Weak links play a very important role in modern superconductivity. In most cases weak links are oxide barriers between two superconducting thin films, but it can also be a crystal boundary (in the case of high-Tc superconductors). A schematic representation is given in Fig. 4. Now consider the ring which is thick everywhere except for a small section where the ring is closed via a weak link (Fig. 3b). The velocity is zero except near the weak link. In these regions the velocity contribution to the total phase change in the loop is given by (with Eq. ())
The line integral is over the contact from one side to the other in such a way that the end points of the line are well inside the bulk of the superconductor where . So the value of the line integral is well-defined (e.g. independent of the choice of the end points). With Eqs. (), (), and ()
Without proof we state that the supercurrent through the weak link is given by the so-called DC Josephson relation
The voltage over the contact is given by the AC Josephson relation
The names of these relations (DC and AC relations) are misleading since they both hold in DC and AC situations. In the steady state (constant ) Eq. () shows that V=0 while a nonzero current flows through the junction. In the case of a constant applied voltage (voltage bias) Eq. () can be integrated easily and gives
Substitution in Eq. () gives
This is an AC current. The frequency
is called the Josephson frequency. One μV gives a frequency of about 500 MHz. By using Eq. () the flux quantum is determined with the high precision as given in Eq. ().
The energy difference of a Cooper pair, moving from one side of the contact to the other, is . With this expression Eq. () can be written as  which is the relation for the energy of a photon with frequency ν.
:The AC Josephson relation (Eq. ()) can be easily understood in terms of Newtons law, (or from one of the London equations). We start with Newton's law 
:Substituting the expression for the Lorentz force  and using the general expression for the co-moving time derivative  gives 
:Eq. () gives  so 
:Take the line integral of this expression. In the end points the velocities are zero so the ∇v term gives no contribution. Using  and Eq. (), with  and , gives Eq. ().