Dear Stefano (and Kåre) ~

Noether’s theorem establishes a relationship between the symmetries of a Lagrangian and conservation laws expressed as continuity equations. It is valid for any field theory whose equations are derivable from a Lagrangian. When the symmetries in question are the invariance of the Lagrangian and hence of its Euler-Lagrange equations under translations in space and time, the conservation laws are those of energy and momentum, expressed as the continuity equations ∂iTij = 0 where Tij is the energy-momentum tensor.

Maxwell’s theory is such a theory. It doesn’t matter that “the tensors were introduced 50 years later than the Maxwell equations.”!

That is the deeper level at which the problem I see arises. Maxwell’s theory in its Lagrangian formulation is a theory of the interaction between the electromagnetic field and the fields of charged particles (in particular, the electron field). The Euler-Lagrange equations are ∂iFij = Jj. The current Jj is the source and charge conservation is ∂iJi = 0. The continuity equation ∂iTij = 0 expresses the conservation of total energy-momentum. In regions where the sources are absent we have ∂iTij = 0 for electromagnetic energy-momentum.

This is not about “waves”, or “static solutions” or “steady state solutions”. It is far more general (and deeper) than that.

That is the mathematics. The generally accepted physical interpretation of any energy-momentum tensor, that gives a physical meaning to the conservation laws, is

Tij − “stresses”

T00 − energy density

Ti0 − energy flux density

T0i − momentum density

Most of the replies in this discussion look to me like “hand-waving” arguments that, in the particular case of electromagnetism, are trying to “wish away” the interpretation of Ti0 as energy flux density in various ill-defined particular circumstances. I don’t buy that!

[ To muddy the waters further: it is worth mentioning that the generally employed symmetric energy-momentum tensor of electromagnetism is not the Noether energy-momentum tensor! (see for example https://www.researchgate.net/publication/241377004_A_theorem_on_stress-energy_tensors). I don’t wish to get into that...(-; ]

Pointing vector gives us the direction of electro-magnetic field travelling in space. But in static electro magnetic fields we don't find and travelling fields. But, even then pointing vector has a significant role in static fields.

As mentioned earlier pointing vector is not confined to give information of direction of fields but also it says about the angular momentum of the field. According to feynman pointing vector is really means a momentum flow. It is needed to maintain conservation of angular momentum.

There are many cases where we can't account for conservation of angular momentum in static fields with out considering pointing vector. For example consider a magnetic material placed in another non magnetic material (assume them in concentric rings(inner magnetic material, outer non magnetic material)). Initially they are at rest (Zero angular momentum). If we supply positive and negative charges to outer and inner layers respectively, there comes a field between them. This field will try to get some momentum in magnetic material.

So, initially we started with Zero magnetic momentum and we are in static fields even then we end up with some non zero moment. This disagree with conservation. Now if we analyse pointing vectors for this case we get its moment conserved. Hence pointing vector is needed for angular momentum conservation in Static electric fields.

I think that question refer to static case and not to steady state where you have Pointing vector such as in FP interferometer

Julien "Would you consider the case of a capacitor in electrostatic condition for your question?"

Yes, that's the simplest example I had in mind. Put a magnetic field across a capacitor, parallel to the plates. Would an object in that space feel a force due to the Poynting vector? And if not, why not?

The Poynting Vector in static fields is related to Electromagnetic Angular Momentum !!!

In fact there exists a static circulating electromagnetic energy (as a circulating mechanical object).

Wikipedia (and Panini and Christos in their answers) consider the "angular" momentum that provides an explanation in very particular situations first discussed by Feynman. What about the more fundamental "momentum density" in a very general static field configuration? "Angular momentum" is only a derivative concept.

In electromagnetism momentum density and energy flux density happen to be identical because of the symmetry of the electromagnetic energy-momentum tensor.

I imagine that a pressure would be exerted on an object placed in the field (even a non-magnetic object with no charge), analogous to radiation pressure, even though there is no radiation. Am I correct? Has this ever been experimentally tested? When the field does work on such an object, where does the energy come from? These questions have troubled me for a very long time. I've never come across any reference to them at all in the literature.

According to the Poynting vector i.e E x H, it shows direction of flow of the electromagnetic energy.

For capacitor we consider electrostatic energy. so poynting vector will be zero.

Now if we keep magnetic field in parallel to capacitor plate it will give be electromagnetic field.

as force is inversely proportional to a distance, there may be a force existing which may be inversely proportional to the distance of magnetic field. but i am not sure.

now i also started thinking by seeing your question.

but i don't know, we should experiment this.

Mathieu ~ thank you for this reference. I've downloaded the whole book. It's not an "easy read" but the discussion on pp16-20 seems to be pointing (no pun intended...) me towards a resolution of my puzzlement. I need to think some more. I'm relieved to learn that the question I raised has a long-standing history of controversy - it wasn't just me being naive!

I would like to recommend you to read our paper "Experimental test of the compatibility of the definitions of the electromagnetic energy density and the Poynting vector" in The European Physical Journal D  09/2004; 31(1):113-120. DOI: 10.1140/epjd/e2004-00135-x

Hope it woul be interesting for you.

Andrew ~

Sorry for the delay in thanking you for the reference (working from home I was unable to get the pdf because Springer charges outrageously for pdfs of their journal articles so I had to make a trip to my institute library to get a xerox. Then it took time to read and understand...). I found it interesting and informative (though I admit to not having followed in detail the experimental setup...).

I'd not heard of Umov before. The Umov continuity equation (1) is simply WRONG - it's not relativistically covariant unless w is a scalar density (like mass or charge density). Energy density is not - it's part of a tensor density. According to Maxwell's theory the rate of flow of energy across a surface is the integral of the normal component of ExB over the surface. It seems to me that the "velocity" of the energy is, in general, not a valid concept. 

I see no way of avoiding the conclusion that the Poyning vector is the correct expression for energy flux, for any configuration of electromagnetic fields.

"

I see no way of avoiding the conclusion that the Poyning vector is the correct expression for energy flux, for any configuration of electromagnetic fields."

Yes, of course, but we avoid the convlision that the Poynting vector in general  is the special case of the Umov vector!  

Agreed!     (-:

Umov's continuity equation doesn't make any sense to me because, as I said, energy is not a scalar density. The continuity equation that follows from Maxwell's theory (modulo choice of units and making allowances for the impossibility of typing a partial derivative in this textbox...) is

(d/dt)W + divP = -J.E

where W is energy density (E2 + B2), P is the Poynting vector ExB and J is charge current density.  P is NOT Wv for some "velocity" v. There's no such velocity - it's a notion in violation of relativistic physics.

This is an interesting question that you can summarize in a simple example as the one of the capacitor with a magnetostatic field orthogonal to the electric one. Are you going to obtain an electromagnetic field which could be independent of the sources as if you had a Lienard-Wiechert potential? Obviously no, because in such a case you would have an emitted antenna without energy cost.

The answer is simple but deep. Let us to divide it:

1. Poynting vector is not the only physical object to give the flow of energy or the momenta. It is only one component of energy-momentum tensor and only in very simple situations can give us this physical information of the flow of energy.

