What really causes light/photons to appear slower in media?

Question

I know that if we solve the Maxwell equation, we will end up with the phase velocity of light being 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.

Photon absorption-emission theory

Some textbooks that I read explain it in a way kind of like this:
In a material the photons are absorbed by atom and then re-emitted a short time later, then they travel 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 apparent speed of light in that material. So the light appears slower because it has a smaller “drift speed”.

Interference theory

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 results in delayed light (advanced in phase), this can easily be 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 travelling 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 detailed explanation 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 occur 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

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

The electromagnetic field, travelling through a medium, induces polarization $\mathbf{P}$ and magnetization $\mathbf{M}$ of this medium. The auxiliary fields $\mathbf{D}$ and $\mathbf{H}$ appearing in the macroscopic Maxwell equations are then defined in terms of true physical quantities as $$ \mathbf{D}=\epsilon_0\mathbf{E} + \mathbf{P},\\ \mathbf{H}=\frac{1}{\mu_0}\mathbf{B}-\mathbf{M} $$ In many cases the polarization and the magnetization can be considered simply proportional to the inducing fields, which allows writing $$ \mathbf{D}=\epsilon \mathbf{E},\\ \mathbf{H}=\frac{1}{\mu}\mathbf{B} $$ Plugging this equations into the Maxwell equations (for simplicity without sources, i.e., $\rho=0,\mathbf{J}=0$) we can derive wave equation as $$ \frac{\partial^2\mathbf{E}}{\partial t^2}-\frac{1}{\epsilon\mu}\nabla^2\mathbf{E}=0 $$ where we immediately identify $$ v_{ph}^2=\frac{1}{\epsilon\mu} $$ In other words, the change in th phase velocity in the medium reflect the fact that the electromagnetic field in the medium is not the same as in vacuum, and importantly, the electric and the magnetic field scale by different factor. Note also that this is the change in the wave vector and not in the frequency (see [this answer][1]).

Remark: Note that the above equally applies to classical and quantum description of the electric and magnetic response, as long as we define the speed of photons via their wave equation. [1]: https://physics.stackexchange.com/a/661720/247642

Answer 5

I will try to give a most comprehensive physical explanation of why light slows down inside a transparent medium like glass and why this slow down in speed is only apparent and the speed of photons inside the medium remain fixed at c, the speed of light in a vacuum, travelling through the vacuum space between the atoms of the glass.

For many years physicists are troubled in physically explaining of why light slows down going from air, refractive index n1~1 to a denser transparent medium like glass, n2~1.5, but with the speed of photons propagating through the medium (glass) still at a fixed speed c?

There are many different explanations given, both classical or quantum [1][2] some are correct and others wrong but I find that they all fail to completely physically and intuitively explain the phenomenon.

Feynman characterizes this drop of the speed of light as apparent [3]:

“Our problem is to understand how the apparently slower velocity comes about”.

I will try herein to give a different physical explanation which compiles however with what we know about light and matter. Before we do that however we have to distinguish between the concepts of phase velocity and group velocity of light in a transparent medium like air, water or glass.

Figure 1 Group velocity vs. Phase velocity for a laser pulse

Simply, as shown in fig.1, the phase velocity is derived from the time it takes the front of a laser pulse (i.e. length of pulse can be few mm), to reach the destination target whereas the group velocity is the actual velocity by which each point of the laser pulse moves through space with all the points having the same speed namely the group velocity. When we refer to the speed of photons c=3x10E8 m/s in the laser pulse we refer only to the group velocity with each point of the laser pulse being a photon. When we refer to the whole laser pulse we can choose either its group velocity or its phase velocity. Keep this descriptions and definitions in mind since they will become important to explain the phenomenon later on.

Also the other point important to understand is that at the quantum scale level there is nothing rigid or solid but a complex sea of interacting fields (QFT) of electromagnetic flux and electromagnetic flux waves as field distortions similar to distortions inside water as a rough analogy.

