Timeline for Equilibrium configurations of ions on n-Dim balls.
Current License: CC BY-SA 4.0
9 events
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| Feb 9, 2022 at 13:02 | history | edited | Glorfindel | CC BY-SA 4.0 |
3 broken links fixed
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| Oct 23, 2011 at 21:09 | comment | added | Pietro Majer | Of course the vertices of the permutohedron $P_n$ makes a minimum configuration for some (unreasonable) potential function of the form $V(x_1,\dots,x_{n!}):=\sum_{i<j}\phi(\|x_i-x_j\|)$ : we may choose a function $\phi\ge0$ vanishing exactly on the finite set of distances between all pairs of veritices of $P_n$. But this hasn't much to do with $P_n$, since it can be done as well for any set of points. | |
| Oct 23, 2011 at 18:45 | vote | accept | Tom Copeland | ||
| Oct 23, 2011 at 17:46 | comment | added | Henry Cohn | I just realized my first paragraph was a little unclear on whether I was comparing cases with the same number of particles, so I've edited it to try to clarify. | |
| Oct 23, 2011 at 17:44 | history | edited | Henry Cohn | CC BY-SA 3.0 |
clarified number of particles
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| Oct 23, 2011 at 17:42 | comment | added | Henry Cohn | For $n=4$ and 120 particles the optimal configuration is a regular 600-cell, so you definitely do not get a permutahedron. For $n=4$ and 42 particles, you can find simulation results at aimath.org/data/paper/BBCGKS2006/4-42.txt. I haven't checked whether it is an associahedron, but I am pretty confident it won't be. You can also get the 3d data from aimath.org/data/paper/BBCGKS2006. | |
| Oct 23, 2011 at 17:15 | comment | added | Tom Copeland | I tried to clarify my question after reading your response. For n=4 the problem is restricted to two separate cases with either 5!=120 identical "ions" or 10!/(5!6!) = 42, i.e., I've constrained the problem to match the number of vertices of either of the two polytopes to be considered separately. Your approach seems to be a more general optimization problem. | |
| Oct 23, 2011 at 15:16 | comment | added | Henry Cohn | P.S. My answer deals with particles on a sphere, but the original question may allow them to be inside the ball as well. For an inverse power law potential $r \mapsto 1/r^s$ with $s \le n-2$, the maximum principle implies that these two variants of the question have the same answer. When $s > n-2$, the questions often have different answers, but it is easier to rule out permutohedra: when $s$ is large enough, one can lower energy by moving one particle to the center of the ball. | |
| Oct 23, 2011 at 14:58 | history | answered | Henry Cohn | CC BY-SA 3.0 |