Why is the Pauli Exclusion Principle not a force?

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  • #76
vanhees71
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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.
 
  • #77
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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.
 
  • #79
Ken G
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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.
 
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  • #80
Ken G
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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.
 
  • #81
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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 ?
 
  • #82
Ken G
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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.
 
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  • #83
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Mentor's note: A number of somewhat digressive posts have been removed.... Everyone, please try to keep on topic here.
 
  • #84
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To compress the fermions into a smaller volume requires raising them to higher energy levels and so requires work. They will resist this.
 
  • #85
zonde
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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?
 
  • #86
zonde
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The original question has transformed into: what pushes back ?
It can be positive charge of metal ions that pushes back.
 
  • #87
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It can be positive charge of metal ions that pushes back.
This idea has already been discussed and discarded in this thread.
 
  • #88
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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.
 
  • #89
zonde
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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.
 
  • #90
Ken G
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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.
 
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  • #91
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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.]
 
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  • #92
PeterDonis
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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).
 
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  • #93
Ken G
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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.
 
  • #94
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[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.
 
  • #95
Ken G
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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.
 
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  • #96
PeterDonis
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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?)
 
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  • #97
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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..
 
  • #98
PeterDonis
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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.
 
  • #99
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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.
 
  • #100
Ken G
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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.
 
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