My syllabus is giving a common proof for relativistic momentum; we consider a symmetric collision, where the two objects (of equal mass) will move with an angle to the opposite direction after the collision. My first question is: how do we know there can be an angle involved? Is it because it keeps the situation symmetrical, so an angle>0 is feasible?

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Now the more pressing issue: using the symmetry of the situation, they get that $\frac{\alpha (V)}{\alpha (w)}= \gamma$. What they do next is taking the limit for w to 0. And what they claim is that we are dealing with a classical collision by taking this limit. I don’t really understand what’s happening then. If w approaches zero, then particle 1 (in the left picture) is stationary, and we have particle 2 coming at particle 1 with velocity u. Now why is this classical? If the velocity u is a significant portion of c, then we’re still dealing with a relativistic collision, I’d say… So could someone explain to me what really happens when we take this limit? If the situation isn’t clear, I will translate the document, but given that it’s just a common example, I reckoned most of you are familiar with this and need the document only for the symbols and directions.

  • $\begingroup$ Not an answer, but there are other ways to tackle this issue. In my modern physics class I have student read an derivation based on this glancing collision methodology you discuss here, but in class I show them an argument simplified from DOI: 10.1088/0143-0807/26/1/005 which is available without paywall as arXiv:0402024. The paper covers an elegant, but very mathematical way of deducing the right formulae starting by assuming that the work-energy theorem holds in it's differential form. $\endgroup$ – dmckee --- ex-moderator kitten Jan 9 '17 at 0:06
  • $\begingroup$ The collision is not classical in the limit of w to zero, however $\alpha(w)$ must equal $m$ in this limit, because the expression for the momentum must coincide with the classical expression in this limit. Particle 1 will, after all, have some momentum that depends on its velocity irrespective of that the other particle is doing, so it can be in the classical range while the other particle is not. $\endgroup$ – Count Iblis Jan 9 '17 at 0:16
  • $\begingroup$ @CountIblis But the particle only has a velocity $w$ in the left frame of reference, so by taking this limit, it doesn't have a velocity at all (this makes p=0, so what about the mass?). And once again, how does it coincide with the classical expression, if we don't know the magnitude of the velocities? What I imagine now is a stationary particle 1 colliding with particle 2 that comes from the right with a certain velocity. $\endgroup$ – Sha Vuklia Jan 9 '17 at 8:38
  • $\begingroup$ There is an angle because in general whenever two bodies collide head-on they may move apart making an angle to their initial trajectories. For example, billiard balls. $\endgroup$ – Abhijeet Melkani Feb 11 '17 at 16:55
  • $\begingroup$ The unstated premise is that mass of a particle at rest is the same as the mass that we have been using "classically". There are no corrections involved for a stationary particle. Now, in the light of relativity as the lengths of rigid bodies and time intervals for periodic events change with change of reference frame, the question asked is: is there also a change in mass? The question is investigated using conservation of momentum as a presumed truth. $\endgroup$ – Abhijeet Melkani Feb 11 '17 at 16:59

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