- #1
yungman
- 5,755
- 293
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
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
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