Q & A: Energy and interference

When two light waves cancel each other out, where does the energy go?
- Jonno
Waltham, MA

Jonno- That's one of the best questions we've had in a while.

When the light waves cancel (or at least partially cancel) in one region, there's less energy there than you would get from adding the energies in the two waves. The missing energy must have gone somewhere else. It turns out that the destructive interference in one place is always balanced by some constructive interference someplace else, where the total energy is bigger than the sum of the two separate energies. In a typical example, two combined beams form interference fringes- alternating regions in which the energy is more or less than the sum of the separate energies. In another type of example, the electric fields in a region can cancel, but the magnetic fields add (or vice versa), so that energy shifts from one type of field to another in the same region.

Mike W.

(published on 10/22/2007)

I understand your explanations so far, but in Physics class we learn that when two identical wave pulses travelling in opposite directions meet, they cancel each other out perfectly. This would mean the particles in the medium do not move. Does this mean there is no energy transfer? What then, causes the next particles to move since the pulses apparently keep travelling unaffected by the temporary cancelling out.
- Jasmine (age 16)
Seattle, WA, USA

Nice question!  I'm really glad that you asked it, because I tried raising exactly the same question in my sophomore physics class a couple of weeks ago, but the students didn't seem very interested.

You're talking about waves made of particles- sound waves, for example. Let's say you're talking about the type of sound ("longitudinal") that travels in air, with the particles moving back and forth along the same direction that the wave travels. 
If the sound pressures from the two waves traveling in opposite directions just cancel, the particle velocities from the two waves add up. If the velocities just cancel, then the pressure changes add up. Either way, although one of those two variables (pressure or velocity) momentarily looks like there's no wave there, the other variable has all the evidence that the wave is still there.

Mike W.


Another way of looking at it is that the two waves produce what is called a "standing wave pattern".    "Nodes" exist where  there is little or no action at all.  In between the nodes are  areas where the air molecules oscillate back and forth at twice the amplitude.  The total energy is the same as the sum of the two individual waves. This is more easily visualized by considering transverse waves on a string.  The stationary pattern of a vibrating violin string can be mathematically decomposed into two traveling waves going in opposite directions.

Lee H

(published on 10/22/2007)

We know that in interference redistribution of energy takes place .Is this redistribution remain constant with time or vary with time?
- Yash (age 17)
India

That depends on how the waves themselves change over time. If they have a big mix of frequencies then the interference patterns will keep shifting around. If they consist of a very narrow range of frequencies, from fixed sources, the interference patterns won't change much. For ordinary light, the interference patterns shift around so much that they usually aren't noticeable. On the other hand, for laser light one can make rather stable interference patterns.

Mike W.

(published on 10/22/2007)

Could you not create two wavetrains of light and send them toward each other. (make one a reflection of the other so that they can interfere destructively). When they are far apart: each wavetrain carries some electric and magnetic energy. At some time, they completely annihilate each other and theres no electric/magnetic field anywhere, so the total energy is zero. Am I missing anything? Also, I can imagine the same thing done with a string. Except in the case of the string, I think the missing energy can be accounted for: the kinetic energy transfers to potential energy between the atoms that make up the string.
- Eugene (age 21)
CA, USA

What you imagine for the string is almost correct. At the moment when the two pulses traveling opposite ways cancel, in the sense of having opposite string displacements, they have the same string velocities. So actually it's the potential energy that's zero then, and all the energy is kinetic.

Something like that happens with the light waves. If the electric fields exactly cancel, the magnetic fields add up, so all the energy is magnetic. That has to work because the Poynting vector, which gives the energy flow, is opposite for the two oppositely moving waves. It's ExB, so if the E's are opposite the B's must be the same.


Mike W

(published on 10/22/2007)

I understand the electric energy is trasfered to magnetic energy when 2 light waves are propagating towards each other. But if the 2 light wave are propagating at same direction and so their electric phase and magnetic phase are both opposite, what does the energy go? I think Michelson Interferometer can do this.
- Keitha (age 28)
China

If the 'two' light waves are really propagating in the same direction, then actually they're one light wave. Whatever the polarization pattern happens to be, it will have 1/2 electric and 1/2 magnetic energy.

