Home » Physics » A Brief Discussion on Participation of \mu-mesons (muons) based upon Relativistic Dynamics

# A Brief Discussion on Participation of \mu-mesons (muons) based upon Relativistic Dynamics

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 The life time of muons { $\mu$-mesons } is $2.2 \times 10^{-6}$ seconds and their speed $0.998c$, so that they can cover only a distance of $0.998c \times 2.2 \times 10^{-6}$ or 658.6 meters in their entire lifetime, and yet they are found in profusion at sea level, i.e., at a depth of 10 kilometers from the upper atmosphere where they are produced. How may this be explained on the basis of (i) Lorentz – Fitzgerald Contraction; (ii) Time Dilation.

• (i) Here, $2.2 \times 10^{-6} s$ being the mean life time of the muons in their own frame of reference, i.e., their proper life-time, the distance covered by them during this time in their own frame of reference, say $l_0= 658.6 m.$

The observed distance, as seen from our reference frame (the earth) will therfore, be, say, $l=\gamma l_0 = \dfrac {l_0}{\sqrt {1-(v^2/c^2)} }$
or $l=658.6/\sqrt {1-{(0.998c/c)}^2} = 658.6/0.064 = 10290 m = 10.29 km$

Thus, the muons are able to cover a distance of 10km or more despite their short life time. This explains their presence at sea level.

•(ii) Here again, the proper life time of the muons is $\tau = 2.2 \times 10^{-6} s$ and therefore their observed life time $\tau '=\gamma \tau = 2.2 \times 10^{-6}/0.064 s$
or
$\tau '= 34.38 \times 10^{-6} s$
i.e., their observed life-time is about 16 times their proper life-time and , in this increased or dilated lifetime, they can cover a distance $0.998c \times 34.38 \times 10^{-6} = 10290 m =10.29 km$ as in (i) above.
Hence their presence at sea level.