Posted by **Tisha** on Sunday, April 24, 2011 at 6:35pm.

Let us assume that the muon production happens at an altitude of about 15 kilometers above the surface of the Earth, and that the produced muons have a velocity of 0.99 c. Thus, in the Earth system, it takes the muon approx. 50500 nanoseconds to reach the surface of the Earth. In a non-relativistic calculation, what is the probability for a muon to reach the surface of the Earth?

Follow up question asks: What is the probability for a muon to reach the surface of the Earth, taking into account time dilation of its mean-life in the Earth system?

mean life:2193 nanoseconds

- Physics -
**drwls**, Sunday, April 24, 2011 at 6:47pm
Here is the relativistic (correct) answer:

In the muon's own coordinate system, the time that it takes to reach the earth is

t' = 50500*10^-9 s * sqrt[1 - (0.99)^2]

= 5*10^-5 s * 0.141

= 7.05*10^-6 s

The fraction that arrive without decaying is

P = exp(-t'/2.193*10^-6) = 0.040

For a nonrelativistic calculation, you would get

P = exp(-50500/2193),

a much lower number.

Note that we are using mean life and not half life numbers for the lifetime. They are not quite the same.

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