The shortest distance between two points on a sphere

Question

The shortest distance between two points $P_{1}(x_1,y_1,z_1)$ and $P_{2}(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$.

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If both $P_1$ and $P_2$ lie on the sphere $(x-x_0)^2+(y-x_0)^2+(z-z_0)^2=r^2$, then what is the length of the shortest path between the two points such that the path lies on the surface of the sphere?

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The formula for distance written above will not work to solve this problem.

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How to express the length of the shortest path that lies on the sphere in terms of $x_0,y_0,z_0,x_1,y_1,z_1,x_2,y_2,z_2,$ and $r$?

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I think the arc length using integration is the right way, but I am struggling with it. Any help will be appreciated.

Answer 1

Consider a unit circle centered at $O = (0,0,0)$ with two points $A$ and $B$ on it. If the arc length between $A$ and $B$ is $\alpha$ (which is equal to the angle between $OA$ and $OB$), then the chord length $d$ satisfies

$$\frac{d}{2} = \sin\left(\frac{\alpha}{2}\right).$$

To find the surface distance between two points $A = (x_1, y_1, z_1)$ and $B = (x_2, y_2, z_2)$ on the unit sphere, note that the shortest path on the sphere between $A$ and $B$ is a great circle arc. By the previous observation, we find that the spherical distance is $$\alpha = 2\sin^{-1}\frac{d}{2} = 2\sin^{-1}\frac{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}{2}.$$

Generalizing this to spheres of radius $r$, we get

$$\alpha = 2r\sin^{-1}\frac{d}{2r} = 2r\sin^{-1}\frac{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}{2r}.$$

Alternatively, we can use the dot product of vectors, using the fact that for any two vectors $a$ and $b$ with an angle of $\gamma$, we have $a\cdot b = \|a\|\cdot\|b\|\cdot\cos\gamma$. If we let $a = A - O$ and $b = B - O$, we get two unit vectors with an angle of $\alpha$, so we get

$$\alpha = \cos^{-1}(a\cdot b) = \cos^{-1}(x_1x_2 + y_1y_2 + z_1z_2).$$

For arbitrary spheres of radius $r$ centered at $(x_0, y_0, z_0)$, we can reduce everything to the $r=1$ case by scaling and translating, and we get

$$\alpha = r\cos^{-1}\left(\frac{(x_1-x_0)(x_2-x_0)+(y_1-y_0)(y_2-y_0)+(z_1-z_0)(z_2-z_0)}{r^2}\right).$$

Answer 2

In spherical coordinates you have: $$x=\rho\cos(\theta)\cos(\phi)$$ $$y=\rho\cos(\theta)\sin(\phi)$$ $$z=\rho\sin(\theta)$$ So, using inverse formula you can find $P_1(\rho_1,\phi_1,\theta_1)$,$P_2(\rho_2,\phi_2,\theta_2)$

The minimum path between two points $P_1(x_1,y_1,z_1)$, $P_2(x_2,y_2,z_2)$ is given by: $$d(P_1,P_2)=\rho\arccos(\cos\Delta\phi\cos(\theta_{P_1})\cos(\theta_{P_2})+\sin(\theta_{P_1})\sin(\theta_{P_2})$$ where: $\Delta\phi=\phi_{P_2}-\phi_{P_1}$