The Pole-in-the-Barn Paradox


    "Does the pole fit in the barn?"
  1. Initially, the front door of the barn is open, and the back door is closed [(a) →].
  2. After the pole enters the front door of the barn, this door is closed and the pole is completly insider the barn [(b) →].
  3. After the front door is closed, the rear door is opened, and the pole exits the barn by the rear door [(c) →].
The paradox: In the reference frame of the pole, the barn is shorter than the pole (length contraction again), so how can the pole fit completly inside the barn as is claimed by an observer in the barn's reference frame?

(Non-)Paradox

The pole-barn paradox must be addressed with the ideas of simultaneity in relativity. The fact that two events are simultaneous in one frame of reference does not imply that they are simultaneous as seen by an observer moving at a relativistic speed with respect to that frame.

To calculate the times for the two frames of reference, consider the pole entering the barn and set t = t' = 0 at that instant and x = x' = 0 to establish the coordinate system. In the events described below, x' and t' refer to the pole frame while x and t refer to the barn frame. The Lorentz factor g = 2.29 and the Lorentz transformation is used to transform quantities from one frame to the other (c = 0.3 m/ns).

Barn frame of reference
═╡   │
  • Front of pole enters: t = 0
  • The back of the pole enters the barn before (4.69 ns) the front of the pole leaves, so a 1 ns gate could be closed on both ends, containing the entire pole. [Partly adapted from]
    ╞═ │
  • Back of pole enters, Front gate closes: t = 8.73m/0.9c = 32.35 ns
  • │ ═╡
  • Back gate opens, Front of pole leaves: t = 10m/0.9c = 37.04 ns
  • │   ╞═
  • Back of pole leaves: t = 32.35ns + 37.04 ns = 69.38 ns
  • Pole frame of reference
    ═╡   │
  • Front of pole enters: t' = 0
  • From the pole point of view, the front gate closes just as the back of the pole enters. The surprising result is that the back gate is seen to open earlier from the pole framework, before the front of the pole reaches it.

    The gate closings are not simultaneous, and they permit the pole to pass through without hitting either gate. [Partly adapted from]

    ╪═╡
  • Back gate opens at t = 37.04 ns, but at x = 10m. It is (almost) simultaneous with the Front gate closing (32.35 ns) in the barn frame, but not in the pole frame. The time for back gate opening in the pole frame is t' = γ (t-vx/c2) = 2.29(37.04 - (0.9c)(10 m)/c2) = 16.12 ns.
  • ╪═╡
  • Front of pole leaves barn: t' = 4.37m/0.9c = 16.14 ns
  • ╞═╪
  • Back of pole enters: t' = 20m/0.9c = 74.07 ns
  • ╞═╪
  • Front gate closes at t = 32.35 ns, but t' = γ t = 2.29(32.35 ns) = 74.07 ns
  • │   ╞═
  • Back of pole leaves: t' = 16.14 ns + 74.07 ns = 90.21 ns
  • Space-time Diagram

    In the following space-time diagram of the situation, the gray rectangle represents the inside of the barn (top view). The front door is on the left and the rear door is on the right. The thick (vertical) lines correspond to events where the door is closed. The diagonal shaded region represents the pole. A horizontal line on this figure represents a "snapshot" in time from the point of view of a ground (or barn) observer.

    reverse Lorentz transformation
    † The axis ct' is the trace ∋ x' = 0 in the reverse Lorentz transformation. Thus, x = γ vt', ct = γ ct' = (c/v)x (∴ slope = 1/β). The axis x' is the trace ∋ ct' = 0. Thus, x = γ x', ct = γ (v/c)x' = (v/c)x (∴ slope = β).

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