Which of the four forces is responsible for degeneracy pressure?

[sokrates]

It is simply amazing that a standard question on Pauli Exclusion principle became all about de Broglie Bohm interpretation and how it makes much more sense and how trivial things just become under that.

My take on the issue is very similar to the standard viewpoint that has been provided here, I think there's no deeper motivation to invoke the Exclusion principle, it is fundamental.

Any question on Quantum Theory can be exploited to popularize one interpretation over the other, but I believe we must pay attention not to make every thread on this forum about highly speculative interpretations..

Maybe we should consider renaming this forum, since 90% of the posts are about how different the invisible variants of the same old theory are among themselves!

 

[zenith8]

I feel that some posters here are ignoring the fact that ordinary main sequence stars, which are not noticeably affected by degeneracy pressure, are still prevented from collapsing by the classical pressure $$P = nk_BT$$.

Well obviously, but we're not talking about main sequence stars - where the thermal pressure is largely a consequence of the heat generated by fusion reactions - we're talking about white dwarfs. The electron degeneracy pressure (which is almost independent of temperature) exists even in main sequence stars but is much less important than the thermal pressure.. When the fusion reactions stop, and the star become a white dwarf, the reverse is true.

The answer is that none of them are. What happens is: if you consider a fixed region in the star, particles are constantly moving in and out of that region, and if there's a pressure gradient the result is that there is a net momentum flow into the region. This flow has dimensions of force, but it's not a 'true force' in the sense that you'll never find it in the equations of motion of an individual particle. In equilibrium, this pressure 'force' exactly balances the gravitational attraction, keeping the star from the collapsing.

I don't understand this. So what's the difference between the 'thermal pressure' in a main sequence star and in the white dwarf, then?

Quoting the 'white dwarf' article in Wikipedia:

The material in a white dwarf no longer undergoes fusion reactions, so the star has no source of energy, nor is it supported against gravitational collapse by the heat generated by fusion. It is supported only by electron degeneracy pressure, causing it to be extremely dense.

I'm not trying to be unhelpful, I'm just not sure I understand the point you're trying to make.

I don't deny that the quantum force described by zenith8 exists, but it is not the only thing which can keep a white dwarf star from collapsing.

So why did you just say that the pressure 'force' exactly balances the gravitational attraction, if 'my' quantum force exists as well?

 

[zenith8]

Okay - i will look into the lecture sheets when i have more time.

Enjoy!

By the way, one quick question before I've read it - does de Broglie-Bohm interpretation destroy quantum computers? Or does it still allowed entangled states. If it doesn't and the BB interpretation turns out to be true, then a lot of people are wasting their time and money!

No, it still works, and of course yes it still allows entangled states. You just can't get away with going on about how all the calculations on a quantum computer are being done simultaneously in all possible parallel universes. But then again that always did sound a bit far-fetched.

Although I'm naturally suspicious, if not skeptical, about alternative, "conspiracy" theories that go against accepted physics theorems, I've been hearing more noise about QM. Not from average joe's, but qualified people like Van het Hooft and others. Maybe QM and GR are so hard to unify because QM isn't quite what it's supposed to be...?

Quite right! I for one certainly suspect that this is the case..

It's not really a conspiracy theory. De Broglie came up with it in 1924-1927, so it actually predates the Copenhagen interpretation. When Bohr made his contribution he'd recently been reading some logical-positivist philosophical books, and in the light of this he decided that quantum theory was not allowed to speculate about what actually exists, and somehow this got moulded into 'things don't exist until you look at them'.

So here's an analogy for the difference between the standard way of looking at things and the so-called 'conspiracy theory':

Take a look around you. You see your computer, your untidy bedroom, your computer books, your old moth-eaten teddy bear, your collection of Playboy, Razzle and Reader's Wives, your half-eaten breakfast, your old boxing trophies. Now close your eyes.. You can't see the stuff any more! So, where's it gone? Here's two possible answers:

(COPENHAGEN): The things in your room have dissolved into a mass of potentialities and no longer have 'positions'.

(de BROGLIE-BOHM): The things are still there, but you've just closed your eyes..

Note both answers use exactly the same equations.

So how the first of these options became the work of a brilliant scientific and philosophical genius, and the second became a 'conspiracy theory' is somewhat difficult to understand.

Ahh, yes.. but could you explain what the quantum potential represents? I mean, in here lies the key to my answer. The Coulomb potential is caused by the Coulomb force, the gravitational potential comes from the gravitational force, etc...

I've answered that question before - see post #15 in https://www.physicsforums.com/showthread.php?p=2369492#post2369492".

Well, i can't really blame them. QM did (and does) produce correct results. In the end, that's what counts. GR also introduces concepts like time and distance dilation which is ridiculous concept intuitively, but turns out to be correct...

Sure. Just remember that deBB QM did (and does) produce correct results as well. It is after all, just QM. But both produce correct results for the statistics of experimental observations. You, however, are not asking about the results of solving the Schroedinger equation, you are asking a conceptual question. This requires you to take a stand about what you believe exists - you must make an ontological commitment - do particles actually exist which can have forces applied to them, and so forth.

I think you mean SR rather than GR, but even so, there are various 'interpretations' of relativity as well, which differ in terms of whether objects really are length-contracted and so on, or whether this is just a sort of perspective effect.

 

[LennoxLewis]

I feel that some posters here are ignoring the fact that ordinary main sequence stars, which are not noticeably affected by degeneracy pressure, are still prevented from collapsing by the classical pressure $$P = nk_BT$$. You may as well ask, "which of the four forces is responsible for thermal pressure?"

The answer is that none of them are. What happens is: if you consider a fixed region in the star, particles are constantly moving in and out of that region, and if there's a pressure gradient the result is that there is a net momentum flow into the region. This flow has dimensions of force, but it's not a 'true force' in the sense that you'll never find it in the equations of motion of an individual particle. In equilibrium, this pressure 'force' exactly balances the gravitational attraction, keeping the star from the collapsing.

In a "normal" gas, the pressure is caused by particles that bounce off the wall on which you measure the pressure. The force that makes them bounce off the wall the is E.M. force. The reason they have the kinetic energy to get to that wall is because of their temperature (in the order of 10 m/s at R.T. if I'm not mistaken), and that temperature they get from somewhere. It's not the same as energy, but certainly related to their energy.

But that's different from degeneracy pressure, where:

-The pressure is not caused by E.M. repulsion (or is it?)
-There is no energy source to keep the particles' energy up, like fusion in a normal sequence star. In a white dwarf/neutron star, there is no energy source, so if normal pressure balances gravity, then wouldn't the star approach absolute zero within no-time, and collapse when the kinetic energy of the gas becomes too small to provide enough pressure to counteract gravity?

It is simply amazing that a standard question on Pauli Exclusion principle became all about de Broglie Bohm interpretation and how it makes much more sense and how trivial things just become under that.

My take on the issue is very similar to the standard viewpoint that has been provided here, I think there's no deeper motivation to invoke the Exclusion principle, it is fundamental.

Any question on Quantum Theory can be exploited to popularize one interpretation over the other, but I believe we must pay attention not to make every thread on this forum about highly speculative interpretations..

Maybe we should consider renaming this forum, since 90% of the posts are about how different the invisible variants of the same old theory are among themselves!

Sorry if i missed the point, but no post was able to explain the concept to me. I don't particularly care for specific Q.M. interpretations, but if it helps explain this effect, then i'll be happy to look into it.

 

[sokrates]

Sorry if i missed the point, but no post was able to explain the concept to me. I don't particularly care for specific Q.M. interpretations, but if it helps explain this effect, then i'll be happy to look into it.

I was just trying to point out that there is really no concept that sits -below- the Exclusion principle. It just is, in this case, as far as our current accumulated knowledge suggests. This was my point, I think, this was first indicated by Dr. Chinese in this thread.

 

[pellman]

No it isn't. Pressure exists even in a gas of completely non-interacting particles.

