1. Jul 9, 2008

Domnu

Problem
Imagine a wheel of radius $$R$$ consisting of an outer rim of length $$2\pi R$$ and a set of spokes of length $$R$$ connected to a central hub. If the wheel spins so fast that its rim is traveling at a significant fraction of $$c$$, the rim ought to contract to less than $$2\pi R$$ in length by length contraction, but the spokes ought not change their lengths at all (since they move perpendicular to their lengths). How do you think this problem is resolved?

Solution
Well, we know that there is no such thing on earth as a "rigid" body (ie, if we take a ladder and run into a wall, where the back of the ladder makes contact with the wall, we know that the front of the ladder keeps moving for a small amount of time, $$t < L/c$$ where $$L$$ is the length of the ladder since the front of the ladder has no way of "knowing" to stop (no information can travel faster than the speed of light).

The same way, let us say that we start rotating the wheel with angular velocity $$\omega$$. The outer rim doesn't start to move immediately... in fact, we will have turned the wheel a full $$L\omega / c$$ radians before the end of the wheel "knows" to start moving. Because of this, the spokes effectively shear slightly, reducing the velocity of the outer rim enough to account for the lessened circumference.

Note that if we accelerate the wheel, this wouldn't apply to special relativity, since special relativity only deals with non-accelerating frames. $$\blacksquare$$

Is this a valid explanation? Also, how can we apply special relativity to this in the first place if an edge element of the wheel is considered to be accelerating? Should we consider the wheel to be extremely large such that the acceleration of each edge element, $$v^2/r$$, is very small, eliminating the need to use general relativity?

2. Jul 9, 2008

Gear300

Interesting question...though I'm not so sure about the answer. What was the point of the black square?

3. Jul 10, 2008

Domnu

The black square was to just signify the end of my solution

4. Jul 10, 2008

dynamicsolo

We don't need to be concerned about how the wheel gets started up, because the "paradox" will still be operative in the "steady-state" situation.

Ah, you're onto it here...

We can't sneak out of it that way -- the effect is real and must be dealt with. (This means, of course, that the effect is present even for the wheels on your bike just riding to and from class -- it's just really tiny. [Don't try using relativistic phenomena to explain why you're late to physics class; they won't buy it...] )

You are correct in noting that the circumference would be contracted, yet the radius would be (essentially, though, I think, not exactly) unchanged. What does this mean for the ratio of circumference to radius? What sort of geometry permits "circles" to have C/R different from pi?! (This is an entree to one of the basic concepts of general relativity.)

5. Jul 11, 2008

Mute

Special relativity handles (at least most...) accelerating reference frames just fine. It's when gravity is involved that General Relativity must be invoked.

http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

Of course, the 'rigid' rotating disk does seem to have led to Einstein's conclusion that not all space can be assumed to be Euclidean...

http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

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Looking into it a little more, it seems like the resolution to the paradox isn't quite all that clear cut (at least as far as one can trust the wikipedia article), and there's still a bit of controversy about it:

If that page can be trusted, then it appears that the two seemingly conflicting statements I made above are resolved by the line that claims

"2000: H. Nikolic points out that the paradox disappears when (in accordance with general theory of relativity) each piece of the rotating disk is treated separately, as living in his own local non-inertial frame."

i.e., point particles (or I guess large objects that can be approximated as point particles for a given approximation) can be handled just fine in special relativity, but the rotating disk is a different case for which special relativity doesn't seem to be quite able to handle it.

Since this all seems a little confusing, if this is a homework assignment, I suspect the desired answer is just the one given by dynamicsolo.

Last edited: Jul 11, 2008
6. Jul 12, 2008

dynamicsolo

Thanks for this page and its references. The subtleties of relativity theory are indeed seemingly endless...

It doesn't surprise me to hear that people still find the implications even of SR to be complicated or surprising: even as straightforward a phenomenon as "Terrell rotation" wasn't found until 1961. (It also took some ways into the 20th Century for enough physicists to come to grips with the intricacies of differential geometry. There's a reason Einstein went to Michele Besso and Marcel Grossmann for help. A lot of the work in this mathematical field was going on mostly in Italy at the time...)

I was a little bit surprised by the suggestion that the circumference might somehow carry a larger number of "meter-sticks" in rotation. I think this sort of confusion comes up when comparing inertial vs. non-inertial frames: many "paradoxes" appear which are due to momentarily misplacing something that belongs only in one frame into the other. You could just as easily imagine a series of marker posts evenly spaced along the circumference of the disk in angle. Surely the number of posts required on the circumference wouldn't increase with angular velocity; it is the unit of arclength that appears to change, as seen from the center.

The problem persists without even the presence of a physical disk. You could simply imagine
a ring of particles orbiting the center at constant angular velocity, without having to be distracted by issues of material stress in the disk and so forth... One certainly encounters a related problem is dealing with the magnetic field around a pulsar, where there is a critical radius (somebody-or-other's Limit, IIRC) at which the tangential velocity would be c, so there would have to be some change in regime for the field as this radius is approached.

One can have a bit more fun with this disk. At the critical radius $$r = \frac{c}{\omega}$$, there would be an "event horizon", with the circumference at that distance approaching zero, the time dilating to infinity (and so light becoming redshifted without limit), the mass of anything there approaching infinity, etc., as seen from the center. (Of course, a "disk-rider" near that radius will say much the same about the individual at the center.) The center observer sees the effects increase continuously with radius out to the "horizon", owing to a potential growing with radius (described classically by $$\frac{1}{2}\omega^{2}r^{2}$$).

Again, thank you for the material on this problem. I'd read a little on it, but had no idea it had quite so much history.

Last edited: Jul 12, 2008
7. Jul 12, 2008

dynamicsolo

It's sometimes known as the "Halmos Q.E.D."; the mathematician Paul Halmos used it to mark the end of proofs in his books.