Question

Many of the particles we see in cosmic rays (and also create in particle accelerators) are unstable. They live only for a short period of time before they decay to other particles. One such unstable particle is called a muon, essentially a heavy version of an electron. Imagine that a muon created in the upper atmosphere is headed straight down towards the Earth at 99.99% the speed of light. This particular muon lives for exactly one millionth of a second as measured in the muon’s rest frame, before decaying into other particles. How far does the muon travel as viewed by a person on the ground? (Show exactly how you set up the calculation, as well as the result.)

Answer 1

According to the concept of time dilation in special relativity, the time experienced by a moving object relative to an observer is dilated or stretched. In this case, the muon is moving at 99.99% of the speed of light, which means its time is dilated. The muon travels a distance of approximately 14.1 nanoseconds as viewed by a person on the ground.

Step-by-step explanation:

The gamma factor, represented by the symbol γ, is given by the equation γ = 1 / sqrt(1 - (v^2 / c^2)), where v is the velocity of the muon and c is the speed of light. First, we need to calculate the gamma factor: γ = 1 / sqrt(1 - (0.9999^2)). This gives us γ ≈ 70.7. Next, we can calculate the time experienced by the muon as viewed by a person on the ground: t' = t / γ, where t is the time measured in the muon's rest frame. Given that t = 1 millionth of a second (1 μs), we can substitute the values into the equation: t' = 1 μs / 70.7 ≈ 14.1 nanoseconds. Therefore, as viewed by a person on the ground, the muon travels a distance of approximately 14.1 nanoseconds.