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The speed of light is the same in all frames of reference. But imagine a scenario in which you passed by a ship which had (B=.78), while your ship had (B=.94). While somehow being able to look through the window of the ship you were passing, you see someone pointing a green laser into a fish tank, and of course the light refracts. Solving for the new angle is simple,(a bit of linear algebra, using the Lorentz matrix with angular components), but we all know that the speed of that light changes after it was refracted. So, would I measure the same velocity of that light as someone who is in the rest frame of the refraction incidence? I mean it is light, so all observers should measure it moving the same speed, correct? But, my gut tells me that I would clearly have to add the velocities using the velocity component Lorentz matrix. Where is the kicker?

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All these dilemmas with different speed of light as observed by different observers are not real. They never occur in reality. They are only imaginary and can never be tested experimentally.

Observers in two different inertial frames cannot see/measure the same ray of light. This is physically impossible, for in order for you to see the light it must come exactly to your eye (measuring device). Light "exists" only for the one to whom it comes directly. If light is "passing you by", you cannot see it.

Therefore, two different observers will never actually measure different speeds of the same light. If one will measure it, the other will not be able to.

If I understand your set-up correctly, the person being in the other spaceship (not the one where the tank is located) will see only the light sent to him from the tank through vacuum separating the two spaceships. Therefore, as the speed of light in vacuum is $c$, than that's exactly what he will measure. (And yet, the person in the spaceship with the tank in it will never see light sent to the other rocket, so he will not be able to make any direct comparisons.)

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  • $\begingroup$ Kick ass conclusion, thank you. I never thought about it like that. $\endgroup$ – Gödel May 20 '14 at 7:57
  • $\begingroup$ You are absolutely excused. The problem is that thousands of highly qualified people never thought about that. So if you want to give that answer to the teacher at your college, you better get prepared for lots of counter-arguments that you might not be able to answer right away due to the lack of experience, like: "OK, but we can install a lot of devices along the path of the light that would split the ray of light in two, therefore sending "samples" to both observers." Or something of the kind. $\endgroup$ – bright magus May 20 '14 at 8:09
  • $\begingroup$ If you are interested about (horrendous) misconceptions in SR, which even physics professors are not immune to, see my answer here (especially the EDIT): physics.stackexchange.com/questions/2554/… $\endgroup$ – bright magus May 22 '14 at 10:11
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You need to distinguish between being able to see and/or measure what happens in different frames and using the Lorentz transformations to tell you what happens in different frames. If you use the Lorentz transformations you'll find that the speed of light in the fish tank is not the same for both sets of observers. In effect the refractive index will be different in the two frames.

A closely related question is How does wind speed affect the velocity of light?. Substitute water for air and it's the same as your question. Another related question is Does the speed of medium affect the path of light?.

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  • $\begingroup$ "If you use the Lorentz transformations you'll find that the speed of light in the fish tank is not the same for both sets of observers." Sure, but the question was about possible measurement (observation) discrepancies, and not about calculation. Calculation will never produce any contradictions in this case (other than those inherent in the theory, i.e. the clock paradox), because nobody will be able to verify it experimentally (in a direct manner). $\endgroup$ – bright magus May 20 '14 at 12:11
  • $\begingroup$ @brightmagus: you could do the measurement using kit like this. Just have a flow of water through the bottle at some known velocity. I'll certainly concede there would be formidable experimental problems, but it could be done in principle. $\endgroup$ – John Rennie May 20 '14 at 12:23
  • $\begingroup$ It's a common misconception. You cannot see the light as it goes all the way through the bottle. You can only see the photons that were redirected to you after having traveled a part of the bottle unseen by you. So the photons you see are those being sent to you after they traveled only locally. If you send a single photon within the bottle (and it makes it through) you will only be able to see it, if it is redirected to you after an entirely local travel inside the bottle. It would be equivalent to not sending a photon through the bottle at all, ... $\endgroup$ – bright magus May 20 '14 at 12:55
  • $\begingroup$ but only sending you a single photon from the end of the bottle after calculating time necessary for the travel inside bottle. It would be like telling you: "Locally the time of the travel measured was t". To which you can apply transforms. Because if you are able to see the photon that made the whole travel it would be only after it completed this travel, but not seen on its way. The same logic applies to all the photons redirected to you on the way that you can see in this video. Each and every of them travels within the bottle unseen, and only after its redirected it becomes visible to you $\endgroup$ – bright magus May 20 '14 at 12:55
  • $\begingroup$ So obviously in your bottle experiment if you measure the velocity of every single photon, the measured speed will be specific for air through which it reaches the measuring device, and not to the water in the bottle. But if you somehow measure intervals between consecutive photons reaching your device from specific points within the bottle, it will be exactly the same intervals as measured locally (plus Lorentz transform for time applied if you are moving relative to the bottle). Effectively you are receiving information how the speed is measured locally and not making independent measurement $\endgroup$ – bright magus May 20 '14 at 13:19

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