Hi Demetrios,

from a 'classical' point of view (no tunneling), a paradox seems to exist in fig. A already: If, for example, in the dark fringes the E fields of the beams cancel each other exactly, the poynting vector would be zero within the fringes. So, the power flow of each beam would hit an impassable, nonreflecting barrier, even without any material sheet.

My suspicion is that there are no realistic beams which could produce dark fringes cutting across both beams completely. After all, the beams cannot be just cutouts of homogeneous plane waves. Did you have a certain geometry of the beam fields in mind?

Thanks Joerg for your reply. I too agree with you that, even without a 'barrier', the physical 'picture' implied by the stationary fringe pattern of these two freely propagating waves is in itself paradoxical, in light of the vanishing Poynting vector. Yes, the 'real' situation would of course not involve infinitely thin and perfectly reflective sheets, or ideal plane waves.......but it does seem that a 'barrier' can be placed across the whole width of the beams, and the proposed 'effect' should remain largely intact under realistic condirions. Please see attached file of a Moire fringe pattern, which is equivalent to the stationary fringe pattern created in my setup (involving mutually coherent and collimated light beams).

Hi, Demetrios,

Your experiment is not so clear to me:

==> "It certainly appears as though we can "cut each of the light beams in two, across an impassable barrier", yet they will persist and continue to freely propagate!"

How can they persist to propagate freely? A perfect conductor is a perfectly reflecting mirror. At its surface the horizontal electric field is null. Each one of your beams should be reflected by the mirror.

To convince yourself try the following: make the two beams different by wavelength. Without the conductor you won't get a stable interference pattern. But with the conductor you'd get on each side of it a stable interference pattern. This pattern appears between the beam comming to the conductor, and the beam reflected.

Hi Sofia.......The point is that we have  stationary fringes in the setup I propose, so that means the two beams have the same wavelength and have a steady phase relation between them such that a 'dark fringe' is present at the center of the "diamond shaped" interference region. Furthermore, the beams must first overlap and the fringe pattern established before we can insert the thin reflective sheet; look at the pic file I uploaded to see how the dark fringe extends across the width of beams; remember, at the center of the dark fringe the light intensity is zero and so is the Poynting vector, so inserting the barrier should not destroy the fringe pattern or affect the light propagation.......meaning that we have put a wall between two parts of a wave.

No, Demetrios, no!

It's obvious that a perfect conductor would reflect perfectly the electromagnetic beam, isn't it? An object that you place on the way of the beams stops them, unless it is transparent to the beam. But a perfect conductor does more, it reflects the beams.

For convincing yourself, I recommend you to think of the experiment with different frequencies.

The fact that for beams of the same frequency, the interference tableau does not change by putting in the middle of the dark fringe a perfect conductor, elludes you. The waves don't pass through the conductor. Inside the conductor the electric field is null. So, each wave is reflected by the conductor and generates an interference pattern between the direct and reflected part.

It seems to me that indeed we could put a sheet of perfectly conducting material into the dark fringe and it would not result in a change of the interference pattern. First, we may reason causally: putting a conductor where the field is zero anyway, cannot have any effect.

Second, we may compare the interference patterns to be expected with and without conductor. Without conductor, we have two propagating waves crossing each other and interfering to produce a standing wave pattern in the direction perpendicular to the fringes. There is still propagation parallel to them and the Poynting vectors of the two propagating waves will add to something parallel to the fringes and zero in the dark fringes, so there is energy flow parallel to the fringes in the bright regions.

With conductor, we see an interference pattern on both sides of the conductor resulting from a wave and its reflection. But this pattern is precisely the same as that of the two crossing waves before! And it has the same Poynting vector.

Classically, the two situations cannot be distinguished. Quantum mechanically, there might be effects from the fact that a single photon can either pass the interference region, in which case it would have to interfere with "another" photon to make a standing wave, or be reflected, in which case it could interfere with itself. 

Klaus, your attention, please,

Demetrios posed a quite subtle problem. (I would also recommend to read what others wrote.)

I would like to stress the subtle part. Without the conductor, the two waves PASS through one another. It is true that the electric vector vanishes at the points of destructive interference, but the two waves EXIST even there. We cannot detect there a particle because of the mutually cancelling ellectric fields, but the two waves EXIST even there. Beyond the interference region they also separate as if they never and nowhere cancelled one another.

