Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy variance, and pair production

  1. Apr 19, 2010 #1


    User Avatar
    Science Advisor
    Gold Member

    Ok, so here's a question. The energy of say photons is frame-dependent. Photons are blue-shifted or red-shifted depending on my velocity towards or away from the source. However, what happens when I apply this to pair-production? For example, if my photons are energetic enough, they may create a positron-electron pair. If my photons are NOT energetic enough they don't create the pair.

    However, since the energy is frame-dependent, it would seem to suggest that whether a pair is created or not is frame-dependent. The existence of particles, however, should not be frame dependent. Paradox! Obviously my logic is faulty somewhere.

    Please resolve this for me. Thanks. =)
  2. jcsd
  3. Apr 19, 2010 #2


    User Avatar

    Staff: Mentor

    A photon all by itself cannot convert into an electron-positron pair and conserve both energy and momentum. There has to be another object in the picture, typically an atomic nucleus which is normally (almost) at rest. In a reference frame in which the photon has a small energy, the nucleus has a lot of kinetic energy which can be contributed towards creating the pair.
  4. Apr 19, 2010 #3


    User Avatar
    Science Advisor
    Gold Member

    What if I replace the atomic nucleus with another photon? Are you saying 1 photon will be blue-shifted and one will be red-shifted or some such?
  5. Apr 20, 2010 #4
    Assume this occurs as the reverse of electron-positron annihilation where two photons are generated.

    Take the annihilation event and work it backwards. In the center-of-mass frame of two leptons, the photons will travel in opposite directions with equal wavelength so that momentum is conserved, as jtbell elucidated.

    If the center of mass has a non zero velocity, the wavelengths will not be equal.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook