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}
$$
Perhaps Related
Best Answer
Some hints: (not a complete solution)
I guess once you think about it you find it clear that $$\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).