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

by BJD
Tags: angle, decay, muon, particle, physics, pion, revision
 P: 2 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?
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,862 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##.
 P: 2 Thank you vela

 Related Discussions High Energy, Nuclear, Particle Physics 5 High Energy, Nuclear, Particle Physics 12 High Energy, Nuclear, Particle Physics 3 Advanced Physics Homework 2 Advanced Physics Homework 10