Building up a simple cubic lattice: Difference between revisions
		
		
		
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| z = k \times (\delta l) &; k=0,1,\cdots, m-1   | z = k \times (\delta l) &; k=0,1,\cdots, m-1   | ||
| \end{array} | \end{array} | ||
| \right. | \right. | ||
| </math> | </math> | ||
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| \right. | \right. | ||
| </math> | </math> | ||
| == Atomic position(s) on a cubic cell == | |||
| Atom 1: X=0, Y=0, Z=0. | |||
| Cell dimensions:  | |||
| *<math> a=b=c </math> | |||
| *<math> \alpha = \beta = \gamma = 90^0 </math> | |||
Revision as of 19:34, 19 March 2007
- Consider:
- a cubic simulation box whose sides are of length
- a number of lattice positions, given by with being a positive integer
- The positions are those given by:
where
Atomic position(s) on a cubic cell
Atom 1: X=0, Y=0, Z=0.
Cell dimensions: