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Which of the four forces is responsible for degeneracy pressure?

  1. Dec 19, 2009 #1
    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?

    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?
     
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  3. Dec 20, 2009 #2
  4. Dec 20, 2009 #3

    Doc Al

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    I'd say that degeneracy pressure is not attributable to any of the four fundamental forces, but is a new quantum effect.
     
  5. Dec 20, 2009 #4

    tiny-tim

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    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). :wink:
     
  6. Dec 20, 2009 #5
    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.
     
  7. Dec 20, 2009 #6
    Mike Towler of Cambridge University addresses precisely that question on p.34 of http://www.tcm.phy.cam.ac.uk/~mdt26/PWT/towler_pilot_waves.pdf" [Broken].

    It's trivial to understand, providing you adopt the de Broglie-Bohm interpretation of QM, which merely involves postulating that particles (e.g. electrons, neutrons..) exist continually - rather than only when you look at them - and therefore have definite trajectories. One can then analyze the dynamics of these particles just by looking at the usual QM probability current. What one finds is that the particles are acted on by a 'quantum force' (over and above the usual classical force due to attraction/repulsion between particles) and that this force has its origin in the wave field (the physical object that is mathematically represented by the wave function). The wave field pushes the particles around. This is why quantum particles do not, in general, follow classical Newtonian trajectories (see p.29 of the above lecture).

    Since de Broglie-Bohm theory is simply an analysis of perfectly ordinary quantum mechanics from a different perspective (i.e. discarding the usual 1920s positivistic philosophical bollocks) then this tells you that QM itself seems to imply the existence of a 'fifth force'. It seems crazy - since the existence of only four forces is hammered into us from birth - but it seems to be true. It is this force which is responsible for 'electron degeneracy pressure', 'Pauli repulsion', the Pauli exclusion principle, the strength of the covalent bond, stopping stars collapsing, digging out 'Fermi holes' in an electron gas, etc.. etc...

    The main reason that this has not been accepted (other than most people being unaware of the possibility) is the usual insistence that the wave function represents 'knowledge' or the 'probability of obtaining particular experimental results' rather than an objectively existing wave field. But today's accumulated experimental knowledge in matter wave optics (which shows us directly that the wave field can be manipulated by essentially optical instruments, and therefore must objectively exist) should be enough evidence to show that people have been barking up the wrong tree for 80 years. Cool, isn't it?
     
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  8. Dec 20, 2009 #7
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  9. Dec 20, 2009 #8

    alxm

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    Nonsense. A fermion will obey the Pauli principle whether it has a magnetic moment or not, period.

    Too weak for what?

    That follows how?
     
  10. Dec 20, 2009 #9

    alxm

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    Elaborating on what was said already: The exclusion principle is not a 'force' in itself. E.g. it's not something that goes into the Hamiltonian, but is rather a boundary condition, a constraint put on the solutions to the Schrödinger equation.

    Now you can model your system without this constraint, but you'll get the wrong energy. You can then look at the difference in energy and turn that into a pseudopotential, which is more or less what you're doing when you're talking about degeneracy 'pressure'. Same goes for density-functional methods and what's known as 'exchange functionals'. (And in fact, the equations for degeneracy pressure in Neutron stars and such turn up again for describing electrons in atoms in books on DFT)

    A somewhat similar thing is using pseudopotentials to model relativistic effects in atoms. While special relativity isn't a force either, you can model its effects as if it were.
     
  11. Dec 20, 2009 #10
    OK. So when a star runs out of gas (as it were) and stops its fusion reactions, and it starts to collapse because of the gravitational interaction between its constituent particles, what stops the collapse at the radius of a white dwarf, if it isn't a force? Forget maths, forget boundary conditions on equations. Just think about the real world. What stops it physically?
     
  12. Dec 20, 2009 #11

    tiny-tim

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    erm :redface: … this is highly non-standard, it's the pilot-wave interpretation of quantum theory.

    Any attempt to understand it without weeks of study is doomed.

    And degeneracy pressure can be explained without it.
     
