Trying to prove the following identity

Question

So I am trying to prove that:

$$\nabla \cdot (f \cdot \vec{r}\,) = \frac{(r^{3}f)'}{r^{2}}$$

where $f=f(r)$ and $\vec{r}=r \hat{r}$ and $\hat{r}$ is the unit vector in the radial direction in spherical coordinates, and del is the del operator.

I have tried to use the definition of the del operator in spherical coordinates and just plug it in the equation, namely:

$$ \Bigl( \frac{\partial}{\partial r} \hat{r} \Bigr) (f(r) \cdot r \hat{r}) = \frac{\partial}{\partial r} (f(r)r) = r \frac{d}{dr}f + 1 \cdot f(r). $$ However this does not match the right hand side of the equation I am trying to prove.

Thanks

Answer 1

$\newcommand{\vect}[1]{{\bf #1}}$ In spherical coordinates

$$ \nabla \cdot \vect{F} = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2F_r) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta A_\theta) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi}A_\phi $$

In your case $\vect{F} = f(r)\vect{r} = f(r)r\hat{\vect{r}}$, so that

$$ \nabla \cdot \vect{F} = \frac{1}{r^2}\frac{{\rm d}}{{\rm d}r}(r^2 f(r)r) = \frac{1}{r^2}\frac{{\rm d}}{{\rm d}r}(r^3 f(r)) $$

Answer 2

I thought it would be instructive to present a coordinate-free approach. To that end, we proceed.

Using the product rule, $\nabla \cdot (\phi \vec A)=\phi (\nabla\cdot \vec A)+(\nabla \phi)\cdot \vec A$, we have with $\vec A=\vec r$ and $\phi=f$

$$\nabla\cdot(f\vec r)=f(\nabla\cdot\vec r)+(\nabla f)\cdot \vec r$$

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Then, noting that $\nabla\cdot\vec r=3$ and $(\nabla f)\cdot \vec r=r\frac{\partial f}{\partial r}$ reveals

$$\nabla\cdot(f\vec r)=3f+r\frac{\partial f}{\partial r}=3f+rf'$$

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Finally, using the product rule, we see that $\frac{(r^3f)'}{r^2}=3f+rf'$ thereby establishing the equality

$$\nabla\cdot(f\vec r)=\frac{(r^3f)'}{r^2}$$

as was to be shown!