I Why is the Pauli Exclusion Principle not a force?

[zonde]

It's just garden variety kinetic pressure if you know E and V, it makes no difference to P what the spin statistics are, or even if the particles are distinguishable or not.

I agree with you about kT and E part you say but I don't agree with that "garden variety kinetic pressure" part. Quantum states for stable configuration are determined by potential well, not simply volume. Another thing is that interacting particle changes it's quantum state. If there are no available quantum states that it can occupy after interaction it can't participate in interaction. And how can we talk about kinetic pressure without interactions between particles?

 

[Ken G]

There is nothing to disagree with, if you understand that degeneracy pressure of a monatomic nonrelativistic gas obeys P = 2/3 E/V, where E is the internal kinetic energy and V is the volume. That's garden variety kinetic pressure, that's all it is. You simply look at the momentum fluxes of all the particles in there, and it ends up giving you that pressure. There are no "potential wells" at all, those are all neglected in what gets called "degeneracy pressure." The only "interactions" between the particles are how they partition that kinetic energy, but we don't care how they partition the kinetic energy if we only need to understand P. We only need that if we want to understand T. The PEP constrains heat transfer, it says nothing about pressure if you have a way of knowing E without reference to T-- as is commonly true in stars. The way I would put it is, the PEP is thermodynamic, but pressure is purely mechanical.

 

[PeterDonis]

if you understand that degeneracy pressure of a monatomic nonrelativistic gas obeys P = 2/3 E/V, where E is the internal kinetic energy and V is the volume.

Is there a way to derive this relationship for a degenerate gas at zero absolute temperature? As I understand it, the usual definition of "temperature" in kinetic theory requires that kinetic energy is directly proportional to temperature, but that obviously can't be the case if E can be nonzero when T is zero.

 

[Ken G]

Is there a way to derive this relationship for a degenerate gas at zero absolute temperature? As I understand it, the usual definition of "temperature" in kinetic theory requires that kinetic energy is directly proportional to temperature, but that obviously can't be the case if E can be nonzero when T is zero.

Yes, the temperature being proportional to kinetic energy is only true in ideal gases, not degenerate ones. But for either type, the pressure is simply the momentum flux density, which has the same units as kinetic energy density. So it is informative to form the ratio of those quantities, for any nonrelativistic monatomic gas that has an isotropic distribution function f(v). The ratio of momentum flux density (i.e., pressure) to kinetic energy density is thus the ratio of the integral over dmu (here by mu I mean the Greek mu, the direction cosine relative to a plane of reference) and dv of mu² mv² f(v) to the integral over dmu and dv of 1/2 mv² f(v), where f(v) is the density of particles in bin dv and it doesn't matter because it is the integrals over dmu that count. (The two powers of mu in the pressure come from the fact that we are considering a flux, through a plane, of momentum perpendicular to that plane, so this is an isotropic stress tensor.) So carrying out the mu integrals gives the ratio of pressure to kinetic energy density is 2/3, and all we used was isotropic f(v) and nonrelativistic momentum and energy. So it is valuable to recognize the core connection between pressure and kinetic energy density that is independent of the temperature and the way the kinetic energy is partitioned over the particles. This helps separate the mechanical from the thermodynamic properties of the gas, an issue which gets terribly muddled in a lot of what is written about degenerate gases.

 

[PeterDonis]

carrying out the mu integrals gives the ratio of pressure to kinetic energy density is 2/3, and all we used was isotropic f(v) and nonrelativistic momentum and energy.

Ah, I see. I assume that the range of $\mu$ must be $0$ to $1$, then? (You used the term "direction cosine", so I assume you mean the magnitude of the cosine without taking into account the sign, since the cosine itself ranges from $-1$ to $+1$.)

 

[Ken G]

Remarkably, you can go 0 to 1, or -1 to 1, it doesn't matter. The kinetic energy is a scalar so doesn't care about the sign of mu, and momentum flux is always positive, even when mu is negative, because you either have a positive flux of positive momentum, or a negative flux of negative momentum.

 

[PeterDonis]

you can go 0 to 1, or -1 to 1, it doesn't matter.

