How does Special Relativity remain valid in the calculation of time period of an $LC$-Circuit?

Question

I have a solenoid of length $l$, cross section $A$ and total turns $N$. So, it has an inductance of $L = \frac{\mu_0 N^2 A}{l}$.

I also have a square parallel-plate capacitor of plate area $A = d^2$ and distance between the plates $d$. So, it has a capacitance of $C = \frac{\epsilon_0 A}{d}$.

If I create an $LC$-Circuit using them, the time period will be $2 \pi \sqrt{LC}$.

Call the end points of the Solenoid as $P$ and $Q$ and that of the capacitor as $R$ and $S$. I create two scenarios:

(1) The LC-circuit is rectangular, with $PQ$ and $RS$ along the same side of the rectangle, and the circuit moves with relativistic speed along $PQRS$. So, Inductance will become $\gamma L$ (as the length of the solenoid will get contracted but cross section will remain same) and Capacitance will become $\gamma C$ (as the plate area will remain same but the plate distance will get contracted). So, the time period will be $2 \pi \gamma \sqrt{LC}$.

(2) The LC-circuit is rectangular, with $PQ$ and $RS$ perpendicular to each other, and the circuit moves with relativistic speed along $PQ$, perpendicular to $RS$. So, Inductance will be $\gamma L$ and Capacitance will become $\frac{C}{\gamma}$ (as one side of the plate will get contracted, but the other side and the plate distance will remain same). So, the time period will be $2 \pi \sqrt{LC}$.

So, how does Special Relativity remain valid in the calculation of time period of the second case?

The equation governing the LC-circuit is : $LC \frac{d^2q}{dt^2} + q = 0$

How does the equation get modified by SR?

Answer 1

Circuit theory is non-relativistic. You cannot use the equations of circuit theory to analyze a relativistically moving circuit. The laws of circuit theory are not laws of physics in the sense meant by the first postulate of relativity.

The only correct way to use circuit theory to analyze a relativistically moving circuit is to do the analysis in the rest frame of the circuit and then Lorentz transform the result into the moving frame.

If the analysis must be done in the frame where the circuit is moving, then you must use the correct relativistic laws, which are Maxwells equations.

Answer 2

You can reframe your setup in a more physical setting. An LC circuit can represent a specific mode of a resonating cavity. Consider a box cavity of dimension $l_x\times l_y\times l_z$ along the $x,y,z$ directions. Say it contains vacuum with permittivity $\epsilon_0$ and permeability $\mu_0$. Consider the mode $(k_x,k_y,k_z)$ with $k_z=0$ and $k_y \gg k_x$ (you cannot take $k_x=0$ hence the surrogate), with the magnetic polarisation along $x$ so that $B_y=B_z=0=E_x=0$ and $E_z\gg E_y$. You can check that: $$ L = \mu_0\frac{l_yl_z}{l_x} \quad C = \epsilon_0\frac{l_xl_y}{l_z} $$ and the frequency is: $$ \omega = ck_y\propto \frac c{l_y}=\frac1{\sqrt{LC}} \quad c = \frac1{\sqrt{\epsilon_0\mu_0}} $$ with $c$ the speed of light. Therefore, it is a physical realisation of your second scenario. You can now consistently boost the box in the $x$ direction according to special relativity. Indeed, you have length contraction only along $x$ so that: $$ l_x\to \frac{l_x}\gamma \quad l_y\to l_y \quad l_z \to l_z $$ The resonant mode is essentially light of bouncing back and forth at the reflecting walls of the cavity along the $y$ direction. After the boost, you would therefore expect a transverse Doppler effect so that: $$ \omega \to \gamma \omega $$ hence the apparent contradiction. From the perspective of the light bouncing, you need to take into account relativistic aberration. After the boost, the wave vector acquires a longitudinal component: $$ k_x=0\to k_x=\frac{\gamma v}{c^2}\omega \quad k_y\to k_y $$ so you still have: $$ \omega = ck $$ Similarly, according to relativistic transformation of fields: $$ \begin{align} E_x=0 &\to E_x=0 & E_y\sim 0 &\to \gamma E_y\sim0 & E_z &\to \gamma E_z \\ B_x &\to B_x & B_y=0 &\to \frac{\gamma v}{c^2} E_z & B_z &\to -\frac{\gamma v}{c^2}E_y \sim 0\\ \end{align} $$ In other words, the solenoid effectively rotates in the $x-y$ plane, maintaining orthogonality with the aberrated wavevector. Therefore, the inductance is not deduced from the same formula. Taking into account the rotation gives you a consistent resonant frequency with the one predicted by Doppler effect.

Your first scenario is not reproducible in this setting. The direction PQ of the solenoid is basically the one of the magnetic field and the direction RS of the capacitor is basically the one of the electric field. Your first scenario would mean that the electric and magnetic field are roughly parallel which is incompatible with Faraday's law.