Helmholtz equation Poynting vector integral

Question

The Maxwell's equation for harmonic time dependent field in vacuum is \begin{align} \nabla \times B + i\omega E &= 0\\ \nabla \times E - i\omega B &= 0 \\ \nabla \cdot B &= 0 \\ \nabla \cdot E &= 0 \end{align} where $\omega$ is a real number, $i$ is the unit pure imaginary number, $E$ and $B$ are both complex valued 3-d vector function of 3-d space variables: $R^3\rightarrow Z^3$. As a consequence, we have the Helmholtz equation \begin{align} (\nabla^2+\omega^2)E&=0 \\ (\nabla^2+\omega^2)B&=0 \end{align} Suppose these holds in the interior of a compact simply connected domain $\Omega$ in $R^3$ with piecewise smooth boundary. Let $n$ be the unit normal vector to the surface at a point of the boundary pointing inward, we have the following boundary condition. \begin{align} n\times E&=0 \\ n\cdot B&=0 \end{align} Now the Helmholtz equation becomes an eigenvalue boundary value problem with eigenvalue $-\omega^2$. Define Poynting vector $S = E\times B^*$, where $^*$ denotes complex conjugation. Is the following true, and if it is, what is the proof? \begin{equation} \int_\Omega SdV=0. \end{equation}

I conjecture this proposition is true due to either the Poynting vector volume integral or the boundary surface inner product integral of the Maxwell energy-stress tensor being zero for a rectangular $\Omega$.

I am well aware of the fact that the time domain expression of this Helmholtz equation is by definition a harmonic function in time, and so the time average of such Poynting vector is always zero. I am also well aware that the field energy in this domain $\Omega$ is a constant in time. The fact of the matter is that these do NOT answer the question whether the total volume integral of the Poynting vector or the field momentum is a constant in time. So please refrain from making comments showing these facts unless they are part of the argument either proving or disproving the conjecture.

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Possible Approach:

What is the variational formulation of the Maxwell's equation above including the boundary condition? Could the Poynting vector volume integral turn out to be one of the invariance under some group transformation via Noether's Theorem?

Answer 1

Allow me to expand on my comment.

We can safely assume that the electric field $\vec{E}$ is real, so $\vec{B}=-(i/\omega)\nabla\times \vec{E}$ is purely imaginary. Then the integrated Poynting vector becomes $$\vec{P}=\int_\Omega \vec{E}\times \vec{B}^\ast\,dV=\frac{i}{\omega}\int_\Omega \vec{E}\times(\nabla\times \vec{E})\,dV$$ $$\quad\quad=-\frac{i}{2\omega}\int_\Omega\nabla E^2\,dV.$$ In the last equality we have used that $\nabla\cdot \vec{E}=0$ within $\Omega$ and that $\hat{n}\times \vec{E}=0$ on the boundary $\delta\Omega$.

On the boundary $\vec{E}=-4\pi\sigma\,\hat{n}$, with $\sigma$ the surface charge density, so we can equivalently write, $$\vec{P}=-\frac{i}{2\omega}\int_{\delta\Omega}E^2 \vec{n}\,dS=\frac{2\pi i}{\omega}\int_{\delta\Omega}\sigma \vec{E}\,dS.$$

The integral over the surface of charge density times electric field is the total electrical force on the cavity walls. The Poynting vector is purely imaginary, which means that this force oscillates in time with frequency $2\omega$ (as explained here). The real part of $\vec{P}$ vanishes, so the time-averaged force is zero ($\vec{E}$ and $\vec{B}$ are $90^\circ$ out of phase).

The OP's conjecture is equivalent, in physical terms, to asking whether the total force on the cavity walls is zero at any time, not just upon averaging over one period. I see no reason for this, neither from a mathematical point of view, nor from a physical point of view --- at least not in the absence of inversion symmetry. For a rectangular $\Omega$ (the example offered by the OP in support of the conjecture) the integral vanishes by symmetry, because $E^2\hat{n}$ is opposite on opposite surfaces.