In a Michelson interferometer, if the phase is just right there will be perfect destructive interference in one output direction and perfect constructive interference in the other, at right angles. Thus the one beam really does have zero energy. All the energy goes into the other.

Mike W.

(published on 08/01/2009)

I couldn't understand how redistribution of energy occurs in interference ??? Our sir said , giving a formula based on intensity that the intensity of the resultant wave increases, nd he gave an xample lyk if 2 waves each with 20 joules show interference the reultant energy inc, say becomes 60 joules, from where does this extra 20 joules come?????
- Deboshree Banerjee (age 17)
India

Perhaps I don't understand your instructor's explanation.  I think most people would say that if you have two interfering waves the energy will add linearly.  Now, the energy spatial distribution might vary from point to point but if you do the averaging correctly you will find the total energy will be the sum of the two individual energies.
 
LeeH 

(published on 09/29/2010)

About your answer to waves and energy... Lets say there are 2 transverse waves 180 out of phase travelling towards each other on a rope. When they interfere destructively you say the energy will be evident in the velocity aspect of the superimposing waves and not in the displacement of that part of the rope, seeing as there would be none. Ok...mechanical waves transmit energy: kinetic, potential or both, the sum of which must remain constant...Got it. But adding the velocities give us zero for 2 waves passing through each other at a node. I know the energy must be all kinetic at a node, but what is the best way understand it? Thanks for the site! Its great! sincerely, shaun
- Shaun Kelly (age 45)
plymouth,ma,us

Shaun- Try drawing the picture. Say that we have an upward bump coming in from the left and a downward bump coming in from the right. If they're both the same symmetrical shape and the same size, when they exactly overlap the displacements cancel.

Now look at the leading (right) edge of the right-going wave. The transverse velocity there is obviously upward. Look at the trailing (right) edge of the left-going wave. The transverse velocity there is also upward. So when these overlap the net upward velocity is doubled.

Mike W.

(published on 03/29/2011)

Hi, I have a thought experiment on this topic. Consider two E.M. waves emitted at nearly the same position at equal amplitude but out of phase by 180deg, maybe two very closely spaced RF sources or spinning dipoles. Far away from this source the magnetic fields cancel another as do the electric fields, even though there is still energy wound up in the individual EM waves. Is there anyway to collect energy from this system given the destructive interference of the two fields? Classical theory would say no, am I missing something?
- Tyler (age 20)
Madison, WI, USA

Hi Taylor,
If the distance between the two antennas is very small, much smaller than the wavelength of the radiation,  then your conjecture that the far away fields will cancel each other is correct and the net radiated power will be close to zero.  But when the distance between them is a good fraction of a wavelength or more, then the two EM fields will cancel in some directions but not in others and so there will be net radiation.  I tried to calculate it exactly but couldn't do the integrals.  You probably have to do it numerically.  If I can find the solution I'll post the answer later.

[If the two out-of-phase dipoles are close, compared to a wavelength, the main surviving radiation follows a quadrupole pattern. For example, if the dipoles are up-down, and located near each other in the east-west direction, you get lobes of radiation going east and west. The dipoles themselves were already not radiating in the up-down directions, and the radiation from the pair cancels in the north-south direction. The radiant power intensity still falls off as 1/r² at a big distance, as required by energy conservation. The far-away power density goes as the square of the separation between the dipoles. /mbw]


There is a video on youtube that shows the effect in two dimensional water waves:
 
Here the 'antennas' are spaced about 3.5 wavelengths apart.  You can easily see the constructive and destructive interference phenomenon.  These antennas are in phase but you will see similar patterns when they are driven out of phase.

LeeH

(published on 12/26/2011)

What happens to the energy when destructive superposition occurs?
- Samiha Nanjiba (age 16)
Dhaka,Bangladesh

I've marked this as a follow-up to a similar question.

Mike W.

(published on 11/11/2012)

Don't know you're still answering these questions, but the first answer quenched one of my questions. So, there's a paper http://arxiv.org/abs/1304.7469 that talks about nested Mach-Zehnder interferometers where the inner interferometer creates destructive interference, but the outer interferometer seemingly undos the destructive interference from the inner one. Is there an explanation for this in classical QM?
- eden
cypress, ca, usa

Hi!

