Relativistic collision and invariant s

1. Sep 25, 2015

carllacan

1. The problem statement, all variables and given/known data
Write the invariant s = (P1+P2)2 as a function of masses amd energies of the process

1+2 → 3+4

in the center of momentum frame and on the lab frame, in which b is at rest. Interpret the result.

2. Relevant equations

3. The attempt at a solution
For the CoM frame I have:
s = (P1+P2)2 = (E1+E2, p1 + p2)2 = (E1+E2)2
(and similar for particles 3 and 4)

In the lab frame (primed symbols are used for variables in this frame)
s' = (P'1+P'2)2 = (E'1+E'2, p'1)2 = (E'1+E'2)2 + |p1|2 = (E'1+E'2)2 + m012-E'12 = 2E'1E'2+m012+m022
(and similar for particles 3 and 4)

Is this right? I still don't feel secure with this problems .Is if is right, can we get any interesting conclusions from the result? We must have s = s', so
2E'1E'2+m012+m022 = 2E1E2+E12+E22

And I get to E'1E'2 = E1E2 + |p1|2. If the energies were being summed I would interpret it as the fact that the energies in the lab frame is the sum of the energies in the CoM frame plus the momentum of the particles, or something like that, but this means nothing to me.

2. Sep 25, 2015

Staff: Mentor

I highlighted a typo (by the way: this forum supports LaTeX).

Before you interpret the result: you know E'2. What happens if you are interested in very large invariant $\sqrt{s}$? 100 or 1000 times the masses, for example.
Compare this to a collider where b is not at rest.

3. Sep 27, 2015

carllacan

If s is very large compared to the masses then I guess the energy E'1 will depend mostly on E'2, but not on the masses or the mometums.
Compare this to a collider where b is not at rest.[/QUOTE]
If I repeat the calculus with particle 2 having momentum I get an expression where the momentums and the angle between them matter. Is that what you mean?

4. Sep 27, 2015

Staff: Mentor

What is E'2 for a particle at rest? This is the first question, everything below relies on that answer.

Can anything depend on that? You can assume that particle masses are fixed.
No. For two particles of equal mass m, what do you need as E'1 to get $\sqrt{s}=100m$ if particle 2 is at rest?
What do you need if they collide head-on with equal energy?