[Pw_forum] Relaxation of low symmetry lattices

[Manoj Srivastava]

Dear users,
 Thank you all for quick reply. Through all the replies I am under the
impression that the forces can never be actually zero, even for perfect
FCC lattice. I dont know the exact reason yet, whether it is just
numerical or something bad on my side. So, I would like to give it one
more shot, and i have some questions on that.

 Duy Le: I will try using crystal and report to the forum. Although
relaxation stops after completing one scf cycle, so atoms actually dont
move, but still nonzero forces.

 S. Baroni: When you use cubic symmetry, do you get forces on all atoms
zero. Did you do the calculation or is this your intution? also, can one
use cubic symmetry for low symmetry unit cell as in my case? I increased
the kinetic energy cutoff to a really high value 150Ry !!  forces are -
     atom   1 type  1   force =    -0.00000074   -0.00000315    0.00000310
     atom   2 type  1   force =     0.00000084   -0.00000391    0.00000354
     atom   3 type  1   force =    -0.00000128    0.00000470   -0.00000401
     atom   4 type  1   force =     0.00000111    0.00000044   -0.00000083
     atom   5 type  1   force =     0.00000008    0.00000193   -0.00000181

which is one order of magnitude smaller than before :) but the KE cutoff
is ridiculously high! I also increase k points with keeping KE cutoff 50
Ry, and that gave me -

     atom   1 type  1   force =    -0.00000103    0.00000086   -0.00000038
     atom   2 type  1   force =     0.00000535   -0.00000155   -0.00000025
     atom   3 type  1   force =     0.00000643    0.00000039   -0.00000244
     atom   4 type  1   force =    -0.00000603   -0.00000141    0.00000310
     atom   5 type  1   force =    -0.00000471    0.00000171   -0.00000003 

 which is also one order of magnitude smaller than before, so i am
guessing there might me some other parameters that are not set up
correctly and causing symmetry breaking? Would someone mind to tell me? I
dont know much about FFT grid, Isn't FFT dimension determined by the KE
cutoff? Is there anyother way of changing it?

Gabriele: Would you mind to tell me how can i get rid of FFT
incompatibility issue?

Andrea: I changed it from e(-8) to e(-10), still no improvement.

Once again, thank you all for your help. 

Regards, 
Manoj Srivastava
University of Florida
Gainesville, FL



On Thu, 8 Oct 2009, Lorenzo Paulatto wrote:

> In data 08 ottobre 2009 alle ore 10:46:13, Gabriele Sclauzero  
> <sclauzer at sissa.it> ha scritto:
> > if you use cubic it should find 2 more, but they are not compatible with  
> > your FFT grid (see at the beginning of the output).
> 
> The code will only use the 2 additional operations if nr3 is a multiple of  
> nr1 and nr2, which is weird as it is, by default, much smaller in your  
> case. Anyway, if you set nr3 equal to nr1 it will use all the symmetry  
> operations.
> 
> For the symmetry experts: could you please check if the code behaviour is  
> correct?
> 
> cheers
> 
> -- 
> Lorenzo Paulatto
> SISSA  &  DEMOCRITOS (Trieste)
> phone: +39 040 3787 511
> skype: paulatz
> www:   http://people.sissa.it/~paulatto/
> 
>      *** save italian brains ***
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