\subsubsection{The dynamics of a collision}
\label{sec:kin_dynamics}

To evaluate the collision integrals for a more complex potential than the hard sphere, a more detailed analysis of the particle collision dynamics than that presented in Section \ref{subsubsec:boltzmann_eq} is required. Here, the relationship between the deflection angle, $\chi$, impact parameter $b$ and dimensionless relative speed, $\sg$ will be derived for an arbitrary spherically symmetric potential that vanishes at infinite separation. Thus making it possible to evaluate the dimensionless collision integral

\begin{equation}
    W_{\ell, r} = 2 \int_0^\infty \int_0^\infty \exp{\lrp-\sg^2\rrp} \sg^{2r + 3} \lsp 1 - \cos^\ell\lrp\chi\lrp \sg, b\rrp\rrp\rsp \pfrac{b}{\sigma} \d \pfrac{b}{\sigma} \d \sg.
    \label{eq:kin_dimless_collision_integral}
\end{equation}

Where $\ell$ and $r$ are integer parameters that enter the equation through the derivation of the summational expressions given in Section \ref{subsubsec:summational_expressions}. The significance of the variables is illustrated in Figures \ref{fig:kin_dyn_red_mass}, \ref{fig:kin_dyn_polar_coords} and \ref{fig:kin_dyn_closest_approach}. Chapman and Cowling give a derivation of the relationship for a purely repulsive potential,\cite{chapman_cowling} but it is not immediately clear what the consequences are if the potential consists of both a repulsive and an attractive part. Therefore, rather than directly using the results from Chapman and Cowling, the derivation given here has been done to ensure that the results are valid for an arbitrary spherical potential. The final results are in accordance with those reported by Chapman and Cowling.

The final evaluation of the results requires a combination of numerical root solving and integration that is described further in Section \ref{sec:kin_numerics}.

The dynamics of a collision between two particles is governed by conservation of mass and momentum. To begin, denote the position of the centre of mass of the particles as $\vr_1$ and $\vr_2$. The intermolecular distance is $\vr_{12} \equiv \vr_2 - \vr_1$. The force acting on the particles is then $\Vec{F}_1 = \pder{u_{12}(\vr_{12})}{\vr_1} = - \pder{u_{12}(\vr_{12})}{\vr_{12}} = - \pder{u_{12}(\vr_{12})}{\vr_{2}} = - \Vec{F}_2 \equiv - \Vec{F}$, where $u$ denotes the intermolecular potential. Newtons second law for the two particles is then
\begin{equation}
    m_1 \Ddot{\vr}_1 = - \Vec{F}, \hspace{1cm} m_2 \Ddot{\vr}_2 = \Vec{F}.
\end{equation}
Where Newton notation, $\Dot{\varphi}$ is used to indicate time-derivatives. To simplify further calculations, recognise that 
\begin{equation}
    \begin{split}
        m_1 m_2 \Ddot{\vr}_{12} &= m_1 m_2 (\Ddot{\vr}_2 - \Ddot{\vr}_1) = (m_1 + m_2) \Vec{F}\\
        &= m_0 \Vec{F}\\
        \Ddot{\vr}_{12} &= \frac{m_0}{m_1 m_2} \Vec{F} = \frac{m_0}{m_1 m_2} \pder{u_{12}}{\vr_{12}},
    \end{split}
\end{equation}
where $m_0 = m_1 + m_2$. That is, the two-particle system behaves equivalently to a single particle of unit mass orbiting a stationary potential $\frac{m_0}{m_1 m_2} u_{12}$, or a single particle of reduced mass $\frac{m_1 m_2}{m_0}$ orbiting a stationary potential $u_{12}$. The former, illustrated in Figure \ref{fig:kin_dyn_red_mass} is arbitrarily chosen as a basis for the remaining analysis in this section, and will be referred to as the single-particle frame of reference (FoR).

\begin{figure}[htb]
    \centering
    \includegraphics[width=.75\textwidth]{kinetic_gas/figures/red_masse.eps}
    \caption{Illustration of two equivalent descriptions of the two particle system. Left: Two particles, each the centre of a potential $u_{12}$. Right: A single particle of unit mass with velocity $\Vec{g}_{12}$, orbiting a stationary potential $\frac{m_0}{m_1 m_2} u_{12}$.}
    \label{fig:kin_dyn_red_mass}
\end{figure}

Due to conservation of angular momentum, the plane of orbit is constant. It is therefore convenient to define a polar coordinate system in this plane, as illustrated in Figure \ref{fig:kin_dyn_polar_coords} to analyse the interaction. Furthermore, in the following analysis, the subscript on $r$ is dropped as the treatment is now that of a one-particle system. Keep in mind that the distance $r$ in the one-particle system is equivalent to $r_{12}$ in the two-particle system. Keep also in mind that $r(t = 0) = \infty$, such that the velocity $\Vec{g}$ is the relative velocity of the particles before they begin to interact.

