Contraction paradoxes and rigidity

Run a pole of proper length L₀ = 5.0 m through a barn of proper length D = 3.0 m with a door at each end, at β = 0.866 so that γ = 2. The barn crew measures the pole at L₀/γ = 2.50 m and sees 0.50 m of room to spare. The pole crew measures the barn at D/γ = 1.50 m and sees 3.5 m of pole hanging out of it. Both statements are correct measurements, and asking which describes the real fit is the wrong question. The claim that the pole fits is shorthand for a claim about two events: the near door shuts, the far door shuts, and no part of the pole is at either door when it does. Events are frame-independent — either a door met wood or it met air, and no boost can change that. Lengths are not. So put the two lengths aside and write the event table.

Put the near door at x = 0 and the far door at x = D = 3.0 m, and slam both at t = 0 with the pole centred, so its tail is at 0.25 m and its nose at 2.75 m. With v = βc = 2.596 × 10⁸ m s⁻¹, the tail crossed the near door at t = −0.25/v = −0.96 ns and the nose reaches the far door at t = +0.96 ns, so the pole is wholly enclosed for a 1.93 ns window. Both doors are shut, both are shut at the same instant, and neither touches the pole. Nothing here is strange, which is why this is the frame everyone trusts — and exactly why it settles nothing on its own. A paradox is only resolved once the other frame's ledger is written too.

Boost the two closings with t′ = γ(t − vx/c²). The near door, at (t = 0, x = 0), gives t′ = 0. The far door, at (t = 0, x = 3.0 m), gives t′ = −γβD/c = −17.3 ns. In the pole's frame the far door shuts first, by 17.3 ns, and reopens long before it reaches the nose. Positions complete the picture. At t′ = −17.3 ns the far door sits at x′ = γD = 6.0 m and the pole's nose at 5.5 m, so the door shuts on 0.50 m of air. Nothing about the pole moves in this frame, so it is the door that eats that gap, arriving at the nose 1.93 ns later. Over the full 17.3 ns the barn sweeps γβ²D = 4.5 m back along the pole — further than its own proper length — and at t′ = 0 the near door shuts at x′ = 0 with the pole's tail at 0.50 m. The clearance γs/2 = 0.50 m is identical at both doors. Two orderings, one event list, no contact anywhere.

A rigid body keeps fixed separations no matter what you do to it. Strike one end of a rod of proper length L and its far end must start moving at the same instant: two events with Δt = 0 and Δx = L. Boost that pair and Δt′ = −γβL/c, so some frame sees the far end move before the hammer landed. Rigidity is therefore not an idealisation that gets better with stiffer steel; it is causally forbidden, and the bound shows up in material constants. The longitudinal signal speed √(B/ρ) can never exceed c, capping B at ρc² = 7.0 × 10²⁰ Pa for steel at ρ = 7.8 × 10³ kg m⁻³. Real steel manages 1.6 × 10¹¹ Pa — more than nine orders short — giving a sound speed of 4.5 km s⁻¹, or 1.5 × 10⁻⁵ c. Strike one end of a 1.00 m bar and the other end learns of it 0.22 ms later, some 66 000 light-crossing times.

What survives is Born rigidity: motion in which every element holds a constant proper distance from its neighbours. It exists, but the constraint bites hard — fix one element's worldline and every other element's is already settled. A rod boosting along its own length must be driven harder at the tail than at the nose, with proper acceleration a(X) = c²/X measured from the Rindler horizon, so a rod of proper length L₀ cannot be pushed Born-rigidly with a nose acceleration above c²/L₀. Bell's spaceships show the cost of ignoring this. Two rockets separated by L in the launch frame run identical acceleration programmes, so their launch-frame separation stays exactly L — but each rocket's own frame measures the gap as γL, which grows without bound, and a taut thread between them snaps. Identical acceleration is not rigid motion; it is a programme that stretches the object.

Contraction is what a ruler reports when both ends are located at one time. A camera does something else: it collects light that arrives together, and light from an object's far parts left earlier. Photograph a cube of side L passing broadside at β. Light from the receding back corner travels an extra L, taking L/c, during which the cube advances βL — so that side face appears with width βL, beside a front face foreshortened to L√(1 − β²). Those two widths are precisely what an unsquashed cube turned through α, with sin α = β, would project. That is the Terrell–Penrose result: a small distant object photographs as rotated rather than flattened, and Penrose showed a sphere's outline stays a perfect circle at any speed. None of it softens contraction: the apparent turn is what finite light-travel time adds on top of a side that really does measure 0.800 L at β = 0.60.