Why is the Pauli Exclusion Principle not a force?

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  • #51
zonde
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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?
 
  • #52
Ken G
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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.
 
  • #53
PeterDonis
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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.
 
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  • #54
Ken G
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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.
 
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  • #55
PeterDonis
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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##.)
 
  • #56
Ken G
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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.
 
  • #57
PeterDonis
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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?
 
  • #58
Ken G
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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 accomodate 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.
 
  • #59
PeterDonis
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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.
 
  • #60
vanhees71
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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.
 
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  • #61
Ken G
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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.
 
  • #62
PeterDonis
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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.
 
  • #63
vanhees71
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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.
 
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  • #64
Ken G
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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.
 
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  • #65
zonde
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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.
 
  • #66
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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.
 
  • #67
zonde
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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?
 
  • #68
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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.
 
  • #69
Ken G
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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.
 
  • #70
vanhees71
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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).
 
  • #71
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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.
 
  • #72
Ken G
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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.
 
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  • #73
Ken G
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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.
 
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  • #74
PeterDonis
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the free electron model for metals
Can you give a reference on which you are basing your understanding of this model?
 
  • #75
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Ashcroft/Mermin chapter 2.
 

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