I need to show that the del operator in 2D polar coordinates is $\nabla=e_r\partial_r+\frac{1}{r}e_r+\frac{1}{r}e_{\phi}\partial_{\phi}$. I try the following approach:
$\nabla=\partial_xe_x+\partial_ye_y=\bigg(\frac{\partial r}{\partial x}\partial_r+\frac{\partial \phi}{\partial x}\partial_{\phi}\bigg)\bigg(\cos(\phi)e_r-\sin(\phi)e_{\phi}\bigg)+\bigg(\frac{\partial r}{\partial y}\partial_r+\frac{\partial \phi}{\partial y}\partial_{\phi}\bigg)\bigg(\sin(\phi)e_r+\cos(\phi)e_{\phi}\bigg) $
and then use $r=\sqrt{x^2+y^2}$, $y=\arctan(y/x)$, $\partial_{\phi}e_r=e_{\phi}$ and $\partial_{\phi}e_{\phi}=-e_r$.
The problem is, I end up with $\nabla=e_r\partial_r+\frac{1}{r}e_{\phi}\partial_{\phi}$, missing a term compared to the expression above. Any ideas?
Best,
Jorgen
EDIT: spelling
Best Answer
As you know, the operator $\nabla$ is defined by $$\nabla\equiv i\partial_x+j\partial_y+k\partial_z$$ It is a vector operator so, it needs to have $\partial_{.}$ terms. Your calculation above is right in 2D polar system and I think something is missing in your original expression where you wrote $\frac{1}{r}e_r$. In fact you don't need to have such this term when you want to calculate del operator in 2D polar coordinates.