Car, wheels and Lorentz contraction of the road: Is this a Paradox?

[LikenTs]

A car travels at low speed on a road from start to finish, and counts the number of turns of the wheel, which gives it a road length of N.2πR, where R is the radius of the wheel.

Then he does the race again at relativistic speed. He sees the road with a Lorentz contraction. However, he has to record the same number of wheel turns N between start and finish. This has to be an invariant.

So he has to measure that the road length is N.2πR/γ, contracted, where γ is the Lorentz factor.

What is happening then, from the pilot's point of view? I think he cannot see his wheel deformed so that it is not a circle of perimeter 2πR, and that he cannot see its radius contracted, and I think that rolling, without slipping, is an invariant. How then does he come to observe the Lorentz factor in the formula and the road contraction?

 

[Dale]

However, he has to record the same number of wheel turns N between start and finish. This has to be an invariant.

Careful. The low speed and high speed scenarios are physically different scenarios that are not related to each other by a Lorentz transform. The number of turns is invariant under the Lorentz transform, but not between the slow and fast scenarios.

I think he cannot see his wheel deformed

In fact, he must see it deformed. There is no Born rigid motion with different angular velocity.

 

[anuttarasammyak]

However, he has to record the same number of wheel turns N between start and finish. This has to be an invariant.

Not invariant for fast and slow cases, the faster car moves the fewer wheel rotates. But N is invariant for any frame of reference. Let us check it.

In pilot's IFR, due to Lorentz contraction of moving road, the number of necessary wheel rotaion is reduced.
$$N=\frac{N_0}{\gamma}<N_0$$ where
$$N_0=\frac{D}{2\pi R}$$

In wheel's FR which is rotation frame of reference the perimeter is
$$2\pi R\gamma$$
By N turn
$$2\pi R\gamma N=2\pi R N_0=D$$
Accumulated moving disance of the contact point of "rotating" road coincides with D.

In road's IFR, let us see Fig.9 of Galilei's New Science.

AB, part of the hexagone periphery, is on the line C. Next the hexagone rotates around B and BC is on the line C, next the hexagone rotates around C, etc with constant pace. There is a moment that both A and B are on the line C in hexagone (I, approximately)FR. But for the line IFR "A is on C" and "B is on C" can not be simultaneous. "A is on C" is past of "B is on C". Distance between A,B marks on C is $\gamma l_0$ where $l_0$ is proper lenght of periphery AB. The result is similar for any n-polygone, circle in limit. Thus we know in N turn
$$2\pi R \gamma N = 2\pi R N_0 = D$$

So in all the three FRs same N rotation of wheel for transfer of distance D in road IFR.

 

[PeterDonis]

I think he cannot see his wheel deformed so that it is not a circle of perimeter 2πR, and that he cannot see its radius contracted

Neither of these are correct. The wheel is rotating in his rest frame; it is not motionless.

I think that rolling, without slipping, is an invariant.

I'm not sure what you mean by "rolling". The invariant is the number of times a given spot on the wheel contacts the road during the trip. But, as has already been pointed out, "invariant" means invariant for the same scenario under a Lorentz transformation. It doesn't mean the same for two different scenarios.

 

[Ibix]

He sees the road with a Lorentz contraction. However, he has to record the same number of wheel turns N between start and finish.

This is a contradiction, so it's where I'd start investigating my premises.

Your reasoning in the car's rest frame seems solid at first glance. The number of revolutions of the wheel needed to complete the course will reduce with speed due to length contraction of the course (in the implausible case that a road wheel does not expand under centrifugal forces as it spins with a rim velocity near $c$, anyway). We are using fairly simple physics here - so let's accept it for the time being.

Now let's look at it from the road rest frame. Wheels are funny things in relativity. When viewed in a frame where they are moving they are not circular:

Source: https://www.spacetimetravel.org/tompkins/7
So to me it is immediately doubtful that those two wheels would need the same number of revolutions to complete the course, since their rim lengths are very different and the rotational behaviour of the right hand one must be very strange. Of note, the speed of the rim varies around its perimeter, from zero at the contact point to $\frac{2v}{1+v^2/c^2}$ at the top. So I'd now be very wary of your assumption that the wheel must turn the same number of times in both low- and high-speed cases.

In fact, the answer is that the number of turns of the wheel will be proportional to $1/\gamma$, as is obvious from the length of the road as measured in the car's rest frame where the wheel is circular and the rim moves with constant speed. As others have noted, you are rather misusing the term "invariant", which may have mislead you.

