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The proper lifetime of an unstable particle is equal to `Deltat_0=10ns`. Find the distance this particle will traverse till its decay in the laboratory frame of reference, where its lifetime is equal to `Deltat=20ns`. |
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Answer» The distance travelled in the laboratory frame of reference is `vDeltat` where v is the velocity of the particle. But by time dilation `Deltat=(Deltat_0)/(sqrt(1-v^2//c^2)` So `v=csqrt(1-(Deltat_0//Deltat)^2)` Thus the distance traversed is `cDeltatsqrt(1-(Deltat_0//Deltat)^2)` |
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