 Solution

Guided electromagnetic waves propagate along a structure whose boundaries select transverse field patterns. At a perfect conductor the tangential electric field vanishes. In a hollow rectangular guide, TE modes have $E_z=0$ and TM modes have $H_z=0$. Their longitudinal wavenumber obeysBelow the waveguide cutoff frequency ${\omega }_c=v{k}_{\perp }$, that mode is evanescent rather than freely propagating. TM modes require both indices positive; TE modes may have one zero but not both. For $a>b$, the lowest cutoff is the TE${}_{10}$ mode, with $f_c=v/(2a)$. A hollow single-conductor guide has no TEM mode, whereas a two-conductor transmission line can support one without this geometrical cutoff.^[ref]

For a propagating TE or TM mode, $v_p=v/\sqrt{1-{\omega }_c^2/{\omega }^2}$ and $v_g=v\sqrt{1-{\omega }_c^2/{\omega }^2}$. The phase velocity can exceed $v$ without superluminal information transfer. Frequency-dependent group delay can broaden pulses; excitation of several modes adds intermodal dispersion.