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  13. Dec 20, 2009 #12
    Isn't that what I said?
    Towler's popular lecture that I referred to above should take you about half an hour to read. If you need the details then you can move on to his http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" [Broken]. That's where I learnt it. Didn't take me that long... You can too!

    Anyway, string theory - for example - would take several years of study to understand. And that's if you've already got a Ph.D. and are a mathematical genius.. That doesn't make it wrong.. Or right, now I come to think of it.

    Oh, I love it. Go on, then.. we're waiting.
     
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  14. Dec 20, 2009 #13

    tiny-tim

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    erm :redface: … no, actually, you didn't.
    I didn't say it was wrong, i said it was unnecessary and non-standard. And most people do find it difficult to understand (if not downright unhelpful).

    It isn't helpful, to most PF members asking questions about quantum theory, to introduce the pilot-wave interpretation.
     
  15. Dec 20, 2009 #14
    Look, the standard term for it (say in http://en.wikipedia.org/wiki/De_Broglie–Bohm_theory" [Broken]) is the de Broglie-Bohm theory/interpretation. If you want to refer to it as pilot-wave theory, then that's fine, but why on earth you are quibbling about me referring to it in the usual way is beyond me...
    If you find it difficult to understand then you haven't spent more than 5 minutes thinking about it. The basic ideas are trivial - again, see the Towler popular lecture.
    So what would you rather I used? The Copenhagen interpretation? The guy is asking a conceptual question. The Copenhagen interpretation is based on the supposition that conceptual questions are 'meaningless'. The 'shut up and calculate' interpretation? I can't, because I have to shut up and calculate. Any other interpretation except de Broglie-Bohm? No, because it is impossible to answer the question in the terms in which it was phrased..

    It is better to use the pilot-wave theory (see, I'm using your language now) because it is clear what it claims exists and is therefore capable of answering conceptual questions. Which of the four known forces is responsible for 'Pauli repulsion' in stars? Answer, none of them. If you analyze quantum theory in terms of forces (and only pilot-wave theory allows you to do that) then it gives you a mathematical formula for the Pauli repulsion force (the negative gradient of the quantum potential), it allows you therefore to postulate that the force objectively exists (if you want), and it allows you to realize that the required characteristics of the force are not at all those of any of the other four forces.
     
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  16. Dec 20, 2009 #15
    And don't think I didn't notice that you didn't explain the degeneracy pressure, as I requested and as you claim to be able to do.
     
  17. Dec 20, 2009 #16

    alxm

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    A pseudo-force, yes.

    A consequence of the properties of fermions. Pilot-wave theory may have a different explanation for spin, but it doesn't make this more or less of a force. I.e. What's the exchange particle for this "force", then?

    I don't see how the pilot-wave theory makes it any simpler either. If you think otherwise, feel free to go ahead and use it to solve a real problem: develop an exact exchange functional. That'll win you a lot more converts than message-board proselytizing.
     
  18. Dec 21, 2009 #17
    Pauli exclusion principle is an empirical theory, isn't it?

    Because the causes and forces of this principle have not been found.
    The original QM and the other theories using the various forces could not predict the Pauli exclusion principle.
    It is based on the spectrum data etc. And it was added to the original QM.
    (Of course, the real spin itself can not be ascertained.)

    The exchange interaction (the exchange integral) is caused by the original form of the wavefunction (antisymmetry etc.)
    If the wavefunction is antisymmetric, this means the two electrons can't be in the same state. Of course this will influence the interaction energy between the electrons because the two electrons can't be in the same state.
     
  19. Dec 21, 2009 #18

    DarMM

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    In quantum theory a type of statistics called Fermi statistics is possible. Without getting into too much detail this means that two particles which obey fermi statistics cannot occupy the same state, or roughly they cannot have the same spin and have the same position as each other at the same time. This applies to particles with half-integer spin.
    So in a neutron star, the neutrons are not allowed to be located near each other by Fermi statistics and this cannot be overcome below a certain mass. This effect keeps the star supported.

    Degeneracy pressure isn't a new force, rather it is just a consequence of the fermi-statistics of half-integer spin particles.

    As for fermi-statistics itself, it is a consequence of the basic properties of quantum field theory, but that would be too much to get into at the moment.
     
