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Relativistic Mechanics

Problem 1.386

How high must be the kinetic energy of a proton striking another, stationary, proton for their combined kinetic energy in the frame of the centre of inertia to be equal to the total kinetic energy of two protons moving toward each other with individual kinetic energies \(T=25.0 \mathrm{GeV} ?\)

Reveal Answer
T′=2T(T+2m0c2)/m0c2=1.43⋅103GeVT^{\prime}=2 T\left(T+2 m_{0} c^{2}\right) / m_{0} c^{2}=1.43 \cdot 10^{3} \mathrm{GeV}
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