Let me start by saying that this is not a question about how to measure the one-way speed of light (OWSOL). It's about the physical implications of the idea that this speed is merely conventional, an idea that I've been strenuously trying to make sense of for weeks now.

From what I can tell, the conventionality of the OWSOL is equivalent to the conventionality of simultaneity (and of clock synchronization), which is a "debate [that] seems far from settled" per this article in the Stanford Encyclopedia of Philosophy and this chapter from John Norton (in which he gives interesting commentary on the debate itself and says he leans towards realism over conventionalism). So while I realize that some prominent members of this community view it as a proven fact, I'll be open to the possibility that it's not until I'm able to make sense of it.

Thought Experiments

If the OWSOL isn't $c$ in every direction then it's necessarily slower than $c$ one way and faster than $c$ the other way (since we know that the two-way SOL is always $c$). So let's say that it's $\frac 23 c$ one way and $2c$ the other way (so that light with an isotropic one-way speed would take $2t$ to travel to a mirror and back but our light will instead take $\frac 12 t$ one way and $\frac 32 t$ the other way). In this case, is our speed limit still $c$ or is it $\frac 23 c$, $2c$, or something else? Does it depend on the direction of travel? The last paragraph of this answer seems to insinuate that it's always simply the OWSOL in the direction of travel, which makes sense to me, but just in case there's any doubt, let's consider both cases:

  • If the speed limit is not $c$ in every direction (presumably, this would mean that it's the OWSOL in the direction of travel), this would seem to have absurd implications. For example, let's imagine a giant accelerator for visible objects, including manned spacecraft (in case we're concerned about the uncertainties of measuring particles), that accelerates them up to $0.99999c$ in a weightless vacuum and then lets inertia carry them. Notice that they would travel around the accelerator in a loop, so 1) the time they take per lap could be measured by a stationary observer with a single clock, so synchronizing clocks wouldn't be necessary and 2) the length of the loop could be measured using the two-way SOL; therefore, the travelers' precise average speed could be determined. Presumably, their two-way speed limit would be $c$, but if the one-way limit changes with direction, wouldn't they be speeding up and slowing down (e.g., between $\frac 23 c$ and $2c$) as they move around the loop? If so, what forces would be acting on them to cause such changes? Wouldn't they feel these forces and notice that things appear to be moving by them at changing speeds?
  • If the speed limit is $c$ in every direction, there would seemingly be other absurd implications. We'd be able to travel at a speed in between the lower OWSOL, $\frac 23 c$ in this case, and $c$. So let's say we speed up to $0.99c$ (either in a straight line or in the giant accelerator—whichever you think makes my point better). When we're moving in the direction of the $\frac 23 c$ speed, we'd be outpacing light (and thus unable to see or be affected by anything behind us). When we're moving in the direction of the $2c$ speed, things would appear to be passing by us at a speed that outpaces light (thus we'd be unable to see or be affected by anything in front of us until it passes by us). And in both cases, I assume that causality violations would be possible.

Additional Questions

These all sound problematic, which suggests to me that the OWSOL can't be—or at least isn't—anything other than $c$ and is therefore not conventional. If this isn't a sound conclusion, what am I missing?

Similarly and perhaps equivalently: If the OWSOL is conventional, is the one-way speed of particles or objects that are looping in an accelerator at $0.99999c$ (on average) also conventional 1) within the reference frame of the accelerator and 2) within that of its travelers? If so, how can we make sense of this physically in light of the issues I've raised?

  • 1
    $\begingroup$ I'm 99% confident that "the one-way speed of light" is not conventional, absolute, real, fictitious, unknown, unknowable, knowable, or even arbitrary; but rather a complete nonsense concept like "the square root of a starfish if the universe is composed entirely of the ratio of Tanzania to bananas." $\endgroup$
    – g s
    Feb 24, 2022 at 4:24
  • 3
    $\begingroup$ @gs It is not quite that bad. There is a legitimate choice you have to make setting up the formalism of special relativity, and "the one-way speed of light" is a legitimate description of this. But, there is a uniformly agreed upon standard convention that makes the physics much clearer than any other choice. The real issue is that some youtubers who shall go unnamed have given much more oxygen to this issue than it deserves and made it seem much more profound than it really is. $\endgroup$
    – Andrew
    Feb 24, 2022 at 4:36
  • 1
    $\begingroup$ As I said here, the one-way speeds cannot be measured independently of a choice of $\epsilon$. So in a sense, those speeds have no physical existence, they're just a mathematical artifact of the $\epsilon$ associated with your synchronisation convention. In contrast, we can measure the round-trip speed of light, and show that it's invariant, and that's the real basis of special relativity. $\endgroup$
    – PM 2Ring
    Feb 24, 2022 at 4:52
  • $\begingroup$ @PM2Ring Thanks. I've read that answer before and just reread it. I've seen you mention John Norton a fair amount but are you aware that his "view is more sympathetic to the non-conventionalist view. I incline towards the realist view of Minkowski geometry [and a] unique notion of simultaneity provided by the spacetime geometry"? This implies a unique set of one-way speeds, so does your view that they "have no physical existence" differ from his? How do you interpret their lack of physical existence? Could there be a wave function that collapses when light arrives back at its starting point? $\endgroup$ Feb 25, 2022 at 10:29
  • $\begingroup$ @GumbyTheGreen Yes, I'm aware of Norton's non-conventionalist leanings, and my viewpoint differs from his. But he does give excellent coverage of the topic, so I'm happy to link to his site. My view is that there is no physical basis for the choice of $\epsilon$, but it's sensible to use $\epsilon=\frac12$ to keep the equations (relatively) simple & intuitive. $\endgroup$
    – PM 2Ring
    Feb 25, 2022 at 10:52

4 Answers 4


Consider ordinary Minkowski spacetime. In standard cartesian coordinates $(t,x^1,x^2,x^3)$ - in which the speed of light is isotropic - the line element takes the form $$\mathrm ds^2 = -c^2 \mathrm dt^2 + \sum_{i=1}^3 (\mathrm dx^i)^2\tag{1}$$ In these coordinates, the fact that a light ray travels along a null worldline implies that along that worldline, $$\mathrm ds^2 = 0 \implies \sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2 = c^2$$ Now we choose other coordinates $(T,x^1,x^2,x^3)$, where the new time coordinate is $T = t - x/c$. The line element now takes the form $$\mathrm ds^2 = -c^2 \mathrm dT^2 + \sum_{i=2}^3(\mathrm dx^i)^2 - 2c\big( \mathrm dT \mathrm dx^1\big)\tag{2}$$

This is the same spacetime and the same metric - just an unconventional choice of coordinates. Furthermore, the null condition now takes a different form. Restricting our attention to motion along the $x^1$ direction, the null condition becomes $-c^2 \mathrm dT^2- 2c\big(\mathrm dT \mathrm dx^1\big) = 0$ If $\mathrm dx^1>0$ (so the ray is moving to the right) then the only possibility is that $\mathrm dT =0$ (so the velocity $\mathrm dx^1/\mathrm dT$ is formally infinite). On the other hand, if the ray is moving to the left then for future-directed null worldlines we have that $\mathrm dx^1/\mathrm dT = -c/2$.

As you can see, whether the velocity (which is a coordinate-dependent quantity, after all) is isotropic or not depends entirely on our choice of time coordinate. When one adopts the Einstein summation convention, they obtain $(1)$, but this is not mandatory.

Presumably, their two-way speed limit would be $c$, but if the one-way limit changes with direction, wouldn't they be speeding up and slowing down (e.g., between $2/3 c$ and $2c$) as they move around the loop? If so, what forces would be acting on them to cause such changes? Wouldn't they feel these forces and notice that things appear to be moving by them at changing speeds?

No forces would be necessary, and nothing out of the ordinary would be felt. If you run through a clock shop and define your velocity as the distance between adjacent clocks divided by the difference in their readings, then you could be moving at what you consider a stationary pace but the numerical value of your velocity would change if the clocks are not synchronized across the shop. This is essentially the same concept - velocity is not measurable in a coordinate-independent way, a central point of special relativity.

Concretely, consider the worldline $(t,x,y)=\big(\lambda,R \cos(\omega \lambda), R\sin(\omega \lambda)\big)$ in coordinate system $(1)$, corresponding to a particle moving in a circle. Its easy to verify that the speed of the particle in these coordinates is constant, and equal to $\omega r$.