But the limits of integration will affect the value of the integral, won't they? Basically you have

$$
\frac{P}{E} = \frac{\int \mu^2 d\mu}{\frac{1}{2} \int d\mu}
$$

If we evaluate the integrals, we have

$$
\frac{P}{E} = \frac{\frac{1}{3} \mu^3}{\frac{1}{2} \mu} = \frac{2}{3} \mu^2 \vert_{\mu_0}^{\mu_1}
$$

So we must have

$$
\mu^2 \vert_{\mu_0}^{\mu_1} = 1
$$

to obtain the result $P/E = 2/3$. If the limits are $0$ to $1$, I see how that is obtained; the lower limit gives $0$ and the upper limit gives $1$. But if the limits are $-1$ to $1$, then the integral should be zero; the lower limit gives a $1$ which is subtracted from the upper limit of $1$ to get a vanishing final value. What am I missing?

 

[Ken G]

Both integrands are even functions of mu, it doesn't matter if they both go from -1 to 1, or 0 to 1. You just have to be consistent top and bottom. Officially, you are going -1 to 1 to accommodate all directions, but it's the same answer if you just go 0 to 1, any half-sphere gives you the right ratio because the other half-sphere just doubles both the momentum flux and the kinetic energy density.

 

[PeterDonis]

Both integrands are even functions of mu, it doesn't matter if they both go from -1 to 1, or 0 to 1.

Ah, I see; I was being too sloppy in taking the ratio.

 

[vanhees71]

Everything can be derived from the grand-canonical partition sum. We start at finite temperature and a Fermi gas of particles with $g=2s+1$ spin degrees of freedom in a large volume $V$. In the thermodynamic limit the sum over the discrete momenta determined by periodic boundary conditions (taking the volume as a cube of edge length $L$) can be approximated with good accuracy by an integral. The final result for the grand-canonical potential, yields
$$\Omega(T,\mu,V)=\ln Z=\mathrm{Tr} \exp[-(\hat{H}-\mu \hat{N})/T]\\ =\frac{g V}{(2 \pi)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \ln[1+\exp(-\vec{p}^2/(2mT)+\mu/T)].$$
I use natural units with $\hbar=1$ and $k_{\text{B}}=1$.

The integral can first be simplified by rewriting it in terms of $E=\vec{p}^2/2m$ via the introduction of spherical coordinates in momentum space. The angular integral just gives a factor $4 \pi$:
$$\Omega(T,\mu,V)=\frac{P V}{T}=\frac{g V}{2 \pi^2} \int_{0}^{\infty} \mathrm{d} p p^2 \ln[1+\exp(-\vec{p}^2/(2mT)+\mu/T)] \\
= \frac{g V (2m)^(3/2)}{4 \pi^2} \int_0^{\infty} \mathrm{d} E \sqrt{E} \ln[1+\exp(-E/T+\mu/T)].$$
Now we make $T \rightarrow 0^+$. For $E>\mu$ we get $0$, and for $E<\mu$ we can neglect the $1$ against the exp under the ln, and this gives
$$\Omega=\frac{g V (2 m)^{3/2}}{4 \pi^2 T} \int_0^{\mu} \mathrm{d} E \sqrt{E}(\mu-E)=\frac{g V (2m)^{3/2} \mu^{5/2}}{15 \pi^2 T}=\frac{p V}{T}.$$
The mean particle number is given by (taking again $T \rightarrow 0^+$ under the integral)
$$N=\frac{\mathrm{d}}{\mathrm{d} \alpha} [\Omega|_{\mu=T \alpha}]=\frac{g V (2m)^{3/2}}{4 \pi^2} \int_0^{\mu} \mathrm{d} E \sqrt{E} = \frac{g V (2m)^{3/2} \mu^{3/2}}{6 \pi^2}.$$
Finally the energy is
$$U=-\frac{\mathrm{d}}{\mathrm{d} \beta} [\Omega_{T=1/\beta,\mu=\alpha/\beta}]=\frac{g V (2m)^{3/2} \mu^{5/2}}{10 \pi^2}.$$
From these relations you can derive
$$U=\frac{3}{5} \mu N=\frac{3}{2} P V.$$
The fact that a Fermi gas at $T=0$ has a finite pressure and energy is due to the Pauli principle. You have to fill the phase space up to the Fermi energy, which in the limit $T \rightarrow 0$ is simply the chemical potential $\mu$. Thus you have Fermi motion and thus a finite pressure and energy.

 

[Ken G]

Yet note that the relation U = 3/2 PV is a mechanical relation that is more general than any particular thermodynamic partition function. In particular, there is no need to involve T, as it is not used in the result. So it's a question of whether you want to get everything, including T, and use the "full Monte," or if you just want to focus on issues that don't involve T. When T gets involved, a lot of strange things get said about degeneracy pressure! I've even heard it said that degeneracy pressure is independent of T, which is an odd statement for a pressure that specifically applies when T=0.

