Is $0\hat{\mathbf{r}}+\frac{\pi}{4}\hat{\mathbf{\theta}}+\frac{\pi}{3}\hat{\mathbf{\phi}}$ a meaningful vector in spherical coordinates?

Question

Is the vector $\displaystyle 0\hat{\mathbf{r}}+\frac{\pi}{4}\hat{\mathbf{\theta}}+\frac{\pi}{3}\hat{\mathbf{\phi}}$ a meaningful vector?

I think it is equal to the zero vector but I just wanted to confirm.

Answer 1

To belabor the point:

Let's consider two copies of $\Bbb R^3$, and call them $S$ and $C$. We then have a map $$ p : S \to C : (r, \theta, \phi) \mapsto (r \cos \theta \cos \phi, r \sin \theta \cos \phi, r \sin \phi) $$ that's sometimes called "polar coordinates" (although in some texts, $\sin \phi$ and $\cos \phi$ are swapped, or the coordinates in $C$ are permuted; let's stick with just one of the various possibilities for now).

Notice that this map is defined for every element $(r, \theta, \phi)$ of the space $S$, i.e., $p$ is actually a function.

As far as I can tell, your question asks, "Is $p(0,\frac{\pi}{4}, \frac{\pi}{3})$ defined?" Once we write it this way, the answer is obviously "yes."

The function $p$ is useful (and widely used), but it's worth understanding that although it's surjective (every element of $C$ is $p(s)$ for some $s \in S$), it's not injective: for many elements $c \in C$, we have multiple distinct elements of $s$, say $s_1, s_2$ such that $p(s_1) = c = p(s_2)$.

These injectivity failures can be thought of as happening in several forms:

1. Negating $r$: $$p(r, \theta, \phi) = p(-r, \theta + \pi, -\phi)$$ for all $r, \theta, \phi$. We can (mostly) get rid of this problem by replacing $S$ with $\Bbb U \times \Bbb R \times \Bbb R$, where $U$ is the set of nonnegative reals, i.e., restricting $r$ to be nonnegative. (But see below).

2. Periodicity in $\theta$: $$ p(r, \theta, \phi) = p (r, \theta + 2\pi, \phi)$$ for all $r, \theta, \phi$. This might be called the "international dateline" problem, for if $r$ is fixed at $1$, then $\theta, \phi$ give longitude-latitude coordinates on the sphere, and the periodicity in $\theta$ corresponds to $180$E longitude being the same as $180$W longitude. This can be addressed by restricting the second factor of $S$ to an interval like $[0, 2\pi)$, or $[-\pi, \pi)$, but that has at least one obvious disadvantage: we have to choose which end of the interval to include/exclude.

3. Periodicity in $\phi$: $$ p(r, \theta, \phi) = p (r, \theta , \phi+ 2\pi)$$ for all $r, \theta, \phi$.

4. A worse "periodicity" in $\phi$: $$ p(r, \theta, \phi) = p (r, \theta + \pi , \pi - \phi)$$ for all $r, \theta, \phi$. I think of this as a "double-bagging" trick: put some dirt in a long thin tubular plastic bag, squeeze out all the air, and spin the bag so as to make a tight closure at the top. Now you can take the remaining free part of the bag and drape it down over the ball of dirt, and do the same thing, forming a second "twist" at the bottom, and then lift the remaining bag back up, and so on. This, and the previous item, can be addressed by limiting the range of $\phi$ to an interval of length $\pi$.

Even if we choose to limit $\phi$ and $\theta$ to intervals of length $2 \pi$ and $\pi$ respectively, and limit $r$ to nonnegative reals, the function $p$ is still not injective. The first injectivity failure remains, for when $r = 0$, we see that $$ p(0, \theta, \phi) = (0,0,0) $$ for all values of $\theta, \phi$.

What this means in practice is that you cannot say, for an arbitrary point of $C$, "what are the spherical polar coordinates" --- the point $(0,0,0) \in C$ has multiple valid sets of spherical coordinates.

It's possible, of course, to only allow one of these. You can take the half-infinite rectangular block defined by $$ r \ge 0\\ \theta \in [0, 2\pi) \\ \phi \in [-\pi/2, \pi/2] $$ and exclude, from the end where $r = 0$, all points except $(0,0,0)$, i.e., you can restrict the function $p$ to the domain $$ D = \{(0,0,0)\} \cup \left( (0, \infty) \times [0, 2\pi) \times [-\pi/2, \pi/2] \right) $$

If you do so, then the point you asked about originally will no longer be in the domain of the function $p$, so the answer to your question will be "no, that's not valid."

There are good reasons not to do this. The main one is calculus: we'd like to be able to do calculus on $C$ by doing calculus on $S$ and transferring results via the chain rule. When you computed "the volume element for spherical coordinates," for instance, you did this. To make that work your functions need to be defined on an open set around a domain point $s \in S$, and once you restrict the domain of $p$, this fails to happen at all the "interesting" points (i.e., those with $r = 0$, and those on the international dateline, etc.)

There is also a good reason to go ahead and make the restriction: it makes polar coordinates "unambiguous". But in doing so, you complicate everything else that you ever choose to do with them; I think that the ambiguity is a small price to pay for the benefits of being able to do calculus properly.