# Some expressions with Del (nabla) operator in spherical coordinates

1. Apr 18, 2012

### lol_nl

Reading through my electrodynamics textbook, I frequently get confused with the use of the del (nabla) operator. There is a whole list of vector identities with the del operator, but in some specific cases I cannot figure out what how the operation is exactly defined.

Most of the problems occur at this single equation:
$\nabla(\frac{\vec{p} \cdot \hat{r}}{r^2})$

which occurs in the calculation of the electric field from a dipole potential, i.e.
$\vec{E_{dip}} = \nabla(V_{dip}) = \nabla(\frac{1}{4\pi\epsilon_{0}}\frac{\vec{p} \cdot \hat{r}}{r^2})$,
where the vector p stands for the dipole moment of the dipole and r hat is the unit vector in the radial direction in spherical coordinates.

Now I know that the nabla operator satisfies the product rule, so this simplifies to:
$\nabla(\frac{\vec{p} \cdot \hat{r}}{r^2}) = \nabla(\vec{p} \cdot \hat{r})\frac{1}{r^2} + (\vec{p} \cdot \hat{r}) \nabla(\frac{1}{r^2})$

Now using the formula for the gradient in spherical coordinates I can easily calculate the last term, but the first term is what is giving me trouble. Using one of the product rules, I see that
$\nabla(\vec{p} \cdot \hat{r}) = \vec{p} \times (\nabla \times \hat{r}) + \hat{r} \times (\nabla \times \vec{p}) + (\vec{p} \cdot \nabla)\hat{r} + (\hat{r} \cdot \nabla)\vec{p}$.

The second term is zero, since p is a single vector, so its curl must be zero.
The first term appears to be zero as well by looking at the curl in spherical coordinates.
The third and fourth terms are where I get stuck. I know that
$(\vec{p} \cdot \nabla) \neq (\nabla \cdot \vec{p})$
and that in (x,y,z) coordinates
$(\vec{p} \cdot \nabla) = (p_{x} \frac{\partial}{\partial x} + p_{y} \frac{\partial}{\partial y}+ p_{z} \frac{\partial}{\partial z})$,
but since I have to let this operator work on r hat, I wish to express it in spherical coordinates. Is this even possible or is the only way to compute this by writing r hat in (x,y,z) coordinates?
A similar problem occurs for the fourth term.

Apparently the result can be written as a relatively simple equation involving only p and r hat.

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Another equation that has caught my attention is this one:
$(\vec{p} \cdot \nabla)\vec{E})$.
The author claims that in his context, where p is a single vector not depending on the position, this can be written as
$\nabla(\vec{p} \cdot \vec{E})$.
However, these are two very different equations, with the first one involving the divergence and the third one involving the gradient. How can they at all be equal?

2. Apr 18, 2012

### sunjin09

The way you attempt to simplify is very difficult, but $\hat{p}\cdot\hat{r}$ is simply $\cos(\theta)$ if $\hat{p}$ is aligned with z axis.

Vector identities are very awkward to work with. There are better ways, don't waste your time on them.