Alec Aivazis edited kinematics_angle.tex  almost 10 years ago

Commit id: 809d36d90f452c0dd06f340a7ef70a7e5d955aaa

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\end{split}  \end{equation}  If we consider the case where the beam has much more energy than the relative masses, then $E_{\pi} \gg m_{\pi}$, $\beta \approx 1$, and $\gamma \gg 1$. In this case, the previous  equation\ref{eq:tan}  becomes \begin{equation}  \tan \theta \approx \frac{E_{\nu}' \sin \theta'}{E_{\nu}}