What happens to an $RC$ circuit moving at relativistic speeds?
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. 483
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. 903