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Allowed and Forbidden Electron-Photon Reactions

  1. Aug 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Show, using spacetime diagrams and also using frame-independent calculations, that the law of conservation of 4-momentum forbids a photon to be absorbed by an electron, e + γ → e.

    2. Relevant equations
    ##\textbf{p}_{e1}+\textbf{p}_{\gamma}=\textbf{p}_{e2}## : ##(E_1,\vec{p_1})+(E_{\gamma},E_{\gamma}\vec{n})=(E,\vec{p_2})##,
    where ##\textbf{p}## is 4-vector, ##\vec{p}## is ordinary 3-vector and ##\vec{n}## is unit vector in photon's direction.
    ##E=\sqrt{p^2+m^2}##

    3. The attempt at a solution
    I take only 3-vecotrs and I (I hope I may) suppose ##\vec{p_1}=0## i.e. is in rest.
    ##\vec{0}+E_{\gamma}\vec{n}=p_2\vec{n}=>\sqrt{E^2-m^2}=E_{\gamma}##
    First time I put there ##E=\frac{1}{2}mv^2## but I feel it is not good probably. Is my begining right? What should I use or think about?
     
  2. jcsd
  3. Aug 10, 2017 #2

    mfb

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    2016 Award

    Staff: Mentor

    You used conservation of momentum so far.
    What about conservation of energy?

    The problem is easier in the reference frame of the electron in the final state.
     
  4. Aug 11, 2017 #3
    Ok, ##(\gamma m, \vec{p}) + (E_{\gamma},E_{\gamma}\vec{n})=(m,\vec{0})##. Right? From energy part than we have ##E_{\gamma}=m(1-\frac{1}{\sqrt{1-v^2}})<0##. And it is this problem, negative photon's energy?
     
  5. Aug 11, 2017 #4

    mfb

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    2016 Award

    Staff: Mentor

    Right.

    Time-reversed, the process would be an electron at rest emitting a photon and gaining kinetic energy at the same time. Written like that, it should be clear that it cannot happen.
     
  6. Aug 14, 2017 #5
    Thank you.
     
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