The train and the gap in the rails paradox

A situation similar to the tunnel one, but with the opposite physical conclusion. A train of proper length $L$ runs at speed $v$ on a track missing a stretch of rail of length $L$ (again, the same length as the train at rest). Call SRA the train’s frame and SRB the rails’ frame. Will the train fall into the gap, or will it pass straight over it?

SRB (rails). The train is contracted: $L_{(B)} = L/\gamma < L$. When the front reaches the first edge of the gap, the train “fits comfortably” inside the gap and should fall like a stone.

SRA (train). The gap is contracted: $l_{(A)} = L/\gamma < L$. Then the train, of length $L$, does not fit into the gap: it straddles it and should pass over undisturbed.

Who is right? Both. The apparent contradiction arises from tacitly assuming that the train is a rigid body: that all of its points “know” instantaneously what happens at the leading edge and react in unison. But rigid bodies do not exist in relativity. No information, not even the internal stress in the train’s structure, can travel faster than $c$.

When the front of the train reaches the edge of the gap and “discovers” that there is no rail beneath it, support ceases: that point begins to fall. But the information “there is no more rail” must propagate backwards along the train, at a speed no greater than $c$. The molecules further back do not yet know they are above the gap: for them the support is still there. They start falling only when the signal arrives.

What really happens?

• In the rails’ frame: as soon as the front is above the gap, it falls. The rear part continues straight for a while longer, remaining above the intact track. As the train advances and the mechanical signal propagates backwards, more and more parts begin to fall. The train is no longer a rigid straight line but a curve tilting downward.
• In the train’s frame: the gap comes “towards” it from the right at speed $-v$. As soon as the first edge of the gap reaches the front, the front starts to fall. The signal propagates from the front towards the rear, always at speed $\leq c$, deforming the train into a curve. There is no need for “the whole train to fall at once”: different points fall at different instants.

In both frames, therefore, the train ends up falling into the gap. The only difference is how the deformation propagates in time, and it is precisely this quantity that depends on the frame, owing to the relativity of simultaneity.

In the rails’ frame: from $F$ (where the front enters the gap) a mechanical signal departs, warning the rear parts of the train; each falls only when the signal reaches it.

In the train’s frame: the gap (tilted grey lines) comes towards the train; from $F$ the same mechanical signal propagates towards the rear at speed $\leq c$. In both frames the train deforms and falls: rigid bodies do not exist.

Why this paradox is instructive

The gap-in-the-rails paradox is subtler than the tunnel one: solving the first requires only the relativity of simultaneity, i.e. pure kinematics; solving the second requires abandoning a notion that seemed obvious, the “rigid body”, and accepting that any internal deformation propagates at finite speed. In real solids the speed of sound is a few thousand m/s, far from $c$: if a train could really be made to travel at $0.9\,c$, it would deform spectacularly well before reaching the gap. Relativity is not just a game of formulae: it forces us to rethink even the macroscopic properties of matter.

Links

Topics: Special relativity Concepts: Relativity of simultaneity Skills: Constructing Minkowski diagrams Methods: Minkowski diagram

Related exercises: Space-type or time-type · Problem — Tunnel at 0.8c · Problem — Simultaneity (true or false)