How can point group symmetry operations be used to reduce the number of independent crystal properties?

Question

How can point group lattice symmetry operations be applied to reduce the full second-rank elasticity tensor (in Voigt notation) from:

to, for example, in the cubic case, this:

A reference would be much appreciated.

Answer 1

This question is better suited for PHYS.SE, but here's the outline of an answer. It is mathematically the easiest way I know.

With no symmetry, there are clearly 21 (7*6/2) independent components in the Voigt elasticity matrix, but symmetry always reduces this number.

By the definition of $[C]$,

$$\{\sigma\}_{6 \times 1} = [C]_{6 \times 6} \{\epsilon\}_{6 \times 1} $$

where the $\{\}$ are column vectors containing the 6 independent components of the stress and strain tensors.

Consider one plane of symmetry, coinciding with one of the coordinate planes, for simplicity. Consider $[\beta]$, a transformation matrix representing a reflection of the coordinate axes in this plane of symmetry, the $\sigma$ and $\epsilon$ tensors transform in the usual way, i.e.

$$ [\sigma'] = [\beta] [\sigma] [\beta]^T \\ [\epsilon'] = [\beta] [\epsilon] [\beta]^T $$

The ' indicates components in the transformed domain.

(Note: The matrix of tensor components $\sigma$ and the column vector $\{\sigma\}$ are completely different entities)

$\beta$ has a very simple form for reflection about a coordinate plane; for the 2-3 coordinate plane it looks like $$ \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Since we assert that the plane is a plane of symmetry, the components of the 4th order $C$ tensor are unchanged, i.e., $$\{\sigma'\}_{6 \times 1} = [C]_{6 \times 6} \{\epsilon'\}_{6 \times 1} $$

Since the components of the stress and strain are known in the reflected basis, one can obtain pairs of equations between identical stress and strain components, from which the coefficients that are 0 are easily deduced. For example, from the first row of $[C]$

$$ \sigma_{11} = C_{11} \epsilon_{11} + C_{12} \epsilon_{22} + C_{13} \epsilon_{33} + 2 C_{14} \epsilon_{23} + 2 C_{15} \epsilon_{13} + 2 C_{16} \epsilon_{12} \\ \sigma'_{11} = C_{11} \epsilon'_{11} + C_{12} \epsilon'_{22} + C_{13} \epsilon'_{33} + 2 C_{14} \epsilon'_{23} + 2 C_{15} \epsilon'_{13} + 2 C_{16} \epsilon'_{12} \\ \Rightarrow \sigma_{11} = C_{11} \epsilon_{11} + C_{12} \epsilon_{22} + C_{13} \epsilon_{33} + 2 C_{14} \epsilon_{23} - 2 C_{15} \epsilon_{13} - 2 C_{16} \epsilon_{12} \\ $$

and from the first and third equations, you can conclude that $C_{15} = C_{16} =0$. If you exhaust all 6 rows, you'll see you have 13, not 21 independent components in the $C$ matrix.

Proceed in this way, imposing as much additional symmetry as required. For instance, with three mutually perpendicular planes of symmetry you have an orthotropic crystal, and with invariance under rotations about three mutually perpendicular axes by $\pi/2$, you'll have cubic symmetry. (For these, the $\beta$ are again quite simple.)

Note: If every plane is a plane of symmetry and every axis is an axis of symmetry, you have a maximally symmetric condition, known as isotropy. The symmetry group then contains the entire rotation group.