Continuum mechanics/Maxwell relations for thermoelasticity

Maxwell relations between thermodynamic quantities For thermoelastic materials, show that the following relations hold: $\frac{\partial \psi }{\partial 𝑬}=\frac{1}{{\rho }_0}\widehat{𝑺}(𝑬,T);\frac{\partial \psi }{\partial T}=-\hat{\eta }(𝑬,T);\frac{\partial g}{\partial 𝑺}=\frac{1}{{\rho }_0}\widetilde{𝑬}(𝑺,T);\frac{\partial g}{\partial T}=\tilde{\eta }(𝑺,T)$ where $\psi (𝑬,T)$ is the Helmholtz free energy and $g(𝑺,T)$ is the Gibbs free energy. Also show that $\frac{\partial \widehat{𝑺}}{\partial T}=-{\rho }_0\frac{\partial \hat{\eta }}{\partial 𝑬}\text{and}\frac{\partial \widetilde{𝑬}}{\partial T}={\rho }_0\frac{\partial \tilde{\eta }}{\partial 𝑺}.$

Proof:

Recall that

$\psi (𝑬,T)=e-T\eta =\overline{e}(𝑬,\eta )-T\eta .$

and

$g(𝑺,T)=-e+T\eta +\frac{1}{{\rho }_0}𝑺:𝑬.$

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

The derivatives are

$\frac{\partial \psi }{\partial 𝑬}=\frac{\partial \overline{e}}{\partial 𝑬}=\frac{1}{{\rho }_0}𝑺;\frac{\partial \psi }{\partial T}=-\eta .$

and

$\frac{\partial g}{\partial 𝑺}=\frac{1}{{\rho }_0}\frac{\partial 𝑺}{\partial 𝑺}:𝑬=\frac{1}{{\rho }_0}𝑬;\frac{\partial g}{\partial T}=\eta .$

Hence,

$\frac{\partial \psi }{\partial 𝑬}=\frac{1}{{\rho }_0}\widehat{𝑺}(𝑬,T);\frac{\partial \psi }{\partial T}=-\hat{\eta }(𝑬,T);\frac{\partial g}{\partial 𝑺}=\frac{1}{{\rho }_0}\widetilde{𝑬}(𝑺,T);\frac{\partial g}{\partial T}=\tilde{\eta }(𝑺,T)$

From the above, we have

$\frac{{\partial }^2\psi }{\partial T\partial 𝑬}=\frac{{\partial }^2\psi }{\partial 𝑬\partial T}⟹-\frac{\partial \hat{\eta }}{\partial 𝑬}=\frac{1}{{\rho }_0}\frac{\partial \widehat{𝑺}}{\partial T}.$

and

$\frac{{\partial }^{2}g}{\partial T\partial 𝑺}=\frac{{\partial }^{2}g}{\partial 𝑺\partial T}⟹\frac{\partial \tilde{\eta }}{\partial 𝑺}=\frac{1}{{\rho }_0}\frac{\partial \widetilde{𝑬}}{\partial T}.$

Hence,

$\frac{\partial \widehat{𝑺}}{\partial T}=-{\rho }_0\frac{\partial \hat{\eta }}{\partial 𝑬}\text{and}\frac{\partial \widetilde{𝑬}}{\partial T}={\rho }_0\frac{\partial \tilde{\eta }}{\partial 𝑺}.$

More Maxwell relations For thermoelastic materials, show that the following relations hold: $\frac{\partial \hat{e}(𝑬,T)}{\partial T}=T\frac{\partial \hat{\eta }}{\partial T}=-T\frac{{\partial }^2\hat{\psi }}{\partial T^2}$ and $\frac{\partial \tilde{e}(𝑺,T)}{\partial T}=T\frac{\partial \tilde{\eta }}{\partial T}+\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}=T\frac{{\partial }^2\tilde{g}}{\partial T^2}+𝑺:\frac{{\partial }^2\tilde{g}}{\partial 𝑺\partial T}.$

Proof:

Recall,

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

and

$\tilde{g}(𝑺,T)=g=-e+T\eta +\frac{1}{{\rho }_0}𝑺:𝑬=-\tilde{e}(𝑺,T)+T\tilde{\eta }(𝑺,T)+\frac{1}{{\rho }_0}𝑺:\widetilde{𝑬}(𝑺,T).$

Therefore,

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

and

$\frac{\partial \tilde{e}(𝑺,T)}{\partial T}=-\frac{\partial \tilde{g}}{\partial T}+\tilde{\eta }(𝑺,T)+T\frac{\partial \tilde{\eta }}{\partial T}+\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}.$

Also, recall that

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

$\tilde{\eta }(𝑺,T)=\frac{\partial \tilde{g}}{\partial T}⟹\frac{\partial \tilde{\eta }}{\partial T}=\frac{{\partial }^2\tilde{g}}{\partial T^2},$

and

$\widetilde{𝑬}(𝑺,T)={\rho }_0\frac{\partial \tilde{g}}{\partial 𝑺}⟹\frac{\partial \widetilde{𝑬}}{\partial T}={\rho }_0\frac{{\partial }^2\tilde{g}}{\partial 𝑺\partial T}.$

Hence,

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

and

$\frac{\partial \tilde{e}(𝑺,T)}{\partial T}=T\frac{\partial \tilde{\eta }}{\partial T}+\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}=T\frac{{\partial }^2\tilde{g}}{\partial T^2}+𝑺:\frac{{\partial }^2\tilde{g}}{\partial 𝑺\partial T}.$

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