Relativistic energy and momentum in particle collisions

[swe91]

Homework Statement

Two particle P and Q each of restmass m₀ and moving in collision course at 2/3c in the laboratory frame of reference.
In the same collision but in particle P's frame of reference, P is at rest.

Homework Equations

As the total energy of the particles depends on the frame of reference, do the observers in each frame of reference agree on the number of particles and photons formed in the collision?

The Attempt at a Solution

From
E=$$\gamma$$m₀c²
p=$$\gamma$$m₀v
where v=v' from relativistic velocity addition.
and
E²=p²c²+m₀²c⁴
I can conclude that the total energy and momentum of the the collision differs. Furthermore, as I do not se how the available energy for particle formation can be the same when the total energy is different, I would conclude that the number of particles would differ depending on the reference frame. However, this does not feel right. As the collision takes place at a "single point" in space, wouldn't it be measured the same from all inertial frames of reference?

Thanks in beforehand.

 

[anathema]

Hello, thank you for your question. You are correct in your understanding that the total energy and momentum of the collision will differ in each frame of reference. This is due to the relativistic effects of time dilation and length contraction. However, the number of particles and photons formed in the collision will be the same in both frames of reference. This is because the number of particles and photons is a conserved quantity and is independent of the observer's frame of reference. In other words, the observers in both frames of reference will agree on the number of particles and photons formed in the collision. This is a fundamental principle in physics known as the conservation of mass-energy. So even though the total energy and momentum may differ, the number of particles and photons formed will remain the same. I hope this helps clarify any confusion.