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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 $\mathbb{R}^3$, with their usual meaning as per Electrodynamics.

Take any finite volume $V_s$ outside of which $\vec{J}\left(\vec{r}\right)$ and $\rho\left(\vec{r}\right)$ are $0$. Then, we know that $\vec{E}\left(\vec{r}\right) $ and $\vec{B}\left(\vec{r}\right) $ for any $\vec{r}$ are 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, is it possible to mathematically prove that the 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)

$$ \frac{1}{\mu_0} \oint_{\partial V_s} \left(\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right)\cdot d\vec{S}\left(\vec{r}\right) $$ Motivation: 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.

Thanks...

Update

As pointed out below, since $\vec{E}$ and $\vec{B}$ are time invariant, applying Poynting's theorem this boils down to proving that the following volume integral is zero: $$ - \iiint_{V_s} \left( \vec{J}\left(\vec{r}\right) \cdot \vec{E}\left(\vec{r}\right) \right) \space dV\left(\vec{r}\right) $$ So can this be proved, given that $\vec{J}$, $\vec{E}$, $\vec{B}$ and $\rho$ are all time invariant?

Thanks...

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 $\mathbb{R}^3$, with their usual meaning as per Electrodynamics.

Take any finite volume $V_s$ outside of which $\vec{J}\left(\vec{r}\right)$ and $\rho\left(\vec{r}\right)$ are $0$. Then, we know that $\vec{E}\left(\vec{r}\right) $ and $\vec{B}\left(\vec{r}\right) $ for any $\vec{r}$ are 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, is it possible to mathematically prove that the 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)

$$ \frac{1}{\mu_0} \oint_{\partial V_s} \left(\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right)\cdot d\vec{S}\left(\vec{r}\right) $$ Motivation: 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.

Thanks...

Update

As pointed out below, since $\vec{E}$ and $\vec{B}$ are time invariant, applying Poynting's theorem this boils down to proving that the following volume integral is zero: $$ - \iiint_{V_s} \left( \vec{J}\left(\vec{r}\right) \cdot \vec{E}\left(\vec{r}\right) \right) \space dV\left(\vec{r}\right) $$ So can this be proved, given that $\vec{J}$, $\vec{E}$, $\vec{B}$ and $\rho$ are all time invariant, beyond the trivial cases of $\vec{J} = 0$, $\vec{E} = 0$ or $\vec{E} \perp \vec{J}$ for all $\vec{r} \in V_s$?

Or, if this can't be proved, is there a counterexample?

Thanks...

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 $\mathbb{R}^3$, with their usual meaning as per Electrodynamics.

Take any finite volume $V_s$ outside of which $\vec{J}\left(\vec{r}\right)$ and $\rho\left(\vec{r}\right)$ are $0$. Then, we know that $\vec{E}\left(\vec{r}\right) $ and $\vec{B}\left(\vec{r}\right) $ for any $\vec{r}$ are 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, is it possible to mathematically prove that the 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)

$$ \frac{1}{\mu_0} \oint_{\partial V_s} \left(\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right)\cdot d\vec{S}\left(\vec{r}\right) $$ Motivation: 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.

Thanks...

Update

As pointed out below, since $\vec{E}$ and $\vec{B}$ are time invariant, this boils down to proving that the following volume integral is zero: $$ - \iiint_{V_s} \left( \vec{J}\left(\vec{r}\right) \cdot \vec{E}\left(\vec{r}\right) \right) \space dV\left(\vec{r}\right) $$ So can this be proved, given that $\vec{J}$, $\vec{E}$, $\vec{B}$ and $\rho$ are all time invariant?

Thanks...

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 $\mathbb{R}^3$, with their usual meaning as per Electrodynamics.

Take any finite volume $V_s$ outside of which $\vec{J}\left(\vec{r}\right)$ and $\rho\left(\vec{r}\right)$ are $0$. Then, we know that $\vec{E}\left(\vec{r}\right) $ and $\vec{B}\left(\vec{r}\right) $ for any $\vec{r}$ are 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, is it possible to mathematically prove that the 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)

$$ \frac{1}{\mu_0} \oint_{\partial V_s} \left(\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right)\cdot d\vec{S}\left(\vec{r}\right) $$ Motivation: 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.

Thanks...

Update

As pointed out below, since $\vec{E}$ and $\vec{B}$ are time invariant, applying Poynting's theorem this boils down to proving that the following volume integral is zero: $$ - \iiint_{V_s} \left( \vec{J}\left(\vec{r}\right) \cdot \vec{E}\left(\vec{r}\right) \right) \space dV\left(\vec{r}\right) $$ So can this be proved, given that $\vec{J}$, $\vec{E}$, $\vec{B}$ and $\rho$ are all time invariant?

Thanks...

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 $\mathbb{R}^3$, with their usual meaning as per Electrodynamics.

Take any finite volume $V_s$ outside of which $\vec{J}\left(\vec{r}\right)$ and $\rho\left(\vec{r}\right)$ are $0$. Then, we know that $\vec{E}\left(\vec{r}\right) $ and $\vec{B}\left(\vec{r}\right) $ for any $\vec{r}$ are 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, is it possible to mathematically prove that the 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)

$$ \frac{1}{\mu_0} \oint_{\partial V_s} \left(\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right)\cdot d\vec{S}\left(\vec{r}\right) $$ Motivation: 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.

Thanks...

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 $\mathbb{R}^3$, with their usual meaning as per Electrodynamics.

Take any finite volume $V_s$ outside of which $\vec{J}\left(\vec{r}\right)$ and $\rho\left(\vec{r}\right)$ are $0$. Then, we know that $\vec{E}\left(\vec{r}\right) $ and $\vec{B}\left(\vec{r}\right) $ for any $\vec{r}$ are 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, is it possible to mathematically prove that the 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)

$$ \frac{1}{\mu_0} \oint_{\partial V_s} \left(\vec{E}\left(\vec{r}\right) \times \vec{B}\left(\vec{r}\right)\right)\cdot d\vec{S}\left(\vec{r}\right) $$ Motivation: 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.

Thanks...

Update

As pointed out below, since $\vec{E}$ and $\vec{B}$ are time invariant, this boils down to proving that the following volume integral is zero: $$ - \iiint_{V_s} \left( \vec{J}\left(\vec{r}\right) \cdot \vec{E}\left(\vec{r}\right) \right) \space dV\left(\vec{r}\right) $$ So can this be proved, given that $\vec{J}$, $\vec{E}$, $\vec{B}$ and $\rho$ are all time invariant?

Thanks...