Distances on the unit sphere

Yuta Okamoto

2026-11-04

Abstract

A sample post that shows every statement environment of the site, with short statements about the great-circle distance and the chordal distance on the unit sphere.

This post shows every statement environment of the site. The statements with a filled background are marked as new only to show the style.

Setting

Let $\left\langle x, y \right\rangle = x_1 y_1 + x_2 y_2 + x_3 y_3$ be the standard inner product on $\mathbb{R}^3$ and $\left\lVert x \right\rVert = \left\langle x, x \right\rangle^{1/2}$ the Euclidean norm. Let $S^2 = \left\lbrace x \in \mathbb{R}^3 : \left\lVert x \right\rVert = 1 \right\rbrace$ be the unit sphere. For $p, q \in S^2$ , let $d_{\mathbb{R}^3}\left(p, q\right) = \left\lVert p - q \right\rVert$ be the chordal distance.

Definition 1 Great-circle distance. For $p, q \in S^2$ , the Cauchy–Schwarz inequality gives $\left\lvert \left\langle p, q \right\rangle \right\rvert \leq 1$ . The great-circle distance between $p$ and $q$ is $d_{S^2}\left(p, q\right) = \arccos \left\langle p, q \right\rangle$ , where $\arccos : \left[-1, 1\right] \to \left[0, \pi\right]$ .

Figure 1: The great circle through

p

and

q

, and the angle

\theta = d_{S^2}\left(p, q\right)

. The thick arc has length

d_{S^2}\left(p, q\right)

and the dashed segment has length

d_{\mathbb{R}^3}\left(p, q\right)

Assumption 2 Hemisphere. The points $p, q \in S^2$ satisfy $\left\langle p, q \right\rangle \geq 0$ .

Chord and arc

Lemma 3. For all $p, q \in S^2$ , $d_{\mathbb{R}^3}\left(p, q\right)^{2} = 2 - 2 \cos d_{S^2}\left(p, q\right)$ holds.

Proof. By the bilinearity and the symmetry of the inner product, $\left\lVert p - q \right\rVert^{2} = \left\lVert p \right\rVert^{2} - 2 \left\langle p, q \right\rangle + \left\lVert q \right\rVert^{2}$ holds. Since $\left\lVert p \right\rVert = \left\lVert q \right\rVert = 1$ , the right-hand side equals $2 - 2 \left\langle p, q \right\rangle$ . By Definition 1, $\cos d_{S^2}\left(p, q\right) = \left\langle p, q \right\rangle$ holds.□

Theorem 4 Chord and arc. For all $p, q \in S^2$ , the following identity holds.

$d_{\mathbb{R}^3}\left(p, q\right) = 2 \sin \frac{d_{S^2}\left(p, q\right)}{2}$ (1)

Proof of Theorem 4. Let $\theta = d_{S^2}\left(p, q\right)$ . By Lemma 3 and the double-angle formula $\cos \theta = 1 - 2 \sin^{2} \left(\theta / 2\right)$ , the following holds.

$d_{\mathbb{R}^3}\left(p, q\right)^{2} = 2 - 2 \cos \theta = 4 \sin^{2} \frac{\theta}{2}$

Since $0 \leq \theta \leq \pi$ , $\sin \left(\theta / 2\right) \geq 0$ holds, and taking the nonnegative square roots gives Equation 1.□

Question 5. For $p, q \in S^2$ with $p \neq q$ , how small can the ratio $d_{\mathbb{R}^3}\left(p, q\right) / d_{S^2}\left(p, q\right)$ be?

Proposition 6. For all $p, q \in S^2$ , the following inequalities hold.

$\frac{2}{\pi} d_{S^2}\left(p, q\right) \leq d_{\mathbb{R}^3}\left(p, q\right) \leq d_{S^2}\left(p, q\right)$

Proof. Let $t = d_{S^2}\left(p, q\right) / 2$ , so that $0 \leq t \leq \pi / 2$ . By Theorem 4, $d_{\mathbb{R}^3}\left(p, q\right) = 2 \sin t$ holds, so it suffices to show $\left(2 / \pi\right) t \leq \sin t \leq t$ . The function $s \mapsto s - \sin s$ vanishes at $s = 0$ and has the nonnegative derivative $1 - \cos s$ , so $\sin t \leq t$ holds. The second derivative $-\sin s$ of $\sin$ is nonpositive for $0 \leq s \leq \pi / 2$ , so $\sin$ is concave on $\left[0, \pi / 2\right]$ . Let $\lambda = 2 t / \pi$ , so that $0 \leq \lambda \leq 1$ . The concavity at the points $0$ and $\pi / 2$ gives the following inequality.

$\sin t = \sin \left(\lambda \frac{\pi}{2} + \left(1 - \lambda\right) 0\right) \geq \lambda \sin \frac{\pi}{2} + \left(1 - \lambda\right) \sin 0 = \frac{2}{\pi} t$ □

Corollary 7. Let $p, q \in S^2$ satisfy Assumption 2. Then $d_{S^2}\left(p, q\right) \leq \pi / 2$ and $d_{\mathbb{R}^3}\left(p, q\right) \leq \sqrt{2}$ hold.

