3.1.7 IBRAV=5 simple trigonal lattice(単純三方格子)

CELLDM(1) $=a$, CELLDM(4) $=\cos{\alpha} (0 < \alpha < 120^{\circ})$,
$b\equiv\sqrt{2} a \sqrt{1-\cos{\alpha}}/\sqrt{3}$, $c\equiv a \sqrt{1+2\cos{\alpha}}/\sqrt{3}$
$\Omega=a^{3}(1-\cos{\alpha})\sqrt{1+2\cos{\alpha}}=3\sqrt{3} b^2 c/2$
$({\bf a}_{1},{\bf a}_{2},{\bf a}_{3})= ({\bf x},{\bf y},{\bf z}) \left( \beg... ...2} & 0 \\ \frac{b}{2} & \frac{b}{2} & -b \\ c & c & c \end{array} \right)$
$({\bf b}_{1},{\bf b}_{2},{\bf b}_{3})= \frac{2 \pi}{a} ({\bf x},{\bf y},{\bf ... ...}b & 0 \\ a/3b & -a/3b & -2a/3b \\ a/3c & a/3c & a/3c \end{array} \right)$
Relationship with a lattice constant of hexagonal lattice and choice of the sub-lattice.
六方格子格子定数との関係および副格子のとり方
$a_{\rm h}= a \sqrt{2(1-\cos{\alpha})}$, $c_{\rm h}= a \sqrt{3(1+2\cos{\alpha})}$
$a=\frac{a_{\rm h}}{3}\sqrt{(\frac{c_{\rm h}}{a_{\rm h}})^{2}+3}$, $\cos{\alpha}=1-\frac{9}{2}\frac{1}{(\frac{c_{\rm h}}{a_{\rm h}})^{2}+3}$
${\bf s}_{1} = 0$
${\bf s}_{2} = \frac{2}{3} {\bf a}_{\rm h} + \frac{1}{3} {\bf b}_{\rm h} + \fr... ...{\bf c}_{\rm h} = \frac{\sqrt{3}b}{2} {\bf x} + \frac{b}{2} {\bf y} + c {\bf z}$
${\bf s}_{3} = \frac{1}{3} {\bf a}_{\rm h} + \frac{2}{3} {\bf b}_{\rm h} + \frac{2}{3} {\bf c}_{\rm h} = b {\bf y} + 2 c {\bf z}$