What happens to the energy when waves completely cancel each other out via destructive interference? It seems like the energy just disappears, but that would violate the law of energy conservation.

My guess is that the kinetic energy is transformed into potential energy. Or maybe what happens to the energy depends on the specific interference scenario? Can someone elaborate on that or correct me if I'm wrong?

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    $\begingroup$ Exactly discussed by MIT youtu.be/SnNmbVH5DAM around 25:00 $\endgroup$
    – Kashmiri
    Commented Dec 28, 2020 at 16:51
  • $\begingroup$ The energy is always stored in the medium as potential energy. Air and water are elastic, the EM field stores energy but we cannot observe it directly. Water wave superposition is easy to observe. $\endgroup$ Commented Apr 2, 2022 at 15:14

15 Answers 15


Waves always travel. Even standing waves can always be interpreted as two traveling waves that are moving in opposite directions (more on that below).

Keeping the idea that waves must travel in mind, here's what happens whenever you figure out a way to build a region in which the energy of such a moving wave cancels out fully: If you look closely, you will find that you have created a mirror, and that the missing energy has simply bounced off the region you created.

Examples include opals, peacock feathers, and ordinary light mirrors. The first two reflect specific frequencies of light because repeating internal structures create a physical regions in which that frequency of light cannot travel - that is, a region in which near-total energy cancellation occurs. An optical mirror uses electrons at the top of their Fermi seas to cancel out light over a much broader range of frequencies. In all three examples the light bounces off the region, with only a little of its energy being absorbed (converted to heat).

A skip rope (or perhaps a garden hose) provides a more accessible example. First, lay out the rope or hose along its length, then give it quick, sharp clockwise motion. You get a helical wave that travels quickly away from you like a moving corkscrew. No standing wave, that!

You put a friend at the other end, but she does not want your wave hitting her. So what does she do? First she tries sending a clockwise wave at you too, but that seems to backfire. Your wave if anything seems to hit harder and faster. So she tries a counterclockwise motion instead. That seems to work much better. It halts the forward progress of the wave you launched at her, converting it instead to a loop. That loop still has lots of energy, but at least now it stays in one place. It has become a standing wave, in this case a classic skip-rope loop, or maybe two or more loops if you are good at skip rope.

What happened is that she used a canceling motion to keep your wave from hitting her. But curiously, her cancelling motion also created a wave, one that is twisted in the opposite way (counterclockwise) and moving towards you, just as your clockwise wave moved towards her. As it turns out, the motion you are already doing cancels her wave too, sending it right back at her. The wave is now trapped between your two cancelling actions. The sum of the two waves, which now looks sinusoidal instead of helical, has the same energy as your two individual helical waves added together.

I should note that you really only need one person driving the wave, since any sufficiently solid anchor for one end of the rope will also prevent the wave from entering it, and so end up reflecting that wave just as your friend did using a more active approach. Physical media such as peacock features and Fermi sea electrons also use a passive approach to reflection, with the same result: The energy is forbidden by cancellation from entering into some region of space.

So, while this is by no means a complete explanation, I hope it provides some "feel" for what complete energy cancellation really means: It's more about keeping waves out. Thinking of cancellation as the art of building wave mirrors provides a different and less paradoxical-sounding perspective on a wide variety of phenomena that alter, cancel, or redirect waves.

  • $\begingroup$ Antireflection coatings(ARC) on solar cells work by facilitating destructive interference between partially reflected waves from the air-ARC interface and ARC-solar cell interface. It is well established that in this scenario, there is minimum reflection and maximum transmission through the ARC. So destructive interference does not carry energy. Quantum mechanically, the EM wave is the wave function for photons. Photons actually carry energy. So, destructive interference=> no waves=>no photons=>no energy. Is it appropriate to compare destructive interference in mechanical waves and EM waves? $\endgroup$
    – user103515
    Commented Jan 19, 2018 at 6:25
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    $\begingroup$ @TerryBollinger , but your answer basically relies on the fact that the 2 waves do not travel in the same direction and thus any cancelation is only instantaneous and not relevant to conservation of energy. But assuming waves can be reflected perfectly without any loss of energy, consider the following scenario: Suppose that we have 2 sources of same frequency light aimed directly at each other call this line l horizontal, and let P be a plane containing l and consider a perfectly reflective square (or cube ) S contained in P oriented 45 degrees to l and intersecting l on 2 adjacent $\endgroup$
    – Hao S
    Commented Dec 25, 2018 at 23:52
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    $\begingroup$ You have to remember, energy is the ability to do work. Work causes displacement along a vector. If wave A causes exactly opposite displacement to wave B, the waves will fully cancel (they do not form a standing wave!). This is not because energy has been lost. Its because you have combined negative and positive energy to create zero energy. Conservation of energy is not violated because -1 + 1 = 0. Equilibriums do not violate the laws of physics... $\endgroup$
    – CommaToast
    Commented Apr 5, 2019 at 19:56
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    $\begingroup$ All physically real waves represent accelerations or real mass or variations of real fields, and so all have positive energy. Be careful no not to confuse convenient mathematical abstractions with the actual flows of energy in the physical world, which for classical physics are always positive. $\endgroup$ Commented Apr 6, 2019 at 1:52
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    $\begingroup$ A monochromatic electromagnetic wave is described by a vector that specifies its wavenumber and propagation direction, and 2 more numbers: amplitude and phase for both polarization. In order to have negative interference between two waves all of these must exactly match except for the phase, which must be shifted by pi. If the other quantities do not match you will NOT have complete cancellation. None of the examples in this answer are relevant to the question as they do not produce complete negative interference. $\endgroup$
    – bkocsis
    Commented Mar 11, 2020 at 19:13

We treated this a while back at University...

First of all, I assume you mean global cancellation, since otherwise the energy that is missing at the cancelled point simply is what is added to points of constructive interference: Conservation of Energy is only global.

The thing is, if multiple waves globally cancel out, there are actually only two possible explanations:

  • One (or more) of the sources is actually a drain and converts wave energy into another form of energy, (e.g. whatever is used to generate the waves in sources, like electricity, and also as Anna said, very often heat)
  • You are calculating with parts of an mathematical expansion which are only valid when convoluted with a weight function or distribution. For example, plane waves physically don't exist (But when used in the Fourier Transform they are still very useful) because their total energy is infinite
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    $\begingroup$ The OP did not ask about a drain nor a plane wave. You can have two wavepackets shifted in phase by pi, and you guide the two wave packets into the same region so that they cancel. This exists in practice, e.g. the laser beam in the LIGO cavities, two beams are combined to have negative interference. $\endgroup$
    – bkocsis
    Commented Mar 11, 2020 at 19:19
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    $\begingroup$ @bkocsis OP asked about waves perfectly cancelling each other out, not just on some locations. Partial/localized negative interference is of course perfectly possible. $\endgroup$ Commented Mar 14, 2020 at 19:09

Just in case anyone (e.g. student) would be interested in the simple answer for mechanical waves:

CASE 1 (global cancellation): Imagine that you have crest pulse moving right and equally large trough pulse moving left. For a moment they "cancel", e.g. there is no net displacement at all, because two opposite displacements cancel out. However, velocities add up and are twice as large, meaning that all the energy in that moment is stored within kinetic energy.

Instructive and opposite situation happens, when crest pulses meet. For a moment, displacements add up and are twice as large, meaning that all the energy in that moment is stored within potential energy, as velocities on the other hand cancel out.

Because wave equation is linear differential equation, you can superpose different waves $\psi_{12} = \psi_1 + \psi_2$. As a consequence, after meeting both crest pulses or pair crest / trough pulses keep traveling as if nothing had happened.

It is instructive, that you can add velocities separately of amplitudes, as $\dot{\psi}_{12} = \frac{\partial}{\partial t} (\psi_1 + \psi_2) = \dot{\psi}_1 + \dot{\psi}_2$. So even if amplitudes do cancel out at a given moment ($\psi_1 + \psi_2 = 0$), speeds do not ($\dot{\psi}_1 + \dot{\psi}_2 \ne 0$).

It is just as if you see that oscillator is in a equilibrium position at a given moment. That does not mean that it is not oscillating, as it still might posses velocity.

If we generalize written above: in any wave you have exchange of two types of energy: kinetic vs. potential, magnetic vs. electrical. You can make such two waves that one of the energies cancels, but the other energy will become twice as big.

