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\title{Superconductivity - Assignment 2}
\author{
	Kees van Kempen (s4853628)\\
	\texttt{k.vankempen@student.science.ru.nl}
}
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\begin{document}

\section{Temperature dependence in Landau model}
In the Landau model, free energy is given as function of order parameter $\psi$ and temperature $T$ as
\[
	\mathcal{F} = a(T - T_c) \psi^2 + \frac{\beta}{2}\psi^4.
\]

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.

\[
	0 = \pfrac{\mathcal{F}}{\psi} = \pfrac{}{\psi} \left[ a(T-T_c)\psi^2 \right] = 2a(T-T_c)\psi + 2\beta\psi^3
\]

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
\[
	\mathcal{F}_0(T \leq T_c) = \frac{-a^2}{\beta}(T-T_c)^2 + \frac{a^2}{2\beta}(T-T_c)^2 = \frac{-a^2}{2\beta}(T-T_c)^2 \leq \mathcal{F}_0(T \geq T_c)
\]
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}

\section{Currents inside type-II superconducting cylinder}

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%\appendix

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