problem in proof for magnetic vec pot is 0

[Tor Fredrik]

 

I have tried to go through a proof for this which is used in physic texts:

 

$$\nabla \cdot \vec{A}=\frac{\mu _0}{4\pi}\nabla \cdot \int     \frac{ J(r')}{|r-r'|} dV'$$

 

again we use

 

$$   \nabla \cdot \frac{ 1}{|r-r'|}=-\nabla' \cdot \frac{ 1}{|r-r'|}$$

 

If you go through the equations you would obtain that you could write:

 

$$\nabla \cdot \vec{A}=-\frac{\mu _0}{4\pi}  \int  J(r') \cdot \nabla' \frac{ 1}{|r-r'|} dV'$$

 

Then they look at the integration as from integration by parts

 

$$\int_{-\infty}^\infty g \frac{\partial f}{\partial x} dx = [gf]_{-\infty}^\infty - \int_{-\infty}^\infty f \frac{\partial g}{\partial x} dx$$

 

initially the integral is over the current which does not go from $$-\infty \:\:   to  \:\:   \infty$$. I could perhaps reason and say that integrating from $$-\infty \:  to \:  \infty$$ would get the same result. But if we extend the reasoning I did get to

 

For the $$-\infty \:\:  to \:\:  \infty$$ we did get:

 

$$\nabla \cdot \vec{A}=-\frac{\mu _0}{4\pi}  \int_{-\infty}^\infty  J(r') \cdot \nabla' \frac{1 }{|r-r'|} dV'$$

 

 

$$ J(r') \cdot \nabla' \frac{1 }{|r-r'|}=J(r') \cdot[\frac{\partial}{\partial x'}\textbf{i}+\frac{\partial}{\partial y'}\textbf{j}+\frac{\partial}{\partial z'}\textbf{k}] \frac{1}{[(x-x')^2+(y-y')^2+(z-z')^2]^{0.5}}$$

 

for the first component:

$$ g=J_x(r')$$    $$f=\frac{1}{[(x-x')^2+(y-y')^2+(z-z')^2]^{0.5}}$$

 

If we would have kept the integration limits from dV' it would be apparent that

$$ [gf] \neq0$$

 

If we increase to from $$-\infty \:\:   to \: \:  \infty$$ it would be apparent that

 

$$\int_{-\infty}^\infty g \frac{\partial f}{\partial x} dx $$

would not change after the increase in integration limits to $$-\infty \: \:   to \: \:  \infty$$

and

$$- \int_{-\infty}^\infty f \frac{\partial g}{\partial x} dx$$

 

would change since  $$\frac{\partial g}{\partial x} $$ is undefied when the current density ends

but with from  $$-\infty \:\:   to \:\:   \infty$$ we would get

 

$$ [gf]_{-\infty}^\infty =0$$

 

which is a change? How does this prove that $$\nabla \cdot A=0$$ ?

 

 

 

 

Additional info:

Here is a proof from a physic text just in case. C below is used as current density above:

 

Edited December 1, 2020 by Tor Fredrik

[swansont]

 

Quote

initially the integral is over the current which does not go from −∞to∞

I don’t see any equation that is an integral over current. The equation to which you refer is an integral over volume.

[Tor Fredrik]

5 hours ago, swansont said:

 

I don’t see any equation that is an integral over current. The equation to which you refer is an integral over volume.

What other components could there be?

[swansont]

7 minutes ago, Tor Fredrik said:

What other components could there be?

.The variable of integration is volume in the equation. Not current. dV, not dJ. You are integrating the current over the volume.

“What other components could there be?” doesn’t make any sense. There can’t be any “other components”. The equation clearly says dV. There’s no wiggle room.

[joigus]

I'm not 100 % sure of what's bothering you, but it's useful to distinguish in these integrals the field point \( x \) and the source point \( x' \).

Now, the sources \( \boldsymbol{J} \) do not extend to spatial infinity. Nor do the fields \( \boldsymbol{A} \) (at the field points: the points at which the field is calculated, \( x \) ).

So,

\[\int\frac{\partial}{\partial x_{i}}\left[f\left(x\right)g\left(x-x'\right)\right]=\left.fg\right|_{\textrm{boundary}}\rightarrow0\]

The boundary is spatial infinity.

I hope that helps.

Edited December 2, 2020 by joigus

[Markus Hanke]

This seems like an awfully complicated way to do this. Why not just use Helmholtz’s Theorem? We know that the curl of the potential field gives the magnetic field (by definition!), so this is already fixed. The potential field is also invariant under certain gauge transformations (I think it’s the addition of a scalar field gradient, but I’d have to check that), hence we will always be free to make the divergence vanish, simply be choosing a suitable gauge, without affecting the curl. So in essence, under Helmholtz’s Theorem, the divergence has no physical relevance at all in this.

[Tor Fredrik]

Edited January 15, 2021 by Tor Fredrik