# I N-spherical coordinate angle intervals

1. Oct 29, 2016

### TubbaBlubba

This is a kind of silly-sounding question I never realized puzzled me until moments ago, when I looked up the algorithm for spherical coordinates in n dimensions.

In two dimensions, we have polar coordinates, consisting of r from 0 to ∞, and θ from 0 to 2π. In spherical coordinates, we have a third angle, from 0 to π, measured from the z-axis. Intuitively the reason for this is clear. So I wondered about n dimensions - is it the case that a 4-sphere requires an angle running only from 0 to π/2? Apparently not - all angles invoked for more than two dimensions run fron 0 to π. What in a mathematical sense, is special about the two-dimensional case that does not hold for higher dimensions?

I'm a final-year physics undergrad; I've taken courses in multilinear algebra, multivariable calculus, and complex caluculus, and I have a casual interest in topology and advanced geometry, but basically I do not have a good grasp of anything beyond the level of, say, Spivak's "Calculus on Manifolds".

EDIT: I realize this is somehow related to there being no straightforward 1d equivalent because you cannot account for parity in a continuous manner unlike higher dimensions, but that's as far as I get.

Last edited: Oct 29, 2016
2. Oct 29, 2016

### Staff: Mentor

The formulas (from Wikipedia) for n dimensions are:

$\displaystyle {\begin{array}{lcr}x_{1}&=&r\ \cos \varphi \ \sin \vartheta _{1}\ \sin \vartheta _{2}\ \cdots \ \sin \vartheta _{n-3}\ \sin \vartheta _{n-2}\\x_{2}&=&r\ \sin \varphi \ \sin \vartheta _{1}\ \sin \vartheta _{2}\ \cdots \ \sin \vartheta _{n-3}\ \sin \vartheta _{n-2}\\x_{3}&=&r\ \cos \vartheta _{1}\ \sin \vartheta _{2}\ \cdots \ \sin \vartheta _{n-3}\ \sin \vartheta _{n-2}\\x_{4}&=&r\ \cos \vartheta _{2}\ \cdots \ \sin \vartheta _{n-3}\ \sin \vartheta _{n-2}\\\vdots &\vdots &\vdots \qquad \qquad \qquad \quad \\x_{n-1}&=&r\ \cos \vartheta _{n-3}\ \sin \vartheta _{n-2}\\x_{n}&=&r\ \cos \vartheta _{n-2}\end{array}}$

where the $\vartheta_i \in [0,\pi]$. I don't think there are many applications for a physicist above $n=3$ but I'm not sure about it.

Last edited: Oct 29, 2016
3. Oct 29, 2016

### TubbaBlubba

Indeed. I mean, in two dimensions you need to use theta to account for the signs one both x AND y, whereas any higher-dimension angles only needs to account for the signs of whatever additional cartesian coordinate you introduce.

But it irks me, because I can't put my finger about what is so different about going from one to two dimensions beyond, "parity, continuity, something something something".

Spherical coordinates are useful because of the properties of the sphere being the set of all points a certain distance from some other points using the Euclidean measure, but this simply a set of two points on the real line. But what is so different about the real line compared to Cartesian coordinates that changes this? It seems so trivial, but I can't put my finger on it.

Last edited: Oct 29, 2016
4. Oct 29, 2016

### Staff: Mentor

I assume you meant the step from two to three (and higher) dimensions.
If we would allow the second angle to be in $[0,2 \pi)$ we could describe coordinates of a torus (if my intuition is correct).
And this points to the reason: $S^2 \text{ sphere } \ncong S^1 \times S^1 \text{ torus }$. So the n-sphere isn't just adding spheres of lower dimension.

5. Nov 13, 2016

### Ben Niehoff

A more symmetric way to think of this is in terms of "direction cosines", which we sometimes call $\mu_i \equiv \cos \theta_i$, where there are as many $\mu_i$ as there are axes; thus, for the $n$-sphere, there are $n+1$ of them. Each direction cosine is the cosine of the angle $\theta_i$ subtended from the positive $i$-th axis, and thus each $\theta_i \in [0,\pi)$. But of course, these are too many coordinates to describe the $n$-sphere! And it so happens that the direction cosines satisfy one constraint:

$$\sum_i \mu_i^2 = 1.$$
So, we can describe the n-sphere more symmetrically, at the expense of introducing an extra coordinate and a constraint. We can explicitly solve the constraint any way we like to obtain exactly $n$ coordinates, but this will break the symmetry.

The standard way to solve the constraint is to pick out two preferred axes, x and y, and then project down onto them to define the angle $\varphi \in [0,2\pi)$. Then $\varphi$ and the remaining $n-1$ of the $\theta_i$ are a good set of coordinates.

But there are many other ways to write spheres! For example, the 3-sphere sits in $\mathbb{C}^2$ described by the set of complex vectors

$$(z_1, z_2) = (\cos \vartheta \, e^{i \phi}, \sin \vartheta \, e^{i \psi}),$$
where the ranges of the coordinates are now $\vartheta \in [0,\pi/2)$, $\phi \in [0, 2\pi)$, $\psi \in [0,2\pi)$.

6. Nov 13, 2016

### TubbaBlubba

That is a very satisfying account! Thank you!