There exists no JUMPING over the dark regions, not even over the darkest line in the middle of the dark fringe.

To convince you, there are experiments with more than one particle, that shed light on this issue, and they are very interesting - but their discussion goes beyong the scope of the present thread - so I think.

Now, in the presence of the conductor the situation changes. Inside the conductor the electric field is ZERO. The waves don't pass through it, they are reflected at its surface. For better clarifying the situation, it is good to do the experiment with two frequences as I recommanded.

With kind regards and best wishes,

Sofia 

Sofia,

I read what others wrote.

So what is your meaning of "the two waves EXIST even there"? The electric and magnetic fields are zero on the surface of exact destructive interference. The same applies, if an infinitely thin conductor is put there -- which therefore is essentially invisible to the wave.

Now, a wave is a nonlocal concept. So clearly, we may say that the wave exists even across surfaces of zero field. However that is true even if the conductor is there. If it does not matter that the field is zero in the case without conductor, why should it matter in the presence of a conductor? If you believe the wave to exist across a region of destructive interference, why don't you believe it to exist across a conductor in that region?

I maintain that the physics is not affected by the - hypothetical - infinitely thin conductor in the classical case, if the latter is put precisely in the region where there is no field. The field distributions are the same, everything that is observable is the same. Since we are talking physics, not magic, we should consider the situations physically equivalent.

Dear Klaus,

==> "So what is your meaning of "the two waves EXIST even there"? The electric and magnetic fields are zero on the surface of exact destructive interference. The same applies, if an infinitely thin conductor is put there -- which therefore is essentially invisible to the wave."

Nooooo! How can the wave pass through a conductor? Klaus, you know electromagnetism. Inside the conductor there is no electric field and no current. Such conductors serve as mirrors. The wave won't penetrate them. So, what the metal does, is to send back by reflection each one of the incoming waves.

As to how I know that the wave EXISTS there where the interference is desctuctive, I can't prove in base of single-particle waves. But I know this from 2-particle interference judging according to different frames of coordinates (Hardy's cellebrated paradox). From the point of view of one frame of reference two 2-particle waves undergo destructive interference, and from the point of view of another frame they do not.

So, it doesn't go abra-cadabra, if the waves ARE THERE from the point of view of one frame, then THEY ARE.

==> "If you believe the wave to exist across a region of destructive interference, why don't you believe it to exist across a conductor in that region?"

Neither I, nor you need beliefs. Even without interference, i.e. if a single wave is in the experiment, the conductor stops it, reflects it back. You see, people sometimes think that certain wonders occur, while simple physics elliminates the wonders. Look what Demetrios says

==> "cut each of the light beams in two, across an impassable barrier", yet they will persist and continue to freely propagate!"

No, there is no continuing of free propagation. That's why I consider a good thing, first of all to tell Demetrios the physics behind apparences. The APPARENT similarity between the pattern with the conductor (supposed) infinitely thin, and the pattern without conductor, is ELLUDING. The physics is different.

With my best wishes

Dear Sofia,

I think that, once again, you are confusing interpretations and reality.

First, it is clear that the two situations are distinguishable in reality, because there are no infinitely thin conductors and no real conductor pushes a field down to zero, there is a finite penetration depth for AC fields (see the skin effect). An infinitely thin conductor is not an impenetrable barrier at all, given that the penetration depth for AC fields is finite. So even a very good 2D conductor such as graphene will be easily penetrated by an electromagnetic wave. If it does so for a propagating wave, it will however perturb that wave, whereas if it is put into the node surface of a standing wave, its effect will be almost zero.

However, there were some idealizations in the experiment, and it was this idealized case that I was discussing. The idealization is that the conductor is infinitely thin. You might also make it ideal, i.e., set its conductivity equal to infinity. This will make its penetration depth zero, so if it had finite thickness, no wave could get through. For zero thickness, the answer is indefinite and depends on the limiting process by which the conductivity was sent to infinity while the thickness was made zero.

