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5.2 Collisions of hard-ball particles
We consider two spherical particles and denote their radii and masses as $R_i$ and $m_i$ with $i \in \{1,2\}$, respectively. At the initial time $t=t_0$ the particles motion is not (yet) subjected to a force such that \begin{align*} \mathbf q_i (t) = \mathbf q_i ( t_0 ) + v_i \: ( t - t_0 ) \, , \quad\text{for}\quad i \in \{1,2\} \end{align*}
5.2.1 Center of mass motion
Analogous to the treatment of the Kepler problem, we decompose the motion of the particles into a center-of-mass motion $\mathbf Q(t)$ and a relative motion $\mathbf r (t)$. Introducing the notion $M = m_1 + m_2$ the former amounts to \begin{align}\label{eq:EXT-MQ} M \; \mathbf Q (t) = m_1 \: \mathbf q_1 (t) + m_2 \: \mathbf q_2 (t) = M \: \mathbf Q (t_0) + \dot{\mathbf Q} ( t_0 ) \: ( t - t_0 ) \end{align} Since there are not external forces the total momentum $M \, \dot{\mathbf Q} (t)$ is conserved (cf.\Thm{Newton-conserveP}) such that \cref{eq:EXT-MQ} applies for all times – even when the particles collide. A collision will therefore only impact the evolution relative to the center of mass. \cref{eq:EXT-MQ} holds for all times.
5.2.2 Condition for collisions
To explore the relative motion we write $\mathbf q_i = \mathbf Q + \mathbf x_i$, and we introduce the momentum $\mathbf p = m_1 \: \dot{\mathbf x}_1 = - m_2 \: \dot{\mathbf x}_2$ and the distance coordinate $\mathbf r = \mathbf x_1 - \mathbf x_2$. With these notations the angular momentum of the relative motion reads $ \mathbf L = \mathbf r \times \mathbf p $, and it is conserved when the collision force is acting along the line connecting the centers of the particles (cf.\Thm{Newton-conserveL} and the discussion of Kepler's problem in \cref{sec:Kepler}). Moreover, $\mathbf r(t)$ is the only time-dependent quantity in this equation because $\mathbf L$ and $\mathbf p$ are preserved. Let us first assume that the particles do not collide, and that the closest approach occurs at some time $t_c$ to a distance $r_c = | \mathbf r(t_c) |$. Then the vectors $\mathbf r(t_c)$ and $\mathbf p$ will be orthogonal, and $|\mathbf L| = r_c \, | \mathbf p |$. By the properties of the vector product the distance of the closest encounter will always be \begin{align*} r_c = \frac{ |\mathbf L| }{ | \mathbf p | } = \frac{ \bigl| m_1 \mathbf q_1(t_0) \times \dot{\mathbf q}_1(t_0) + m_2 \mathbf q_2(t_0) \times \dot{\mathbf q}_2(t_0) - M \mathbf Q (t_0) \times \dot{\mathbf Q} (t_0) \bigr| } { m_1 \bigl| \dot{\mathbf q}_1(t_0) - \dot{\mathbf Q} \bigr| } \end{align*} and there will be no collision if $r_c > R_1 + R_2$.
5.2.3 The collision
Conservation of angular momentum implies that the relative motion of the particles proceeds in a plane.
When they collide they approach until, at time $t_c$,
they reach a position $\mathbf r (t_c)$ where their distance is $|\mathbf r (t_c)| = R_1+R_2$.
We denote the direction of $\mathbf r$ at this time as $\hat{\boldsymbol \beta}$ and augment it by an orthogonal direction $\hat{\boldsymbol \alpha}$
such that $( \hat{\boldsymbol \alpha}, \hat{\boldsymbol \beta}, \hat{\boldsymbol L} = \mathbf L/\left\lvert \mathbf L \right\rvert )$ form an orthonormal basis.
We select the origin of the associated coordinate system such that
\[
\mathbf p
= ( \mathbf p \cdot \hat{\boldsymbol \alpha} ) \: \hat{\boldsymbol \alpha}
+ ( \mathbf p \cdot \hat{\boldsymbol \beta} ) \: \hat{\boldsymbol \beta}
\]
At the collision there is a force $\mathbf F = F \, \hat{\boldsymbol \beta}$ acting on the particles,
that acts in the direction of the line $\mathbf r (t_c)$ connecting the particles.
Hence,
1. the momentum component in the $\hat{\boldsymbol \alpha}$ direction is preserved during the collision
because there is no force acting in this direction
2. the collision in $\hat{\boldsymbol \beta}$ direction proceeds like a one-dimensional collision,
\Example{1dCollision}, with the exception
that one must retrace the argument using the center-of-mass frame,
as discussed in \cref{quest:ODE-1dCollision}.
Consequently, we obtain the following momentum $\mathbf p'$ after the collision
\[
\mathbf p'
= ( \mathbf p \cdot \hat{\boldsymbol \alpha} ) \: \hat{\boldsymbol \alpha}
- ( \mathbf p \cdot \hat{\boldsymbol \beta} ) \: \hat{\boldsymbol \beta}
= \mathbf p
- 2 \; ( \mathbf p \cdot \hat{\boldsymbol \beta} ) \: \hat{\boldsymbol \beta}
\]
5.2.4 Self Test
Collision of two hard-ball particles with radii $R_1$ and $R_2$: (top) Trajectory shape. The labels denote the ratios $(\mathbf p \cdot \hat{\boldsymbol \alpha}) / (\mathbf p \cdot \hat{\boldsymbol \beta})$. (bottom) Scattering angle $\theta$. }
Problem 5.2: Scattering angle for hard-ball particles
In\cref{fig:HardBallScatteringAngle} we show shows the trajectory shape and the scattering angle for hard-ball scattering.
- What is the dimensionless length scale adopted to plot the trajectory shapes?
- What is the impact of the angular momentum on the trajectory shape?
What is the impact of the energy?
- Verify that
\begin{align} \label{eq:HardBallScatteringAngle} \sin^2\theta = \frac{ L^2 }{2\mu \, E \, (R_1+R_2)^2} \end{align} and that this dependence is plotted in the lower panel of\cref{fig:HardBallScatteringAngle}.
- What happens when $L^2 > 2\mu \, E \, (R_1+R_2)^2$?
Which angle $\theta$ will one observe in that case?
- $\star$ Show that\cref{eq:HardBallScatteringAngle,eq:CoulombScatteringAngle} agree when one identifies
the length scale $R_1+R_2$ of the hard-ball system with the distance $R_{\text{eff}}$ of symmetry point of the cone section from the origin, i.e., with the mean value of the two intersection points with the $\hat x$-axis \begin{align*} R_{\text{eff}} = \frac{1}{2} \; \left( \frac{R_0}{ 1+\epsilon } + \frac{R_0}{1-\epsilon} \right) = \frac{\epsilon \, R_0}{1-\epsilon^2} \end{align*} Can you provide a physical argument why that must be true?
Problem 5.3: Reflection from a wall
Show that a particle reflected at a flat wall follows the same trajectory
as a particle that collides with a particle of the same mass
and at a position obtained as mirror image of the particle.
Problem 5.4:
Collisions on a billiard table
The sketch to the right shows a billiard table.
The white ball should be kicked (i.e. set into motion with velocity $\mathbf v$),
and hit the black ball such that it ends up in pocket to the top right.
What is tricky about the sketched track?
What might be a better alternative?