2. Notice that 4-momentum as a Noether current associated to space-time symmetries is a quite general object which needs some physical conditions to exist. And the angular momenta much more and even is not possible to give a general expression containing the spin and orbital components in a straightforward form.

3. The equation of motion is not only the one that you have chosen Eric. This is just the time projection but you have also the space projection which contain the Maxwell stress tensor beside the time derivative of the Poynting vector. But even assuming that this term is zero because in this case is with static fields and also it happens the same in your equation with the density of energy w, pair of differential equations are not with trivial solutions.

4. Fortunately there is a simple answer if you consider that 4-momentum is equivalent to the product of the charge by the 4-potential. And if you assume that you work within the Coulomb gauge with zero scalar potential, then notice that the Poynting vector is reduced to eA(r) whose time derivative is zero.

Thus there are not flux of electromagnetic energy in any configuration that you can think for static or even stationary fields!

Daniel ~ Thank you for your answer. I'm aware that the Poynting vector is part of the four-dimensional stress-energy-momentum tensor. Electromagnetism is only a particular case - in general an energy-momentum tensor contains flux and density of momentum and of energy. According to this generally accepted prescription energy flux for electromagnetism is the Poynting vector. So I cannot see how you can conclude that " there are no flux of electromagnetic energy in any configuration that you can think for static or even stationary fields". Yes, in a static configuration, the time derivative of the Poynting vector is, of course, zero. But the vector itself is not, in general, zero.  If it's not an energy flux, what is it?

Dear Eric,

The answer in my four point. Poynting vector is equivalent to eA(r) into Coulomb gauge and with the other mentioned conditions. The time derivative of this quantity gives you zero and therefore this is the value of the electric force or the energy produced in a certain space. 

    Notice that you cannot do this simple deduction with the usual Poynting vector even if you assume the same physical conditions. The time derivative of the Poynting vector (made directly) can be only interpreted through the conservative laws, but these too complex in their physical interpretation because they include more physical quantities as stress Maxwell tensor or electromagnetic sources. In the canonical electromagnetic momentum you do not need to do it and you have only a physical possible energy to interprete.

"... conservative laws, but these are too complex in their physical interpretation because they include more physical quantities as stress Maxwell tensor or electromagnetic sources"

Physical interpretation of the conservation laws arising from the stress-energy tensor is not complicated, as you are making it out to be. The generally accepted interpretation is that the Poynting vector is energy flux (or momentum density, which is the same thing since the energy-momentum tensor is symmetric). Maxwell's equations imply divP = -E.j in the static case. Thus -E.j acts as a source of the Poynting vector P (analogous to charge as a source of E). It need not be zero in a static situation.

You can make many different interpretations of an incomplete equation as you use. The only thing that you have is that the density of energy lost in producing electric currents is due to divergence of the Poynting vector. But this is not solving your general question of what happens with the flow of energy for static fields which have to involve all the electrodynamics. Notice that instead of J you can substitute it by the fields only (using Ampere's law) and you can try to see what happens with the fields. This is what I have tried to tell you with the sentence"...conservative laws, but these are too complex in their physical interpretation... ".

Perhaps for a deeper discussion of this issue you could go to the book:

D.Baldomir and P.Hammond, Geometry of Electromagnetic Systems, Clarendon Press (1996).

I'm sorry Daniel but I'm lost now -  I have to admit that I cannot understand what you are trying to say. You seem to me to be be making an unnecessarily complex issue out of a simple implication of Maxwell's theory.

The equation divP = -E.j is not an "incomplete equation". 

"The only thing that you have is that the density of energy lost in producing electric currents is due to divergence of the Poynting vector".

That's right. But where does the energy go? Isn't the obvious interpretation of the equation that energy is lost because there is an energy flux P outwards from j into the static region??

The question "If P is not an energy flux, what is it?"  shouldn't require a "deep" discussion. The interpretation of P as energy flux is quite generally accepted. It's not just my personal idiosyncracy!

Incidentally, I disagree with your statement that "conservative laws ... are too complex in their physical interpretation". A continuity equation and the result of integrating it are not rocket science...

  The full equation of conservation is 

   daTba  =-FbcJc

   where usually the a,b,c indices are greek letters for taking the values 0,1,2,3. Your equation is only for b=0. That is to say, you have lost the other 3 equations of conservation in Electrodynamics. 

  Sorry, for clearness and accuracy,I was thinking only in energy and linear momentum when I wrote that remain three conservation laws (translations). In fact in Electrodynamics there are eleven laws of conservation.

  Incidentally, there are no problems with your interpretation of the Poynting vector

Thank you for the clarification. daTba =-FbcJc and daJa = 0 are five conservation laws. (The six for angular momentum conservation can be derived from them, I believe.) It's not necessary to consider all these equations at once, though (which is what you seemed to be implying). It's quite legitimate to look at particular components, such as energy conservation divP + dw/dt = -E.j. In a region away from the sources (currents and charge density),  P = ExB can be non-zero even in a static situation. That is obvious. And that is what motivated my original question. If P is not energy flux, which it should be according to the canonical interpretation of Tba , then what is it?

I have understood your question about the Poynting vector. But I think that you are fixed only in one wrong idea: if you have a Poynting vector, i.e. the existence of the cross product of ExH doesn´t means that you have a flow of energy. Notice that flow means dependence of time and in the case of static fields this should create energy instead to transform it. This is a non trivial statement in Electrodynamics and it was what I have tried to prove you in previous answers.

1. Electro-dynamics allows to propagate the Poynting vector only if the fields depends of the time in such a form that Maxwell equations work. That is to say, electric field transforms in magnetic field or viceverse.

2. I have shown you that they can move with respect to an inertial frame without transforming energy. Say, if you have an external observer seeing your static Poynting vector as depending of time, even in such a case there is not transfer of energy or momentum.

3. By the way, the six conservation laws associated to the angular momentum are absolutely independent of the other 5 that you properly identified.

4. Finally, trying to answer you in a very simple form, the producto ExH for static fields is a Poynting vector, but without propagating energy or momentum at all. 

If I've understood correctly, you are saying that if energy density is not changing, it cannot be flowing. Only in circumstances where E and B are  time-dependent, as in an electromagnetic wave, the energy density (E2 + B2) flows and the flux is then the Poyning vector P = ExB.

Sounds reasonable. But here is an analogy:

The continuity equation for charge is (d/dt)rho + divJ = 0. The normal interpretation of current density J is that it is a charge flux. A wire carrying a current may have total charge density rho = 0 because the density of - and + charges cancel out in a continuum level. But the current flows. There is no charge density but there is nevertheless a flow of charge!

I asked "if the Poynting vector is not a flux of energy, what is it?" The analogous question is now, "if J is not a flux of charge, what is it?" The answer in this case is clear: It is always a charge flux density even when charge density is zero! The elementary + and - charges cancel to give rho = 0, but there is a charge flux density J because the elementary minus particles are moving faster than the elementary plus particles.

Unfortunately this analogy throws no light on the electromagnetic energy situation. (There are no elementary units of positive and negative energy!!)

