Regarding atomic positions in Al2o3

Dear Lammpser,

I am also going to create Al2O3 by “lattice command” but I am still unsuccessful.

unfortunately, as you know LAMMPS does not do symmetry operations so I need to type in all 30 basis atoms: 12 Al (Wyckoff position 12c) and 18 O (18e) for one unit cell of R-3c Al2O3.

So, I tried to get these 30 positions from an existing .cif file attached. but actually there is 46 atomic positions based on this .cif file. I checked several .cif file for Al2O3 but all of them have 46 atom positions!

So, any help is highly appreciated regarding calculating such 30 Wyckoff positions.

Best regards,

Al2O3-corundum.cif (1.89 KB)

The best solution is to acquire some software that read CIF files, such as Jmol.

46 atoms does not seem to be reasonable. For starter, 46 is not a multiple of 5, so how does it work for Al2O3?

The first Google search of “al2o3 cif” came up with a cif file with 30 atoms: . May be you should try again?

Hope this helps.