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Another divergence and curl.

  1. Aug 13, 2013 #1
    Let ##\vec {F}(\vec {r}')## be a vector function of position vector ##\vec {r}'=\hat x x'+\hat y y'+\hat z z'##. I want to find ##\nabla\cdot\frac {\vec {F}(\vec {r}')}{|\vec {r}-\vec{r}'|}##.

    My attempt:
    Let ##\vec {r}=\hat x x+\hat y y+\hat z z##. Since ##\nabla## work on ##x,y,z##, not ##x',y',z##'
    [tex]\Rightarrow\;\nabla\cdot\vec {F}(\vec{r}')=\nabla\times\vec {F}(\vec{r}')=0[/tex]
    Then ##\frac {1}{|\vec {r}-\vec{r}'|}## is not even a vector. How can I perform ##\nabla\cdot\frac {\vec {F}(\vec {r}')}{|\vec {r}-\vec{r}'|}##?




    BUT
    according to http://faculty.uml.edu/cbaird/95.657%282012%29/Helmholtz_Decomposition.pdf. It uses:
    [tex]\nabla\cdot (\Phi\vec A)=\vec A\cdot\nabla\Phi+\Phi\nabla\cdot\vec A\;\hbox { Where }\;\Phi=\frac{1}{|\vec {r}-\vec{r}'|}\;\hbox {and }\;\vec A=\vec {F}(\vec {r}')[/tex]
    You can see this from the bottom of page 2 to the top of page 3. I cannot agree with this, ##\vec A=\vec {F}(\vec {r}')## is function of ##x',y',z'##, it's a constant respect to ##x,y,z## as indicated above.

    Even the Wikipadia use the same method:http://en.wikipedia.org/wiki/Helmholtz_decomposition. You can see the 5th equation under "Proof".

    Please tell me what's going on. Thanks
     
    Last edited: Aug 13, 2013
  2. jcsd
  3. Aug 13, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    With respect to which vector do you want to take the derivatives? It's important to say, whether you want to calculate [itex]\partial/\partial \vec{x}[/itex] or [itex]\partial/\partial \vec{x}'[/itex]!
     
  4. Aug 13, 2013 #3
    I think I got the answer, please verify my finding:

    [tex]\nabla\cdot\frac {\vec {F}(\vec {r}')}{|\vec {r}-\vec{r}'|}=\frac{\partial}{\partial x}\left(\frac{1}{|\vec {r}-\vec{r}'|}F_{x'}\right)+\frac{\partial}{\partial y}\left(\frac{1}{|\vec {r}-\vec{r}'|}F_{y'}\right)+\frac{\partial}{\partial z}\left(\frac{1}{|\vec {r}-\vec{r}'|}F_{z'}\right)[/tex]
    [tex]=\nabla\left(\frac{1}{|\vec {r}-\vec{r}'|}\right)\cdot \vec {F}(\vec {r}')+\left(\frac{1}{|\vec {r}-\vec{r}'|}\right)(\nabla\cdot\vec{F}(\vec{r}'))[/tex]
    [tex]\hbox{Because }\;\nabla\cdot \vec {F}(\vec {r}')=0[/tex]
    [tex]\Rightarrow\;\nabla\cdot\frac {\vec {F}(\vec {r}')}{|\vec {r}-\vec{r}'|}=\vec {F}(\vec {r}')\cdot\nabla\left(\frac{1}{|\vec {r}-\vec{r}'|}\right)[/tex]

    Even though ##\vec {F}(\vec {r}')## is not a function of ##(x,y,z)## but you still need to tread it as a vector and use the product rule. Am I correct.
     
    Last edited: Aug 13, 2013
  5. Aug 13, 2013 #4

    king vitamin

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    Gold Member

    Indeed, even though [itex]\vec{F}(\vec{r}')[/itex] is independent of [itex]\vec{r}[/itex], the vector [itex]\frac{\vec{F}(\vec{r}')}{|\vec{r}-\vec{r}'|}[/itex] is not and it will have nonzero divergence with respect to the unprimed coordinates. The formula you gave in your first post is correct, we should have

    $$ \nabla \cdot \left( \frac{\vec{F}(\vec{r}')}{|\vec{r}-\vec{r}'|} \right) = \frac{1}{|\vec{r}-\vec{r}'|} \left( \nabla \cdot \vec{F}(\vec{r}') \right) + \vec{F}(\vec{r}') \cdot \left( \nabla \frac{1}{|\vec{r}-\vec{r}'|} \right) $$

    where the last term involves a gradient of a scalar. As you say, the first term on the right hand side is zero, so only the last term contributes. The above formula shouldn't be too hard to derive yourself, try writing it out in components.
     
  6. Aug 13, 2013 #5
    Thanks for the reply, I just updated the last post, please check my work. I think it's agreeing with you.

    Thanks
     
  7. Aug 13, 2013 #6

    king vitamin

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    Gold Member

    Yes, your reasoning in the post looks perfect!
     
  8. Aug 13, 2013 #7
    Thanks
     
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