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

  1. Feb 26, 2012 #1

    BJD

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    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 + [itex]\nu[/itex]. If the neutrino emerges at 90° to the original pion direction at what angle does the muon come off?
    [ Answer: tanθ = ( 1 - m[itex]_{\mu}[/itex][itex]^{2}[/itex] / m[itex]_{\pi}[/itex][itex]^{2}[/itex] ) / ( 2βγ[itex]^{2}[/itex] ) ]

    2. Relevant equations
    → using particle physics (pp) units:
    E[itex]_{\pi}[/itex] = E[itex]_{\mu}[/itex] + E[itex]_{\nu}[/itex] → energy conservation.
    [itex]\bar{p_{\pi}}[/itex] = [itex]\bar{p_{\mu}}[/itex] + [itex]\bar{p_{\nu}}[/itex] → momentum conservation. (3 vector)
    βγm[itex]_{\pi}[/itex] = |[itex]\bar{p_{\pi}}[/itex]| (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[itex]_{\pi}[/itex][itex]^{2}[/itex] = ( E[itex]_{\mu}[/itex] + E[itex]_{\nu}[/itex] )[itex]^{2}[/itex] - ( [itex]\bar{p_{\mu}}[/itex] + [itex]\bar{p_{\nu}}[/itex] )[itex]^{2}[/itex]

    →m[itex]_{\pi}[/itex][itex]^{2}[/itex] = E[itex]_{\mu}[/itex][itex]^{2}[/itex] + E[itex]_{\nu}[/itex][itex]^{2}[/itex] + 2E[itex]_{\mu}[/itex]E[itex]_{\nu}[/itex] - { [itex]\bar{p_{\mu}}[/itex][itex]^{2}[/itex] + [itex]\bar{p_{\nu}}[/itex][itex]^{2}[/itex] + 2[itex]\bar{p_{\mu}}[/itex][itex] \cdot[/itex][itex]\bar{p_{\nu}}[/itex]}

    substituting ( m[itex]^{2}[/itex] = E[itex]^{2}[/itex] - p[itex]^{2}[/itex] ) into:
    →m[itex]_{\pi}[/itex][itex]^{2}[/itex] = E[itex]_{\mu}[/itex][itex]^{2}[/itex] - p[itex]_{\mu}[/itex][itex]^{2}[/itex] + E[itex]_{\nu}[/itex][itex]^{2}[/itex] - p[itex]_{\nu}[/itex][itex]^{2}[/itex] + 2E[itex]_{\mu}[/itex]E[itex]_{\nu}[/itex] - 2|[itex]\bar{p_{\mu}}[/itex]||[itex]\bar{p_{\nu}}[/itex]|cos ( 90°+θ )
    gives:
    →m[itex]_{\pi}[/itex][itex]^{2}[/itex] = m[itex]_{\mu}[/itex][itex]^{2}[/itex] + ( m[itex]_{\nu}[/itex][itex]^{2}[/itex] = 0 ) + 2E[itex]_{\mu}[/itex]E[itex]_{\nu}[/itex] - 2|[itex]\bar{p_{\mu}}[/itex]||[itex]\bar{p_{\nu}}[/itex]|( - 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. jcsd
  3. Feb 26, 2012 #2

    vela

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    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##.
     
  4. Feb 26, 2012 #3

    BJD

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    Thank you vela
     
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