ass2: This seems like a pretty thorough Landau theory solution.

superconductivity_assignment2_kvkempen.tex

@@ -49,18 +49,34 @@ In the Landau model, free energy is given as function of order parameter $\psi$
 	\mathcal{F} = a(T - T_c) \psi^2 + \frac{\beta}{2}\psi^4.
 \]
 
-To find the equilibrium value with respect to the order parameter $\psi_0(T)$,
-we need to equate the derivatives with respect to both $T$ and $\psi(T)$ to zero.
+The equilibrium state as function of temperature $T$ is the state of minimal free energy with respect to the order parameter $\psi(T)$.
+This point we call $F_0(T)$ with order parameter $\psi_0(T)$.
+For this, we will take the derivative of $F$ with respect to $\psi$ and equate it to zero.
 
 \[
-	0 = \frac{\delta\mathcal{F}}{\delta\psi} = \frac{\partial \mathcal{F}}{\partial \psi} - \nabla \cdot \frac{\partial \mathcal{F}}{\partial (\nabla \psi)}
+	0 = \pfrac{\mathcal{F}}{\psi} = \pfrac{}{\psi} \left[ a(T-T_c)\psi^2 \right] = 2a(T-T_c)\psi + 2\beta\psi^3
 \]
 
-Now we seek the Q = TdS, C = dQ/dT = T dS/dT
-For the entropy, we know
+Extreme points are found at $\psi = 0$ and $\psi = \pm\sqrt{\frac{-a}{\beta}(T-T_c)}$.
+
+For $T \geq T_c$, $\psi_0(T \geq T_c) = 0$ gives the minimum, i.e. $\mathcal{F}_0(T \geq T_c) = 0$.
+
+For $T \leq T_c$, $\psi_0(T \leq T_c) = \sqrt{\frac{-a}{\beta}(T-T_c)}$ is the minimum,
+giving free energy
 \[
-	S = -\pfrac{}
+	\mathcal{F}_0(T \leq T_c) = \frac{-a^2}{\beta}(T-T_c)^2 + \frac{a^2}{2\beta}(T-T_c)^2 = \frac{-a^2}{2\beta}(T-T_c)^2 \leq \mathcal{F}_0(T \geq T_c)
 \]
+where we chose the positive of the $\pm$ as the order parameter is understood to increase from finite at the phase transition.
+	
+For the specific heat, we find
+\[
+	C(T) = -T\pfrac{^2\mathcal{F}}{T^2} =
+	\begin{cases}
+		0 & T > T_c \\
+		\frac{a^2}{\beta}T & T < T_c
+	\end{cases}.
+\]
+There is thus a discontinuity in $C(T)$ at $T = T_c$ with size $\Delta C(T) = \frac{a^2}{\beta}T_c$.
 \section{Type-I superconducting foil}
 
 \section{Type II superconductors and the vortex lattice}