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Photon-photon collision - electron-anti electron pair creation

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data
    two photons collide head-on and create an electron and an anti electron. The energy of photon 1 is given, but not for photon 2.
    Find the velocity β of the systems rest frame (where the total momentum is zero.)

    2. Relevant equations
    β=v/c
    γ=1/√1-β^2)
    β=pc/E

    3. The attempt at a solution
    I have tried this every way I could think of, by adding the two momentum four vectors and using a lorentz transform on them, by using the formula for speed of the rest from β=pc/E, but no matter what, I always need the energy of the second photon. How do I get around needing this?!

    Thank you for your help
     
  2. jcsd
  3. Feb 17, 2012 #2

    vela

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    Do you have any other constraints? You can't solve the problem without more information.
     
  4. Feb 17, 2012 #3
    You can find the energy converted because the total rest mass of the electron and positron is known and I guess that any KE they gain can be considered as negligible.
     
  5. Feb 17, 2012 #4
    Well that's the thing; I'm pretty sure now that the problem just wants the speed β in terms of E1 and E2, because a later part of the question asks to find the minimum energy required for the second photon. The only problem is that now that β=(E1-E2)/(E1+E2), that makes both γ and βγ very messy constants, which makes solving for E2 impossible (at least for me and maple) The minimum energy would be 1.022E6 eV (2m_eC^2) and the first photon has an energy of 1E12 eV, and by doing the lorentz transform on the added momentum frames, you get an equation for E' (which is shown below in the attachments.) Then E'=1.022E6 eV, E1=1E12 eV, and I have no idea how to solve for E2... haha
     

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  6. Feb 18, 2012 #5

    vela

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    You made an algebraic error. The units on the RHS is energy cubed. You should have
    $$E' = \frac{4E_1E_2}{(E_1+E_2)\sqrt{1-\frac{(E_1-E_2)^2}{(E_1+E_2)^2}}}.$$ That will simplify down quite a bit.
     
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