# Magnetic field tangential surface integral

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.
• Welcome to Physics Stack Exchange. We generally leave out the "thanks for reading" pleasantries in questions here. The goal is clarity above all else. I edited that stuff out. Have fun on the site :-) – DanielSank Jul 25 '15 at 19:00
• Can you give more details about your question? One derives that $E=-\partial_t A-\nabla\phi$ from the fact that $E+\partial_t A$ is irrotational. – Enredanrestos Jul 26 '15 at 2:08
• Sure, I added the main steps to the question. – user3586384 Jul 26 '15 at 11:48

I do not know whether I understood exactly your question. In any case, I do not think that you need to use Helmholtz decomposition to establish that $E=-\nabla\phi-\partial_t A$ (using $c=1$). Particularly, the fields do not have to fulfill the hypotheses of Helmholtz decomposition in all space for this identity to be true. Maybe we could work in infinitesimal volumes, but then the full Helmholtz decomposition needs to be taken (the one that includes the border terms). I use all space, so that is a disadvantage of the following answer. I noted also that you missed a number of $4 \pi$s.
We have $E=-\nabla\phi-\nabla\times F$, where $$F=-\int \frac{\nabla'\times E(r')}{4\pi |r-r'|}dV'~~.$$ Therefore, let's calculate \begin{align} \nabla\times F &=-\nabla\times\int \frac{\nabla'\times E(r')}{4\pi |r-r'|}dV'\\ &=\nabla\times\int \frac{\partial_t B(r')}{4\pi |r-r'|}dV'\\ &=\partial_t\int\left(\nabla\frac{1}{|r-r'|}\right)\times\frac{B'}{4\pi}dV'\\ &=\partial_t\int\left(-\nabla'\frac{1}{|r-r'|}\right)\times\frac{B'}{4\pi}dV'\\ &=\partial_t\int\left(-\nabla'\times\frac{B(r')}{4 \pi|r-r'|}\right)+\frac{\nabla'\times B(r')}{4\pi|r-r'|}dV'\\ &=\partial_tA+\partial_t\int-\nabla'\times\frac{B(r')}{4 \pi|r-r'|}dV'\\ \end{align}
So, we need to prove that $\partial_t\int-\nabla'\times\frac{B(r')}{4 \pi|r-r'|}dV'=0$, but we are assuming that $B$ has Helmholtz decomposition in all space. For this to happen, we need that $|B(r)|r^2$ to be bounded. This may be related with the Sommerfeld condition.
Therefore, \begin{align} |\int-\nabla'\times\frac{B(r')}{4 \pi|r-r'|}dV'|&=\lim_{R\rightarrow\infty}|\int_{\textrm{B}(R)}-\nabla'\times\frac{B(r')}{4 \pi|r-r'|}dV'|\\ &=\lim_{R\rightarrow\infty}|\int_{\partial \textrm{B}(R)}\hat{r}^\prime\times\frac{B(r')}{4 \pi|r-r'|}dS'|\\ &\le \lim_{R\rightarrow\infty} (\max_{\partial\textrm{B}(R)}{|B|})*4\pi R^2*\frac{1}{4\pi R} \end{align} But since $(\max_{\partial\textrm{B}(R)}{|B|})*4\pi R^2$ is bounded in $R$, then the limit is zero.
• Great, thank you! I was missing the bounded $|B(r)|r^{2}$ requirement - I disregarded the boundary conditions \ convergence entirely. I wasn't trying to establish the relation, but to get an expression for the potentials using the decomposition, which will consist while deriving from both fields. Your answer was very helpful. – user3586384 Jul 27 '15 at 23:40