# Homework Help: Pion Decay & Lorentz Transforms

1. Feb 10, 2010

### ian2012

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

A pi+, produced by a cosmic ray, has an energy of 280MeV in the laboratory frame S. It undergoes a decay: pi+ ---> muon+ & muon-neutrino. The muon produced continues to travel along the same direction as the pi+ and has a mean lifetime of 2.2mirco seconds in the muon's rest frame (S''). Calculate the expected distance it would have travelled in the laboratory frame S before decaying.

2. Relevant equations

In the rest frame of the pion (S'):
$$E^{'}_{\mu}=\frac{(m^{2}_{\pi}+m^{2}_{\mu})c^{2}}{2m_{\pi}}$$
$$P^{'}_{\mu}=\frac{(m^{2}_{\pi}-m^{2}_{\mu})c}{2m_{\pi}}$$
and other relativistic relations and Lorentz Transforms

$$m_{\pi}=139.6\frac{MeV}{c^{2}}, m_{\mu}=105.7\frac{MeV}{c^{2}}$$

3. The attempt at a solution

I have attempted the problem, I don't know if i am correct.
The obvious thing to do is find the mean lifetime in the frame S' and the distance in S' in order to solve the equation:
$$x_{\mu}=\gamma(x^{'}_{\mu}+vt^{'}_{\mu})$$

To find t' I have used the fact that time dilates in moving frames therefore t' = gamma x t'' (2.2 mircoseconds)
I then used the relativistic momentum equation to find the velocity of muon in the pion rest frame S': v' = 0.28177/gamma
Then i used LT to find the distance travelled in the pion frame = gamma x v x t'' = 0.28177 x t''.
Then I used the energy of the pion in the lab frame to find the velocity of the pion in the lab frame using the relativistic energy equation and the momentum equation to give = 1.739/gamma.
Subbing these into the above equation I got the expected distance = gamma x 4.45x10^-6 metres.
Is this correct? Is the gamma supposed to be there?

2. Feb 10, 2010

### vela

Staff Emeritus
Looks okay so far.

So just to be clear: this gamma is for the pion moving relative to S, not the gamma above.

I don't follow what you did here. Could you provide more detail, like exactly what quantities you plugged into what equations?

The answer isn't right. You left out a factor of c, for one thing. Even after multiplying your answer by c, it's still off. You should get 1565 meters.

3. Feb 13, 2010

### ian2012

Sorry, but I have been away for a while. Looking back at this question, I've realised what mistakes I've made:
1. I didn't realise gamma can be calculated using Energy and Mass (very trivial).
2. I didn't realise there are two different gamma's, since there are three different frames.
3. I also missed out the speed of light in the final answer, as you pointed out.

The final answer I get is: 1551 metres with minimal rounding of answers. (x' = 179m, v[lab/pion] = 0.867c, t' = 2.29 microseconds). This is a little bit off your 1565 metres.

4. Feb 13, 2010

### vela

Staff Emeritus
You miscalculated x'. It should be 186 m.

5. Feb 14, 2010

### ian2012

Oh right yeah of course. I divided by an extra gamma term. Yeah i get x' = 186m and x = 1.56km with rounding. Thanks vela