How do we distinguish, in general,  cases in which P is an energy flux from cases where it is not? And when it is not the energy flux, what is the energy flux? (Those are rhetorical questions, though if you can answer them, please do....)

Thanks for your patience (-:

Dear Eric,

The analogies are just analogies, but it is a very dangerous thing to carry them far away of the reasonable. The continuity equation associated to the local conservation of the charge is easy to understand because the electric charges are easy to imagine. But the energy in electrodynamics is very difficult to imagine for me, even in simple cases and less the linear or angular momenta. All this issue is quite complex when you put it real situations, at least for me and my intelligence.

If I have understood you well, your analogy is rho by w and J by P. Then you would have:

                             P=w.v

being v the velocity that the density of energy associated to the fields. Therefore if you have fields no dependent of time and also their density of energy, this means that the velocity had to be zero and in such a case the Poynting vector would be also. In other cases this could give you a Poynting vector different of zero and with a flux of energy.

This is pure analogy and trying to assume your model. In a first glance this could be reasonable but unfortunatelly the things a much more complicated in electrodynamics, for instance, it exist stress on matter, also energy waste in heat or in electronic transport.

Yes, of course there is no "velocity" v such that electromagnetic energy flux is wv. That was the mistake made by Umov, cited in a paper* by an earlier contributor to this thread - the mistake of treating energy density as a scalar density.

(An analogue even simpler than my charge conservation example, illustrating the fallacy, is the flow of a liquid - it flows even though its density is uniform and constant...)

I've collected and looked into quite a lot of published papers on the topic of electromagnetic energy and the Poynting vector since I started this discussion. Mostly they pursue particular contrived "thought experiments" involving capacitors etc. and are not very enlightening about general principles.

I agree, electromagnetic energy is not easy to visualize, even when the complications you mention (stress on matter, energy waste in heat or in electronic transport etc) can be dispensed with by considering only electromagnetism in "free space" - i.e, in regions not containing sources and "matter". Then there are only configurations of the fields E and B in space and time to think about.

If indeed, as you have convinced me, the Poynting vector describes energy flux in some situations but not all, then there ought to be some mathematical entity that does express this flux correctly for all possible solutions of Maxwell's "free space" equations. But I'm at a loss to know what it might be!

Anyway, thank you for a stimulating discussion.

_______________________________________________________

* Experimental test of the compatibility of the definitions of the electromagnetic energy density and the Poynting vector in The European Physical Journal D 09/2004; 31(1):113-120. DOI: 10.1140/epjd/e2004-00135-x

Dear Eric,

I also happened to ask the same question as you just now. After putting the question I found that there has been so much activity on this point. Very interesting indeed. Are you satisfied with the replies you got?

Regards,

Rajat

Hello Rajat ~

I was surprised to see this discussion thread come alive again. I thought it had run its course.

You ask “Are you satisfied with the replies you got?”

The brief answer is "no". I found the discussion interesting and thought-provoking (especially my discussion with Daniel Baldomir). However, in the end, I was left as much in the dark as when I posted the question (and somewhat more confused). It surely should not be necessary to resort to deep considerations, strange “thought experiments” and convoluted arguments, to deal with what is really a very elementary question arising from Maxwell’s theory!

In all branches of physics, whenever we have a continuity equation

       divP = ∂W/∂t,

it implies that “something” is conserved. W is the density of that “something", and P is its flux density.

The energy density of an electromagnetic field is W = ½(E2 + B2). (No controversy there, I hope!). Maxwell’s theory tells us, unequivocably, that

       divP = ∂W/∂t

where P is the Poynting vector E×B. Nothing could be clearer. Maxwell's equations are telling us that P is a flux of energy, for any solution whatsoever of those equations. There are many solutions of Maxwell’s equations in free space with E and B static and P non-zero. Maxwell’s equations imply that there can be energy flow when E and B are static.

Why do people try to negate this conclusion by saying that divP = ∂W/∂t is “incomplete” because W and P are parts of a larger entity, the “stress-energy-momentum tensor? That’s a red-herring and completely irrelevant. No-one talks like that in other branches of physics. Why try to claim that P is energy flux in some circumstances but not in others? No-one talks like that in other branches of physics. Energy flow can be detected by the pressure it exerts on an object placed in it’s way, so why have I never heard of any such experiment? Those are the questions I’m left with.

Rajat ~

I've just been over to your posting of the same question. I see that Hanno Krieger has brought up the cylindrical capacitor example in which there is energy flux (or momentum density, which is the same thing because of the symmetry of the electromagnetic energy-momentum tensor) in closed loops even though E and B are static. I believe this example was first put forward by Feynman. I wonder why no-one has set this up and actually tried to detect the energy flow experimentally...

Had I been an experimentalist rather than a theoretician I'd have tried it myself! (-: 

Dear Eric,

I thought also that the this question was solved and it is very curious that some people have also asked it independently. I shall try to help you again:

1. The continuity equation of energy that you use, div P = ∂W/∂t, is only a very simple case of conservation energy in electrodynamics. The simplest physical case that I can imagine now, is an electromagnetic wave in vacuum without any material close enough to interacting on it. Notice that the density of energy in such a case is dependent of time and the the electromagnetic fields transform each other using Maxwell equations, i.e. no stationary or static fields at all.

2. If we imagine a system static or stationary which have to follow the above continuity equation, them we had

div P=0

and choosing a closed region of surface S of the space we can integrate such expression for obtaining the surface integral of P equal to zero and you have two possibilities:

a- The P that it enters in the volume surrounded by S is the same that the one that it leaves it.

b- P is zero, which is the reasonable if you think that we have non determined any condition for the surface as how big it is or so on.

I hope that everything is clear now with this example that you have asked 

Daniel ~

Thanks for your input. I'm afraid I still don't get it. Yes, of course, divP = 0 tells us that the P that enters a volume is balanced by the P that leaves, it doesn't mean that P is zero! You say that "P is zero is reasonable if you think that we have not determined any condition for the surface." It's not reasonable: I'm posulating a static B and static E for which P isn't zero. There's nothing unreasonable there, its a configuration that's easily set up in a lab. It has nothing to do with any "surface". Let's forget about closed surfaces. Take a surface element dS with unit normal n. Then according to Maxwell's theory P·ndS should be interpreted as energy that crosses dS per unit time. If dS is an element of an actual material surface, it should feel a force. I see no way of getting around that without violating well-established physical principles.

   Eric,

   By Gauss the integral of volume of a divergence is always a closes surface which surround it, and if you take in general all the surfaces that you want, it seems reasonable (for a vectorial magnitude as P) that it must be zero. Notice that the Poynting vector is not with spherical symmetry.

  Eric,

  Being more explicit. At least that you had an external source of energy which were making to pass a continuous flux of electromagnetic energy, On the other hand, given that the Poynting vector depends of r^(-4) if you can choose the surface as far as you want, then it must be zero.

Dear Eric,

I have added another answer more to your question in the Rajat forum.

Daniel ~

"By Gauss the integral of volume of a divergence is always a closed surface which surround it, and if you take in general all the surfaces that you want, it seems reasonable (for a vectorial magnitude as P) that it must be zero."

This argument doesn't work. A vector can be divergenceless but not zero.

    Eric,

    Obviously all constant vectors and many others can be divergenceless without being zero. I  was spoken about a very different thing. If you want to solve your continuity equation, you need to integrate it with respect to volume space. By Gauss you transform the integral of the divergence and volume in other of Poynting vector integrated on a closed surface. That integral must be zero because w doesn't depends of time, isn't?. Now as Poynting vector is proportional to one divided by r raised to four it means that the electromagnetic moment is zero. Notice that I assumed always that we speak of the electromagnetic moment because the Poynting vector is only a mathematical product that needs to be interpreted through the motion equation.

Why do you insist on integration over a closed surface? Integration over an open surface is what's relevant here. That's how a flow of anything is defined.

"Poynting vector is proportional to one divided by r raised to four". Where on earth has that come from??

Forget electromagnetism for a moment. Just think about two time-independent vector fields E and B with ExB not zero. Then ask yourself, in the context of the time-independent Maxwell equations and their energy-momentum tensor: what does ExB mean?

    If you integrate a divergence field on a volume always have the integral of such a field on a closed surface which surrounds it. This is a fundamental statement that if you do not follow it then we are wasting our time: we are speaking about conservative laws. Have a good weekend!

You misunderstand. I'm not talking about ∫div PdV = ∮P·dS = 0

I’m taking about ∫P.dS over an open surface patch

You are not paying attention to what i'm writing - we are indeed "wasting our time".

    Dear Eric,

   I know  that you like open surfaces of integration, but that is one your mistakes. Please, look for the Gauss theorem and also look for the Poynting vector interpretation as DENSITY of momentum, not momentum. But if you want to follow what you want with mathematics, it is obvious that you are going to have troubles as your question says.

   I wish you the best and please look for the concepts that I told you.

You are very much mistaken, Daniel,  in thinking I don't know and understand these very elementary physical and mathematical principles. It's you, not I, who needs to revise them. No hard feelings, Daniel. We did have an interesting discussion earlier and I appreciated that. 'Bye, and best wishes,

     The same for you Eric, although I am afraid that you are not going to solve your question. It is worse if you know the basic Gauss theorem and you do not apply it. Bye

Dear Eric and Baldomir,

I went through the hot and cold discussions between you both and I hope that no ill-feelings persist. This is a very important question and has the potential of overthrowing some of our most cherished and valued understanding of fields and of momentum. I think we should get down to find a solution rather than quarreling over trifles like Gauss law over closed surfaces etc.

Eric has asked a perfectly valid question and Feynman also points out the same difficulties on the topic.

Probably we have to modify our understanding of momentum, if Poynting is right. Or may be we have to modify our idea of fields if the understanding of momentum as a concept associated with motion is right.

Regards,

Rajat

Rajat ~

i feel that what you've just said provides at last some real insight this question - that our usual understanding of momentum density (or energy flux) in a field theory such as EM is incorrect and that in fact it is not necessarily associated with motion. Rather, it is simply an aspect of the field that is capable of exerting a force on an object.

I hope Daniel will give some more thought to the question - he appears to have misled himself into a fallacy (which I tried in vain to point out). Of course Gauss's theorem doesn't prove that divP = 0 implies P = 0!

   Dear Eric,

   You are right that if you think as P of a general vector, it is absolutely true that the Gauss theorem only cannot prove that div P=0, mens that P=0. That is obviously true and I have told you in another part of our discussion.

   There is an ingredient more that you do not tell it. The static Poynting vector P depends of the inverse of fourth power of the distance and therefore if you integrate on a closed surface that surrounds the sources of this field, it is reasonable to take zero for P. This is the answer for trying to tell it in simple words and no going to calculate integrals. But if you want to do it try to choose concrete geometry, for example to electric charged parallel plates with a solenoid into these plates, in such a form that you have E(r) and B(r) orthogonal. Now choose several points and calculate there divergence of such a field and after that you can see what is linear momentum associated to them, i.e. the integral of surface which surrounds them. This is all!

   Dear Rajat.

  I think that Classical Electrodynamics doesn't need to change anything on this basic concepts as Poynting vector interpreted as a density of linear momentum. In fact this pseudo-vector plays two different roles in the conservation laws of Electrodynamics but we have always forgotten space projection of the motion equation and we have only concentrate in the time-space projection.

  I have tried to give my answer to Eric and I think that is very clear, at least what I said, if you think that it is wrong; please go ahead and prove it. The linear momentum of the Poynting vector associated to static fields is always zero!

Daniel ~

If P is any "momentum density" (not necessarily electromagnetic – think for example of the flow of a liquid) the momentum of a volume V is ∫PdV. It isn’t ∮P·dS. Momentum density is usually thought of in terms of something actually flowing, and then for a divergenceless P, ∮P·dS = 0 just means that what flows inwards across the closed surface is equal to what flows outward. It's a conservation law – it doesn’t usually mean that nothing is flowing.

I think that Rajat is on the right track when he says that in some situations we need to abandon our habitual assumption that “momentum” is necessarily associated with motion. In particular, in electromagnetism the Poynting vector doesn't necessarily always denote a flow of energy. However, it is not correct to presume that in a static EM field the Poynting vector doesn't mean anything at all! And it's certainly not correct to try to prove that it's zero – it isn’t.

I have no more to say.

   Eric.

   Really I haven´t to say more too. I do not understand you and all these integrals are wrong:

   of a volume V is ∫PdV. It isn’t ∮divP·dS. The usual way we think of momentum density is in terms of something actually flowing. Then for a divergenceless P, ∮divP·dS = 0

    Gauss says

      ∫divPdV.=∮P·dS 

    You cannot define the integral of

    ∮divP·dS = 0

   That doesn´t exits in mathematics.

Daniel ~

Sorry, it was a typing error. I corrected it within seconds but you were too quick for me!. Look again  (-:

  Eric.

   It does't matter.

I think that you make many analogies but you don't think on the properties of the electric and magnetic fields. Please forget the fluids that they haven't anything like the Poynting vector.

By the way the Poynting vector has not an only physical meaning in Electrodynamics and also it can be defined in other forms because it enters in the energy-momentum tensor within a differential operator.

     Eric.

    ∮divP·dS = 0

     this is not an equation it is one identity.

I understand what you are saying: I know that an energy-momentum tensor Tij can be changed according to Tij →Tij + ∂kKkij (Kkij satisfing Kkij = - Kikj but otherwise arbitrary) without affecting ∂iTij = 0. That amounts to redefining the Poynting vector by adding the curl of an arbitrary vector. But that doesn't make any difference to what I've said - you can't eliminate ExB that way.

    Perfect. It was exacly what I wanted to say, it is a kind of gauge which keeps invariant the motion equations of Electrodynamics and tell us that fixed numerical value is no important. On the other hand you have it sequivalence with eA(r, t), where ii is usualy  definend the gauge condition for making unique the famaly of solutions or independent the electromagnetic equations.

   But you are right that this not the answer to your question which is much more basic and just a condition of the field momentum for static fields.

Dear Eric and others on this thread,

The following very interesting answer of mine  I put here from a related thread started by me. Hopefully you'll all enjoy the argument.

I want to make the following rather pedestrian observation:

We know that the Electric field can accelerate a charge. Now, what can possibly accelerate something i.e change the velocityof something? Ans: A force i.e. a quantity that can change momentum .

How can a force lead to a change in momentum unless it itself is endowed with the ability to suck in or impart momentum?

Thus the force or the field must be a storehouse of momentum that can act as a source as well as a sink of momentum.

It means that the static electric field also (not just the Poynting vector) must carry momentum.

For that matter, any force field must be such a storehouse of momentum, since it can give only momentum to another and that is what it has and that is what it does i.e. change momentum.

We may call it the potential form of momentum, to distinguish it from the kinetic form (massx velocity) that we are familiar with.

Regards,

Rajat

Rajat ~

Thanks for your answer. What you say is very like the conclusion I eventually arrived at: If E and B are static nothing is happening (nothing 'flows') until an object is placed in the field. Then a force will be exerted on the object - analogous to radiation pressure. I'm surprised that I know of no experimental attempt to test this. Your analogy with the static E-field is an interesting one. We have divE = ρ − charge is the source of E.  The static Maxwells equations imply divP = −E.j; the 'source' of the Poynting vector is provided by the currents j required to set up the static field.

Dear Colleagues! Too shy to ask, what is the debate? In the attached file you will find the derivation of  Poynting Theorem and see if the time derivatives are zero, then the Poynting vector is zero. In other words!!! Poynting vector, as well as the transfer of energy, exists only for the alternating fields.

Dear M L Shur ~

In the static case Poynting's theorem (equtn 3) reduces to ∮(E×B)·dS = 0. That implies div(E×B) = 0. It doesn't imply (E×B) = 0. There is, in general, even in a static configuration, a non-vanishing vector field E×B. The question was simply: "what is the correct physical interpretation of E×B in a static configuration?"

   The answer is very simple and basic and it has been said in our previous discussion. Poynting vector is defined into an integral and the ExH is only a density of momentum. Thus if you have static field moments into integrals of surface (2D)that have the limits of spatial integration depending of 1/(d^4) ,i.e. D^(-4), obviously gives zero.

   Notice that this seems to be in contradiction to have electromagnetic momentum for the radiation fields. No!, because the radiation fields depends of r^(-1).

Dear Baldomir ~

I don't wish to open up this ancient discussion again. I felt satisfied in the end that I understood.

Best wishes ~ Eric

   Dear Eric,

I do not want also to enter again in this old basic question. I strongly agree.

 Greetings. Daniel

                                Dear Eric,

      That is the question I answered!  In Static electromagnetic fields, of course, not necessarily (E × B) = 0. However, no physical meaning, this expression does not matter.

     Poytting asked the question: "What is the amount of electromagnetic energy  that is transferred ..." and found the answer: it is proportional to the amount of (E × B).

    But, as follows from the paper that I sent to you, this is true only for the variable components of EM fields.

      Static electric fields are  created by motionless electric charges.Static magnetic fields are created by constant currents. As we can see from the Maxwell's equations, these fields are independent.

   I will express a stronger assertion: no function, depending on the vector and scalar products of E and H can not have a physical meaning in the static case. But the functions of the form P (E) + T (H) may be .......

Dear Eric,

"what is the correct physical interpretation of E×B in a static configuration?"

The interpretation is that it does not have one. In stationary conditions the fields are not related at all,  that vector product cannot be defined.

   Dear Stefano,

   The product can be defined even if the fields are static; i.e., they depend of the position only. The problem is that the electromagnetic moment associated to this product is given within one spatial integral on a surface, while the static fields depend as the fourth power of the position. Thus the moment is always zero. The problem is not associated with the mathematics or the physics, the problem is simply the value of one integral.

   Notice that if you have radiation fields, then they depend inversely of the distance ( no square as the static ones) and therefore the integral gives a constant value because there is one surface integral of surface on a physical magnitude that depends of the square of the distance.

    Sorry, it is badly explained. Where is written:

    the static fields depend as the fourth power of the position

     obviously the dependence is for the product of the electric and magnetic fields which

   depend r^(-2) each and therefore its product r^(-4). Thus the surface integral of this magnitude for long distances gives zero.

    Sorry for trying to write too quickly and without seeing what is written. 

 In case it still could be of some interest to the present discussion, I would like to stress that ( at least as far as I am aware) the Poynting Vector Field (with its meaning related with both energy and momentum) could be applied in all cases, apparently never coming into conflict neither with experience nor  conservation laws. Besides the known examples of steady fields (for example a battery and a conducting wire in steady state), there are examples ot its application to static fields like the ones in Pugh and Pugh article "Physical significance of the Poynting Vector in static fields" (in part attached). 

After I mentioned in another conversation that momentum related kinetic energy seems to remain adiabatically induced in electrons while translationally immobilized in resonance states in atomic orbitals, Eric drew my attention to this thread, wondering if this might have a relation with what Rajat Pradhan and himself mentioned regarding the possibility that momentum energy might not necessarily be associate with motion. 

This idea to possibly be applied to the case of a charged capacitor statically resting in an ambient magnetic field.

Please read back Rajat and Eric contribution on this issue on page 5 of this thread.

I know of one case of apparent macroscopic momentum induction by ambient magnetic field that might be connected with this case even if in the confirming experiment, the momentum was expressed as motion.

It is the Einstein-de Haas effect that I analyze in the paper below and where references are given.

Since this effect has also been noted in materials other than ferromagnetic, It struck me that if some or all of the electrons in excess on the negative plate of the capacitor are unpaired, they are likely to be force aligned magnetically in parallel spin with respect to each other (all all of them antiparallel to any ambient magnetic field, such as the Earth magnetic field), but since the capacitor most likely is not free to move physically, then the related macroscopic momentum related motion observed for the freely rotating Einstein-de Haas cylinder by alignment in the same direction of the translational half of the magnetically aligned carrying energy of the unpaired electrons would be prevented from being expressed.

Note that this is not momentum energy added to the system, but the normal momentum component of the carrying energy adiabatically induced by the Coulomb force in each electron due to the distance each electron stabilizes at with respect to the atomic nucleus, momentum energy which is not usually expressed in the same direction at the level of the electrons but that cancel each other's motion because when not forced to magnetically align by an external magnetic field, they would naturally settle by default into least action anti-parallel spin orientation with respect to each other in the material.  

I think that this could be experimentally verified by having a capacitor made free to frictionlessly move, maybe in a circle about some axis.

Since so few electrons would be involved with respect to the mass of the capacitor, the momentum expressed as macroscopic motion would be minute.

If this is a case of inhibited Einstein-de Haas effect of unpaired electrons momentum kinetic energy alignment, and if the capacitor material is chosen so that the electrons in excess on the negative plate do not pair up (maybe they usually don't, which is an aspect of capacitor charging that I am unfamiliar with), then the capacitor should start moving as it is being charged, like the Einstein-de Haas cylinder.

http://ijerd.com/paper/vol6-issue12/B06120711.pdf

We have considered the significance of the Poynting vector in M.Ribarič and L.Šušteršič, Conservation Laws and Open Questions of Classical Electrodynamics, World Scientific, Singapore 1990, https://www.researchgate.net/publication/319111561_Conservation_

Laws_and_Open_Questions_of_Classical_Electrodynamics

Book Conservation Laws and Open Questions of Classical Electrodynamics

Thank you Marijan ~

I've downloaded the ebook. It appears to be relevant to many of my interests...

I will appreciate your comments.

Dear Eric, your question is excellent, and unanswered by most standard books on electricity and magnetism.

Please, read the article by Galili and Goihbarg, see

http://sites.huji.ac.il/science/stc/staff_h/Igal/Research%20Articles/Pointing-AJP.pdf

This article describes a simple example of static electromagnetic fields: a circuit that consists of a battery, two wires and a linear resistor. Electromagnetic energy flows from the plus and minus battery terminals to the resistor, which dissipates the electric power. Learn why the energy flow is described by the Poynting vector. Hint: the metalic wires that connect the battery and the resistor must have a non-zero electrostatic charge, such that the two wires source a radial static electric field, perpendicular to the wire surface. One wire is charged positively, and the other one is charged negatively.

It is not surprising that both wires conduct energy, that flows from the battery to the resistor, expressed by the Poynting vector. The Poynting vector is directed towards the 'center of the resistor (the resister converts all the 'static' electromagnetic energy into heat radiation, so it is the 'end station' of the static electromagnetic energy flow).

It is surprising that the article by Galili and Goihbarg is not part of the standard educational literature on Electricity and Magnetism (classical electrodynamics, electrostatics, magnetostatics).

Some extra comments. The paper "Energy transfer in electrical circuits" by Galili and Goihbarg is very good. Their theory should be standard education material, and they explained something essential in a simple way. How many PhD physicists/engineers understand that static fields can perform Poynting energy flow? How many electrical engineers understand that it is essential to have non zero net charge on metallic wires in order to establish wired energy flow from an electric energy source to a load (DC circuits, and even 50/60 Hz AC circuits)? Not many, I suppose. Before I read this paper, I did not know this either :), so, despite my masters degree in electrical engineering I did not even understand how/why electromagnetic energy flows in the simplest of electric circuits.

For example, double the battery voltage and also double the resistor's Ohmic value. The electric current in Amps remains the same (the rotational magnetic fields close to the wires are unchanged), however, the static net charge in the wires has doubled, and the radial electric field strength has doubled, so the Poynting flow has doubled as well, and also the power dissipation in the resistor has doubled. It should be possible to calculate the net charge on the wires, for a given wire diameter, battery voltage and Ohmic resistor value, which "is an exercise for the interested reader".

 I would like to remind my former answer: "In case it still could be of some interest to the present discussion, I would like to stress that ( at least as far as I am aware) the Poynting Vector Field (with its meaning related with both energy and momentum) could be applied in all cases, apparently never coming into conflict neither with experience nor  conservation laws. Besides the known examples of steady fields (for example a battery and a conducting wire in steady state), there are examples ot its application to static fields like the ones in Pugh and Pugh article "Physical significance of the Poynting Vector in static fields".

The battery and wires case for the Poynting vector field in a steady state appears already explained in the second volume (Elektrodynamik) of W. Döring, Einführung in die Theoretische Physik (Sammlung Göschen; fünf Bände: Mechanik, Elektrodynamik, Optik, Thermodynamik, Statistische Mechanik), Berlin, 1957. I don't know an English translation of this five volume textbook.

However, in the paper of Galili and Goihbarg recommended by Dr. van Vlaenderen ( "Energy transfer in electrical circuits: A qualitative account" Am. J. Phys. 73 (2), February 2005) the development of the battery and wires case is more detailed, complete and in English!

Dear Eric, sorry but I really miss the point. In a PURE STATIC regime the electric field is not coupled with any magnetic one and viceversa. Thus, even the definition of electromagnetic wave is missed. As a consequence, the Poynting vector is not defined , at least in a pure formal way.

As soon as any transient condition is induced, electric and magnetic fields start to be coupled together and the Poynting vector is then set up.

de gustibus non est disputandum

First you have to define what is the real meaning of the electrostatic charge.Next step you will be able to define the electric permittivity and magnetic permeability of empty space.Then you are ready to define electric and magnetic field strengths.Then you can make discussions about Poynting's vector.Without knowing the nature of elementary charge all you do is to repeat what is happening for the last 150 years .Talking about invisible fields,ascribing them unnatural properties and giving them names like Coulomb, Ampere, Henry, Weber, Tesla etc.Do you really know what is all about???Do you?I am 55 years old mechanical engineer totally unsatisfied for the empty knowledge deliberately given to scientists about electromagnetism.Yes it works ,but you don't know what and why is working and you take this as a given fact to make discussions about details of the details without trying to find out about the essence of it .Have you ever thought that Poynting's vector in static fields firstly must derive from a combined electromagnetic field and not from two irrelevant stationary magnetic and electric fields that happened to be in a volume in proper configuration and secondly that particular Poynting's vector changes the spacetime curvature around the conductor that is vertically directed to??Gentlemen you represent universities , research centers, and you call yourselves researchers, scientists etc.You don't impress me a bit about your superiority of handling tensor calculus,theorems continuity laws ,differential equations etc.I watched Phd instructors ,professors etc repeating like parrots the same and the same in lectures every year participating obligatorily in research teams knowing absolutely nothing about what they are up to and making careers like this .Enough.Do something useful and tell us what is all about E/M fields that our senses cannot receive but it is there and it is working.Spare us from useless long discussions.I am so dissapointed reading the above.I really thought that I could find something interesting

Dear Stelyo ~

I see you have resuscitated this discussion thread, which I thought (and hoped...) had faded away long ago (-;

You are quite correct in your assertion that physicists simply don’t know what the aspects of Nature that Physics deals with actually ARE. Kant understood that long ago (https://en.wikipedia.org/wiki/Thing-in-itself).

"It is important to realize that in physics today, we have no knowledge what energy is" − The Feynman Lectures (1961–1963) Vol 1

You ask what is the “real meaning” of electric charge, electric and magnetic field strengths, permittivity and permeabilty of “empty space”, etc, and you complain that professional physicists cannot tell you what is the “essence” of those quantities. Of course they can’t! They are not obliged to because Physics is the science of measurement; it is a pragmatic endeavour that seeks patterns of correlation in the relationships between various results of measurement . It’s not about “essences” or Kant’s “Ding an sich”.

“A physical quantity is defined by the series of operations and calculations of which it is the result” − A S Eddington

The appropriate language for expressing the relationships among measured quantities is mathematics.

"Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true." − Bertrand Russell

I take that to mean that mathematics deals with relationships, not with the "essences" or "real meanings" of “things in themselves”.

Maxwell studied the experimental results of Faraday, Ampère and others and unified them in an elegant set of differential equations. The experimental verification of those equations has withstood the test of time and has led to the miraculous advances in communications engineering and electronics that we enjoy (and take for granted) today.-

My original question was a simple straightforward request for clarification concerning an aspect of Maxwell’s equations that I had not understood. It had nothing to do with metaphysical questions of “essences”. Let me express it differently: “what experimental predictions, if any, can be deduced from the measured Poynting vector, in general, when the electric and magnetic fields may or may not be wholly propagating?” I see nothing mystical or metaphysical about that question. I have found no satisfactory answer in the literature. And, like you, “I am so disappointed reading the above. I really thought that I could find something interesting.”

PS: incidentally, you say, about electromagnetism: “...our senses cannot receive”. We have eyes, don’t we?? We can feel the pull of a magnet, can we not??

Dear Eric

Thank you so much being kind enough to write the above long text as an answer to my protestation not to you personally but generally to the established mentality about E/M fields.Many things changed according to me to the worst after the acceptance (I accept the results of it totally too) of STR that refers to empty space as really empty( I don't accept it).The author of it later on ,set up GTR as a pure aether theory but unfortunately the concept of empty vacuum is still on no matter what zero -point field theory tells us.What I say is that with my senses I can understand mass, length, time and the derivatives of them.But I fail to understand electric charge(yes I can feel the repel or attraction but I would like to know what is between them as density elastic modulus if any and other natural properties) .Is it property of the particle or the reaction to it of the surrounding "empty space"?And what is bothering me a lot is the magnetic field :closed loops, always vertical to electric field,one produces the other but without try to be philosophical I really would like to know which is the essence of the one and the other before my life is over in this planet .Are they part of the surrounding fabric of vacuum?We need to know more.What is given in research papers are insufficient .You are different and this is the mentality what the rest of us need to keep hoping for a better science from people that took over from the previous to advance the research related to E/M fields.Nice interacting with you.I am sure you will keep going wondering to the proper direction and not taking things for granted.We need you

Dear Sydney

Thank you for your text.I read it carefully.With your permission I would like to address the following remarks about your thoughts that obviously are the outcome of many years of study and research:

If the phenomenological physics is our subjective truth and the modern quantum field is the objective reality that we cannot conceive with our senses, if we are so much advanced to even consider such a field, then scientists owe to the rest of humanity (not from now on but since long time ago)to search the properties as you mentioned " The properties of the underlying field structure – the basic quantum fields – are responsible for the creation..............".This is what I am looking for :

Give me properties of the quantum field.Don't even dare to tell me that the local

fluctuations is the product of nothing since the outcome is something that our poor senses can conceive.If on average, as some assertions claim, the total energy

of universe is zero that doesn't mean zero locally.Over there at last, researchers must be able to measure properties as quality and quantity.For example it is ridiculous still to measure the local property of space called electrostatic charge

in Coulomb units to honour Coulomb.It could be in Grimm units.These are not units.These are the proof of knowledge absence.Everything must be expressed as combination of the units of phenomenological physics that really works in practice

mass , length, time.Except observing them we have the apparatus to measure them.Electrostatic charge could be the derivative of mass with respect time

dm/dt.Immediate the question that is raised : dm/dt of what?You can answer that if you know the properties of the medium that is responsible for the local quantum fluctuations. Somebody with the status of university professor in a discussion open to public having as a subject " gravity" dared to address a statement that :

"the universe exists because we can observe it" obviously read it from somewhere.

What an arrogance to declare something like this.Unbelievable.These are the pioneers in research and science.Do the good job.We need scientists like you and Eric .Kind regards .

Stelyo

Dear Sydney ~

Your final paragraph implies that the original question is unanswerable without considering the quantum nature of the fields. It even seems to imply that it is not a valid question because “represents phenomenological physics”!

It is a question about the interpretation (ie, the experimental implications) of Maxwell’s theory − a purely classical (ie. “phenomenological”) theory. Since the theory is mathematically self-consistent the question must have an answer expressible in terms of classical physics. No-one in this lengthy discussion has provided one. Physics is the science of observations and measurements; it is “phenomenological” by definition!

“A physical quantity is defined by the series of operations and calculations of which it is the result” − A S Eddington

Dear Eric,

The poynting vector is conceived for field which vary in time and can be correlated through the maxwell equation. For static field the E and M are separated do not propagate variations so the poynting vector is zero.

Dear Stefano ~

That is an answer others have given, earlier in the discussion. But it doesn’t answer the question because there are many static solutions of Maxwell’s equations that make sense physically and for which E × B is not zero (a trivial example: a stationary charged object and a stationary magnet placed side-by side). Electromagnetic radiation is only one aspect of how electromagnetism can behave; it doesn’t have to propagate! In a very general EM field how can one separate the “propagating component” from the “non-propagating component”? It seems to me that one cannot − the idea doesn’t make sense. It follows that if E × B is physically meaningful it must be meaningful for all electromagnetic field configurations.

OK Sydney, I agree - my question was “about a model, not reality”. What is your answer to the question?

Another question: what is "reality?"

Dear Eric,

it does not make sense to make that product of uncorrelated E and B, they have to be part of the same "source" and to be part of the same "source" they have to be time variant.

I totally agree. EM radiation is just a result of the transitories of the EM fields

the zero frequency component of the FIELD is the non propagating component, can be separated from the non-zero frequency components according to the Fast Fourier Transform.

>

it does not make sense since in general they may not belong to the same EM field.

In other words take a current through a wire and static electrons. You cannot combine the B of the current and the E of the static electron, it does not make sense, you get nothing...

Dear Stefano,

I'm not sure if I understand you here:

I totally agree. EM radiation is just a result of the transitories of the field.

The radiation electromagnetic field is very different than the static field because it depends a r-1 instead of r-2, explained by the Lienard-Wichert potentials. Thus when you calculate the momentum associated to the Poynting vector you must integrate on a surface. In the case of a radiation field you obtain a constant while in the case of the static fields you get zero. Thus if you apply two static fields you always cannot obtain directly radiation emittion for such you need to have accelerated charges.

Dear Daniel,

so EM radiation is just one result of the transitories of EM fields....

in other words in a transmitting antenna there are E and M time variant on the surface and till at a certain distance non negligible, radiation may be generated or not depending on how the antenna is built, usually there is..

Dear Stefano,

One antenna has an oscillating current which is therefore made by accelerated electrons, for instance a dipolar antenna. What is important is that there are two kind of fields the static ones E and H depending of r-2 and the radiation ones which depends as r-1 . if you calculate the associated momentum p to both Poynting vectors it is inmediate to see that only the surface integral sourranding them survive for the radiation fields. The other gives zero.

I'm used to employ the word transitory for the currents when you cut them or you put them till the permanent current is stablished. It is true that during the transition of currents there are also accelerations and therefore radiation fields, but the most important for one antenna are produced for such aim. But perhaps we are just with a question of language, in any case both are understanding how the Poynting vector of static fields must always gives a zero linear momentum p.

Dear Daniel,

a transitory is the definition in control system, a transient, a non stationary behavior.

An antenna at its emission is intrinsically non stationary.

actually there are 3 kind of fields

a) Stationary B and static E (which in principle are negligible in the case of an active dipole with a sinusoidal excitation),

b) non stationary B(w) and E(w) which are the time variable version of a) and are the fields of the accelerated charges which I call a transitory of the field

c) emission of a EM wave in space or in a medium which is a consequence of B(w) and E(w)

from b) it is not at all guaranteed c), this means that two time variable fields may not be able to emit EM waves...

hence an EM wave is a consequence of the transients of B and E, an EM wave is just a consequence of the transient of the EM field itself.

