core38 min studyLesson 17 in Physics

Waveguides and cavity modes

Use conductor boundary conditions to select modes, calculate cutoff, and separate phase velocity from energy transport.

Before you begin

On this page

The question for today

What you’ll understand

Relate transverse confinement to a mode's cutoff frequency.
Calculate propagation and group velocity for the rectangular TE₁₀ mode.
Explain why cavity frequencies are discrete and mode labels need boundary constraints.

Metal walls can direct electromagnetic energy without acting like mirrors for a narrow geometric ray. The field must fit the walls everywhere, and only certain transverse patterns satisfy those conditions. Each pattern defines a mode with its own propagation law. Closing the guide at both ends imposes another standing-wave condition, turning a continuum of propagation frequencies into discrete cavity resonances.

The wall selects allowable field patterns

Begin with a hollow rectangular guide filled with vacuum and ideal perfectly conducting walls. The tangential electric field vanishes at each wall. A mode traveling along z can vary sinusoidally across the rectangular cross-section while retaining a definite frequency and longitudinal wave number. These boundary restrictions quantize transverse wave numbers even though the guide can be arbitrarily long.

A transverse electric mode has no longitudinal electric component, while its magnetic field generally does have one. A transverse magnetic mode instead has no longitudinal magnetic component. A simply connected hollow metal guide does not support a nontrivial TEM mode with both fields entirely transverse; a two-conductor transmission line can. Mode names encode longitudinal components, not a claim that every vector component is perpendicular to every wall.

k_z^2=\frac{\omega^2}{c^2}-k_\perp^2,\qquad k_\perp^2=\left(\frac{m\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2

a and b are the cross-sectional side lengths. For rectangular TE modes m,n are nonnegative and not both zero; for TM modes both are positive. These index restrictions are essential when using the dispersion relation.↔ Scroll wide equations horizontally.

Cutoff separates propagation from evanescence

For $a>b$ , the lowest-cutoff mode is TE₁₀. Below its cutoff, longitudinal wave number becomes imaginary and an excited field decays exponentially along a uniform section. This is evanescence, not a propagating wave with a very small real speed. A finite below-cutoff section can couple fields between its ends, so the infinite-guide statement should not be mistaken for an absolute prohibition on finite-device transmission.

Above cutoff, phase velocity exceeds the vacuum speed because a constant-phase pattern advances along z faster than its energy envelope. For a narrow-band packet away from cutoff in this lossless guide, group velocity is smaller than c and equals the energy-transport speed. Near cutoff, dispersion becomes strong; a broad pulse changes shape, and one group-velocity number no longer describes its entire evolution.

f_c=\frac{c}{2a},\qquad v_p=\frac{c}{\sqrt{1-(f_c/f)^2}},\qquad v_g=c\sqrt{1-(f_c/f)^2}\quad\text{(TE}_{10}\text{, }f>f_c\text{)}

These formulas assume an ideal uniform vacuum guide. vₚ = ω/kz and v_g = dω/dkz obey vₚv_g = c². A superluminal phase pattern does not transmit a new signal faster than c.↔ Scroll wide equations horizontally.

Closing the guide quantizes longitudinal structure

Conducting end plates require a compatible standing wave along the cavity length. The resonant frequencies then depend on the full three-dimensional geometry. An index triplet alone is insufficient: the associated TE or TM field must obey the boundary conditions and cannot be an identically zero configuration. For example, a rectangular TE₁₀₁ cavity mode has a legitimate electric pattern varying across its width and along its length.

Perfect walls imply no ohmic dissipation and infinitely sharp isolated resonances. Real cavities have conductor losses, dielectric absorption, radiation through ports, and loading by coupled devices. Their quality factor relates stored energy to loss per cycle, and a measured linewidth reflects these mechanisms. The ideal mode frequency remains useful as a design estimate, while precision work needs corrections rather than more digits from the ideal formula.

f_{101}=\frac c2\sqrt{\frac{1}{a^2}+\frac{1}{d^2}},\qquad Q=\frac{\omega U}{P_{\mathrm{loss}}}

The frequency is for the allowed TE₁₀₁ mode of an ideal rectangular vacuum cavity of width a and length d. Q uses cycle-averaged stored energy U and average dissipated power in a weakly damped resonator.↔ Scroll wide equations horizontally.

Work it through

A ten-gigahertz guide

An ideal rectangular vacuum guide has a = 2.0 cm and b = 1.0 cm. At 10 GHz, determine whether TE₁₀ propagates and estimate its guided wavelength and group velocity. Take c = 3.0 × 10⁸ m/s.

Its TE₁₀ cutoff is $f_c=c/(2a)=7.5\,\mathrm{GHz}$ . The 10 GHz operating frequency exceeds it, while TE₂₀ and TE₀₁ each cut off at 15 GHz, so only the lowest mode can propagate among these candidates.
Calculate $s=\sqrt{1-(7.5/10)^2}=0.6614$ . The free-space wavelength is $c/f=0.030\,\mathrm m$ , while the guided wavelength is $\lambda_g=(c/f)/s=0.0454\,\mathrm m$ .
The group velocity is $v_g=cs=1.98\times10^8\,\mathrm{m/s}$ . The phase velocity is $c/s=4.54\times10^8\,\mathrm{m/s}$ ; their product is $c^2$ , providing a consistency check.

The result

TE₁₀ propagates with a guided wavelength of approximately 4.54 cm and a narrow-band energy-envelope speed of 0.661c. The higher phase velocity does not violate causality.

The living notebook

Explore the model

Interactive lab

Interactive lab · spectral

Inspect a finite collection of spatial modes

Change the number N and decay exponent p in the finite sine sum Σ sin(nx−t)/nᵖ. It illustrates how combining allowed spatial scales changes a pattern. Its modes share an illustrative time dependence; it does not impose metal boundaries or calculate waveguide dispersion, TE/TM fields, or cavity frequencies.

Finite sumHighest included mode

Sum at x = π/2: 0.867
Smallest wavelength: 2π / 5
Highest-mode coefficient: 0.2

Mode count N5Coefficient decay p1

t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA modal expansion needs both an allowed spatial basis and the correct physical frequency for each mode.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Below cutoff, kz² is negative, so the longitudinal dependence is exponential rather than oscillatory. This does not exclude finite evanescent coupling through a short section or fields near the excitation port.

Your turn

Try it yourself

An ideal rectangular cavity has width 2.0 cm and length 3.0 cm. Estimate the TE₁₀₁ resonant frequency and explain why the height does not enter this particular formula.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The second transverse mode index is zero; insert the two nonzero inverse-length terms.

Compare with the solution

The frequency is $(3.0\times10^8/2)\sqrt{(1/0.020)^2+(1/0.030)^2}=9.01\times10^9\,\mathrm{Hz}$ , or 9.01 GHz. The TE₁₀₁ mode has no variation associated with the height index, so height does not enter this ideal frequency. Height still affects field normalization, stored energy, other mode frequencies, and real conductor losses.

Keep the idea

Take it with you

Boundary conditions select field modes and exclude invalid index combinations.
Cutoff produces evanescence; phase speed is distinct from energy transport.
Real cavity linewidths require loss mechanisms beyond ideal-conductor eigenfrequencies.

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