Not so. It is true that pressure in a calculated sense can exist if the particles of the gas are non-interacting. But that gas cannot actually produce a force-per-unit-area on a surface unless it interacts with that surface. If the particles do not interact with the walls of the container, then they would not bounce off--they would just pass right through the container walls. The force that makes them bounce off is the EM force.

The EM force is what makes two billiard bounce off each other and is what keeps you from falling through the floor.

 

[DrChinese]

So how the first of these options became the work of a brilliant scientific and philosophical genius, and the second became a 'conspiracy theory' is somewhat difficult to understand.

It is because the "baggage" of BM - that being the non-local interactions - adds nothing useful to the theory. At least, so far it hasn't. Perhaps one day it will.

In the meantime, how do you justify the ad hoc Bohmian hypotheses to students? I mean, as far as I can see, the functional dependence on positions of other particles is just one way to have non-locality included. I would think there could be any number of other ways (i.e. other hidden non-local relationships, perhaps momentum or even new and unknown attributes), so why this one in particular? You could be silent about it and be just as effective, so that is what is usually taught.

Don't get me wrong, I am not "against" Bohmian theory. But it is a gross mischaracterization to claim that it is somehow objectively superior to any other interpretation. The only reason to consider it "better" is for one's own personal/subjective reasons. I have not seen even one single new scientific point that arose from it in the past 50 years. (Earlier than that, just becomes a matter of useless historical debate.)

As I have said before, we owe it to posters here to respond with generally accepted theory first. We all have personal speculations and suspicions about the "true" underlying nature of reality, but that doesn't really qualify as science.

 

[kakaz]

Actually, i have two questions:
1. Because of the Pauli exclusion principle, there can be degeneracy pressure, for instance in neutron stars, but also in electron gasses (and any fermion cluster?). What force causes this pressure?

Reality is stranger than fiction ;-) So let's begin with some hints.

As You've may seen, fermions has spin which is multiple of $$1/2$$, whilst bosons has spin given by natural numbers. What is that mean? When You consider wavefunction of composed system which consists of many particles ( bosons, or fermions exclusively) then You may notice that every two particles are indistinguishable. This means that such system has special symmetry: You may exchange any number of particles among themselves and solutions of equations of dynamic of such system should not change. This means that wavefunctions of such systems has some symmetry over "permutations". How can You physically realize such permutation? Lest say by rotation of chosen pair of particles. You take a pair, and then rotate its placement above chosen (freely) axis. Result should not depend whether You turn this pair assuming that one particle replaces in the end other.

So such rotation is realized on the wavefunction space by means of representation of SO(3) - general group of rotation. There are several representations of such group: some of them has paradoxical properties, some not. The most known one is typical matrix realization of general rotation in 3dimensional coordinate system. Other, are more sophisticated. Consider simple example: choose one dimensional system, and one-dimensional rotations. Simple representations are:
$$1 ... R(\phi) = \exp(i k \phi)$$
$$2 ... R(\phi) = \exp(i \frac{1}{2} (2k+1) \phi)$$
k is natural number, both has property that $$R(\phi) R(\phi') = R(\phi + \phi')$$
which means we satisfy (in one dimension) group law of SO(3).

Look! When You insert $$\phi = 2*\pi$$ into equation 1 and 2 You will see that first expression is the same as for $$\phi = 0$$ whilst second one will change its sign. We have then situation that anything which we transform by representation 1. has the same mathematical shape before and after rotation, and this is what we are used to. But the second one turns objects into the same objects but wits opposite sign.

What objects? Wave functions. Vectors. Whatever. You have to rotate again in order to obtain first sign of expression with this representation.

First one - this are bosons, second one - this are fermions. There is mathematical theorem called Spin-Statistics Relation which precisely qualifies what I wrote. In three dimensions there is no other representations of rotational symmetry than this two kind of. So every pure quantum multi-particle state which is symmetric over particle exchange has to be invariant over 1 or 2 case. In 2 dimensions there are surprises which I only mention about: anyons..

Now, From that picture You have symmetrical or anti symmetrical wavefunctions according to representations of rotational group in 3 dimensions.

Now You want to change by pressure state of such multi particle fermionic system. You press, but there is no free states - every state where there is "symmetrization" of antisymmetric wavefunction should vanish, and other ones, pure anti-symmetrical, has only one particle in certain quantum state. It depends on spin, as internal state of freedom, but in fact there is quite possible to have two fermions in the same spin state if they differ by other quantum numbers, for example momentum ( in BCS for example).

As long as You nave not reach any other "way of grouping" of particles ( I mean different than simple accessible energetic sates) there is no way to change the behavior of the system - when You press, gas takes higher temperature etc. because You lower volume, but there is no real change in the states.
There is no forces needed at all to explain what happened: in order to change state of a system, You have to find new energetic state for it. When You find it ( pressure so high that some nuclear reactions may happen) then system will follow that way.
I have hope I wrote it in not very mysterious way...

2. According to the Pauli Exclusion principle, no two fermions can have the same state in the same position. Now, by his formula, you can calculate delta x if you insert delta p, but states are integers. So, at what "range" does this principle work? How far away must a fermion be from the other, in order to still be in the same quantum state?

This may be obtained by Wigner functions for example, but I do not know it very much. Particles are in the same state as long as they are indistinguishable according to involved potentials, forces etc.

 

[LennoxLewis]

I was just trying to point out that there is really no concept that sits -below- the Exclusion principle. It just is, in this case, as far as our current accumulated knowledge suggests. This was my point, I think, this was first indicated by Dr. Chinese in this thread.

In other words, you don't know what force causes the Exclusion principle? It just is?

Isn't that a bit strange for a physicist to say? 80 years of science ever since, there must have been some publications about the principle that drives the Pauli Exclusion?

Reality is stranger than fiction ;-) So let's begin with some hints.

As You've may seen, fermions has spin which is multiple of $$1/2$$, whilst bosons has spin given by natural numbers. What is that mean? When You consider wavefunction of composed system which consists of many particles ( bosons, or fermions exclusively) then You may notice that every two particles are indistinguishable. This means that such system has special symmetry: You may exchange any number of particles among themselves and solutions of equations of dynamic of such system should not change. This means that wavefunctions of such systems has some symmetry over "permutations". How can You physically realize such permutation? Lest say by rotation of chosen pair of particles. You take a pair, and then rotate its placement above chosen (freely) axis. Result should not depend whether You turn this pair assuming that one particle replaces in the end other.

So such rotation is realized on the wavefunction space by means of representation of SO(3) - general group of rotation. There are several representations of such group: some of them has paradoxical properties, some not. The most known one is typical matrix realization of general rotation in 3dimensional coordinate system. Other, are more sophisticated. Consider simple example: choose one dimensional system, and one-dimensional rotations. Simple representations are:
$$1 ... R(\phi) = \exp(i k \phi)$$
$$2 ... R(\phi) = \exp(i \frac{1}{2} (2k+1) \phi)$$
k is natural number, both has property that $$R(\phi) R(\phi') = R(\phi + \phi')$$
which means we satisfy (in one dimension) group law of SO(3).

Look! When You insert $$\phi = 2*\pi$$ into equation 1 and 2 You will see that first expression is the same as for $$\phi = 0$$ whilst second one will change its sign. We have then situation that anything which we transform by representation 1. has the same mathematical shape before and after rotation, and this is what we are used to. But the second one turns objects into the same objects but wits opposite sign.

What objects? Wave functions. Vectors. Whatever. You have to rotate again in order to obtain first sign of expression with this representation.

First one - this are bosons, second one - this are fermions. There is mathematical theorem called Spin-Statistics Relation which precisely qualifies what I wrote. In three dimensions there is no other representations of rotational symmetry than this two kind of. So every pure quantum multi-particle state which is symmetric over particle exchange has to be invariant over 1 or 2 case. In 2 dimensions there are surprises which I only mention about: anyons..

Now, From that picture You have symmetrical or anti symmetrical wavefunctions according to representations of rotational group in 3 dimensions.

