A large plastic ball full of holes is given. (So the holes are in a plastic shell.) Straight threads connect these holes randomly, by passing through the interior of the ball/shell.

For a big ball or shell, say a meter in size, with thousands of holes, this makes (1/2 times) thousands of straight threads inside it. (Each hole has the diameter of the thread, so that each hole can only have one string passing through it.)

Now the question: Inside the ball/shell (assumed to be large), is the density of the random threads homogeneous, or does it depend on the radius?

  • $\begingroup$ Do you mean holes in a spherical shell? Can a string go from hole 1 to hole 2 and another one from hole 1 to hole 3? On other words, is more than one string per hole allowed? $\endgroup$ – G. Smith Mar 26 at 5:27
  • $\begingroup$ I see you added the cosmology tag. Does this model have anything to do with cosmic strings? $\endgroup$ – G. Smith Mar 26 at 6:55
  • 4
    $\begingroup$ I think the answer will depend on how you choose your random placement of threads. See Bertrand paradox for a continuous 2D example of what I mean. $\endgroup$ – Ruslan Mar 26 at 9:49
  • 1
    $\begingroup$ @Ruslan: you have 10,000 holes, say. When you put the thread through a hole you choose a random empty hole to connect it to. There really isn't any ambiguity about this probability distribution. $\endgroup$ – Peter Shor Mar 26 at 13:14
  • 1
    $\begingroup$ @Ruslan The Bertrand paradox applies to randomly selecting an element from an infinite set. With a finite set, you can assign each element the same, finite, probability. With an infinite set, you can't assign elements the same finite probability, so you have to have some notion of "measure", and there paradox comes in when there is more than one reasonable measure. There are a finite number of holes, so Bertrand's paradox doesn't apply to choosing the holes for a string. $\endgroup$ – Acccumulation Mar 26 at 14:57

I believe the density is homogeneous throughout the ball.

I did a numerical simulation of this in Mathematica. I assumed the sphere had radius 1 and generated 100,000 pairs of random points on it, each pair to be connected with string. Then I analyzed this set of random strings to see how much total mass (i.e., length of string) lay between $r$ and $r+dr$ in various spherical shells with radii (0.1, 0.2, …, 0.9), using a bit of geometry. Then I divided by the square of the radius of the spherical shell to get the volume density and plotted it. The 9 points lay almost on a horizontal line:

enter image description here

The horizonal axis is the radial coordinate and the vertical axis is the mass density.


Here is an analytic proof that the density is homogeneous, based on @Gec's answer. I agree with his approach but not his former result.

Take the sphere to have unit radius and the strings to have unit linear mass density so that the mass of a small segment is just the length of that segment.

As Gec points out, a string can be characterized by the angle it subtends, which I'm going to call $\theta$. A string has a minimum radial distance of $\cos{(\theta/2)}\equiv a$ and a length of $2\sin{(\theta/2)}=2\sqrt{1-a^2}$.

Introduce a linear coordinate $s$ along the string, measured from its midpoint. Then one has $a^2+s^2=r^2$ so


Differentiating with respect to $r$, we find


This tells us that the mass of this string that lies within a spherical shell between $r$ and $r+dr$ is


(The string passes through the shell on both sides of its center.)

We can check that this is correct by integrating it over $r$ from $a$ to $1$:


which agrees with the length of the string.

Now we need to integrate over strings between random points on a sphere.

As Gec pointed out, the spherical symmetry means that we can consider just strings with one endpoint at the north pole, and the other end at polar angle $\theta$ and azimuthal angle $\phi$. To randomly average a quantity $f$ over the randomly placed other end, we compute $\langle f \rangle=\frac{1}{4\pi}\iint f\,\sin{\theta}\,d\theta\,d\phi$. By azimuthal symmetry, this simplifies to $\frac{1}{2}\int f\,\sin{\theta}\,d\theta$.

