I Energy of Electromagnetic Waves in Destructive Interference

[MartinG]

Hello !

As we know by definition that:

"Constructive interference occurs when the phase difference between the waves is an even multiple of π (180°), whereas destructive interference occurs when the difference is an odd multiple of π."

But my question is in the case of destructive interference, what happens to the energy carried by the two electromagnetic waves that annihilate, the energy carried by the electromagnetic waves also disappears, or is transformed into some other type of energy.

Because according to the law of conservation of energy, energy can neither be created nor destroyed, it can only be transformed into other types of energy. This is why I ask you what happens to the energies of photons or electromagnetic waves in a destructive interference.

I thank you for the response and I send you my regards.

 

[Charles Link]

In ideal interference problems, the energy is always conserved, so that it doesn't disappear, but winds up at another location, where constructive interference occurs.

The conservation of energy in these interference problems is something that is really fundamental. You might find it of interest to read an Insights article which I authored a couple of years ago. The textbooks seem to come up short on this topic, and it took me a very long time (=years) to finally figure out, that more importantly than multiple reflections, the Fabry-Perot effect involves the interference that results from two beams incident on an interface from opposite directions. The result surprised me when I first figured it out. Here is the "link": https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/

 

[berkeman]

But my question is in the case of destructive interference, what happens to the energy carried by the two electromagnetic waves that annihilate, the energy carried by the electromagnetic waves also disappears, or is transformed into some other type of energy.

That's a good question, and it's good that you are asking it.

As a first step, have you learned about waves in general, and their propagation and interference? The first time that I became comfortable with destructive wave interference was when I studied it for waves on strings. The energy in the tension components helped me to become more comfortable.

 

[Charles Link]

It is also the case with two mutually coherent radiating point sources with spherically symmetric patterns from each, that the total radiated power is the sum of the powers of the two sources. There will be regions where destructive interference occurs, but this is exactly offset by regions of constructive interference.

 

[Charles Link]

One additional comment on my post 2: @vanhees71 had mentioned to me a while back that my approach to the interference that takes place with two beams incident on a dielectric interface is not completely new, and that apparently J. Schwinger introduced this same approach in the solution of the beamsplitter problem.

The beamsplitter problem, e.g. in the Michelson interferometer, is a case where you can have complete destructive interference (=no signal) at one of the receivers, (depending on the relative phases of the two beams incident on the beamsplitter), and completely constructive interference at the other receiver, so that energy is conserved, and all of the energy emerges at the site of the constructive interference.

 

[Quarker]

An even more interesting question is what happens to the pattern of the EM wave at the point of destructive interference. How is information about the structure of the interference pattern transmitted across the dark gaps, when the wave is no longer electro or magnetic?

 

[tech99]

An even more interesting question is what happens to the pattern of the EM wave at the point of destructive interference. How is information about the structure of the interference pattern transmitted across the dark gaps, when the wave is no longer electro or magnetic?

When two waves meet they travel across each other without interaction. One wave simply rides up on top of the other. Then they each emerge unaltered. When we say cancellation, we are considering that an antenna placed at a certain position receives opposing fields and so couples zero energy from its environment.

 

[Charles Link]

I think @tech99 in post 7 tries to oversimplify the interference process and thereby misses some key features. Over the years I've had a couple discussions with a couple intelligent and older physics Ph.D.'s who were puzzled how such interference phenomena could occur when Maxwell's equations are completely linear. It took me a while to resolve this puzzle, but the reason is that the energy equations are second order in electric field amplitude, so that although Maxwell's equations are linear in the electric field amplitude, to obtain the complete picture it is necessary to include the energy equation. It is with this energy part, e.g. the intensity $I \propto E^2$, that the interference appears. With this second order dependence of the energy, linear principles no longer need to apply. For example, the energy pattern of two antennas is not the sum of patterns of the individual antennas.

 

[Delta2]

I am puzzled here, someone please help:
We consider two waves of same frequency and phase and same amplitude $A$. Then the amplitude of the resultant wave is $2A$, hence the energy it carries proportional to $4A^2$.
However the energy of each wave is proportional to $A^2$ hence the total energy of two waves, proportional to $A^2+A^2=2A^2\neq 4A^2$

What's going on here?