 

[LikenTs]

So I'd now be very wary of your assumption that the wheel must turn the same number of times in both low- and high-speed cases.

My reasoning that the wheel must turn the same number of times at any speed is the following:

Suppose the wheel is a gear and the road is a gear track. The number of teeth of the road between start and finish has to be the same at any speed, and consequently also the number of turns of the wheel. So I assume that in the rolling case, without gears, the number N of turns must be conserved. In that sense I meant invariant.

In the case of gears, at first glance, the driver sees no Lorentz contraction, Road Length = Number of teeth x Nominal Distance between teeth, except that if he measures the distance between teeth, perhaps with lasers and clocks, he sees a Lorentz contraction . Then it concludes that L = Number of teeth x Nominal Distance between teeth/γ

However in the case of rolling, which I think we can imagine as gears with teeth of infinitesimal spacing, the driver thinks: The number of wheel turns N is the same, the wheel is circular, its radius R remains the same, and pi is pi, therefore L = N.2πR, and there is not contraction, what of his assumptions is wrong?

 

[Dale]

So I assume that in the rolling case, without gears, the number N of turns must be conserved.

You are further assuming that the same gear which meshes with the track at low speeds will mesh with the track at high speeds. That, mathematically, cannot be the case. There is no Born rigid transformation between the low speed gear and the high speed gear. The gear must unavoidably undergo strain in going from the low speed to the high speed scenario.

 

[anuttarasammyak]

Say car speed v is almost light speed, in pilot's IFR wheel turns infinitesimal angle of less than one tooth of gear in wheel which should correspondent to many teeth of line. It is impossible. Teeth of line is so densed by Lorentz contraction to pinch teeth of wheel between. Gear track does not seem to work for relativistic speed.

[EDIT] To make it work we can apply dense-sparce asymmetric design to teeth of wheel and to teeth of line. Line teeth width and distance to next tooth should be expanded by factor $\gamma$.

 

[Nugatory]

My reasoning that the wheel must turn the same number of times at any speed is the following:

The teeth-and-gear model is suspect because it contains a bunch of unstated assumptions about the rigid geometry of the wheel and these assumptions fail at relativistic speeds. We can fix this by using @PeterDonis's formulation of the invariant (" the number of times a given spot on the wheel contacts the road during the trip") and yes, that is indeed the same in both frames.

The cause of the apparent paradox is the bogus assumption that the circumference of the wheel is $2\pi R$ where $R$ is the radius of the wheel when it is not turning. It's not, although the effect is completely insignificant except at relativistic speeds so is usually ignored, and it is different in different frames.

Relativistic rotating motion is a surprisingly non-trivial problem, and your paradox has challenged more than one competent physicist. Googling for "Ehrenfest paradox" and "Relativistic rolling wheel" will bring up many good references.

 

[PeroK]

@LikenTs Your basic assumption about the shape of things being unchanged at any relative speed ignores length contraction in the first place.

 

[Dale]

We can fix this by using @PeterDonis's formulation of the invariant (" the number of times a given spot on the wheel contacts the road during the trip") and yes, that is indeed the same in both frames

That is the same in different frames, but it may not be the same in the slow vs fast scenarios. The wheel must strain, so the circumference may be different

I know you are aware, I just want to make sure that the OP doesn’t misunderstand this statement

 

[Ibix]

In the case of gears

Have a look at the spokes of the relativistic wheel I posted above. It's not even clear that the teeth would mesh with the rack at relativistic speeds. So there's quite a lot of work you need to do before you can reason from that analogy. (In fact it will mesh, but you'll need to change the pitch of the rack or cog for the given speed in order to achieve it.)

A much better approach is to start from the car's rest frame where things are simple and work out what the description in the road frame ought to be. It's going to be a mess and probably require numerical solution, but will give you a consistent answer where trying to work from analogy will just keep biting you in different ways.

 

[Nugatory]

from the car's rest frame where things are simple

"Less complicated", but relativistic rotation is never simple

 

[Vanadium 50]

Tooth and notch. Pole and barn. To-may-to, To-mah-to.

 

[Sagittarius A-Star]

View attachment 329825

However in the case of rolling, which I think we can imagine as gears with teeth of infinitesimal spacing, the driver thinks: The number of wheel turns N is the same, the wheel is circular, its radius R remains the same, and pi is pi, therefore L = N.2πR, and there is not contraction, what of his assumptions is wrong?

