Question #266247

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 (πœ‡βˆ’ β†’ π‘’βˆ’ + 𝜈 + πœˆΜ…).  

(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? 


Expert's answer

Lorentz factor:\text {Lorentz factor:}

v=0.99cv = 0.99c

Ξ³=11βˆ’v2c2=11βˆ’0.992β‰ˆ7.09\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-0.99^2}}\approx7.09

for Earthbound observer:\text{for Earthbound observer:}

sE=4.6kms_E = 4.6km

vE=0.99cv_E= 0.99c

cβ‰ˆ3βˆ—108mcc \approx 3*10^8\frac{m}{c}

tE=sEvE=46000.99βˆ—3βˆ—108β‰ˆ1.54βˆ—10βˆ’5 st_E= \frac{s_E}{v_E}= \frac{4600}{0.99*3*10^8}\approx1.54*10^{-5}\ s

for muonbound observer:\text{for muonbound observer:}

tm=tEΞ³=1.54βˆ—10βˆ’57.09β‰ˆ2.18βˆ—10βˆ’6st_m = \frac{t_E}{\gamma}=\frac{1.54*10^{-5}}{7.09}\approx 2.18*10^{-6}s

sm=vtm=vβˆ—tEΞ³=sEΞ³β‰ˆ648.8ms_m = vt_m= v* \frac{t_E}{\gamma}=\frac{s_E}{\gamma}\approx648.8m

Answer:\text{Answer:}

a)1.54βˆ—10βˆ’5 sa)1.54*10^{-5}\ s

b)648.8 mb)648.8\ m


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