Talk:Electron degeneracy pressure

Formatting error

In the paragraph containing the formula for "P" I think that somebody tried to insert either a reference, or a footnote (or both), but somehow forgot an opening tag. I would correct it myself but, since I'm not a specialist on the topic, I'll ask somebody knowledgeable do it, please. — Preceding unsigned comment added by 109.100.69.106 (talk) 20:20, 30 September 2012 (UTC)[reply]

Untitled

I think the problem with this article is self-evident. As the above tag says, it needs more background (or more accurately, some background) to allow ordinary readers to have some idea of the general principle being discussed.

The degenerate matter article seems to cover electron degeneracy pretty well; do we realy need a separate article here? - Warren Dew

I think the introduction is just fine the way it is. I expected to walk into this article having to reread it many times to understand the concept, but the introduction pretty much explained the whole article to me.

"pressure due to Heisenberg speed exceeds the pressure due to thermal motion of electrons" -not very clear. can one seperate an electron's motion into Heisenberg section and pure kinetic section? If yes how did they do it? -NA

"Heisenberg speed" ?

Smoke crack much? --76.209.50.222 22:46, 12 February 2007 (UTC)[reply]

Anyone looking up "electron degeneracy" will already have sufficient background and will find the article to be just a basic overview. I hardly find the article too technical, but that's a very subjective statement. Leave it alone..

Heisenberg Speed? Never read or heard of this... Looks and feels fishy to me. Also with the equation of pressure, wouldn't it be better to have a derivation of how the conclusion was reached (or at least the assumptions made to get there) —Preceding unsigned comment added by Ajihood (talk • contribs) 09:50, 2 September 2010 (UTC)[reply]

Since when a a pressure is a force? "Electron degeneracy pressure is a force" The definition of "degeneracy pressure" is given by dW/dV = P. Please, read griffiths chapter 5 (solids). Also, including the formula would not be a bad idea. —Preceding unsigned comment added by 128.227.139.68 (talk) 02:33, 22 January 2008 (UTC)[reply]

The opening statement discusses the Pauli exclusion principle and the electron degeneracy pressure. The immediate sentence following is ambiguous at best to which concept it is further referring to.

eg ". The force provided by this pressure sets a limit on how much matter can be squeezed together without it collapsing into a black hole."

Someone can mistake that statement as reference to the electron degeneracy pressure, when in fact it is the neutron degeneracy pressure that determines if a neutron star, already exceeding the electron degeneracy pressure, will collapse into a black hole. —Preceding unsigned comment added by 72.244.203.224 (talk) 19:49, 13 September 2008 (UTC)[reply]

How is this pressure computed?

I mean, the exclusion principle of Pauli seems to imply that beyond a certain pressure there are no longer particles of a certain type... I'd really like to understand, if only someone pointed be to the right book.

Or is it something like: Beyond a certain pressure, the indeterminacy principle of Heisenberg tells that there's no way of seeing particles of that type. Therefore, I can assume they don't exist any more. And this is cool, because this interpretation predicts well some astronomical observations (white dwarves, neutron stars, blacks holes). Dpotop (talk) 21:46, 5 April 2008 (UTC)[reply]

Astronomy Stub

This subject comes up in astronomy, but its really a subject from quantum mechanics. This should be a physics stub instead of an astronomy stub. —Preceding unsigned comment added by Pulu (talk • contribs) 00:43, 22 May 2009 (UTC)[reply]

Unwarranted original research – editors please review

This edit is effectively a revert of my revert. In my opinion, it is simply WP:OR, unsourced, and downright wrong. Please would some other editors look into this. — Quondum 08:35, 20 January 2013 (UTC)[reply]

Additional Equation Needed?

The discussion I'm following regarding degenerate matter and degeneracy pressure is the question of when the core of a large star converts to a neutron star, or a black hole. The usual description is that 'if the pressure exceeds the degeneracy pressure, then the core collapses.' However, the equation here for degeneracy pressure is not a limit, but appears to be a continuous function of pressure as a function of density.

