Another way of looking at the guided waves
We want now to show you another way of understanding why a waveguide attenuates the fields rapidly for frequencies below the cutoff frequency $\omega_c$. Then you will have a more “physical” idea of why the behavior changes so drastically between low and high frequencies. We can do this for the rectangular guide by analyzing the fields in terms of reflections—or images—in the walls of the guide. The approach only works for rectangular guides, however; that’s why we started with the more mathematical analysis which works, in principle, for guides of any shape. For the mode we have described, the vertical dimension (in $y$) had no effect, so we can ignore the top and bottom of the guide and imagine that the guide is extended indefinitely in the vertical direction. We imagine then that the guide just consists of two vertical plates with the separation $a$. Let’s say that the source of the fields is a vertical wire placed in the middle of the guide, with the wire carrying a current that oscillates at the frequency $\omega$. In the absence of the guide walls such a wire would radiate cylindrical waves. Now we consider that the guide walls are perfect conductors. Then, just as in electrostatics, the conditions at the surface will be correct if we add to the field of the wire the field of one or more suitable image wires. The image idea works just as well for electrodynamics as it does for electrostatics, provided, of course, that we also include the retardations. We know that is true because we have often seen a mirror producing an image of a light source. And a mirror is just a “perfect” conductor for electromagnetic waves with optical frequencies. Now let’s take a horizontal cross section, as shown in Fig. 24–15, where $W_1$ and $W_2$ are the two guide walls and $S_0$ is the source wire. We call the direction of the current in the wire positive. Now if there were only one wall, say $W_1$, we could remove it if we placed an image source (with opposite polarity) at the position marked $S_1$. But with both walls in place there will also be an image of $S_0$ in the wall $W_2$, which we show as the image $S_2$. This source, too, will have an image in $W_1$, which we call $S_3$. Now both $S_1$ and $S_3$ will have images in $W_2$ at the positions marked $S_4$ and $S_6$, and so on. For our two plane conductors with the source halfway between, the fields are the same as those produced by an infinite line of sources, all separated by the distance $a$. (It is, in fact just what you would see if you looked at a wire placed halfway between two parallel mirrors.)  For the fields to be zero at the walls, the polarity of the currents in the images must alternate from one image to the next. In other words, they oscillate $180^\circ$ out of phase. The waveguide field is, then, just the superposition of the fields of such an infinite set of line sources. We know that if we are close to the sources, the field is very much like the static fields. We considered in Section 7–5 the static field of a grid of line sources and found that it is like the field of a charged plate except for terms that decrease exponentially with the distance from the grid. Here the average source strength is zero, because the sign alternates from one source to the next. Any fields which exist should fall off exponentially with distance. Close to the source, we see the field mainly of the nearest source; at large distances, many sources contribute and their average effect is zero. So now we see why the waveguide below cutoff frequency gives an exponentially decreasing field. At low frequencies, in particular, the static approximation is good, and it predicts a rapid attenuation of the fields with distance. Now we are faced with the opposite question: Why are waves propagated at all? That is the mysterious part! The reason is that at high frequencies the retardation of the fields can introduce additional changes in phase which can cause the fields of the out-of-phase sources to add instead of cancelling. In fact, in Chapter 29 of Vol. I we have already studied, just for this problem, the fields generated by an array of antennas or by an optical grating. There we found that when several radio antennas are suitably arranged, they can give an interference pattern that has a strong signal in some direction but no signal in another. Suppose we go back to Fig. 24–15 and look at the fields which arrive at a large distance from the array of image sources. The fields will be strong only in certain directions which depend on the frequency—only in those directions for which the fields from all the sources add in phase. At a reasonable distance from the sources the field propagates in these special directions as plane waves. We have sketched such a wave in Fig. 24–16, where the solid lines represent the wave crests and the dashed lines represent the troughs. The wave direction will be the one for which the difference in the retardation for two neighboring sources to the crest of a wave corresponds to one-half a period of oscillation. In other words, the difference between $r_2$ and $r_0$ in the figure is one-half of the free-space wavelength: \begin{equation} r_2-r_0=\frac{\lambda_0}{2}.\notag \end{equation}  The angle $\theta$ is then given by \begin{equation} \label{Eq:II:24:33} \sin\theta=\frac{\lambda_0}{2a}. \end{equation}  There is, of course, another set of waves travelling downward at the symmetric angle with respect to the array of sources. The complete waveguide field (not too close to the source) is the superposition of these two sets of waves, as shown in Fig. 24–17. The actual fields are really like this, of course, only between the two walls of the waveguide. At points like $A$ and $C$, the crests of the two wave patterns coincide, and the field will have a maximum; at points like $B$, both waves have their peak negative value, and the field has its minimum (largest negative) value. As time goes on the field in the guide appears to be travelling along the guide with a wavelength $\lambda_g$, which is the distance from $A$ to $C$. That distance is related to $\theta$ by \begin{equation} \label{Eq:II:24:34} \cos\theta=\frac{\lambda_0}{\lambda_g}. \end{equation}  Using Eq. (24.33) for $\theta$, we get that \begin{equation} \label{Eq:II:24:35} \lambda_g=\frac{\lambda_0}{\cos\theta}= \frac{\lambda_0}{\sqrt{1-(\lambda_0/2a)^2}}, \end{equation}  which is just what we found in Eq. (24.19). Now we see why there is only wave propagation above the cutoff frequency $\omega_0$. If the free-space wavelength is longer than $2a$, there is no angle where the waves shown in Fig. 24–16 can appear. The necessary constructive interference appears suddenly when $\lambda_0$ drops below $2a$, or when $\omega$ goes above $\omega_0=\pi c/a$. If the frequency is high enough, there can be two or more possible directions in which the waves will appear. For our case, this will happen if $\lambda_0<\tfrac{2}{3}a$. In general, however, it could also happen when $\lambda_0<a$. These additional waves correspond to the higher guide modes we have mentioned. It has also been made evident by our analysis why the phase velocity of the guided waves is greater than $c$ and why this velocity depends on $\omega$. As $\omega$ is changed, the angle of the free waves of Fig. 24–16 changes, and therefore so does the velocity along the guide.  Although we have described the guided wave as the superposition of the fields of an infinite array of line sources, you can see that we would arrive at the same result if we imagined two sets of free-space waves being continually reflected back and forth between two perfect mirrors—remembering that a reflection means a reversal of phase. These sets of reflecting waves would all cancel each other unless they were going at just the angle $\theta$ given in Eq. (24.33). There are many ways of looking at the same thing.