I Without degeneracy, when would Solar cores collapse?

[PeterDonis]

a device for locating the "go no further" sign is quite different from a physical description of what is actually going on.

I disagree. Forces are just as "actual" as energies, and a description in terms of forces is just as much a description of "what is actually going on" as a description in terms of energies. Both can be true and valid at the same time. One or the other might be preferred for a particular purpose or by a particular person, but that doesn't make the other one any less "actual" or valid.

 

[Ken G]

I disagree. Forces are just as "actual" as energies, and a description in terms of forces is just as much a description of "what is actually going on" as a description in terms of energies. Both can be true and valid at the same time.

Yes, they are just as actual, but that's not the issue at all. It just doesn't have to do with forces. But wait on that, forget the critique of pedagogy, this whole thread is to avoid subjective critiques. We have a real question on the table, which again is, why can you point to a 2 or 3 solar mass iron core that is ideal gas and see that it is just having no problem at all, at this moment, maintaining force balance in a star with a mass of like 30 solar masses, but you will never be able to do that with a highly degenerate core. What's more, if you enter weirdo classical electron world, you can start with a 1 solar mass highly degenerate core, add a bunch of weirdo electrons, and have no problem going past 1.4 solar masses (if you do it fast enough), but you'll never be able to do that with highly degenerate real electrons. Why is that?

 

[PeterDonis]

to understand why degenerate gas can never go above 1.4 solar masses, while ideal gas can

I still don't understand what comparison you are trying to make here or why you think it's so important. Yes, there are a variety of objects that astronomers look at, some of which have degeneracy as a significant factor in their behavior, and some of which don't. But I don't see what particular pair of such objects are "similar" enough to make the kind of comparison you are making in the quote above.

 

[PeterDonis]

why can you point to a 2 or 3 solar mass iron core that is ideal gas and see that it is just having no problem at all, at this moment, maintaining force balance in a star with a mass of like 30 solar masses, but you will never be able to do that with a highly degenerate core. Why is that?

The ideal gas core in this case is nowhere near white dwarf densities, so again I don't see why I would even want to compare it with a degenerate core that is at white dwarf densities in some other star with some other mass.

 

[PeterDonis]

forget the critique of pedagogy

I don't see how I can, since the main problem you are having in this thread is one of pedagogy. It's not a matter of me disagreeing with any actual physics you are expounding (leaving aside the issue I have already pointed out about thought experiments that violate the laws of physics). I just don't get why you are concentrating on the particular things you are concentrating on or why I should think they are important. That's a problem of pedagogy.

 

[Ken G]

The ideal gas core in this case is nowhere near white dwarf densities, so again I don't see why I would even want to compare it with a degenerate core that is at white dwarf densities in some other star with some other mass.

Then consider the situation where the densities are the same, the real electrons and the weirdo electrons. That is the device to understand what degeneracy is actually doing, because the only difference in the two situations is the particle distribution functions. Answer in that situation, why can the weirdo electrons go past 1.4 solar, but the highly degenerate ones cannot? What would actually happen to those populations as you quickly (but in force balance) add mass from 1 to 1.4? (That's not a pedagogical question, it's a physics question, albeit with hypothetical elements.) And by the way, it will be important that both these situations involve not just electrons, but also ions. (And it's not the charges that matter.)

 

[Vanadium 50]

I'm still unclear on what will happen when the magic want is waved.

Electrons become distinguishable, so the emtropy goes away up. The internal energy initially stays the same, because the system is virialized, and so the temperature goes way up. That surely cranks up the fusion rate, so you end up again with very different stars.

 

[PeterDonis]

consider the situation where the densities are the same, the real electrons and the weirdo electrons.

In this situation, as I have already said, the real electrons will stop contracting due to degeneracy, whereas the weirdo electrons won't. The weirdo electron core will keep contracting, so if, say we have both cores at 1 solar mass and a typical white dwarf density for that mass, then by the time both cores have grown to, say, 1.2 solar mass, the weirdo electron core will be significantly more dense than the real electron core. And that, to me, means the chances of endothermic processes like electron capture will be significantly higher in the weirdo electron core (since the rates for such processes become significant at densities right at the high end of the white dwarf density range), meaning that the weirdo electron core will likely start a catastrophic collapse at a mass below the Chandrasekhar limit, whereas the real electron core, whose density will stay lower, will not catastrophically collapse until it exceeds the limit.

why can the weirdo electrons go past 1.4 solar

Can they? Under what circumstances? See above.

 

[Ken G]

I'm still unclear on what will happen when the magic want is waved.

We start with two identical cores of highly degenerate electrons (and ideal iron ions). Both are at 1 solar mass, so they look a lot like white dwarfs. We wave the wand over one of them, and strip them of their PEP, that is, we label each electron so they are no longer indistinguishable, but all this does is release them from the PEP, nothing else.

Electrons become distinguishable, so the emtropy goes away up. The internal energy initially stays the same, because the system is virialized, and so the temperature goes way up. That surely cranks up the fusion rate, so you end up again with very different stars.

You are correct in all those statements, except the fusion. This is iron.

 

[Ken G]

In this situation, as I have already said, the real electrons will stop contracting due to degeneracy, whereas the weirdo electrons won't.

Yes, I said that also, from the very start of the thread. But remember, the degenerate electrons won't actually stop contracting, because mass is being added. We completely agree that the weirdo electrons will have a very high T and may start losing heat prodigiously, even if that is somewhat limited by that fusing silicon shell around them. But there is a timescale involved there, and it should be rather long because energy transport timescales are generally much longer than momentum transport (i.e., free fall) timescales, until you get into the core collapse phase.

The weirdo electron core will keep contracting, so if, say we have both cores at 1 solar mass and a typical white dwarf density for that mass, then by the time both cores have grown to, say, 1.2 solar mass, the weirdo electron core will be significantly more dense than the real electron core.

Yes, but the key point is this takes time, and we have a silicon burning shell adding mass quite fast. It takes only a few weeks to significantly enhance the core mass. This is the key player in all core collapse scenarios, the core mass is rising rapidly, and the new material is generally coming in "undervirialized", meaning it is not pulling its own weight in terms of the kinetic energy it shows up with. The core sags under the new weight, until it recovers virialization. However, the weirdo electrons are also losing heat at the same time, and this is where the timescale competition comes in that I mentioned early in the thread. The key is that the mass has to be added faster than the heat loss is happening to the ideal core, so the main contraction is from the rising mass and not the contraction from heat loss. That is what happens, for example, in a regular core collapse in a real star, the core collapse is ushered in more by adding mass than by waiting for the core to lose heat. I grant you that we may have to add the mass pretty quickly, but so be it, that's the scenario under consideration. We are asking why adding mass to a degenerate core makes it collapse, when it does not cause collapse of an ideal core with all the same properties except the particle distribution functions, i.e., it has the same density and energy but the kinetic energy is distributed differently. In the latter case, you can point to it and say, look, that same core we had a bit ago is now less contracted than the degenerate one, even though the degenerate one is supposed to have all that extra quantum mechanical pressure force! The remaining question is, why does this happen? There is a very specific reason. You are still wondering if it happens, I'm saying, add mass fast enough so that it does.