CASE 2 (local cancellation): In case of spatial interference of two continuous waves there are areas of destructive and areas of constructive interferences. Energy is no longer uniformly distributed in space, but in average it equals added up energies of two waves. E.g. looking at standing waves, there is no energy at nodes of the standing waves, while at crests energy is four times the energy of one wave - giving a space average of twice the energy of one wave.

More engineer-like explanations can be found here:



Maybe the question can simply be answered by the observation that a wave like

$$\Psi(x,t)=A \cos(x)-A \cos(x+\omega\ t),$$

where the two cosines cancel at periodic times $$t_n=\frac{2\pi}{\omega}n\ \ \longrightarrow\ \ \Psi(x,t_n)=0,$$ still has nonvanishing kinetic energy, if it looks something like $$E=\sum_\mu\left(\frac{\partial \Psi}{\partial x^\mu} \right)^2+\ ...$$

You really would have to construct an example.

Since non-dissipative waves whose equations of motions can be formulated by a Lagrangian will have an energy associated to them, as you say, you'd have to find a situation/theory without an energy quantity. The energy is related to the wave by its relation to the equation of motion. So if the energy is defined as that which is constant because of time symmetry and you don't have such a thing, then there is no question.

Also don't make the mistake and talk about about two different waves with different energy. If you have a linear problem, the wave will be "one wave" in the energy expression, wherever its parts may wander around.

edit: See also the other answer(s) for a discussion of a more physical reading of the question.


I think a good way to approach this question is with a Mach-Zehnder interferometer:Mach-Zehnder interferometer

The field landing on detector 1 is the interference between two waves, one from the lower path, and one from the upper path. Let's suppose the field in each arm is a collimated beam of coherent light, well-approximated as a plane wave, and the interferometer is well-aligned, so the two outputs are almost perfectly overlapped. By changing the thickness of the sample, we can change the relative phase between the two waves, changing our interference from destructive (less energy on detector 1) to constructive (more energy on detector 1). Where did this energy come from? If the beams are nicely matched, this interference can even be completely destructive, and detector 1 will register zero signal. Where does the energy go?

The short answer is: detector 2. The total energy hitting the two detectors is constant, as you vary the phase shift caused by the sample. Constructive interference at detector 1 goes hand-in-hand with destructive interference at detector 2.

If you only look at one detector or the other, it might seem like energy is created or destroyed by interference, but as other answers mention, we must consider the whole system.

  • $\begingroup$ If you have negative interference you do not have any energy flux left for detector 2 to receive. $\endgroup$
    – bkocsis
    Commented Mar 11, 2020 at 19:25
  • $\begingroup$ I'm not sure your comment makes any sense, @bkocsis. Can you try to express yourself more clearly? $\endgroup$
    – Andrew
    Commented Mar 15, 2020 at 1:20
  • $\begingroup$ If there's no sample, or if the same thickness sample would also be placed on RB then wouldn't both detectors register the same interference, be it constructive or destructive? $\endgroup$
    – mireazma
    Commented Dec 5, 2020 at 11:12

I had prepared this answer for a question that was made duplicate, so here it comes, because I found an instructive MIT video. (the second link) This answer is for electromagnetic waves mainly

Have a look at this video to get an intuition how interference appears photon by photon in a two slit experiment.

It comes because the probability distribution for the photons, as accumulated on the screen, has destructive and and constructive patterns, ruled by the underlying quantum mechanical solution of "photon + two slits".

The classical electromagnetic wave emerges from a great plethora of photons which have phases and such that they build up the electric and magnetic fields. The nu in the E=h*nu of the photon is the frequency of the electromagnetic wave that emerges from the confluence of the individual photons. In order to get an interference pattern the photons have to react with a screen, or some some matter, as in the laser experiments.

The reason that matter is needed for light interference phenomena is due to the very small electromagnetic coupling constant. Photon photon interactions due to the 1/137 end up having a probability of interaction of order of ~10^-8 . With respect to photon electron interactions, which to first order is ~ 10^-2,( and is the main photon-matter interaction) there are 6 orders of magnitude. To all intents two laser beams crossing will go through each other without any measurable interaction,( interference pattern may exist , but they are not photon photon interactions but quantum mechanical superpositions). (Keep this in mind when you reach the last question at the end of the next video.)

