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Regarding pair production and annihilation

  1. Apr 30, 2012 #1
    Hi, i have a few questions about these two processes. Now, im only 16 years old, in my last year of school, so im not so familiar with physics, if you could put terms simple enough for an average 16 year old to understand, i'd much appreciate it :)
    my question is:
    If e‾ + e+ → γ + γ and γ → e‾ + e+, wouldn't that mean its theoretically possible to produce almost an infinite amount of positron and electron pairs? What i mean is annihilate a pair, force both gamma rays to undergo pair production, keep one e-/e+ pair, and annihilate the other?

    on a side note, why exactly are photons the force carriers for EM force, and W/Z bosons the force carriers for weak force?
  2. jcsd
  3. Apr 30, 2012 #2
    Throughout all these reactions, energy must be conserved. In order for pair creation to happen, the photon must have at least enough energy to create the two particles (2 times 0.511 MeV). So you can certainly turn the kinetic energy of the original pair into additional pairs of particles (if it exceeds the mass threshold), in fact, that is the principle behind some calorimeters (detectors that measure energy) used in particle physics, for example at the LHC.
  4. Apr 30, 2012 #3


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    The force carries for the EM force are called "photons", and the force carriers for the weak force are called "W+, W-, Z". The names are just definitions.

    If you ask why they have their properties (and why the weak force has 3 particles): Well, first, it is an observation. Based on this observation, the theory of the electroweak force was developed, and with this theory it is possible to calculate the properties of these particles (more properties than the theory needed as input ;)).
  5. Apr 30, 2012 #4
    In order to create an electron-positron pair with a photon, BOTH energy AND momentum have to be conserved. This not possible in vacuum, but it is possible in the Coulomb field of a nucleus, because the heavy nucleus can carry away a lot of momentum with only a small amount of energy loss. It is much easier to create electron positron pairs with photons over about 2 MeV (threshold 1.02 MeV) near high-Z nuclei, like lead, for example, than copper.
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