Now if you first create the interference pattern and then put the conductor of zero thickness in the region where the field is zero, i.e. in the nodal plane of the standing wave, then standard physical arguments suggest that there is no effect of the conductor. A conductor put in a field-free region does not have any effect. There was no energy current through the nodal plane before and after the conductor was put in the system. Observation confirms this interpretation, because there is no change in the interference pattern.

The other interpretation is that the conductor indeed stops both propagating waves and reflects them in a way producing the same interference pattern between incident and reflected wave as there was before between the two propagating waves. This assumes that no wave goes through the conductor, something not necessarily to be expected from the properties of an infinitely thin conductor. Nevertheless, this interpretation also is confirmed by observation, so it may also be considered admissible.

The important point to realize here is that interpretations are not reality. So they should not be considered true or false nor correct or wrong but rather admissible or implausible. The closest we have to what is real is the mathematical description (in classical theories). That gives the same result in both interpretations, so their reality content is the same. While the two interpretations seem to contradict each other, this does not render either of them useless. In their context, both have plausibility. (No field -> plausibility of no effect of conductor; conductor being obstacle to propagating wave -> plausibility of conductor having strong effect on wave.) There are other examples where we have contradicting interpretations and nobody cares. In elementary particle theories, fermionic antiparticles may be interpreted as particles propagating to the past or else as holes in a sea of occupied negative energy states. These interpretations are both compatible with the math, they contradict each other, but particle physicist are well-versed in quantum mechanics and use both of them ad libitum, knowing that these are just interpreations, not reality (nobody understands quantum reality).

Now, if you do additional experiments, one of the interpretations may become much more plausible than the other. For example, if you switch off one of the two counterpropagating waves with the conductor in place, you will get a completely different field pattern than without, if the conductor is really blocking the wave (i.e., if it has zero penetration depth). Then you will see an interference pattern on one side and no electromagnetic waves on the other side of the conductor and just a propagating wave, if there is no conductor. This will clearly render your interpretation more plausible -- it works in two cases, while the other works in one only. (It does not invalidate the other interpretation for the first experiment, but it strongly limits its scope.) On the other hand, if the conductor has an infinitesimally positive penetration depth, it will be only a weak perturbation to the field, which will essentially propagate through. In that case, the situations with and without conductor will still look very similar albeit not identical. Nevertheless, in this case the first interpretation will be more plausible, because it gives correct predictions for both experiments and your interpretation will be implausible, because it incorrectly predicts strong reflection of a wave by a conductor, which in fact has imperfect reflectivity.

Dear Klaus,

I don't know what you mean by interpretation, we are not in QM here.

Your explanations about realistic cases are O.K., but Demetrios was speaking of an idealized case, not about the realistic one.

==> "Suppose we insert a very thin (compared to the fringe width) and, ideally, perfectly conducting 'sheet' across, say, the central 'dark fringe'(fig. 'B')."

You don't have the right to change the data of the problem. And my intention was to stress the physics for the problem posed by him. I repeat, the central point in my explanation is that the waves don't flow through the conductor as Demetrios thinks.

Now, your statement describe APPARENCES

==> "A conductor put in a field-free region does not have any effect. There was no energy current through the nodal plane before and after the conductor was put in the system. Observation confirms this interpretation, because there is no change in the interference pattern."

The dark fringe is not exactly a field-free region Klaus, YOU KNOW that the two beams pass through it. For removing any doubt, please see a simple argument: in the figure B presented by Demetrios, place an absorber on the way of the wave going upwards toward the conductor foil. (I repeat it's a perfectly conducting foil as Demetrios says, s.t. the electric field on buth upper and lower surface is zero.) Thus, no wave would touch the conductor from below, and the interference pattern below the conductor would disappear. However, above the conductor the interference pattern would remain intact. Therefore, the reflected wave in the upper plane is not a transmitted wave (through the conductor) of the incident beam in the lower plane.

Mathematically, without the conductor we describe the interference pattern as

(1) ψ1 + ψ2.

In the presence of the conductor we have to write

(2) (ψ+,in + ψ+,r) (ψ-,in + ψ-,r)

where by ψ+  I mean the wave in the upper half-plane and by ψ-  I mean the wave in the lower half-plane. The subscript "in" means incident and "r" means reflected. The waves ψ+ and ψ- evolve independently knowing NOTHING of one another.

