Problem — Lorentz factor of the muon

Problem

A muon moves relative to the laboratory with velocity $v = (\sqrt{3}/2)\,c \approx 0{,}866\,c$. Calculate the Lorentz factor $\gamma = 1/\sqrt{1 - v^2/c^2}$.

Solution

Given. $v = (\sqrt{3}/2)\,c$, so $\dfrac{v^2}{c^2} = \dfrac{3}{4} = 0{,}75$.

Lorentz factor.

$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0{,}75}} = \frac{1}{\sqrt{0{,}25}} = \frac{1}{0{,}5}$

$\ev{\gamma = 2{,}0}$

The muon “lives” in the laboratory for twice its proper time: it is precisely this dilation that allows atmospheric muons to reach the ground.

Links

Topics: Special relativity Concepts: Time dilation Skills: Symbolic setup