Speed of Muon: Conservation of Momentum/Energy

[Tony11235]

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

[tex] \frac{u}{c} = \frac{ (m_\pi/m_\mu)^2 - 1}{ (m_\pi/m_\mu)^2 + 1} [/tex]I've tried [tex] m_\pi c^2 = \gamma m_\mu c^2 [/tex]. No success with that. I've also tried the relationship [tex] \beta = pmuc/E [/tex]. 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?

 

[Tide]

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

 

[Tony11235]

So did I leave out the energy from the neutrino?
[tex] \pi^+ \rightarrow \mu^+ +v[/tex]

[tex] m_\pi c^2 = \gamma m_\mu c^2 + p_vc[/tex]

 

[Tide]

Yes, you left out the neutrino energy.

You should arrive at

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

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