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P: 321
 Quote by DrGreg Note that the reason why it makes sense to consider $$\Sigma E$$ and $$\Sigma \textbf{p}$$ is because of conservation of energy and momentum. We want to replace our multi-particle system with a single particle which behaves like the system, as far as possible. For each particle, the four dimensional vector $$\left(E, \textbf{p}c \right)$$ is a "4-vector", which means that it obeys the Lorentz transform $$E' = \gamma_u \left(E - u p_x \right)$$ $$p_x' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$ $$p_y' = p_y$$ $$p_z' = p_z$$
Yes, you were right all along about $$p_y' = p_y$$
$$p_z' = p_z$$, I had to do the calculations myself, it wasn't obvious.
$$E'^2-(p'_xc)^2=E^2-(p_xc)^2$$
$$E'^2-(\vec{p'}c)^2=E^2-(\vec{p}c)^2$$