Building up a simple cubic lattice
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- Consider:
- a cubic simulation box whose sides are of length \(\left. L \right. \)
- a number of lattice positions, \( \left. M \right. \) given by \( \left. M = m^{3} \right. \) ; with \( m \) being a positive integer
- The \( \left. M \right. \) positions are those given by:
\[ \left\{ \begin{array}{ll} x = i \times (\delta l) &; i=0,1,\cdots, m-1 \\ y = j \times (\delta l) &; j=0,1,\cdots, m-1 \\ z = k \times (\delta l) &; k=0,1,\cdots, m-1 \end{array} \right. \]
where \( \left. \delta l = L/m \right. \)
[edit] Atomic position(s) on a cubic cell
- Number of atoms per cell: 1
- Coordinates:
Atom 1: \( \left( x_1, y_1, z_1 \right) = \left( 0, 0, 0 \right) \)
Cell dimensions:
- \( a=b=c \)
- \( \alpha = \beta = \gamma = 90^0 \)