We have one equation that gives the pressure as a function of density, and an inequality that shows that the product of uncertainty in momentum and position has a minimum value. What's missing is a relationship between (P,rho) and (p,x). I think with such a conversion, there may be enough information here to turn the Heisenberg inequality into a statement about the maximum density of electrons, and/or the maximum pressure the electron cloud could withstand. JDoolin (talk) 15:55, 21 November 2013 (UTC)[reply]

I expect you're right. Though if you interpret the statement carefully, it does not imply a limit. There is a missing explanation, though, that at the balance between the two becomes unstable at the "limit". It would be nice and informative to have this point developed. —Quondum 05:40, 22 November 2013 (UTC)[reply]

Should they not be talking about atomic nuclei instead of protons? JRSpriggs (talk) 07:24, 22 November 2013 (UTC)[reply]

Yes, the equation seems to completely ignore the contribution to the density from the neutrons or other particles. I think that the equation is misguided in trying to relate the degeneracy pressure to the mass density, as this is irrelevant and confusing, and is used only to calculate the electron density. The equation should be written to use the electron particle density directly. Surely there must be a more reliable source that gives the formula in simpler form without involving non-electrons? —Quondum 14:59, 22 November 2013 (UTC)[reply]

I looked in Griffiths Section 5.3 on solids, equation 5.46 says $P={\frac {(3\pi ^{2})^{2/3}\hbar ^{2}}{5m}}\rho ^{5/3}$ . After some algebra, this agrees with the equation given in the article, except that $m_{p}$ and $\mu _{e}$ are not there. His description deals with the "number of free electrons per unit volume" inside a solid. But the density in Griffiths is the "number of free electrons per unit volume." I think the Wikipedia article is defining $\rho$ as the actual mass density, but maybe $m_{p}$ should be replaced with the mass of a nucleon (or nucleus), and $\mu _{e}$ should be the number of nucleons (or nuclei) per free electron, then the quantity ${\frac {\rho }{\mu _{e}m_{p}}}$ would be equal to the number of free electrons per unit volume.

Griffiths goes on to say this is "a partial answer to the question of why a cold solid object doesn't simply collapse: There is a stabilizing internal pressure that has nothing to do with electron repulsion" ... "or thermal motion" ... "but is strictly quantum mechanical." JDoolin (talk) 22:22, 22 November 2013 (UTC)[reply]

The way you describe Griffiths here is exactly what I had in mind and would be an excellent replacement for and significant improvement on what is there. I imagine bound electrons play a part, but am not sure of the detail. —Quondum 06:12, 23 November 2013 (UTC)[reply]

I think: Long before you reach the Chandrasekhar limit, the distinction between valence electrons and inner electrons has disappeared. The extra space which allowed valence electrons to settle into bonding orbitals rather than populate anti-bonding orbitals has been squeezed out. So the valence electrons have become like inner electrons in that they exclude other electrons from their space. But the inner electrons have become like valence electrons in that they are no longer firmly attached to one nucleus because the inner orbitals are now overlapping for different nuclei. JRSpriggs (talk) 15:22, 23 November 2013 (UTC)[reply]

Yes, I expect this in the high densities of a white dwarf. However, degeneracy pressure is a valid concept under less extreme pressures. Take metallic hydrogen at low temperature (and suitable pressure), where the electrons are not bound. Essentially all the pressure should be electron degeneracy pressure, and should match the formula. Now add a percentage of atoms of high atomic number, which will result in many electrons in bound state around these atoms, whether these bond to neighbouring hydrogen atoms or not. Assuming a sufficiently low density of the added atoms, the only pressure is clearly still due to the free electrons, not the bound electrons. What is unclear to me is whether the bound electrons reduce the effective available volume available to the free electrons, thus making a change to a formula that uses only free electron density, defined in terms of total volume. This is what I was referring to. For the purposes of this article, the Griffiths statement should be adequate. JDoolin's analysis above seems correct (assuming all mass arises from nucleons). It shows that the current formula in the article needs change to even have a chance of being correct, and that it is the number of free electrons per unit volume that is what we should be using without reference to total density. —Quondum 17:56, 23 November 2013 (UTC)[reply]

