Posted by Chelsea on Wednesday, April 16, 2008 at 7:40pm.
The muon does not have a fixed lifetime, it just has a fixed probability to decy per unit time. The mean lifetime of the muon is 2.197 micro seconds in the rest frame of the muon.
Suppose that the muon decays after T = 2.197 microseconds in its rest frame. To compute how much time this is for the observer, you can use the formula for distances in Space-Time which generalizes Pythagoras's theorem:
S^2 = (distance in time)^2 - (distance in space)^2
S is the space time-distance between two points in space-time. Distance in time is defined as the speed of light times the time difference. It does not matter in which coordinate system you evaluate it (similarly when you compute the distance between two points using Pythagoras the answer does not depend on which coordinate system you use to compute that distance).
Let's evaluate S^2 between the space-time points where the muon starts and where it decays. In the frame of the muon, only the muon's time coordinate changes. So we find:
S^2 = c^2 T^2
If we evaluate S^2 in the frame ogf the observer, then
S^2 = c^2 T'^2 - V^2 T'^2
Here T' is the time it takes for the muon to decay according to the observer. Since both computations for S^2 must yield the same answer, you find:
c^2 T'^2 - V^2 T'^2 = c^2 T^2 ----->
T' = T/sqrt(1-v^2/c^2)
v is the speed of 0.7 c
I am trying to answer this question too, but plugging in doesnt work. Can you tell me what numbers were used to answer this? Observed lifetime in microseconds is what is wanted, i am probably doing this wrong.
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