# Number of wheel rotations relativistic car

1. Jan 1, 2016

### lightarrow

A car has wheels with exactly 1m of circumference and travels at constant relativistic speed with $\gamma = 2$ in an horizontal rectilinear road between two points A and B marked on the road itself and separated from a distance of 8 m, as measured from a frame of refence which is still with respect of the road (so it's the "proper distance" between A and B).
Assuming that the wheels can resist the centrifugal forces and that they don't expand because of those forces (it's not a realistic car, of course), how many rotations do the wheels make from the mark A to the mark B?

To measure that number of rotations one can simply mark a red point on a wheel's circumference and see how many times that point comes in contact with the road, assuming it's in contact with the mark A at t = 0.

Edit: the wheel have exactly 1 m of circumference at car stationary on the road, before starting the engine.

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lightarrow

Last edited: Jan 1, 2016
2. Jan 1, 2016

### Staff: Mentor

I assume that you mean that the wheel has exactly 1 m circumference in the inertial reference frame of the wheel. Then the wheels make 4 rotations.

3. Jan 1, 2016

### lightarrow

Sorry not to have specified it, the wheel has exactly 1 m circumference when the car is still with respect to the road ( I mean, before starting the engine). I've edited my first post.

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lightarrow

4. Jan 1, 2016

### pervect

Staff Emeritus
Well, the circumference of the wheel will, in general, change when you spin it up. So one needs more details on what the wheel is made of to answer the questions. The Ehrnfest paradox tells us there is no way to spin up a wheel rigidly. Without getting overcomplicated, one could reasonably ask what the answer was if the wheel was spun up without changing it's circumference, or one could ask what happens if one spin up the wheel without changing it's radius, or one could ask what happens if the wheel is made up of some specific material. For any known material the answer to the last question is that the wheel fails, a figure of merit for the strength of a wheel is the "characteristic velocity", closely related to the speed of sound in the wheel. And even for buckytubes, the speed of sound is a lot less than the speed of light. I won't give a detailed calculations, but see for instance http://hyperphysics.phy-astr.gsu.edu/hbase/sound/souspe2.html, and the following article on the design of "space tethers", http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19870000736.pdf

So the characteristic velocity Vc of the material of the wheel is a good predictor of how fast a tether cab be spun up before it fails (and, generally speaking, a wheel as well, though some of the calculational details will have different constants)

If one is really interested in the Ehrnfest paradox itself, the best approach is to address the issue directly, where there is some (perhaps too much) literature on the topic, both in the professional literature and here on PF.

Last edited: Jan 1, 2016
5. Jan 1, 2016

### Staff: Mentor

Then the problem has no easy solution. The wheel cannot retain its shape when spinning up since Born rigid motion cannot involve angular acceleration. Therefore you need a material model, like a relativistic version of Hooke's kaw to determine the strains and stresses in the wheel.

Once the circumference has been determined in the axle's frame, then it is an easy matter to determine the number of rotations in the ground frame. But the first part of the calculation is beyond me.

6. Jan 2, 2016

### lightarrow

Thanks to both for the replies.
Does the question changes if I specify that, in some way, realistic or not, the wheel's points on the border, stays at the same constant radius r, in the axle's frame, where r is the wheel radius at car stopped?

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lightarrow

7. Jan 2, 2016

### Staff: Mentor

Then you have 4 rotations.

8. Jan 2, 2016

### lightarrow

So you assume that the circumference lenght does not vary as effect of relativistic lenght contraction/dilatation.
Is this proved?

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lightarrow

9. Jan 2, 2016

### A.T.

Without length contraction you would have 8 rotations.

10. Jan 2, 2016

### Ibix

In the inertial frame where the car is at rest, how could it be otherwise? This means that the rim of the wheel is under ridiculous stress due to the atoms being further apart in their own instantaneous rest frames than they would if the wheel were not rotating.

11. Jan 2, 2016

### lightarrow

Why? Furthermore, does length contraction happens or not?

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lightarrow

12. Jan 2, 2016

### lightarrow

Einstein initially believed the geometry of the relativistic rotating disk couldn't be Euclidean, then he seemed to have changed idea, or maybe he first believed the circumference had to contract, then he believed the opposite, that it had to expand; anyway I don't know the background of the rotating disk problem. I simply ask if, assuming r is constant, the circumference lenght stays the same or not, that is, if the wheel's geometry stays Euclidean or not, in the inertial frame where the car is at rest.
Sorry I haven't understood this part.

lightarrow

13. Jan 2, 2016

### Staff: Mentor

No, you just told me to assume it:

Last edited: Jan 2, 2016
14. Jan 2, 2016

### Ibix

In the rest frame of the car its wheels are circular at rest and are circular when spun up. This frame is Euclidean (well, Lorentzian), so the circumference is $C=2\pi r$ - as long as both C and r are measured in the car's rest frame. As DaleSpam noted, you said that C=1m - so problem solved.

