I Why is the Pauli Exclusion Principle not a force?

[my2cts]

Summing up, when I hit a hammer on a nail, the hammer exerts a force to the nail.
This force is transmitted by the PEP, which is not a force.
That may be correct but is very counterintuitive and deserves further study.

 

[Nugatory]

Summing up, when I hit a hammer on a nail, the hammer exerts a force to the nail.
This force is transmitted by the PEP, which is not a force.
That may be correct but is very counterintuitive and deserves further study.

Indeed it is, and this may be true of force transmission in general.

It's a lot easier to describe the force that A exerts on B than to describe the the mechanism by which that force is transmitted. It's been this way for a long time too - ask a seventeenth-century student of natural philosophy or a 21st-century layperson how it is that the hammer blow affects the nail, and you'd get an answer that reduces to "because they're solid objects and solidity means they can't both occupy the same space at the same time".

 

[Collin237]

The PEP says that a doubly occupied fermion state is equal to zero. That in itself is merely math. You could have two fermions approaching the same state, and their combined state approaching zero, and it would be valid as a mathematical statement. The physics comes in when you mention the empirical fact that matter doesn't approach zero; that it has measures, such as total energy, that evidently do not decrease. In other words, the question isn't why the hammer and the nail remain separate from each other, but why they don't fade from existence when they touch.

If a possibility of two fermions approaches the same state and thus decreases in norm, then another possibility must increase in norm to compensate. So there is a "force", in the generalized sense of a conserved quantity being transferred. Something transfers an amount of norm from one possibility to another. But what? Has anyone attempted to describe interactions at this level?

 

[Khashishi]

With forces, you can overcome some force with greater force. For example, the Coulomb force between two protons could be overcome by the strong force.The Pauli exclusion principle is absolute. The Pauli exclusion principle cannot be overcome by greater force. Two electrons are not allowed to overlap in phase space, which is different than saying they repel.

There are several other differences. Forces are essentially a gradient in some potential, and so they have a direction. PEP has no direction. It's just a restriction.

The PEP does keep you from falling into your chair or sliding between atoms or slipping off. Without PEP, Earnshaw's theorem would prevent stable arrangements of charges in your butt and the chair. Actually, without PEP, all molecules would collapse.

 

[Collin237]

The PEP applies only to identical fermions. So the restriction that a phase space domain is occupied by a fermion can be "overcome" by changing the fermion's identity.

For example, it would be theoretically possible that one of an atom's electrons gets hit by a proton, and they change into a neutron which goes into the atom's nucleus, and a neutrino which is emitted. Then where that electron was, there's room for another electron to take its place.

 

[bhobba]

The chair thing is definitely coulomb force. PEP plays a role in determining the electro structure but that isn't what is holding you up.

Thats a common misconception - but its wrong as discovered by Dyson:
https://en.wikipedia.org/wiki/Electron_degeneracy_pressure

The reason its not a force is, as you stated, its really a symmetry thing that prevents electrons being in the same state.

Thanks
Bill

 

[Simon Bridge]

Thats a common misconception - but its wrong as discovered by Dyson:
https://en.wikipedia.org/wiki/Electron_degeneracy_pressure

That was a while ago - thanks for correcting it here. I was confusing models.
Note: we usually advise to avoid taking wikipedia's word for things, the Dyson Paper on the stability of matter is here:
http://scitation.aip.org/content/aip/journal/jmp/8/3/10.1063/1.1705209

There's another discussion on these forums:
https://www.physicsforums.com/threads/matters-solidity.15605/ ... some of the respondents may benefit from reviewing the argument though.

The reason its not a force is, as you stated, its really a symmetry thing that prevents electrons being in the same state.

Thanks.

 

[Collin237]

http://scitation.aip.org/content/aip/journal/jmp/8/3/10.1063/1.1705209

It says I can rent it for $4, but when I click on that, it says to join "DeepDyve" for $40 a month.

Where can I get the article without such a trick?

 

[my2cts]

Everybody successfully taking quantum mechanics I understands the Pauli exclusion principle. I.

So anyone who is puzzled by PEP would fail "quantum mechanics I".
I do not support this approach.

 

[Vanadium 50]

So anyone who is puzzled by PEP would fail "quantum mechanics I".
I do not support this approach.

Well, you should. If you don't understand Newton's laws, you should not pass Physics 1. If you don't understand what a derivative is or how to take one, you should not pass Calc 1. If you don't understand the Pauli Exclusion Principle, you should not pass QM 1.

 

[my2cts]

Only if you define understanding the PEP as passing QM1.
What is taught is the starting point, but not the end point of understanding.
Consider what Erik Verlinde is doing. His aim is to understand gravity.
Did he fail his courses on GRT ? I don't think so.

