Relativistic Elastic Collision

Question

I am having trouble getting my head around the transfer of energy in a relativistic elastic collision. My understanding of a relativistic elastic collision is one in which the total rest mass on each side of the equation is unchanged, i.e:

$$\sum_{\text{before}} m_{i}=\sum_{\text{after}} m_{j}$$

If we have a particle of rest mass $M$ and relativistic energy $E$ colliding with a particle at rest with rest mass $m$, then we have by conservation of 4-momentum:

$$\mathsf{P}_{1}\cdot\mathsf{P}_{2}=\mathsf{P}_{1}'\cdot\mathsf{P}_{2}'$$

Initially we have: $\mathsf{P}_{1}=(\frac{E}{c},p,0,0)$ and $\mathsf{P}_{2}=(mc,0,0,0)$, and therefore:

$$\mathsf{P}_{1}'\cdot\mathsf{P}_{2}'=\frac{E'E_{2}}{c^{2}}-p_{1}'p_{2}=mE$$

However, this leaves us with a lot of unknowns $E'$, $E_{2}$ and $p_{2}$, so I'm not sure how I'd go about reducing this to one unknown $E'$ (assuming it's possible)?

Answer 1

Energy-momentum conservation is a stronger statement than the statement* that the inner product $p_\mu p'^\mu$ is conserved. It states that the sums are conserved individually/coordinatewise - $P_1+P_2=P_1'+P_2'$.

As I see it, the conservation of the inner product is a statement about change in reference frames, whereas the conservation of energy and momentum is a physical principle*.

So then, we have (in the 1D case), separately: $$E_1+E_2=E_1'+E_2'$$ $$p_1+p_2=p_1'+p_2'$$

If the masses before and after are conserved, then these are two equations in two unknowns (the two velocities), which can be solved for exactly. If only the sum of the masses is conserved, there's still a degree of freedom and we have to decide what happens based on some other information involving the dynamics of the system.

*Let me know if these statements are wrong, because I'm far from an expert in the 4-vector approach.