Time dilation and length contraction

The standard-configuration boost, ct′ = γ(ct − βx) and x′ = γ(x − βct), relates the coordinates of one event in two frames. Hand it a pair of events and it returns Δt′ and Δx′ — and that is all. Nothing in it yet mentions a clock running slow or a rod being short, because a measurement is not simply a pair of events: it is a pair of events chosen by a rule. Time dilation chooses the rule Δx′ = 0, meaning the two events happen where one clock is sitting. Length contraction chooses Δt = 0, meaning the two ends are marked at one instant of the measuring frame. Different rule, different answer, out of the same transformation. Write the constraint on the page before the algebra and the γ can never come out on the wrong side.

Let a clock ride in S′. Two of its ticks are two events at the same place in S′, so Δx′ = 0. The inverse boost, Δt = γ(Δt′ + vΔx′/c²), then collapses to Δt = γΔt′ = γΔτ, where Δτ is the proper time — the interval read by that one clock. Since γ ≥ 1, every other inertial frame assigns more. At β = 0.866, γ = 2.000, so a 1.00 s tick aboard occupies 2.00 s of lab time. Notice the asymmetry hiding inside the procedure: the clock's own frame needs one clock, while the lab needs two clocks at different places, synchronised beforehand. The dilation is not a fault in the moving clock. It is a comparison between one clock and a pair, and the pair's synchronisation is exactly what depends on the frame.

A moving rod cannot be measured by laying a ruler beside it at leisure; you have to fix where both ends are, and 'both ends at once' is where the frame dependence enters. Why that is the right definition of a moving length is argued in Special Relativity II; here it is simply the constraint Δt = 0. Put the rod at rest in S′ with its ends at x′ = 0 and x′ = L₀. In S, mark the ends at one lab time t. Apply x′ = γ(x − vt) to each mark: 0 = γ(x_A − vt) and L₀ = γ(x_B − vt). Subtract, and the vt cancels: L₀ = γ(x_B − x_A) = γL, so L = L₀/γ. Take L₀ = 15.0 m at β = 0.800, where γ = 1.667: the lab marks 9.00 m. Had you instead marked the ends simultaneously in S′, you would have recovered 15.0 m, because that is the rod's own rule. The rod did not change. The rule did.

Each frame says the other's clocks run slow and the other's rods are short, and both are right. The apparent paradox comes from imagining a single symmetric comparison when there are two different comparisons in play. When the lab says the ship's clock is slow, it is checking one ship clock against a succession of lab clocks it has synchronised. When the ship says the lab's clocks are slow, it is checking one lab clock against a succession of ship clocks it has synchronised. The two procedures use different event pairs, and the word doing the work — synchronised — is precisely what the two frames disagree about. Nothing here is a race with a winner. A real asymmetry needs something more, such as one twin changing frames, which is Unit 3's business.

The interval delivers both results without trusting any one frame. For two ticks of a clock moving at v, the lab has Δx = vΔt, so (cΔτ)² = (cΔt)² − (vΔt)² = (cΔt)²(1 − β²), and Δt = γΔτ follows with no boost written down. That is what makes Δτ the invariant of the pair: every frame computes the same number from its own Δt and Δx, and only the frame with Δx′ = 0 reads it straight off a single clock. The two marking events of a length measurement are the spacelike counterpart. In the lab they have Δt = 0 and Δx = L, so (Δx)² − (cΔt)² = L². In the rod's frame they lie on its two ends, Δx′ = L₀, and the simultaneity offset puts them βL₀/c apart in time, so the same invariant reads L₀² − (βL₀)² = L₀²/γ², which is L² again. With L₀ = 20.0 m at β = 0.600 the lab marks 16.0 m, the rod's frame sees a 40.0 ns offset, so cΔt′ = 12.0 m, and 20.0² − 12.0² = 16.0² in both frames. How the rod's frame turns that offset into metres of marker drift is worked through in Length contraction and the measurement it rests on, in Special Relativity II.

Four limits. First, both were derived between inertial frames in constant relative motion; an accelerating clock needs the proper time integrated along its worldline, τ = ∫√(1 − β²) dt, which is Unit 3's tool. Second, contraction acts along the boost axis only, because y′ = y and z′ = z; what that means for transverse size, volume and density is taken up in Length contraction and the measurement it rests on, in Special Relativity II. Third, contraction is not a squeeze: no force acts, no stress appears, and the rod's proper length is exactly what it was. Fourth, it is not what a camera records. A photograph gathers light that left different parts of an object at different times, so a small fast object images as rotated rather than flattened, and a fast sphere keeps a circular outline — the Terrell–Penrose effect. Contraction is a statement about coordinates assigned by a synchronised frame, not about an image.