Étendue measure of the set of lines between two Euclidean balls

Question

Let $d>0$ and $r_1,r_2>0$ such that $r_1+r_2 < d$. Consider two (say, closed) balls $B_1,B_2$ in $\mathbb{R}^m$ having radii $r_1,r_2$ and whose centers are at distance $d$. Let $C$ be the set of lines intersecting both $B_1$ and $B_2$.

Is there an exact formula for the étendue measure of $C$? To be clear, the “étendue” of a Borel set of lines $C$ in $\mathbb{R}^m$ is defined as one half the integral over unit vectors $u$ in $\mathbb{R}^m$ of the (“area”) measure $\mathcal{A}(C,u)$ of the intersection of an arbitrary hyperplane perpendicular to $u$ with the union of all lines in $C$ having direction $u$. (The factor ½ is to avoid doubly counting each direction.) In other words, this is, up to normalization, the translation-and-rotation invariant measure on the Grassmannian of affine lines in $\mathbb{R}^m$.

For $r_1,r_2$ small with respect to $d$ we should get the asymptotic $V_{n-1}^2 \, r_1^{n-1} \, r_2^{n-1} / d^{n-1}$, where $V_{n-1}$ refers to the volume of the unit $(n-1)$-ball, because we are seeing a small disk of area $V_{n-1} \, r_1^{n-1}$ so solid angle $V_{n-1} \, r_1^{n-1} / d^{n-1}$ from one of are $V_{n-1} \, r_2^{n-1}$ (or vice versa, the situation is symmetric). What I wonder is whether an exact formula can be found.

Answer 1

$\newcommand\R{\mathbb R}\renewcommand{\th}{\theta}\newcommand{\om}{\omega}\newcommand{\Ga}{\Gamma}$Here is a solution for any dimension $m\ge2$:

Without loss of generality (wlog), we have two open balls: $B_1=B_0(r_1)$ and $B_2=B_D(r_2)$, for some $D\in\R^m$ with $\|D\|=d$, where $B_x(r)$ is the open ball in $\R^m$ of radius $r$ centered at $x$ and $\|\cdot\|$ is the Euclidean norm.

Take any unit vector $u$. Take any $x\in B_1$ such that $x\cdot u=0$, where $\cdot$ denotes the dot product. Then the condition that the line through $x$ in the direction of $u$ intersects $B_2$ is that $x\in B_D(R_u)$, where $R_u:=\sqrt{(D\cdot u)^2+r_2^2}$; this can be obtained by minimizing $\|x-tu-D\|^2$ in real $t$ and taking into account the condition $x\cdot u=0$.

So, $$\mathcal A(C,u)=L_{m-1}(B_0(r_1)\cap B_D(R_u)\cap H_u),$$ where $H_u:=\{x\in\R^m\colon\, x\cdot u=0\}$ and $L_{m-1}$ is the $(m-1)$-dimensional Lebesgue measure over the hyperplane $H_u$. So, $\mathcal A(C,u)$ is the $(m-1)$-volume of the intersection of two balls, $B_0(r_1)\cap H_u$ and $B_D(R_u)\cap H_u$, in $H_u$.

So, $\mathcal A(C,u)$ is the $(m-1)$-volume of the intersection of two balls in $\R^{m-1}$, of radii $r_1$ and $r_2$, with the centers of the two balls at distance $d_u:=\sqrt{d^2-(D\cdot u)^2}$ from each other; this conclusion can be probably be obtained in a simpler manner.

So, $\mathcal A(C,u)$ is an ordinary integral of the ($(m-2)$-dimensional) volumes of $(m-2)$-dimensional balls, which is actually known -- but is expressed via the incomplete beta function, rather than in elementary functions.

Now it remains to integrate $\mathcal A(C,u)$ over all $u$ on the $(m-1)$-dimensional unit sphere $S^{m-1}$. When doing that, by the spherical symmetry we can, and will, assume that $D=d\,e_1$, where $e_1$ is the first standard basis vector in $\R^m$.

