Massive Meets Massless: Compton Scattering Revisited

[neilparker62]

Introduction
In a previous article entitled “Alternate Approach to 2D Collisions” we analyzed collisions between a moving and stationary object by defining the coordinate axes as being respectively parallel and perpendicular to the post-collision direction of motion of the stationary object. In this article, we will be adopting the same approach to analyze the well-known physical phenomenon known as Compton scattering in which a photon collides with a stationary electron, imparts momentum to the latter, and hence loses momentum and energy which manifests in a change of direction (‘scattering’ angle) and wavelength (of the photon).
The established physics of Compton scattering relates the scattering angle (angle of deflection) to the change in wavelength of the incident photon according to the following expression...

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[azntoon]

\begin{equation}\Delta \lambda = \frac{h}{m_e c}\left(1 - \cos(\theta)\right)\end{equation}where $\Delta \lambda$ is the change in wavelength of the incident photon, $h$ is Planck’s constant, $m_e$ is the mass of the electron, and $\theta$ is the scattering angle.To determine the scattering angle using the alternate approach, let us consider a point $(x,y)$ on the co-ordinate plane representing the photon's pre-collision position and direction of motion. The post-collision position and direction of motion can be represented by $(x',y')$. The co-ordinate axes are parallel and perpendicular to the post-collision direction of motion of the electron. Therefore, the components of the momentum of the photon before and after the collision can be written as:\begin{equation}\vec p_{pre} = (px,py)\end{equation}\begin{equation}\vec p_{post} = (p'x,p'y)\end{equation}where $px$ and $py$ are the components of the momentum of the photon in the pre-collision direction of motion and $p'x$ and $p'y$ are the components of the momentum of the photon in the post-collision direction of motion.Since the momentum of the photon is conserved during the collision, we have:\begin{equation}px + py = p'x + p'y\end{equation}Also, since the change in momentum of the electron is equal and opposite to the change in momentum of the photon, we can write:\begin{equation}p'x - px = -(p'y - py)\end{equation}Rearranging the equations for $px$ and $py$, we get:\begin{equation}px = \frac{p'y - p'x}{2}\end{equation}\begin{equation}py = \frac{p'x + p'y}{2}\end{equation