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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?

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    $\begingroup$ 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. $\endgroup$ – KF Gauss Feb 15 at 9:15
  • $\begingroup$ @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... $\endgroup$ – Jon Custer Feb 15 at 13:54
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This was studied by mathematician Gareth A. Jones in the paper Geometric and Asymptotic Properties of Brillouin Zones in Lattices, Bulletin of the London Mathematical Society 16, 241 (1984). Non-paywalled link here.

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.

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