In our new coordinates $(2)$, this becomes $$(T,x,y)=\big(\lambda-\frac{r}{c}\cos(\omega\lambda),r\cos(\omega\lambda),r\sin(\omega\lambda)\big)$$ $$\implies \frac{dx}{dT} = -\frac{\omega r \sin(\omega\lambda)}{1+\frac{\omega r}{c}\sin(\omega\lambda)} \qquad \frac{dy}{dT} = \frac{\omega r \cos(\omega\lambda)}{1+\frac{\omega r}{c}\sin(\omega\lambda)}$$ $$\implies \sqrt{\left(\frac{dx}{dT}\right)^2 + \left(\frac{dy}{dT}\right)^2} = \frac{\omega r}{1+\frac{\omega r}{c}\sin(\omega\lambda)} = \frac{\omega r}{1+\frac{\omega y}{c}}$$ Therefore, the speed as calculated in these coordinates is not constant.

It should be obvious that nothing has physically changed here - we're just using new coordinates. The trajectory of the particles as observed by a human looking at it with their eyes in a laboratory is unchanged. However, speed is by definition a coordinate dependent notion, and using unconventional coordinates yields to unconventional results like this.

In this case, is our speed limit still $c$ or is it $2/3 c$, $2c$, or something else? Does it depend on the direction of travel?

The "speed limit" is a manifestation of the condition that the worldline of a massive particle must be timelike. In coordinate system $(1)$, this means that $\sqrt{\sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2} < c$, making the term speed limit a reasonable one.

In coordinate system $(2)$, the same condition means that $$-2c\big(\mathrm dT \mathrm dx^1) + \sum_{i=2}^3 \big(\mathrm dx^i\big)^2 < c^2 \mathrm dT$$ This is harder to cast into the form of a speed limit in general. However, if $\mathrm dx^1=0$ so the motion is occuring in the $(x^2,x^3)$-plane, then we obtain the same limit as before, while if the motion is occurring along the $x^1$ axis only then $\mathrm dx^1/\mathrm dT \in (-c/2, \infty)$.

  • $\begingroup$ 1) Could you add an algebraic step or two to show how equation (2) is obtained? 2) Why doesn't $dT = 0$ when the ray is moving to the left as well? 3) If "no forces would be necessary" and nothing unusual would be experienced, then speed isn't really changing in any physical, meaningful sense—only our definition of it is changing (since our definition of time coordinates is changing) as we travel around the loop—right? 4) Is it possible for the OWSOL to $\neq c$ without a coordinate system change? Coordinate systems are constructs. My question is about what's physically happening in nature. $\endgroup$ Feb 24, 2022 at 6:11
  • 1
    $\begingroup$ @GumbyTheGreen 1) $g^{(y)}_{\mu\nu} = \frac{dx^\alpha}{dy^\mu} \frac{dx^\beta}{dy^\mu} g^{(x)}_{\alpha\beta}$. The calculation is straightforward, and ultimately amounts to setting $\mathrm dt = \mathrm dT + \mathrm dx/c$ in $(1)$. 2) $\mathrm dx^1<0$ and $\mathrm dT=0$ would correspond to a past-directed null curve - see here. Note that if you transform back to the original coordinates (1), then $\mathrm dT=0\implies \mathrm dt=\mathrm dx^1/c$, and so $\mathrm dx^1<0 \implies \mathrm dt<0$. $\endgroup$
    – J. Murray
    Feb 24, 2022 at 6:31
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    $\begingroup$ 3) You're almost there. The point is that speed itself is based on our choice of coordinates. Speed is the rate of change of our coordinates with respect to one another; there is no coordinate-independent way to define it, which is a central theme in relativity. All of the intuition you have about speed is based on the assumption that you've adopted the Einstein synchronization convention; once that goes out the window, your intuition does as well. 4) See point 3. You are under the impression that "speed" is a property of an object independent of a choice of coordinates; this is wrong. $\endgroup$
    – J. Murray
    Feb 24, 2022 at 6:33
  • $\begingroup$ Thanks, I'm following it all now. It looks like movement in $x$ affects movement in $T$ such that as an object moves in a circle in the $xy$-plane, there's an oscillating trade off between speed through space and speed through time. The problem I'm seeing with this is that time is a different kind of dimension, so the laws of physics distinguish it from the spatial dimensions. For example, momentum is only a function of velocity through space, which is changing in the direction of motion, so momentum is changing, which requires force since $F = \frac{dp}{dT}$, correct? So where's the force? $\endgroup$ Feb 24, 2022 at 16:44
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    $\begingroup$ @GumbyTheGreen I've added the explicit calculation for your example to my answer. Note that force is not $\frac{dp}{dT}$, but rather $F^\mu =\frac{dp^\mu}{d\tau} = m \frac{d^2 u^\mu}{d\tau}$ where $\tau$ is the proper time and $u^\mu = \frac{dx^\mu}{d\tau}$ is the 4-velocity (note that the Christoffel symbols are still zero in this coordinate system). Remember that all I've done here is perform a coordinate change, so nothing is physically different; the point is that whether the speed of light is isotropic or not is merely a question of which coordinates you're using. $\endgroup$
    – J. Murray
    Feb 24, 2022 at 17:23

