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The tree-level Cooper-pair scattering amplitude is simply $V$. Define the density of states at the Fermi surface, in the normalization of the question, byIn the one-loop Cooper diagram the two internal fermions have opposite momenta. After integrating over internal frequency, the remaining normal-energy integral containsConsequentlyand summing the leading geometric series gives the running BCS theory couplingFor repulsive $V>0$ it flows logarithmically toward zero. For attractive $V<0$, its denominator vanishes at the Cooper instability scaleAt this scale the normal Fermi liquid becomes unstable: opposite-momentum fermions form Cooper pairs, a superconducting or superfluid condensate develops, and a gap opens.

Superconductivity is a phase of matter that supports steady electric current with zero bulk electrical resistivity and exhibits the Meissner effect: equilibrium screening of a sufficiently weak, static magnetic field from the interior. It occurs below a material-dependent superconducting transition temperature, provided the electric current and applied magnetic field are also sufficiently small. ^[ref] Bulk expulsion means that a sample much thicker than its London penetration depth has an almost field-free interior; the magnetic field need not vanish at its surface. Conventional BCS theory explains superconductivity through correlated Electrons forming Cooper pairs. ^[ref]

The two observations answer different questions. A resistance-versus-temperature experiment for superconductivity asks whether a steady electric current requires a measurable electric potential difference. A Field-cooled Meissner experiment asks whether a magnetic field already present in the material is expelled when it enters superconductivity. An ideal perfect conductor with zero electrical resistivity can preserve that pre-existing magnetic field; zero electrical resistance alone therefore does not establish the Meissner effect.

The usefulness follows directly from Joule heating, $P=I^{2}R$: maintaining a steady electric current through a segment with zero electrical resistance produces no resistive heating in that segment. This makes persistent electric currents attractive for producing steady magnetic fields. The beauty is that one phase has both a transport property and an equilibrium magnetic response, and the second cannot be inferred just by making an ordinary electrical resistance very small. The London equations encode this extra magnetic response. ^[ref]

Weak coupling means a suitably dimensionless interaction strength is small. In BCS theory, ${\nu }_{s}V\ll 1$ gives a gap exponentially small compared with the pairing-shell energy cutoff; the expansion parameter is not the dimensional quantity $V$ alone.