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