Griffiths has a problem in the same section, to show how 3/5 of bulk modulus of a copper is close to the degeneracy pressure. Assuming I did the problem correctly, if 2 free electrons per copper atom are assumed, the relation becomes almost exact. I wonder if in the Chandrasekhar limit, ALL the available electrons are contributing to the degeneracy pressure, and when that pressure is exceeded by external compression, Carbon detonation occurs. — Preceding unsigned comment added by JDoolin (talk • contribs) 18:07, 23 November 2013 (UTC)[reply]

Griffiths's point is probably to convince people that degeneracy pressure is a significant factor in the structure of many metals. In covalent bonding, e.g. diamond, the same momentum transfer via electrons in the lowest available energy state will still be a significant pressure mechanism, but with the difference that the confinement volume is now determined by the potential well of the orbital; it would be less than the total volume.

With reference to your statement, I gather from carbon detonation that a white dwarf remains in pressure equilibrium (between degeneracy pressure and gravity) during the detonation; this equilibrium is barely affected by the very energetic detonation, which says that the degeneracy pressure is not exceeded by external pressure and is still strongly stable at the time that the atomic nuclei reach a fusion threshold. And yes, I expect that all the electrons in a white dwarf have become valence electrons; at the high densities, the orbital overlap and closely spaced nuclei have essentially reduced the bonding energy, probably creating a metallic plasma in which all electrons are valence electrons.

The Chandrasekhar limit will correspond to substantially higher density, where compression instability occurs: where an increase of density increases the compressive pressure due to gravity faster than it increases the degeneracy pressure. You will see that the density climbs rather sharply just before the limit, but that the equilibrium remains stable up to the limit. —Quondum 19:40, 23 November 2013 (UTC)[reply]

This appears to say that the Chandrasekhar limit is an environment that creates an infinite density object, where I was under the impression that the Chandrasekhar limit was a much lower density environment where Type 1a supernova occurred. So I was under the impression that the Chandrasekhar limit was a non-relativistic limit. What is a relativistic fermi gas? If you have the radius going down to zero, that means that the ground state particles would have zero wavelength. Is this relevant--that each time you produce enough energy to excite the last particle in the ground state, it changes the minimum deBroglie wavelength, and allows the body to shrink? — Preceding unsigned comment added by JDoolin (talk • contribs) 17:41, 24 November 2013 (UTC)[reply]

Not necessarily of infinite density, even if the graph does appear to rapidly approach zero radius. Nevertheless, it should still be nonzero at the Chandrasekhar limit, i.e., if you magnify the graph, the curve should end away from the axis. It is determined by total mass and number of free electrons, and so technically has nothing to do with supernova mechanisms. Various nuclear reactions presumably occur at points along the density scale. The nonrelativistic Fermi gas is evidently a Newtonian extrapolation of the low-pressure graph, whereas the relativistic graph is corrected for special relativity (see Degenerate matter#Concept). Basically, just ignore the non-relativistic extrapolation as inaccurate. A relativistic Fermi gas is a Fermi gas in which the highest occupied energy levels imply energies and hence momentum spreads associated with velocities approaching the speed of light. This apparently leads to the pressure increasing less rapidly with electron density, hence the difference. The "minimum" de Broglie wavelength is not really applicable: as the body is compressed, higher energy levels get occupied, reducing the smallest wavelength. When more energy is generated through gravitational collapse than is taken by squeezing electrons into higher energy levels, the configuration is unstable, and the Chandrasekhar limit has been reached.

Would you like to try your hand at inserting Griffiths's version, noting that this is only valid in nonrelativistic contexts (low Fermi energy)? —Quondum 20:20, 24 November 2013 (UTC)[reply]

I added a couple of lines, with the alternate equation, and a brief note on how this was derived. I'm not exactly clear on where the relativistic limit comes into play, but I left the statement that was already there. — Preceding unsigned comment added by JDoolin (talk • contribs) 23:54, 24 November 2013 (UTC)[reply]

Thanks for doing that. I've now removed the previous formula due to its failings, leaving only this simpler, more correct formula. —Quondum 02:02, 25 November 2013 (UTC)[reply]