==> "Now if you first create the interference pattern and then put the conductor of zero thickness in the region where the field is zero, i.e. in the nodal plane of the standing wave, then standard physical arguments suggest that there is no effect of the conductor."

I address all the time Demetrios' question, i.e. whether "they will persist and continue to freely propagate!". THIS WAS THE QUESTION.

With kind regards from me!

Thanks Sofia and Klaus for the good discussion of this scenario.

So far, I agree with Klaus regarding the description of the setup and the implications. In this light, consider the following modification:

Instead of an idealized reflective sheet (of perfect conductivity and infinitesimal width), lets say we have arranged for a very dense 'sheet' of atoms to stream through the nodal plane of the central 'dark fringe' formed by the two intersecting light beams. Suppose we have chosen the atomic species and the wavelength of the light beams so that, when taking into account effects related to the velocity of the atomic stream, the light is strongly absorbed by the atoms--in other words, the dense flux of streaming atoms acts like a strongly absorbing sheet that can be, literally, only a few nanometers 'thick' (due to atomic dimensions). In such a scenario, if, say, we change the phase of either beam so that a 'bright fringe' is shifted into the plane of the streaming atoms, then BOTH light beams are absorbed; if we "shut off" either one of the beams, the other beam will be absorbed; if we do nothing, then the light beams effectively "pass 'over' or 'through' the wall of absorption". So, in this scenario, we have a distinguishable effect  (on the light beams and atoms) if we perform any of the two above-mentioned actions, when comparing the case where the atomic stream propagates in the 'dark' region of the intersecting beams and the case where the atomic stream isn't present at all.

Dear Demetrios

==> "a very dense 'sheet' of atoms to stream through the nodal plane of the central 'dark fringe' formed by the two intersecting light beams."

In the "Divina comedia" (Divine comedy) of Dante Aligheri is told that on the gate of the Hell was written "Lasciate ogni speranza voi che'ntrate" (Abandon any hope you that enter). What you propose is playing with the uncertainty principle. It's quantum mechanics (QM) here.You can't arrange a beam of atoms of infinitely narrow diameter, because you'd get by the uncertainty principle total indetermination in the atoms' velocities. You won't be able to keep your beam thin.

==> "the dense flux of streaming atoms acts like a strongly absorbing sheet . . . if we do nothing, then the light beams effectively "pass 'over' or 'through' the wall of absorption" "

No!!! First of all the atom-light interaction has a couple of channels: absorption, ellastic scattering, inellastic scattering. You can't have ONLY absorption. Second, both the absorption and the scattering damage the interference pattern that you think that is preserved. The absorption of the beam ψ- "eats" this beam, doesn't let it continue upwards and meet the beam ψ+. The absorption of the beam ψ+ "eats" this beam, doesn't let it go downwards and meet the beam ψ-. Thus, the atoms extinguish part of the interference tableau by eating the photons. The photons not eaten, but ellastically scattered, would fly in all the directions. As to inellastic scattering, the photon energy is partly absorbed (for raising the electrons in the atom onhigher levels), so the photon continues its flight with longer wavelength. So the pattern of interference is destroyed.

Dear Sofia,

"I don't know what you mean by interpretation, we are not in QM here."

And you really believe there are no interpretations in classical physics? Very strange idea. Are you not aware that physics arose from natural philosophy and the question of how much we can trust our senses as regards reality (a question that can then be extended to experimental apparatus) was predominant in philosophy for some time? (Descartes, Hume, Kant)

Anyway, it was not me who changed the data of the experiment suggested by Demetrios. It was you. You were the one who said that an ideal conductor is always reflecting. I showed that this is not true. It depends on how you take the limit in going to "ideal". If you first take the limit of zero thickness and then take conductivity to infinity, your conductor will not reflect at all. If you first take conductivity to infinity and then thickness to zero, your conductor will reflect perfectly. And you can have any finite reflectivity different from zero by taking the limits of zero thickness and infinite conductivity in the appropriate way together. This follows from standard electrodynamics, which you, however, did not know well enough to realize that an ideal conductor has both possibilities of being opaque and transparent in the limit of zero thickness. In the setup, Demetrios only talked of a seemingly impassible barrier. I discussed that it need not be impassible in reality and that the question which of the two interpretations that both describe the same reality is more plausible (i.e. accounts for a larger set of experiments) depends on the level of impassibility.

