Divergence question

1. Feb 6, 2013

matematikuvol

I see identity in one mathematical book
$$div \vec{A}(r)=\frac{\partial \vec{A}}{\partial r} \cdot grad r$$
How? From which equation?

2. Feb 6, 2013

Simon Bridge

what does $gradr$ mean?

Do you mean: $$\text{div}(\vec{A}(r)) = \frac{\partial\vec{A}(r)}{\partial r}\text{grad}(r)$$ ... still not sure it makes sense.

But it looks a bit like it might be a special case of the grad operator in spherical or polar coordinates.
I'm afraid you'll have to provide the reference - the book could simply be wrong.

3. Feb 6, 2013

joeblow

Is r a vector or not? Either way it seems that there is a problem with that equation.

4. Feb 6, 2013

HallsofIvy

Staff Emeritus
Assuming that r is the distance from (0, 0) to (x, y) then that equation is correct and is just the chain rule (although, strictly speaking, that partial derivative ought to be an ordinary derivative since A is assumed to be a function of r only).

5. Feb 6, 2013

matematikuvol

Yes. But I'm not sure why is that correct? Could you explain me that?

6. Feb 6, 2013

HallsofIvy

Staff Emeritus
As I said, it is the chain rule. We have $r= \sqrt{x^2+ y^2+ z^2}$ so that $\partial r/\partial x= x(x^2+ y^2+ z^2)^{-1/2}= x/r$, $\partial r/\partial y= y(x^2+ y^2+ z^2)^{-1/2}= y/r$, $\partial r/\partial z= z(x^2+ y^2+ z^2)^{-1/2}= z/r$. So $grad r= (xi+ yj+ zk)/r$.

If we write $\vec{A(r)}= A_1(r)i+ A_2(r)j+ A_3(r)k$ then $d\vec{A(r)}/dr= (dA_1/dr) i+ (dA_2/dr)j+ (dA_3/dr)k$ and $(d\vec{A})/dr\cdot grad r= (x(dA_1/dr)+ y(dA_2/dr)+ z(dA_3/dr))/r$

On the left, $div \vec{A(r)}= dA_1/dr+ dA_2/dr+ dA_3/dr= [(\partial A_1/\partial x)(\partial r/\partial x)+ (\partial A_1/\partial y)(\partial r/\partial y)+ (\partial A_1/\partial z)(\partial z/\partial r)]+ [(\partial A_2/\partial x)(\partial r/\partial x)+ (\partial A_2/\partial y)(\partial r/\partial y)+ (\partial A_2/\partial z)(\partial z/\partial r)]+ [(\partial A_3/\partial x)(\partial r/\partial x)+ (\partial A_3/\partial y)(\partial r/\partial y)+ (\partial A_3/\partial z)(\partial z/\partial r)]$
Now use the fact that $\partial x/\partial r= 1/(\partial r/\partial x)= r/x$, $\partial y/\partial r= r/y$, and $\partial z/\partial r= r/z$.