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 $\left.M=4m^{3}\right.$,

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

• $0\leq i_{a}<2m$
• $0\leq j_{a}<2m$
• $0\leq k_{a}<2m$,
• 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:

• $a=b=c=l$

• $\alpha =\beta =\gamma =90^{0}$