And that, to me, means the chances of endothermic processes like electron capture will be significantly higher in the weirdo electron core (since the rates for such processes become significant at densities right at the high end of the white dwarf density range), meaning that the weirdo electron core will likely start a catastrophic collapse at a mass below the Chandrasekhar limit, whereas the real electron core, whose density will stay lower, will not catastrophically collapse until it exceeds the limit.

That would certainly be true if the ideal core had enough time to lose heat and contract more than the degenerate core does. But if the mass is added quickly, the degenerate core will contract much faster, because it must reach zero radius by the time the Chandra mass is reached, if it didn't have all those endothermic processes kick in first. (The other timescale of interest is the force balance timescale, but that's the free fall time and is very fast indeed, so we will always assume force balance until collapse kicks in.)

Can they? Under what circumstances? See above.

If you don't give the heat loss enough time to act. Heat transport is generally slow, though I admit we have a lot of processes like neutrino escape and ultimately neutronization and photodisintegration. But those latter really kick in when you have core collapse, and we are analyzing the time before that, when energy transport timescales are still slow.

 

[PeterDonis]

If you don't give the heat loss enough time to act.

When I say the ideal core will keep contracting while the degenerate one will stop, I'm not just talking about heat loss at a fixed mass (or more precisely a fixed number of baryons and electrons). I am talking about the effect of adding more mass. As mass is added, the ideal core will compress more than the degenerate core.

Here's why I think that: for an ideal Maxwell-Boltzmann gas, the kinetic energy, and hence the pressure, goes like $1 / R$ always. So it compresses the same way in all regimes.

But for a non-relativistic Fermi gas, the kinetic energy, and hence the pressure, goes like $1 / R^2$. That means that the pressure increases faster as the gas is compressed.

Once the Fermi gas becomes relativistic, the kinetic energy, and hence, the pressure, goes like $1 / R$, as with the ideal gas, so in that regime the gas will compress the same as an ideal gas.

But from the above, we can see that a real electron gas will have a "bump" in its pressure behavior, while it is in the non-relativistic degenerate regime, whereas the weirdo electron gas will not. And while the real core is in the "bump" regime, it will compress less than the weirdo core does. This is true regardless of what is causing the compression, mass being added or heat loss. So once the "bump" regime is entered, the real core will be less compressed than the weirdo core, for the rest of the "experiment".

 

[Ken G]

When I say the ideal core will keep contracting while the degenerate one will stop, I'm not just talking about heat loss at a fixed mass (or more precisely a fixed number of baryons and electrons). I am talking about the effect of adding more mass. As mass is added, the ideal core will compress more than the degenerate core.

This is indeed the crucial issue, let us see where it leads.

Here's why I think that: for an ideal Maxwell-Boltzmann gas, the kinetic energy, and hence the pressure, goes like $1 / R$ always. So it compresses the same way in all regimes.

But for a non-relativistic Fermi gas, the kinetic energy, and hence the pressure, goes like $1 / R^2$. That means that the pressure increases faster as the gas is compressed.

This is a very important point. If indeed the degenerate gas stayed nonrelativistic, which we would normally assume for simplicity, then it is true that the degenerate gas would not contract as much. But the reason there is this 1.4 solar mass limit is that the degenerate gas goes relativistic. So that's why the radius actually goes to zero, in force balance, for completely degenerate gas. All you have to do is add mass to get to 1.4, maintain force balance and complete degeneracy, and the degenerate core is gone, it has fallen into core collapse.

To expound a bit, the reason highly relativistic gases contract to zero radius is that their virial theorem has a qualitatively different character. If you add undervirialized mass, a nonrelativistic gas can contract, release gravitational energy, and it only takes half that released energy to maintain the previous level of virialization, the remainder is available to improve the degree of virialization (if that excess kinetic energy is not lost as heat, so this is why we have to add the mass fast). But highly relativistic gas has no such margin for revirializing itself, it needs all the released gravitational energy just to maintain whatever level of virialization it had before. Hence adding undervirialized gas is catastrophic for highly relativistic gas, and this is a crucial ingredient of core collapse.

But the ideal gas core does not suffer that fate. This is the punchline. Why doesn't the ideal gas core also go relativistic, and also contract to zero radius? Why can it sail past 1.4 solar masses and not be relativistic, even these weirdo electrons that started out at the same energy as their degenerate cousins in the scenario where they were both at the same total energy and density when they had 1 solar mass? (And here is where the ions matter.)

Once the Fermi gas becomes relativistic, the kinetic energy, and hence, the pressure, goes like $1 / R$, as with the ideal gas, so in that regime the gas will compress the same as an ideal gas.

Ah yes, I see you have already realized the key point. But something is missing.

But from the above, we can see that a real electron gas will have a "bump" in its pressure behavior, while it is in the non-relativistic degenerate regime, whereas the weirdo electron gas will not. And while the real core is in the "bump" regime, it will compress less than the weirdo core does. This is true regardless of what is causing the compression, mass being added or heat loss.

Everything you are saying is correct, but forgetting one critical point: the ions!

 

[PeterDonis]

the reason highly relativistic gases contract to zero radius is that their virial theorem has a qualitatively different character

Yes, because the ratio of kinetic energy to (absolute value of) potential energy goes from $1/2$ in the non-relativistic limit to $1$ in the relativistic limit. But this will be true whether the relativistic gas is an ideal gas or a Fermi gas. So it will affect both cases (real electrons and weirdo electrons) the same, as soon as each case reaches the density where it becomes relativistic. And, as already noted, the weirdo electrons will be compressed more so they will reach that density sooner.

the ions!

The ions are an ideal gas in both cases (they don't become degenerate until neutron star densities are reached), so I don't see any difference in behavior there between the two cases.

 

[Vanadium 50]

Ah yes, I see you have already realized the key point. But something is missing.

I really, really, really hope you are not playing Twenty Questions here.

 

[Vanadium 50]

You are correct in all those statements, except the fusion. This is iron.