 

[PeterDonis]

I've even heard it said that degeneracy pressure is independent of T, which is an odd statement for a pressure that specifically applies when T=0.

Degeneracy pressure doesn't go away for $T > 0$, though, does it? It stays the same.

 

[vanhees71]

Of course, the gas stays "degenerate" (which is a somewhat misleading name for the fact that at low temperatures quantum effects become important at the macroscopic scale) also at not too high temperatures. At higher temperatures the gas can be described by classical statistics. This happens, because then in each phase-space cell doesn't contain too much particles on average. That's clear from the phase-space distribution function of the ideal gas, which reads
$$f(\vec{x},\vec{p})=\frac{g}{\exp[(E(\vec{p})-\mu)/T] \pm 1},$$
where $T$ is the temperature (measured in energy units, i.e., setting $k_{\text{B}}=1$), $\mu$ the chemical potential, $g=2s+1$ the spin-degeneracy factor, and $E(\vec{p})=\vec{p}^2/(2m)$ for a non-relativistic gas. The upper (lower) sign is for fermions (bosons).

The case for bosons must be handled with special care in the low-temperature limit since there $\mu \geq 0$, and if for a fixed temperature the particle number in the volume cannot be fixed for any $\mu \geq<0$, Bose-Einstein condensation must be taken into account. At $T=0$ for an ideal Bose gas all particles occupy the lowest energy state $E=0$, which is the ground state of any fixed number of non-interacting bosons. The reason, why this is a bit complicated is that I immediately wrote down the phase-space distribution function in the "thermodynamic limit", i.e., where formally $V \rightarrow \infty$ and the momentum sum of the finite volume was approximated by an integral, where the zero-mode contribution is cut away, because even for $\mu=0$ and for $E_{\vec{p}}\ll T$ the integral over $\mathrm{d}^3 \vec{p} f(\vec{p})$ stays finite, because $f(\vec{p}) \approx \frac{T}{E}$ and $\mathrm{d}^3 \vec{p}=p^2 \mathrm{d} p \mathrm{d}^2 \Omega$.

The classical Boltzmann distribution is reached if $f \ll 1$, i.e., if $-\beta \mu \gg 1$, and then for both fermions and bosons
$$f \simeq g \exp[-(E(\vec{p})-\mu)/T].$$
Then the gas behaves like a classical ideal gas as expected. The reason, of course, is that if the occupation number per phase-space cell becomes small (on average) the Bose/Fermi nature of the particles becomes negligible.

 

[Ken G]

Right, there is extreme ambiguity about the phrase "degeneracy pressure." Some people treat it like you could somehow add it to the "thermal pressure," as if a gas has two sources of pressure, one normal and one quantum. But there is only one kind of pressure here, it's perfectly mundane kinetic pressure. The difference between "degeneracy pressure" and "thermal pressure" are simply the low- and high-T limits of the mundane kinetic pressure (for fermions). So that's all those terms mean, whether we are going to apply a low T limit or a high T limit when we simplify the equation of state. But all of that only comes into play if you want to know T, if all you care about is P, and you know E and V, then you don't need to know anything about T or anything about the particle statistics (other than that you may assume a nonrelativistic monatomic gas). Not realizing this leads to a lot of confusion, and you hear strange statements like "degeneracy pressure doesn't depend on temperature." Of course it depends on temperature, it's the low-T limit for the equation of state, so it only applies when kT << E, so how does that "not depend on temperature"? All they mean is that P isn't zero at T=0, that doesn't mean you have two types of pressure, it just means the P(T) function isn't always the ideal gas law.

 

[zonde]

Pressure is defined as force (per unit area of surface perpendicular to surface). But PEP is not a force. So phrases like "electrons can be squeezed into smaller volume by adding energy" do not make sense. Electrons can be squeezed into smaller volume by squeezing potential well i.e. squeezing ions (protons) into smaller volume. And squeezing ions of course can be analyzed in terms of force and energy. But then it's electrostatic repulsion of ions that creates the opposite force not degeneracy of electrons. Electrostatic repulsion of course is relieved to the level as much electrons "follow" ions into smaller volume but higher energy electron states will tend to occupy bigger volume.

