The cosmic-ray muon

Problem

A muon created by cosmic rays in the upper atmosphere has a mean lifetime $\tau_0\approx 2{,}2\;\mu$s in its rest frame. Yet many muons reach the Earth’s surface after travelling $\sim 10$ km. How does this square with $c\tau_0 \approx 660$ m? Explain from both points of view (Earth and muon).

Solution

The classical problem. Travelling at the speed of light during its mean lifetime, a muon would cover only

$c\,\tau_0 \approx (3\cdot 10^8)(2{,}2\cdot 10^{-6}) \approx 660\ \text{m},$

well short of the $10$ km of atmosphere. Without relativity, almost no muon would reach the ground. Instead, many do: the explanation is the same from both points of view, but with different mechanisms.

Earth’s point of view — time dilation. The muon travels at $v\approx 0{,}998\,c$, with $\gamma \approx 15$. In Earth’s frame, its mean lifetime appears dilated:

$\tau = \gamma\,\tau_0 \approx 15 \times 2{,}2\ \mu\text{s} \approx 33\ \mu\text{s}.$

In this time it covers $v\,\tau \approx (3\cdot 10^8)(33\cdot 10^{-6}) \approx 10$ km: it reaches the ground.

The muon’s point of view — length contraction. In its own frame the muon lives only $\tau_0 = 2{,}2\ \mu$s, but it is the atmosphere that moves towards it at $0{,}998\,c$, and its thickness is contracted:

$L = \frac{L_0}{\gamma} \approx \frac{10\ \text{km}}{15} \approx 660\ \text{m}.$

These $660$ m are exactly $\approx c\,\tau_0$: the muon has just enough time to cross them.

$\ev{\text{Earth: } \tau=\gamma\tau_0;\quad \text{muon: } L=L_0/\gamma;\quad \text{same physical result}}$

The two observers disagree on what “the time” and “the distance” are, but agree on the one objective fact: the muon reaches the surface.

Links

Topics: Relatività ristretta Concepts: Dilatazione dei tempi · Contrazione delle lunghezze Skills: Cambio di sistema di riferimento