# Collision of electrons, kinetic energy?

1. Nov 11, 2013

### skrat

1. The problem statement, all variables and given/known data
What is the minimum amount of kinetic energy of electron that hits another electron, so ectra electron-positron pair is produced.

2. Relevant equations
$p^{\mu }=(E/c,\vec{p})$

3. The attempt at a solution

Before the collision:

$p^{\mu }=(\frac{E_e+m_ec^2}{c},p)$

and after: (not sure but... i guess?)

$p^{\mu }=(\frac{E_1+E_2+2m_ec^2}{c},p_1cos\varphi +p_2cos\theta ,p_1sin\varphi+p_2sin\theta )$

For index 1 the first electron, index 2 the second one which had no kinetic energy before the collision and pair electron-positron with no kinetic energy...

Is that right? Probably not, because I can't find a way to answer the question: minimum amount of kinetic energy....

2. Nov 11, 2013

### Staff: Mentor

First you'll have to find out how that collision will look like. What is the configuration of minimal energy? Hint: consider the center of mass frame.
With that knowledge, your equations simplify a lot.

3. Nov 11, 2013

### skrat

Well that is my biggest problem: going in center of mass frame!

Let's try.. The way I understand it, what will happen is: $e^{-}+e^{-}\rightarrow e^{-}+e^{-}+(e^{-}+e^{+})$

Before the collision in the center of mass frame:

$p^{*\mu }=(\frac{E_1+E_2}{c},p_{1}^{*}+p_{2}^{*})$ where index 1 indicates electron that has some kinetic energy in the first frame and index 2 for electron that has no kinetic energy in first frame.

so $p_{1}^{*}=-p_{2}^{*}$

After the collision in the center of mass frame: Let's put " ' " on everything after the collision

$p_{x1}^{'}=p_3^{'}cos\varphi ^{'}$

$p_{y1}^{'}=p_3^{'}sin\varphi ^{'}$

$p_{x2}^{'}=-p_4^{'}cos\varphi ^{'}$

$p_{y2}^{'}=-p_3^{'}sin\varphi ^{'}$

Ok, now to be completely honest with you... I simply copied the last 4 equations from my notes... I have got absolutely no idea what do they mean, why would they be useful or why do I need them and I haven't got a clue on how to continue from here.

Notes continue:

$p_3^{'}=mc\gamma ^*=mc\sqrt{\frac{\gamma-1}{2}}$ therefore

$p_{x1}^{'}=p_3^{'}cos\varphi ^{'}=mc\gamma ^*=mc\sqrt{\frac{\gamma -1}{2}}cos\varphi ^{'}$ and

$p_{x1}^{'}=-p_3^{'}cos\varphi ^{'}=mc\gamma ^*=-mc\sqrt{\frac{\gamma -1}{2}}cos\varphi ^{'}$

Using Lorentz transformation back the primary frame:

$p_{x1}=\frac{mc}{2}\sqrt{\gamma ^2-1}(1+cos\varphi ^{'})$

$p_{x2}=\frac{mc}{2}\sqrt{\gamma ^2-1}(1-cos\varphi ^{'})$

If I understand correctly, these two are now moments after the collision in the first frame.

And that is where everything stops.. It' would probably be the easiest if somebody knows a website where these things are well explained?

4. Nov 11, 2013

### Staff: Mentor

Stop. You are doing this way too complicated.

What happens in the center of mass system when the energy is just enough to let the process happen? What can you know about the kinetic energies of the particles then?
Do not calculate anything. This is not necessary at this step.

5. Nov 12, 2013

### skrat

If the energy is just enough to let the process happen, than after the process they wouldn't have any kinetic energy.

6. Nov 12, 2013

### Staff: Mentor

Right.

In the lab frame momentum conservation does not allow that, but what about their relative motion in the lab frame? In particular, do you need all those angles and different momenta?