 

[my2cts]

it's electrostatic repulsion of ions that creates the opposite force not degeneracy of electrons.

In the free electron model both electron and compensating positive charge repulsion are neglected but there still is "degeneracy" pressure coming from the PEP.

 

[zonde]

In the free electron model both electron and compensating positive charge repulsion are neglected but there still is "degeneracy" pressure coming from the PEP.

Do you say that PEP is force?

 

[my2cts]

The PEP is the reason why the energy increases with electron density in the free electron model.
Yet its formulation does not involve a potential and I don't know how to see it as a force.
Something important is missing in my opinion.

 

[Ken G]

What is missing is the recognition that energy does not appear at the behest of the PEP. What the PEP actually is is an interdiction on heat transport out of a system as that system approaches its quantum mechanical ground state, where the PEP also constrains, in participation with other factors, what that ground state is. The energetics of the ground state are controlled by something else, some true force with some actual potential energy associated with it. Indeed, whether the ground state, and the PEP, is associated with high or low energy and pressure depends entirely on other aspects of the system. If you take a box filled with a gas, and simply remove heat, while disallowing any forces between the particles that could lead to phase changes away from a gaseous state, then the ground state kinetic energy and its associated "degeneracy pressure" are both minima of kinetic energy and pressure for that system, and are both very low indeed.

 

[vanhees71]

What do you mean by "What the PEP actually is is an interdiction on heat transport out of a system as that system approaches its quantum mechanical ground state"? That's a bit misleading. I guess what you refer to here is Nernst's Theorem, according to which the entropy at $T=0$ is given by $S_0=\ln g_0$, where $g_0$ is the degeneracy of the ground state (provided there's an energy gap between the ground state and the excited states). This implies that the heat capacity goes to 0:

https://en.wikipedia.org/wiki/Third_law_of_thermodynamics

This however is valid also for bosons, not only fermions.

The Pauli exclusion principle is of course indeed not a force but the fact that by definition many-fermion systems are described in the antisymmetric Fock space of many-body theory (or in the case that the particle number is conserved the antisymmetric $N$-body Hilbert space, where antisymmetric refers to the behavior of wave functions under exchange of two identical particles within a pure many-body state, which operation for fermions necessarily flips the sign of the corresponding state vector).

 

[my2cts]

The Pauli exclusion principle is of course indeed not a force but the fact that by definition many-fermion systems are described in the antisymmetric Fock space of many-body theory ...

Somewhere there has to be a force in the PEP story.
The example of the free electron model for metals makes that clear.
On the other hand, even a single electron in a box pushes back if the volume is reduced.

 

[Ken G]

It is indeed a fascinating application of Nernst's theorem, thanks for that, but I just mean it in the more general sense that any system whose temperature is being suppressed somehow will be worse at transporting heat out of that system, and that's the important thing the PEP is doing. I am imagining a gas of noninteracting particles, appropriate for the usual applications like white dwarfs, and asking what the Pauli exclusion principle is doing to the global properties of that gas. The way it acts is by driving down the temperature, that's the global impact of degeneracy. If you look more closely at the microphysics, you see that the way this happens is that the distribution of kinetic energy over the particles start to look a lot different due to the PEP (the Fermi-Dirac distribution starts to look a lot different from the Maxwell-Boltzmann distribution as the PEP starts to matter), but that microphysical distribution has no effect on anything global except heat transport via its effect on temperature. It doesn't directly affect the density, pressure, volume, or any other global properties but temperature and heat transport, because there are no forces associated with the PEP that can do anything to density, pressure, volume, etc. The forces there are just the usual ones, and the pressure there comes from the kinetic energy density, which is controlled by the history of energy transport (including heat loss). So that's it, the PEP impacts the system exclusively because it reduces heat loss by driving down the temperature, in any situation where you are tracking the energy transport history.

A key point about the PEP is that it involves very low entropy, as you say, so the way to get it to appear (reversibly) is to make a system lose heat. As the ground state is approached, the normal way the PEP functions is by suppressing the temperature, but a lot of misconceptions appear when people try to think of the situation as having the temperature externally imposed, say by a heat reservoir. When you do that, the PEP takes on a very different character, because it cannot suppress T, so you don't notice the tendency to suppress heat loss when you are in a situation that fixes T. Instead, the PEP appears in the form of requiring a strangely high amount of work to squeeze out a given amount of heat, because the internal kinetic energy is way higher than what it would be for an ideal gas at the same T. But fixing T like that is not the usual astrophysical application, so it tends to be a false lead when thinking about things like white dwarfs. There, you just track the energy history, and all the PEP is doing globally is suppressing T and heat loss.

