Time dilation, proper time & length contraction

Build a clock from two mirrors a distance d apart, with a light pulse bouncing between them and d set perpendicular to the direction of motion. In the frame where the clock is at rest one tick is Δτ = 2d/c. Now watch the same tick from a frame in which the clock moves at v. Between emission and return the bottom mirror advances vΔt, so the pulse travels two slanted legs, each the hypotenuse of a right triangle with sides d and vΔt/2. The second postulate says the pulse still moves at c in this frame, so each leg has length cΔt/2, and Pythagoras gives (cΔt/2)² = d² + (vΔt/2)². Solving, Δt = 2d/√(c² − v²) = γ(2d/c) = γΔτ with γ = 1/√(1 − v²/c²). At v = 0.6c the triangle is 3-4-5: sides 0.6 and 0.8 in units of the hypotenuse, so Δt/Δτ = 1/0.8 = 1.25. Two assumptions did the work and both are checked later in the lesson: the perpendicular spacing d is the same in both frames, and c is.

A tick of the moving clock is two events: the pulse leaves the bottom mirror, and the pulse returns to it. In the clock's own frame those events happen at the same place, so one clock reads both, and its reading is the proper time Δτ. In the lab the events are vΔt apart, so timing them takes two lab clocks, one at each place, synchronised beforehand. That is the asymmetry: the proper time is the shortest interval any inertial observer assigns to the pair, and whoever needs two clocks measures γ times more. It also settles the hardest first-year question: the moving observer is entitled to say the same about lab clocks, so each frame finds the other's clocks slow. There is no contradiction: they are timing different pairs of events, and the lab's two clocks, synchronised in the lab, are out of step as seen from the ship, by the relativity of simultaneity from the previous topic. A ship at 0.6c whose clock advances 8.0 y between passing Earth and passing a beacon at rest relative to Earth is timed at 10.0 y by Earth's clocks: the ship clock attended both events, so 8.0 y is proper.

Let a rod of proper length L₀ rest in the lab and let a clock fly along it at v. Lab observers time the clock from one end to the other, Δt = L₀/v, using a clock at each end. The flying clock attends both events, passing end A and passing end B, so it reads the proper time Δτ = Δt/γ. Now tell the story from the clock's frame: the rod moves past at the same speed v, since relative speed is symmetric and both frames agree on it, and takes Δτ to pass, so its length is L = vΔτ = vΔt/γ = L₀/γ. No new postulate went in; contraction is time dilation read from the other side, plus the shared value of v. What shrinks is always the length measured in the frame where the object moves, and only along the motion. A 1000 m accelerator hall passed at 0.8c, where γ = 5/3, is 600 m long to the particle, whose own transit time is 600 m ÷ 0.8c = 2.50 μs against 1000 m ÷ 0.8c = 4.17 μs on hall clocks. Both frames agree that the hall's clocks record the longer time; they disagree about why.

Contraction acts only along v, and the argument is symmetry. Send two identical metre sticks past each other sideways, each with a brush at its tip that paints the other stick as they pass. If motion shortened transverse lengths, A would find B shorter, so B's brush would mark A below A's tip while A's brush passed clean over B; B, by the same reasoning, would find a mark on B below B's tip and none on A. Where a mark sits is a fact about paint on wood that every frame must agree on, so both claims cannot hold; with neither frame preferred, the marks land exactly at the tips. That secures the light clock's d. Now turn the clock through 90° so the light runs along the motion. In the lab the pulse closes on the front mirror at c − v and returns at c + v, so for a mirror spacing L the tick is Δt = L/(c − v) + L/(c + v) = 2Lγ²/c. If L equalled the rest spacing d, this clock would tick at a different rate from the transverse one and comparing them would reveal absolute motion, which the first postulate forbids. With L = d/γ the tick is 2dγ/c = γΔτ, the same tick. Contraction is what makes the two clocks agree: it is the Michelson-Morley null result.

Cosmic-ray muons are created near 15 km altitude with rest-frame mean life τ₀ = 2.20 μs and speeds around 0.995c, so γ ≈ 10. Without dilation a muon's mean decay length would be cτ₀ ≈ 660 m and the fraction surviving 15 km would be e^{(−15000/660)} ≈ 10⁻¹⁰: none would reach the ground. With Δt = γτ₀ = 22 μs the decay length is 6.6 km and the surviving fraction is e^(−15/6.6) ≈ 0.10, which is what counters record. The muon's own account differs and is equally valid: its life is 2.2 μs, but the atmosphere is contracted to 1.5 km and passes in 5.0 μs, so e^(−5.0/2.2) ≈ 0.10 survive. The survival ratio is a frame-independent count, and both frames compute it. The cleanest laboratory test used a storage ring at CERN in 1977, where muons at γ = 29.3 circulated with a measured mean life of 64.4 μs against 2.197 μs at rest, matching γτ₀ to within 0.2% despite an acceleration in the muon's own frame near 10¹⁸ g. The speed, and so γ, was constant, and Δt = γΔτ held as the model says it should.

First the classical limit. Expanding γ = (1 − β²)^(−1/2) gives γ ≈ 1 + β²/2 + 3β⁴/8, so both effects are second order in v/c. A GPS satellite at 3.87 km s⁻¹ has β = 1.29 × 10⁻⁵ and γ − 1 = 8.3 × 10⁻¹¹; over 86 400 s its clock loses 7.2 μs to motion alone, which the system corrects (gravity adds a larger, opposite term). Newton is recovered, and the correction is measurable. Second the caveat. Δt = γΔτ compares two inertial frames at constant relative speed. What is always true is the rate: in any inertial frame a clock moving at instantaneous speed v(t) advances dτ = dt √(1 − v²/c²), so over a trip Δτ = ∫ dt/γ(t). When v is constant the integral is Δt/γ; when it is not, you must integrate, and in an inertial frame, because the accelerating clock's own frame is not one and cannot simply turn the formula round. That is why a traveller who goes out and comes back is genuinely younger: the Earth clock stayed in one inertial frame and the traveller's did not. Constant speed on a curved path, as in the storage ring, keeps γ constant, so the simple relation survives there.