To compute the averaged mass $dM$ in a spherical shell between $r$ and $r+dr$, we integrate $dm$ over $\theta$, but only between $2\cos^{-1}r$ and $\pi$. For smaller angles, the string would not pass through the shell and thus would not contribute any mass. So

$$\frac{dM}{dr}=\int_{2\cos^{-1}r}^\pi \frac{r\sin{\theta}\,d\theta}{\sqrt{r^2-\cos^2{(\theta/2)}}}$$

The substitution $u=\cos{(\theta/2)}$ simplifies this integral to


To get the volume mass density $\rho=dM/dV$, we divide by the area of the spherical shell, $4\pi r^2$, to get a homogeneous density of


My numerical simulation gave $2$ rather than $1/\pi$ because (1) I didn’t multiply by 2 to take into account that a string passes through a shell on both sides of its midpoint, and (2) at the end I divided by $r^2$ rather than $4\pi r^2$.

  • $\begingroup$ Awesome - even though I expected something different, namely a decrease with radius. $\endgroup$ – frauke Mar 26 at 7:03
  • $\begingroup$ Well, the question is whether I did the geometry right. But it seems unlikely that if I got it wrong the density would have turned out constant. The calculation was nontrivial. $\endgroup$ – G. Smith Mar 26 at 7:05
  • $\begingroup$ I first calculated the midpoint of the string. Call its distance from the center $a$. That string only contributes mass to shells at $r>a$. Then I calculated the points where the string passed through radius $r$ and $r+dr$ and took the distance between them. I had found analytically that this was $r\,dr/\sqrt{r^2-a^2}$. So this is proportional to the mass $dm$ of this string in this shell. $\endgroup$ – G. Smith Mar 26 at 7:13
  • $\begingroup$ So you connected random points on the sphere? But what if you use a "random midpoint" method (from Bertrand paradox): choose a random point inside the ball, and then generate a chord with this point being the midpoint of the chord. I think you'll get a different result. $\endgroup$ – Ruslan Mar 26 at 9:55
  • $\begingroup$ Once you figure out how much mass lies between $r$ and $r+dr$, don't you have to divide by the radius squared to get the volume density? $\endgroup$ – Peter Shor Mar 26 at 13:33

Added. Now, I think that G. Smith gave the right answer to the initial question. And I was solving a different problem. My former solution implies that we chose any thread with equal probability and then uniformly chose the point of this thread. This procedure is not equivalent to the finding of mass distribution. To find the distribution of mass we should choose threads with probabilities proportional to their length. Just because of the lengthy thread contains more mass. Doing like this one obtains a mass distribution with constant density.

Former solution. I obtained the following expression for the density of "matter" inside the sphere of radius 1 $$ \rho(r) = \frac{A}{r}\log\left(\frac{\sqrt{2}(1+r)}{\sqrt{|2r^2+\cos(\varphi(r))-1|}+\sqrt{2}\cos(\varphi(r)/2)}\right). $$ Here $A$ is constant and $\varphi(r) = 2\arcsin(r)$. The valiue of this density at $r=0$ is equal to $A$, and it diverges as $r$ tends to 1.

Upd. This expression is obtained in the following way. For any pair of holes let's draw z-axis through one of them and the center of sphere. Then a position of the second is defined by a polar angle $\varphi\in[0,\pi]$. The angle is random and the corresponding pdf is $w_1(\varphi)=\sin(\varphi)/2$. Uniform distribution of "matter" along the line connecting two holes leads to the following distribution of radius: $$ w_2(r|\varphi) = \frac{r}{\cos(\varphi/2)\sqrt{r^2-\sin^2(\varphi/2)}}, $$ where $r\in[\sin(\varphi/2),1]$. The minimal value of radius along the line is equal to $\sin(\varphi/2)$, hence the definition of $\varphi(r)$. Averaging with respect to angles gives the radius pdf: $$ w_3(r) = \int_0^{\varphi(r)} w_1(\varphi)w_2(r|\varphi)d\varphi . $$ And the density of "matter" is proportional to $w_3(r)/r^2$.

