Waveguides and Cavities | Physics

A rectangular waveguide with dimensions $a$ (width) and $b$ (height) supports electromagnetic waves propagating in the $z$-direction. Two families of modes exist: TE (transverse electric, $E_z = 0$) and TM (transverse magnetic, $B_z = 0$).

TE$_{mn}$ modes. The longitudinal field is $B_z = B_0\cos(m\pi x/a)\cos(n\pi y/b)\,e^{i(kz-\omega t)}$.

The transverse fields are determined from $B_z$ via:

$E_x = \frac{i\omega}{k_c^2}\frac{\partial B_z}{\partial y}, \quad E_y = -\frac{i\omega}{k_c^2}\frac{\partial B_z}{\partial x}$

$B_x = \frac{-ik}{k_c^2}\frac{\partial B_z}{\partial x}, \quad B_y = \frac{-ik}{k_c^2}\frac{\partial B_z}{\partial y}$

Where $k_c^2 = (m\pi/a)^2 + (n\pi/b)^2$ is the cutoff wavenumber.

Cutoff frequency: Waves propagate only when $\omega > \omega_{c,mn}$ where:

$f_{c,mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}$

The dominant (lowest frequency) mode is TE$_{10}$ with $f_{c,10} = c/(2a)$ (for $a > b$).

Dispersion relation:

$k = \sqrt{\frac{\omega^2}{c^2} - k_c^2}, \quad v_{\text{phase} = \frac{\omega}{k} = \frac{c}{\sqrt{1 - (\omega_c/\omega)^2}} > c}$

$v_{\text{group} = \frac{d\omega}{dk} = c\sqrt{1 - \left(\frac{\omega_c}{\omega}\right)^2} < c}$

The product $v_p \cdot v_g = c^2$.

The wave impedance for TE modes:

$Z_{\text{TE} = \frac{E_x}{H_y} = \frac{\omega\mu_0}{k} = \frac{Z_0}{\sqrt{1 - (f_c/f)^2}}}$

Where $Z_0 = \sqrt{\mu_0/\varepsilon_0} \approx 377\,\Omega$ is the impedance of free space.

The time-averaged power carried by TE$_{10}$ mode:

$\langle P \rangle = \frac{ab}{4}E_0^2\frac{\beta}{\omega\mu_0} = \frac{ab}{4Z_{\text{TE}}E_0^2}$

Where $\beta = k$ is the propagation constant and $E_0$ is the peak electric field.

A rectangular cavity of dimensions $a \times b \times d$ supports standing waves at resonant frequencies:

$f_{mnp} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 + \left(\frac{p}{d}\right)^2}$

Where $m, n, p$ are non-negative integers (not all zero). For TM modes, $p \geq 1$; for TE modes, $m$ and $n$ cannot both be zero.

Quality factor:

$Q = \frac{\omega \times \text{energy} stored}{\text{power} dissipated} = \frac{2\pi \times \text{energy} stored}{\text{energy} lost per cycle}$

For a cavity with conducting walls of conductivity $\sigma$:

$Q \approx \frac{V}{S\,\delta} \cdot \frac{3}{2}$

Where $V$ is the cavity volume, $S$ is the surface area, and $\delta$ is the skin depth.

• Assuming TEM modes exist in hollow waveguides: TEM modes require at least two separate conductors (e.g., coaxial cables). Hollow rectangular and circular waveguides cannot support TEM modes; they only support TE and TM modes.
• Confusing cutoff frequency with zero propagation: At $\omega = \omega_c$, $k = 0$ and the wave does not propagate. Below cutoff, $k$ becomes imaginary and fields decay exponentially (evanescent mode), carrying no net power.
• Forgetting that $m$ or $n$ can be zero in TE modes but not in TM modes: For TE$_{mn}$ modes, $m$ and $n$ cannot both be zero, but one may be zero. For TM$_{mn}$ modes, both $m$ and $n$ must be non-zero, meaning the lowest TM mode is TM$_{11}$.
• Misapplying the quality factor formula: The $Q$ of a cavity depends on the specific mode, as different field distributions produce different wall currents and hence different ohmic losses. The approximate formula $Q \approx (V/S\delta) \cdot 3/2$ is for the dominant mode only.

