Question in divergence and curl

[yungman]

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

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

 

[CompuChip]

Yes. By

Let $\vec {F}(\vec {r}')$ be a vector function of position vector $\vec {r}'=\hat x x'+\hat y y'+\hat z z'$.

we mean that it does not depend on x, y or z, so $$\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial z} = 0.$$

 

[voko]

LBut what if
$$\vec {F}(\vec {r}')=\frac {1}{4\pi r r'}\;\hbox { where }\; r=\sqrt{x^2+y^2+z^2}$$

This equality is impossible because the right hand side is not vectorial.

But your general line of thinking is correct.

 

[yungman]

Thanks for verifying this.