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I've been interested in a thought experiment that a friend proposed to me. If we take a huge block of glass through which a beam of light passes, it results that the velocity of the beam in the medium is slower following the ratio of the refractive index $n= \frac{c}{v}$ due to the interaction of light with the material.

Then if my friend runs next to the block at a speed of $v=\frac{c}{n}$ (which is a valid velocity since it is slower than c, how does he see the beam of light? In his inertial frame of reference the light stays static? By my basic knowledge of special relativity, it is clear that light moves at the same speed regardless of the inertial reference system in which we are immersed, so I haven't found the way to solve this problem. My best guess is that he still sees the beam at the speed of c but I don't know how to justify my answer.

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This experiment must be done in deep, deep space!

Be aware that the disturbance is not light, but a quantum superposition of light states and excited matter states. There is no problem with the disturbance as a whole moving at a velocity other than $c$, although freespace light states always move at $c$.

Your friend would see:

  1. A stationary region wherein the medium is in an excited state
  2. The medium is anisotropic, becoming optically denser along the direction of motion, with both $\epsilon$ and $\mu$ tensors reflecting this anisotropy.
  3. The medium also picks up, from your friend's frame of reference, further to $\epsilon$ and $\mu$, a nonzero magnetoelectric tensor coupling the electric field with magnetic induction and the magnetic field with electric displacement and proportional to $v\times$ (i.e. the matrix corresponding to the cross product with $v$ - I can't find a reference right now woth the exact proportionality constants). The "cross coupling" is the same in both directions.

It is 3. in combination with effect 2. that actually "stops" the pulse. A finite rapidity Lorentz transformation cannot yield infinite refractive index.

Alternatively to (2) and (3), and less theoretically awkwardly, (but practically tedious) you would work out the full details as follows. One defines a rank four susceptibility tensor $\chi$ mapping the Faraday tensor:

$$F=\left(\begin{array}{cccc}0&E_x&E_y&E_z\\-E_x&0&-B_z&B_y\\-E_y&B_z&0&-B_x\\-E_z&-B_y&B_x&0\end{array}\right)$$

to the displacement tensor:

$$G=\left(\begin{array}{cccc}0&D_x&D_y&D_z\\-D_x&0&-H_z&H_y\\-D_y&H_z&0&-H_x\\-D_z&-H_y&H_x&0\end{array}\right)$$

as $G_{\mu\,\nu}=\chi_{\mu\,\nu}^{\alpha\,\beta}\,F_{\alpha\,\beta}$ and then reads points 2. and 3. above from the usual transformation of $\chi$

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Well for a start, in your reference frame you are already seeing the light travel at a speed less than $c$, so you shouldn't expect your friend to see the beam at $c$.

The crux of this question is "How is light slowing down?" Answer: it's not. That is, there is never a photon which travels at a speed less than $c$. Hence if you were to look at any individual photon, it would have the speed $c$. However, as the photon travels through the material, it is able to interact with the material. It could for example be absorbed, and re-emitted, with a small delay between the two events. It's this small delay that makes it appear as though light is travelling slower. Really the beam travels at $c$, but it occasionally makes pit-stops along the way.

An analogy would be, if I could run at some velocity $v$, along a street with a a length $L$. But let's say there are donut shops along this street, and everytime I come across a donut shop, I am compelled to go in, and buy a donut, before resuming running at $v$. Then if it takes me $\tau$ time in a donut shop, and there is a density of donut shops $\sigma$, my travel time will now be $L(\sigma\tau +1/v)$. So although I never travel at a speed other than $v$, it's taking me a longer time to get along the street, giving me an effective speed of $$v_{\text{eff}}=\frac{1}{\sigma\tau +1/v}=\frac{v}{v\sigma\tau + 1},$$ where you can see the effective speed must be smaller, as $\sigma\tau > 0$. Identifying ${v \sigma \tau + 1}=n$, you can compare this to the case for light.

Returning to your question now, if your friend saw any photon, it would have a speed $c$. However by moving at the effective speed, he would be see the regions of the material that were absorbing and re-emitting as stationary. This is sort of like saying every time I went into a donut shop, he would be there beside the shop, but when I got out, I would travel at my unimpeded speed.$^{12}$ Then when I was in the next donut shop, he would catch up, until he was right there beside me again.

$^1$Donut running analogy does break down a little, as he running next to me of course wouldn't measure me as travelling at $v$, because my velocity is not the same in all reference frames, but he could tell when I was travelling at my unimpeded velocity, $v$.

$^2$Note in the material, the refractive index is a statistical treatment of a many d.o.f. system.

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a short non-mathematical answer would be that what we call "the speed of light" used in relativity is a mathematical constant, and not necessarily the apparent speed of your beam (but the individual speed of individual photons is c).

As far as relativistic calculations go, you can do as if the beam of light was just an object moving at speed $v=\frac{c}{n}$

but of course when you go into details the beam is not a singular objects: photons are constantly being absorbed and reemited, and travelling at the speed of light between those events.

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