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Though the physical concepts and mathematics are highly rudimentary, I often try to comprehend exactly why light wants to keep it's frequency, yet alter it's wavelength as it travels through a medium, while at the same time, alter both it's wavelength and frequency while traveling through a vast series of gravitational fields.

My question is this:

If possible, try to address the situation without poking at the obvious holes in it.

Imagine that there exists a uniform $[2*2(ly^2)]$ body of fluid in the vacuum of space with an index of refraction $N$. Now imagine a massive wall of light enters this fluid (ignore any intensity fading/ dispersion), and doesn't exit to the other end for two years. Would the light wave exit the fluid with the same wavelength that it entered with? Why ?

Gödel
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3 Answers3

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The size or path-length that the wave has nothing to do with it.

There are two explanations on offer for why the frequency of the radiation remains unchanged as it crosses the boundary between different media.

(a) If a medium is considered to be lots of driven oscillators, then they will oscillate at the frequency of the driving force - the electric field of the wave.

(b) The components of the E- and H-fields that are parallel to the boundary of the medium, must be the same immediately either side of it. Equating the E-fields either side of the boundary and demanding this is true at all times results in the constancy of frequency.

Even if the medium were dissipative in the sense of having a conductivity, the light would still propagate, with decreasing amplitude, at the same frequency.

The only thing that can change the frequency would be inelastic scattering events - things like Compton scattering. Frequently in astrophysics you hear about light being "Comptonised" when it has a significant optical depth to Compton scattering. Here, both the frequency and wavelength of light are altered. Obviously such scattering has a cross section - the more scatterers there are, the more likely it is to happen.

ProfRob
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$\lambda = c/nf$ While in the medium, $n$ is greater than one, in vacuum $n=1$.

The medium responds to whatever is driving it. The molecules of the medium oscillate at whatever frequency shakes them. Each molecule is then a tiny radiator, generating light at that same frequency. The light that the polarized, oscillating, molecule produces interferes with the light generated by all of the other molecules and the initial radiation. But all of those fields are oscillating at the frequency of the incident light. The result of this interference is that the locations of the peaks and troughs of the radiation field are not where they would be if the medium were not there. If the medium is linear and transparent, then the amplitude remains constant, and the shape of the wave in the medium is sinusoidal.

The result: the wavelength in the medium is different than the wavelength in the vacuum, yet all of the molecules and all of the radiation are still oscillating at the frequency of the incident light.

So upon entering the medium the wavelength is reduced. The light that emerges from the other end is generated by the oscillating dipoles in the medium, which, of course, oscillate at that same frequency as everything else. But now in the vacuum, there are no molecules to polarize, no generated fields, no interference ... the light propagates as it normally does in a vacuum, with $\lambda = c/f$

garyp
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I often try to comprehend exactly why light wants to keep it's frequency, yet alter it's wavelength as it travels through a medium

Have you not heard the standard (macroscopic, continuum approximation) explanation?

To whit: the electric and magnetic fields must both be continuous at the boundary, and must therefore have the same time variation on both sides of the boundary.

That really doesn't leave room for any other behavior at this scale.

Note that this kind of consideration is a complemet to the microscopic understanding offered by garyp's answer.

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dmckee --- ex-moderator kitten
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