Particle collision problem (accelerator experiment)

[leonmate]

Homework Statement

An accelerator experiment collides a beam of electrons head-on with a beam of positrons. The particles in each beam have energy E_e 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 2E_e > m_π, where m_π is the mass of the pion show that the sped v_π of the pion in the lab frame is given by

v_π = 4E_e² - m_π² / 4E_e² + m_π²

Homework Equations

The invariant:
E² - p²c² = m²c⁴

Conservation of energy
Conservation of momentum

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:

2E_e = 2√(m_e² + p_e²) = √(m_π² + p_π²) + cp_γ

Then conservation of momentum:

p_e^- + p_e⁺ = p_π + p_γ

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

p_π = - p_γ

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

2E_e = 2√(m_e² + p_e²) = √(m_π² + p_π²) + cp_π

Now, I figured I should solve this for p_π, 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.

 

[mfb]

There are missing brackets in the first equation.

2E_e = ... = √(m_π² + p_π²) + cp_π

Get cp_π on the left side, square and you have a quadratic equation you can solve.

 

[leonmate]

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

p_π = 4E_e² - m_π² / 4E_e

Now, I was hoping to get E = 4E_e² + m_π² / 4E_e

Using E² - p² = m²

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

 

[leonmate]

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

 

[mfb]

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

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

 

[leonmate]

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

I feel like I'm overlooking something.

I need E_π = 4E_e² + m_π / 2m_π

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

E_π² = m_π² + p_π²

Only problem is when I square out my value for p_π 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 textbook I must be missing something

 

[mfb]

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?

 

[Dick]

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

p_π = 4E_e² - m_π² / 4E_e

Now, I was hoping to get E = 4E_e² + m_π² / 4E_e

Using E² - p² = m²

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

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.

 

[leonmate]

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