Brillouin Zone for Graphene

So the Brillouin Zone can be described as the set of points that are closer to the origin than any other lattice point in reciprocal space. So one way to get the shape of the Brillouin Zone would be to first find the reciprocal lattice. Using the Bravais Lattice generating vectors we found before: \(  \vec{b_{1}} = ( \frac{3}{2}b , \frac{ \sqrt{3}}{2} b) \)  and  \( \vec{b_{2}} = ( \frac{3}{2}b , -\frac{ \sqrt{3}}{2} b) \) ,  we can find the reciprocal vectors \( \vec{a_{1}} \)  and \( \vec{a_{2}}\) by the relation \( \vec{b_{1}} \cdot \vec{a_{1}} = \vec{b_{2}} \cdot \vec{a_{2}} = 2\pi  \) . More simply we can find the reciprocal vectors by:
 \( \vec{a_1} = 2 \pi \frac{R \vec{b_2}}{\vec{b_1} \cdot R \vec{b_2}} \) and \( \vec{a_2} = 2 \pi \frac{R \vec{b_1}}{\vec{b_2} \cdot R \vec{b_1}} \) where \( R\) is the  \( 90^{\circ} \) rotation matrix \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \)

Following this through yields \( \vec{a_1} = ( \frac{ 2 \pi} { 3 b} , \frac{ 2 \pi \sqrt{3}}{3b} ) \) and \( \vec{a_2} = ( \frac{ 2 \pi} { 3 b} , -\frac{ 2 \pi \sqrt{3}}{3b} ) \).

So our reciprocal lattice looks like:

To get the Brillouin Zone from the earlier definition, we'll draw in the perpendicular bisectors of the reciprocal vectors, along with the vertical line through the midpoint between the nearests neighbor on the  \( k_x \)-axis to get something like:

Through some geometry you can find this boundary intersects the \( k_y \) -axis at \( k_y = \frac{4 \pi }{3 \sqrt{3}b} \) and the \( k_x \) - axis at \( k_x = \frac{2 \pi}{3b} \).