# [lammps-users] About the new "compute heat/flux"

``````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