# Why is the number of independent elements of a stiffness tensor 21?

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?

• What is "deviatoric strain"?
– nasu
Commented Mar 29, 2021 at 10:50
• See, paragraph "Strain deviator tensor" of entry "Infinitesimal strain theory" in Wikipedia. Commented Mar 31, 2021 at 5:07
• Please note that the tensor in elasticity =/= the tensor in fluid mechanics, they are functions of different things, this had me stumped for years until I tried to implement my own shallow water equation implementation.
– Emil
Commented Jan 30, 2022 at 7:35

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 $$e_{ij}$$.
This will depend upon your constitutive equation. Under generalized hooks law you begin with a total of 81 constants. If the strain energy function has the commutative property this reduces to 45 due to the fact that $$C_{ijkl}=C_{klij}$$ (diagonal). If the loading is non polar that is $$\sigma_{ij}=\sigma_{ji}$$ and infinitesimal strain $$\varepsilon_{ij}=\varepsilon_{ji}$$ and use of $$C_{ijkl}=C_{klij}$$ then it becomes 21 constants. Look at Reddy Continuum mechanics