If superconductors have no resistance, what prevents you making the cross-section as small as you can handle, if there will be no power dissipated anyhow?

Is there a limit on the magnetic field that the superconductor can contain, if there is no limit on the current itself?

  • $\begingroup$ I hope this question gets the quality of answer it deserves. An answer which combines the theoretical aspect of the phase gradient with some kind of physical picture would be really nice. $\endgroup$ – DanielSank Jan 23 '16 at 22:10
  • $\begingroup$ Whilst this question could be a substantial part of a PhD thesis, I thoroughly recommend the highly accessible paper by the late David Dew-Hughes, w2agz.com/Library/Classic%20Papers%20in%20Superconductivity/…, which provides a historically informed discussion. $\endgroup$ – Landak Jan 24 '16 at 0:26

The super-current is carried by the gradient of the phase of the condensate, and there is a finite energy cost associated with this. If the gradient energy is larger than the BCS condensation energy (the energy gained by forming Cooper pairs), then superconductivity will disappear. This is the critical current.

In order to be more quantitative, consider the Landau-Ginzburg functional $$ {\cal F} = \frac{1}{2m}|(k-qA)\psi|^2+\alpha|\psi|^2+\frac{1}{2}\beta|\psi|^4 $$ where $k$ is the momentum, $q$ is the charge, $A$ is the gauge potential, $\psi$ is the condensate wave function, and $\alpha,\beta$ are parameters ($\alpha<0$ in the BCS phase). We can compute the super-current $$ j= \frac{q}{m}(k-qA)|\psi|^2 $$ and define the superconducting velocity $v_s=(k-qA)/m$. Vary the free energy with respect to $\psi$ to determine the ground state at finite $v_s$. We get $$ j = 2qn_s^0 v_s \left ( 1-\frac{mv_s^2}{2|\alpha|}\right) $$ where $n_s^0$ is the density of paired electrons in the absence of a current. You see that there is a maximum. $\alpha$ and $\beta$ are related to other quantities, so you can express the critical current in terms of properties like the critical field and the penetration depth.

  • $\begingroup$ "... carried by the gradient of the phase..."? $\endgroup$ – Daniel Griscom Jan 24 '16 at 3:19
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    $\begingroup$ In quantum mechanics, in order to get a current, we need $\psi=\psi_0 e^{i\phi}$ with $\phi=kx$. Then $j \sim k |\psi_0|^2$. This is the sense in which the current $j$ is proportional to the gradient $k$ of the phase $\phi$. The Hamiltonian has a term $|\nabla\psi|^2$, so there is a cost $k^2|\psi_0|^2$ to having a current, even in a superconductor. $\endgroup$ – Thomas Jan 24 '16 at 3:25
  • $\begingroup$ ... I was trying to point out a typo; I can't do a 2-character edit to your post, but I believe you can. $\endgroup$ – Daniel Griscom Jan 24 '16 at 3:26
  • $\begingroup$ So this means a superconductor can never asymptote the relativistic limit of current at $J = nec$ where $n$ is the number of free electrons per unit volume? $\endgroup$ – busukxuan Jan 24 '16 at 3:36
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    $\begingroup$ @busukxuan Indeed, see my answer to this question physics.stackexchange.com/questions/36053/… $\endgroup$ – Thomas Jan 24 '16 at 4:12

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