The twin paradox

While both twins move inertially, each is entitled to say the other's clock runs slow by the same factor — at β = 0.80 each says the other ticks at 0.60 of their own rate — and both are right. There is no contradiction, because the two statements are not about the same pair of events. Each twin is comparing one clock of their own against a pair of separated clocks in the other frame, and those pairs are synchronised differently. The paradox only bites when the twins meet a second time, because who is older at the reunion is decided at one place at one moment, and no frame can dodge it. So the question is never which twin's dilation is real. It is which twin's account of the whole journey can be written in a single inertial frame.

Strip the frames out and the problem is geometry. Along any worldline dτ² = dt² − dx²/c², and the elapsed proper time is the integral of dτ — the Minkowski length of the path between two events. Bob never moves in his own frame, so dx = 0 and his clock records the coordinate time itself, 10.0 yr. Alice's outbound leg covers Δt = 5.0 yr and Δx = 4.0 ly, so Δτ = √(5.0² − 4.0²) = √9 = 3.0 yr, and the return leg gives another 3.0 yr: 6.0 yr in all. The minus sign is what makes this unlike Euclidean geometry. There the straight line is the shortest path; here the straight timelike worldline between two events is the one with the most proper time, and every bend costs ageing. Alice returns younger because her path through spacetime was bent — a statement that names no frame at all.

Alice may still tell the story from her side, provided she admits she is using two frames. On the outbound leg she computes Bob's clock at 1/γ of hers: over her 3.0 yr he ages 3.0 × 0.60 = 1.8 yr, and the inbound leg credits him with another 1.8 yr. That accounts for only 3.6 of his 10.0. The missing 6.4 yr appears the moment she changes frames. In the outbound frame the event on Bob's worldline simultaneous with her turnaround is t = 5.0 − βD/c = 5.0 − 3.2 = 1.8 yr; in the inbound frame the same turnaround is simultaneous with t = 5.0 + 3.2 = 8.2 yr. Her present has swung across 6.4 yr of his life, which is the Δtⱼᵤₘₚ = 2vD/c² of the formula card, and 1.8 + 6.4 + 1.8 = 10.0. Nothing crossed the gap: no signal, no force, no influence on Bob's clock. What changed is which of Bob's birthdays Alice counts as happening now. Give the turnaround a finite duration and the slice sweeps instead of jumping — smoother, and the same 6.4 yr.

The cleanest resolution never mentions simultaneity. Let each twin send one pulse per year of their own proper time and use the Doppler factor k = √((1 + β)/(1 − β)), which is √(1.80/0.20) = 3 at β = 0.80: a receding observer receives at 1/k, an approaching one at k. Alice's telescope therefore delivers 1 pulse across her outbound 3.0 yr and 9 across her inbound 3.0 yr, and 10 is Bob's age at the reunion. Bob's own tally runs the other way and returns 6, which is Alice's. Every pulse emitted is received, and an arrival is a single event, so those totals carry over unchanged into any frame and no synchronisation convention is ever selected. What separates the twins is when the rate flips: Alice sees 1/k become k at the instant she turns, while the news of that turn has to cross 4.0 ly of space before it reaches Bob. That delay, and not a g-force, is the asymmetry in a form each twin can point at.

Replace Alice with two probes. An outbound probe passes Bob at 0.80c and zeroes its clock; at the star it passes an inbound probe running the other way and hands over its reading of 3.0 yr; the inbound probe adds its own 3.0 yr and calls out 6.0 yr as it passes Bob, whose clock reads 10.0. Nobody in that version ever accelerates, and the answer does not move. What the handoff keeps is the bend: the relay's combined worldline is still not straight. Acceleration is simply how one traveller changes frames, which makes it a reliable marker of who bent — but it is not what sets the size. Δage = T(1 − 1/γ) is built from β and T alone, so the gap tracks how fast she cruised and for how long, and is deaf to how sharply she turned.

Muons stored in a ring at CERN circulated at γ = 29.3 on a closed loop — out and back, with the laboratory playing the stay-at-home twin. Their proper lifetime of 2.197 µs appeared in the laboratory as 29.3 × 2.197 = 64.4 µs, matching γτ₀ to about one part in a thousand. The circular path means their proper acceleration was around 10¹⁸ g, beyond anything a spacecraft could survive, and it shifted the decay rate by nothing measurable. That is the clock hypothesis in raw form: the rate of an ideal clock depends on its instantaneous speed and not on its acceleration. It is an experimental result rather than a theorem of the two postulates, and it is exactly what licenses treating Alice's turnaround as a corner rather than as a separate physical effect.