot like introducing them at the surface with very little kinetic energy).