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I have some expressions with grad and div operators and would like to re-write the expressions so that there are no grads or divs. Basically, I have been told that $\nabla \phi$, where $\phi = 1/4 \pi |\textbf{r}|$, can be re-written as

$\nabla \phi = \frac{\textbf{r}}{|\textbf{r}|} \frac{\text{d}\phi}{\text{d}|\textbf{r}|}$.

I might need to refresh my knowledge of vector calculus, but I suppose this just follows from the chain rule?

Edit:

$\nabla \nabla \phi = \frac{\nabla (\textbf{r})}{|\textbf{r}|}\frac{\partial}{\partial r}\phi + \hat{r} \nabla \bigg( \frac{\partial}{\partial r} \phi \bigg)=\frac{\textbf{r}}{|\textbf{r}|} \frac{\text{d} \phi}{\text{d} |\textbf{r}|} + \frac{\textbf{r}}{|\textbf{r}|}\frac{\textbf{r}}{|\textbf{r}|} \frac{\partial}{\partial r} \bigg( \frac{\partial}{\partial} \phi \bigg) = \frac{\delta_{ij}}{|\textbf{r}|}\frac{\text{d} \phi}{\text{d}|\textbf{r}|} + \frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} , $

$\nabla \nabla \nabla \phi = \nabla \bigg( \frac{\delta_{ij}}{|\textbf{r}|}\frac{\text{d} \phi}{\text{d}|\textbf{r}|} + \frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} \bigg)=\nabla \bigg(\frac{\delta_{ij}}{|\textbf{r}|} \bigg) \frac{\text{d}\phi}{\text{d} |\textbf{r}|} + \frac{\delta_{ij}}{|\textbf{r}|} \nabla \bigg(\frac{\partial \phi}{\partial r} \bigg) + \nabla \bigg(\frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \bigg) \frac{\partial^2 \phi}{\partial r^2} +\frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \nabla \bigg( \frac{\partial^2 \phi}{\partial r^2} \bigg) = \nabla \bigg(\frac{\delta_{ij}}{|\textbf{r}|} \bigg) \frac{\text{d}\phi}{\text{d} |\textbf{r}|} + \frac{\delta_{ij} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} + \frac{\delta_{ij} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} + \frac{\textbf{r} \otimes \delta_{ij} }{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} + \frac{\textbf{r} \otimes \textbf{r} \otimes \textbf{r}}{|\textbf{r}|^3} \frac{\text{d}^3 \phi}{\text{d} |\textbf{r}|^3} , $

$\nabla \cdot \nabla \phi = \nabla^2 \phi = \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} $.

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Your expression for the gradient will work with any function that is a function purely of the radius (i.e., no angular dependencies). In spherical coordinates:

$\displaystyle \nabla\Phi = \hat{r}\frac{\partial}{\partial r}\Phi + \hat{\theta}\frac{\partial}{\partial \theta}\Phi+\hat{\phi}\frac{1}{\sin{\theta}}\frac{\partial}{\partial \phi}\Phi $

If $\Phi$ is a function of $r$ only,

$\displaystyle \nabla\Phi = \hat{r}\frac{\partial}{\partial r}\Phi$

But $\hat{r}=\bf{r}/|\bf{r}|$ and $\partial/\partial r = d/d|\bf{r}|$. Make those substitutions and you get your expression.

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  • $\begingroup$ And for example, if I then want to $\nabla \nabla \phi$ it's just repeat the same rule? What would $\nabla \cdot \nabla \phi$ be? $\endgroup$ – Tom Jun 19 '19 at 17:25
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    $\begingroup$ @Tom Again, if $\Phi$ depends only on $r$, it would just be $\partial^2\Phi/\partial r^2=d^2\Phi/dr^2$ $\endgroup$ – bob.sacamento Jun 19 '19 at 17:28
  • $\begingroup$ I've tried a few more as practice, I'm not sure if these are correct. One thing that I'm not sure about in the expression is that you end up with $\nabla \delta_{ij}$, what happens when you have the grad operator act on a second rank tensor (obviously I know it will have to be a third rank tensor). $\endgroup$ – Tom Jun 19 '19 at 19:35

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