I can't tell what's wrong with it. Imagine the wheels of a car, jack the car, put a sensor on one wheel, a laser on another, start rotating, with enough speed the sensor will start to detect the laser on another spot than when it was stationary, and by calculating the difference taking into account the RPM and distance, you get the speed of light. And, in my view, the one-way speed of light. So, how is this a two way experiment?

Of course the wheels of a car is just an example to visualize the experiment.


  • $\begingroup$ How do you know the two wheels are spinning at the same RPM? $\endgroup$
    – RC_23
    Commented Sep 23, 2023 at 3:53
  • $\begingroup$ Same axle for both wheels, single engine positioned at the center rotating axle. $\endgroup$
    – Hudson ST
    Commented Sep 23, 2023 at 4:14
  • 1
    $\begingroup$ if they are on the same axle the laser would be moving at the same rate , so the light would always fall at the same spot. $\endgroup$
    – anna v
    Commented Sep 23, 2023 at 4:44
  • $\begingroup$ There's a distance between the laser and the sensor. $\endgroup$
    – Hudson ST
    Commented Sep 23, 2023 at 5:02
  • 1
    $\begingroup$ but if the wheels are fixed to the same axle this distance is fixed. $\endgroup$
    – anna v
    Commented Sep 23, 2023 at 6:59

2 Answers 2


The same issue as every other one-way speed of light measurement- you need to make assumptions about synchronizing the "clocks" (in this case, wheels) at both ends in order to make the measurement.

If you make the assumption that both wheels start spinning at the same time, you will end up with the usual two-way speed of light $c$. But you can also assume that one wheel starts some amount of time after the other and end up with any other result.

Assume one wheel is at $x$ and starts turning at time $t$, while the other wheel is at $x^\prime$ and starts turning at $t^\prime$, both measured in an inertial frame where neither wheel is moving. Saying the two wheels start moving at the same time means that you are stating that $t=t^\prime$. But $t$ is a time measured on a clock at $x$, and $t^\prime$ is a time measured on a clock at $x^\prime$. You can't relate these two clocks without a synchronization convention.

Assuming that $t=t^\prime$ is assuming that the clocks are synchronized via Einstein synchronization (or an equivalent method), or equivalently that the one-way speed of light is isotropic.

  • 1
    $\begingroup$ "If you make the assumption that both wheels start spinning at the same time, you will end up with the usual two-way speed of light $c$." Why? And what is being assumed for rigid wheels on an axle to not begin turning at the same time? Does this tie into the fact that nothing can be perfectly rigid? $\endgroup$ Commented Sep 23, 2023 at 4:59
  • 1
    $\begingroup$ I don't get which assumption is being made on this example. It's the same axle for both wheels, being rotated by a engine positioned at the center of the axle. If there's any inconsistencies in the axle construction that would make one wheel start spinning first, you could simply swap sides of the laser/sensor and do another measurement to account for it. $\endgroup$
    – Hudson ST
    Commented Sep 23, 2023 at 5:24
  • 1
    $\begingroup$ @HiddenBabel The OP assumes that the laser moving on one wheel relative to its initial spot is due to the time delay of the light. But the other wheel starting earlier would have the same effect. There is nothing unusual about the wheels not starting at the same time. Indeed, they don't start at the same time in most reference frames! But it's more than that. What does starting at the same time mean? It means that if you had a clock next to each wheel and you marked off the time that the wheel started, you get the same time for each wheel. But then how do you synchronize those clocks? $\endgroup$
    – Chris
    Commented Sep 23, 2023 at 5:24
  • 2
    $\begingroup$ @HudsonST It's not an inconsistency with the axel. To say they start at the "same time" is meaningless without a synchronization convention. If you assume that turning a rigid axel turns both its wheels at the same time, you are tacitly assuming Einstein synchronization. $\endgroup$
    – Chris
    Commented Sep 23, 2023 at 5:27
  • 1
    $\begingroup$ @Chris OH! It clicked now. Thanks! $\endgroup$
    – Hudson ST
    Commented Sep 23, 2023 at 5:39

I recommend that you look into the relativistic concept called Born rigidity. (Named after the physicist Max Born.)

As a variation on the setup you are proposing: let's imagine floating in space a very, very long cylinder, and we will take the end caps of that cylinder to correspond to the wheels of the setup you described.

We will assume special measures are in place so ensure that the long cylinder remains straight, even during a process of spinning it up.

We start a rotation of that cylinder around its long axis by causing angular acceleration at the middle of the cylinder.

That angular acceleration then propagates out to the ends of the cylinder.

