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Vector field (rotors and nabla operators)

  1. Jun 16, 2014 #1
    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. jcsd
  3. Jun 16, 2014 #2
    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}## ?
     
  4. Jun 16, 2014 #3
    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##.
     
  5. Jun 16, 2014 #4
    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?
     
  6. Jun 16, 2014 #5
    I meant that it should be ##3r^{-p}n^2## instead of ##r^{-p}n^2##. How do you get ##r^{-p}n^2##?
     
  7. Jun 16, 2014 #6
    Well

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

    ##\vec{n}r^{-p}## and finally ##\vec{n}^2r^{-p}=n^2r^{-p}##
     
  8. Jun 16, 2014 #7
    No, this is not correct. How do you conclude this?
     
  9. Jun 16, 2014 #8
    ##\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?
     
  10. Jun 16, 2014 #9
    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

    [tex]\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}[/tex]
     
  11. Jun 16, 2014 #10
    I was a bit lazy and didn't write terms that disappear after derivation. :)
     
  12. Jun 16, 2014 #11

    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
     
  13. Jun 16, 2014 #12

    CAF123

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    Gold Member

    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.
     
  14. Jun 16, 2014 #13
    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: )
     
  15. Jun 16, 2014 #14
    I made this calculation here breaking into components and I got the same result. [itex]p=0[/itex] and [itex]\alpha =1/3[/itex]. How do you know this is wrong?

    edit: excuse my blindness, that's not the result you found.
     
    Last edited: Jun 16, 2014
  16. Jun 16, 2014 #15
    Mainly because I got ##\alpha =-\frac{1}{3n}## and you haven't.

    No other reason actually. I just doubt that I did right.
     
  17. Jun 16, 2014 #16
    Nonononononono!

    I have a sign error in my notes. You are right, I also get ##\alpha = \frac 1 3 ##
     
  18. Jun 16, 2014 #17
    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?
     
  19. Jun 16, 2014 #18

    CAF123

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    Gold Member

    $$\vec r \cdot \nabla = (xe_x + ye_y + ze_z) \cdot \left\{e_x\frac{\partial}{\partial x} + e_y\frac{\partial}{\partial y} + e_z\frac{\partial}{\partial z} \right\}$$ 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.
     
  20. Jun 16, 2014 #19
    Understood, thanks a lot! :smile:
     
  21. Jun 16, 2014 #20
    I assume you mean finding [itex]\varphi[/itex] from the definition
    [tex]-\nabla \varphi = \vec A [/tex]

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