The non-Euclidean stuff comes in if you try to work out what a fly clinging for dear life to the rim of the wheel would observe. Look up the Ehrenfest paradox.

My comment about stresses was simply that, in the car frame, the fly is moving with $\gamma=2$ and is therefore length contracted. We can fit twice as many flies around the rim as we could when it was stationary. As with flies, so with atoms. But there is no way to introduce more atoms, so the existing ones must move apart in the tangential direction if the circumference is to remain constant in the car's rest frame. That implies stresses over and above the obvious centripetal forces.

15. Jan 2, 2016

### Staff: Mentor

@lightarrow. You are being excessively confusing about this. There is a simple and a complicated part of your question. You are making the whole thread confusing by jumping around and not being clear and consistent in your assumptions.

Simple part:

Once you have the circumference in the axle's frame while rotating then you simply apply length contraction to the ground to determine the distance in the axle's frame. Divide the circumference by the length to get the number of rotations. It doesn't matter if the circumference in this frame is given or calculated via some other assumptions. Once you have it, this part is easy.

Complicated part:

If you are only given the circumference in the wheel's frame when not rotating then you have to do a lot of work to determine its shape while rotating. You need some additional assumptions like a relativistic version of Hooke's law, the stress and strain relationship in the material, etc. This part is really difficult and it is not possible to assume a rigid wheel to make it easier.

16. Jan 2, 2016

### lightarrow

Thanks, I see now where I was confused: I was reasoning in the wheel's non inertial frame, not in the car's inertial one... We actually "are" in pure SR here.
It's because I had looked it up that has confused me here :-).
If I've taken something of this, you mean that it's something the sort of Bell's spaceship paradox: there the thread lenght is held constant because of how the spaceships moves, but it should be Lorentz length contracted because of the relativistic speed, so this mean that the thread is stretched. Is the same here with the wheel's atoms at the rim?

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lightarrow

17. Jan 2, 2016

### lightarrow

But you don't have it, because I wrote the circumference is 1m when the car is stopped, that is when the wheel doesn't spin, not that this value is kept when it spins, in the car's frame. Maybe is a language problem of mine.
Now I (think to) have understood that the circumference have 1m length even when it spins, but this is what I didn't know and that I was asking.
Certainly.
Agreed.
I believed it was obvious that I was asking this! How did you interpret "...when the car is still (engine not working)"?
So it's not enough to assume that every atom of the wheel stays at the same distance to the axle when the wheel doesn't spin and when does?

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lightarrow

18. Jan 2, 2016

### TSny

Why not use FBR wheels? You know, where FBR stands for “Forget Born Rigidity”. See picture below. A top of the line model would have many more (smaller) shoes more closely spaced than the one shown. Before the wheel is set spinning, the shoes are equally spaced around the circumference.

We imagine the car is initially floating at rest in deep outer space in some inertial frame and the wheels are not yet spinning. We next get the wheel spinning by using the BS mechanism, where BS stands for…you guessed it… “Bell Spaceship”. That is, at t = 0 in this frame each shoe begins accelerating tangentially to the circumference (by a small rocket thruster attached to the shoe, not shown.) In this frame, the shoes accelerate with the same acceleration for the same time until they obtain $\gamma = 2$ relative to our car frame. Consecutive shoes have therefore not changed their relative separation distance according to the frame of the car.

Of course, in an instantaneously comoving frame of one of the shoes, the distance to the next shoe has about doubled. But since the shoes are not connected to one another along the circumference, we have eliminated most of the Born stress that would occur in a standard wheel. FBR!

Now that the wheels are spinning, we can imagine another inertial frame that is moving relative to the car frame at $\gamma = 2$. This frame has an 8 m stretch of road lined up with the relative motion of the car. In this frame, the car is going to zoom past with the wheels of the car just touching the road. The shoes of the FBR wheel will be “running” along the road with zero relative velocity between the road and any shoe that is touching the road.

Now we can ask how many revolutions the wheels make in traveling the length of road.

[Ok. OK. For you serious types, this was meant only as an amusement. Happy New Year to everyone!]

#### Attached Files:

• ###### FBR wheel.png
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19. Jan 2, 2016

### Ibix

Actually you were reasoning in a mixture of the two, which is what was really giving you problems. If you were reasoning purely in the rotating frame you'd have described the road as spinning (and curved, I suspect, although I haven't worked that out in detail), which is how it looks from a rotating reference frame.

Pretty much, yes.

20. Jan 2, 2016

### Staff: Mentor

This is why I am confused, maybe it is just a language issue, but it seems more than that to me. First you don't specify which condition:
Then you specify the non rotating condition:
Then you specify in the rotating condition:
Then you specify the non rotating condition:
Then you specify the rotating condition:
If you specify the circumference in the rotating condition then the circumference in the non rotating condition is irrelevant to the question. It would only become relevant if you want to know the strain in the wheel.

In your most recent specification the strain is purely tangential and is an expansion. But again, that part does not impact the answer to the question about the number of rotations.