﻿ Solution: Classical 2D Elastic Scattering | Primer on Geometric Algebra | David Hestenes

## Solution: Classical 2D Elastic Scattering

Here we'll derive the formulas used for the collision of billiard balls shown in the interactive diagram for Compton scattering shown on the previous page. We'll use lab coordinates.
The CCW unit bivector is i, so multiplying a vector by this on the right rotates the vector in the CCW direction. We're using the same scattering angle (chosen from the diagram slider) as for the Compton scattering, but since the energy and momentum formulas are different, we will get a different target recoil angle (call it ϕ here for simplicity of the derivation, but it's called ϕ2 in the diagram) than for the photon scattering.

μ m = mass of target ball,  m = mass of incident cue ball, the slider allows (1 ≤ μ ≤ 20).

v = incident cue ball speed,  V = scattered exit cue ball speed,  u = target ball exit recoil speed.

θ  = scattering angle of cue ball above the incident axis,  ϕ = target ball recoil angle below the incident axis.

Target ball initially at rest.  Given all else, find formulas for Vu and ϕ:

$\Large{\begin{array}{l}{\rm{Momentum:}}\,\,\,\,mv{\bf{\hat v}}\,\, + \,\,0\,\,\, = \,\,\,mV{\bf{\hat v}}{e^{{\bf{i}}\theta }}\,\, + \,\,\mu mu{\bf{\hat v}}{e^{ - {\bf{i}}\phi }}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v\,\,\, = \,\,\,V{e^{{\bf{i}}\theta }}\,\, + \,\,\mu u{e^{ - {\bf{i}}\phi }}\,\,\, = \,\,\,V{e^{ - {\bf{i}}\theta }}\,\, + \,\,\mu u{e^{{\bf{i}}\phi }}\\\\{\rm{Eliminate}}\,\,\phi :\,\,\,\,\left( {v\,\, - \,\,V{e^{{\bf{i}}\theta }}} \right)\,\left( {v\,\, - \,\,V{e^{ - {\bf{i}}\theta }}} \right)\,\,\, = \,\,\,{\mu ^2}{u^2}\,\,\, = \,\,\,{v^2}\,\, + \,\,{V^2}\,\, - \,\,2vV\cos \theta \\\\{\rm{Energy:}}\,\,\,\,\frac{1}{2}m{v^2}\,\, + \,\,0\,\,\, = \,\,\,\frac{1}{2}m{V^2}\,\, + \,\,\frac{1}{2}\mu m{u^2}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mu ^2}{u^2}\,\,\, = \,\,\,\mu {v^2}\,\, - \,\,\mu {V^2}\\\\{\rm{Eliminate}}\,\,u:\,\,\,\,\mu {v^2}\,\, - \,\,\mu {V^2}\,\,\, = \,\,\,{v^2}\,\, + \,\,{V^2}\,\, - \,\,2vV\cos \theta \\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\mu \,\, + \,\,1} \right)\,{V^2}\,\, - \,\,\left( {2v\cos \theta } \right)\,V\,\, - \,\,\left( {\mu \,\, - \,\,1} \right)\,{v^2}\,\,\, = \,\,\,0\\\\\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\, \le \,\,\,\left( {\frac{{\mu \,\, - \,\,1}}{{\mu \,\, + \,\,1}}} \right)v\,\,\, \le \,\,\,{\bbox[8px,border:2px solid red]{V\,\,\, = \,\,\,\left( {\frac{v}{{\mu \,\, + \,\,1}}} \right)\,\,\left( {\cos \theta \,\, + \,\,\sqrt {{\mu ^2}\,\, - \,\,{{\sin }^2}\theta } } \right)}}\,\,\, < \,\,\,v\\\\{\rm{Then,}}\,\,{\rm{from}}\,\,{\rm{the}}\,\,{\rm{energy}}\,\,{\rm{equation}}\,\,{\rm{again:}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\, < \,\,\,{\bbox[8px,border:2px solid red]{u\,\,\, = \,\,\,\sqrt {\frac{{{v^2}\,\, - \,\,{V^2}}}{\mu }}}} \,\,\, \le \,\,\,\left( {\frac{2}{{\mu \,\, + \,\,1}}} \right)\,v\\\\\\\\\\\\{\rm{Finally,}}\,\,{\rm{go}}\,\,{\rm{back}}\,\,{\rm{to}}\,\,{\rm{the}}\,\,{\rm{momentum}}\,\,{\rm{equation}}\,\,{\rm{to}}\,\,{\rm{get}}\,\,\phi :\\\\{e^{ - {\bf{i}}\phi }}\,\,\, = \,\,\,\cos \phi \,\, - \,\,{\bf{i}}\,\sin \phi \,\,\, = \,\,\,\frac{{v\,\, - \,\,V{e^{{\bf{i}}\theta }}}}{{\mu u}}\,\,\, = \,\,\,\frac{{v\,\, - \,\,V\cos \theta }}{{\mu u}}\,\, - \,\,{\bf{i}}\,\frac{{V\sin \theta }}{{\mu u}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\, \le \,\,\,{\bbox[8px,border:2px solid red]{\phi \,\,\, = \,\,\,{\rm{Arc}}\cos \left( {\frac{{v\,\, - \,\,V\cos \theta }}{{\mu u}}} \right)}}\,\,\, < \,\,\,\frac{\pi }{2}\end{array}}$

When playing with the interactive Compton Effect diagram on the previous page, you may notice
two peculiarities in the billiard ball collision when they have the same masses (μ=1) not present in the photon/electron scattering.

1. The angle between the balls' paths after the collision is always a right angle (θ + ϕ = π/2).
The equations are much easier to solve in this case.
Also, notice our main boxed formulas become
V = v cosθ, u = v sinθ, and cosϕ = sinθ.

2. When we choose the scattering angle
θ to be greater than π/2, V and ϕ2 are always zero and
u is always equal to the incident speed of v. The cue ball can never bounce in any backwards direction!
That's a consequence of conserving both momentum and energy with equal masses.
When masses are equal, in the most direct head on collision,
the incoming ball stops and the target ball
moves on with the original incident speed

The analog for this case cannot happen in the photon/electron collision. That's not too surprising since the governing
equations are pretty different anyway, but notice also that this would correspond to
the photon stopping (in other
words being absorbed) and transferring all of its energy to the electron's kinetic energy, which, as is shown at the
bottom of the previous page,
cannot happen.