Proper time and time dilation

Dilation problems go wrong at the first line, when γ is written before deciding what is being dilated. Fix that by naming the event pair. An event is a point in space and time: the muon is created at (t₁, x₁); the muon decays at (t₂, x₂). Every frame agrees these two things happened, and disagrees only about the coordinates. Now ask a mechanical question — is there an inertial frame in which both events occur at the same place? For a particle that is born, moves, and dies, yes: its own rest frame, where Δx′ = 0 and one clock carried along is present at both. The reading of that clock is the proper time Δτ between those two events, and the phrase 'the proper time' means nothing until the pair is named. It is not a property of a clock in general; the same clock has a different Δτ for a different pair. Two events joined by a light ray have no such frame at all: the invariant gives Δτ = 0 for them, and no clock, however fast, can be present at both.

The boost from the laboratory S to the particle frame S′ reads t′ = γ(t − vx/c²) and x′ = γ(x − vt). Work in that direction and you must supply Δx, the laboratory separation, which is usually one of the unknowns. Use the inverse instead — swap the primes and reverse the sign of v — giving t = γ(t′ + vx′/c²). Take differences between the two events: Δt = γ(Δt′ + vΔx′/c²). Now impose the condition that defines the proper frame, Δx′ = 0. The second term vanishes, Δt′ is Δτ, and Δt = γΔτ. That is the whole derivation. Notice what did the work: not a claim that moving clocks are defective, but the fact that in S the two events are separated in space, so the laboratory interval has to be read off two clocks rather than one. With β = 0.60, γ = 1.25, and a proper interval of 4.00 s appears in the laboratory as 5.00 s.

Each frame finds the other's clock slow, and that sounds self-defeating until you notice the two measurements are not the same measurement. When S times a clock at rest in S′, it compares one moving clock against a pair of stationary, pre-synchronised clocks standing at different places. When S′ times a clock at rest in S, it does the same thing with its own pair. The asymmetry hides in the synchronisation: two clocks that S has set to agree are, in S′, offset by vΔx/c², the relativity of simultaneity. So the two frames are running different experiments, each internally consistent, and neither result contradicts the other. Every clock face read at an actual meeting is agreed by both. The moment you insist on bringing one clock back to meet the other a second time, something must accelerate, the situation stops being symmetric, and you have the twin problem rather than this one.

Because γ ≥ 1 for every real v, Δt = γΔτ says the proper time is the smallest interval any inertial frame will assign to the pair. That is not an accident of the algebra; it follows from an invariant. Square the boost and the combination c²Δt² − Δx² comes out the same in every inertial frame. In the proper frame Δx′ = 0, so that invariant is exactly c²Δτ²; in any other frame it is c²Δt² − Δx² with Δx ≠ 0, and subtracting a positive quantity forces Δt > Δτ. Two consequences matter in practice. First, Δτ is a property of the event pair, so there is no 'really right' frame to hunt for. Second, the relation is one-directional numerically: an astronaut's 4.00 s at β = 0.60 is 5.00 s in the laboratory, and a laboratory interval of 5.00 s between those same two events is 4.00 s of proper time. You never multiply by γ twice.

Cosmic-ray muons are produced high in the atmosphere and decay with a rest-frame mean lifetime τ₀ = 2.20 μs. At 0.995c a muon covers only 0.995 × 3.00 × 10⁸ × 2.20 × 10⁻⁶ = 657 m in one mean lifetime if dilation is ignored, so almost none should survive the roughly 2000 m from a mountain summit to sea level: the surviving fraction would be e^{(−2000/657)} = 0.048. With γ = 10.0 the laboratory mean lifetime is 22.0 μs, the mean decay length becomes 6.57 km, and the surviving fraction is e^{(−2000/6570)} = 0.74 — about fifteen times more muons at the bottom. Rossi and Hall first measured a survival ratio of this kind between two altitudes in Colorado in 1941, and Frisch and Smith repeated it two decades later between the summit of Mount Washington and sea level with muons selected near 0.995c; both results landed on the dilated side. Ride along with the muon and nothing odd happens to its clock: its mean life is still 2.20 μs. What has changed in its frame is the mountain, contracted from 2000 m to 200 m, which at 0.995c takes 0.67 μs to pass — and e^{(−0.67/2.20)} is the same 0.74.

Δt = γΔτ is a statement about one pair of events and one constant relative velocity, and three restrictions follow. It does not apply to a clock whose speed changes: for that you integrate, τ = ∫√(1 − v(t)²/c²) dt, which is the next topic. It says nothing about what an observer sees, because light from the two events takes different times to arrive; the visual Doppler factor √((1 + β)/(1 − β)) is a separate calculation from γ, and confusing the two is a standard source of wrong answers about the twin problem. And it is not a claim about the mechanism of any particular clock. A caesium transition, a muon, and a pendulum all stretch by the same γ, because what dilates is the coordinate interval between two events, not the workings of a device. Whether acceleration itself alters a clock's rate — the clock hypothesis — is a separate experimental question, taken up with the worldline integral.