Answer to Question #266247 in Mechanics | Relativity for Dinesh priyananda

Question

A muon formed high in the Earth’s atmosphere is measured by an observer on the Earth’s surface to travel at speed 𝑣 = 0.990𝑐 for a distance of 4.60 km before it decays into an electron, a neutrino, and an antineutrino (𝜇− → 𝑒− + 𝜈 + 𝜈̅).  

(a) For what time interval does the muon live as measured in its reference frame?  

(b) How far does the Earth travel as measured in the frame of the muon? 

Answer 1

$\text {Lorentz factor:}$

$v = 0.99c$

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-0.99^2}}\approx7.09$

$\text{for Earthbound observer:}$

$s_E = 4.6km$

$v_E= 0.99c$

$c \approx 3*10^8\frac{m}{c}$

$t_E= \frac{s_E}{v_E}= \frac{4600}{0.99*3*10^8}\approx1.54*10^{-5}\ s$

$\text{for muonbound observer:}$

$t_m = \frac{t_E}{\gamma}=\frac{1.54*10^{-5}}{7.09}\approx 2.18*10^{-6}s$

$s_m = vt_m= v* \frac{t_E}{\gamma}=\frac{s_E}{\gamma}\approx648.8m$

$\text{Answer:}$

$a)1.54*10^{-5}\ s$

$b)648.8\ m$