Electron-positron creation from colliding photons

[chem_heather]

Homework Statement

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 (E_e- and E_e+) and velocities (v_e- and v_e+) 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.)

Homework Equations

ε₁ = 2 MeV
ε₂ = 3 MeV
m_e = 0.511 MeV/c²
p = ϒmv
E = ϒmc² = mc² + E_K = mc² + mc²(ϒ-1)
E = √[(pc)² + (mc²)²]
ϒ = 1/√1-(v²/c²)

The Attempt at a Solution

I set up four frames of reference (FORs):
E: lab frame, before collision
E' : lab frame, after collision
E_cp: Center-of -mass frame before collision
E'_cp: Center-of -mass frame after collision

E = c(p₁ + p₂) = 5MeV
E'_cp = 2m_ec²
E' = 2m_ec² + E_K

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.

 

[mfb]

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.

 

[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 = (2m_ec²,0,0,0)? When I use these, P² = 25 MeV².

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

 

[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?​

 

[mfb]

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 = (2m_ec²,0,0,0)? When I use these, P² = 25 MeV².

No, and guessing does not help.

Or do the momentums need to be subtracted since they're going in opposite directions?

They have to be added, but one photon needs a negative momentum as it is going in the opposite direction (-x direction).

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

There is, but that needs significantly more work.