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Calculate the energy of the neutrino in the following decay

  1. May 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Stopped pions provide a useful mono-energetic source of neutrinos. For a pion at rest, calculate the energy of the neutrino in the decay $$\pi^+\rightarrow \mu^++\nu_{\mu}$$ You do not need to consider the subsequent decay of the ##\mu^+## and you can assume that the neutrinos are massless.

    The answer is 30 Mev/c^2.

    2. Relevant equations

    ##m_{\pi^+}=140 \text{ Mev/c}^2;~~~~~~~m_{\mu^+}=106 \text{ Mev/c}^2##

    3. The attempt at a solution

    In the rest frame of the pion, ##\vec{p}=\vec{p}_{\mu}=-\vec{p}_{\nu}## and in the massless approxiamation ##E_{\nu}|=\vec{p}_{\nu}|\implies |\vec{p}_{\mu}|=E_{\nu}##.

    The invariant mass is ##m_{\pi^+}^2##.
    ##m_{\pi^+}^2 = ((E_{\mu},\vec{p}_{\mu})+(E_{\nu},\vec{p}_{\nu}))^2##
    ##~~~~~~~=E_{\mu}^2+E_{\nu}^2+2E_{\mu}E_{\nu}-p_{\mu}^2-p_{\nu}^2-2\vec{p}_{\mu}\cdot\vec{p}_{\nu}##
    ##~~~~~~~=E_{\mu}^2+E_{\nu}^2+2E_{\mu}E_{\nu}-E_{\nu}^2-E_{\nu}^2-0##
    ##~~~~~~~=E_{\mu}^2-E_{\nu}^2+2E_{\mu}E_{\nu}##
    I know that ##|\vec{p}_{\mu}|^2=E_{\nu}^2=E_{\mu}^2-m_{\mu}^2\implies E_{\nu}=\frac{m_{\pi}^2-m_{\mu}^2}{2E_\mu}##

    So now I just need to get ##E_{\mu}## in terms of things I know. I know ##E_{\mu}=\sqrt{E_{\nu}^2+m_{\mu}^2}##, but now this is seeming cat and mouse.

    I tried this:
    ## E_{\nu}=\frac{m_{\pi}^2-m_{\mu}^2}{2E_\mu}##
    ##2\sqrt{E_{\nu}^2+m_{\mu}^2}E_{\nu}=(m_{\pi}^2-m_{\mu}^2)## then square both sides and expand to get:
    ##4E_{\nu}^2+4m_{\mu}^2E_{\nu}^2-(m_{\pi}^2-m_{\mu}^2)^2=0##

    I can then make the substitution ##x=E_{\nu}^2## which results in a quadratic. I can solve this and find the roots using the quadratic equation, but this gives ##E_{\nu}=52.64 ## Mev, which isn't right. I'm not sure why this doesn't give the correct answer and i'm not sure how to get the correct answer, there is probably an easier way.
     
  2. jcsd
  3. May 12, 2016 #2
    By direct conservation of energy, don't you just have
    [tex]m_{\pi} = E_{\nu} + \sqrt{m_{\mu}^2 + E_{\nu}^2}?[/tex]
    There's no need to always resort to using invariant quantities and four-vectors when a direct approach is available.

    The mistake you made in your calculations is that [itex]\vec{p}_{\mu} \cdot \vec{p}_{\nu} \neq 0[/itex].
     
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