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In a solid crystal, for example, copper, if we treat the electrons as a free electron gas we can obtain that the pressure exerted by the gas at absolute zero is about $10^5$ atm or $10$ GPa. What balances the electron degeneracy pressure in a solid?

Furthermore, if we apply external pressure greater than $10$ GPa, sometimes it will only lead to a phase transition rather than destroying the crystal. What is the force that balances the external pressure? Is it still the electron degeneracy pressure that balances the external pressure?

Paul T.
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Yuan Fang
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    atmospheric pressure is about $10^5$Pa – Yuan Fang Mar 30 '20 at 11:03
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    The obvious answer is the electric attraction between electrons and nuclei, as otherwise electrons would repel via electric forces and prefer to be far apart to minimize degeneracy pressure. Are you looking for something else? – KF Gauss Mar 30 '20 at 11:42
  • Also, to whoever down-voted. Its a bit much, don't you think? I mean, nothing seems to bring out the passion to press the down-vote button (and there is quite a lot of down-vote worthy content on this site. There is a lot of useful stuff too, which is why I keep coming back). But anyway. – insomniac Mar 30 '20 at 12:28
  • For the record (really, at this point, not much else to do) : 2nd downvote. I dislike controversies. They lead to hairloss. So I'll take my leave. My comments are left to read below. – insomniac Mar 30 '20 at 13:34
  • Also, for the record : @JonCuster's comment deserves a reply . I quote "...Considering electrons as a fluid really doesn't get you anywhere except in a very hand waving way..." I would suggest that the author of this comment look up Fermi-liquid theory on Google scholar. It is pretty big (at least, it provides my bread). Anyway. Point is that for phenomena at the lattice-scale, it is (as I have pointed out in several comments below), the tight-binding model is the appropriate description. This statement is absolutely true. – insomniac Mar 30 '20 at 13:38
  • However, at the length scale at which ANY fluidic analogy is valid, There Is No Lattice. This is the modern, emergent picture, that followed the works of Kadanoff, Wilson, etc, that depending on the length scale we are describing, we have an effective theory appropriate to that length scale. – insomniac Mar 30 '20 at 13:41
  • OP's question, as I have said several times below, rests on an analogy between an electron fluid, and what we understand with our classical intuitions as fluid behaviour (which, in addition to other things, is fluid in a macroscopic way). Explaining this conundrum with any reference to the underlying lattice is, I repeat, misleading. – insomniac Mar 30 '20 at 13:43
  • Finally, reading makes us all wiser : May I suggest (IMHO) two very nice articles on what I understand is the modern view of understanding how fluidic behavior emerges from electrons on a lattice. $(1)$ https://arxiv.org/abs/hep-th/9210046 (Effective Field theory and Fermi surface, Polchinski) and $(2)$ https://arxiv.org/abs/cond-mat/9307009 (RG approach to interacting fermions, Shankar). Have a nice day. – insomniac Mar 30 '20 at 13:52
  • @insomniac, I think you are really misguided here, none of those (excellent) works are agreeing with you. If you treat the electrons as a fluid, you must treat the ionic background as a counterbalancing positively charged fluid to get overall neutrality. This is the jellium model. You seem to be thinking that you can think of the electron as a fluid and somehow tight-binding lets you forget about the positively charged background completely. – KF Gauss Mar 30 '20 at 14:27
  • One day, a band-theorist met one who uses boltzmann equations to calculate transport coefficients, and both wondered How on earth are we both part of the same umbrella called Condensed Matter Physics? – insomniac Mar 30 '20 at 14:31
  • I mean, I do not doubt your wonderment at having self-evident (to you) truths questioned. It mirrors my own. Its called Perspective – insomniac Mar 30 '20 at 14:33
  • I don't know who you are talking to anymore, but as far as I can tell your answer runs into a contradiction when you demand it satisfies charge neutrality – KF Gauss Mar 30 '20 at 14:41
  • Let us continue this discussion in chat. – KF Gauss Mar 30 '20 at 14:46
  • Sorry, no. I have said what I had to say (including on HBar, the main chat room). I am usually not here to chat, and besides, I feel any intellectual debate, if it cannot be settled here, needs to be left to the audience. Neither of us knows everything, though the certainty with which I have been downvoted shows that someone here is less certain about their own understanding than at least I am. Good for them. Nevertheless, I do not respond to meanness very well. These answers will stay in place. Rest is upto others. It was a pleasure talking to you. See you. – insomniac Mar 30 '20 at 14:52
  • @YuanFang Yes, should have gone to SpecSavers – jim Mar 31 '20 at 12:23

3 Answers3

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The obvious answer here is that the degeneracy pressure is balanced by electrostatic attraction forces due to the positively charged nuclei. This is same idea for any bound fermion system, the degeneracy pressure that makes fermions stick together always is balanced by whatever is attracting them in the first place.

For protons and neutrons this is the (residual) strong force.

For electrons it is the EM force.

For neutron stars it is gravity.

