Landau's Maximum Mass Limit Derivation in Shapiro & Teukolsky (1983)

In summary, the maximum mass limit for white dwarfs and neutron stars is derived using a variational method. The energy limit is determined when the number of fermions present is less than a critical value.
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TL;DR Summary
Shapiro & Teukolsky describe a simple derivation by Landau of the maximum mass limit for white dwarfs and neutron stars. They say that the result is the same for both types of object. But is that correct?
In Section 3.4 of Shapiro & Teukolsky (1983), a simple derivation, due to Landau, of the maximum mass limit for white dwarfs and neutron stars is given. I will briefly describe it here and then pose my question.

The basic method is to derive an expression for the total energy (excluding rest mass energy--this is not mentioned in Shapiro & Teukolsky, but I take it that it is assumed to be constant and so drops out of the variational method being used) of the object, and then ask under what conditions that energy will have a minimum, indicating a stable equilibrium. The energy is the sum of two terms: a positive term coming from the Fermi energy of the fermions that are producing the degeneracy pressure, and a negative term coming from the gravitational potential energy.

The object is assumed to contain ##N## fermions that produce degeneracy pressure, and some number of baryons of mass ##m_B## that produce the mass of the object. For an object of radius ##R##, the number density of fermions is ##n = N / R^3##, and the momentum of a fermion is proportional to ##\hbar n^{1/3}##. So the Fermi energy per fermion is the momentum times ##c##, or

$$
E_F \approx \frac{\hbar c N^{1/3}}{R}
$$

The gravitational potential energy per fermion is given by

$$
E_G \approx - \frac{G M m_B}{R}
$$

where ##M = N m_B##. The total energy is then ##E = E_F + E_G##, and an argument is then given (which I am not questioning) that there will only be a minimum of ##E## if ##N## is less than a critical value, which denotes the maximum mass.

My question is about the formula ##M = N m_B##. For a neutron star, that is obviously true, since the neutrons play both roles: they provide the degeneracy pressure and they provide the mass. For a white dwarf, however, it does not seem correct: ##N## is the number of electrons, but ##M## comes from the baryons, and the number of baryons will be larger than the number of electrons (since not all of the baryons are protons). So it seems like the formula for white dwarfs should be ##M = (N / f) m_B##, where ##f < 1## is the fraction of baryons that are protons (so ##N / f## is the total number of baryons).

I don't see this mentioned anywhere in Shapiro & Teukolsky, so I'm wondering if I've missed something, or if they just left it out, presumably because they were only giving a rough order of magnitude argument and weren't concerned about numerical factors of order unity. The cube root of the factor ##f## would be what would appear in the final formula for the maximum mass [Edit--this turns out not to be correct, it should be f^2, see post #14], so the effect on the result would not be large. But it would mean that the results, even at this rough order of magnitude level, would be different for white dwarfs and neutron stars; as far as I can tell from the description in Shapiro & Teukolsky, Landau's original 1932 paper in which he published his result said they were the same to this level of approximation.
 
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  • #2
Isn't this statistical physics?
 
  • #3
The maximum mass of compact stars depend on the equation of state of the matter they are made of. The standard way to model this is to use general relativistic hydrostatics, i.e., you have the equations describing energy-momentum and conserved-charge conservation. To close the system of equations you need an equation of state, i.e., pressure as a function of density.

In the case of neutron stars this is a challenging research problem for decades. One way to figure out the equation of state of nuclear matter are heavy-ion collisions, and the problem still is that there's a plethora of equations of state, trying to describe what's observed in heavy-ion collisions to find the phase diagram of strongly interacting matter with various transitions/cross overs (the confinement-deconfinement transition from a hadron-gas state to a (strongly coupled) quark-gluon plasma; the chiral phase transition), the search for a first-order transition line ending in a critical point, etc. etc. The observation of neutron-star masses and (more challenging) mass-radius relations provide important constraints on the equation of state, and it's an open question, whether there are neutron stars with a quark-matter core or not. Another very new possibility is the observation of the gravitational-wave signals and also the electromagnetic signals of the same source in neutron-star-neutron-star mergers or black-hole-neutron-star mergers. There one can hope to find further constraints on the equation of state from the gravitational-wave signal (and, if available the electromagnetic signals too). It's a very active field of research.

An example for a recent publication from my colleagues at the Goethe University, Frankfurt is here:

https://doi.org/10.3847/2041-8213/ac8674 (open access)
 
  • #4
malawi_glenn said:
Isn't this statistical physics?
Not really, at least not at the rough order of magnitude level that Landau's argument is made at. The Fermi energy could be sort of thought as a "statistical" quantity since it depends on Fermi-Dirac statistics for fermions. But the argument does not involve any statistical calculations.
 
  • #5
vanhees71 said:
The maximum mass of compact stars depend on the equation of state of the matter they are made of.
To get a more accurate limit, yes, this is true, one needs to have an equation of state.

