Length contraction and proper length

For an object at rest you can measure at leisure: read one end this morning and the other after lunch, and because neither end has moved the answer is the same. That leisure is what makes proper length L₀ well defined, and it is why L₀ belongs to the object's rest frame alone. Set the object moving and the leisure is gone. Take a 100 m rod passing a lab at β = 0.600, which the lab will find to be 80.0 m long: read its nose at t = 0 and its tail 100 ns later, by which time the tail has advanced 18.0 m, and you report 62.0 m. Read the tail first and the nose 100 ns later and you report 98.0 m. The reading has become a story about your delay rather than about the rod. So the procedure must be pinned down: mark both ends at the same time in the frame doing the measuring, Δt = 0 there. A length is therefore a pair of events, not a property — and the moment simultaneity stops being universal, that pair stops being universal too.

Special Relativity I did the algebra, and it fits on one line: with the rod at rest in S′ and both ends marked at one lab time, Δt = 0 turns x′ = γ(x − vt) into L₀ = γL, so L = L₀/γ = L₀√(1 − β²). What matters here is which Δt vanished. It is the lab's, because the lab's apparatus fired both markers together, and that is the whole of the physical input. The common slip is to start instead from the inverse form x = γ(x′ + vt′) and set Δt′ = 0. That is a legal calculation, but it answers a different question — the lab separation of two events simultaneous in the rod's frame — and it delivers γL₀, which for the 100 m rod at β = 0.600 is 125 m where the lab's own rule gives 80.0 m. Choose the form of the transformation whose Δt is the one your apparatus actually zeroed. One check catches the error every time: γ ≥ 1, so the largest number always belongs to the frame in which the object is at rest.

The boost leaves y and z alone: y′ = y, z′ = z. This is not an aesthetic choice, and a symmetry argument forces it. Let a peg and a hole have equal proper diameters and send the peg through, its motion along the axis. If transverse lengths contracted, the hole's frame would say the peg is narrow and it passes cleanly, while the peg's frame would say the hole is narrow and it jams. Whether the peg scrapes is a single physical fact that no change of frame can rewrite, so the only consistent transverse factor is 1. Consequences follow at once. A sphere of proper radius r measures as an oblate spheroid, squashed to r/γ along v while staying r across; a volume becomes V₀/γ and, since mass is invariant, a density γρ₀. Charge is invariant while the volume holding it is not, so charge density transforms as well — which is exactly why a wire that is neutral in the lab is not neutral in the electrons' frame, and why the force one frame calls magnetic another calls electric.

Run the lab's measurement through the boost and follow the two marking events rather than the rod. Rod of proper length 100 m, β = 0.600, γ = 1.250: the lab marks the ends at t = 0, so the events are R = (x = 0, t = 0) and F = (x = 80.0 m, t = 0). In the rod's frame they sit at x′ = 0 and x′ = 100 m, on the rod's two ends, so both frames agree which atoms were marked. They disagree only about when: t′ = 0 for R and t′ = −200 ns for F, the offset Δt′ = βL₀/c that Special Relativity I derived, leading end first. Now convert that offset into a length with the invariant interval. The lab's pair has Δt = 0, so its spacelike separation is simply L² = (80.0 m)² = 6400 m². The rod's frame must return the same number, and it does, because cΔt′ = βL₀ = 60.0 m and (100 m)² − (60.0 m)² = 6400 m². In general L² = L₀² − β²L₀² = L₀²/γ², and with the offset set to zero the lab would get L₀ itself. The entire shortfall is timing and none of it is material: at every t′ the rod's ends sit at x′ = 0 and x′ = 100 m, before, during and after the marking. Special Relativity I spends the same 200 ns as distance instead — the lab's markers, 64.0 m apart in the rod's frame, slide 36.0 m between firings — and that is the account the figure draws.

Contraction is reciprocal: two 100 m rods passing at β = 0.600 each measure the other at 80.0 m. That sounds like a contradiction until you notice that 'shorter than' is not a frame-free relation between rods — it is a claim about a pair of events, and the two crews use different pairs. Follow the frame-free facts instead, which are coincidences. Let the two noses meet at t = 0 as B overtakes A. In A's frame B is 80.0 m long, so B's tail already lies 20.0 m past A's tail, and the tails met 111 ns before the noses did. In B's frame A is the short one, A's tail falls 20.0 m short of B's tail, and the tails meet 111 ns after the noses. The two frames reverse the order of the same two events — which is permitted, because with Δx = 100 m against cΔt = 33.3 m the pair is spacelike separated, and only spacelike pairs can be reordered. No signal outruns light, no causal chain is disturbed, and no contradiction has been created.

The evidence is routine, and the clearest case has already been met: Special Relativity I's cosmic-ray muon at β = 0.995 reaches the ground because, in its own frame, the atmosphere is contracted by γ ≈ 10, the same survival count the ground frame gets from a dilated lifetime. At the LHC a 7 TeV proton has γ ≈ 7460, so a metre of straight beam pipe measures 0.134 mm in the proton's rest frame; the rule belongs to inertial frames, which is why it is applied to a straight metre and not to the whole curved 26.7 km ring. None of this is a squeeze. Lorentz and FitzGerald first proposed contraction as a real dynamical deformation produced by motion through the aether, and that version returns the same measured lengths, so no single measurement refutes it. What retired it is universality. A stress produces a strain set by the material's stiffness, so steel and rubber, whose Young's moduli differ by a factor of thousands, would shorten by different amounts; contraction gives both exactly 1/γ, and gives the same factor to the gap between two unconnected probes coasting one behind the other, where there is no material to squeeze at all. That makes it a statement about spacetime rather than about the forces holding any one material together. Accelerating a rod up to speed does involve genuine internal forces, but that is a separate, dynamical question.