Speed of the muon

1. Feb 2, 2006

Tony11235

The positive pion decays into a muon and a neutrino. The pion has rest mass $$m_\pi$$, the muon has $$m_\mu$$ , while the neutrino has $$m_v = 0$$. Assuming the original pion was at rest, use conservation of momentum and energy to show that the speed of the muon is given by:

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

I've tried $$m_\pi c^2 = \gamma m_\mu c^2$$. No success with that. I've also tried the relationship $$\beta = pmuc/E$$. Still no. I know it SAYS to use conservation of energy and momentum, but I have yet to put together a correct relationship. Any help?

Last edited: Feb 2, 2006
2. Feb 2, 2006

Tide

HINT: Since the neutrino is massless its energy and momentum are related by E = pc.

3. Feb 2, 2006

Tony11235

So did I leave out the energy from the neutrino?
$$\pi^+ \rightarrow \mu^+ +v$$

$$m_\pi c^2 = \gamma m_\mu c^2 + p_vc$$

Last edited: Feb 3, 2006
4. Feb 3, 2006

Tide

Yes, you left out the neutrino energy.

You should arrive at

$$\frac { m_{\pi}} {m_{\mu}} = \gamma (1 +u/c)$$

after combining the energy and momentum equations and the rest is simple algebra.

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