A muon lives only 2.2 µs, so even at near-lightspeed it should decay within ~660 m of where it's born — yet muons made ~15 km up in the upper atmosphere reach sea level in force. How do they survive a fall that should cost them ~50 lifetimes?
▶ Launch the interactive simulationForward-model the Frisch–Smith 1963 experiment (Mt Washington, h≈1907 m, β=0.9952, muon mean lifetime tau0=2.197 µs). A muon's decay clock ticks in its PROPER time; the lab sees it run slow by γ=1/√(1−β²), so survival to a detector h below is S=exp(−(h/βc)/(γ·tau0)). Run a SEEDED Monte-Carlo of millions of individual muons (proper lifetimes drawn from Exp(tau0)), measure the surviving fraction S, and invert γ=(h/βc·tau0)/(−ln S) — a formula that never contains γ. Repeat across five β, and run a Galilean control (no dilation, the clock ticks in lab time) as a falsification test. ?world=muon.
the special-relativistic Lorentz factor γ=1/√(1−β²) — recovered as γ=10.218 at β=0.9952 purely from how many muons survive the fall, and tracking the law to ~1e-3 across β∈[0.99,0.999] (γ from 7.1 to 22.4). The result is decisive because the Galilean alternative is RULED OUT: with no time dilation the same muons survive with probability S_gal≈5.5% — about 13.8× fewer than the relativistic S≈75% — and fed the real summit flux (563/hr) it predicts ~31/hr at sea level against the measured ~408/hr, while relativity predicts ~423/hr (within ~4%). The muon's reprieve is exactly the time dilation Einstein's 1905 kinematics demands; the lab's first special-relativity world, the SR companion to ?world=schwarzschild's general relativity.