# Massless Particles and the algebra involved

1. Feb 1, 2010

### forrealfyziks

1. The problem statement, all variables and given/known data
The positive pion decays into a muon and a neutrino. The pion has rest mass m=140 MeV/c^2, the muon has m=106 MeV/c^2 while the neutrino is about m=0 (assume it is). Assuming the original pion was at rest, use conservation of energy and momentum to show that the speed of the muon is given by

U/c = [(m$$\pi$$/m$$\mu$$)2 - 1] / [(m$$\pi$$/m$$\mu$$)2 + 1]

2. Relevant equations
For massless particles, E=pc u=c $$\beta$$=1
p=$$\gamma$$mu E=$$\gamma$$mc2

3. The attempt at a solution (Note $$\gamma$$ and u are for mu, since it's the only particle using them)
Alright, I know that P$$\pi$$=0 and P$$\mu$$+P$$\nu$$=0. I found that P$$\nu$$=-$$\gamma$$m$$\mu$$u$$\mu$$
I replace p in m$$\pi$$c2=P$$\nu$$c + $$\gamma$$m$$\mu$$c2 and reduced it down and got
m$$\pi$$c=$$\gamma$$m$$\mu$$(c-u)
I then unraveled the gamma, moved some stuff around, and squared both sides and got (m$$\pi$$/m$$\mu$$)2=(c-v)2/(1-v2/c2)
I don't know if this is the right path, but I have tried many different methods from this point and nothing seems to get any closer to what I need. Thanks for any help.

2. Feb 1, 2010

### gabbagabbahey

You seem to have dropped a factor of $c^2$;

$$m_{\pi}c=\gamma m_{\mu}(c-u)=\left(1-\frac{u^2}{c^2}\right)^{-1/2}m_{\mu}(c-u)$$

$$\implies m_{\pi}^2c^2= \left(1-\frac{u^2}{c^2}\right)^{-1}m_{\mu}^2(c-u)^2$$

$$\implies c^2-u^2=\left(\frac{m_{\mu}}{m_{\pi}}\right)^2(c-u)^2$$

From here, just expand $(u-c)^2$, group terms in powers of $u$ and solve via the quadratic formula (studying university physics is no excuse for forgetting highschool algebra!)

3. Feb 2, 2010

### Count Iblis

I think you started out with way too complicated algebra, by considering the 3-mometum seperately. Also, you should use c = 1 units to avoid making mistakes with factors of c.

The simple way to soleve these sorts of problems is to use the energy momentum relation in the form:

p^2 = m^2

where p is the four-momentum, the square denotes the Lorentz inner product of the momentum with itself and m is the mass. By bringing the four-momentum of the particle in which you are not interested in to one side and squaring, you eliminate its momentum and energy in one go. So, in this case, you would proceed like:

p_neut = p_pi - p_mu

0 = m_pi^2 + m_mu^2 - 2 m_pi E_mu

Then solving for E_mu and dividing by m_mu gives you the gamma factor and you're done.

4. Feb 2, 2010

### forrealfyziks

eh I was up all night doing this so it was no doubt the fatigue, but I got to that point once and gave up(talking to gabba). I haven't been taught the four-momentum process so I can't use it unfortunately.

Thanks you!