Ladder and Barn Paradox

The ladder-and-barn paradox

A ladder of proper length L0 slides at speed v through a barn of proper length Lb with a door at each end. In the barn frame the ladder is length-contracted, L = L0/γ with γ = 1/√(1 - v²/c²), so if L0/γ < Lb there is a time interval in which the whole ladder is inside and both doors can be slammed shut at the same instant. In the ladder frame the barn is contracted instead, Lb/γ < L0, and the ladder never fits. There is no contradiction because the doors do not close simultaneously in the ladder frame.

Resolution with the Lorentz transformation

x' = γ(x - v t), t' = γ(t - v x / c²). Two events that are simultaneous in the barn frame, closing the entry door at x = 0 and the back door at x = Lb at time ts, happen at t'entry = γ ts and t'back = γ(ts - v Lb/c²). The back door closes (and, if the ladder's front were there, is already passed) earlier by γ v Lb/c². The two events are spacelike separated, so their time order is frame-dependent while the interval s² = c²Δt² - Δx² is invariant.

Critical speed

The ladder fits in the barn frame only above vcrit = c √(1 - (Lb/L0)²) (for L0 > Lb). The window during which both ends are inside lasts Δt = (Lb - L0/γ)/v.

Honest simplifications

• The ladder is perfectly rigid in each frame's description; a real ladder is not, and information about the front being stopped cannot travel faster than light, so a real ladder would compress or break rather than stay straight. Closing the doors "for an instant" never exerts a force.
• No acceleration phase, no Terrell-Penrose rotation or light-travel-time visual distortion: the drawing shows the geometry of simultaneous positions, not what a camera would photograph.
• The 3D scene is stretched in the x direction only for the contraction; the units are metres and nanoseconds (c = 0.2998 m/ns), a replay speed slows the action down.