Alec Aivazis edited kinematics_flux.tex  almost 10 years ago

Commit id: d26a02ba1f754d92b37ac7a2aaa23809dc3d9cca

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\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, To understand the behavior of the nuetrino flux, notice that  \begin{equation}  \cos \theta & = \sqrt{1 - \sin^2 \theta}   \end{equation}  and from  equation \ref{eq:tan} \begin{equation}  \sin \theta' \approx \frac{E_{\nu}}{E_{\nu}'} \tan \theta  \end{equation}  Therefore,  \begin{equation}  \begin{split}  \cos \theta' & \approx \sqrt{1-\frac{E_{\nu}^2}{E_{\nu}^2}\tan^2 \theta} \\  & = \sqrt{1-\frac{E_{\nu}^2}{E_{\nu}^2} \left(\frac{1}{\cos^2 \theta}-1\right)}  \end{split}  \end{equation}