MSc particle physics revision question - angle of muon from pion decay

[BJD]

I am trying to revise for PhD, going over MSc work. Could anyone help me with this question?

Homework Statement

A pion traveling at speed β(=v/c) decays into a muon and a neutrino, π→μ + \nu. If the neutrino emerges at 90° to the original pion direction at what angle does the muon come off?
[ Answer: tanθ = ( 1 - m_{\mu}^{2} / m_{\pi}^{2} ) / ( 2βγ^{2} ) ]

Homework Equations

→ using particle physics (pp) units:
E_{\pi} = E_{\mu} + E_{\nu} → energy conservation.
\bar{p_{\pi}} = \bar{p_{\mu}} + \bar{p_{\nu}} → momentum conservation. (3 vector)
βγm_{\pi} = |\bar{p_{\pi}}| (speed of light c not included as pp units)

The Attempt at a Solution

invariant mass squared from decay of the moving pion: m_{\pi}^{2} = ( E_{\mu} + E_{\nu} )^{2} - ( \bar{p_{\mu}} + \bar{p_{\nu}} )^{2}

→m_{\pi}^{2} = E_{\mu}^{2} + E_{\nu}^{2} + 2E_{\mu}E_{\nu} - { \bar{p_{\mu}}^{2} + \bar{p_{\nu}}^{2} + 2\bar{p_{\mu}}\cdot\bar{p_{\nu}}}

substituting ( m^{2} = E^{2} - p^{2} ) into:
→m_{\pi}^{2} = E_{\mu}^{2} - p_{\mu}^{2} + E_{\nu}^{2} - p_{\nu}^{2} + 2E_{\mu}E_{\nu} - 2|\bar{p_{\mu}}||\bar{p_{\nu}}|cos ( 90°+θ )
gives:
→m_{\pi}^{2} = m_{\mu}^{2} + ( m_{\nu}^{2} = 0 ) + 2E_{\mu}E_{\nu} - 2|\bar{p_{\mu}}||\bar{p_{\nu}}|( - sin (θ) ) (the mass of the neutrino is taken as zero here)

also as: cos (90+θ) = cos(90) cos(θ) - sin(90)sin(θ) = - sin (θ)

I got stuck a few lines after this, can anyone who understands this help? Am I on the right track with the methodology?

 

[vela]

You'll find it more convenient to work with four-vectors. Let $p_\pi^\alpha = (E_\pi, \vec{p}_\pi)$, $p_\mu^\alpha = (E_\mu, \vec{p}_\mu)$, and $p_\nu^\alpha = (E_\nu, \vec{p}_\nu)$ be the four-momentum of the pion, muon, and antineutrino respectively, where $\vec{p}$ denotes three-momentum. Conservation of energy and momentum gives you
$$p_\pi^\alpha = p_\mu^\alpha + p_\nu^\alpha.$$ Squaring this yields
$$m_\pi^2 = m_\mu^2 + m_\nu^2 + 2p_\mu^\alpha {p_\nu}_\alpha = m_\mu^2 + m_\nu^2 + 2(E_\mu E_\nu - \vec{p}_\mu \cdot \vec{p}_\nu),$$ which is the same thing you got with a bit more algebra.

Often, the trick to these problems is to rearrange the original equation so that the product of the various four-vectors results in the conveniently placed zero. For example, you know that $\vec{p}_\pi$ and $\vec{p}_\nu$ are perpendicular to each other, so their dot product will vanish. This suggests you try squaring $p_\pi^\alpha - p_\nu^\alpha = p_\mu^\alpha$.

 

[BJD]

Thank you vela