0
$\begingroup$

Find the probability that given two random points on a sphere of radius $k$, their distance is at most $d,$ where $0\leq d \leq 2k.$

Obviously the probability function is increasing in $d$. By scaling, we may assume WLOG that the sphere has radius $1.$ So we want to find the probability of the distance between the two points being at most $\frac{d}k.$ But I have no idea how to compute it. Maybe considering the probability a point is of the form $(x,0,0)$ given that it is a distance $r$ from the center of the sphere might be useful? I know some integral will definitely be necessary here, perhaps involving some variable $r$ that could represent the distance from the first point to the center of the sphere, which ranges from $0$ to $r$.

$\endgroup$
5
  • 3
    $\begingroup$ Just compute the area of the spherical cap as defined by the distance and divide it by the total area of the sphere. $\endgroup$ Sep 19 '20 at 1:20
  • $\begingroup$ Do you have $k=r=x$ here? $\endgroup$
    – Henry
    Sep 19 '20 at 1:21
  • $\begingroup$ @Henry no. At the beginning, I already mentioned scaling the circle. $\endgroup$
    – user747916
    Sep 19 '20 at 1:51
  • $\begingroup$ Also, @DavidG.Stork what do you even mean? I don't even know what the "spherical cap" you're talking about is. $\endgroup$
    – user747916
    Sep 19 '20 at 1:51
  • $\begingroup$ en.wikipedia.org/wiki/Spherical_cap $\endgroup$
    – Henry
    Sep 19 '20 at 2:03
0
$\begingroup$

Not a full answer, but a graphical response to the OP:

enter image description here

Take one of the two points on the sphere and arbitrarily rotate the sphere so that point lies on the top. This then defines a (blue) "spherical cap" whose radius is your distance. If the second point lies in the cap, it is closer than the distance, otherwise, it is outside that distance.

That second point is equally likely to be anywhere on the sphere.

The probability it is closer than your distance is given by the ratio of the area of the blue spherical cap to the area of the sphere... do you see that?

$\endgroup$
2
  • $\begingroup$ Just ONE cap ?. $\endgroup$ Sep 19 '20 at 6:43
  • $\begingroup$ Of course. One point is at its center, the other point can be anywhere within the cap to be closer than the stated distance. Imagine this problem were done on a flat field. How would you solve that? Then the transfer to a sphere is trivial. $\endgroup$ Sep 19 '20 at 15:46
0
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\iint_{S}{1 \over 4\pi a^{2}} \iint_{S}{1 \over 4\pi a^{2}} \bracks{\vphantom{\Large A}\verts{\vec{r} - \vec{r}\, '} < d}\dd^{2}\vec{r}\, '\dd^{2}\vec{r}} \\[5mm] = &\ {1 \over 16\pi^{2}a^{4}}\iint_{S}\int_{0}^{2\pi}\int_{0}^{\pi} \bracks{\vphantom{\Large A}\root{a^{2} -2a^{2}\cos\pars{\theta'} + a^{2}} < d}\ \times \\[2mm] &\ \phantom{{1 \over 16\pi^{2}a^{4}}} a^{2}\sin\pars{\theta'}\dd\theta'\dd\phi' \dd^{2}\vec{r} \\[5mm] = &\ {1 \over 8\pi a^{2}}\iint_{S}\int_{0}^{\pi} \bracks{\vphantom{\Large A}\sin\pars{\theta' \over 2} < {d \over 2a}} \sin\pars{\theta'}\dd\theta'\dd^{2}\vec{r} \\[5mm] = &\ {1 \over 8\pi a^{2}}\iint_{S}\ \underbrace{\int_{0}^{2\arcsin\pars{d/\bracks{2a}}} \sin\pars{\theta'}\dd\theta'} _{\ds{d^{2} \over 2a^{2}}}\dd^{2}\vec{r} = {1 \over 8\pi a^{2}}\,4\pi a^{2}\,{d^{2} \over 2a^{2}} \\[5mm] = &\ \bbx{\pars{d \over 2a}^{2}} \\ &\ \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy