# Compton scattering equations

In my book, when it discusses Compton Scattering it provides the following momentum equations, where $$\gamma$$ is Lorentz factor, $$m$$ is the mass, $$u$$ is the velocity, $$h$$ is Plank constant, and $$\lambda$$ is the wavelength:

$$\text{x-component: }\frac{h}{\lambda} = \frac{h}{\lambda'}\cos\theta + \gamma m u\cos\phi$$ $$\text{y-component: }0 = \frac{h}{\lambda'}\sin\theta - \gamma mu\sin\phi.$$

However, it does not elaborate on why it divides the equation into an x-compoenent and a y-compoenent. What exactly is those two equations telling us in regards to the geometry?

• The fact that we have defined the $x$ direction to be the direction of propagation of the incoming photon (with the $e^{-}$ at rest). If you see, the initial y-momentum = $0$ – insomniac Mar 20 '20 at 9:16

Look at this drawing

For our convenience we have put the $$xyz$$ coordinate frame in such a way, that the $$x$$-axis is in the direction of the incident photon, and the scattered electron and photon are moving in the $$xy$$-plane.

From the drawing above we can read off the various momentum vectors (each with $$x$$, $$y$$ and $$z$$ component):

• The incident photon has momentum with an $$x$$ component only: $$\vec{p}_{\gamma,\text{incident}} =\frac{h}{\lambda}\begin{pmatrix}1\\0\\0\end{pmatrix} \tag{1a}$$
• The initial electron is at rest and therefore has zero momentum: $$\vec{p}_{e,\text{initial}} =\begin{pmatrix}0\\0\\0\end{pmatrix} \tag{1b}$$
• The scattered photon has momentum with an $$x$$ and $$y$$ component: $$\vec{p}_{\gamma,\text{scattered}} =\frac{h}{\lambda'}\begin{pmatrix}\cos\theta\\\sin\theta\\0\end{pmatrix} \tag{1c}$$ (Here we used the geometric definition of $$sin$$ and $$cos$$.)
• The scattered electron has momentum with an $$x$$ and $$y$$ component: $$\vec{p}_{e,\text{scattered}} =\gamma m u\begin{pmatrix}\cos\phi\\-\sin\phi \\0\end{pmatrix} \tag{1d}$$ (Here again we used the geometric definition of $$sin$$ and $$cos$$.)

During Compton scattering the total momentum is conserved. So we have the vector equation: $$\vec{p}_{\gamma,\text{incident}} + \vec{p}_{e,\text{initial}} = \vec{p}_{\gamma,\text{scattered}} + \vec{p}_{e,\text{scattered}} \tag{2}$$

Plugging in the four momenta from (1a-d) we get $$\frac{h}{\lambda}\begin{pmatrix}1\\0\\0\end{pmatrix} + \begin{pmatrix}0\\0\\0\end{pmatrix} = \frac{h}{\lambda'}\begin{pmatrix}\cos\theta\\\sin\theta\\0\end{pmatrix} + \gamma m u\begin{pmatrix}\cos\phi\\-\sin\phi\\0\end{pmatrix} \tag{3}$$

Writing vector equation (3) in components $$(x,y,z)$$ we get: \begin{align} \frac{h}{\lambda} &= \frac{h}{\lambda'}\cos\theta + \gamma mu\cos\phi \\ 0 &= \frac{h}{\lambda'}\sin\theta - \gamma mu\sin\phi \\ 0 &= 0 \end{align} \tag{4}

Momentum is a vector, so 'conservation of momentum' implies 3 equations.

These can be taken along any set of axes you choose. You can call them $$x,y,z$$ or $$e_1,e_2,e_3$$, whatever.

However it makes life simple if you choose the $$x$$ axis to be the initial direction of the photon. And then the $$y$$ axis to be the orthogonal axis that gives the scattering plane. So, by construction, neither the electron nor the scattered photon has any $$z$$ component of momentum. The third equation, for $$z$$, is just $$0=0$$ which is automatically satisfied and you only have to bother with two.