change in unit radial vector and unit polar angle due to change in polar angle

Question

In cylindrical coordinates, the set B of basis vectors is

$B=\left \{ \vec{e}_{r},\vec{e}_{\theta} \right \}$.

It is clear, geometrically, that

$\frac{\partial e_{r}}{\partial r}=0=\frac{\partial e_{\theta}}{\partial r}$

However, I am unable to convince myself that

$\frac{\partial e_{r}}{\partial \theta}=e_{\theta}$ and $\frac{\partial e_{\theta}}{\partial \theta}=-e_{r}$

Any help to shed light on this is appreciated.

Thanks in advance

Answer 1

This is for polar coordinates, cylindrical would have a third basis vector $e_z$, but regardless, as the angle $\Delta\theta$ goes to $0$ in the ccw direction, the differential angle makes an appearance resulting in a differential change in the direction of the radial vector

So wherever $e_r$ was pointing, it now has to rotate over an amount of $d\theta$ (ccw) to point in the new correct direction, and the direction of where it should go is already accounted for by $e_\theta$

Likewise, $e_\theta$ is perpendicular to $e_r$, so when going counterclockwise, the new $e_\theta$ is more flattened horizontally than the original $e_\theta$, so the direction $e_\theta$ must go when rotated by $d\theta$ is in the direction of $-e_r$

This is to say

$$\underline{e_r}=\cos\theta\underline{i}+\sin\theta\underline{j}$$

By definition, they are orthogonal

$$\underline{e_r}\cdot\underline{e_\theta}=0$$

And we want $e_\theta$ to have positive orientation sense, so

$$\underline{e_\theta}=-\sin\theta\underline{i}+\cos\theta\underline{j}$$

And we can now clearly see

$$\frac{d\underline{e_r}}{d\theta}=-\sin\theta\underline{i}+\cos\theta\underline{j}=\underline{e_\theta}$$

$$\frac{d\underline{e_\theta}}{d\theta}=-\cos\theta\underline{i}-\sin\theta\underline{j}=-\underline{e_r}$$

Answer 2

The quickest and easiest way I know how to handle this one is via the realization that

$\vec e_r(r, \theta) = (\cos \theta, \sin \theta), \tag 1$

and, if $\theta$ increases in a counter-clockwise direction, that

$\vec e_\theta(r, \theta) = (-\sin \theta, \cos \theta); \tag 2$

(1) is very nearly self-evident, and (2) follows from (1) by a counter-clockwise rotation of the unit vector $\vec e_r$ by $\pi/2$, which takes

$\theta \to \theta + \dfrac{\pi}{2}, \tag 3$

whence

$\cos (\theta + \dfrac{\pi}{2}) = (\cos \theta)(\cos \dfrac{\pi}{2}) - (\sin \theta)(\sin \dfrac{\pi}{2}) = -\sin \theta; \tag 4$

$\sin(\theta + \dfrac{\pi}{2}) = (\cos \theta)(\sin \dfrac{\pi}{2}) + (\cos \dfrac{\pi}{2})(\sin \theta) = \cos \theta. \tag 5$

It is then clear that

$\dfrac{\partial \vec e_r}{\partial r} = 0 = \dfrac{\partial \vec e_\theta}{\partial r}, \tag 6$

since neither $\vec e_r$ nor $\vec e_\theta$ depends on $r$; it is equally clear that

$\dfrac{\partial \vec e_r}{\partial \theta} = (-\sin \theta, \cos \theta) = \vec e_\theta, \tag 7$

and

$\dfrac{\partial \vec e_\theta}{\partial \theta} = (-\cos \theta, -\sin \theta) = -\vec e_r. \tag 8$

Note: Of course this is all based upon the fact that any unit vector may be written $(\cos \phi, \sin \phi)$ for appropriate $\phi$. End of Note.