I could get the dispersion relation for a wave traveling in a rectangular waveguide of width $a$

$$\omega=c\sqrt{ \left( \frac{\pi}{a}\right)^2 + k_{z}^2 }$$

with $k_z = \dfrac{2\pi}{\lambda_g}\,.$

where z is the direction along the guide. From here I got the group velocity

$$ \dfrac{\partial\omega}{\partial k_z} = \dfrac{\lambda_0}{\lambda_g}c$$

First of all, is this process correct? And second, I've searched and apparently the guided wavelength is longer than the wavelength in free space, which agrees with the result I got, but I really can't understand how this could happen?


The expression of the group velocity $$v_g=\frac {\partial \omega}{\partial k_z}$$ for the waveguide with wave vector $k_z$ in propagation direction is correct. And the phase velocity is $$v_{ph}=\frac{\omega}{k_z}$$ According to the dispersion relation of the wave guide, when $\omega$ approaches the cut-off frequency $\omega_c=c\frac{\pi}{a}$, the wave vector $k_z$ and and group velocity $v_g$ go to zero and the wavelength approaches infinity.

This can be understood by the fact that the guided wave can be decomposed into plane waves with an inclination angle $\theta$ to the z-direction which are reflected at the upper and lower waveguide walls. If $theta$ approaches 90 degrees, the superposition gives wavelength in z-direction approaching infinity and there is only a standing plane wave reflected betweeen the upper and lower waveguide wall.

  • $\begingroup$ Thanks for your answer. I understand this concept, but what I can't wrap my head around is the fact that the wavelength of the component of the wave propagating along the waveguide (in the z direction in this case) $\lambda_g$ seems to be larger than the wavelength of the original wave propagating in free space $\lambda_0$. How can this happen? $\endgroup$ – Julian Nov 6 '16 at 12:58
  • $\begingroup$ @Julian - The plane wave with wavelength $\lambda_0$, inclined at an angle $\theta $ to the z-axis and its reflection on the side wall inclined at an angle $-\theta$ produces the considered wave mode of the wave guide. When the plane waves with $\lambda_0$ are directed at an angle $±\theta$ to the z-direction their crests intersect the z-axis at a distance $\lambda_ =\frac{\lambda_0}{cos{\theta}}$ which gives the wavelength in z-direction of the waveguide mode. You can also see this from the projection of the wave vector $ k_{0z} =\vec k_0 \cos{\theta}$ of the plane waves on the z-axis. $\endgroup$ – freecharly Nov 6 '16 at 15:31
  • $\begingroup$ Ohh I see! I was looking at it in a wrong way, thanks a lot! $\endgroup$ – Julian Nov 6 '16 at 15:46

Good question. Guided wavelenght is a somewhat tricky concept. Let's see if we can get some intuition for it.

The group velocity is the straight line velocity of propagation of the wave down the center-line of the waveguide. Its value is always less than the propagation velocity of the wave in free space ($c$), because the wave propagates along the guide by a series of oblique reflections from one side of the guide to the other, thus increasing the total path length to a value greater than the straight line path. The relationship between $c$ and $v_g$ is:

$$v_g = c \sin \alpha$$

Where $\alpha$ is the angle of incidence in the waveguide.

On the other hand, we call phase velocity the velocity of propagation of the spot on the waveguide wall where the wave impinges. This velocity is actually faster than both $v_g$ and $c$. It is helpful to consider the “beach analogy” to understand the relationship between the phase and group velocities:

Consider an ocean beach that waves will arrive from offshore at an angle other than 90°, meaning the arriving wave fronts will not be parallel to the shore line. The arriving waves at $v_g$ as it hits the shore will strike a point down the beach first, and this “point of strike” races up the beach at a faster phase velocity, $v_p$, that is faster than $v_g$. This is why the phase velocity in a waveguide can be greater than $c$. Finally, consider that the definition of wavelength is the distance between points of equal phase along the wave.

You'll get the following relationship:


Let me know if something is unclear and I'll try to improve my answer.

  • 1
    $\begingroup$ Thanks for the interesting analogy, that was very clear, I appreciate it! $\endgroup$ – Julian Nov 6 '16 at 21:41
  • 1
    $\begingroup$ @Julian My pleasure. Feel free to upvote the answer if it was useful! $\endgroup$ – dahemar Nov 6 '16 at 21:51

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