Landau's argument described in the OP, however, does not involve any specific equation of state. It does assume, though, that the only significant energies in the problem are the Fermi energy (due to the Pauli exclusion principle) and the gravitational potential energy. At the time Landau made his argument, the strong interaction was not at all well understood. Now, however, we know that the strong interaction becomes repulsive at short ranges. This doesn't really affect white dwarfs, but it does affect neutron stars; the effect of this for neutron stars, as far as Landau's argument is concerned, would be to add another positive energy term due to the repulsive potential. As long as this energy scales as ##1/R##, it would not affect Landau's overall argument, but it would affect the numerical value of the estimated maximum mass limit that the argument gives.
 
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  • #6
I'd say Landau's arguments are typical "back-of-the-envelope estimates", using dimensional arguments, and of course he gets the principle "mechanism" right: You need the balance between "thermodynamic pressure" (in the case of white dwarfs and neutron stars indeed "degeneracy pressure" due to the Pauli principle) and gravitation.
 
  • #7
PeterDonis said:
For a white dwarf, however, it does not seem correct: ##N## is the number of electrons, but ##M## comes from the baryons, and the number of baryons will be larger than the number of electrons (since not all of the baryons are protons).
One way to do the order-of-magnitude is to say that we encounter the N limit when p ~ mc, so we can just use that in the deBroglie wavelength, and also say the total gravitational potential energy in the system, GM^2/R, is of order N times mc^2, where m is the electron or neutron mass. When I do this, I get that the critical N is of order (m_P / m_B)^3 times f^3, where m_P is the Planck mass. So that's about 10^57 times f^3, that last bit being what you are talking about-- except I get f cubed, not cube root, which sounds a lot more important!

I concur that in a white dwarf, f =1/2, because the ions generally have equal protons and neutrons, so gram for gram, the white dwarf has half as many electrons to carry the load as a neutron star has neutrons. But with f^3 scaling we should be talking about a factor of 8 smaller critical mass for a white dwarf than a neutron star, which is quite a different mass, even at the level of a rough estimate. So I'm agreeing with your point, except the calculation I did gets an even larger discrepancy than you reported, making your point all the more significant. So please check your result that it only scales like f^(1/3) instead of f^3, so we can see just how large the issue is.

I think all this implies that if the scaling is like f^3, then the actual neutron star maximum mass is considerably less than we get if we ignore GR and EOS corrections, so then we should be asking, not how do GR and strong-force corrections cause the neutron star to have a higher mass limit than the white dwarf, but rather why do those corrections move down the NS mass limit to make it closer to the WD. Landau may have missed an opportunity to make that point, though apparently his only goal was to say the two maximum masses should not be vastly different.
 
  • #8
Ken G said:
One way to do the order-of-magnitude
I'm not concerned about just an order of magnitude estimate; I already said in the OP that it might be that Shapiro & Teukolsky left out the factor ##f## I described simply because they were doing just an order of magnitude estimate and weren't concerned about numerical factors of order unity.

Ken G said:
I concur that in a white dwarf, f =1/2
For most white dwarfs, yes, I would expect that to be a reasonable value.

Ken G said:
please check your result that it only scales like f^(1/3) instead of f^3
The ##f^{1/3}## scaling is obvious from the Fermi energy term, which goes like ##N^{1/3}##, so if ##N = M f / m_B##, the term will scale like ##(M f / m_B)^{1/3}##. The gravitational potential energy term already has ##M## in it so no factor of ##f## appears there (since the goal is to express everything in terms of ##M##).
 
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  • #9
PeterDonis said:
I'm not concerned about just an order of magnitude estimate; I already said in the OP that it might be that Shapiro & Teukolsky left out the factor ##f## I described simply because they were doing just an order of magnitude estimate and weren't concerned about numerical factors of order unity.
Yet the simple scaling laws can get to the question of the impact of omitting the f factor, because it speaks to whether the actual maximum mass difference of about a factor of 2 is due to f, or if only part of it is due to f (as you maintain), or if it actually produces too much of a difference that has to be corrected back down by GR and EOS issues.
PeterDonis said:
The ##f^{1/3}## scaling is obvious from the Fermi energy term, which goes like ##N^{1/3}##, so if ##N = M f / m_B##, the term will scale like ##(M f / m_B)^{1/3}##. The gravitational potential energy term already has ##M## in it so no factor of ##f## appears there (since the goal is to express everything in terms of ##M##).
It seems we both made a mistake, which are easily corrected. Your approach is to look at energy per fermion, so you can use the Fermi energy, but then the gravitational energy also has to be expressed as per fermion. You are correct that the Fermi energy characterizing the electrons scales like f^(1/3), but you are not correct that the gravitational energy per electron has no f dependence because it is per electron, not per baryon. So the gravitational energy per electron is GM^2/R times 1/f, that last term being the per electron correction. This leads to M proportional to f^2, not f^(1/3).