the conventionality of the OWSOL is equivalent to the conventionality of simultaneity (and of clock synchronization), which is a "debate [that] seems far from settled" per this article in the Stanford Encyclopedia of Philosophy and this chapter from John Norton

Note that this is a statement in the philosophy literature. Indeed, philosophers continue to debate this sort of thing endlessly. Among physicists this is settled. The OWSOL is purely conventional based on your choice of synchronization convention. This is well established by Reichenbach and later by Anderson (R. ANDERSON, I. VETHARANIAM, G.E. STEDMAN, "CONVENTIONALITY OF SYNCHRONISATION, GAUGE DEPENDENC AND TEST THEORIES OF RELATIVITY", Physics Reports 295 (1998) 93-l80) in a more complete form. At this point, any debate in the physics community is limited to people who are unfamiliar with the literature on the topic.

In this case, is our speed limit still c or is it 2/3c, 2c, or something else? Does it depend on the direction of travel?

The "speed limit" is the speed of light, which is no longer equal to c in these anisotropic coordinates.

Presumably, their two-way speed limit would be c, but if the one-way limit changes with direction, wouldn't they be speeding up and slowing down (e.g., between 2/3c and 2c) as they move around the loop?

Yes, the anisotropic synchronization convention affects all one-way speeds, not just that of light. If we use Anderson's convention then $\frac{2}{3}\ c$ one direction and $2\ c$ the other direction corresponds to his $\kappa = 0.5$

So in standard isotropic coordinates with natural units we can define a loop by $\vec r=(t,x,y,z)=(\lambda, r \cos(\lambda \omega), r \sin(\lambda \omega), 0)$ for $-\pi < \lambda \omega \le \pi$ and then we can transform to the Anderson coordinates with $$t \rightarrow T + \kappa X$$ $$x \rightarrow X$$ $$y \rightarrow Y$$ $$z \rightarrow Z$$ to get $$\vec r = (T,X,Y,Z) = (\lambda -\kappa r \cos (\lambda \omega ),r \cos (\lambda \omega ),r \sin (\lambda \omega ),0) $$ so the one way speed around the loop is $$\frac{\sqrt{dX^2+dY^2}}{dT^2}=\frac{r \omega}{1+\kappa r \omega \sin(\lambda \omega)}$$ which is anisotropic for any $\omega \ne 0$ and gives the correct 2/3 c and 2 c for $\kappa=0.5$ and $r\omega=1$

If so, what forces would be acting on them to cause such changes?

The forces from the rails. Indeed, if you work out the geodesics in the Anderson coordinates you see that an inertially moving object in the isotropic coordinates moves in a straight line at constant velocity. However, an object travelling in a loop is not moving inertially. It is experiencing a centripetal force which accelerates it in the loop. This centripetal force does not change the speed in the isotropic coordinates, but the Anderson coordinates are anisotropic. The centripetal force in the Anderson coordinates changes speed as well as direction.

Wouldn't they feel these forces and notice that things appear to be moving by them at changing speeds?

Yes, they feel the centripetal force, though it won't feel any different than centripetal force feels in any other coordinates. And yes, in those Anderson coordinates things move by at changing speeds, even though radar measurements will not detect it (radar measurements being "distorted" by the anisotropic speed of light).

In fact, in Anderson coordinates the four-acceleration from the centripetal force is: $$A^\mu = \left(\frac{\kappa r \omega ^2 \cos (\lambda \omega )}{\sqrt{1-r^2 \omega ^2}},-\frac{r \omega ^2 \cos (\lambda \omega )}{\sqrt{1-r^2 \omega ^2}},-\frac{r \omega ^2 \sin (\lambda \omega )}{\sqrt{1-r^2 \omega ^2}},0\right) $$ where the non-zero time component of the centripetal force indicates precisely that the force causes the speed to change in these coordinates.

If so, how can we make sense of this physically in light of the issues I've raised?

