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I know a lot of people already asked about this, but i still can't figure out why it's impossible to measure the one-way speed of light. Sorry if it's a dumb question, but I'd really like understand.

Let's say we have to clocks, both at 0.00s, or off (so we don't need to sync them) (1).

We fire a beam of light, so the clock 1 sensor is triggered, the clock 2 is still at 0.00 (2).

The beam of light reaches the second clock and triggers it (3).

Now if we subtract, we have clock1 - clock2 = t (time for light to travel d meters), so c = d / t. Of course, this is limited by the precision of the sensors, but fundamentally, why wouldn't this work?

One-way speed of light experiment

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    $\begingroup$ This question gets asked a lot. As a result, I like to ask questioners why they care about it. It obviously is not a obstruction to the usefulness of SR. So, why does this question matter to you? $\endgroup$
    – m4r35n357
    Commented Sep 28, 2021 at 9:32
  • $\begingroup$ How do you get the synchronised readings on the two clocks at step 3? The light beam starts the two clocks, but you don't mention anything stopping the clocks. $\endgroup$
    – PM 2Ring
    Commented Sep 28, 2021 at 13:48
  • $\begingroup$ Like everything in sciences or maths, it's all about the will to understand things. The fact that light travels at 2c in one direction and goes infinitely fast in the other does not matters at all. I just saw that veritasium video (like everyone else asking about this), and wondered about it. That's how great discoveries are made ; by doubting and wondering "what if"? @m4r35n357 $\endgroup$
    – FoxYou
    Commented Sep 28, 2021 at 17:36
  • $\begingroup$ If we simply put a button in the middle of the clocks, and send an electric signal to stop them simultaneously, wouldn't it work? @PM2Ring $\endgroup$
    – FoxYou
    Commented Sep 28, 2021 at 19:17
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    $\begingroup$ How do you know the speed of the electrical signal in each direction? Relativity says that any signal travels at a speed $\le c$, so this one-way speed of light issue affects all possible signals, not just light beams. $\endgroup$
    – PM 2Ring
    Commented Sep 28, 2021 at 19:39

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Of course, this is limited by the precision of the sensors, but fundamentally, why wouldn't this work?

In your diagram you have 4.33 on one clock at the same time as 0.8 on the other. The only way to know that 4.33 goes with 0.8 and not some other time is to assume a simultaneity convention. That assumption defines the one way speed of light, not the experiment.

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  • $\begingroup$ What do you mean by "simultaneity convention" ? The difference between the two clocks will always remain the same, so, by looking at the clocks, we can compare them and deduce the difference. @Dale $\endgroup$
    – FoxYou
    Commented Sep 27, 2021 at 19:31
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    $\begingroup$ @FoxYou whenever you have spatially separated things you need a simultaneity convention to say this event here happened at the same time as that event over there. Looking at the clocks won’t help because what you see is delayed by the speed of light. $\endgroup$
    – Dale
    Commented Sep 27, 2021 at 19:39
  • $\begingroup$ So the problem is to get the measurements, not to make them at all? The fact that what i see is delayed by the speed of light doesn't seem to be an issue to me, since the difference between the clocks remain the same. What am I missing? @Dale $\endgroup$
    – FoxYou
    Commented Sep 27, 2021 at 19:56
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    $\begingroup$ @FoxYou The fact that the difference remains constant is irrelevant. To do the comparison you need to know what the difference is, not just that it is constant. To know that requires a simultaneity convention. There simply is no way around it. If you wish to have a back and forth discussion then I recommend going to a forum that encourages discussion, such as physicsforums.com. Here they strongly discourage discussions in the comments. $\endgroup$
    – Dale
    Commented Sep 27, 2021 at 20:06
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    $\begingroup$ @FoxYou You might like to take a look at an earlier answer by Dale on this topic: physics.stackexchange.com/a/591436 $\endgroup$
    – PM 2Ring
    Commented Sep 28, 2021 at 13:42

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