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Ok, this is (probably) my last question about this topic. My other two (here and here) had some misguided assumptions, but I think this one is better. (There is probably something else I am neglecting, but I'll give it a shot anyway.)

Set up a rotating apparatus with two toothed wheels, one attached to each end of a rod. (You would know the distance between the wheels because it is just the length of the rod.) Then, you rotate the apparatus, let's say, counterclockwise. You would then offset the rotation of the wheel farthest from the light source by some known angle (clockwise) so that the light has a possibility of traveling through both wheels (as the light travels from the closer to the farther wheel, the farther wheel would "catch up" and the light would pass through it.) Since they are both attached to the same rod, I don't think you would need to worry about clock synchronization because both wheels would have to spin at the same rate. Then, you could adjust the rotation rate of the apparatus until the light passes through both wheels, at which point you would be able to tell how long it took for the light to pass between the two wheels, which would allow you to calculate the one-way speed of light. (I'm not sure if the location at which the torque is applied to the rod matters, but, if you were to generate the rotation of the apparatus [apply the torque to it] at the center of the rod, I don't think clock synchronization would matter here.)

As a little diagram:

SOURCE --- WHEEL #1 --- ROD --- WHEEL #2 --- DETECTOR

As with my other questions, I'm guessing there is something I haven't thought of. Please let me know what that thing is (since there most likely is one). Thank you!

EDIT: Also, I think that applying the torque to the rod at the center of the rod might be super important here. Since it takes time for the "signal" that the rod is rotating to travel to each end of the rod, having the torque in the middle would mean that the "signal" would reach both ends at the same time.

EDIT #2: The math of it all (or at least what I believe is the correct math):

$$v_{\text{one-way}}=\frac{\Delta x}{\Delta t}$$ $$\omega_{\text{rod}}=\omega_{\text{w1}}=\omega_{\text{w2}}=\frac{\Delta \theta}{\Delta t}\text{ so }\Delta t=\frac{\Delta \theta}{\omega_{\text{rod}}}$$ $$v_{\text{one-way}}=\frac{\Delta x}{\frac{\Delta \theta}{\omega_{\text{rod}}}}=\frac{\omega_{\text{rod}}\Delta x}{\Delta \theta}$$ $$\omega_{\text{rod}}=\frac{2\pi}{T}$$ $$v_{\text{one-way}}=\frac{2\pi\Delta x}{T\Delta \theta}$$

where $T$ is the period of rotation for the rod/wheels, $\Delta x$ is the length of the rod (which is also the distance between the wheels), and $\Delta \theta$ is the angle that you offset the farther rod (this is the amount that the farther wheel would then have to rotate during the time $ \Delta t$, which is the amount of time it would take for the light to pass between the two wheels).

EDIT #3:

I wasn't clear when I was talking about applying the torque to the rod. I meant to apply the torque before the experiment, not during.

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  • $\begingroup$ @SolomonSlow I agree. I don't think I was clear when I said to apply a torque. What I meant was to apply the torque before the experiment, not to apply the torque while performing the experiment. Thanks for pointing that out. I'll edit it now. $\endgroup$
    – user276997
    Nov 3, 2020 at 21:48
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    $\begingroup$ In case you didn't see it already, you should check out what John Norton has to say on this topic, which I linked in this comment. Norton is a physicist who is also a science historian & philosopher, specialising in relativity. $\endgroup$
    – PM 2Ring
    Nov 4, 2020 at 3:01
  • $\begingroup$ @PM2Ring Thanks a lot for the link. I did CMD + F to find "one-way," and I read through the parts where he talks about it. It was quite interesting; he basically just said there are no physical changes that different one-way speeds would cause, so it doesn't really matter. He said it was just a convention. If you want, you can post this as an answer (maybe pull some specific quotes) and I'd be glad to accept it! Thank you! $\endgroup$
    – user276997
    Nov 4, 2020 at 4:27
  • $\begingroup$ "Since they are both attached to the same rod" - ah, but rods can bend. In fact rods can only transmit information at the speed of sound, which is even slower than the speed of light. When you start turning one end, there's a speed-of-sound delay before the other end starts turning. $\endgroup$ Nov 5, 2020 at 10:06
  • $\begingroup$ And if the speed of light is faster in one direction, so is the speed of sound, because sound is based on electromagnetic forces. $\endgroup$ Nov 5, 2020 at 10:07

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This question doesn't mention the Veritasium video, but I'll assume it's about that video since the others were. As I said in another answer, Veritasium's variable-speed-of-light model is just standard special relativity in different coordinates. The variant in which the speed of light ranges from $c/2$ to $\infty$ is special relativity in coordinates like $(t',x,y,z)$ where $t'=t-x/c$ and $(t,x,y,z)$ are standard inertial coordinates.

The easiest way to analyze any experiment in the variable-speed coordinates is to analyze it in the standard coordinates and then transform. I'll assume your rod is parallel to the $x$ axis. With respect to standard coordinates, a symmetry argument shows that the rod should have no net twist at any $t$ in its equilibrium rotating configuration (if the torque is applied at the center, or if it's rotating without torque and without friction). Therefore, it is twisted at every $t'$. The amount of the twist is $ω_\text{rod}Δx/c$. This is just enough such that the light moving at $c/2$ or $\infty$ will make it through.

Note that the rod is also twisted with respect to an isotropic inertial frame that is moving in the $x$ direction with respect to the lab frame, due to the relativity of simultaneity, and this "explains" why the light makes it through in both directions in that frame even though the speed of the light relative to the apparatus is $c\pm v$ in that frame. Whether the primed coordinates are inertial or not doesn't matter.

Explaining the twist more directly in either of these primed coordinate systems would be rather difficult. It comes down to the fact that the rigidity of the rod is enforced by electromagnetic interactions between atoms, and those interactions are effectively anisotropic in the primed frames. But what I want to stress is that you don't need to do this analysis. The fact that you'll get the same answer in any coordinates is not a physical assumption; it's purely a statement about the internal consistency of the mathematics.

(In the case of inertial frames, it is a nontrivial physical assumption that the laws of physics are the same with respect to any of them, but that assumption is already built into the physical laws that you're using to do this calculation. You aren't performing an independent test of the assumptions of the theory by calculating the shape of the rod in the primed frame. It's constructed such that the answer must be consistent.)

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  • $\begingroup$ I'm sorry, but I don't understand what your answer is explaining. I am comfortable with special relativity (I was never "formally" taught it, but I have done fairly extensive online research about it), and I understand that the laws of physics hold in all inertial frames. Still, I don't understand why you are talking about the "twist" (nor do I know what you mean by "twist" -- torque? angular velocity?). Would you mind explaining further? Also, when you are talking about a frame moving in the x-direction, how does the speed of light move at $c\pm v$ in that frame? Wouldn't it just be $c$? $\endgroup$
    – user276997
    Nov 4, 2020 at 1:19

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