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In a rectangular waveguide with sides of length $a$ and $b$, the dispersion relation is

$$\beta^2 =\omega^2\mu\epsilon=\beta_z^2+\beta_x^2+\beta_y^2=\beta_z^2+\beta_s^2.$$

So we have

$$\beta_z = \sqrt{\omega^2\mu\epsilon-\beta_x^2-\beta_y^2}$$

which gives us the cutoff frequency if the wave propagates in $z$.

My confusion arises with using these modes in practice - what are we referring to as the frequency, and analogously the wavelength? When we talk about the wavelength, are we talking about $2\pi/\beta_z$ or are we talking about $\omega/c$?

Or are the dimensions usually so small of the waveguides that $\omega/c$ is roughly equal to $2\pi/\beta_z$?

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2 Answers 2

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The frequency is $f = \omega/(2 \pi)$ and specifies the oscillation in time. The wavenumber is $\beta_z = (2\pi)/\lambda_z$ and specifies the oscillation in space. Thus the electric wave is given by $E \cos(\beta_z z + \omega t)$.

The dispersion relation describes the relation between $\omega$ and $\beta$. In vacuum, this becomes $\beta \omega = c$ or $\lambda f = c$. This is a linear dispersion relation. In the rectangular waveguide, you have a changed dispersion because of the confinement (which gives rise to the additional terms $\beta_x$ and $\beta_y$) and material (which changes the speed of the wave $c' = 1/\sqrt{\mu_r \mu_0 \epsilon_r \epsilon_0}$).

The wavelength in the z-direction in the waveguide is given by $2\pi/\beta_z$. When the frequency ($\omega$) is sufficiently high, the terms $\beta_x$ and $\beta_y$ can be neglected in the dispersion relation. In this case (which is usually the case in practice), we end up with the linear dispersion relation. Thus, in practice when the frequency is sufficiently high, the real wavelength $\lambda = 2\pi/\beta$ can be approximated by $f/c' = \omega/(2\pi c')$.

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First, let me just point out that the relationship between the angular frequency $\omega$ and the propagation constant $\beta_z$ is not in general as simple as indicated by the second expression. For instance, in an optical fibre, which is a widely used waveguide, the relationship is quite complicated and generally need numerical methods to determine.

Now, when a wavelength is associated with such a mode (of instance 1550 nm for fibre optic telecoms), it is usually the one related to the angular frequency via the vacuum dispersion relation $$ \omega = \frac{2\pi c}{\lambda} . $$

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  • $\begingroup$ Yes, I understand - the question is moreso in regards to, in practice, what is the wavelength. Like in the other answer, $\lambda = \lambda_z$ because the frequency is high and the transverse dimensions are small. $\endgroup$
    – lumicoh
    Nov 9, 2019 at 22:46

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