It will induce fusion in the envelope, and it will increase the rate of fusion before you get to iron. But no matter - let's start the clock with an iron core where electrons suddenly behave classically.

As we agreed, the temperature goes way, way up. So the radiation emitted also goes way, way up. As energy is lost, the temperature goes up (the specific heat is negative) and the radiation increases more. My copy of S&T is in a box at the moment, but I think this is similar to their argument. This in turn causes the core to contract more. This slows the process, but cannot stop it - it doesn;t stop until the neutron degeneracy kicks in.

So if you turn off electron degeneracy, you skip the whit dwarf phase (and the core of a red giant is white dwarf-like) and go straight to the neutron star phase. As expected.

 

[Ken G]

Yes, because the ratio of kinetic energy to (absolute value of) potential energy goes from $1/2$ in the non-relativistic limit to $1$ in the relativistic limit. But this will be true whether the relativistic gas is an ideal gas or a Fermi gas. So it will affect both cases (real electrons and weirdo electrons) the same, as soon as each case reaches the density where it becomes relativistic. And, as already noted, the weirdo electrons will be compressed more so they will reach that density sooner.

Right, another way to say the same thing is that when you contract undervirialized nonrelativistic particles, half the gravitational energy released goes into maintaining the previous level of virialization, and half goes into making up the previous shortfall, so in that way contraction brings the gas closer to being virialized. But for highly relativistic particles, essentially all of the liberated gravitational energy goes into simply maintaining the previous degree of virialization, and no "catching up" can occur, so at that point adding any more mass causes complete contraction to zero. When you mix highly nonrelativistic particles with highly relativistic particles, you can derive that the half of the released energy that would have gone into catching up gets reduced to just the fraction of the kinetic energy that is in the nonrelativistic population.

The weirdo electrons do lose more heat, so would contract faster if losing heat was the only issue. But if revirialization after mass increase is the only issue, then both groups contract similarly. So you correctly point out that when we take that into account and tack on the extra heat loss, the loser would be weirdo electrons, but only if they had the same degree of relativity as the degenerate case.

The ions are an ideal gas in both cases (they don't become degenerate until neutron star densities are reached), so I don't see any difference in behavior there between the two cases.

The ions don't go relativistic. I might be overstressing the importance of this, given that there is only one iron ion for twenty six electrons (it's more important for carbon in type Ia supernovae, where you have one ion for six electrons), but a very important thing that happens when you mix ions and degenerate electrons is that the degenerate electrons steal almost all of the kinetic energy of the ions (for very high degeneracy). This is the thing that degeneracy does that rarely gets mentioned, it is the reason that the valence electrons in a metal fork have way more kinetic energy than the ions in there (but they don't burn us because they are degenerate). So this is the way that degeneracy actually assists collapse: the electrons get all the kinetic energy, and when they go relativistic, there is no nonrelativistic fraction of the kinetic energy to produce any "catching up" of virialization when mass is added. But if the gas is ideal, then the ions get their fair share of the kinetic energy, and when that stays nonrelativistic, it allows for some protection against contraction via revirialization. Given that in core collapse, we have only one such ion for twenty six electrons, it might not be enough to induce less contraction in the ideal gas, so the ideal gas might not win after all in an actual simulation of this scenario rather than this idealized one.

Still, we can use this to understand what actually does happen in core collapse. We have a core where two things are happening at the same time, it is contracting for two reasons at once. One is that it is losing heat, because it is probably above the SC limit so must have a temperature gradient (it is not yet fully degenerate so shares some ideal gas behavior). The other is that undervirialized mass is being added to it, producing the need for "catching up" the virialization. All this contraction is causing the energy scales to rise. But because it is losing heat, its degree of degeneracy is rising, so there is one more thing happening: the degenerate electrons are stealing kinetic energy from the highly nonrelativistic ions, making everything more relativistic, and providing less ability to "catch up" whenever the virialization falls behind due to the gravitational load of the added mass. So degeneracy is not always helping prevent contraction, it only helps when interdicting heat loss is more important than increasing degree of relativity.

There is nothing like that in any explanation that says gravity is overcoming some added type of pressure in the core collapse process. But if you simply track energy, and say where the energy is going, the picture is crystal clear.

 

[PeterDonis]

the degenerate electrons steal almost all of the kinetic energy of the ions

Or, to put this another way, the ion temperature goes way down (because it is trying to equilibrate with the electron temperature, which is way down because the electrons are degenerate--their kinetic energy is not thermal). So another difference between the "real electron" core and the "weirdo electron" core is that the "real electron" core will be much colder than the surrounding shell in which fusion is taking place, while the "weirdo electron" core will be at least as hot and likely hotter.

This means that, while the "weirdo electron" core will be losing heat, the "real electron" core will not--it will be gaining heat (as well as mass in the form of iron ash) from the surrounding shell. This seems like yet another effect that would cause the "real electron" core to contract more slowly than the "weirdo" electron core, and therefore to take longer to become relativistic. Remember that the only regime where there is a difference between the pressure behavior of the two cores is the regime in which the electrons are non-relativistically degenerate. And in that regime, the real electron core contracts more slowly.

 

[Ken G]

It will induce fusion in the envelope, and it will increase the rate of fusion before you get to iron. But no matter - let's start the clock with an iron core where electrons suddenly behave classically.

Yes, you have a point, rising the core temperature will likely bump up the shell fusion a bit, and I'm not including that. So yes, there is always the issue of how far down the road you want to track the counterfactuality!

As we agreed, the temperature goes way, way up. So the radiation emitted also goes way, way up.

Probably neutrino emission would be the biggest deal, given that it doesn't have to wait like radiative diffusion.

As energy is lost, the temperature goes up (the specific heat is negative) and the radiation increases more. My copy of S&T is in a box at the moment, but I think this is similar to their argument. This in turn causes the core to contract more. This slows the process, but cannot stop it - it doesn;t stop until the neutron degeneracy kicks in.

That is the endothermic runaway that causes core collapse, so when thermal timescales start to compete on the free fall time, that's core collapse. You are saying that might be ushered in by the weirdo electron case, which is also what @PeterDonis is worried about. You're not wrong. But the energy losses are not just a function of temperature, many of the endothermic processes care more about the electron energy than the temperature. For those, like electron capture, what matters is the gravitational energy scale, so those are the same for real and weirdo electrons as long as they are at the same stage of contraction. I grant you that the ones that depend on temperature will be more active in the weirdo case, this all falls under the heading of heat transport. Remember we are considering the situation where mass is being added faster than heat transport timescales, which are generally pretty slow compared to virialization timescales (the latter being the free fall time). But I'm not disputing that these timescales are not being carefully tracked in this hypothetical scenario, it's not clear that the mass can be added fast enough to override the effects you and @PeterDonis are talking about.

