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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?

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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?

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# Homework Help: Relativity, decay

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