In the rest frame of the driver, the circumference of the rotating gear moves with relativistic speed and is length-contracted compared to the circumference $2\pi R\gamma$ in the rotating frame, which causes mechanical stress along the circumference. If a rubber band would be used as a ruler, it would also provide a wrong measurment result, if it would be stretched by the factor $\gamma$.

Assumption: The gear was constructed without internal mechanical stress, when it was not rotating.

Source (see under "Einstein and general relativity"):
https://en.wikipedia.org/wiki/Ehrenfest_paradox

 

[LikenTs]

You are further assuming that the same gear which meshes with the track at low speeds will mesh with the track at high speeds. That, mathematically, cannot be the case. There is no Born rigid transformation between the low speed gear and the high speed gear. The gear must unavoidably undergo strain in going from the low speed to the high speed scenario.

I assume gears mesh at any speed if they mesh at low speed, because in road IFR the contact point of whell-Gear is at zero speed, and in the car IFR it is at same speed than track, so any contraction should be the same.

Relativistic rotating motion is a surprisingly non-trivial problem, and your paradox has challenged more than one competent physicist. Googling for "Ehrenfest paradox" and "Relativistic rolling wheel" will bring up many good references.

"Ehrenfest paradox" is very interesting. Happy that I am not the only one who has doubts about this problem or paradox. However, I don't see or understand a clear resolution of the problem on the wikipedia page. At the moment I don't care about the point of view of a rotating observer, which I think is outside the scope of the SR, but the point of view of the driver and an observer on the ground, particularly the first one for symmetry and apparent greater simplicity.

@LikenTs Your basic assumption about the shape of things being unchanged at any relative speed ignores length contraction in the first place.

I don't quite understand why. I focus on the driver's IFR. The center of the wheel is at rest. The entire contour of the wheel goes at the same speed. Due to symmetry, it maintains its circular shape, and the radius, perpendicular to the rotation, maintains its same length as at rest. He doesn't see anything unusual about his wheel.

Rigid solids have been discussed, but this is an ideal experiment. It does not seem to me that resorting to theories of elasticity serves as an argument to resolve apparent paradoxes of SR. The driver can change the wheels for the relativistic high-speed test, so that at that high speed of rotation they maintain the same radius (and number of teeth in the case of gear-wheels) as the low-speed wheels.

Tooth and notch. Pole and barn. To-may-to, To-mah-to.

I feel that my command of English does not allow me to understand what you mean.Above all, I am almost convinced that the number of wheel turns must be conserved in tests at different speeds.

 

[Sagittarius A-Star]

I focus on the driver's IFR. The center of the wheel is at rest. The entire contour of the wheel goes at the same speed. Due to symmetry, it maintains its circular shape, and the radius, perpendicular to the rotation, maintains its same length as at rest. He doesn't see anything unusual about his wheel.

Replace the rotating gear by a disk made of very elastic rubber, and only the circumference is made of a thin layer of metal. This would allow the metal-circumference to length-contract to $2\pi R/\gamma$ while spinning up. Then the metal-circumference would compress the rubber to have a shorter radius $r=R/\gamma$. In this case you would have no apparent paradox in the inertial frame of the driver.

 

[Ibix]

The driver can change the wheels for the relativistic high-speed test, so that at that high speed of rotation they maintain the same radius (and number of teeth in the case of gear-wheels) as the low-speed wheels.

Right - but the road will be length contracted in the driver's frame, so the same number of teeth on the same radius wheel will not mesh with the shorter pitch of the teeth on the fast moving road.

 

[PeterDonis]

Your reasoning in the car's rest frame seems solid at first glance.

No, it isn't, because the wheel is rotating in this frame, i.e., it is moving. The OP assumed the wheel was at rest in the car's rest frame.

 

[Ibix]

No, it isn't, because the wheel is rotating in this frame, i.e., it is moving. The OP assumed the wheel was at rest in the car's rest frame.

I don't see where he assumed that. And in any case, it only invalidates his analysis if the radius of the wheel changes as its speed varies. Which it would in reality, but for a sufficiently strong material you could neglect that. Or change wheels, as proposed in his last post. The problem in the OP is using the same $N$ for both low and high speed cases.

 

[PeterDonis]

I don't see where he assumed that.

He assumed that the wheel's circumference is $2 \pi R$ in the car's rest frame. That requires the wheel to be at rest in that frame, not rotating. In the non-relativistic case, of course the correction for wheel rotation is negligible; but in the relativistic case, it most certainly isn't, as has already been pointed out.

