Find Threshold Energy to Generate a Pion

[sebb1e]

Homework Statement

A high speed proton of rest mass M collides with a proton at rest in such a way that not only do the two protons emerge from the collision, but also a pion of mass m. Find the threshold kinetic energy of the incoming proton for such a pion to be generated.

Show that the efficiency of the interaction is never better than a half.

The Attempt at a Solution

I've done very similar questions before but usually it would be the minimum kinetic energy required to produce one particle from two colliding ones, so there is only one variable on the RHS of the equation. Quite simply, can I just take the centre of mass of the two protons and the pion on the RHS so I don't have to have three separate velocities? If this is wrong what should I do instead, also how do I deal with the efficiency bit?

Mgamma(u)[c,u]+M[c,0]=(2M+m)gamma(w)[c,w]

This leads to an answer of c^2(2m+m^2/2M) which looks neat enough to be reasonable.

mc^2/(c^2+2m+m^2/2M) can never be more than a half, so is the efficiency the rest energy of the new particle over the energy used.

 

[mark726]

Hello there!

First of all, you are correct in using the center of mass frame to simplify the equations. This will make your calculations much easier.

To find the threshold kinetic energy of the incoming proton, you need to set the total energy of the system equal to the rest energy of the two protons and the pion. This will give you the minimum kinetic energy required for the pion to be generated.

As for the efficiency of the interaction, you are on the right track. The efficiency can be defined as the ratio of the rest energy of the new particle (in this case, the pion) to the total energy used in the collision. So, in this case, the efficiency would be (mc^2)/(c^2+2m+m^2/2M). And as you correctly pointed out, this can never be more than a half. This means that at most, only half of the energy used in the collision can be converted into the rest energy of the new particle.

I hope this helps! Let me know if you have any other questions.