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If we consider a localized current distribution and the following natural boundary condition : $\vec A(\vec r \rightarrow 0)$ for $\vec r \rightarrow \infty$

If $$\vec A(\vec r)=\int_V G_0(\vec r, \vec r')j(\vec r')dV'=\frac {\mu_0}{4\pi}\int_V \frac{j(\vec r')}{|\vec r-\vec r'|}dV'$$

In my script we consider the divergence of this gauged vector potential (Lorenz gauge):

$$\nabla_{r}\cdot \vec A(\vec r)= \frac {\mu_0}{4\pi} \int_V j(\vec r')\nabla_{r} \left(\frac{1}{|\vec r-\vec r'|}\right) dV' \\\nabla_{r}\cdot \vec A(\vec r)= -\frac {\mu_0}{4\pi}\int_V j(\vec r')\nabla_{r'}\cdot \left(\frac{1}{|\vec r-\vec r'|}\right)dV' \\\nabla_{r}\cdot \vec A(\vec r)=-\frac {\mu_0}{4\pi}\int_V \nabla_{r'}\cdot \left(\frac{j(\vec r')} {|\vec r-\vec r'|}\right)dV'+\frac {\mu_0}{4\pi}\int_V \frac{1}{|\vec r-\vec r'|} \nabla_{r'}j(\vec r')dV'$$

Applying Gauss Law here:

$$\nabla_{r}\cdot \vec A(\vec r)=-\frac {\mu_0}{4\pi}\int_{(V)}\left(\frac{j(\vec r')} {|\vec r-\vec r'|}\right)d\vec f'+ \frac {\mu_0}{4\pi}\int_V \frac{1}{|\vec r-\vec r'|} \nabla_{r'}j(\vec r')dV'$$

The second term here is zero because as written in the title, we are in magnetostatics.

I don't understand the reason that was given as to why the first term was zero. The reason is:

The first term vanishes after applying Gauss' theorem (current density distribution is said to be localized at finite and thus vanishes on the surface of a boundary at infinity)

I don't understand this part "distribution is said to be localized at finite and thus vanishes on the surface of a boundary at infinity".

I understand that the current distribution is localized, meaning we have moving charges to a certain region in space, I don't understand the 2nd part of this reasoning. Can anyone help me with further clarification ?

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1 Answer 1

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The integral "$\int_{(V)}\ldots $" is over a surface "at infinity" i.e far from where the current $j$ is located. So $j=0$ on the surface and the integrand vanishes.

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  • $\begingroup$ But the integral, the surface integral should be over the surface of the localized volume in which the current is flowing? Where is this infinity coming from? When you use the gauss law for a charge inside a spherical shell, the surface integral is over the surface of the enclosing volume. SHouldn't the same apply here? $\endgroup$
    – imbAF
    Feb 6, 2022 at 16:35
  • $\begingroup$ The region $V$ can be any volume as long at it encloses all the current. Its surface can therefore be as far away as one likes. $\endgroup$
    – mike stone
    Feb 6, 2022 at 17:23

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