 

[Charles Link]

I am puzzled here, someone please help:
We consider two waves of same frequency and phase and same amplitude $A$. Then the amplitude of the resultant wave is $2A$, hence the energy it carries proportional to $4A^2$.
However the energy of each wave is proportional to $A^2$ hence the total energy of two waves, proportional to $A^2+A^2=2A^2\neq 4A^2$

What's going on here?

If you consider two or more photons in the same mode in a cavity, the photons need to have phases that are not all the same, in order to have energy conservation. One possibility would be that the phases differ by 90 degrees. Random phases would also work for a large population. I think that might be what has you puzzled here.
One other case of interest is two identical beams getting combined by a beamsplitter. For a dielectric beamsplitter, the Fresnel coefficients when the energy reflection coefficient $R=1/2$ are $\rho= \pm 1/\sqrt{2}$ and $\tau= 1/\sqrt{2}$. For a symmetric type beamsplitter (e.g. a silvered layer in a glass block), there necessarily must be a 90 degree phase shift between the transmitted and reflected waves. Otherwise, if you have two beams getting combined by the beamsplitter that are $\pi$ out of phase, the energy would disappear, and would lead to all kinds of contradictions.

I don't want to get away from the topic of the thread, but it could be this last beamsplitter case that had puzzled the OP. You can not have a case where you have destructive interference, unless there is also some form of constructive interference.

 

[vanhees71]

Don't speak about photons, if the classical theory is not yet understood!

The linearity of Maxwell's equations tells you that if there are two sources $(\rho_1,\vec{j}_1)$ and $\rho_2,\vec{j}_2)$, the solution with both sources present together are given by
$$\vec{E}=\vec{E}_1+\vec{E}_2, \quad \vec{B}=\vec{B}_1+\vec{B}_2.$$
Each solution is given by the retarded potentials (or equivalently directly the fields by Jefimenko's solutions), and the field of the total source is the sum of the solutions of the individual sources. That's because Maxwell's equations are linear partial differential equations with the sources as inhomogeneities.

The energy density is given by (in Heaviside-Lorentz units which avoids the awful $\epsilon_0$ and $\mu_0$ factors I always mess up ;-)).
$$u=\frac{1}{2} (\vec{E}^2+ \vec{B}^2).$$
Now we have
$$u=\frac{1}{2} (\vec{E}_1+\vec{E}_2)^2 + \frac{1}{2}(\vec{B}_1+\vec{B}_2)^2 = u_1 + u_2 + (\vec{E}_1 \cdot \vec{E}_2 + \vec{B}_1 \cdot \vec{B}_2).$$
So the total energy density is NOT the sum of the energies of the fields from the individual sources but there are the extra "mixing terms", the so called interference terms, and this is crucial for all wave theories.

 

[tech99]

I agree that the power flux density where waves constructively interfere involves the square of the fields. If we consider a 1 metre cube in space, with two identical waves entering from opposite sides, then the aggregate power flowing in the cube will be zero, because there is as much power entering as leaving. On the other hand, the energy within the cube is the sum of that in the two waves. So in effect the cube is an energy store. The peak of the fields will be twice those of a single wave and the power flux density, or intensity, will be four times.
A similar situation occurs with reflection from a metal, in a cavity resonator, in a waveguide, and in a short circuited transmission line. The peaks of the electric and magnetic fields are in quadrature and occur in different positions, being a quarter of a wavelength apart.
In these structures it is convenient to consider both energy which is traveling and energy which is stored. If the waves are unequal in amplitude we then see a traveling wave and the transport of energy, or in other words a power flow. At the same time we see stored energy, which has no aggregate power flow and which has no progressive phase shift along the cavity. For a cavity or transmission line the ratio of the two energies gives us the resonant Q of the system.

 

[DaveE]

However the energy of each wave is proportional to A2 hence the total energy of two waves, proportional to A2+A2=2A2≠4A2

No, you must add the wave amplitudes first, before you find the intensity. So in your example, you would have $(A+A)^2$, not $A^2 + A^2$. This is key to understanding interference.

 

[sandy marcus]

SandyM
If you restrict your analysis to the classical domain energy is not conserved. Thus you could use this superposition effect as an energy multiplier. If you take the view that EM is an extraction from the background vacuum energy, this would result in energy conservation.