 

[PeterDonis]

it only invalidates his analysis if the radius of the wheel changes as its speed varies.

This is not correct; it invalidates his analysis for the obvious reason that you yourself point out:

The problem in the OP is using the same $N$ for both low and high speed cases.

And he is deducing this from his assumption that the wheel's circumference in the car's rest frame is the same ($2 \pi R$) for both cases. Which it isn't.

 

[Nugatory]

At the moment I don't care about the point of view of a rotating observer,

You should, because the teeth of your hypothetical gear are rotating observers. Which means….

Due to symmetry, it maintains its circular shape, and the radius, perpendicular to the rotation, maintains its same length as at rest.

Yes, but…

He doesn't see anything unusual about his wheel.

is not right. The circumference of the wheel is equal to $2\pi R$ only when the wheel is not rotating. All other times, the circumference is a fairly complicated function of the rotation rate as well as the radius, and the effect is to exactly cancel out the effects of length contraction along the distance traveled by the car.

(And be aware that because of relativity of simultaneity and the different parts of the wheel moving in different directions so not at rest relative to one another in any inertial frame, “circumference” in this context is a slippery concept, it doesn’t work like you expect. You can’t trust your intuition here, a lifetime of experience with non-relativistic speeds will steer you wrong. You have to do the math).

 

[Sagittarius A-Star]

It does not seem to me that resorting to theories of elasticity serves as an argument to resolve apparent paradoxes of SR.

That's wrong. The length of a (part of a) body follows the length contraction exactly, if it has no internal mechanical stress. If you stretch a body by using a force, you can for example cancel-out a length contraction effect.

 

[Nugatory]

Above all, I am almost convinced that the number of wheel turns must be conserved in tests at different speeds.

At any given speed of the car relative to the ground, the number of wheel turns between two given marks on the ground will be the same for all observers. However, it will be different for different speeds of the car relative to the ground.

 

[Nugatory]

It does not seem to me that resorting to theories of elasticity serves as an argument to resolve apparent paradoxes of SR.

In any problem involving the change of speed of anything except point particles, elasticity will be an essential part of the analysis. Perfect rigidity of anything is an approximation only valid under non-relativistic conditions, and this isn't just because we lack ideal rigid materials. Classical rigidity is DEFINED as all parts of the rigid body changing speed at the same time if any part does, and that definition is incompatible with the relativity of simultaneity.

The driver can change the wheels for the relativistic high-speed test, so that at that high speed of rotation they maintain the same radius (and number of teeth in the case of gear-wheels) as the low-speed wheels.

No he cannot. He change out his wheels for a new set that has the same number of teeth and the same circumference once spun up, but those wheels will have a different radius and diameter than the wheels that work in the low-speed test.

 

[Ibix]

He assumed that the wheel's circumference is $2 \pi R$ in the car's rest frame. That requires the wheel to be at rest in that frame, not rotating. In the non-relativistic case, of course the correction for wheel rotation is negligible; but in the relativistic case, it most certainly isn't, as has already been pointed out.

That surely depends upon your material model, though. There is a sense in which a spun-up wheel wants to contract radially, because that would reduce the stress caused by its circumference length contracting. Of course that imposes radial stresses on the wheel, which will resist being compressed. And the wheel will try to expand like a centrifuge. I can't see how one can make a categorical statement that this never nets to (near) zero change in radius without specifying how the material behaves. Other than the obvious "any known material would have disintegrated long since".

This is not correct; it invalidates his analysis for the obvious reason that you yourself point out:

The use of the same $N$ invalidates his analysis in one frame or other. If you take $2\pi RN$ to be the length of the road in its rest frame then $2\pi RN/\gamma$ is the correct contracted length, whether $N$ and $R$ are relevant to the wheel in the car rest frame or not.

 

[LikenTs]

I have to read and think more carefully about the answers, but as a quick summary.

According to @Sagittarius A-Star there is a contraction of the radius due to a mixed elasticity and lorentz effect, so the formula for the road length I think would be:

N.2π(R/γ)

According to @ibix, and several others, the number of wheel turns would change with speed, formula:

(N/γ).2πR

And according to "Ehrenfest paradox", also a change in pi is suggested, which would give the formula:

N.2(π/γ)R

Without ruling out any other intermediate alternative.

Clearly all the formulas give the same result, the measurement of a Lorentz contraction, as expected in SR, however they are very different physical scenarios. In one it can be seen that the number of turns of the wheel changes according to the speed, in another one the height at which the vehicle circulates changes due to wheel radius, and in another one it is due to a non-Euclidean geometry.

