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Let us consider (in spherical coordinates) the expression

Great arc distance between two points on a unit sphere

$$d({\bf v}_1,{\bf v}_2)=\cos^{-1}\left(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\cos\left(\varphi_1-\varphi_2\right)\right) \qquad (1)$$ that gives the geodesic distance between two points $${\bf v}_i=(\sin\theta_i\cos\varphi_i,\sin\theta_i\sin\varphi_i,\cos\theta_i)$$ on the unit sphere $S$.

A question:

Assume $\,{\bf v}_i\in C\,$ where $\,C\,$ is the circle $\, \left\{\theta=\pi/4 \right\}\cap S\,$. The pole (the origin of the spherical coordinate system) is still the center of the sphere of course.

Then $\,\theta_1=\theta_2=\frac{\pi}{4}\,$ and, by (1), we have $$d({\bf v}_1,{\bf v}_2)=\cos^{-1}\left(\frac{1}{2}+\frac{1}{2}\cos\left(\varphi_1-\varphi_2\right)\right)=\cos^{-1}\left(\frac{1}{2}+\frac{1}{2}\cos\left(\varphi_1-\varphi_2\right)\right)=\\ =\cos^{-1}\left(\cos^{2}\frac{\varphi_1-\varphi_2}{2}\right)$$.

On the other hand, the radius of $C$ is $\frac{1}{2}$. So the arc length between $\,{\bf v}_1,\,{\bf v}_2\,$ should also be $$\text{radius of}\;\;C\times |\varphi_1-\varphi_2|= \frac{|\varphi_1-\varphi_2|}{2}$$

My problem is $$\cos^{-1}\left(\cos^{2}\frac{\varphi_1-\varphi_2}{2}\right)\not\equiv \frac{|\varphi_1-\varphi_2|}{2}$$ Take for instance $\varphi_1=\pi/2$, $\varphi_2=0$. Then $\cos^{-1}\left(\cos^{2}\frac{\varphi_1-\varphi_2}{2}\right)=\frac{\pi}{3}$ while $\frac{\varphi_1-\varphi_2}{2}=\frac{\pi}{4}$.

Where am I mistaken ?

Thanks

Medo
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  • The geodesic between the points you are describing is not the circle $C$. Remember the geodesic is always a great circle (i.e. the intersection of a plane passing through the origin and the sphere). – Hamed Jun 25 '18 at 01:27

2 Answers2

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This is not an answer. This is a comment in reply to the great answer of @Chappers above. I am using the answer environment because I do not know how insert a figure inside a comment.

@Chappers Thank you for the great answer.

It does not really matter, but the radius of the circle $C$ is $\frac{1}{2}$ I still believe. Consider the isosceles right triangle with the vertices at the origin $O$, the north pole $\,B (\theta=0, \varphi=0),\,$ and $\,A (\theta=\frac{\pi}{2}, \varphi=0).\,$ Let $A,B\in S$ so that $OA=OB=1$. Let the horizontal line segment $\overline{DC}$ represent a radius in $C$. Since $\theta=\angle COD=\frac{\pi}{4}$ then all the acute inner angles in the figure below are $\pi/4$. So $BC=OC=CD=\frac{1}{2}$.enter image description here

Medo
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You've got two problems here.

  • Firstly, the radius of the circle $\theta = \pi/4$ is not $1/2$, it's $\cos{(\pi/4)} = 1/\sqrt{2}$.
  • Secondly, the circle $\theta=\pi/4$ is not a great circle, so it does not minimise the distance between the two points on the sphere. This is very clear in the case where $\phi_1-\phi_2=\pi$, because then the great circle passing through both points goes through the north pole of the sphere, and the distance along this is $\pi/2$, whereas the distance along $\theta=\pi/4$ is $\pi/\sqrt{2}$ which is rather larger. There's a comparison image of this sort of situation here, with example distances given. And here is a comparison of the great-circle distance (blue) with the "latitudinal" distance (orange) for the case you're considering:

enter image description here

Clearly for small distances the latitudinal distance is a reasonable approximation, but it gets worse and worse for large differences in angle.

Chappers
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