The tensor operation shown in the red box is used in the textbook to prove that there are only 9 independent constants for orthotropic materials:
[![enter image description here][1]][1]
I want to use MMA to reproduce the operation of $C_{pqmn}=l_{ip}\;l_{jq}\;l_{km}\;l_{ln}\;C_{ijkl}$ (Where $C_{ijkl}$ is the stiffness tensor), but at present, I have no specific idea. I will continue to update the details to make it perfect.
**Additional details:**
Thanks to the help of `xzczd`, I updated some details of the problem. I want to know how to get 5 independent constants of transversely isotropic materials in the textbook when $l_{ij}$ is the rotation matrix of any angle below:
$$l_{ij}=\left(
\begin{array}{ccc}
\cos (\alpha ) & \sin (\alpha ) & 0 \\
-\sin (\alpha ) & \cos (\alpha ) & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$$
[![enter image description here][2]][2]
[![enter image description here][3]][3]
**Other details are being added...**
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**Anisotropic material:**
The following formula is the constitutive relation of anisotropic materials:
$${\displaystyle {\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _ {6}\end{bmatrix}}\,=\,{\begin {bmatrix} C_ {11}& C_ {12}& C_ {13}& C_ {14}& C_ {15}& C_ {16} \\ C_ {12}& C_ {22}& C_ {23}& C_ {24}& C_ {25}& C_ {26} \\ C_ {13}& C_ {23}& C_ {33}& C_ {34}& C_ {35}& C_ {36} \\ C_ {14}& C_ {24}& C_ {34}& C_ {44}& C_ {45}& C_ {46} \\ C_ {15}& C_ {25}& C_ {35}& C_ {45}& C_ {55}& C_ {56} \\ C_ {16}& C_ {26}& C_ {36}& C_ {46}& C_ {56}& C_ {66}\end {bmatrix}}{\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _ {6}\end{bmatrix}}}$$
It can be seen from the stiffness matrix that there are 21 independent elastic constants.
----
**Orthotropic material:**
Orthotropic materials have three elastic symmetries which are perpendicular to each other(Orthotropic materials have three orthogonal planes of symmetry). In other words, if the coordinate axes are perpendicular to the three material symmetry planes of orthotropic materials, the elastic properties of the materials will not change after 180 degrees of rotation around these axes.
In the transformation process of the coordinate axis, the elastic symmetry of the material requires the fourth-order tensor to meet the following conditions:
$$C_{pqmn}=l_{ip}\;l_{jq}\;l_{km}\;l_{ln}\;C_{ijkl}$$
Where $l_{ij}$ is the symmetric transformation tensor.
For example, first of all, consider the material symmetry of orthotropic materials rotating 180 degrees around the Z axis, so $l_{ij}$ is :
$$l_{\text{ij}}=\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$$
From formula $C_{pqmn}=l_{ip}\;l_{jq}\;l_{km}\;l_{ln}\;C_{ijkl}$, we can get the following relations in this case:
$$C_{1311} = -C_{1311},\quad C_{1322} = -C_{1322} $$
$$C_ {1333} = -C_ {1333}, \quad C_ {1313} = -C_ {1313}$$
So,
$$C_ {1311} = C_ {1322} = C_ {1333} = C_ {1313} = 0$$
the following relationships can also be obtained:
$$C_ {2311} = C_ {2322} = C_ {2333} = C_ {2312} = 0 \\
C_ {1213} = C_ {1223} = C_ {1123} = C_ {1113} = 0 \\
C_ {2223} = C_ {2213} = C_ {3323} = C_ {3313} = 0$$
Similarly, a similar relationship can be obtained by rotating around the x-axis or y-axis; thus, the symmetry of these subscripts reduces the independent elastic constants to 12.
--------
**Transversely isotropic material:**
In this case, the material exhibits rotational elastic symmetry about a certain coordinate axis. If the z-axis is an elastic symmetry axis, then the isotropic plane is `xoy` plane. The symmetry condition of transverse isotropy can be obtained from orthotropic materials. That is, for any rotation transformation $l_{ij}$ about the z-axis, equation x must satisfy $C_{pqmn}=l_{ip}\;l_{jq}\;l_{km}\;l_{ln}\;C_{ijkl}$:
$$l_{ij}=\left(
\begin{array}{ccc}
\cos (\alpha ) & \sin (\alpha ) & 0 \\
-\sin (\alpha ) & \cos (\alpha ) & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$$
where $\alpha$ is any rotation angle about elastic plane axis (z axis).
---------
**Current problems to be solved:**
`xzczd` has answered the solution of the number of independent constants of orthotropic materials when $l_{\text{ij}}=\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$. But now I want to know how to solve the number of independent constants of transversely isotropic materials when $l_{ij}=\left(
\begin{array}{ccc}
\cos (\alpha ) & \sin (\alpha ) & 0 \\
-\sin (\alpha ) & \cos (\alpha ) & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$.
Because at this time, $l_{ij}$ is not a fixed number matrix, but a matrix related to the rotation angle $\alpha$, which increases the difficulty of solving. I need to change the existing code of `xzczd` to solve this problem.
[1]: https://i.sstatic.net/ncHj0.png
[2]: https://i.sstatic.net/8Lgs7.png
[3]: https://i.sstatic.net/2cXbw.png
[4]: https://i.sstatic.net/eJBd8.png