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[Anglade Pierre-Matthieu <anglade@gmail.com>]

Hi Andrei,

Usually when one wants to make a surface calculation the procedure is the following :
1) choose the surface to be calculated (your case : 111)
2) find the smallest cell that based on a vector normal to the surface (in your case likely 111  will fit :-), and two vectors normals to the previous one (you may find the first by imposing the scalar product with 111 to be zero and the second by doing the vector product between the two previous vectors). 
3) find the coordinates of the atoms within this new cell. From the cell volume you can guess how many atoms you must find.
4) Input in abinit a cell based on the one you have just derived but with many time the (111) axis.
5) within this cell put a few layers of your atoms translated along the 111 axis. Don't forget that in order to get a surface you need at least one empty layer along 111.
6) Check the convergence of your results with respect to the number of vacuum and cristal layers. 

Let's proceed on an example: let'st take a simple cubic cristal.
1) we want a surface along 111
2) the smallest cell correponding to our criterions is
(111)(11-2)(-110)
3) the volume of this cell is 6 (if -6 we must change the vectors order). We must find six node/atoms matching this cell. For instance two of them are
at1 : (010)  -> (1/3,1/6,1/6)
at2 : (11-1) -> (1/3,2/3,0)
...
4) In this case all 3 vectors are orthogonal but not normalized it is then easy to input
rprim 1 0 0 0 1 0 0 0 1
acell sqrt(3)*a*N sqrt(6)*a sqrt(2)*a
(where a is your original cell parameter and N the number of layers you want to use)
5) supposed you have three layers (N=3) among which one is empty you have to input 
to coordinate per atoms in your cell :
at1 gives two sets of coordinates just like at2  and you may input
xcart
1/3*sqrt(3)*a  1/6*sqrt(6)*a 1/6*sqrt(2)*a
4/3*sqrt(3)*a  1/6*sqrt(6)*a 1/6*sqrt(2)*a
1/3*sqrt(3)*a  2/3*sqrt(6)*a  0 
4/3*sqrt(3)*a  2/3*sqrt(6)*a  0
...

Obviously I have supposed that you know what cristobalite is. 
Note that, since cristobalite is of the "I" kind (assuming you are correct) you may use a smaller cell than that of my example by taking (111)/2 ...

Hope this helps.
Regards

PMA

On Mon, Feb 16, 2009 at 4:43 AM, Andrey Chibisov <andreichibisov@yandex.ru> wrote:

Hello!
How construct of (111) surface of cristobalite with I42d structure?

--
Andrey Chibisov
Structural Ceramics Laboratory
Institute of Geology and Nature Management
Far Eastern Branch of Russian Academy of Sciences

-- 
Pierre-Matthieu Anglade