When working with spherical coordinates, one has the basis vectors $\hat{e}_{r}$, $\hat{e}_{\theta}$ and $\hat{e}_{\varphi}$. And I'm following the physics convention, where $\theta$ is the angle between the position vector and the z-axis and $\varphi$ is, therefore, the azimuthal angle.
Well, I was wondering if
$$\hat{e}_{r} = \frac{\vec{r}}{r} = \sin(\theta)\cos(\varphi)\, \hat{\mathbf{x}} + \sin(\theta)\sin(\varphi) \, \hat{\mathbf{y}} + \cos(\theta) \,\hat{\mathbf{z}},$$
is well-defined at the origin (r=0).
For example, the position vector $\vec{r}$ goes to zero in the limit $r\to 0$. However, the unit radial vector does not have the "$r \,\times$ trig. functions" anymore. So, how is the behavior of $\hat{e}_{r}$ when $r\to 0$?
EXTRAS: I will insert below what I've thought so far.
I know that I have three possible limits to take in this case, and they are of the form:
$$\lim_{(x,y,z)\to(0,0,0)} \frac{x}{\sqrt{x^2 + y^2 + z^2}},$$
while the other two are an exchange of $x$ by $y$ and $z$ in the numerator. So, I will concentrate on the one above (the others are no new information).
If I choose a path along the x-axis ($y = 0 = z$) to take the limit, then
we get
$$\lim_{\substack{x\to 0 \\ y = 0 = z}} \frac{x}{\sqrt{x^2 + y^2 + z^2}} = \lim_{x\to 0} \frac{x}{\sqrt{x^2}}.$$
When I got to this expression, I noticed I should be more careful and think about $x$ going to zero by $\lim_{x \to 0^+}$ and $\lim_{x \to 0^-}$. This of course has distinct values $+1$ and $-1$ respectively. So, as far as I know, this limit does not exist!
One could also think about a path where $y = mx$ and $z = nx$ with $m,n \in \mathbb{R}$. This one would gives us another values.
So, my conclusion is that $\hat{e}_{r}$ is not defined at the origin. And for me, this agrees with the fact that those trig functions are also ill-defined at the origin, which follows from the fact that the origin does not uniquely specify $\theta, \varphi$.
So, does my argument make sense? Could you please tell me if there are any more problems/subtleties I haven't noticed?
Thanks in advance!
Best Answer
You are correct. the limit $$\lim_{(x,y,z)\to(0,0,0)} \frac{x}{\sqrt{x^2 + y^2 + z^2}},$$is path dependent and the basis vectors in spherical coordinates are not well defined for $r \to 0$