Picking a point on a sphere uniformly at random

Question

I wanted to read a bit about the Borel-Kolmogorov paradox on Wikipedia, but I don't understand the following: to me it seems that picking a point uniformly at random on the sphere amounts to pick a longitude $\lambda$ uniformly at random from $[-\pi,\pi]$ and picking a latitude $\phi$ uniformly at random from $[-\pi/2,\pi/2]$. However, on Wikipedia it's written we pick the latitude $\phi$ from $[-\pi/2,\pi/2]$ with density $\frac 1 2 \cos(\phi)$. Why is that?

Answer 1

We have this density on the latitude essentially because there is more land near the equator, and thus we want to give the latitudes near the equator more weight.

To find the density yourself, I like to use $[0,\pi]$ for the latitude coordinate, where $\phi = 0$ or $\phi = \pi$ corresponds to the $z$-axis, since it corresponds with the polar coordinate system I was taught in multivariable calculus. Then you'll recall from multivariable calculus that $\rho^2\sin\phi\,d\rho\,d\phi\,d\lambda$ is the volume element of $\mathbb R^3$, written in these coordinates, where $(\rho,\phi,\lambda)\in (0,\infty)\times[0,\pi]\times [-\frac\pi2,\frac\pi2]$.

Evaluating at $\rho = 1$ gives us $\sin\phi\,d\phi\,d\lambda$. Dividing this by $4\pi$, which is the area of the sphere, we get the uniform distribution on the sphere: $\frac{1}{4\pi}\sin\phi\,d\phi\,d\lambda = (\frac{1}{2}\sin\phi\,d\phi)(\frac{1}{2\pi}\,d\lambda)$. Or in other words, to sample a point $(\phi,\lambda)$ uniformly at random from the sphere, we sample the latitude $\phi$ from $[0,\pi]$ from a distribution with density $\frac{1}{2}\sin\phi$ and we sample the longitude $\lambda$ from $[-\pi,\pi]$ uniformly, i.e., from a distribution with density $\frac{1}{2\pi}$.

• When $\phi \approx \pi/2$, which corresponds to points near the equator, $\sin\phi\approx 1$, so these points have the most weight.
• When $\phi\approx 0$, or $\phi\approx \pi$, $\sin\phi\approx 0$, so these points have much less weight.

Both of the last two bullet points make sense if you think about how there is so much more land in a latitudinal strip of width $d\phi$ near the equator, versus near the poles.

Answer 2

A picture is worth a thousand words:

The density in the "naive" approach is proportional to $\frac{1}{\cos \phi}$...

Answer 3

Visualize the points on the sphere given by every combination of $\lambda \in\{\pi, \pi -\frac{2\pi}{n}, \pi - 2\frac{2\pi}{n}, ..., -\pi + \frac{2\pi}{n}\}$ and $\phi\in\{\frac{\pi}{2},\frac{\pi}{2}-\frac{\pi}{m}, \frac{\pi}{2} - 2\frac{\pi}{m},...,-\frac{\pi}{2} + \frac{\pi}{m}\}$. According to your hypothesis, these points are distributed uniformly across the sphere (i.e, equal density per unit area). However, as you approach the poles, the latitudinal distance between the points will remain constant, but the longitudinal distance will decrease, meaning there will be more point density at the poles.

To offset this, we will need to make points more sparse towards the poles. It is not hard to see that the number of points on a given circle of latitude will have to be proportional to the circumference of that circle. However, a circle at latitude $\phi$ has circumference $2\pi r\cos(\phi)$ (where $r$ is the radius of the sphere), leading to the factor of $\cos(\phi)$ in the latitude distribution.