# Does the sequence of the $n$th Brillouin zones converges to any particular geometrical form?

I am currently studying solid state physics and we just saw the construction of the Brillouin zones. My question is the following:

Does the sequence of the $$n$$th Brillouin zones converges to any particular geometrical form? From some figures I have seen, it seems to tend to a circle ( for different lattices ) but if this is true, I don't have any idea why and whether it is convergent whatever the form of the lattice at the beginning.

Does somebody have an idea?

• This is an interesting geometry question, but I guess the point is that a high enough order polygon is bound to look like a circle, especially with the symmetries of Brillouin zones enforced. – KF Gauss Feb 15 '19 at 9:15
• @KFGauss - I agree, with the caveat that you start with a high symmetry crystal. Books rarely have nice pictures of the Brillouin zones or Fermi surface of a triclinic crystal... – Jon Custer Feb 15 '19 at 13:54

The relevant conclusion is that, as $$n\rightarrow \infty$$, the $$n$$-th Brillouin zone approaches the shape of a thin spherical shell of radius $$O(n^{1/d})$$, where $$d=1,2,3,\dots$$ is the dimension of the lattice. This fact implies that $$\bigcup\limits_{i=1}^{n}B_n$$ approaches the shape of the $$d-1$$-sphere $$S^{d-1}$$. I.e. in two dimensions you get a circle $$S^1$$ as $$n\rightarrow \infty$$, and in three dimensions you get the regular sphere.
A possibly relevant follow-up paper is Jones. and Lansberg Brillouin Zones and the Fundamental Regions, Physica Status Solidi b 128, 619 (1985). I have not read it, but from its abstract it appears to describe some properties of the $$n$$-th Brillouin zone.