Virtual laboratory/Physics/University-level physics/Atmospheric muons — proof of time dilation

Atmospheric muons — proof of time dilation

Muons form high in the atmosphere and live on average only 2.2 μs — without relativity they would have no chance of reaching the surface. Set their speed and formation height, and compare how many muons arrive with time dilation and without it. This is the Rossi–Hall experiment — a direct proof of special relativity straight from nature.

Atmospheric muons — proof of time dilation
Physics · Theory of relativity · university level
Muon speed β=v/c\beta = v/c0.9950
Formation height hh15 km
Without time dilation — how many muons would reach the surface?
factor γ
10.0
reaches with dilation
10%
reaches without dilation
10⁻¹⁰
🎯 Challenge: hit the factor γ
streak: 0

I’ll give you a target Lorentz factor γ — tune the speed β to reach it. γ = 1/√(1−β²).

What just happened?

Muons form high in the atmosphere and live on average only τ0=2.2 μs\tau_0 = 2.2\,\mu s. Even at nearly the speed of light they would travel barely ~660 m — yet they have more than ten kilometres to cover. And still they reach the surface in huge numbers. Why?

Time dilation:

τ=γ τ0,γ=11−β2\tau = \gamma\,\tau_0, \qquad \gamma = \frac{1}{\sqrt{1 - \beta^2}}

For an observer on Earth the “clock” of a fast muon runs γ\gamma times slower, so the muon lives correspondingly longer and travels a greater distance. Fraction reaching the ground: N/N0=e−h/(βc γ τ0)N/N_0 = e^{-h/(\beta c\,\gamma\,\tau_0)}.

Invariant — two equivalent descriptions:

In the muon’s own frame its lifetime is normal (τ0\tau_0), but the atmosphere is contracted: L=L0/γL = L_0/\gamma. It can cover the shorter distance in time. Time dilation and length contraction are two sides of the same phenomenon — they give an identical result.

Why this matters:

This is an academic extension of special relativity. The Rossi–Hall experiment (1941) with muons is “tangible” proof that time dilation is not an abstraction but the everyday reality of particles from space.

Remember: τ=γτ0\tau = \gamma\tau_0. Muons reach Earth only thanks to time dilation (or equivalently — contraction of the atmosphere).

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Theory & formulas

The muon problem

A muon lives on average τ0=2.2 μs\tau_0=2.2\ \mathrm{\mu s}. Classically, even at a speed close to cc, it would travel only hundreds of metres — too little to reach the surface from a dozen-odd km.

Time dilation (Earth frame)

τ=γτ0,γ=11−v2/c2\tau = \gamma\tau_0, \qquad \gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} A fast muon lives γ\gamma times longer, so it has time to arrive.

Length contraction (muon frame)

For the muon it is the atmosphere that is contracted: L=L0/γL=L_0/\gamma. Both descriptions give the same result. The Rossi–Hall experiment (1941) confirmed it.

Notation: τ0\tau_0 — rest lifetime, τ\tau — in motion, γ\gamma — Lorentz factor, vv — speed, cc — speed of light, L0,LL_0,L — proper and contracted length.

Frequently asked questions

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Why do more muons arrive than they “should”?

Because for an observer on Earth a fast muon lives γ times longer (time dilation). Its internal “clock” runs slower, so it has time to travel a much longer path than its own 2.2 μs lifetime would allow.

›

Length contraction or time dilation?

They are two descriptions of the same phenomenon. In the Earth frame we speak of the muon’s time dilation; in the muon’s frame it is the atmosphere that is contracted (L = L₀/γ), so it has a shorter path to cover. Both give an identical result — how many muons arrive.

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Has this really been measured?

Yes — the Rossi–Hall experiment (1941) compared the muon flux at a mountain top and at its base. The measured drop was much smaller than the classical lifetime would imply — in exact agreement with the time-dilation prediction.

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