Base-centred orthorhombic (ORCC)#
Pearson symbol: oS
Constructor: ORCC()
It is defined by three parameters \(a\), \(b\) and \(c\) with \(a < b\). Standardized primitive and conventional cells in the default orientation are
Transformation matrix from standardized primitive cell to standardized conventional cell is
K-path#
\(\mathrm{\Gamma-X-S-R-A-Z-\Gamma-Y-X_1-A_1-T-Y\vert Z-T}\)
Point |
\(\times\boldsymbol{b}_1^s\) |
\(\times\boldsymbol{b}_2^s\) |
\(\times\boldsymbol{b}_3^s\) |
|---|---|---|---|
\(\mathrm{\Gamma}\) |
\(0\) |
\(0\) |
\(0\) |
\(\mathrm{A}\) |
\(\zeta\) |
\(\zeta\) |
\(1/2\) |
\(\mathrm{A_1}\) |
\(-\zeta\) |
\(1-\zeta\) |
\(1/2\) |
\(\mathrm{R}\) |
\(0\) |
\(1/2\) |
\(1/2\) |
\(\mathrm{S}\) |
\(0\) |
\(1/2\) |
\(0\) |
\(\mathrm{T}\) |
\(-1/2\) |
\(1/2\) |
\(1/2\) |
\(\mathrm{X}\) |
\(\zeta\) |
\(\zeta\) |
\(0\) |
\(\mathrm{X_1}\) |
\(-\zeta\) |
\(1-\zeta\) |
\(0\) |
\(\mathrm{Y}\) |
\(-1/2\) |
\(1/2\) |
\(0\) |
\(\mathrm{Z}\) |
\(0\) |
\(0\) |
\(1/2\) |
Variations#
There are no variations for base-centered orthorombic.
One example is predefined: orcc with
\(a = \pi\), \(b = 1.3\pi\) and \(c = 1.7\pi\).
Examples#
Brillouin zone and default kpath#
import wulfric as wulf
cell = wulf.cell.get_cell_example("ORCC")
backend = wulf.visualization.PlotlyBackend()
backend.plot(cell, kind="brillouin-kpath")
# Save an image:
backend.save("orcc_reciprocal.png")
# Interactive plot:
backend.show()
Primitive, Wigner-Seitz and conventional cells#
Click on the legend to hide a cell.
import wulfric as wulf
cell = wulf.cell.get_cell_example("ORCC")
backend = wulf.visualization.PlotlyBackend()
backend.plot(cell, kind="primitive", label="primitive", color="black")
backend.plot(cell, kind="conventional", label="conventional", color="blue")
backend.plot(cell, kind="wigner-seitz", label="wigner-seitz", color="green")
# Save an image:
backend.save("orcc_real.png")
# Interactive plot:
backend.show()
Cell standardization#
Length of third vector of the primitive cell has to be different from the lengths of the first two vectors of the primitive cell. Together with the \(a < b\) we arrive at the following condition of the angles of the primitive cell in a standard form: \(\alpha^s = \beta^s = 90^{\circ}\) and \(\gamma^s > 90^{\circ}\). Wulfric uses angles of the primitive cell for standardization.
If \(\alpha = \beta = \frac{\pi}{2}\) and \(\gamma > \frac{\pi}{2}\) 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 \(\alpha = \beta = \frac{\pi}{2}\) and \(\gamma < \frac{\pi}{2}\) 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 \(\beta = \gamma = \frac{\pi}{2}\) and \(\alpha > \frac{\pi}{2}\) 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 \(\beta = \gamma = \frac{\pi}{2}\) and \(\alpha < \frac{\pi}{2}\) then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (\boldsymbol{a}_3, -\boldsymbol{a}_2, \boldsymbol{a}_1)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \qquad \boldsymbol{S}^{-1} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\end{split}\]If \(\alpha = \gamma = \frac{\pi}{2}\) and \(\beta > \frac{\pi}{2}\) 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 \(\alpha = \gamma = \frac{\pi}{2}\) and \(\beta < \frac{\pi}{2}\) then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (\boldsymbol{a}_1, -\boldsymbol{a}_3, \boldsymbol{a}_2)\]and
\[\begin{split}\boldsymbol{S} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} \qquad \boldsymbol{S}^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}\end{split}\]
Note
All six changes of the cell preserve handiness of the original one.
Edge cases#
If \(a = b\), then the lattice is Tetragonal (TET).
If \(a = b = \sqrt{2} c\), then the lattice is Cubic (CUB).