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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, 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...

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3 Answers

up vote 2 down vote accepted

So, as noted, we use Poynting's theorem to get: $$\frac{1}{\mu_0} \oint_{\partial V} (\vec{E} \times \vec{B}) \cdot d\vec{S} = -\int_V \vec{E} \cdot \vec{J} dV$$ [The static version of Poynting's theorem is just: divergence theorem, $\nabla \cdot (\vec{E} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{E}) - \vec{E} \cdot (\nabla \times \vec{B})$, then $\nabla \times \vec{E} = 0$ and $\nabla \times\vec{B} = \mu_0\vec{J}$]

The electric field is just the negative gradient of electric potential. We can use integration by parts in higher dimensions:

$$-\int_V \vec{E}\cdot\vec{J} dV = \int_V (\nabla \phi) \cdot \vec{J} dV = \oint_{\partial V} \phi \vec{J} \cdot d\vec{S} - \int_V \phi(\nabla \cdot \vec{J}) dV$$

The current density doesn't diverge. So it's just the first term, basically, the surface integral of voltage times current.

Of course your boundary condition says there's no current out of the surface, still, the previous statement holds in static situations even without that boundary condition.

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First of all, substitute in the integral expressions for $\vec{E}$ and $\vec{B}$ and pull the volume integrals out of the surface one. This way you do the calculation for fixed $\vec{r}_s$ in both integrals, which amounts to doing it for a point-charge source for $\vec{E}$ and a point current element (which is unphysical but can be used in this case) source for $\vec{B}$.

You then need to use the identities $$\nabla_{\vec{r}} \frac1{|\vec{r}-\vec{r}_s|}=\frac{\vec{r}-\vec{r}_s}{|\vec{r}-\vec{r}|^3}$$ and $$\nabla_{\vec{r}}\times \frac{\vec{J}(\vec{r}_s)}{|\vec{r}-\vec{r}_s|}=-\vec{J}(\vec{r}_s)\times\frac{\vec{r}-\vec{r}_s}{|\vec{r}-\vec{r}|^3}$$ where singular terms vanish as $\vec{r}$ and $\vec{r}_s$ occupy different domains.

Using this the surface integral will look like $$ \oint\left[\nabla_{\vec{r}} \frac1{|\vec{r}-\vec{r}_s|}\times\left(\nabla_{\vec{r}}\times \frac{\vec{J}(\vec{r}_s')}{|\vec{r}-\vec{r}_s'|}\right)\right]\cdot d\vec{S}(\vec{r}). $$ Now this is not patently zero but it looks ripe for some fancy vector integration by parts. If necessary, you can also trade the $\vec{r}$ derivatives for $\vec{r}_s$ ones if you are careful.

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From Maxwell's equations you use the many derivations to get Poynting's theorem:enter image description here

Since the fields are static, the second term is zero. The rate at which mechanical work is done on each charge is $e\mathbf E \cdot \mathbf V= \mathbf E\cdot \mathbf J$. This needs to be zero so that the $\mathbf {J}$s don't change, keeping $\mathbf B$ static. So you're left with your expression equaling zero.

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Thanks @LarryHarson. It was obvious to me that the second term is zero, but I can't quite get to grips with why the third must be too. Can you help me with a physical interpretation? I mean if you can elaborate why it is necessary to ensure a constant $\vec{J}$? –  Avijit Feb 16 '13 at 3:24
    
@Avijit is the physical interpretation any better now? –  Larry Harson Feb 17 '13 at 1:28
    
Yes @LarryHarson, it is all clear now: both physically and mathematically. Thank you and everyone else for the help! –  Avijit Feb 17 '13 at 3:37
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