So if you turn off electron degeneracy, you skip the whit dwarf phase (and the core of a red giant is white dwarf-like) and go straight to the neutron star phase. As expected.

It was never disputed that the ultimate endpoint will be a neutron star or black hole, the issue was always if there will come a point where you can point at, say, a 2 solar mass core that is still in pretty good force balance in the case of the weirdo electrons. That is what you know for certain you will never be able to do for the highly degenerate real electrons. @PeterDonis pointed out that the weirdo electrons are always more apt to contract as heat is lost, so that 2 solar mass situation should never occur if it doesn't for the degenerate electrons starting out from the same state of contraction. I pointed out that the weirdos have one ace in the hole: their kinetic energy is shared (somewhat) by nonrelativistic gas, making them better at revirializing when mass is added and potentially allowing them to sail through the 1.4 solar mass limit. But yeah, the timescales are not completely clear, and there are not a lot of iron nuclei, so this might not happen.

Still, in summary, the whole enterprise was just a device to bring us into a conversation that is focusing where it needs to be: on the energy, and its transport, and how that energy is shared between low temperature, high temperature, relativistic, and nonrelativistic particles. Not forces, or weird quantum mechanical additions to them. I am very pleased that in all of the well posed challenges you both have placed on the scenario I was advocating, neither of you ever found any use in citing any kind of modification to a force! I'm not surprised by that, because degeneracy is a thermodynamic effect, not a mechanical one.

 

[PeterDonis]

the weirdos have one ace in the hole: their kinetic energy is shared (somewhat) by nonrelativistic gas

The ions being non-relativistic only matters if the electrons are relativistic--but we have already agreed that degeneracy doesn't make a difference if the electrons are relativistic, the kinetic energy/pressure goes like $1 / R$ in both cases. It is only if the electrons are non-relativistic that degeneracy changes the kinetic energy/pressure behavior to $1 / R^2$, which slows the contraction as mass is added.

And since we are considering cores with the same mass (adding mass to both at the same rate since they are both surrounded by the same kind of shell in which fusion is taking place), the electrons will become relativistic at the same size range for both, so the only question is which one reaches that size range first. So far, both differences between the cores--degeneracy changing the pressure behavior, and the degenerate core being colder--look to me to cause the contraction to be slower in the degenerate case. That means the "weirdo electron" core should reach the size range at which the electrons become relativistic first.

 

[PeterDonis]

I am very pleased that in all of the well posed challenges you both have placed on the scenario I was advocating, neither of you ever found any use in citing any kind of modification to a force!

Not quite. I cited a difference in the pressure behavior in the regime where the electrons are non-relativistically degenerate, a $1 / R^2$ dependence instead of $1 / R$. You can of course equally well describe this as a difference in the kinetic energy behavior. As I have already commented, both descriptions, the force/pressure description and the energy description, are equally valid. You prefer the energy description so that has been the description we have been mainly using in this discussion. But that doesn't make the force/pressure description invalid or wrong. Both things will coexist in any scenario.

 

[Ken G]

Not quite. I cited a difference in the pressure behavior in the regime where the electrons are non-relativistically degenerate, a $1 / R^2$ dependence instead of $1 / R$. You can of course equally well describe this as a difference in the kinetic energy behavior. As I have already commented, both descriptions, the force/pressure description and the energy description, are equally valid. You prefer the energy description so that has been the description we have been mainly using in this discussion. But that doesn't make the force/pressure description invalid or wrong. Both things will coexist in any scenario.

There is no problem with using the degeneracy condition to determine the energy environment, that is a standard way to find the mass/radius relation. That's never been what I was talking about, because that is actually an energy argument simply framed in force units, which has nothing to do with any "additional quantum mechanical forces."

 

[PeterDonis]

that is actually an energy argument simply framed in force units, which has nothing to do with any "additional quantum mechanical forces."

I disagree, not with the physics, but with your ordinary language description of it. Classical physics cannot explain how the degenerate electron gas in this scenario can have so much pressure and kinetic energy while being at such a low temperature. Quantum mechanics is required to explain that. Both the pressure and the kinetic energy are present. Yes, they are related the same way as thermal pressure and thermal kinetic energy would be, but that doesn't mean that only the energy is really there or that only the energy can be appealed to to give a valid explanation of what is going on. The kinetic energy of a Fermi gas is "quantum mechanical energy", and the pressure that gas exerts is a "quantum mechanical force" because quantum mechanics is required to explain how they can be there. I don't see any justification for calling such statements "wrong". You have a certain preference for how to describe such things, but that doesn't make other descriptions wrong.

 

[PeterDonis]

There is no problem with using the degeneracy condition to determine the energy environment

What you call "the energy environment" can equally well be called "the pressure environment". And since we are talking about the core resisting compression due to having more mass piled on it, talking about pressure seems perfectly natural. You might prefer a different description, but that doesn't make the pressure description wrong.

 

[Ken G]

What you call "the energy environment" can equally well be called "the pressure environment". And since we are talking about the core resisting compression due to having more mass piled on it, talking about pressure seems perfectly natural. You might prefer a different description, but that doesn't make the pressure description wrong.

It's not about whether we call it pressure or energy, it's where the energy comes from. All the usual places, no special quantum mechanical forces or energy sources.

 

[PeterDonis]

It's not about whether we call it pressure or energy, it's where the energy comes from. All the usual places, no special quantum mechanical forces or energy sources.

I disagree; as I've already said, quantum mechanics is required to explain how a gas that is so cold can still have so much kinetic energy and pressure. You prefer not to focus on this, but that doesn't make it any less true.

 

[Ken G]

I disagree, not with the physics, but with your ordinary language description of it. Classical physics cannot explain how the degenerate electron gas in this scenario can have so much pressure and kinetic energy while being at such a low temperature. Quantum mechanics is required to explain that.

Yes, exactly, we need QM to explain the temperature. It's thermodynamic, not mechanical. All the mechanical issues stem from energy, not force interactions. Yes pressure is always that way, so I don't mind saying there is a pressure there, but it's not some quantum mechanical force.

Both the pressure and the kinetic energy are present. Yes, they are related the same way as thermal pressure and thermal kinetic energy would be, but that doesn't mean that only the energy is really there or that only the energy can be appealed to to give a valid explanation of what is going on. The kinetic energy of a Fermi gas is "quantum mechanical energy", and the pressure that gas exerts is a "quantum mechanical force" because quantum mechanics is required to explain how they can be there. I don't see any justification for calling such statements "wrong". You have a certain preference for how to describe such things, but that doesn't make other descriptions wrong.