Then $D\cdot u=u_1d$, where $u_1$ is the first coordinate of the unit vector $u$, and hence $d_u=d(1-u_1^2)$. So, \begin{equation*} \mathcal A(C,u)=g(u_1^2), \tag{00}\label{00} \end{equation*} for a certain function $g=g_{r_1,r_2,d}$, expressed explicitly via an incomplete beta function. So, the étendue measure of $C$ is \begin{multline} E=E_m(r_1,r_2,d)=\frac12\,\int_{S^{m-1}}du\,\mathcal A(C,u)= \frac12\,\int_{S^{m-1}}du\,g(u_1^2) \\ =\frac12\,|S^{m-1}|Eg(U_1^2)=b_m\,\int_0^1 dt\, g(t)t^{1/2-1}(1-t)^{(m-1)/2-1}, \tag{10}\label{10} \end{multline} where $|S^{m-1}|$ is the $(m-1)$-dimensional "surface area" of $S^{m-1}$ and $U_1^2$ is a (real-valued) random variable with the beta distribution with parameters $1/2,(m-1)/2$, and \begin{equation*} b_m:=\frac{\pi^{(m-1)/2}}{\Ga((m-1)/2)}. \tag{20}\label{20} \end{equation*}

Let us detail the latter expression in \eqref{10}. By \eqref{00}, $g(t)$ is the $(m-1)$-volume of the intersection of two balls, say $B_{1,t}$ and $B_{2,t}$, in $\R^{m-1}$, of respective radii $r_1$ and $r_2$, with the centers of the two balls at distance $d_u=d\sqrt{1-t}$ from each other. Here and in what follows, $t\in(0,1)$. Let $L_k$ denote the Lebesgue measure over $\R^k$.

Without loss of generality, $r_1\ge r_2$.

There are three possible cases here:

• Case 1: $t\le t_-:=1-\dfrac{(r_1+r_2)^2}{d^2}$

• Case 2: $t\ge t_+:=1-\dfrac{(r_1-r_2)^2}{d^2}$

• Case 3: $t_-<t<t_+$

In Case 1, $r_1+r_2\le d\sqrt{1-t}=d_u$ and hence the interiors of the balls $B_{1,t}$ and $B_{2,t}$ do not intersect, so that
\begin{equation*} g(t)=0 \tag{g: Case 1}\label{cs1} \end{equation*}

In Case 2, $r_1-r_2\ge d\sqrt{1-t}=d_u$ and hence $B_{1,t}\supseteq B_{2,t}$, so that \begin{equation*} g(t)=L_{m-1}(B_{2,t})=\om_{m-1}r_2^{m-1} \tag{g: Case 2}\label{cs2} \end{equation*} where \begin{equation} \om_k:=\frac{\pi^{k/2}}{\Ga(k/2+1)}, \tag{30}\label{30} \end{equation} the $k$-volume of the unit ball in $\R^k$.

In Case 3, $B_{1,t}\cap B_{2,t}$ is the union of two spherical segments $S_1$ and $S_2$ in $\R^{m-1}$ with disjoint interiors. More specifically, let us identify $\R^{m-1}$ with $\R\times\R^{m-2}$ and assume, wlog, that the balls $B_{1,t}$ and $B_{2,t}$ are centered at the respective points $(0,0)$ and $(d_u,0)=(d\sqrt{1-t},0)$ in $\R\times\R^{m-2}$. Then, for $m\ge3$,
\begin{equation*} B_{1,t}\cap B_{2,t} \\ =\{(z,y)\in\R\times\R^{m-2}\colon |y|^2<\min[r_1^2-z^2,r_2^2-(d\sqrt{1-t}-z)^2] \\ =S_1\cup S_2, \end{equation*} where \begin{equation*} S_1:=\{(z,y)\in\R\times\R^{m-2}\colon z_1<z<r_1,|y|^2<r_1^2-z^2\}, \end{equation*} \begin{equation*} S_2:=\{(z,y)\in\R\times\R^{m-2}\colon d\sqrt{1-t}-r_2<z\le z_1,|y|^2<r_2^2-(d\sqrt{1-t}-z)^2\}, \end{equation*} \begin{equation*} z_1:=z_1(r_1,r_2)(t):=\frac{d^2 (1-t)+r_1^2-r_2^2}{2 d \sqrt{1-t}}. \end{equation*} Using the reflection $z\leftrightarrow d \sqrt{1-t}-z$, we see that spherical segment $S_2$ is congruent to \begin{equation*} S'_2:=\{(z,y)\in\R\times\R^{m-2}\colon z_2\le z<r_2,|y|^2<r_2^2-z^2\}, \end{equation*} where \begin{equation*} z_2:=d \sqrt{1-t}-z_1=\frac{d^2 (1-t)-r_1^2+r_2^2}{2 d \sqrt{1-t}} =z_1(r_2,r_1)(t). \end{equation*}

