# Is the speed of light in all media independent of reference frame?

We know that, the speed of light in vacuum is independent of reference frame. I read the reason for this fact from Why and how is the speed of light in vacuum constant, i.e., independent of reference frame?. Similarly, is the speed of light in a medium of refractive index $$n \neq 1$$ also independent of reference frame?

I tried to frame my question visually. So, let us consider the following image:

The entire setup is in a medium of refractive index not equal to unity. The light source is at rest with respect to the medium. The violet coloured squares below the light beam numbered $$1,2,$$ and $$3$$ are speed detectors which we can use to determine the speed of light beam with respect to the reference frame attached to them. $$1$$ is at rest with respect to the light source. $$2$$ moves parallel to the light beam and towards the source with a speed of $$v$$ and $$3$$ moves away with the same speed. Now my question is, is the speed of light beam detected by the three detectors equal? If it's vacuum we know they are equal. But what happens in this case?

Image Courtesy: My Own Work :)

• I looked for a similar question within the linked questions to the one mentioned in my question using this but I couldn't find one. If there exists a similar question, kindly provide the links for them. – Guru Vishnu Dec 11 '19 at 7:16
• All of physics is independent of (inertial) reference frame. – WillO Dec 11 '19 at 7:18
• If the speed of light in (say) water were different in different frames, you could measure it in your own frame to determine whether you were in motion. But no experiment can reveal that you're in motion. – WillO Dec 11 '19 at 7:21
• The speed could be different when travelling relative to water without violating the postulates of relativity – bemjanim Dec 11 '19 at 7:43
• @WillO But the speed of light in a medium is not a fundamental quantity in physics like the speed of light in a vacuum is. As Qmechanic states in their answer, the frame where the medium is at rest is a "preferred" reference frame. You can tell whether you're moving relative to the medium by measuring the speed of light in that medium (possibly in multiple directions). This doesn't break the equivalence principle, though, because you still can't tell whether, when you're in the rest frame of the medium, both you and the medium are moving in any absolute sense. – probably_someone Dec 11 '19 at 11:38

Let's use the Lorentz transformation to calculate this. Let S be a reference frame, in which a coordinate system $$(t,x)$$ is used. Let S$$'$$ be a frame moving in the $$x$$ direction relative to S at speed $$v$$. The Lorentz transformation asserts that if event A is at $$(t,x)$$ then its coordinates in S$$'$$ are given by $$\begin{eqnarray} t' &=& \gamma(t - v x / c^2) \\ x' &=& \gamma(x - v t) \end{eqnarray}$$ where $$\gamma = (1-v^2/c^2)^{-1/2}$$. Now consider a pulse of light that starts from $$(0,0)$$ and propagates at the speed $$c/n$$ relative to S. For example, there could be some water at rest in S, and $$n$$ is the refractive index of this water. At time $$t$$ in S, such a pulse will have reached the event $$(t,x) = (t, ct/n)$$. The coordinates of this event relative to S$$'$$ are given by $$\begin{eqnarray} t' &=& \gamma(t - v (ct/n) / c^2) \\ x' &=& \gamma((ct/n) - v t) \end{eqnarray}$$ Now the coordinates of the starting point are $$(0,0)$$ in both frames, so we can find the speed of this pulse relative to S$$'$$ by using the distance traveled divided by the time elapsed: $$\mbox{speed relative to S}' = \frac{x' - 0}{t' - 0} = \frac{c/n - v}{1 - v (c/n) / c^2} = \frac{c/n - v}{1 - v / n c}.$$ If you are familiar with the formula for addition of velocities, you could find this same result by applying it. We now find that when $$n=1$$ the speed relative to S$$'$$ is equal to $$c$$, but when $$n \ne 1$$ the speed relative to S$$'$$ is not equal to $$c/n$$.

The above formula gives the speed that will be measured by your detector number 3, if we take it that the detector works in the usual way by measuring distances and times in its own rest frame. The result for detector number 2 will be $$\frac{c/n + v}{1 + v / n c}.$$

Another question that arises is the speed of the light relative to the water. That is just $$c/n$$. As soon as one says "relative to the water" then to calculate it you must use the rest frame of the water. End of story. But someone might ask instead, what is the rate of change of the distance between the light pulse and something floating in the water? If, relative to some given frame, the water is flowing at speed $$w$$ and the light is moving at speed $$u$$, then the answer to this question is $$u-w$$.

