Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I know that if we solve the maxwell equation, we will end up with the phase velocity of light is related to the permeability and the permittivity of the material. But this is not what I'm interested in, I want to go deeper than that. We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing. The rare atoms will disturb the light in some way. So I am interested in how the atoms affect the light.

Some textbooks that I read explain it in a way kind of like this: In a material the photons is absorbed by atom and then re-emitted a short time later, it then travels at a short distance to the next atom and get absorbed&emitted again and so on. How quickly the atoms in a material can absorb and re-emit the photon and how dense the atoms decides the speed of light in that material. So the light appears slower because it has a smaller “drift speed”.

But recently I realize an alternative explanation: Atoms respond to the light by radiating electromagnetic wave. This “new light” interferes with the “old light” in some way that result in delayed light(advanced in phase),this can easily shown by using simple phasor diagram. Consequently effectively the light covers a smaller phase each second, . Which gives the impression of a lower phase velocity . However the group velocity is changing in a complicated way.

I think that the first explanation does not explain the change in phase velocity of light. if we consider light traveling into a slab of negative refractive index non-dispersive material, let’s say the light is directed perpendicular to the slab. The phase velocity’s direction will be flipped, but group velocity’s direction in the material will not change. Only the second explanation can explain the flipped phase velocity direction. I guess that the velocity that we get in the first explanation is actually belongs to the group velocity. It makes sense to me that the front most of the photon stream determines the first information that the light delivers.

So the question is What really cause the phase velocity of light to be decreased?

1."drift velocity" of photons ( they aren't the same photons, they are re-emitted all the time) 2. phase difference between absorbed and emitted light 3. something else

And also, I still don't really understand about the detail of the absorption-emission process for small light's wavelength ( for large lambda compare to the atoms spacing, the photons will be absorbed by the phonons). The dispersion relation that we know is continuous and also some material is non-dispersive, therefore the absorption process must occurs in all frequency for a certain range. So definitely it doesn't involve the atomic transition, otherwise it will be quantized. My guess is that the relevant absorption process gets smooth out by the dipole moment. What makes the spectrum continuous?

EDIT: link for dispersion relation: http://refractiveindex.info/?group=CRYSTALS&material=Si

share|improve this question
    
I actually read a paper that took this approach at some point. That was decades ago, and I haven't a clue where I found it. –  dmckee Jul 2 '11 at 22:34
    
@Emitabsorb I do not think that the atoms can be absorbing the light because the frequency and phase would change as there is no guarantee that the de-excitation will come with the same frequency ( unless it is a laser setup). IMO it must be higher order QED diagrams playing ball with the electric field of the atoms. Now for x rays, which have a lamda smaller than the crystal/glass spacings, do you have a link that a difference between phase velocity and group velocity has been measured? –  anna v Jul 4 '11 at 7:02
    
@anna v "...there is no guarantee that the de-excitation will come with the same frequency" why not? en.wikipedia.org/wiki/File:Ramanscattering.svg –  pcr Jul 4 '11 at 16:57
    
@pcr Inelastic scattering, which is Raman scattering, means change of frequency –  anna v Jul 4 '11 at 17:12
    
@ anna v I don't think it is necessary to be spontaneous emission, if it is spontaneous emission, then the emission will be directed in random direction right. If the emission is random, then the light would be heavily diffused, which is not what usually happen in daily life. In the real life case e.g light emitted by a lamp, the light's intensity can be considered as high. therefore it is still possible that the absorption emission process is primarily governed by stimulated absorption/emission. Is it true? –  Emitabsorb Jul 4 '11 at 18:11
show 6 more comments

3 Answers

up vote 5 down vote accepted

Looks like you are already familiar with the classical explanation but are still curious about the quantum version of it.

2.phase difference between absorbed and emitted light

Yeah, this is essentially the lowest order contribution to the phase shift in the photon-electron scattering. Here is the sloppy way to visualize it continuously (this is basically the 'classical EM wave scattering' point of view): you can imagine that the "kinetic energy" (-> frequency) of the "photon" increases as it approaches the atom's potential well and then it goes back to its normal frequency upon leaving the atom. This translates to a net increase in the phase ($(n-1)\omega/c$).

  1. "drift velocity" of photons ( they aren't the same photons, they are re-emitted all the time)

By "drift velocity" do you mean a pinball-like, zigzag motion of the photon? This won't contribute that much because it requires more scattering (basically it is a higher order process).

And also, I still don't really understand about the detail of the absorption-emission process.

Yes the absorption will still occur in all range of the frequency. The hamiltonian of the atom will be modified by the field (by $- p \cdot E$ where p is the dipole moment of the atom and E is the electric field component of the light). This will give us the required energy level to absorb the photon momentarily, which will be re-emitted again by stimulated+spontaneous emission.

edit: clarification, the term 'energy level' is misleading, since the temporarily 'excited' atom is not in an actual energy eigenstate.

