1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Which processes are posible and why?

  1. Feb 27, 2008 #1
    1. Photon runs into motionless electron and gives the electron all of its energy
    2. Fast positron interacts with motionless electron producing one photon

    1. I know that if that electron were in atom, photon of some energy could give its energy, and electron would jump into a higher energy state. But what about free electron, as in this case? And why not if it's not possible?

    2. positron and electron should anihilate producing photon, but I'm not sure is it 1 or two photons, and I don't know the argument why.
  2. jcsd
  3. Feb 27, 2008 #2
    If you calculate the momenta and energy using Conservation laws, you should be able to see whether they are possible or not.
  4. Feb 27, 2008 #3


    User Avatar
    Science Advisor

    1. Are p and E conserved?
    2. Calculate E^2-p^2 for the initial and final state.
  5. Feb 27, 2008 #4
    p=(E(photon)+E(electron), p1+p2=0) before, after I should have E of electron that exceeds rest mass of electron, and no momentum, so the answer should be that process is not possible.

    It should not be possible because photon should have some momentum (three component momentum) and since it doesn't have any mass, it can't be possible.

    Are these correct explanations?
  6. Feb 27, 2008 #5


    User Avatar
    Science Advisor

    Neither is possible, but your reasoning is flawed.
    For #1, look at the barycentric (or "center of mass") system where the total momentum is zero. The final state would have total energy mc^2, while the initial system would have a higher total energy.
    For #2, one photon cannot be produced because there is no center of mass system for a single photon.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook