2

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?

Gumby The Green
  • 235
  • 1
    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." – g s Feb 24 '22 at 04:24
  • 3
    @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. – Andrew Feb 24 '22 at 04:36
  • 1
    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. – PM 2Ring Feb 24 '22 at 04:52
  • @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? – Gumby The Green Feb 25 '22 at 10:29
  • @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. – PM 2Ring Feb 25 '22 at 10:52
  • OTOH, maybe the universe is trying to give us a hint that there is a physical basis for $\epsilon=\frac12$. ;) But at this stage, I think it's best to say that the round-trip speed is physical, and the one-way speeds are just mathematical artifacts. – PM 2Ring Feb 25 '22 at 10:53

4 Answers4

7

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

J. Murray
  • 69,036
  • 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.
    • – Gumby The Green Feb 24 '22 at 06:11
  • 1
    @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 $\mathrm dx^1<0 \implies \mathrm dt<0$. – J. Murray Feb 24 '22 at 06:31
  • 2
  • 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.
    • – J. Murray Feb 24 '22 at 06:33
  • 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? – Gumby The Green Feb 24 '22 at 16:44
  • 2
    @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. – J. Murray Feb 24 '22 at 17:23
  • I see, thanks for the addition. It sounds like you're implying that the OWSOL must be $c$ in the coordinates we actually tend to use and therefore cannot differ physically? What I think is confusing me (and many others) is that we aren't just doing any arbitrary coordinate change here; we're doing a very specific one of the form $T=t-kx/c, k\in(-1,1)$ that's designed to fit within a window of uncertainty that some believe to physically exist in the simultaneity of two events, which makes it seem intended to have physical implications. If that's not intended, what's the point of all this? – Gumby The Green Feb 24 '22 at 20:31
  • Let us continue this discussion in chat. – J. Murray Feb 24 '22 at 20:44