In terms of newtonian mechanics we will expect that once you stop the angular acceleration at some final angular velocity the ends of the cylinder will over time catch up to the angular velocity of the middle (allow for extra time so that the energy of any oscillation that might have build up dissipates). In terms of newtonian mechanics: after a sufficiently long time for the system to settle down all of the parts of the system will be in synchrony with each other.

As we know: in terms of special relativity things are more complicated.

There is the question: as seen from an inertial point of view, what is the shape of a rotating cylinder? That is a tricky problem, because we need to take relativity of simultaneity into account.

Let's say we would attempt to verify that the end caps are in sync by exchanging light signals. Well, that's already Einstein synchronization procedure.

So we would have to assume that since the angular acceleration was applied to the middle of the long cylinder the angular accelerations traveled out to the end caps in the same way, which would mean the end caps must be synchronized with respect to each other (while not necessarily synchronized with respect to the middle of the cylinder).

I don't have a definitive answer, but these are the complications that I can come up with right now.

More generally, the idea of rigidity that is applicable in terms of newtonian mechanics does not carry over to special relativity. The best known example of this is the Ehrenfest paradox

In special relativity: combination of rotation and spatial extent gives interesting phenomena.


I now have definitive answer.

It is along lines that I have discussed in previous answers I posted on physics.stackechange. (Which of course means I should have come up with the definitive answer right away - but it took me a while.)

Also, judging by your "It clicked now" remark in a comment to the answer by contributor Chris you have already figured it out by yourself. I'm adding this discussion for completeness .

I start with discussing the following:
In Minkowski spacetime all forms of dissemination of time will give the same result. Examples of what I mean: you can disseminate time by sending out electromagnetic signals, or by shooting out particles, or by transporting clocks; all of those procedures will produce mutually consistent results.

Example: dissemination of time with pulses of, say, protons.
Let's say we have two particle accelerators with storage rings, with fast protons, having a precisely known velocity relative to the facility. Let's say there is some device in place that allows release of precisely timed pulses of protons towards a target. The other facility detects the incoming pulses of protons, and in return releases timed pulses of protons (with the same velocity) back to the first facility. The facilities use that procedure to synchronize clocks.

Let another observer have a velocity relative to the two particle accelerator facilities. As represented in the the coordinate system of that third oberver the two proton beams do not have the same velocity , as per relativistic velocity addition. Hence as represented in the coordinate system of the third observer the time-of-flight is not the same in the two opposite directions. In the end: you arrive at the same relativity of simultaneity as when Einstein synchronization procedure is applied.

Next I discuss the implications of relativity of simultaneity for a long cylinder, rotating along its long axis.

Let a fleet of spaceships be in motion parallel to the long axis of the long cylinder. The ships of that fleet of spaceships use Einstein synchronization procedure to maintain synchronized fleet time.

That means that as represented in the coordinate system of the fleet of spaceships the rotating cilinder has along its long axis a helical twist, in accordance with relativity of simultaneity.

Further down I will abbreviate 'the fleet of spaceships' to 'the fleet'.

Finally I come to the process of setting the long cilinder into rotation.

The setup: the torque to get the rotation going is applied at the midpoint along the length of the cylinder. That twisting action then starts propagating along the length of the cylinder, towards the end caps. The stiffer the cylinder, the faster the velocity of propagation of the twisting action.

For an observer who is stationary with respect to the length of the cylinder the twisting action is propagating in both directions at the same velocity. Hence for that observer the end caps, once they are up to speed, will be synchronized with the middle, where the angular velocity state commenced.

We shift to the perspective of the fleet, which has a velocity relative to the cylinder, parallel to the length of the cylinder. For the velocity of the propagation-of-the-twisting-action relative to the fleet relativistic velocity addition must be used. Consequently, as represented in the coordinate system of the fleet the twisting action is not in both directions propagating at the same velocity. As represented in the coordinate system of the fleet the end state of the cylinder is that it has along its long axis a helical twist, consistent with relativity of simultaneity.

General statement:
For any arrangement where the time keeping is along the length of a straight line there is no way to avoid relativity of simultaneity.

  • $\begingroup$ That was a great reading. Thanks! I'll be taking a look into Born rigidity. $\endgroup$
    – Hudson ST
    Commented Sep 23, 2023 at 6:16
  • $\begingroup$ @HudsonST In case you have set up that you will be notified of a comment addressed to you: I have edited my answer, adding what I believe is a definitive answer. (Judging by your "It clicked now" remark you have already figured it out. For completeness I added the discussion anyway.) $\endgroup$
    – Cleonis
    Commented Sep 24, 2023 at 7:20

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