For a set of spherical coordinates ($\rho, \theta, \phi$), we'll define $\rho= \sqrt{x^2 + y^2 + z^2}$, $\theta \in [0, 2\pi]$ as the rotation away from the $x$-axis (forward-back horizontal axis), and $\phi \in [0, \pi]$ as the rotation down from the $z$-axis (vertical axis). So, routine stuff, but just to dispel any ambiguities.

Consider the point with Cartesian/ rectangular coordinates (0,0,10); really we could generalize this to some sort of (0,0,$z$) for $z \in \mathbb{R}$. Suppose we want to convert this into spherical coordinates.

What are we supposed to do here?

From some self-teaching, I've essentially just thought of going from rectangular to spherical coordinates as solving this "system":

$\begin{cases} x = \rho \sin\phi \cos \theta \\ y = \rho \sin \phi \sin \theta \\ z = \rho \cos \phi \end{cases}$

It's quickly clear that we run into an issue, and I haven't yet found a definitive answer anywhere; only tangentially related answers and wiki's. $\phi = 0 \implies \sin \phi = 0$, so we cannot determine what $\theta$ is.

Thinking geometrically, wouldn't we be able to give any theta considering we're going straight up the $z$-axis? Would that not indicate the same point? Testing my hypotheses with some online calculators yielded nothing: the ones I tried just didn't spit out an answer, so the other possibility I can imagine is just that there is... well, no spherical representation for these such points. I assume this system is rigorous, well-defined, etc. (and that, therefore, there is an answer to this), but this seems to be quite an oversight.

TLDR; how do we handle points on the $z$-axis with the spherical coordinate system?

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    $\begingroup$ Even when we take $\theta \in [0, 2\pi)$ we are faced with a violation of one-to-one correspondence: the plane $r = 0$ of the space $r\phi \theta$ is mapped to the origin of the coordinates $x = y = z = 0$, a line $\phi=0(\pi), r=r$ is mapped to one point: $x=y=0,z=r$. $\endgroup$
    – zkutch
    Jan 24, 2021 at 17:22

1 Answer 1


It is hard to answer this because it is hard to know what you are asking precisely. You haven't yet run into any difficulties but you are suspicious of the idea that the change of coordinates map is not a bijection.

This can be a problem when you are trying to push around jacobians near the z axis or something.

The answer is that in 3d minus a neighborhood around the z axis everything will work well. And in the case where you have to do something fancy around the z axis you will want to do a change of coordinates so that you are working with a different set of spherical coordinates near that problem area.

This is why we have the machinery of atlases and tangent bundles, etc ..in differential geometry. To deal with these problems that might arise in a general way by looking locally, however you may be best to not worry too much about it while you go through your earlier classes.

  • $\begingroup$ I've put together without saying it that these transformations are not bijective, but, with the TLDR in mind, what do we do with these points? That's my question. Do we accept the ambiguities? Or is there just not a spherical representation for these points? [Upon rereading a few times, seems that the problem that originally sparked this is a bit deeper than a course titled "Honors Intermediate Calculus" should cover. I've been somewhat self-didactic with analysis/ interests lie in applied math, but Differential Geo is on the list for what it's worth] $\endgroup$
    – user838358
    Jan 24, 2021 at 17:32

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