According to Gauss's theorem the integral of $\vec \nabla⋅(\phi \vec E)$ taken over the whole space is equal to 0?

Question

The following excerpt is from Space—Time—Matter by Hermann Weyl, starting on page 101. Among the things I'm not understanding is Weyl's assertion that the integral of $\nabla\cdot(\phi\mathbf{E})$ vanishes due to Gauss's theorem. Why should I believe that?

The potential $\phi$ is defined by $\mathbf{E}=\nabla\phi.$ I've noticed the current German language edition has some of the signs reversed, so that might correctly be $-\phi.$ But I don't think that matters for my question.

A system of discrete point-charges $e_{1}, e_{2}, e_{3},\dots,$ has potential energy $$ U = \frac{1}{8\pi} \sum_{i \neq k} \frac{e_{i} e_{k}}{r_{ik}} $$ in which $r_{ik}$ denotes the distance between the two charges $e_{i}$ and $e_{k}$. This signifies that the virtual work which is performed by the forces acting at the separate points (owing to the charges at the remaining points) for an infinitesimal displacement of the points is a total differential, viz. $\delta U$. For continuously distributed charges this formula resolves into $$ U = \iint \frac{\rho(P) \rho(P')}{8\pi r_{PP'}}\, dV\, dV' $$ in which both volume integrations with respect to $P$ and $P'$ are to be taken over the whole space, and $r_{PP'}$ denotes the distance between these two points. Using the potential $\phi$ we may write $$ U = -\tfrac{1}{2} \int \rho\phi\, dV. $$ The integrand is $\phi \nabla\cdot\mathbf{E}$. In consequence of the equation $$ \nabla\cdot(\phi\mathbf{E}) = \phi \nabla\cdot\mathbf{E} + \mathbf{E}\cdot \nabla\phi $$ and of Gauss's theorem, according to which the integral of $\nabla\cdot(\phi\mathbf{E})$ taken over the whole space is equal to $0$, we have $$ -\int \rho\phi\, dV = \int (\mathbf{E} \cdot\nabla\phi)\, dV = \int |E|^{2}\, dV; $$ i.e. $$ U = \int \frac{1}{2} \left|\mathbf{E}\right|^{2} dV . $$

Answer 1

Gauss's theorem is another name for the divergence theorem, which says that if you have some region $M$ with boundary $\partial M$ then

$$\int_M \mathrm{div}(\mathbf F) \mathrm dV = \oint_{\partial M} \mathbf F \cdot \mathrm d\mathbf S$$

Let $M$ be a ball of radius R centered at the origin. We would then have that

$$\int_{|\mathbf r|<R} \mathrm{div}(\mathbf F) \mathrm dV = \oint_{|\mathbf r|= R} \mathbf F \cdot \mathrm d\mathbf S \leq 4\pi R^2 \cdot \sup_{|\mathbf r|=R}|\mathbf F(\mathbf r)|$$

As a result, if $|\mathbf F(\mathbf r)|\rightarrow 0$ faster than $1/|\mathbf r|^2$ as $|\mathbf r|\rightarrow \infty$, then we can take the limit as $R\rightarrow \infty$ (thereby integrating over all space) to obtain $$\int_{\mathbb R^3} \mathrm{div}(\mathbf F) \mathrm dV = 0$$

All that remains is to show that this is true for $\phi \mathbf E$. As long as the number of point charges in question is finite, then this is not difficult to see (in the limit of large $R$, $\phi \mathbf E \sim R^{-3}$).

Answer 2

The physical meaning for that, I'm not sure if this is something you already know, there is no potential difference at the same surface. There's potential difference between one surface and another, but at the same surface the potential difference is zero, and this is what we call an equipotential surface. The definition of an equipotential surface is a surface composed of all those points having the same value of potential. So, there is no potential difference between any two points on this surface. therefore, No work is involved in moving a unit charge around on an equipotential surface, or any closed path.