Asked by Hercules
An unstable high-energy particle enters a detector and leaves a track 1.35 mm long before it decays. Its speed relative to the detector was 0.992c. What was its proper lifetime? That is, how long would the particle have lasted before decay had it been at rest with respect to the detector? please use units for final answer
Answers
Answered by
drwls
The apparent lifetime of the particle in the laboratory/detector coordinate system is (track length)/speed =
1.35*10^-3 m/[(0.992)(3*10^8) m/s]
= 4.5*10^-12 s
In its own frame of reference, the lifetime is shorter by the relativistic factor sqrt[1 - (v/c)^2]= 0.126
The high speed in the labratory frame of reference slows the observed decay process.
Its actual lifetime is 5.5*10^-13 s
1.35*10^-3 m/[(0.992)(3*10^8) m/s]
= 4.5*10^-12 s
In its own frame of reference, the lifetime is shorter by the relativistic factor sqrt[1 - (v/c)^2]= 0.126
The high speed in the labratory frame of reference slows the observed decay process.
Its actual lifetime is 5.5*10^-13 s
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