Superconductivity resistance I'm confused about the resistance of superconductors. It is known that the resistance at the the critical temperature goes exactly to zero (the resistance is not very small, its really zero, right?). But in a superconducting state, not all electrons form Cooper pairs right? so why then is the resistance exactly equal to zero?
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 A: Its because of the band structure, in a regular metal if we plot the band structure we see that the spectrum is gapless. There is a conduction band and this band intersects the Fermi level. Thus very small energy is needed to excite the electrons on Fermi surface. However, mostly due to phonons these excited electrons scatter from those phonons and that's why we have finite resistance.
On the superconducting phase however we can diagonalize Hamiltonian by using Bogoliubov transformation.  The spectrum for Bogoliubov quasi particles will be gapped and the gap is $\Delta$ the superconducting order parameter.  This gap causes 0 resistance, because it blocks all scattering channels. That is when a charge carrier  wants to scatter away from impurity, phonon etc it can't find a state to be scattered to. All allowed states need much higher energy because of the gap thus charge carrier moves freely in superconductor.
A: It is absolutely true that the resistance of a superconductor goes to zero at at absolute zero, but just below the critical temperature, most of the conduction electrons are still in the normal state.  The current density has two parts:  J=...A+...E, the coefficients being proportional to the numbers of electrons in the paired and normal states.  The DC resistance vanishes because it is possible to have current without any electric field, but there can still be some AC resistance.
