Proof divergence of vector potential = 0

[grantdenbrock]

Homework Statement

I need to show that $$\del*\vec{A(\vec{r})}=\frac{\mu}{4\pi}\int{\frac{\vec{J{vec\r'}}}{\vec{R}}}d\tau=0$$
where A is the vector potential and R refers to "script r" or (r-r') where r is source point of charge and r' is the measurement point. tau refers to a volume integral. I have tried many times now to show this by bringing del into the integrand using product rules and the fact that $$\delR=-\del'R'$$ but cannot make it equal zero. I am not sure if there is something I have overlooked or another method to use but any help or suggestions are much appreciated!

Homework Equations

del*R=-del'R' must be used at some point[/B]

The Attempt at a Solution

My solution thus far goes like this (Sorry My latex is awful so I will just write out my method)
1) bring Del into the integrand
2) using product rule rule of dot products expand into 2 terms each with its own dot product
3) del*J' =0 since del operates on unprimed coordinates
4) J' del*1/R does not equal zero therefore integrand does not equal zero :/[/B]

 

[grantdenbrock]

I should add that its divergence of A, meaning del dot A =0 . Also, I am aware of the other thread regarding this problem however that thread does not contain any useful information or attempt a solution.

 

[fzero]

From what you've said, you should be able to show that
$$ \nabla \cdot \mathbf{A}(\mathbf{x}) = - \frac{\mu}{4\pi} \int \mathbf{J}(\mathbf{x}') \cdot \nabla' \left(\frac{1}{|\mathbf{x}- \mathbf{x}'|} \right) d\mathbf{x}'.$$
The next step would be to integrate by parts and argue that both terms vanish independently subject to reasonable assumptions.

 

[rude man]

You write "... r is source point of charge ..." but don't you mean "source point of current density"? The integral certainly looks like the magnetic vector potential, especially with the μ in it ... based on that assumption I offer the following:

∇⋅A = 0 is not an identity. It is chosen as a convenience.
We define A as H = ∇ x A.
By Maxwell, ∇ x H = j (j = current density).
So j = ∇ x (∇ x A).
But by a mathematical identity, this can be rewritten as
j = ∇(∇⋅A) - ∇²A
So we choose ∇⋅A = 0, giving
∇²A = -j
i.e. the Poisson equation, the solution for which is your integral. BTW the denominator in the integrand should read |R|. You can't divide one vector by another, at least not to my knowledge.