The Barn and the Pole
updated 4-AUG-1992 by SIC
original by Robert Firth
These are the props. You own a barn, 40m long, with automatic doors at either end, that can be opened and closed simultaneously by a switch. You also have a pole, 80m long, which of course won't fit in the barn.
Now someone takes the pole and tries to run (at nearly the speed of light) through the barn with the pole horizontal. Special Relativity (SR) says that a moving object is contracted in the direction of motion: this is called the Lorentz Contraction. So, if the pole is set in motion lengthwise, then it will contract in the reference frame of a stationary observer.
You are that observer, sitting on the barn roof. You see the pole coming towards you, and it has contracted to a bit less than 40m. So, as the pole passes through the barn, there is an instant when it is completely within the barn. At that instant, you close both doors. Of course, you open them again pretty quickly, but at least momentarily you had the contracted pole shut up in your barn. The runner emerges from the far door unscathed.
But consider the problem from the point of view of the runner. She will regard the pole as stationary, and the barn as approaching at high speed. In this reference frame, the pole is still 80m long, and the barn is less than 20 meters long. Surely the runner is in trouble if the doors close while she is inside. The pole is sure to get caught.
Well does the pole get caught in the door or doesn't it? You can't have it both ways. This is the "Barn-pole paradox." The answer is buried in the misuse of the word "simultaneously" back in the first sentence of the story. In SR, that events separated in space that appear simultaneous in one frame of reference need not appear simultaneous in another frame of reference. The closing doors are two such separate events.
SR explains that the two doors are never closed at the same time in the runner's frame of reference. So there is always room for the pole. In fact, the Lorentz transformation for time is t'=(t-v*x/c^2)/sqrt(1-v^2/c^2). It's the v*x term in the numerator that causes the mischief here. In the runner's frame the further event (larger x) happens earlier. The far door is closed first. It opens before she gets there, and the near door closes behind her. Safe again - either way you look at it, provided you remember that simultaneity is not a constant of physics.
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