What is the expansion of the operation $\nabla(r^2\phi(r))$ , where $r$ is $ |r|$

Question

What is the expansion of the operation $\nabla(r^2\phi(r))$ , where $r$ is $|\mathbf{r}|$?

Is it $(r(\mathrm{d}\phi/\mathrm{d}r)+2\phi)\mathbf{r}$ ?

Answer 1

Indeed the identity holds:

$\mathbf{\text{grad}} (r^2\phi(r))=(2\phi(r)+r\phi'(r))\mathbf{r}$ [1]

where $\mathbf{\text{grad}}$ is the gradient operator, $\phi(r)$ a scalar function, $\mathbf{r}$ a three dimensional vector and $r=|\mathbf{r}|$. I just modified a bit the notation of the OP using bold notation for vector properties.

To prove [1], just perform the calculation using:

• The i component of the gradient $(\mathbf{\text{grad}}f(\mathbf{r}))_i$ is by definition $\partial_{r_i} f(\mathbf{r})$ for every function $f(\mathbf{r})$
• $\partial_{r_i} r = r_i/r$ (in case this is not known, try to prove it :) )

Answer 2

By the chain rule$$\frac{\partial}{\partial r_j}f(r)=\frac{1}{2r}\frac{\partial r^2}{\partial r_j}\frac{\partial}{\partial r}f(r)=\frac{r_j}{r}f^\prime(r)\implies\nabla f=\frac{f^\prime}{r}\mathbf{r}.$$You asked about the case $f=r^2\phi$,$$\nabla(r^2\phi)=(r\phi^\prime+2\phi)\mathbf{r}.$$