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Question in divergence and curl

  1. Aug 12, 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'##.

    Question is why:
    [tex]\nabla\cdot\vec {F}(\vec{r}')=\nabla\times\vec {F}(\vec{r}')=0[/tex]
    I understand ##\nabla## work on ##x,y,z##, not ##x',y',z'##. But what if
    [tex]\vec {F}(\vec {r}')=\frac {1}{4\pi r r'}\;\hbox { where }\; r=\sqrt{x^2+y^2+z^2}[/tex]

    My answer is if ##\vec {F}(\vec {r}')=\frac {1}{4\pi r r'}## , then it is a vector function of both ##r,r'##..........##\vec{F}(\vec {r},\vec {r}')## not just ##\vec {F}(\vec {r}')##. So ##\vec {F}(\vec {r}')## cannot have variable of ##x,y,z##. Am I correct?

    Thanks
     
    Last edited: Aug 12, 2013
  2. jcsd
  3. Aug 12, 2013 #2

    CompuChip

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    Yes. By
    we mean that it does not depend on x, y or z, so $$\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial z} = 0.$$
     
  4. Aug 12, 2013 #3
    This equality is impossible because the right hand side is not vectorial.

    But your general line of thinking is correct.
     
  5. Aug 12, 2013 #4
    Thanks for verifying this.
     
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