Shouldn't length contraction cause friction under certain circumstances?

Question

I was thinking about the ladder paradox, which if I understand correctly involves a combination of length contraction and relativity of simultaneity. And with that it is resolved.

But there's this example with a train fitting inside a tunnel. Which got me thinking:

If the length contracted train rapidly slows down a lot and stops (from relativistic speeds), wouldn't that cause friction between the wheels and rails because of the "undoing" of the length contracted train?

Someone told me that it's like Bell's spaceship paradox, but in reverse:

The train and rails are initially at rest in S′. The train is then given an acceleration in the negative x′ direction (I am assuming that the front and rear of the train have the same acceleration relative to S′), but the rails remain at rest in S′.

Assuming that the acceleration of the train relative to S' is due to static friction between the wheels and the rails, i.e., the wheels initially don’t slip on the track, the shortening of the train relative to the track, due to length contraction of the train, must be compensated for by a tensile stretching of the train by the static friction forces on the front and back wheels. Assuming these static friction forces are not strong enough to cause the train to break apart, this will eventually cause the maximum static friction force between the wheels and the rails to be exceeded. There will then be kinetic friction (sliding friction) between the rails and wheels. This is accounted for in S by the relativity of simultaneity. In S the front and back of the train do not slow down at the same rate. Again this will cause sliding of the wheels on the rails.

It’s just like Bell’s spaceship paradox. The cord connecting the two spaceships is like the train cars between the engine and the caboose. Just as the requirement that the two spaceships have the same acceleration in the rest frame leads to the cord breaking, the requirement that the engine and caboose have the same acceleration in the rest frame of the tracks, would cause the train coupling to break. But I am assuming the coupling is stronger than the maximum force of static friction, which would, therefore, cause the wheels to slide on the rails instead of rotating without slipping.

So I get that, but then I thought .. for the observers in the train the rails are length contracted, so to them the frictional forces and sliding is in the opposite way?

That can't be right. This is so confusing to me, I really hope someone can help me out here, because I can't get a full, satisfying answer anywhere.

It's very likely that I'm misunderstanding something big time, but I have no idea what.

Answer 1

I don't think it's all that complicated.

Let’s assume that in your train problem there is a rack and pinion action between the train wheels and the track instead of friction.

Assume that the speeding up of the train relative to the initial rest frame of the train is accomplished by the interaction of the wheels (pinions) with the track (rack). By hypothesis, the rear of the train lags behind the front of the train by a constant number of teeth in the rack. (The rack acts as a coordinate axis for the initial inertial rest frame of the train.) This is because by the design of the problem the rear and front of the train have the same acceleration relative to the rail.

Assume further, that there is a clock at each tooth of the rail, and that these clocks are synchronized in the initial rest frame of the train (which is also the continuing rest frame of the track). Simultaneous measurements of the location of the front wheel and back wheel, using these synchronized clocks always gives the same number of teeth of the rack (rail), between the front and rear, regardless of the speed that the train has achieved. Because of length contraction, this means that the proper length of the train has increased, i.e., the train has been stretched by a difference in contact force between the front and rear wheels. This will become more pronounced as the speed of the train, relative to its initial rest frame, increases. Eventually, teeth of the rack or pinion will not be strong enough to maintain the same acceleration of the front and rear of the train, because to do so would require continued stretching of the train to increase its proper length. (Remember that the proper length must increase to give the same measured contracted length in the initial rest frame of the train as the length of the train when it was initially at rest.) This means that the teeth of the rack and pinion must break.

If we use the same reasoning in a circumstance where friction provides the acceleration of the train, rather than a rack and pinion mechanism, there must come a point at which the wheels skid on the track and the continued equal acceleration of the front and rear cannot be maintained.

Answer 2

This question is complicated by using accelerated frames (not just one but two differently accelerated frames, one for the front and one for the back of the train) plus the unintuitive weirdness that is the relativistic wheel. As a result I am much less certain of the correctness of this answer than I would like.

Simultaneous accelerations of the back and front of the train in the station frame:

Measured from the station: the train slows down and tries to get longer according to the Lorentz boost, while the distance between front and back stay the same. As a result, the train buckles under compressive stress. The front and back wheels rotate at the same rate.

Measured from the train: the front of the train accelerates first while the back of the train keeps going for a moment, then begins to accelerate. There is a relative velocity between the front and back of the train with the front being pushed towards the back, so the train buckles under compressive stress. The front of the train is moving towards the back of the train, so the wheels would, on a track of constant length, therefore need fewer revolutions. However, the track under the train is getting longer and longer as the train accelerates, and since the track itself is moving backwards, the rear wheels cover a greater distance of track per revolution than the front wheels.$^1$ Consequently, the wheels in front have the same number of revolutions as the wheels in the back.$^2$

Eventually the acceleration ends first at the front of the train, then at the rear (allowing the rear of the now-buckled train to match the velocity of the front of the train).

The different frame observers disagree about the geometry of spacetime. They do not agree about in what order events transpired, where, when, and how far apart they were in space and time. But they do agree about which events happened, which events were in the causal past of other events, and, with a bit of math, what everyone else must have seen.

Simultaneous accelerations of the back and front of the train in the train frame:

Measured from the station: the back of the train accelerates first, so the train moving train gets longer as though pulled from the back - the back slows down first, creating space for the train to get longer (without any tensile stress, since the length is changing according to the Lorentz boost). The front and back wheels rotate at different rates to accommodate the fact that the back of the train is slower than the front of the train.

Measured from the train: the train accelerates simultaneously, staying the same length. The moving, slowing track gets longer under it (coordinate accelerating forward opposite its direction of motion). As in the other example, the time-dependent lengthening of the track affects the back wheels more than the front wheels. The front and back wheels rotate at different rates to accommodate the lengthening track, with the back wheels rotating fewer times than the front wheels to cover the same section of track.

Again: The different frame observers disagree about the geometry of spacetime, but they agree about which events happened, which events were in the causal past of other events, and what everyone else must have seen.

---

1: There is also a contributing factor from the fact that an observer at the front wheels, who has a smaller relative velocity to the track than an observer at the rear, will measure a slightly longer total track than an observer at the rear wheels, until the front stops accelerating and the back continues to accelerate for a moment. So an observer on the train will not measure the front and back wheels rotating simultaneously, although she will count the same number of total revolutions during the acceleration for front and back.

2: Note how if the track doesn't in every measurable way actually get longer, the front wheels, which spend some time moving towards the rear of the train in the train frame, would have fewer revolutions than the back wheels over the total elapsed time of the acceleration. The station universe would observe revolutions that never happened in the train universe! Lorentz transformed observables can be usefully thought of as 'illusions' having to do with information transfer, but information transfer is no illusion. It is the whole substance of measurable reality. The information that the track transfers to the wheels includes the fact that it is longer when the relative velocity is smaller.