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["waigen zhang" <waigen.zhang@gmail.com>]

Dear Lallet,
Thank you for your replies.
Have you tried to calculate the elastic constant? There are several ways to define the structural geometry for alpha-Al2O3 and it seems the nonzero elements in the elastic tensor are different if we choose different geometry.

W.G.Zhang

On 6/9/07, Lallet François <francois.lallet@umontreal.ca> wrote:

Dear Zhang,
I followed the instruction on webpage http://cst-www.nrl.navy.mil/lattice/struk/d5_1.html. and I studied the relaxation of the alpha Al2O3 rhombohedral unit cell with ionmov=2; optcell=2, ngkpt=2 2 2 and ecut was fixed to 15Ha (I used PAW pseudopotentials produced with atompaw) in LDA and GGA. The energy was converged within 0.2eV/atom and the forces relaxed below 10^-4Ha/(Bohr.atom), which was sufficient to produce relaxed lattice parameters and internal ionic positions (see Wyckoff "Crystal Structure" is you have any doubts) in very good agreement with experiments. I calculated also the bulk modulus by fitting E(V) with a Murnaghan equation of state and the results are very good... but I don't know whether the hexagonal unit cell is better than the rhombohedral one for this purpose...
Best,

F. Lallet

-----Original Message-----
From: waigen zhang [mailto:waigen.zhang@gmail.com]
Sent: Fri 6/8/2007 3:52 PM
To: forum@abinit.org
Subject: [abinit-forum] structural geometry of alfa-Al2O3

Dear all,
Dose anybody have idea about the structural geometry about alfa-Al2O3 ?
Should we use the rhombohedral or hexagonal primitive cell for ABINIT
calculation of elastic constant?
I followed the instruction on webpage
http://cst-www.nrl.navy.mil/lattice/struk/d5_1.html.
But still not sure about that. Your suggestion is welcome.

Best,
W.G.Zhang