Help needed with partial derivatives and polar coordinates, missing term.

Question

I have a missing $\frac{1}{r}\partial_r$ -term (notice the question mark) but cannot see why, could someone hint where I am doing mistake.

Answer 1

$$\partial_\varphi e_r=e_\varphi\implies\frac{1}{r}e_\varphi\partial_\varphi \cdot e_r \partial_r =\frac{1}{r}(e_\varphi\cdot e_\varphi)\partial_r=\frac{1}{r}\partial_r$$

You evaluated this to zero. If you write this out fully (ignoring the $1/r$ in front), you have

$$\begin{pmatrix}-\sin\varphi \; \frac{\partial}{\partial\varphi} \\ \cos\varphi \; \frac{\partial}{\partial\varphi}\end{pmatrix} \cdot\begin{pmatrix} \cos\varphi \; \frac{\partial}{\partial r} \\ \sin\varphi \frac{\partial}{\partial r}\end{pmatrix} = -\sin\varphi \; \frac{\partial}{\partial \varphi}\left(\cos\varphi \; \frac{\partial}{\partial r}\right)+\cos\varphi \; \frac{\partial}{\partial\varphi}\left(\sin\varphi\frac{\partial}{\partial r}\right)$$

$$=-\sin\varphi \left(-\sin\varphi\frac{\partial}{\partial r}+\cos\varphi\frac{\partial^2}{\partial \varphi \partial r}\right)+\cos\varphi\left(\cos\varphi\frac{\partial}{\partial r}+\sin\varphi \frac{\partial^2}{\partial \varphi \partial r}\right)=\frac{\partial}{\partial r}.$$