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Can anyone explain to me qualitatively, what kind of field exists inside a waveguide at EXACTLY the cut-off frequency?
I would be grateful if an explanation was given about waveguides with metalic boundaries and dielectric waveguides (optical fibers).
One way of looking at it (not thinking too hard about the differences between TE, TM, and hybrid modes) is to consider the momentum vector, or k-vector, of the mode.
Like any vector, a k-vector can be decomposed as \$\vec{k}=k_x \hat{x} + k_y \hat{y} + k_z \hat{z}\$. If the waveguide is aligned in the z-direction, then \$k_x\$ and \$k_y\$ tell us how much of the wave's momentum is (speaking crudely) bouncing back and forth between the walls of the waveguide, and \$k_z\$ tells us how much of the momentum is devoted to propagating the wave along the guide.
The boundary conditions for the (n,m)'th mode to propagate can be expressed as $$k_x=\frac{n\pi}{\ell_x}$$ and $$k_y=\frac{m\pi}{\ell_y}$$
As the waveguide dimensions \$\ell_x\$ and \$\ell_y\$ shrink, more of the k-vector must be used to fulfill these boundary conditions and less is available for the \$k_z\$ component.
When \$|\vec{k}|^2 = k_x^2 + k_y^2\$, and therefore \$k_z=0\$, we have reached the cut-off condition.
Basically we have a wave that is resonating in the x and y directions (transverse to the guide) and simply a plane wave (no variation) in the z direction.
