# Particle collision problem (accelerator experiment)

1. Mar 3, 2015

### leonmate

1. The problem statement, all variables and given/known data

An accelerator experiment collides a beam of electrons head-on with a beam of positrons. The particles in each beam have energy Ee as measured in the lab frame. Suppose one electron-positron pair collide to form a photon and neutral pion particle:

e- = e+ ---> γ + π0

Assuming 2Ee > mπ, where mπ is the mass of the pion show that the sped vπ of the pion in the lab frame is given by

vπ = 4Ee2 - mπ2 / 4Ee2 + mπ2
2. Relevant equations

The invariant:
E2 - p2c2 = m2c4

Conservation of energy
Conservation of momentum

3. The attempt at a solution

The first place I've been getting confused is that the electron and positron have the same momenta but in opposite directions. Not entirely sure if I should be including their momentum in the equations as it sums to zero. In a collision their momentum towards each other would surely contribute to the rest energy, kinetic energy and photon energy of the collision products.

I've been attempting to use conservation of energy:

2Ee = 2√(me2 + pe2) = √(mπ2 + pπ2) + cpγ

Then conservation of momentum:

pe- + pe+ = pπ + pγ

We know that the electron have opposite momentum so their sum equals zero and ..

pπ = - pγ

So this gets rid of any photon terms from the equation,

2Ee = 2√(me2 + pe2) = √(mπ2 + pπ2) + cpπ

Now, I figured I should solve this for pπ, use that to find the pion energy and then it's velocity but it doesn't solve nicely and doesn't look anything like the equation given in the question.

Been working on this for a while but I'm not coming up with anything.

2. Mar 3, 2015

### Staff: Mentor

There are missing brackets in the first equation.
Get cpπ on the left side, square and you have a quadratic equation you can solve.

3. Mar 3, 2015

### leonmate

That's helpful, I've used that to get:

pπ = 4Ee2 - mπ2 / 4Ee

Now, I was hoping to get E = 4Ee2 + mπ2 / 4Ee

Using E2 - p2 = m2

The text book I have does that, but it doesn't show the steps and I don't know how to get there! Is it even possible??

4. Mar 3, 2015

### leonmate

I've uploaded the example question from my textbook which shows what I'm trying to do

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5. Mar 3, 2015

### Staff: Mentor

There are brackets missing again.
$E^2=m^2+p^2$ looks useful.

The pion energy in the textbook example corresponds to your 2 Ee.

6. Mar 3, 2015

### leonmate

I keep trying that, but I can't get the maths to work

I feel like I'm overlooking something.

I need Eπ = 4Ee2 + mπ / 2mπ

I'm a little confused as to how plugging in 2Ee will help me get there. Surely I need to find:

Eπ2 = mπ2 + pπ2

Only problem is when I square out my value for pπ I can't simplify it to what I want. Been trying to work this out all day, I can't even follow how it was achieved in the text book I must be missing something

7. Mar 4, 2015

### Staff: Mentor

Brackets!

There is nothing to plug in for $E_e$. I think the equation you want to get for E follows quite naturally if you use the equation of post 5. Can you show what you did?

8. Mar 4, 2015

### Dick

Since you already have $p_\pi$, how is that related to the energy of the photon? I think that's what mfb is hinting at using the post 5 equation for. Now just thinking about energy conservation should give you $E_\pi$ pretty fast.

9. Mar 8, 2015

### leonmate

Yeah, I managed to figure it out. Think I was overtired, I couldn't do math!
Thanks for the help guys