Compton Scattering: Klein-Nishina formula derivation

Question

I'm following a derivation of the Klein-Nishina formula from scratch and this is what I have so far: $P_{e,i} = m_0\gamma_u[1,u]$ and $P_{\gamma,i} = [\frac{\hbar\omega_i}{c^2}, \frac{\hbar\omega_i}{c}n_{\gamma,i}]$,

where $i=1,2$ is before and after and I'm assuming $n_\gamma$ is the number density.

Firstly, four momentum is $[E/c, p]$, so how has $P_{\gamma,i}$ got a $c^2$ and $n_\gamma$ in it?

So, if we accept all that. We can conserve the momenta before and after, square it and reduce it to

$$P_{e1}.P_{\gamma1} = P_{e2}.P_{\gamma2}$$

Secondly, multiplying by $P_{\gamma2}$ gives

$$P_{e1}.P_{\gamma2} +P_{\gamma1}.P_{\gamma2} = P_{e2}.P_{\gamma2} + P_{\gamma2}.P_{\gamma2}$$ . How does that work?

Then finally, how do I get

$$P_{e1}.P_{\gamma2} = \gamma_1m_e\hbar\omega_2(1-\frac{v_1.n_{\gamma2}}{c})$$ from the definitions on the second line?

So confused!

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user Lucidnonsense

Comments

$\hat{n}$ should denote the direction of propagation. If you're trying to find the Klein-Nishina formula, Compton scattering with the electron at rest will only allow the electron to acquire energy. The full expression is quite complicated.

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user auxsvr

Answer 1

The original publication is of some use

http://adsabs.harvard.edu/abs/1929ZPhy...52..853K

But, this derivation is even clearer:

http://isites.harvard.edu/fs/docs/icb.topic521209.files/QFT-Schwartz.pdf

It is quite envolved though ... Hope this helps.