I Without degeneracy, when would Solar cores collapse?

[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.

 

[PeterDonis]

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

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

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

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

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

 

[Ken G]

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

Agreed.

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

Agreed.

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

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

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

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

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

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

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

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

 

[Ken G]

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

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

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

Yes, I understand.

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

Yes.

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

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

 

[PeterDonis]

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

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

 

[Ken G]

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

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