This MIT video is instructive and a real experiment that shows that in destructive interference set up with interferometers there is a return beam, back to the source, as far as classical electromagnetic waves go. So the energy is balanced by going back to the source.

What is happening at the photon level? If the laser emitted photons one by one as in the two slit video? I will hand wave as there is no corresponding video to show:

The quantum mechanical solution with the complicated boundary values of the interferometer allows the elastic scattering ( not small, that is how we get reflections) of photons also back to the source. You can see in the video that there always exists a beam going back to the source, that beam is carried by individual photons scattering elastically backwards through the system of the optics of the interferometer. In total destructive interference all the energy is reflected back ( minus some due to absorption and scattering in the matter of the optical system).

In essence this experiment is a clear demonstration that the system laser-optical-bench is in a coherent quantum mechanical state, the returning photons joining the ensemble of photons within the laser action , which also includes reflections to be generated.

In this video, the first beam carries the information of the phases such that in space interference patterns will form if a screen or other matter intervenes. The energy of the final beam after it leaves the interferometer system and falls on the screen and is redistributed according to the pattern of the interference. The amount of energy carried by the beam there depends on the proportion of energy that manages to leave the interferometer/laser system, i.e. if all of the energy is returned to the laser (destructive interference) , or a part of it goes out of the lasing system to impinge on the screen.

In the case of waves in matter, as sound waves or water waves:

In the case of two sound waves interfering destructively, the temperature of the medium will go up and energy is conserved because it turns into incoherent kinetic energy of the molecules of the medium.

For two water waves, ditto.


Explaining this problem using classical equations is simple, well known and well understood. The field values become zero in destructive interference and there is no energy as a result- full stop. The QM explanation is equally simple.. the probability of finding a photon there is zero and no more discussions is needed. So where is the problem, and clearly there is one in the logic- judging by the length and number of replies.

We first need to realize that radiation is something special.. it is the advance of a force field in vacuum- but as if this vacuum is a medium, and when we can't see or feel any. This medium is homogeneous, has constant properties of permeability, permittivity and even an electric resistance 376.73 ohms. It has a constant speed of propagation as a result of that, and the speed is as in matter, given by; the square root of the ratio of Bulk modulus of elasticity divided by the density (get this by dividing the two sides of E=m c^2 by the volume. B=E/V, and ρ=m/V).

When we deal with waves in a matter medium, we find the we can't transfer energy without the existence of a sink that absorbs this energy. This is the basis of the Newman-Feynman absorber theory. You can establish a force all the time in all the space- with or without an absorber, but not energy. So, matter sitting in intense radiation field at points of total darkness is under intense stress, but does not receive any energy. Energy is force times distance and matter needs to move to absorb energy. For this reason if you have a perfect microwave oven, you don't spend any electrical energy on it if it is running empty-despite being full of intense radiation with regions of destructive and constructive interference- accept the small amount needed to establish these from zero.

Thus the simple answer is not that the energy is going back to the source which sounds ridiculous in my opinion, but by the source not giving its energy in the first place, because there is no absorber to take it. There is one mysterious exception in this however.. it is that it is possible to send energy to an infinite space even if we don't see any obsorber there- as in antenna sending to outer space. We need to resort here to the distant masses of Mach to provide an absorber.

  • $\begingroup$ This is wrong. An RF power amplifier that is connected to an antenna with a transmission whose far-end is mismatched because the antenna is not equal to the wave impedance $R_w$ of the transmission line may blow up even if the amplifier's own output impedance is equal to $R_w$. The effective load impedance can be anything complex between $-\infty \to 0 \to \infty$ and if it appears essentially reactive nothing actually will come out of the amplifier after the first reflection. Don't experiment with this on your home magnetron, it can be a very unpleasant surprise! $\endgroup$
    – hyportnex
    Commented Jan 14 at 15:57
  • $\begingroup$ You give a nice practical example but it doesn't prove anything wrong. It seems to me that your connection piece is working in a standing mode,, and you are pumping more energy without a way to go out to outer space because of the mismatch, Like pumping a laser or microwave cavity without letting any laser out, or like pushing a child swing more than the opposing friction losses. $\endgroup$
    – Riad
    Commented Jan 15 at 17:57

This figure shows two common situations.

