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pondzo
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Homework Statement
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.
Homework Equations
##m_{\pi^+}=140 \text{ Mev/c}^2;~~~~~~~m_{\mu^+}=106 \text{ Mev/c}^2##
The Attempt at a Solution
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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.