# 54 Waveguides I

## Introduction

As detailed in the previous lecture, the idea of confining a
propagating electromagnetic wave in a metal tube was first proposed by J. J.
Thomson in 1893, and mathematically analyzed by Lord Rayleigh in 1897. Rayleigh discovered that waves below a certain
cutoff frequency *would not propagate*.

The first attempt at actually *constructing* waveguides for possible commercial use was by George Southworth
at Bell Labs in the 1930's. Unaware of Rayleigh's work, the group there was at
first bewildered by their lack of success in transmitting low-frequency
waves. Fortunately, a mathematician colleague
who had escaped from Russia in 1917 recreated Rayleigh’s analysis, and the
experimenters raised the frequency to one that *would* propagate (the first message transmitted was “send money”.)

Here is Southworth demonstrating

an early waveguide in 1938. (Notice the field diagrams above the transmitter.)

The development of radar in the Second World War led to a rapid development of waveguide technology. Waveguides were used to feed the power from the oscillator (a magnetron, invented at the University of Birmingham) to the antenna. This work was done at the MIT Rad Lab, a team led by Purcell, and including Schwinger, Bethe and others.

Nowadays, waveguides are mainly used for high frequency power transmission (typically in the gigahertz range) in radar, of course, (see picture) but also in accelerators. The waveguide transverse dimensions are comparable to the wavelength being transmitted. For some accelerator installations, this is roughly the same as a 2"x4" wooden beam. The similarity proved useful on one occasion: a granting agency checking progress of accelerator construction was bamboozled by "waveguides" that were in fact two by fours painted with copper paint (this from an anonymous source).

Waveguides are also used in microwave transmitters: these were the original underwater long lines, carrying most of the telephone traffic, now largely replaced by fiber optics.

*In preparing this
lecture, I used Feynman’s Lectures in
Physics, Volume 2, Chapter 24. I highly recommend that as an
introduction. I also used Griffiths, Introduction
to Electrodynamics, Third Edition, Section 9.5, which gives more detail
than Feynman on TE, TM; **ω**, k
relationships, etc. Of course, much of the lecture follows Jackson.*

## Preliminaries: Boundary Conditions

### First: for a Perfect Conductor

In thinking about electromagnetic waves being guided, in other words directed down some kind of tube, the first thing to establish is how does the tube constrain the wave, in other words, what are the boundary conditions?

The second practical question is: how quickly are the currents induced in the boundary conducting material draining energy from the wave?

We'll begin by considering the wall to be a *perfect* conductor: there can be no
electric field inside this conductor, it would instantly be shielded out by the
charges adjusting on the surface, so *the
electric field inside the guide, on approaching the wall, is normal to the wall
surface*, and terminates at a layer of surface charge. (Any electric field at the surface parallel
to the surface would create a large current in the conductor.)

What about the magnetic field? There can be a *static *magnetic field inside a conductor,
a magnet will work through a copper sheet.
But you can't have an *oscillating*
field inside a *perfect *conductor,
since it necessarily has an accompanying electric field. There can’t be a normal component to the
magnetic field, that would require a monopole layer, but there can be a *parallel *component terminated by a layer
of surface current.

### Second: A Real Conductor: Skin Depth

(*Note: this section is
a reminder of material already covered in lectures **37** and **47*.)

*Notation*: We will
use $z$ in* this
section* for depth into the conducting material, the notation Jackson
himself used in the earlier treatment of skin depth (p 220), where the same
equations were derived. (Doubtless he
uses $\xi $ here since he now wants to reserve $z$ for distance along the wave guide, but I want
to make very clear that we’ve already done this problem.)

For a *real*
conductor, like copper, the idealized delta function layers of charge and
current on the “perfect conductor” obviously spread to some finite (if small)
depth, so the electric field to some extent penetrates the conductor,
generating currents and consequently heat, and therefore *attenuating the wave*.

The depth of penetration is called the *skin depth*, labeled $\delta ,$ the currents generated are *eddy currents*.

The *displacement*
current (the $\partial \overrightarrow{E}/\partial t$ term) is *negligible*
compared to the free electron current in a good conductor, so the appropriate
equations are (including Ohm’s law, and $\mu ,\text{\hspace{0.33em}}\sigma $ refer to values for the conductor)

$\overrightarrow{\nabla}\times \overrightarrow{H}=\overrightarrow{j}=\sigma \overrightarrow{E},\text{\hspace{1em}}\overrightarrow{\nabla}\times \overrightarrow{E}=-\mu \frac{\partial \overrightarrow{H}}{\partial t},$

so for fields having time dependence ${e}^{-i\omega t},$

${\nabla}^{2}\overrightarrow{H}=-i\mu \sigma \omega \overrightarrow{H}.$

We take the surface to be the $\left(x,y\right)$ plane, and so measure depth into the conductor as $z$ (the equations then coincide with those in lecture 47, on impedance of a metal reflector).

Taking the field outside the conductor to be $\overrightarrow{H}=\left({H}_{x},0,0\right)$ then inside the conductor, $z\ge 0,$

${H}_{x}\left(z,t\right)=h\left(z\right){e}^{-i\omega t},$

and $h\left(z\right)$ satisfies

$\left(\frac{{d}^{2}}{d{z}^{2}}+i\mu \sigma \omega \right)h\left(z\right)=0,$

assuming field variation in the $z$-direction is far more rapid than in other directions, so we can take ${\nabla}^{2}={d}^{2}/d{z}^{2}.$

A solution ${e}^{ikz}$ works provided ${k}^{2}=i\mu \sigma \omega ,$ or $k=\pm \left(1+i\right)\sqrt{\frac{\mu \sigma \omega}{2}}.$

The solution has to be the sign giving exponential decay of
the field into the conductor, that gives a decay length, the *skin depth*:

$\delta =\sqrt{\frac{2}{\mu \sigma \omega}}.$

In terms of $\delta ,$ the magnetic field inside the conductor is

${H}_{x}\left(z\right)={H}_{x}{e}^{-\left(1-i\right)z/\delta},$

neglecting here the possible slow variations parallel to the surface.

Now $\overrightarrow{\nabla}\times \overrightarrow{H}=\overrightarrow{j}=\sigma \overrightarrow{E},$ from which the *electric* field inside the conductor

$\overrightarrow{E}=\left(1/\sigma \right)\left(1-i\right)\left(1/\delta \right)\left(\overrightarrow{n}\times \overrightarrow{H}\right){e}^{-\left(1-i\right)z/\delta}=\sqrt{\mu \omega /2\sigma}\left(1-i\right)\left(\overrightarrow{n}\times \overrightarrow{H}\right){e}^{-\left(1-i\right)z/\delta}.$

This can also be written in terms of the impedance, as in lecture 47. In this regime, the impedance $Z=\sqrt{\mu /\epsilon}\approx \sqrt{\mu \omega /i\sigma},$ and this internal electric field is ${E}_{y}\left(z\right)=Z{H}_{x}\left(z\right).$

This field is generated purely by the oscillating magnetic
field, and is much smaller than the electric field outside the conductor in
general. But, unlike the outside
electric field, this internal field is tangential to the surface, and therefore
there is also a nonzero tangential electric field infinitesimally *above* the surface, in the tube, and so a
*nonzero Poynting vector*, an energy
flow, into the surface from the wave:

$-{\scriptscriptstyle \frac{1}{2}}\mathrm{Re}\left[\overrightarrow{n}\cdot \overrightarrow{E}\times {\overrightarrow{H}}^{\ast}\right]={\scriptscriptstyle \frac{1}{4}}\mu \omega \delta {\left|{H}_{x}\right|}^{2}={\scriptscriptstyle \frac{1}{4}}\sqrt{\frac{2\mu \omega}{\sigma}}{\left|{H}_{x}\right|}^{2}={\scriptscriptstyle \frac{1}{2\sqrt{2}}}\left|Z\right|{\left|{H}_{x}\right|}^{2}.$

From Ohm's law inside the conductor $\overrightarrow{J}=\sigma \overrightarrow{E}=\left(1/\delta \right)\left(1-i\right)\left(\overrightarrow{n}\times \overrightarrow{H}\right){e}^{-\left(1-i\right)z/\delta}$, it's straightforward to show this energy all goes into Joule heating.

