Understanding the Compton Effect: Derivation of the Equation

The_ArtofScience
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The Compton effect does make sense conceptually but I'm having trouble with the actual derivation.

The equation starts out from the principles of conservation of energy and mass: E_{\gamma}+E_{e}=E_{\gamma'}+E_{e'}
which strangely proceeds into the form: \ E-E'}+m=\sqrt{p^2+m^2}} The \sqrt{p^2+m^2}} actually expands into (E-E'cos\theta)^2{}+m^2 when the RHS is squared

So my question is where does the \sqrt{p^2+m^2}} come from and why does it expand to (E-E'cos\theta)^2{}+m^2?
 
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The_ArtofScience said:
where does the \sqrt{p^2+m^2}} come from

The relativistic relationship between (rest) mass, momentum and energy is E^2 = (pc)^2 + (mc^2)^2. Your source must be using units that make c = 1. I assume this refers to the energy of the electron after the interaction (because the electron is usually assumed to be stationary before the interaction) so it really should read

E_{\gamma} - E^{\prime}_{\gamma} + m = \sqrt {(p^{\prime}_e)^2 + m^2}


and why does it expand to (E-E'cos\theta)^2{}+m^2?

I assume the E and the E' refer to the outgoing photon (gamma). I'm sorry, different textbooks use different routes to derive the Compton-scattering equation and I'm too tired to try to guess which route your source uses. Does your source give any more details?
 
Looking at your second question again, that step probably comes from conservation of momentum. Usually in this derivation one let's the direction of the incoming photon be along the x-axis. Then the outgoing electron and photon both have x- and y-components of momentum. The x- and y-components are each conserved.

Also, for a photon, E = pc which in your units (with c = 1) becomes E = p. So that E^{\prime} \cos \theta is probably the x-component of the outgoing photon momentum: p_x^{\prime} = p^{\prime} \cos \theta = E^{\prime} \cos \theta.
 
Thank you jtbell for clearing that up. I'm still curious where that radical came from, but for now I'm very satisfied with the explanations. =D
 
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