# Ball ricochetting from a plane of close-packed spheres

Suppose the lower $z \le 0$ halfspace of $\mathbb{R}^3$ is filled with a rigid close-packing of unit-radius spheres. (I don't think it matters much for my purposes if it is an FCC or an HCP packing.) Shoot in from $z>0$ another sphere/ball $S$ of radius $R$ along a random ray at a random angle $\phi$ (w.r.t. the $xy$-plane). Perhaps $R>1$, maybe even $R \gg 1$; but also perhaps $R<1$.

(Sphere packing image from Ed Pegg Mma Demo.)
$S$ bounces from the rigid unit-spheres halfspace boundary via elastic collision, perhaps ricochetting from several in the top layer—or near-top layers—before emerging free into the upper halfspace $z>0$.

Q1. Is it true that, as $R \to \infty$, the angle of reflection from the $xy$-plane approaches the angle of incidence $\phi$, despite the (inevitable) ricochetting?

Q2. Is it true that, for $R<1$, the distribution of the angles of reflection, for a random ray at incidence angle $\phi$, approaches a diffuse reflection cone independent of $\phi$?

Q3. Is it true that, for $R \ll 1$, some trajectories would never emerge from the lower halfspace?

For small $R$, one might instead think of lightrays bouncing from mirrored spheres.1

1(Distantly) related: "Light reflecting off Christmas-tree balls".

• For Q1, you might shift perspective and "throw" the cube at the sphere. Ignoring angular momentum, it seems the answer would be yes, as for large R the sphere looks like a plane surface. Gerhard "Backwards Thinking For Problem Solving" Paseman, 2016.05.07. – Gerhard Paseman May 7 '16 at 23:58
• I think for Q3 with HCP you can construct a trajectory that would penetrate infinitely deeply: There are triangle shaped "tunnels" of empty space perpendicular to the hexagonal planes. If you take a small ball (or light ray) that bounces off a packed ball in each even hexagonal plane and at angle of $30^\circ$ from the normal of the packed ball, then I think it would follow a triangle shaped helix-like path downward. – Jaap Eldering May 8 '16 at 13:18
• If there is no rotation of the ball, then you might instead imagine a light ray bouncing off reflective overlapping balls of radius $1+R$. – Douglas Zare May 8 '16 at 16:04

An answer to Q1 only, and I'm ignoring angular momentum considerations.

As $R \to \infty$ the maximum penetration depth of the ball $S$ into the half-space $z < 0$ goes to zero. Hence, if the ball bounces off one of the packed balls only, then the tangent plane to both balls at the collision point must be close to horizontal. The reflected velocity vector is mirrored in this plane from the incident velocity vector, hence close to reflection in the horizontal plane. This also shows that multiple bounces can only occur when the incident angle smaller than the maximum angle $\varphi_m$ of the collision tangent plane away from horizontal, and by symmetry the reflection angle as well, hence their difference is smaller than $2\varphi_m$. This only measures the angle to the normal, but if you can bound the number of bounces, then it also provides a bound on the deviation of the direction tangent to the plane.

Q3: If $r<\sqrt{2}-1$, there are trajectories that enter the lower half-space and don't come out.

For example, let position $A_n$ be at $(\frac{1+r}{\sqrt{2}},\frac{1+r}{\sqrt{2}}, -2n)$. Let $B_n$ be at $(2- \frac{1+r}{\sqrt{2}},2-\frac{1+r}{\sqrt{2}},-2n)$. A ball centered at $A_{-1}$ directed toward $B_0$ will bounce toward $A_1,B_2,A_3,...$ while its center stays in the plane $y=x$.

You can think of this as covering a closed reflecting path in a $3$-torus that is nontrivial in $H_1$. There are other paths that enter and don't come out that are aperiodic, even with the center confined to $y=x$.

• Nice! Could you explain the phrase "nontrivial in $H_1$"? Thanks. – Joseph O'Rourke May 11 '16 at 11:39
• If the closed reflecting path in the torus is null-homotopic, then it lifts to a closed path in the universal cover $\mathbb{R}^3$. On the other hand, if it is a nontrivial loop then it lifts to an infinite path $\mathbb{R} \to \mathbb{R}^3$ which has a last time it crosses some planes. It becomes a trajectory that enters a half-space and doesn't come out. – Douglas Zare May 11 '16 at 13:52

Q2: The distribution of reflection angles certainly depends on the incident angle. For example, if the incident angle is nearly parallel to the surface, the particle will make a small number of reflections and exit in another nearly parallel direction.