# Observed lifetime of muons (relativity)

1. Feb 5, 2009

### Ravenatic20

1. The problem statement, all variables and given/known data
The muon is an unstable particle that spontaneously decays into an electron and two neutrinos. If the number of muons at t = 0 is $$N_{o}$$, the number at time t is given by $$N = N_{0}e^{-t/\tau}$$, where $$\tau$$ is the mean lifetime, equal to 2.2 $$\mu s$$. Suppose the muons move at a speed of 0.95c and there are $$5.0 X 10^{4}$$ muons at t = 0. (a) What is the observed lifetime of the muons? (b) How many muons remain after traveling a distance of 3.0 km?

2. Relevant equations
$$N = N_{0}e^{-t/\tau}$$
$$t = d/v$$
$$\tau ' = \gamma \tau$$
$$\gamma = 1/ \sqrt{1 - v^{2}/c^{2}}$$

3. The attempt at a solution
$$t = d/v => 3 km/0.95c = 1.05 X 10^{-5}$$ where c is $$3 X 10^{8}$$

In the Muon frame: $$\tau = 2.2\mu$$
In the Earth frame: $$\tau ' = \gamma \tau$$
$$\gamma = 1/ \sqrt{1 - v^{2}/c^{2}} => 1/ \sqrt{1 - (0.95c/c)^{2}}$$
$$\gamma = 3.2 \mu s$$
Plug this back into the equation for $$\tau '$$...
$$\tau ' = 7.04 \mu s$$
In the back of my book, the solution for part (a) is $$7.1 \mu s$$. This answer ($$\tau ' = 7.04 \mu s$$) doesn't have anything to do with the answer to (a), does it?

Next, I need to plug in the numbers I have into N, which is $$N = N_{0}e^{-t/\tau}$$
If I read the problem right, $$N_{o} = 5.0 X 10^{4}$$. We know t, and we know the new value of tau. So it's plug and chug, right?
If so, this is what I got:
$$N = N_{0}e^{-t/\tau}$$
$$N = 5.0 X 10^{4}e^{-1.05 X 10^{-5}/7.04 \mu s}$$
$$N = 4.9 X 10^{-4}$$
This is not the same answer in the back of the book, nor does it make sense. The answer in the back of the book is $$1.1 X 10^{-4}$$

Where am I going wrong? Please put me on track, thanks!

2. Feb 5, 2009

### Ravenatic20

I apologize for the shameless 'thread bumping', but I'm really stuck on this problem and could use some help. It's due in about 12 hours so that's why I'm rushing now. I don't like turning in HW I know is wrong.

3. Feb 5, 2009

### Dick

I think your book answers have some problems. Sure, the gamma factor is about 3.2 (no units, it's dimensionless). So the observed lifetime is about 7.04 microseconds. Why the book says 7.1, I don't know. Now travelling for 3km at 0.95c takes about 10.5 microseconds. There should be a decrease in the number muons by a modest factor. Certainly not from 5*10^4 to 1.1*10^(-4). No, it doesn't make much sense.

4. Feb 6, 2009

### Ravenatic20

Dick, thanks for catching that gamma is dimensionless. If anyone has any last minute ideas please post them. Thank you.

5. Feb 6, 2009

### Ravenatic20

I found out what I was doing wrong. It's really stupid. When solving for N, my value for tau(7.04 microseconds), I didn't punch it into my calculator right. Was entering 7.04 instead of 7.04X10^-6. The answer I got then was right, about 1.13X10^4.

Mark this problem as solved. Thanks anyways guys.