I just found out, all neutron stars in binary systems have also been found to be right around 1.4 solar masses. I had been under the impression that a Type 1a supernova leaves nothing behind, but it sounds like it leaves behind a neutron star. JDoolin (talk) 14:53, 26 November 2013 (UTC)[reply]

Type Ia supernova seems to suggest that a neutron star is the result if the white dwarf does not become a Type Ia supernova, which (reading between the lines) leaves nothing behind. The difference in evolution is determined by the chemical composition of the white dwarf. —Quondum 15:35, 26 November 2013 (UTC)[reply]

It occurs to me that the total gravitational binding energy might be calculated from a sum

\Sigma G{\frac {MM}{r}}

representing the energy of chopping the star in two repeatedly. One could use such a formula to estimate how many times a white dwarf could potentially be ripped in two by an energy of

10^{44}

Joules. Or, by using some different assumptions, an integral could figure the sum of energies of several differential masses being cast off from a central body .

E={\frac {G}{r}}\int MdM

JDoolin (talk) 15:59, 28 November 2013 (UTC)[reply]

The first sum yields

{\frac {GM^{2}}{R}}\Sigma \left({\frac {2*.5*.5}{\sqrt[{3}]{0.5}}}\right)^{n}

. The ratio per split is less than one, so the infinite series of splits converges at finite energy. Using M = 1.44 solar masses, and R = 0.01 Solar radii, the series converges around

2.15\times 10^{44}

Joules. So if only gravitational potential were involved, this would be enough to atomize the star, leaving nothing behind. So that still leaves somewhat of a mystery of why neutron stars are almost exactly the same size. JDoolin (talk) 20:14, 28 November 2013 (UTC)[reply]

Mechanism of collapse

When the collapse occurs (because heat is no longer being generated by fusion and electron degeneracy pressure is insufficient to counteract gravity), most of the electrons will be elevated to such high energy levels that it is advantageous to combine with a proton and form a neutron and electron-neutrino. This burst of electron-neutrinos blows away the outer layers of the star in a supernova and leaves behind a neutron star. The neutron star may have enough neutron degeneracy pressure to avoid further collapse or, perhaps after cooling down a bit, it may collapse into a black hole. JRSpriggs (talk) 02:20, 26 November 2013 (UTC)[reply]

Which is really saying that the following statement in the article may be inaccurate: "For stars that are sufficiently large, electron degeneracy pressure is not sufficient to prevent the collapse of a star and a neutron star is formed." The point of electron–proton combination may occur before or after the balance between electron degeneracy pressure versus gravitational force becomes unstable. It would be interesting to discover which happens first. —Quondum 07:01, 26 November 2013 (UTC)[reply]

Incorrect citations

The lead paragraph says In 1967, Freeman Dyson showed that solid matter is stabilized by quantum degeneracy pressure rather than electrostatic repulsion., with three references, none of which use the words "quantum degeneracy". Neither does a review ten years later: Lieb, Elliott H. "The stability of matter." Reviews of Modern Physics 48.4 (1976): 553. Johnjbarton (talk) 14:26, 26 July 2023 (UTC)[reply]

Griffith's textbook cited here says that it is "sometimes called degeneracy pressure, although 'exclusion pressure' might be better term." Jähmefyysikko (talk) 15:02, 26 July 2023 (UTC)[reply]

Thanks. Neither of the 2 remaining references use "electron degeneracy pressure". Ashcroft and Mermin say: "It is absurd to expect the free electron gas pressure alone to completely determine the resistance of a metal to compression, but ... at least as important as other effects". page 40. The context here is Sommerfeld metal theory, free electron gas. Johnjbarton (talk) 15:13, 26 July 2023 (UTC)[reply]

Do you think it is a problem that the references do not use this specific terminology? Although I haven't read most of the references carefully, to me it seems clear that they are talking about the same thing. Lieb uses the term Fermi pressure in his Rev. Mod. Phys. to refer to this effect. Jähmefyysikko (talk) 18:29, 26 July 2023 (UTC)[reply]

I think you are correct but I also think that a sentence with a reference should be expressed in the terms used by the authors.

