The rectangular waveguide
The next thing we want to talk about seems, at first sight, to be a striking phenomenon: if the central conductor is removed from the coaxial line, it can still carry electromagnetic power. In other words, at high enough frequencies a hollow tube will work just as well as one with wires. It is related to the mysterious way in which a resonant circuit of a condenser and inductance gets replaced by nothing but a can at high frequencies. Although it may seem to be a remarkable thing when one has been thinking in terms of a transmission line as a distributed inductance and capacity, we all know that electromagnetic waves can travel along inside a hollow metal pipe. If the pipe is straight, we can see through it! So certainly electromagnetic waves go through a pipe. But we also know that it is not possible to transmit low-frequency waves (power or telephone) through the inside of a single metal pipe. So it must be that electromagnetic waves will go through if their wavelength is short enough. Therefore we want to discuss the limiting case of the longest wavelength (or the lowest frequency) that can get through a pipe of a given size. Since the pipe is then being used to carry waves, it is called a waveguide. We will begin with a rectangular pipe, because it is the simplest case to analyze. We will first give a mathematical treatment and come back later to look at the problem in a much more elementary way. The more elementary approach, however, can be applied easily only to a rectangular guide. The basic phenomena are the same for a general guide of arbitrary shape, so the mathematical argument is fundamentally more sound. Our problem, then, is to find what kind of waves can exist inside a rectangular pipe. Let’s first choose some convenient coordinates; we take the $z$-axis along the length of the pipe, and the $x$- and $y$-axes parallel to the two sides, as shown in Fig. 24–3. We know that when light waves go down the pipe, they have a transverse electric field; so suppose we look first for solutions in which $\FLPE$ is perpendicular to $z$, say with only a $y$-component, $E_y$. This electric field will have some variation across the guide; in fact, it must go to zero at the sides parallel to the $y$-axis, because the currents and charges in a conductor always adjust themselves so that there is no tangential component of the electric field at the surface of a conductor. So $E_y$ will vary with $x$ in some arch, as shown in Fig. 24–4. Perhaps it is the Bessel function we found for a cavity?  No, because the Bessel function has to do with cylindrical geometries. For a rectangular geometry, waves are usually simple harmonic functions, so we should try something like $\sin k_xx$. Since we want waves that propagate down the guide, we expect the field to alternate between positive and negative values as we go along in $z$, as in Fig. 24–5, and these oscillations will travel along the guide with some velocity $v$. If we have oscillations at some definite frequency $\omega$, we would guess that the wave might vary with $z$ like $\cos\,(\omega t-k_zz)$, or to use the more convenient mathematical form, like $e^{i(\omega t-k_zz)}$. This $z$-dependence represents a wave travelling with the speed $v=\omega/k_z$ (see Chapter 29, Vol. I). So we might guess that the wave in the guide would have the following mathematical form: \begin{equation} \label{Eq:II:24:12} E_y=E_0e^{i(\omega t-k_zz)}\sin k_xx. \end{equation}  Let’s see whether this guess satisfies the correct field equations. First, the electric field should have no tangential components at the conductors. Our field satisfies this requirement; it is perpendicular to the top and bottom faces and is zero at the two side faces. Well, it is if we choose $k_x$ so that one-half a cycle of $\sin k_xx$ just fits in the width of the guide—that is, if \begin{equation} \label{Eq:II:24:13} k_xa=\pi. \end{equation}  There are other possibilities, like $k_xa=2\pi$, $3\pi$, $\dotsc$, or, in general, \begin{equation} \label{Eq:II:24:14} k_xa=n\pi, \end{equation}  where $n$ is any integer. These represent various complicated arrangements of the field, but for now let’s take only the simplest one, where $k_x=\pi/a$, where $a$ is the width of the inside of the guide. Next, the divergence of $\FLPE$ must be zero in the free space inside the guide, since there are no charges there. Our $\FLPE$ has only a $y$-component, and it doesn’t change with $y$, so we do have that $\FLPdiv{\FLPE}=0$. Finally, our electric field must agree with the rest of Maxwell’s equations in the free space inside the guide. That is the same thing as saying that it must satisfy the wave equation \begin{equation} \label{Eq:II:24:15} \frac{\partial^2E_y}{\partial x^2}+ \frac{\partial^2E_y}{\partial y^2}+ \frac{\partial^2E_y}{\partial z^2}- \frac{1}{c^2}\,\frac{\partial^2E_y}{\partial t^2}=0. \end{equation}  We have to see whether our guess, Eq. (24.12), will work. The second derivative of $E_y$ with respect to $x$ is just $-k_x^2E_y$. The second derivative with respect to $y$ is zero, since nothing depends on $y$. The second derivative with respect to $z$ is $-k_z^2E_y$, and the second derivative with respect to $t$ is $-\omega^2E_y$. Equation (24.15) then says that \begin{equation*} k_x^2E_y+k_z^2E_y-\frac{\omega^2}{c^2}\,E_y=0. \end{equation*}  Unless $E_y$ is zero everywhere (which is not very interesting), this equation is correct if \begin{equation} \label{Eq:II:24:16} k_x^2+k_z^2-\frac{\omega^2}{c^2}=0. \end{equation}  We have already fixed $k_x$, so this equation tells us that there can be waves of the type we have assumed if $k_z$ is related to the frequency $\omega$ so that Eq. (24.16) is satisfied—in other words, if \begin{equation} \label{Eq:II:24:17} k_z=\sqrt{(\omega^2/c^2)-(\pi^2/a^2)}. \end{equation}  The waves we have described are propagated in the $z$-direction with this value of $k_z$. The wave number $k_z$ we get from Eq. (24.17) tells us, for a given frequency $\omega$, the speed with which the nodes of the wave propagate down the guide. The phase velocity is \begin{equation} \label{Eq:II:24:18} v=\frac{\omega}{k_z}. \end{equation}  You will remember that the wavelength $\lambda$ of a travelling wave is given by $\lambda=2\pi v/\omega$, so $k_z$ is also equal to $2\pi/\lambda_g$, where $\lambda_g$ is the wavelength of the oscillations along the $z$-direction—the “guide wavelength.” The wavelength in the guide is different, of course, from the free-space wavelength of electromagnetic waves of the same frequency. If we call the free-space wavelength $\lambda_0$, which is equal to $2\pi c/\omega$, we can write Eq. (24.17) as \begin{equation} \label{Eq:II:24:19} \lambda_g=\frac{\lambda_0}{\sqrt{1-(\lambda_0/2a)^2}}. \end{equation}  Besides the electric fields there are magnetic fields that will travel with the wave, but we will not bother to work out an expression for them right now. Since $c^2\FLPcurl{\FLPB}=\ddpl{\FLPE}{t}$, the lines of $\FLPB$ will circulate around the regions in which $\ddpl{\FLPE}{t}$ is largest, that is, halfway between the maximum and minimum of $\FLPE$. The loops of $\FLPB$ will lie parallel to the $xz$-plane and between the crests and troughs of $\FLPE$, as shown in Fig. 24–6.