# Homework Help: Proton-electron collision, particle produced?

1. Oct 17, 2009

### jeebs

Hi,
Got what seems like it should be a basic particle physics problem that I can't quite get my head around, I could do with some guidance. Here's the problem:

I have electrons with 4GeV energy striking target protons and producing some neutral particle Q. I'm supposed to calculate the maximum mass of Q that could possibly be produced, and all I am told is the electron mass is 0.511MeV/c2 and the proton mass is 938.3Mev/c2. I assume these are rest masses.

So, attempt at solution...
I first tried thinking of things in the proton's rest frame. The proton momentum Pp1 = 0, and the electron has all the momentum in this system, Pe1. Once the electron collides with the proton, it will lose energy/momentum and Q will be produced using some of this.
So, before collision, (total momentum) Ptot = Pp1 + Pe1 = Pe1
and after collision, Ptot = Pe2 + Pp2 + Pq

Then I tried thinking of things in the centre of mass frame, ie. the frame where the proton and electron momenta are equal and opposite: Ptot' = Pp' + Pe' = 0, or Pe' = -Pp'
Since I am looking for the most masive Q possible, I thought that all the proton and electron momenta should go into producing Q, so after the collision there would be 3 stationary particles just sitting there, ie. Ptot' = Pp2' + Pe2'+ Pq' = 0

This is where I start to get a bit confused.

If I've got E2 = p2c2 + m2c4
or if c=1, then E2 = p2 + m2

then I can use these to work out the electron's initial momentum in the proton frame, but I'm not too sure where I can go from there. Do I need to somehow get what the electron's momentum (and therefore the proton's momentum) is in the centre of mass frame?
This is confusing me a lot, I've been trying to do this for ages now. I appreciate it someone could break it down for me.

Thanks.

2. Oct 18, 2009

### fantispug

I'm going to suggest that a better reaction (i.e. one that would produce a more massive Q) would be if the electron and proton annihilated; if they are still around they are hogging rest and kinetic energy that Q would otherwise have. (Of course we could have more particles produced, but again this would detract from the rest energy of Q).

This should make your analysis much simpler. Using conservation laws and your mass-energy relation can you work out what the mass of Q must be in this process?

3. Oct 18, 2009

### jeebs

ah thanks man, I can't believe I didn't think of them annihilating each other... suppose I have only been doing particle phys for 2 weeks cheers.

4. Oct 18, 2009

### jeebs

actually they do not annihilate, I just looked at the question and the reaction is

e + p --> e + p + Q

so I am back to square one.

5. Oct 20, 2009

### mattias

interesting.
i've got the exact same problem to do, and came across this here.

6. Oct 20, 2009

### turin

From the point of view that Q is some hypothetical lepto-baryon resonance such that:

e,p -> Q

What is the energy of Q? What is the momentum of Q? Then, use m=sqrt(E2-p2). This calculation should be straightforward in any frame of reference.

From the point of view of, say, electron capture:

e,p -> Q,nu

where nu is the electron neutrino, and Q is some new baryon (similar to a neutron). This is more complicated. I find it most convenient to work in the CM frame. This fixes the energy and magnitude of momentum of Q, and then you can use the same formula as above. (hint: you can make the mass of Q a function of the neutrino energy, for instance.)

From the point of view of:

e,p -> e,p,Q

This is complicated. Again, I would use the CM frame, and treat the final-state e,p as a pseudoparticle. (hint: you can make the mass of Q a function of two kinematic variables of the final-state e,p, say, invariant mass and energy.)

Even if this last most complicated scenario is the relevant one, you should work through the first two, since they are a bit easier, and they will give you ideas for how to approach these kinds of problems.

Last edited: Oct 20, 2009
7. Oct 22, 2009

### fantispug

That makes the problem slightly more difficult, but not much more so (depending on your approach). Consider the reaction

e + p -> e + p + Q

Try to solve it in the centre of mass frame; can you straight away deduce the final momenta of the particles such that the mass of Q is maximised? (If you can't you can always find the mass as a function of the final momenta of the electron and proton and use differential calculus to find it, but it's a little bit of work).

Now all you need to do is boost your frame to a CM frame, or boost your CM solution back to your original frame and plug in the relevant numerical values.

8. Oct 23, 2009

### turin

No boosting is required. That is the beauty of using invariants. You need to determine the invariant mass of the initial state, and this is frame-independent.

9. Oct 23, 2009

### fantispug

That's true, but the mass of the particle will depend on the energies (say) of the particles in the initial system. So when you change frames you need to account for this.

(Well, if you solved the problem generally in an arbitrary frame you wouldn't need to - as you say it's a Lorentz invariant quantity - but when you use the symmetry of the centre of mass frame some of the terms vanish. You could try to guess the missing terms, but this would be the same as just boosting your solution to a general frame.)

10. Oct 26, 2009

### turin

fantispug, I don't understand your concern. Since this problem has nothing to do with dynamics, but rather with endpoints, then it is a strictly kinematical question. The important initial-state quantity is the invariant mass. The kinematics of the final-state do not care how the energy-momentum of the initial state is divided among the various initial-state particles. And, since the question is again about mass (in the final state), the frame of reference is actually irrelevant. The only concession that I would make is this:

Calculate the invariant mass of the initial state in the lab frame, and then boost to the CM frame to do the extremization. The point that I want jeebs (and anyone else) to understand is that the boost is utterly trivial, because the invariant mass is invariant.