# 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 ∞ P: 12,369 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,293 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 $\partial z/\partial r= r/z$.