# Homework Help: Energy Threshold for 2 proton -> 2 proton 1 Pion

1. Jan 30, 2012

### Goatsenator

1. The problem statement, all variables and given/known data
When a beam if high-energy protons collides with protons at rest in the laboratory, pions are produced by the reaction p + p --> p + p + ∏. Compute the threshold energy of the protons in the beam for this reaction to occur.

2. Relevant equations

(mc^2)^2 = E^2 - (pc)^2
Einitial = Efinal

3. The attempt at a solution

I tried to do an energy balance but can't seem to get it right. I don't understand how to account for the momentum of the proton if I'm not given its velocity. How can the momentum of the first stage be equal to the momentum of the second stage if none of the particles are moving?

I think:

2x(rest energies of protons) + (Kinetic Energy of proton) = 2x(rest energies of protons) + (rest energy of pion)

Is the threshold energy the total energy (i.e. the rest energy and kinetic energy) of the proton that is moving, or is it just its kinetic energy? My professor went over this very quickly and wrote down that zero is equal to something which makes no sense to me. I'm pretty confused and the book isn't helping either. Can anyone clarify this?

2. Jan 31, 2012

### vela

Staff Emeritus
It can't, so the two protons and the pion must be moving after the interaction.

Try analyzing the problem in the center-of-mass frame, and then transform back to the lab frame.

3. Jan 31, 2012

### Goatsenator

OH! Okay I think I get it now! So when I view it from the COM frame both of the protons appear to have equal kinetic energies when they collide so that is the threshold energy? And then after the collision there appears to be no movement of the particles relative to one another?

4. Jan 31, 2012

### vela

Staff Emeritus
No. You're trying to find the energy of the incident beam as measured in the lab frame. It's related to the threshold energy as measured in the COM frame, but they are not the equal.

This is true in both frames. Perhaps you didn't write exactly what you meant?

5. Mar 16, 2012

### LeePhilip01

I'm also kind of stuck on this.

So far I've got:

BEFORE the collision:

Using the invariant (E2 - c2p2) = (2mpc2)2 - c2(pp)2

Then AFTER the collision in the COM frame:

(E2 - c2p2) = (mpc2+mpc2+mc2)2

Then to get the threshold I'm equating the two equations and solving for mpc2. However, I can't seem to the the momentum term to cancel. Or I can't simplify it to get an equation for the energy.

Any help?

6. Mar 16, 2012

### vela

Staff Emeritus
I'm not sure what you did here.

7. Mar 16, 2012

### LeePhilip01

I just took the E of the two protons ( = 2 x mpc2 ) and subtracted the product of c2 and the momentum of the incoming proton, all squared. As in the invariant.

The second part in the COM frame has no momentum so is just the sum of the E components.

Is this wrong?

8. Mar 16, 2012

### vela

Staff Emeritus
Since the incoming proton is moving, its energy isn't equal to its rest energy, which is what you seem to be saying.