I got proportional to f^3, but that's because I solved for the number of electrons N, not the maximum mass M. So my claim that it should be 8 times more mass for the NS should have been stated as just 4 times, since we are talking M not N. That's actually a good thing, correcting 8 all the way down to 2-3 was pretty severe, now we see we only have to get 4 down to 2-3 using GR and EOS corrections. I therefore conclude that the f effect you are drawing attention to is pretty close to the major effect here, at least as important as GR and EOS corrections. More importantly, it can be used to explain why NS exist at all, since if GR and EOS corrections reduce the maximum M, then NS could not exist without the f effect, they'd all be white dwarfs instead.
 
  • #10
PeterDonis said:
The cube root of the factor would be what would appear in the final formula for the maximum mass
Actually, this is a misstatement. The factor of ##f^{1/3}## appears in the formula for the Fermi energy. In the final maximum mass formula, that factor appears in the formula for ##M^{2/3}##, so it has to be raised to the ##3/2## power to give the factor that affects the maximum mass, which will then be ##f^{1/2}##.
 
  • #11
PeterDonis said:
Actually, this is a misstatement. The factor of ##f^{1/3}## appears in the formula for the Fermi energy. In the final maximum mass formula, that factor appears in the formula for ##M^{2/3}##, so it has to be raised to the ##3/2## power to give the factor that affects the maximum mass, which will then be ##f^{1/2}##.
Not quite, there's still the problem of the gravitational energy being the energy per electron, not per baryon. So it's really f^2, and this is an important difference-- it means your post here is far from a minor issue, it is likely the reason neutron stars exist (it's hard to be more quantitative with a simple scaling law, but I suspect the effect is large enough to justify this statement).
 
  • #12
Ken G said:
you are not correct that the gravitational energy per electron has no f dependence because it is per electron, not per baryon
The formula in Shapiro & Teukolsky does it per electron. Perhaps it will help to follow their analysis directly instead of trying to express everything in terms of ##M## as I was trying to do before.

The relevant equations in S&T are 3.4.1 through 3.4.5. If we do everything in terms of ##N##, the number of fermions (in the white dwarf case, electrons), then a factor of ##f## does appear in the gravitational term--but it is the only factor of ##f## that appears, since the Fermi energy term is now expressed directly in terms of ##N##, so no adjustment there is required. The Fermi energy per fermion is taken directly from S&T equation 3.4.1, and the gravitational energy per fermion is given by S&T equation 3.4.2, which has ##M## in it, so we have to substitute ##M = N m_B / f## to get everything in terms of ##N##. So S&T equation 3.4.3 for the total energy becomes:

$$
E = E_F + E_G = \frac{\hbar c N^{1/3}}{R} - \frac{G N m_B^2}{f R}
$$

Setting ##E = 0## to find the maximum for ##N##, as S&T do to obtain their equation 3.4.4, gives a modified equation 3.4.4 as follows:

$$
N_\text{max} = \left( \frac{\hbar c f}{G m_B^2} \right)^{3/2}
$$

and S&T equation 3.4.5 then becomes

$$
M_\text{max} = \frac{N_\text{max} m_B}{f} = \left( \frac{\hbar c}{G m_B^2} \right)^{3/2} m_B \frac{f^{3/2}}{f} = \left( \frac{\hbar c}{G m_B^2} \right)^{3/2} m_B f^{1/2}
$$

So the final factor in the formula for the maximum mass is ##f^{1/2}##, in agreement with the correction I made to my statement in the OP in my post #10.

Ken G said:
there's still the problem of the gravitational energy being the energy per electron, not per baryon
See above.
 
  • #13
PeterDonis said:
$$
E = E_F + E_G = \frac{\hbar c N^{1/3}}{R} - \frac{G N m_B^2}{f R}
$$
Then they have clearly made the same mistake you did! I know it will make me unpopular to point out an error in S&T, but it is obviously an error. The equation above has the wrong expression for the gravitational potential energy per electron. This is clear, we can agree the total gravitational potential energy is GM^2/R (I will assume I need no reference for that), so per electron is
$$
\frac{GM^2}{RN}= \frac{GN m_B^2}{ f^2 R} .
$$
Note the f^2 in the denominator, which corrects the result to what I'm saying, N ~ f^3. I believe the error in the textbook stems from thinking in terms of "per electron" is a bit tricky, that's why I did it in terms of total energies.
 
  • #14
Ken G said:
Then they have clearly made the same mistake you did!
Their analysis in section 3.4 does not include ##f## or any similar factor, any mistake is mine, not theirs. The equations in my post #12 are my modifications to the S&T equations to take ##f## into account. See further comments below.

Ken G said:
Note the f^2 in the denominator
Yes, I see that, and I see where the missing piece was in my previous post--the factor of ##1 / f## baryons per electron providing the gravitational potential well has to be included twice because the gravitational potential goes like the product of the masses.

