The free energy difference between normal metal and superconductor can be then approximated ignoring the small metal susceptibility as
{$$G_n(T,0)-G_s(T,0)\approx Gs(T,H)-Gs(T,0) = \mu_0 \frac 1 2 H_c^2$$}
This is called condensation energy density, and it will be shown that it corresponds to the condensation of composite bosons (formed by pairs of electrons) into their unique ground state.
A simple calculation for Nb, with {$B_c=0.17$} T, gives a condensation energy corresponding to 0.02 K per Nb atom, a minute energy scale compared to that set by the Fermi energy of the metal, of order 10000 K. This was a long standing conundrum in the early understanding of superconductivity.
Index
Thermodynamics
The Landau free energy density is a series expansion in {$|\psi|^2$}, valid as long as the order parameter is small, i.e for {$T\lesssim T_c$}. Introducing a second order term in free energy, at {$H=0$}
{$$\delta f = -a(T-T_c)|\psi|^2 + \frac b 2 |\psi|^4$$}
and minimizing the order parameter {$|\psi|^2$} one gets its temperature dependence
{$$|\psi|^2 = \frac a b (T_c-T)$$}
In this interval the free energy difference, obtained by putting back the order parameter in the free energy density, to obtain
{$$\delta f = - \frac 1 2 \frac {a^2} b (T_c-T)$$}
is equal to the condensation energy, therefore
{$$B_c(T) =\left(\mu_0 \frac {a^2} b (T_c-T)^2\right)^{\frac 1 2} = \sqrt{\frac {\mu_0} b} a(T_c-T), \qquad T<T_c$$}
The change in entropy and heat capacity can be also computed. The former is {$\delta s= -\frac {\partial \delta f}{\partial T} = \frac {a^2} 2b (Tc-T)$} for {$T<T_c$} and vanishes above. The latter is {$\delta c= -T\frac {\partial \delta s}{\partial T}$}, i.e.
{$$c_s = c_n + \frac {a^2} b T,\qquad T<T_c$$}
that has a jump at {$T_c$} equal to {$\frac {a^2} b T_c$}. The experimental data for Al in Fig. 3 show the jump at {$T_c$}
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