 

[Charles Link]

@vanhees71 in post 11 does a very good job of showing the non-linear (second order in the electric field amplitude) nature of the energy. I do believe under most normal conditions the interference term will average to zero when integrated over all space, so that energy will be conserved. I don't agree with post 14.

 

[Delta2]

No, you must add the wave amplitudes first, before you find the intensity. So in your example, you would have $(A+A)^2$, not $A^2 + A^2$. This is key to understanding interference.

Sorry I don't agree, the point is that the sum of the energies of the constituent waves, does not equal the energy of the resultant wave.

 

[Delta2]

@vanhees71 in post 11 does a very good job of showing the non-linear (second order in the electric field amplitude) nature of the energy. I do believe under most normal conditions the interference term will average to zero when integrated over all space, so that energy will be conserved. I don't agree with post 14.

Well, it doesn't average to zero if you take $\vec{E_1}=\vec{E_2},\vec{B_1}=\vec{B_2}$ which seems a perfect "normal" condition to me.

 

[hutchphd]

Sorry I don't agree, the point is that the sum of the energies of the constituent waves, does not equal the energy of the resultant wave.

What point? You are saying things that are either very incorrect or perhaps I misunderstand. Can you elaborate?

 

[Delta2]

What point? You are saying things that are either very incorrect or perhaps I misunderstand. Can you elaborate?

I say in the sentence you quoted what is the point. If you find this very incorrect, I don't know what to say, seems perfectly correct to me...

 

[hutchphd]

Is energy conserved according to your calculations?

 

[Delta2]

No.

 

[Charles Link]

Well, it doesn't average to zero if you take $\vec{E_1}=\vec{E_2},\vec{B_1}=\vec{B_2}$ which seems a perfect "normal" condition to me.

Good point. If the sources are physically separated, then the energy is conserved. When one source is overlaid on the other, it gives what may seem to be an inconsistency, but because the energy is second order in the electric field amplitude, it does not need to obey linear principles. We see this with the Fabry-Perot effect also, that I mentioned in post 2 and the "links" there.

 

[Delta2]

If the sources are physically separated,

Yes sorry didn't think of that case.

 

[Charles Link]

Yes sorry didn't think of that case.

I think you might find the "link"in post 2, an Insights that I authored on the Fabry-Perot effect, of some interest. One thing it explains is how you can have a beamsplitter with a single incident beam, and the energy reflection coefficient $R=1/2$ works just fine. When a second beam is introduced, this $R$ is no longer a good number=linear principles simply don't work for the energy when there is a second beam from the other direction, even though the Fresnel coefficients for the E-field remain good numbers.

 

[DaveE]

Good point. If the sources are physically separated, then the energy is conserved. When one source is overlaid on the other, it gives what may seem to be an inconsistency, but because the energy is second order in the electric field amplitude, it does not need to obey linear principles. We see this with the Fabry-Perot effect also, that I mentioned in post 2 and the "links" there.

But he's stipulated that the frequency and phase are equal.

We consider two waves of same frequency and phase and same amplitude A. Then the amplitude of the resultant wave is 2A, hence the energy it carries proportional to 4A2.

If you add that they come from the same place, then you really don't have two separate waves, do you? Nor would you have interference. This leaves the trivial result that the intensity of a wave is the square of its amplitude.

 

[Delta2]

If you add that they come from the same place, then you really don't have two separate waves, do you? Nor would you have interference.

Yes ok I was thinking of the waves without thinking of their sources, as we do quite often in physics. But yes when sources come into play it turns that we can't have interference in this case.

 

[vanhees71]

SandyM
If you restrict your analysis to the classical domain energy is not conserved. Thus you could use this superposition effect as an energy multiplier. If you take the view that EM is an extraction from the background vacuum energy, this would result in energy conservation.

In classical electrodynamics energy is strictly conserved. It's a Poincare-invariant theory, and time-translation invariance implies the strict conservation of energy. There is no way to extract energy from some ominous "background vacuum energy", whatever you think that might be.

 

[vanhees71]

@vanhees71 in post 11 does a very good job of showing the non-linear (second order in the electric field amplitude) nature of the energy. I do believe under most normal conditions the interference term will average to zero when integrated over all space, so that energy will be conserved. I don't agree with post 14.

