Module 4 : Thermodynamics of the superconducting transition
Lecture 2 : Application to the superconducting transition followed by problem solving
 
Application to superconductors
We will now calculate the Gibbs free energy difference between the normal state and superconducting state. At a fixed temperature, the applied magnetic field is varied to a critical field $H_{\mathrm{c}}$, so that superconductivity is destroyed and the substance becomes normal. Therefore, we integrate the $dG$ equation from $0$ to $H_{\mathrm{c}}$. This yields
  \begin{displaymath} G_{\mathrm{s}}(T,H_{\mathrm{c}})-G_{\mathrm{s}}(T,0)=-V\intop\overrightarrow{M}\cdot d\overrightarrow{H} \end{displaymath} (16)
Since, in the Meissner state, $H=-4\pi M$ , the integral gives $V\frac{H_{\mathrm{c}}^{2}}{8\pi}$. At the critical field, of course, the free energy of the superconductor is equal to that of the normal phase i.e., $G_{\mathrm{s}}(T,H_{\mathrm{c}})=G_{\mathrm{n}}(T,H_{\mathrm{c}})$. Also, since the normal state has negligible magnetisation, $G_{\mathrm{n}}(T,H_{\mathrm{c}})=G_{\mathrm{n}}(T,0)$. Therefore, in zero magnetic field, the superconducting state is lower in energy (Gibbs free energy) than the normal state by an amount $V\frac{H_{\mathrm{c}}^{2}}{8\pi}$. Clearly in zero field, the Gibbs and the Helmholtz free energy are the same and so the above statement is also valid for Helmholtz free energy. The above difference in the energy density between the normal and the superconducting state (the normal state being of higher energy) is called the condensation energy. The entropy difference can be calculated by taking the derivative of the Gibbs free energy with temperature. The equation below gives the difference in the entropy per unit volume between the superconducting and the normal state.
  \begin{displaymath} s_{s}(T,H_{c})-s_{n}(T,H_{c})=-\frac{H_{c}}{4\pi}\frac{dH_{c}}{dT} \end{displaymath} (17)
Therefore, when the external field is zero, the entropy difference is zero and there is no latent heat. The transition is then second-order. On the other hand, in the presence of a field there is a non-zero entropy change and hence the transition is of first-order.