So, in Case 3, \begin{equation*} g(t)=L_{m-1}(B_{1,t}\cap B_{2,t})=L_{m-1}(S_1)+L_{m-1}(S'_2) \\ =V_{r_1,r_2}(t)+V_{r_2,r_1}(t) \tag{g: Case 3}\label{cs3} \end{equation*} for $m\ge3$, where \begin{equation*} V_{r_1,r_2}(t):=\om_{m-2}\int_{z_1(r_1,r_2)(t)}^{r_1} dz\,(r_1^2-z^2)^{(m-2)/2} =\om_{m-2}r_1^{m-1} I_{r_1,r_2}(t), \end{equation*} \begin{equation*} I_{r_1,r_2}(t):=\int_{c_1(r_1,r_2)(t)}^1 dc\,(1-c^2)^{m/2-1}, \end{equation*} \begin{equation*} c_1(r_1,r_2)(t):=\frac{z_1(r_1,r_2)(t)}{r_1} =\frac{d^2 (1-t)+r_1^2-r_2^2}{2 r_1 d \sqrt{1-t}} \\ =\frac{d}{2 r_1}\,(1-t)^{1/2} +\frac{r_1^2-r_2^2}{2 r_1 d }\,(1-t)^{-1/2}. \tag{40}\label{40} \end{equation*} One can similarly see that \eqref{cs3} holds for $m=2$ as well.

Collecting \eqref{10}, the descriptions of Cases 1--3, \eqref{cs1}, \eqref{cs2}, and \eqref{cs3}, we get the following expression for the étendue measure of $C$: \begin{equation*} E=b_m\,(J_2+J_{3;1,2}+J_{3;2,1}) \tag{!}\label{!} \end{equation*} where \begin{equation*} J_2:=\om_{m-1}r_2^{m-1}\int_{t_+}^1 dt\, t^{-1/2}(1-t)^{m/2-3/2}, \end{equation*} \begin{equation*} J_{3;i,j}:=\om_{m-2}r_i^{m-1}\int_{t_-}^{t_+} dt\, t^{-1/2}(1-t)^{m/2-3/2}\, \int_{c_1(r_i,r_j)(t)}^1 dc\,(1-c^2)^{m/2-1}, \end{equation*} $b_m$ is given by \eqref{20}, $\om_k$ is given by \eqref{30}, $t_-$ and $t_+$ are defined in the descriptions of Cases 1 and 2, and $c_1(r_1,r_2)(t)$ is defined in \eqref{40}.

Using the binomial expansions of $(1-c^2)^{m/2-1}$, of the latter expression for $c_1(r_1,r_2)(t)$ in \eqref{40}, and of $(1-t)^p$, one can further express the expression \eqref{!} of the étendue measure $E$ of $C$ as the ordinary integral of $dt\,t^{-1/2}$ times a triple hypergeometric-like series in powers of $t$.

If $m$ is even, this triple series is a finite sum, and then the integral expressing $E$ is an elementary function.

