## Empty lattice approximation for a simple monoclinic crystal

a/b:    a/c:    $\alpha$: (degrees)

 Symmetry points ($u,v,w$) [$k_x,k_y,k_z$] Γ: (0, 0, 0) [0, 0, 0] Y: (0, 1/2, 0) [0, $\frac{\pi}{b \, \sin(\alpha)}$, 0] B: (1/2, 0, 0) [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}$, 0] A: (1/2, 1/2, 0) [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}+\frac{\pi}{b \, \sin(\alpha)}$, 0] Z: (0, 0, 1/2) [0, 0, $\frac{\pi}{c}$] C: (0, 1/2, 1/2) [0, $\frac{\pi}{b \, \sin(\alpha)}$, $\frac{\pi}{c}$] D: (1/2, 0, 1/2) [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}$, $\frac{\pi}{c}$] E: (1/2, 1/2, 1/2) [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}+\frac{\pi}{b \, \sin(\alpha)}$, $\frac{\pi}{c}$]
$$$\vec{b}_1=\frac{-2\pi}{a}(\hat{k}_x + \frac{1}{\tan(\alpha)}\hat{k}_y), \hspace{1cm} \vec{b}_2=\frac{2\pi}{b\, \sin(\alpha)}\hat{k}_y, \hspace{1cm} \vec{b}_3=\frac{2\pi}{c}\hat{k}_z.$$$