Difference between revisions of "Talk:Lattice:Hexagonal diamond"

Revision as of 10:03, 9 January 2018

$4\,\mathrm {bottom\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$ $4\,\mathrm {bottom\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
- $\left(0,0,0\right),(a,0,0),\left({\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},0\right),\left(a+{\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},0\right)$
$4\,\mathrm {mid\,\,layer} :\,{\frac {1}{6}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{3}}=1$ $4\,\mathrm {mid\,\,layer} :\,{\frac {1}{6}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{3}}=1$
- $\left(0,0,{\frac {5c}{8}}\right),\left(a,0,{\frac {5c}{8}}\right),\left({\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},{\frac {5c}{8}}\right),\left(a+{\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},{\frac {5c}{8}}\right)$
$2\,\mathrm {internal\,\,strut} :\,1+1=2$ $2\,\mathrm {internal\,\,strut} :\,1+1=2$
- $\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {c}{8}}\right),\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {4c}{8}}\right)$
$4\,\mathrm {top\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$ $4\,\mathrm {top\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
- $\left(0,0,c\right),(a,0,c),\left({\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},c\right),\left(a+{\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},c\right)$

For a particle-particle bond-length of $l$ :

a={\frac {2{\sqrt {6}}}{3}}l

b=a={\frac {2{\sqrt {6}}}{3}}l

c={\frac {8}{3}}l

{\frac {a}{c}}={\frac {\sqrt {6}}{4}}\approx 0.61237

{\frac {c}{a}}={\frac {4}{\sqrt {6}}}\approx 1.63299

$4\,\mathrm {bottom\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$ $4\,\mathrm {bottom\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
- $\left(0,0,0\right),\left({\frac {2{\sqrt {6}}}{3}}l,0,0\right),\left({\frac {\sqrt {6}}{3}}l,{\sqrt {2}}l,0\right),\left({\sqrt {6}}l,{\sqrt {2}}l,0\right)$
$4\,\mathrm {mid\,\,layer} :\,{\frac {1}{6}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{3}}=1$ $4\,\mathrm {mid\,\,layer} :\,{\frac {1}{6}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{3}}=1$
- $\left(0,0,{\frac {5}{3}}l\right),\left({\frac {2{\sqrt {6}}}{3}}l,0,{\frac {5}{3}}l\right),\left({\frac {\sqrt {6}}{3}}l,{\sqrt {2}}l,{\frac {5}{3}}l\right),\left({\sqrt {6}}l,{\sqrt {2}}l,{\frac {5}{3}}l\right)$
$2\,\mathrm {internal\,\,strut} :\,1+1=2$ $2\,\mathrm {internal\,\,strut} :\,1+1=2$
- $\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {c}{8}}\right),\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {4c}{8}}\right)$
$4\,\mathrm {top\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$ $4\,\mathrm {top\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
- $\left(0,0,{\frac {8}{3}}l\right),\left({\frac {2{\sqrt {6}}}{3}}l,0,{\frac {8}{3}}l\right),\left({\frac {\sqrt {6}}{3}}l,{\sqrt {2}}l,{\frac {8}{3}}l\right),\left({\sqrt {6}}l,{\sqrt {2}}l,{\frac {8}{3}}l\right)$

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 ** <math>\left(0,0,0\right),\left(\frac{2\sqrt{6}}{3}l,0,0 \right),\left(\frac{\sqrt{6}}{3}l,\sqrt{2}l,0\right),\left(\sqrt{6}l,\sqrt{2}l,0\right)</math>
 * <math> 4 \, \mathrm{mid\,\, layer}: \, \frac{1}{6} + \frac{1}{3} + \frac{1}{6} + \frac{1}{3} = 1</math>
-** <math>\left(0,0,\frac{5c}{8}\right),\left(a,0,\frac{5c}{8}\right),\left(\frac{b}{2},\frac{\sqrt{3}b}{2},\frac{5c}{8}\right),\left(a+\frac{b}{2},\frac{\sqrt{3}b}{2},\frac{5c}{8}\right)</math>
+** <math>\left(0,0,\frac{5}{3}l\right),\left(\frac{2\sqrt{6}}{3}l,0,\frac{5}{3}l \right),\left(\frac{\sqrt{6}}{3}l,\sqrt{2}l,\frac{5}{3}l\right),\left(\sqrt{6}l,\sqrt{2}l,\frac{5}{3}l\right)</math>
 * <math> 2 \, \mathrm{internal\,\, strut}: \, 1 + 1 = 2</math>
 ** <math>\left(\frac{a}{2},\frac{b}{2\sqrt{3}},\frac{c}{8}\right),\left(\frac{a}{2},\frac{b}{2\sqrt{3}},\frac{4c}{8}\right)</math>
 * <math> 4 \, \mathrm{top\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}</math>
 ** <math>\left(0,0,\frac{8}{3}l\right),\left(\frac{2\sqrt{6}}{3}l,0,\frac{8}{3}l \right),\left(\frac{\sqrt{6}}{3}l,\sqrt{2}l,\frac{8}{3}l\right),\left(\sqrt{6}l,\sqrt{2}l,\frac{8}{3}l\right)</math>