Advanced elasticity/Maxwell relations for thermoelasticity

From Wikiversity

Jump to: navigation, search

[edit] 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, η, \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}} ~. }

[edit] 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}~. }
Retrieved from "http://en.wikiversity.org/wiki/Advanced_elasticity/Maxwell_relations_for_thermoelasticity"