• This effect was discovered experimentally in 1851 and was among the earliest evidences for Einstein's relativity, even though this was half a century before Einstein's theory: en.wikipedia.org/wiki/Fizeau_experiment – Mark H Dec 12 '19 at 7:52

The media itself constitutes a preferred reference frame in which the speed of light is $$v=c/n$$. The speed of light in a reference frame that moves relative to the media will change anisotropically.

• Thank you for your answer. Could you please explain this statement "will change anisotropically"? Does it mean, in my (edited) question, the speed of light measured by the three detectors are different and is dependant on the speed and direction of motion of the detectors? – Guru Vishnu Dec 11 '19 at 11:29
• The speed and reference frame of the light source are irrelevant. – Qmechanic Dec 11 '19 at 13:20
• So, all the three detectors will measure the same speed of light which is unique for the medium. Could you please clarify this? – Guru Vishnu Dec 11 '19 at 13:23
• The speed of light depends on the motion of the detector relative to the media. – Qmechanic Dec 11 '19 at 13:28

The formula for adding velocities is $$\frac {v_1 + v_2}{1+\frac{v_1v_2}{c^2}}$$. Plugging $$c$$ in for $$v_1$$ yields $$c$$ for any $$v_2$$. Plugging in a value of $$v_1$$ other than $$c$$ and a nonzero value for $$v_2$$ results in a value other than $$v_1$$.

In other words, anything observed to be travelling at $$c$$ in one reference frame will be observed to be travelling at $$c$$ in all reference frames. Anything observed to be travelling at a velocity other than $$c$$ in one reference frame will be observed to be travelling at other velocities (and by "other", I mean different from its observed velocity in the first reference frame, not just different from $$c$$) in other reference frames.

In particular, anything travelling at a velocity other than $$c$$ has a rest reference frame in which it will be observed to have zero velocity. If Detector 3 in your diagram is travelling at $$\frac c n$$, then the light wave will appear to be a standing wave.

• +1: Thank you for your answer. Actually detectors 2 and 3 in my question are travelling at a different speed $v$ which is much less than the speed of light in the medium. I understand, if the third detector travels with the speed of light in the medium then the photons appear to be at rest (if at all we are able to see them) as mentioned in your third paragraph. But here it's not that case, so, I think you may have to make a slight modification to your last paragraph. – Guru Vishnu Dec 12 '19 at 6:27
• @M.GuruVishnu The fact that you are not imagining it traveling at $\frac c n$ is irrelevant to my point. This shows that there are reference frames where the light travels at a velocity other than $/frac c n$. This then shows that the velocity is not independent of reference frame. Which means that the answer to the question in the title is "no". – Acccumulation Dec 12 '19 at 6:30
• Fine. I understood your point. Actually, I neglected that case because, I thought travelling with speed equal to that of light even in a denser medium is too large to be approachable. – Guru Vishnu Dec 12 '19 at 6:37
• @M.GuruVishnu There are media with extremely high n, to the point that it is possible for someone to travel at $\frac c/n$. – Acccumulation Dec 12 '19 at 6:40
• Thanks for letting me know that. This article says the slowest speed attained is $17$m/s. That's low enough to be easily attainable. – Guru Vishnu Dec 12 '19 at 6:44

The speed of light only differs when comparing it in two mediums with a different refraction index. With the added changes we can say that the detector with $$v=v+$$ will detect the light as going $$-v$$ slower than $$c$$ and vice versa.

The other answers give some good results how to arrive at the correct result from the reference frame of the water: We just relativistically add the two speeds to see that our speed of light measured changes with the detector movement.

However, relativity means the same result (up to time dilation, obviously) should come out when we look at it from the detector's reference frame. So what does that look like?

We have our light pulse, starting at some point away from the detector, then moving towards it until it reaches it and is registered (and its speed measured somehow). While doing so, it moves through a refractive medium (the water) that itself moves relative to the reference frame.

Therefore, the water (from the perspective of the detector) undergoes Lorentz contraction, meaning it becomes "shorter" in the direction of movement. For a flow of water, this means it becomes denser: The molecules along the direction of flow appear squashed and tighter together.

While the exact processes of how the interaction of light in matter results in a refractive index are complicated, it suffices for our explanation that it's dependent on density: If there's more matter in a volume to interact with, the light will be slowed down more. Thus, what changes from the perspective of the detector is $$n$$: Moving water has a different refractive index compared to water at rest.