Is 4-momentum conservation used in this answer?

Question

In Dale's answer to the collision rest mass problem, how is he concluding that the momentum 4-vector of the resulting particle is equal to the sum of the 4-momentums of the incoming particles. I am specifically wondering about this step in his analysis: $$R^\mu = P^\mu + Q^\mu$$

I agree with the final answer, it just seems to me that in regular collision problems we always use some form of momentum conservation (and maybe energy if collision is elastic) to find final velocities. How is Dale getting away without using conservation here? What is different to regular mechanics?

Related fact: Momentum 4-vector is a conserved quantity in special relativity.

Answer 1

How is Dale getting away without using conservation here?

I am not using the conservation of the four momentum in that question because that question is not about conservation. That question is about the fact that the mass of the whole is not the sum of the masses of the parts.

In that answer $R^\mu$ is the four-momentum of the system. Because four-momentum is conserved it is the same before and after the collision. Let me repeat that concept for emphasis: $R^\mu$ is the four momentum of the system which is the same before the collision and after.

So the mass of the system, even before the collision, is given by the expression in the answer, $m^2 c^2 = 2(1+\gamma) \ m_0{}^2 c^2$ even before the collision and therefore without invoking the conservation law before and after the collision.

Note, I am not claiming that the conservation of four-momentum is violated in my answer. The four-momentum is indeed conserved. However its conservation is simply not part of the answer. It was not needed and not used for this question, although it would have been a valid property to use if it had been needed.

Answer 2

In a collision, the total 4-momentum is conserved. This is true whether or not the collision is elastic.

For the decay, the conservation of total 4-momentum is the equation $$R^\mu = P^\mu + Q^\mu,$$ which says (in component form) the total relativistic-energy is conserved and the total relativistic-spatial momentum is conserved.

However, the decay is not elastic.

In special relativity, an elastic collision is characterized by having the particles retain their rest-masses. In special relativity, this implies that the total relativistic-kinetic energy is also conserved.