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\subsubsection{Neutrino Flux}  Since NO$\nu$A is an off-axis neutrino detector, of most interest to this paper is the total flux of neutrinos at a given energy and angle.  According to \cite{McDonald_2001} since the pion is a spin zero particle, the decay is isotropic in the pion rest frame. Therefore, the volume elements can be seen as concentric cylinders to integrate over.   Assuming the total neutrino flux is proportional to the $\cos$ of the angle with which they are created and the energy of the beam,  \begin{equation}  \begin{split}  & d^2 N \propto d \cos \theta d E_{\nu} \\  \implies & \frac{d^2 N}{d \cos \theta' d E_{\nu}} \propto 1  \end{split}  \end{equation}  From the point of view of an observer in the lab, the total flux at a given angle and energy can by found using the relation  \begin{equation}  \frac{d^2 N}{d \cos \theta d E_{\nu}} = \frac{d^2 N}{d \cos \theta' d E_{\nu}} \frac{d \cos \theta'}{d \cos \theta}  \end{equation}  From, equation \ref{eq:tan}   \begin{equation}  \sin \theta' \approx \frac{E_{\nu}}{E_{\nu}'} \tan \theta  \end{equation}