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It is known that there are five Platonic solids:

enter image description here

If, lets say, there are 4 particles with the same electricity charge and whose movement is constrained to be on a sphere, resulting forces will eventually move particles to form a tetrahedron.

I have two questions here:

1) If number of particles are 8, 6, 20, and 12, would resulting solid (in similar experiment) be hexahedron, octahedron, dodecahedron, and icosahedron, respectively?

2) What would happen if number of particles is not 4, 8, 6, 20, or 12? Would an equilibrium be reached ever? Would the equilibrium shape be unique or not? Are those solids (resulting from such experiments) widely known? What would be their other properties?

I attempted some simulations, and researche the internet but didn't reach any sound conclusion.

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    $\begingroup$ Possibly related to math.stackexchange.com/questions/66365/… $\endgroup$ Commented Apr 30, 2014 at 13:22
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    $\begingroup$ Electrostatic energy should be proportional to the sum of the inverse distance of all the pairs of points. So to turn this into a mathematical problem, you are trying to minimize $$\sum_{i=1}^{n-1}\sum_{j=i+1}^n\lVert p_i-p_j\rVert^{-1}$$ with the sphere constraint $\lVert p_i\rVert=1$. This makes the problem a mathematical one, while we leave it to physics.stackexchange.com to decide whether electrons actually reach and maintain that energy minimum. $\endgroup$
    – MvG
    Commented Apr 30, 2014 at 13:38
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    $\begingroup$ See this web page for an investigation of this problem. $\endgroup$
    – Jim Belk
    Commented Apr 30, 2014 at 13:59
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    $\begingroup$ Cross reference to Physics SE: Shape of electric charges on sphere in equilibrium state $\endgroup$
    – MvG
    Commented Apr 30, 2014 at 16:23

1 Answer 1

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This is known as the Thomson problem. For $4$, $6$, and $12$ particles, the minimum energy configuration is the corresponding Platonic solid, but not for $8$ or $20$ particles. In particular, for $8$ particles a square antiprism has lower energy than a cube. Minimum energy configurations of course exist for other $n$, but finding them is a difficult numerical problem; as with most packing problems, there appears to be no general pattern for large $n$.

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  • $\begingroup$ Thanks for answer, @Rahul! But is there any formal proof that any distribution of particles would eventually lead to the one with minimum energy? In particular, if particles were initially at vertices of a cube, would they move to vertices of an antiprism? $\endgroup$
    – VividD
    Commented Apr 30, 2014 at 14:14
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    $\begingroup$ No, in general if you start from an arbitrary distribution of particles and perform, say, gradient descent, you will probably only land on a local minimum, not the global minimum. This is why the problem is hard for large $n$ if you care about global minima. // Starting from a cube, actually, you may not end up anywhere because the cube is a critical point with zero gradient -- sort of like a stationary bicycle that's perfectly balanced vertically. But if you perturb the particles slightly away from the cube randomly, you should end up at the antiprism. $\endgroup$
    – user856
    Commented Apr 30, 2014 at 15:25
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    $\begingroup$ Is it known that the energy function on the space of particle distributions is a Morse function? If so, then if you start with not any distribution but with almost any distribution (with respect to Lebesgue measure on the sphere) then gradient descent will converge to a global minimum. $\endgroup$
    – Lee Mosher
    Commented May 1, 2014 at 12:51
  • $\begingroup$ @Lee: Can't Morse functions have multiple local minima? $\endgroup$
    – user856
    Commented May 1, 2014 at 15:19
  • $\begingroup$ @LeeMosher: Min-Energy Configurations of Electrons On A Sphere says that N = 16 electrons has 2 local minima, but only one of them is the unique global minimum. $\endgroup$
    – David Cary
    Commented May 2, 2014 at 4:04

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