See the diagram here: http://en.wikipedia.org/wiki/Raman_scattering

share|improve this answer
    
ok seems like I don't have any problem with the classical explanation, but I'm still pretty confused about the absorption process. Would u mind to elaborate? where does the phase shift comes from quantum mechanically? and how do u explain for the case of negative index of refraction? Is it due to the same effect but the decrease in phase is larger than the phase that the light gains phase by travelling between atoms. –  Emitabsorb Jul 3 '11 at 19:06
add comment

I don't believe it is generally helpful to try and analyze these things in terms of photons, so I'm going to try and point out a few things about the classical picture.

The big difficulty from the mathematical perspective is that you're working in a continuous medium where the phase of the wave is changing continuously. It makes the visualisation much easier to start off with if we restrict ourselves to a thin slab, where "thin" means small with respect to the wavelength.

We know that there is a dielectric constant which represents the tendency for charges to displace themselves in response to an external field. But how fast to the charges respond? Is it a quasi-static case, where the maximum field strength coincides with the maximum charge displacement? I think we will find that this is the case, for example, when light is travelling through glass.

Note that in this case the displacement current is leading the incident field by 90 degrees. This makes sense: as the frequency of light approaches the resonant frequency of the material, the phase lags more and more; when the phase difference goes to zero, you have resonant absorption. (EDIT: To be more clear, I choose to define the phase difference in terms of its far-field relation to the incident field!) In the case of the thin slab, you can see that the transmitted wave is the sum of the incident wave and a wave generated by the displacement current. Because you are absorbing, the phase in the far field must be opposite so that energy is removed from the incident wave.

It is instructive to do the energy balance. Let's say the displacement current generates a wave equal to 2% of the incident wave. Then the amplitude of the reflected wave is 2%, and the transmitted wave is 98%. It is easy to calculate (by squaring amplitudes) that almost 4% of the energy is missing. Where does it go? It continuously builds up the amplitude of the displacement current until the resistive losses in the material are equal to the power extracted from the incident wave.

Let's now go back to the case of the transparent medium. Take the same value for the displacement current, namely 2%. The reflected wave is the same, but the transmitted wave is different because now you are adding phasors that are at 90 degrees to each other, so the amplitude of the transmitted wave is, to the first order, unchanged.

It's the phase that's confusing. Because we are in the quasi-static regime, the phase is leading. In any case it must be leading in comparison to the absorptive case. Don't we want a lagging phase in order to slow down the wave? This is where you have to be very careful. Because we are adding a leading phase, the wave peaks occur sooner than otherwise...in other words, they are close together. This is indeed the condition for a wave to travel slower. It's all very confusing, which is why I took the case of a thin slab so the math would be simpler. Let the incident wave be

sin(kx-wt)

Then the wave generated by the slab will be

0.02*cos(kx-wt)

Note the cosine wave leads the sine wave by 90 degrees. If you draw these two waves on a graph and add them together, you can see that the peaks of the sine wave are pushed slightly to the left. This makes the wave appear slightly delayed.

The continous case is harder to do mathematically but you can see that it ought to follow by treating it as a series of slabs.

share|improve this answer
    
Actually I'm already using that point of view too, we can understand it more easily with phasor diagram. If the "new wave" is pi/2 degrees lag then if we add the old phasor with the new one, the result is like the old phasor gets rotated a little bit. So how do u explain for the case of negative index of refraction? Is it due to the same effect but the decrease in phase is larger than the phase that the light gains by travelling between atoms? –  Emitabsorb Jul 3 '11 at 19:28
add comment

In addition to everything said, I'd like to comment on the following:

We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing.

1) The speed of light does change. It's the speed of light in vacuum that doesn't.

2) A very short answer to your question: Light is, so to say, ‘larger’ than the inter-atomic space. Therefore I wouldn't speak of it travelling in the spacing.

This situation is similar to a human running through bushes as opposed to running in a forest. Because you are larger than the individual branches of a bush, you interact with the bush in a different manner than with trees in the forest.

May be an illustration with a subwavelength-diameter optical fibre could help to see the problem from another side. This fibre is thinner than the wavelength of light. For example, half a micron diameter for a 1 um light. As the light propagates through such a fibre (which can be done with basically 100 % transmission), the light field doesn't ‘fit’ into the fibre, and about half of energy propagates actually outside the fibre. However, it is not the case that the outer part of light goes faster than the one inside the fibre. The wave remains single, travelling with the speed of

$$ v = \frac{c}{n_\text{eff}}, $$

where $ n_\text{eff}$ is the effective refractive index, which is somewhere between 1 (the apparent refractive index for the outer part of the wave) and $n_\text{glass}$ (for the inner one).

So, light is ‘larger’ than the inter-atom distance, and therefore it will continuously ‘see’ the atoms.

share|improve this answer
    
I am actually interested in every wavelength, what you are saying is for long light's wavelength or small atoms spacing. In that case, I know that the photons will be scattered by atoms as a group(phonons). But for small enough wavelength, the light will only collide with one atoms at a time. And yet in practice, we didn't see any quantized the efect. I would like to hear your explanation for the case of short wavelength, a quantum mechanics explanation. –  Emitabsorb Jul 3 '11 at 19:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.