Looking at S&T section 3.3, where they include a factor they call ##\mu_e## to take into account chemical composition, that factor looks like the reciprocal of what we have been calling ##f## (you have to go back to chapter 2 to see how they define it). Their equation 3.3.17, for the maximum mass in the extreme relativistic limit, is

$$
M =1.457 \left( \frac{2}{\mu_e} \right)^2 M_S
$$

(where ##M_S## is the mass of the Sun). This would equate to an ##f^2## dependence of the maximum mass. In terms of the formulas in post #12, we would have an extra factor of ##f^{3/2}## in the formulas for ##N_\text{max}## and ##M_\text{max}##, leading to a total factor of ##f^2## in ##M_\text{max}##.

This explains why S&T didn't bother taking into account chemical composition in section 3.4; they had already shown the dependence on chemical composition in section 3.3, and section 3.4 was simply an order of magnitude estimate by a different route (based on a 1932 paper by Landau).
 
  • #15
Ken G said:
it is likely the reason neutron stars exist
S&T Equation 3.3.17 would indeed suggest a factor of ##4## increase in ##M_\text{max}## for a neutron star, with ##\mu_e = 1##, vs. a white dwarf, with ##\mu_e \approx 2##. Taking into account that the Oppenheimer-Volkoff estimate for a neutron star, which took into account GR correction factors in the TOV equation that are negligible for white dwarfs but not for neutron stars, but AFAIK did not take into account any details of chemical composition, was about half the S&T Equation 3.3.17 value, that would indicate a rough estimate of twice the white dwarf maximum mass for neutron stars, or about 3 solar masses. That seems to be about where current estimates are, although there is still significant uncertainty (due to our lack of understanding of the detailed behavior of the strong interaction at these densities).
 
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  • #16
OK, then it seems equation 3.4.3 in S&T was not wrong after all, it was just the nontrivial interpretation of their definitions of quantities. So I believe we are back to the fundamental point here, which is that by glossing over the electron-per-baryon effect, S&T (and Landau) were able to get the useful result that neutron stars should have about the same maximum mass as white dwarfs, but they missed the opportunity to make the point that the electron-per-baryon difference is the reason neutron stars are able to exist at all. That lost opportunity has now been filled in!

Put in more physical terms, white dwarfs lose a little bit of pressure efficiency from the fact that they have only half as many energy-carrying particles, as they start to go degenerate, as neutrons stars of the same mass would have. This makes it harder for electrons to go degenerate than neutrons in objects of the same mass, because for electrons the virialized energy has to be put into fewer particles, which holds off the degeneracy (by shortening their deBroglie wavelengths), and achieving degeneracy is essential for preventing continued heat loss and the contraction it produces. That difference creates room for a second type of object, the neutron star, to exist later in the evolution of cores, in the mass regime where electrons could not reach degeneracy but neutrons can. (Short way to say this: having more neutrons allows them to reach degeneracy where electrons could not.)
 
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  • #17
Ken G said:
it seems equation 3.4.3 in S&T was not wrong after all, it was just the nontrivial interpretation of their definitions of quantities
I don't think it's a matter of interpretation; S&T's equation 3.4.4 simply didn't include any effect of chemical composition, I assume because they had already treated that in the previous section, as I said before.

Ken G said:
white dwarfs lose a little bit of pressure efficiency from the fact that they have only half as many energy-carrying particles
Yes, in the sense that there is more mass, and hence more weight for pressure to resist, at a given Fermi energy.

Ken G said:
achieving degeneracy is essential for preventing continued heat loss and the contraction it produces
I'm not sure what you mean here. A white dwarf can still radiate heat.

Ken G said:
That difference creates room for a second type of object, the neutron star, to exist later in the evolution of cores, in the mass regime where electrons could not reach degeneracy but neutrons can.
I'm not sure what you mean here either. First, AFAIK neutron star matter does not exist in the cores of stars that are still burning nuclear fuel, so I'm not sure what you're referring to by "the evolution of cores". Second, in the mass regime between the white dwarf maximum mass and the neutron star maximum mass, the issue is not that electrons "could not reach degeneracy but neutrons can"--the densities involved are several orders of magnitude larger than white dwarf densities, and a Fermi electron gas at such densities would be extremely relativistically degenerate if it were to exist. The issue is that the equation of state of degenerate matter goes through a transition in between white dwarf densities and neutron star densities, in which any electrons present combine with protons to form neutrons, so there simply aren't any significant number of electrons (or protons) left: the chemical composition has changed to be virtually all neutrons. This is discussed in Chapter 8 of Shapiro & Teukolsky.
 