It's the difference between "what" and "why." The "what" is the energy, the "why" is the quantum mechanics. The quantum mechanics controls the thermodynamics, which controls the energy, which determines the pressure. But the pressure is not quantum mechanical, it's not mysterious, it's just pressure. What's mysterious is the nature of the heat transport (and why it gets shut off), that's where the student should be pointed, that's where the quantum mechanics lives. Your posts in this thread were just like that, always pointing to the energy, and where it was going. That's exactly the right way to do it.

 

[Ken G]

The ions being non-relativistic only matters if the electrons are relativistic--but we have already agreed that degeneracy doesn't make a difference if the electrons are relativistic, the kinetic energy/pressure goes like $1 / R$ in both cases. It is only if the electrons are non-relativistic that degeneracy changes the kinetic energy/pressure behavior to $1 / R^2$, which slows the contraction as mass is added.

I'm not sure what you are saying here. When the Chandra mass is approached, you have a bunch of very relativistic electrons, and you have a gas that is obeying the relativistic virial theorem, that's why it collapses. But there are ions there, and they are nonrelativistic. That would prevent the collapse by altering the nature of the virial theorem, except for one thing: the ions don't get a vote, because kinetic energy is what votes on the nature of the virial theorem, and the ions have had all their kinetic energy robbed from them by the degenerate nature of those electrons. That's what wouldn't happen if electrons were distinguishable, so the ideal case does not have the collapsing relativistic virial theorem (but it does have all that extra heat loss, so you are correct that the winner of that competition becomes reliant on the details).

And since we are considering cores with the same mass (adding mass to both at the same rate since they are both surrounded by the same kind of shell in which fusion is taking place), the electrons will become relativistic at the same size range for both, so the only question is which one reaches that size range first. So far, both differences between the cores--degeneracy changing the pressure behavior, and the degenerate core being colder--look to me to cause the contraction to be slower in the degenerate case. That means the "weirdo electron" core should reach the size range at which the electrons become relativistic first.

What you have correctly surmised is that the ideal version will always lose more heat. What is not correct is this means it will have contracted more. The connection between the heat lost, and the amount of contraction, is controlled by whether the virial theorem is acting more relativistically or more nonrelativistically. That depends on how the kinetic energy is partitioned between relativistic electrons, and nonrelativistic ions. Degeneracy affects that, it puts more kinetic energy in the relativistic electrons.

 

[PeterDonis]

we need QM to explain the temperature.

If that's what you choose to emphasize, yes. You are basically saying: we have this system that has to have a certain kinetic energy, but it's too cold, so where does that energy come from?

But I could equally well say: we have this system that is at a very cold temperature, but it's also in force balance when, according to classical physics, it shouldn't be, so where does the force come from?

I understand you don't like the latter way of talking, but, once more, that doesn't make it wrong. That is why I keep objecting when you say it is wrong. If you want to post about how the energy viewpoint you take would analyze a scenario, that's great! As long as you just do that. But as soon as you start saying that any other viewpoint is wrong, you're going to get pushback. At least if you do it here.

 

[Ken G]

If that's what you choose to emphasize, yes. You are basically saying: we have this system that has to have a certain kinetic energy, but it's too cold, so where does that energy come from?

Yes.

But I could equally well say: we have this system that is at a very cold temperature, but it's also in force balance when, according to classical physics, it shouldn't be, so where does the force come from?

You will always end up back at the energy when you answer that. You will need to see the heat that came into that system, which explains where that force came from. Otherwise, you will never know where that force came from, it will seem like some "mysterious quantum mechanical force." It's like when con men try to show mysterious behaviors of systems that are plugged into the wall, if you don't see the plug, it looks very mysterious. (Yes, it may be mysterious how the electricity works, and why current is coming in, but if you see that current is bringing energy in, then you at least understand the "what" if not the "why".)

I understand you don't like the latter way of talking, but, once more, that doesn't make it wrong. That is why I keep objecting when you say it is wrong. If you want to post about how the energy viewpoint you take would analyze a scenario, that's great! As long as you just do that. But as soon as you start saying that any other viewpoint is wrong, you're going to get pushback. At least if you do it here.

I accept that your way of looking at it is completely valid. My concern is misconceptions that are often fostered in people with a less deep understanding. The only way to see that is to look for those misconceptions in action. I see them all the time, I saw one in the context of the exchange interaction recently. If you look out for them, you will see them too.

 

[PeterDonis]

When the Chandra mass is approached, you have a bunch of very relativistic electrons

Yes, but by that time, the real electron core has contracted less than the "weirdo" electron core, because of the different pressure behavior during the phase when the electrons were not yet relativistic. That was my point.

the ideal case does not have the collapsing virial theorem

In the relativistic regime the pressure behavior is the same for both cores. So is the virial theorem.

In the "weirdo" electron core, the electron temperature is much higher, so the ion temperature will be much higher as well, yes. You appear to be saying that the increased "re-virialization" available from the ions in this core will outweigh the increased heat loss due to its higher temperature. I'm not sure how we would figure that out either way, but at best, it would mean the degenerate core might "catch up" somewhat during the relativistic phase.

The connection between the heat lost, and the amount of contraction, is controlled by whether the virial theorem is acting more relativistically or more nonrelativistically.

And while the electrons are not yet relativistic, this is the same for both cores. What is not the same is the pressure behavior--as the degenerate electron core is compressed, its pressure goes up faster than the ideal gas electron core. That means the degenerate core compresses less for a given amount of mass added, during the phase when the electrons are non-relativistic. This is independent of heat loss; it is a consequence of the adding mass process, not the heat loss process.

 

[PeterDonis]

My concern is misconceptions

I understand that. But that still does not justify expounding your entire position in detail in order to correct a single misconception. For example:

I saw one in the context of the exchange interaction recently.

Yes, and it has now been corrected by me in a single one-paragraph post. As I pointed out in our PM conversation on this, you yourself could have made a similar post and left the issue there. But you posted a lot more than that. That resulted in a thread hijack that I had to take the time to deal with. That is what needs to stop.

 

[PeterDonis]

You will always end up back at the energy when you answer that.

This is your viewpoint, I understand that. My viewpoint is that all of these things are connected, and trying to point to just one as where one has to "end up" is a waste of time. Use whatever analysis works.

 

[Ken G]

Yes, but by that time, the real electron core has contracted less than the "weirdo" electron core, because of the different pressure behavior during the phase when the electrons were not yet relativistic. That was my point.

