4 ms·

Wigner Crystals (2017)

45 points by leephillips 5y ago

c1ccccc1 5y ago
So the 2d crystals make sense to me, but how do the 3d crystals form? Charge is usually concentrates on the boundaries of a conductor, right? Shouldn't the same apply to an empty region of space?
- contravariant 5y ago
  Your objection applies equally to the 2D crystal. As far as I can tell the fact that charges reside on the boundary only really applies to induced charges. When the conductor itself has a net charge then you'd normally expect these to configure themselves so they're as far apart from each other as possible. If the conductor doesn't have any kind of weird shape this should normally result in a roughly uniform distribution of charge (if it has spikes then most charges will move there to be away from the bulk).
  - c1ccccc1 5y ago
    I think for my objection to apply in the 2D case, you'd need for the force law to be 1/r instead of 1/r^2. That would happen in a truly 2D world, but not in this case where we have charges confined to a 2D plane, but with 3D electric fields extending above and below that plane. I think charge still goes to the surface, even if not induced. A spherical conductor with a net charge would have a uniform distribution of charge on its surface, and 0 charge density inside the bulk. Any net charge density inside the conductor would create a diverging electric field around itself, which would cause a current to flow, dissipating that net charge in the process. The same argument should apply to a bunch of free electrons, shouldn't it?
    - contravariant 5y ago
      Ah you're right by definition a conductor does have a current provided there's any electrical field inside it (divergent or not). So indeed they can't have charges inside them in equilibrium. However if you induce an electrical field in a vacuum then no charges at all will flow because a vacuum is the perfect insulator. And even if there is nonzero electric field in and around an individual electron they won't move as long as things cancel out at their exact position. Really the theory of conductors and charge densities seems to break down somewhat once you get to the point of individual electrons near absolute zero in a vacuum. One way to put this to the test would be to charge a conductor to its absolute limit, basically removing all free charges from it and seeing if the rest will crystalize. This might require an impractical amount of energy.
- avsteele 5y ago
  The actual distribution will depend on the confining potential. The arrangement that minimizes the total potential energy of the system is just some regular (crystalline) structure. This is an ionic crystal in a harmonic potential well https://youtu.be/Pe3_IosfaTo https://youtu.be/Pe3_IosfaTo They are regularly spaced but taper to the top-bottom and foreword-backward because the 'k' is larger along those axis. It is weaker laterally so more fit along that dimension.
jhoechtl 5y ago
> I’d like to explain a conjecture about Wigner crystals, which we came up with in a discussion on Google+. There is a link to the Google+ discussion. No, you can't read what has been discussed there, another abandoned and folded Google service.
contravariant 5y ago
The author seems to get quite close to a proof that there must be 6 'defects' but then gives up for some reason. To get the result all you really need is the additional requirement that the vertices on the edge each have 4 neighbours. Once you have that then you can just glue it to its mirror image. This now functions as a triangulation of the sphere so it must have 12 more vertices with 5 neighbours than vertices with 7 neighbours, since it's composed of 2 equal halves each half must have exactly 6 more vertices with 5 neighbours. You can probably generalize this by letting vertices on the edge have one neighbour more or fewer as well, but things get a bit messy.
- kmill 5y ago
  The answer to the puzzle is in the comments on that page (Prof. Baez gives your argument in one of them, too). It would be interesting knowing why there are roughly six clusters of defects. The figure seems to technically have seven clusters, but it seems plausible that low-energy configurations tend to have at most six. One point of view is that in a triangulation, the degree d of a vertex corresponds to something like the curvature there, where the curvature is concentrated into a cone angle of 60 * d. The total cone angle defect (the sum of 360 - 60 * d over all vertices) is 720 degrees for a sphere, either by Euler characteristic or Gauss-Bonnet, so if the only vertex degrees you see are 5 and 6, there are exactly 12 vertices of degree 5. So, fancifully, it seems like the potential function makes the space "want" to be flat (i.e., have vertices of degree 6 for cone angle defect of 0), so it seems plausible that these defects, some kind of fundamental particle of curvature, would "want" to be concentrated in packets of total-defect-60-degrees, and that these would "want" to be relatively far apart from one another.
deleted 5y ago
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45 points by leephillips 5y ago

