Special Relativity; 3 Momentum Conservation

In SR, I understand you can use 4 momentum conservation, but what are the special cases where you can use 3 momentum/energy conservation?

An example I have seen is with $$P_1=(M_1, 0) \\ P_2=(M_2,0) \\ P_3=\left(\sqrt{(m_3^2+p^2)},p\right) \\ P_4 =(|p|,-p)$$

where $$P_1$$ is the initial state and $$P_2$$, $$P_3$$, $$P_4$$ are the final. From here the example said

$$M_1 = M_2 + \sqrt{(m_3^2+p^2)} + |p|$$

Could somebody explain why we are allowed to say that, I thought that energy and momentum became "intertwined" and energy and momentum conversation were coupled into four momentum conservation.

2 Answers

The fact that the four-momentum vector $$p^\mu=(E,p_x,p_y,p_z)$$ is conserved,

$$p^\mu_\text{after}=p^\mu_\text{before},$$

simply means that $$E$$ is conserved, $$p_x$$ is conserved, $$p_y$$ is conserved, and $$p_z$$ is conserved. A vector equation is simply a set of equations for each component. The equation that puzzles you is simply the conservation of energy.

Energy and momentum are not "intertwined" because they are conserved quantities. They are intertwined because under Lorentz boosts they transform into linear combinations of each other. This is why they compose a Lorentz four-vector. How they transform has nothing to do with whether they are conserved or not.

• That makes perfect sense. Thank you – Tp123 May 13 '19 at 21:51

Four-momentum is conserved as a four-vector, i.e. each component is conserved. Two inertial observers, moving relative to each other, observing the same particle, would see the particle to have different 3-momentum and different energy, but both would agree that the three-momenum and the energy (of the particle) does not change as a function of time (be it time of observer One or observer Two)