I Energy of Electromagnetic Waves in Destructive Interference

[keep_it_simple_silly]

The problem with this is that the Power in the wave depends on the impedance as well as the E field. P0 is only P/2 when the sources are far enough apart to ignore interactions. This expression:
View attachment 315986
doesn't include the impedances involved - there's more to it than just the spacing. Both E and H fields count and also the source impedance, self Impedances of the antennae and also the mutual impedance. So that expression 'just' a system that may produce the right answers under some circumstances.

I'd be very suspicious of any work that suggests non-conservation of Energy. In the case of radiating systems, calculations are always of limited accuracy and they always require numerical solutions so I'd suggest that there will be a flaw that remains to be found.

1. There is absolutely no violation of the conservation of energy. The only difference is the consideration of work done which is usually ignored.
2. The expression doesn't contain the terms of wave impedances, but the impedance is already taken care of in the equation.
For example, a source will experience the highest impedance when it experiences a wave of the same amplitude, frequency and phase, which also means when the 2 sources are put at the same point. Now the power radiated in this case is 4P_0 by both sources, so this already includes the additional power dissipation by the source to overcome the resistance. You can see the same with destructive interference and other interferences in between.
3. This formula hasn't considered parameters like heating effects or others.

 

[Charles Link]

I'm very glad to see a somewhat thorough discussion of the energy/power interplay in the case of the two radiating sources as presented by @sophiecentaur and @keep_it_simple_silly . Meanwhile please see the "Edits" that I added to post 50 above.

 

[sophiecentaur]

3. This formula hasn't considered parameters like heating effects or others.

So your sources are approximations, which is fine, and they do the job that's required, apparently but you don't need to suggest departure from that basic conservation of energy principle to explain why the energy is apparently disappearing.

I always try to reach for the simplest wave systems available and RF antenna fill the bill well. They are pretty good omni radiators (in one plane) and they radiate coherent signals. Also, the Impedances can be calculated with some accuracy (a lot better than narrow slits, for instance). When some Power appears to be missing then the realities of non perfect transmitter amplifiers is somewhere to look. The same thing will apply to acoustic experiments with multiple speakers.

 

[Charles Link]

This thread is a couple months old, but I thought of another case that the OP and others might be interested in that involves interference and diffraction, and it also illustrates the conservation of energy that occurs in the formation of constructive and destructive interference: When a collimated beam (parallel rays) is focused to a small spot by a paraboidal mirror=[not a point as geometric (ray trace) optics would yield, but rather a diffraction limited spot], the energy is conserved as shown by calculations using diffraction theory for computing the intensity pattern in the focal plane. The (effective) area of the focused spot is finite, and is inversely proportional to the cross-sectional area of the beam/focussing optic. Meanwhile the intensity of the focused spot is proportional to the second power of the cross-sectional area of the beam/focusing optic, thereby conserving energy, as more detailed calculations show very precisely.
See https://www.physicsforums.com/insights/diffraction-limited-spot-size-perfect-focusing/

Edit: It should be noted that constructive interference occurs at and near the focal point of the paraboidal mirror because the optical path distance of every ray to the focal point is identical, (with each ray reflecting off the mirror with angle of incidence =angle of reflection).

 

[Charles Link]

@MartinG , @sophiecentaur, @keep_it_simple_silly You might find the above post of interest. The energy conservation that occurs when light is brought to a focus by a lossless focusing optic (diffraction theory of optical imaging) is another example of the conservation of energy that occurs in optical interference as was also illustrated by post 44 above.

 

[sophiecentaur]

and the spacial arrangement. How is information about the structure of the interference pattern transmitted across the dark gaps, when the wave is no longer electro or magnetic?

The 'information' that governs the interference pattern is contained in the waves themselves (wavelengths and amplitude in every direction) and the spatial layout of the experiment such as the positions of the sources and their size / widths. No other information is needed to calculate the value of the fields across the pattern; the pattern doesn't 'communicate amongst itself.

 

[Quarker]

The 'information' that governs the interference pattern is contained in the waves themselves (wavelengths and amplitude in every direction) and the spatial layout of the experiment such as the positions of the sources and their size / widths. No other information is needed to calculate the value of the fields across the pattern; the pattern doesn't 'communicate amongst itself.

An Airy Disc’s rings get brighter if the point source of light is increased. How is that extra energy transferred to the outer rings?

 

[Dale]

An Airy Disc’s rings get brighter if the point source of light is increased. How is that extra energy transferred to the outer rings?

The same way the extra energy is transferred to the inner rings. The energy is transferred from the source to both sets of rings, not from one set of rings to the other.

 

[Ibix]

The same way the extra energy is transferred to the inner rings. The energy is transferred from the source to both sets of rings, not from one set of rings to the other.

Just to add to the above, note that the eye's response to illumination is not linear so your direct perception may be that the relative brightness changes. The same may be true with an electronic sensor, depending on how much you paid for it and how well you used it.

 

[sophiecentaur]

How is that extra energy transferred to the outer rings?

Energy is not 'transferred'. The Airy Disc is just a diffraction pattern which is formed by contributions from all parts of what would be the image. We start by considering simple 'interference' between ideal point sources but the image of a star (for example) is due to contributions from all over the objective lens. The resulting Airy pattern is given by an Integral of all contributions. The amplitudes in all directions all scale according to the brightness of the source (star). The energy is spread over an infinite range (in theory) but in practice it's confined to the usual disc image.

