Dear Prof. Stefaan Cottenier
I want to make a struct file for a rhombohedral compound but with hexagonal
symmetry. How can i do that?
the positions in rhomboherdral are (0 0 0) (2/9 2/9 2/9) (7/9 7/9 7/9) with
166-R3m
the equivalent positions by using rhom2hex are (0 0 0) (0 0 0.22) (0 0 0.77)
with H space group, with this positions WIEN2K suggest 183 (P 6 m m) space
group which seems it is not correct because i think in regard to
crystallography, the number of atom during the conversion from rhombohedral to
hexagonal may change.
On Thursday, 6 February 2014, 9:50, negin kamali <negin_kamali6...@yahoo.com>
wrote:
Dear Wien2K Users,
I want to run the sm-element in hexagonal symmetry. Sm-element is a
rhombohedral structure with 166(R-3m space group) and a=8.966A and alpha=23.13
and the atomic position are (0 0 0) (2/9 2/9 2/9) (7/9 7/9 7/9).
I know, In WIEN2K, when we want to run rhombohedral structure, we must to
change the lattice constant to hexagonal which is in sm case a=b=3.621 and
c=26.25 A (aphlfa=beta=90 and gamma=120), but i could not determine the
positions in hexagonal axes.As you know in crystallography the R-3m space group
has two sub-state first r-axes an second h-axes.
So, i want to know;
First: how the number of atoms are changed in this conversion (from
rhombohedral to hexagonal symmetry)? (it means, is it necessary to change
number of atom when i want to convert a rhombohedral symmetry to hexagonal)
Second: What are the atomic positions in h-axes?
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