5

How can I find an average distance between two points lying on surface of a sphere of a certain radius?

More importantly : can knowing the average distance between two points on surface of a disk ( this question has already an answer on MSE) be useful to answer the question about average distance between two points on surface of sphere? Or there are no immediate obvious relationship/generalization between the two question?

jimjim
  • 9,675
  • 2
    If you mean what I think you mean, then you just want to consider a fixed point and average the distances along a great circle passing through that point. At that point, you can just drop down to the 2D problem of the average distance along a circle. – Cameron Williams Apr 27 '15 at 14:17
  • @CameronWilliams : How can I generalise that to spheres of n dimesion? That is the motivation for asking this question. – jimjim Apr 27 '15 at 14:28
  • I could be wrong but if you make a slice along a great circle of a sphere in $\Bbb R^n$, I think you get a sphere in $\Bbb R^{n-1}.$ So, I think you can just keep reducing the problem all the way down to the case of the circle. I could be wrong though. – Cameron Williams Apr 27 '15 at 14:35
  • 1
    I don't think the fixed-point-and-single-great-circle method generalizes well to even a sphere in three-dimensional space. If the points are uniformly distributed on the sphere, the probability density of the angular separation is greater at $\pi/2$ than at $0$ or $\pi$. But if both points are uniformly distributed, then one fixed point and one uniformly distributed point works. I'd try to find $dA/dr$ where $A$ is the measure of the portion of the sphere within $r$ units of the fixed point. – David K Apr 27 '15 at 14:37
  • By distance, I assume you mean the geodesic distance on the sphere. Consider the case the fixed point is the north pole. If you have a point at a distance $r$ from the north pole, its mirror image wrt to the $xy$-plane is at a distance $\pi - r$. Since the reflection wrt $xy$-plane is an isometry, .... – achille hui Apr 27 '15 at 14:50
  • @achillehui : just a curiosity , is there a generalisation that generalises geodesic distance to n dimensions? – jimjim Apr 27 '15 at 15:00
  • @Arjang, for $n$-dim sphere, it is essentially the same thing as the one on a $2$-dim sphere, you construct the great circle between the two points and the geodesic distance is the length of the shorter arc joining those two points along the great circle. – achille hui Apr 27 '15 at 15:02
  • @DavidK: I would agree with you that you should not take the points as being uniformly distributed on a great circle, for the reason you give, but there is an argument that they are symmetrically distributed, making the mean surface distance between two points $\pi r / 2$. – Henry Apr 27 '15 at 15:57
  • @CameronWilliams: The harder (and more analogous) question is, what is the average distance between two points in the unit ball? – Brian Tung Apr 27 '15 at 16:48
  • @BrianTung $\frac{36}{35}$, see answers of this. – achille hui Apr 27 '15 at 17:16
  • Right, I'm directing OP to a question I think is more analogous. – Brian Tung Apr 27 '15 at 17:37
  • 1
    @Henry I was not assuming geodesic distance. On second thought, it is clear that the "great circle" method did assume geodesic distance, and not clear at all that the OP did not so assume. The answer really depends on what the desired kind of distance measurement is; Euclidean distance in $\mathbb R^n$ gives a different result, for example. – David K Apr 27 '15 at 18:23

1 Answers1

3

Without loss of generality, assume the first point is at the "north pole"; also without loss of generality, assume the second point is along the "prime meridian." Then the probability of being at "latitude" $x$ degrees north is equal to the probability of being at "latitude" $x$ degrees south (and is proportional to $\cos x$). Therefore, the average latitude is at the "equator," and the average distance is $\pi r/2$, as stated by Henry and achille hui.

Brian Tung
  • 34,160