The same may be true with an electronic sensor, depending on how much you paid for it and how well you used it.

And how long you are prepared to wait for an exposure. The outer rings can in fact coincide with another very faint star and that can determine the resolution between to stars.

 

[Quarker]

Energy is not 'transferred'. The Airy Disc is just a diffraction pattern which is formed by contributions from all parts of what would be the image. We start by considering simple 'interference' between ideal point sources but the image of a star (for example) is due to contributions from all over the objective lens. The resulting Airy pattern is given by an Integral of all contributions. The amplitudes in all directions all scale according to the brightness of the source (star). The energy is spread over an infinite range (in theory) but in practice it's confined to the usual disc image.

And how long you are prepared to wait for an exposure. The outer rings can in fact coincide with another very faint star and that can determine the resolution between to stars.

Assuming an Airy disc with one ring, does the dark ring around the central disc have the same area as the disc and bright ring? Do the bright and dark areas balance out exactly? Is that how energy is conserved?

 

[Charles Link]

Assuming an Airy disc with one ring, does the dark ring around the central disc have the same area as the disc and bright ring? Do the bright and dark areas balance out exactly? Is that how energy is conserved?

The energy (total power) incident onto the aperture is the same as the energy (total power) in the diffraction pattern.

 

[Dale]

Is that how energy is conserved?

No. The preceding two questions have nothing to do with the conservation of energy.

 

[sophiecentaur]

Do the bright and dark areas balance out exactly? Is that how energy is conserved?

What would they 'balance out' to except the total energy? 'Balancing out' implies cancelling out to a zero; that's not what happens. Conservation of energy is a 'law' which you can use to check on your results when you calculate or measure the diffraction pattern. You would expect the Integral of the energy all over the (two dimensional) pattern to equal the total energy entering the aperture / lens. Put a large area thermal detector to measure the energy and it would (should) give the same answer whatever distance the detector is placed and whatever area is covered by the visible pattern (a small dot or a wide disc + rings).

 

[tech99]

It is interesting that where we have an interference pattern, such as from a knife edge or large aperture, individual bright zones can have greater intensity than the average. But due to the smaller area of the bright zone, energy is conserved.

 

[sophiecentaur]

But due to the smaller area of the bright zone, energy is conserved.

I'd put it the other way round, rather. Conservation of energy forces the bright zones to be narrower than you'd maybe expect. There's nowhere else for energy in the pattern to come from or to go.

 

[Quarker]

It is interesting that where we have an interference pattern, such as from a knife edge or large aperture, individual bright zones can have greater intensity than the average. But due to the smaller area of the bright zone, energy is conserved.

Could the central disc of an Airy disc be “brighter” than the surrounding ring? Or could the central disc itself have a higher intensity in the very center of the disc?

 

[Dale]

Could the central disc of an Airy disc be “brighter” than the surrounding ring? Or could the central disc itself have a higher intensity in the very center of the disc?

Certainly. But this has nothing whatsoever to do with conservation of energy. You could also arrange lenses so that the center had a low or zero brightness. Again, with nothing to do with conservation of energy.

The energy flows from the light source to the disks and from the source to the rings, not from the disk to the rings.

 

[Charles Link]

The airy disc is very similar to a single slit diffraction pattern in two dimensions with a circular aperture being very similar to the square that results from the two-dimensional single slit. See also: https://www.physicsforums.com/threa...erfect-focusing-comments.907511/#post-5749159

The peak of the pattern appears in the center.

There is Poisson's bright spot that can get generated when the center of the aperture is blocked, but there is no Poisson's dark spot for an open aperture.

Edit: I need to study this further because, ignoring any obliquity factor, the phasor diagram for a circular aperture is a circle, (that returns to the starting point=zero intensity), if I recall correctly from previous calculations. That would indicate you can get something of a dark spot in the center of the pattern from an open aperture if you have just the right number (=even number) of Fresnel zones.

and I think I just answered this question: For the far field pattern which appears in the focal plane of the lens, the entire aperture all has zero phase angle at the center of the diffraction pattern. The result is you always get the bright spot in the center if you sample in the focal plane of the lens.

 

[sophiecentaur]

But this has nothing whatsoever to do with conservation of energy.

I wouldn't say that; it's more a matter of cause and effect. COE always applies but the actual patterns are more. obviously due to vector additions of a multiplicity of fields. Energy, being a scalar, is difficult (inconvenient) to use as a way of predicting the energy distribution in a diffraction pattern.

Both descriptions of the process will apply and measurement of the accuracy of a pattern can be verified / checked by integrating what you get over a sphere (or whatever). With radio antennae, power can be 'lost' in directions that the designer ignored. Summing the measured powers around the direction of interest could give, say half the expected main beam power and COE would instantly lead you to smell a rat.

 

[Dale]

Could the central disc of an Airy disc be “brighter” than the surrounding ring?

But this has nothing whatsoever to do with conservation of energy.

I wouldn't say that

The reason I said it is that energy is conserved if you arrange for a bright center or if you arrange for a dark center. Energy flows from the source to the disk and from the source to the rings. It doesn’t flow from the rings to the disk or vice versa. So it doesn’t matter which is darker or brighter

 

[sophiecentaur]

energy is conserved if you arrange for a bright center or if you arrange for a dark center.

Yes. Whatever diffraction pattern you manage to produce with your source (and you can design a source to produce any pattern you want - even a hologram), the total energy (integral) over the whole pattern (the hemisphere) will be the energy from the source.