Derivatives of Unit Vectors in Spherical and Cartesian Coordinates

I noticed something interesting yesterday. If I consider the unit vectors in spherical coordinates expressed in terms of the Cartesian unit vectors:

$\hat{\textbf{r}} = \sin\theta \cos\phi \, \hat{\textbf{x}} + \sin\theta \sin\phi \, \hat{\textbf{y}} + \cos\theta \, \hat{\textbf{z}}$

$\hat{\boldsymbol{\theta}} = \cos\theta \cos\phi \, \hat{\textbf{x}} + \cos\theta \sin\phi \, \hat{\textbf{y}} -\sin\theta \, \hat{\textbf{z}}$

$\hat{\boldsymbol{\phi}} = -\sin\phi \, \hat{\textbf{x}} + \cos\phi \, \hat{\textbf{y}},$

I see that one can get the expression for $\hat{\boldsymbol{\theta}}$ by taking the derivative of $\hat{\textbf{r}}$ with respect to $\theta$. Similarly, one can get the expression for $\hat{\boldsymbol{\phi}}$ by taking the derivative of $\hat{\textbf{r}}$ with respect to $\phi$ and then setting $\theta$ to $\pi/2$. (I'm not completely sure what the second part of this means. I know that $\hat{\boldsymbol{\phi}}$ cannot depend on $\theta$, but why that particular value? I guess $\hat{\boldsymbol{\phi}}$ "lives" in the $x$-$y$ plane or something.) Except for this parenthetical remark, this makes sense to me, as the unit vectors in curvilinear coordinates are functions of the coordinates, and their derivatives with respect to the coordinates should be easily related to the other unit vectors in an orthogonal coordinates system. This is particularly useful, because I don't have the expressions for $\hat{\boldsymbol{\phi}}$ and $\hat{\boldsymbol{\theta}}$ in terms of the Cartesian unit vectors memorized, but I can easily derive them from the expression for $\hat{\textbf{r}}$

However, I also noticed a similar relationship for the Cartesian unit vectors:

$\hat{\textbf{x}} = \sin\theta \cos\phi \, \hat{\textbf{r}} +\cos\theta \cos\phi \, \hat{\boldsymbol{\theta}} -\sin\phi \, \hat{\boldsymbol{\phi}}$

$\hat{\textbf{y}} = \sin\theta \sin\phi \, \hat{\textbf{r}} + \cos\theta \sin\phi \, \hat{\boldsymbol{\theta}} +\cos\phi \, \hat{\boldsymbol{\phi}}$

$\hat{\textbf{z}} = \cos\theta \, \hat{\textbf{r}} -\sin\theta\, \hat{\boldsymbol{\theta}}$

If one takes the $\textit{partial}$ derivative of $\hat{\textbf{x}}$ with respect to $\phi$ (ignoring the $\phi$-dependence of the spherical unit vectors), one gets the expression for $-\hat{\textbf{y}}$. Similarly, taking the partial derivative of $\hat{\textbf{x}}$ with respect to $\theta$ and setting $\phi$ to $0$, yields the expression for $\hat{\textbf{z}}$. However, since Cartesian coordinates are not curviliear, taking their derivatives with respect to the coordinates doesn't really make sense.

What's going on here? Is this just a coincidence or something meaningful?

You seem to have raised two questions here. The first is why is $\hat{\boldsymbol\phi} = \dfrac{\partial\hat{\mathbf r}}{\partial\phi}$ only true for $\theta=\pi/2$. The second is why does the same derivative trick seem to work for Cartesian coordinates as well when it shouldn't.

The answer to question 2 is that it doesn't work. When you take the $\phi$ derivative of the expression $\hat{\mathbf x} = \sin\theta\cos\phi\hat{\mathbf r} + \cos\theta\cos\phi\hat{\boldsymbol \theta} -\sin\phi\hat{\boldsymbol \phi}$, you cannot "ignore the $\phi$-dependence of the spherical unit vectors", since they are explicitly dependent on the coordinates. The extra terms containing the $\dfrac{\partial\hat{\mathbf r}}{\partial\phi}$, $\dfrac{\partial\hat{\boldsymbol\theta}}{\partial\phi}$, $\dfrac{\partial\hat{\boldsymbol\phi}}{\partial\phi}$ derivatives will eventually cancel out all the other derivatives and give you $0$.