Still, it sounded like you were still claiming the weirdos would always have to have contracted more than the real ones, after any given time. That's what is not necessarily true, it would depend on the details of how fast the heat transport was, versus how fast was mass being added.

In the relativistic regime the pressure behavior is the same for both cores. So is the virial theorem.

The "relativistic regime" is what is not the same for both, the weirdos will be less in that regime than the real ones, because of the ions.

In the "weirdo" electron core, the electron temperature is much higher, so the ion temperature will be much higher as well, yes. You appear to be saying that the increased "re-virialization" available from the ions in this core will outweigh the increased heat loss due to its higher temperature. I'm not sure how we would figure that out either way, but at best, it would mean the degenerate core might "catch up" somewhat during the relativistic phase.

Exactly, the degenerate core might catch up as the real electrons go deeper into the relativistic regime. This is a very interesting element of degenerate behavior, which works toward greater contraction than ideal gases, as a twist on the norm.

And while the electrons are not yet relativistic, this is the same for both cores. What is not the same is the pressure behavior--as the degenerate electron core is compressed, its pressure goes up faster than the ideal gas electron core.

Not sure what you mean by its pressure going up faster. The pressure only depends on the radius since they have the same mass. The amount of compression when both are nonrelativistic is the same for the same heat loss, so if the weirdos are losing heat faster, they will always be more contracted at any given time, I agree with that point. This is still true if mass is being added, if we assume the mass is being added with the same degree of "undervirialization." For simplicity, let us imagine that the mass comes in with very little kinetic energy, so it is highly undervirialized.

That means the degenerate core compresses less for a given amount of mass added, during the phase when the electrons are non-relativistic. This is independent of heat loss; it is a consequence of the adding mass process, not the heat loss process.

This claim seems like a perfect example of the importance of separating the mechanical aspects from the thermodynamic ones. If one imagines an adiabatic situation, then there is no thermodynamics, it's all mechanical, and there will never be any difference in the two cases, the PEP doesn't matter at all. How could it, it's just mass and initial energy, that's it! Partical distribution functions are of no importance (if nonrelativistic). You just have a given amount of initial kinetic energy, and a given new higher mass, and you must revirialize this to get force balance. The PEP does absolutely nothing there. I am not seeing how you conclude the degenerate version contracts less. (I could see getting the weirdos to contract less if you imagine the added mass comes in at the same temperature as the core gas, in which case the weirdo added mass comes in with all kinds of more kinetic energy. But that would be unfair to the real electrons, there is no reason for the weirdo added mass to have any different properties, so let's just bring it in with very little kinetic energy in both cases, as the shell is a kind of an atmosphere on the core.)

If you are not treating it adiabatically, so you are taking account of the weirdo heat loss, then the weirdos will contract more in the same time, and will be at higher density and higher energy scale. But that's not independent of heat loss, it's all about heat loss.

 

[Ken G]

As I pointed out in our PM conversation on this, you yourself could have made a similar post and left the issue there. But you posted a lot more than that. That resulted in a thread hijack that I had to take the time to deal with. That is what needs to stop.

I understand your perspective there and will respect it, this is your right to stipulate.

 

[PeterDonis]

Not sure what you mean by its pressure going up faster.

Just what I said: when the degenerate electrons are non-relativistic, the pressure as the core gets compressed goes up as $1 / R^2$. In all other cases (non-relativistic ideal gas, or relativistic anything), the pressure only goes up as $1 / R$.

 

[PeterDonis]

The amount of compression when both are nonrelativistic is the same for the same heat loss

But not necessarily for the same mass added. Or more precisely, for the same number of baryons and electrons added. That is the point I have been making about the pressure behavior.

 

[Ken G]

But not necessarily for the same mass added. Or more precisely, for the same number of baryons and electrons added. That is the point I have been making about the pressure behavior.

And the point I have been making is that this cannot be correct, because there is nothing thermodynamic in it, and the PEP is pure thermodynamics. By that I mean, it is about the distribution of energy over the particles, not the total energy in the system. The pressure depends only on the latter, as pressure is 2/3 the kinetic energy divided by the volume. So we start with two systems at the same mass and volume, and the same internal kinetic energy. We add the same mass to both, with no additional kinetic energy (for simplicity). So we now have two systems of the same mass, volume, and kinetic energy. They are both undervirialized, so they must both contract to recover virialization. That will happen when the gravitational energy is twice the kinetic energy, which fixes the new radius for both. This is all the mechanics of kinetic energy and gravity, it is essentially nothing but Newtonian physics. There is zero quantum mechanics in it, expressly because we never asked about the particle distribution functions, because we never needed to in order to get the adiabatic mechanics right.

 

[PeterDonis]

So we now have two systems of the same mass, volume, and kinetic energy. They are both undervirialized, so they must both contract to recover virialization. That will happen when the gravitational energy is twice the kinetic energy, which fixes the new radius for both.

I see what you're saying, but what I have been struggling with is how to reconcile this with the fact that, for degenerate electrons, the Fermi energy goes up as $1 / R^2$ instead of $1 / R$, so the same amount of contraction should result in more total kinetic energy for the real core than for the "weirdo" electron core. Since the total kinetic energies must be the same, that led me to think that the real core must contract less.

However, I think I might have found a way to reconcile the two. In the real core, with degenerate electrons, the electrons have all the kinetic energy; whereas in the "weirdo" electron core, the electrons only have part of it; the ions have the rest. So in the degenerate core, the kinetic energy per electron does end up higher--but the kinetic energy per ion is zero to compensate. In the "weirdo" core, the kinetic energy per electron and ion is the same, and both are smaller, for the same total kinetic energy in both cases.

I'm still not entirely comfortable with this because it doesn't seem to me to guarantee that exact equality will hold at the same value of $R$. I'll have to think about it some more when I have time.

 

[Ken G]

What you and I always know from the virial theorem, for nonrelativistic gases ideal or degenerate alike, is that if we know the total energy E (including gravitational potential energy, so E is negative), and keep it conserved (so adiabatic contraction under gravity), the kinetic energy will always be -E whenever force balance is achieved, and the potential energy will always be 2E. That latter criterion sets the equilibrium radius. So we know R(M) in equilibrium, if we know E, and the PEP never enters because we know E and the rest is all basic mechanics.

OK so you see that already, but you are wondering what went wrong with the calculation of the Fermi energy. The problem there is that the Fermi energy is just a device, it's not connected to any real energies in the problem unless you use the correct R. But you are using R like a variable to find the correct R, which is a fine device, but the other Rs and the other Fermi energies have no physical significance, they are like points on a curve that only matters when it crosses some other curve.

