# What happens to the energy when waves perfectly cancel each other?

What happens to the energy when waves perfectly cancel each other (destructive interference)? It appears that the energy "disappear" but the law of conservation of energy states that it can't be destructed. My guess is that the kinetic energy is transformed into potential energy. Or maybe it depends on the context of the waves where do the energy goes? Can someone elaborate on that or correct me if I'm wrong?

-

1
– Tobias Kienzler Apr 18 '12 at 7:12

3
You can't make waves perfectly cancel each other everywhere, because then energy isn't conserved. – Ron Maimon Jun 25 '12 at 19:47
This whole question and it's answers are frightening and saddening. – Colin K Jun 29 '12 at 3:34

## 10 Answers

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
-

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 would really 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 sitation/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 symmetriy 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, whereever its parts may wander around.

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

-

1
@Pygmalion: $\partial_\mu$ is supposed to be just a derivative like $\frac{\partial}{\partial t}$. If $\Phi(t)=\sin(t)$, then at $t=0$ you have $\Phi=0$, but $(\frac{\partial \Phi(t)}{\partial t})^2=1$. And yeah, you can extend my example to $\Psi(x,t)=\cos(k_1x-\omega_1t)+\cos(k_2x-\omega_2t)$ and not much will change but the expression for $t_n$. And as I said in the answer, if you describe the waves by one theory with supperposition, then you usually need to talk only about one field. There are not two different electric field energies. It's a linear theory. PS: You can edit comments btw. – Nick Kidman Apr 17 '12 at 19:43
@Pygmalion: And as I said before, if you take a linear theory like eletrodynamics, you add up all parts of the field and then compute the kinetic equation of that. I.e. $(\partial(\psi_1+\psi_2))^2$, not $(\partial\psi_1)^2+(\partial\psi_2)^2$. (side note: Since the form of the fields, related to the kinetic energy by the equation of motion, are not arbitrary you can't just write down a kinetic energy with a nonlinear additive partition propery and choose an example like $\cos$. These are phases which, in physics typically come from a spectral analysis of a linear differential operator.) – Nick Kidman Apr 17 '12 at 20:05
I understand your answer and notation now, but I had to cleared it up in my head my way. Also, maybe I was too sleepy around midnight yesterday ;) – Pygmalion Apr 18 '12 at 8:01

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.

-

Wherever the +100 came from (I'm assuming a person, not a bot), thanks! I am deeply appreciative of such a nice bit of positive feedback! – Terry Bollinger Jul 2 '12 at 3:14

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 though 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 / though pulses keep traveling if nothing has 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: http://van.physics.illinois.edu/qa/listing.php?id=1891

-

When two waves interfere totally destructively the energy turns into heat.

Two sound waves, 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.

Two water waves, ditto.

I have been trying unsuccessfully to find a reference about two electromagnetic waves with total destructive interference and energy conservation.

If it is an interference pattern, as observed in a detecting medium, screen etc. and one gradually goes to total destructive interference then again the medium will be heated.

If it is two electromagnetic waves in vacuum, I expect the energy to go into soft infrared invisible photons, as there is no ether to take it up, but may be wrong. One would need QED to see how this could happen, imo. Figure 6.8 allows for photon photon scattering and the change in frequency but the probability is small. Or maybe there cannot be complete destructive interference in vacuum, the Heisenberg uncertainty principle ( delta(E)*delta(t) always allowing for constructive ones. Or the truth is in between that the energy propagates and we can only see interference if the beams impinge on a medium, which will take care of energy conservation .

Edit Thanks to @HelderVelez for the link of the antilazer. Saves me thinking of an experiment.

When the alignment was right, the light waves canceled each other out. The silicon absorbed the light and converted it to another form of energy, like heat or electrical current.

I had not thought of current, but heat is right there.

-

2
I guess it's worth pointing out that this sort of thing can only happen in nonlinear theories. – Mark Eichenlaub Apr 17 '12 at 20:27

4
Two EM waves which interfere destructively will also interfere constructively in another region, such that energy is conserved globally. – user2963 Apr 17 '12 at 21:14
@zephyr Would it be fair to say generally that it is impossible to create two particles with exactly opposite wave functions for the whole universe for some, say, exclusion principle? Then it is absolutely necessary that somewhere there is a constructive interference. – Pygmalion Apr 17 '12 at 22:01
Downvote -- As Mark says, wave energy can't just turn into heat just because you want it to, there has to be a mechanism--a nonlinear interaction. There is no reason to expect that nonlinear interactions of exactly the correct strength will magically appear in all situations to transfer exactly the appropriate amount of energy. – Steve B Apr 18 '12 at 0:14
If destructive interference just shifted light to infrared, then all those interferometers I've got wouldn't work very well... – Colin K Apr 18 '12 at 3:36
show 16 more comments

http://www.opticsinfobase.org/josaa/abstract.cfm?uri=josaa-27-11-2468

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.

-

thanks, an interesting calculation behind a paywall – anna v Apr 25 '12 at 4:38

I did some reaserch into this and after a lot of reading I came acrose a site with a good description of what you are talking about so here's a link. http://skullsinthestars.com/2010/04/07/wave-interference-where-does-the-energy-go/ If you go to this site it should answer your question and I hope it helps.

-

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).

-

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!)

-

What about momentum conservation? Do you have a link for an experiment on this? – anna v Apr 19 '12 at 14:19

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9430378079414368, "perplexity_flag": "middle"}