Continuum mechanics/Maxwell relations for thermoelasticity

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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}} ~. $$
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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}} ~. } $$

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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}~. $$
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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}~. } $$