PH 4433/6433 Homework 4, Problem 2

Mikhail Gaerlan
30 September 2015

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Introduction

A potential field is given by the equation
$\displaystyle{V(x,y)=x^2y^2e^{-\left(x^2+y^2\right)}.}$
The path of a particle interacting with this field can expressed by a system of ODE's.
By using numerical methods, the path of of a particle can be plotted in the potential field.


Code

Repulsive

  1. Main
  2. Deriv
  3. RK4

Attractive

  1. Main
  2. Deriv
  3. RK4

Results

The lines represent the path of the particles. The color of the background represents the strength of the potential field.

Motion Motion Motion

Motion Motion Motion

Scattering

Scattering

Discussion

$\displaystyle{\text{For all values of }b\text{ at both }t=0\text{ and }t\gg t _0\text{, }KE\gg PE\text{ and }KE _{t=0}=KE _{t\gg t _0}\Rightarrow E\approx KE\text{ and thus }E _{before}=E _{after}\text{.}}$

$\displaystyle{\text{The scattering angles for }b=-1\text{ and }b=0\text{ are }\frac{\pi}{2}\text{ (except repulsive, }v _i=0.5\text{), which make sense since the particles}}$
$\displaystyle{\text{are entering perpendicular to the field and will continue to move in straight lines. For repulsive }v _i=0.5\text{, the}}$
$\displaystyle{\text{particle does not have enough kinetic energy to overcome the field and thus goes back in the opposite direction.}}$
$\diaplsystyle{\text{For all other values of }b\text{, the scattering angle is seemingly random for low }v _i\text{. For higher }v _i\text{, most of the}}$
$\displaystyle{\text{scattering angles are around }\frac{\pi}{2}\text{ since most of the particles pass through the field mostly unchanged.}}$