Waveguides & Resonant Cavities

Step 1: Assume a waveguide mode with z-dependence

For a wave propagating in the $\hat{z}$ direction inside a hollow metallic waveguide:

$$\mathbf{E}(x,y,z,t) = \mathbf{E}_0(x,y)\,e^{i(k_z z - \omega t)}, \quad \mathbf{B}(x,y,z,t) = \mathbf{B}_0(x,y)\,e^{i(k_z z - \omega t)}$$

Step 2: Decompose into transverse and longitudinal parts

$$\mathbf{E}_0 = \mathbf{E}_T(x,y) + E_z(x,y)\,\hat{z}, \quad \mathbf{B}_0 = \mathbf{B}_T(x,y) + B_z(x,y)\,\hat{z}$$

Step 3: Start from the curl equations

From $\nabla \times \mathbf{E} = i\omega\mathbf{B}$, write out components. Since $\partial/\partial z \to ik_z$:

$$\frac{\partial E_z}{\partial y} - ik_z E_y = i\omega B_x$$

$$ik_z E_x - \frac{\partial E_z}{\partial x} = i\omega B_y$$

Step 4: Similarly from Ampere's law

From $\nabla \times \mathbf{B} = -i(\omega/c^2)\mathbf{E}$:

$$\frac{\partial B_z}{\partial y} - ik_z B_y = -\frac{i\omega}{c^2} E_x$$

$$ik_z B_x - \frac{\partial B_z}{\partial x} = -\frac{i\omega}{c^2} E_y$$

Step 5: Solve for transverse fields in terms of longitudinal components

Define $\gamma^2 = (\omega/c)^2 - k_z^2$. Solving the four equations above simultaneously:

$$E_x = \frac{i}{\gamma^2}\left(k_z\frac{\partial E_z}{\partial x} + \omega\frac{\partial B_z}{\partial y}\right)$$

$$E_y = \frac{i}{\gamma^2}\left(k_z\frac{\partial E_z}{\partial y} - \omega\frac{\partial B_z}{\partial x}\right)$$

$$B_x = \frac{i}{\gamma^2}\left(k_z\frac{\partial B_z}{\partial x} - \frac{\omega}{c^2}\frac{\partial E_z}{\partial y}\right)$$

$$B_y = \frac{i}{\gamma^2}\left(k_z\frac{\partial B_z}{\partial y} + \frac{\omega}{c^2}\frac{\partial E_z}{\partial x}\right)$$

Step 6: The waveguide equation for TE and TM modes

All transverse fields are determined by $E_z$ and $B_z$. These satisfy the 2D Helmholtz equation:

$$\nabla_T^2 E_z + \gamma^2 E_z = 0, \qquad \nabla_T^2 B_z + \gamma^2 B_z = 0$$

where $\nabla_T^2 = \partial^2/\partial x^2 + \partial^2/\partial y^2$ is the transverse Laplacian.

TE modes: $E_z = 0$, solve for $B_z$ with $\partial B_z/\partial n|_S = 0$ (Neumann).

TM modes: $B_z = 0$, solve for $E_z$ with $E_z|_S = 0$ (Dirichlet).

Step 1: Write the Helmholtz equation

$$\frac{\partial^2 B_z}{\partial x^2} + \frac{\partial^2 B_z}{\partial y^2} + \gamma^2 B_z = 0$$

Step 2: Separate variables

Let $B_z(x,y) = X(x)Y(y)$. Substitute and divide by $XY$:

$$\frac{X''}{X} + \frac{Y''}{Y} + \gamma^2 = 0$$

$$\frac{X''}{X} = -k_x^2, \quad \frac{Y''}{Y} = -k_y^2, \quad k_x^2 + k_y^2 = \gamma^2$$

Step 3: General solutions

$$X(x) = A\cos(k_x x) + B\sin(k_x x), \quad Y(y) = C\cos(k_y y) + D\sin(k_y y)$$

Step 4: Apply Neumann boundary conditions for TE

For TE modes, $\partial B_z/\partial n = 0$ at the walls. At perfectly conducting walls, the tangential E must vanish, which translates to:

$\partial B_z/\partial x = 0$ at $x = 0$ and $x = a$: requires $B = 0$ and $k_x = m\pi/a$.

