# Determine velocity of muon

1. Jan 28, 2010

### abalmos

1. The problem statement, all variables and given/known data

Particle Physicists use particle track detectors to determine the lifetime of short-lived particles. A muon has a mean lifetime of 2.2 microseconds and makes a track of 9.5 cm long before decaying into a electron and two neutrinos. What was the speed of the muon?

2. Relevant equations

I believe:

$$L = L_0\sqrt{1 - \frac{v^{2}}{c^{2}}$$
$$T = \frac{T_0}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

and the standard transformations as well

3. The attempt at a solution

I suppose I understand the process but this might fundamentally be my issue. My understanding of what needs to be done is I need to determine what the length of time and space which the muon sees from its inertial frame (i.e. the muon is at rest). From there it is trivial to determine the speed the muon is traveling.

My biggest issue is I don't understand how I can put any of these equations together to get any type of solution. No matter how I manipulate them I am left with more variables then independent equations.

Any advise and guidance you could provide would be greatly appreciated. I seemed to be doing very well with this topic until this question albeit this is the last and hardest.

Thank you,
- Andrew Balmos

2. Jan 29, 2010

### Lok

Classically the speed would be v=x/t=0.095/2.2*10^-6=43181 m/s = 43.181 km/s (v<<c) so the relativity effects are barely present.

http://en.wikipedia.org/wiki/Muons

"When a cosmic ray proton impacts atomic nuclei of air atoms in the upper atmosphere, pions are created. These decay within a relatively short distance (meters) into muons (the pion's preferred decay product), and neutrinos. The muons from these high energy cosmic rays, generally continuing essentially in the same direction as the original proton, do so at very high velocities. Although their lifetime without relativistic effects would allow a half-survival distance of only about 0.66 km at most, the time dilation effect of special relativity allows cosmic ray secondary muons to survive the flight to the earth's surface. Indeed, since muons are unusually penetrative of ordinary matter, like neutrinos, they are also detectable deep underground and underwater, where they form a major part of the natural background ionizing radiation. Like cosmic rays, as noted, this secondary muon radiation is also directional."

An exemple of relativity on earth. I've found a better example somewhere (but this wil have to do). Basically the muon would travel only 0.7 km not enough to reach our surface but it does. So because of time dilation it's time life is longer from our POV
(around 6.6us as it's speed is close to c), but it is the same in it's referance frame (2.2us). The exact values for this example are somewhere but i cannot fin them now.

3. Jan 29, 2010

### abalmos

Of course the obvious answer... Thank you very much for helping me out with this.

For my own curiosity and interest in knowing I fully understand this topic, is it possible to come to this answer using the transformations? I suppose it must, but the real question is does this problem contain enough information?

Thank you again for all your help, you and this community are always so helpful.

- Andrew Balmos

4. Jan 31, 2010

### Lok

There is no other way to come to a solutiuon other that using the transformations.

The problem does contain enough info, and usually the speed is the observed speed, so not in the muons reference frame.

Again the speed is very slow in comparison to c, so you will have only a small time or lenght difference.

5. Jan 31, 2010

### abalmos

Right, I suppose my confusion was trying to find the muons proper time.

Next time I need to first check if relativity factors will have any effect in the first place.

Thanks again!

- Andrew Balmos