  • #18
PeterDonis said:
Yes, in the sense that there is more mass, and hence more weight for pressure to resist, at a given Fermi energy.
That's one way to frame it, I was taking the approach of fixing the mass, and asking why, for that same mass, a neutron star succeeds in going degenerate where a white dwarf fails.
PeterDonis said:
I'm not sure what you mean here. A white dwarf can still radiate heat.
Not well. It's true that nothing is every truly degenerate, which would require zero temperature (and hence no further loss of heat). We simply idealize white dwarfs as if they had reached that point (that's also what you do when you use the formula for "degeneracy pressure"), when in fact they're not quite there yet. Still, they are getting there-- and their ability to lose heat is interdicted as a result. This greatly slows, and ultimately stops, their contraction, because as the virial theorem so clearly shows, it is always heat loss that produces contraction of any object in force balance. This is the story of any star.
PeterDonis said:
I'm not sure what you mean here either. First, AFAIK neutron star matter does not exist in the cores of stars that are still burning nuclear fuel, so I'm not sure what you're referring to by "the evolution of cores".
I mean what evolution does to them. Yes of course a star undergoing fusion has not yet been neutronized, it hasn't lost enough net heat yet because fusion keeps replacing it. Evolution is caused by two things-- net loss of heat (if they are not fusing), or chemical change (if they are). The latter process is self-limited, so the former process is the road to neutronization-- unless the electrons succeed in going degenerate first.
PeterDonis said:
Second, in the mass regime between the white dwarf maximum mass and the neutron star maximum mass, the issue is not that electrons "could not reach degeneracy but neutrons can"--
Actually, that is the issue, this is the insight I'm attempting to convey because I think it is quite valuable to understand. The key point is, you take exactly your equations, but do not assume the gas is already degenerate and ask if it can be in force balance, assume it is in force balance (i.e., has the virial kinetic energy) and ask if it can be degenerate. The comparison is precisely the same, it's that Fermi energy versus gravitational potential energy that you've been using all along, just framed the way the star is actually constrained to work. Degeneracy is not a requirement, virialization is (because the sound crossing times are much shorter than heat-loss timescales). When you see the Fermi energy must be below the virial energy whenever M > 1.4, you see that the electrons cannot go degenerate, because their necessary higher kinetic energy forces a smaller deBroglie wavelength, so they're just not degenerate. That's all there is to it, no different equations at all.

Now note the real purpose of this reframing: the clear fact that your own equations, taken with force balance as the requirement and degeneracy as the thing being tested instead of the other way around, implies something quite important. If the electrons cannot go degenerate, they cannot interdict heat loss (their entropy is not getting small enough to cling to that heat, via the laws of thermodynamics), and the virial theorem tells us what heat loss will do to those electrons: it will raise their kinetic energy scale. First they will go relativistic, which is bad enough because now the smallest heat loss causes drastic rises in their virial kinetic energy, and then they will reach the neutronization scale, and that's the end of them. If you personalize the electrons, you can say they are trying to go degenerate as that is their only way to save their skins, but for M>1.4, they can't pull it off because their virial energy always stays ahead of their Fermi energy. (That's in your equations.)

By the way, this is the same point I was making in your insights post, but I apologize for trying to make it there-- that was indeed a kind of hijack of the point you were making there. But it's quite apropos here, as we are talking about what happens to stars with M > 1.4 vs < 1.4.

PeterDonis said:
the densities involved are several orders of magnitude larger than white dwarf densities, and a Fermi electron gas at such densities would be extremely relativistically degenerate if it were to exist.
Yes, that is why we use the relativistic Fermi energy whenever we are talking about a relativistic virial energy. I hope it is clear what I mean by the virial energy-- it's just the gravitational potential energy that you have in all your order-of-magnitude expressions above.

PeterDonis said:
The issue is that the equation of state of degenerate matter goes through a transition in between white dwarf densities and neutron star densities, in which any electrons present combine with protons to form neutrons, so there simply aren't any significant number of electrons (or protons) left: the chemical composition has changed to be virtually all neutrons. This is discussed in Chapter 8 of Shapiro & Teukolsky.
I realize that, this is the whole point. If the electrons wish to survive against an inexorable background of net heat loss, they must achieve degeneracy, and interdict that heat loss, else they will inexorably reach the energy of neutronization. My point is, assuming degeneracy to test if there can be force balance is a valid way to find the critical mass where this fails, but actual stars are in force balance and are in essence asking themself the question, am I degenerate yet so I can cease this process of heat loss and contraction?
 
  • #19
Ken G said:
I was taking the approach of fixing the mass, and asking why, for that same mass, a neutron star succeeds in going degenerate where a white dwarf fails.
That's not what the respective maximum mass limits show. In fact they show the opposite: a given mass that is below both mass limits, say 1 solar mass, is closer to the white dwarf mass limit than to the neutron star mass limit, so the degeneracy in a white dwarf at that mass is more relativistic (since the relativistic limit is what determines the maximum mass limit) than in a neutron star.

At a mass above the white dwarf mass limit, you can't even compare the two cases because a white dwarf can't even exist at that mass.