So the 2d crystals make sense to me, but how do the 3d crystals form? Charge is usually concentrates on the boundaries of a conductor, right? Shouldn't the same apply to an empty region of space?

contravariant 5y ago
Your objection applies equally to the 2D crystal. As far as I can tell the fact that charges reside on the boundary only really applies to induced charges. When the conductor itself has a net charge then you'd normally expect these to configure themselves so they're as far apart from each other as possible. If the conductor doesn't have any kind of weird shape this should normally result in a roughly uniform distribution of charge (if it has spikes then most charges will move there to be away from the bulk).
- c1ccccc1 5y ago
  I think for my objection to apply in the 2D case, you'd need for the force law to be 1/r instead of 1/r^2. That would happen in a truly 2D world, but not in this case where we have charges confined to a 2D plane, but with 3D electric fields extending above and below that plane. I think charge still goes to the surface, even if not induced. A spherical conductor with a net charge would have a uniform distribution of charge on its surface, and 0 charge density inside the bulk. Any net charge density inside the conductor would create a diverging electric field around itself, which would cause a current to flow, dissipating that net charge in the process. The same argument should apply to a bunch of free electrons, shouldn't it?
  - contravariant 5y ago
    Ah you're right by definition a conductor does have a current provided there's any electrical field inside it (divergent or not). So indeed they can't have charges inside them in equilibrium. However if you induce an electrical field in a vacuum then no charges at all will flow because a vacuum is the perfect insulator. And even if there is nonzero electric field in and around an individual electron they won't move as long as things cancel out at their exact position. Really the theory of conductors and charge densities seems to break down somewhat once you get to the point of individual electrons near absolute zero in a vacuum. One way to put this to the test would be to charge a conductor to its absolute limit, basically removing all free charges from it and seeing if the rest will crystalize. This might require an impractical amount of energy.
avsteele 5y ago
The actual distribution will depend on the confining potential. The arrangement that minimizes the total potential energy of the system is just some regular (crystalline) structure. This is an ionic crystal in a harmonic potential well https://youtu.be/Pe3_IosfaTo https://youtu.be/Pe3_IosfaTo They are regularly spaced but taper to the top-bottom and foreword-backward because the 'k' is larger along those axis. It is weaker laterally so more fit along that dimension.

jhoechtl 5y ago

> I’d like to explain a conjecture about Wigner crystals, which we came up with in a discussion on Google+. There is a link to the Google+ discussion. No, you can't read what has been discussed there, another abandoned and folded Google service.

contravariant 5y ago

The author seems to get quite close to a proof that there must be 6 'defects' but then gives up for some reason. To get the result all you really need is the additional requirement that the vertices on the edge each have 4 neighbours. Once you have that then you can just glue it to its mirror image. This now functions as a triangulation of the sphere so it must have 12 more vertices with 5 neighbours than vertices with 7 neighbours, since it's composed of 2 equal halves each half must have exactly 6 more vertices with 5 neighbours. You can probably generalize this by letting vertices on the edge have one neighbour more or fewer as well, but things get a bit messy.

kmill 5y ago
The answer to the puzzle is in the comments on that page (Prof. Baez gives your argument in one of them, too). It would be interesting knowing why there are roughly six clusters of defects. The figure seems to technically have seven clusters, but it seems plausible that low-energy configurations tend to have at most six. One point of view is that in a triangulation, the degree d of a vertex corresponds to something like the curvature there, where the curvature is concentrated into a cone angle of 60 * d. The total cone angle defect (the sum of 360 - 60 * d over all vertices) is 720 degrees for a sphere, either by Euler characteristic or Gauss-Bonnet, so if the only vertex degrees you see are 5 and 6, there are exactly 12 vertices of degree 5. So, fancifully, it seems like the potential function makes the space "want" to be flat (i.e., have vertices of degree 6 for cone angle defect of 0), so it seems plausible that these defects, some kind of fundamental particle of curvature, would "want" to be concentrated in packets of total-defect-60-degrees, and that these would "want" to be relatively far apart from one another.

deleted 5y ago

[deleted]