```
Dear Steve,
Could you please replace the current equation:
lamda (KCal/(mol fmsec Ang K)) = V/(k_B*(T^2)) x "thermo" output frequency x timestep
to:
kappa (KCal/(mol fmsec Ang K)) = V/(k_B*(T^2)) x "thermo" output frequency x timestep x Integral Value
Sorry for the inconvenience.
Thanks.
Mario
Dear Jianxin,
1> Lambda is the thermal conductivity, which is represented in the equation as kappa.
2> Regarding the value of the integral:
The example is more to do with how to go from a given value of the integral
to the final vale of the thermal conductivity.
The integral does not stabilize to a precise value.
(See for example the attached plot, posted by German some time ago on the mailing list).
The value is indicative.
3> As per the example given in the doc page:
timestep 4.0
thermo 10
-these are the timestep and thermo output frequency.
4> From the doc page:
lamda (KCal/(mol fmsec Ang K)) = V/(k_B*(T^2)) x "thermo" output frequency x timestep x Integral Value
where
V = 10213.257 Angs^3
k_B = 1.98721e-3 KCal/(mol K)
T = ~70K
Therefore, lamda = 3.7736e-6 (KCal/(mol fs A K)).
Converting to W/mK gives:
3.7736e-6 * (4182 / (1e-15 * 1e-10 * N_Avogradro)) =
3.7736e-6 * 69443.837 =
~0.26 W/mK
![LJ_90K.jpg|720x504](upload://5L33DkCqb5mvCLJDLwaOxEpSMp9.jpeg)
```

doc page updated

Steve