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?


$\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} $.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.