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., $e=e(𝑭,T)$. Show that, for thermoelastic materials, the Clausius-Duhem inequality $\rho (\dot{e}-T\dot{\eta })-\mathit{\sigma }:\mathrm{\nabla }𝐯\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}$ can be expressed as $\rho (\frac{\partial e}{\partial \eta }-T)\dot{\eta }+(\rho \frac{\partial e}{\partial 𝑭}-\mathit{\sigma }\cdot {𝑭}^{-T}):\dot{𝑭}\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}.$

Proof:

Since $e=e(𝑭,T)$, we have

$\dot{e}=\frac{\partial e}{\partial 𝑭}:\dot{𝑭}+\frac{\partial e}{\partial \eta }\dot{\eta }.$

Therefore,

$\rho (\frac{\partial e}{\partial 𝑭}:\dot{𝑭}+\frac{\partial e}{\partial \eta }\dot{\eta }-T\dot{\eta })-\mathit{\sigma }:\mathrm{\nabla }𝐯\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}\text{or}\rho (\frac{\partial e}{\partial \eta }-T)\dot{\eta }+\rho \frac{\partial e}{\partial 𝑭}:\dot{𝑭}-\mathit{\sigma }:\mathrm{\nabla }𝐯\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}.$

Now, $\mathrm{\nabla }𝐯=𝒍=\dot{𝑭}\cdot {𝑭}^{-1}$. Therefore, using the identity $𝑨:(𝑩\cdot 𝑪)=(𝑨\cdot {𝑪}^T):𝑩$, we have

$\mathit{\sigma }:\mathrm{\nabla }𝐯=\mathit{\sigma }:(\dot{𝑭}\cdot {𝑭}^{-1})=(\mathit{\sigma }\cdot {𝑭}^{-T}):\dot{𝑭}.$

Hence,

$\rho (\frac{\partial e}{\partial \eta }-T)\dot{\eta }+\rho \frac{\partial e}{\partial 𝑭}:\dot{𝑭}-(\mathit{\sigma }\cdot {𝑭}^{-T}):\dot{𝑭}\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}$

or,

$\rho (\frac{\partial e}{\partial \eta }-T)\dot{\eta }+(\rho \frac{\partial e}{\partial 𝑭}-\mathit{\sigma }\cdot {𝑭}^{-T}):\dot{𝑭}\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}.$