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Electron-positron creation from colliding photons

  1. Sep 9, 2015 #1
    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. jcsd
  3. Sep 9, 2015 #2


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    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.
  4. Sep 9, 2015 #3
    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?
  5. Sep 9, 2015 #4
    Hmm, I just found this on another thread:
    "Since m=0, combining those two gives the four-momentum of a photon as:
    or equivalently for a photon traveling in the x direction:

    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?​
  6. Sep 10, 2015 #5


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    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.
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