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Let $\mathbb{S}^n$ be the round sphere, $p_0$ be the north-pole and $T$ be the equator whose center is $p_0$. Given a point $p$ on the sphere, how do we calculate the (spherical, intrinsic) distance of $p$ to $T$? It certainly has to do with the inner product $\langle p, p_0 \rangle$, but I am unsure about the explicit formula. If $p$ is next to $\pm p_0$, $d(p, T)$ is next to $\pi/2$, and if $p$ is next to $T$, obviously $d(p, T)$ is next to $0$. And it is easy to see that $d(p, T)$ is even with respect to $\langle p, p_0 \rangle$. I suspect $d(p, T) = |\arcsin \langle p, p_0 \rangle|$. Is this correct?

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Let $p$ and $q$ be two points on the surface of the sphere, of radius $R$, subtending a central angle of $\theta$ (along any great circle through $p$ and $q$). Then $$ \frac{\langle p,q \rangle}{||p|| \, ||q||} = \frac{\langle p,q \rangle}{R^2} = \cos \theta \text{.} $$

Any great circle through $p_0$ meets $T$ perpendicularly. So the angle $\theta$ satisfying $$ \frac{\langle p_0,p \rangle}{R^2} = \cos \theta $$ is the angle between $p_0$ and $p$ along a great circle perpendicular to $T$. If $0 \leq \theta\leq \pi/2$, the (smallest) angle between $p$ and $T$ is $\varphi = \frac{\pi}{2} - \theta$. If $\pi/2 \leq \theta \leq \pi$, then $\varphi = \theta - \pi/2$. (It can help to draw a picture here, where in one case $p$ is in the same hemisphere bounded by $T$ as is $p_0$ and in the other case, $p$ is in the opposite hemisphere.) Combining all this, $$ \varphi = \left| \frac{\pi}{2} - \theta \right| = \left| \frac{\pi}{2} - \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \right| \text{.} $$

Finally, the distance along the great circle through $p$ perpendicular to $T$ is $\varphi R$, so $$ d(p,T) = R \left| \frac{\pi}{2} - \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \right| \text{.} $$

If this is the unit sphere, $R = 1$ and this simplifies a bit.


Suppose we now want to solve that for $\langle p_0,p \rangle$. \begin{align*} \frac{d(p,T)}{R} &= \left| \frac{\pi}{2} - \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \right| \\ &\in \pm \left( \frac{\pi}{2} - \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \right) \text{.} \end{align*} Then \begin{align*} \cos\left( \frac{d(p,T)}{R} \right) &= \cos \pm \left( \frac{\pi}{2} - \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \right) \\ &= \cos \left( \frac{\pi}{2} - \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \right) && \text{cosine is even} \\ &= \cos \frac{\pi}{2} \cos \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) + \sin \frac{\pi}{2} \sin \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \\ &= \sin \cos^{-1}\left( \frac{\langle p_0,p \rangle}{R^2} \right) \text{.} \end{align*} The range of arccosine is $[0, \pi]$, and sine is nonnegative for all those angles, so \begin{align*} \cos\left( \frac{d(p,T)}{R} \right) &= \frac{+\sqrt{R^4 - \langle p_0,p \rangle^2}}{R^2} \\ &= \sqrt{1 - \left( \frac{\langle p_0,p \rangle}{R^2} \right)^2} \text{,} \end{align*} so $$ \left( \frac{\langle p_0,p \rangle}{R^2} \right)^2 = 1 - \cos^2 \left( \frac{d(p,T)}{R} \right) $$ and then \begin{align*} \langle p_0,p \rangle &= \pm R^2 \sqrt{1 - \cos^2 \left( \frac{d(p,T)}{R} \right)} \\ &= \pm R^2 \sqrt{\sin^2 \left( \frac{d(p,T)}{R} \right)} \\ &= \pm R^2 \sin \left( \frac{d(p,T)}{R} \right) \text{.} \end{align*} This is the best we can do, because $d(p,T)$ doesn't contain information about which hemisphere $p$ is in. (This missing information was first indicated by the "$\pm$".) We choose the positive root if $p$ is in the same hemisphere as $p_0$ and the negative root if $p$ is in the other hemisphere (and the square root has the value zero if $p$ is on $T$, so either choice of sign is fine).

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  • $\begingroup$ Can we solve for $\langle p,p_0 \rangle$? $\endgroup$ – Eduardo Longa Dec 1 '17 at 14:38
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    $\begingroup$ @EduardoLonga : THat would be a new question. However, I'll edit in a second answer. $\endgroup$ – Eric Towers Dec 1 '17 at 14:51
  • $\begingroup$ we can further simplify the last square root to get a sine, right? $\endgroup$ – Eduardo Longa Dec 1 '17 at 20:12
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    $\begingroup$ @EduardoLonga : Yes we can and I have (now, together with a dropped square of $R$ near the end). The result shouldn't be so surprising; The angle between $p$ and $p_0$ is complementary to the angle between $p$ and $T$, so while $\langle p, p_0 \rangle$ is proportional to $\cos \theta$, it is also proportional to the complementary cosine (that is, sine) of the complementary angle. $\endgroup$ – Eric Towers Dec 1 '17 at 21:37

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