# 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

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

I think this argument only leads to

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

In my opinion one should argue that ${\displaystyle {\boldsymbol {E}}}$ is symmetric and therefore
${\displaystyle \left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)_{ij}={\frac {\partial e}{\partial {\boldsymbol {E}}_{ij}}}={\frac {\partial e}{\partial {\boldsymbol {E}}_{ji}}}=\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)_{ji}}$