How Does Muon Decay Provide Evidence for Time Dilation?

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Because atmospheric muons are typically produced several kilometres above Earth, classical physics predicts that very few should reach detectors at the surface. Experiments instead detect many more.

In Earth’s reference frame, the moving muon’s internal decay process runs more slowly. The observed lifetime is

For $v=0.998c$, $\gamma \approx 15.8$, so

The muon can therefore travel approximately $0.998c(35\,\mu\text{s}) \approx 10\,\text{km}$, allowing many muons to reach the surface. The measured survival rates agree with relativistic predictions.

More precisely, radioactive decay is statistical: the number remaining follows $N=N_0e^{-t/(\gamma\tau_0)}$ in Earth’s frame. Increasing the measured lifetime therefore increases the predicted fraction that survives a given journey. Measurements comparing muon count rates at different altitudes match this exponential prediction, rather than the much smaller survival fraction expected without relativity.

In the muon’s frame, its lifetime remains $2.2\,\mu\text{s}$, but the atmosphere is length-contracted. Both explanations are consistent because proper time and proper length belong to different reference frames.

A common misconception is that high speed physically makes muons more stable. It does not: observers in relative motion measure different time intervals.

Exam technique

For an IB Explain question, identify the proper lifetime, calculate or describe the dilated lifetime using $\gamma$, and connect the increased survival rate directly to experimental evidence for special relativity. Show clear substitution and state a sensible conclusion.