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I think that the dilatation of an elastic material is proportional to the hydrostatic pressure.
That is, ${\rm{tr}}(\boldsymbol\sigma)=3k\ {\rm{tr}}(\boldsymbol\epsilon)$ for small strains.
If so, there is a stiffness tensor of the constitutive relationship between deviatoric strain and stress.
However, the stress deviator tensor satisfies ${\rm{tr}}(\boldsymbol s)=0$ for arbitrary deviatoric strain of ${\rm{tr}}(\boldsymbol e)=0$.
[Edited]
If the deviation stiffness tensor is represented by a block matrix of four 3x3 blocks, the number of independent elements is decreased from 6 to 4 in the upper diagonal block $C'_{11}$ and from 9 to 6 in off-diagonal blocks $C'_{12},\ C'_{21}$.
The strain deviation tensor can be diagonalized to the principal stress state by rotating the coordinate system, and the stress deviation tensor for the new coordinate system can be obtained from the above two blocks. Then the stress deviation tensor of the original coordinate system can be obtained by re-rotating the coordinate system.
As a result, the number of independent elements of a stiffness tensor is $1+4+6=11$.
Is the above correct?

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  • $\begingroup$ What is "deviatoric strain"? $\endgroup$
    – nasu
    Mar 29 '21 at 10:50
  • $\begingroup$ See, paragraph "Strain deviator tensor" of entry "Infinitesimal strain theory" in Wikipedia. $\endgroup$
    – tossy
    Mar 31 '21 at 5:07
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I am answering the question in the title of your post rather than the text, as I do not understand what you are getting at in the with the ${\rm tr}(e)$ stuff.

I'm also not sure what a "stiffness tensor" is, but the tensor of elastic constants $c_{ijkl}$ such that $\sigma_{ij}=c_{ijkl} e_{kl}$ has the symmetries $$ c_{ijkl}= c_{klij}, \quad c_{ijkl}= c_{jikl}=c_{ijlk}. $$ In the absence of these symmetries there would be $3^4=81$ independent coefficients. Taking the symmetries into account one has 6 possible equivalent pairs $ij$, similarly for $kl$, so the total number of independent coefficients is the number of independent entries in a 6 by 6 symmetric matrix. This is 6 for the diagonal entries and $(36-6)/2=15$ for the off-diagonal entries, making a total of 6+15=21.

This count is due to George Green of the "Green function" fame, a codiscoverer of the Cauchy-Green strain tensor $e_{ij}$.

I image that the "stiffness" tensor mught be an inverse of the elastic constants tensor, so it will have the same symmeties and hence the same number of independent entries.

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  • $\begingroup$ Thank you for your answer. When an elastic solid loses shear stiffness, it turns into a fluid. Then elastic anisotropy can be thought of as due to anisotropic shear stiffness. It was my mistake to think that pure-shear strain or divergence-free deformation causes divergence-free stresses. I forgot that pure-shear strain can cause dilatancy or hydrostatic pressure can causes shear strain. However, I think that there is a linear relationship of the first invariants between the stress and the strain tensor, and the number of wave velocities in a linear elastic solid is much less than 21. $\endgroup$
    – tossy
    Apr 4 '21 at 10:38
  • $\begingroup$ There is an account of the waves in a general linear elastic solid in Imaging Phonons (Acoustic Wave Propagation in Solids) by James P. Wolfe. Lots of pretty pictures too. The theory is similar to, but more complicated than Fresnel's theory of crystal optics: en.wikipedia.org/wiki/Wave_surface $\endgroup$
    – mike stone
    Apr 4 '21 at 11:32

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