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!

  • $\begingroup$ $e_{\theta}$ is a function of position. $\endgroup$ – NicNic8 Jul 7 '18 at 19:36
  • $\begingroup$ Ok, @NicNic8. I know this. But my question is about $\hat{e}_{r}$. $\endgroup$ – MLPhysics Jul 7 '18 at 19:37

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 $r \to 0$

  • $\begingroup$ Thank you!!! So, just to confirm... If instead I have a projector (in cartesian coordinates) given by $P^{ia} \equiv \frac{x^{i} x^{a}}{r^2}$, where $x^{i}$ (i=1,2,3) stands for $x,y,z$... this would also be ill-defined at the origin, right? Since when I take limits by different paths, I get different results. For instance, $P^{12} = \frac{xy}{x^{2}+y^{2}+z^{2}}$ would have similar problems. $\endgroup$ – MLPhysics Jul 7 '18 at 20:55

Note $\hat{e}_{r}$ is independent of of $r$ . So $r\to 0$ does not affect anything.

Other way look away it is that $\hat{e}_{r}$ is a parametric representation of a unit sphere . This means you have already specified the radius as $1$. So no question of changing radius here .

  • $\begingroup$ Yes, I know that $\hat{e}_{r}$ is independent of r. And I'm also aware this is a parametrization of a unit sphere. However, I don't see how this answers my question. $\endgroup$ – MLPhysics Jul 7 '18 at 19:54
  • $\begingroup$ Think about the geometry, if you are taking z=0 and y=0 , then what you have left out is the x . The possibilities for x is (on co-ordinate system) (1,0,0) and (-1,0,0) . Thats why your limit x/r does not exist and gives answer +1 or -1 . Taking another path suppose y=mx then z=0 would give some kind of circle on XY plane and points will be according to that . $\endgroup$ – Chinmaya mishra Jul 7 '18 at 20:14

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