= Equal-angle relativistic elastic scattering
{title2=$\cos^2\theta=(T+2Mc^2)/(T+4Mc^2)$}
For two identical particles of <rest mass> $M$, one initially stationary and the other with <kinetic energy> $T$, symmetric final paths at $\pm\theta$ force equal final <energies> $Mc^2+T/2$. The <relativistic energy-momentum relation> and longitudinal <momentum conservation> give the title formula. The angle between the paths is $2\theta$, tending to $90^\circ$ at low <kinetic energy> and to zero at high <kinetic energy>.
Back to article page