Notice that for a given field strength, power loss goes up with frequency as $\sqrt{\omega}.$ It also goes as $1/\sqrt{\sigma},$ because it goes as $\sigma {E}^{2},$ but $E\sim H/\sqrt{\sigma},$ and the volume of dissipation per unit area is given by the penetration depth $\sim 1/\sqrt{\sigma}.$

### Summary: Fields Near the Surface

The key is that the surface can have charge density and current density, spread through a depth of order $\delta .$

There can be a strong perpendicular electric field *outside* the surface, from the layer of
charge. Such a field *inside* would dissipate quickly, moving
charge to or from the surface layer. As discussed above, there is a *small* tangential electric field,
continuous across the surface.

There can be a *strong*
tangential *magnetic* field outside the
surface, generated by surface currents. As we have seen, tangential fields
inside the conductor attenuate at the decay length.

Any outside *perpendicular
*magnetic field would continue into the conductor (there are no sheets of
magnetic charge!), and decay rapidly from eddy currents (this is of course an
oscillating field) so any such field must be very weak.

## Waveguides

*Notation**: from now on, **$z$** measures
distance along the waveguide. *

*The first waveguide
was Heaviside’s coaxial cable, discussed in the previous lecture.*

### Coaxial Cable

Commercial electrical power is AC, with a frequency of 50 or
60 Hz, depending on country, it is transmitted by wires, and therefore radiates
power. But the loss of power by radiation is trivial compared with the ohmic
heating loss. However, AC radiation goes
as the *fourth power of frequency*
(we’ll be discussing this a little later in the course), so going up from 50Hz
to megahertz makes all the difference. The most common solution to transmitting
high frequency signals is the coaxial cable, two concentric cylindrical
surfaces, with equal and opposite currents and charges at any point on the
cable, so no external fields, and hence no radiation. (Image from Wikipedia.)
This is hooked up to some oscillator generating the field, so evidently waves
go down the cable.

## Hollow Cylindrical Waveguide

(*Jackson 8.2*. *Note*:
*cylindrical just means the cross section
is always the same, not necessarily circular*.)

### Deriving Transverse Differential Equations for *E*_{z}, *B*_{z}

The coaxial cable and the two-wire model discussed in the
previous lecture are easy to visualize in terms of local capacitance and
inductance. But in fact, as J. J.
Thomson suggested and Lord Rayleigh analyzed, we can send waves down a *single hollow conductor*: it's
straightforward to write down Maxwell's equations and solve them.

We take it that the oscillating source generates waves with a time dependence ${e}^{-i\omega t}.$

Inside the waveguide, Maxwell’s equations are

$\begin{array}{l}\overrightarrow{\nabla}\times \overrightarrow{E}=i\omega \overrightarrow{B}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\overrightarrow{\nabla}\cdot \overrightarrow{B}=0\\ \overrightarrow{\nabla}\times \overrightarrow{B}=-i\mu \epsilon \omega \overrightarrow{E}\text{\hspace{1em}}\text{\hspace{0.33em}}\overrightarrow{\nabla}\cdot \overrightarrow{E}=0\end{array}$ .

It follows that

$\begin{array}{l}\left({\nabla}^{2}+\mu \epsilon {\omega}^{2}\right)\overrightarrow{E}=0,\\ \left({\nabla}^{2}+\mu \epsilon {\omega}^{2}\right)\overrightarrow{B}=0.\end{array}$

We’ll now assume that all components of $\overrightarrow{E}$ vary in the $z$ direction, down the guide, as ${e}^{ikz},$ so

$\left(\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}-{k}^{2}+\mu \epsilon {\omega}^{2}\right)\overrightarrow{E}\left(x,y\right)=0.$

Evidently we’re now looking at a 2-D ${\nabla}^{2}$ problem: Jackson writes

$\left({\partial}^{2}/\partial {x}^{2}+{\partial}^{2}/\partial {y}^{2}\right)={\nabla}_{t}{}^{2},$

*$t$** *for
*transverse*, meaning in the plane
perpendicular to the direction of the wave guide.