\begin{figure}[htb]
    \centering
    \includegraphics[width=.75\textwidth]{kinetic_gas/figures/polar_coords.eps}
    \caption{The polar coordinate system, and corresponding cartesian coordinate system in the plane of orbit.}
    \label{fig:kin_dyn_polar_coords}
\end{figure}

Denoting the position in cartesian coordinates as $\Vec{s} = \lsp r \sin \theta, r \cos \theta \rsp$, and the velocity as\\$\Dot{\Vec{s}} = \lsp \Dot{r} \sin \theta + r \Dot{\theta} \cos \theta, \Dot{r} \cos \theta - r \Dot{\theta} \sin \theta \rsp$, the angular momentum of the particle is
\begin{equation}
    \begin{split}
        L &= \vert \Vec{s} \times \Dot{\Vec{s}} \vert\\
        &= \left| r \sin \theta \lrp \Dot{r} \cos \theta - r \Dot{\theta} \sin \theta \rrp - r \cos \theta \lrp \Dot{r} \sin \theta + r \Dot{\theta} \cos \theta \rrp \right|\\
        &= r^2 \Dot{\theta}.
    \end{split}
\end{equation}
At $t = 0$ the angular momentum is
\begin{equation}
    \begin{split}
        L &= \vert \vr \times \Vec{g} \vert \\
        &= \vert \vr \vert \vert \Vec{g} \vert \sin \lrp \angle \lcp \vr, \Vec{g} \rcp \rrp\\
        &= gb
    \end{split}
\end{equation}
Where the final equivalency is acquired by noting that $r(0)\sin \lrp \angle \lcp \vr, \Vec{g} \rcp \rrp = b$, as can be seen in Figure \ref{fig:kin_dyn_closest_approach}. The governing equation for angular momentum is thus
\begin{equation}
    r^2 \Dot{\theta} = gb.
    \label{eq:kin_dyn_govern_momentum}
\end{equation}

\begin{figure}
    \centering
    \includegraphics[width=.55\textwidth]{kinetic_gas/figures/closest_approach.eps}
    \caption{The distance of closest approach $R$, impact parameter $b$ and deflection angle $\chi$ illustrated in the single-particle FoR.}
    \label{fig:kin_dyn_closest_approach}
\end{figure}

Conservation of energy must also be obeyed. At any time, the kinetic energy of the particle is $\frac{1}{2} \vert \Dot{\Vec{s}} \vert ^2$, the total mechanical energy is thereby
\begin{equation}
    \begin{split}
        E_{tot}^{mec} &= \frac{1}{2} \vert \Dot{\Vec{s}} \vert ^2 + \frac{m_0}{m_1 m_2} u_{12}(r) \\
        &= \frac{1}{2}\lrp \Dot{r}^2 + r^2 \Dot{\theta}^2\rrp + \frac{m_0}{m_1 m_2} u_{12}(r).
    %\Dot{r}^2 \sin^2 \theta + r \Dot{r} \Dot{\theta} \sin 2 \theta + r^2 \Dot{\theta}^2 \cos^2 \theta  + \Dot{r}^2 \cos^2\theta - r \Dot{r} \Dot{\theta} \sin 2 \theta + r^2 \Dot{\theta}^2 \sin^2 \theta
    \end{split}
\end{equation}
At $t = 0$ the velocity of the particle is $\Vec{g}$ and the potential vanishes at infinite distance, such that the governing equation for conservation of energy is
\begin{equation}
    \frac{1}{2}\lrp \Dot{r}^2 + r^2 \Dot{\theta}^2\rrp + \frac{m_0}{m_1 m_2} u_{12}(r) = \frac{1}{2}g^2.
    \label{eq:kin_dyn_govern_energy}
\end{equation}

