Partial derivatives and orthogonality with polar-coordinates We are stuck with this question here because I cannot understand the following results. I find it hard to visualize this, let alone deduce from that. How to do it?
Objective to Attack The closely Related Problems with Orthogonal Basis and Dot Products In Polar-coordinates

  
*
  
*$\left(\hat{e}_{r}\partial_{r}\right) \cdot \left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right)= 0$
  
*$\left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) 
     \cdot \left(\hat{e}_{r}\partial_{r}\right) 
         = \frac{1}{r} \partial_r$
  
*$\partial_\theta \hat e_r = \hat e_\theta$
  
*$\partial_\theta \hat e_\theta = -\hat e_r$

Trials

  
*
  
*I have some errors there, related to 3-4 apparently.
  
  
  $$ \left(\hat{e}{r}\partial{r}\right) \cdot
   \left(\frac{1}{r}\hat{e}{\theta}\partial{\theta}\right) 
   =  

\left(\hat{e}{r}\partial{r}\right) \cdot \frac{1}{r}+
\left(\hat{e}{r}\partial{r}\right) \cdot
\left( \hat{e}{\theta}\partial{\theta} \right)
\not
\frac{-\hat{e}_r}{r^2}+
\left(\hat{e}{r}\cdot\hat{e}\theta \right) \partial_{r}
\partial_{\theta}
$$
Perhaps Related


*

*Explain Dot product with Partial derivatives in Polar-coordinates

*Orthonormal vectors in Polar coordinates, show $\hat{e}_R=\frac{(x,y,z)}{r}$

*Visual Ways to Remember Cross products of Unit vectors? Cross-product in $\mathbb F^3$?
 A: Some hints: (not a complete solution)


*

*I guess once you think about it you find it clear that $$\hat{e}_{r}\cdot\hat{e}_\theta=0$$
or in words the unit vector along the radial direction is orthogonal to the unit vector along the angular direction.

*I guess the confusion you have originates from the following fact: the unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ themselves depend on the coordinate $(r,\theta)$. This dependence is usually not made explicit but you should always keep that in mind. If you think about it then I guess it is clear. For $\theta=0$ the radial unit vector points along the $x$-axis whereas for $\theta=\pi/2$ the unit vector points along the $y$-axis.

*The last two points hold for an arbitrary rectangular coordinate system. What is special for the polar coordinate system is that even though $\hat{e}_{r,\theta}$ depend on $\theta$ they do not depend on $r$, i.e., $$\hat{e}_{r,\theta} = \hat{e}_{r,\theta} (\theta).$$

*The important relations $\partial_\theta \hat e_r(\theta) = \hat e_\theta (\theta)$ and $\partial_\theta \hat e_\theta(\theta) = -\hat e_r(\theta)$, you can check by taking the explicit expressions
$$\begin{array}\ \hat{e}_r(\theta)&= \begin{pmatrix} \cos \theta  \\ \sin \theta \end{pmatrix} &
\hat{e}_\theta(\theta)&= \begin{pmatrix} -\sin \theta  \\ \cos \theta \end{pmatrix}
\end{array}$$
With these remarks, it is easy to show that (here, I make the dependence of the unit vectors on the coordinates explicit)
$$\left(\hat{e}_{r}(\theta)\partial_{r}\right) \cdot \left(\frac{1}{r}\hat{e}_{\theta}(\theta) \partial_{\theta}\right)
= \underbrace{\hat{e}_{r}(\theta) \cdot \hat{e}_{\theta}(\theta)}_{=0} 
 \,\left(\partial_{r} \frac{1}{r} \partial_{\theta} \right) =0  .$$
The other results follow similarly (but I will leave the proof up to you).
