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- From: mmikami@rc.m-kagaku.co.jp
- To: forum@abinit.org
- Subject: Re: distorted BaTiO3
- Date: Wed, 15 Oct 2003 10:57:23 +0900
Hello,
(I will redirect my reply to ML, not in the direct correspondence ...
the reference is :
http://www.abinit.org/wws/arc/forum/2003-10/msg00030.html)
OK, you are trying on the basis of GS calculation.
And I overlooked the slight differences among xcart/acell.
(But I am not sure of the objective of the ten datasets without comments.)
My suggestions are:
1) the 2nd derivative equation will be, for example,
(Etot(delta_x) - 2Etot(0)+Etot(- delta_x))/(delta_x)^2
Here, Etot(0) means the total energy at the equilibrium position, and delta_x means the deviation from the equilibrium position ...
The above equation may be transformed using forces (I mean, e.g., F(0)=(E(delta_x/2)-E(-delta_x/2))/delta_x, so the above equation can be transformed by using "F"). Numerically, the equation based on forces may be preferred. (There should be references ... )
2) the deviation "delta_x" could be carefully tested.
Please redirect to the forum ML ... a concise/specific summary would be welcome.
(e.g. which xcart corresponds to Etot(0) in the above, and which xcart corresponds to Etot(delta_x), ... etotal also could be cited ... etc.)
Regards,
Masayoshi
On 2003.10.14, at 21:01 Asia/Tokyo, fenglan hu wrote:
Hello,
I am sorry the xcart7,xcart8,xcart9,xcart10 is error. They are as follows:
xcart7
0 0 0
3.72460921843687 3.64960921843688 3.72460921843687
0 3.64960921843688 3.72460921843687
3.72460921843687 0 3.72460921843687
3.72460921843687 3.64960921843688 0
xcart8
0 0 0 &nb! sp;
3.72460921843687 3.72460921843687 3.72460921843687
0 3.72460921843687 3.72460921843687
3.72460921843687 0 3.72460921843687
3.72460921843687 3.72460921843687 0
xcart9
0 0 0
3.72460921843687 3.79960921843688 3.72460921843687
0 3.79960921843688 3.72460921843687
3.72460921843687 0 3.72460921843687
3.72460921843687 3.79960921843688 0
xcart10
0 0 0
3.72460921843687 3.83960921843688 3.72460921843687
0 3.83960921843688 3.72460921843687
3.72460921843687 0 3.72460921843687
3.72460921843687 3.83960921843688 0
I have used tolvrs 1.0d-20 ,this also appears the problem,
as toldfe 1.0d-7
the second derivatives of total energy with respect to atomic displacements equals to
[total energy (atomic displacements is not zero)- total energy (atomic displacements is not zero)]
multiply two then divide the square of the atomic displacements .
for xcart6 ecut=45
rms dE/dt= 0.0000E+00; max dE/dt= 0.0000E+00; dE/dt below (all hartree)
1 0.000000000000 0.000000000000 0.000000000000
2 0.000000000000 0.000000000000 0.000000000000
3 0.000000000000 0.000000000000 0.000000000000
4 0.000000000000 0.000000000000 0.000000000000
5 0.000000000000 0.000000000000 0.000000000000
cartesian forces (hartree/bohr) at end:
1 0.00000000000000 0.00000000000000 0.00000000000000
2 0.00000000000000 0.00000000000000 0.00000000000000
3 0.00000000000000 &nbs! p; 0.00000000000000 0.00000000000000
4 0.00000000000000 0.00000000000000 0.00000000000000
5 0.00000000000000 0.00000000000000 0.00000000000000
frms,max,avg= 0.0000000E+00 0.0000000E+00 0.000E+00 0.000E+00 0.000E+00 h/b
Cartesian components of stress tensor (hartree/bohr^3)
sigma(1 1)= -1.15995004E-05 sigma(3 2)= 0.00000000E+00
sigma(2 2)= -3.03387139E-04 sigma(3 1)= 0.00000000E+00
sigma(3 3)= -1.15995004E-05 sigma(2 1)= 0.00000000E+00
for xcart6 ecut=50
rms dE/dt= 0.0000E+00; max dE/dt= 0.0000E+00; dE/dt below (all hartree)
1 0.000000000000 0.000000000000 0.000000000000
2 0.000000000000 0.000000000000 0.000000000000
3 0.000000000000 0.000000000000 0.000000000000
4 0.000000000000 0.000000000000 0.000000000000
5 0.000000000000 0.000000000000 0.000000000000
cartesian forces (hartree/bohr! ) at end:
1 0.00000000000000 0.00000000000000 0.00000000000000
2 0.00000000000000 0.00000000000000 0.00000000000000
3 0.00000000000000 0.00000000000000 0.00000000000000
4 0.00000000000000 0.00000000000000 0.00000000000000
5 0.00000000000000 0.00000000000000 0.00000000000000
frms,max,avg= 0.0000000E+00 0.0000000E+00 0.000E+00 0.000E+00 0.000E+00 h/b
Cartesian components of stress tensor (hartree/bohr^3)
sigma(1 1)= -3.56049023E-06 sigma(3 2)= 0.00000000E+00
sigma(2 2)= -2.93177716E-04 sigma(3 1)=&nb! sp; 0.00000000E+00
sigma(3 3)= -3.56049023E-06 sigma(2 1)= 0.00000000E+00
&nbs ! p; & nbsp;
for xcart1 ecut=45
rms dE/dt= 6.0808E-03; max dE/dt= 1.4538E-02; dE/dt below (all hartree)
1 0.000000000000 -0.002120910278 0.000000000000
2 0.000000000000 0.014538059644 0.000000000000
3 0.000000000000 0.001349765952 0.000000000000
4 0.000000000000 -0.018306719901 0.000000000000
5 0.000000000000 0.001349765952 0.000000000000
cartesian forces (hartree/bohr) at end:
1 0.00000000000000 0.00020541040071 0.00000000000000
2 0.00000000000000 -0.00210217594924 0.00000000000000
3 0.00000000000000 -0.00027534471990 0.00000000000000
4 0.00000000000000 0.00244745498835 0.00000000000000
5 0.00000000000000 -0.00027534471990 0.00000000000000
frms,max,avg= 8.4075373E-04 2.4474550E-03 0.000E+00 8.838E-05 0.000E+00 h/b
Cartesian components of stress tensor (hartree/bohr^3)
sigma(1 1)= -1.16069878E-05 sigma(3 2)= 0.00000000E+00
sigma(2 2)= -3.07659430E-04 sigma(3 1)= 0.00000000E+00
sigma(3 3)= -1.16069878E-05 sigma(2 1)= 0.00000000E+00
for xcart1 ecut=50
rms dE/dt= 5.2491E-03; max dE/dt= 1.3501E-02; dE/dt below (all hartree)
1 0.000000000000 0.002135000854 0.000000000000
2 0.000000000000 -0.014934690746 0.000000000000
3 0.000000000000 -0.001304869299 0.000000000000
4 0.000000000000 0.013501114188 0.000000000000
5 0.000000000000 -0.001304869299 0.000000000000
cartesian forces (hartree/bohr) at end:
1 0.00000000000000 -0.00034860611795 0.00000000000000
2 0.00000000000000 0.00201587304958 0.00000000000000
3 0.00000000000000 0.00012788177091 0.00000000000000
4 0.00000000000000 -0.00192303047345 0.00000000000000
5 0.00000000000000 0.00012788177091 0.00000000000000
frms,max,avg= 7.2645367E-04 2.0158730E-03 0.000E+00 5.287E-05 0.000E+00 h/b
Cartesian components of stress tensor (hartree/bohr^3)
sigma(1 1)= -4.19905181E-06 sigma(3 2)= 0.00000000E+00
sigma(2 2)= -2.98402143E-04 sigma(3 1)= 0.00000000E+00
sigma(3 3)= -4.19905181E-06 sigma(2 1)= 0.00000000E+00
input file for the RF calculation
ndtset 3
kptopt1 1
tolvrs1 1.0d-20
iscf1 3
rfelfd2 2
rfdir2 1 0 0
nqpt2 1
qpt2 0.0 0.0 0.0
getwfk2 -1
kptopt2 2
iscf2 -3
tolwfr2 1.0d-24
rfphon3 1
rfatpol3 1 5
rfelfd3 3
rfdir3 1 1 1
nqpt3 1
qpt3 0.0 0.0 0.0
getwfk3 -2
getddk3 -1
kptopt3 2
tolvrs3 1.0d-10
iscf3 3
acell 3*7.4492
rprim 1.0 0.0 0.0
0.0 1.0 0.0
0.0 0.0 1.0
ixc 3
ntype 3
znucl 56 22 8
natom 5
type 1 2 3 3 3
xred
0.0 0.0 0.0
0.5 0.5 0.5
0.0 0.5 0.5
0.5 0.0 0.5
0.5 0.5 0.0
ecut 45.0
nband 20
kptrlatt
6 0 0
0 6 0
0 0 6
nstep 300
diemac 5.0
Thank you ,
hfl
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- distorted BaTiO3, kxyhu, 10/13/2003
- Re: [abinit-forum] distorted BaTiO3, mmikami, 10/14/2003
- <Possible follow-up(s)>
- Re: distorted BaTiO3, mmikami, 10/15/2003
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