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# There is a narrow beam of negative pions with kinetic energy T equal to the rest energy particles. Find the ratio of fluxes at the sections of the beam separated by a distance $$l = 20\,m$$. The proper mean lifetime of these pions is $$\tau_{0} = 25.5\,ns$$.

Solution
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#### Here $$\eta = \dfrac{T}{mc^{2}} = 1$$ so the life time of the pion in the laboratory frame is $$\eta = (1 + \eta) \tau_{0} = 2 \tau_{0}$$ The law of radioactive decay implies that the flux decreases by the factor. $$\dfrac{J}{J_{0}} = e^{-t/\tau} = e^{-l/ \nu\ \tau} = e^{-l/c \tau_{0}} \sqrt{\eta (2 + \eta)}$$ $$= exp \left (- \dfrac{m\ c\ l}{\tau_{0} \sqrt{T (T + 2m\ c^{2})}} \right ) = 0.221$$

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