In particular, for $m=2$ we get the same result as the one found in the other answer.

For $m=4$, we get \begin{align*} &E \\ &= \frac{4}{3} \pi ^2 \Big((r_1^3+r_2^3) \sin^{-1}(\frac{\sqrt{d^2-(r_1-r_2){}^2}}{d}) \\ &-(r_1^3+r_2^3) \sin ^{-1}(\frac{\sqrt{d^2-(r_1+r_2){}^2}}{d}) +2r_2^3 \sin ^{-1}(\frac{r_1-r_2}{d})\Big) \\ &-\frac{4 \pi ^2\sqrt{d^2-(r_1+r_2){}^2} (2 d^4-2 d^2 (7 r_1^2-r_2 r_1+7 r_2^2)-3 (r_1+r_2){}^2 (r_1^2-3 r_2 r_1+r_2^2))}{45 d^2} \\ &+\frac{4 \pi ^2\sqrt{d^2-(r_1-r_2){}^2} (2 d^4-2 d^2 (7 r_1^2+r_2 r_1+7 r_2^2)-3 (r_1-r_2){}^2 (r_1^2+3 r_2r_1+r_2^2))}{45 d^2}. \end{align*}

Answer 2

Regarding this comment and this comment: Here is a solution for $m=2$:

Without loss of generality, the disk of radius $r_1$ is centered at $(0,0)$, the disk of radius $r_2$ is centered at $(d,0)$, and $r_1>r_2>0$. We also are given the condition $d>r_1+r_2$.

Then the line $y=kx+b$ intersects both (say open) disks iff exactly one of the following three cases takes place:

• $-k_2< k\le-k_1\ \&\ b_{11}<b<b_{12}$
• $-k_1< k\le k_1\ \&\ b_{21}<b<b_{22}$
• $k_1< k<k_2\ \&\ b_{31}<b<b_{32}$

where $$k_1:=\frac{r_1-r_2}{\sqrt{d^2-(r_1-r_2){}^2}}, \quad k_2:=\frac{r_1+r_2}{\sqrt{d^2-(r_1+r_2){}^2}}$$

• $b_{11}:=-dk-\sqrt{1+k^2}\,r_2$, $b_{12}:=\sqrt{1+k^2}\,r_1$
• $b_{21}:=-dk-\sqrt{1+k^2}\,r_2$, $b_{22}:=-dk+\sqrt{1+k^2}\,r_2$
• $b_{31}:=-\sqrt{1+k^2}\,r_1$, $b_{32}:=-dk+\sqrt{1+k^2}\,r_2$

-- see the Detail at the end of this answer.

So, writing $k=\tan t$ for $t\in(-\pi/2,\pi/2)$ and letting $t_j:=\arctan k_j$ for $j=1,2$, we see that the quantity of interest is \begin{align} &\int_{-t_2}^{-t_1}dt\,(b_{12}-b_{11})\cos t \\ &+\int_{-t_1}^{t_1}dt\,(b_{22}-b_{21})\cos t \\ &+\int_{t_1}^{t_2}dt\,(b_{32}-b_{31})\cos t \\ &=2\int_{t_1}^{t_2}dt\,(b_{32}-b_{31})\cos t \\ &+2\int_0^{t_1}dt\,(b_{22}-b_{21})\cos t \\ &=2\sqrt{d^2-(r_1+r_2)^2} \\ &-2\sqrt{d^2-(r_1-r_2)^2} \\ &+2(r_1+r_2) \tan^{-1}\frac{r_1+r_2}{\sqrt{d^2-(r_1+r_2){}^2}} \\ &-2(r_1-r_2) \tan^{-1}\frac{r_1-r_2} {\sqrt{d^2-(r_1-r_2)^2}}. \end{align}

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Detail on the three cases: That the consideration reduces to those three cases was obtained by solving a rather simple system of algebraic equalities and inequalities. I think it is possible to do this by hand, but I did it with Mathematica: