3
$\begingroup$

I was trying to explain the dynamics of a DC current in a superconductor -- for example, a ring -- and was asked, essentially,

Cooper pairs are the charge carriers, are all in the same state, and cannot scatter. Well how do they "change direction" to travel in a circle then? Equivalently, why doesn't their inertia cause them to collide with the superconductor edge? What happens in a superconducting wire with a sharp angle?

After some thought, this is what I came up with:

The superfluid/Cooper pairs, since they are all in the same state, can be thought of like a single (rigid?) body with a total charge and total mass, in the "shape" of the superconductor. The initial applied external electric or magnetic field gives this body a "push" and it moves without dissipation. So for example a magnetic field sets it rotating, while an electric field gives it a linear momentum (with Andreev reflection/creating and destroying Cooper pairs at the ends). Therefore there are no worries about how individual Cooper pairs find their direction, they are "pulled along" by the Cooper pair state as a whole.

Leaving aside flux quantisation, is there anything wrong with this analogy, can you add more detail, or can you do better?

$\endgroup$
  • $\begingroup$ "Normal charge carriers" aren't changing direction because of "scattering", either, an electron beam in vacuum can, for instance, form a nice ring in the presence of a magnetic field just fine. Charges are mainly following electromagnetic fields and the scattering part only causes dissipation. The wave function of a superconductor does the same, but it can avoid the dissipation altogether. Sounds a bit like a trick question... ? $\endgroup$ – CuriousOne Mar 17 '16 at 19:33
  • $\begingroup$ Yes, I agree with you, maybe the explanation is same as in a normal conductor. But there is no magnetic field inside the superconductor -- so if we are only thinking about the microscopic dynamics, it's not the external magnetic field that's affecting the Cooper pairs -- it's the field created by the displacement/movement of the Cooper pairs around them? Hence they are "dragged along" with the current direction as a whole? Does that make any sense (it would seem to apply to normal conductors as well)? $\endgroup$ – Tejas Guruswamy Mar 18 '16 at 13:38
  • $\begingroup$ That there is no field in superconductors is incorrect, see en.wikipedia.org/wiki/London_penetration_depth. It's analogous to the skin effect in conductors. $\endgroup$ – CuriousOne Mar 18 '16 at 21:31
  • $\begingroup$ Is it true that the current is always within the penetration depth? $\endgroup$ – Tejas Guruswamy Mar 21 '16 at 13:11
  • $\begingroup$ I don't think there is a complete theory that can model every possible situation. Self-organizing superconducting and non-superconducting domains will form near the critical field, I believe, and one would have to include fluctuations into the calculation. I have no experience with the more complex scenarios. My aim was to give you the most trivial case... that's not enough to discuss all possible thermodynamic configurations. $\endgroup$ – CuriousOne Mar 22 '16 at 0:15
3
$\begingroup$

Phil Anderson used to refer to persistent currents as a "generalized rigidity phenomenon". Nevertheless, I think it is misleading to think of this in classical terms. All Cooper pairs are in the same wave function. If the wave function acquires an angular factor $\exp(in\phi)$, then the current is going around a ring. If the wave function has a linear phase $\exp(ikx)$, then all electrons are moving at the same speed. These phases are exactly what suitable external fields will create.

$\endgroup$
  • $\begingroup$ Thanks, I do know this is not correct or complete. Just trying out an analogy for the behaviour of Cooper pairs in terms of more "familiar" dynamics. The Anderson reference was actually very helpful. $\endgroup$ – Tejas Guruswamy Mar 18 '16 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.