Solving a Divergence Question: What Equation Do I Use?

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Discussion Overview

The discussion revolves around the mathematical identity involving the divergence of a vector field, specifically the expression div \vec{A}(r) and its relation to the gradient of the radial distance r. Participants explore the validity of the equation, its derivation, and the implications of treating r as a scalar or vector.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the meaning of "grad r" and suggests that the equation might be a special case in spherical or polar coordinates.
  • Another participant raises a concern about whether r is a vector, indicating a potential issue with the equation.
  • A different participant asserts that if r is defined as the distance from the origin, the equation is correct and relates it to the chain rule, although they note that the partial derivative should be an ordinary derivative.
  • Further elaboration is provided on the chain rule, detailing how to derive the gradient of r and its components in terms of x, y, and z.
  • One participant expresses uncertainty about the correctness of the equation and requests clarification on the derivation process.
  • A detailed breakdown of the divergence operation is presented, including the relationships between partial derivatives and the radial distance r.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the equation or the interpretation of r. Multiple competing views and uncertainties remain regarding the mathematical treatment of the terms involved.

Contextual Notes

There are unresolved questions about the definitions of r and its treatment as a scalar or vector, as well as the assumptions underlying the derivation of the divergence expression.

matematikuvol
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I see identity in one mathematical book
div \vec{A}(r)=\frac{\partial \vec{A}}{\partial r} \cdot grad r
How? From which equation?
 
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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.
 
Is r a vector or not? Either way it seems that there is a problem with that equation.
 
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).
 
Yes. But I'm not sure why is that correct? Could you explain me that?
 
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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