Cavities with rectangular boundaries

Consider a vacuum region totally enclosed by rectangular conducting walls. In this case, all of the field components satisfy the wave equation

$$\nabla^2\psi - \frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}=0,$$ (1047)

where $\psi$ represents any component of ${\bfm E}$ or ${\bfm H}$. The boundary conditions (6.4) require that the electric field is normal to the walls at the boundary whereas the magnetic field is tangential. If $a$, $b$, and $c$ are the dimensions of the cavity, then it is readily verified that the electric field components are

$$E_x = E_1 \cos(k_1 x)\sin(k_2 y) \sin(k_3 z)\, {\rm e}^{-{\rm i}\,\omega t},$$ (1048)

$$E_y = E_2 \sin(k_1 x)\cos(k_2 y) \sin(k_3 z)\, {\rm e}^{-{\rm i}\,\omega t},$$ (1049)

$$E_z = E_3 \sin(k_1 x)\sin(k_2 y) \cos(k_3 z)\, {\rm e}^{-{\rm i}\,\omega t},$$ (1050)

where

$$k_1 = \frac{l\,\pi}{a},$$ (1051)

$$k_2 = \frac{m\,\pi}{b},$$ (1052)

$$k_3 = \frac{n\,\pi}{c},$$ (1053)

with $l$, $m$, $n$ integers. The allowed frequencies are given by

$$\frac{\omega^2}{c^2} = \pi^2 \left(\frac{l^2}{a^2} + \frac{m^2}{b^2} + \frac{n^2}{c^2}\right).$$ (1054)

It is clear from Eq. (6.9) that at least two of the integers $l$, $m$, $n$ must be different from zero in order to have non-vanishing fields. The magnetic fields obtained by the use of $\nabla\wedge{\bfm E}={\rm i}\,\omega{\bfm B}$ automatically satisfy the appropriate boundary conditions, and are in phase quadrature with the electric fields. Thus, the sum of the total electric and magnetic energies within the cavity is constant, although the two terms oscillate separately.

The amplitudes of the electric field components are not independent, but are related by the divergence condition $\nabla\!\cdot\!{\bfm E} = 0$, which yields

$$k_1\,E_1 + k_2 \,E_2 + k_3\, E_3 = 0.$$ (1055)

There are, in general, two linearly independent vectors ${\bfm E}$ that satisfy this condition, corresponding to two polarizations. (The exception is the case that one of the integers $l$, $m$, $n$ is zero, in which case ${\bfm E}$ is fixed in direction.) Each vector is accompanied by a magnetic field at right angles. The fields corresponding to a given set of integers $l$, $m$, and $n$ constitute a particular mode of vibration of the cavity. It is evident from standard Fourier theory that the different modes are orthogonal (i.e., they are normal modes) and that they form a complete set. In other words, any general electric and magnetic fields which satisfy the boundary conditions (6.4) can be unambiguously decomposed into some linear combination of all of the various possible normal modes of the cavity. Since each normal mode oscillates at a specific frequency it is clear that if we are given the electric and magnetic fields inside the cavity at time $t=0$ then the subsequent behaviour of the fields is uniquely determined for all time.

The conducting walls gradually absorb energy from the cavity, due to their finite resistivity, at a rate which can easily be calculated. For finite $\sigma$ the small tangential component of ${\bfm E}$ at the walls can be estimated using Eq. (6.5):

$${\bfm E}_\parallel = \frac{1-{\rm i}}{\sqrt{2}} \sqrt{\frac{\mu_0\omega}{\sigma}}\, {\bfm H}_\parallel \wedge{\bfm n}.$$ (1056)

Now, the tangential component of ${\bfm H}$ at the walls is slightly different from that given by the ideal solution. However, this is a small effect and can be neglected to leading order in $\sigma^{-1}$. The time averaged energy flux into the walls is given by

$$\overline{\bfm N} = \frac{1}{2} \, {\rm Re}\,({\bfm E}_\para... ...,{\bfm n} = \frac{H_{\parallel 0}^{~2}}{2\sigma d}\, {\bfm n},$$ (1057)

where $H_{\parallel 0}$ is the peak value of the tangential magnetic field at the walls predicted by the ideal solution. According to the boundary condition (6.4)(d), $H_{\parallel 0}$ is equal to the peak value of the surface current density $K_0$. It is helpful to define a surface resistance,

$$\overline{\bfm N} = \overline{K^2} R_s\,{\bfm n},$$ (1058)

where

$$R_s = \frac{1}{\sigma d}.$$ (1059)

This approach makes it clear that the dissipation of energy is due to ohmic heating in a thin layer, whose thickness is of order the skin depth, on the surface of the conducting walls.