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I was lightly tickled by Veritasium video Why no one has measured the speed of light in which the author says that a one-way measurement of speed of light is impossible (or, rather, that $c$ could vary depending on direction).

So riddle me this:

You have a mirror (M), light source (LS) and light detector (LD) at a certain distance apart (either d/2 or d).

M ←d/2→ LS ←d→ LD

Let's periodically shoot two photons from the light source in two directions:

  1. one towards the detector
  2. one towards the mirror

In first case, it travels distance 'd' (LS to LD). In second case it travels distance 2d (LS to M, M to LS, LS to LD).

If lightspeed is isotropic, you will see the photons that traveled directly arrive with some frequency, and the others bouncing off the mirror with half of that frequency. If it isn't isotropic, they will de-sync (since it traveled 0.5d in one direction and 1.5d in the other).

Measurement is done by a single device at a single place, no sync needed. No moving parts either.

So what am I missing?

Dale
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Wejn
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    well isn't michelson moorley experiment exactly proves that? – physshyp Oct 31 '20 at 23:33
  • Related question about the same video: https://physics.stackexchange.com/questions/590825/on-measuring-velocity-of-light-in-vacuum – G. Smith Oct 31 '20 at 23:40
  • Also https://physics.stackexchange.com/q/590790/123208 and a related question on Astronomy: https://astronomy.stackexchange.com/q/39626/16685 – PM 2Ring Nov 01 '20 at 00:03
  • so what about this https://www.youtube.com/watch?v=EtsXgODHMWk – physshyp Nov 01 '20 at 06:23
  • Ok, what I'm missing is that the round-trip to the mirror and back has no chance to move the photon out of sync, because the delta in speed will cancel out. – Wejn Nov 01 '20 at 08:07
  • So originally I modified this question based on feedback, because I realized my mistake (frequency stays the same, but phase of the two streams of photons would maybe change). But apparently doing edits like that is against site policy, so it was reverted. As such, I'm dropping this question, because I don't want to spam this site further. – Wejn Nov 01 '20 at 15:36

1 Answers1

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If lightspeed is constant in all directions, you will see the photons that traveled directly arrive with some frequency, and the others bouncing off the mirror with half of that frequency. If it isn't constant in all directions, they will de-sync (since it traveled 0.5d in one direction and 1.5d in the other).

This is incorrect. Regardless of the path of the signal and the speed of the signal as long as the geometry and speed are constant the frequency of arrival will be constant. If the emitter and receiver are at the same gravitational potential then the frequency received will be equal to the frequency emitted, again regardless of the speed of propagation.

Dale
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  • the energy levels depend on $c$, or on $\epsilon_0$, does this change when c is not a constant?, or are they a function of the averaged $c$? –  Nov 01 '20 at 00:54
  • @Wolphram jonny I assume you mean e.g. the energy levels for atomic transitions. The measured energy levels depend on the fine structure constant, not c. The fine structure constant is isotropic, so that would not change even with a variable c. I don’t know if the expression for the fine structure constant in terms of c would change or if one of the other constants would also become anisotropic – Dale Nov 01 '20 at 03:08