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Relativity, decay

  1. Nov 27, 2005 #1
    Hello,

    1. Show that a free electron in a vacuum at velocity v cannot emit a single photon.

    My ideas:

    Momentum is conserved. Energy is conserved.

    Hence (E(before)=E(after)): gamma(before)*m(e)*c^2 = E(photon) + gamma(after)*m(e)*c^2

    where E(photon)=hc/lamda

    (P(before)=P(after)): gamma(before)*m(e)*v(after) = E(photon)/c + gamma(after)*m(e)*v(after), where P and v are vectors.

    This is where I get stuck. Do I have to worry about the emisssion angle - i.e. does the electron change direction? How do I proceed?

    --

    Also, where can I find information about two body decay of a moving particle?
    I have computed the energy and velocity of a particle (pi meson), which now decays into two photons. I imagine the photon energies depend on the emission angles - but how?
     
  2. jcsd
  3. Nov 27, 2005 #2

    Physics Monkey

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    Yes, in principle you should account for the possiblity that the photon is emitted at some angle relative to the electron's initial velocity. This gives you two equations for momentum conservation and one for energy conservation and you can show from these equations that the emission is impossible.

    An easier way to approach to problem is to describe the emission of the photon in the initial rest frame of the electron. You can then make an argument based on the principle of relativity.
     
  4. Nov 27, 2005 #3
    "An easier way to approach to problem is to describe the emission of the photon in the initial rest frame of the electron. You can then make an argument based on the principle of relativity."

    Hmm... the electron will see the photon as moving away with c. I don't follow you.
     
  5. Nov 27, 2005 #4

    Physics Monkey

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    Hint: if you consider the emission in the frame where the electron is initially at rest, the momentum of the system is zero.
     
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