Continuum mechanics/Clausius-Duhem inequality for thermoelasticity

Proof: Since $$e = e(\boldsymbol{F}, T)$$, we have

\dot{e} = \frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} + \frac{\partial e}{\partial \eta}~\dot{\eta} ~. $$ Therefore,

\rho~\left(\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} + \frac{\partial e}{\partial \eta}~\dot{\eta}    - T~\dot{\eta}\right) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} \qquad\text{or}\qquad \rho\left(\frac{\partial e}{\partial \eta} - T\right)~\dot{\eta} + \rho~\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$ Now, $$\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{l} = \dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}$$. Therefore, using the identity $$\boldsymbol{A}:(\boldsymbol{B}\cdot\boldsymbol{C}) = (\boldsymbol{A}\cdot\boldsymbol{C}^T):\boldsymbol{B}$$, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\sigma}:(\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}) = (\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} ~. $$ Hence,

\rho\left(\frac{\partial e}{\partial \eta} - T\right)~\dot{\eta} + \rho~\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} - (\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} $$ or,

\rho~\left(\frac{\partial e}{\partial \eta} - T\right)~\dot{\eta} + \left(\rho~\frac{\partial e}{\partial \boldsymbol{F}} - \boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}\right):\dot{\boldsymbol{F}} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$