In Landau's Statistical Physics's analysis of solids, he begins with the remark that solids are caracterized by their atoms' small oscillations about equilibrium positions. However, he states that these positions' configurations must be "distinguished from all other configurations and must therefore be regular", or crystalline. This surprised me a bit, because I would expect the equilibrium state of a solid to be less ordered (that is, I wouldnt think a regular configuration is the most entropic one). Additionally, I don't understand his argument very well. Would someone cast some light on this?
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I wouldn't think a regular configuration is the most entropic one
It is a common mistake to believe that a regular configuration corresponds to a low entropy value.
First, thermodynamic and statistical mechanics entropies are not properties of a single configuration but of a macrostate, i.e., of all the microstates corresponding to a fixed set of thermodynamic variables and a given macroscopic order parameter. Therefore, the meaningful question is why the ordered configurations corresponding to a crystalline structure should have higher entropy than disordered configurations.
The reason is the presence of a short-range, strongly repulsive interaction. When the interaction effect is important, there are many more configurations in the neighborhoods of a crystalline spatial configuration than those of a disordered configuration. Therefore, the entropy per particle of the ordered system is higher than that of a disordered system.
We can intuitively understand such an issue by considering what happens when we fill a suitcase. If we have few objects, the most frequent configurations will correspond to the disordered positions of those objects. However, if we have many objects, we have to put them close to the ordered configurations, or we do not fit them in the suitcase. I.e., a disordered macrostate may correspond to no microstate, while an ordered macrostate allows the presence of many microstates.
answered Jun 14 at 13:19
GiorgioP-DoomsdayClockIsAt-90GiorgioP-DoomsdayClockIsAt-90
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1$\begingroup$ Could you elaborate on "...there are many more configurations in the neighborhoods of a crystalline spatial configuration than those of a disordered configuration"? This seems to be the crux of my question. Additionally, I did not understand your suitcase analogy. $\endgroup$ Commented Jun 14 at 14:01
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3$\begingroup$ The suitcase analogy is that sometimes entropy is higher simply by virtue of a state being possible compared to hypothetical alternatives. In certain crystal systems there is no sensible way to form a disordered lattice in the first place, and so the crystal lattice is, by default, the highest in entropy simply because it actually exists. $\endgroup$ Commented Jun 14 at 15:11
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2$\begingroup$ @LourencoEntrudo The suitcase analogy corresponds to the extreme case where the overlap of objects would correspond to an infinite cost in energy. Since configurations with overlap are forbidden, too much disorder would leave no room for fitting the suitcase without overlap. I.e., no or very few microscopic configurations would be allowed. More ordered average positions allow more possible configurations and consequently correspond to higher entropy (according to Statistical Mechanics). $\endgroup$ Commented Jun 14 at 15:38
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$\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 I think I see your point regarding the suitcase analogy. But then, if I were to have a certain lattice, and displace one of its points by a certain ammount so that the structure becomes irregular, I have a bit of difficulty seeing how that has lowered the entropy. Why has the number of microstates decreased? Is it because there is now "less spacing" between the displaced point and its neighbours and as such only more specific configurations of atoms can correspond to this macrostate? If that is the reason, it's still a little fuzzy to me $\endgroup$ Commented Jun 15 at 17:45
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1$\begingroup$ @LourencoEntrudo if you displace one-quarter of the regular lattice positions randomly, part of the remaining regular positions would correspond to a situation where less free space is available. Of course, this is a handwaving argument and not a proof. Proof comes from the comparison of the entropy per particle of a crystalline solid and a fluid of hard spheres at the melting point. The fluid has a lower enropy than the solid. See for instance sciencedirect.com/science/article/abs/pii/S0378437198005019 $\endgroup$ Commented Jun 16 at 16:50
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I think the issue lies with the definition of a solid. If you define a solid to be a crystalline material, all solids will be crystalline materials. However, if you define a solid via its elastic properties (such as non-zero shear modulus) you will find a large variety of solids everywhere around you that are not crystalline, ranging from glass to plastics or even wood. These are examples of disordered solids that do not have periodic structure, and understanding their properties is a very active field of research. Even more generally, there is a class of materials called viscoelastic, which have properties of both solids and liquids!
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3$\begingroup$ Landau's definition of solid is that of a collection of atoms that oscillate about equilibrium position. The question is not that all solids are crystalline, but that all solids in equilibrium are crystalline. Landau also comments on amorphous solids, but remarks that such solids are "thermodynamically metastable and must ultimately become crystalline", even though he points out that their relaxation times are so long that for all practical purposes they can he considered in equilibrium. My question still stands as to why that is not a real equilibrium. $\endgroup$ Commented Jun 14 at 14:59