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 $\omega = ck$.


1 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.

  • $\begingroup$ Is this a roundabout way of saying that $k_{3}$ is indeterminate and so the minimum frequency is for $k_{3}=0$ so that $k = k_{2}$ and $\omega = k_{2}c$? If so, are we saying that the wavelength in the $z$ direction is then infinite? If so, are we saying that the wave is not actually propagating down the guide? It is basically a standing wave? Thanks. $\endgroup$ Commented Jan 13 at 1:36
  • $\begingroup$ @PoissonAerohead Yes, any value of $k_3 > 0$ yields a valid propagating mode, for which $\omega=c\sqrt{k_1^2+k_2^2+k_3^2}$. $k_3 = 0$ thus yields the minimum possible frequency of traveling waves. $k_3=0$ is a special case that marks the border between propagating and evanescent waves. It isn't really a propagating mode, the Poynting vector has no component down the waveguide in this mode. It is not really useful in in waveguides in practice as far as I know, normally $\omega$ is chosen (or the waveguide is designed) so that $k_3 > 0$ by a safe margin. $\endgroup$
    – Puk
    Commented Jan 13 at 1:57
  • 1
    $\begingroup$ Yup, got it, it is truly the "cut off" frequency, and not really useful for sending anything down the guide. Anything less is going to decay and anything more will propagate at some rate. The boundary is basically standing. To send anything down the guide at any sort of "reasonable rate" you would need some "safe margin" like you said. Thanks. $\endgroup$ Commented Jan 13 at 2:14
  • $\begingroup$ should write: $k_3^2 >0$ for a propagating mode, $k_3 <0$ just propagates in the opposite direction to $k_3 >0$ . Below cutoff, also called evanescent mode, waveguides are used to design frequency selective structures, filters and some such, Regarding both their performance (insertion loss/selectivity) and size they are between lumped element and resonant transmission line structures. $\endgroup$
    – hyportnex
    Commented Jan 13 at 3:29
  • $\begingroup$ @hyportnex Yes, I should have written $k_3^2 > 0$, thanks. $k_3^2 < 0$ (for which $k_3$ is imaginary) gives the evanescent modes. $\endgroup$
    – Puk
    Commented Jan 13 at 3:44

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