What is the RC time constant in a superconductor?

Question

In conventional conductors, the RC time constant is the time required to charge or discharge a capacitor through a resistor by ≈ 63.2 percent of the difference between the initial value and final value:

$$\tau = R \cdot C $$

However, in a superconductor, the resistance is exactly zero.

$$\tau = 0 \cdot C = 0$$

Which would require infinite current because the capacitor is charged in 0 time. So the RC time constant equation above must not be valid for superconductors. How should the RC time constant be defined in a superconductor such that it does not require infinite current?

That is, what is the expression for the time constant of charging a superconducting capacitor by ≈ 63.2 percent?

Edit: I re-framed the question because we were getting bogged down about signal propagation time. Now instead of using special relativity to point out the flaw in the equation, I note that it would require infinite current, which would not be physically realizable.

Answer 1

The RC constant is not fundamentally related to the speed of signals. It is derived from an equation which already assumes the speed of signals in your circuit is effectively infinite. Therefore there is no relativistic constraint on RC time. To arrive at a formula which takes into account the propagation time you would need to account for the impedance of the wires at least.

Superconductors display a variety of complicated behaviors when put into larger circuits. To give you a more concrete formula would require a more concrete circuit. The simplest possible behavior is that of a perfect inductor.

Also to correct errors in the comments, the speed of signals in a conductor has almost no relation to the "speed of electrons".

Answer 2

Short answer: The RC-time is not applicable for a superconducting capacitor, because it ignores the inductance $L$ that the capacitor has. When $L$ is bigger than $R$, you should use an LC-circuit to describe your system.

Long answer: There are many reasons why superconductors don't carry infinite current:

1. Pracitcal Limit No matter what method you use to put a current $I$ into a superconductor, there is a natural limit, which is given by the critical current $I_c$. You'll probably want to design your circuit in such a way that you don't go over this limit.

2. AC Limit The resistance of a superconductor is only zero in the stationary case, that is when neither voltages nor currents change. For any AC-signals, it definitely has some resistance! There are several effects that have to be taken into account when minimizing losses in, for example, superconducting LC tank circuits for signal pickup, as it is done in single-ion detectors. You mentioned RC-time constants, this often implies that we are talking about AC-signals, or at least something non-stationary, so there will be some resistance. But it will be low.

3. DC limit Even in the (almost) stationary case of non-changing currents and voltages, real superconductors have some resistance. The superconducting magnet (a superconducting coil with 30 A of current in it) that I use has a field that decays with a time-constant of a few hundred thousand years, probably due to the weld that joins the superconducting wire to a loop. There seems to be an ongoing debate on whether or not a superconducting loop emits synchrotron radiation, but in any case this would be an extremely small effect that leads to decay constants that are longer than the age of the universe. I guess in real life, cosmic rays knocking the occasional cooper-pair out of order play a much bigger effect.

In the case of a properly designed superconducting capacitor, the resistance of the superconductor is negligible compared to its inductance, which is why an RC description is not a good way to model the system. (Note that even a straight wire has self-inductance, and so does any capacitor, be it superconducting or not.) If you use a more appropriate LC-model to describe your system, you quickly see why the current is limited:

As you start charge your superconducting capacitor, the inductance limits the rate of the current change to $dI/dt = U/L$, so the current increases until the capacitor is charged. But the current keeps going (due to the inductance!) so that the voltage of the capacitor will be higher than what the voltage source applies. The voltage source will work against the inductance to reverse the current, so that the voltage goes down. Again, as the capacitor is at nominal voltage that the voltage source applies, the current is still going. The voltage across the capacitor will go down further, as the current is reversed again. Basically, by switching on the voltage source you excited a parallel LC circuit, and it will ring forever at its resonance frequency $\omega = \sqrt{L C}$.

Really forever? No, the AC-losses will cause the oscillation to die down within $10^3 - 10^6$ periods.

Answer 3

Which is in disagreement with special relativity because information could be transmitted along a superconductor instantaneously.

For even a theoretical ideal wire with exactly zero resistance, signals do not propagate with infinite speed.

The reason is that, even without resistance, there is associated inductance and capacitance that cannot be eliminated. In fact, we have, from transmission line theory, that the signal propagation speed is given by:

$c = \dfrac{1}{\sqrt{LC}}$

where $L$ and $C$ are the inductance and capacitance per unit length.