The electron degeneracy pressure seems to trace to work by Subrahmanyan Chandrasekhar. In his paper he calls the "degenerate case" requires electron density

>6\times 10^{29}

per cc, applicable to white dwarf stars. Is this degenerate matter in the same realm as atoms treated by Lenard and Dyson? I suspect from Leib's Bulletin article that the answer is yes, but he never talks about a "degenerate case" or degeneracy directly.

I end up with many more questions that it seems to me our articles ought to answer. Who created the term "electron degeneracy pressure". Is it in fact the same as "Fermi pressure"? Is the model a Fermi gas? A result of Thomas-Fermi model? Johnjbarton (talk) 23:17, 26 July 2023 (UTC)[reply]

Looks like Shapiro and Teukolsky 1983 Black Holes, White Dwarfs, and Neutron Stars: The Physics of Compact Objects is the right place to look.

Koester, Detlev, and Ganesar Chanmugam. "Physics of white dwarf stars." Reports on Progress in Physics 53.7 (1990): 837. is a review that talks about degeneracy pressure. Johnjbarton (talk) 20:04, 28 July 2023 (UTC)[reply]

I added the Koester ref and removed the content related to Dyson and Lieb (which is not cited by Koester FWIW).

I'm not against these references coming back in a separate section that clearly explains how they are related to the topic.

Resolved

Johnjbarton (talk) 23:20, 28 July 2023 (UTC)[reply]

Renaming and repurposing this article?

How about we hijack this article? What if we redirect "electron degeneracy pressure" to degenerate matter and call the content here "Quantum stability of matter"? The same references but building out their point of view rather than using them to kinda sorta explain electron degeneracy pressure.

I'll bring the idea to Physics Talk. Johnjbarton (talk) 18:54, 27 July 2023 (UTC)[reply]

I again weakly oppose this change, "quantum stability of matter" is not a very popularly searched term, contrary to degeneracy pressure that is studied in many physics textbooks. Please bring this and the discussion in degenerate matter to the WikiProject to have other user perspectives. Maybe both articles can be merged into a single "degeneracy pressure" article but again effort is needed so we need to find consensus here.--ReyHahn (talk) 10:54, 28 July 2023 (UTC)[reply]

We also have the article stability of matter, which takes the quantum aspect into account. What would be the difference in the scope of the articles? Jähmefyysikko (talk) 11:24, 28 July 2023 (UTC)[reply]

Oh, hah, of course we do! Thanks. Johnjbarton (talk) 13:57, 28 July 2023 (UTC)[reply]

"degeneracy pressure that is studied in many physics textbooks" Yes I know that is what Wikipedia implies, but if so why is it that this article does not have any references? Why is there no primary or review articles on "electron degeneracy pressure"? Science isn't about popular searches. Johnjbarton (talk) 14:01, 28 July 2023 (UTC)[reply]

Because anything about it is so old and established. Wikipedia is not about primary sources is about secondary or tertiary sources, and textbooks are the best for that. Do you think stability of matter is any better? I just noticed that Stability of matter exists and its terrible, I will take a look at that can of worms.--ReyHahn (talk) 17:16, 28 July 2023 (UTC)[reply]

The references in stability of matter actually discuss stability of matter. By consulting those references we can improve the article. The references for electron degeneracy pressure here actually discuss stability of matter and white dwarf stars, so we have no way of checking anything said here. The article is essential unreferenced or worse, because the references are off-topic.

At this point I believe that the models discussed under stability of matter -- incompressible waves (uncertainty principle) and Pauli-exclusion (Fermi-Dirac statistics) -- explain the stability of white dwarf stars. The references say exactly that. They also explain "degeneracy pressure", a T->0 pressure effect. But "degeneracy pressure" does not explain stability of white dwarf stars. These are two results, one does not explain the other. Johnjbarton (talk) 18:56, 28 July 2023 (UTC)[reply]

Well I still don't understand what "degeneracy pressure" means, but I removed the refs that do not talk about it and added one that (almost) does. I'll be looking for better explanations, or at least ones I can understand. In the meantime I'll withdraw my suggestion.

Resolved

Johnjbarton (talk) 23:17, 28 July 2023 (UTC)[reply]

Completely new qualitative description added.

