I Without degeneracy, when would Solar cores collapse?

[Ken G]

"Lose some heat" only in the sense that kinetic energy will be converted to potential energy as the core expands to virialize.

Yes, I should have said "lose some heat to virialize and be completely degenerate at the end of the process."

The expansion could be adiabatic. To carry the analysis of that case to completion, if we define $U_1 = G M_1^2 / R^1$ and we have $W_1 = - U_1$, $K_1 = U_1 / 2$, and $E_1 = - U_1 / 2$ by the virial theorem, then we have

$E_1 = E_0$ by the adiabatic condition, and therefore $U_1 = U_0$.

$U_1 = U_0 \left( 1 + 2 f \right) R_0 / R_1$ by simple algebra.

And, combining the two above conditions:

$R_1 = \left( 1 + 2 f \right) R_0$.

So the radius increases by a factor which is the square of the mass increase factor. The core could then contract by heat loss, but in the degenerate case, as I think we've already commented, the core will be colder than the surrounding shell, so it will be gaining heat, not losing it.

Actually, I'll bet it would be hotter than the shell if it contracted adiabatically. I think what happens if you drop the mass from infinity and have it virialize adiabatically, is it expands and its temperature rises (because even though the average kinetic energy per particle will drop, the degree of degeneracy will also drop, and the T will rise). So it should still lose heat to the shell, not gain heat from it. Another complication is there are always neutrino losses, based on the electron energy, even at low temperature (I believe that's true, which is also what @Vanadium 50 seemed to be saying). So even if there is heat coming in from the shell, there could be net heat loss due to the neutrinos.

 

[Vanadium 50]

Neutrinos: I doubt very much that neutrinos thermalize. When a core (of real electrons, not weirdo electrons) collapses, the nutronization pulse duration is comparable to the core radius (divided by c). That means they can't scatter too often and thermalize.

Further, the core is cold on neutrino scales - keV. Neutrinos would heat it, not cool it.

Derivation: @PeterDonis you may find it easier to work with differentials than deltas. That opens up the full power of calculus, and can make dependencies more obvious, It doesn't always help, but usually does. i.e. R + dR might be the better starting point.

Often if you get two different answers depending on your assumptions means your starting point is unphsyical.

 

[Ken G]

This raises a couple of other questions:

First, how does this relate to the Fermi energy? The Fermi energy in the non-relativistic degenerate regime goes like $M^{2/3} / R^2$. So the Fermi energy in this case will change by a factor $\left( 1 + f \right)^{2/3} / \left( 1 + f \right)^4 \approx 1 - \left( 10 / 3 \right) f$.

This seems odd.

You are using two things that are inconsistent. The behavior of R when M is added comes from keeping the total energy constant, which means you are dropping the mass from infinity. But then you are taking that resulting expanded adiabatic radius, and putting it into a formula for what the Fermi energy would be at that M and R. But the electrons won't have the Fermi energy, because dropping the mass from infinity spoils the degeneracy. If you let it go degenerate, then it will have the Fermi energy, but a smaller R than what you are using.

Second, if we look at the white dwarf equilibrium solutions in S&T section 3.3, we find that they obey the proportionality $M \propto R^{-3}$, or $R \propto M^{- 1/3}$. Whereas in our analysis here we found $R \propto M^2$. In other words, if our core were an isolated white dwarf and we added mass to it, we would expect the new equilibrium to have a smaller radius; but here the new equilibrium has a larger radius. Why does our degenerate core not act like a white dwarf?

Because it had mass dropped onto it, which must lose heat to recover degeneracy. The minimum energy you can add is place it at the surface with no kinetic energy, that will be much closer to degenerate.

Of course we could change our assumptions about the kinetic energy of the added mass, and the solution $R \propto M^{- 1/3}$ is within the range of the assumptions we considered. But that still leaves the question of how the total energy of the core can be lower after mass is added, when there has not been time for any heat loss to take place.

When you drop mass onto a white dwarf, you keep its energy the same but raise the total kinetic energy too much, so like you said, it will expand. Then the potential energy eats up the extra kinetic energy and it revirializes, but it won't have the Fermi energy. If you want to use the white dwarf relation between M and R, you have to let it lose heat to get degenerate again, and then it will have the Fermi energy, but it will have contracted to a smaller R by that point. So the total energy of the core will not be lower until you let heat be lost, you just can't use the Fermi energy for anything.

The point is, if you are considering an adiabatic process, why bring in the Fermi energy at all? It is not physically relevant, you already know the M, the R, and the energy situation, you have nothing left to do but revirialize it. The Fermi energy is a red herring, unless you also stipulate that you are letting it lose heat and go to its minimum possible energy state.

 

[PeterDonis]

Why does our degenerate core not act like a white dwarf?

I realized on re-reading this that I phrased that post and this question badly. Let me try to rephrase.

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. 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.

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. 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.

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?

 

[Ken G]

Neutrinos: I doubt very much that neutrinos thermalize. When a core (of real electrons, not weirdo electrons) collapses, the nutronization pulse duration is comparable to the core radius (divided by c). That means they can't scatter too often and thermalize.

The issue was not if neutrinos thermalize to some T, it is if their emission mechanisms refer to the kT of the electrons, or to the energy of the electrons (which is much higher if they are degenerate). The processes themselves look like it would be the electron energy that matters, but that can't be right because a degenerate system is in its ground state and cannot lose energy except via a process that changes the composition of the star, like electron capture. We know that doesn't happen much until you get to core collapse. So it must end up being the kT of the electrons that matter for neutrino cooling. Hence I agree that the weirdos might have a significant problem with neutrino cooling (by which I mean energy loss, not T drop, that term is always ambiguous that way!), I'm really not including any of the detailed cooling timescales in this simplistic analysis, I'm wondering what the added mass is doing if it is added very quickly on those timescales.

Further, the core is cold on neutrino scales - keV. Neutrinos would heat it, not cool it.

No, neutrinos cool (in the sense of removing energy from) the cores of all late stages of stellar evolution. The energy scale of electrons in an iron core is much higher than that, almost the MeV range even before core 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).