This gets even more interesting. For given M, there is only one R where the Fermi energy will actually be achieved, so there is only one E. If E is larger than that, heat must be lost to recover the degeneracy, and if E is below that, the initial condition is probably impossible. So we can assume for simplicity that as mass is added, the new E always gives the correct Fermi energy for that M at the equilibrium R. We can assume the same for the weirdos, but all we need to know is that we have the same E and the same M, and the mechanics will result in the same R. What this also means is,, since the R that obeys the degenerate mass-radius relation is R is proportional to M to the -1/3 power, the ideal gas version must obey that same relation as mass is added adiabatically and is given the right E to match the Fermi energy of the real electrons. But if we gave them any other E, as long as it was all adiabatic, they would both always come to the same R in force balance, the real electrons just might not be degenerate any more.

 

[PeterDonis]

you are wondering what went wrong with the calculation of the Fermi energy

Not what went wrong with it, but just how to reconcile the radial dependence of the Fermi energy with the radial dependence of the total energy. We have a system with a given $M$ (after adding the mass) that has to contract to a given $R$ (to recover virialization), causing an increase in total kinetic energy that goes like $1 / R$ (because that's what the virial theorem tells us). That will produce a given increase in the Fermi energy per electron in the degenerate case, but this increase has a $1 / R^2$ dependence in the non-relativistic regime. There's no way around that: there's nowhere else for the energy to go, because the electrons have all the kinetic energy in this case. It's the same change in $R$ in both cases (it has to be, it's the same object), so the change in Fermi energy per electron is larger than the change in total energy per particle. The only way I can see to reconcile those facts is to take into account that the total energy per particle includes both electrons and ions, but the kinetic energy per particle only includes electrons.

 

[Ken G]

So following up on that, we can get back to what would happen to the real degenerate electrons and the weirdos. Perhaps the real electrons lose whatever heat they need to sustain degeneracy, and the weirdos lose even more heat because of their high temperature, so the weirdos reach a smaller radius in the nonrelativistic regime. But later on, they may have the chance to catch up, when the real electrons go relativistic, whereas the weirdos can give kinetic energy to their nonrelativistic ions to help them avoid the nastiest areas of the relativistic virial theorem. What happens then depends on the details of how fast the mass is being added, and how much contraction occurs before the runaway endothermic processes kick in. I haven't looked too closely at that last bit because it would require knowing all the details and probably do a full simulation, but I do expect, on further reflection, that there might not be enough nonrelativistic iron nuclei to allow the weirdo radius to hang on longer than the real electron radius does. The core collapse is just looming too close to the edge of the relativity, and only one iron to twenty six electrons is not too promising for the weirdos! (But for type Ia supernovae, you have one carbon to six electrons, so the weirdo prospects there are more interesting.)

 

[Ken G]

Not what went wrong with it, but just how to reconcile the radial dependence of the Fermi energy with the radial dependence of the total energy. We have a system with a given $M$ (after adding the mass) that has to contract to a given $R$ (to recover virialization), causing an increase in total kinetic energy that goes like $1 / R$ (because that's what the virial theorem tells us). That will produce a given increase in the Fermi energy per electron in the degenerate case, but this increase has a $1 / R^2$ dependence in the non-relativistic regime. There's no way around that: there's nowhere else for the energy to go, because the electrons have all the kinetic energy in this case. It's the same change in $R$ in both cases (it has to be, it's the same object), so the change in Fermi energy per electron is larger than the change in total energy per particle. The only way I can see to reconcile those facts is to take into account that the total energy per particle includes both electrons and ions, but the kinetic energy per particle only includes electrons.

That's not the way to reconcile it, because nothing in your calculation even requires that there are any ions present. The kinetic energy per particle in force balance will scale like M/R, and the Fermi energy will scale like M to the 2/3 over R squared. But that latter is completely hypothetical if we are treating the adiabatic case (as per the claim that heat loss is not important to this), because the kinetic energy won't be the Fermi energy. The only time it will actually be the Fermi energy is if we make sure the mass comes in with the right kinetic energy, and also if we wait for the gas to contract to its equilibrium radius (which will then obey the white dwarf radius relation, R proportional to M to the -1/3 power). No other Fermi energy is physically realized in this situation, regardless of how it scales with R.

 

[PeterDonis]

the kinetic energy won't be the Fermi energy

Ah, I see; to put it another way, the "degree of degeneracy" of the electrons does not have to remain constant.

 

[Ken G]

Ah, I see; to put it another way, the "degree of degeneracy" of the electrons does not have to remain constant.

Right, there are implicit assumptions there. The go-to thing to remember is that if you already know what is happening to the energy, you always have a purely mechanical calculation on your hands, and the PEP never matters a whit to the overall scaling of things like the contraction radius, unless you are worrying about how the kinetic energy is distributed between relativistic vs. nonrelativistic particles.

 

[Vanadium 50]

Probably neutrino emission would be the biggest deal, given that it doesn't have to wait like radiative diffusion.

Doubt it. Neutrino emission is not thermal. Stellar cores are way too cold. (Like a million or more)

 

[PeterDonis]

if we know the total energy E (including gravitational potential energy, so E is negative), and keep it conserved (so adiabatic contraction under gravity), the kinetic energy will always be -E whenever force balance is achieved, and the potential energy will always be 2E. That latter criterion sets the equilibrium radius.

I've been working through the math for this and have uncovered a couple of possible complications.

First, the conditions: we have a core of initial mass $M_0$ and initial radius $R_0$. We add some more mass, which I will express as a fraction $f M_0$ of the initial mass, where $f$ is much less than $1$; so we have a new mass $M_1 = \left( 1 + f \right) M_0$. Then we ask what the new radius $R_1$ will be once virialization is re-established.

First, we define the useful quantity $U_0 = G M_0^2 / R_0$. Then, in the initial state, we have the total energy $E_0 = K_0 + W_0$, the sum of kinetic and gravitational potential energies, which gives $E_0 = U_0 / 2 - U_0 = - U_0 / 2$.

Now comes the first possible complication. Consider the state immediately after mass has been added, but before the radius has a chance to change. What is the total energy at this point?

The gravitational potential energy now is $W_i = - G M_1^2 / R_0 = - U_0 \left( 1 + f \right)^2 \approx - U_0 \left( 1 + 2 f \right)$ (the subscript $i$ is to designate that this is an intermediate state between adding the mass and completing virialization). If we assume that the added mass has zero kinetic energy, which was what I understood to be the proposal, then we have $K_i = K_0$ and $E_i = W_i + K_0 = - U_0 \left( 1/2 + 2 f \right)$. Note that this is less than $E_0$ by a quantity we'll call $\Delta = 2 f U_0$. That doesn't seem right, because we haven't included any way for energy to be lost from the core in the adding mass process.