Recognise that due to the symmetry of the intermolecular potential, $\chi(g, b) = \pi - 2 \theta(R; g, b)$, where $R$ is the distance of closest approach, as illustrated in Figure \ref{fig:kin_dyn_closest_approach}, and the semicolon indicates a parametric dependency. Now that the governing equations for the orbit have been developed, $\chi(g, b)$ will be determined by obtaining an explicit expression for $\theta(R; g, b)$. As $R$, and thereby $\chi$, is not time-dependent, time can be eliminated from the governing equations without losing relevant information. This is accomplished by rewriting Equation \ref{eq:kin_dyn_govern_energy} as 
\begin{equation}
    \frac{1}{2}\lsp \lrp \frac{\d r}{\d \theta} \rrp^2 + r^2 \rsp \Dot{\theta}^2 + \frac{m_0}{m_1 m_2} u_{12}(r) = \frac{1}{2}g^2.
\end{equation}
Inserting for $\Dot{\theta}$ from Equation \ref{eq:kin_dyn_govern_momentum} yields
\begin{equation}
    \frac{1}{2}\frac{g^2 b^2}{r^4}\lsp \lrp \frac{\d r}{\d \theta} \rrp^2 + r^2 \rsp + \frac{m_0}{m_1 m_2} u_{12}(r) = \frac{1}{2}g^2.
\end{equation}
To find the distance of closest approach, solve this equation as
\begin{equation}
    \begin{split}
        \frac{\d r}{\d \theta} &= \pm \lrp \frac{r^4}{b^2}\lsp 1 - \frac{2}{g^2}\frac{m_0}{m_1 m_2} u_{12}(r)\rsp - r^2 \rrp^{\frac{1}{2}}\\
        &= \pm \lrp \frac{r^4}{b^2}\lsp 1 - \frac{u_{12}(r)}{k_B T \sg^2}\rsp - r^2 \rrp^{\frac{1}{2}}.
    \end{split}
    \label{eq:kin_dyn_drdtheta}
\end{equation}
Where $\sg = \pfrac{m_1 m_2}{2 m_0 k_B T}^{\frac{1}{2}}\vec{g}$ is the dimensionless relative speed introduced in Section \ref{sec:theory_kingas}. It is evident that $R$ is found at the point where $\frac{\d r}{\d \theta} = 0$, or
\begin{equation}
    \begin{split}
         \frac{R^4}{b^2}\lsp 1 - \frac{u_{12}(R)}{k_B T \sg^2}\rsp - R^2 = 0.
    \end{split}
    \label{eq:kin_dyn_R_root}
\end{equation}
For some general collision, this equation may have several roots, corresponding to stable orbits. However, because we consider a situation where the particles start at infinite separation, simply requiring that the particles rotational or vibrational energy cannot change is enough to ensure that no such stable orbit exists. Due to conservation of energy, even a particle starting with $g = 0$ at infinite distance will always remain exactly at the escape velocity of the potential. Any particle with $g > 0$ will always have a velocity greater than the escape velocity.

To obtain an expression for $\theta$, we invert Equation \eqref{eq:kin_dyn_drdtheta} such that 
\begin{equation}
    \frac{\d \theta}{ \d r} = \pm \lrp \frac{r^4}{b^2}\lsp 1 - \frac{u_{12}(r)}{k_B T \sg^2}\rsp - r^2 \rrp^{-\frac{1}{2}}.
\end{equation}
From the geometry of the encounter it is evident that 
\begin{equation}
    \frac{\d \theta}{\d r} = \lrp \frac{r^4}{b^2}\lsp 1 - \frac{u_{12}(r)}{k_B T \sg^2}\rsp - r^2 \rrp^{-\frac{1}{2}} \times 
    \begin{cases}
    -1,& \theta < \theta(R; \sg, b, T)\\
    1,& \theta > \theta(R; \sg, b, T)\\
    \end{cases}
    \label{eq:kin_dyn_dthetadr}
\end{equation}
As $\theta$ is strictly increasing as a function of time, while $r$ is decreasing until $\theta = \theta(R)$, after which $r$ increases. Integrating Equation \eqref{eq:kin_dyn_dthetadr} now gives an explicit expression for $\theta$,
\begin{equation}
    \begin{split}
        \theta(r') &= \int_\infty^{r'} \pfrac{\d \theta}{\d r} \d r\\
        \int_0^{\theta(R; \sg, b, T)} \d \theta &= \int_\infty^R  - \lrp \frac{r^4}{b^2}\lsp 1 - \frac{u_{12}(r)}{k_B T \sg^2}\rsp - r^2 \rrp^{-\frac{1}{2}} \d r\\
        \theta(R; \sg, b, T) &= \int_R^{\infty} \lrp \frac{r^4}{b^2}\lsp 1 - \frac{u_{12}(r)}{k_B T \sg^2}\rsp - r^2 \rrp^{-\frac{1}{2}} \d r.
    \end{split}
    \label{eq:kin_dyn_theta_integral}
\end{equation}

The dimensionless collision integral may now be evaluated numerically, the specifics of the numerics are presented in Section \ref{sec:kin_numerics}. In essence, for a given $\sg$, $b$, $T$ the value of $\chi$ is given as 
\begin{equation}
    \chi(\sg, b; T) = \pi - 2 \theta(R; \sg, b, T),
\end{equation}
as shown geometrically in Figure \ref{fig:kin_dyn_closest_approach}. To evaluate the integrand of Equation \eqref{eq:kin_dimless_collision_integral} at a single point one must first solve Equation \eqref{eq:kin_dyn_R_root} for $R$, then evaluate the integral of Equation \eqref{eq:kin_dyn_theta_integral} to obtain $\theta$ from which $\chi$ is computed.