Hi Luke,
See the comments below.
2010/6/7 JhonY. I. <[email protected]...>:
Dear Dongshan:
Hello, Dongshan?
First many thanks for all your pevious helps really.
I've seen your previous post in LAMMPS
archive(http://sourceforge.net/mailarchive/message.php?msg_id=e6628c5a1003032314s7b89b946p210864050a89bbe2%40mail.gmail.com)
and I'd like to indicate on some fault of your calculation and also ask some
questions on it.1. The result of that post was as follows.
step flux
700000 0
...
4268000 11355.837You calculated kappa in the following way.
kappa= dJ/dt / (dT/dz) = 3.18eV/ps / (0.3K/A) / (42.6*3.4A^2) ~ 118 W/K/m
But if your unit is metal and time step is 1 fs, I think 3.18 eV/ps should
be
changed to 31.8 eV/ps since it is 11355.837 eV/((4268000-700000)*0.001 ps),
and your thermal conductivity will be increased ten times, I suppose.
What do you think on it?In addition, I think that the equation your derived is correct in principle.
Is it right?
How you can get 11355.837 eV/((4268000-700000)*0.001 ps) equal to 31.8
eV/ps? It's actually 3.18eV/ps.
2. Incidientally asking, I've got the tmp.profile as follows.
101000 25
1 -0.7175 0 0
2 0.7175 23.69 291.21
3 2.1525 24.31 299.481
4 3.5875 16 315.053
5 5.0225 16 303.967
6 6.4575 16 298.175
7 7.8925 16 308.913
8 9.3275 23.84 281.964
9 10.7625 24.16 281.042
10 12.1975 16 290.864
11 13.6325 16 307.152
12 15.0675 16 309.009
13 16.5025 16 304.191
14 17.9375 23.35 339.291
15 19.3725 24.65 311.915
16 20.8075 16 313.163
17 22.2425 16 304.582
18 23.6775 16 300.64
19 25.1125 16 282.925
20 26.5475 23.56 280.165
21 27.9825 24.44 292.552
22 29.4175 16 298.56
23 30.8525 16 295.138
24 32.2875 16 289.97
25 33.7225 16 271.865
...These values are "Layer Coord Ncount v_temp" in order.
I suppose that v_temp is the average temperature of corresponding layer.
The z length was 34.439999 A and I determine delta as 34.439999/24 ~ 1.435
A.
I think the first 1 layer is the same with 25 layer (34.439999 - 0.7175 ~
33.7225) and the 1 layer may be ignored. (It should appear by a certain bug
of LAMMPS, I guess)I think we can obtain temperature gradient as follows.
dT/dz = (339.291 K(the hottest 14th layer)-271.865 K (the coldest 25 th
layer, the same with 1th layer) / (33.7225-17.93765)To get more statistically meaningful value, finally, I should average all
dT/dz of analyses performed every 1000 steps (101000 step, 102000 step
etc.), e.g. by coding program for it, and there may be no way to produce
this average value automatically in LAMMPS.In addition, Ncount is "counted atom number" to calculate average
temperature of corresponding layer and it is not so signficant.Is there any fault in my description, would you indicate it, please?
To get the correct temperature gradient, there are two things to must
be assured:
(1) Plot the whole system's temperature variation with your simulation
time when you perform thermal conductivity calculation to check the
ssystem temperature whether or not conserved. If you see a increase of
temperature with the simulation time, it will indicate the chosen
Nevery is too small and you should increase it.
(2) Plot the temperature along the heat flux direction say z direction
to check there are two symmetry linear regions (with opposite slope)
in the curve. ONLY after you can observe the linear regions, you can
estimate the temperature gradient dT/dz. To get the T vs z data, you
need do a time average over the data after the linear regions have
appeared from the tmp.profile file.
For point (2), it's my personal experience and I'm not very sure, so I
post this email in the mailing list to hope other people also to give
comments.
Best,
Dongshan