Building up a face centered cubic lattice

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A face centered cubic lattice

• Consider:

1. a cubic simulation box whose sides are of length \(\left. L \right. \)
2. a number of lattice positions, \( \left. M \right. \) given by \( \left. M = 4 m^3 \right. \),

with \( m \) being a positive integer

• The \( \left. M \right. \) positions are those given by:

\[ \left\{ \begin{array}{l} x_a = i_a \times (\delta l) \\ y_a = j_a \times (\delta l) \\ z_a = k_a \times (\delta l) \end{array} \right\} \]

where the indices of a given valid site are integer numbers that must fulfill the following criteria

• \( 0 \le i_a < 2m \)
• \( 0 \le j_a < 2m \)
• \( 0 \le k_a < 2m \),
• the sum of \( \left. i_a + j_a + k_a \right. \) must be, for instance, an even number.

with \( \left. \delta l = L/(2m) \right. \)

Atomic position(s) on a cubic cell

• Number of atoms per cell: 4
• Coordinates:

Atom 1: \( \left( x_1, y_1, z_1 \right) = \left( 0, 0, 0 \right) \)

Atom 2: \( \left( x_2, y_2, z_2 \right) = \left( 0 , \frac{l}{2}, \frac{l}{2}\right) \)

Atom 3: \( \left( x_3, y_3, z_2 \right) = \left( \frac{l}{2}, 0, \frac{l}{2} \right) \)

Atom 4: \( \left( x_4, y_4, z_2 \right) = \left( \frac{l}{2}, \frac{l}{2}, 0 \right) \)

Cell dimensions:

• \( a=b=c = l \)

• \( \alpha = \beta = \gamma = 90^0 \)