Building up a simple cubic lattice: Difference between revisions

Revision as of 19:32, 19 March 2007

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

\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 $\left.\delta l=L/m\right.$

@@ Line 1: / Line 1: @@
 * Consider:
-# a Cubic Simulation box of length <math>\left. L  \right. </math>
+# a cubic simulation box whose sides are of length <math>\left. L  \right. </math>
-# a numer of lattice positions, <math> \left. M \right. </math> given by:
+# 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
-: <math> \left. M = m^3    \right. </math>
+* The <math> \left. M \right. </math> positions are those given by:
+:<math>
-: with <math> m </math> being a positive integer
-* The <math> \left. M \right. </math> positions are those given by:
-<math>
 \left\{ \begin{array}{ll}
 x = i \times (\delta l) &; i=0,1,\cdots, m-1 \\
@@ Line 18: / Line 13: @@
 </math>
-with:
+where
 <math>
 \left.