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    Relativistic Collision

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    This is problem 12.34 from Griffiths' third edition of Electrodynamics:

    In the past, most experiments in particle physics involved stationary targets: one particle (usually a proton or an electron) was accelerated to a high energy E, and collided with a target particle at rest (Fig12.29a). Far higher relative energies are obtainable (with the same accelerator) if you accelerate both particles to energy E, and fire them at each other. Classically, the energy E' of one particle, relative to other, is just 4E(why?)-not much of a gain (only a factor of 4). But relativistically the gain can be enormous. Assuming the two particles have the same mass m, show that E' = 2E^2/mc^2 - mc^2

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    Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here.
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    (For Ebar I am using E')

    Assume that the particles approach each other with velocities v. (Fig. b)

    Classically, the velocity of one particle with respect to the other is v + v = 2v

    Therefore the ...

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