Table of Contents
5.1 Motivation and outline: How do particles collide?
In order to get a first impression about this idea we consider the case of two particles at the positions $\mathbf q_i$, $i \in \{1,2\}$ that interact by a repulsive Coulomb force that derives from a potential $\Phi_C(\left\lvert \mathbf R \right\rvert)$ with $\mathbf R = \mathbf q_2 - \mathbf q_1$, \begin{align*} \Phi_C (\left\lvert \mathbf R \right\rvert) = \frac{ C }{ \left\lvert \mathbf R \right\rvert } \quad \Rightarrow \quad \mathbf F_c( \mathbf q_i ) = - \nabla_{\mathbf q_i} \Phi_C( \left\lvert \mathbf q_i - \mathbf q_{2-i} \right\rvert ) = \frac{ C \; \bigl( \mathbf q_i - \mathbf q_{2-i} \bigr) }{ \left\lvert \mathbf q_i - \mathbf q_{2-i} \right\rvert^3 } \end{align*} Here, $2-i$ is the index of the other particle ($1$ for $i=2$ and $2$ for $i=1$), and the constant $C$ is the product of the permittivity of the vacuum and the particle charges. For charges of opposite signs this force has agrees with the gravitational force when one substitutes $C \to -m_1\, m_2\, G$. This results in the same dimensionless equations of motion as obtained for the Kepler problem, with the important difference that the length and time units adopted to defined the dimensionless units take vastly different values. When the two particles carry charges of equal signs the force is repulsive, giving rise to the EOM \begin{align*} 0 = w'' ( \theta ) + w(\theta) + \frac{ \mu \, C }{L^2} \; w( \theta) \end{align*} such that \begin{align*} R (\theta) = \frac{1}{W(\theta)} = \frac { R_0 }{ -1 + \epsilon \: \cos\bigl( \theta - \theta_0 \bigr) } \quad \text{where} \quad R_0 = \frac{L^2}{\mu \, C} \end{align*} agrees with Equation 4.9.3 up to a change of the sign of the one in the denominator and the length unit $R_0$.
Remark 5.1. It is illuminating to adopt a different perspective on the origin of the minus sign in front of the one. Let us write the force on particle $1$ as $\mathbf F_1 = F_1 \: \hat{\boldsymbol e}(\theta)$ where $\hat{\boldsymbol e}(\theta)$ is the vector pointing from particle $1$ to particle $2$. The strength of the scalar force $F_1$ will be positive for an attractive force and negative for a repulsive force. In the dimensionless force $F t_0^2 / \mu\, R_0$ the change of sign is taken into account by the sign of $C$ in $R_0 = L^2 / \mu \, C$ and the solution takes the form of Equation 4.9.3. In order to obtain a positive length scale $\left\lvert R_0 \right\rvert = \pm R_0$ we multiply the numerator and denominator of the solution by the $\pm 1$ and absorb the factor in front of $\epsilon$ in a rotation of the angle by $\pi$ such that the polar coordinates are always aligned with the direction of the force. Hence, one finds \begin{align*} R (\theta) = \frac{1}{W(\theta)} = \frac {\left\lvert R_0 \right\rvert }{ \pm 1 + \epsilon \: \cos\bigl( \theta - \theta_0 \bigr) } \quad \text{where} \quad \pm 1 = \text{sign}\! \left( C \right) \end{align*} At this point dimensionless units play out their strength. We obtain the solution of the nontrivial EOM by an analysis of the ODE an