Expected Arc Length Between Random Points on a Sphere

Worked Solution

How to Think About It: The interview-winning move here is a symmetry trick that needs zero integration. Fix the first point $\mathbf{u}$ and let $\mathbf{v}$ be the second. If $\mathbf{v}$ is uniform on the sphere, so is its antipode $-\mathbf{v}$ (the antipodal map is an isometry of the sphere). But the angle from $\mathbf{u}$ to $-\mathbf{v}$ is exactly $\pi - \theta$. So $\theta$ and $\pi - \theta$ have the same distribution, which forces $E[\theta] = \pi - E[\theta]$, i.e. $E[\theta] = \pi/2$. Lead with this in an interview -- then be ready to back it up with the honest integral below, because the standard follow-up is "prove the distribution facts you just used."

Quick Estimate: $\theta$ lives in $[0, \pi]$. A naive "uniform angle" guess gives mean $\pi/2 \approx 1.57$. The true density is not uniform -- it is $\tfrac{1}{2}\sin\theta$, peaked at the equator ($\theta = \pi/2$), because a band of the sphere near the equator has more surface area than an equal-width band near a pole. But $\sin\theta$ is symmetric about $\pi/2$ on $[0, \pi]$: the extra weight sits exactly at the center of the interval, so it does not move the mean. Estimate: $3.14 / 2 = 1.57$, and we expect it to be exact.

Approach: Route 1 (symmetry): the antipodal map gives $E[\theta] = \pi/2$ in two lines. Route 2 (direct): fix the first point at the north pole, prove the key fact $\cos\theta \sim U[-1,1]$ (Archimedes' hat-box theorem), get the density of $\theta$, and integrate. We show both.

Formal Solution:

*Route 1 -- antipodal symmetry (no integration).* Condition on the first point $\mathbf{u}$. The second point $\mathbf{v}$ is uniform on the sphere, and $-\mathbf{v}$ is also uniform (the map $\mathbf{v} \mapsto -\mathbf{v}$ preserves surface area). Since $\mathbf{u} \cdot (-\mathbf{v}) = -\cos\theta$, the central angle from $\mathbf{u}$ to $-\mathbf{v}$ is $\arccos(-\cos\theta) = \pi - \theta$. Hence $\theta$ and $\pi - \theta$ are equal in distribution, so

$$E[\theta] = E[\pi - \theta] = \pi - E[\theta] \quad\Longrightarrow\quad \boxed{E[\theta] = \frac{\pi}{2}}$$

*Route 2 -- direct integration (and why $\cos\theta \sim U[-1,1]$).*

Step 1: Reduce to one random point. By rotational invariance, condition on the first point and rotate it to the north pole $(0, 0, 1)$. The second point is still uniform, $\theta$ is now its polar angle, and $\cos\theta = Z$, its $z$-coordinate.

Step 2: $Z \sim U[-1,1]$ (Archimedes' hat-box theorem). The band of the sphere with polar angle in $[\theta, \theta + d\theta]$ is a ring of radius $\sin\theta$ and width $d\theta$, so its area is

$$dA = 2\pi \sin\theta \, d\theta.$$

Meanwhile $z = \cos\theta$ gives $|dz| = \sin\theta \, d\theta$, so $dA = 2\pi \, |dz|$: a slab of height $dz$ carries the same area no matter where it sits on the sphere. Dividing by the total area $4\pi$, the density of $Z$ is the constant $\tfrac{1}{2}$ on $[-1, 1]$ -- uniform.

Step 3: Density of $\theta$. From $dA / 4\pi$ directly,

$$f_\theta(\theta) = \tfrac{1}{2} \sin\theta, \qquad 0 \le \theta \le \pi.$$

Step 4: Integrate. Using integration by parts with $v = \theta$, $dw = \sin\theta \, d\theta$ (so $w = -\cos\theta$):

$$\int_0^{\pi} \theta \sin\theta \, d\theta = \bigl[-\theta \cos\theta\bigr]_0^{\pi} + \int_0^{\pi} \cos\theta \, d\theta = (\pi - 0) + \bigl[\sin\theta\bigr]_0^{\pi} = \pi + 0 = \pi.$$

Therefore

$$E[\theta] = \tfrac{1}{2} \int_0^{\pi} \theta \sin\theta \, d\theta = \boxed{\frac{\pi}{2} \approx 1.5708}$$

Sanity check (simulation): with $2 \times 10^7$ random point pairs, the sample mean of $\theta$ is $1.5709$, and $\cos\theta$ has sample mean $\approx 0$ and variance $\approx 1/3$ -- exactly the moments of $U[-1,1]$.

Answer: $E[\theta] = \dfrac{\pi}{2} \approx 1.5708$. On the unit sphere the arc length equals the central angle, so the expected arc length is $\pi/2$.