How can I express $\nabla^2(f(r)\mathbf{r})$ with index notation?

Question

I am trying to get an expresion of \begin{equation*} \nabla^2(f(r)\mathbf{r}), \end{equation*} where $\mathbf{r} = \mathbf{r}(x_1,x_2,x_3)$ and $|\mathbf{r}| = r$, using index notation. For example \begin{align*} \nabla^2f(r) &= \nabla\cdot\nabla f(r) \\ &= (\mathbf{e}_i\cdot\mathbf{e}_j)\partial_i(\partial_j(f(r))) \\ &= \delta_{ij}\partial_i(\partial_jf(r)) \\ &= \partial_i\partial_if(r) \\ &= \partial_1^2f(r) + \partial_2^2f(r) + \partial_3^2f(r). \end{align*} But I have not been able to it, because I don't know how continue from here \begin{align*} \nabla^2(f(r)\mathbf{r}) &= \nabla\cdot\nabla(f(r)\mathbf{r}) \\ &= \mathbf{e}_i\partial_i\cdot(\mathbf{e}_j\partial_j(\mathbf{e}_k f(r)r_k)). \qquad\longleftarrow\textbf{Here} \end{align*} I know it is a tensor, but, what is the notation for: $\mathbf{e}_i\mathbf{e}_j$.

Answer 1

Before we do any detailed calculations, let's see roughly what to expect.

1. We're applying a spherically symmetric vector-to-vector operator to (a spherically symmetric scalar times $\mathbf{r}$), so the answer must be a spherically symmetric scalar times $\mathbf{r}$.
2. Dimensional analysis tells us the only legal terms are $f^{(n)}r^{n-2}\mathbf{r}$. The Leibniz rule's generalization to higher-order derivatives tell us only the cases $0\le n\le2$ can come up. The case $n=0$ would be surprising too, since it means putting two derivatives on $\mathbf{r}$, which should give $\mathbf{0}$.
3. While you asked about the $3$-dimensional case, the $1$-dimensional case is easy, giving $(f(r)r)^{\prime\prime}=(f^{\prime\prime}+2f^\prime/r)r$. We'll find that's more or less what we get in $3$ dimensions, except the $r$ at the end becomes $\mathbf{r}$, while the coefficient $2$ will change... let's see to what.

Write $\nabla^2=\partial_i\partial_i$. Your expression is$$\partial_i\partial_i(f(r)r_j\mathbf{e}_j)=\left([\partial_i\partial_i f(r)]r_j+2\partial_if\partial_ir_j+f(r)\partial_i\partial_ir_j\right)\mathbf{e}_j.$$Since $\partial_ir_j=\delta_{ij}$, the above simplifies to $\left([\partial_i\partial_i f(r)]r_j+2\partial_jf\right)\mathbf{e}_j$. But$$\partial_kf(r)=\frac{1}{2r}f^\prime(r)\partial_k(r^2)=\frac{1}{2r}f^\prime(r)\partial_k(r_lr_l)=\frac{r_kf^\prime}{r},$$and so (we'll need this special case in a moment)$$\partial_i\frac1r=-\frac{r_i}{r^3}.$$So we can simplify again to$$\left([\partial_i(r_i f^\prime/r)]r_j+2r_jf^\prime/r\right)\mathbf{e}_j=(4f^\prime/r+f^{\prime\prime})r_j\mathbf{e}_j=(4f^\prime/r+f^{\prime\prime})\mathbf{r}.$$I've cancelled/collected many terms, but I recommend you write out all the details yourself. Note the $2$ became $4$; in $d$ dimensions, it would be $d+1$.