Frequency and Wavenumbers in Rectangular Waveguide for TE01

Question

In Chapter 9 of Principles of Electrodynamics by Schwartz, which is on waveguides and cavities, there is an analysis of a simple rectangular guide and a calculation of the lowest frequency that can pass through the guide, either $TE_{01}$ or $TE_{10}$ depending on whether $a$ (the $x$ dimension) is greater or less than $b$ (the $y$ dimension). It should be clear, but $TE_{m,n}$ is the transverse electric wave for eigen mode $m,n$.

From the standard wave equation $(\nabla^{2} - \frac{1}{c^2}\frac{\partial^{2}}{\partial t^{2}})\vec{E}$ and likewise for $\vec{B}$, which are clearly true on a component by component basis for each, there is a standard linear PDE analysis as:

$$ E_{x}(x,y,z,t) = E_{x1}(x)E_{x2}(y)E_{x3}(z)E_{x4}(t) $$

$$ \frac{1}{E_{x1}(x)}\frac{d^{2}E_{x1}}{dx^2}+\frac{1}{E_{x2}(y)}\frac{d^{2}E_{x2}}{dy^2}+\frac{1}{E_{x3}(z)}\frac{d^{2}E_{x3}}{dz^2}-\frac{1}{c^{2}E_{x4}(t)}\frac{d^{2}E_{x4}}{dt^2}=0 $$

These are then equal to the four eigenvalues, which give equations

$$ \frac{d^{2}E_{xi}}{du^{2}} = -k_{i1}^{2}E_{xi} $$

for $1 <= i <= 3$ and $u = x,y,z$ respectively, and

$$ \frac{d^{2}E_{x4}}{dt^{2}} = -c^{2}k^{2}E_{x4} $$

The author defines $\omega = ck$ and points out that it must be true that $k_{11}^{2}+k_{21}^{2}+k_{31}^{2}=k^{2}$. There is a bunch of math that eliminates some of the trig functions and also shows that the various $k_{i}$ must be equal, as $k_{11}=k_{12}=k_{13}=k_{1}$, (note the index order is reversed here from the above). The final answer is

$$ E_{x} = E_{0x}\cos(k_{1}x)\sin(k_{2}y)e^{-i(k_{3}z-\omega t)} $$ $$ E_{y} = E_{0y}\sin(k_{1}x)\cos(k_{2}y)e^{-i(k_{3}z-\omega t)} $$ $$ E_{z} = E_{0z}\sin(k_{1}x)\sin(k_{2}y)e^{-i(k_{3}z-\omega t)} $$

where all remaining constants have been absorbed into $E_{0x}$, etc. and by Maxwell's first law

$$ k_{1}E_{0x}+k_{2}E_{0y}+ik_{3}E_{0z}=0 $$

The $k_{1}$ and $k_{2}$ values are fixed by boundary conditons:

$$ k_{1} = \frac{m\pi}{a} $$ $$ k_{2} = \frac{n\pi}{b} $$

There is then an analysis of the transverse electric and transverse magnetic waves, and it is determined that one of the 0 order transverse electrics is the minimum frequency. Under the assumption that it is $E_{01}$, the final answer is:

$$ E_{x} = E_{0x}\sin(\frac{\pi}{b}y)\cos(k_{3}z-\omega t) $$ $$ B_{y} = \frac{k_{3}}{k}E_{0x}\sin(\frac{\pi}{b}y)\cos(k_{3}z-\omega t) $$ $$ B_{z} = -i\frac{k_{2}}{k}E_{0x}\cos(\frac{\pi}{b}y)\cos(k_{3}z-\omega t) $$

I copied these equations exactly as written, but the $k_{2}$ in $B_{z}$ could have been substituted with $\pi/b$ as it is in the trigonometric.

I understand all this derivation, but I do not see how $k_{3}$ is unambiguously defined, and therefore how $k$ is, since $k$ is given by $k_{11}^{2}+k_{21}^{2}+k_{31}^{2}=k^{2}$. I could see how $k_{1}E_{0x}+k_{2}E_{0y}+ik_{3}E_{0z}=0$ could have been used, but for the TE wave, $E_{0z} = 0$. Without knowing how $k$ is calculated, I don't know what $\omega$ is, since $\omega = ck$.

Answer 1

Just as the waveguide supports infinitely many modes, it also supports waves of infinitely many frequencies. The value of $k_3$ or $\omega$ isn't determined exclusively by the waveguide, one of these can be considered a free parameter. In practice, $\omega$ can be determined by the source generating the wave, for example.

In other words, there is a valid propagating solution for any value of $\omega$ above the cut-off frequency of a given mode, and $k_3$ can be calculated from $\omega$ (or vice versa) via $k_1^2 + k_2^2 + k_3^2 = \omega^2/c^2$. In this respect $k_3$ and $\omega$ differ from $k_1$ and $k_2$, which are dictated exclusively by the waveguide geometry.