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.