Exercise 7.8.2: A cube and what happens when it is squeezed

$c/2a=3\mathrm{GHz}$, and

• $\sqrt{2}\times 3=4.243\mathrm{GHz}$: TE${}_{101}$, TE${}_{011}$, TM${}_{110}$, three modes,
• $\sqrt{3}\times 3=5.196\mathrm{GHz}$: TE${}_{111}$, TM${}_{111}$, two modes,
• $\sqrt{5}\times 3=6.708\mathrm{GHz}$: TE${}_{102}$, TE${}_{201}$, TE${}_{012}$, TE${}_{021}$, TM${}_{120}$, TM${}_{210}$, six modes.

TE${}_{101}$ does not involve $b$ and stays at $4.243\mathrm{GHz}$. TE${}_{011}$ and TM${}_{110}$ both contain $(1/b{)}^2$ and move up together:

The triple splits into a single mode at $4.243\mathrm{GHz}$ and a pair $43\mathrm{MHz}$ higher. TE${}_{111}$ and TM${}_{111}$ rise from $5.196$ to $5.232\mathrm{GHz}$ and stay degenerate.

At $4.243\mathrm{GHz}$ $\delta =1/\sqrt{\pi f{\mu }_0{\sigma }_c}=1.01\mu \mathrm{m}$. The squeezed box is still almost a cube, so $Q\approx a/3\delta =0.05/(3\times 1.01\times 10^{-6})=16400$ and $\Delta f=4.243\mathrm{GHz}/16400=0.26\mathrm{MHz}$. The $43\mathrm{MHz}$ split is $170$ bandwidths: the modes are fully resolved.

A vertical probe couples to $E_y$. TE${}_{101}$ has $E_y=E_0\sin \frac{\pi x}{a}\sin \frac{\pi z}{d}$, maximal at the probe. TE${}_{011}$ has only $E_x$ ($m=0$), and TM${}_{110}$ only $E_z$ ($p=0$). Only TE${}_{101}$ is excited.

The relative shift of the pair, $43/4243=1.0\%$, is half of the $2\%$ change of $b$, as expected: $b$ contributes half of $f^2$ for these two modes.