I was at the conference when this paper was first presented, and I really enjoyed the ensuing chaos and confusion. A lot of smart people were thoroughly confused; the results of that experiment are indeed very counterintuitive. 

The author's interpretation of his experiment is that photons travel on discontinuous trajectories; that they can appear inside a region of space without ever entering or leaving that region.

Your interpretation is that destructive interference can be undone at a later time or position. 

Both of these interpretations seem to be supported by the experimental evidence, but they are radical departures from standard classical or even quantum theory. As such, they should only be accepted if there is no simpler explanation within our conventional framework. 

It turns out that this experiment can be understood completely using basic quantum mechanics. In fact, it can even be understood using completely classical electromagnetism! So the two interpretations above are unjustified and unnecessary.

Here's a sketch of the proper understanding in quantum mechanics. The reason this understanding wasn't immediately obvious is that the experiment cleverly wrapped several separate tricks into one cool and sneaky demonstration.

The experiment uses a small interferometer inside a larger interferometer. The inner interferometer has the cool physics, an effect called "weak value amplification." Basically, a clever arrangement of interference effects allows for any small changes made at certain mirrors to be amplified, while small changes made at other mirrors are not amplified. Specifically, the vibrations at A and B were designed to be amplified, while those at E and F were not. This effect is well understood even at a classical level; e.g. http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.66.1107
or http://arxiv.org/abs/0906.4832

Meanwhile, the outer interferometer is simply a "local oscillator"; basically, it is a strong field which interferes with any small fields coming from the inner interferometer, amplifying them until they are easy to measure. This too is a well-understood interference effect.

To put the entire picture in terms of interfering waves, it is important to realize that small amplitudes of waves are leaking through the interferometers, even when they are aligned for ALMOST destructive interference. These small bits of amplitude pick up tiny vibrations from all of the mirrors. The tiny vibrations from A and B are amplified by weak interference, and then the tiny vibrations from A, B, E, and F are all amplified by interference with the strong wave from the outer interferometer. Because vibrations from A and B get amplified twice, they are visible on the detector; E and F remain below the experimental noise.

The sneakiest part of this experiment is that the authors claim no vibrations from mirrors E and F were observed. In fact, such vibrations should have been detected, albeit at a much smaller level than those from mirrors A and B. Once you realize this, it is clear that tiny amplitudes of the wave ARE in fact interfering constructively, and it is those amplitudes that later are registered on your detector in an amplified fashion. There is NOT any "undoing of the destructive interference", nor are the photons traveling in discontinuous paths.

All of these statements agree with quantitative calculations and are equally true in quantum and classical wave mechanics. As beautiful as the experiment is, it doesn't force one to drastically change one's worldview, and it certainly doesn't challenge standard quantum mechanics.

Hope that helps!

David Schmid

(published on 12/03/2015)

I have a few problems with the answer to the question where does the energy go when two waves are added. This answers the problem beautifuly if the waves arn’t completely out of phase or the waves are different but what happen if two identical waves are added when compleetly out of phase, then logicaly you get no wave at all. So where does all the energy go?
- ben (age 17)

Nice question. How can you make the two waves be out of phase everywhere? Their sources would have to have just the same spatial distribution, so the waves had just the same shape. Let's make an approximation to that.

Say you have two speakers facing each other, hooked up out of phase to the same amplifier. (That can easily happen by accident.) Then the total amount of sound energy emitted (at least for long wavelengths) will be much less than either one would emit separately, for a given speaker displacement. However, the speakers would be doing much less work on the air than they would out in the open. In that case, the work done by the amplifiers on the speakers would also be reduced. So the energy flowing in is actually much less than would be found if the speakers were far apart.

So in this case, because of the effect of the waves from each source on the other source, less work is done than would be done on the separate sources. If the sources are moved far enough apart to avoid effects like that, then our previous argument about how the waves interfere destructively in some places but constructively in others becomes applicable.

Mike W.

(published on 10/22/2007)