Ken G said:
It's true that nothing is every truly degenerate, which would require zero temperature (and hence no further loss of heat).
I don't know where you're getting this definition of "truly degenerate". Degeneracy doesn't require zero temperature. It just requires that the main thing determining which states are occupied is Fermi-Dirac statistics instead of thermal distribution. Any temperature much less than the Fermi energy will satisfy that requirement. "Zero temperature" in the math is simply a useful approximation for this regime, not an exact claim.

Ken G said:
their ability to lose heat is interdicted as a result
Please give a reference for this, since I don't know where you are getting it from or what comparison it is based on.

Ken G said:
The latter process is self-limited, so the former process is the road to neutronization-- unless the electrons succeed in going degenerate first.
Again I'm not sure what you mean. If you mean that, for an object that is below the white dwarf mass limit, it will stop collapsing at the white dwarf stage--which is at a density a few orders of magnitude smaller than the density at which the onset of neutronization occurs--then yes, I agree.

But in an object above the white dwarf mass limit, which will not stop collapsing at the white dwarf stage (because there is no stable equilibrium there), if you are claiming that the electrons will not be degenerate at the point where neutronization occurs, that is false. That point in the equation of state is a couple of orders of magnitude smaller in radius and higher in density than the white dwarf stage, so the electrons will be extremely relativistically degenerate--it's just that electron degeneracy will have failed to stop the collapse process.

Ken G said:
it is always heat loss that produces contraction of any object in force balance
More precisely, heat loss will produce contraction of any object in force balance whose structure is primarily determined by hydrostatic equilibrium--outward pressure balancing weight due to gravity, with the matter in the object being well described as a fluid.

But objects such as collapsing supernova remnants are not in force balance as they collapse--they will only reach a force balance if their masses are within the white dwarf or neutron star mass limits, in which case the collapse process will (probably) stop at the corresponding range of density and radius.

I'll comment on the rest of your post separately since it raises another issue.
 
  • #20
Ken G said:
this is the insight I'm attempting to convey because I think it is quite valuable to understand
Ken G said:
this is the same point I was making in your insights post
This seems like something that you have come up with personally and which is not currently reflected in the mainstream literature on the topic. If it is, you will need to provide a mainstream reference. If it is not, then it is out of bounds for discussion here at PF, because it is something that needs to be published in the literature and subjected to review and criticism by other experts in the field before it can get to the point where it is suitable for discussion here. The purpose of PF is not to try to resolve contested scientific questions. It is to help people understand science that is already mainstream, i.e., not contested.

Specifically, this claim of yours does not seem consistent with the scientific literature on this topic (such as Shapiro & Teukolsky):

Ken G said:
The comparison is precisely the same, it's that Fermi energy versus gravitational potential energy that you've been using all along, just framed the way the star is actually constrained to work. Degeneracy is not a requirement, virialization is (because the sound crossing times are much shorter than heat-loss timescales). When you see the Fermi energy must be below the virial energy whenever M > 1.4, you see that the electrons cannot go degenerate
In Shapiro & Teukolsky, the point at which the Fermi energy equals the (absolute value of) the gravitational potential energy is the point that determines the maximum mass limit, which is not the point at which the fermions "go degenerate". They are already degenerate. The maximum mass limit point is the point beyond which there is no longer a stable equilibrium.

For the mainstream view on when the fermions actually "go degenerate", see, for example, Shapiro & Teukolsky section 3.2, The Onset of Degeneracy. Note that their treatment takes into account the things you mention, such as the virial theorem and the role of gravitational potential energy. But their criterion for the onset of degeneracy is that the volume occupied by an electron in phase space becomes of order ##\hbar^3##. Equation 3.2.16 gives the equation for phase space volume and makes clear that there is a substantial possible range of parameters corresponding to the onset of degeneracy, most of which come nowhere near satisfying the condition you claim is required for degeneracy, that the Fermi energy must equal the (absolute value of) the gravitational potential energy.
 
  • #21
Ken G said:
actual stars are in force balance and are in essence asking themself the question, am I degenerate yet so I can cease this process of heat loss and contraction?
This claim also does not appear to be consistent with the mainstream scientific literature on this topic, at least not for the regime under discussion here. Obviously main sequence stars are in force balance, as are stable white dwarfs and neutron stars, but we are not just discussing those cases here. We are discussing cases in intermediate regimes between these--for example, a star of one solar mass with a radius significantly smaller than that of our current Sun but significantly larger than the radius of a one solar mass white dwarf; or a star of, say, two solar masses with a radius much smaller than that of a main sequence star of that mass but significantly larger than the radius of a two solar mass neutron star.

The key point of the equation of state discussed in Shapiro & Teukolsky is that there are no stable equilibrium states in these regimes. That means that it is impossible for, say, a main sequence star of one solar mass to slowly contract to a white dwarf of that mass by a quasi-static heat loss process. There are no intermediate equilbrium states for it to be in during such a contraction. The contraction must be a non-equilibrium collapse process in which the star's surface is more or less in free fall, until it reaches white dwarf size (and therefore density) and the collapse is halted by electron degeneracy pressure. Similarly, there is no way for, say, a white dwarf which accretes enough mass to go just over the Chandrasekhar limit to slowly collapse to a neutron star of that mass by a quasi-static heat loss process; again, there are no intermediate equilibrium states for it to be in. The collapse must be a catastrophic one until neutron star densities are reached and the collapse is stopped by neutron degeneracy pressure (and, as I said earlier, the electrons present will have become degenerate well before the onset of neutronization, given the relevant densities for each to occur).

