[Wien] In the calculation of elastic properties why ​​​​​rhombohedral lattice parameter needs same k-mesh as used in pristine case

[Dr. K. C. Bhamu]

Dear Wien2k users,

I am running a cubic (F) system having a=b=c=10.446600Bohr. The default
k-mesh size is 14 14 14 (3000kpt in FBZ).
I applied tetragonal and rhombohedral strain on optimized lattice
parameters (on above a,b,c) and the resultant strained a/b/c are 10.429247/
10.429247/10.481393 Bohr for tetragonal strain and 7.386862/7.386862/
18.184512 Bohr for rhombohedral strain.

My script advise me to use 20 20 8 kmesh size for the rhombohedral case and
"14 14 14" (as there is only slight change in a/b/c) for tetragonal case.

But for the rhombohedral case I am not able to use 20 20 8 mesh size and
the kgen program suggest me to use equal mesh in x and z direction (see
below in blue).

Could you someone please advice me why I can not use the mesh size  20 20 8
for the rhombohedral case?

I may be convinced about my doubt because I am having  equal "length of
reciprocal lattice vectors:   1.041   1.041   1.041".  Am I right? If so,
then also I need to know why a lattice parameter which is almost double in
one direct wrt original cubic latice need the same k-points as used in case
of cubic case. The answer may be somewhere in symmetry operations but I
could not completely understand it.



           6  symmetry operations without inversion
 inversion added (non-spinpolarized non-so calculation)
  NUMBER OF K-POINTS IN WHOLE CELL: (0 allows to specify 3 divisions of G)
0
 length of reciprocal lattice vectors:   1.041   1.041   1.041   0.000
0.000   0.000
  Specify 3 mesh-divisions (n1,n2,n3):
20 20 8
 Lattice symmetry requires equal mesh in x and z direction
  Specify 3 mesh-divisions (n1,n2,n3):


regards
Bhamu
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