Now You want to change by pressure state of such multi particle fermionic system. You press, but there is no free states - every state where there is "symmetrization" of antisymmetric wavefunction should vanish, and other ones, pure anti-symmetrical, has only one particle in certain quantum state. It depends on spin, as internal state of freedom, but in fact there is quite possible to have two fermions in the same spin state if they differ by other quantum numbers, for example momentum ( in BCS for example).

As long as You nave not reach any other "way of grouping" of particles ( I mean different than simple accessible energetic sates) there is no way to change the behavior of the system - when You press, gas takes higher temperature etc. because You lower volume, but there is no real change in the states.
There is no forces needed at all to explain what happened: in order to change state of a system, You have to find new energetic state for it. When You find it ( pressure so high that some nuclear reactions may happen) then system will follow that way.
I have hope I wrote it in not very mysterious way...

Thanks for the extended and well written post... not to come across as unthankful, but i know what fermions and bosons are, and their wavefunctions. But what you don't talk about is which force acts to make sure no same states are occupied and cause degeneracy pressure!

 

[zenith8]

Hi Dr. Chinese,

In the meantime, how do you justify the ad hoc Bohmian hypotheses to students? I mean, as far as I can see, the functional dependence on positions of other particles is just one way to have non-locality included. I would think there could be any number of other ways (i.e. other hidden non-local relationships, perhaps momentum or even new and unknown attributes), so why this one in particular? You could be silent about it and be just as effective, so that is what is usually taught.

How do you justify it to students? Here's a daring strategy - try actually telling them about it and let them make their own minds up. I've seen Towler do it at Cambridge and the students just lap it up. The positive response that young people give to the de Broglie-Bohm quantum mechanics is astonishing, particularly if they've only been told about the so-called 'standard view' recently and can compare and contrast their relative merits. For them the clarity and lack of paradoxes works wonders.

Don't get me wrong, I am not "against" Bohmian theory. But it is a gross mischaracterization to claim that it is somehow objectively superior to any other interpretation. The only reason to consider it "better" is for one's own personal/subjective reasons. I have not seen even one single new scientific point that arose from it in the past 50 years. (Earlier than that, just becomes a matter of useless historical debate.)

OK - we know that we can lose all the 'weirdness' and the paradoxes and we can 'explain' the reality of any quantum event. How do we know this is the true explanation? We don't but you could say that about anything. One can certainly adopt it as a reasonable working hypothesis that allows one to visualize things, and for that reason one should definitely teach it in addition to the regular "all conceptual questions are meaningless" view.

As for new scientific points, how about (just to pick the first four that occurred to me)..

(1) Bell's theorem(!).
(2) The realization that the sums over an infinite number of paths in Feynman path-integral theory can be done over a single path (the one that the particle actually follows according to the Schroedinger equation/de Broglie-Bohm theory).
(3) Valentini's testable predictions about the possibility of quantum non-equilibrium states and its effect on e.g. the cosmic background radiation.
(4) And er.. a coherent explanation for why white dwarfs don't undergo gravitational collapse (rather than saying, effectively, they just don't).

As I have said before, we owe it to posters here to respond with generally accepted theory first. We all have personal speculations and suspicions about the "true" underlying nature of reality, but that doesn't really qualify as science.

Well indeed, so I did let the orthodox crowd respond first. I waited for well over a day after Lennox posted his question before answering, during which time four responses were received. Let's review:

Lennox himself: [begging someone to answer the question]

Doc Al: "I'd say that degeneracy pressure is not attributable to any of the four fundamental forces, but is a new quantum effect." Fair enough answer from a very clever guy- effectively says no-one knows the answer to the question.

tiny-tim: "No, it's that no two fermions can have the same state in the same "ball-park" … eg two electrons "orbiting" the same nucleus cannot occupy the same orbit (with the same spin) … their positions do not matter, only their orbits. The ball-park can actually be quite large (like the region round a nucleus)." A roundabout way of saying that fermions can't be in the same state, which the OP clearly already knows. Doesn't answer the question.

ytuab: "The spin up and down electrons have the opposite magnetic moments. So it seems that the magnetic force is related to the Pauli exclusion principle. But for example, in the helium atom, the magnetic force of spin is too weak in comparison to the Coulomb force. So the Pauli exclusion principle is not related to the magnetic force. To be precise, if the two electrons are apart, in all areas except in the part at just the same distance from the two electrons, the magnetic fields are theoretically produced." I have no idea what this means. Doesn't answer the question.

So, having concluded that the orthodox crowd have had their turn, I decide to answer the question in my own way, stating in the very first sentence that the answer is clear if you adopt the de Broglie-Bohm interpretation. I don't see why many people here think this is such a bad thing; the OP has stated that he found the explanation interesting and thanked me for providing it. I really don't understand why everyone has to get on my back all the time about it.

Seriously, can you honestly tell me that the responses above (or indeed any of the later ones) were more informative than the one I gave? As I have said to you many times before, the interpretation of QM is an open question, and nowadays Copenhagen has no right whatsoever to be considered the 'standard one.' And despite what has been stated by some people the OP is clearly not a beginner and doesn't need his hand holding.

 

[Astronuc]

Locked pending moderation. Dec 28, 2009, 06:28 PM EST (2328 UTC)

Unlocking: Dec 29, 2009, 07:18 EST (0018 UTC)

Please keep comments on-topic and avoid personal slights/attacks on other members.

================================================================================
Some notes on White Dwarfs and Electron Degeneracy

Degenerate electron gases
http://farside.ph.utexas.edu/teaching/qmech/lectures/node65.html

White-dwarf stars
http://farside.ph.utexas.edu/teaching/qmech/lectures/node66.html
http://farside.ph.utexas.edu/teaching/qmech/lectures/node67.html (problems)

Eventually, the mean separation becomes of order the de Broglie wave-length of the electrons, and the electron gas becomes degenerate. Note, that the de Broglie wave-length of the ions is much smaller than that of the electrons, so the ion gas remains non-degenerate. Now, even at zero temperature, a degenerate electron gas exerts a substantial pressure, because the Pauli exclusion principle prevents the mean electron separation from becoming significantly smaller than the typical de Broglie wave-length (see the previous subsection). Thus, it is possible for a burnt-out star to maintain itself against complete collapse under gravity via the degeneracy pressure of its constituent electrons. Such stars are termed white-dwarfs. Let us investigate the physics of white-dwarfs in more detail.

The physical universe: an introduction to astronomy By Frank H. Shu

http://universe-review.ca/R08-04-degeneracy.htm

Electron degeneracy pressure and white dwarfs
http://www.astro.psu.edu/users/niel/astro130/powerpoint/watson/watson-Lect_11.ppt

. . . there is an additional quantum-mechanical repulsion of electrons by each other, which sets in at very small distances, such that wave properties are displayed.
If the separation is small enough that this quantum repulsion is bigger than the electric repulsion, the electrons are said to be degenerate.
Note for those who have taken physics or chemistry before: you may know this quantum repulsion as the Pauli exclusion principle.
Protons can confine each other in a similar fashion; so can neutrons. Because electrons are less massive, though, they become degenerate with less confinement (a space roughly 1800 times larger, as we have seen).

Implications of confinement arising from the wave properties of elementary particles

If one confines an electron wave to a smaller space, its wavelength is made shorter. The shorter the wavelength, the greater the energy.