So, what would be the real solution?

 

[Dale]

I assume gears mesh at any speed if they mesh at low speed, because in road IFR the contact point of whell-Gear is at zero speed, and in the car IFR it is at same speed than track, so any contraction should be the same.

It is a wrong assumption and the reasoning is not valid. The issue is that the gear cannot be rigid. It must strain. Strain means that its shape changes, it deforms. Strain gauges mounted on the gear will detect this change in shape.

This is not about length contraction. Length contraction does not produce strain. Strain gauges placed on an object do not detect length contraction. This is a completely different effect from length contraction. This is the shape of the gear changing in its own frame.

A gear whose shape changes will not mesh, and we are guaranteed by relativistic kinematics that the gear’s shape must change.

It does not seem to me that resorting to theories of elasticity serves as an argument to resolve apparent paradoxes of SR.

Then you cannot have both the high speed and the low speed scenarios. You must discard one or the other because you cannot go from one to the other without discussing elasticity and strain.

 

[Nugatory]

So, what would be the real solution?

For any given choice of wheels, the number of turns (this is best defined as @PeterDonis did, the number of times that a given point on the wheel touches the road between two given points on the road) will change with the speed of the vehicle relative to the road. The choice of frame (car at rest, road at rest, road and car both moving relative to some third party watching the experiment, ...) will not affect this result.

This is the real solution, the frame invariant and paradox-free fact of what actually happens.
The various explanations you quote above are different approaches to calculating that answer.

 

[PeterDonis]

That surely depends upon your material model, though.

I don't think there can be any material model that has the wheel having the same shape in the relativistic as in the non-relativistic case. Relativity imposes limits on the properties of all materials. And relativistic kinematics imposes limits on what can and cannot be held constant as the spin rate of a wheel is varied; in particular, there is no Born rigid way to spin up a wheel from lower to higher frequency.

 

[PeterDonis]

The use of the same $N$ invalidates his analysis in one frame or other.

Not for the same relative speed (low or high). The issue is that $N$ cannot be the same for both relative speeds. But if we hold the relative speed fixed, then $N$, the number of times a given point on the wheel contacts the road from start to finish, is frame invariant.

 

[A.T.]

I assume gears mesh at any speed if they mesh at low speed, because in road IFR the contact point of whell-Gear is at zero speed, and in the car IFR it is at same speed than track, so any contraction should be the same.

Let's assume you build the gear already rotating at a given rate from tooth-segments that you accelerate tangentially and connect. The number of teeth you can fit on a circumference of a given radius depends on the rotation rate, because with greater tangential speed the segments are more length contracted.

So if you want your rack to mesh with the gear (no deformation of the gear-teeth), then you must build different wheels with a different number of teeth for any speed.

But if you use the same gear and enforce a constant radius, then the circumference will deform differently at different speeds, and it won't mesh with the rack. Not because their kinetic length contractions are different at the contact, but because the gear teeth are actually physically deformed, as is measurable with stain gauges.

 

[Dale]

@LikenTs one other thing to recognize is that any contradiction between the slow scenario and the fast scenario will not be due to relativity because relativity does not govern the difference between the two scenarios. The fast and slow scenarios are related by material laws of elasticity and stress and strain. So any inconsistency would point to your laws of elasticity being incorrect (probably non-relativistic).

In the fast scenario, relativity governs the relationship between the axle frame and the road frame. If you found a contradiction there, the same scenario in two different frames, then that would be a relativity paradox.

Since you have said that you don’t want to resort to theories of elasticity then I would recommend the following approach. Simply specify the geometry in the fast scenario and don’t even consider the slow scenario. Analyze the fast scenario in the axle and ground frames and check for any contradictions that interest you.

 

[Nugatory]

Since you have said that you don’t want to resort to theories of elasticity then I would recommend the following approach. Simply specify the geometry in the fast scenario and don’t even consider the slow scenario. Analyze the fast scenario in the axle and ground frames and check for any contradictions that interest you.

@LikenTs This is really good advice.... And another step that you might want to consider is to ignore the wheel, instead consider a device that puts a drop of paint on the rail once every second (according to the frame in which the car is at rest) - this is basically what you've been expecting to get out of one turn of the rotating wheel, but allows you to avoid the seriously non-trivial analysis of the wheel. Once you have this relativistic analysis under your belt, you will be much better positioned to take on the wheel.