In advance I apologize for any bad English.
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.
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."
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.
It's very likely that I'm misunderstanding something big time, but I have no idea what.
Thanks in advance!!
