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}
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</math>
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== 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 20:34, 19 March 2007

  • Consider:
  1. a cubic simulation box whose sides are of length
  2. 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: