Question

Muons are formed high in the atmosphere by the interaction of cosmic rays with atomic nuclei in the upper atmosphere. They have an average lifetime of about two millionth of a second and according to classical physics should decay before reaching the sea level. However, measurements show a large number of muons reach the Earth’s surface. Give an explanation how this is possible with respect to the earth and muon’s reference frames using ideas of time dilation and length contraction.

Answer 1

Time dilation-

Muon lifetime,$\tau$  = 2.2 $\times$ 10^-6 S, muon speed, v = 0.9999 C.

Muon created 15 km above earth.

Non-relativistic answer, Distance travel by muon =  0.9999 C $\times$ 2.2 $\times$ 10^-6 S = 660 m (<< 15km)

But, relativistically for earth observer- Muon life time t = $\gamma$ $\tau$

  

2.2 x 10-6 = 70 x 2.2 x 10-6 71 - (v/c)2

= 70 v.

So, distance travel by muon = 70 $\times$ 660 = 46200 m = 46.2 km. So, we can say muon reached the earth.

Length Contraction-

Now from Muon frame, muon will see that earth is coming towards it with velocity = v =0.9999 C

So, distance from earth to muon will contract relativistically by length contraction formula.

1=LXV1 - (U/C)2 = 15000 X V1 - (0.9999)2 = 212 m

Which is less than 660 m. So Muon will reach the earth's surface.