What happens to an $RC$ circuit moving at relativistic speeds?

Question

Suppose I have a capacitor of capacitance $C$ connected to a resistor of resistance $R$. In a frame where this circuit is stationary, the time it takes to discharge $63\%$ or $(1-1/e)$ of charge on the capacitor is $RC$. I have effectively created a "clock" with which I can measure time, assuming I can measure the amount of charge left on the capacitor plates.

Now suppose we are in a frame where the circuit is moving at some speed $v$, my "lab" frame. Moving clocks run slow, so I will observe any process on the capacitor to take longer by a factor of $\gamma=(1-v^2/c^2)^{-1/2}$. This suggests that my circuit now has an effective time constant of $\gamma RC$, where $R$ and $C$ are the resistances and capacitances measured in the circuit's frame.

What do I observe in the lab frame as the cause of this increase in time constant? Further, does $RC$ actually retain meaning as "time to discharge to 1/e" if the circuit is moving?

Possible thoughts might include the following. Suppose the capacitor is a simple parallel plate capacitor, and it is moving in a direction parallel to the normal of the plates. One could argue that in the lab frame the plates appear to get closer together by a factor of $\gamma$ (length contraction) and so the capacitance as measured in the lab frame is $\gamma C$, potentially leading to time constant $\gamma RC$.

I have a problem with the above argument though. What if the capacitor is moving in a direction perpendicular to the normal of the plates? Then the area of the capacitor will shrink by a factor of $\gamma$, but the distance between the plates remains the same. Since $C=A\varepsilon_0/d$ the capacitance measured in the lab frame is now $C/\gamma$.

I can think of two possible routes to fix this. Does consideration of the resistor play a role? e.g. for a material with resistivity $\rho$, the resistance is $R=\rho L/A$, and $L$ and $A$ could change by factors of $\gamma$ depending on the orientation of the resistor. Does the $\bf B$-field that will be present in the lab frame also have anything to do with the answer? The $\bf B$-field will appear because of the way you Lorentz boost EM fields.

Answer 1

Circuit theory is inherently non-relativistic. You simply cannot use circuit theory to analyze a circuit moving at relativistic velocity. When a circuit is moving relativistically you need to use Maxwell’s equations to analyze it and cannot take the usual shortcuts afforded by circuit theory.

The issue is the following. Circuit theory rests on three assumptions:

1. all lumped elements have no net charge
2. there is no inductive coupling between lumped elements (a mutual inductance is considered a single lumped element with four terminals)
3. the circuit is small enough that c can be ignored and all effects are assumed to happen instantaneously

Assumption 3) is expressly non-relativistic, and assumption 1) is also violated for current-containing devices, as described by Purcell. Only with all three of these assumptions can circuit theory be derived from Maxwell’s equations. So without them any attempt to use circuit theory will inevitably lead to contradictions.

Circuit theory must be applied in the rest frame of the circuit, or at least in a frame where it is moving non-relativistically. Once you have done that, then you can transform the results into a different frame if you need

Answer 2

It sounds like a cop out but whatever the details are, the observed time for anything in the moving inertial frame must reduce by the factor $\gamma$.

I can think of another experiment. You could have a weight vibrating in simple harmonic motion on a spring. If the spring was horizontal (in the direction of motion), it would be shorter but just as thick so you might conclude that it was stiffer so the frequency should be greater. Whereas, if the spring was perpendicular to the direction of motion, it would be the same length but narrower ans so would have less stiffness and consequently the vibrations should have a lower frequency.

On the other hand, if you were in the moving frame, you would measure all the distances and times to be the same as you would if they were at rest. Similarly the capacitor and resistor would appear to you to be performing as they would at rest, because you could consider them to be at rest. You, at rest, might be able to measure the increased time that things take in the moving frame and the shortenings in the direction of motion but you can't do calculations on them as if they were in your world.

Answer 3

I admit that I am not familiar with Maxwell's equations in matter taken to relativistic speeds. But I strongly guess that this would be the right approach to answer your question in principle, and I feel that all the quantities characterizing matter (namely permittivity, permeability and conductivity) should then be reinterpreted as components of properly transforming 4-tensors.

However this Wikipedia article writes the constitutive equations of matter a little differently, involving the 4-velocity of the experiment. But, to be honest, this does not look very convincing to me because you might also have a dynamic material arrangement (think of a conductor moving relative to the magnetic field of a coil) that is translated as a whole at relativistic speeds. Should the (local) Maxwell's equations really depend on the determination of a global property like the 4-velocity of the experiment?

I have found other references that seem to indicate that material electrodynamics at relativistic speeds is not very well investigated and even ambiguous (especially with respect to the relativistic form of Ohm's law). I can hardly believe that, but if it were the case this might probably be due to lack of applications: nobody has ever had the opportunity to send an electric circuit on a relativistic flight.

Anyways, those kinds of ugly transformation details (including also the elastic properties of the electric parts for example) are most probably the reason why special relativity is used for macroscopic problems instead of the (intrinsically relativistic) Maxwell's equations.

Answer 4

Moving resistor

Like cable squeezed so that it becomes thinner, resistance does not change.

Capacitor moving in a direction perpendicular to the normal of the plates

The area of the capacitor will shrink by a factor of γ. But like in the case of the the resistor, that does not matter. The reason being that the fields shrink the same way as the plates.

But the force that an electron feels when moving from one plate to another is reduced, so the energy released is reduced by a factor of γ. We can say the force is reduced because the electron is moving according to us and there is a B-field according to us because the plates are moving according to us, so there is a Lorentz-force pointing to the opposite direction compared to the electrostatic force.

As the energy released is reduced by a factor of γ, the potential difference is reduced by a factor of γ, so capacitance is increased by a factor of γ.

Capacitor moving in a direction parallel to the normal of the plates

The force that an electron feels when moving from one plate to another is unchanged, but distance is reduced, so the energy released is reduced by a factor of γ.

As the energy released is reduced by a factor of γ, the potential difference is reduced by a factor of γ, so capacitance is increased by a factor of γ.