Ok I sorted out my thinking and wrote a new description of this phenomena at the level I think appropriate.

Please review while I look for references. Johnjbarton (talk) 14:40, 29 July 2023 (UTC)[reply]

I rewrote the two paragraph discussing the origin of the effect quite extensively, trying to make it simpler. For qualitative description, I believe it is enough to refer to the nonvanishing kinetic energy of the particles, this is what distinguishes quantum mechanics from classical one at T=0, and it also gives a hand-waving, but quite intuitive, connection to pressure. Quantitative description is then given in the body. The part about infinite potential well I did not find from the references. That sounds dubious to me, as it would mean that the system becomes an insulator at low temperatures. Jähmefyysikko (talk) 09:20, 30 July 2023 (UTC)[reply]

I understand trying to make it simple but:

"The strong electromagnetic forces between the negatively charged electrons are balanced by the positive nuclei, and can be neglected in simple models."

How can something "strong" also be "neglected"? It is true that the form of the potential is -- surprisingly -- unimportant so this is in the right direction. But is not about "simple". We need to figure out how so say this.

If many electrons are confined to a small volume, the highest occupied state has a large kinetic energy, and a large pressure is exerted.

I don't believe this is correct. The change in energy with volume comes from all levels, not the just the highest level.

I wonder if we can avoid "pressure is exerted" because it seems too imply "pushing" at least to me, whereas in the metal case the pressure is "pulling".

The infinite potential is discussed in Fermi's original (though he used harmonic in his analysis). On the other hand he points out repeatedly that the potential is completely irrelevant. Confinement and Pauli are all that matter.

I would like to have a more complex but not mathematical description, either here or in degenerate matter. My original problem with both pages is that the physics discussion was incomplete. Having hand waving followed by a math derivation which no one reads is the combo I want to get away from. Johnjbarton (talk) 14:34, 30 July 2023 (UTC)[reply]

Perhaps can is a too strong word. They are neglected would be more accurate. For metals in low temperatures, it is not usually a good approximation, as there tends to be some phase transition or another. It is indeed tricky.

You are of course right about 'highest level'. I simply wanted to avoid speaking about averages, but perhaps it is necessary.

I need to understand the details of the negative pressure better to take a stand on pushing and pulling. Did you have some reference which describes the mechanism in metals?

There is definitely is room in the article for more nuanced discussion, both within the Fermi gas and with respect to realistic systems, white dwarf stars and metals. Jähmefyysikko (talk) 15:16, 30 July 2023 (UTC)[reply]

So far I only find references that compute a "pressure" but don't explain what it means. That was part of my confusion when I first read this article.

Equilibrium requires that the energy of solid metal must increase if you reduce or increase the interatomic distances. If you take away the electrons you have positive ion cores. Clearly the the energy due the ion cores will increase with decreasing distance and decrease with increasing distance. What then of the electrons? They must have the opposite behavior.

The situation in metals is opposite the white dwarf star case, because in the star the nuclei have switched from electromagnetic repulsion to gravitational attraction, while the electrons have gone from enjoying freedom relative to atomic distances to compression to near nuclear distances.

It is exactly this kind of material that I would like to include. Maybe we can have a "Qualitative physics" or "Qualitative models" and keep a short lead?

By the way, I now realize why the Lieb/Dyson/Lenard work is confusing in the context of electron degeneracy pressure. They are computing bulk stability: why the atoms or solids don't collapse on the nuclei. Confinement + Pauli explains this broadly, for solids and dwarf stars. They are not computing bonding, that is they are not computing the energy of solids relative to atoms. Johnjbarton (talk) 21:10, 30 July 2023 (UTC)[reply]

Well as soon as I said that I found a reference that claims the degeneracy pressure prevents metallic compression.

Taylor, John Robert; Zafiratos, Chris D. (1991). Modern physics for scientists and engineers. Englewood Cliffs, N.J: Prentice Hall. ISBN 978-0-13-589789-8

https://openlibrary.org/works/OL3232611W/Modern_physics_for_scientists_and_engineers?edition=key%3A/books/OL1866571M

Section 13.9 "Degeneracy pressure".

