Peek behind a radar dish, inside a satellite dish's feed, or into the plumbing of a particle accelerator, and you will find something that looks oddly like a length of rectangular metal pipe. No wire runs down the middle. It is hollow. Yet it is carrying gigawatts of microwave power from one end to the other with almost no loss. This hollow pipe is a waveguide, and it is one of the most elegant devices in all of electromagnetism: it herds an electromagnetic wave down a tube the way a canyon channels a river.

Here is the single surprising fact this whole page is built to explain: a hollow conducting pipe will only carry a wave if the wave's frequency is high enough. Feed in a signal below a certain cutoff frequency and nothing comes out the far end — the wave does not travel, it simply decays away within a few centimetres. Above the cutoff it sails through. A waveguide is not a wire; it is a high-pass filter with walls. And the wave inside cannot take just any shape: it is forced into a discrete set of patterns called modes, each with its own cutoff. Everything below is the story of where those modes and that cutoff come from.

Why a pipe is picky: the walls set the rules

The metal walls are the whole reason a waveguide behaves so differently from free space. A good conductor cannot support a tangential electric field at its surface: any parallel field would drive currents that instantly cancel it. So the boundary condition is uncompromising — the component of \mathbf{E} along each wall must vanish everywhere on that wall:

Now think about what that demands of the field across the guide. Take a rectangular pipe of width a (in x) and height b (in y), with the wave running along z. The transverse field cannot be uniform: it has to rise from zero at one wall, do something in the middle, and fall back to zero at the opposite wall. The only patterns that fit are standing waves that have a whole number of half-wavelengths across the guide — exactly like the vibrations of a rectangular drum skin clamped at its edges. That "whole number" requirement is what quantises the field into discrete modes.

For the transverse-electric \text{TE}_{mn} modes (the field has no E_z component, only transverse electric field), fitting m half-waves across the width and n across the height pins the two transverse wavenumbers to the discrete values

The integers m and n count the humps of the field pattern in the two transverse directions. You cannot have a fraction of a hump, because a fractional hump would not meet the wall at zero. This is the same quantisation you meet for a particle in a box or a string clamped at both ends — the box just happens to be a copper pipe.

The dispersion relation: how the pieces of the wavevector must add up

Inside the empty guide the field still obeys the ordinary wave equation, which ties the frequency \omega to the full wavevector. The wave oscillates as \cos(k_x x)\cos(k_y y)\,e^{i(k_z z - \omega t)}, and plugging that into the wave equation forces the three wavenumber pieces to satisfy

Here k_z is the only piece that describes actual travel down the guide; k_x and k_y are locked to the standing pattern across it. It is tidy to bundle the two transverse pieces into a single cutoff wavenumber

so the relation collapses to the clean, central equation of the whole topic:

Read the middle equation like a right triangle: the "hypotenuse" \omega/c must be long enough to cover the fixed transverse side k_c before there is anything left over for the travelling side k_z. Drop the frequency until \omega/c = k_c and k_z hits zero — the wave stops advancing and just sloshes side to side. Below that, (\omega/c)^2 - k_c^2 goes negative, its square root is imaginary, and propagation dies.

The dominant mode and its cutoff

Which mode has the lowest cutoff? For a guide that is wider than it is tall (a > b), the smallest non-zero k_c comes from m = 1, n = 0 — one hump across the wide dimension, none across the short one. This is the celebrated dominant mode \text{TE}_{10}, and its cutoff frequency depends only on the width:

This is why practical guides are run in a band just above the \text{TE}_{10} cutoff but below the next mode's cutoff: in that window only one mode can propagate, so the signal keeps a single clean shape. The width a of the pipe is essentially chosen to place c/2a just below your operating frequency.

A tangible check: your kitchen microwave oven runs at 2.45\ \text{GHz}. Its magnetron feeds the cooking chamber through a short rectangular guide whose wide wall is about a \approx 8\ \text{cm}, giving f_c = c/2a \approx 1.9\ \text{GHz} — safely below 2.45\ \text{GHz}, so the wave sails through. Shrink that wall to 5\ \text{cm} and the cutoff climbs to 3\ \text{GHz}; now 2.45\ \text{GHz} is below cutoff and the same oven would go dead.

See it: the dispersion curve and the light line

The dispersion relation is best seen. The chart plots \omega/c against the guide wavenumber k_z. The straight dashed light line \omega/c = k_z is how a wave would travel in free space. The curved line is the guided mode \omega/c = \sqrt{k_z^2 + k_c^2}: drag the slider to change the cutoff k_c and watch it lift off.

Notice three things. The guided curve never touches the origin: at k_z = 0 it starts at \omega/c = k_c — that is the cutoff, the minimum frequency the guide will pass. For large k_z the curve hugs the light line from above, because at high frequency the transverse structure barely matters and the wave behaves almost as in free space. And crucially the guided curve always sits above the light line, which — as we are about to see — is exactly why the phase velocity in a guide exceeds c.

Phase velocity, group velocity, and a beautiful identity

Two speeds live inside a propagating mode. The phase velocity v_p = \omega/k_z tracks the crests; the group velocity v_g = d\omega/dk_z tracks the energy and the information. Starting from \omega = c\sqrt{k_z^2 + k_c^2} and using k_c = \omega_c/c, a little algebra gives both in terms of how far above cutoff you are running, f/f_c:

Multiply them together and the square roots annihilate, leaving one of the prettiest results in the subject:

The crests moving faster than light also stretch the wavelength inside the guide. The guide wavelength \lambda_g = 2\pi/k_z is always longer than the free-space wavelength \lambda = c/f:

Worked example: an X-band guide

The workhorse of radar and microwave labs is the WR-90 rectangular guide, with inner dimensions a = 2.29\ \text{cm} and b = 1.02\ \text{cm}. Let's run it at f = 10\ \text{GHz} (the "X band"). Throughout, use the handy fact that in these units c \approx 30\ \text{cm}\cdot\text{GHz}, so c/2 = 15\ \text{cm}\cdot\text{GHz}.

Step 1 — dominant-mode cutoff. For \text{TE}_{10},

Since 10\ \text{GHz} > 6.55\ \text{GHz}, the mode propagates. Good.

Step 2 — is it single-mode? The next modes up are \text{TE}_{20} at c/a = 13.1\ \text{GHz} and \text{TE}_{01} at c/2b \approx 14.7\ \text{GHz}. Both are above 10\ \text{GHz}, so at 10\ \text{GHz} only \text{TE}_{10} lives — single-mode operation, as designed.

Step 3 — the velocities. The ratio is f_c/f = 6.55/10 = 0.655, so \sqrt{1 - (f_c/f)^2} = \sqrt{1 - 0.429} = \sqrt{0.571} \approx 0.756. Hence

Check the identity: v_p v_g = (1.32)(0.76)\,c^2 \approx 1.00\,c^2. The energy crawls down the guide at about three-quarters of light speed while the crests overtake light by a third — and their product is exactly c^2.

Step 4 — guide wavelength. Free-space \lambda = c/f = 30/10 = 3.0\ \text{cm}, so

Nearly a centimetre longer than in open air — the stretch that phase velocity above c buys you.