  20. Dec 21, 2009 #19
    So - to translate into English -- what you're saying is that because they are fermions the particles repel each other, and yet this is not because there is a force, but because they just do.

    OK. Nice logic. I'd love to hear what the Original Poster thinks of your answer to his question.
     
  21. Dec 21, 2009 #20

    DarMM

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    Well, first of all, it's not my answer. I'd be a prize winning physicist if Fermi-statistics was my idea.

    Also, the fermions do not repel each other, in the sense that they cause changes in each other's momentum. In fact they do not repel at all. Rather for the star to keep collapsing any two given fermions would have to occupy the same position. However this isn't a possible two particle state because of the half-integer spin of the particles. So there is no repulsion, it's just that for the star to keep collapsing it would have to move into a state which doesn't exist, hence it doesn't and so remains as it is. Which means the star is supported.
     
  22. Dec 21, 2009 #21
    Ok, so the way I understand it (limited as that is) is that the reason we call it degeneracy "pressure" is because the way it affects the (thermodynamic) equation of state is like an extra pressure term. If we increase the strength of gravity--and hence how much total force is acting on the star--the predicted decrease in volume is consistent with the usual gas pressure plus this extra degeneracy pressure. (Think of a balloon. You squeeze it and the volume decreases only a certain amount because the pressure is pushing back. ) At high densities, the degeneracy pressure becomes very significant.

    Now we say it is not due to a force. Rather it is because once the lowest energy states are filled, that's it. They can't give up energy and go to any lower state. The volume they take up is fixed (in a sense). Additional electrons will have to occupy higher energy states, and that requires more energy. It gets complicated from there and i've probably already said something incorrect. But one straightforward question...

    Can the degeneracy pressure do work?

    If so, that energy has to come from somewhere, either from a force field or from converting matter to energy. I presume it has to be former, so wouldn't it have to be the EM field?
     
  23. Dec 21, 2009 #22
    OK - so they don't repel each other, in the sense that they don't cause changes in each other's momentum. Fine. Let's ask QM itself to confirm this, shall we?

    You're talking about fermions as if they are continuously existing particles, so they must have trajectories. If the classical force is the only force acting on them (as you effectively state) then they must be following Newtonian trajectories, right?

    Since we don't know precisely where the particles are, we'll have to work with a statistical distribution [tex]\rho[/tex] of particles with unknown positions. They're obeying ordinary Newtonian dynamics, but just for convenience instead of F=ma I will use the entirely equivalent Hamilton-Jacobi equation [tex]-\frac{\partial S}{\partial t} = \frac{(\nabla S)^2}{2m} + V[/tex] - where [tex]S[/tex] is related to the 'action' - to calculate the trajectories.

    Furthermore, the probability distribution of the particles must obey the usual continuity equation - [tex]\partial \rho / \partial t = -\nabla\cdot(\rho {\bf v})[/tex] - in order that it remains normalized as it changes shape over time (here [tex]v[/tex] is the velocity of the particle).

    Just to play a trick, let's combine the classical Hamilton-Jacobi equation and the continuity equation (two real equations, note) into a single complex equation. To do this, we introduce a general complex function [tex]\Psi = r e^{i\theta}=\sqrt{\rho} e ^{\frac{iS}{\hbar}}[/tex] with [tex]\hbar[/tex] an arbitrary constant giving a dimensionless exponent. After a small amount of algebra, we find that the complex equation that results is:

    [tex]i\hbar \frac{\partial \Psi}{\partial t} = \left( -\frac{\hbar^2}{2m}\nabla^2 + V - Q\right)\Psi[/tex] .

    Well, well - that's the time-dependent Schrodinger equation - straight out of quantum mechanics - with something ('[tex]Q[/tex] ') that looks like a potential subtracted off the Hamiltonian, and [tex]\Psi[/tex] has the same interpretation as in QM: a probability density of particle positions. So what this is telling us is that if we want the particles to follow Newtonian trajectories, we need to subtract off an extra 'quantum force' [tex]-\nabla Q[/tex] - the negative gradient of the `quantum potential' - from the usual dynamic equation of quantum mechanics. This extra force is what makes quantum systems different from classical ones. Is doing this, the momentum of the particle is clearly changed, in contradiction to what you state.

