In studying simple drift based conductivity in metals and semiconductors, we follow the simplistic drift based model wherein, $$J = e(n\mu_n + p\mu_p)E = e(nV_{d,n} + pV_{d,p})$$ This is a completely intuitive model used for explaining behavioural modelling of semi-conductors and metals at least at the device-physics levels.
Here in $n$ and $p$ scale as $n = N_c e^{(E_c - E_F)/(k_B T)}, \, p=N_p e^{(E_F - E_V)/(k_B T)}$. Of course the derivation oversees itself from an electron-gas model(no lattice potential included)
My question is, is this approach also valid for relating to superconducting behaviour? I mean, if it were, then for having a $\rho = 0$, we would need $\sigma \to \infty$?
From what I can see is that, $\sigma \propto n\mu_n + p\mu_p$. Can this be resolved by saying $n,p \to \infty$, but this doesn't make much sense. Or is it resolved by saying $\mu \to \infty$, again this is problematic because then we would have $V_d \to \infty$. Does this model completely break down for superconductors?
What I do understand is that, since cooper pairs are bosons, we are not handicapped by Pauli's exclusion principle hence, every possible pair can exist at the same energy level thus overpopulating the values of $n$ or $p$ intuitively. But, even so we would be limited to the total no. of electrons in the system to form cooper pairs again setting up a hard limit for $n$ (though extremely large). Also relativistically the $V_d < c$. So, is there a hard limit to how low $\rho$ can be?