  • $\begingroup$ This answer could be simplified greatly by using the formula $\cos(2\arcsin(x)) = 1-2x^2\,.$ $\endgroup$ – Peter Shor Mar 26 at 15:13
  • $\begingroup$ I think $w_2$ is wrong. If I integrate it over $r$ from $\sin(\varphi/2)$ to 1, which should give the total mass of a string with polar angle $\varphi$, I get 1. But the strings should have different masses. I’m also totally confused because the minimal radius should be $\cos{\varphi/2}$, not $\sin{\varphi/2}$. $\endgroup$ – G. Smith Mar 26 at 17:19
  • $\begingroup$ I think $w_2=r/\sqrt{r^2-\cos^2{(\varphi/2)}}$. $\endgroup$ – G. Smith Mar 26 at 17:27
  • $\begingroup$ @G. Smith, I think I was solving a different problem. And you gave the right answer to the initial question. $\endgroup$ – Gec Mar 26 at 18:26

While the number of strings passing nearby the shell is higher than that of the strings passing through the center, also the constant radius surfaces near the shell are larger than those near the center of the sphere.

We can define the string density $\rho $ through

$$4\pi r^2\rho(r)= N (r) $$

where $N (r) $ is the number of times that the strings intersect the constant radius $r$ surface.

Note that, assuming the holes follow a homogeneous distribution, in the limit of large number of holes you are just connecting random points of the sphere with lines that cross the sphere.

Fixed a point from which the line is drawn, you have equal probability of connecting it to any other point of the sphere.

The line (string) will pass through the center only if the opposite point is chosen.

Conversely every line will pass through the sphere surface and almost every line will pass at a slightly smaller radius.

You can compute the number of lines of a certain length $L $ that can be drawn from a chosen point; even better you can express this using the angle formed by the two points connected and the center of the sphere: $$L=2Rsin\theta\;,\qquad N_L= 2\pi R sin 2\theta$$ The segments corresponding to an angle $\theta$ will contribute to the density for radiai in the range $[Rcos\theta, R]$ with 2 points each except in the case of the minimum radius value (here the string passes only once). Now $N (r) $ will be proportional to

$$ \int_{\theta*}^{\pi/2} 2\pi R sin (2\theta)d\theta $$

where $cos\theta*=\frac {r}{R}$. The proportionality constant is basically the number of the endpoints since you integrate their distribution on the $R $ shell (you also have a factor 2 because each string is counted twice almost everywhere and a factor 1/2 to avoid the overcounting when integrating over endpoints).

The integral gives $$2\pi R \left(\frac {r}{R}\right)^2$$

so that when you compute $\rho (r)$ you indeed get a term which is independent from $r $.

If we were to stop here, the density would be uniform.

One could think that we still need to remove the overcounting of pieces of string in the minimum radius each segment reaches: do we have to subtract from $N (r)$ one counting of the intersection at minimum radius, i.e. the quantity $$2\pi R sin (2\theta*)$$

This would give a part which is dependent on $r $ in the distribution: $$\rho (r)\sim const+\frac {\sqrt{1-(r/R)^2}}{r}$$

The truth is that the term must be subtracted into the integral of $N (r) $ and there gives zero contribute, since its a modification on a set of zero measure.

So in conclusion there is no term to be subtracted and the density is indeed constant. It would be nice to see if there are other endpoints distributions that are mimicked by the string density..

  • $\begingroup$ You seem to have neglected the fact that, while a shell of radius $r$ around the midpoint of the sphere will indeed have less threads passing through it as $r$ decreases, it also has correspondingly less area. A priori, without calculating the expected number of threads passing through the shell, it doesn't seem obvious which effect will dominate -- and, in fact, G. Smith's numerical results seem to suggest that they might cancel out exactly! $\endgroup$ – Ilmari Karonen Mar 26 at 10:28
  • $\begingroup$ Yeah @Ilmari Karonen you are right! Indeed the effect precisely cancels the radial growth. I will soon add the explicit computations $\endgroup$ – france95 Mar 26 at 14:13
  • $\begingroup$ Why not simplify $\sin(\arccos(x))$? $\endgroup$ – Peter Shor Mar 26 at 15:15
  • $\begingroup$ @Peter Shor yes, but more importantly there is no need to subtract that term I think. (See how I edited) $\endgroup$ – france95 Mar 26 at 15:35
  • $\begingroup$ Why is the string density proportional to number of times the string crosses the shell of radius r? Doesn't it depend on the angle between the string and the local slope of the shell? Eg the string could be perpendicular and the intersection would be a circle, or close to parallel and the intersection would be a stretched out elipse. @G.Smith seems to have taken this into account, and got the same answer as you. $\endgroup$ – jwg Mar 30 at 0:06

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.