For a circular waveguide of radius $R$, the TE modes have cutoff wavenumbers $k_c = p'_{mn}/R$ where $p'_{mn}$ is the $n$-th root of $J'_m(x) = 0$. The TM modes have $k_c = p_{mn}/R$ where $p_{mn}$ is the $n$-th root of $J_m(x) = 0$.

The dominant mode in a circular waveguide is TE$_{11}$, with cutoff $f_c = \frac{cp'_{11}}{2\pi R} = \frac{1.841c}{2\pi R}$. Circular waveguides are used in rotating joints and polarisation-sensitive applications because TE$_{11}$ maintains polarisation orientation.

Problem. Design a rectangular cavity ($a = 3$ cm, $b = 1.5$ cm, $d = 2$ cm) that resonates at approximately 10 GHz. Which mode should be used?

Solution. The resonant frequency formula is $f_{mnp} = \frac{c}{2}\sqrt{(m/a)^2 + (n/b)^2 + (p/d)^2}$.

For TE$_{101}$: $f = \frac{3 \times 10^8}{2}\sqrt{(1/0.03)^2 + (1/0.02)^2} = 1.5 \times 10^8 \times \sqrt{1111.1 + 2500} = 1.5 \times 10^8 \times 60.09 = 9.01$ GHz.

For TE$_{102}$: $f = 1.5 \times 10^8 \times \sqrt{1111.1 + 10000} = 1.5 \times 10^8 \times 105.4 = 15.8$ GHz.

For TE$_{011}$: $f = 1.5 \times 10^8 \times \sqrt{(1/0.015)^2 + (1/0.02)^2} = 1.5 \times 10^8 \times \sqrt{4444.4 + 2500} = 1.5 \times 10^8 \times 83.33 = 12.5$ GHz.

TE$_{101}$ at 9.01 GHz is closest to 10 GHz. Fine-tuning the dimensions or inserting a dielectric can adjust the resonant frequency upward to exactly 10 GHz.

• Electromagnetic Waves — The wave equation and dispersion relation for free-space propagation provide the starting point for waveguide mode analysis.

• Special Relativity and Electromagnetism — The phase velocity exceeding $c$ in waveguides is consistent with special relativity because only the group velocity carries information.

• Lasers — Laser cavities are optical resonators governed by the same standing-wave and quality-factor principles as microwave cavities.

• Calculus

• Linear Algebra

• Vector Calculus

• Cutoff frequency determines single-mode operation: For a waveguide with $a > b$, the TE$_{10}$ mode has the lowest cutoff. Operating between $f_{c,10}$ and the next higher cutoff ensures only one mode propagates, avoiding modal dispersion.
• Phase velocity exceeds $c$ while group velocity is below $c$: This is consistent with special relativity because no information travels at the phase velocity; signal velocity is bounded by $v_g \leq c$.
• The product $v_p \cdot v_g = c^2$ is universal for all waveguide modes in a lossless rectangular guide, a direct consequence of the dispersion relation.
• Quality factor $Q$ increases with cavity size: Larger cavities store more energy relative to wall losses, giving higher $Q$. This is why microwave cavities in particle accelerators are large.
• Skin depth decreases with frequency: Higher frequency means thinner current-carrying layer on walls, reducing resistive losses but also reducing the effective conductor cross-section.

• Microwave communication: Rectangular waveguides (e.g., WR-90 for X-band) carry radar signals with low loss, as the confined mode avoids radiation losses.
• Particle accelerators: Resonant cavities (e.g., RF cavities in synchrotrons) accelerate charged particles by sustaining strong oscillating electric fields at precise frequencies.
• Microwave ovens: The magnetron generates microwaves at 2.45 GHz that propagate into the oven cavity, where standing waves heat food.
• Fibre optics: Although optical fibres are dielectric waveguides rather than metallic, the same concepts of modes, cutoff, and dispersion apply.
• Radar systems: Waveguide components (bends, twists, directional couplers) route microwave signals between the transmitter, antenna, and receiver with minimal loss.

flowchart TD
    A[9_Waveguides And Cavities] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

A waveguide is a metallic pipe that channels electromagnetic waves like a flute channels sound. The cutoff frequency acts like a minimum note: waves below this frequency cannot propagate and decay exponentially. Phase velocity exceeding c is not a paradox because it carries no information; the group velocity, which does, stays below c. Cavities are like organ pipes for microwaves, resonating at specific frequencies determined by their dimensions. The quality factor measures how long a cavity rings, like how long a bell sustains its tone after being struck.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

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