 

[Ken G]

Somewhere there has to be a force in the PEP story.

The force is garden variety gas pressure, 2/3 the kinetic energy density for a nonrelativistic monatomic noninteracting gas.

On the other hand, even a single electron in a box pushes back if the volume is reduced.

Right, because it's garden variety gas pressure.

 

[PeterDonis]

the free electron model for metals

Can you give a reference on which you are basing your understanding of this model?

 

[my2cts]

Ashcroft/Mermin chapter 2.

 

[vanhees71]

A very good treatment of the theory of metals can be found in Sommerfeld's theory textbook, vol. 5, which is no surprise, because he and Bethe wrote a famous review about the topic before.

 

[my2cts]

The force is garden variety gas pressure, 2/3 the kinetic energy density for a nonrelativistic monatomic noninteracting gas.

This statement has a good point in it.
Good point:
The contribution to the pressure of each electron in a Fermi gas is P=-dE/dV=2/3 E/V, as for a thermal particle.
So the pressure can be interpreted as to result from some inelastic quantum bumping into the wall.
Less good points:
1) Electrons in a degenerate Fermi gas have velocities much higher than thermal and depend on density, while garden variety gases have energy per electron equal to 1.5kT independent of density. For a metal the energy per electron is of the order of 10000 K. PEP is the root cause of this high energy.
2) A single electron in a box at T=0 has P=0. Garden variety gases at T=0 have P=0.

 

[PeterDonis]

Ashcroft/Mermin chapter 2.

Ok, thanks!

 

[Ken G]

A very good treatment of the theory of metals can be found in Sommerfeld's theory textbook, vol. 5, which is no surprise, because he and Bethe wrote a famous review about the topic before.

It's probably worth distinguishing how the PEP works in metals, where you have ions that undergo a phase transition and are electrostatically bound, versus in stars, where the ions remain a gas and are gravitationally bound. The volume responds rather differently, a fact that gets pretty mutilated in a lot of astrophysical explanations.

 

[Ken G]

This statement has a good point in it.
Good point:
The contribution to the pressure of each electron in a Fermi gas is P=-dE/dV=2/3 E/V, as for a thermal particle.
So the pressure can be interpreted as to result from some inelastic quantum bumping into the wall.

Yes, it's pressure due to kinetic energy, so in a fluid description, it's really all the same thing, degenerate or not.

Less good points:
1) Electrons in a degenerate Fermi gas have velocities much higher than thermal and depend on density, while garden variety gases have energy per electron equal to 1.5kT independent of density.

That's why I never said anything about "garden variety gases", by which you mean gases that obey the ideal gas law. I spoke only of "garden variety gas pressure," which is the pressure that is there expressly because of the presence of kinetic energy in the gas particles. Efforts to distinguish degeneracy pressure from "thermal pressure" really muddle the situation. When one means "ideal gas pressure", one should say so, there is no such thing as "thermal pressure" that means anything different from "the pressure of an ideal gas." But the more general meaning of gas pressure in these contexts is simply 2/3 the kinetic energy density, there is no need to bring in temperature unless one is interested in issues of heat transport. Which one is, of course, but that's the thermodynamics, not the pressure. What is meant by "degeneracy pressure" is simply the garden variety gas pressure that a fermionic gas has at T=0, but the reason it has the pressure is all about the kinetic energy content, not the temperature. If you are tracking the energy content, say via the virial theorem, you don't even need to know the temperature, so you don't care if the gas is ideal or degenerate.

For a metal the energy per electron is of the order of 10000 K.

Yes, but the temperature is much lower because of the PEP.

PEP is the root cause of this high energy.

Only in situations where you are fixing the T, as I said above. That's not generally the way things work in astronomy though, there you are fixing the energy history, not the T. So solid-state concepts are a poor guide there, since solid-state thinking is so set around having a fixed T. That's rather backwards in astrophysical applications, where you don't need to know T to understand P, you only need it to understand the heat transport.

2) A single electron in a box at T=0 has P=0. Garden variety gases at T=0 have P=0.

Again you are talking about "garden variety gases," a term I never use. I use "ideal gases." But I do use "garden variety gas pressure," by which I mean pressure that is present entirely due to the presence of kinetic energy in a known volume. That type of pressure is completely ambivalent to whether the gas is ideal or degenerate, which is the point of the concept.