 

[vanhees71]

I don't care about the requirements for passing QM 1. What I wanted to say is that anybody who has taken the introductory course lecture on quantum mechanics should have understood the Pauli principle. It's the statement that for fermions (bosons) the Hilbert space is the Fockspace of antisymmetric (symmetric) many-body states (antisymmetric (symmetric) meant with respect to exchanging two particles within the state) or said differently: Any physicist with a BSc is understanding the Pauli principle contrary to the belief expressed in the posting I was answering to.

I also don't know whether Verlinde failed a GR exam. I'd guess not ;-)).

 

[my2cts]

Erik Verlinde understands gravity in this sense, but yet he searches an explanation of gravity.
In the same way I understand PEP, I know how to state it and use it and I know what its consequences are.
Still I miss an explanation of it.

 

[vanhees71]

Well, the explanation is that it describes the observations. That's the usual explanation science has to offer: It describes what's objectively observable in nature. A mathematical explanation why in spaces with dimension $\geq 3$ there are only bosons or fermions, while in 2D there are also anyons possible, can be found in the paper

M. G. G. Laidlaw and C. M. DeWitt, Feynman Functional Integrals for Systems of Indistinguishable Particles, Phys. Rev. D, 3 (1970), p. 1375.
http://link.aps.org/abstract/PRD/v3/i6/p1375

 

[ddd123]

There's not only the degeneracy pressure as an example of PEP manifesting as a force. There's also the exchange interaction.

Oh, never mind, it's already been said.

 

[Collin237]

Huh? Since when is a description the same as an explanation?

Q: Explain how a car works.
A: You get in, turn the key, put your foot on the accelerator, and it moves.

 

[Ken G]

The connection between spin and statistics has been proved
so we know that it is indeed half integer spin that implies increased repulsion.
Yet my gut feeling is that there is something important missing.
There is a force, as V50 argues, but there is no force field, no force carrier.

I think what a lot of people miss about "degeneracy pressure" is that it is nothing other than completely mundane kinetic pressure. As such, it is no more of a force than gas pressure is-- when you take into account the kinetic energy enclosed, and treat what is inside as a fluid, then you have something in there that acts as a momentum flux. It doesn't even require collisions, only that fluid averages are appropriate. The reason you can tell the "degeneracy pressure" is actually nothing but mundane kinetic pressure is that it obeys P = 2/3 E/V for kinetic energy E contained in volume V, for any nonrelativistic monatomic particles.

So what is degeneracy doing, if it is not producing any "extra" pressure (as is so often erroneously claimed)? Simple-- degeneracy repartitions the kinetic energy in such a way that drastically lowers the ratio kT/E, ultimately all the way to 0 when the gas reaches its ground state. So degeneracy is all about the T given the E, and is nothing at all about P. The only reason there is an expression for "degeneracy pressure" is that there is an expression for the completely mundane kinetic pressure when degeneracy drives kT/E to 0. That's all "degeneracy pressure" in a gas ever meant-- it's not any special kind of pressure, it's just garden variety kinetic pressure.

 

[Collin237]

nothing but mundane kinetic pressure is that it obeys P = 2/3 E/V

But if the gas is behaving non-classically, how can there be only one formula? If all the atoms are identical, isn't the behavior described by one of two entirely different sets of forumlas depending on whether the atomic spin is an integer?

 

[Nugatory]

But if the gas is behaving non-classically, how can there be only one formula? If all the atoms are identical, isn't the behavior described by one of two entirely different sets of forumlas depending on whether the atomic spin is an integer?

This is a thread about the Pauli exclusion principle, so we're talking about a "gas" made up of particles with half-integral spin. The integral-spin case isn't under consideration here.

 

[Ken G]

What is behaving differently is how the kinetic energy is partitioned among the particles-- you have a Fermi-Dirac distribution instead of the more familiar Maxwell-Boltzmann. But none of that affects the pressure-- that depends only on the kinetic energy content, and the volume, via P = 2/3 E/V. The problem is, often people like to track the temperature, and degeneracy alters kT/E dramatically. But if you simply track E instead of kT, you don't need to even know if the gas is fermionic or not-- you still know P from E and V, even if it gets called "degeneracy pressure" when kT/E is driven way down. 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. What's more, very often you do know E and V, such as when you apply the virial theorem to a star of given mass and radius, or more generally, when you know the P(V) function of some kind of containment vessel, and then you only need to read off V. You will then know the E of the gas, and will understand its pressure perfectly, without even knowing if it is fermionic or not. People have some strange ideas about "degeneracy pressure"! It's really a kind of limiting pressure at which point you should not be able to extract any more heat, but if you don't care what T is, you can always get P from E and V, that's why it's such a mundane form of pressure.

 

[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.