Another point is the although the bare photon is regarded in the literature as a dimensionless point its actual physical shape when inside a condensed matter medium is dressed with a field o electromagnetic flux energy resembling strongly not a point but a small sphere. The size of the dressed photon spherical field varies according to its wavelength. Larger wavelengths translate to larger size dressed photon spheres and short wavelength to small size dressed photon spheres. In fig.2 we demonstrate and explain physically the whole phenomenon using a green laser pulse. Wavelength of photons in green laser is λ=630nm on air, represented as the perimeter size of the green spheres shown in fig.1. Laser pulse length is a few mm represented as the length of the red line in fig.1. Fig.1 is not to be scaled:

Figure 2 Photons traveling at fixed group velocity c inside the transparent medium like glass, are shank in size (i.e. wavelength) therefore the total length but also phase velocity of the laser pulse is reduced by 40% compared to the group velocity of the photons.

As best explained and illustrated in fig.2, the photons transitioning from the much less dense matter field of air to the much more dense matter field of glass as indicated accordingly by their refraction index, undergo a wavelength reduction and shrink in size by 40%. Therefore, their phase velocity and also total length of the laser pulse and its phase velocity are effectively reduced by the same amount 40%.

Using a rough analogy for example, it is like in the case of a sprinter horse (in analogy to the laser pulse) running a very relative short lane to the finish line, just after passing the starting line suddenly shrinks in size and although its speed is unchanged it has now to travel more distance until its front cuts the finish line. Therefore, its time to finish increases thus its effective phase velocity recorded has decreased compared to its actual speed running thus its group velocity.

There are two last points to cover to explain to complete our explanation of the phenomenon.

There are some views expressed by others that the photon will strongly interact with the atoms of the glass causing phenomena like scattering, absorption-reemission etc. This is however not supported by our everyday life experience. The laser beam comes straight right through, out of the glass with negligible loss in intensity and without any spreading of the beam. These have been dismissed by us but also others 4 5 . The 600 nm in size green light photon in the case of visible light shown in fig.2 does not interact with the ~100 picometers Si atoms of the glass (photon is 6,000 larger than the atom) and cuts right through them, traveling through the vacuum space between atoms at fixed speed group velocity c. A minimum giggling in the atom’s field is of course present by the passing of the photon and recorded as a moving relative small distortion in the matter field of the glass.

The next point we have to clarify is that according to the following equation:

$$V_{P}=\lambda f$$

But also what it is known in physics, that the frequency f of the photons (i.e. number of photons passing by per unit of time, thus, how intense the laser beam pulse is) of the laser pulse (see fig.1) will not be affected by the different two media shown in fig.2 and will remain unchanged. This is illustrated in fig.2, where separation distance of the photons in the air and in the glass remains the same as shown. Therefore the reduction of the total length of the laser pulse and consequently the reduction in the phase velocity $V_{P}$ are directly proportional, only to the reduction of the wavelength λ, size of the photon.

Other researchers suggest that the photon is not a massless particle anymore when it passes through a dense matter field like that of glass and becomes something different called polariton [4] which has mass and therefore would slow down its group velocity lower than c. Offered as an alternative explanation of a deformed photon (i.e. polariton) actually having a group velocity less than c. Although this quasiparticles are real in some semiconductor materials I dismiss this explanation regarding transparent matter like glass, since this would also suggest a stronger interaction of the laser beam on its path with the atoms in the glass which we don’t observe since a continuous laser beam comes out practically unchanged in its characteristics, frequency, intensity and thickness. This is very evident at 0° incident angle of the beam to the glass where there is no refraction present and photon’s electric and magnetic vectors are uniformly compressed.

