# Limiting shape for Brillouin zones

Is it true that the limiting shape for Brillouin zones (for any lattice) is a circle? You can find the definition and the step by step construction of Brillouin zones here. This picture is taken from the Graphics Gallery for The Mathematica GuideBook for Graphics.

It is known that all Brillouin zones have the same area, so if we have a picture with $n$ zones and we want to see it in a unit square then it is necessary to multiply it by $1/\sqrt{n}$. But they could be close to the circle of radius $c\sqrt{n}$ even without rescaling.

edit by j.c.: This question is stated in 2D but makes sense in arbitrary dimensions.

Suppose we have a lattice $\Gamma$ in $d$-dimensions (i.e. a discrete subgroup of $\mathbb{R}^d$ that spans $\mathbb{R}^d$) (in crystallography terms, $\Gamma$ is the reciprocal lattice of some Bravais lattice). A Bragg plane of $\Gamma$ is a hyperplane that perpendicularly bisects a line segment in $\mathbb{R}^d$ between the origin and any other element of $\Gamma$. Then the $n$th Brillouin zone of $\Gamma$ is defined to be the closure of the set of points $P$ in $\mathbb{R}^d$ such that the line segment between $P$ and the origin intersects exactly $n-1$ Bragg planes.

• The limit as what converges to what? – Liviu Nicolaescu Oct 23 '15 at 13:54
• As the number of Brillouin zone tends to infinity. The pictures from the question are just the union of large number of Brillouin zones. The last zone is a thin lace along the border, and it looks like a circle. – Alexey Ustinov Oct 23 '15 at 13:56
• I assume that there is also a rescaling involved. Could you include these details in your question? – Liviu Nicolaescu Oct 23 '15 at 14:11
• on which lattice? for any lattice? – Carlo Beenakker Oct 23 '15 at 14:18
• Alexey: I really think that you should make an effort to define the terms that you use. The link that you provided does not help me understand your question. Please explain your figures: did you make them? Is so, how did you make them? Did somebody else make them? If so: in what context did they occur? – André Henriques Oct 23 '15 at 14:54

## 1 Answer

Yes, it is. Take a point P, and let us check if it belongs to Nth Brillouin zone. The Bragg planes (rather, lines, as we are in $R^2$) that we have to cross while going from the origin $O$ to $P$, correspond to the points $L$ of the lattice $\Gamma$ such that $P$ is closer to $L$ than to $O$. In other words, we are looking for the number of the points $L$ of the lattice $\Gamma$ that belong to the $P$-centered ball of radius $|OP|$.

This number is approximately equal to $\pi \cdot |OP|^2/\mathop{\mathrm{covol}} \Gamma$. Hence, $N$th Brillouin zone is very close to a circle of radius $\sqrt{N/\pi \cdot \mathop{\mathrm{covol}} \Gamma}$.

• Victor, I understand from you answer that we have to cross (while going from the origin $O$ to $P$) $\pi \cdot |OP|^2/\mathop{\mathrm{covol}} \Gamma$ Bragg planes. But why it implies that we know the distance from $P$ to $O$? It is clear for symmetrical picture but not for arbitray lattice. – Alexey Ustinov Oct 23 '15 at 15:25
• Assume that you know the distance from $P$ to $O$. Then you (approximately) know, to which zone does it belong: it is exactly the number of lines to cross. Now, $N$th zone is where this number is equal to $N$, and thus the distance should be approximately equal to the above square root. Well, formally, it implies that $N$th zone is contained in a neighborhood of the above-described circle. But it should separate the "inner" part from the "outer" (any ray from $O$ to infinity intersects it), so there are no gaps. – Victor Kleptsyn Oct 23 '15 at 15:29
• Oh, great! I've missed this argument. It means that the distance from $P\in B(N)$ is $c\sqrt{N}+O(N^{\frac{k-1}{2}}),$ where $k$ is the exponent from Gauss circle problem $\pi R^2+O(R^k)$. We know that $k<1$, so Nth zone lies in at most $O(N^{-1/6})$ (for $k=2/3$) neighbourhood of corresponding circle of radius $c\sqrt{N}.$ – Alexey Ustinov Oct 23 '15 at 15:51
• For different lattices there's some discussion of Gauss circle counting in the answer to this question mathoverflow.net/questions/109515 . Is it obvious that the points in the Brillouin zone become equidistributed on the circle? And does this say something about the asymptotics of the "outer perimeter" of the Brillouin zone as well? – j.c. Oct 23 '15 at 16:20
• @j.c. I think that you questions are interesting and should be asked separately. Will you do it? – Alexey Ustinov Oct 24 '15 at 1:40