Chemistry, Physics, and Mathematics behind the Soccer Ball Design Peter Dragnev Or How to Distribute Dictators on a Planet?

Chemistry of the Soccer Ball Fullerenes (Buckyballs) Large carbon molecules discovered in 1985 by Richard Smalley (1996 Chemistry Noble Prize winner) Richard Smalley C 60 C 70

C 60, C 70, C 1,000,000

Nanotechnology NanowireNanowire -- “a giant single fullerene molecule”, “ a truly metallic electrical conductor only a few nanometers in diameter, but hundreds of microns (and ultimately meters) in length”, “expected to have an electrical conductivity similar to copper's, a thermal conductivity about as high as diamond, and a tensile strength about 100 times higher than steel”.

Physics of the Soccer Ball N electrons orbit the nucleus Electrons repel Equilibrium will occur at minimum energy Electrons in Equilibrium Problem: If Coulomb’s Law is assumed, then we minimize the sum of the reciprocals of the mutual distances, i. e. Σ i  j |X i - X j | -1 over all possible configurations of N points X 1, …, X N on the sphere.

32 Electrons122 Electrons In Equilibrium

Minimum Energy Problem on the Sphere Given an N-point configuration  N ={X 1, …, X N } on the sphere we define its generalized energy as E  (  N )= Σ i  j |X i - X j | . We then minimize this energy over all possible configurations when  0, and when  =0 we minimize the logarithmic energy (or maximize the product of the mutual distances). This extremal energy we denote by E( ,N). Mathematics of the Soccer Ball

 =1 Except for small N a long standing open problem in discrete geometry (L. Fejes Toth )  =-1 Known as J. J. Thompson problem - dates back to Hilbert. Due to the recent discovery of fullerenes it has attracted the attention of researchers in chemistry, physics, crystallography.  =0 The problem was raised by Steven Smale, a Fields medallist for The answer known only for N=1-4,8,12.Steven Smale   -  Best packing (of spherical caps) problem -- to maximize the minimum distance between any pair of points. Known for N= 1-12 and 24.

Dirichlet Cells (School Districts) D 1, …, D N The D-cells of 32 electrons at equilibrium are the tiles of the Soccer Ball. Is it a coincidence that the D-cells are made out of pentagons and hexagons? Call such a tiling of the sphere a soccer ball design. D j :={xє S 2 : |x-x j |=min k |x-x k |} j=1,…,N

Q1. Is it possible to tile a soccer ball with only hexagons? NO! Why Not? Euler’s Formula F + V - E = 2 F= # of faces, V= # of vertices, E= # of edges Example: Cube F=6, V=8, E=12

Suppose we could… Cut edges in half. Each vertex has three half-edges. 3V=2E Each face has six edges. 6F=2E Thus 3V=6F and V=2F. Then F+V-E=F+2F-3F=0  2, which contradicts the Euler’s formula. QED

Q2. How many pentagons are needed for a soccer ball design? (We shall assume that exactly three edges emanate from each vertex). Answer: Exactly 12 Proof: Let F p = # of pentagons, F q = # of hexagons. Then F=F p +F q. How many pentagons does a soccer ball have?

There is a new formula 5F p +6F q =2E. Still 3V=2E. Substitute in Euler’s formula F+V-E=F p +F q +1/3(5F p +6F q )-1/2(5F p +6F q )=2 6(F p +F q )+2(5F p +6F q )-3(5F p +6F q )=12 ( )F p +( )F q =12 1F p =12 QED

C Buckminster Fulerene

Numerical Fact: For large number of points (N   ) all but exactly 12 of the D-cells of an optimal configuration are hexagons. The exceptional cells are pentagons. The proof of this Conjecture is a very difficult problem. Numerical computations for large N are also very important aspect of the problem, especially with the development of Nanotechnology. But this is also very difficult problem, because of the exponentially increasing number of local minima.

Buckyball under the Lion’s Paw

The problem for small N Best Packing problem The solution is known for N=1,2, …, 12, 24. N=4 - regular tetrahedron N=5,6,7 - south, north and the rest on the equator N=8 - skewed cube N=12 - centers of icosahedron (12 pentagons) Minimum logarithmic energy problem N=4 - regular tetrahedron N=12 - centers of icosahedron (12 pentagons)

Minimum LE problem - N=4 Choose A, B, C, D on the unit sphere, such that AB · AC · AD · BC · BD · CD  max Proof: First, optimal choice exists (S 2 is compact). Let A, B, C, D, be an optimal configuration. Let {x=0}=(OAB) and AB  {y=0} and let C and D spin in horizontal orbits. Then maximum occurs when CA=CB and DA=DB.

We now can write CA 2 ·CB 2 =[(rcos  ) 2 +(rsin  +y) 2 +(s-t) 2 ] ·[(rcos  ) 2 +(rsin  -y) 2 +(s-t) 2 ] =[r 2 +y 2 +(s-t) 2 + 2rysin  ] ·[r 2 +y 2 +(s-t) 2 - 2rysin  ] which achieves maximum when sin  =0. The same will be true for DA 2 ·DB 2 when sin  =0, therefore CA=CB and DA=DB.Fortunately, when  =0 and  =  CD also achieves maximum. Therefore, CA=CB and DA=DB is a necessary condition for optimization. By symmetry all six edges must be equal. Indeed, in cylindrical coordinates A=(0,-y,t) C=(rcos ,rsin ,s) B=(0,y,t) D=(lcos ,lsin ,u)

For N=5,6,7 the problem has been posed by Rakhmanov. Computational examples suggest that the solution is the north pole, the south pole and regular figure on the equator. Nice open problem! GOOD LUCK! THE END