Divergence operator in polar coordinates

Question

I'm an absolute beginner and I'm having some problems with the del operator in polar coordinates, any help would be appreciated.

The del operator in polar coordinates is defined as: $$ \nabla = \left(\frac{d}{dr}, \frac{1}{r} \frac{d }{d \theta}\right) $$

Assuming a vector field $F=(u,v)$ then the divergence of $F$ should be:

$$\begin{align} \nabla \cdot F &= \left(\frac{d}{dr}, \frac{1}{r} \frac{d }{d \theta}\right) \cdot (u,v) \\ &= \frac{du}{dr}+\frac{1}{r} \frac{d v }{d \theta} \\ &= \frac{1}{r} \left(r\frac{du}{dr}+ \frac{d v }{d \theta}\right) \end{align}$$

But everywhere I see: $$ \nabla \cdot F =\frac{1}{r}\left(\frac{d(ru)}{dr}+ \frac{d v }{d \theta}\right) $$

My question is, is something wrong with my product, or perhaps it shouldn't be treated as a dot product (I read some people telling that divergence is not a real dot product)?

Hint . You should use the chain rule to express $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ as linear combinations of $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$. The "del" operator in polar coordinates is not what you were thinking. — Kurt G., Apr 17 '22 at 15:40
You have to note that the basis vectors change in polar coordinates. So $\mathrm{div}:F =\nabla\cdot F$ is only true in cartesian coordinates. You can see the general formula here. — Silas, Apr 17 '22 at 16:24
Does this answer your question? How to Derivate Divergence in Cylindrical Coordinates? — Kurt G., May 17 '24 at 10:56

Divergence operator in polar coordinates

0 Answers0