# Compton effect

1. Jun 26, 2008

### The_ArtofScience

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 $$(E-E'cos\theta)^2{}+m^2$$?

2. Jun 26, 2008

### Staff: Mentor

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

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

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?

3. Jun 27, 2008

### Staff: Mentor

Looking at your second question again, that step probably comes from conservation of momentum. Usually in this derivation one lets 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: $p_x^{\prime} = p^{\prime} \cos \theta = E^{\prime} \cos \theta$.

4. Jun 27, 2008

### The_ArtofScience

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