Consider a fourth oder stiffness tensor $\mathbb{C}$. The components $C_{ijkl}$ are given with respect to a global coordinate system. Is there a way to determine the orthotropic axes from this representation?
In Mandel-Kelvin-Notation we can express the tensor as $6\times6$-Matrix \begin{equation} \mathbb{C} = \begin{pmatrix} \tilde{C}_{11} & \tilde{C}_{12} & \tilde{C}_{13} & \tilde{C}_{14} & \tilde{C}_{15} & \tilde{C}_{16} \\ & \tilde{C}_{22} & \tilde{C}_{23} & \tilde{C}_{24} & \tilde{C}_{25} & \tilde{C}_{26} \\ & & \tilde{C}_{33} & \tilde{C}_{34} & \tilde{C}_{35} & \tilde{C}_{36} \\ & & & \tilde{C}_{44} & \tilde{C}_{45} & \tilde{C}_{46} \\ & \text{sym} & & & \tilde{C}_{55} & \tilde{C}_{56} \\ & & & & & \tilde{C}_{66} \end{pmatrix} \tilde{\boldsymbol{B}}_\alpha \otimes \tilde{\boldsymbol{B}}_\beta = \begin{pmatrix} C_{11} & C_{12} & C_{13} & 0 & 0 & 0 \\ & C_{22} & C_{23} & 0 & 0 & 0 \\ & & C_{33} & 0 & 0 & 0 \\ & & & C_{44} & 0 & 0 \\ & \text{sym} & & & C_{55} & 0 \\ & & & & & C_{66} \\ \end{pmatrix} \boldsymbol{B}_\alpha \otimes \boldsymbol{B}_\beta \end{equation} with the global basis $\lbrace \tilde{\boldsymbol{B}}_\alpha \rbrace$: \begin{align} \tilde{\boldsymbol{B}}_1 &= \boldsymbol{e}_x \otimes \boldsymbol{e}_x \\ \tilde{\boldsymbol{B}}_2 &= \boldsymbol{e}_y \otimes \boldsymbol{e}_y \\ \tilde{\boldsymbol{B}}_3 &= \boldsymbol{e}_z \otimes \boldsymbol{e}_z \\ \tilde{\boldsymbol{B}}_4 &= \frac{1}{\sqrt{2}}\left(\boldsymbol{e}_y \otimes \boldsymbol{e}_z + \boldsymbol{e}_z \otimes \boldsymbol{e}_y \right) \\ \tilde{\boldsymbol{B}}_5 &= \frac{1}{\sqrt{2}}\left(\boldsymbol{e}_x \otimes \boldsymbol{e}_z + \boldsymbol{e}_z \otimes \boldsymbol{e}_x \right) \\ \tilde{\boldsymbol{B}}_6 &= \frac{1}{\sqrt{2}}\left(\boldsymbol{e}_x \otimes \boldsymbol{e}_y + \boldsymbol{e}_y \otimes \boldsymbol{e}_x \right) \\ \end{align}
and the orthotropic basis $\lbrace \boldsymbol{B}_\alpha \rbrace$
\begin{align} \boldsymbol{B}_1 &= \boldsymbol{e}_1 \otimes \boldsymbol{e}_1 \\ \boldsymbol{B}_2 &= \boldsymbol{e}_2 \otimes \boldsymbol{e}_2 \\ \boldsymbol{B}_3 &= \boldsymbol{e}_3 \otimes \boldsymbol{e}_3 \\ \boldsymbol{B}_4 &= \frac{1}{\sqrt{2}}\left(\boldsymbol{e}_2 \otimes \boldsymbol{e}_3 + \boldsymbol{e}_3 \otimes \boldsymbol{e}_2 \right) \\ \boldsymbol{B}_5 &= \frac{1}{\sqrt{2}}\left(\boldsymbol{e}_1 \otimes \boldsymbol{e}_3 + \boldsymbol{e}_3 \otimes \boldsymbol{e}_1 \right) \\ \boldsymbol{B}_6 &= \frac{1}{\sqrt{2}}\left(\boldsymbol{e}_1 \otimes \boldsymbol{e}_2 + \boldsymbol{e}_2 \otimes \boldsymbol{e}_1 \right) \\ \end{align}
The orthonormal-system $\lbrace \boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3 \rbrace$ would represent the orthtropic axes, I am looking for.
Help would be really much appreciated. TIA