Title: Chemistry, Physics, and Mathematics behind the Soccer Ball Design
1Chemistry, Physics, and Mathematics behind the
Soccer Ball Design
Or How to Distribute Dictators on a Planet?
2Chemistry of the Soccer Ball
Large carbon molecules discovered in 1985 by
Richard Smalley (1996 Chemistry Noble Prize
winner)
C60
C70
3C60, C70, C1,000,000
4Nanotechnology
Nanowire -- 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.
5Physics of the Soccer Ball
Electrons in Equilibrium
N electrons orbit the nucleus
Electrons repel
Equilibrium will occur at minimum energy
Problem If Coulombs Law is assumed, then we
minimize the sum of the reciprocals of the mutual
distances, i. e. Si?jXi - Xj-1 over all
possible configurations of N points X1 , , XN
on the sphere.
632 Electrons
122 Electrons
In Equilibrium
In Equilibrium
7Mathematics of the Soccer Ball
Minimum Energy Problem on the Sphere Given an
N-point configuration ?N X1 , , XN on the
sphere we define its generalized energy as Ea (?N
) Si?j Xi - Xj?. We then minimize this energy
over all possible configurations when ?lt0,
maximize it when ?gt0, and when ?0 we minimize
the logarithmic energy (or maximize the product
of the mutual distances). This extremal energy we
denote by E(?,N).
8?1 Except for small N a long standing open
problem in discrete geometry (L. Fejes Toth -
1956)
?-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 1965. The answer known only
for N1-4,8,12.
?? -? Best packing (of spherical caps) problem --
to maximize the minimum distance between any pair
of points. Known for N 1-12 and 24.
9Dirichlet Cells (School Districts) D1, , DN
Djx? S2 x-xjmink x-xk j1,,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.
10Q1. Is it possible to tile a soccer ball with
only hexagons?
Why Not?
NO!
Eulers Formula F V - E 2 F of faces, V
of vertices, E of edges
Example Cube F6, V8, E12
11Suppose we could Cut edges in half. Each vertex
has three half-edges. 3V2E Each face has six
edges. 6F2E
Thus 3V6F and V2F. Then FV-EF2F-3F0
?2, which contradicts the Eulers formula.
QED
12Q2. How many pentagons are needed for a soccer
ball design? (We shall assume that exactly three
edges emanate from each vertex).
How many pentagons does a soccer ball have?
Answer Exactly 12
Proof Let Fp of pentagons, Fq of
hexagons. Then FFpFq .
13There is a new formula 5Fp6Fq2E. Still 3V2E.
Substitute in Eulers formula FV-EFpFq1/3(5Fp
6Fq)-1/2(5Fp6Fq)2 6(FpFq)2(5Fp6Fq)-3(5Fp6Fq)
12 (610-15)Fp(612-18)Fq12 1Fp12
QED
14C60 -- Buckminster Fulerene
15Numerical 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.
16Buckyball under the Lions Paw
17The problem for small N
- Best Packing problem
- The solution is known for N1,2, , 12, 24.
- N4 - regular tetrahedron
- N5,6,7 - south, north and the rest on the
equator - N8 - skewed cube
- N12 - centers of icosahedron (12 pentagons)
- Minimum logarithmic energy problem
- N4 - regular tetrahedron
- N12 - centers of icosahedron (12 pentagons)
18Minimum LE problem - N4 Choose A, B, C, D on the
unit sphere, such that AB AC AD BC BD
CD ? max
Proof First, optimal choice exists (S2 is
compact). Let A, B, C, D, be an optimal
configuration. Let x0(OAB) and AB?y0 and
let C and D spin in horizontal orbits. Then
maximum occurs when CACB and DADB.
19Indeed, in cylindrical coordinates A(0,-y,t)
C(rcos?,rsin?,s) B(0,y,t) D(lcos?,lsin?,u)
We now can write CA2 CB2 (rcos?)2(rsin?y)
2(s-t)2 (rcos?)2(rsin?-y)2(s-t)2
r2y2(s-t)2 2rysin? r2y2(s-t)2-
2rysin? which achieves maximum when sin?0. The
same will be true for DA2 DB2 when sin?0,
therefore CACB and DADB.Fortunately, when ?0
and ?? CD also achieves maximum. Therefore,
CACB and DADB is a necessary condition for
optimization. By symmetry all six edges must be
equal.
20 For N5,6,7 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