For some phase transitions, symmetry allows the Landau free energy to be written as $$ F(\eta)=\frac{1}{2} a\left(T-T_{\mathrm{c}}\right) \eta^{2}+\frac{1}{4} \eta^{4}+\frac{1}{2} \lambda \epsilon_{\mathrm{s}} \eta+\frac{1}{2} C_{\mathrm{el}} \epsilon_{\mathrm{s}}^{2} $$ where \(\epsilon_{\mathrm{s}}\) is the spontaneous strain. Show (a) that the equilibrium condition \(\partial F / \partial \epsilon_{\mathrm{S}}=0\) gives \(\epsilon_{\mathrm{s}}=-\lambda \eta / 2 C_{\mathrm{el}}\); (b) that this coupling leads to a new transition temperature at \(T_{\mathrm{c}}+\lambda^{2} / 4 a C_{\mathrm{el}} ;\) (c) that the corresponding elastic constant falls to zero at the phase transition temperature.

To proceed, we start by partially differentiating the given Landau free energy expression with respect to \(\epsilon_{\text{S}}\): \[ F(\eta, \epsilon_{\text{s}}) = \frac{1}{2} a \left(T-T_{\text{c}}\right) \eta^{2} + \frac{1}{4} \eta^{4} + \frac{1}{2} \lambda \epsilon_{\text{s}} \eta + \frac{1}{2} C_{\text{el}} \epsilon_{\text{s}}^{2} \] Let's calculate the partial derivative \(\frac{\partial F}{\partial \epsilon_{\text{S}}} \): \[ \frac{\partial F}{\partial \epsilon_{\text{S}}} = \frac{1}{2} \lambda \eta + C_{\text{el}} \epsilon_{\text{s}} \]

According to the equilibrium condition, \(\frac{\partial F}{\partial \epsilon_{\text{s}}}=0\). Therefore, we set the partial derivative equal to zero: \[ \frac{1}{2} \lambda \eta + C_{\text{el}} \epsilon_{\text{s}} = 0 \]

From the equation \(\frac{1}{2} \lambda \eta + C_{\text{el}} \epsilon_{\text{s}} = 0\), solve for \(\epsilon_{\text{s}}\): \[ \epsilon_{\text{s}} = -\frac{\lambda \eta}{2 C_{\text{el}}} \]

Substitute \(\epsilon_{\text{s}} = -\frac{\lambda \eta}{2 C_{\text{el}}}\) back into the Landau free energy expression: \[ F(\eta) = \frac{1}{2} a \left(T-T_{\text{c}}\right) \eta^{2} + \frac{1}{4} \eta^{4} + \frac{1}{2} \lambda \left(-\frac{\lambda \eta}{2 C_{\text{el}}}\right) \eta + \frac{1}{2} C_{\text{el}} \left(-\frac{\lambda \eta}{2 C_{\text{el}}}\right)^{2} \] Simplify the expression: \[ F(\eta) = \frac{1}{2} a \left(T-T_{\text{c}}\right) \eta^{2} + \frac{1}{4} \eta^{4} - \frac{1}{4} \frac{\lambda^{2} \eta^{2}}{C_{\text{el}}} + \frac{1}{8} \frac{\lambda^{2} \eta^{2}}{C_{\text{el}}} \] Combine terms: \[ F(\eta) = \left( \frac{1}{2} a \left(T-T_{\text{c}}\right) - \frac{1}{8} \frac{\lambda^{2}}{C_{\text{el}}} \right) \eta^{2} + \frac{1}{4} \eta^{4} \] The new transition temperature is obtained by setting the coefficient of \eta^{2} to zero: \[ \frac{1}{2} a \left(T-T_{\text{c}}\right) - \frac{1}{8} \frac{\lambda^{2}}{C_{\text{el}}} = 0 \] Hence, the new transition temperature is: \[ T'_{\text{c}} = T_{\text{c}} + \frac{\lambda^{2}}{4 a C_{\text{el}}} \]

To demonstrate that the elastic constant falls to zero at the new transition temperature, evaluate the second derivative of the Landau free energy with respect to \(\eta\): \[ C_{\text{el}} \left.\frac{\partial^{2} F}{\partial \eta^{2}}\right|_{T=T'_{\text{c}}} = \frac{1}{2} a \left(T'_{\text{c}}-T_{\text{c}}\right) - \frac{1}{8} \frac{\lambda^{2}}{C_{\text{el}}} = \frac{1}{2} a \left( \frac{\lambda^{2}}{4 a C_{\text{el}}} \right) - \frac{1}{8} \frac{\lambda^{2}}{C_{\text{el}}} = 0 \]