For the usual observations with light you average rather over time, leading to the observered "intensity". It then depends on the "coherence" of the sources and thus the coherence between the partial fields resulting from them. If you have "natural" (thermal) light, usually this averaging cancels the interference terms, because the phases of the partial waves are fluctuating randomly.

 

[Delta2]

There is no way to extract energy from some ominous "background vacuum energy"

That is according to mainstream physics.

I sense that the view of mainstream physics has something wrong regarding this topic, but since I can't argue in more detail (and I don't want to get banned if I start talking about non mainstream physics) I ll accept it for the time being.

 

[vanhees71]

This is out of the range of topics to be discussed in this forum then. There is not the slightest evidence for something you are suggesting, i.e., "mainstream physics" works extremely well for everything concerning electromagnetism (in its quantum-theoretical formulation, quantum electrodynamics, it's even among the best confirmed theories ever discovered).

 

[Orodruin]

That is according to mainstream physics.

I sense that the view of mainstream physics has something wrong regarding this topic, but since I can't argue in more detail (and I don't want to get banned if I start talking about non mainstream physics) I ll accept it for the time being.

There is nothing wrong with mainstream physics. What fails here is your assumption that energy quadrupling with the field doubling is somehow giving you energy non-conservation (it doesn’t).

The fields are linear in the source and the energy is quadratic in the fields. This means that ultimately energy is quadratic in the source as well. If you double the source, then you also quadruple the energy input into the fields. If, for example, you look up the power radiated by an oscillating dipole, you will find that it is quadratic in the magnitude of the dipole moment.

 

[Charles Link]

I want to follow-up post 24, especially for @Delta2 : It is very interesting to me that you can have mutually coherent beams incident onto a beamsplitter, e.g. with a Michelson interferometer, where when you have a single beam, you get a 50-50 energy split, but with two beams present, the way the energy redistributes itself depends on the relative phases of the beams, with complete energy conservation. You can have complete destructive interference on one arm, and the other arm (to the receiver) gets completely constructive interference. The Fresnel coefficients still apply for the electric field, and that is how the interference is computed. The $R=1/2$ which was computed for a single beam can no longer be used with both beams, (any redistribution of the energy is possible), even though the Fresnel reflection coefficient $\rho=\pm 1/\sqrt{2}$, which is essentially computed from the $R$ still applies to compute the result, (along with the composite $\tau=1/\sqrt{2}$).

This is a very classic case, where we can have complete destructive interference on one arm, as the OP asked about, but it is counteracted by constructive interference on the other arm.

 

[sophiecentaur]

Good point. If the sources are physically separated, then the energy is conserved. When one source is overlaid on the other, it gives what may seem to be an inconsistency,

That "inconsistency" is due to the fact that the two 'sources' will have to interact because they are in the same place. The apparently lost energy will actually be flowing from one of the sources into the other source. Imagine two radio transmitters connected to the same antenna. If they are both fed from the same oscillator (identical frequencies) and their phases adjusted then, for zero power to be radiated, then their two outputs must be an odd number of half cycles out of phase. They will be 'fighting' each other and dissipating the total transmitter powers internally somewhere (i.e. conserving it).
That is only when they are actually in the 'same place'. If they are even slightly apart (two antennae) then there will be some direction in which there is non-zero radiated power. In practice, there may be horrific mis-matches between the two transmitters and their individual antennae if they are 'very close'. But that's very practical and not an ideal situation. You will never get two optical sources that are much less than one wavelength apart.

 

[Charles Link]

When two solutions of Maxwell's equations are overlaid on one another, because of the linearity of the equations, the sum of the solution is also a solution. It is an interesting feature that the energy does not superimpose in the same way, and instead what we often get is interference patterns. The quadratic dependence of the energy density on the electric field amplitude makes this to be the case.

Meanwhile when a single source has its electric field amplitude doubled, the resulting source has 4 times the energy instead of just two times. This is how two sources (in phase with each other) can be overlaid or superimposed, (you double the electric field amplitude), but it can't be done using two physical sources that have fixed energy.

 

[hutchphd]

The most recent example is the alignment of the mirrors on the James Webb Space Telescope. In particular the final distance ("piston alignment") for each mirror segment had to be equal to within something like a tenth of a wavelength get the nice coherent center spot. They did okay I guess!

