The relevant decay for the NuMI beam is
\[\pi^{+} \rightarrow \mu^{+} \nu_{\mu}\]
Which must conserved energy and momentum according to the 4 vector equation:
\[\label{eq:energy-momentum} \mathbf{\pi^{+}} = \mathbf{\mu^{+}} + \mathbf{\nu_{\mu}}\]
Where \(\mathbf{\pi^{+}}\), \(\mathbf{\mu^{+}}\), and \(\mathbf{\nu_{\mu}}\) are the energy-momentum 4-vectors.
Rearranging equation \ref{eq:energy-momentum} as
\[\mathbf{\mu^{+}} = \mathbf{\pi^{+}} - \mathbf{\nu_{\mu}}\]
And squaring both sides, we get
\[\label{eq:energy-momentum-expanded} \mathbf{\mu^{+}}^2 = \mathbf{\pi^{+}}^2 - 2 ( \mathbf{\pi^{+}} \cdot \mathbf{\nu_{\mu}} ) - \mathbf{\nu_{\mu}}^2\]
It is important to note that since the magnitude of a 4-vector is a Lorenz invariant, for any energy-momentum 4 vector \(\mathbf{p}\), where
\[\mathbf{p} = (E, \vec{p} )\]
one can boost to the rest frame of the particle where \(\vec{p} = 0\) and a find the value for \(\mid \mathbf{p} \mid ^2\). Since these particles are moving close to the speed of light, in natural units,
\[E^2 = \vec{p}^2 + m^2\]
Then in the rest frame of the particle,
\[E^2 = m^2\]
Therefore,
\[\begin{split} \mid \mathbf{p} \mid^2 &= E^2 + p^2 \\ &= m^2 \end{split}\]
The neutrinos mass is many orders of magnitude smaller than that of the other particles \cite{Robertson_2008}; to good approximation,
\[\begin{split} &\mathbf{\nu_{\mu}}^2 = m_{\nu}^2 \\ \implies &\mathbf{\nu_{\mu}}^2 \approx 0 \end{split}\]
Plugging these results into equation \ref{eq:energy-momentum-expanded} we get
\[\label{eq:energy-momentum-expanded-more} m_{\mu}^2 = m_{\pi}^2 - 2(\mathbf{\pi^{+}} \cdot \mathbf{\nu_{\mu}})\]
In the rest frame of the pion, with z as the direction of travel for the nutrino
\[\begin{split} \mathbf{\pi^{+}} &= (m_{\pi} , 0, 0, 0) \\ \mathbf{\nu_{\mu}} &= (E_{\nu}', E_{\nu}' \sin \theta, 0, E_{\nu}' \cos \theta) \end{split}\]
Therefore, the dot-product is given by
\[( \pi \cdot \nu ) = m_{\pi} E_{\nu}'\]
Plugging this into equation \ref{eq:energy-momentum-expanded-more},
\[\begin{split} m_{\nu}^2 = m_{\pi}^2 - 2 m_{\pi}E_{\nu}' \end{split}\]
and solving for the energy of the neutrino in the rest frame of the pion,
\[\begin{split} E_{\nu}' &= \frac{m_{\pi}^2 - m_{\mu}^2}{2 m_{\pi}} \\ &= 29.8 MeV \end{split}\]
These derivations are adapted from \cite{McDonald_2001}↩