 

[my2cts]

There is always need for good questions.
Without PEP the pressure of a degenerate electron gas at T=0 and metallic density would be P=0.
Such a free electron metal would be highly compressible.
With PEP it the pressure is comparable to that of a gas at several 10.000 K.
The original question has transformed into: what pushes back ?

 

[Ken G]

There is always need for good questions.

And a good answer makes the question go away.

Without PEP the pressure of a degenerate electron gas at T=0 and metallic density would be P=0.

Or, one could say that without the PEP, the temperature at nonzero pressure would always be nonzero. See the difference? It does not have to be stated like a consequence to the pressure, this depends on other details of the situation. So it always is with equations of state, they should never be regarded as the physics that sets the pressure, but merely a constraint that involves pressure. Pressure never comes from the equation of state, it always comes from forces and equations of state are not forces, they are constraints that take into account the actual forces. A very simple example of this principle is the pressure in the ideal-gas air all around you-- that pressure comes from the work done on that air by a history of compressing that gas by the weight of the air above you, it certainly does not come from the ideal gas law. Indeed, given the column of air above you, the only thing the ideal gas law is responsible for controlling is the volume of that column, not its pressure. But we wouldn't say that the ideal gas law causes volume, any more than we should say it causes pressure.

Such a free electron metal would be highly compressible.

How compressible is a gas of noninteracting nonrelativistic monatomic particles always depends on exactly one thing: the kinetic energy density.

With PEP it the pressure is comparable to that of a gas at several 10.000 K.

Which as I said, can be viewed as easily as an effect of the PEP on the temperature as on the pressure. It always depends on other aspects of the apparatus, not the PEP itself.

The original question has transformed into: what pushes back ?

That depends on the situation. In a white dwarf, what pushes back is garden variety gas pressure (by which I mean, the pressure that stems from the momentum flux density of the particles, which is directly related to the kinetic energy density). In solid state applications, the electrostatic repulsion of the ions is largely screened by the electrons, especially in metals, so the dominant "push back" is also from kinetic energy density, but only that of the "gas" of electrons in the conduction band. So the solid state applications are more complicated than in white dwarfs, but the dominant "push back" in a metals stems from a similar source: kinetic energy of effectively freely moving particles. There is no value in thinking that the PEP is a source of that pressure, instead it is simply a constraint on the energy conditions.

 

[Nugatory]

Mentor's note: A number of somewhat digressive posts have been removed... Everyone, please try to keep on topic here.

 

[Jilang]

To compress the fermions into a smaller volume requires raising them to higher energy levels and so requires work. They will resist this.

 

[zonde]

To compress the fermions into a smaller volume requires raising them to higher energy levels and so requires work. They will resist this.

How do you imagine compressing fermions into a smaller volume? Do you imagine some kind of piston in cylinder that creates square potential well that can be made smaller by pushing the piston? Or what?

 

[zonde]

The original question has transformed into: what pushes back ?

It can be positive charge of metal ions that pushes back.

 

[my2cts]

It can be positive charge of metal ions that pushes back.

This idea has already been discussed and discarded in this thread.

 

[my2cts]

And a good answer makes the question go away. ... There is no value in thinking that the PEP is a source of that pressure, instead it is simply a constraint on the energy conditions.

I have not seen that "good answer" here or elsewhere.
"Simply a constraint ..."
This problem is anything but "simple".
Also, thermodynamics is off-topic, since degeneracy pressure is a ground state property.
"garden variety gas pressure"
If this is supposed to explain degeneracy pressure then it is mistaken,
as pointed out before.

 

[zonde]

This idea has already been discussed and discarded in this thread.

I can't find it. Maybe you are misunderstanding the arguments that it is not coulomb force between electrons that is responsible for solidity of matter?

Anyways it was clearly stated in this thread many times that PEP does not produce force.

 

[Ken G]

I have not seen that "good answer" here or elsewhere.

Well, you have seen it here.

"Simply a constraint ..."
This problem is anything but "simple".

