The symmetric thermal conductivity tensor κ can be written in components as,
\kappa = \Biggl(\begin{matrix} κ_{11} & κ_{12} & κ_{13} \\ κ_{12} & κ_{22} & κ_{23} \\ κ_{13} & κ_{23} & κ_{33} \end{matrix} \Biggr) .
The rotation matrix \Re_θ describing a rotation of angle θ = π/6 in the horizontal plane around the vertical axis that leaves the thermal conductivity tensor invariant and its inverse \Re^{−1}_θ read,
\Re _\theta = \frac{1}{2} \Biggl(\begin{matrix} 1 & -\sqrt{3} & 0 \\ \sqrt{3} & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \Biggr) .
and
\Re^{-1} _\theta = \Re _\theta = \frac{1}{2}\Biggl(\begin{matrix} 1 & \sqrt{3} & 0\\ -\sqrt{3} & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \Biggr) .
Since the physical properties of the crystal are invariant under the rotation \Re _\theta , we rotate Fourier’s law (11.26) by an angle θ,
j_Q = −κ (s, n_A, q) ·∇ T . (11.26)
\Re _\theta · j_Q = −\Re _\theta · κ ·∇T = −\Re _\theta · κ · \Re _{-\theta} · \Re _\theta ·∇T .
where \Re _{-\theta} · \Re _\theta = 1 Since Fourier’s law (11.26) is invariant under such a rotation,
\Re _\theta · j_Q = j_Q and \Re _\theta ·∇T = ∇T
and
\Re _\theta · κ · \Re _{-\theta} = κ or \Re _\theta · κ = κ · \Re _\theta
which is written in components as,
\Biggl(\begin{matrix} 1 & -\sqrt{3} & 0 \\\sqrt{3} & 1 & 0 \\ 0 & 0 & 1\end{matrix} \Biggr) \Biggl(\begin{matrix} κ_{11} & κ_{12} &κ_{13} \\ κ_{12} & κ_{22} & κ_{23} \\ κ_{13} &κ_{23} & κ_{33} \end{matrix} \Biggr) = \Biggl(\begin{matrix} κ_{11} & κ_{12} &κ_{13} \\ κ_{12} & κ_{22} & κ_{23} \\ κ_{13} &κ_{23} & κ_{33} \end{matrix} \Biggr) \Biggl(\begin{matrix} 1 & -\sqrt{3} & 0 \\\sqrt{3} & 1 & 0 \\ 0 & 0 & 1\end{matrix} \Biggr)
The solutions of this matrix system are,
κ_{12} = κ_{13} = κ_{23} =0 and κ_{11} = κ_{22}
With the identifications,
κ_\parallel= κ_{33} and κ_⊥ = κ_{11} = κ_{22}
the thermal conductivity tensor κ reduces to,
κ = \Biggl(\begin{matrix} κ_⊥ & 0 & 0 \\ 0 & κ_⊥ & 0 \\ 0 & 0 & \kappa _\parallel \end{matrix} \Biggr).