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Divergence in Polar Coordinates

  1. Sep 20, 2008 #1
    Why is
    [tex]\nabla\cdot\vec{A}=\frac{1}{r}\frac{\partial}{\partial r}(rA_{r})+\frac{1}{r}\frac{\partial}{\partial \theta}(A_{\theta})[/tex]

    [tex]\nabla=\hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}[/tex]
    Instead of just:

    [tex]\nabla\cdot\vec{A}=\frac{\partial}{\partial r}(A_{r})+\frac{1}{r}\frac{\partial}{\partial \theta}(A_{\theta})[/tex]
    Last edited: Sep 20, 2008
  2. jcsd
  3. Sep 20, 2008 #2
    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.
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