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Let's take a standard ACME Thought Experiment Division car with a max speed of a leisurely, constant 0.75c in a straight line.

So, for an external observer, the body of the car behaves like a typical Thought Experiment Car, it's shorter than at rest, the dashboard clock goes slower than the clock in the observer's pocket, and so on.

Thing is, the car uses oft-neglected in thought experiments, wheels.

The surface of the wheel in contact with the road surface remains at rest relative to the road (friction etc). But from the classical mechanics point of view, the upper surface of the wheel would be moving at 1.5c relative to the ground (and the external observer). Obviously this is impossible.

What gives? What shape will the wheels be to the external observer? Will they shatter as the shearing forces grow towards infinity? What shape will they have to the driver of the car?

Just to avoid trivial answers ( e.g. "wheels explode in all directions torn apart by centrifugal force") let's impose some limits:

The wheels are extremely (though, let's say, not infinitely) durable, and quite (though not infinitely) flexible and stiff: you need a very strong force to distort them - certainly more than the their centrifugal force alone. And once that force is exerted it doesn't instantly break them - they distort flexibly for an observable while before an eventual critical failure.

They move over the surface without slipping, and the car's motor provides just enough torque (if any) to keep the car moving at constant speed despite any flexibility/friction/other forces the wheels might exert on it; the car retains its 0.75c no matter what the cost to the wheel integrity or such.

SF.
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    This is just another way to see that there are no rigid bodies in relativity. Your "trivial" answer is the answer - any time you derive a contradiction from special relativity + rigid body (such as the wheel), the solution is that the body doesn't stay rigid, but is torn apart. – ACuriousMind Jan 24 '16 at 16:38
  • @ACuriousMind: Hey, the wheel is NOT a Rigid Body. If it's undergoing a force strong enough, it will bend/twist/stretch/whatever. It's just hard enough that it won't do it trivially, before relativistic effects kick in, and flexible enough that you can observe how it stretches/bends/twists before it eventually shatters as the distortion goes towards large values. – SF. Jan 24 '16 at 16:42
  • An edit was suggested tagging this as a specific relativity problem. But the wheels' reference frame definitely isn't inertial. The centripetal force definitely causes acceleration of the wheel surface towards the axis. – SF. Jan 24 '16 at 16:44
  • @ACuriousMind I disagree. I dont think this has anything to do with the rigidity of the wheels. The wheels would be limited to going less than the speed of light, thus COM of the car can only approach half the speed of light. – DilithiumMatrix Jan 24 '16 at 16:48
  • The car cannot go at .75c through the wheels, you yourself state it in the problem. Anyway long before the car reaches .1C they will melt due to the friction, to say the least. – anna v Jan 24 '16 at 16:48
  • @DilithiumMatrix: or the car will stretch/unroll the wheels and rip through them eventually. Treat them not as means of propulsion but a kind of decorative/thought-device attachments. Their movement, from point of view of passenger of the car doesn't violate laws of GR (mostly) - no point of the wheel moves faster than 0.75c relative to the car. I believe from the driver's point of view they would twist spiral galaxy style as the outer rim trails in angular velocity behind the axis (and eventually shatter torn by that force)... – SF. Jan 24 '16 at 16:54
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  • Special relativity means flat metric, not inertial frame. You don't need general relativity unless you have gravity, so that edit was actually good. 2. I'm now confused what your actual question is. If you think one can predict how exactly the wheels will meet their demise, you can't. How exactly they bend or tear part is dependent on the specific wheel and its weak points, not on some general principle.
    • – ACuriousMind Jan 24 '16 at 16:58
  • @ACuriousMind ... special relativity only works in inertial systems, which also requires a flat metric. Even in GR one can always find a locally flat metric. – DilithiumMatrix Jan 24 '16 at 17:00
  • @DilithiumMatrix: What, exactly, do you call "special relativity" that it only works for inertial systems? To me, special relativity is essentially kinemetics in Minkowski space - if you want to consider non-inertial coordinate system on Minkowski space, you're free to do so, that doesn't mean you're doing GR. – ACuriousMind Jan 24 '16 at 17:04
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    @ACuriousMind you cannot perform a Lorentz Transformation between non-inertial frames. The comments of OP's question is not a good place for this discussion --- a good reason why you should post your answers as answers, instead of as comments. – DilithiumMatrix Jan 24 '16 at 17:07
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    @ACuriousMind: In GR acceleration is indistinguishable from gravity... though I might be wrong here. 2. This is not a material engineering question. Assume whatever construction is comfortable to you but tough enough that it can survive a couple microseconds needed to make an observation of what goes on with it, with the rim moving at 0.75c relative to the car. It can shatter and explode all you want if it exceeds 0.8c. – SF. Jan 24 '16 at 17:15
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    Possible duplicate of Rotate a long bar in space and get close to (or even beyond) the speed of light $c$ – Daniel Griscom Jan 24 '16 at 17:20
  • @DanielGriscom: This assumes a bar that would necessitate its end to exceed c from point of view of the observer at the axis of the bar. These wheels don't exceed c from the point of view of the driver (and are not rigid). – SF. Jan 24 '16 at 17:24
  • They do exceed C from the point of view of the ground, which is what is pushing the top of the wheels to go faster than C relative to the ground observer. – Daniel Griscom Jan 24 '16 at 17:40
  • @DanielGriscom: So? All reference frames are equal. Can a situation be valid in one and invalid in another? Observer 1 sees two spaceships traveling in two opposite directions, each at 0.75c. Observer 2 flies one of these spaceships. Replace 2 spaceships with armada where each flies slightly slower than the previous, you have the wheel. Measure distance between nearest two spaceships, you have the local distortion. – SF. Jan 24 '16 at 17:48
  • @DanielGriscom: Take another thought experiment. Take a long, flexible bar and rotate it at a speed that is well below c and completely within confines of reasonable material engineering, barely exhibiting any relativistic effects at all, say, 0.01%c. Then observe it from a point of view of a very fast-traveling observer, traveling at 99.99% c. Will that invalidate the bar somehow? – SF. Jan 24 '16 at 18:06
  • Or let's reduce my original problem to the old "car on a train". Train travels at 0.9c (as seen by an external observer). Car on the train travels at 0.1c relative to the train (as seen by driver). What happens to the wheels of the car? It's the same old worn problem, except instead of observing a single point moving in a straight line I want to observe an area, a rotating shape. – SF. Jan 24 '16 at 18:13
  • The point that matters, as you say in the question, is the top of the wheel. But maybe you're right, and it isn't a duplicate of the rotating bar, but instead it's a duplicate of http://physics.stackexchange.com/q/113818. Either way you're right: it's the "same old worn problem". – Daniel Griscom Jan 24 '16 at 18:55
  • @DanielGriscom: More like extension of that into a plane/field of particles. I've got a fairly good idea what happens to any single particle when translating linear motion between two frames of reference, but I both fail to grasp a larger group moving together at varied speeds and imagine any more complex motion, like rotary. – SF. Jan 24 '16 at 23:02
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    This is not a duplicate of the rotating bar problem. Here we have a relativistic rotation, AND a relativistic translation. It is much more similar to Gamow's cyclist http://www.spacetimetravel.org/tompkins/node7.html a bicycle moving at 0.93 c. The link covers nicely what happens to the wheels, or rather that the wheels must actually be assembled while they are rotating because the circumference of the spinning wheel is much higher than that of the stationary wheel. What happens to the wheel is also very involved in the Ehrenfest paradox, look it up. – mwengler Jan 25 '16 at 16:33
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