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 nonrelativistic 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 meanlife 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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