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

Question

Introduction

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?

Answer 1

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.

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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 $\mathrm dx^1/\mathrm dT \in (-c/2, \infty)$.

Answer 2

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.

Answer 3

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.

Answer 4

"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.