When given two vectors $\mathbf{A}$ and $\mathbf{B}$, the curl of the cross product of these two is given by
$$\nabla\times(\mathbf{A}\times\mathbf{B})=(\mathbf{B}\cdot\nabla)\mathbf{A}-\mathbf{B}(\nabla\cdot\mathbf{A})-(\mathbf{A}\cdot\nabla)\mathbf{B}+\mathbf{A}(\nabla\cdot\mathbf{B}).$$
Using this relation, we can write
$$\nabla\times(\boldsymbol{\mu}_I\times\mathbf{r})=(\mathbf{r}\cdot\nabla)\boldsymbol{\mu}_I-\mathbf{r}(\nabla\cdot\boldsymbol{\mu}_I)-(\boldsymbol{\mu}_I\cdot\nabla)\mathbf{r}+\boldsymbol{\mu}_I(\nabla\cdot\mathbf{r}).$$
In a lecture course that I'm reading, it is stated that this can actually be rewritten as
$$\nabla\times(\boldsymbol{\mu}_I\times\mathbf{r})=-(\boldsymbol{\mu}_I\cdot\nabla)\mathbf{r}+\boldsymbol{\mu}_I(\nabla\cdot\mathbf{r}),$$
which means that the first two terms on the right hand side cancel. These can be written out, which results in
$$\begin{array}{l@{\;}l}
(\mathbf{r}\cdot\nabla)\boldsymbol{\mu}_I-\mathbf{r}(\nabla\cdot\boldsymbol{\mu}_I)&=\left(y\partial_y\mu_x-x\partial_y\mu_y+z\partial_z\mu_x-x\partial_z\mu_z\right)\boldsymbol{\hat{\textbf{i}}}\\
&+\left(x\partial_x\mu_y-y\partial_x\mu_x+z\partial_z\mu_y-y\partial_z\mu_z\right)\boldsymbol{\hat{\textbf{j}}}\\
&+\left(x\partial_x\mu_z+y\partial_y\mu_z-z\partial_x\mu_x-z\partial_y\mu_y\right)\boldsymbol{\hat{\textbf{k}}}.
\end{array}$$
(Sorry for the ugly ihat, SE doesn't support \imath for some reason.) In order for this to be zero, each term should be zero. Taking for example the $\boldsymbol{\hat{\textbf{i}}}$ component, this means that
$$y\partial_y\mu_x-x\partial_y\mu_y+z\partial_z\mu_x-x\partial_z\mu_z=0.$$
In this case, $\boldsymbol{\mu}_I$ is proportional to an angular momentum operator. Is this a special case, or is the above equation always true? And if so, why?
[Physics] The curl of a special cross product
angular momentumcalculusvectors
Related Solutions
To show that $$ \left(\sigma\cdot\mathbf{n}\right)^2=\mathbf n\cdot\mathbf n+i\sigma\cdot\left(\mathbf n\times\mathbf n\right)\tag{1} $$ consider writing the above as \begin{align} \left(\sigma\cdot\mathbf a\right)\left(\sigma\cdot\mathbf b\right)&=\sum_j\sigma_ja_j\sum_k\sigma_kb_k\\ &=\sum_j\sum_k\left(\frac12\{\sigma_j,\,\sigma_k\}+\frac12[\sigma_j,\,\sigma_k]\right)a_jb_k\\ &=\sum_j\sum_k\left(\delta_{jk}+i\epsilon_{jkl}\sigma_l\right)a_jb_k\tag{2} \end{align} where the 2nd line arises from using the anti-commutating and commutating relation for the matrices. In the third line, we have the Kronecker delta and Levi-Civita symbol. The result (1) follows from completing the math from (2) (that is, writing it in vector notation and replacing $\mathbf a$ and $\mathbf b$ with $\mathbf n$).
The remainder is to show that this is equal to 1. For that, the following two hints should suffice:
- Note that for two vectors $\mathbf a$ and $\mathbf b$, $\mathbf a\times\mathbf b=-\mathbf b\times\mathbf a$. What requirement is needed if $\mathbf b=\mathbf a$: $\mathbf a\times\mathbf a=?$
- For the unit vector, e.g. $\mathbf n=(1,\,0)^T$, what is the dot product?
As ACuriousMind has already noted, you can geometrically interpret the length of the cross product of two vectors as the area of the parallelogram (or as twice the area of the triangle) spanned by them, and (the absolute values of) its components as the areas of the projections of that parallelogram onto the coordinate planes.
As for the dot product of two vectors, based on the law of cosines, you can interpret it as half the difference between the sum of their squares and the square of their difference:
$$\|\vec a - \vec b\|^2 = \|\vec a\|^2 + \|\vec b\|^2 - 2(\vec a \cdot \vec b).$$
In other words, taking the vectors to be two sides of a triangle, the dot product measures (half) the amount by which Pythagoras' law fails for this triangle.
Another way to geometrically interpret (the absolute value of) the dot product is as half the area of the triangle formed by rotating one of the vectors by 90° in their common plane, and then taking the resulting vectors as two sides of a triangle:
This follows from the well known dot product formula $\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos \gamma$, where $\gamma$ is the angle between $\vec a$ and $\vec b$, from the triangle area formula $T = \frac12 \|\vec a'\| \|\vec b\| \sin \gamma'$, where $T$ is the area of the triangle formed by the vectors $\vec a'$ and $\vec b$ and $\gamma'$ is the angle between them, and the fact that the angles $\gamma$ and $\gamma'$ are complementary, and so $|\cos \gamma| = |\sin \gamma'|$.
Note the similarity with the cross product here. In fact, we always have $\|\vec a \times \vec b\| = |\vec a' \cdot \vec b|$, where $\vec a'$ is $\vec a$ rotated by 90° in their common plane (or in any of the planes, if there are several)!
Ps. I did notice (after posting this answer) that you asked specifically about the units of the products and "not about geometric interpretations." Even so, these examples should at least show that both the dot and the cross product of two length vectors can, in fact, be meaningfully interpreted as areas, and it should therefore not be surprising that, if the original vectors have units of, say, meters, then their product will be measured in square meters.
Best Answer
Since $\mu_I$ is proportional to the angular momentum, I'm guessing that $\mu_I$ is the magnetic moment of some particle. If so, $\mu_I$ is a constant vector; any derivative hitting on it will vanish.
Recall that magnetic moment is defined for a localized current distribution, which is the following integral, \begin{equation} {\bf m } = \frac{1}{2}\int {\bf x'} \times J({\bf x'}) dV' \end{equation} just like the total charge of a localized charge distribution, it has no spatial dependence.