Some subtleties of spherical coordinates

Review of Spherical Coordinates

Recall that any point \((x, y, z) \in \mathbb{R}^3\) can be described by its distance from the origin, \(\rho = \sqrt{x^2 + y^2 + z^2}\), its polar angle \(\phi\)¹, and its azimuthal angle \(\theta\).

From this picture we have \(\rho \in [0, \infty)\), \(\phi \in [0, \pi]\), and \(\theta \in [0, 2\pi)\). Despite the care we take with these possible values, we have still introduced several degeneracies: various choices of “coordinates” that represent the same point. For example, at \(\rho = 0\), we identify \[(0, \phi, \theta) \sim (0, \phi', \theta').\] Similarly, on the \(z\)-axis, \[(\rho, 0, \theta) \sim (\rho, 0, \theta') \quad\text{ and }\quad (\rho, \pi, \theta) \sim (\rho, \pi, \theta').\] As a result, spherical coordinates only really describe a coordinate system in the precise sense (as charts on a topological manifold) provided \[(\rho, \phi, \theta) \in (0, \infty) \times (0, \pi) \times [0, 2\pi).\] This covers \(\mathbb{R}^3 \setminus (\text{$z$-axis})\), and we can only guarantee the existence of \(d\rho\), \(d\phi\), and \(d\theta\), on this domain.

The problem of the 1-form

From the equation for \(z\) we have \[\phi = \arccos\left( \frac{z}{\rho} \right) = \arccos\left( \frac{z}{\sqrt{x^2 + y^2 + z^2}} \right).\] The domain for \(\arccos\) is \([-1, 1]\) and the inequality \(|z| \leq \rho\) ensures that \(\phi\) is a globally defined smooth function on \(\mathbb{R}^3 \setminus (\text{$z$-axis})\). We have more, since this expression for \(\phi\) actually shows that it’s well-defined on \(\mathbb{R}^3 \setminus \{0\}\). This makes it tempting to conclude that \(d\phi\) exists on \(\mathbb{R}^3 \setminus \{0\}\). It turns out that this is not the case, \(d\phi\) is not even a 1-form on \(S^2\).

Direct computation

We can argue that \(d\phi\) fails to exist by a direct computation. For \(\rho > 0\) we have \[\begin{align*} d\phi & % alignment on newline for padding = d \arccos\left( \frac{z}{\rho} \right) % & \paren*{ \text{Explanation} } % \\ & = -\frac{1}{\sqrt{1 - \frac{z^2}{\rho^2}}} d\left(\frac{z}{\rho}\right) % & \paren*{ \text{Next} } = -\frac{1}{\sqrt{\frac{x^2 + y^2}{\rho^2}}} \frac{\rho dz - zd\rho}{\rho^2} % & \paren*{ \text{Next} } % \\ & \\ & = -\frac{1}{\sqrt{\frac{(\rho\sin\phi)^2}{\rho^2}}} \frac{\rho dz - zd\rho}{\rho^2} % & \paren*{ \text{Next} } % \\ & = \frac{1}{\sin\phi} \left( \frac{z}{\rho^2}d\rho - \frac{1}{\rho}dz \right) % & \paren*{ \text{Next} } . \end{align*}\] Computing \(d\rho\), \[2\rho d\rho = d\rho^2 = 2( xdx + ydy + zdz ).\] Combining these we have \[\begin{align*} d\phi & % alignment on newline for padding = \frac{1}{\sin\phi} \left( \frac{z}{\rho^2}d\rho - \frac{1}{\rho}dz \right) % & \paren*{ \text{Explanation} } \\ & = \frac{1}{\sin\phi} \left( \frac{z}{\rho^3}( xdx + ydy + zdz ) - \frac{1}{\rho}dz \right) % & \paren*{ \text{Next} } \\ & = \frac{1}{\sin\phi} \left( \frac{xzdx + yzdy - (x^2 + y^2)dz }{\rho^3} \right) % & \paren*{ \text{Next} } . \end{align*}\] Since we’re away from the origin, the factor on the right is not an issue. So the only problem for \(d\phi\) happens at \(\phi \in \{0, \pi\}\). In this case, the right vanishes as \(\phi \to 0\) and \(\phi \to \pi\) but computing limits shows that the derivative still diverges.

While this calculation shows why the derivative does not exist on the \(z\)-axis, our goal now is to go over a shorter and more geometric argument for why it does not exist.

A geometric argument

We work on \(S^2\), for arbitrary \(\rho\) changing \(\phi\) does not change the distance of the point so the argument will transfer over identically. Focusing on the positive \(z\)-axis, let’s see what happens as \(\phi\) gets close to \(0\).

The distance any given point on \(S^2\) (\(z > 0\)) has from the \(z\)-axis is \(\sin\phi\). For small values of \(\phi\) the small angle approximation \(\sin\phi \approx \phi\) (justified by Taylor expansion) says that the point is approximately \(\phi\)-distance away from the axis. On the other hand, spherical coordinates also tell us that \[\sin\phi = r = \sqrt{x^2 + y^2}.\] Our small angle approximation tells us \(\phi \approx r\) and the problem of differentiating \(\phi\) near the \(z\)-axis is exactly the problem of differentiating \(r\) in polar coordinates. In particular \[\phi \approx \sqrt{x^2 + y^2},\] and the derivative of this explodes at 0, as seen by \[dr = \frac{1}{\sqrt{x^2 + y^2}} d(x^2 + y^2).\]

Consequences

The observation \(\phi \approx r = \sqrt{x^2 + y^2}\) for small values of \(\phi\) offers a suggested way to smooth it over by instead considering \(d\phi^2\). In this case \[d\phi^2 = 2\phi d\phi\] and a direct computation shows that \(\phi d\phi \to 0\) as \(\phi \to 0\) for the same reason \(rdr \to 0\) as \(r \to 0\).

Notice that this still fails to fix the issue on the negative \(z\)-axis since here \(\phi \to \pi\) which does not remove the singular behavior of its derivative given by the same picture. To justify the small-angle approximation, we need to instead consider the complementary angle \(\psi = \pi - \psi\). Now \(\phi \to \pi\) is the same as considering \(\psi \to 0^+\) and our previous argument holds. This means that, away from the positive \(z\)-axis, \[d\psi^2 = (\pi - \phi)d(\pi - \phi) = -(\pi - \phi)d\phi\] is well-defined.