Anyway, the first problem is solved for me. About the second, I have yet to think.

Klaus,

I said again and again that what I discuss is the question of Demetrios whether dispite the object placed in the darkmost part of the fringe the waves freely propagate.

==> "standard electrodynamics, which you, however, did not know well enough"

Did I ever talk with you in this style? Did you see me talking with somebody in this style?

Well, let's say so, you are right in whatever you say, I am wrong because I don't know enough. Are you pleased? Is this what you want? If it is, then, from my part you can HAVE IT.

Dear all,

A perfect conductor reflects back EM waves, full stop. There is noting to invent or interpret. We are not reinventing Physics here.

@Quattrini

No. As I have discussed, a "perfect conductor" is not defined well enough to say whether it will reflect, if it has zero thickness in addition. It depends on the order of limits to perfect conductivity and zero thickness, starting from a conductor with finite conductivity and thickness. If you first take the limit thickness to zero, your conductor will be transparent, because it always has a finite penetration length. If you then take the limit of conductivity to infinity, that will not change anything. 100% transparency will remain at all thicknesses. If you first take the limit of conductivity to infinity, then the transparency will be zero at all thicknesses, and so it will be zero at thickness zero as well.

Anyway. Since we can make perfect conductors, but not objects of zero thickness, it is certainly sound physical intuition to assume that a perfect conductor is reflecting at 100%. The discussion could have remained brief, if instead of perfect conductor the notion of perfect reflector would have been used.

And of course, it remains true that a perfect conductor introduced in a stationary zero-field region will not change the interference pattern. It does not change the boundary conditions of the electromagnetic field, hence, it will not affect it.

As to the second problem proposed by Demetrios, an absorbing sheet, that is more interesting.

First, I do not think that the uncertainty relation enters the problem in any severe way. Atoms are far smaller than a nanometer, so keeping them in a sheet the thickness of a nanometer will not be beyond experimental possibilities. Scientists can make crystal surfaces that are smooth on the subnanometer scale and they can produce silicon spheres (for the redefinition of the kg) that are smooth on the 10 nm scale, having a diameter of 10 cm. So a sheet of atoms, say, 2 nm thick and collimated across a distance of a mm, should be feasible.

Second, light has a wavelength of a few hundred nanometers. Let us assume 500 nm for the purpose of consideration and take the angle of the two plane waves at 45°. Then the distance between the fringes of zero field strength will be 250 nm x sqrt(2), i.e. roughly 350 nm. So the sheet when put into a nodal plane of the interference pattern will occupy 2/350 of a half-wavelength of the standing-wave pattern. In that range, the sine is essentially linear, so the field strength will be about pi/2/175 of the field strength in the fringes of maximal brightness, let us say 1%. That means the intensity of the field felt by the atoms is about 1/10000 in the dark fringes in comparison with the bright fringes. So under normal circumstances (finite absorption probabilitly, impossibility of an atom to absorb a second quantum as long as it has not reemitted the first one, etc.) the absorbing sheet of atoms will be a weak perturbation in the dark fringes and may be a strong one in the bright fringes. So we still may expect the sheet to have little effect, if put into the region of weak field. It will essentially reduce the visibility of the interference pattern, while in the strong-field region, it may completely destroy it.

Now, a factor of 10000 may not be sufficient, if we make our atoms super-absorbant. Or if we assume, for simplicity, a perfectly absorbing sheet of material put into the field-free region. Assuming that sheet to have zero thickness would be stretching credibility. Contrary to reflection, absorption cannot be described by a simple boundary condition on a surface. To model absorption in electrodynamic problems, one usually assumes a whole layer, the thickness of which is adapted so that the reflectivity becomes zero (there are nice problems for students of the calculation of the thickness of an anti-reflection coating considering three dielectric layers). With a single surface, zero reflectivity means 100% transparency, so to have absorption, you need a bit of bulk behind the surface, with dissipative properties. Of course, one may imagine that this bulk can be made arbitrarily thin, but that should not mean zero here but rather infinitesimally positive :-). Anyway, the effective boundary conditions of such an absorber, introduced in the nodal plane will not be Dirichlet nor Neumann. Maybe Robin boundary conditions can approximate the situation. At exactly zero field, these may be satisfied by the original interference pattern, but since I have argued that the thickness should be slightly positive, they will not be satisfied slightly off the nodal plane. Then the field configuration will be changed by the absorber. Intuition suggests that the change will be so that we have two propagating plane waves on both sides of the absorber that enter the sink constituted by the latter, without reflections and without appearance on the other side. So the interference pattern will go away and after a transient, we will have nothing but two plane waves hitting the absorber and being consumed by it.

If the absorber is put into a bright fringe, the final result will be the same, but the transient may be much shorter.

Just my two cents.

@Kassner

"A perfect conductor or perfect electric conductor (PEC) is an idealized material exhibiting infinite electrical conductivity or, equivalently, zero resistivity (cf. perfect dielectric). "

it is perfectly defined and it reflects back anything. A non ideal or non perfect conductor is influenced by its thickness . 

K. Kassner says

==> "Atoms are far smaller than a nanometer, so keeping them in a sheet the thickness of a nanometer will not be beyond experimental possibilities."

Who says that a nanometer is enough thin? Demetrios didn't say what is the wavelength of the beams.

However, what Demetrios wanted was the effect that

==> "the dense flux of streaming atoms acts like a strongly absorbing sheet"

Before someone says that a sheet of a nanometer thickness can be a strong absorber, "a perfectly absorbing sheet", I'd advise to ask Demetrios what is the wavelength, and then look in cross section tables. Absorption is only one channel in scattering. There also exists "scattering cross section". 

Stefano explains

==> "A perfect conductor . . . is perfectly defined and it reflects back anything. A non ideal or non perfect conductor is influenced by its thickness . "

Moreover, the resistence of a piece of metal is proportional to its thickness. Thus, an infinitely thin sheet of metal is a bad candidate for highly conducting.

But, there is another issue here, and it is systematically avoided in this talk. Demetrios' question wasn't if a shhet of metal can become really transparent. He asked if the beams flow through the barrier, given that outside the metal, the pattern looks as if the metal is not there. The answer is NO, and I think that Stefano's words are clear enough. In the opposite case, if the material placed on the way of the beams is transparent, then there is no barrier there, and no question.

I want to sincerely thank Klaus for taking the time to provide us with truly expert commentary on the proposed scenarios.......I think we have all learned from his descriptions of both the conductor and the absorption cases. Klaus focused on the 'essentials' of the scenarios and, in my opinion, properly put aside the imperfections and non-ideal character of the elements involved, in any realistic setting, when those imperfections and non-ideal characteristics did not interfere with the 'in principle' generation of the effect that I proposed.......I thought we all tacitly assumed that the effect of 'cutting the light beams' , if at all possible, would only be observed in an incomplete manner, as in any realistic description of an idealized 'proof of principle' proposal.......but, in any case, I agree with the conclusions reached by Klaus regarding the conductor and the atomic absorption. Of course, I also thank Sofia and Stefano for their input, even though I disagree.

As Weschler and Kassner say, with a perfectly reflecting sheet and two identical beams at the same frequency, the reflected beam cannot be distinguished from the transmitted beam that was there before, so there is no observable difference in the pattern with and without the conducting sheet.  No power crosses the conducting plate, but the pattern does not change.  If the beams are at different frequency then the nulls will move very fast.  I'm not sure if a moving plate could be used.  I think it could, but I also suspect the null lines would be tilted.  Perhaps there is a moving reference frame in which both beams have the same frequency - in this frame the conducting plate would be stationary!

In practice, the situation of two intersecting beams is what exists in rectangular waveguide, in which the pattern of energy is the same (inside the waveguide) as a slice cut out of an interference pattern between two crossing plane waves by two conducting plates running along two adjacent electric field nulls (or not adjacent for higher order modes).  The copper, brass or aluminium used is not a perfect conductor but the skin depth (1 micron at microwave frequencies) is small enough so that the transmission through 1/2 mm is millions of times lower than the incident field.  The reflection is not quite 100% but is very close.  The use of rectangular waveguide (hollow rectangular pipes) is commonplace in microwave engineering.