Solving a Divergence Question: What Equation Do I Use?

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In summary: So\begin{align*}div \vec{A(r)} &= (x(dA_1/dr)+ y(dA_2/dr)+ z(dA_3/dr))/r\\&= (x(r/x)+ y(r/y)+ z(r/z))/r\end{align*}
  • #1
matematikuvol
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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?
 
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  • #2
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
Is r a vector or not? Either way it seems that there is a problem with that equation.
 
  • #4
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
Yes. But I'm not sure why is that correct? Could you explain me that?
 
  • #6
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].
 

Related to Solving a Divergence Question: What Equation Do I Use?

1. What is a divergence question?

A divergence question is a type of mathematical problem that involves finding the rate at which a vector field is expanding or contracting at a given point.

2. What equation do I use to solve a divergence question?

The equation used to solve a divergence question is known as the divergence theorem, which states that the divergence of a vector field in a region is equal to the flux of the vector field through the boundary of that region.

3. How do I know when to use the divergence theorem?

The divergence theorem is used when you need to find the overall behavior of a vector field, such as whether it is expanding or contracting, rather than just the behavior at a specific point.

4. Are there any simplifications or special cases for using the divergence theorem?

Yes, there are a few simplifications and special cases for using the divergence theorem. For example, if the vector field is constant, the divergence is zero and the flux is equal to the surface area of the region. Additionally, if the region is closed, the flux through the boundary is equal to zero.

5. Can the divergence theorem be used in multiple dimensions?

Yes, the divergence theorem can be used in multiple dimensions. In two dimensions, it is known as Green's theorem, and in three dimensions, it is known as the divergence theorem. The formula for each version may vary slightly, but the basic concept remains the same.

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