# Divergence in Polar Coordinates

1. Sep 20, 2008

### neutrino2063

Why is
$$\nabla\cdot\vec{A}=\frac{1}{r}\frac{\partial}{\partial r}(rA_{r})+\frac{1}{r}\frac{\partial}{\partial \theta}(A_{\theta})$$

Where
$$\vec{A}=A_{r}\hat{r}+A_{\theta}\hat{\theta}$$
And
$$\nabla=\hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}$$

$$\nabla\cdot\vec{A}=\frac{\partial}{\partial r}(A_{r})+\frac{1}{r}\frac{\partial}{\partial \theta}(A_{\theta})$$

Last edited: Sep 20, 2008
2. Sep 20, 2008

### nathan12343

Because the unit vectors are actually functions of position in cylindrical coordinates. This means all the derivative in the gradient operator act not only on the components of a particular vector, but also the unit vectors themselves.