This problem grew out of my playing with car tire circumferences. It is possible to show that tire circumferences, as measured by the distance between contact points, are non-euclidean. That is, Euclid would expect the distance to be 2$\pi$ times the radius, as at low velocities, but as velocities grow, that distance grows. (A simple way to see that is that for fast moving vehicles of tire rpm X, we roadside observers see that X slows by X/$\gamma$, so rpm's decrease yet velocity doesn't, implying that the distance between contact points grows to 2$\pi \gamma$r).

Anyway, my simple thinking created a model that works well to ~0.75c, but faster than that, I get a divergence in time versus the time I would expect for one rotation. Analysis shows that the reason is that a point on the rim of tire goes "too slow" near the top. Here is an extreme example considering a car traveling at 0.9999c, which is 0.9999*299.792458 meters/$\mu$second.

From the driver's point of view, his tires rotate with edge velocity = 0.9999c and radius 1 meter. The time for one rotation is 2$\pi$/(.9999*299.792458) = 0.021 $\mu$seconds to the driver.

$\gamma$ is 70.7. So we see one rotation occurring in time .021*70.7 = 1.48 $\mu$seconds

However - at the top of the rotation, we roadside observers see the velocity of a point on the rim by relativistic velocity addition of (.9999+.9999)/(1+.9999*.9999) - .9999 = 9.9995e-5c relative to the hub (the top point overtakes the center at this modest velocity)

So for the top 1/100th of a revolution, the time to rotate 1/100ths of a rotation (3.6 degrees) is: 2$\pi$/(100 * 9.9995e-5*299.792458) = 2.1 $\mu$seconds

That is, it takes longer for that top point to move 1/100th of a rotation than we would predict the wheel to do one full rotation!

Obviously, I am doing something wrong. Is this a kind of restatement of the Erhenfest paradox (in which disks separate at these rotational velocities)? But this would apply to just a spinning rod on the side of a passing spaceship as well. The rod would be rotating at allowable velocities to the pilot, but we watching the ship go by, can't reconcile the motion with our expectation of time of one rotation versus the top 1/100th of rotation using the velocity addition formula.

What am I doing wrong? How can the velocity addition formula fail to reconcile with $\gamma$ times rotation time for a circular path whose center is moving past us at near light speed?

*** Resolved! Both md2perpe and my chihuahua gave me the right answer - it's not a circular path, it's a highly squished ellipse. My simulation is now giving me perfect agreement with expected rotation time and "lorentz-stretched" circumference so that everything works out for cars going any speed. I'll write it up and arxiv.org -it if anyone wants to see the details. It is really interesting because the tire tread both gets shorter (as seen by an ant going round and round) and longer (as seen by a roadside observer) at the same time. And (although the numbers are tiny) it applies to automobiles at any speed, including 65 mph freeways - tires are non-Euclidean.

  • $\begingroup$ "a car traveling at 0.9999c, which is 0.9999*299.792458 $\mu$meters/second" Don't you mean "megameters/second"? $\endgroup$
    – md2perpe
    Jul 7, 2021 at 19:11
  • $\begingroup$ "So for the top 1/100th of a revolution, the time to rotate 1/100ths of a rotation (3.6 degrees) is: 2$\pi$/(100 * 9.9995e-5*299.792458) = 2.1 $\mu$seconds" But the wheel is length contracted and not circular in the frame of the ground. So that length $2\pi/100$ meters is not correct. $\endgroup$
    – md2perpe
    Jul 7, 2021 at 19:21
  • $\begingroup$ @Ralph 0.9999 c is very annoying to work with. Why would you choose such a terrible speed? $\endgroup$
    – Dale
    Jul 7, 2021 at 21:29
  • $\begingroup$ I solved it! Sometimes it just takes writing your question down, and then taking the dogs for a walk. I had an aha moment, and of course - the resolution is exactly as md2perpe says. It isn't a circular path, it's a very squished ellipse! And md2perpe is also correct about my units, it is meters per $\mu$second. I've got my simulation fixed. As to Dale, I wanted my simulation to work for every speed from 1 mph to 0.9999c, and now it does. The tangential velocity slows way down at the top, and the time step would grow, but the path lenght is shorter and offsets perfectly! $\endgroup$ Jul 7, 2021 at 21:35

1 Answer 1


Congratulations- you beat me to it!

Yes, the RPM of the wheel is effectively a clock frequency, so from the stationary frame the rotation is time dilated and slows. But also from the stationary frame, the wheel is length contracted in the direction of motion, so the circumference is not longer circular. If you imagine the tire as being made of small segments, the portion of the tire in contact with the ground is stationary in the ground frame, while the diametrically opposite patch at the top of the tire is moving even faster than the centre, so the relativistic effects are quite complicated.


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