Some expressions with Del (nabla) operator in spherical coordinates

Click For Summary
The discussion focuses on the complexities of using the del (nabla) operator in spherical coordinates, particularly in the context of calculating the electric field from a dipole potential. The main confusion arises from the expression \nabla(\frac{\vec{p} \cdot \hat{r}}{r^2}), where the application of the product rule leads to difficulties in evaluating the first term, \nabla(\vec{p} \cdot \hat{r}). The author highlights the challenge of expressing the operator in spherical coordinates and questions whether it's necessary to revert to Cartesian coordinates for clarity. Additionally, a separate equation involving (\vec{p} \cdot \nabla)\vec{E} raises concerns about the validity of equating it to \nabla(\vec{p} \cdot \vec{E}), emphasizing the distinct nature of divergence and gradient operations. Overall, the discussion underscores the intricacies of vector identities and the challenges they present in electrodynamics calculations.
lol_nl
Messages
40
Reaction score
0
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.

-----------------------------
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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K