Simply realize that the whole discussion about the OWSOL is not physical. It is purely a perverse coordinate system that nobody ever uses for any reason other than to discuss the OWSOL. You don't make sense of it physically because it is purely mathematical and has no physical significance whatsoever.

  • $\begingroup$ To be fair, John D. Norton isn't only a philosopher of science. He's a specialist in the history of relativity, and regularly teaches a relativity course (admittedly with a philosophical / historical focus). See sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/index.html $\endgroup$
    – PM 2Ring
    Feb 25, 2022 at 4:00
  • $\begingroup$ Fair enough, but that doesn’t change my statement. What I said is entirely accurate. His quoted statement is in the philosophy literature and this debate currently is among philosophers not physicists. $\endgroup$
    – Dale
    Feb 25, 2022 at 4:27
  • $\begingroup$ Also fair enough. :) I just wanted to make it clear to future readers that Norton isn't just a philosopher who happens to know some relativity. $\endgroup$
    – PM 2Ring
    Feb 25, 2022 at 4:49
  • $\begingroup$ Wow, thanks! Now, it sounds like you're saying that the forces that change the speed are there in one coordinate system but not in the other, but aren't forces invariant across reference frames and coordinate systems? The travelers are either subject to them (able to "feel" them) or they aren't, right? $\endgroup$ Feb 25, 2022 at 10:07
  • $\begingroup$ Please see the "chat" link under J. Murray's answer for a list of reasons why I've thought that this notion of conventionality has physical significance. In addition to those, from what I've read, Reichenbach came up with it because he thought that any two spacelike events—presumably within a given inertial reference frame and coordinate system—could optionally be considered simultaneous, giving them a range of acceptable simultaneity relations (and thus a range of acceptable OWSOL values). You seem to instead take the view that each frame and system has a unique relation. Is that the case? $\endgroup$ Feb 25, 2022 at 10:12

Your question contains a number of objections that I too have held against the idea that the OWSOL is arbitrary between infinity and c/2. However, I have become content to accept that it may be the case because it is simply a matter in effect of convention about how to synchronise clocks.

Suppose you had a car with a broken speedometer and wanted to check how fast it went at 1000rpm in top gear. You could drive it to a local town and back and see how long it took. If the local town was ten miles away and the round trip took 2 hours you could say the car went at ten miles an hour. However, if the clocks at the local town were all set an hour ahead of the clocks in your town, you would find that if you left at noon you would arrive at the local town at 2pm, so the trip would have taken 2 hrs at 5mph. Travelling back you would find, having left the town at 2pm local time, you arrived back home at 2pm, the trip having been accomplished at infinite speed.

In a sense, the 'real' speed of the car was always 10mph, but the measured speed was not. I believe that the arguments about the OWSOL are simply a more complicated (ie having to take account of the hyperbolic geometry of spacetime) analog of what I have just described.

  • $\begingroup$ Thanks for the input. Your example is harmless enough but I was trying to push the idea to its limits to see if it would break down. I understand the connection between the OWSOL, clock synchronization, and simultaneity relations. Apparently, any anisotropy in these things would be indistinguishable from a coordinate change and therefore be non-physical. What I'm still not 100% clear on is whether it would be non-physical literally or just effectively, i.e., whether it could happen without a coordinate change. If it can't then I'm not sure how this even became a topic in physics. $\endgroup$ Mar 4, 2022 at 7:36

"Would paradoxes arise if the one-way speed of light weren't $c$ ? [...]"

There are (at least) two distinct ways to understand and answer this question:

(1): Does the (internal, a priori) consistency of the theory of relativity,
as a system of

  • obvious notions (foremost the notion of "coincidence", and thus of "coincidence events"),

  • declarations of specific definitions in terms of the obvious notions (notably, the declarations of how to determine "mutual rest", a.k.a. "joint membership in the same inertial frame", of given participants; followed by how to determine, alias compare, "distances" between pairs of participants who were and remained at rest wrt. each other; along with how to determine/compare "durations"; and subsequently: how to determine/compare values of "speed", etc.), and

  • all theorems which can be derived from these definitions, and how they depend of each other,

depend on the conventional use of the letter "c" for symbolizing what's more explicitly called "signal front speed", alias "speed of light in vacuum" ?? -- Certainly not.

(2): Does consistency of the RT require that signal fronts are unambiguous; or in other words:
if two events are identified as lightlike wrt. each other then they cannot be identified as timelike wrt. each other as well, nor as spacelike wrt. each other ?? -- Surely yes.


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