Dear Lammps users,
Does anyone have a clue on how to decide on the appropriate box size for a rotated unit cell?
I am applying the rotation of
orient x 1 0  1 orient y 1 1  1 orient z 1 2 1
to a periodic system containing FccFe with the lattice parameter of 3.65 Angstroms.
Finding the lattice spacing calculated by either LAMMPS or hand, is a piece of cake. When I, however, use a multiplicative of the lattice spacing as the box size, I still don’t get a correct arrangement of atoms and I can see lots of internal interfaces and overlapping of atoms. Using the minimization of energy also proves that I don’t have a fcc at hand.
I would be appreciative if someone could give some advices.
regards,
– Hamid

You need to read a solidstate physics book on lattices and figure out
the periodic replication distance for your orientation. In 2d you can
generally draw it on a piece of paper. For orientaion [abc[, the interplane
distance is something like 1/(a^2 + b^2 + c^2), and you need to
figure out that every Nth plane is identical. But I’m not looking
at a text book …
Steve
Hi,
Your periodic length for this particular orientation in X would be 2^1/2 similar Y would be 3^1/2 and Z would be 6^1/2.
The unit cell volume would now be 6a^3 , ‘a’ is lattice parameter.
Regards
Ilaksh