Building up a square lattice: Difference between revisions

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m (New page: * Consider: # a cubic 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...)
 
mNo edit summary
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* Consider:
* 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:  
* The <math> \left. M \right. </math> positions are those given by:  
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x = i \times (\delta l) &; i=0,1,\cdots, m-1 \\
x = i \times (\delta l) &; i=0,1,\cdots, m-1 \\
y = j \times (\delta l) &; j=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}
\end{array}


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</math>
</math>


== Atomic position(s) on a cubic cell ==
== Atomic position(s) on a square cell ==


* Number of atoms per cell: 1  
* Number of atoms per cell: 1  
* Coordinates:
* 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>

Revision as of 14:16, 20 March 2007

  • Consider:
  1. a square 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 square cell

  • Number of atoms per cell: 1
  • Coordinates:

Atom 1: