3
$\begingroup$

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.


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?

$\endgroup$
  • 1
    $\begingroup$ physics.stackexchange.com/questions/131509/… --- researchgate.net/post/… $\endgroup$ – Carlo Beenakker Aug 30 '14 at 8:55
  • $\begingroup$ I like to think about this in mechanical terms: the domain $\Omega$ is a box filled with photons, sloshing around periodically with period $2\pi/\omega$; their total momentum oscillates with the same period, why would the amplitude of this oscillation be zero? Momentum conservation just says that forces on the boundary of the box have to appear if the momentum oscillates, so there will be some net force on the boundaries that oscillates with the same period. Why do you see a conflict with conservation of momentum? $\endgroup$ – Carlo Beenakker Aug 30 '14 at 13:16
  • $\begingroup$ @CarloBeenakker: You are right. Conservation of momentum per se does not give zero field momentum. I will edit the last statement in the question. Thank you. $\endgroup$ – Hans Aug 30 '14 at 17:12
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This constitute neither a proof nor a counterexample but a speculation. Indeed a very week speculation at that. Seeing no reason is no reason for claiming nonexistence. More importantly, you did not even bother to check the rectangular example that I have mentioned which agrees at least my conjecture. We do not need this derivation to know Poynting vector oscillates with frequency $2\omega$. It comes straight from the set up and direct computation of $\vec E\times \vec B$. Also, the pure imaginarity for $\vec P$ is just a phase shift. It does not implies the oscillation and frequency. $\endgroup$ – Hans Sep 15 '14 at 23:39
  • $\begingroup$ Please excuse my bluntness, but this answer not only contributes nothing but distracts from the question, because by blithely claiming "see no reason for this,neither from a mathematical point of view, nor from a physical point of view" it even, perhaps deliberately, ignores reasons and examples, which I clearly spelled out in the last sentence of my question, that indeed support the conjecture. $\endgroup$ – Hans Sep 15 '14 at 23:46
  • $\begingroup$ Carlo. I appreciate your opinion. The fact remains though your edited answer still offers neither a proof nor a counterexample, which is what I am seeking with my hard earned bounty. I can not accept a non-answer. According to the rule, if there is no accepted answer after the bounty period expires, the answer with the highest point will be awarded with the bounty. If you do not have a proof or counterexample in the next 12 hours, would you please kindly remove this answer temporality until after the bounty period? It will be unfair to award the bounty to a wrong answer. Thank you in advance ! $\endgroup$ – Hans Sep 16 '14 at 17:01
  • $\begingroup$ at your service. $\endgroup$ – Carlo Beenakker Sep 16 '14 at 17:40
  • $\begingroup$ Carlo, I hope you don't mind I undeleted this answer. The bounty is no longer an issue. However, if you sincerely want to delete this answer I am happy to oblige. $\endgroup$ – François G. Dorais Sep 20 '14 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.