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Pair Annihilation

  1. Jul 21, 2008 #1


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    I'm a little unsure about a certain part of this shown in a book.
    There is an electron and a positron moving in the +x direction. They annihilate each other and release 2 photons. to conserve momentum 1 moves in the -x and one in the +x direction.
    So then the first step is to do conservation of momentum: [tex]p_{1} - p_{2}[/tex] (photon momentum 1 - photon momentum 2). It's a (-) because the second photon moves in the -x direction.

    Then we find conservation of energy: [tex]p_{1}c + p_{2}c[/tex]
    (obviously we set these equations equal to the electron values, but I'm leaving that out.)

    Ok, so, they find values for both of those equations. But, what I'm not sure of is the next step. It says
    "Now we add the two results and solve for [tex]p_{1}[/tex] and [tex]p_{2}[/tex]
    so it looks like:
    [tex](p_{1} - p_{2}) + (p_{1} + p_{2})[/tex]

    Then I understand the rest, I just don't know why they add the 2 values. Total Energy and net Momentum.
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  3. Jul 22, 2008 #2


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    Hi jk4! :smile:

    You have p1 + p2 = A, p1 - p2 = B.

    If you add: 2p1 = A + B, and 2p2 = A - B.

    So that gives you p1 and p2. :smile:

    What worries you about that? :confused:

    (How else would you solve it?)
  4. Jul 22, 2008 #3


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    That's why I asked. I like to learn things in a way that I don't have to do much memorizing, but so that it will be obvious to me if I ever come across it.
    So I'm just not sure this will be that obvious to me.. Might have to actually consult my notes :(

    Sorry, It just threw me because I'm not sure what is the significance of momentum+totalEnergy
  5. Jul 22, 2008 #4


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    Your initial question is confusing because you don't say what information was provided to you. I am assuming that they gave you the total momentum and energy of the e-e+ pair, right?

    Forget about particle physics for a second. It's just algebra. Let's say you have to solve

    x+y = 10

    x-y = 6

    How would you solve that? There is not a single way. You could isolate x from the first equation and plug in the second and then solve for y. But the quick way is of course to add them up to get rid of y.

    well, this is basically exactly the same type of algebra problem you are dealing with here except that your unknowns are p1 and p2. That's all there is to it.
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