Calculate the energy of the neutrino in the following decay

[pondzo]

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

[/B]
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.

 

[Fightfish]

By direct conservation of energy, don't you just have
m_{\pi} = E_{\nu} + \sqrt{m_{\mu}^2 + E_{\nu}^2}?
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 \vec{p}_{\mu} \cdot \vec{p}_{\nu} \neq 0.