Building up a face centered cubic lattice: Difference between revisions

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== Atomic position(s) on a cubic cell ==
* Number of atoms per cell: 4
* Coordinates:
Atom 1: <math> \left( x_1, y_1, z_1 \right) = \left( 0, 0, 0 \right) </math>
Atom 2: <math> \left( x_2, y_2, z_2 \right) = \left( 0 , \frac{l}{2}, \frac{l}{2}\right) </math>
Atom 3: <math> \left( x_3, y_3, z_2 \right) = \left( \frac{l}{2}, 0, \frac{l}{2} \right) </math>
Atom 4: <math> \left( x_4, y_4, z_2 \right) = \left( \frac{l}{2}, \frac{l}{2}, 0  \right) </math>
Cell dimensions:
*<math> a=b=c = l </math>
*<math> \alpha = \beta = \gamma = 90^0 </math>

Revision as of 18:36, 20 March 2007

  • Consider:
  1. a cubic simulation box whose sides are of length L
  2. a number of lattice positions, M given by M=4m3,

with m being a positive integer

  • The M positions are those given by:
{xa=ia×(δl)ya=ja×(δl)za=ka×(δl)}

where the indices of a given valid site are integer numbers that must fulfill the following criteria

  • 0≤ia<2m
  • 0≤ja<2m
  • 0≤ka<2m,
  • the sum of ia+ja+ka must be, for instance, an even number.

with δl=L/(2m)

Atomic position(s) on a cubic cell

  • Number of atoms per cell: 4
  • Coordinates:

Atom 1: (x1,y1,z1)=(0,0,0)

Atom 2: (x2,y2,z2)=(0,l2,l2)

Atom 3: (x3,y3,z2)=(l2,0,l2)

Atom 4: (x4,y4,z2)=(l2,l2,0)

Cell dimensions:

  • a=b=c=l
  • α=β=γ=900