In addition if you consider different sources, it is not defined at all a pointing vector for a static current through a wire B and a static field of one electron E outside and at rest with the wire.

If you just want to discuss the energy part, a static electric field can polarize the charges in a material medium, and this requires energy.

What can flow in crossed static fields are charges, if there are any.

Dear All ~

The Poynting vector is part of a tensor in Minkowskian spacetime, the stress-energy tensor of the electromagnetic field (derived from Noether’s therem). It is observer-dependent. It can be zero for some inertial observers and not for others. The attached file gives the transformation law of E and H under a Lorentz transformation.

Consider the simple example of a uniform electrostatic field in the direction of the x axis, in the absence of a magnetic field. E = (E, 0, 0), H = (0, 0, 0) (eg, the region inside a large stationary capacitor). The Poynting vector is of course zero. Now transform to the inertial frame of an observer moving with uniform velocity along the z axis. We have E′ = ( γE, 0, 0), H′ = (0, - γβE, 0), (where β = v/c). The Poynting vector P = (0, 0, E2β/(1− β )) is not zero. For the moving inertial observer the electromagnetic field remains uniform and static. There is no radiation. (radiation comes from acceleration of the sources, not from their steady motion).

It makes nonsense of the concept of the stress-energy tensor to claim that in this case (and other more complex cases) the Poynting vector has to be ignored as meaningless!

Dear Eric,

In your question there are many subtle things which allow to answer it in several different forms. The tensor energy-momentum contains twice the Poynting vector playing different roles in Electrodynamics. Let us write the equation of motion

∂βTαβ=-FαɣJɣ

In the case of being without sources, i.e. Jɣ=0 and you have conservation of the energy-momentum Tαβ which translates the generators of traslations symmetries of the space-time as conserved Noether's currents. Thus this is a physical conserved quantity, this means that you cannot made disappear choosing a transformation of coordinates.

On the other hand, as you have made, the energy-momentum tensor depends of the electric and magnetic fields. You can transform them, yes, but that is not enough you need to transform the whole energy-momentum tensor for seeing how the different terms transform among them: density of energy(00), density of momentum(0i), density of momentum(j0) and Maxwell stress tensor (ij).

Answering again your initial question, the Poynting vector for static fields always gives zero due to its definition because it is a surface integral on a function depending of 1/r4, being r the distance from the sources which produced such fields.

Dear Daniel ~

“You can transform them, yes, but that is not enough you need to transform the whole energy-momentum tensor for seeing how the different terms transform among them.”

Yes, of course. I can do that. It’s no big deal. It makes no difference to what I said.

A Noether energy-momentum tensor is in general is not symmetic. Conservation of energy is contained in the continuity equation ∂iTi0 + ∂T00/∂t = 0. Energy flow across a surface element dS with unit normal n is niTi0dS. Thus the 3-vector Ti0 is energy flux. (Why do you keep on insisting on integrating over a closed surface? And on including the sources inside that surface? A flow − of anything − is understood to be a flow across a chosen open surface patch and is not concerned with the sources of whatever is flowing...).

The 3-vector T0i on the other hand is momentum density. The momentum contained within a (3-dimentional) volume element dV is T0idV.

The generally adopted symmetrical energy-momentum tensor for electromagnetism is not the Noether tensor it is the symmetrized Belinfante tensor (see, for example https://www.researchgate.net/publication/241377004_A_theorem_on_stress-energy_tensors); The subtle distinction between ‘energy flux’ and ‘momentum density’ is then obscured. Hence, as you say, “The tensor energy-momentum contains twice the Poynting vector playing different roles in Electrodynamics”. However, I cannot see how that affects the fact that the Poynting vector can be non-zero in non-propagating EM fields. Nor can I see how it affects my (very simple and basic) question about its interpretation. Yes, momentum density and energy flux are observer-dependent by definition, and can be “transformed away” even in Newtonian physics, so I've been surprised at all the irrelevant obfuscation that has been thrown at me in this discussion thread since I posted the question long ago. I’m at a loss to see what more I can say.

My own answer to my original question is that yes, a static electromagnetic field does contain a hidden (non-manifest) momentum density or energy flux, which can exchange energy and momentum with objects introduced into the region of the field, in a manner analogous to “radiation pressure” in the case of propagating electromagnetism. All that is lacking is experimental evidence; the relevant experiments have, to the best of my knowledge, simply never been done!

Dear Eric,

You are only forgetting a fundamental detail: the action is a volume integral of the Lagrangian, i.e. the Lagrangian is in fact a density of Lagrangian. The same the energy-momentum tensor, thus if you want to obtain the physical quantities you need to do the integrals. For instance, the oo- component of the energy density and no the energy, which means that if you want to have the energy you need to make the integral with respecto the space. The same happens if you want to have the momentum respect to the Poynting vector.

Look at the eq.27.3 in the Feynman book

http://www.feynmanlectures.caltech.edu/II_27.html

If I understand you well, when you says:

All that is lacking is experimental evidence; the relevant experiments have, to the best of my knowledge, simply never been done!

It is not true. This is simple to measure and surely it was measured but if with static fields you were obtaining radiation this would be a good form to avoid the usual antennas. This is not possible due to the reasons that I have tried to explain you in the past post.

Dear Eric,

Let me to prove it just taking the time projection of the conservation of the energy-momentum that you have written

∂iTi0 + ∂T00/∂t = 0

If the fields are static ( they do not depend of time), you have then that

∂T00/∂t=0

Therefore you must have

∂iTi0 =0

which means that the Poynting vector must be also zero. This is another form to see that your effect couldn't exist for conserving the energy. Do you agree?

Dear Daniel ~

"Therefore you must have

∂iTi0 = 0

which means that the Poynting vector must be also zero."

No. It means only that the Poynting vector Pi = Ti0 is divergeless. Integrating it over a closed surface − the boundary of a volume − gives zero which means only that the energy flowing into the volume enclosed by the surface is equal to the energy that flows out. P itself is energy flux density which means that the rate of flow of energy across any surface element ndS is P · ndS. That is not, in general, zero and should in principle be measureable. (Think of divergeless fluid flow as an analogy....)

(I have a sense of deja vu − haven't you and I had this argument before?)

Dear Eric,

It is true that this is only the conservation of energy and momentum, by the way the momentum is not as you write it, Pi = Ti0, it is with on integral on the space

Pi=∫ T0i d3r=∫S/c2 d3r

And what you obtain is that the derivative with respect to the time for the density of energy is zero because the fields don't depend of time, thus the space derivative of the momentum must be zero, i.e. the surface integral of the Poynting vector must be zero using Gauss theorem. Thus if the integral of a closed surface is zero for one E and H uniques it means that such pseudovector S is zero. I don't know other possibility.

Your mistake is always the same, you don't consider that you are speaking of densities. And I can tell you that if this were different than zero the production of electromagnetic radiation would be very different than it is: only with acceleration of charges never crossing fields.