    If you analyze the electron-degeneracy pressure situation (in a star, say) then one finds that the gravitational field is balanced precisely by a term involving the quantum force at the radius of the white dwarf, and this force has none of the characteristics of the four regular forces. So if you actually want to answer the question in the terms that the guy asked it, then that's the answer.

    If you want, you can say "The particles are fermions and thus they can't be in the same state" but that's simply restating the question in different words. What the above analysis is doing is showing why they can't be in the same state. Because the force exerted by the wave field on the particles keeps them apart.

    Just for the record, the formula for [tex]Q[/tex] turns out to be :

    [tex]Q=-\sum_{i}\frac{\hbar^{2}}{2m_{i}}\frac{\nabla_{i}^{2}\left\vert \Psi\right\vert }{\left\vert \Psi\right\vert }[/tex]

    and the natural interpretation of this is that the wave field (the objectively existing field represented by the wave function [tex]\Psi[/tex]) is 'pushing' the particles, and this force is larger at points in configuration space where the curvature of the wave field is large.

    Look, I'm only trying to be clear about this, and in the end, I'm the one doing the mathematics, and you guys are the ones doing the philosophy. You probably think it's the other way around. Well, so be it. But at the end of the day saying that the answer to the guy's question is 'Fermi-Dirac statistics' is doing no better than the chaps in Brazil who Feynman was complaining about, when in answer to the question 'what causes sugar crystals to emit light when you crush them with a pair of pliers?' they said 'triboluminescence makes them emit light'. Any the wiser? I thought not.
     
    Last edited: Dec 21, 2009
  24. Dec 21, 2009 #23

    DrChinese

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    It is fundamental, so as far as we know there is no explanation or cause. I would not consider a theory to be any weaker because we used facts to help us form it. That pretty much describes all theories. :smile:

    Whether you consider degeneracy pressure something distinct (as a force) from the electromagnetic force is a question which is conventionally resolved as a NO. It is more of a consequence, and is not absolute as degenerate matter can still collapse to a black hole. With the strong force, you have a loosely similar analogy with the asymptotic freedom of quarks. The attractive force becomes larger as distance increases. Is that is separate force? No, because it is associated with strong force rules (especially the effects of gluons).
     
  25. Dec 21, 2009 #24
    I see what you mean, and I think you are right. :smile:

    As far as I know, the concrete force which causes Pauli exclusion principle is not known.
    Of course, it depends on the meaning of "concrete".
    (Relativistic QFT has more abstract and "mathematical" property, I think.)

    First, the spin itself has strange properties (such as the spinning speed which is much faster than the speed of the light).

    So in 1920's Pauli opposed to the idea of "spin" strongly, and tried to make the others abandon the idea of "spin".

    Schrodinger first thought that the Schrodinger wave function is a "real matter wave". But later his idea was denied and replaced by the idea of "probability amplitude". He was very disappointed.
    QM itself was a very strange thing for them.

    If you say we should accept the Pauli exclution principle as it is (without further inquiring), that's an idea and I won't criticize your idea.

    But similarly we have to accept "the spin" as it is, though the spin has the very strange property which returns the spinning electrons to its original configuration by the 4 pi rotaion (not 2 pi).
     
    Last edited: Dec 21, 2009
  26. Dec 21, 2009 #25
    Shrug. OK - I just derived a mathematical formula for it directly from ordinary quantum mechanics, gave it a name, and told you what (superficially at least) appears to cause it, but - of course - this is not even slightly interesting.

    OK -- I'll give it one last try. I'm going to go outside now, and I'm going to sacrifice a chicken. Then you'll believe me.. won't you?
    Or, you could adopt the de Broglie-Bohm viewpoint, and one finds the electron particle doesn't have a property called spin it all - it turns out to be part of the angular momentum of the wave field, thus becoming instantly and directly comprehensible. See my earlier post #26 in https://www.physicsforums.com/showthread.php?p=2332306#post2332306".

    Squawk..
     
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