In the Tevatron accelerator/storage ring at the Fermi National Accelerator Laboratory, two beams of protons travel in opposite directions each with a total energy of 1 TeV and interact. Since these beams have momenta of equal magnitude but opposite direction, they interact in their center of momentum inertial frame. Hence s2 = 4TeV2. If one wished to achieve the same total energy in the center of momentum with a target proton at rest and a very high energy beam proton, what kinetic energy must this beam proton have?

1 TeV = 106 MeV

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The explanation of the calculation is in the attached pdf file.

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centerline{bf Tevatron}

Recall that
$$
s^2 = (p_1+p_2)^2,
eqno(1)
$$
where $p_1$ and $p_2$ are the 4-momenta in the center of mass frame,
$$
p_1 = mc^2gamma(1,beta,0,0), ~~~p_2 = mc^2gamma(1,-beta,0,0),
eqno(2)
$$
where $m = 938.3 ...

Solution Summary

This Solution contains calculations to aid you in understanding the Solution to this question.

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