The approach is to write out Maxwell’s equations in terms of
transverse and parallel (to $z$ ) components, then write the *transverse* $\overrightarrow{E}$ and $\overrightarrow{B}$ fields just in terms of ${E}_{z},{B}_{z}$ as follows (I decided to follow Griffiths’ presentation, which is a bit more transparent than Jackson’s here. I'll connect with Jackson's approach in the next lecture.):

$\begin{array}{l}\overrightarrow{\nabla}\times \overrightarrow{E}=\dot{\overrightarrow{B}}=i\omega \overrightarrow{B}:\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\overrightarrow{\nabla}\times \overrightarrow{B}={\mu}_{0}{\epsilon}_{0}\dot{\overrightarrow{E}}=-\frac{i\omega}{{c}^{2}}\overrightarrow{E}\\ \frac{\partial {E}_{y}}{\partial x}-\frac{\partial {E}_{x}}{\partial y}=i\omega {B}_{z}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\frac{\partial {B}_{y}}{\partial x}-\frac{\partial {B}_{x}}{\partial y}=-\frac{i\omega}{{c}^{2}}{E}_{z}\\ \frac{\partial {E}_{z}}{\partial y}-ik{E}_{y}=i\omega {B}_{x}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\frac{\partial {B}_{z}}{\partial y}-ik{B}_{y}=-\frac{i\omega}{{c}^{2}}{E}_{x}\\ ik{E}_{x}-\frac{\partial {E}_{z}}{\partial x}=i\omega {B}_{y}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}ik{B}_{x}-\frac{\partial {B}_{z}}{\partial x}=-\frac{i\omega}{{c}^{2}}{E}_{y}.\end{array}$

Note now that we can choose *two* equations for ${E}_{x}:$

$\begin{array}{c}-\frac{i\omega}{{c}^{2}}{E}_{x}=\frac{\partial {B}_{z}}{\partial y}-ik{B}_{y},\\ ik{E}_{x}=\frac{\partial {E}_{z}}{\partial x}+i\omega {B}_{y}.\end{array}$

*The strategy is to get
**${E}_{x},{E}_{y},{B}_{x},{B}_{y}$** all
in terms of only **${E}_{z},{B}_{z}.$** *

So, we need to eliminate ${B}_{y}$ between the two equations above, multiplying the first by $\omega ,$ the second by $k$ and adding:

$\begin{array}{l}\frac{-i{\omega}^{2}}{{c}^{2}}{E}_{x}+i{k}^{2}{E}_{x}=\omega \frac{\partial {B}_{z}}{\partial y}+k\frac{\partial {E}_{z}}{\partial x},\\ {E}_{x}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(\omega \frac{\partial {B}_{z}}{\partial y}+k\frac{\partial {E}_{z}}{\partial x}\right).\end{array}$

Similarly,

$\begin{array}{l}{E}_{y}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(k\frac{\partial {E}_{z}}{\partial y}-\omega \frac{\partial {B}_{z}}{\partial x}\right),\\ {B}_{x}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(k\frac{\partial {B}_{z}}{\partial x}-\frac{\omega}{{c}^{2}}\frac{\partial {E}_{z}}{\partial y}\right),\\ {B}_{y}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(k\frac{\partial {B}_{z}}{\partial y}+\frac{\omega}{{c}^{2}}\frac{\partial {E}_{z}}{\partial x}\right).\end{array}$

We have already shown that

$\left[\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}+{\left(\frac{\omega}{c}\right)}^{2}-{k}^{2}\right]{E}_{z}=0.$

and similarly

$\left[\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}+{\left(\frac{\omega}{c}\right)}^{2}-{k}^{2}\right]{B}_{z}=0.$

*So, we have 2-D **${\nabla}^{2}$** equations for **${E}_{z},{B}_{z}$** and
from the earlier equations, it’s straightforward to compute **${E}_{x},{E}_{y},{B}_{x},{B}_{y}$** from **${E}_{z},{B}_{z}.$** So in
principle we’ve solved the problem.*

### Electric and Magnetic Boundary Conditions

Of course, to actually find solutions of the differential equations for ${E}_{z},{B}_{z}$, we need the boundary conditions!

Let’s start with ${E}_{z}$: this is easy, we need ${E}_{z}=0$ at the boundary, since the metal is taken to be a perfect conductor (and even if it isn't, the skin depth must be far smaller than the waveguide size, or it wouldn't work).