With this increase in energy, each electron exerts itself harder on the walls of its “cell;” this is the same as an increase in pressure.
So:
•squeeze a lot of matter from a very small space into an even smaller space...
•electrons are more tightly confined...
•thus the electrons have more energy and exert more pressure against their confinement.
This extra pressure from the increase in wave energy under very tight confinement is degeneracy pressure. (Fowler, 1926)

WHITE DWARFS (DEGENERATE DWARFS)
http://www.astro.princeton.edu/~burrows/classes/514/wd.pdf

White Dwarfs & Neutron Stars [Revision : 1.1]
http://www.astro.wisc.edu/~townsend/resource/teaching/astro-310-F08/38-white-dwarfs.pdf

More general comments on white dwarf properties
http://imagine.gsfc.nasa.gov/docs/science/know_l2/dwarfs.html


The bottom line seems to be that the electron degeneracy pressure is a QM effect, and not due to one of the 4 fundamental forces. It's certainly not gravity, and in fact degeneracy pressure is opposing gravity and preventing collapse of the white dwarf.
------------------------------------------------------------------------------------------------

The interior of a white dwarf of one solar mass has a density about 200000 that of the mean density of the earth. The atomic density of a typical solid on Earth is of the order of 10²² atoms/cm³ which puts the atomic density of a white dwarf on the order of 10²⁷-10²⁸ atoms/cm³. The interatomic spacing of solids on Earth is on the order of angstroms (look at lattice parameters for solids) and based on 200000 times greater density, the interatomic spacing for a white dwarf would be approximately 1/60 or ~1/100 of a typical terrestrial solid, i.e. ~ 0.01 angstrom.

The nuclei are still not degenerate.

 

[zenith8]

Reality is stranger than fiction ;-) So let's begin with some hints.

As You've may seen, fermions has spin which is multiple of $$1/2$$, whilst bosons has spin given by natural numbers. What is that mean? When You consider wavefunction of composed system which consists of many particles ( bosons, or fermions exclusively) then You may notice that every two particles are indistinguishable. This means that such system has special symmetry: You may exchange any number of particles among themselves and solutions of equations of dynamic of such system should not change.

OK, this is a nice post as far as it goes, but as Lennox has said you're not actually answering his question, you're merely restating it. You're saying that fermions can't be in the same state. He's effectively asking why.

One of the reasons that this may not be clear to you is because, as a serious physicist, you've been taught that philosophical questions are not important (whereas in reality the boundary between physics and philosophy is very unclear). This allows you to use language in a very loose way. For example, these 'particles' you talk about that are 'indistinguishable'. What are they, actually? Is a particle a tiny point-like thing? Is a particle a wave function? What is it? And if you don't know what it is, how can you say they are indistinguishable? (For example, if they were actually point-like particles, as your wording seems to imply, then surely they would be distinguishable by their trajectories?). Do you think that indistinguishability implies that fermions have to have antisymmetric wave functions and therefore that they cannot be in the same state? By itself, no it doesn't. Unless you give clear answers to these questions, then your answers to questions like Lennox's will have no clear meaning. In particular, if quantum mechanics describes only probabilities, how can you use it to talk about forces and (dynamical equations of) motion?

If the thing is actually a point-like particle, then it has a wave function in position space (for example), and the square of this gives the probability of the particle being found in the various positions. So saying that two fermions 'can't be in the same state' is equivalent to saying that their probability distributions of positions can't have the same shape - which is presumably because they 'repel' each other. Lennox is asking what causes that repulsion? So in some sense your answer (which boils down to 'fermions can't be in the same state') is not an answer to Lennox's question, it is a logical consequence of it.

If you believe strictly in the orthodox interpretation then the actual (and perfectly correct) answer to a conceptual question like this is 'I don't know, since I don't know what exists', or (because it sounds less silly) 'It is a quantum effect' (thanks, Astronuc!). For some reason most believers in orthodox QM have difficulty saying this, because they don't want to appear like they don't know everything (even though this approach is philosophically perfectly respectable).

So it is an important thing to realize that questions like this cannot be answered by means of the orthodox interpretation of QM, and you shouldn't be afraid of saying so. There is nothing wrong with that, but when you - despite this - attempt to answer such questions you need to be very aware of the limitations your philosophical viewpoint imposes on you.

To answer a question of this nature, you need to make an ontological commitment - a statement of your belief in what exists. All I have been pointing out is that if you state that particles exist and (necessarily) have trajectories - as in the deBB interpretation - then the answer to the question about forces is straightforward, both mathematically and philosophically (see my post #22). One need not, in fact, get annoyed about this, as others have been doing. The mere fact that one can do this should be interesting in itself. You should be asking yourself 'Is zenith's (i.e. Nobel prize winner de Broglie's and almost Nobel prize winner Bohm's) force equation unique? Does it depend on one's philosophical assumptions? What are the properties of this force?'. You should not be asking yourself 'Why is this idiot allowed to post on this forum when everyone knows that philosophy has no place in physics and that Bohr sorted all this out in 1928?'. A discussion of interpretational questions is fundamental in answering a question such as this, and is no way 'off-topic'.

This may be obtained by Wigner functions for example, but I do not know it very much. Particles are in the same state as long as they are indistinguishable according to involved potentials, forces etc.

Note also my answer to this in post #29 - do you agree?

 

[cortiver]

I don't know what astronuc considers "on-topic", but I feel it is important to point out, as I said before, that degeneracy pressure is really not any different from classical thermal pressure. Both are simply a result of the kinetic energy of the gas particles.

To specify more clearly what I am saying: consider a region of gas. Now, particles are constantly passing through the surface of this region in either direction. Particles moving into the region carry momentum which should be added to the total momentum contained in the region. Conversely, particles moving out of the region should have their momenta subtracted from the total momentum in the region.

The flow of momentum into the region due to particles passing through an element $d\mathbf{A}$ of the surface (oriented outwards), turns out to be (I am ignoring relativistic effects here):
$$-N \langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle d\mathbf{A}$$,
where N is the density of particles, $\mathbf{n}$ is the unit vector pointing in the direction of $d\mathbf{A}$, $\mathf{v}$ is the velocity of the particles, and the angle brackets denote averaging over all nearby particles. Now, assuming isotropy, $\langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle$ will be the same regardless of $\mathbf{n}$, so we can define the pressure by

$$P = N \langle m(\mathbf{v}\cdot{\mathbf{n})^2\rangle$$

In particular we have
$$\begin{align*} P &= N \langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle \\ &= \frac{N}{3}\left(\langle mv_x^2 \rangle + \langle mv_y^2 \rangle + \langle mv_z^2 \rangle\right) \\ &= \frac{2N}{3} \left\langle \frac{1}{2}m|\mathbf{v}|^2\right\rangle \\ &= \frac{2}{3} u \end{align}$$
where u is the kinetic energy density.

Now, for a classical gas in thermal equilibrium the equipartition theorem gives $u = 3/2NkT$, leading immediately to the ideal gas law [itex]P = NkT[/tex]. For a degenerate gas the equipartition theorem doesn't hold, but if you extract the correct formula for the energy density of a degenerate gas from the Fermi-Dirac distribution in the limit $T \to 0$, you will get the standard formula for degeneracy pressure - see the first result in http://books.google.com.au/books?ei...ectron+degeneracy+pressure&btnG=Search+Books", for example.

Well obviously, but we're not talking about main sequence stars - where the thermal pressure is largely a consequence of the heat generated by fusion reactions - we're talking about white dwarfs.

I raised the issue of main sequence stars since there is clearly no actual force supporting them, contrary to your claim that a quantum force is necessary to keep a white dwarf star from collapsing.

I would also make the distinction, though probably you know this already, that the thermal pressure is not caused directly by the fusion reactions - it is a result of the kinetic energy of the gas particles, which is itself maintained by the fusion reactions. In a white dwarf star the degeneracy pressure is also caused by the kinetic energy of the gas particles, but this doesn't need to be "maintained" because it is forced to exist by the exclusion principle.

So why did you just say that the pressure 'force' exactly balances the gravitational attraction, if 'my' quantum force exists as well?

Maybe 'exactly' wasn't the right word. But I was talking about main sequence stars, where the quantum force is going to be negligible.

I don't know what the relative contributions of the quantum force and the pressure force is in real white dwarf stars. But if the characteristic length scale of variation of the star is much greater than the de Broglie wavelength of the particles, then it should be valid to use a 'semiclassical approximation' (where you use the Fermi-Dirac distribution instead of the Maxwell-Boltzmann distribution, but otherwise you treat the particle motions classically). In that limit your quantum force is going to disappear.

In a "normal" gas, the pressure is caused by particles that bounce off the wall on which you measure the pressure. The force that makes them bounce off the wall the is E.M. force.