The top is an example where the waves are coming from different directions--one from "S1", one from "S2". Then there is destructive interference in some areas ("nodes") and constructive interference in others ("hot spots"). The energy has been redistributed but the total amount of energy is the same.

The bottom is an example where the two sources S3 and S4 are highly directional plane-wave emitters, so that they can destructively interfere everywhere they overlap. For that to happen, the source S4 itself has to be sitting in the field of S3. Then actually what is happening is that S4 is absorbing the energy of S3. (You may think that running the laser S4 will drain its battery, but ideally, the battery can even get recharged!)

  • $\begingroup$ What about momentum conservation? Do you have a link for an experiment on this? $\endgroup$
    – anna v
    Commented Apr 19, 2012 at 14:19
  • $\begingroup$ physics discovers impedance mismatch ... $\endgroup$
    – hyportnex
    Commented Dec 9, 2019 at 13:26
  • $\begingroup$ How about when the light from S3 and S4 pass through a (same) beamsplitter, (S3 transmission and S4 90 degree reflection)? $\endgroup$
    – mireazma
    Commented Dec 5, 2020 at 14:04


Oblique superposition of two elliptically polarized lightwaves using geometric algebra: is energy–momentum conserved? Michelle Wynne C. Sze, Quirino M. Sugon, Jr., and Daniel J. McNamara JOSA A, Vol. 27, Issue 11, pp. 2468-2479 (2010)

We added the two elliptically polarized waves and computed the energy–momentum density of their sum. We showed that energy and momentum are not generally conserved, except when the two waves are moving in opposite directions. We also showed that the momentum of the superposition has an extra component perpendicular to the propagation directions of both waves. But when we took the time-average of the energy and momentum of the superposition, we found that the time-average energy and momentum could also be conserved if both waves are circularly polarized but with opposite handedness, regardless of the directions of the two waves. The non-conservation of energy and momentum of the superposition of two elliptically polarized plane waves is not due to the form of the plane waves themselves, but rather to the accepted definitions of the electromagnetic energy and momentum. Perhaps we may need to modify these definitions in order to preserve the energy–momentum conservation. In our computations, we restricted ourselves to the superposition of two waves with the same frequency.


Think of light as photons, then black spots means that no photons were detected and bright spots means lots of photon detection. Since the energy of photons are always quantized: $E=h\nu$ there would be no problem on the energy conservation.

One question arises, what makes a photon "arrives" here and not there? The answer is in the wave amplitude of probability associated to the photon. It is the wave amplitude of probability that interferes, in consequence the probability density becomes the intensity pattern for large amount of light quanta.

It is often said that a photon 'interferes with itself', but it is the wave amplitude of probability which interferes. In this sense the statement is ok when you think of the 'wave function' as the photon itself.

PD: wave function for photons is still an arguable issue, but some progress has been made. You may want to check Galuber's theory of photodetection.


The Poynting_vector

In physics, the Poynting vector represents the directional energy flux density (the rate of energy transfer per unit area, in Watts per square metre, W·m−2) of an electromagnetic field.

If the antilaser antilaser experiment is performed in the vacuum there is no thermal dissipation, and the Poynting vectors are opposed, and cancel, for the same field intensity and with the fields out-of-phase. For plane waves (WP, link above):
"The time-dependent and position magnitude of the Poynting vector is" : $\epsilon_0cE_0^2\cos^2(\omega t-\mathrm{k\cdot r})$ and the average is different of zero for a single propagating wave, but, for two opposing plane waves of equal intensity and 100% out-of-phase the instantaneous Poynting vector, that measures the flux of energy, is the vector $\vec{S}(t)=\mathrm{\vec{0}}$.

If you have one electromagnetic beam at a time then work can be done. If you have two in the above conditions then no work can be extracted. (Energy is canceled, destroyed, ;)

BUT, things can be more complicated then described by the eqs, because a physical emmiter antenna also behaves as a receiving antenna that absorbs and reradiates etc, ... changing and probably trashing my first oppinion.