 

[sophiecentaur]

The most recent example is the alignment of the mirrors on the James Webb Space Telescope. In particular the final distance ("piston alignment") for each mirror segment had to be equal to within something like a tenth of a wavelength get the nice coherent center spot. They did okay I guess!

In this example - as for all optics - the intention is to get diffraction to give you enhancement of all light from one direction and cancellation from all other directions. Your 'main lobe' of directivity is pretty much the same for all reasonable reflectors (optical or radio) or lenses but the higher quality design will provide better and better cancellation of signals arriving off-axis (shaping and spreading 'side lobes' as much as possible to acceptable levels). It's a bit like an air mattress; you squash one part down and another part springs up but the volume of air inside stays the same.

 

[keep_it_simple_silly]

Hello Martin!

Many of the answers say that the conservation of energy is maintained because energy is 'redistributed' or goes to other location of constructive interference. This is an explanation which only roughly fits in certain cases but fails in some other.

The real explanation is not as straightforward as it seems. Although the energy of the whole system (source + receiver medium) is conserved, the energy radiated to the medium is not always the same or conserved!

In the case of YDSE or Young’s Double Slit Experiment, we observe that the average intensity of the fringes to be 2 times the Intensity of a single source. So far so good, then where’s the problem?

Now consider the following situations. Consider 2 sources are radiating waves of same frequency, same amplitude and no phase difference. :-

1. In case 1, the distance between these sources is λ/2, as shown in Figure 1.
   
2. then it is reduced to 0 in Figure 2.
   
   
3. If we calculate the power dissipated in these 2 cases, it turns out to be equal to 2*P in case of Figure 1, and 4*P in case of Figure 2. This clearly shows that by changing the distance, the net energy dissipated in medium is changing, not conserved.
   
   We can qualitatively see this in Figure 1 and Figure 2. In Figure 1, The waves are absent in the sides, and in Figure 2, waves are present everywhere. This gives us the clear idea that more energy is being spent in case of Figure 2 The question is how?
   
   • In order to give an explanation to this phenomenon, which fits all the cases and doesn't fail at any case, it is necessary to consider the ‘work done' by the source or the ‘wave impedance’. For example, do you think that a source generating a wave of amplitude A in a vacuum and a source generating the same wave in the environment when there is a varying electromagnetic field present at the source requires same amount of energy?
   • This situation can be compared to pushing a rock on a plane surface vs pushing it on a hill. To get the same displacement, more energy needs to be transferred in case of pushing rock on the hill. This is because there is an ‘opposing force’ present in the latter case.
   • In a similar way, when a source is generating a wave of amplitude 𝐴A when there is already a wave of amplitude 𝐴A present at the source, it needs to provide more energy to the system. The additional energy it is providing, in this case, is KA^2. So total energy provided by one source becomes 2KA^2, and when we add the energy from both sources, the net energy becomes 4KA^2.
   • The case of 2 sources of opposite phases present at the same point, can be compared to pushing a rock down the hill. Because of the opposite nature of the field present at the source, the total work done by the source will become 0, and that is why there is no net radiation present, radiation being the transfer of energy to the medium by the source.
   • 
   • The concept of work done by the source was explained using term wave impedance by Levine in 1980 Ref, it’s explained in ref written by Authors Robert Drosd, Leonid Minkin, and Alexander S. Shapovalov
     
     [Link to personal web blog redacted by the Mentors]
   • Hope this helps! thank you!

 

[sophiecentaur]

Now consider the following situations. Consider 2 sources are radiating waves of same frequency, same amplitude and no phase difference. :-

That is only possible if the two sources are in exactly the same place. There will always be regions where two real sources are in anti phase. So there's no need to worry that Physics got this wrong.

In that situation, the two sources have to be in phase or power will flow from one source into the other, which can make a lot of smoke at times, and accounts for your apparent energy imbalance

So I think your argument fails on the grounds that you haven't considered all contributions to your energy equation.

 

[hutchphd]

The real explanation is not as straightforward as it seems.