Okay, of course the force the particles exert on the wall is electromagnetic. But that's not what I'm talking about. Pressure is a local property of the individual locations in the gas (take an infinitesimal surface element at a particular point - how much momentum flows through it in unit time?). It doesn't depend on what's happening at the exterior of the gas.

-There is no energy source to keep the particles' energy up, like fusion in a normal sequence star. In a white dwarf/neutron star, there is no energy source, so if normal pressure balances gravity, then wouldn't the star approach absolute zero within no-time, and collapse when the kinetic energy of the gas becomes too small to provide enough pressure to counteract gravity?

Well, the whole point of degeneracy pressure is that the exclusion principle forbids this. At high densities, the only way you can avoid having two particles in the same state is by filling up lots of very high energy states. So even at absolute zero the particles continue to have lots of kinetic energy.

Not so. It is true that pressure in a calculated sense can exist if the particles of the gas are non-interacting. But that gas cannot actually produce a force-per-unit-area on a surface unless it interacts with that surface.

True. It can, however cause a momentum flow into a region of a star, counteracting the gravitational force - which is all that is needed to keep a star from collapsing (note that a star doesn't have any container walls). Actually I think the particles in a real main sequence star can be treated as non-interacting (except via gravity, of course) to a good approximation for the purposes of hydrostatic equilibrium calculations.

 

[ytuab]

Pauli exclution principle is troublesome for the energy calculation (the variational methods, etc.)

For example, For the Lithium (Li) atom, if we don't consider the Pauli exclusion principle, the result of the variational methods shows about 30 eV lower energy (-230 eV) than the experimantal value(-203.5 eV).

So we need to restrict the wavefunction artificially to incorporate this effect.

Does it mean "choosing a basis set is an art, not science" ?

 

[DarMM]

To everybody who keeps asking why fermions can't be in the same state or why the wavefunction needs to be restricted to antisymmetric wavefunctions the answer is the spin and statistics theorem. Which is a result from quantum field theory, you can't get it from quantum mechanics.

Basically in quantum field theory particles can only have spins 0, 1/2, 1. Also a prior there can only be two types of statistics Fermionic or Bosonic. You can show that using the properties of quantum field theory like locality and causality that spin 1/2 particles must be fermionic and spin 0 and 1 must be bosonic. You can also show that no other types of statistics are possible in four dimensions.

If you want an intuitive picture of the proof, although very loose, remember that a spin 1/2 particle picks up a minus sign upon a 360 degree rotation. So if $$\mathcal{R}(2\pi)$$ is this rotation, then:
$$\mathcal{R}(2\pi)\Psi = -\Psi$$. This is simply group theory.

Now if you have two fermions two the left and right of a common center and then do a 180 degree rotation on both of them about the common center they will have swapped places. So the state is no longer $$\left | ab \right \rangle$$, but $$\left | ba \right \rangle$$. However since you have performed two 180 degree rotations, one on each particle, this has the same effect as a single 360 degree rotation, hence the new state is the old one with a minus sign. Hence:
$$\left | ab \right \rangle = - \left | ba \right \rangle$$
So,
$$\left | aa \right \rangle = - \left | aa \right \rangle$$

Which implies,
$$\left | aa \right \rangle = 0$$

So spin 1/2 particles cannot be in the same state.

You're saying that fermions can't be in the same state. He's effectively asking why.

This question as formulated is easily answerable. Fermions are particles which obey Fermi-Dirac statistics and in those statistics you cannot occupy the same state, since that is part of the definition of these statistics.
What you might ask instead is, "Why to some particles obey Fermi-Dirac statistics?". Basically I've given the answer above. In four dimensions you can show that only two statistics are possible: Fermi-Dirac and Bose-Einstein. So the question reduces as to why particles don't all obey Bose-Einstein. The asnwer is that locality and causality require spin-1/2 fields to anti-commute so their particles must have Fermi-Dirac statistics.
Although the more intuitive explanation above probably works better.

 

[kakaz]

In other words, you don't know what force causes the Exclusion principle? It just is? Isn't that a bit strange for a physicist to say? 80 years of science ever since, there must have been some publications about the principle that drives the Pauli Exclusion?

I will wrote for You once more what I wrote last time: You do not have any forces in order to have pressure in system. You push with some force system with fermion particles. You want to lover of volume in such gas for example. But as the whole energetic states are occupied, then after You "press the wall " particles has to look for new energetic states. But there is no one available except higher energetic ones, because lower energies are taken by other particles. As there is no other states, the higher energetic ones will be occupied, and energy for this is given by You, when You press the wall. So: You feel lake You energy is given somewhere, and You think it was to break some force: in fact You pump additional energy into system. You may call it "breaking Pauli force" although it was only pumping additional energy for higher energetic states occupied.

Thanks for the extended and well written post... not to come across as unthankful, but i know what fermions and bosons are, and their wavefunctions. But what you don't talk about is which force acts to make sure no same states are occupied and cause degeneracy pressure!

Ha! Here You are: there is no force needed: wavefunctions has some symmetries, and as usually there is no way to destroy it (symmetry), until we do not reach new ways of particle reactions for example nuclear one. So: there is no force needed: as long as there is no other way for wavefunction to behave, it has internal symmetry or antisymmetry and there is degeneracy pressure. Possible ways of breaking symmetry usually are located much higher than normal energetic levels available for system. In fact such types of behaviors usually are connected with destroying system at all!

In order to destroy this internal symmetry You have to pump as much energy into system as there will be another possibilities of decay for system accessible.

For example for gas of nucleons You have to break nuclear forces which contract against nuclear fusion of particles in system. In this case You may think that answer for "which force acts" the answer is "nuclear forces". But in other cases there may be other forces: when as fermionic particles You will consider ions the answer would be "molecular forces" etc. So the proper answer is: there is no one force at all, it depends on case.

 

[kakaz]

OK, this is a nice post as far as it goes, but as Lennox has said you're not actually answering his question, you're merely restating it. You're saying that fermions can't be in the same state. He's effectively asking why.

I understand. In 3-dimensional space, there are 2 different families of representation od SO(3) - gerenal group of rotation. One family has spin which is natural number whils second one family has multiplicity of 1/2. This is mathematics, not physics.
Question of Your friend may be, rephrased acordingly to situation, changed as follows:
1. why there are two families of representations SO(3)?
2. Why real objects has to be governed by one of them?

Answer to first one is unimportant: as long as w accept that SO(3) is important for us in describing real world objects. It is feature os group of rotation. No more o no less.

Answer to second question is interesting one: as long as system has some symmetry, we may choose our description in such way, that states of that system are pure eigenstates of some representations of group of symmetry which system has.

For example, as system has transnational symmetry, You may choose to use momentum states as basis in Hilbert space of wavefunctions.

Other way, as we have symmetry of translation in time, our systems states has well defined energies ( and are in states which are eigenstates for some energy operators which in fact gives us translation in time etc.)

This is the real meaning of symmetry in system. In order to look as it works please take any Quantum Mechanics book with chapter about Representation Group Theory and its relation to Noether Theorem.

 

[LennoxLewis]

The flow of momentum into the region due to particles passing through an element $d\mathbf{A}$ of the surface (oriented outwards), turns out to be (I am ignoring relativistic effects here):
$$-N \langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle d\mathbf{A}$$,
where N is the density of particles, $\mathbf{n}$ is the unit vector pointing in the direction of $d\mathbf{A}$, $\mathf{v}$ is the velocity of the particles, and the angle brackets denote averaging over all nearby particles. Now, assuming isotropy, $\langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle$ will be the same regardless of $\mathbf{n}$, so we can define the pressure by

$$P = N \langle m(\mathbf{v}\cdot{\mathbf{n})^2\rangle$$

In particular we have
$$\begin{align*} P &= N \langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle \\ &= \frac{N}{3}\left(\langle mv_x^2 \rangle + \langle mv_y^2 \rangle + \langle mv_z^2 \rangle\right) \\ &= \frac{2N}{3} \left\langle \frac{1}{2}m|\mathbf{v}|^2\right\rangle \\ &= \frac{2}{3} u \end{align}$$
where u is the kinetic energy density.