So "force balance" is irrelevant in analyzing the collapse processes involved.
 
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  • #22
PeterDonis said:
the volume occupied by an electron in phase space becomes of order ##\hbar^3##.
This can also, of course, be stated as a criterion for the density of the object; the specifics will of course depend on chemical composition. Per Shapiro & Teukolsky (chapters 2 and 8), for typical white dwarf chemical compositions, the onset of degeneracy is at a density of roughly ##10^5## g/cm^3, with the degeneracy becoming relativistic at around ##10^7## g/cm^3; also at around that point, the chemical composition of the equilibrium nuclide starts to shift to a higher proportion of neutrons. The "neutron drip" onset (i.e., the point at which neutrons start to "drip" out of the nuclei and electrons and protons start to combine into neutrons) is at about ##3 \times 10^{11}## g/cm^3; neutron degeneracy is already occurring by the time stable states become possible again at about ##10^{12}## g/cm^3, and the neutron degeneracy becomes relativistic at about ##10^{15}## g/cm^3. By that point, however, the analysis is complicated by the fact that the strong interaction has become significant but we don't understand it very well in this regime.
 
  • #23
PeterDonis said:
That's not what the respective maximum mass limits show. In fact they show the opposite: a given mass that is below both mass limits, say 1 solar mass, is closer to the white dwarf mass limit than to the neutron star mass limit, so the degeneracy in a white dwarf at that mass is more relativistic (since the relativistic limit is what determines the maximum mass limit) than in a neutron star.
I'm not sure why you think that is relevant to my statement, which was simply a fact: if we watch a 2 solar mass ball of self-gravitating hydrogen gas lose net heat, it will definitely start out with protons and electrons in force balance, and end up as neutrons in force balance, and the electrons will definitely never go degenerate, and the neutrons definitely will. This is all I said, and it's simply true.
PeterDonis said:
At a mass above the white dwarf mass limit, you can't even compare the two cases because a white dwarf can't even exist at that mass.
The limit is an awkward mass, as it is perched between two behaviors. I am looking at M > 1.4, and contrasting to M < 1.4. I don't know what you mean by "can't compare the two cases", those are the two cases I'm comparing.
PeterDonis said:
I don't know where you're getting this definition of "truly degenerate".
Then I shall clarify: the limit of complete degeneracy, i.e., zero temperature, i.e., when the Fermi-Dirac distribution looks like a step function. This is certainly a common idealization, and it is certainly the only time the formula called "degeneracy pressure" formally applies.
PeterDonis said:
Degeneracy doesn't require zero temperature. It just requires that the main thing determining which states are occupied is Fermi-Dirac statistics instead of thermal distribution.
Yes, I know all that. However, you did use what gets called "degeneracy pressure", and I presume you realize that formula does not apply for Fermi-Dirac statistics, except in the limit of zero temperature. Why are you splitting hairs here, we are using the limit of complete degeneracy as a useful idealization for thinking about the pressure in the gas. A real gas never obeys that pressure, because it never has zero temperature, but we can still use the benchmark, and we do. Similarly, we can notice that a gas approaching complete degeneracy is experiencing the interdiction of its heat loss, expressly because its temperature is being suppressed. That's what the Fermi-Dirac distribution does as it looks more and more like a step function, it has a low temperature and weak heat loss.

Think of a box of fixed volume containing gas, with no gravity to simplify. Let the gas lose heat. Its distribution will go from Maxwell-Boltzmann to Fermi-Dirac (it's actually always Fermi-Dirac, but we don't bother with it until we need it). Eventually it will look a lot like a step function. That's degeneracy, that's when its pressure equals the formula called "degeneracy pressure."
PeterDonis said:
Any temperature much less than the Fermi energy will satisfy that requirement. "Zero temperature" in the math is simply a useful approximation for this regime, not an exact claim.
Yes, this is obvious, and not relevant to my point.
PeterDonis said:
Please give a reference for this, since I don't know where you are getting it from or what comparison it is based on.
You need a reference that objects reaching a Fermi-Dirac distribution that looks like a step function cannot spontaneously lose heat? All right, here's Nernst's heat theorem: (https://en.wikipedia.org/wiki/Nernst_heat_theorem).
Relevant expression:
\lim _{T\to 0}\Delta S=0

