# Packing spheres on the boundary of a larger sphere

Consider the following problem, which is a variation of the sphere packing problem and is somehow related to the kissing number problem. For a dimension $n\ge 2$ and a natural $k$, let $r=r(n,k)$ be the maximal radius of $k$ spheres (in $\mathbb{R}^n$) that can be packed into the boundary ring (is that the name of it?) of radius $2r$ of the unit sphere; that is, into the set of all points whose distance from 0 is in the interval $[1-2r, 1]$.

It is easier to explain this using examples:

• $r(2,1)$ is 1, as the maximal radius $r$ of a single sphere that should be embedded in the subset of the unit sphere containing all points whose distance from 0 lies in the interval $[1-2r,1]$ is simply 1 (in which case, the said subset is the unit sphere itself). In fact, $r(n,1)=1$ for any $n\ge 2$.
• $r(2,2)$ is 0.5, as one can pack 2 0.5-radius spheres inside the unit sphere. I guess $r(n,2)=0.5$ for any $n\ge 2$.
• $r(2,3)$ is ... well, I already don't know that.

What I do know (know might be too harsh here) is that: $$\lim_{k->\infty}k\cdot r(2,k)=\pi$$

And that makes me guess $k\cdot r(2,k)$ is an increasing function, going from 1 to $\pi$. A more general limit can be described as follows: the limit of ratio of the sum of all surface areas of all packed spheres, and the surface area of the unit sphere. If we denote that ratio with $R(n,k)$, we obtain (I think) $$\lim_{k\to\infty}R(2,k)=\pi$$ and $$\lim_{k\to\infty}R(3,k)=\pi$$ but this is where it stops, as $$\lim_{k\to\infty}R(4,k)=\frac{\pi^2}{4}$$ and I think that $$\lim_{k\to\infty}R(9,k)=\frac{\pi^4}{840}$$

So my question actually consists of three sub-questions:

• Can you imagine how $r(n,k)$ (or $R(n,k)$) looks like? Is it monotone, does it always have a limit for $k\to\infty$, can you estimate that limit with respect to $n$, etc.
• I there any clear relation between $k\cdot r(n,k)$ and $R(n,k)$?
• Is $$\lim_{n,k\to\infty}R(n,k) = 0,$$ and if so, do you have any intuition about it?
• If I understood correctly, $r(2,k)$ should be equal to $1-1/\left(1+\sin(\pi/k)\right)$ for $k\geq 2$ (confirming your first limit being equal to $\pi$). – Peter Košinár Feb 21 '14 at 2:14
• Also, the relation between $R(n,k)$ and $k\cdot r(n,k)$ should be just $R(n,k)=k\cdot r(n,k)^{n-1}$, since the surface area of an $n$-dimensional ball is proportional to its radius raised to $(n-1)$-th power. – Peter Košinár Feb 21 '14 at 10:25

Let $\mathcal{R}_n = \lim\limits_{k\to\infty} R(n,k)$. Your numbers on $\mathcal{R}_n$ doesn't feel right for $n \ge 3$.

For any fixed $n$ and large $k$, the centers of the small spheres of radius $r$ should be constrained to a sphere of radius $1-r$. To the small spheres, the space between two spherical shell of radius $1$ and $1-2r$ will look flat. So the optimal $(r,k)$ configuration will correspond to some sort of close packing of $S^{n-2}$ in a $\mathbb{R}^{n-1}$.

Let $\rho_{n}$ be the optimal packing density of $S^{n-1}$ in $\mathbb{R}^n$. Let $\displaystyle\sigma_n = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}$ be the volume of the unit $n$-ball in $\mathbb{R}^n$. We know the "surface area' of the unit $n$-ball is given by $n\sigma_n$. This leads to \begin{align} & \sigma_{n-1} k\,r^{n-1} \approx n\sigma_n (1-r)^{n-1} \times \rho_{n-1} \\ \implies & \mathcal{R}_n = \lim_{k\to\infty} k\,r^{n-1}(n,k) \approx n\frac{\sigma_n}{\sigma_{n-1}}\rho_{n-1} = \frac{2\sqrt{\pi}\,\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\rho_{n-1} \end{align}

It is known that $\rho_1 = 1$, $\displaystyle \rho_2 = \frac{\pi}{2\sqrt{3}}$, $\displaystyle \rho_3 = \frac{\pi}{\sqrt{18}}\color{blue}{^{[1]}}$ and $\displaystyle \rho_8 \approx \frac{\pi^4}{384} ( 1 + O(10^{-14}) )\color{blue}{^{[2]}}$. This give us an estimate

$$\mathcal{R}_n \approx \begin{cases} \pi,&n = 2\\ \\ \frac{2\pi}{\sqrt{3}},& n = 3\\ \\ \frac{\pi^2}{4\sqrt{2}}, & n = 4\\ \\ \frac{2\pi^4}{105},&n = 9 \end{cases}$$ Something very different from your guess when $n \ge 3$.

About what happens to $\mathcal{R}_n$ for large $n$, we know that for $n \ge 115$, there is an upper bound for $\rho_n$ of the form$\color{blue}{^{[3]}}$:

$$\rho_n \le 2^{-(0.5990\ldots + o(1))n}$$ This means $\mathcal{R}_n$ converges to $0$ as $n$ tends to infinity.

Notes

• $\color{blue}{[1]}$ T.C. Hales, A proof of the Kepler conjecture, Ann. of Math. (2) 162 (2005), 1065-1185. MR2179728 doi:10.4007/annals.2005.162.1065.

• $\color{blue}{[2]}$ the packing density $\frac{\pi^4}{384}$ is achieved by regular packing of spheres on an $E_8$ lattice in $\mathbb{R}^8$. The $O(10^{-14})$ error bound is given by H.Cohn and A. Kumar but I don't know the exact reference.

• $\color{blue}{[3]}$ G.A.Kabatyanskii and V.I.Levenshetin. Bounds for packings on a sphere and in space (Russian), Problemy Peredaci Informacii 14 (1978), 3-25; English translation in Problems of Information Transmission 14 (1978), 1-17, MR0514023.

• What do you mean by To the small spheres, the space between two spherical shell of radius $1$ and $1−2r$ will look flat? In addition, I am not totally sure you got me right. There's only one layer of small spheres, and they "touch" the boundary of the bigger sphere. Or perhaps I didn't get you right; you've used terms I am not familiar with (close packing, packing density...) – Bach Mar 13 '14 at 10:52
• @Bach, On the human length scale, the earth looks flat instead of a sphere. If you want to pack basket balls on a basket ball field effectively, you will just pack them as if the earth is a plane (at least locally). The same thing happens to the small spheres when $r$ is sufficiently small. About the number of layers of small spheres. Yes, there is only one layer. However, for the configuration when $r$ reaches its minimal (for given $n, k$), one will expect their center are all at a distance $1-r$ from the center and hence lies between 2 concentric spheres with radii $1$ and $1-2r$. – achille hui Mar 13 '14 at 11:20