Vector field (rotors and nabla operators)

[skrat]

Homework Statement

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.

Homework Equations

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 :(

 

[skrat]

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

 

[Saitama]

$=(r^{-p}\vec{n}-pr^{-p-2}(\vec{r}\vec{n})\vec{r})\vec{n}-3\alpha n^2 r^{-p}+p\alpha n^2r^{-p-1}$

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 $p=0$.

 

[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?

 

[Saitama]

hmmm.

First term only:

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

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

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

 

[skrat]

Well

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

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

 

[Saitama]

Well

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

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

 

[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?

 

[diegzumillo]

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

is it not?

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

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

 

[skrat]

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

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

 

[Saitama]

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

is it not?

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

 

[CAF123]

Any idea what's wrong with the above?

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.

 

[Saitama]

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.

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: )

 

[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.

 

[skrat]

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?

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

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

 

[skrat]

Nonononononono!

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

 

[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?

 

[CAF123]

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: )

$$\vec r \cdot \nabla = (xe_x + ye_y + ze_z) \cdot \left\{e_x\frac{\partial}{\partial x} + e_y\frac{\partial}{\partial y} + e_z\frac{\partial}{\partial z} \right\}$$ 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.

 

[Saitama]

$$\vec r \cdot \nabla = (xe_x + ye_y + ze_z) \cdot \left\{e_x\frac{\partial}{\partial x} + e_y\frac{\partial}{\partial y} + e_z\frac{\partial}{\partial z} \right\}$$ 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.

Understood, thanks a lot!

 

[diegzumillo]

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?

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 $-\frac{\partial \varphi}{\partial x} = A_x \Rightarrow \varphi = \int dx A_x$

 

[diegzumillo]

Just be careful when performing an integration on x, for example, since its integration constant could very well be a function of y and z. Yeah, you'll have to do the other integrations as well.

 

[skrat]

Ok, I get it.

$\varphi=\int A_xdx+C(y,z)$ now the idea is to find what $C(y,z)$ really is.

Will do that in my next post.

 

[CAF123]

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

Given the supposition $\vec A = \nabla \varphi$, you could find $\varphi$ by doing a line integral over an arbritary path between some reference point and some point in space. Analogous to the definition of potential energy in physics.

 

[skrat]

Ok,now second part.

$\vec{A}=[\vec{n}(\vec{r}\vec{n})-\alpha n^2\vec{r}]r^{-p}=[\vec{n}(\vec{r}\vec{n})-\frac{n^2}{3}\vec{r}]$ if $p=0$ and $\alpha = 1/3$.

$[\vec{n}(\vec{r}\vec{n})-\frac{n^2}{3}\vec{r}]=[n_x(xn_x+yn_y+zn_z)-\frac{n^2}{3}x,n_y(xn_x+yn_y+zn_z)-\frac{n^2}{3}y,n_z(xn_x+yn_y+zn_z)-\frac{n^2}{3}z]$

So $U=\int A_xdx=\int (n_x(xn_x+yn_y+zn_z)-\frac{n^2}{3}x)dx=$
$=\int (n_x(xn_x+yn_y+zn_z)-\frac{n^2}{3}x)dx=\int (xn_x^2+yn_yn_x+zn_zn_x-\frac{n^2}{3}x)dx=$
$=n_x^2\frac{x^2}{2}+(yn_yn_x+zn_zn_x)x-\frac{n^2}{6}x^2+C(z,y)$

Now $\frac{\partial U}{\partial y}=xn_yn_x+\frac{\partial }{\partial y}C(z,y)$ which should be equal to $A_y=n_y(xn_x+yn_y+zn_z)-\frac{n^2}{3}y$ meaning that $\frac{\partial }{\partial y}C(z,y)=yn_y^2+zn_zn_y-\frac{n^2}{3}y$.

Therefore $C=yzn_yn_z+\frac{n^2}{2}z^2-\frac{n^2}{6}z^2+D(z)$

Now if you don't mind... I think you already got the idea how I calculated this so if you don't mind, I will just write what Is my final expression for $U$

$U=n_x^2\frac{x^2}{2}+(yn_yn_x+zn_zn_x)x-\frac{n^2}{6}x^2+C(z,y)=$
$=n_x^2\frac{x^2}{2}+(yn_yn_x+zn_zn_x)x-\frac{n^2}{6}x^2+2zyn_zn_y-\frac{n^2}{6}(y^2+z^2)+\frac{n_y^2}{2}y ^2+\frac{n_z^2}{2}z^2=$
$=(yn_yn_x+zn_zn_x)x+2zyn_zn_y-\frac{n^2}{6}(x^2+y^2+z^2)+\frac 1 2 (x^2n_x^2+y^2n_y^2+z^2n_z^2)$

Hopefully..