Quote by DrGreg
Note that the reason why it makes sense to consider [tex]\Sigma E[/tex] and [tex]\Sigma \textbf{p}[/tex] is because of conservation of energy and momentum. We want to replace our multiparticle system with a single particle which behaves like the system, as far as possible.
For each particle, the four dimensional vector [tex]\left(E, \textbf{p}c \right) [/tex] is a "4vector", which means that it obeys the Lorentz transform
[tex] E' = \gamma_u \left(E  u p_x \right) [/tex]
[tex] p_x' = \gamma_u \left(p_x  \frac{u E}{c^2} \right) [/tex]
[tex] p_y' = p_y [/tex]
[tex] p_z' = p_z [/tex]

Yes, you were right all along about [tex] p_y' = p_y [/tex]
[tex] p_z' = p_z [/tex], I had to do the calculations myself, it wasn't obvious.
This leads to:
[tex]E'^2(p'_xc)^2=E^2(p_xc)^2[/tex]
and ultimately to:
[tex]E'^2(\vec{p'}c)^2=E^2(\vec{p}c)^2[/tex]
Now I am very happy.