Continuum mechanics/Balance of energy for thermoelasticity

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Balance of energy for thermoelastic materials
Show that, for thermoelastic materials, the balance of energy

\rho~\dot{e} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 ~. $$ can be expressed as

\rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. $$
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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} ~. $$ Plug into energy equation to get

\rho~\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} + \rho~\frac{\partial e}{\partial \eta}~\dot{\eta} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 ~. $$ Recall,

\frac{\partial e}{\partial \eta} = T \qquad\text{and}\qquad \rho~\frac{\partial e}{\partial \boldsymbol{F}} = \boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T} ~. $$ Hence,

(\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} + \rho~T~\dot{\eta} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 ~. $$ Now, $$\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}} ~. $$ Plugging into the energy equation, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \rho~T~\dot{\eta} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 $$ or,

{    \rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. } $$

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Rate of internal energy/entropy for thermoelastic materials
For thermoelastic materials, the specific internal energy is given by

e = \bar{e}(\boldsymbol{E}, \eta) $$ where $$\boldsymbol{E}$$ is the Green strain and $$\eta$$ is the specific entropy. Show that

\cfrac{d}{dt}(e - T~\eta) = - \dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} \qquad\text{and}\qquad \cfrac{d}{dt}(e - T~\eta - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}) = - \dot{T}~\eta - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} $$ where $$\rho_0$$ is the initial density, $$T$$ is the absolute temperature, $$\boldsymbol{S}$$ is the 2nd Piola-Kirchhoff stress, and a dot over a quantity indicates the material time derivative.
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Taking the material time derivative of the specific internal energy, we get

\dot{e} = \frac{\partial \bar{e}}{\partial \boldsymbol{E}}:\dot{\boldsymbol{E}} + \frac{\partial \bar{e}}{\partial \eta}~\dot{\eta} ~. $$ Now, for thermoelastic materials,

T = \frac{\partial \bar{e}}{\partial \eta} \qquad \text{and} \qquad \boldsymbol{S} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}} ~. $$ Therefore,

\dot{e} = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} + T~\dot{\eta} ~. \qquad \implies \qquad \dot{e} - T~\dot{\eta} = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} ~. $$ Now,

\cfrac{d}{dt}(T~\eta) = \dot{T}~\eta + T~\dot{\eta} ~. $$ Therefore,

\dot{e} - \cfrac{d}{dt}(T~\eta) + \dot{T}~\eta = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} \qquad \implies \qquad {  \cfrac{d}{dt}(e - T~\eta) = -\dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} ~. } $$ Also,

\cfrac{d}{dt}\left(\cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}\right) = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} + \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} ~. $$ Hence,

\dot{e} - \cfrac{d}{dt}(T~\eta) + \dot{T}~\eta = \cfrac{d}{dt}\left(\cfrac{1}{\rho_0} \boldsymbol{S}:\boldsymbol{E}\right) - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} \qquad \implies \qquad {  \cfrac{d}{dt}\left(e - T~\eta - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}\right) = - \dot{T}~\eta - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} ~. } $$

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Energy equation for thermoelastic materials
For thermoelastic materials, show that the balance of energy equation

\rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s $$ can be expressed as either

\rho~C_v~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s +\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} $$ or

\rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) ~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} $$ where

C_v = \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} \qquad \text{and} \qquad C_p = \frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} ~. $$ For the special case where there are no sources and we can ignore heat conduction (for very fast processes), the energy equation simplifies to

\rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{\alpha}\right) ~\dot{T} =     -\cfrac{\rho}{\rho_0}~T~\boldsymbol{\alpha}:\dot{\boldsymbol{S}} $$ where $$\boldsymbol{\alpha} := \frac{\partial \tilde{\boldsymbol{E}}}{\partial T}$$ is the thermal expansion tensor which has the form $$\boldsymbol{\alpha} = \alpha\boldsymbol{1}$$ for isotropic materials and $$\alpha\,$$ is the coefficient of thermal expansion. The above equation can be used to calculate the change of temperature in thermoelasticity.
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Proof:

If the independent variables are $$\boldsymbol{E}$$ and $$T$$, then

\eta = \hat{\eta}(\boldsymbol{E}, T) \qquad \implies \qquad \dot{\eta} = \frac{\partial \hat{\eta}}{\partial \boldsymbol{E}}:\dot{\boldsymbol{E}} + \frac{\partial \hat{\eta}}{\partial T}~\dot{T} ~. $$ On the other hand, if we consider $$\boldsymbol{S}$$ and $$T$$ to be the independent variables

\eta = \tilde{\eta}(\boldsymbol{S}, T) \qquad \implies \qquad \dot{\eta} = \frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}}:\dot{\boldsymbol{S}} + \frac{\partial \tilde{\eta}}{\partial T}~\dot{T} ~. $$ Since

\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} = -\cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}   ~; \frac{\partial \hat{\eta}}{\partial T} = \cfrac{C_v}{T} ~; \frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~;\text{and} \frac{\partial \tilde{\eta}}{\partial T} = \cfrac{1}{T}\left(C_p - \cfrac{1}{\rho_0}    \boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) $$ we have, either

\dot{\eta} = -\cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \cfrac{C_v}{T}~\dot{T} $$ or

\dot{\eta} = \cfrac{1}{\rho_0}~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \cfrac{1}{T}\left(C_p - \cfrac{1}{\rho_0}    \boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right)~\dot{T} ~. $$ The equation for balance of energy in terms of the specific entropy is

\rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. $$ Using the two forms of $$\dot{\eta}$$, we get two forms of the energy equation:

-\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \rho~C_v~\dot{T} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s $$ and

\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \rho~C_p~\dot{T} - \cfrac{\rho}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}~\dot{T} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. $$ From Fourier's law of heat conduction

\mathbf{q} = - \boldsymbol{\kappa}\cdot\boldsymbol{\nabla} T ~. $$ Therefore,

-\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \rho~C_v~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s $$ and

\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \rho~C_p~\dot{T} - \cfrac{\rho}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s ~. $$ Rearranging,

{ \rho~C_v~\dot{T}  = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s +\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} } $$ or,

{ \rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) ~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} ~. } $$