KF Gauss
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  • Then what to balance the externally exerted pressure? – Yuan Fang Mar 30 '20 at 12:22
  • How exactly is the fact that the electron fluid does not flow out of the crystal (which is what, as I understand, OP's concern) related to the "electrostatic attraction of the positively charged nuclei", when at the length scales at which we observe said fluid, the details of the lattice are completely averaged out? – insomniac Mar 30 '20 at 13:19
  • The point is this: Yes, there is an underlying lattice of positive ions with its attractive potential. We write the tight-binding hamiltonian (which is the valid description of the system at the length scale of the lattice constant) using this fact. However, at this scale, There Is No Electron Fluid. The scale at which the fluid behaviour emerges is much larger than this lattice-scale. The Fluid Experiences No attraction from an underlying lattice of ions. – insomniac Mar 30 '20 at 13:27
  • Hence, to explain the fact that the electron fluid does not flow out on account of its degeneracy pressure using the attraction of the underlying lattice of ions is simplistic at best, and misleading at worst. But, that is just an opinion. I always have an open mind. – insomniac Mar 30 '20 at 13:29
  • @YuanFang it is the same as the case of any equilibrium state in mechanics. Total force is zero so that $P_{Coul.}(r) + P_{Degen.}(r) + P_{ext.}=0$. If the external pressure is zero we have $P_{Coul.}(r) = -P_{Degen.}(r)$, otherwise if it is not then we get $P_{Coul.}(r) + P_{Degen.}(r) = - P_{ext.}$. One must solve for $r$ to get the density either way – KF Gauss Mar 30 '20 at 14:03
  • @insomniac If you have a charge neutral system it is not possible to have the lattice completely screened and leftover electrons to form an "electron fluid" as you state. You are double counting charge and/or are using a definition for the electron fluid that is not the normal meaning in the solid state. – KF Gauss Mar 30 '20 at 14:06
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"...What balances the electron degeneracy pressure in a solid?..."

The problem with this statement is that it assumes that the "electron gas" is exactly analogous to, say, Helium in a balloon. We need to (in my opinion) find slightly different ways of thinking of Pressure of a solid, as against pressure of a fluid.

Helium in a balloon is a fluid. What that means is that if we say that the Pressure of a Helium balloon in equilibrium with its surroundings is $P$, then, we can conclude that the pressure in the "atmosphere" surrounding the balloon must also be $P$. The fact that we are dealing with something that is able to flow implies that unless balanced by an equal and opposite pressure, the fluid will "spread" till its pressure equals that of its surroundings.

Let us now compare this with the situation you describe. That is, a lattice (a rigid solid, definitely cannot flow); On this background, a "sea" of nearly free electrons. Obviously, this whole setup is in equilibrium as we see it, notwithstanding any Pressure it has.

Note that we may bring in atmospheric pressure into this, but I think that is misleading. I am pretty sure a piece of metal (which is an example of a crystalline solid with a degenerate electron gas) won't disintegrate if it is placed in a vacuum.

So how should we think of pressure in a solid? Strictly speaking, using the thermodynamic relation $dE = -PdV$ ; ie., Pressure is the rise in Internal energy of the solid per unit volume of compression. If we compress/stretch the solid, then energy rises/falls by $P$ per unit volume. Compare this with the physical implication of pressure of a fluid above. Solids deform via intentionality. Fluids, in contrast, deform spontaneously, just because they can flow.

Seen in this way, it is obvious why the electron-gas should have a "Pressure". If we compress the system, we are changing the boundary conditions for the electronic states, (occupied by the electrons forming the gas) and hence its internal energy. In short, Pauli exclusion principle ; the energies of the levels change, but their occupation numbers can't (at absolute $0$), so the total energy changes.

Also, note that this is not the only source of "Pressure" in this system. Obviously, the rigid lattice has its own associated pressure as well. But none of this means that this needs to be balanced by external pressures in order for it to be in equilibrium.

Addendum, on nomenclature: The exchange of comments below also made me realize, that the nomenclature of electron gas or fluid may be confusing. An electron fluid is so called because it is fluid with respect to transport of energy, charge, spin, etc (whatever the electrons can transport). But an electron fluid is definitely not to be thought of as fluid with respect to mechanical deformations. If I had to say what the other answer is saying (from what I understand), is that the reason the electron fluid does not require an external pressure to hold it together is the attraction of the ionic lattice. Frankly, I find this a little confusing, because the fluidity of a so called electronic fluid is very much an emergent phenomenon specific to this special situation of electrons on a lattice. What I mean is, normally, one starts with a tight-binding model, then takes the long wavelength (low energy) limit ; then, a fluid emerges. I don't know how to make sense of this outside of this specific context.

So, the safest thing to say, it seems, is that the electron fluid is not fluid in a mechanical sense because it is tied to a mechanically rigid elastic medium (the lattice). How exactly it is tied is not at all a factor in the behaviour of the electron fluid. Remember, that the tying of free electrons to the lattice happens on the length scale of the lattice. The fluid emerges at a much larger lengthscale, by which point, the exact nature of the underlying rigid lattice is irrelevant. All that matters is the fluid is tied to some elastic solid, so its mechanical deformations are constrained to this unspecified background solid. That is all that is required to explain why the degeneracy pressure of the electron fluid is not to be thought of as something that needs to be counterbalanced by an equal and opposite pressure to prevent the electron fluid from flowing out.