$\partial B_z/\partial y = 0$ at $y = 0$ and $y = b$: requires $D = 0$ and $k_y = n\pi/b$.

Step 5: The TE_$_{mn}$ mode solution

$$B_z = B_0\cos\left(\frac{m\pi x}{a}\right)\cos\left(\frac{n\pi y}{b}\right), \quad m,n = 0,1,2,\ldots \text{ (not both zero)}$$

Step 6: Derive the cutoff frequency

$$\gamma^2 = k_x^2 + k_y^2 = \pi^2\left[\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2\right]$$

Since $\gamma^2 = (\omega/c)^2 - k_z^2$ and we need $k_z^2 > 0$ for propagation:

$$\omega > \omega_{mn} = c\gamma = c\pi\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}$$

For $a > b$, the lowest cutoff is TE$_{10}$: $\omega_c = c\pi/a$, i.e., $f_c = c/(2a)$.

Step 1: The Helmholtz equation for E_z

$$\frac{\partial^2 E_z}{\partial x^2} + \frac{\partial^2 E_z}{\partial y^2} + \gamma^2 E_z = 0$$

Step 2: Separate variables (same as TE)

$$E_z(x,y) = X(x)Y(y) = [A\cos(k_x x) + B\sin(k_x x)][C\cos(k_y y) + D\sin(k_y y)]$$

Step 3: Apply Dirichlet boundary conditions

For TM modes, $E_z = 0$ at the walls (tangential E vanishes on a perfect conductor):

$E_z = 0$ at $x = 0$: requires $A = 0$ (sine only in $x$).

$E_z = 0$ at $x = a$: requires $k_x = m\pi/a$, $m = 1, 2, 3, \ldots$

$E_z = 0$ at $y = 0$: requires $C = 0$ (sine only in $y$).

$E_z = 0$ at $y = b$: requires $k_y = n\pi/b$, $n = 1, 2, 3, \ldots$

Step 4: The TM_$_{mn}$ mode solution

$$E_z = E_0\sin\left(\frac{m\pi x}{a}\right)\sin\left(\frac{n\pi y}{b}\right), \quad m, n = 1, 2, 3, \ldots$$

Note: both $m \geq 1$ and $n \geq 1$ are required (if either is zero, $E_z = 0$ everywhere).

Step 5: Same cutoff frequency formula

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

The lowest TM mode is TM$_{11}$, which has the same cutoff as TE$_{11}$. This is always higher than TE$_{10}$.

Step 6: Key difference from TE modes

TE modes use cosines (Neumann BC), TM modes use sines (Dirichlet BC). TE modes allow one index to be zero; TM modes require both indices nonzero. Therefore TE$_{10}$ (or TE$_{01}$) exists but TM$_{10}$ does not.

Step 1: Write the waveguide dispersion relation

$$k_z^2 = \frac{\omega^2}{c^2} - \gamma^2 = \frac{\omega^2}{c^2} - \frac{\omega_c^2}{c^2} = \frac{\omega^2 - \omega_c^2}{c^2}$$

where $\omega_c = c\gamma$ is the cutoff frequency for the mode in question.

Step 2: Solve for k_z

$$k_z = \frac{1}{c}\sqrt{\omega^2 - \omega_c^2} = \frac{\omega}{c}\sqrt{1 - \left(\frac{\omega_c}{\omega}\right)^2}$$

This is real only when $\omega > \omega_c$ (propagating); imaginary when $\omega < \omega_c$ (evanescent).

Step 3: Compute the phase velocity

$$v_p = \frac{\omega}{k_z} = \frac{\omega}{\frac{\omega}{c}\sqrt{1 - (\omega_c/\omega)^2}} = \frac{c}{\sqrt{1 - (\omega_c/\omega)^2}}$$

Since $\sqrt{1 - (\omega_c/\omega)^2} < 1$ for all $\omega > \omega_c$, we have $v_p > c$ always.

