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I've done a derivation showing that the wave length of light in a waveguide is longer than that of free space light. But I don't really have an intuitive understanding of why this is. I think my prof gave an example but i'm not sure I get it. Basically, two waves are propagating down the wavelength. But their angle of propagation is slightly off center. So the intersection of these waves now corresponds to the wave speed? And this is slower than in free space so to make up for that the wave length must increase?

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  • $\begingroup$ Without seeing your calculation, it's hard to answer this. It's not even clear to me what you're asking, since of course you can make the wavelength of an EM wave in a waveguide that has a longer wavelength than some wave in vacuum. Are the vacuum wave and the guided wave meant to be related in some way? $\endgroup$ – d_b Mar 20 at 0:50
  • $\begingroup$ Say you had a microwave emitter. The light has a wave length of 3cm. If you send that light down a wave guide its wave length will be increased to 4 cm. Why does this happen? $\endgroup$ – David Mar 20 at 1:21
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It's better to understand this in the wavenumber domain. The dispersion relation (relating frequency and wavenumber) in a wave-guide is:

$$ \omega_{WG} =\sqrt{(ck_{WG})^2 + \omega_0^2}$$

where $\omega_0$ is the cutoff frequency.

In free space, $\omega_0=0$ so:

$$ \omega_{FS} =ck_{FS}$$

At fixed frequency, $\omega$, you have:

$$ k_{FS}=\frac{\omega} c $$

and

$$ k_{WG}=\frac{\sqrt{\omega^2-\omega_0^2}} c $$

Clearly:

$$ k_{WG} < k_{FS} $$

which means:

$$ \frac{2\pi}{k_{WG}} = \lambda_{WG} > \lambda_{FS}= \frac{2\pi}{k_{FS}}$$

Note that the cutoff frequency means that the wavelength goes to infinity at finite frequency.

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  • $\begingroup$ That answer is great and I follow it. But can this be explained without equations? I think that will give me the best intuitive understanding. What is happening in the wave guide that makes the wave length longer? $\endgroup$ – David Mar 20 at 2:00
  • $\begingroup$ Does this mean the phase velocity $v_{phase}$ in the waveguide is faster than light? $v_{phase} = \frac{\partial \omega_{WG}}{\partial k_{WG}} = \frac{c}{\sqrt{1 + \left( \frac{\omega_{WG}}{c k_{WG}} \right)^2}} > c$ $\endgroup$ – A. P. Mar 20 at 11:27
  • $\begingroup$ @A.P. yes, phase velocity is fast than light. $\endgroup$ – JEB Mar 20 at 15:38
  • $\begingroup$ @David The same way mass means finite energy at zero momentum (thanks to a global interaction with the Higgs field), a wave-guide creates an "effective mass" (via global interaction with the conducting surfaces) so that there is non-zero energy at zero wavenumber. Check out: desy.de/~njwalker/uspas/coursemat/notes/unit_2_notes.pdf $\endgroup$ – JEB Mar 20 at 15:47
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Here is a way of thinking of this a little more intuitively, which might help.

The waveguide has a characteristic impedance which is different from that of free space. Therefore we expect that the propagation velocity of an EM wave in a waveguide will be different from its free-space value. And indeed it is; it will always be a bit slower in the waveguide than it would be in free space. As a consequence, the wavelength of an EM wave propagating down a waveguide will be slightly longer than it would be in free space.

This has important practical consequences.

The original SLAC beam tube was a waveguide 10,000 feet long, at the head end of which electrons were injected at very nearly the speed of light. From there, the electrons rode the crests of a wave train of microwave radiation propagating down the beamline, which did little to increase their velocity but increased their energy enormously. To do this, it was necessary to precisely trim the impedance of the waveguide so the propagation velocity of the EM waves in the tube would properly match that of the electrons, in every segment of that 10,000 foot long tube. This was done essentially by hand on every subsegment of the tube after it was fitted together but before it was installed in the beam line.

The result? Nobel prizes!

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  • $\begingroup$ Thank you, this does help a lot. But could you speak more to impedance? As far as I know its just the ratio of electric over magnetic field ratio. Does a lower impedance mean a faster wave? Are there any other reasons we would want low impedance? $\endgroup$ – David Mar 20 at 5:36
  • $\begingroup$ Have a look at the wikipedia article on waveguide velocity factor, it explains this stuff much better than I can. $\endgroup$ – niels nielsen Mar 20 at 5:44

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