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Electron-positron collisions

  1. Mar 23, 2005 #1
    I was told that when electrons and positrons collide they produce 2 photons of 0.511MeV each. But what happens to the kinetic energy of the electrons and positrons? It just feels to me that the energy is not conserved.

    I was told somewhere that the relative KE of the particles to each other is 0, but I don't understand this.

    Any help would be greately appreciated.
     
  2. jcsd
  3. Mar 23, 2005 #2
    All the energy from is positrons and electrons are 'converted' into two photons. Energy is very much conserved. This is easy to prove when just writing down the energy and momentum conservation laws.

    Regards
    marlon
     
  4. Mar 23, 2005 #3
    I think in the other thread anti_crank mentioned that the 0.511MeV photons produced is only when the collision occurs with the positron + electron at rest. But then how do they "collide." Don't they have to move close to each other, therefore they have KE just before collision?
     
  5. Mar 23, 2005 #4
    "At rest" in this context means that the velocities are significantly smaller than the speed of light, so that the kinetic energy is small compared to the rest mass.

    You are correct that the photon energy will be shifted due to the kinetic energy in the collision. So instead of seeing a spike at exactly 511 keV, there will be a distribution with some width. Again, it's a small effect.

    Note that when you see actual data looking for the 511 line from astrophysical sources there are many other effects that distort the line-shape as well (doppler broadening, etc.... There's a long list).
     
  6. Mar 24, 2005 #5
    All that's required is a partial overlap of their wavefunctions to count as a "collision". Think of the positronium atom: in the ground state the electron's wave fcn vanishes at the positron "nucleus" and there is no "motion" as defined clasically - yet the system still decays.
     
  7. Mar 25, 2005 #6

    nrqed

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    Actually, the wavefunction does NOT vanish at the origin. The decay rate is indeed proportional to the square of the wavefunction at the origin.

    Pat
     
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