# Continuum mechanics/Balance of energy for thermoelasticity

Proof:
Since $e=e({\boldsymbol {F}},T)$, we have
    ${\dot {e}}={\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}+{\frac {\partial e}{\partial \eta }}~{\dot {\eta }}~.$
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,
    ${\rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}$
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,
    ${\dot {e}}-{\cfrac {d}{dt}}(T~\eta )+{\dot {T}}~\eta ={\cfrac {d}{dt}}\left({\cfrac {1}{\rho _{0}}}{\boldsymbol {S}}:{\boldsymbol {E}}\right)-{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}\qquad \implies \qquad {{\cfrac {d}{dt}}\left(e-T~\eta -{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}\right)=-{\dot {T}}~\eta -{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}~.}$
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,
    ${\rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}\right)~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s-{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}~.}$