Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Divergence question

  1. Feb 6, 2013 #1
    I see identity in one mathematical book
    [tex]div \vec{A}(r)=\frac{\partial \vec{A}}{\partial r} \cdot grad r[/tex]
    How? From which equation?
     
  2. jcsd
  3. Feb 6, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  4. Feb 6, 2013 #3
    Is r a vector or not? Either way it seems that there is a problem with that equation.
     
  5. Feb 6, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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).
     
  6. Feb 6, 2013 #5
    Yes. But I'm not sure why is that correct? Could you explain me that?
     
  7. Feb 6, 2013 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

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

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

    On the left, [itex]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)][/itex]
    Now use the fact that [itex]\partial x/\partial r= 1/(\partial r/\partial x)= r/x[/itex], [itex]\partial y/\partial r= r/y[/itex], and [itex]\partial z/\partial r= r/z[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Divergence question
  1. Question on Divergence (Replies: 5)

Loading...