# Moving from the exclusion principle to degenercy pressure/forces

1. Aug 29, 2010

### zhermes

I understand the basics of the pauli exclusion principle emerging from antisymmetric wavefunctions being invalid for two fermions with the same quantum numbers. I've also seen how, empirically, you can add a term to the potential energy, of (e.g.) the ionic bond between K+ and Cl-, that incorporates the effects of an exclusion force preventing the atoms from merging (or coming closer than some optimal distance).

How can one transition from a wavefunction simply not being allowed, to the existence of a new potential energy term (and therein force)? Some moderately-rigorous mathematics might help. Similarly, how do you derive something like the degeneracy pressure in a neutron star from the pauli exclusion principle?

Thanks!

2. Aug 29, 2010

### zenith8

See the lecture "Exchange, antisymmetry and Pauli repulsion" under 'Lecture and Slides' on http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" [Broken].

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3. Aug 29, 2010

### zhermes

This cite looks really cool, I'll check it out--thanks!

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4. Aug 30, 2010

### Zarqon

5. Aug 30, 2010

### alxm

I'm not an astrophysicist, I'm a chemical physicist. But the question you're asking is basically one of the central issues of Density Functional Theory: How do you calculate the energetic contribution of the Pauli principle (better-known as the exchange energy) as an external potential (functional) of the particle density?

The first attempt at that was Dirac's exchange correction to the Thomas-Fermi model (1930), which gives you essentially the same expression that's used for degeneracy pressure: $$\left({3\over\pi}\right)^{2/3} \rho^{5/3}$$.

This isn't empirical; DFT has rigorous mathematical underpinnings in the http://prola.aps.org/abstract/PR/v136/i3B/pB864_1" [Broken] (1964) (showing that 1) Such a functional exists and 2) It is variational), and the Kohn-Sham proofs (showing that you could turn this functional into an external potential acting on a non-interacting reference system of particles). Walter Kohn got the Nobel prize for this work.

The central problem here is that it is, indeed, very very difficult to derive exact expressions for the exchange potential. Basically there's only a handful of systems that have exact expressions, such as the homogeneous electron gas (a horrible model for atoms and molecules, which have anything but a homogeneous density!), the Hooke atom, and some others. But for the astrophysicists (and to a lesser extent, solid state physicists), where they're dealing with dense and fairly homogenous matter, the existing expressions are pretty good (or at least usable. The TFD model cannot describe chemistry at all.) Any DFT book will tell you all about it. A classic is Parr and Yang's book. A more rigorous treatment of the mathematical underpinnings of DFT is found in Dreizler and Gross's book.

(If nothing else, the H-K paper is worth a read, it's really a quite simple proof.)

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6. Aug 30, 2010

### zhermes

I appreciate the link, but these slides are essentially just propaganda, they aren't very informative.

Unfortunately I read through almost all of this, and some of the other threads (basically the same as this OP) linked therein---they're all entirely useless :( I don't understand why so many people need to make responses when they clearly have NO IDEA what the answer to the OP is, or what any body else is talking about (a problem from which i suffer as-well).
The frequent "examples" which show that people don't understand the effects of E&M forces is especially painful....

Thanks for the link, I'll check this out as soon as I get back to my institution (and therefore APS access), hopefully it isn't too far above my head.

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7. Aug 31, 2010

### peteratcam

Well as one of the responders to the "entirely useless" thread, maybe this will not help: But to me the trouble is in "...to the existence of a new potential energy term (and therein force)?". You don't need a potential energy term to be able to measure a force. Turn the handle of statistical mechanics and out pops thermodynamic quantities like pressure. At the level of the free energy or the entropy, there might be some generalized potentials associated with generalized forces, but that doesn't mean there is a microscopic potential energy. Maybe I haven't understood your question, in that I don't know what sort of "transition" you're looking for.

8. Aug 31, 2010

### zenith8

The http://www.tcm.phy.cam.ac.uk/~mdt26/PWT/lectures/towler_pauli.pdf" [Broken] describe precisely how - if you assume the wave function is 'allowed', how the potential energy term/force arises (using completely rigorous mathematics) and explains how the degeneracy pressure in a neutron star arises from the Pauli exclusion principle, using the only known method for doing so. If you don't think that answers your question, then well, we can't help you. Not my problem.

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9. Sep 1, 2010

### alxm

It doesn't explain anything. No calculation is done from deBB theory to arrive at the expressions for the degeneracy pressure. And as I already said in this thread, those equations were arrived at well before Bohm's work anyway. What you have there is a reinterpretation of the existing theory, reworking the boundary conditions on the mysterious $$\Psi$$ so you now have a mysterious $$Q$$ instead. Maybe you prefer that as an 'explanation', but from the standpoint of actually calculating it, I don't see how that's useful, as no way is shown on how to arrive at $$Q$$ without first calculating $$\Psi$$, at which point the calculation is already 'done' from any practical standpoint.

You're right though, I don't see how deBB can help me, or anyone else interested in practical results. When recent papers using deBB for atomic problems (doi:10.1007/s10701-009-9317-6) have to invoke relativistic theory, spin-dependence and do computer calculations in order to arrive at results which, from the practical perspective, were surpassed 80 years ago using 'orthodox' theory without relativity or invoking spin, calculated on a mechanical calculator, then I don't see any practical point.

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10. Sep 7, 2010

### zhermes

Well, thats an excellent question zhermes. Consider two pairs of fermions (A1 & A2, and B1 & B2) each pair in the ground state, with opposite spins (i.e. A1-up, A2-down; B1-up, B2-down). To bring the pairs together (i.e. compress them), work must be done to increase the energy of one of the pairs such that the two pairs no longer occupy the same quantum state (i.e. A1-A2=excited, B1-B2=ground-state or visa-versa). From the volume-dependence (via density, etc) of the total energy of the system, a pressure arises. For such a derivation, see, e.g: http://www.physics.thetangentbundle.net/wiki/Statistical_mechanics/Fermi_gas [Broken]

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