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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

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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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  • $\begingroup$ "Typical values for μH0 is in range of 0.01-0.1 Tesla." Assuming 10mT critical field for a normal superconductor wire of radius 5mm. It would take 250A through the wire before the wire reaches the 10mT critical value and stop behaving like a superconductor!! hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html Therefore the benefits of using superconductive wires for power transmission are enormous. $\endgroup$
    – Markoul11
    Commented May 17, 2022 at 15:22
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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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    $\begingroup$ 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) $\endgroup$
    – diffeomorphism
    Commented Oct 21, 2015 at 15:27
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Commented Oct 21, 2015 at 16:31
  • $\begingroup$ @Thomas How the magnetic field get induced? Cooper pairs are pairs with opposite magnetic dipole moments and for this the magnetic field cancels out? If so, not paired electrons involved too? They move in circle (in the coil) and induce the magnetic field? Somehow the EM radiation from this accelerated electrons could not radiate? Where are the mistakes? Never before thought about this until there was this question about physics.stackexchange.com/questions/226250/… and my - perhaps flippantly - answer. $\endgroup$
    – HolgerFiedler
    Commented Dec 29, 2015 at 18:57
  • $\begingroup$ @HolgerFiedler 1) There is no B field inside a superconductor. In the case of vortices the B field goes through the vortex core, where the superfluid density is small. The field and current distribution in a superconducting wire or coil are complicated, but the main observation is that the current is concentrated near the surface of the wire. 2) There is no radiation (just like in an atom), the electrons are in a quantum state with fixed L. $\endgroup$
    – Thomas
    Commented Dec 30, 2015 at 2:40
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    $\begingroup$ @Thomas Happy new year. For permanent magnets it is obvious today that the aligned magnetic dipole moments are responsible for the magnetic field. The thoughts, that a current, and not a deeper explanation, is responsible for electrodynamic fields, is how old? Are we sure that the measurement of a stream of electrons does not induce (influence) a magnetic filed by alignment of the magnetic dipole moments of the involved electrons? Thinking in such a way I have doubts about that Cooper paired electrons are responsible for the magnetic field. $\endgroup$
    – HolgerFiedler
    Commented Jan 1, 2016 at 8:14

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