# Electron-positron creation from colliding photons

Tags:
1. Sep 9, 2015

### chem_heather

1. The problem statement, all variables and given/known data
Consider two photons, one with energy ε1 = 2MeV traveling to the right, and the other with energy ε2 = 3MeV moving tot he left. The two photons collide head-on and produce a positron-electron pair. Suppose the the electron and positron move along the same axis as the photons.

What are the final energies (Ee- and Ee+) and velocities (ve- and ve+) of the positron and the electron?

(Hint: it is easier to do this problem by first switching to a frame of reference where the two photons have the same energy (and thus same momentum); in this frame, after the collision the center of mass is at rest.)

2. Relevant equations
ε1 = 2 MeV
ε2 = 3 MeV
me = 0.511 MeV/c2
p = ϒmv
E = ϒmc2 = mc2 + EK = mc2 + mc2(ϒ-1)
E = √[(pc)2 + (mc2)2]
ϒ = 1/√1-(v2/c2)

3. The attempt at a solution
I set up four frames of reference (FORs):
E: lab frame, before collision
E' : lab frame, after collision
Ecp: Center-of -mass frame before collision
E'cp: Center-of -mass frame after collision

E = c(p1 + p2) = 5MeV
E'cp = 2mec2
E' = 2mec2 + EK

I know that the norms of the energies will be equal from one FOR to another, so:
E2 = (E'cp)2
(E'cp)2 = 25 MeV2

And this is where I'm stuck. We've covered four-vectors, and I think I might be getting confused on whether or not I use them here, and if so, how to set up the components for the photons.

2. Sep 9, 2015

### Staff: Mentor

Your notation is confusing. If E are energies, they cannot be frames. Also, squaring the energy does not give anything invariant. You'll need the 4-vectors of the particles involved.

3. Sep 9, 2015

### chem_heather

That's where I'm getting confused. Would the 4-vector for the lab frame before the collision be P = (0,c(p1 + p2),0,0)? I used 0 in the A0 spot because the photons don't have any rest-mass. And for the center-of-mass frame after the collision be P'cp = (2mec2,0,0,0)? When I use these, P2 = 25 MeV2.

Is there a way to do this without using 4-vectors?

4. Sep 9, 2015

### chem_heather

Hmm, I just found this on another thread:
"Since m=0, combining those two gives the four-momentum of a photon as:
(|p|,p)
or equivalently for a photon traveling in the x direction:
(E/c,E/c,0,0)"

So, if that's the case, the 4-vector for the lab-frame before the collision would be
P = (c(p1 + p2),c(p1 + p2),0,0), correct? Or do the momentums need to be subtracted since they're going in opposite directions?​

5. Sep 10, 2015

### Staff: Mentor

No, and guessing does not help.
They have to be added, but one photon needs a negative momentum as it is going in the opposite direction (-x direction).
There is, but that needs significantly more work.