Gruber P. Convex and Discrete Geometry

158 Convex Bodies
Wulff’s Theorem in Crystallography
Why do crystals have such particular forms? Since the late eighteenth century it was
the belief of many crystallographers that underlying each crystal there is a point
lattice, where any lattice parallelotope contains a certain set of atoms, ions or molecules,
and any two such parallelotopes coincide up to translation. Compare also the
discussion in Sect. 21. Any facet of the crystal is contained in a 2-dimensional lattice
plane. The free surface energy per unit area of a facet (whatever this means) is small
if the corresponding lattice plane is densely populated by lattice points and large otherwise.
Thus, only for a small set of normal directions, the free surface energies per
unit area in the corresponding lattice planes are small. Since real crystals minimize
their total free surface energy according to Gibbs, Curie and Wulff, this explains why
crystals have the form of particular polytopes. More precisely, we have the following
theorem of Wulff [1031]. A first proof of it is due to Dinghas [268]. In our version
of Wulff’s theorem, no compatibility condition for the free surface energies per unit
area is needed. For this reason the proof is slightly more difficult than otherwise.
There are various other versions of Wulff’s theorem, for example those of
Busemann [180], Fonseca [339] and Fonseca and Müller [340]. See also the surveys
of Taylor [990], McCann [702] and Gardner [360] and the book of Dobrushin,
Koteck´y and Shlosman [275]. For applications to statistical mechanics and combinatorics,
compare the report of Shlosman [932].
Theorem 8.13. Let u1,...,un ∈ S d−1 (the exterior unit normal vectors of the facets
of the crystal) be such that E d ={α1u1 +···+αnun : αi ≥ 0}. Let ε1,...,εn > 0
(the free surface energies per unit area of the facets). Let K be the convex polytope
K = � x : ui · x ≤ εi for i = 1,...n �
(the crystal). Then, amongst all convex polytopes with volume equal to V (K ), and
such that the exterior normal vectors of their facets belong to {u1,...,un}, those
with minimum total free surface energy are precisely the translates of K .
The proof of Dinghas is difficult to understand. Possibly, our proof is what Dinghas
had in mind. It is related to Minkowski’s proof of the isoperimetric inequality, see
Theorem 8.7. Let v(·) denote the (d − 1)-dimensional area measure.
Proof. We may assume that each of the halfspaces {x : ui · x ≤ εi}, i = 1,...,m,
contributes a ((d − 1)-dimensional) facet to K ,say
Fi = � �
x : ui · x = εi ∩ K,
while each of the halfspaces {x : ui · x ≤ εi}, i = m + 1,...,n, has the property that
Fi = � �
x : ui · x = δiεi ∩ K
is a face of K with dim Fi < d − 1, where 0 ≤ δi ≤ 1 is chosen such that the
hyperplane {x : x · ui = δiεi} supports K .