Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Doc Al

    User Avatar

    Staff: Mentor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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?
    Last edited by a moderator: May 4, 2017
  8. Dec 20, 2009 #7
    Last edited by a moderator: Apr 24, 2017
  9. Dec 20, 2009 #8


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
    Last edited by a moderator: May 4, 2017
  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.
    Last edited by a moderator: May 4, 2017
  14. Dec 20, 2009 #13


    User Avatar
    Science Advisor
    Homework Helper

    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.
    Last edited by a moderator: May 4, 2017
  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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook