next up previous
Next: Relativity and electromagnetism Up: Electromagnetic radiation Previous: Reflection at a dielectric


A wave-guide is a hollow conducting pipe, of uniform cross-section, used to transport high frequency electromagnetic waves (generally, in the microwave band) from one point to another. The main advantage of wave-guides is their relatively low level of radiation losses (since the electric and magnetic fields are completely enclosed by a conducting wall) compared to transmission lines.

Consider a vacuum-filled wave-guide which runs parallel to the $z$-axis. An electromagnetic wave trapped inside the wave-guide satisfies Maxwell's equations for free space:

$\displaystyle \nabla\cdot{\bf E}$ $\textstyle =$ $\displaystyle 0,$ (1280)
$\displaystyle \nabla\cdot{\bf B}$ $\textstyle =$ $\displaystyle 0,$ (1281)
$\displaystyle \nabla\times{\bf E}$ $\textstyle =$ $\displaystyle - \frac{\partial {\bf B}}{\partial t} ,$ (1282)
$\displaystyle \nabla\times{\bf B}$ $\textstyle =$ $\displaystyle \frac{1}{c^2}\frac{\partial {\bf E}}{\partial t}.$ (1283)

Let $\partial/\partial t\equiv -{\rm i} \omega$, and $\partial/\partial z\equiv
{\rm i} k$, where $\omega$ is the wave frequency, and $k$ the wave-number parallel to the axis of the wave-guide. It follows that
$\displaystyle \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + {\rm i} k E_z$ $\textstyle =$ $\displaystyle 0,$ (1284)
$\displaystyle \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + {\rm i} k B_z$ $\textstyle =$ $\displaystyle 0,$ (1285)
$\displaystyle {\rm i} \omega  B_x$ $\textstyle =$ $\displaystyle \frac{\partial E_z}{\partial y} - {\rm i} k E_y,$ (1286)
$\displaystyle {\rm i} \omega B_y$ $\textstyle =$ $\displaystyle -\frac{\partial E_z}{\partial x} + {\rm i} k E_x,$ (1287)
$\displaystyle {\rm i} \omega B_z$ $\textstyle =$ $\displaystyle \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y},$ (1288)
$\displaystyle {\rm i} \frac{\omega}{c^2} E_x$ $\textstyle =$ $\displaystyle - \frac{\partial B_z}{\partial y} + {\rm i} k B_y,$ (1289)
$\displaystyle {\rm i} \frac{\omega}{c^2} E_y$ $\textstyle =$ $\displaystyle \frac{\partial B_z}{\partial x}
- {\rm i} k B_x,$ (1290)
$\displaystyle {\rm i} \frac{\omega}{c^2} E_z$ $\textstyle =$ $\displaystyle -\frac{\partial B_y}{\partial x}+
\frac{\partial B_x}{\partial y}.$ (1291)

Equations (1287) and (1289) yield
E_x = {\rm i}\left(\omega \frac{\partial B_z}{\partial y} +...
...{\partial x}\right)\left(\frac{\omega^2}{c^2}-k^2\right)^{-1},
\end{displaymath} (1292)

B_y = {\rm i} \left(\frac{\omega}{c^2}\frac{\partial E_z}{\...
...{\partial y}\right)\left(\frac{\omega^2}{c^2}-k^2\right)^{-1}.
\end{displaymath} (1293)

Likewise, Eqs. (1286) and (1290) yield
E_y = {\rm i}\left(-\omega \frac{\partial B_z}{\partial x} ...
...{\partial y}\right)\left(\frac{\omega^2}{c^2}-k^2\right)^{-1},
\end{displaymath} (1294)

B_x = {\rm i} \left(-\frac{\omega}{c^2}\frac{\partial E_z}{...
...{\partial x}\right)\left(\frac{\omega^2}{c^2}-k^2\right)^{-1}.
\end{displaymath} (1295)

These equations can be combined to give
$\displaystyle {\bf E}_t$ $\textstyle =$ $\displaystyle {\rm i}\left(\omega \nabla B_z\times\hat{\bf z} + k \nabla
E_z\right)\left(\frac{\omega^2}{c^2}-k^2\right)^{-1},$ (1296)
$\displaystyle {\bf B}_t$ $\textstyle =$ $\displaystyle {\rm i}\left(-\frac{\omega}{c^2} \nabla E_z\times\hat{\bf z} + k \nabla
B_z\right)\left(\frac{\omega^2}{c^2}-k^2\right)^{-1}.$ (1297)

Here, ${\bf E}_t$ and ${\bf B}_t$ are the transverse electric and magnetic fields: i.e., the electric and magnetic fields in the $x$-$y$ plane. It is clear, from Eqs. (1296) and (1297), that the transverse fields are fully determined once the longitudinal fields, $E_z$ and $B_z$, are known.

Substitution of Eqs. (1296) and (1297) into Eqs. (1288) and (1291) yields the equations satisfied by the longitudinal fields:

$\displaystyle \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\!
E_z + \left(\frac{\omega^2}{c^2}-k^2\right)\!E_z$ $\textstyle =$ $\displaystyle 0,$ (1298)
$\displaystyle \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\!
B_z + \left(\frac{\omega^2}{c^2}-k^2\right)\!B_z$ $\textstyle =$ $\displaystyle 0.$ (1299)

The remaining equations, (1284) and (1285), are automatically satisfied provided Eqs. (1296)-(1299) are satisfied.

We expect ${\bf E} = {\bf B} = {\bf0}$ inside the walls of the wave-guide, assuming that they are perfectly conducting. Hence, the appropriate boundary conditions at the walls are

$\displaystyle E_{\parallel}$ $\textstyle =$ $\displaystyle 0,$ (1300)
$\displaystyle B_{\perp}$ $\textstyle =$ $\displaystyle 0.$ (1301)

It follows, by inspection of Eqs. (1296) and (1297), that these boundary conditions are satisfied provided
$\displaystyle E_z$ $\textstyle =$ $\displaystyle 0,$ (1302)
$\displaystyle \hat{\bf n}\cdot\nabla B_z$ $\textstyle =$ $\displaystyle 0,$ (1303)

at the walls. Here, $\hat{\bf n}$ is the normal vector to the walls. Hence, the electromagnetic fields inside the wave-guide are fully specified by solving Eqs. (1298) and (1299), subject to the boundary conditions (1302) and (1303), respectively.

Equations (1298) and (1299) support two independent types of solution. The first type has $E_z=0$, and is consequently called a transverse electric, or TE, mode. Conversely, the second type has $B_z=0$, and is called a transverse magnetic, or TM, mode.

Consider the specific example of a rectangular wave-guide, with conducting walls at $x=0, a$, and $y=0, b$. For a TE mode, the longitudinal magnetic field can be written

B_z(x,y) = B_0 \cos(k_x x) \cos(k_y y),
\end{displaymath} (1304)

The boundary condition (1303) requires that $\partial B_z/\partial x =0$ at $x=0, a$, and $\partial B_z/\partial y=0$ at $y=0, b$. It follows that
$\displaystyle k_x$ $\textstyle =$ $\displaystyle \frac{n \pi}{a},$ (1305)
$\displaystyle k_y$ $\textstyle =$ $\displaystyle \frac{m \pi}{b},$ (1306)

where $n=0, 1, 2, \cdots$, and $m=0, 1, 2, \cdots$. Clearly, there are many different kinds of TE mode, corresponding to the many different choices of $m$ and $n$. Let us refer to a mode corresponding to a particular choice of $m, n$ as a ${\rm TE}_{mn}$ mode. Note, however, that there is no ${\rm TE}_{00}$ mode, since $B_z(x,y)$ is uniform in this case. According to Eq. (1299), the dispersion relation for the ${\rm TE}_{mn}$ mode is given by
k^2 c^2 = \omega^2 - \omega_{mn}^{ 2},
\end{displaymath} (1307)

\omega_{mn} = c \pi \sqrt{\frac{n^2}{a^2} + \frac{m^2}{b^2}}.
\end{displaymath} (1308)

According to the dispersion relation (1307), $k$ is imaginary for $\omega < \omega_{mn}$. In other words, for wave frequencies below $\omega_{mn}$, the ${\rm TE}_{mn}$ mode fails to propagate down the wave-guide, and is instead attenuated. Hence, $\omega_{mn}$ is termed the cut-off frequency for the ${\rm TE}_{mn}$ mode. Assuming that $a>b$, the TE mode with the lowest cut-off frequency is the ${\rm TE}_{10}$ mode, where

\omega_{10} = \frac{c \pi}{a}.
\end{displaymath} (1309)

For frequencies above the cut-off frequency, the phase-velocity of the ${\rm TE}_{mn}$ mode is given by

v_p = \frac{\omega}{k} = \frac{c}{\sqrt{1-\omega_{mn}^{ 2}/\omega^2}},
\end{displaymath} (1310)

which is greater than $c$. However, the group-velocity takes the form
v_g = \frac{d\omega}{d k} = c \sqrt{1-\omega_{mn}^{ 2}/\omega^2},
\end{displaymath} (1311)

which is always less than $c$. Of course, energy is transmitted down the wave-guide at the group-velocity, rather than the phase-velocity. Note that the group-velocity goes to zero as the wave frequency approaches the cut-off frequency.

For a TM mode, the longitudinal electric field can be written

E_z(x,y) = E_0 \sin(k_x x) \sin(k_y y),
\end{displaymath} (1312)

The boundary condition (1302) requires that $E_z=0$ at $x=0, a$, and $y=0, b$. It follows that
$\displaystyle k_x$ $\textstyle =$ $\displaystyle \frac{n \pi}{a},$ (1313)
$\displaystyle k_y$ $\textstyle =$ $\displaystyle \frac{m \pi}{b},$ (1314)

where $n=1, 2, \cdots$, and $m=1, 2, \cdots$. The dispersion relation for the ${\rm TM}_{mn}$ mode is also given by Eq. (1307). Hence, Eqs. (1310) and (1311) also apply to TM modes. However, the TM mode with the lowest cut-off frequency is the ${\rm TM}_{11}$ mode, where
\omega_{11} = c \pi \sqrt{\frac{1}{a^2}+\frac{1}{b^2}}>\omega_{10}.
\end{displaymath} (1315)

It follows that the mode with the lowest cut-off frequency is always a TE mode.

There is, in principle, a third type of mode which can propagate down a wave-guide. This third mode type is characterized by $E_z=B_z=0$, and is consequently called a transverse electromagnetic, or TEM, mode. It is easily seen, from an inspection of Eqs. (1286)-(1291), that a TEM mode satisfies

\omega^2= k^2 c^2,
\end{displaymath} (1316)

$\displaystyle {\bf E}_t$ $\textstyle =$ $\displaystyle - \nabla\phi,$ (1317)
$\displaystyle {\bf B}_t$ $\textstyle =$ $\displaystyle c^{-1} \nabla\phi\times\hat{\bf z},$ (1318)

where $\phi(x,y)$ satisfies
\nabla^2\phi = 0.
\end{displaymath} (1319)

The boundary conditions (1302) and (1303) imply that
\phi = {\rm constant}
\end{displaymath} (1320)

at the walls. However, there is no non-trivial solution of Eqs. (1319) and (1320) for a conventional wave-guide. In other words, conventional wave-guides do not support TEM modes. In fact, it turns out that only wave-guides with central conductors support TEM modes. Consider, for instance, a co-axial wave-guide in which the electric and magnetic fields are trapped between two parallel concentric cylindrical conductors of radius $a$ and $b$ (with $b>a$). In this case, $\phi=\phi(r)$, and Eq. (1319) reduces to
\frac{1}{r}\frac{\partial}{\partial r}\left( r \frac{\partial \phi}{\partial r}\right) = 0,
\end{displaymath} (1321)

where $r$ is a standard cylindrical polar coordinate. The boundary condition (1320) is automatically satisfied at $r=a$ and $r=b$. The above equation has the following non-trivial solution:
\phi(r) = \phi_b \ln(r/b).
\end{displaymath} (1322)

Note, however, that the inner conductor must be present, otherwise $\phi\rightarrow\infty$ as $r\rightarrow 0$, which is unphysical. According to the dispersion relation (1316), TEM modes have no cut-off frequency, and have the phase-velocity (and group-velocity) $c$. Indeed, this type of mode is the same as that supported by a transmission line (see Sect. 7.7).
next up previous
Next: Relativity and electromagnetism Up: Electromagnetic radiation Previous: Reflection at a dielectric
Richard Fitzpatrick 2006-02-02