# Energy of Electromagnetic Waves in Destructive Interference

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• MartinG
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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.

• Pushoam and Delta2

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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/

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• Delta2 and MartinG
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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.

• MartinG
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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.

• Delta2 and MartinG
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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.

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• MartinG and vanhees71
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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?

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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.

• berkeman
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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.

• • • DaveE, vanhees71, Delta2 and 1 other person
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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?

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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 and Delta2
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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.

• • Pushoam, dextercioby, weirdoguy and 2 others
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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.

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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.

• Pushoam, weirdoguy and vanhees71
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.

• • weirdoguy and Delta2
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@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 and DaveE
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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.

• davenn and weirdoguy
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@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.

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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?

• • davenn, weirdoguy and Delta2
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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...

• davenn
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Is energy conserved according to your calculations?

• vanhees71
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No.

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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.

• vanhees71 and Delta2
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If the sources are physically separated,
Yes sorry didn't think of that case.

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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.

• • vanhees71 and Delta2
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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.

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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.

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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.

• • dextercioby, weirdoguy, Orodruin and 3 others
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@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
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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.

• weirdoguy
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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).

• Staff Emeritus
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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.

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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.

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• • sophiecentaur, vanhees71 and Delta2
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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.

• hutchphd and vanhees71
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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.

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• sophiecentaur and vanhees71
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