Electron and positron annihilation

In summary, both methods give the same result, but the momentum of the photon is opposite of what is predicted by conservation of momentum.
  • #1
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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]

then

[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!
 
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  • #2
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.
 
  • #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!
 

1. What is electron-positron annihilation?

Electron-positron annihilation is a process in which an electron and a positron (the antiparticle of an electron) collide and are converted into energy in the form of gamma rays.

2. How does electron-positron annihilation occur?

When an electron and a positron collide, they can either annihilate directly or form a short-lived positronium atom before annihilating. In both cases, the particles are converted into energy in the form of gamma rays.

3. What is the significance of electron-positron annihilation in physics?

Electron-positron annihilation is an important process in particle physics as it allows scientists to study the properties of subatomic particles and their antiparticles. It is also used in medical imaging techniques such as positron emission tomography (PET).

4. Can electron-positron annihilation produce other particles besides gamma rays?

Yes, in some cases, electron-positron annihilation can produce other particles such as muons, pions, and neutrinos. This occurs when the energy of the annihilation is high enough to create these particles.

5. How is electron-positron annihilation related to the Big Bang theory?

In the early universe, it is believed that there were equal amounts of matter and antimatter. As the universe cooled and expanded, particles and antiparticles annihilated, leaving behind a small excess of matter. Electron-positron annihilation played a role in this process, helping to shape the universe as we know it today.

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