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

1. Feb 26, 2012

### BJD

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

1. The problem statement, all variables and given/known data
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$^{2}$ ) ]

2. Relevant 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)

3. 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?

2. Feb 26, 2012

### vela

Staff Emeritus
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 $p_\pi^\alpha - p_\nu^\alpha = p_\mu^\alpha$.

3. Feb 26, 2012

### BJD

Thank you vela