IBRION = 6 elastic moduli from ionic relaxartion is negative
Posted: Wed Oct 28, 2015 5:28 pm
I am wondering if the negativity of elastic constant from ionic relaxation contribution is reasonable? If it is, does it mean the ionic relaxation mode is unstable?
I use M-P 888 Kpoints. The material is cubic BaSnO3, which is stable at 0 K.
Here is my INCAR and and OUTPUTS:
NSW = 1
ISIF = 3
IBRION = 6
ENCUT = 520
EDIFF = 1E-5
EDIFFG = -1E-8
ISYM = 0
ISMEAR = -1
SIGMA = 0.1
LPLANE = .TRUE.
NSIM = 4
LVTOT = .T.
KPAR = 4
SYMMETRIZED ELASTIC MODULI (kBar)
Direction XX YY ZZ XY YZ ZX
--------------------------------------------------------------------------------
XX 2467.6439 711.4288 708.2952 -3.4668 2.3529 0.1122
YY 711.4288 2461.1179 709.1010 -3.4270 0.3129 2.3419
ZZ 708.2952 709.1010 2461.9007 1.4224 0.0945 0.3244
XY -3.4668 -3.4270 1.4224 910.0718 -4.0215 -0.0116
YZ 2.3529 0.3129 0.0945 -4.0215 900.1764 -4.0591
ZX 0.1122 2.3419 0.3244 -0.0116 -4.0591 900.1012
--------------------------------------------------------------------------------
ELASTIC MODULI CONTR FROM IONIC RELAXATION (kBar)
Direction XX YY ZZ XY YZ ZX
--------------------------------------------------------------------------------
XX -0.2350 0.0134 0.0415 0.0068 0.0028 0.0152
YY 0.0134 -0.1952 0.0590 0.0003 0.0075 -0.0393
ZZ 0.0415 0.0590 -0.1998 0.0026 0.0006 0.0327
XY 0.0068 0.0003 0.0026 -0.0078 -0.0000 -0.0001
YZ 0.0028 0.0075 0.0006 -0.0000 -0.0078 0.0004
ZX 0.0152 -0.0393 0.0327 -0.0001 0.0004 -0.1603
--------------------------------------------------------------------------------
TOTAL ELASTIC MODULI (kBar)
Direction XX YY ZZ XY YZ ZX
--------------------------------------------------------------------------------
XX 2467.4089 711.4422 708.3368 -3.4599 2.3558 0.1274
YY 711.4422 2460.9227 709.1600 -3.4267 0.3205 2.3025
ZZ 708.3368 709.1600 2461.7010 1.4250 0.0950 0.3570
XY -3.4599 -3.4267 1.4250 910.0640 -4.0215 -0.0117
YZ 2.3558 0.3205 0.0950 -4.0215 900.1686 -4.0587
ZX 0.1274 2.3025 0.3570 -0.0117 -4.0587 899.9409
--------------------------------------------------------------------------------
I use M-P 888 Kpoints. The material is cubic BaSnO3, which is stable at 0 K.
Here is my INCAR and and OUTPUTS:
NSW = 1
ISIF = 3
IBRION = 6
ENCUT = 520
EDIFF = 1E-5
EDIFFG = -1E-8
ISYM = 0
ISMEAR = -1
SIGMA = 0.1
LPLANE = .TRUE.
NSIM = 4
LVTOT = .T.
KPAR = 4
SYMMETRIZED ELASTIC MODULI (kBar)
Direction XX YY ZZ XY YZ ZX
--------------------------------------------------------------------------------
XX 2467.6439 711.4288 708.2952 -3.4668 2.3529 0.1122
YY 711.4288 2461.1179 709.1010 -3.4270 0.3129 2.3419
ZZ 708.2952 709.1010 2461.9007 1.4224 0.0945 0.3244
XY -3.4668 -3.4270 1.4224 910.0718 -4.0215 -0.0116
YZ 2.3529 0.3129 0.0945 -4.0215 900.1764 -4.0591
ZX 0.1122 2.3419 0.3244 -0.0116 -4.0591 900.1012
--------------------------------------------------------------------------------
ELASTIC MODULI CONTR FROM IONIC RELAXATION (kBar)
Direction XX YY ZZ XY YZ ZX
--------------------------------------------------------------------------------
XX -0.2350 0.0134 0.0415 0.0068 0.0028 0.0152
YY 0.0134 -0.1952 0.0590 0.0003 0.0075 -0.0393
ZZ 0.0415 0.0590 -0.1998 0.0026 0.0006 0.0327
XY 0.0068 0.0003 0.0026 -0.0078 -0.0000 -0.0001
YZ 0.0028 0.0075 0.0006 -0.0000 -0.0078 0.0004
ZX 0.0152 -0.0393 0.0327 -0.0001 0.0004 -0.1603
--------------------------------------------------------------------------------
TOTAL ELASTIC MODULI (kBar)
Direction XX YY ZZ XY YZ ZX
--------------------------------------------------------------------------------
XX 2467.4089 711.4422 708.3368 -3.4599 2.3558 0.1274
YY 711.4422 2460.9227 709.1600 -3.4267 0.3205 2.3025
ZZ 708.3368 709.1600 2461.7010 1.4250 0.0950 0.3570
XY -3.4599 -3.4267 1.4250 910.0640 -4.0215 -0.0117
YZ 2.3558 0.3205 0.0950 -4.0215 900.1686 -4.0587
ZX 0.1274 2.3025 0.3570 -0.0117 -4.0587 899.9409
--------------------------------------------------------------------------------