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Spherical coordinates are inherently asymetric. One way to see that is in spherical coordinates $(r, \theta, \phi)$, lines of constant $\theta$ are great circles whereas those of $\phi$ are not. (Using the convention that azimuth is $\theta$ and polar angle is $\phi$).

Is there anything similar to spherical coordinates, but with two angles symmetric? If not: why not?

Also: Without designating a fixed "north pole", how does $\theta$ "know" to be great circles and $\phi$ not to? What in the definition makes $\theta$ operate differently than $\phi$? The references I've seen for this all use a fixed, known "north pole" to identify this. Understood that the cartesian mapping proves this, but why does making $\theta$ azimuth and $\phi$ polar angle result in that mapping? Without talking about "up" and "down", what's different about azimuth vs polar?

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  • $\begingroup$ As far as "knowing" which angle traces out great circles and which doesn't, this should be apparent from a coordinate transition map to cartesian coordinates: $(x, y, z) = (r\sin\theta\cos\varphi, r\sin\theta\sin\varphi, r\cos\theta)$. Then curves of constant $\varphi$ satisfy $x^2 + y^2 + z^2 = r^2$ and $x\sin\varphi - y\cos\varphi = 0$, which is a vertical plane passing through the origin intersecting the sphere, giving a great circle. $\endgroup$ Commented Dec 1, 2024 at 18:07
  • $\begingroup$ There is a representation via direction cosines, effectively determining a point by its distance from the origin and the angles it determines with positive coordinate axes. (The cosines associated with $(x,y,z)$ are simply $x/r$, $y/r$, $z/r$, where $r:=\sqrt{x^2+y^2+z^2}$.) This provides a natural conceptual symmetry, with the trade-off that the angles aren't independent. $\endgroup$ Commented Dec 2, 2024 at 6:22
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    $\begingroup$ If the coordinates don't have to be angles the most symmetric version of spherical coordinates is given by stereographic projection. $\endgroup$ Commented Dec 2, 2024 at 9:11
  • $\begingroup$ @TobErnack But why does making $\theta$ azimuth and $\phi$ polar angle result in that mapping? Without talking about "up" and "down", What's different about azimuth vs polar? $\endgroup$ Commented Dec 2, 2024 at 11:28

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You could draw the longitude lines on a globe, and then rotate the globe about a horizontal axis so that the former N + S poles are now E and W. Then draw a second set of longitude lines from where the N+S poles used to be. That's 'symmetric' in the sense that $q = C$ lines are great circles regardless of which coordinate is "q". It's also not symmetric, because the density of coordinate lines at the front and back poles is small, while at the other four poles it's large.

In coordinates:

$$ \alpha (x, y, z) = atan2(x, y) \\ \beta(x, y, z) = atan2(x, z) $$ gives you those two coordinates (starting from xyz coordinates on a unit sphere. Oh...and you can have a radial coordinate, too, at no extra cost.

As for an intrinsic way to identify coordinates on a sphere...there's nothing, really, because one of the key things about a sphere is that it's homogeneous; indeed, for any two points $P$ and $Q$ on the sphere, there's an isometry $u_{PQ}: S^2 \to S^2$ such that $u_{PQ}(P) = Q$. (There are actually infinitely many such isometries for any pair $P$ and $Q$.)

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You could force the two angles to play "symmetric" roles by admitting a radius with negative values. Indeed, instead of the usual spherical coordinates $(r,\theta,\phi) \in \Bbb{R}_+ \times [0,\pi] \times [0,2\pi]$, you can consider a radius $\rho \in \Bbb{R}$, as well as two angles $\alpha,\beta \in [0,\pi]$ for instance, in such a way that one hemisphere is parametrized as usual, when the other one is described from the antipodal points and a negative radius, hence $$ \begin{cases} r = |\rho | \\ \theta = \alpha \\ \phi = \beta + \pi(1-H(r)) \end{cases}, $$ where $H$ denotes the Heaviside function.

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  • $\begingroup$ The angles still do not play symmetric roles when you do this. Sure, the ranges are both $0$ to $\pi$; however, it's still true that when the azimuthal angle is set to a constant, we get a great circle, and when the polar angle is set to a constant, we get a non-geodesic small circle (and in the extreme case, a point). $\endgroup$ Commented Dec 2, 2024 at 5:53

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