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$\varphi(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\cdot\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'-\frac{1}{4\pi}\int_{S}\frac{\mathbf{F}(\mathbf{r}')\cdot\mathbf{\mathrm{d}S}'}{\left|\mathbf{r}-\mathbf{r}'\right|},$ $\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\times\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'+\frac{1}{4\pi}\int_{S}\frac{\mathbf{F}(\mathbf{r}')\times\mathbf{\mathrm{d}S}'}{\left|\mathbf{r}-\mathbf{r}'\right|}.$ If $V$ is $\mathbb{R}^3$ itself (unbounded), and $F$ vanishes sufficiently fast at infinity, then the second component of both scalar and vector potential are zero. That is, $\varphi(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\cdot\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V',$ $\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\times\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'.$

Why is it true that second components of these two potentials are zero?

user66077
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1 Answers1

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Informal explanation: because $S$ is the boundary of $V$, and when $V=\mathbb R^3$, the boundary of $V$ is empty.

More precisely: assume $\mathbf F(\mathbf r')=o(|\mathbf r'|^{-1})$. Apply the decomposition with $V$ being the ball of radius $R$. In either integral over $S$, the function to be integrated is estimated from above by $|\mathbf F(\mathbf r')|/|\mathbf r-\mathbf r'|$, which is $o(R^{-2})$ as $R\to\infty$. Since the area of $S$ is $4\pi R^2$, the integral over $S$ is $o(1)$. In the limit $R\to\infty$ the integral vanishes.

ˈjuː.zɚ79365
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