# Vector field (rotors and nabla operators)

1. Jun 16, 2014

### skrat

1. The problem statement, all variables and given/known data
Find $\alpha$ and $p$ so that $\nabla \times \vec{A}=0$ and $\nabla \cdot \vec{A}=0$, where in $\vec{A}=r^{-p}[\vec{n}(\vec{n}\vec{r})-\alpha n^2\vec{r}]$ vector $\vec{n}$ is constant.

2. Relevant equations

3. The attempt at a solution

$\nabla \times \vec{A}=0$

$\nabla \times \vec{A}=\nabla \times [r^{-p}\vec{n}(\vec{n}\vec{r})-r^{-p}\alpha n^2\vec{r}]=$
$=r^{-p}(\vec{r}\vec{n})\nabla \times \vec{n}+\nabla(r^{-p}(\vec{r}\vec{n}))\times \vec{n}-r^{-p}\alpha n^2\nabla \times \vec{r}-\nabla(r^{-p}\alpha n^2)\times \vec{r}=$
$=0+[(\vec{r}\vec{n})\nabla r^{-p}+\nabla(\vec{r}\vec{n})r^{-p}]\times \vec{n}-0-[r^{-p}\alpha n^2\nabla \times \vec{r}+\alpha n^2\nabla r^{-p}\times \vec{r}]=$
$=(\vec{r}\vec{n})(-p)r^{-p-2}\vec{r}\times \vec{n}$

$\nabla \cdot \vec{A}=0$

$\nabla \cdot \vec{A}=\nabla [r^{-p}\vec{n}(\vec{n}\vec{r})-r^{-p}\alpha n^2\vec{r}]=$
$=\nabla (r^{-p}(\vec{n}\vec{r}))\vec{n}-r^{-p}\alpha n^2\nabla \vec{r}-\nabla(r^{-p}\alpha n^2)\vec{r}=$
$=(r^{-p}\vec{n}-pr^{-p-2}(\vec{r}\vec{n})\vec{r})\vec{n}-3\alpha n^2 r^{-p}+p\alpha n^2r^{-p-1}$

I know something is wrong, I just don't know what and where :(

2. Jun 16, 2014

### skrat

Does this mean that from $\nabla \times \vec{A}=0$ we know that $p=0$ and accordingly from the second equation that $\alpha =-\frac{1}{3n}$ ?

3. Jun 16, 2014

### Saitama

There should be a factor of 3 in the first term and the last term should have $r^{-p}$ instead of $r^{-p-1}$.

I think you are correct about $p=0$.

4. Jun 16, 2014

### skrat

hmmm.

First term only:

$\nabla r^{-p}(\vec{n}\vec{r})\vec{n}=$
$=r^{-p}(\vec{n}\vec{r})\nabla\vec{n}+(\nabla r^{-p}(\vec{n}\vec{r}))\vec{n}=$
$=0+[(\vec{n}\vec{r})\nabla r^{-p}+(\nabla(\vec{n}\vec{r}))r^{-p}]\vec{n}=$
$=[-p(\vec{n}\vec{r})r^{-p-1}\frac{\vec{r}}{r}+r^{-p}\vec{n}]\vec{n}=$
$=-pr^{-p-2}(\vec{n}\vec{r})^2+r^{-p}n$

I don't know where would factor 3 come from?

5. Jun 16, 2014

### Saitama

I meant that it should be $3r^{-p}n^2$ instead of $r^{-p}n^2$. How do you get $r^{-p}n^2$?

6. Jun 16, 2014

### skrat

Well

$\nabla(\vec{r}\vec{n})=\vec{n}$

$\vec{n}r^{-p}$ and finally $\vec{n}^2r^{-p}=n^2r^{-p}$

7. Jun 16, 2014

### Saitama

No, this is not correct. How do you conclude this?

