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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 ?

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  • $\begingroup$ It isn't clear what you are asking. What do you think the size of the medium has to do with the question? If you are thinking of GR effects, you need to specify who is observing the wavelength, from where and when. $\endgroup$
    – ProfRob
    Commented Oct 31, 2014 at 20:03
  • $\begingroup$ It doesn't matter what the actual size of the medium is here, as long as it is greater than 1.0 (vacuum). But I am not asserting any GR effects either. I used it as an analogy, but my question is simple: Would light lose intrinsic energy after traveling through a medium that slowed it down (increased it's wavelength)for an extended amount of time? $\endgroup$
    – Gödel
    Commented Nov 1, 2014 at 2:16
  • $\begingroup$ By an extended amount of time I mean years (which we would never be able to observe). $\endgroup$
    – Gödel
    Commented Nov 1, 2014 at 2:19
  • $\begingroup$ I know that the frequency remains constant in all practical cases, but could that really hold up with light working directly against a medium for so long? $\endgroup$
    – Gödel
    Commented Nov 1, 2014 at 2:21

3 Answers 3

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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.

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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 $\lambda = c/f$

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    $\begingroup$ One of the many reasons for spending a lot of time on the behavior of driven harmonic oscillators in mechanics. That the response has the driving frequency and not the systems frequency or some combination comes as a surprise without it. $\endgroup$ Commented Oct 31, 2014 at 20:49
  • $\begingroup$ How could molecules shake at visible light frequencies. Seems like it too fast. $\endgroup$
    – Lambda
    Commented Jul 2, 2018 at 16:37
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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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  • $\begingroup$ Yes,I wrote that I knew both the physical and mathematical reason, that was not my main question. My question was basically : If a wave of light travels through a medium with an index of refraction, N, for a very long period of time (say, like 2 years, that is why I made the fluid body 2 ly across), if the wavelength would exit at the numerical value that it entered, just as with short distances. Or, would the light wave eventually lose potential, and become stretched, analogous to traveling through the vacuum against dark energy. $\endgroup$
    – Gödel
    Commented Oct 31, 2014 at 22:20
  • $\begingroup$ Or through gravitational fields, where lensing occurs. $\endgroup$
    – Gödel
    Commented Oct 31, 2014 at 22:22

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