How Far Does a Muon Travel Before Decaying?

[spidey007]

i have a question
i have to do this problem
i know somewhat how it sets up but what do u take from the initial condition? do u take the distance the muon travels?
can someone help me set this bad boy up

Suppose a muon produced as a result of a cosmic ray colliding with a nucleus in the upper atmosphere has a velocity v = 0.9500c. Suppose it travels at constant velocity and lives 1.52 µs as measured by an observer who moves with it (this is the time on the muon's internal clock). It can be shown that it lives for 4.87 µs as measured by an earth-bound observer. Use c=2.997E+8 m/s and give all answers correct to 4 significant figures.
(a) How long would the muon have lived as observed on Earth if its velocity was 0.0900c?
(b) How far would it have traveled as observed on earth?
(c) What distance is this in the muon's frame?

 

[Doc Al]

This is a basic relativity problem; it probably should be moved to the Homework Help forum. (It sure doesn't belong in the Quantum Physics forum.)

Here are some tips:

Originally posted by spidey007

(a) How long would the muon have lived as observed on Earth if its velocity was 0.0900c?

Consider relativistic time dilation; you have the proper time.

(b) How far would it have traveled as observed on earth?

Distance = speed x time.

(c) What distance is this in the muon's frame?

Consider relativistic length contraction. (Or just use "Distance = speed x time" again.)

 

[R0man]

To solve this problem, we first need to understand the initial condition, which is the velocity of the muon (v=0.9500c) and its lifespan (1.52 µs on its internal clock and 4.87 µs as measured by an earth-bound observer). From this, we can calculate the distance the muon travels using the formula d=vt, where d is the distance, v is the velocity, and t is the time.

(a) To find the time the muon would have lived as observed on Earth if its velocity was 0.0900c, we can use the time dilation formula t'=t/√(1-v^2/c^2), where t' is the time measured by the earth-bound observer, t is the time measured by the muon, v is the velocity, and c is the speed of light. Plugging in the values, we get t'=4.87 µs/√(1-0.0900^2)=4.88 µs. This means that the muon would have lived for 4.88 µs as observed on Earth.

(b) To find the distance the muon would have traveled as observed on Earth, we can use the formula d=vt. Plugging in the values, we get d=(0.0900c)(4.88 µs)=1.31 km. This means that the muon would have traveled 1.31 km as observed on Earth.

(c) To find the distance in the muon's frame, we can use the Lorentz contraction formula x'=x√(1-v^2/c^2), where x' is the distance measured in the muon's frame, x is the distance measured on Earth, v is the velocity, and c is the speed of light. Plugging in the values, we get x'=1.31 km√(1-0.0900^2)=1.31 km. This means that the distance traveled in the muon's frame is also 1.31 km. This makes sense because the muon is traveling at a constant velocity, so there is no relative motion between the two frames.