However this reference makes me more convinced that this makes no sense. If degeneracy pressure explains incompressibility of metals, what explains the incompressibility of say rocks?

IMO this all comes down to the confusion of "pressure" in what is really energy vs radius analysis. Johnjbarton (talk) 21:35, 30 July 2023 (UTC)[reply]

At least in ionic compounds, the Coulomb force provides a long-range attraction, and the degeneracy pressure provides a short range repulsion, making the ions into largely incompressible spheres. It seems that the size and compressibility of ionic compounds is determined largely by the degeneracy pressure. My guess is that it is important for other bond types also. Ashcroft and Mermin have a good discussion about the physics of chemical bonding. Jähmefyysikko (talk) 11:45, 31 July 2023 (UTC)[reply]

Yes, in the limit of a simple ionic model electron repulsion including Pauli principle balances Coulomb attraction. (There is no mention of degeneracy pressure in ionic compounds article however.)

The case I neglected for metals is: the degeneracy pressure is positive and stronger than the core repulsion and the Coulomb attraction -- unexplained in the free electron model -- is the balancing force.

Ashcroft and Mermin:

In Chapter 2 we noted that the pressure of a free electron gas at the density of the alkali metals gave their observed compressibilities to within a factor of two or less. To go from this to a crude theory of cohesion in the alkali metals, we must add to the electron gas kinetic energy the total electrostatic potential energy. This contains, among other things, the energy of attraction between the positively charged ions and the negatively charged electron gas, without which the metal would not be bound at all.

I will repair the article. Johnjbarton (talk) 17:34, 31 July 2023 (UTC)[reply]

Two meanings of degeneracy

@ReyHahn, what do you mean by "of course electrons are degenerate"? For example, in Fermi gas, there is some degree of degeneracy due to symmetry, but most of the levels are not degenerate. Level degeneracy is anyway irrelevant for the degeneracy pressure. I did not find the section non-sensical. Jähmefyysikko (talk) 11:26, 31 July 2023 (UTC)[reply]

@Jähmefyysikko: I have not been able to follow all conversations so excuse me if I have missed something. But as you point out there is a degree of degeneracy, hence its name. If it is irrelevant or not the paragraph removed was very confusing about it. --ReyHahn (talk) 11:46, 31 July 2023 (UTC)[reply]

The name does not refer to level degeneracy at all. It comes from 'degenerate electron gas', which is simply electron gas at low temperature or high pressure. Degenerate in this context refers to a limiting case. Compare to Degeneracy (mathematics). Jähmefyysikko (talk) 11:55, 31 July 2023 (UTC)[reply]

The paragraph had an online text book reference. I had been looking for this kind of reference for some time. This is another point that confused me about the original article: what is the meaning of "degeneracy" in this context? The typical discussions about degeneracy are 1) the levels are energetically identical 2) they mix and split. None of that appears in discussion of degeneracy pressure. Eventually I worked out that "degeneration" was used by Nerst before Schrodinger's equation, its term derived from gas behavior as T->0.

Anyway I included this paragraph in the History section of degenerate matter. If you have concerns or suggestions on how to improve it please check there. Johnjbarton (talk) 13:31, 31 July 2023 (UTC)[reply]

Degeneracy pressure is caused by the Heisenberg uncertainty principle

The article states 'it is not possible for *all* the electrons to have zero kinetic energy'.

Isn't it more accurate to say that *no* electrons can have zero kinetic energy? (due to the uncertainty principle)

_____

It also says 'the confinement makes the allowed energy levels quantized'.

Doesn't quantum mechanics make the allowed energy levels quantized, regardless of confinement?

_____

It then says 'and the electrons fill them [the energy states] from the bottom upwards. If many electrons are confined to a small volume, on average the electrons have a large kinetic energy, and a large pressure is exerted'.