Step 4: Compute the group velocity

Differentiate the dispersion relation $\omega^2 = c^2 k_z^2 + \omega_c^2$ implicitly:

$$2\omega\,d\omega = 2c^2 k_z\,dk_z \implies v_g = \frac{d\omega}{dk_z} = \frac{c^2 k_z}{\omega}$$

$$v_g = c^2 \cdot \frac{k_z}{\omega} = c^2 \cdot \frac{1}{c}\frac{\sqrt{\omega^2 - \omega_c^2}}{\omega} = c\sqrt{1 - \left(\frac{\omega_c}{\omega}\right)^2}$$

Since $\sqrt{1 - (\omega_c/\omega)^2} < 1$, we have $v_g < c$ always.

Step 5: Prove v_p * v_g = c^2

$$v_p \cdot v_g = \frac{c}{\sqrt{1 - (\omega_c/\omega)^2}} \cdot c\sqrt{1 - (\omega_c/\omega)^2} = c^2$$

This is an exact relation, valid at all frequencies above cutoff.

Step 6: Limiting behavior

As $\omega \to \omega_c^+$: $v_p \to \infty$ and $v_g \to 0$ (wave barely propagates).

As $\omega \to \infty$: $v_p \to c$ and $v_g \to c$ (behaves like free space).

The guide wavelength is $\lambda_g = 2\pi/k_z = \lambda/\sqrt{1 - (\omega_c/\omega)^2} > \lambda$.

Step 1: Define the quality factor

$$Q \equiv 2\pi\frac{\text{Energy stored}}{\text{Energy dissipated per cycle}} = \omega_0\frac{W}{P_{\rm loss}}$$

where $W$ is the total electromagnetic energy stored in the cavity and $P_{\rm loss}$ is the time-averaged power dissipated.

Step 2: Energy stored in the cavity

The time-averaged electromagnetic energy in the volume $V$:

$$W = \frac{1}{2}\int_V \left(\epsilon_0|\mathbf{E}|^2 + \frac{|\mathbf{B}|^2}{\mu_0}\right)dV$$

At resonance, the time-averaged electric and magnetic energies are equal: $W_E = W_B = W/2$.

Step 3: Power loss in the walls

For walls with finite conductivity $\sigma$, currents flow in a skin depth $\delta = \sqrt{2/(\mu_0\sigma\omega)}$. The power dissipated per unit area:

$$\frac{dP}{dA} = \frac{1}{2}\frac{|\mathbf{K}|^2}{\sigma\delta} = \frac{1}{2\sigma\delta}|\hat{n} \times \mathbf{H}|^2$$

where $\mathbf{K}$ is the surface current density and $\hat{n}$ is the outward normal.

Step 4: Total power loss

$$P_{\rm loss} = \frac{1}{2\sigma\delta}\oint_S |\hat{n} \times \mathbf{H}|^2\,dA = \frac{R_s}{2}\oint_S |H_{\rm tan}|^2\,dA$$

where $R_s = 1/(\sigma\delta) = \sqrt{\mu_0\omega/(2\sigma)}$ is the surface resistance.

Step 5: Q for a rectangular cavity (TE_$_{101}$ mode)

For a rectangular cavity $a \times b \times d$ operating in the TE$_{101}$ mode:

$$W = \frac{\epsilon_0 E_0^2}{4}(abd)$$

The surface integral over all six walls gives:

$$P_{\rm loss} = \frac{R_s E_0^2}{4\mu_0^2\omega^2}\left[\frac{2b(a^2 + d^2)}{a^2 d^2}\cdot ad + \ldots\right]$$

Step 6: General scaling of Q

The Q factor scales as the ratio of volume to surface area times the skin depth:

$$Q \sim \frac{\text{Volume}}{\text{Surface area} \times \delta} \sim \frac{V}{S\delta}$$

For a cubic cavity of side $L$: $Q \sim L/(3\delta)$. Since $\delta \propto 1/\sqrt{\omega}$, higher frequency cavities can achieve higher Q with smaller size.

Step 7: Physical interpretation and examples

Q measures how many oscillations occur before the energy decays to $1/e$:

$$W(t) = W_0\,e^{-\omega_0 t/Q}$$

The bandwidth of the resonance is $\Delta\omega = \omega_0/Q$.

Copper cavity at 10 GHz: $Q \sim 10^4$. Superconducting Nb cavity at 1.3 GHz: $Q \sim 10^{10}$.