Building up a square lattice: Difference between revisions

Revision as of 15:16, 20 March 2007

a square simulation box whose sides are of length $L$
a number of lattice positions, $M$ given by $M = m^{2}$ ; with $m$ being a positive integer

{\begin{matrix} x = i \times (δ l) & ; i = 0, 1, \dots, m - 1 \\ y = j \times (δ l) & ; j = 0, 1, \dots, m - 1 \end{matrix}

where $δ l = L / m$

Atom 1: $(x_{1}, y_{1}) = (0, 0)$

@@ Line 1: / Line 1: @@
 * Consider:
-# a cubic simulation box whose sides are of length <math>\left. L  \right. </math>
+# a square simulation box whose sides are of length <math>\left. L  \right. </math>
-# a number of lattice positions, <math> \left. M \right. </math> given by <math> \left. M = m^{3}    \right. </math> ; with <math> m </math> being a positive integer
+# a number of lattice positions, <math> \left. M \right. </math> given by <math> \left. M = m^{2}    \right. </math> ; with <math> m </math> being a positive integer
 * The <math> \left. M \right. </math> positions are those given by:
@@ Line 8: / Line 8: @@
 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}
@@ Line 21: / Line 20: @@
 </math>
-== Atomic position(s) on a cubic cell ==
+== Atomic position(s) on a square cell ==
 * Number of atoms per cell: 1
 * Coordinates:
-Atom 1: <math> \left( x_1, y_1, z_1 \right) = \left( 0, 0, 0 \right) </math>
+Atom 1: <math> \left( x_1, y_1\right) = \left( 0, 0 \right) </math>