# Question about compute heat/flux documentation page.

Dear all I have a confusion regarding the compute heat_flux command.

According to the documentation page http://lammps.sandia.gov/doc/compute_heat_flux.html,

To obtain the thermal conductivity, one must divide the values obtained by compute_heatflux by the volume, which would give Jx, Jy, Jz.

And in the equation on the document page to obtain kappa, it is shown that it is the integral of J multiplied by some scaling factor where

scale = V/(kB*T^2)

I have looked around textbooks and understand that this is a standard result to obtain the thermal conductivity from the heat flux vector, J. The volume V is in the numerator.

What confuses me in the documentation is that, in the Solid Argon example, the scaling is given by

scale = {convert}/{kB}/\$T/\$T/\$V*s*{dt}

Ignoring {convert} and {s}, I noticed that the volume V is actually in the denominator.

I recognize that if we do the scaling this way, it is actually the scaling in the equation to find the thermal conductivity if we have the heat current vector S, where S = J/V. (as defined in Eqn1 in this paper http://li.mit.edu/A/Papers/99/Kaburaki99.pdf except he uses the symbol ‘J’ as the heat current vector instead of S as defined by me).

But didn’t we just calculate J instead of S?

Regards,
J.

Not sure what you are asking. Note this sentence on the compute heat/flux doc page:

Note that as discussed below, the 1/V scaling factor in the equation for J is NOT included in the calculation performed by this compute; you need to add it for a volume appropriate to the atoms included in the calculation.

Kappa is intensive (independent of system size).

If you run the script on the doc page for different

system sizes you should be able to verify

that the resulting Kappa’s do not scale with volume

or the inverse of volume.

Steve