No.
P6/mmc is NOT a rhombohedral space group, but a hexagonal one.
Thus NO conversion of positions .....
So just give a,a,c,alpha,beta,gamma (120) and
1/3,2/3,0.25 as position of an atom.
The second atom at 2/3,1/3,.75 will be created automatically if you use
the spacegroup symbol.
PS: make sure you enter 1/3 and not just 0.3333
On 01/27/2014 09:17 AM, saurabh singh wrote:
I am making struct file for Mg (magnesium) using wien2k_13.
I have following information of Mg from mincryst
(http://database.iem.ac.ru/mincryst/s_carta.php?MAGNESIUM+2671)
structure : Hexagonal
Spacegroup: P6(3)/mmc with spacegroup No. 194.
lattice parameter a=b= 3.2095, alpha=beta= 90, gama = 120.
lattice parameter I am giving of hexagonal structure, where as atomic
positions I have given of rhombohedral as suggested in wien2k manual.
when I am making struct file by using the spacegroup or spacegroup
number i.e. 194.
then the struct file is not making properly. and while initialising the
struct file its spacegroup no. is changing from 194 to 191 i.e.
correspond to spacegroup P6/mmm.
If I am using lattice type setting (L) while making struct file i.e.
when it ask :
would you like to enter Spacegroup or Lattice (S/L)(def=S)?
then it make the proper struct file and also give the same spcegroup
while initialising the struct file.
Therefore I want to know why it is showing problem of changing
spacegroup while using the spacegroup setting, where as it gives proper
struct file and iitialisation in Lattice type setting.
Is it due to Hexagonal structure of system or due to any other problem.
If you need any further details please let me know.
Thanks & Regards
Saurabh Singh IIT mandi
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