I have no idea what you are saying here but I fear it is largely not relevant. Suppose I have two identical point sources separated by $\vec d$, and I consider the intensity at x where $\vec x \cdot \vec d =0$ i.e. perpendicular to $\vec d$. If the sources are out of phase there will be destructive interference for all wavelengths (and at any value for d). It goes elsewhere.

 

[Charles Link]

Many of the answers say that the conservation of energy is maintained because energy is 'redistributed' or goes to other location of constructive interference. This is an explanation which only roughly fits in certain cases but fails in some other.

I disagree with this statement. I think the "answers" that were given in the previous discussion are very accurate, and the mathematics clearly shows the energy gets redistributed in cases of interference where there is energy conservation.

 

[Charles Link]

It might be worth mentioning that, when two sinusoidal point sources are used, (each with a given amplitude $E_o$, and each radiating spherically), there will be an increase in the radiated power of the system as the distance between them becomes smaller. If the separation is fairly large, then to a very good approximation the total radiated power stays the same regardless of the separation, and the interference pattern will result in a redistribution of the energy.

(If I have analyzed it correctly, there isn't a sudden discontinuity when the sources are superimposed, but rather an increase in power occurs as the separation decreases, and then the power nearly doubles when the separation is very small. [Edit: The relative phases of the sources is also going to play a role. I believe this one might get a little complicated, and I haven't analyzed it in its entirety.])

I agree with @sophiecentaur in post 38 that the energy needs to be considered very carefully. e.g. When the point sources get too close together, the power needs to be re-computed, and you can't simply assume that the power is what it was when the sources were further apart.

@keep_it_simple_silly may also have been trying to make a statement to that effect, but it wasn't totally clear on first reading.

 

[keep_it_simple_silly]

Hello Martin!

Many of the answers say that the conservation of energy is maintained because energy is 'redistributed' or goes to other location of constructive interference. This is an explanation which only roughly fits in certain cases but fails in some other.

The real explanation is not as straightforward as it seems. Although the energy of the whole system (source + receiver medium) is conserved, the energy radiated to the medium is not always the same or conserved!

In the case of YDSE or Young’s Double Slit Experiment, we observe that the average intensity of the fringes to be 2 times the Intensity of a single source. So far so good, then where’s the problem?

Now consider the following situations. Consider 2 sources are radiating waves of same frequency, same amplitude and no phase difference. :-

1. In case 1, the distance between these sources is λ/2, as shown in Figure 1.View attachment 315822
2. then it is reduced to 0 in Figure 2.
   View attachment 315823
3. If we calculate the power dissipated in these 2 cases, it turns out to be equal to 2*P in case of Figure 1, and 4*P in case of Figure 2. This clearly shows that by changing the distance, the net energy dissipated in medium is changing, not conserved.
   
   We can qualitatively see this in Figure 1 and Figure 2. In Figure 1, The waves are absent in the sides, and in Figure 2, waves are present everywhere. This gives us the clear idea that more energy is being spent in case of Figure 2 The question is how?
   • In order to give an explanation to this phenomenon, which fits all the cases and doesn't fail at any case, it is necessary to consider the ‘work done' by the source or the ‘wave impedance’. For example, do you think that a source generating a wave of amplitude A in a vacuum and a source generating the same wave in the environment when there is a varying electromagnetic field present at the source requires same amount of energy?
   • This situation can be compared to pushing a rock on a plane surface vs pushing it on a hill. To get the same displacement, more energy needs to be transferred in case of pushing rock on the hill. This is because there is an ‘opposing force’ present in the latter case.
   • In a similar way, when a source is generating a wave of amplitude 𝐴A when there is already a wave of amplitude 𝐴A present at the source, it needs to provide more energy to the system. The additional energy it is providing, in this case, is KA^2. So total energy provided by one source becomes 2KA^2, and when we add the energy from both sources, the net energy becomes 4KA^2.
   • The case of 2 sources of opposite phases present at the same point, can be compared to pushing a rock down the hill. Because of the opposite nature of the field present at the source, the total work done by the source will become 0, and that is why there is no net radiation present, radiation being the transfer of energy to the medium by the source.
   • View attachment 315824
   • The concept of work done by the source was explained using term wave impedance by Levine in 1980 Ref, it’s explained in ref written by Authors Robert Drosd, Leonid Minkin, and Alexander S. Shapovalov
     
     [Link to personal web blog redacted by the Mentors]
   • Hope this helps! thank you!