8. Jun 16, 2014

### skrat

$\nabla(\vec{n}\vec{r})=\nabla(xn_x+yn_y+zn_z)=(\frac{\partial }{\partial x}(xn_x),\frac{\partial }{\partial y}(yn_y),\frac{\partial }{\partial z}(zn_z))=(n_x,n_y,n_z)=\vec{n}$

is it not?

9. Jun 16, 2014

### diegzumillo

Weird. I was about to correct you because you seem to consider each term as a vector component, but when I wrote the thing down it resulted in the exact same thing :P

$$\nabla(\vec{n}.\vec{r})= \nabla(xn_x+yn_y+zn_z)=\left( \frac{\partial }{\partial x}(xn_x+yn_y+zn_z),\frac{\partial }{\partial y}(xn_x+yn_y+zn_z),\frac{\partial }{\partial z}(xn_x+yn_y+zn_z) \right)=(n_x,n_y,n_z)=\vec{n}$$

10. Jun 16, 2014

### skrat

I was a bit lazy and didn't write terms that disappear after derivation. :)

11. Jun 16, 2014

### Saitama

Hmm.....that looks correct although the following identity I found on wiki doesn't seem to agree.

$$\nabla (\vec{f}\cdot \vec{g})=(\vec{f}\cdot \nabla)\vec{g}+(\vec{g}\cdot \nabla)\vec{f}+\vec{f}\times(\nabla \times \vec{g})+\vec{g}\times(\nabla \times \vec{f})$$

In the prsent case, $\vec{f}=\vec{n}$ and $\vec{g}=\vec{r}$. Three terms in the above identity are zero and $\nabla\cdot r=3$ and this gives the factor of 3. Any idea what's wrong with the above?

http://en.m.wikipedia.org/wiki/Vector_calculus_identities

12. Jun 16, 2014

### CAF123

The first term is the only non-vanishing term. $\nabla \cdot r \neq r \cdot \nabla$. The first is a scalar (divergence of r) and the latter is a differential operator.

13. Jun 16, 2014

### Saitama

Thanks CAF!

I haven't yet seen that in the book I am currently following. How do you calculate $\vec{r}\cdot \nabla$? (Sorry if this is an idiotic question, I have only a 2-days experience with vector calculus :tongue: )

14. Jun 16, 2014

### diegzumillo

I made this calculation here breaking into components and I got the same result. $p=0$ and $\alpha =1/3$. How do you know this is wrong?

edit: excuse my blindness, that's not the result you found.

Last edited: Jun 16, 2014
15. Jun 16, 2014

### skrat

Mainly because I got $\alpha =-\frac{1}{3n}$ and you haven't.

No other reason actually. I just doubt that I did right.

16. Jun 16, 2014

### skrat

Nonononononono!

I have a sign error in my notes. You are right, I also get $\alpha = \frac 1 3$

17. Jun 16, 2014

### skrat

This problem has one more question I have no idea how to answer:

How can scalar potential of this vector field with calculated $p$ and $\alpha$ be written?

18. Jun 16, 2014

### CAF123

$$\vec r \cdot \nabla = (xe_x + ye_y + ze_z) \cdot \left\{e_x\frac{\partial}{\partial x} + e_y\frac{\partial}{\partial y} + e_z\frac{\partial}{\partial z} \right\}$$ Then doing the dot product gives $$x \frac{\partial}{\partial x} + y \frac{\partial}{\partial y} + z \frac{\partial}{\partial z}$$ which vanishes when acted on a constant vector. So I do not mean to say $\vec r \cdot \nabla$ vanishes identically.

19. Jun 16, 2014

### Saitama

Understood, thanks a lot!

20. Jun 16, 2014

### diegzumillo

I assume you mean finding $\varphi$ from the definition
$$-\nabla \varphi = \vec A$$

I don't remember any elegant way of solving this besides solving three (or even less) integral equations like $-\frac{\partial \varphi}{\partial x} = A_x \Rightarrow \varphi = \int dx A_x$