As for the first question, if you express the derivative $\dfrac{\partial\hat{\mathbf r}}{\partial\phi}$ in terms of $\hat{\mathbf r}$, $\hat{\boldsymbol \theta}$, $\hat{\boldsymbol \phi}$ again (by substituting the expressions for $\hat{\mathbf x}$, $\hat{\mathbf y}$, $\hat{\mathbf z}$ into the expression after you take the derivative, for example), then it becomes $\sin\theta\hat{\boldsymbol \phi}$. This then equals $\hat{\boldsymbol \phi}$ when you set $\theta$ to $\pi/2$. Now knowing that the algebra checks out, how is this to be interpreted geometrically? Well, consider at a point $(r, \theta, \phi)$, we move a tiny bit in the local $\hat{\boldsymbol\phi}$ direction, which amounts to a $\delta\phi$ rotation about the $z$ axis. The tip of the $\hat{\mathbf r}$ vector will have also rotated after this rotation. However, the amount of rotation depends on the "lattitude" of the point $(r, \theta,\phi)$, or the value of $\theta$. We can see that near the "north pole", a small change in the azimuthal angle only results in a very small change in the $\hat{\mathbf r}$ direction, but the same change will result in a large change at "low lattitudes." and is maximal at the "equator." This is the effect of the $\sin\theta$ factor.

• Thanks. The stuff on $\theta = \pi/2$ makes sense. But for the derivative, wouldn't taking the partial derivative still work? That's what I meant when I said "ignore the $\phi$-dependence of the spherical unit vectors." Taking the partial derivative would involve the explicit dependence on the coordinates. – Klein Four Nov 4 '16 at 6:17
• Yes, taking partial derivative does involve all explicit dependence on the coordinates, which is exactly why you cannot ignore the $phi$ dependence of the spherical unit vectors, because they explicitly depend on the coordinates as well. Your derivative of $x = f_1(r, \theta, \phi)\hat{r}(r,\theta,\phi)+ f_2(r, \theta, \phi)\hat{\theta}(r,\theta,\phi)+f_3(r, \theta, \phi)\hat{\phi}(r,\theta,\phi)$ will then contain 6 terms, instead of 3. – Elliot Yu Nov 4 '16 at 6:25
• Though technically, in the framework of differential geometry, the coordinate dependence of the unit vectors is a very hairy business. Just as OkThen's answer pointed out, the unit vectors can be thought of as derivatives. More precisely, they live in tangent spaces at each point. So to compare the unit vectors at each point, one would need to relate the tangent spaces somehow. This issue is solved by covariant derivatives, which tells you how to properly take derivatives of unit vectors. – Elliot Yu Nov 4 '16 at 6:39
• Everything simplifies a bit in Euclidean space, so we can still take derivatives of these unit vectors, which is good news for our scenario here. – Elliot Yu Nov 4 '16 at 6:41

There is a clever way to look at vectors. They are differential operators, for example:

$$\mathbf{x} = \frac{\partial}{\partial x}.$$

So, in a Cartesian basis, we would have

$$\mathbf{r} = x \frac{\partial}{\partial x} + y \frac{\partial}{\partial y} + z \frac{\partial}{\partial z}.$$

It also follows that the usual change of basis is given by the chain rule:

$$\mathbf{x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} + \frac{\partial}{\partial \phi} \frac{\partial}{\partial \phi} = \frac{\partial r}{\partial x} \mathbf{r} + \frac{\partial \theta}{\partial x} \mathbf{\theta} + \frac{\partial \phi}{\partial x} \mathbf{\phi}.$$

This result is stablished in every differential geometry book. I recommend you take a look at some general relativity texts.

p.s.: Watch out that I did not care to normalize my vectors. But it should explain the phenomenon that you found.