No, it actually is pretty simple, especially for the free particles-- it gets a bit more complicated in the atomic bonds. But the concept of a ground state is not all that complicated, and what the PEP does is alter the ground state, and the states that are allowed as that ground state is approached. That's it, that's what it does. It's all you need, because all the forces are already there, they go into the energetics of the allowed states. Looking for the PEP itself to produce a "push back" is a mistake, because all the push-back is already there in the actual forces. For example, in a fluid description, particle momentum flux counts as a force, and that's "garden variety gas pressure," the momentum flux density of the particles. The PEP didn't give the particles those momentum fluxes, it merely constrains what momentum fluxes they are allowed to be given by the actual forces present.

Also, thermodynamics is off-topic, since degeneracy pressure is a ground state property.

The ground state is very much a thermodynamics topic, it is the state at T=0. That's thermodynamics.

"garden variety gas pressure"
If this is supposed to explain degeneracy pressure then it is mistaken,
as pointed out before.

Again, the only form of pressure that appears with the PEP for free particles is the garden variety, and in metals the useful idealization is that the conduction band is just a subset of effectively free particles that are acting like a gas. Atomic and molecular bonds interact with the PEP in more complicated ways, but then they are more complicated to begin with, with or without the PEP. The PEP just alters the state that is associated with zero T, and the approach to that state as heat is lost from a system. It controls the partition of energy between the particles, but is not a source of that energy, and is also not a source of force. Even the "exchange energy," mentioned in the OP question, requires an actual source of that energy that is not the PEP. What the PEP does, as always, is determine what states are possible, the energy of those states comes from somewhere else.

 

[my2cts]

Well, you have seen it here.

No I have not. I have seen an explanation of degeneracy pressure in terms of momentum, but that is not the full answer.
That was a good point.
The question that remains is why the momentum is so high. There is no answer to that question here.

[Moderator's note: off topic comments deleted.]

 

[PeterDonis]

The question that remains is why the momentum is so high. There is no answer to that question here.

The answer is: Because the fermions are confined to a finite region of space by some potential. But the only way to know what specific potential, and therefore what specific form the momentum will take, is to look at the details of the specific system (e.g., a hydrogen atom will be very different from a metal).

 

[Ken G]

No I have not. I have seen an explanation of degeneracy pressure in terms of momentum, but that is not the full answer.

Yes, it really is. Some real forces give momentum to the particles, and that momentum flux density is what we call pressure. All the PEP does is constrain how that momentum can be given to the particles, it is not a source of any momentum so that answers the OP question. The OP also asked about exchange energy, but again, that is not energy that the PEP provides, it is provided by the actual forces, but the PEP helps constrain how much energy those forces provide by controlling the allowed states.

The question that remains is why the momentum is so high. There is no answer to that question here.

Sure there is, the answer is that the momentum is not necessarily high. It is only high if the forces that are present in some specific situation deposit a huge amount of momentum, as always happens where there is a huge degree of compression.

 

[Jilang]

[QUOTE="my2cts, post: 5647534, member: 488871
The question that remains is why the momentum is so high. There is no answer to that question here.[/QUOTE]
That was the point of my answer in post #84. With fermions when the lower energy levels are filled only the higher ones are available. If the volume increases the number of possible ground states decreases. So to reduce the volume further some will need to be promoted to higher energy levels (wth higher momentum). To do this requires energy. When you talk about what pushes back, it is the same as what resists acceleration...inertia.

 

[Ken G]

I think what a lot of people don't understand is what would happen if you waved a magic wand over a white dwarf and made all the electrons distinguishable, so the PEP no longer applied. Those under the mistaken impression that the PEP is responsible for some mystery force might imagine the star would suddenly collapse. But what would actually happen is this: almost nothing! Globally and right away, anyway. That's because the pressure doesn't care if the PEP is active or not, it only cares about the momenta the particles have. If you wave away the PEP, all that would immediately happen is that the momenta and kinetic energy would start repartitioning into the more likely configuration of the Maxwell-Boltzmann distribution, with no effect at all on pressure. So much for "degeneracy pressure" versus "ideal gas pressure" as two different causes of pressure! But there would be a huge effect on temperature, revealing the fundamentally thermodynamic character of the PEP. So the energy transport would change drastically, and the white dwarf would become much brighter on a radiative diffusion time, but none of these changes would appear on the sound crossing time because degeneracy isn't a force.

 

[PeterDonis]

If you wave away the PEP, all that would immediately happen is that the momenta and kinetic energy would start repartitioning into the more likely configuration of the Maxwell-Boltzmann distribution, with no effect at all on pressure. So much for "degeneracy pressure" versus "ideal gas pressure" as two different causes of pressure! But there would be a huge effect on temperature, revealing the fundamentally thermodynamic character of the PEP. So the energy transport would change drastically, and the white dwarf would become much brighter on a radiative diffusion time, but none of these changes would appear on the sound crossing time because degeneracy isn't a force.

