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Is there a formula for the effective speed of electron currents inside superconductors?

The formula for normal conductors is:

$$ V = \frac{I}{nAq}$$

I wonder if there are any changes to this formula for superconductors.

Is there any regime for existing superconductors where the electrons will be flowing at speeds near light speed? Or more precisely, is it possible to have carrier currents that produce drift velocities that are relativistic, while maintaining the superconducting phase?

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2 Answers 2

This formula is derived using conservation of charge principle and so it's valid for the superconductors as well. There's a critical magnetic field that above which a superconductor becomes normal conductor and it's a function of temperature.

If a large current is to pass through a superconductor, a magnetic field will be produced that disrupts superconductivity when exceeds this critical magnetic field, so you can't have arbitrarily large currents and drift speeds will be well below relativistic speeds.

This is an approximate formula for dependence of this critical magnetic field on temperature: $H_c(T) = H_0[1-({T \over T_c})^2]$

In which $T_c$ is the critical temperature at zero field and $H_0$ is the critical field at zero temperature. Typical values for $\mu H_0$ is in range of 0.01-0.1 Tesla.

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I think there is a misunderstanding underlying your question. The fact that the material is a superconductor does not imply that electrons are flowing at relativistic speeds.

1) In a superconductor the current is carried by the condensate, not by random motion of electrons. In that sense, there is no drift velocity. The formula for the current density is $$ \vec\jmath = \frac{e n_s}{2m}\hbar\vec\nabla\phi $$ where $n_s$ is the superfluid density, and $\phi$ is the phase of the condensate wave function. Note that $$ v_s = \frac{\hbar}{2m}\vec{\nabla}\phi $$ is the velocity of the supercurrent.

2) At $T=0$ we have $n_s\to n$, and the first formula is essentially the same as the usual formula for the current density, $\vec\jmath=en\vec{v}$, so a naive estimate of the drift velocity from the known current and density of electrons give the correct supercurrent velocity (even though the physical mechanism is quite different).

3) The current density in a superconductor is not larger than what can be achieved in ordinary conductors (in fact, it is often smaller). Superconductors have a critical current (roughly, the kinetic energy of the supercurrent cannot exceed the condensation energy). As a result the supercurrent velocity is limited to speeds much smaller than $c$.

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No "misunderstanding", I just completely ignore if that was possible or not, I asked if there was a regime where that would happen, or if it would be possible for any sort of conductor to have relativistic velocities in the charge carriers (graphene electron velocities are about 3000 km/s in certain circumstances) – diffeomorphism Oct 21 at 15:27
Very good. No, in ordinary superconductors the velocity of the supercurrent is nowhere near c. Even in relativistic superconductors (e.g. paired quarks in neutron stars), getting close to c is not esay. – Thomas Oct 21 at 16:31

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