Divergence of a function of r

  • Thread starter avidfury
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Main Question or Discussion Point

Why is this true?

[tex]
\vec \bigtriangledown \cdot \vec f ( \vec r ) = \frac {\partial}{\partial r} (r^2 | \vec f ( \vec r ) | )
[/tex]
 

Answers and Replies

  • #2
Mute
Homework Helper
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It's not, at least not without some conditions on [itex]\mathbf{f}(\mathbf{r})[/itex] that you haven't given us. Take [itex]\mathbf{f}(\mathbf{r}) = \hat{\mathbf{x}}[/itex], for example. The divergence is zero, and the magnitude is just one, so

[tex]\frac{\partial}{\partial r}(r^2 |\mathbf{f}(\mathbf{r})|) = \frac{\partial}{\partial r}(r^2) = 2r[/tex]

[itex]\nabla \cdot \mathbf{f}(\mathbf{r}) = \partial_r(r^2|\mathbf{f}(\mathbf{r})|)[/itex] looks more like a differential equation to be solved rather than an identity.
 

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