Without degeneracy, when would Solar cores collapse?

[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.

 

[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.