# Divergence question

by matematikuvol
Tags: divergence
 P: 192 I see identity in one mathematical book $$div \vec{A}(r)=\frac{\partial \vec{A}}{\partial r} \cdot grad r$$ How? From which equation?
 Homework Sci Advisor HW Helper Thanks ∞ PF Gold P: 11,009 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.
 P: 71 Is r a vector or not? Either way it seems that there is a problem with that equation.
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P: 38,885

## Divergence question

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).
 P: 192 Yes. But I'm not sure why is that correct? Could you explain me that?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,885 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 $\partial z/\partial r= r/z$.

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