Let me try to rephrase this in more physical terms. The immediate effect of the magic wand is to open up a large number of low energy states for electrons that were not previously available (because of the PEP constraint). But that doesn't mean the electrons will instantaneously occupy those states; it just means the states are available. So there will have to be a process whereby electrons that are in higher energy states give up energy and occupy lower energy states. This process is a process of heat transport--the energy has to be radiated away from the white dwarf as a whole, out into empty space--and so its characteristic time scale is the radiative diffusion time, which is much longer than the sound crossing time. On the time scale of the sound crossing time, nothing significant will happen, because on that time scale not much energy has actually been given up by the electrons, so the pressure hasn't significantly changed. Once the energy has been given up, the pressure of the white dwarf will be much lower and it will have to contract significantly in order to re-establish a new equilibrium. (Actually this will have been happening as the energy was radiated away.)

Is this a fair summary of what you are saying?

(Btw, although the white dwarf will become brighter for a while, it will ultimately end up dimmer, won't it?)

 

[my2cts]

But what would actually happen is this: almost nothing!

What a mistake. I bet you don't have a reference for this to a real scientific paper or textbook.
What would happen is that all the extra energy that was required by PEP now becomes available as thermal energy.
The star would suddenly become much hotter, gravitationally implode under emission of huge amounts of radiation and explosively reject its outer parts. It would go supernova.
What would happen is the extreme opposite of "almost nothing".
This is what actually happens when a neutron star forms: the matter turns from degenerate into non-degenerate..

 

[PeterDonis]

This is what actually happens when a neutron star forms: the matter turns from degenerate into non-degenerate..

A neutron star is degenerate. So is the core of a star that goes supernova and forms a neutron star in the remnant. (These are different types of degeneracy--electrons in the star core vs. neutrons in the neutron star--but they're both degenerate.) I don't know what you're talking about here.

 

[my2cts]

A neutron star is degenerate. So is the core of a star that goes supernova and forms a neutron star in the remnant. (These are different types of degeneracy--electrons in the star core vs. neutrons in the neutron star--but they're both degenerate.) I don't know what you're talking about here.

It is too bad that you don't know what I am talking about, for this is really an interesting thought experiment, and on-topic as well..

When the electrons and protons are converted into neutrons the neutrons are not degenerate.
Indeed, a neutron gas at the same density as a degenerate electron gas plus nondegenerate protons
has a 2000 times smaller fermi energy that the electron gas, namely the ratio of neutron over electron mass.
It will have to shed this excess energy to become degenerate.
Its volume will shrink so it will also have to get rid of a lot of gravitational potential energy.
Boom. Supernova.
In the thought experiment of Ken G the end product will be a degenerate proton gas
embedded in a see of electrons that are no longer subject to PEP due to the "magic wand".
Such a body could be similar to a neutron star.

 

[Ken G]

What a mistake. I bet you don't have a reference for this to a real scientific paper or textbook.

So are you saying you dispute that P = 2/3 E/V for both degenerate and ideal gases? Goodness, I thought we had established that a long time ago. So if you realize that expression is correct, just ask yourself this: does the "magic wand" that makes the electrons distinguishable change the E, or the V, or neither? Of course, the answer is neither, so nothing happens to P at first. It's just obvious, it's very hard to find references for statements of basic math. Of course, I already said that things will begin to evolve over a radiative diffusion time.

What would happen is that all the extra energy that was required by PEP now becomes available as thermal energy.

What happens is just what I said-- nothing at all to the pressure, but a big jump in temperature.

The star would suddenly become much hotter, gravitationally implode under emission of huge amounts of radiation and explosively reject its outer parts.

If you reread my last post, you will find a more careful accounting of the timescales here.

It would go supernova.

Yes-- eventually. But not for quite a long time, and certainly not on a sound crossing time, as I said above. The star will need to wait for many radiative diffusion times in order to lose that energy that is now freed up. But this is all what I stressed above-- the primary effects of the PEP are on heat transport. You are only repeating what I already said, but exaggerating how quickly it would happen.

What would happen is the extreme opposite of "almost nothing".

Again, if you would quote me, it is important to quote my entire statement, or you will miss the important parts, as you have done here. You have missed the timescales involved.