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This shows a 2-dimensional 'electron crystal', formed when very cold electrons at low density are confined to a disk. The vertices are the electrons. The edges are just to help you see the pattern.
To minimize energy, the pattern wants to be a lattice of equilateral triangles. But since such a lattice doesn’t fit neatly into a disk, there are also some red and blue ‘defects'.
5 triangles meet at each red vertex. 7 meet at each blue vertex.
Now it's time to make some conjectures!
(1/n)
Take N points 𝑥ᵢ on the unit disk, arranged so as to minimize the energy
I conjecture that when N is large, when we form the Delauney triangulation of these points, most points will be connected to 6 others, while those on the boundary will be connected to 4. There will also be 'red and blue defects' - points connected to 5 or 7 others.
And I conjecture that there will be 6 more red defects than blue defects!
(2/n)
I don't have much evidence for this conjecture. But I have a hand-wavy argument for why there should be 6 more red defects than blue ones. It's here:
https://johncarlosbaez.wordpress.com/2017/12/07/wigner-crystals/
The easiest way forward would be to take N points on the unit disk, and use some algorithm to move them around until their energy is approximately minimized, and then work out the Delauney triangulation:
https://en.wikipedia.org/wiki/Delaunay_triangulation
and count the defects of various kinds. If you give it a try, let me know!
(3/n, n = 3)
@johncarlosbaez Maybe Euler's formula can be used here? A related well-known fact about e.g. 3D meshes (triangulations of closed surfaces) is that on average every vertex has degree 6 (the exact value is 6-\chi(X)/12V or something like that).
@lisyarus - yes, if you read my article about this you'll see I use Euler's formula to heuristically argue for 6 more red defects than blue ones:
https://johncarlosbaez.wordpress.com/2017/12/07/wigner-crystals/
But it's not a proof!
@johncarlosbaez Ah, I see!
Now I'm also incredibly tempted to write some simulation that would do the thing with any number of points...
@johncarlosbaez Interestingly, I consistently get results that look more like this, i.e. density gets higher closer to the disk edge
@johncarlosbaez Oh, I've read the wiki article and they say they have an extra parabolic potential added to the system. Gonna try that as well.
@lisyarus - Very nice result! Sorry, I forgot to mention the quadratic potential. If it's easy to run 100 simulations with different number of points and see if there are always 6 more red defects than blue ones, that would be cool.
@johncarlosbaez @lisyarus So the conjecture is that only these two types of defects occur? I think that the Euler characteristic argument goes through exactly once you assume that. (If you say that every interior vertex _should_ have six neighbors and every boundary vertex _should_ have four neighbors and you sum the total surplus/deficit over all vertices, you always get -6 for a triangulated disk, right?)
@eruonna @lisyarus - You're right, Euronna.
For a triangulated sphere where each vertex has 5, 6, or 7 edges incident to it, Euler characteristic shows that there must be 12 more vertices with 5 edges than vertices with 7. But this implies that for a triangulated disk where each boundary vertex has just 4 incident edges, there must be 6 more vertices with 5 edges than vertices with 7. (Why? Because we can take two copies of such a disk and glue them together to get a sphere.)
@johncarlosbaez @eruonna With many particles (5000 here) I get even funnier results: large clusters of red (degree-5) particles on the boundary of the disk. Not sure if this is an actual phenomenon or just the algorithm didn't converge yet, though.
@lisyarus wrote: ") I get even funnier results: large clusters of red (degree-5) particles on the boundary of the disk."
Oh no! I'm hoping you just need to wait longer for the algorithm to converge. Even for just 100 points it took a long time for the red points on the boundary to all go away. I felt sure they should eventually go away... and they eventually did.
@johncarlosbaez That was my impression as well! Sometimes it helps to increase step size to "push" the system in the right direction, sometimes decreasing it makes it stabilize, similar to how heating up & cooling down a system moves it towards a more efficient state. Global minimization is hard, I'm not that good at it :)
@lisyarus - sometime when I'm working at my computer for a long time I'll try to run a big example and push the step size up and down to try to get all the defects at the boundary to go away. I guess it''s possible that defects naturally lie on the boundary since that's where the equilateral triangle lattice can't possibly work. But I'm still hoping they don't.
@lisyarus @johncarlosbaez @eruonna Love it. With simple gradient descent, mine tends to stall with lots of degree-5 points around the edge as well, but if it gets past those, it does tend toward six… https://observablehq.com/d/833399641ea39c4e (Y'all really shouldn't be nerd-sniping people like this though. I have work to do!)
@rreusser @johncarlosbaez @eruonna Cool! What's that "AABB" thing? I guess you're doing something smarter than a basic quadtree for Barnes-Hut?
@lisyarus @johncarlosbaez @eruonna Ah, yes. The acceleration data structure I had handy was a bounding volume hierarchy, so I'm using aabb's as the state vector, but really they're just collapsed to a point, i.e. aabb = (x, y, x, y). I've modified the tree display to not just show the leaf nodes now. (sorry the implementation is quite thin on details!)
@johncarlosbaez Do you know of the Thomson problem :) https://en.wikipedia.org/wiki/Thomson_problem
I have thought about it, made some progress, and realized I didn't even _know_ if the solution would be stable in other cases :)
BTW: Do you know if the electrons on the exterior/boundary of your disk are in general uniform for a solution?
@rrogers - I know a nice webpage listing solutions to the Thomson problem for up to 75 points. The webpage is defunct, but it's still available on the Wayback Machine:
https://web.archive.org/web/20160224233855/http://www.csun.edu/~hcmth007/electrons/points.html
Here is the solution for 25 points:
(1/2)
@rrogers wrote: "Do you know if the electrons on the exterior/boundary of your disk are in general uniform for a solution?"
You mean are the points on the boundary of the disk exactly equally spaced? That seems incredibly unlikely, except in some special cases.
@johncarlosbaez That's what I thought.
@johncarlosbaez Thanks! I have added it to my collection. Since I have fermented it in my mind (subconscious) for more than a couple of years, maybe I'll try some experiments to test my ideas; so far it's just been thoughts that seem reasonable. It is nice to have a reasonable number of figures (with precalculated points!) available.
@johncarlosbaez I just want to point to the reference https://arxiv.org/pdf/cond-mat/9707204.pdf (section V and Appendix A), where they discuss the same.
@The_Correlator - excellent, thanks! This topic is mainly a hobby for me so I haven't been digging into the literature. I'm glad to know that experts are thinking about this.