Isn't that the flow of kinetic energy? I know that doesn't really make sense as kinetic energy isn't a vector, but momentum is m.v, while m.v.v, which you are using, is related to kinetic energy?

Okay, of course the force the particles exert on the wall is electromagnetic. But that's not what I'm talking about. Pressure is a local property of the individual locations in the gas (take an infinitesimal surface element at a particular point - how much momentum flows through it in unit time?). It doesn't depend on what's happening at the exterior of the gas.

I'm not used to that definition, but for the moment i'll accept it as being true.


Maybe my question becomes more clear when i re-phrase it:
Pressure is related to kinetic energy of the particles. In a normal sequence star, nuclear fusion provides this kinetic energy, leading to a pressure that opposes gravitational collapse. When fusion runs out, the gas loses kinetic energy and with it the pressure needed to retain size. The star collapses until degeneracy pressure equals the gravitational pressure. Where does the energy needed for this pressure come from? The Pauli exclusion principle provides an endless source of energy when inter atomic distances become small enough?

Well, the whole point of degeneracy pressure is that the exclusion principle forbids this. At high densities, the only way you can avoid having two particles in the same state is by filling up lots of very high energy states. So even at absolute zero the particles continue to have lots of kinetic energy.

So where does this energy come from? It never depletes, if the Pauli Principle is universal and timeless (not taking big crunches etc into account).

 

[cortiver]

Isn't that the flow of kinetic energy? I know that doesn't really make sense as kinetic energy isn't a vector, but momentum is m.v, while m.v.v, which you are using, is related to kinetic energy?

The flow of kinetic energy would be
$$N\left\langle \frac{1}{2}mv^2 \mathbf{v}\cdot d\vect{A} \right\rangle$$
(though for an isotropic velocity distribution this just evaluates to zero). To calculate the flow of a quantity through a surface, you have to take into account that the faster particles are moving perpendicular to the surface, the more particles pass through per unit time, i.e. you multiply the quantity you're calculating the flow for by the velocity of the particles perpendicular to the surface. What I've shown is that the flow of momentum and the density of kinetic energy are closely related. There are similar results in other parts of physics, e.g. in electromagnetism the http://en.wikipedia.org/wiki/Poynting_vector" tells you both the flow of energy and the density of momentum.

Maybe my question becomes more clear when i re-phrase it:
Pressure is related to kinetic energy of the particles. In a normal sequence star, nuclear fusion provides this kinetic energy, leading to a pressure that opposes gravitational collapse. When fusion runs out, the gas loses kinetic energy and with it the pressure needed to retain size. The star collapses until degeneracy pressure equals the gravitational pressure. Where does the energy needed for this pressure come from?

The original main sequence star had kinetic and gravitational potential energy. When the core collapses to form a white dwarf, some of the energy is radiated but some of it remains in the white dwarf. This is where the kinetic energy of the particles in a white dwarf 'comes from'. The energy then remains in the white dwarf forever - there is no need for a source of energy. (Obviously since white dwarfs are 'white', they are still radiating energy. This is surplus energy, on top of the minimum energy mandated by the exclusion principle. Eventually this surplus energy is exhausted, and the white dwarf becomes a http://en.wikipedia.org/wiki/Black_dwarf" )

 

[LennoxLewis]

The flow of kinetic energy would be
$$N\left\langle \frac{1}{2}mv^2 \mathbf{v}\cdot d\vect{A} \right\rangle$$
(though for an isotropic velocity distribution this just evaluates to zero). To calculate the flow of a quantity through a surface, you have to take into account that the faster particles are moving perpendicular to the surface, the more particles pass through per unit time, i.e. you multiply the quantity you're calculating the flow for by the velocity of the particles perpendicular to the surface. What I've shown is that the flow of momentum and the density of kinetic energy are closely related. There are similar results in other parts of physics, e.g. in electromagnetism the http://en.wikipedia.org/wiki/Poynting_vector" tells you both the flow of energy and the density of momentum.

Fair enough. When seeing the entire derivation, it would probably make more sense, but when i see a factor proportional to v squared, my instinct tells me it's related to energy, regardless of whether it's per unit time and/or multiplied by a unit vector.

The original main sequence star had kinetic and gravitational potential energy. When the core collapses to form a white dwarf, some of the energy is radiated but some of it remains in the white dwarf. This is where the kinetic energy of the particles in a white dwarf 'comes from'. The energy then remains in the white dwarf forever - there is no need for a source of energy. (Obviously since white dwarfs are 'white', they are still radiating energy. This is surplus energy, on top of the minimum energy mandated by the exclusion principle. Eventually this surplus energy is exhausted, and the white dwarf becomes a http://en.wikipedia.org/wiki/Black_dwarf" )

I don't understand why there is no need for a energy source to provide degeneracy pressure. It can withstand a gigantic amount of gravitational force, for eternity... I understand that the gas obtains some, or even a lot of kinetic energy upon the collapse to a white dwarf, but the merely provides a temporary, "normal" pressure that clearly doesn't stop the collapse. It's the pressure from the Pauli Exclusion principle that holds a stop to the collapse, and this has little to do with the kinetic energy of the gas.

In fact, the formula for degeneracy pressure:

Does not contain a temperature term, or even an energy term! It is dependent only on the density and the ratio of electrons to neutrons.

Quoting Wikipedia, about the formula i wrote above:
"This degeneracy pressure is omnipresent and is in addition to the normal gas pressure P = nkT / V."

Continueing, on the Heisenberg Uncertainty principle:
"A material subjected to ever increasing pressure will become ever more compressed, and for electrons within it, the uncertainty in position measurements, Δx, becomes ever smaller. Thus, as dictated by the uncertainty principle, the uncertainty in the momenta of the electrons, Δp, becomes larger. Thus, no matter how low the temperature drops, the electrons must be traveling at this "Heisenberg speed," contributing to the pressure. "

So, where does this gain in momentum come from? What happened to conservation of momentum? Or energy??

 

[cortiver]

I don't understand why there is no need for a energy source to provide degeneracy pressure.

I don't understand why you think there is a need for an energy source... you realize that the degeneracy pressure isn't doing any work? The energy of the white dwarf star (apart from the radiation of surplus energy which I mentioned before) is constant. There is no source of new energy.

I understand that the gas obtains some, or even a lot of kinetic energy upon the collapse to a white dwarf, but the merely provides a temporary, "normal" pressure that clearly doesn't stop the collapse. It's the pressure from the Pauli Exclusion principle that holds a stop to the collapse, and this has little to do with the kinetic energy of the gas.

No, the degeneracy pressure has everything to do with the kinetic energy of the gas. The point of my derivation above was that both degeneracy pressure and thermal pressure come from the kinetic energy, via the relation $P = \frac{2}{3}u$. The only difference between the two is that thermal pressure is the pressure you get in a classical gas, whereas degeneracy pressure is the pressure you get in a extremely degenerate (i.e. very high density or very low temperature) gas. Because classical and degenerate gas have different thermodynamic properties, the equation of state for the pressure is different in the two cases. In particular, the pressure in a high-density gas is much higher than the pressure you would get if classical mechanics still worked at such high densities.

Does not contain a temperature term, or even an energy term! It is dependent only on the density and the ratio of electrons to neutrons.

Well, the equation $P = \frac{2}{3}u$ still holds. The kinetic energy density of a degenerate gas is itself dependent only on the density and the ratio of electrons and neutrons (because the kinetic energy of a degenerate gas is determined by the exclusion principle - all the lowest possible energy states are filled with one particle each).

Quoting Wikipedia, about the formula i wrote above:
"This degeneracy pressure is omnipresent and is in addition to the normal gas pressure P = nkT / V."

I think this is not really true. If $T \neq 0$, then the pressure will not be exactly given by the degeneracy pressure formula you quoted above (because that is derived by setting $T = 0$ in the Fermi-Dirac distribution), but it won't simply be the sum of the P = NkT pressure and the degeneracy pressure - it's more complicated than that.