This says that a system with zero entropy also has zero temperature. Then we have the zeroth law of thermodynamics:
(https://en.wikipedia.org/wiki/Zeroth_law_of_thermodynamics)
This says that if a system at zero temperature cannot lose heat to anything with finite temperature, such as the rest of universe. Ergo, a Fermi-Dirac distribution that looks like a step function, which most people call "complete degeneracy", is a state of complete interdiction of heat loss. This is the state that white dwarfs, for example, are headed toward. It is not surprising their T is dropping before they get there, and it is not surprising the rate that they lose heat is also dropping. There is a simple physical interpretation that is seen everywhere: as electrons go degenerate, those in the Fermi sea become inert to heat loss, and only the ones above the sea can lose heat. This strongly interdicts the ability of the system to lose heat, which is not surprising since it is also an effect that lowers its temperature.
PeterDonis said:
Again I'm not sure what you mean. If you mean that, for an object that is below the white dwarf mass limit, it will stop collapsing at the white dwarf stage--which is at a density a few orders of magnitude smaller than the density at which the onset of neutronization occurs--then yes, I agree.
I'm certainly saying that obvious thing, but I'm also saying something else: the reason it stops contracting (many authors prefer "contract" to "collapse", as the latter generally implies being out of force balance, which white dwarfs are not) is that is has lost the ability to lose heat. For proof, I cite the virial theorem, which I trust you already know. (And as we've said, of course the white dwarf never actually gets to this state, as it is a zero temperature state, but it is approaching that state so we idealize it that way-- exactly like we use the "degeneracy pressure" formula. I trust you know that formula is only actually true in the zero temperature limit.)
PeterDonis said:
But in an object above the white dwarf mass limit, which will not stop collapsing at the white dwarf stage (because there is no stable equilibrium there),
That's not correct, which is my whole point-- you have the misconception I'm correcting. Of course the object will be in force balance, it is a ball of gas. What it won't be is degenerate. So there is no "white dwarf stage" above the mass limit, it will still be an ideal gas, or somewhere between.

All this follows directly from your own equations, simply don't assume the gas is degenerate. Remember, you added that part, as your own conceptual device to find the limiting mass. A standard device, I've used it myself many times, but it is designed to find the limiting mass, not as a physical rule that the object must obey regardless of its mass.

Look at your textbook, I guarantee you it never says a 2 solar mass ball of ion and electron gas will ever actually be degenerate (in the sense of the electrons having the Fermi energy, or exhibiting the "degeneracy pressure" formula). It's a very different question to ask "what if it's degenerate," and find that something breaks-- the question is, what breaks? You think the answer is force balance, but that's what's wrong-- the sound crossing time is much shorter than heat-loss timescales, so what breaks is the assumption that enough heat has been lost to create degeneracy (and yes, degeneracy requires heat loss, that's dS = dQ/T where dS leads to degeneracy since degeneracy is a low S state).

PeterDonis said:
if you are claiming that the electrons will not be degenerate at the point where neutronization occurs, that is false.
That's exactly the crucial place where you are wrong, which is the value of this discussion. Remember, we are talking about a specific situation, a ball of, say, 2 solar masses of ions and electrons. Such a ball will not be degenerate when neutronization occurs, that's my whole point. This is proven from simply requiring force balance, equating the electron kinetic energy to the gravitational potential energy, and noticing by your own equations that this requires the electron energy to be above the Fermi energy. Do it, it's right there in front of you.

PeterDonis said:
That point in the equation of state is a couple of orders of magnitude smaller in radius and higher in density than the white dwarf stage, so the electrons will be extremely relativistically degenerate--it's just that electron degeneracy will have failed to stop the collapse process.
No, you can see that is wrong from your equations. Use the relativistic versions, you will still see that since M > 1.4, the electrons will have a virial kinetic energy is above the Fermi energy. It's in your equations, indeed it is just another way to interpret the meaning of those same equations-- a way that respects what is actually happening, i.e., force balance, rather than just a way to find the boundary value of the mass.
PeterDonis said:
More precisely, heat loss will produce contraction of any object in force balance whose structure is primarily determined by hydrostatic equilibrium--outward pressure balancing weight due to gravity, with the matter in the object being well described as a fluid.
Yes, exactly. And that happens on a sound-crossing time. Let's have a look at a typical sound crossing time in a white dwarf-- the escape speed is like 10,000 km/s, the radius is like 10,000 km, so it's like 1 second. That's how long it takes such an object to find force balance.
PeterDonis said:
But objects such as collapsing supernova remnants are not in force balance as they collapse
I thought we were talking about white dwarfs.
PeterDonis said:
--they will only reach a force balance if their masses are within the white dwarf or neutron star mass limits, in which case the collapse process will (probably) stop at the corresponding range of density and radius.
No, they are in force balance right up until their final collapse, the very last step. That happens when exothermic processes appear, including photodisintegration of nuclei, neutronization of electrons, and neutrino generation. Any reference on core collapse supernovae will feature those processes prominently, for this reason.
PeterDonis said:
I'll comment on the rest of your post separately since it raises another issue.
OK I will continue the discussion there.
 
  • #24
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