In many books, I've seen equations that describe the conduction of charge across a material. It seems all of them define some drift velocity as proportional to the electric field inside the conductor, which makes sense to me given that charges move from atom to atom instead of movent in free space, so as long as the field does not change with time the velocity is constant. But it got me thinking about superconductors. If there is no resistance, charges can move across the material as if it was free space so an electric field could accelerate the charges and the drift velocity would increase with time. Then the relationship that makes this velocity proportional to the field does not apply to superconductors. I also know that there is a limit for the current inside a superconductor and it will be resistive if the limit is reached. Taking that reasoning it would mean that if we want to maintain a current in a loop made of a superconductor material we must first get charges moving by means of an electric field (generated by magnetic induction for example) and then stop the acceleration so we don't reach the current limit. Is that reasoning correct? If not, would that mean that drift velocity is constant even with no resistance and a steady electric field?

  • $\begingroup$ I don't know the full answer to this question, but amazingly, the phenomenology is correct. A piece of superconductor in an other wise normal conducting circuit can be modeled like an inductor. It doesn't store charge, and it doesn't have resistance. It's state at a certain time is specified by how much current runs through it, and that has some energy associated with it in the magnetic field it produces. So like an inductor, that bit of superconductor has a current whose rate of change is proportional to the voltage you put across the superconductor. It's is exactly what you're proposing. $\endgroup$
    – AXensen
    Apr 2, 2023 at 17:20

1 Answer 1


you got most of the ideas right, I'll try to complete your reflexion the best I can. So buckle up, I'll try to make this as short as possible, but I believe it will be long.

You already know that a type I superconductor (SC) has a $\chi=-1$ (Meissner effect) up to a $T_c$ and/or $H_c$. Type II is a wee bit more complicated, but has the same $\chi$ under a certain range under $H_c$ and $I_c$ (read more about this type that produces vortex of current to protect its state, really interesting in my humble opinion). Magnetization of SC type I (left) and type II (right)

In the case of a time depend field, (out of London's limit where B=0), there is induction of an alternative current in a conductor (by Faraday's law). Using the two fluid model, where there is the normal ($n$ and over $T_c$) and superconducting electrons ($n_s$), a number given by $n_s (T) = n\left\lbrack 1 - \frac{T}{T_c} \right\rbrack^4 $ enter image description here

, one can write the superoconducting electron mouvement as: $m \dot{\mathbf{v}}_s =-e \mathbf{E}$, where there is no friction terme as there is no resistance. The current density yields $\mathbf{j}_s = -n_s e \mathbf{v}_s$ that we can derive to get $\dot{\mathbf{j}}=-n_s e \dot{\mathbf{v}}_s = \frac{n_s e^2}{m} \mathbf{E}$.
This then implies that one can change the current within a SC only when the current density varry through time, which differs from a metal. In other words, you need a potential difference between the extremities of the superconductor to change the current, but do not require it to maintain it.

The supeconducting electrons behavior can be compared to the one of a capacitor. In both case, a difference of potential $V=-\frac{L d I}{dt}$ is required to change the current, but in the case of a null potential, the current stay steady and non zero. This implies that SC have an intrinsic induction that is absent in metals: kinetic inductance. So, in a way, yes the electric field can accelerate the superconducting electrons up to a certain point, which I'm not sure when and how.

In the case of a stationnary field, $\dot{\mathbf{j}}_s = \mathbf{E} =0$. Also,one can use, surprisingly well, Maxwell's equations as follow : $\dot{\mathbf{B}}=-\nabla \times \mathbf{E}$. But since in the SC $\dot{\mathbf{B}}=0$ because $\mathbf{E}=0$. In this case, the magnetic field in the SC is time independent.
Starting from $\nabla \times \dot{\mathbf{j}}_s = \frac{n_s e^2}{m} \mathbf{\nabla}\times \mathbf{E}$ and $\dot{\mathbf{B}}=-\frac{m}{n_s e^2}\mathbf{\nabla}\times \dot{\mathbf{j}}_s$, we can extracte a penetration depth in the SC (mind you there are a few steps before reaching that point that I'm just avoiding to write) $\lambda_L =\left(\frac{m}{\mu_0 n_s e^2} \right)^{1/2} $. However, all this seems in contradiction with the first paragraphs and what was said a bit earlier. However, by intergrating $\dot{\mathbf{B}}$, one obtain: $\dot{\mathbf{B}}= - \frac{m}{n_s e^2} \nabla \times \mathbf{j}_s$ and realise that the magnetic field is actually related to a current density, even if on a minuscule depth of the SC. By using the fact that $\nabla\times \mathbf{B}= \mu_0 \mathbf{j}_s$, we obtain: $\nabla^2 \mathbf{B} = \mathbf{B}/\lambda_L^2$, which is London's equation. Solving it usually yields an exponential solution, as shown in the next figure, for a $\mathbf{B}$ along y.

London's equation representation in a SC

So there's always a penetration depth in a SC, even if extremely small. The closer the SC temperature gets to $T_c$, the deeper the fields can be coherent in the SC without "destroying" the superconducting state (of my understanding). Here's a graph of it

Penetration depth according to London's equation

To complete this long answer, another detail I won't dive into is the coherence length (or depth) which details a surface layer on the SC where there are different behavior than the rest of the SC (worth a read), is linked to the impuritites and the phenomological theorem of Ginzburg-Landau. All this culminates into the BCS theory which is crucial to understand most of the mechanism of electron displacement in a SC.

Hope this helps you out in your reflection, cheers


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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