# Trying to derive compton scattering using 4-vectors [closed]

I'm trying to derive the energy of a photon after compton scattering:

$$E_\gamma' = \frac{E_\gamma}{1 + \frac{E_\gamma}{m_ec^2} (1-\cos \theta)}$$

where $E_\gamma'$ is the photon energy after scattering, $E_\gamma$ is the energy before scattering, $m_e$ is the mass of the electron and $\theta$ is the scattering angle.

It seems like I'm running into some mistake, because I get an additional $-m_e c^2$ term.

Let $\textbf{p}_\gamma = (E_\gamma/c, \vec{p}_\gamma)$ and $\textbf{p}_\gamma' = (E_\gamma'/c, \vec{p}_\gamma')$ the 4-vectors of the photon before and after the scattering. Let $\textbf{p}_e = (m_e c, \vec{0})$ and $\textbf{p}_e' = (E_e'/c, \vec{p}_e')$ the 4-vectors of the electron before and after the scattering. The electron is at rest before the scattering.

Then because of 4-momentum conservation, we get

$$\textbf{p}_\gamma + \textbf{p}_e = \textbf{p}_\gamma' + \textbf{p}_e'$$ $$\textbf{p}_\gamma - \textbf{p}_\gamma' = \textbf{p}_e' - \textbf{p}_e$$

$$\textbf{p}_\gamma^2 - 2\textbf{p}_\gamma\textbf{p}_\gamma' + \textbf{p}_\gamma'^2 = \textbf{p}_e'^2 -2\textbf{p}_e'\textbf{p}_e + \textbf{p}_e^2$$

Using $\textbf{p}_\gamma^2 = 0$ and $\textbf{p}_e^2 = m_e^2c^2$ gives

$$-2\textbf{p}_\gamma\textbf{p}_\gamma' = 2m_e^2c^2 - 2\textbf{p}_e'\textbf{p}_e$$ $$\textbf{p}_\gamma\textbf{p}_\gamma' = \textbf{p}_e'\textbf{p}_e - m_e^2c^2$$ $$\frac{E_\gamma E_\gamma'}{c^2} - \vec{p}_\gamma \vec{p}_\gamma' = \frac{E_e'}{c} m_e c - m_e^2c^2$$ $$\frac{E_\gamma E_\gamma'}{c^2} - |\vec{p}_\gamma| |\vec{p}_\gamma'| \cos \theta = E_e' m_e - m_e^2c^2$$

Because for photons $|\vec{p}| = \frac{E}{c}$, we get:

$$\frac{E_\gamma E_\gamma'}{c^2} (1 - \cos \theta) = E_e' m_e - m_e^2c^2$$

Because of energy conservation ($E_\gamma - E_\gamma' = E_e'$), we get:

$$\frac{E_\gamma E_\gamma'}{c^2} (1 - \cos \theta) = (E_\gamma - E_\gamma') m_e - m_e^2c^2$$

$$E_\gamma' = \frac{E_\gamma - m_ec^2}{1 + \frac{E_\gamma}{m_ec^2} (1-\cos \theta)}$$

Where is my mistake?

You say energy conservation gives

$$E_{\gamma}' + E_e' = E_{\gamma}$$

I think it should be

$$E_{\gamma}' + E_e' = E_{\gamma} + E_e$$

where $E_e = m_e c^2$.

This would make your second to last equation

$$\frac{E_\gamma E_\gamma'}{c^2} (1 - \cos \theta) = (E_\gamma + E_e - E_\gamma') m_e - m_e^2c^2 = E_\gamma - E_\gamma'$$