Mode Selection for a gyrotron cavity

The microwave is reflected at the wall of the cavity to form the standing wave in a closed cavity. If the plane boundary surfaces are at z=0 and z=d, the boundary conditions can be satisfied at each surface only if

$d=p \frac{\lambda_g}{d}\,\,\,\, p=1,2,3,…$

where $lambda_g$ is the waveguide length, p is the mode number in the z-axial direction. However, if the cavity boundary is open, there is no clear walls at the end of the cavity. The microwave energy is stored mainly in the cavity and a little part of the energy goes into the tapper section. So the cavity length is slightly smaller than a half of the waveguide wavelength. It means,

$k=\frac{2 \pi f_0}{c}=\sqrt{k_z^2+(\frac{\chi_{mn}}{R_0})^2}$

\lambda_g=\frac{2 \pi}{k_z}>L_2

where, k is the wave number in vacuum, $f_0$ is the frequency, $k_z$ is the wave number in the z-axial direction, c is the speed of light, $\chi_{mn}$ is the n-th root of the equation $J_m'(x)=0$.

The main mode is found by compare the waveguide wavelength for all the TE modes and the length of the uniform section in the high power cavity.