Alec Aivazis edited kinematics.tex  almost 10 years ago

Commit id: bc646c532ad2daef9301bac3cad3ecb559e7d162

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\begin{equation}  \begin{split}  &\mathbf{\nu_{\mu}}^2 = m_{\nu} m_{\nu}^2  \\ \implies &\mathbf{\nu_{\mu}} &\mathbf{\nu_{\mu}}^2  \approx 0 \end{split}  \end{equation}  Plugging this result back into \ref{eq:energy-momentum-expanded}, we get  \begin{equation}  \mathbf{\pi^{+}} = ( m_{\pi} , 0, 0, 0 )  \end{equation}