Hello,

Dear Bruce，

Thanks for your quick reply.

I noticed that in the previous email,you said that

**>You cannot assume that qy = qz = 0 here, if the off-diagonal elements of the conductivity tensor is nonzero and the temperature gradient is in the x direction.**

if qy and qz !=0 here, then what about ∂T / ∂y and ∂T / ∂z, do they not need to be zero here?

if you fix the temperature gradient, then the heat flux is fixed according to Fourier’s law;

if you fix the heat flux, then the temperature gradient is also fixed accordingly.

You cannot prescribe both in an inconsistent way. You have at most three independent inputs and then you have 3 outputs. In steady state, there should be no contradiction.

**>A better way to treat anisotropic systems is to first work with principal axes, where the conductivity tensor only has nonzero diagonal elements: k11, k22 and k33.**

But at sometimes ,it is very hard to first know the princinple axes.

One can use symmetry arguments to identify the principal axes, or just calculate the conductivity tensor in one coordinate system and figure out how it can be transformed into a diagonal form. Then the principal axes can be identified from the transformation matrix. But this might be overkill. I think symmetry arguments are enough in most cases.

I noticed that in the paper you referenced to me, it is possible to get the thermal conductivity tensor by using HNEMD

method.Then what about emd method?Emd method can get the three direction thermal conductivity value but not the tensor, is it possiable to get the thermal conductivity tensor?

HNEMD is physically equivalent to EMD. So you can also calculate the full conductivity tensor using the EMD (Green-Kubo) method. Check early papers (1980s or 1990s) by D. J. Evans et al… For example, the xy component of the conductivity tensor can be calculated from the time integral of <Jx(0) * Jy(t)>, where Jx and Jy are the heat flux in the x and y directions, respectively.

Best,

Bruce