When deriving: $\vec{E}=-\frac{\partial\vec{A}}{\partial t}-\nabla\phi$ from Helmholtz decomposition of $\vec{E}$, It seems necessary to show: $$\iiint \nabla' \times \frac{\vec{B}(\vec{r')}}{|\vec{r} - \vec{r'}|} d^{3}r'= 0 \, .$$
I'm then using the identity $$\iiint(\nabla \times \vec{A})\, dv \:=\: \oint_s(\hat{n} \times\vec{A}) \, dS \, .$$
Is there a reason why, in the general case, we can state that $$\oint_{s}\frac{\vec{B}(\vec{r')}}{|\vec{r}-\vec{r'}|}\times dS'=0 \, .$$
Is it somehow related to Sommerfeld Radiation Condition?
** Main steps of derivation:
- We need to show: $\nabla\times\iiint\frac{\nabla'\times\vec{E}(\vec{r'})}{|\vec{r}-\vec{r'}|}d^{3}r'=-\frac{\partial\vec{A}}{\partial t}$.
- We can use Faraday's law, take the time derivative out, and the spatial curl into the integral.
- We then use: $\nabla\times\varphi\vec{F}=\nabla\varphi\times\vec{F}+\varphi\nabla\times\vec{F}$, and take $\nabla'=-\nabla$ on the scalar term $\frac{1}{|\vec{r}-\vec{r'}|}$
- By using the same identity again, we get: $-\iiint\frac{\nabla'\times\vec{B}(\vec{r}')}{|\vec{r}-\vec{r}'|}d^{3}r'$ which we know is $\vec{A}$, plus the term stated above which we need to show is zero.