Orthorhombic (ORC)#
Pearson symbol: oP
Constructor: ORC()
It is defined by three parameters \(a\), \(b\) and \(c\) with \(a < b < c\). Standardized primitive and conventional cells in the default orientation are
Transformation matrix from standardized primitive cell to standardized conventional cell is
K-path#
\(\mathrm{\Gamma-X-S-Y-\Gamma-Z-U-R-T-Z\vert Y-T\vert U-X\vert S-R}\)
Point |
\(\times\boldsymbol{b}_1^s\) |
\(\times\boldsymbol{b}_2^s\) |
\(\times\boldsymbol{b}_3^s\) |
|---|---|---|---|
\(\mathrm{\Gamma}\) |
\(0\) |
\(0\) |
\(0\) |
\(\mathrm{R}\) |
\(1/2\) |
\(1/2\) |
\(1/2\) |
\(\mathrm{S}\) |
\(1/2\) |
\(1/2\) |
\(0\) |
\(\mathrm{T}\) |
\(0\) |
\(1/2\) |
\(1/2\) |
\(\mathrm{U}\) |
\(1/2\) |
\(0\) |
\(1/2\) |
\(\mathrm{X}\) |
\(1/2\) |
\(0\) |
\(0\) |
\(\mathrm{Y}\) |
\(0\) |
\(1/2\) |
\(0\) |
\(\mathrm{Z}\) |
\(0\) |
\(0\) |
\(1/2\) |
Variations#
There are no variations for orthorhombic lattice.
One example is predefined: orc with
\(a = \pi\), \(b = 1.5\pi\) and \(c = 2\pi\).
Examples#
Brillouin zone and default kpath#
import wulfric as wulf
cell = wulf.cell.get_cell_example("ORC")
backend = wulf.visualization.PlotlyBackend()
backend.plot(cell, kind="brillouin-kpath")
# Save an image:
backend.save("orc_reciprocal.png")
# Interactive plot:
backend.show()
Primitive and Wigner-Seitz cells#
Click on the legend to hide a cell.
import wulfric as wulf
cell = wulf.cell.get_cell_example("ORC")
backend = wulf.visualization.PlotlyBackend()
backend.plot(cell, kind="primitive", label="primitive", color="black")
backend.plot(cell, kind="wigner-seitz", label="wigner-seitz", color="green")
# Save an image:
backend.save("orc_real.png")
# Interactive plot:
backend.show()
Cell standardization#
Lengths of the lattice vectors have to satisfy \(\vert\boldsymbol{a}_1^s\vert < \vert\boldsymbol{a}_2^s\vert < \vert\boldsymbol{a}_3^s\vert\) for the primitive cell in the standard form.
If \(\vert \boldsymbol{a}_3\vert > \vert \boldsymbol{a}_2\vert > \vert \boldsymbol{a}_1\vert\), then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad \boldsymbol{S}^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\end{split}\]If \(\vert \boldsymbol{a}_3\vert > \vert \boldsymbol{a}_1\vert > \vert \boldsymbol{a}_2\vert\), then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (-\boldsymbol{a}_2, -\boldsymbol{a}_1, -\boldsymbol{a}_3)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \qquad \boldsymbol{S}^{-1} = \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}\end{split}\]If \(\vert \boldsymbol{a}_2\vert > \vert \boldsymbol{a}_3\vert > \vert \boldsymbol{a}_1\vert\), then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (-\boldsymbol{a}_1, -\boldsymbol{a}_3, -\boldsymbol{a}_2)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix} \qquad \boldsymbol{S}^{-1} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}\end{split}\]If \(\vert \boldsymbol{a}_2\vert > \vert \boldsymbol{a}_1\vert > \vert \boldsymbol{a}_3\vert\), then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (\boldsymbol{a}_3, \boldsymbol{a}_1, \boldsymbol{a}_2)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \qquad \boldsymbol{S}^{-1} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\end{split}\]If \(\vert \boldsymbol{a}_1\vert > \vert \boldsymbol{a}_3\vert > \vert \boldsymbol{a}_2\vert\), then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (\boldsymbol{a}_2, \boldsymbol{a}_3, \boldsymbol{a}_1)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \qquad \boldsymbol{S}^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\end{split}\]If \(\vert \boldsymbol{a}_1\vert > \vert \boldsymbol{a}_2\vert > \vert \boldsymbol{a}_3\vert\), then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (-\boldsymbol{a}_3, -\boldsymbol{a}_2, -\boldsymbol{a}_1)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix} = \boldsymbol{S}^{-1} = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix}\end{split}\]
Note
All six changes of the cell preserve the handiness of the original one.
Edge cases#
If \(a = b \ne c\) or \(a = c \ne b\) or \(b = c \ne a\), then the lattice is Tetragonal (TET).
If \(a = b = c\), then the lattice is Cubic (CUB).