Train and Tunnel Paradox

This thought experiment is usually called the ladder paradox — told with a ladder or pole carried at speed through a barn with a door at each end — but the train version is the same puzzle. A train, parked, is a little longer than a tunnel. Now run it through at a good fraction of $c$ . Each door opens just as the front of the train reaches it and slams shut the instant the rear clears it. From the ground the moving train is shortened by length contraction, and past a certain speed it is shorter than the tunnel — so there is a moment when the entrance has already shut and the exit has not yet opened: the train is sealed inside. But from the train it is the tunnel that is moving, so the tunnel is the one that shrinks. It is now shorter than the train was to begin with. The exit must open before the entrance shuts; the train pokes out of both ends and is never enclosed.^[1]

From the ground

the train is contracted to 0.79 of the tunnel's length

On the train

the tunnel is contracted to 0.55 of the train's length

Ground: the entrance shuts 0.28 light-crossing times before the exit opens — the train is sealed inside. Train: the exit opens 0.72 light-crossing times before the entrance shuts — right now the train sticks out of both ends of the tunnel.

Train speed = 75% of light (γ = 1.51) — train is 1.2× the tunnel at rest, 0.79× from the groundScrub time

Top: the same passage from the two frames, each running on its own clock. Below: the spacetime diagram that reconciles them. The tunnel is the grey strip and its doors are the thick lines (dashed while open); the train is the blue band. A is the entrance shutting, B the exit opening. The ground's "now" is horizontal and can slice between A and B, where the red block marks the train sealed inside. The train's "now" is tilted and meets B before A — from the train, the exit opens first. Neither view is wrong: B lies outside A's light cone (the yellow lines), so nothing can fix which came first.

Both accounts are correct, and the thing that makes them compatible is the relativity of simultaneity. "The entrance shuts before the exit opens" is a claim about the order of two events at different places, and the spacetime diagram shows that order is not something the two frames share. The ground's line of "now" is horizontal; the train's is tilted, and the tilt is enough to swap the two door events. There is no experiment that could settle which door acted first, because the two events lie outside each other's light cone — no signal, not even light, can get from one to the other in time to matter.^[2] Length contraction is not the train "really" being short; it is what you get when you measure both ends at the same time, and "at the same time" is exactly what the frames disagree about.

Deep dive · The numbers

Take $c = 1$ , tunnel rest length $L = 1$ , train rest length $L_0 = 1.2$ . From the ground the train has length $L_0/\gamma$ , which is shorter than the tunnel once $\gamma > L_0/L = 1.2$ , i.e. $\beta > \sqrt{1 - (L/L_0)^2} \approx 0.55$ .

Put the tunnel at $x \in [0, L]$ with the train's front at $x = \beta t$ and its rear at $\beta t - L_0/\gamma$ . The two events that matter are

\begin{aligned} A\ (\text{entrance shuts}):&\quad t_A = \frac{L_0}{\gamma \beta}, \quad x_A = 0 \\[4pt] B\ (\text{exit opens}):&\quad t_B = \frac{L}{\beta}, \quad x_B = L. \end{aligned}

On the ground $t_A < t_B$ whenever $L_0/\gamma < L$ — the sealed interval. Now apply the Lorentz transformation $t' = \gamma\,(t - \beta x)$ :

t'_A = \frac{L_0}{\beta}, \qquad t'_B = \gamma\!\left(\frac{L}{\beta} - \beta L\right) = \frac{L}{\gamma\beta}.

Since $L_0 > L > L/\gamma$ , we always have $t'_B < t'_A$ : on the train the exit opens first, at any speed. The $\beta x$ term in $t'$ is doing the work — it shifts the exit event (at $x = L$ ) earlier by $\gamma \beta L$ relative to the entrance event (at $x = 0$ ).

Why can the order flip at all? The separation between A and B is $\Delta x = L$ in space and $\Delta t = t_B - t_A$ in time, and $|\Delta x| > |\Delta t|$ whenever the train fits. So the interval is spacelike: B is outside A's light cone. Only for spacelike pairs can a boost reverse the order — and for such pairs nothing physical can depend on it, which is why no door ever actually collides with a train that "wasn't there yet."

The version with a crash. Suppose instead the ground crew shuts both doors at the same ground instant while the train is inside, then re-opens them. On the train those two slams are not simultaneous: the exit shuts first, the front of the train hits it, and the train is compressed — the rear keeps moving because news of the collision cannot outrun light — until the entrance shuts behind it. Nothing in a rigid body is truly rigid in relativity; the train genuinely fits, with a crumpled nose to show for it.