Number of wheel rotations relativistic car

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Thus, if we compute the circumference of the wheel by breaking the wheel up into many segements (we can imagent the segments are delimited by spokes), the number of revolutions of the wheel multiplied by the circumference of the wheel is the distance traveled in the road frame. The circumference is cacluated by adding together the distance between spokes in the limit as the spokes are closely spaced, calculated in a series of frames co-moving with the ends of the spokes. You cal alternatively think of this as the non-inertial rotating frame of the wheel. This is a standard textbook calculation - while there are some confused papers on the topic, unfortunately (mostly the confused papers are rather old), the modern textbook answer for the circumference of the wheel is well understood to be ##2 \pi r / \sqrt{1 - \omega^2 r^2/c^2}##

Surely is a silly question, by why should ##\omega r## be equal to the car's speed v?

 

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It would seem possible that in the car's reference frame, the road distance, A to B could be contracted to less than 1m, so less than one wheel rotation would occur;
But this problem is too complex for me. For example, the top surface of the wheel, if it stayed round, is advancing forward at twice the speed of the axial - more than the speed of light? (I think not.)

No, you have to use the relativistic composition of velocities: V = (v₁+v₂)/(1+v₁*v₂/c²).
At ##\gamma = 2##: v/c = sqrt(3)/2 ~= 0.866. Using the composition of velocities with v₁ = v₂ = v, you get:
V/c = 4sqrt(3)/7 ~=0.987 which means ##\gamma' = 7##.

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[Dale]

It's not necessarily a bad idea to work it out in both frames and make sure you get the same answer.

Yes, that is true. Also, just doing it in the car frame you never notice the weird shape of the spokes in the road frame, which is interesting.

 

[Laurie K]

Just to clarify that the car frame is centered on the axle frame and as such moves parallel to the road frame in the direction of motion of the wheel.

In "Space geometry in rotating reference frames: A historical appraisal" Øyvind Grøn presents an interesting compendium of various different approaches to relativistic rotation. He contends that Born rigidity problems can be avoided by making the wheel zero thickness (i.e. x, y and t only with z = 0) and provides some interesting structural insights into other SR based aspects of relativistically rolling wheels. Figures 8 shows how the spokes look in each frame while Figure 9 show the differences between the locations of 16 equidistant points in the various frames.

http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf

Some complementary figures showing the appearance of the rotating disk in accordance with two operational procedures performed in the rotating rest frame of the disk, have been presented by K. MacFarlane [110]. The positions of points on a rolling ring at retarded points of time were calculated with reference to K0 by Ø. Grøn [111]. The result is shown in Fig. 9. Part C of the figure shows the “optical appearance” of a rolling ring, i.e. the positions of emission events where the emitted light from all the points arrives at a fixed point of time at the point of contact of the ring with the ground. In other words it is the position of the points when they emitted light that arrives at a camera on the ground just as the ring passes the camera.

While not exactly the OP question his Figure 9 part C provides some interesting structural insights into problems of this nature. I have seen Grøn's Figure 9 part C “optical appearance” plot verified (at 0.86c) on another forum and agree that the standard Euclidean geometry of the plot plays a major part. While the plots of the emission points in the PDF show the physical location of the discrete points at the time of emission, the length of a straight line drawn between each retarded time point shown and the camera (1) also equals the respective distance traveled by the photons (at c) between emission and observation.

Further, as the x and y position of any emission point on the circumference of the wheel, at any time during one complete rotation, depends explicitly on the wheels velocity and the location of the wheels axle at a particular time, the final results can be cross checked again by comparing emission time and axle location on the x-axis between sequential emission points and times. In the verification I saw that the distance traveled along the x-axis by the axle between each emission event divided by the time between emission events should equal the velocity of the wheel (and the angular velocity of the point on the circumference) if your calculations are correct.