Question

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 $$ \tau_{0} = 25.5\,ns$$.

Answer 1

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.
$$ \dfrac{J}{J_{0}} = e^{-t/\tau} = e^{-l/ \nu\ \tau} = e^{-l/c \tau_{0}} \sqrt{\eta (2 + \eta)} $$
$$ = exp \left (- \dfrac{m\ c\ l}{\tau_{0} \sqrt{T (T + 2m\ c^{2})}} \right ) = 0.221 $$