I do not understand your question.

What do you mean by "integrate over a cavity"?

Do you mean a surface integral over the whole surface of the PEC?

Kengo:

I have edited the question. The Poynting vector is integrated over the volume of the cavity, not the surface.

I am sorry  I am not familiar with the Poynting vector integrated over the volume.

What would you get?

Like I said in the question, that is the total linear momentum of the electromagnetic field of the said space. In physical sense, the sum of the field momentum and the conductor material momentum should conserve. Since the wave is standing, there should be no net instantaneous momentum transfer (force) between the field and the enclosing conductor. Therefore, the volume integral of the Poynting vector should be zero at all time. However, there is pressure on almost every point locally on the boundary surface, only that it is completely cancelled globally elsewhere.

The question is, how we prove this mathematically from the perspective of the field equation, particularly the boundary value problem.

Now  I understand your question. I am interested.

Sorry I do not have the answer right now.

Do you want a general proof for arbitrary cavity shapes? I don't have any good ideas there. On the other hand, for specific cavities with certain geometries (and time-harmonic excitation), the fields can be written as an infinite sum of modes. For example, inside a rectangular cavity the solution is the sum of products of sinusoids. Similarly, Bessel functions of the radius and sinusoids of the angles result from solving the problem in a spherical cavity. I think prolate and oblate spherioidal cavities also have closed form solutions of a similar modal form in terms of Mathieu functions. Once you have these forms, it would just be a long exercise to form the cross-product of electric and magnetic fields to form the Poynting vector. Integrating over the volume likely gives zero (or a constant?) for each mode, the sum of which you could probably directly show is equal to zero.

Raj Bhattacharjea: I want this for an arbitrary shape, with boundary at least piecewise smooth if not any continuous boundary. Once an explicit solution, say in special function, for a symmetric cavity is known, it is just a matter of computation as you said, tedious perhaps.

I think your question may be based on a misleading premise. The Poynting Vector in a cavity in a standing wave is *not* constant with respect to time unless you are looking at a time average quantity. See Kaiser, G., “Electromagnetic inertia, reactive energy and energy flow velocity,” J. Phys. A: Math. Theor. 44, 8 August 2011. Link: http://bit.ly/1w51bAB In general, the Poynting Vector will be a function of space and time.

Dr. Schantz, the original post is about showing that a *spatial* average value (spatial integral scaled by the volume) over the cavity of the Poynting vector is a constant in time, not the showing that the Poynting vector itself is a constant in time.

My point is that there's no reason to expect that the spatial average will be constant with respect to time. For instance, consider a short impulse on a 1-D transmission line as a simple model of a signal in a cavity, The 1-D transmission line is shorted at both ends. As the impulse propagates forward, the spatially averaged Poynting vector is positive. After it reflects and propagates in reverse the spatially averaged Poynting vector is negative. At any given moment the spatial average of the Poytning Vector integrated along the transmission line may be positive, negative, or during the reflection, any value in between. So I believe the question is setting out to demonstrate something that is not in general true.

Hans G. Schantz: I am asking about a STANDING wave, NOT a wave packet. Your example is an impulse or a wave packet which is NOT a standing wave and thus will certainly be carrying net nonzero momentum and imparting momentum to the conductor body enclosing the cavity. So it is NOT a counterexample.

OK then, here's a counter example using harmonic waves.

Consider a 1-D transmission line shorted at both ends and lying in the x-y plane with the harmonic standing wave of interest to you. If the transmission line is straight and linear, the flow of energy in each quarter-wave long segment and in any quarter wave period is in an equal and opposite direction and the spatial average Poynting vector is constant (zero). Now bend the transmission line each quarter wave section with a right angle kink every quarter wavelength so that the first quarter wave section is parallel to the +x-direction, the second parallel to the +y-direction, the third parallel to the +x-direction again, etc. The linear transmission line now looks like a right angle kinky diagonal along the x = y line. Consider the Poynting vector. For a quarter period, all the energy flow will be in the +x or +y direction for a net flow in the 1/Sqrt(2)(x-hat+y-hat) direction. For the next quarter period, all the energy flow will be in the -x or -y direction for a net flow in the -1/Sqrt(2)(x-hat+y-hat) direction. The time average is zero. The spatial average is not constant with respect to time.

I hope this example is helpful in saving you time trying to prove something that is not true. Again, you might want to look at the paper I cited by Gerald Kaiser.

Hans G. Schantz:

I have browsed the standing wave section of the Kaiser paper you cited. I did not see any content pertaining to the spatial integral being nonzero with perfect conductor boundary condition. I would appreciate it if you would pinpoint the part you think is specifically relevant to our discussion.

I know the point you are trying to make. However, I do not think the argument stands. It is not even clear what specific Maxwell equation boundary value problem your example corresponds to. I have two groups of questions regarding that.

1) Are you suggesting that the geometry (aside from an overall size scalar) of a cavity has no effect on the eigenfunction of a Helmholtz equation (the spatial part of the standing wave, or simply the eigen mode of a standing wave) so long as the geometry is topologically isomorphic? 

2) Consider your one dimensional example. In the straight line case, does your cavity have transversal dimension. If not, what form does the electromagnetic eigenfunction take on, for what boundary condition? I suppose you first solve for the finite dimensional eigenfunction then take the transversal dimension to zero, right? In the kinky case, are you sure that the kinky line eigenfunction is simply the bending of the straight line wave function and not changed by the changed geometry? Are you sure the right angle corner, acting as a boundary condition when the transversal dimension approaches zero, does not force a node in either electric or magnetic field and halving the wavelength so that the spatial integral of each segment is still zero?

Hans G. Schantz:

Have you considered my enquiry regarding your example? Without answering these questions which seek to specify the Helmholtz equation boundary value problem your two examples, particularly the kinky 1-d one, attempt to characterize, your example serves at most a caricature and is not even a valid (well posed) problem, much less a counterexample. Not every caricature has a factual correspondence. We can perhaps interpret the straight line segment example by taking a cavity with infinite transversal size, but what do we do with the kinky example?

Even the caricature itself does not hold water. The same reasoning applied to the following mechanical scenario leads to a reductio ad absurdum.

Take a 1-d (or infinitesimally thin) spring encased and compressed in the longitudinal direction in a 1-d (or infinitesimally thin) tube. At first the tube is a straight line. There is equal and opposite directional pressure on both ends of the tube and none on the side. There is then no net force on the tube. Now bend the tube in the middle. This time we make it more extreme than the kinky wave example, and bend not 90 degree but 180 and fold it onto itself. Now apply your implicit logic of the kinky wave example. There should be topological invariance of the force relative to the geometry of the tube and we should have double the pressure on the ends of the tube in only one longitudinal direction and none on the tube sides. Voila, we get ourselves a accelerating perpetual motion tube running all by itself!

We can apply the same argument with the spring replaced by gas.

In answer to your questions: (1) No, but it can be made negligible if you are careful (see below), (2) Requires further explanation.

First off - the archived version of Kaiser's paper at ResearchGate appears truncated. Look at section 4 of this version: http://arxiv.org/pdf/1105.4834.pdf Kaiser demonstrates that although the average energy velocity in a standing wave is zero, the instantaneous velocity is not.

I've attached a figure from a forthcoming publication of my own showing the spacetime energy flow in an ideal standing wave. The units are half-wavelength in distance and half-periods in time. If your cavity (i.e. 1-D transmission line) has an open on one side and a short on the other side, that would correspond to an individual half-wavelength section. There, it's clear that you have energy oscillating back and forth and forth in each half wavelength section.

I understand that you want to insist on the same (PEC/short) boundary conditions on both sides. That means you have to have an even number of half-wavelength sections and - so long as the line is linear - there will always be a balance between forward propagating energy in one half wavelength section and reverse propagating energy in the next.

If you prefer not to consider an ideal 1-D transmission line, assume a transmission line with a negligible cross-sectional area - like a coaxial line. The bends may be configured to aligned with the nodes or anti-nodes in the standing wave so as to minimize any interactions with the corners. The kind of interactions you're nervous about in the bend may be significant for microwave signals, so consider the limit of long wavelength signals propagating in skinny coax. Think HF band or lower and RG-58 cable, if you want a concrete example. You will not fundamentally change the standing wave pattern in a coaxial cable at these frequencies by including a gentle bend.

By the way, my example was flawed - I should have been thinking in terms of half-wavelength sections, not quarter wavelength. And your argument regarding my "caricature" is certainly valid - but only on a time average basis. Granted, you cannot implement a perpetual motion machine, but you can make a system that will vibrate or oscillate about some center of mass. I think your 180deg example is much better than the one I thought of. If you have a cable an integral number of wavelengths long (electrical length, of course), with a 180degree bend at each half wavelength, then at any given moment in time all the energy is moving in the same direction. It's not a perpetual motion machine, but rather the system as a whole would oscillate back and forth at the RF frequency ever so slightly given that the effective mass of the EM energy is vastly smaller than the mass of the cable.

The problem with your pipe analogy is that there is a force couple at the middle of the pipe at which you want to bend the pipe. In a straight pipe, the fluid from one side presses against the fluid from the other. When you bend it, the two fluid forces exert a double reaction against the pipe wall. You'd have the same effect taking place at each bend in the coax. All the bends on one side would be opens and all the bends on the other side would be shorts. So if you want to dismiss the example on those grounds, you'd have a point. Keep in mind, however that you can always enclose a local open in a much larger scale closed conducting termination.

Perhaps a different example would make my point more clear. You seemed comfortable with the concept that a transient impulse bouncing back and forth along the line would yield a non-constant Poynting Vector. Note that transient may be decomposed into a superposition of harmonic standing waves.

Another way I express my concern is that given your approach of insisting on standing waves, you are employing a time harmonic analysis. The harmonic approach deals in time average quantities which are inherently constant. Effectively, I'm concerned that you may be setting out to prove a tautology. Time average quantities are inherently constant with respect to time. From the point of view of a harmonic analysis, you may be setting out to prove what you have already assumed.

Don't let my concerns stop you if you are determined to attempt your proof. It's an interesting problem, and I'll be looking forward to your sharing your results. But I really recommend you approach the problem from the time domain to avoid the risk of fooling yourself.

Hans G. Schantz:

There could be some concern as to what we mean by standing wave in an arbitrarily shaped vacuum cavity. So, I have edited the question and specialized the problem to an eigenvalue problem for the Helmholtz equation boundary value problem. Please check and let me know if the question statement is mathematically clear.

Let me emphasize that the question is clearly asking about the volume INTEGRAL of the Poynting vector over the whole cavity at any time, not the value of the Poynting vector at any spatial point at any time.

I did read that very version of Kaiser’s paper you provided the link for. Both this paper and your figure concern with the planar harmonic wave, which give no clue as to how to resolve the problem in arbitrary geometry. You say “the instantaneous velocity is not (zero)”. This is wrong. The Poynting vector spatial integral of this standing wave between two perfect conductor is zero instantaneously and all the time. Incidentally, your citation prompted me, thanks to you, to email Gerald Kaiser this question. Unfortunately, he responded he had no idea how to tackle it.

I am not familiar with your nomenclature of short and open. I suppose by short you mean a perfectly conducting boundary surface and by open you mean there is no boundary? If that is the case, I agree with your discussion about 1-d straight line.

You say

“assume a transmission line with a negligible cross-sectional area - like a coaxial line. The bends may be configured to aligned with the nodes or anti-nodes in the standing wave so as to minimize any interactions with the corners. The kind of interactions you're nervous about in the bend may be significant for microwave signals, so consider the limit of long wavelength signals propagating in skinny coax. Think HF band or lower and RG-58 cable, if you want a concrete example.”

This is not, a concrete example which implies a valid claim, but a speculation, without a shred of mathematical justification and unproven (which I actually think is wrong albeit I do not have a proof for the contrary --- had I had one I would not have posed the question ) In this logical process, you are assuming you can select arbitrary eigen (standing) wave wavelength in a cavity. Quite the contrary, in an enclosed space the eigenvalue of the Helmholtz equation (or Laplacian operator) is countably discrete. “The bends may be configured to aligned with the nodes or anti-nodes in the standing wave so as to minimize any interactions with the corners.” Minimizing all you want, it does not necessarily make the objective value zero or even close to zero. What if the minimal value is huge? It is not even clear what objective function it is you are trying to minimize. It is all very fuzzy, vague and ambiguous. My previous question is exact whether you could ever succeed. What makes you think the bending 90 degrees or 180 degrees a very skinny coax cable with both ends perfect conductor does not cut the wave length in half? “You will not fundamentally change the standing wave pattern in a coaxial cable at these frequencies by including a gentle bend.” First of all that is an leap of logic assertion without proof. Second, the gentleness of the bend would seem to run counter to your objective to minimize the "interaction with cable". How would you align with the nodes or antinodes, so instead of “minimizing” the interaction with the corner now wouldn’t you maximizing it? Third, it is a gentle bend, what prevents the integral of the effects to add up to a big enough one counteracting the pressure exerted on the cable ends? In fact, there is nothing in your argument that could refute the counter assertion that the very skinniness forces the halving of the original wavelength when bent 90 or 180 degrees and thus exact counteract the pressure at the two ends at all time and produces no movement or oscillation whatsoever.

Overall, your argument is too hand waving, sleight of hand, to serve as a valid counterexample. This is very dangerous. These kind of magical arguments are prevalent and characteristic in perpetual motion machine paradoxes. The seemingly insignificant forces always add up large enough to counteract the “main” locomotive force.

You say:

“The problem with your pipe analogy is that there is a force couple at the middle of the pipe at which you want to bend the pipe. In a straight pipe, the fluid from one side presses against the fluid from the other. When you bend it, the two fluid forces exert a double reaction against the pipe wall. You'd have the same effect taking place at each bend in the coax. All the bends on one side would be opens and all the bends on the other side would be shorts. So if you want to dismiss the example on those grounds, you'd have a point. Keep in mind, however that you can always enclose a local open in a much larger scale closed conducting termination.”

I do not see the problem with my fluid pipe analogy. On the contrary, I think my reductio ad absurdum has succeeded in leading you to find the logical flaw in your 1-d kinky example. Specifically, in your words “there is a double reaction against the wall” at the bend in both the pipe and the coax. The reaction at the bend in the coax is forced by the shorting at the bend thus halving the wavelength.

In specific logic derivation, I am applying your logic in the 1-d transmission line (close ended) to the pipe, which would have stipulated that there would be no change to the pressure at the bend and remains at zero as it was when the bending point was on the straight line. This leads to absurdity. This is exactly what I want to demonstrate. Now, what the very reason that you described here, i.e. the double reaction on the fluid pipe and that, as you say, “All the bends on one side would be opens all the bends on the other side would be shorts” is precisely what actually happens with my fluid pipe reductio ad absurdum example and your bending 1-d transmission line example (90 degree or 180 degree), respectively. This very change of open and short forces the halving of the wavelength and exerts equal and opposite pressure on the bending parts, sharp corn or gentle curve to precisely counteract the pressure at the two ends so that the whole pipe and the whole transmission line remain static all the time not just on average time.

You say:

“Another way I express my concern is that given your approach of insisting on standing waves, you are employing a time harmonic analysis. The harmonic approach deals in time average quantities which are inherently constant. Effectively, I'm concerned that you may be setting out to prove a tautology. Time average quantities are inherently constant with respect to time. From the point of view of a harmonic analysis, you may be setting out to prove what you have already assumed.”

That is precisely the very point of which I am trying to go beyond, but you have been trying to argue against. I am trying to prove that not only the spatial INTEGRAL of field momentum is time average zero but is time INSTANTANEOUSLY zero. It has been you who have been trying to argue that only the trivial or what you now fear to be "tautological" result, i.e. time average is zero, is true and nothing beyond. Now you are saying I am trying to prove too little, while previously, in fact merely a couple of paragraphs ago, you have been saying I am trying to prove too much. I am confused by your self-contradictory concern. It is not clear what you are arguing against.

Hans G. Schantz:

I want to separate out the response to your last statement.

You say "I really recommend you approach the problem from the time domain to avoid the risk of fooling yourself."

I do not quite understand what you are suggesting. The one scenario I can think of is the following which is a much harder problem.

Suppose the mass of the cavity wall is infinite, and we have a very concentrated wave packet shooting from one point of the cavity and scattered off of the wall. What can we say about the total momentum evolution? I have no idea. Most probably the resulting wave would be chaotic. I suppose asymptotically at time infinity, the spatial momentum integral would be zero and the wave would be "uniformly distributed" (which needs to be rigorously defined) over the whole cavity space. This would probably be in the realm of ergodic theory which I know nothing about. I would think this is a much harder problem and completely different problem.

@Hansen Chen: The Poynting vector is defined as the directional energy flux density (the rate of energy transfer per unit area, in units of watts per square metre (W/m²)) of an electromagnetic field. If you have a cavity surrounded with a perfect conductor, the energy inside can not pass the boundary and leave (and no "outside" energy can enter the cavity). The energy flux is zero. Is this an acceptable solution?

@Herbert Weidner: Unfortunately this is not a solution.

The area integral of the Poynting vector is the field energy flux through the area. The volume integral is the field momentum. I am asking for the volume integral of the cavity. Energy flux is zero or energy is conserved does not imply the momentum is conserved. One can have a billiard ball perfectly reflecting off of surrounding walls of infinite mass and with no energy loss but momentum varying all the time. So you are not answering the question.

an other attempt: consider a plane, dividing the cavity into two parts. The Poynting vector is defined as S=(energy)/(area*time). In words: How much energy flows through the plane in a given time?

At time A, the energy in the left part of the cavity is WL, in the right part you have WR.

Wait (time 1/f), until one oscillation is finished. Then nothing has changed, you have again the energies WL and WR, the total energy the net flow of energy through the plane is zero. Therefore, the integrated Poynting vector over one oscillation is zero.

You can place that plane everywhere in the cavity and you can wait many multiples of 1/f, since there is no energy loss in the cavity. The result is always S=0.

@Herbert Weidner:  What does that say about the total volume integral of S or the total field momentum at any time?