Continueing, on the Heisenberg Uncertainty principle:
"A material subjected to ever increasing pressure will become ever more compressed, and for electrons within it, the uncertainty in position measurements, Δx, becomes ever smaller. Thus, as dictated by the uncertainty principle, the uncertainty in the momenta of the electrons, Δp, becomes larger. Thus, no matter how low the temperature drops, the electrons must be traveling at this "Heisenberg speed," contributing to the pressure. "

It says the uncertainty in the momentum becomes larger. The momentum itself is conserved. Anyway this explanation is not very good. The fact that the particles still have kinetic energy even at absolute zero is due to the exclusion principle, and can't be satisfactorily explained by the uncertainty principle alone, which applies equally to fermions and bosons (the latter of which don't obey the exclusion principle).

 

[DrChinese]

1. How do you justify it to students? Here's a daring strategy - try actually telling them about it and let them make their own minds up.

2. Scientific points...
(1) Bell's theorem(!).

1. That is not what I meant. I meant: the specific hypothesis that the positions of remote particles completely determines all we see is an ad hoc hypothesis. It could be other attributes, and there could be different (amounts of) influences for different remote particles. And this is clouded by the additional hypothesis that the effect is not observable (QEH). Again, I am not arguing the merits of BM - of which I believe there are many - just the fact that it is additional to standard accepted science. That is the criteria for this board, and hypotheses related to interpretations should be properly labeled as such. If you think BM is the best, great, say so, but don't advance it to others in such a way that they think it is generally accepted.

2. Bell's Theorem was not developed as a result of Bohmian Mechanics. Bell obviously knew of it, but there is nothing in his proof that relates to any element of BM. Further, Bell's work around that time clearly was focused on a variety of mathematical proofs associated with the realistic side of the EPR paradox (such as K-S). There has been a lot of revisionist history associated with BM in recent years, and I think that really speaks poorly for the position in general.

 

[bcrowell]

Now if you have two fermions two the left and right of a common center and then do a 180 degree rotation on both of them about the common center they will have swapped places. So the state is no longer $$\left | ab \right \rangle$$, but $$\left | ba \right \rangle$$. However since you have performed two 180 degree rotations, one on each particle, this has the same effect as a single 360 degree rotation, hence the new state is the old one with a minus sign.

It's very alluring to have such a conceptually straightforward sketch of the spin-statistics theorem. I don't follow you, though, when you say, "since you have performed two 180 degree rotations, one on each particle, this has the same effect as a single 360 degree rotation." If we think of the case of a vector field, which can be treated purely in terms of three-dimensional geometry, a rotation by 180 degrees causes a sign flip, but a 90-degree rotation can't simply be treated as a change of sign or phase. So I don't see why it's valid to cut a 360-degree rotation in half to get a 180-degree one, and then assume that the 180-degree one can just be treated as a change of phase.

 

[DarMM]

It's very alluring to have such a conceptually straightforward sketch of the spin-statistics theorem. I don't follow you, though, when you say, "since you have performed two 180 degree rotations, one on each particle, this has the same effect as a single 360 degree rotation." If we think of the case of a vector field, which can be treated purely in terms of three-dimensional geometry, a rotation by 180 degrees causes a sign flip, but a 90-degree rotation can't simply be treated as a change of sign or phase. So I don't see why it's valid to cut a 360-degree rotation in half to get a 180-degree one, and then assume that the 180-degree one can just be treated as a change of phase.

It's not valid. The full proof of the spin-statistics theorem requires heavy mathematical machinery. My post above is meant as a rough guide. The full proof is given in the monograph of Streater and Wightman "PCT, Spin and Statistics and all that". There are other proofs involving other approaches to QFT, but all are equally difficult.

Remember the Spin and Statistics theorem is a theorem of QFT, it does not follow from QM. So any proof will involve the local and causal properties of QFT.

 

[LennoxLewis]

I don't understand why you think there is a need for an energy source... you realize that the degeneracy pressure isn't doing any work? The energy of the white dwarf star (apart from the radiation of surplus energy which I mentioned before) is constant. There is no source of new energy.


No, the degeneracy pressure has everything to do with the kinetic energy of the gas. The point of my derivation above was that both degeneracy pressure and thermal pressure come from the kinetic energy, via the relation $P = \frac{2}{3}u$. The only difference between the two is that thermal pressure is the pressure you get in a classical gas, whereas degeneracy pressure is the pressure you get in a extremely degenerate (i.e. very high density or very low temperature) gas. Because classical and degenerate gas have different thermodynamic properties, the equation of state for the pressure is different in the two cases. In particular, the pressure in a high-density gas is much higher than the pressure you would get if classical mechanics still worked at such high densities.


Well, the equation $P = \frac{2}{3}u$ still holds. The kinetic energy density of a degenerate gas is itself dependent only on the density and the ratio of electrons and neutrons (because the kinetic energy of a degenerate gas is determined by the exclusion principle - all the lowest possible energy states are filled with one particle each).


I think this is not really true. If $T \neq 0$, then the pressure will not be exactly given by the degeneracy pressure formula you quoted above (because that is derived by setting $T = 0$ in the Fermi-Dirac distribution), but it won't simply be the sum of the P = NkT pressure and the degeneracy pressure - it's more complicated than that.


It says the uncertainty in the momentum becomes larger. The momentum itself is conserved. Anyway this explanation is not very good. The fact that the particles still have kinetic energy even at absolute zero is due to the exclusion principle, and can't be satisfactorily explained by the uncertainty principle alone, which applies equally to fermions and bosons (the latter of which don't obey the exclusion principle).

Okay - i don't dispute that anything you say is not true.

But i still don't get it. I agree that there is no inflow of energy and maybe no work is done, but i still think there must be a force or energy that prevents gravitational collapse. Maybe you can make me understand via this example:

Let's say i am suspended on a bungee jumping rope, hanging from a bridge. I'm not talking about the situation when you are falling, but after you fell. You're not moving; the axial force of the rope is in equilibrium with the opposite gravitational force. No work is being done.

Now, if my rope is cut, the only way to stop me from falling down is to introduce an opposite force. For instance, a rocket on my back... or by falling onto the ground, where the EM force will stop (and probably kill) me from moving. This last scenario does not involve adding energy, but in every situation, there is an opposing force. There has to be; Newton's law.

So what force in a white dwarf opposes the gigantic gravitational pull? I guess you've already answered that there can be pressure without any force, but i don't understand this. By definition, if there is acceleration, then there is a force... how can Newton's law be violated?

 

[cortiver]

Okay - i don't dispute that anything you say is not true.

But i still don't get it. I agree that there is no inflow of energy and maybe no work is done, but i still think there must be a force or energy that prevents gravitational collapse. Maybe you can make me understand via this example:

Let's say i am suspended on a bungee jumping rope, hanging from a bridge. I'm not talking about the situation when you are falling, but after you fell. You're not moving; the axial force of the rope is in equilibrium with the opposite gravitational force. No work is being done.

Now, if my rope is cut, the only way to stop me from falling down is to introduce an opposite force. For instance, a rocket on my back... or by falling onto the ground, where the EM force will stop (and probably kill) me from moving. This last scenario does not involve adding energy, but in every situation, there is an opposing force. There has to be; Newton's law.

So what force in a white dwarf opposes the gigantic gravitational pull? I guess you've already answered that there can be pressure without any force, but i don't understand this. By definition, if there is acceleration, then there is a force... how can Newton's law be violated?

The thing is, Newton's law, F = ma, fundamentally applies to single particles. It can also be generalized to apply to groups of particles whose membership is constant, i.e. the same particles are in the group forever. The set of particles making up a person is, to all intents and purposes, constant, so you can apply Newton's law to a bungee-jumper.

I guess the problem you're having then, is you're trying to apply Newton's Law to a little section of a star, as if it were a little block of solid. But a star is not a solid, and the set of particles contained within a given region does not remain the same over time - because of the kinetic energy of the particles, they are moving around and there will always be particles entering and leaving the region. Hence Newton's Law cannot be applied to a region of gas!

