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Electron and positron annihilation

  1. Apr 21, 2007 #1
    Problem:An electron flies toward +x direction with velocity of 0.9c, while a positron flies toward -y direction with the same velocity. Assuming their speed is so fast that they collide and annihilate at the origin, what will be the magnitude and the direction of the wave vector of the generated photon, k?

    I have two kinds of solution. One is base on conservation of linear momentum, one is base on conservation of energy, but they don't have the same result! Where went wrong??

    Method 1
    By conservation of momentum,

    [tex] P_i = \frac{m_e v}{\sqrt{1-v^2/c^2}} \hat{i} - \frac{m_e v}{\sqrt{1-v^2/c^2}} \hat{j}[/tex]

    since the magnitude is the same in the two direction, we have

    [tex] P_{photon} \cos \frac{\pi}{4} = \frac{m_e v}{\sqrt{1-v^2/c^2}} [/tex]

    [tex] p = \gamma m_ev \sqrt{2} = \frac{h}{\lambda} [/tex]

    so [tex] k = \frac{2\sqrt{2}\pi m_e v}{h\sqrt{1-v^2/c^2}}[/tex]

    substitute v=0.9c

    [tex] k = \frac{1.8\sqrt{2} \pi m_e c}{h \sqrt{1-(0.9)^2}}[/tex]

    Method 2
    By conservation of energy

    [tex] 2 \times \gamma m c^2 = \frac{hc}{\lambda} [/tex]


    [tex] k = \frac{2 \pi}{\lambda} = \frac{4 m_e c \pi}{h\sqrt{1-v^2/c^2}} = \frac{4 m_e c \pi}{h \sqrt{1-0.9^2}}[/tex]

    why the two method turn out different result??

    thanks for help!!
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 21, 2007 #2


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    Gold Member

    You can't find the direction using energy conservation. This will give you the magnitude of the resulting wave. The momentum of the outgoing wave will the opposite of the vector sum of the incoming photons.
  4. Apr 21, 2007 #3
    I know I can't find the direction of the produced photon by only using conservation of energy, I also need conservation of momentum. But I don't see why the magnitude won't be the same by two different methods. I don't understand your last sentence. What do you mean of "outgoing" wave and "imcoming" wave, it seems to me that there is only "outgoing" wave in the situation.

    Thanks for help!
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