1
$\begingroup$

I've just watched Why No One Has Measured The Speed Of Light by Veritasium, but I haven't really understood why this debate exists.

The reason for my misunderstanding, I believe, is that the whole video hinges on examples of light moving between 2 points, i.e. making the way forward and the way along the single line connecting the two points.

So ok, I get that that means that a possible alternative to $c_{forward} = c_{backward} = c$ is $c_{forward} = c/2 \,\,\wedge\,\, c_{backward} = \infty$.

But what if make light travel along a square, with 45° mirrors? If two sides are in the same absolute direction as before, then the speed of light will be $c/2$ and $\infty$ along those sides. $$ c = \frac{4}{\frac{1}{c'} + \frac{1}{c''} + \frac{1}{c'''} + \frac{1}{c''''}} = \frac{4}{\frac{2}{c} + \frac{1}{c''} + \frac{1}{\infty} + \frac{1}{c''''}} $$

Which leaves the speed of light unknown across the other two sides, as long as $1/c'' + 1/c'''' = 2/c$

But so we have that same problem again, just along two directions which are not that of light going away or towards me but... from left of me or from right?? (I'm imagining myself ortogonal to the plane of the screen.)

                       c/2
                 /----------->\
                 ^            |
                 |            |
    me      c''''|            |c''
                 |            |
                 |            v
                 \<-----------/
            ∞

Sure a solution would be $c'' = \infty\,\,\wedge\,\,c'''' = c/2$, but $c'' = c/2\,\,\wedge\,\,c'''' = \infty$ would be a solution as well, and why should I prefer one to the other, given the symmetry?


Is the trick in that none of these 4 sides is on the line of my sight, hence light will... vary speed along the sides in a symmetric way such that all fits with $\frac{1}{c'} + \frac{1}{c''} + \frac{1}{c'''} + \frac{1}{c''''} = \frac{4}{c}$?

$\endgroup$
2
  • $\begingroup$ Please read Dale's excellent answer about the Einstein synchronisation convention, physics.stackexchange.com/a/591436/123208 $\endgroup$
    – PM 2Ring
    Sep 17, 2023 at 11:13
  • 2
    $\begingroup$ "I haven't really understood why this debate exists." - There is no debate; there is just a popular YouTuber who made a video about a subject that he doesn't really understand. The [one-way-speed-of-light] tag was created to organize the questions that showed up here in the wake of that video's publication. It didn't exist before. $\endgroup$
    – benrg
    Sep 17, 2023 at 18:18

2 Answers 2

2
$\begingroup$

This question is very closely related to this, that, and the other questions as well as all of the questions in the one-way-speed-of-light tag.

However, I think that the focus in this question on the OWSOL in multiple spatial dimensions is something that has not been covered explicitly in previous answers that I am aware of. I will focus on that aspect as the rest has already been thoroughly discussed elsewhere here.

Anderson deals with this concept explicitly in his paper Conventionality of synchronisation, gauge dependence and test theories of relativity. In that paper he begins with a fairly typical (although high quality) treatment in a single spatial dimension. Then in section 2.3 he extends his analysis to multiple spatial dimensions. He promotes his synchrony parameter $\kappa$ to a vector field $\vec \kappa$ which allows the OWSOL to vary in every direction, and also from place to place. The corresponding coordinate transform is $$t'=t-\vec \kappa \cdot \vec x/c$$$$\vec x'=\vec x$$ where $\vec x=(x,y,z)$ is the spatial position. And the OWSOL in the direction $\hat n$ is $$c(\hat n)=\frac{c \hat n}{1-\vec \kappa \cdot \hat n}$$ where $\hat n$ is a unit vector.

But so we have that same problem again, just along two directions which are not that of light going away or towards me but... from left of me or from right??

It is not a problem, but it is an additional degree of freedom in specifying our OWSOL convention, or equivalently our synchronization convention. The fact that we have even more freedom to choose our convention simply further strengthens the idea that the OWSOL is a matter of convention.

why should I prefer one to the other, given the symmetry?

In this specific case we set $x$ to be the forward direction and $y$ the left direction. Then since $c_{forward}=c((1,0,0))=c/2$ you have $\vec \kappa = (\kappa_x,\kappa_y,\kappa_z)$ where $\kappa_x=-1$, but $\kappa_y$ is undetermined. However, in this case we can insist that $\kappa_y=\kappa_z=0$. This, in turn, comes from insisting that $|\vec \kappa|\le 1$. From $$c(\hat n)=\frac{c \hat n}{1-\vec \kappa \cdot \hat n}$$ if $|\vec \kappa|>1$ then $c(\hat n)$ would be negative in some direction, but speed is never negative. But in general we could choose $\vec \kappa$ such that none of its components are zero.

$\endgroup$
3
  • $\begingroup$ Since you mention in this specific case, I think you're referring to my example. Would you mind clarifying what (x, y, z) are with resepct to that? Also, what is the x vector? $\endgroup$
    – Enlico
    Sep 17, 2023 at 15:40
  • $\begingroup$ I updated the answer with explanations $\endgroup$
    – Dale
    Sep 17, 2023 at 16:24
  • $\begingroup$ People have been doing physics in arbitrary coordinate systems since 1915. Why did Anderson write a paper about a less general treatment of the issue in 1998? $\endgroup$
    – benrg
    Sep 17, 2023 at 17:57
0
$\begingroup$

The whole point, in my opinion, is that to measure the speed of anything, including light, we need a pre-synchronized set of clocks. These clocks can be evenly spaced (say, separated by 1 meter) and should remain at rest w.r.t. each other.

If we fulfill this condition, then we are ready to measure the one-way speed of light. However, it is light itself that we use to synchronize these clocks in the first place. Now, if you think about it, you will see that there is no way to synchronize two clocks that are separated by a non-zero spatial distance without requiring that the transmitted light be received back by the source. But in doing this, we average out the speed of light in both the direction and hence lose the information of the one-way speed of light.

This paper by Einstein explaining this himself might be helpful (see chapter 8). Here is a quote from the same chapter.

"That light requires the same time to traverse the path $A \rightarrow B$ as for the path $B \rightarrow C$ is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity.”

$\endgroup$
11
  • $\begingroup$ Rømer famously measured the speed of light without transmitting anything. $\endgroup$
    – John Doty
    Sep 17, 2023 at 15:13
  • $\begingroup$ I don't see why you can not synchronize clocks in a one-way sense. Just assume you have several clocks in a line recording the arrival time of the light signal. If all the clocks would already be synchronized, we would get an exactly linear plot of the arrival times vs. distance (assuming obviously that the speed of light is constant throughout the path and also that all clocks are identical). So we just need to have a look at the arrival time for each clock and adjust then their times such that we get a linear plot. All clocks will then be synchronized. $\endgroup$
    – Thomas
    Sep 17, 2023 at 16:14
  • $\begingroup$ @Thomas you assume all clocks are already synchronized and that is the problem. $\endgroup$
    – S.G
    Sep 17, 2023 at 21:28
  • $\begingroup$ Rømer didn't synchronize his clocks (the rotation of Earth and the orbit of Io). $\endgroup$
    – John Doty
    Sep 17, 2023 at 21:52
  • $\begingroup$ Are you referring to this page:= (en.wikipedia.org/wiki/…)? As I understand, this is pre-relativity which means no questions were asked regarding the meaning of simultaneity. The relativity of simultaneity changes things. $\endgroup$
    – S.G
    Sep 18, 2023 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.