Now, you can apply Newton's Law to an individual gas particle. In this case, since the only force acting on it is gravity, it will indeed be accelerating towards the center. But this does not mean that, averaged over long periods of time, it has to get any closer to the center, as you can see by considering the orbit of the Earth around the sun. So you have lots and lots of particles, all accelerating towards the center, but the overall density of the star at every point is unchanged, so it doesn't collapse.

 

[zenith8]

Now, you can apply Newton's Law to an individual gas particle. In this case, since the only force acting on it is gravity, it will indeed be accelerating towards the center. But this does not mean that, averaged over long periods of time, it has to get any closer to the center, as you can see by considering the orbit of the Earth around the sun. So you have lots and lots of particles, all accelerating towards the center, but the overall density of the star at every point is unchanged, so it doesn't collapse.

Hi Cortiver,

OK, but I showed in my post #22 that, under the sole assumption that particles exist and that they have trajectories, then the equations of quantum mechanics themselves predict the existence of an extra force $$-\nabla Q$$ on each particle. It is quite clear in your above posts that you are making the same implicit assumption that particles exist. For your argument to stand up, then, you need to get rid of this force. How do you propose to do that, other than by ignoring it?

Cheers,
Zenith

 

[LennoxLewis]

The thing is, Newton's law, F = ma, fundamentally applies to single particles. It can also be generalized to apply to groups of particles whose membership is constant, i.e. the same particles are in the group forever. The set of particles making up a person is, to all intents and purposes, constant, so you can apply Newton's law to a bungee-jumper.

I guess the problem you're having then, is you're trying to apply Newton's Law to a little section of a star, as if it were a little block of solid. But a star is not a solid, and the set of particles contained within a given region does not remain the same over time - because of the kinetic energy of the particles, they are moving around and there will always be particles entering and leaving the region. Hence Newton's Law cannot be applied to a region of gas!

Good points and i agree.

Now, you can apply Newton's Law to an individual gas particle. In this case, since the only force acting on it is gravity, it will indeed be accelerating towards the center. But this does not mean that, averaged over long periods of time, it has to get any closer to the center, as you can see by considering the orbit of the Earth around the sun. So you have lots and lots of particles, all accelerating towards the center, but the overall density of the star at every point is unchanged, so it doesn't collapse.

But the Earth around the Sun is a special case of free fall. You can't tell me that every particle in a region as dense as a white dwarf (or even a neutron star) is constantly in free fall? Because two particles can't have the same state when their wave functions overlap, they are repulsed (?) "by", or as a consequence of the Pauli Exclusion principle?


One unrelated sidenote: what about the Pauli Exclusion principle behind the event horizon of a black hole? Does our physics break down at a singularity?

 

[cortiver]

zenith8 and LennoxLewis, you both seem to have misunderstood my post. I was trying to clarify how a star can be supported by pressure without the need for an actual force. To avoid quantum complications, I was considering a star made up of classical particles. Of course, the motion of quantum particles is governed by the Schrödinger equation, and Newton's Law doesn't apply directly, at least not without the introduction of the quantum force.

However, for a real white dwarf star, the de Broglie wavelength of the particles (which, in a degenerate gas, is on the order of the average distance between neighbouring particles) is much smaller than the characteristic length scale of variation of the star, and hence the semiclassical approximation (use Fermi-Dirac distribution, but otherwise treat particles as classical) gives correct results for macroscopic quantities. Hence, under this picture (which admittedly is not really the true one), we can talk about the particles as moving under Newton's Law without a quantum force.

Edit: the following argument is wrong! Please ignore it.

I can also show directly that the quantum force should be negligible compared to the pressure 'force' in such a situation:
The quantum force is given by
$$\begin{align*} F_q &= \sum_{i}\nabla_i \left(\frac{\hbar^2}{2m} \frac{\nabla_i^2 |\Psi|}{|\Psi|}\right) \\ &\sim \sum_{i}\frac{\hbar^2}{2m} \frac{1}{R^3} \\ &\sim N\frac{\hbar^2}{2m} \end{align}$$
where R is the radius of the star, N is the average density of particles in the star, and the last line follows because $R^3$ is on the order of the volume of the star. The quantum force density is therefore
$$f_q = NF_q \sim N^2\frac{\hbar^2}{2m}$$

On the other hand, the degeneracy pressure is on the order of the kinetic energy density. So $$P \sim NE$$, where E is the average kinetic energy of particles in the star, and the pressure 'force' density is
$$f_p = -\nabla P \sim \frac{NE}{R}$$
The ratio of the two forces is therefore
$$\begin{align*} \frac{f_p}{f_q} &\sim \frac{N\hbar^2 R}{2mE} \\ &\sim N\lambda_B^2 R \end{align*}$$
where $\lambda_B$ is the average de Broglie wavelength of the particles. Since in a degenerate gas, $\lambda_B \sim N^{-1/3}$, where $N^{-1/3}$ is approximately the average separation between neighbouring particles, we have
$$\frac{f_p}{f_q} \sim N^{1/3} R$$
Since there are obviously many particles in a white dwarf star, we have $R \gg N^{-1/3}$, and hence:
$$f_p \gg f_q$$
So the quantum force is negligible compared to the pressure 'force'.

End of wrong argument.

I note also that the definition of the quantum force is the same for bosons and fermions, so how could it explain repulsion between fermions?

 

[LennoxLewis]

Isn't that fq over fp that you calculated? fp over fq should, other terms aside, be proportional to one over N (before simplifying the terms).

 

[cortiver]

Isn't that fq over fp that you calculated? fp over fq should, other terms aside, be proportional to one over N (before simplifying the terms).

Oops, you're right. Clearly I composed that argument in too much of a hurry - looking at it now, I see it doesn't make much sense. Consider it retracted. I still maintain that the semiclassical approximation should be applicable for white dwarf stars, so the quantum force should be negligible. I'll keep thinking whether it is possible to show this directly.

 

[zenith8]

zenith8 and LennoxLewis, you both seem to have misunderstood my post.

I can also show directly that the quantum force should be negligible compared to the pressure 'force' in such a situation

Your maths is just wrong, as Lennox has already pointed out.

I note also that the definition of the quantum force is the same for bosons and fermions, so how could it explain repulsion between fermions?

Remember that the quantum force is supposed to be a force exerted by the objectively existing wave field (mathematically represented by the wave function) on the particles. It is not directly an interparticle force.

Remember also that the wave fields for fermions and bosons have very different characteristics, the most important of which is the existence of a nodal surface in the fermionic case (i.e. a set of points in configuration space where the wave field has the value zero). Particles trajectories are actively repelled from these surfaces, and effectively, cannot pass through them. This is exactly what is required to produce 'fermionic repulsion' in the sense you state.

 

[cortiver]

Remember that the quantum force is supposed to be a force exerted by the objectively existing wave field (mathematically represented by the wave function) on the particles. It is not directly an interparticle force.

Remember also that the wave fields for fermions and bosons have very different characteristics, the most important of which is the existence of a nodal surface in the fermionic case (i.e. a set of points in configuration space where the wave field has the value zero). Particles trajectories are actively repelled from these surfaces, and effectively, cannot pass through them. This is exactly what is required to produce 'fermionic repulsion' in the sense you state.

Fair enough. I had got the impression somehow that you were claiming that the quantum force could explain the exclusion principle without any additional boundary conditions, but looking back at the thread it looks like I was mistaken and you never made any such claim.

Anyway, all treatments I've seen of the equation of state in a white dwarf star only mention the standard degeneracy pressure
$$P = \frac{(3\pi^2)^{2/3}}{5} \frac{\hbar^2}{m} N^{5/3}$$
(or its relativistic generalization), which comes simply from the kinetic energy via the relation P = (2/3)u, where u is calculated from the Fermi-Dirac distribution at T = 0. This must surely mean that the additional effect of the quantum force is considered negligible. I don't know enough to prove this myself, but I do not believe that all those authors are wrong.