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Four Momentum in Different Frames

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A pion of mass m moves with a speed u in the positive y direction in frame S with parametrs BETAu = u/c and GAMMAu = (1 - BETAu^2)^-1/2. The pion decays into two gammay rays. An observer S' moves in the positive x direction with speed v = (BETA)(c) [this is a different BETA and GAMMA than before] and detects the two photons.

A. Compute the sum of the x components of momentum in S' for the photons.
ANSWER: Px' = -(GAMMA)(GAMMAu)(BETA)mc
B. Compute the combined energy of the two photons in S'.
ANSWER: E' = (GAMMA)(GAMMAu)mc^2

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Solution Summary

The solution shows how to compute the sum of different momentums and combined energies in a system of four momentums.

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Since this decay of a pion into two photons conserves the 4-momentum, the sum of the 4-momenta of the photons (and therefore the sum of their energies as well as the sum of any component of their 3-momenta) are exactly the same as the 4-momentum of the original pion.

Therefore the answers to these questions are exactly the same as if they were asked for the original pion

In frame S, the 4-momentum of the original pion is

p1 = m1*gamma1*c*(1, 0, beta1, 0)

The 4-momentum of the observer (we give it a temporary mass m2 which will not enter the answer anyway) is
...

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