Proof. Since $\arccos$ is decreasing and $\left\langle p, q \right\rangle \geq 0$ , $d_{S^2}\left(p, q\right) = \arccos \left\langle p, q \right\rangle \leq \arccos 0 = \pi / 2$ holds. Since $\sin$ is increasing on $\left[0, \pi / 4\right]$ , Theorem 4 gives $d_{\mathbb{R}^3}\left(p, q\right) = 2 \sin \left(d_{S^2}\left(p, q\right) / 2\right) \leq 2 \sin \left(\pi / 4\right) = \sqrt{2}$ .□

Examples

Example 8. Let $p = \left(1, 0, 0\right)$ and $q = \left(-1, 0, 0\right)$ . Then $\left\langle p, q \right\rangle = -1$ , so $d_{S^2}\left(p, q\right) = \pi$ and $d_{\mathbb{R}^3}\left(p, q\right) = 2$ hold. Both sides of the left inequality of Proposition 6 equal $2$ for this pair, so the coefficient $2 / \pi$ cannot be replaced by a larger number.

Example 9. Let $p = \left(1, 0, 0\right)$ and $q = \left(0, 1, 0\right)$ . Then $\left\langle p, q \right\rangle = 0$ , so the pair satisfies Assumption 2, and $d_{S^2}\left(p, q\right) = \pi / 2$ and $d_{\mathbb{R}^3}\left(p, q\right) = \sqrt{2}$ hold. Hence both bounds of Corollary 7 are attained.

Check 10. Let $p = \left(1, 0, 0\right)$ and $q = \left(1 / \sqrt{2}, 1 / \sqrt{2}, 0\right)$ . Compute both sides of Equation 1.

Solution. Since $\left\langle p, q \right\rangle = 1 / \sqrt{2}$ , $d_{S^2}\left(p, q\right) = \pi / 4$ holds. The coordinates and the formula $\sin^{2} \left(\theta / 2\right) = \left(1 - \cos \theta\right) / 2$ for $\theta \in \mathbb{R}$ give the following.

$d_{\mathbb{R}^3}\left(p, q\right)^{2} = \left(1 - \frac{1}{\sqrt{2}}\right)^{2} + \left(\frac{1}{\sqrt{2}}\right)^{2} = 2 - \sqrt{2} = 4 \sin^{2} \frac{\pi}{8}$

Hence both sides equal $\left(2 - \sqrt{2}\right)^{1/2}$ .

Check 11. For $p \in S^2$ , compute $d_{S^2}\left(p, p\right)$ and $d_{\mathbb{R}^3}\left(p, p\right)$ .

Solution. Since $\left\langle p, p \right\rangle = \left\lVert p \right\rVert^{2} = 1$ , $d_{S^2}\left(p, p\right) = \arccos 1 = 0$ holds. By definition, $d_{\mathbb{R}^3}\left(p, p\right) = \left\lVert p - p \right\rVert = 0$ holds.

Problems

Question 12. For which pairs $p, q \in S^2$ does $d_{\mathbb{R}^3}\left(p, q\right) = d_{S^2}\left(p, q\right)$ hold?

Exercise 13. Show that $d_{\mathbb{R}^3}\left(p, q\right) \leq 2$ holds for all $p, q \in S^2$ , with equality if and only if $q = -p$ .

Hint. Use Lemma 3 and compute $\left\lVert p + q \right\rVert^{2}$ .

Exercise 14. Let $A$ be a real $3 \times 3$ matrix with $A^{\top} A = I$ , where $A^{\top}$ is the transpose of $A$ and $I$ is the identity matrix. Show that $A p \in S^2$ and $d_{S^2}\left(A p, A q\right) = d_{S^2}\left(p, q\right)$ hold for all $p, q \in S^2$ .

Remark 15. The function $\theta \mapsto 2 \sin \left(\theta / 2\right)$ is strictly increasing on $\left[0, \pi\right]$ . Hence, by Theorem 4, for all $p, q, r, s \in S^2$ , $d_{S^2}\left(p, q\right) \leq d_{S^2}\left(r, s\right)$ holds if and only if $d_{\mathbb{R}^3}\left(p, q\right) \leq d_{\mathbb{R}^3}\left(r, s\right)$ holds.

Open Problem 16 Tammes problem. Determine, for every integer $N \geq 2$ , the maximum of $\min_{1 \leq i < j \leq N} d_{S^2}\left(p_i, p_j\right)$ over $p_1, \dots, p_N \in S^2$ .

Open Problem 17 Thomson problem. Determine, for every integer $N \geq 2$ , the minimum of $\sum_{1 \leq i < j \leq N} 1 / d_{\mathbb{R}^3}\left(p_i, p_j\right)$ over pairwise distinct $p_1, \dots, p_N \in S^2$ .

Solutions

Solution of Exercise 13. By Lemma 3, $d_{\mathbb{R}^3}\left(p, q\right)^{2} = 2 - 2 \cos d_{S^2}\left(p, q\right) \leq 4$ holds, so $d_{\mathbb{R}^3}\left(p, q\right) \leq 2$ . Equality holds if and only if $\left\langle p, q \right\rangle = \cos d_{S^2}\left(p, q\right) = -1$ . Since $\left\lVert p + q \right\rVert^{2} = 2 + 2 \left\langle p, q \right\rangle$ , this holds if and only if $p + q = 0$ , that is, $q = -p$ .

Solution of Exercise 14. Since $A^{\top} A = I$ , $\left\langle A x, A y \right\rangle = \left\langle x, A^{\top} A y \right\rangle = \left\langle x, y \right\rangle$ holds for all $x, y \in \mathbb{R}^3$ . Taking $x = y = p$ gives $\left\lVert A p \right\rVert = \left\lVert p \right\rVert = 1$ , so $A p \in S^2$ . Taking $x = p$ and $y = q$ gives $d_{S^2}\left(A p, A q\right) = \arccos \left\langle A p, A q \right\rangle = \arccos \left\langle p, q \right\rangle = d_{S^2}\left(p, q\right)$ .