Continuum mechanics/Clausius-Duhem inequality for thermoelasticity

Clausius-Duhem inequality for thermoelasticity For thermoelastic materials, the internal energy is a function only of the deformation gradient and the temperature, i.e., $e=e({\boldsymbol {F}},T)$. Show that, for thermoelastic materials, the Clausius-Duhem inequality $\rho ~({\dot {e}}-T~{\dot {\eta }})-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}$ can be expressed as $\rho ~\left({\frac {\partial e}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.$

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,

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

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,

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

or,

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

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