Answer to Question #98251 in Mechanics | Relativity for AbdulRehman

Question

A muon is a particle that has a lifetime of about 10−6
seconds. In other
words, when you make a muon, it lives for that long (as measured in its
own rest frame) and then decays (disintegrates) into other particles.
a) The atmosphere is about 30km tall. If a muon is created in the upper
atmosphere moving (straight down) at 1
2
c, will it live long enough to
reach the ground?
b) Suppose that a muon is created at the top of the atmosphere moving
straight down at .999999c. Suppose that you want to catch this muon
at the surface and shoot it back up at .999999c so that it decays just
when it reaches the top of the atmosphere. How long should you hold
onto the muon?

Answer 1

Lifetime in this conditions for observer reference frame.

$t_0 = 10^{-6} (s); v = 0.12c (\frac {m}{s}); c = 3\cdot10^8 (\frac{m}{s})$

$$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

We have

$$t=1. 01\cdot10^{-6} (s)$$

but we need a time

$$t=\frac{L}{v}=\frac{30000}{0.12\cdot 3\cdot 10^8}=8.3\cdot 10^{-4} (s)$$

answer a) he can't reach the ground

Lifetime in this conditions for observer reference frame.

$t_0 = 10^{-6} (s); v = 0.999999c (\frac {m}{s}); c = 3\cdot10^8 (\frac{m}{s})$

$$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

We have

$$t=7.1\cdot10^{-4} (s)$$

but we need a time for muon which moving down and up

$$t=2\frac{L}{v}=\frac{30000}{0.999999\cdot 3\cdot 10^8}=2\cdot 10^{-4} (s)$$

So, we hold it for a time 

$$t'=t_0-t=5\cdot 10^{-4} (s)$$