$\nabla \times \left(\frac{\mathbf{A \times r}}{r^3}\right)$, where $\mathbf{A}$ is independent of $\displaystyle\nabla \times$

Question

The curl is just over $\mathbf{r}$ and $r$. I've been trying to pull the vector $\displaystyle \mathbf{A}$ out of the way, in order to get a expression much easier to deal with, but I have no idea how to do it without expanding the whole expression (I'm not sure whether what I did is correct either).

This is my attempt:

$$\nabla \times \left(\frac{\mathbf{A \times r}}{r^3}\right) = \frac{\nabla \times (\mathbf{A \times r})r^3+ (\mathbf{A \times r)}\nabla r^3}{r^6}$$

In the second term of the numerator:

$\displaystyle \nabla r^3 = 3r^2\mathbf{\hat{r}}$

For the nastiest part of the first term:

\begin{align*} \nabla \times (\mathbf{A \times r}) &= \mathbf{A}(\nabla \cdot \mathbf{r}) - \mathbf{r}(\nabla \cdot \mathbf{A})+ (\mathbf{r}\cdot \nabla ) \mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{r}\\ &= 3\mathbf{A}- 0 + (\mathbf{r}\cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{r} \end{align*}

According to me, $\displaystyle \nabla \cdot \mathbf{A} = 0$. And I'm stuck there, I don't know what to do with the directional derivatives, but it doesn't seem to be correct so far.

Thanks!!

Answer 1

not sure what you mean about the curl, but the following thoughts, along the lines you have indicated, may (or may not) be of some help

for a scalar function $f$ and vector field $\mathbf{v}$ we have: $$ \nabla \times f\mathbf{v} = \nabla f \times \mathbf{v} + f \nabla \times \mathbf{v} $$ setting $f=\frac1{r^3}$ and $\mathbf{v}=\mathbf{A \times r}$ gives $$ \nabla \times \left(\frac{\mathbf{A \times r}}{r^3}\right) = \nabla \frac1{r^3} \times (\mathbf{A \times r}) + \frac1{r^3} \nabla \times (\mathbf{A \times r}) \tag{1} $$ we have $$ \nabla \frac1{r^3} =-\frac3{r^5}\mathbf r $$ and $$ \mathbf{r \times (A \times r)} = r^2(\mathbf {A-A_r}) $$ where $\mathbf{A_r} = (\mathbf{\hat r} \cdot \mathbf {A) \hat r}$ is the projection of $\mathbf{A}$ onto the radius vector. thus for the first summand in (1): $$ \nabla \frac1{r^3} \times (\mathbf{A \times r}) = \frac3{r^3}(\mathbf {A_r-A}) $$ following your analysis of the second summand of (1), (but leaving in the term you set to zero), we first observe that $(\mathbf{A} \cdot \nabla)\mathbf{r} = \mathbf A$. so $$ \frac1{r^3} \nabla \times (\mathbf{A \times r}) = \frac1{r^3} \left( 2\mathbf A - \mathbf{r}(\nabla \cdot \mathbf{A})+ (\mathbf{r}\cdot \nabla ) \mathbf{A} \right) $$ if $\mathbf A$ is constant this reduces to: $$ \nabla \times \left(\frac{\mathbf{A \times r}}{r^3}\right) = \frac1{r^3}(3\mathbf A_r - \mathbf A) $$

Answer 2

We are given a constant vector $\vec A$ and a scalar $\frac{\vec r}{r}$ and asked to find

$$\nabla \times\left(\vec A\times \frac{\vec r}{r}\right)$$

We have the Vector Cross Product identity

$$\nabla \times\left(\vec A\times \frac{\vec r}{r^3}\right)=\vec A\left(\nabla \cdot \frac{\vec r}{r^3}\right)-\frac{\vec r}{r^3}(\nabla \cdot \vec A)+\left(\frac{\vec r}{r^3}\cdot\nabla\right)\vec A-\left(\vec A\cdot\nabla\right)\frac{\vec r}{r^3}\tag 1$$

Inasmuch as $\vec A$ is a constant, both the second and third terms on the left-hand side of $(1)$ are zero.

Since $\frac{\vec r}{r^3}$ has only a radial component in spherical coordinates, its divergence vanishes for $r\ne 0$. And so the first term is also zero for $r\ne 0$.

Therefore, we have only the fourth term to evaluate. To do this, we will use tensor notation with summation implied for repeated indices.

$$-\left(\vec A\cdot\nabla\right)\frac{\vec r}{r^3}=-A_i\frac{\partial }{\partial x_i}\frac{\hat x_j x_j}{r^3}=\hat x_jA_i\frac{3x_ix_j-r^2\delta_{ij}}{r^5}$$

Putting it all together we have

$$\bbox[5px,border:2px solid #C0A000]{\nabla \times\left(\vec A\times \frac{\vec r}{r^3}\right)=\frac{3(\vec A\cdot \hat r)\hat r-\vec A}{r^3}}$$