Follow the paper "Interference and the Law of Energy Conservation" by
Robert Drosd, Leonid Minkin, Alexander S. Shapovalov to know more

 

[sophiecentaur]

I disagree with this statement. I think the "answers" that were given in the previous discussion are very accurate, and the mathematics clearly shows the energy gets redistributed in cases of interference where there is energy conservation.

If you take a number of radio antennae and attempt to combine them with a given phase to produce a highly directional pattern by feeding them according to 'calculations', when the antennae are close together then the Mutual Impedance between them will, in the end, foil any attempt to make the array too directional. This is due to the effect that I mentioned higher up - they interact mutually and basically waste a lot of energy. You could say that that wasted energy has been directed out of the pattern.

It might be worth mentioning that, when two sinusoidal point sources are used, (each with a given amplitude Eo, and each radiating spherically), there will be an increase in the radiated power of the system as the distance between them becomes smaller.

Isn't that just because the local Impedance is not that of free space so Eo is not sufficient to work out the Energy flow. Actually, that seems to be an alternative to the Mutual Impedance I ,mentioned. Power will be dissipated within the sources. This is not an effect that kicks in at more than a half wave or so.

 

[keep_it_simple_silly]

That is only possible if the two sources are in exactly the same place. There will always be regions where two real sources are in anti phase. So there's no need to worry that Physics got this wrong.

In that situation, the two sources have to be in phase or power will flow from one source into the other, which can make a lot of smoke at times, and accounts for your apparent energy imbalance

So I think your argument fails on the grounds that you haven't considered all contributions to your energy equation.

***That is only possible if the two sources are in exactly the same place. There will always be regions where two real sources are in anti phase. So there's no need to worry that Physics got this wrong.***

- not necessarily. Consider 2 sources in phase with a distance, say λ/8. In this condition, the combined energy dissipation is still greater than the sum of individual power dissipation for isolated sources. The formula which represents all cases is (formula is given by Vanderkooy and Lipshitz in their paper Power response of loudspeakers with noncoincident drivers – The influence of crossover design) :-

Where,

1. P is total power radiated by both sources
2. P0 is the power radiated by single source in absence of other source
3. k=2π/λ, λ is wavelength of source
4. d is the distance between 2 sources
5. θ is the phase difference between 2 sources

If you plot graph for this particular equation, it looks like following for a phase difference of 0:-

So for any distance lesser than λ/2, power dissipation will be greater than 2P_0.
Check out this paper :-
Interference and the Law of Energy Conservation" by
Robert Drosd, Leonid Minkin, Alexander S. Shapovalov to know more
It makes things much clearer.

 

[vanhees71]

The link to this nice paper is (unfortunately not open access though)

https://doi.org/10.1119/1.4895362

 

[keep_it_simple_silly]

If you take a number of radio antennae and attempt to combine them with a given phase to produce a highly directional pattern by feeding them according to 'calculations', when the antennae are close together then the Mutual Impedance between them will, in the end, foil any attempt to make the array too directional. This is due to the effect that I mentioned higher up - they interact mutually and basically waste a lot of energy. You could say that that wasted energy has been directed out of the pattern.

Isn't that just because the local Impedance is not that of free space so Eo is not sufficient to work out the Energy flow. Actually, that seems to be an alternative to the Mutual Impedance I ,mentioned. Power will be dissipated within the sources. This is not an effect that kicks in at more than a half wave or so.

I disagree with this statement. I think the "answers" that were given in the previous discussion are very accurate, and the mathematics clearly shows the energy gets redistributed in cases of interference where there is energy conservation.

The 'redistribution of energy' explanation fits well when the distance between the sources, d>>λ but fails for cases when
d<<λ . Check out paper Interference and the Law of Energy Conservation as mentioned by @vanhees71.

 

[keep_it_simple_silly]

It might be worth mentioning that, when two sinusoidal point sources are used, (each with a given amplitude $E_o$, and each radiating spherically), there will be an increase in the radiated power of the system as the distance between them becomes smaller. If the separation is fairly large, then to a very good approximation the total radiated power stays the same regardless of the separation, and the interference pattern will result in a redistribution of the energy.