I think it's explained better with reference to the Hiesenberg uncertainty principle, as it is this source (http://physics.mq.edu.au/~orsola/MQTeaching/ASTR377/RydenPetersonStellarRemnants.pdf / https://assets.cambridge.org/97811088/31956/frontmatter/9781108831956_frontmatter.pdf), which states:

'... electrons become degenerate when they are packed closely enough that the Pauli exclusion principle produces an additional form of pressure to keep them apart*. The electron degeneracy pressure is a consequence of the Heisenberg uncertainty principle, which states that you can't simultaneously specify the position x and momentum p of a particle to arbitrary accuracy. There is always an uncertainty in each such that ΔxΔp ≥ ħ,

where ħ is the reduced Planck constant... Suppose that the degenerate electrons have a number density n_e. In their cramped conditions, each electron is confined to a volume V~n_e^-1. Thus the location of each electron is determined with an uncertainty Δx ~ V^1/3 ~ n_e^-1/3. From the uncertainty principle, the minimum uncertainty in the electron momentum is Δp ~ ħ/Δx ~ ħn_e^1/3.

If the electrons are nonrelativistic,

Δv = Δp/Δm_e ~ ħn_e^1/3/m_e,

where m_e is the mas of the electron.

Thanks to the uncertainty principle, degenerate electrons are zipping around with a speed v_e ∝ n_e^1/3 regardless of how low the temperature drops. These "Heisenberg speeds" contribute to the pressure, just as the thermal speeds do. For ordinary thermal motions, the electron speeds are

v_th ~ (kT/m_e)^1/2,

and the pressure contributed by thermal motions of electrons is

P_th = n_ekT ~ n_em_ev_th².

By analogy, the "Heisenberg speeds" contribute a pressure

P_degen ~ n_em_e(Δv)² ~ n_em_e(ħn_e^1/3/m_e)² ~ ħ².n_e^5/3/m_e.

We label a population of electrons as "degenerate" when P_degen < P_th.

*Note: Electrons, neutrons, and protons are all fermions, particles with half-integral spin, to which the Pauli exclusion principle applies. Photons are examples of bosons, particles with integral spin, to which the exclusion principle does not apply.' _____ @User:Johnjbarton, @User:Hellacioussatyr, I'd prefer someone else take on the task of incorporating this perspective into the article if it's agreed it should be edited. MathewMunro (talk) 15:27, 6 January 2024 (UTC)[reply]

Here is my understanding:

Kinetic energy is a system property, it depends upon relative velocities within the system. So one electron can not have kinetic energy (undefined).
Confinement typically called boundary conditions: ideally free particles in QM are not quantized in energy.
"electrons become degenerate when they are packed closely..." this phrase is not sensible. Electrons cannot become degenerate: that is what the Pauli principle says. (Well you can put two in a level and they will be approximately degenerate). Energy levels can be degenerate, meaning only that more than one level has the same energy (eigenvalue).

An explanation of degeneracy pressure based on the uncertainty principle is insufficient because bosons also obey the uncertainty principle.

The uncertainty principle above seems to be used in this way: when that the levels are filled according to Pauli exclusion, they have energy above the level they would have without Pauli. That energy can be expressed as a pressure using the uncertainty principle reasoning above.

So the uncertain principle isn't being use to "explain" the cause of the pressure, but rather to calculate the value caused. Johnjbarton (talk) 16:37, 6 January 2024 (UTC)[reply]

Yes, the uncertainty principle provides a heuristic way to get a value for the pressure given the more fundamental idea of Pauli exclusion, but it isn't the explanation itself. A full-blown derivation in a statistical physics book wouldn't even invoke the uncertainty principle here, most likely. XOR'easter (talk) 20:30, 6 January 2024 (UTC)[reply]

I was going to say that! Well, not exactly but not so far off. In any case, yes, it is Pauli exclusion. As for, farther up, about confinement and quantization: As the system gets larger, the quantization spacing decreases. So, in the infinite (unconfined) limit, quantization goes away. Gah4 (talk) 23:27, 6 January 2024 (UTC)[reply]

@User:Johnjbarton, you wrote: 'An explanation of degeneracy pressure based on the uncertainty principle is insufficient because bosons also obey the uncertainty principle.'

My understanding is that at white-dwarf densities, the uncertainty principle/Pauli exclusion principle for bosons doesn't contribute significantly to pressure or kinetic energy due to bosons having a much shorter matter-wavelength than electrons. MathewMunro (talk) 03:20, 7 January 2024 (UTC)[reply]