${B}_{z}$ is more tricky, we know that ${B}_{\perp}=0$ at the surface, but ${B}_{z}$ isn’t in general zero. However, we can use $\overrightarrow{\nabla}\times \overrightarrow{B}={\mu}_{0}{\epsilon}_{0}\left(\partial \overrightarrow{E}/\partial t\right)$ and look near the surface in the direction tangential to the surface but also perpendicular to $z$ $\u2014$call it $y$, so $x$ is normal to the surface (see figure). Thus, the $y$ component of $\overrightarrow{\nabla}\times \overrightarrow{B}={\mu}_{0}{\epsilon}_{0}\dot{\overrightarrow{E}}$ is

${\left(\overrightarrow{\nabla}\times \overrightarrow{B}\right)}_{y}=\frac{\partial {B}_{z}}{\partial n}-\frac{\partial {B}_{n}}{\partial z}={\mu}_{0}{\epsilon}_{0}\frac{\partial {E}_{y}}{\partial t}.$

In this equation, we already know that close enough to the perfectly conducting (in our approximation) surface, the tangential ${E}_{y}$ and the normal ${B}_{n}={B}_{x}$ are both identically zero, so we conclude that the boundary condition we need is:

$\partial {B}_{z}/\partial n=0.$

### Boundary Conditions for Field *z*-Components

We now see that ${E}_{z}\left(x,y\right),\text{\hspace{0.33em}}{B}_{z}\left(x,y\right)$ satisfy *exactly*
the same differential equation but with *different
boundary conditions*:

$$

The differential equation is a 2D eigenvalue equation for ${E}_{z}$ (or for ${B}_{z}$ ):

$\left[\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}\right]{E}_{z}\left(\text{or}{B}_{z}\right)=\left({k}^{2}-{\left(\frac{\omega}{c}\right)}^{2}\right){E}_{z}\left(\text{or}{B}_{z}\right),$

and *different boundary
conditions will give in general different allowed eigenvalues* (think
standing waves in an organ pipe with both ends open or one end closed, one
open).

**Therefore, in
general, for consistency, at least one of ****${E}_{z},{B}_{z}$**** must
be zero. This leads to three classes of modes. **

Actually
there are exceptional (termed *hybrid*)
modes in some geometries with $z$ -direction electric *and* magnetic fields, the most important being for rectangular cross-section waveguides. See the full discussion in the next lecture.

### Possible Modes: TM, TE, TEM

The possibilities are:

**TM** (transverse magnetic) solutions: zero
longitudinal (parallel to the cable) magnetic field.

That is, ${B}_{z}=0$ *everywhere*.

( Of course ${E}_{z}=0$ at the surface, as always.)

Now ${E}_{x}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(\omega \frac{\partial {B}_{z}}{\partial y}+k\frac{\partial {E}_{z}}{\partial x}\right)$ and ${E}_{y}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(k\frac{\partial {E}_{z}}{\partial y}-\omega \frac{\partial {B}_{z}}{\partial x}\right).$

Therefore for ${B}_{z}\equiv 0,$ the transverse electric field

$$

and correspondingly

${\overrightarrow{B}}_{t}=\left(\omega /k\right)\widehat{\overrightarrow{z}}\times {\overrightarrow{E}}_{t}.$

**TE **(transverse
electric) solutions: ${E}_{z}=0$ everywhere, $\partial {B}_{z}/\partial n=0$ at surface.

From ${B}_{x}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(k\frac{\partial {B}_{z}}{\partial x}-\frac{\omega}{{c}^{2}}\frac{\partial {E}_{z}}{\partial y}\right)$ and ${B}_{y}=\frac{i}{{\left(\omega /c\right)}^{2}-{k}^{2}}\left(k\frac{\partial {B}_{z}}{\partial y}+\frac{\omega}{{c}^{2}}\frac{\partial {E}_{z}}{\partial x}\right),$

we find for ${E}_{z}\equiv 0$ that

$$

and

${\overrightarrow{E}}_{t}=-\left(\omega /c\right)\widehat{\overrightarrow{z}}\times {\overrightarrow{B}}_{t}.$

**TEM**:
${B}_{z}={E}_{z}=0$ everywhere. As mentioned above in discussing
the coaxial cable, this is equivalent to a 2-D electrostatics problem, the
boundary cross-section needs a minimum of two separate components, at different
potentials in the 2-D problem.

### Solving for the Modes

In the next lecture, we’ll find these modes in detail for the solvable (and important) case of a rectangular cross section pipe, and also analyze the analogous modes in a rectangular cavity.