Will the wheels of a relativistically moving train, according to an observer at rest, show friction with the tracks, when decelerating?

Question

Imagine a train moving relativistically according to an observer at rest wrt to the tracks it's moving on. The train seems contracted in the direction of motion.

At the points where the wheels make contact with the tracks, the wheels have zero velocity, so the distance between these points is not Lorentz contracted (like the caterpillars of a tank are at rest).

Now what will happen when the train suddenly decelerates, simultaneously at all parts, as seen from the observer at rest? Obviously, the wagon will expand, but what will happen to the wheels? Will they follow the expanding motion, which leads to friction between the wheels and the track?

On the other hand, when the train starts to accelerate from zero velocity, the distance between the wheels will not Lorentz contract, so the wheels seem to have to slide with friction, dragging behind the contracting wagon, so to speak. Will the reverse happen when decelerating?

Answer 1

Now what will happen when the train suddenly decelerates, simultaneously at all parts, as seen from the observer at rest?

What will happen is that the train will get crushed. This is not a statement regarding the strength or stiffness of the train, but rather a statement about the description of the motion.

You have specified that it decelerates "simultaneously at all parts, as seen from the observer at rest". Due to the relativity of simultaneity, in the moving frame this means that the front decelerates before the rear. Because the front is decelerated and the rear is not decelerated the train gets crushed.

This crushing will occur regardless of how gentle the deceleration is. The crushing is caused by the stipulation "simultaneously at all parts as seen from the observer at rest" and not the "suddenly decelerates". In other words, the only way to get it to decelerate simultaneously for the ground observer is to physically crush the train. If you do not crush the train in exactly the right way the deceleration will not be simultaneous in the ground frame.

None of the rest of your analysis follows because the rest of your analysis assumes that the train is rigid. But a rigid train is incompatible with your description of the motion.

For further information on what types of acceleration are compatible with a rigid train, please see the Wikipedia entry on Born Rigidity. For Born rigid deceleration the deceleration must be simultaneous in the train frame. If at any point the velocity of one part of the train is different from another in the momentarily co-moving inertial frame then the train is being physically deformed, crushed in this case.

Answer 2

If all parts of the train are decelerated the same from the viewpoint of an outside observer, then Dale already explained what happens: the train gets physically compressed/crushed, but stays the same size from the viewpoint of the outside observer, with the same apparent distance between the wheels.

Instead, we consider the scenario where the train is decelerated the same from the train's viewpoint(s), that is, when you start to apply the same break power to each wheel at the same time in the train's initial frame. In that case, the outsider observer will see the back wheel start to decelerate earlier than the front wheel, the train appears to grow larger, and the distance between the wheels grows larger.

In both scenarios, the wheels will only slip if the breaking force is larger than the friction between wheel and rails.

Answer 3

You've written that "obviously the wagon will expand" (in the track frame) whereas in fact obviously it won't expand.

Why this is obvious: Each end of the train has the same initial velocity $v$. You've assumed that the two ends decelerate simultaneously, which I take to mean that the acceleration of either end is given by the same function of time $a(t)$. The two ends also have the same initial velocity. It follows by elementary calculus that the two ends follow paths $s_0(t)$ and $s_1(t)$ that differ by a constant; that constant is the length of the train both before and after the deceleration.

In a now-deleted comment thread you said that this argument was correct but the conclusion was wrong. If you believe that a correct argument can have an incorrect conclusion, then of course you can believe anything at all, so in that case the answer to your question is whatever you care to believe. But in a world that makes sense, "deceleration simultaneously at all parts" must imply no change in length.

Answer 4

The stipulated train does not expand. The definition of coordinate acceleration doesn't change with velocity.

There are some problems with this thought experiment which I address at the end related to trains don't work like this, but I attempt to proceed with the least-impossible interpretation for now.

You'll find a non-length-contracted train that has been destructively crushed into the amount of length as a length-contracted train by the forces you applied to achieve the desired acceleration.

If the back of the train and the front of the train accelerate simultaneously in the station frame, they stay the same distance apart in the station frame.

In the train frame, the front of the train accelerated before the back, giving it a relative velocity towards the rear of the train by the time the back starts to accelerate. See: relativity of simultaneity. A non-constrained train would equilibrate under the internal forces that hold solids in their shape, but our train is constrained to continue accelerating simultaneously in the station frame, so the front keeps a relative velocity towards the back.

Both frames measure a force acting on the back of the train opposite the electric forces transmitting the rearward force applied on the front of the train. They have a slight disagreement about when the forces were started and what they're doing. In the station frame, the forces are constraining the length of the train. (Imagine trying to inflate a beach-ball under a car.) In the train frame, they're maintaining the velocity difference between the front of the train and the back of the train against the restoring forces applied by the train's structure. (Imagine compressing one end of a spring towards the other.)

Having all parts of the train accelerate simultaneously in the station frame is impossible unless your train is actually an cunningly disguised spring and you carefully tune your acceleration forces to its material properties. Trains (and rockets) buckle when you crush them, they don't smoothly compress. But trains also don't travel at relativistic speeds on train tracks, or have massless, frictionless wheels.... Just check your intuitions at the door, since this is a deeply non-train-like train.

Answer 5

There are some complications at work here, because the train will decelerate simultaneously in its own rest frame, so its behaviour in the frame of the track will appear differently. However, if you leave those complications aside then yes, the train will appear to have one length when it is moving and another, greater, length when it is at rest. Bear in mind, though, that if a train was travelling at the kind of speed at which length contraction becomes significant, to the observer at the trackside the train would come and go in less than the blink of an eye, so you would never be able to witness the expansion in the sense your question implies.

Answer 6

The previous answer is correct. The train will lengthen as it slows down to a stop. A strange thing does happen for a stationary observer. The back of the train slows down faster than the front. This is because of the time difference between the front and rear of the train for the stationary observer.