Surface integral of Poynting vector for time-invariant E & B fields

Question

I've got this question from electrodynamics, but I have a feeling that this is probably purely a maths question. I'm using standard terminology, but for folks not into electrodynamics here is the background:

Consider fields $\rho \left( \vec{r} \right)$, $\vec{J} \left( \vec{r} \right)$, $\vec{E} \left( \vec{r} \right)$ and $\vec{B} \left( \vec{r} \right)$ in $\mathbb{R}^3$.

Take any finite volume $V_s$ outside of which $\vec{J}\left(\vec{r}\right)$ and $\rho\left(\vec{r}\right)$ are $0$. Then define $\vec{E}\left(\vec{r}\right) $ and $\vec{B}\left(\vec{r}\right) $ for any $\vec{r}$ as follows:

$$ \vec{E} \left( \vec{r} \right) = \frac{1}{4\pi \epsilon_0} \iiint_{V_s} \frac{\rho\left(\vec{r}_s\right)}{\left|\vec{r}-\vec{r}_s\right|^3} \left(\vec{r}-\vec{r}_s\right) \space dV\left(\vec{r}_s\right) $$ $$ \vec{B} \left( \vec{r} \right) = \frac{\mu_0}{4\pi} \iiint_{V_s} \frac{\vec{J}\left(\vec{r}_s\right)}{\left|\vec{r}-\vec{r}_s\right|^3} \times \left(\vec{r}-\vec{r}_s\right) \space dV\left(\vec{r}_s\right) $$ Now my question is, given just this information, is it possible to mathematically prove that following surface integral will always evaluate to zero? If so, what is the proof? (To clarify, $\partial V_s$ is the bounding surface of the volume $V_s$ mentioned earlier)

$$ \frac{1}{\mu_0} \oint_{\partial V_s} \left(\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right)\cdot d\vec{S}\left(\vec{r}\right) $$ The meaning in physics is, I'm looking for proof that time invariant sources (static charges and constant currents confined to a volume) cannot radiate any energy, and I'm trying to do that without invoking the Hertzian dipole and Fourier analysis.

Update

I've summarized the consolidated solution in my answer below. Thanks to all in stackexchange Mathematics & Physics who helped.

Answer 1

$\vec E\cdot\vec J$ is the work exerted by the field on the currents. Its integral over all space vanishes if the currents form closed loops, as they must in a static scenario. Since $\vec E$ is static, it's conservative, so the integral of $\vec E\cdot\vec J$ along a closed loop vanishes.

More formally, consider the charge conservation equation $\displaystyle\frac{\partial\rho}{\partial t}+\vec\nabla\cdot\vec J=0$: Since $\displaystyle\frac{\partial\rho}{\partial t}=0$ we also have $\vec\nabla\cdot\vec J=0$, and thus $\vec J=\vec\nabla\times\vec Y$ for some $\vec Y$; substituting this into the integral, integrating by parts and using $\vec\nabla\times\vec E=0$ yields the result.

Answer 2

To get this done, we also need to spell out the condition $\nabla \cdot \vec{J}\left(\vec{r}\right)=0$ which is the continuity equation $-\dot{\rho}\left(\vec{r}\right) = \nabla \cdot \vec{J}\left(\vec{r}\right)$ as applicable in time invariant conditions.

From the definition or $\vec{E}$ and $\vec{B}$, we can apply a few vector identities and rewrite them as: $$ \vec{E} \left( \vec{r} \right) = -\nabla\Phi\left(\vec{r}\right),\quad \Phi\left(\vec{r}\right) = \frac{1}{4\pi \epsilon_0} \iiint_{V_s} \frac{\rho\left(\vec{r}_s\right)}{\left|\vec{r}-\vec{r}_s\right|} \space dV\left(\vec{r}_s\right) \\ \vec{B} \left( \vec{r} \right) = \nabla \times \vec{A}\left( \vec{r} \right), \quad \vec{A}\left( \vec{r} \right) = \frac{\mu_0}{4\pi} \iiint_{V_s} \frac{\vec{J}\left(\vec{r}_s\right)}{\left|\vec{r}-\vec{r}_s\right|} \space dV\left(\vec{r}_s\right) $$ From this definition, we can straightaway infer $\nabla \times \vec{E}\left(\vec{r}\right)=0$. With a bit more work, we can derive that $\nabla \times \vec{B}\left(\vec{r}\right)=\mu_0 \vec{J}\left(\vec{r}\right)$.

To begin the derivation, we apply the divergence theorem, to get: $$ \frac{1}{\mu_0} \iiint_{V_s} \nabla \cdot \left[\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right] \space dV\left(\vec{r}\right) $$ Applying a vector identity, this further becomes: $$ \frac{1}{\mu_0} \iiint_{V_s} \vec{B} \left( \vec{r} \right) \cdot \left[ \nabla \times \vec{E} \left( \vec{r} \right) \right] - \vec{E} \left( \vec{r} \right) \cdot \left[ \nabla \times \vec{B} \left( \vec{r} \right) \right] \space dV\left(\vec{r}\right) $$ Further substituting $\nabla \times \vec{E}\left(\vec{r}\right)=0$ and $\nabla \times \vec{B}\left(\vec{r}\right)=\mu_0 \vec{J}\left(\vec{r}\right)$ this becomes: $$ - \iiint_{V_s} \vec{E} \left( \vec{r} \right) \cdot \vec{J} \left( \vec{r} \right) \space d V\left(\vec{r} \right) $$

So now we substitute $\vec{E} \left(\vec{r}\right) = -\nabla\Phi\left(\vec{r}\right)$, and get: $$ \iiint_{V_s} \vec{J} \left( \vec{r} \right) \cdot \nabla\Phi \left( \vec{r} \right) \space d V\left(\vec{r} \right) $$ Now, we apply vector integration by parts to get: $$ \oint_{\partial V_s} \Phi \left( \vec{r} \right) \space \vec{J} \left( \vec{r} \right) \cdot \space d \vec{s}\left(\vec{r} \right) - \iiint_{V_s} \Phi \left( \vec{r} \right) \space \left[ \nabla \cdot \vec{J} \left( \vec{r} \right)\right] \space d V\left(\vec{r} \right) $$ Substituting $\nabla \cdot \vec{J} \left( \vec{r} \right) = 0$ this becomes: $$ \oint_{\partial V_s} \Phi \left( \vec{r} \right) \space \vec{J} \left( \vec{r} \right) \cdot \space d \vec{s}\left(\vec{r} \right) $$ And further, since by definition of $V_s$ we have $\vec{J}\left(\vec{r}\right) = 0$ for all $\vec{r} \in \partial V_s$, we find the integral evaluates to zero.