# Continuum mechanics/Stress-strain relation for thermoelasticity

Proof:
Recall that the Cauchy stress is given by
    ${\boldsymbol {\sigma }}=\rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}\qquad \implies \qquad \sigma _{ij}=\rho ~{\frac {\partial e}{\partial F_{ik}}}F_{kj}^{T}=\rho ~{\frac {\partial e}{\partial F_{ik}}}F_{jk}~.$
The Green strain ${\boldsymbol {E}}={\boldsymbol {E}}({\boldsymbol {F}})={\boldsymbol {E}}({\boldsymbol {U}})$ and $e=e({\boldsymbol {F}},\eta )=e({\boldsymbol {U}},\eta )$. Hence, using the chain rule,
    ${\frac {\partial e}{\partial {\boldsymbol {F}}}}={\frac {\partial e}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial {\boldsymbol {F}}}}\qquad \implies \qquad {\frac {\partial e}{\partial F_{ik}}}={\frac {\partial e}{\partial E_{lm}}}~{\frac {\partial E_{lm}}{\partial F_{ik}}}~.$
Now,
    ${\boldsymbol {E}}={\frac {1}{2}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})\qquad \implies \qquad E_{lm}={\frac {1}{2}}(F_{lp}^{T}~F_{pm}-\delta _{lm})={\frac {1}{2}}(F_{pl}~F_{pm}-\delta _{lm})~.$
Taking the derivative with respect to ${\boldsymbol {F}}$, we get
    ${\frac {\partial {\boldsymbol {E}}}{\partial {\boldsymbol {F}}}}={\frac {1}{2}}\left({\frac {\partial {\boldsymbol {F}}^{T}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {F}}}}\right)\qquad \implies \qquad {\frac {\partial E_{lm}}{\partial F_{ik}}}={\frac {1}{2}}\left({\frac {\partial F_{pl}}{\partial F_{ik}}}~F_{pm}+F_{pl}~{\frac {\partial F_{pm}}{\partial F_{ik}}}\right)~.$
Therefore,
    ${\boldsymbol {\sigma }}={\frac {1}{2}}~\rho ~\left[{\frac {\partial e}{\partial {\boldsymbol {E}}}}:\left({\frac {\partial {\boldsymbol {F}}^{T}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {F}}}}\right)\right]\cdot {\boldsymbol {F}}^{T}\qquad \implies \qquad \sigma _{ij}={\frac {1}{2}}~\rho ~\left[{\frac {\partial e}{\partial E_{lm}}}\left({\frac {\partial F_{pl}}{\partial F_{ik}}}~F_{pm}+F_{pl}~{\frac {\partial F_{pm}}{\partial F_{ik}}}\right)\right]~F_{jk}~.$
Recall,
    ${\frac {\partial {\boldsymbol {A}}}{\partial {\boldsymbol {A}}}}\equiv {\frac {\partial A_{ij}}{\partial A_{kl}}}=\delta _{ik}~\delta _{jl}\qquad {\text{and}}\qquad {\frac {\partial {\boldsymbol {A}}^{T}}{\partial {\boldsymbol {A}}}}\equiv {\frac {\partial A_{ji}}{\partial A_{kl}}}=\delta _{jk}~\delta _{il}~.$
Therefore,
    $\sigma _{ij}={\frac {1}{2}}~\rho ~\left[{\frac {\partial e}{\partial E_{lm}}}\left(\delta _{pi}~\delta _{lk}~F_{pm}+F_{pl}~\delta _{pi}~\delta _{mk}\right)\right]~F_{jk}={\frac {1}{2}}~\rho ~\left[{\frac {\partial e}{\partial E_{lm}}}\left(\delta _{lk}~F_{im}+F_{il}~\delta _{mk}\right)\right]~F_{jk}$
or,
    $\sigma _{ij}={\frac {1}{2}}~\rho ~\left[{\frac {\partial e}{\partial E_{km}}}~F_{im}+{\frac {\partial e}{\partial E_{lk}}}~F_{il}\right]~F_{jk}\qquad \implies \qquad {\boldsymbol {\sigma }}={\frac {1}{2}}~\rho ~\left[{\boldsymbol {F}}\cdot \left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)^{T}+{\boldsymbol {F}}\cdot {\frac {\partial e}{\partial {\boldsymbol {E}}}}\right]\cdot {\boldsymbol {F}}^{T}$
or,
    ${\boldsymbol {\sigma }}={\frac {1}{2}}~\rho ~{\boldsymbol {F}}\cdot \left[\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)^{T}+{\frac {\partial e}{\partial {\boldsymbol {E}}}}\right]\cdot {\boldsymbol {F}}^{T}~.$
From the symmetry of the Cauchy stress, we have
    ${\boldsymbol {\sigma }}=({\boldsymbol {F}}\cdot {\boldsymbol {A}})\cdot {\boldsymbol {F}}^{T}\qquad {\text{and}}\qquad {\boldsymbol {\sigma }}^{T}={\boldsymbol {F}}\cdot ({\boldsymbol {F}}\cdot {\boldsymbol {A}})^{T}={\boldsymbol {F}}\cdot {\boldsymbol {A}}^{T}\cdot {\boldsymbol {F}}^{T}\qquad {\text{and}}\qquad {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{T}\implies {\boldsymbol {A}}={\boldsymbol {A}}^{T}~.$
Therefore,
    ${\frac {\partial e}{\partial {\boldsymbol {E}}}}=\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)^{T}$
and we get
    ${{\boldsymbol {\sigma }}=~\rho ~{\boldsymbol {F}}\cdot {\frac {\partial e}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}$