Answer 6

Photons travel more slowly in different media ultimately because of conservation of energy. A photon is a fixed quantum of energy which expresses itself as an oscillating EM field travelling through space. Thus to conserve energy when travelling through a dielectric medium with a permittivity lower than the vacuum the rate of oscillation must be reduced and correspondingly the speed at which the light travels. So it is a logical constraint imposed by the nature of light as an oscillating EM field and conservation of energy.

No absorption and emission is involved - the same photons continue through the medium.

Answer 7

The electric field of the photon couples weakly to the dielectric field of the constituent molecules of a medium. This weak interaction drags the photon giving it an effective mass m* and a slower speed v as it travels through it.High frequency photons have stronger electric fields than low frequency photons and therefore the coupling is much stronger for high frequency photons. Momentum is conserved since hk/2pi= m*v upon entering or leaving the medium.

Answer 8

A very simple answer: the photon is absorbed by the constituents of the medium (which is not continuous), after which it's reemitted for the major part in the same direction.

This process costs time, so the effective speed of light in the medium is reduced, while the speed of light between the absorption and the reemission stays equal to the speed of light in vacuum.

The alternative is NOT true!

See also here on Wikipedia.

Answer 9

I see that the answers do not cover a few things.

1. Speed of light is always c in empty space when measured locally. If you measure the speed of light next to the sun (viewed from Earth), you will see it is less then c, that is caused by the Shappiro effect, which consists of basically two things. #1 clocks run slower next to the sun (because the sun's gravity) , viewed from the earth, so (speed=distance/time) speed will become slower, because you divide distance by a larger amount of time (you divide by the time elapsed on your clock here on Earth, which ticks faster). #2 the distance light travels becomes a little bit shorter (viewed from Earth) because the spacetime continuum becomes bent by the sun's gravity, so you divide a shorter distance by a larger time, so you get a speed that is less then c.

2. The speed of light (EM wave, which consists of photons) is always c when measured locally in empty space. If you measure the speed of the EM wave inside a material, even air, you will get a speed that is less then c.

3. The speed of photons is always c when measured locally. The EM wave consisting of photons, will however have a speed that is less then c inside a medium. That is because the EM wavefront will itself slow down in medium. Photons on the other hand, will travel in space inside the medium with speed c always. They will either travel in empty space inside the medium or get absorbed and re-emitted. The electron field around the atoms nucleus will absorb and re-emit the photons, that is the interaction photons will have with the medium, other than that, the photons will travel in empty space.

4. Now in terms of QM, the absorption and re-emission itself is instantaneous. Why is the wavefront itself then slowing down in medium?

5. It is the EM interaction that needs time. For the H atom it is on average 10^-8 sec. The EM interaction itself is the absorption and the re-emission (and the electron field's excited state, and the return to the normal state).

6. How do you measure the speed of the EM wave inside the medium? You measure the time elapsed between when the first photons of the EM wave (the wavefront) enters the medium, and when the first photons (the wavefront) exits the medium.

7. Because the EM interaction needs time, the denser the medium is, the more the EM wave will slow down. Because the denser the medium is, the more atoms per a certain thickness of the medium there are, and the more EM interaction the photons will have to make to get through the medium. The more interaction, the more time the EM wavefront will need to travel.

So your question, why the phase velocity of light will be slower, can be answered by clearing up that :

1. the slowing is in terms of the EM wavefront only, and because we measure the EM wave's speed by measuring the time between when the wavefront enters and exits the medium.

2. Photons will not slow down. Because we measure photons speed when they are traveling. We do not count the time when they are transformed into energy in the electron field (between absorption and re-emission). And although the absorption and re-emission is instantaneous, the EM interaction still needs time. But we do not count that time into the calculation of the photon's speed.

3. When we measure the EM wave's speed as a herd of photons, we do count the time when the herd of photons are transformed into energy in the electron fields.

4. That is the difference between the speed of the herd (and the wavefront) and the individual photons.

So the question you are asking,

"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"

The answer is that it is all three. They aren't the same photons. There is a phase difference. And there is the time that the EM interaction needs.