Continuum mechanics/Maxwell relations for thermoelasticity

Maxwell relations between thermodynamic quantities For thermoelastic materials, show that the following relations hold: ${\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)~;~~{\frac {\partial \psi }{\partial T}}=-{\hat {\eta }}({\boldsymbol {E}},T)~;~~{\frac {\partial g}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~;~~{\frac {\partial g}{\partial T}}={\tilde {\eta }}({\boldsymbol {S}},T)$ where $\psi ({\boldsymbol {E}},T)$ is the Helmholtz free energy and $g({\boldsymbol {S}},T)$ is the Gibbs free energy. Also show that ${\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}=-\rho _{0}~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}\qquad {\text{and}}\qquad {\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=\rho _{0}~{\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}~.$

Proof:

Recall that

$\psi ({\boldsymbol {E}},T)=e-T~\eta ={\bar {e}}({\boldsymbol {E}},\eta )-T~\eta ~.$

and

$g({\boldsymbol {S}},T)=-e+T~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}~.$

(Note that we can choose any functional dependence that we like, because the quantities $e$, $\eta$, ${\boldsymbol {E}}$ are the actual quantities and not any particular functional relations).

The derivatives are

${\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~;\qquad {\frac {\partial \psi }{\partial T}}=-\eta ~.$

and

${\frac {\partial g}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {E}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {E}}~;\qquad {\frac {\partial g}{\partial T}}=\eta ~.$

Hence,

${{\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)~;~~{\frac {\partial \psi }{\partial T}}=-{\hat {\eta }}({\boldsymbol {E}},T)~;~~{\frac {\partial g}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~;~~{\frac {\partial g}{\partial T}}={\tilde {\eta }}({\boldsymbol {S}},T)}$

From the above, we have

${\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial T}}\qquad \implies \qquad -{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}~.$

and

${\frac {\partial ^{2}g}{\partial T\partial {\boldsymbol {S}}}}={\frac {\partial ^{2}g}{\partial {\boldsymbol {S}}\partial T}}\qquad \implies \qquad {\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$

Hence,

${{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}=-\rho _{0}~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}\qquad {\text{and}}\qquad {\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=\rho _{0}~{\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}~.}$

More Maxwell relations For thermoelastic materials, show that the following relations hold: ${\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}=-T~{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}$ and ${\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=T~{\frac {\partial ^{2}{\tilde {g}}}{\partial T^{2}}}+{\boldsymbol {S}}:{\frac {\partial ^{2}{\tilde {g}}}{\partial {\boldsymbol {S}}\partial T}}~.$

Proof:

Recall,

${\hat {\psi }}({\boldsymbol {E}},T)=\psi =e-T~\eta ={\hat {e}}({\boldsymbol {E}},T)-T~{\hat {\eta }}({\boldsymbol {E}},T)$

and

${\tilde {g}}({\boldsymbol {S}},T)=g=-e+T~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}=-{\tilde {e}}({\boldsymbol {S}},T)+T~{\tilde {\eta }}({\boldsymbol {S}},T)+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~.$

Therefore,

${\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}={\frac {\partial {\hat {\psi }}}{\partial T}}+{\hat {\eta }}({\boldsymbol {E}},T)+T~{\frac {\partial {\hat {\eta }}}{\partial T}}$

and

${\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=-{\frac {\partial {\tilde {g}}}{\partial T}}+{\tilde {\eta }}({\boldsymbol {S}},T)+T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$

Also, recall that

${\hat {\eta }}({\boldsymbol {E}},T)=-{\frac {\partial {\hat {\psi }}}{\partial T}}\qquad \implies \qquad {\frac {\partial {\hat {\eta }}}{\partial T}}=-{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}~,$

${\tilde {\eta }}({\boldsymbol {S}},T)={\frac {\partial {\tilde {g}}}{\partial T}}\qquad \implies \qquad {\frac {\partial {\tilde {\eta }}}{\partial T}}={\frac {\partial ^{2}{\tilde {g}}}{\partial T^{2}}}~,$

and

${\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)=\rho _{0}~{\frac {\partial {\tilde {g}}}{\partial {\boldsymbol {S}}}}\qquad \implies \qquad {\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=\rho _{0}~{\frac {\partial ^{2}{\tilde {g}}}{\partial {\boldsymbol {S}}\partial T}}~.$

Hence,

${{\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}=-T~{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}}$

and

${{\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=T~{\frac {\partial ^{2}{\tilde {g}}}{\partial T^{2}}}+{\boldsymbol {S}}:{\frac {\partial ^{2}{\tilde {g}}}{\partial {\boldsymbol {S}}\partial T}}~.}$

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