If we instead assume that $E_i = E_0$, i.e., that the total energy (exclusive of rest mass energy) remains the same as the mass is added, then we must have $K_i = K_0 + \Delta$. In other words, the mass that is added must have kinetic energy $\Delta$--presumably this is the minimum kinetic energy it must gain in the process of falling onto the core from the surrounding shell region where iron ash is being produced.

If we adopt the above as more reasonable, then we get to the second possible complication. We now have the intermediate state $W_i = - U_0 - \Delta$ and $K_i = U_0 / 2 + \Delta$. However, this state is not "under-virialized"; instead it is "over-virialized"--we have $K_i > | W_i | / 2$, instead of $K_i < | W_i | / 2$. So to recover virialization in this state, the core will need to expand, not contract, so that it loses kinetic energy (at least if we require the change to be adiabatic--the alternative of course would be for the core to radiate away the excess kinetic energy).

And we could, of course, adopt the intermediate assumption about how much kinetic energy the added mass contains, and give it kinetic energy $\Delta / 2$--in which case the new state would be exactly virialized and no change in radius would occur at all. But this would still mean the total energy $E_i$ would be less than $E_0$, so the core would have to lose energy somehow in the course of adding the mass.

In short, in order to have a well-defined specification of what will happen to the core when mass is added, we need to make an additional assumption about how much kinetic energy the added mass has, and it's not clear to me either that the proposed "zero kinetic energy" assumption is physically reasonable, or what the most physically reasonable assumption is.

 

[Ken G]

Doubt it. Neutrino emission is not thermal. Stellar cores are way too cold. (Like a million or more)

Yes it likely scales with the electron energy, moreso than temperature, in which case it might happen at a similar rate for the weirdos as for the regular electrons, though there can be some issues with the electrons finding final states if there's a Fermi sea. Not sure just how the neutrino losses would scale, but they get very high near core collapse energies.

 

[Ken G]

Now comes the first possible complication. Consider the state immediately after mass has been added, but before the radius has a chance to change. What is the total energy at this point?

This is what I meant by having to take account of the kinetic energy of that added mass. But I think we can use as a basic simplifier that it comes in with the kinetic energy it needs for the equilibrium state to still be as degenerate as the core was before, i.e., highly so. If it comes in with more kinetic energy than that, it will reduce the degree of degeneracy, but we can get to the same place by just letting the core lose some heat during equilibration.

The gravitational potential energy now is $W_i = - G M_1^2 / R_0 = - U_0 \left( 1 + f \right)^2 \approx - U_0 \left( 1 + 2 f \right)$ (the subscript $i$ is to designate that this is an intermediate state between adding the mass and completing virialization). If we assume that the added mass has zero kinetic energy, which was what I understood to be the proposal, then we have $K_i = K_0$ and $E_i = W_i + K_0 = - U_0 \left( 1/2 + 2 f \right)$. Note that this is less than $E_0$ by a quantity we'll call $\Delta = 2 f U_0$. That doesn't seem right, because we haven't included any way for energy to be lost from the core in the adding mass process.

Probably what is happening there is, if we are giving it zero initial kinetic energy, we have to add the mass at the surface, which is generally less deep in the well than the (1+2f) factor is assuming. So we have two choices, we can either give it zero kinetic energy and at the surface, in which case the gravitational potential energy has less magnitude than 1+2f (it is spoiling the homology that this kind of scaling assumes), or we can spread it over the interior to support the homology, in which case we have to give it some kinetic energy to avoid being quantum mechanically disallowed. Probably the simplest thing is to just give it whatever kinetic energy it needs to produce the Fermi energy when it equilibrates, and maintain the homology. This is what I meant above that the initial condition is probably impossible if the total energy of the configuration is too low to reach the Fermi energy at force balance.

If we instead assume that $E_i = E_0$, i.e., that the total energy (exclusive of rest mass energy) remains the same as the mass is added, then we must have $K_i = K_0 + \Delta$. In other words, the mass that is added must have kinetic energy $\Delta$--presumably this is the minimum kinetic energy it must gain in the process of falling onto the core from the surrounding shell region where iron ash is being produced.

Yes, one could also "drop" the mass in from infinity to keep the total energy the same, and then it will need to lose some heat as it virializes, if you want it to be degenerate at the end. It's the same final result either way.

If we adopt the above as more reasonable, then we get to the second possible complication. We now have the intermediate state $W_i = - U_0 - \Delta$ and $K_i = U_0 / 2 + \Delta$. However, this state is not "under-virialized"; instead it is "over-virialized"--we have $K_i > | W_i | / 2$, instead of $K_i < | W_i | / 2$. So to recover virialization in this state, the core will need to expand, not contract, so that it loses kinetic energy (at least if we require the change to be adiabatic--the alternative of course would be for the core to radiate away the excess kinetic energy).

Yes, that's all true. It's just another example of the importance of "following the energy" rather than trying to directly follow the forces.

And we could, of course, adopt the intermediate assumption about how much kinetic energy the added mass contains, and give it kinetic energy $\Delta / 2$--in which case the new state would be exactly virialized and no change in radius would occur at all. But this would still mean the total energy $E_i$ would be less than $E_0$, so the core would have to lose energy somehow in the course of adding the mass.

In short, in order to have a well-defined specification of what will happen to the core when mass is added, we need to make an additional assumption about how much kinetic energy the added mass has, and it's not clear to me either that the proposed "zero kinetic energy" assumption is physically reasonable, or what the most physically reasonable assumption is.

And this is very much the situation in real core collapse analyses. If you look at papers by people like Woosley, you see that he tends, in his analysis of simulations, to assume the entropy per particle is what stays the same as mass is added. I've never seen a particularly good explanation of why that is what is taken as fixed, but one does have to choose something, if one cannot closely follow the energy transport. (I presume the simulations do include the energy transport, so the after the fact analyses choose some kind of simplifying assumption like fixed entropy per particle.) Since we are doing an even simpler analysis, we are also fixing the entropy per particle, we are just saying it is "small enough to act as if it's zero," i.e., fully degenerate. That's where the Fermi energy device comes in.

 

[PeterDonis]

one could also "drop" the mass in from infinity to keep the total energy the same, and then it will need to lose some heat as it virializes

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

 

[PeterDonis]

So the radius increases by a factor which is the square of the mass increase factor.

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.

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?

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.