How to find divergence of $f(\vec{r})\vec{r}$

Question

I have to find $$\mathrm{div}f(\vec{r})\vec{r}$$ I only know that $\mathrm{r}$ is a motion vector so it should be $\vec{r} = \{x;y;z\}$ and also that $\mathrm{div} = \left(\vec{\nabla},f(r)\vec{r}\right)$ but I do not know how to handle $f(r)\vec{r}$ there properly to substitute it into determinant.

Answer 1

Assumptions:

1. $\vec{r}=\{x,y,z\}$, in other words, $\vec{r}$ is a position vector.

2. $f(\vec{r})$ is a real-valued function.

3. $f(\vec{r})\vec{r}$ is a vector-valued function, where $\vec{r}$ is scaled by $f(\vec{r})$. In other words, $f(\vec{r})\vec{r}=(f(\vec{r})x,f(\vec{r})y,f(\vec{r})z)$.

4. The question asks for $\operatorname{div}(f(\vec{r})\vec{r})$ (there are no parentheses in the original post).

Since $f(\vec{r})\vec{r}=(f(\vec{r})x,f(\vec{r})y,f(\vec{r})z)$, \begin{align*} \operatorname{div}(f(\vec{r})\vec{r})&=\frac{\partial}{\partial x}f(\vec{r})x+\frac{\partial}{\partial y}f(\vec{r})y+\frac{\partial}{\partial z}f(\vec{r})z\\ &=3f(\vec{r})+f_x(\vec{r})x+f_y(\vec{r})y+f_z(\vec{r})z \end{align*}

Answer 2

Here is a standard vector calculus identity: $$\nabla.(f \vec A) = (\nabla f). \vec A + f (\nabla . \vec A).$$

You can use this to get $$ \nabla (f\vec r) = (\nabla f). \vec r + f (\nabla . \vec r).$$

But $\nabla . \vec r = 3$, so this reduces to $$ \nabla (f \vec r) = (\nabla f) . \vec r + 3f.$$

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It might be good to rederive the same result in spherical polar coordinates. First, you write $\vec r$ as $r \vec e_r$, where $r$ is the radial coordinate in your spherical coordinate system, and $\vec e_r$ is a unit vector pointing radially outwards. So $f(\vec r) \vec r = rf(r,\theta, \phi) \vec e_r$.

Using the formula for del in spherical polars, we have $$ \nabla . (rf(r, \theta, \phi) \vec e_r) = \frac 1 {r^2} \frac{\partial (r^3 f(r,\theta, \phi))}{\partial r},$$ and you should be able to verify that this agrees with the earlier result.

In fact, since $f$ only depends on the radial coordinate $r$ in your example, you can change the partial derivative to an ordinary derivative: $$ \nabla . (rf(r) \vec e_r) = \frac 1 {r^2} \frac{d(r^3 f(r))}{dr}.$$