Let's first do an example without any interference at all - imagine an electron trapped in a box with some total energy $E$. At some point it releases a photon and we capture that new particle in a box.

By conservation of energy wavefunction of the two particles looks like $\sum\limits_{E_i+E_n = E} a_{in} e^{-iEt} | E_i \rangle | E_n \rangle$

If we measure box with the photon and find an energy of $E_n$, then by the usual copenhagen interpretation the electron immediately collapses into the eigenstate $|E_i \rangle$. Now one has to be careful, because quantum mechanics doesn't really deal with the creation of particles - but we can analogously think of the situation where we dont measure any energy in the photon box - we would immediately know that the electron has energy $E$. The fact that there is no photon in the box does not violate the conservation of energy, it simply tells us that no photon was released in the first place.

This should tell us that even if perfect destructive interference were possible, energy would still be conserved. We can only definitely say that energy has been transferred once we do a measurement - if there is no light to measure then there is no energy that has been transferred.

Now, is perfect destructive interference actually possible to have? Let's look at the wavefunction of two particles $\Psi(x_1,x_2,t)$. Generally we force the norm of $\Psi$ to equal 1 for normalization, but lets relax that condition and take $\Psi(x_1,x_2,t_1)=0$ We then have that $\Psi(x_1,x_2,t_2) = U(t_2,t_1) \Psi(x_1,x_2,t_1) = 0$ And so the wavefunction is 0 for all time - the wavefunction can't be 0 unless it was always 0, assuming unitarity. Now, if we think of a system where destructive interference happens, at time $t$ our wavefunction is non-zero, and at time $t + dt$ our wavefunction is exactly 0. Since time-evolution is given by the flow of the hamiltonian, this means that our hamiltonian has a delta-function spike at time $t$, which we can think of as a change in the energy of our system. This is really an indicator that we are not looking at a large enough system - that part of the energy we thought was in our system was really coming from somewhere else. Note that this argument actually works for any system whose time-evolution is described by hamiltonian flow.

This also partially explains why the answers to this question are so varied. Destructive interference is common but perfect destructive interference is the same as nothing happening. We can say that perfect destructive interference does not happen and that would be correct. We could also say that two waves perfectly destructively interfering never described a system with energy and that would also be correct. If two waves cancel out in a forest and nobody sees them, were they ever there?

(Note that destructively interfering waves can also lose their energy to another system, such as heat. It is not always the case that the energy lies within the system that created the waves. For example, one could absorb a sound wave to gain energy, which we could model by destructive interference - but also another way to destructively interfere a sound wave is to turn it into heat.)

  • $\begingroup$ "two waves perfectly destructively interfering give their energy back to the source" ... is not likely possible. Consider a laser diode, the electron is excited to its state by the E field in the depletion region of the semiconductor ... it can not be re-excited by a returning photon. Consider 2 water waves in the middle of the ocean ... they certainly superimpose destructively but it has nothing to do with the source, which is the wind. The wave functions above have no term for energy held in the medium .... all wave functions concern themselves with the observables. ...E and M. $\endgroup$ Commented Oct 21, 2023 at 23:24
  • $\begingroup$ I should have phrased myself better - its not that the energy "goes back to the source", its that the energy was never there in the first place. If I consider a perfectly still pond, I can model that pond as a series of overlapping perfectly destructive waves. The enery from this destructive interference isn't lost - it was never there in the first place. $\endgroup$ Commented Oct 23, 2023 at 0:10
  • $\begingroup$ You can take a perfectly still pond and have 10 frogs jump in (one after the other) from the east bank and, at the same time, 10 frogs jump OUT on the west side ... a few minutes late when the wave trains meet in the middle the pond will look perfectly still for a brief moment .... but a moment later the waves reemerge and continue on their merry way ... only to eventually "cancel" when they crash on the shore leaving the pond still again. The point being you have 2 very different types of "still" pond ... one full of potential energy (water is elastic) and one not. $\endgroup$ Commented Oct 23, 2023 at 0:40
  • $\begingroup$ I was thinking perfect destructive interference in the sense that the waves are zero for all time and space. Any standing wave will be uniformly zero at some times, right? There isnt anything to explain energy-wise there because the potential energy is all kinetic. I wouldnt call the pond perfectly still there since the water has a velocity up ot down? $\endgroup$ Commented Oct 23, 2023 at 2:04
  • $\begingroup$ The point to make here is that if the wave is zero on some open set and nonzero elsewhere, then there must be something not modeled by the wave equation - such as energy lost to heat $\endgroup$ Commented Oct 23, 2023 at 2:20

