Number of wheel rotations relativistic car

[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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[Dale]

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?

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.

 

[lightarrow]

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.

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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[pervect]

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

Usable specific strength can be expressed in various ways. Three waysare shown at right, Vc, Lc and Llg

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.

 

[Dale]

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.

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.

 

[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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[Dale]

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?

Then you have 4 rotations.

 

[lightarrow]

Then you have 4 rotations.

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

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[A.T.]

So you assume that the circumference length does not vary as effect of relativistic length contraction/dilatation.

Without length contraction you would have 8 rotations.

 

[Ibix]

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

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.

 

[lightarrow]

Without length contraction you would have 8 rotations.

Why? Furthermore, does length contraction happens or not?

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[lightarrow]

In the inertial frame where the car is at rest, how could it be otherwise?

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 length 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.

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.

Sorry I haven't understood this part.

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[Dale]

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

No, you just told me to assume it:

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

 

[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.

 

[Dale]

@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.

 

[lightarrow]

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.

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.

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.

It's because I had looked it up that has confused me here :-).
Clearly I hadn't understood much about it.

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.

If I've taken something of this, you mean that it's something the sort of Bell's spaceship paradox: there the thread length 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]

@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

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.

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.

Certainly.

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.

Agreed.

Complicated part:
If you are only given the circumference in the wheel's frame when not rotating

I believed it was obvious that I was asking this! How did you interpret "...when the car is still (engine not working)"?

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.

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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[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!]

 

[Ibix]

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...

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.

If I've taken something of this, you mean that it's something the sort of Bell's spaceship paradox: there the thread length 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?

Pretty much, yes.

 

[Dale]

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:

A car has wheels with exactly 1m of circumference

Then you specify the non rotating condition:

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

Then you specify in the rotating condition:

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?

Then you specify the non rotating condition:

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

Then you specify the rotating condition:

assume that every atom of the wheel stays at the same distance to the axle when the wheel doesn't spin and when does?

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.

 

[pervect]

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.

So which option are you taking again? I'm confused.

 

[lightarrow]

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

Dale, If I write that "the radius" r is constant I have not written "the circumference" C is constant, if I don't know that C = 2*pi*r *even when the wheel rotates at relativistic speed* and that it's what I was asking!
I don't know how else to explain it to you...

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[lightarrow]

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.

So which option are you taking again? I'm confused.

Are you asking me? Then it's "what happens if one spin up the wheel without changing it's radius". About the circumference I have never written that it stays the same when you spin up the wheel, I've written that has 1 m value when it doesn't spin. You are at rest in the ground, in your box, car stopped, nothing is moving relative to anything, you take a measuring tape and you measure the wheel's circumference: 1m. Then you go with the car, accelerating it (gently if you want) till you reach $\gamma = 2$ and you make the radius stays the same, in some way, during the acceleration and after, at constant speed.
question 1: does the circumference stays at 1m value in the car frame at relativistic speed?
question 2: how many rotations (or spins? don't know the language well) makes the wheel from mark A to mark B on the road where the proper distance between A and B is 8m?
Thanks.

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[Dale]

I don't know that C = 2*pi*r *even when the wheel rotates at relativistic speed* and that it's what I was asking!

Ah, that makes more sense.

Spacetime is flat in all inertial frames and furthermore the simultaneity convention results in spatial submanifolds with a Euclidean flat metric. So ALL of the standard Euclidean geometry results hold in anyone frame at anyone time. Specifying the radius in an inertial frame is equivalent to specifying the circumference.

 

[lightarrow]

Ah, that makes more sense. Spacetime is flat in all inertial frames and furthermore the simultaneity convention results in spatial submanifolds with a Euclidean flat metric. So ALL of the standard Euclidean geometry results hold in anyone frame at anyone time.

Thanks.

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[lightarrow]

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.

Certainly.

lightarrow:
If I've taken something of this, you mean that it's something the sort of Bell's spaceship paradox: there the thread length 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?

Pretty much, yes.

Thanks.

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[A.T.]

You are at rest in the ground, in your box, car stopped, nothing is moving relative to anything, you take a measuring tape and you measure the wheel's circumference: 1m. Then you go with the car, accelerating it (gently if you want) till you reach $\gamma = 2$ and you make the radius stays the same, in some way, during the acceleration and after, at constant speed.

question 1: does the circumference stays at 1m value in the car frame at relativistic speed?

