A graduate student discovers that an elementary particle produced in his experiment travels 0.250 mm through the lab at a speed of 0.950c before it decays (becomes another particle). What is the lifetime of the particle measured in its rest frame (proper lifetime). A) 8.77 x 10-13 s B) 7.99 x 10-13 s C) 2.74 x 10-13 s D) 2.24 x 10-13 s E) 2.81 x 10-12 s
L = L0 / ϒ
t = ϒt0
γ = 1/(1 - (v^2/c^2))^(1/2)
The Attempt at a Solution
I found ϒ to be equal to 3.2025 which I believe is correct. Given in the problem, it says that the observer sees the particle travel through 0.250 mm. That would be considered to be in the frame that is "moving" with respect to the particle. To find the proper lifetime, I would need to find the distance observed by someone who is a rest with respect to the particle (usually I imagine someone riding on the particle). The distance they would observe would be "L0," because that is the length observed by a someone at rest. L0 = L(ϒ) = 0.250 x 10^-3)(3.2025) = 8 x 10^-4 m.
With that I took the speed of the particle to be 0.95c and I divided that by 8 x 10^-4 to get the time observed by someone at rest with respect to the particle. This gave me 2.81 x 10^-12, which is not correct.
I got the right answer by solving first for the time observed by the graduate student ("moving" with respect to the particle). That time would be considered to be in the primed frame, and then using the second equation listed above I solved for t0. So, v = 0.95c, d = 0.250 x 10^-3... t = 0.250 x 10^-3/(0.95c) = 8.77 x 10^-13 s. Then t0 = t/γ = 8.77 x 10^-13/3.2025 = 2.74 x 10^-13. Shouldn't I be getting the same answer no matter which way I decide to do it?