(If I have analyzed it correctly, there isn't a sudden discontinuity when the sources are superimposed, but rather an increase in power occurs as the separation decreases, and then the power nearly doubles when the separation is very small. [Edit: The relative phases of the sources is also going to play a role. I believe this one might get a little complicated, and I haven't analyzed it in its entirety.])

I agree with @sophiecentaur in post 38 that the energy needs to be considered very carefully. e.g. When the point sources get too close together, the power needs to be re-computed, and you can't simply assume that the power is what it was when the sources were further apart.

@keep_it_simple_silly may also have been trying to make a statement to that effect, but it wasn't totally clear on first reading.

Yes, there is no sudden discontinuity in the power radiation as the distance between sources decreases.
The formula which represents all cases is (formula is given by Vanderkooy and Lipshitz in their paper Power response of loudspeakers with noncoincident drivers – The influence of crossover design) :-

Where,

1. P is total power radiated by both sources
2. P0 is the power radiated by single source in absence of other source
3. k=2π/λ, λ is wavelength of source
4. d is the distance between 2 sources
5. θ is the phase difference between 2 sources

If you plot graph for this particular equation, it looks like following for a phase difference of 0:-

you can see for larger distances(d>>λ), it is converging to 2P_0, and this is where the 'energy redistribution' argument roughly fits. Why rough? because you can see the graph still oscillating, but with minimum amplitude.
Im not sure if external links will be allowed by moderators but if they allow:- https://hackmd.io/@sharada10/wave_interference_energy_conservation

 

[Charles Link]

It may be worth mentioning that for the optics case, with a double slit experiment or with a diffraction grating, the sources are in the forward direction and the energy conservation and redistribution is a little simpler than with the r-f antenna. In any case, even with the optics case, we don't always get the simple and ideal case of energy conservation and redistribution. One case comes to mind that for very narrow slits the diffraction pattern can't expand any further (than $\pi$ radians) as the slit narrows, (to compute the power which needs to be the same as the input power, the peak of the pattern is multiplied by the effective angular spread=the mathematics becomes problematic when the angular spread can't increase any further as the slit width narrows), and thereby the simple textbook diffraction theory for the case of very narrow slits is only a fairly good approximation.

 

[sophiecentaur]

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:

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.

 

[Charles Link]

Now that there has been a considerable addition and perhaps correction to what I said in post 4, I would like to also have you take another look at what I said in post 2, which I think is indeed correct, and where the optics texts often come up short in this interference matter:
In many textbooks, the Fabry-Perot effect is treated as an interference that is the result of the multiple reflections that occur on a slab of thickness $d$. Changing the thickness changes the phases of the multiple-reflected beams as they add to interfere with each other. Meanwhile, the Michelson interferometer is sometimes taught as the interference of the original source superimposed on its mirror image. Both of these are sort of ok, but they both can be presented in the same and more fundamental way as the result of the interference that occurs from two beams incident on a single interface from opposite directions. There is an interference here, and the energy redistribution will depend on the relative phases of the two beams.
[Edit: Note for each incident beam, there are exiting transmitted and reflected E-field sinusoidal components, (each with their own phase), that are computed using the Fresnel coefficients].
I attempted to present this more fundamental concept in the Insight that I authored a couple years ago (the link is in post 2), but perhaps you can follow my short explanation here of the interference that occurs that indeed conserves energy when two sinusoidal sources are incident on a single interface. The energy redistribution depends upon their relative phases.

With this "new" way of looking at it, the Fabry-Perot effect and Michelson interferometer, and a beamsplitter with two incident beams are all explained by the same fundamental concept.

[Edit: Note that many of the textbooks do present the topic with the Fresnel coefficients, e.g. with the Fabry-Perot effect, but they do not emphasize or show in detail what results if we simply consider the interference (i.e. energy redistribution) of the simplest case of two beams incident on a single interface from opposite directions.]

[Edit: and note that for this interference with a single interface, we use the Fresnel coefficients that still work when two beams are present, but we no longer can use the energy reflection coefficient $R$ which works for a single beam incident on the interface. When two beams are present, the energy redistribution can be anything, (the energy no longer obeys linear principles even though the E-fields still obey linear principles), and is determined by the relative phases of the incident beams.]