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Let's have a stiffness tensor $$ a^{ijkl}: a^{ijkl} = a^{jikl} = a^{klij} = a^{ijlk}. $$

It has a 21 independent components for anisotropic body.

How does body symmetry (cubic, hexagonal etc.) changes the number of independent components of the tensor? For example, for absolutely isotropic body tensor has 2 independent components, and for hexagonal symmetry $C_{6}$ (with an z-axis symmetry) it has five components. How to explain it?

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up vote 2 down vote accepted

Feynmann's answer for the case of isotropic bodies is as follows (I copy his argument because I cannot find any online edition or snippet in google books):

The components can only be independent of direction if they can be written as a function of the tensor $\delta_{ij}$ (this needs some thinking, of course). Now, because there are only two possible expressions involving $ \delta_{ij} $ and $\delta_{kl}$ with the necessary symmetries, namely $ \delta_{ij} \delta_{kl}$ and $ \delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk} $, each $a_{ijkl}$ must be a linear combination of these, i.e.

$$ a_{ijkl} = a(\delta_{ij} \delta_{kl}) + b(\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk})$$

The case of cubic crystals (3 components) should be similar.

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