# Does thermal conductivity calculated by NEMD method make any meaning if the system is anisotropic?

Dear all,
When i was comparing the thermal conductivity of an anisotropic system calculated by emd(gree-kubo) method and nemd method, a question occured to me that does the nemd method calculated thermal conductivity has any meaning for the anisotropic system?

As we all know that the fourier heat transform formular has the form
[qx] [k11 k12 k13] [ ∂T / ∂X]
[qy]= -[k21 k22 k23] * [∂T / ∂y] (1)
[qz] [k31 k32 k33] [∂T / ∂z],
Imagining that under a coordinate system ,the thermal conductivity tensor component k12,k13,k31,k32 not equal zero, and under this coorinate system ,along the x axis, we perform the nemd method.
For performing the nemd method in x axis,so the heat flux along y,z ,that is qy and qz euqal 0, also the temperature gradient in y,z direction ∂T / ∂y and ∂T / ∂z equal 0 as well.Bring these into the equation(1)
,then the equation (1)take the form :

[qx] [k11 k12 k13] [ ∂T / ∂X] [k11* ∂T / ∂X]
= - [k21 k22 k23] *  =[k21∂T / ∂X]
 [k31 k32 k33]  [k31
∂T / ∂X],

for ∂T / ∂X not equal zero,so k21=k31=0,which is not consistent with the premise we make before.So does the thermal conductivity λ calculated by using nemd methods make any means if the system is anisotropic?
Sincerely,
XiaonengRan

Hello,

Dear all,
When i was comparing the thermal conductivity of an anisotropic system calculated by emd(gree-kubo) method and nemd method, a question occured to me that does the nemd method calculated thermal conductivity has any meaning for the anisotropic system?

Yes, it is meaningful.

As we all know that the fourier heat transform formular has the form
[qx] [k11 k12 k13] [ ∂T / ∂X]
[qy]= -[k21 k22 k23] * [∂T / ∂y] (1)
[qz] [k31 k32 k33] [∂T / ∂z],
Imagining that under a coordinate system ,the thermal conductivity tensor component k12,k13,k31,k32 not equal zero, and under this coorinate system ,along the x axis, we perform the nemd method.

Fine.

For performing the nemd method in x axis,so the heat flux along y,z ,that is qy and qz euqal 0, also the temperature gradient in y,z direction ∂T / ∂y and ∂T / ∂z equal 0 as well.Bring these into the equation(1)
,then the equation (1)take the form :

[qx] [k11 k12 k13] [ ∂T / ∂X] [k11* ∂T / ∂X]
= - [k21 k22 k23] *  =[k21∂T / ∂X]
 [k31 k32 k33]  [k31
∂T / ∂X],

for ∂T / ∂X not equal zero,so k21=k31=0,which is not consistent with the premise we make before.So does the thermal conductivity λ calculated by using nemd methods make any means if the system is anisotropic?

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.

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. After obtaining the diagonal elements within the principal axes, you can obtain the conductivity tensor in any coordinate system by performing a coordination transformation. The conductivity is a second-rank tensor and will transform like a tensor. For an example, check equation (10) and related discussion in this paper:

Thermal transport properties of single-layer black phosphorus from extensive molecular dynamics simulations
https://iopscience.iop.org/article/10.1088/1361-651X/aae180/meta

Best,

Bruce