a question about the thermal conductivity tensor

[XiangXing96]

Hi, everyone,

I am using alamode to calculate the thermal conductivity of 2d materials. After I get results, I am confused about the thermal conductivity tensor. Are the directions in thermal conductivity tensor along real x, y and z or the lattice vectors, a, b, and c?
My 2d material is MoS2. It has the same a and b with a angle of 120 degrees. Correspondingly, the atomic arrangements are the same along a and b, while different along x (zigzag) and y (armchair) directions. In thermal conductivity tensor, kxx and kyy are the same, which seem different from the intuition. Therefore, I write this to ask the question.
Thank you very much.

Best regards

[ttadano]

The thermal conductivity tensor shown is in the Cartesian coordinate, $\kappa_{xx}$ and $\kappa_{yy}$.

The thermal conductivity tensor $\kappa$ is defined by the Fourier's law:

$$ \boldsymbol{j} = -\kappa \nabla T. $$

In the Peierls-Boltzmann theory, we do not consider the phonon Hall effect, so the off-diagonal elements of $\kappa$ are zero:

$$ \begin{pmatrix} j_x \\ j_y \\ j_z \end{pmatrix} = - \begin{pmatrix} \kappa_{xx} & 0 & 0 \\ 0 & \kappa_{yy} & 0 \\ 0 & 0 & \kappa_{zz} \end{pmatrix} \begin{pmatrix} \nabla_{x}T \\ \nabla_{y}T \\ \nabla_{z}T \end{pmatrix}. $$

In the monolayer MoS2, three-fold rotational symmetry exists along the Cartesian $z$ axis.

If we set the $\vec{a}$ axis to be the same as $\vec{x}$, we have $\kappa_{aa} = \kappa_{xx}$.
When $\vec{b}$ axis is chosen to be

$$ \vec{b} = \vec{x}\cos{\theta} + \vec{y}\sin{\theta}, $$

with $\theta=\frac{2}{3}\pi$.

Since the $a$ and $b$ directions are symmetrically equivalent, we should have $\kappa_{aa}=\kappa_{bb}$.

Since the heat flux and temperature gradient transform in the same way as the lattice vector, we have

$$ \begin{align} j_{b} = j_x \cos{\theta} + j_y \sin{\theta} = -(\kappa_{xx}\nabla_x T \cos{\theta} + \kappa_{yy} \nabla_y T \sin{\theta}), \end{align} $$

and

$$ \begin{align} -\kappa_{bb}\nabla_{b}T = -\kappa_{bb}(\nabla_x T \cos{\theta} + \nabla_y T \sin{\theta}). \end{align} $$

Since the above two equations must be equal to each other (according to the Fourier's law), we have

$$ \begin{align} \kappa_{xx}\nabla_x T \cos{\theta} + \kappa_{yy} \nabla_y T \sin{\theta} = \kappa_{bb}(\nabla_x T \cos{\theta} + \nabla_y T \sin{\theta}). \end{align} $$

Thus, for $(\kappa_{xx}=) \kappa_{aa}=\kappa_{bb}$ to hold, it is necessary to satisfy $\kappa_{yy}=\kappa_{xx}$.

Moreover, when you have $\kappa_{xx}=\kappa_{yy}$, you can show that the thermal conductivity value is the same as $\kappa_{xx}$ in arbitrary directions of the $xy$ plane.

[XiangXing96]

Hi, Tadano,

I get it. Thank you very much.

Best regards,