Yes. But this implies that the proper length of the rim elements has doubled, so let's assume the rim is build of telescopic elements, which can extend that much.

question 2: how many rotations (or spins? don't know the language well) makes the wheel from mark A to mark B on the road where the proper distance between A and B is 8m?

In the car frame AB is just 4m, so it takes 4 rotations for a circumference that is still 1m.

In the ground frame the wheel is distorted due to the differential velocities. as shown here:

Source: http://www.spacetimetravel.org/tompkins/node7.html

The lowest rim element, which is at rest in the ground frame, has now twice the length it had before the car started moving. So you need only half as many of such elements (and thus half as many rotations) to cover AB=8m, compared to a slow rolling wheel with 1m cirumference. So it's again 4 rotations.

 

[lightarrow]

Yes. But this implies that the proper length of the rim elements has doubled, so let's assume the rim is build of telescopic elements, which can extend that much.In the car frame AB is just 4m, so it takes 4 rotations for a circumference that is still 1m.

In the ground frame the wheel is distorted due to the differential velocities. as shown here:

Source: http://www.spacetimetravel.org/tompkins/node7.html

The lowest rim element, which is at rest in the ground frame, has now twice the length it had before the car started moving. So you need only half as many of such elements (and thus half as many rotations) to cover AB=8m, compared to a slow rolling wheel with 1m cirumference. So it's again 4 rotations.

Thanks A.T.

P.S. In a more correct English and clearer formulation, how would you have expressed my initial question?

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[A.T.]

Thanks A.T.

P.S. In a more correct English and clearer formulation, how would you have expressed my initial question?

You're welcome. The important thing (just like with Bell's paradox) is that Lorentz contraction relates lengths in the same situation, but in different frames. It does NOT relate lengths in different situations, like before and after acceleration.

 

[BillyT]

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.)

Same sort of problem as the relativistic tank. I.e. It would seem, that the top tread of the tank is advancing at twice the speed of the axis driving that tread forward as it turns. Thus I suspect a correct understanding of either problem does not have much to due with the question of "Does the wheel still have circular shape, or a constant circumference." I suspect one must understand what is the top speed of the wheel (or tank tread) wrt the inertial frame of car or tank before much progress can be made.

I admit I do not understand either problem.

 

[pervect]

The easiest reference frame to work things out is the inertial reference frame of the road. The wheel can't be described by a single inertial reference frame, but the part of the wheel touching the road has a velocity of zero in the road frame. The velocity of the wheel where it touches the road is the sum of the velocity v of the axle, and the velocity omega*radius due to the rotation of the wheel, the sum is zero when the wheel rolls. That's why wheels roll, if there was no friction wheels would slide, but the friction acts to make the part of the wheel touching the road have the same velocity as the road does.

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 $2 \pi r / \sqrt{1 - \omega^2 r^2/c^2}$

I could dig up a few references (I'd recommond one of Gron's papers, he has several, and one of Ruggiero's papers (he also has a lot). But the papers may be too advanced, and it doesn't make sense to bother to dig up the references if nobody is able (and also interested enough) to read them.

 

[A.T.]

The easiest reference frame to work things out is the inertial reference frame of the road.

The car frame seems simpler to me, where the wheel is still a circle of known circumference. See post #27.

 

[Dale]

The car frame seems simpler to me, where the wheel is still a circle of known circumference. See post #27.

I agree. The wheel is a circle, and all of the parts of the rim are moving at the same velocity.

 

[pervect]

If you do it in the car frame, I suppose you can argue that the length of the road is shorter in that frame, the wheel is circular (and has a circumference of 2 pi r in that frame). You get the same answer for the number of wheel revolutions (4), thought the explanation is different, one attribute the lower number of revolutions to a shorter road rather than to the non-Euclidean geometry of the rotating wheel. I suppose one could argue that this is simpler, though it's not really the way I think about the problem. It's not necessarily a bad idea to work it out in both frames and make sure you get the same answer.

 

[lightarrow]

If you do it in the car frame, I suppose you can argue that the length of the road is shorter in that frame, the wheel is circular (and has a circumference of 2 pi r in that frame). You get the same answer for the number of wheel revolutions (4), thought the explanation is different, one attribute the lower number of revolutions to a shorter road rather than to the non-Euclidean geometry of the rotating wheel. I suppose one could argue that this is simpler, though it's not really the way I think about the problem. It's not necessarily a bad idea to work it out in both frames and make sure you get the same answer.

I agree, to compare the answers in both frames it's always a good idea.

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