From my answer here-PSE-anti-laser-how-sure-we-are-that-energy-is-transported

The Poyinting vectors, and the momenta vectors as the E, B fields are symetric. When we do 'field shaping' with antenae aggregates we simply use Maxwell eqs and go with waves everytime. When we got near a null in energy in some region of space we dont get infrared radiation to 'consume' the canceled field. E,B vectors additive: Light+Light=0

Antenae in sattelites (vaccum) work the same way as the ones at Earth surface to shape the intensity of the field.
Because the "Poyinting vectors" add to null there is no doubt, imo, that energy vanish.

See the antilaser experiment.

We dont have theory? Then we must rethink.
IMO energy is not transported. What is propagating is only an excitation of the medium (we call it photons) and energy is already 'in site' (vacuum, or whatever name we call the medium).


the waves will obliterate each other but they will still exist, they just won't be moving they would just change form (energy cannot be destroyed it can only change form) so when the waves meet they will cancel each other so sound will change to potential and kinetic will change to sound or whatever


When there is a complete destructive interference of two light beams, Maxwell’s equations predict that the energy becomes zero. Let’s have the case of two coherent collinear beams, out of phase 180 degrees, like the case of the antilaser.

\begin{align} E_1 = E_m \sin (kx - \omega t);\quad E_2 = E_m \sin (kx - \omega t + p) \\ B_1 = B_m \sin (kx - \omega t); \quad B_2 = B_m \sin (kx - \omega t + p) \end{align}

$E = E_1 + E_2$ and $B = B_1 + B_2$ \begin{align} E &= E_m \sin (kx - \omega t) + E_m \sin (kx - \omega t + p) \\ B &=B_m \sin (kx - \omega t) + B_m \sin (kx - \omega t + p) \end{align} But, $\sin (kx - \omega t + p) = - \sin (kx – \omega t)$ , Then, $E = 0$ and $B = 0$ and,

\begin{align} UT &= U_E + U_B \\ &= \frac12ԑ_0E^2 + \frac1{2\mu_0}B^2 \\ &= 0 \end{align} This is the classical interpretation of the waves electromagnetism during total destructive interference, following Maxwell. Maxwell's description of the energy of the light wave is of an undulating energy that predictably reaches a maximum and later becomes zero. The proposed solution to this problem is to calculate the mean of the energy when the fields are maxima.

What is the physical meaning of an energy that have to be averaged in order to have the real magnitude. If the principle of conservation of energy is to be applied to this phenomenon, the energy must be constant, have an unique value for each instant during the movement of the wave. What is the meaning of that situation that has not been recognized for more than a century?

What almost nobody want to admit is that electromagnetism is incomplete, because cannot describe the electromagnetic radiation adequately, and generate a violation of the principle of conservation of energy.

As Helder Velez said: “We don’t have theory?”. NO Then we must rethink.” He has a proposition: EM energy is not transported, only is an excitation of the medium, the quantum vacuum, or the quantum plenum as I prefer to call it. But this is only an idea, an intuition, without support or evidence.

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    $\begingroup$ Lots of copy-pasting from poster's other (flawed) answers. $\endgroup$
    – Kyle Kanos
    Commented Feb 13, 2015 at 21:21
  • $\begingroup$ I made one and only one reference: Helder Velez. The fact that his comment have negative punctuation does not mean that he is incorrect; only that they disagree with him. I support him in the violation of the conservation of energy. I support the "EMG theory of the foton" by Diogenes Aybar that can be found at journaloftheoretics.com/Links/Papers/EMG%20III.pdf. He thinks that the photon has another inherent field, the gravitational field, where the energy goes from the two electromagnetic fields to the gravitational field, keeping then, at every moment the energy constant. $\endgroup$ Commented Feb 15, 2015 at 18:39

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