Talk:Continuum mechanics/Stress-strain relation for thermoelasticity

First of all I have to apologize for a probably not conforming style, this is my first own-written discussion so please be patient.

Problem:
The argumentation for

    ${\frac {\partial e}{\partial {\boldsymbol {E}}}}=\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)^{T}$ because of $~{\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{T}\Rightarrow {\boldsymbol {A}}={\boldsymbol {A}}^{T}$
doesn't sound reasonable to me.

I think this argument only leads to

    $\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)^{T}+{\frac {\partial e}{\partial {\boldsymbol {E}}}}={\frac {\partial e}{\partial {\boldsymbol {E}}}}+\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)^{T}$

what is obviously not helpful.

Proposal for Solution:

In my opinion one should argue that ${\boldsymbol {E}}$ is symmetric and therefore

    $\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)_{ij}={\frac {\partial e}{\partial {\boldsymbol {E}}_{ij}}}={\frac {\partial e}{\partial {\boldsymbol {E}}_{ji}}}=\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)_{ji}$

what gives the demanded symmetry.