Addendum (2): A comment points out below, the "nearly free electrons" is a poor approximation : It is more subtle that that ; In most normal metals, the nearly free electron model IS a very good model of the macroscopic properties of what we see (with renormalized parameters, of course). That is why the Boltzmann equation works so well in calculating transport coefficients in normal electron fluids. (leading to the Fermi - liquid theory). Of course, this fails sometimes, in quantum critical systems, which is when we have various kinds of Non-Fermi liquid behaviour. These phases are marked by the absence of quasiparticle excitations of any kind. Hence, it is objectively true in these situations that the model of nearly free electrons is a poor one to explain the macroscopic properties we see. But normally (in the aptly named normal fermi liquid phase) , the nearly free-electron model is an astonishingly good one (again, to explain macroscopic properties of the fluid)

insomniac
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    I don't think the solid vs fluid distinction is necessary here, electrons in liquid mercury have essentially the same degeneracy pressure as in a solid piece of copper. The point about compressibility is true in either case. – KF Gauss Mar 30 '20 at 11:59
  • But it is, right? The point you are making in your comment above is essentially that this electronic fluid is essentially tied to a rigid lattice of ions. THAT is the reason we do not need to worry about what will stop this electron fluid from spreading (which is what seems to be OP's concern, IMO of course) – insomniac Mar 30 '20 at 12:03
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    The reason we don't need to worry about spreading is the electrostatic Coulomb interaction, that's it. – KF Gauss Mar 30 '20 at 12:16
  • The 'nearly free electrons' is a poor statement - all that means in solid state physics is that energy is roughly equal to momentum squared. A real solid has band structure, and the point to remember is that all of those energy states are bound states of the crystal. Considering electrons as a fluid really doesn't get you anywhere except in a very hand waving way. – Jon Custer Mar 30 '20 at 13:27
  • @JonCuster: It is meant to mirror the wording of OP's question (v1). I quote : "...In a solid crystal, for example, copper, if we treat the electrons as a free electron gas we can obtain...". – insomniac Mar 30 '20 at 13:31
  • Which just underlines why that is a poor way to look at the situation. – Jon Custer Mar 30 '20 at 13:39
  • This answer presupposes its conclusion. The Fermi gas model assumes a genuine fluid. In reality, it doesn't expand simply because the degeneracy pressure is balanced by electrostatic attraction (outside the model's scope; recall $\rho_e$ is assumed). Mechanical compression causes volume reduction until the degeneracy pressure balances the applied force, which we observe as the bulk modulus $K=-\frac{dP}{d \ln{V}}$. – alexchandel Nov 03 '20 at 16:55
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tl;dr (rigidity vs positivity of background):

Several comments on this post have pointed out the background of positive ions forming the lattice. Few points about this:

$(1)$ At the scale at which we observe an electron liquid, this appears as a uniform positive background. In an isolated crystal (which is electrically neutral) this constrains the density of the electron fluid to be the density of the background. But this is a trivial fact that cannot really be used to explain anything. $(3)$

$(2)$ But first, let us reiterate that definitely, invoking "positively charged nuclei" to explain why the electron fluid does not flow out under the influence of its own pressure is a gross simplification (that may, as I have mentioned several times on this post, be very misleading). There are no discrete centers of positive charge (nuclei) that the electron fluid sees.

$(3)$ But might we at least use the fact of the uniform positive background to explain the fact that the electronic fluid does not flow out on account of its own pressure? Answer is again, no. For one, A Uniformly Charged Background Exerts no Forces (Obviously).

If that is not convincing, consider that a crystal that is Not Isolated. We are usually allowed to tune the chemical potential of the crystal as well. So we can create a situation where the (uniform background + electron fluid) system is definitely NOT neutral ; still, nothing flows out because of "excess pressure"

$(4)$ Why not? If we think in terms of what the other answer (and other comments) are suggesting, For each configuration of electrons, we would have to think of a source of equal and opposite pressure. The reason this seems tedious is that it is wrong. It seeks to explain an inherently quantum phenomenon in terms a classical toy model.

$(5)$ A normal metal (system in OP's question) (normal $\implies$ quasi-particles exist) is very much a quantum fluid. It has a degeneracy pressure, relating to the fact that (as I mentioned in my other answer): $(I)$ the quasi-particle excitation energies are related to the dimensions of the crystal and $(II)$ Pauli Exclusion principle

This does imply, as I referred to in my other answer, that we have an operational definition of $Pressure.$ However it does NOT imply that we need to be worried about the electron fluid spreading out unless counterbalanced by an equal and opposite pressure (to reiterate).

$(6)$ So what is the key operative reason why the electron fluid does not flow out? As we saw above, it is that the electron fluid is an emergent phenomenon on a Rigid background.

This seems to really be the theme of the whole discussion that has gone before. Rigidity of background vs Positivity of background.

I hope I have demonstrated that correct reason is the former.

insomniac
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