# Destructive interference - where does the energy go?

Discussion in 'Physics & Math' started by James R, May 11, 2011.

1. ### przyksquishyValued Senior Member

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I was actually puzzling over what happened for two counterpropagating waves in the electromagnetic case even before RJ asked. The reason is that, as Alphanumeric pointed out, the energy density associated with the electromagnetic field is $\sim E^{2} + B^{2}$. It only depends on the magnitudes of the electric and magnetic components and not on their time derivatives, so the resolutions you've been given for two counterpropagating waves on a rope don't apply here. In this case the only way energy can be conserved is if the electromagnetic field never gets cancelled out.

3. ### RJBeeryNatural PhilosopherValued Senior Member

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Przyk (or anyone), do you have a response to the above? I understand that the RHR is just a convention of sorts, but is there a way to construct an anti-wave which is left-handed wave to cancel out both components of an EM wave?

Also, not directly relevant, but do circularly polarized light waves necessitate two and only two superposed linear waves, and are they mathematically required to be 90 degrees out of phase?

5. ### arfa branecall me arfValued Senior Member

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What happens when a plane electromagnetic wave is reflected back onto itself?
Does the phase angle between the E and B fields change?

7. ### przyksquishyValued Senior Member

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I've never heard of this, but I'd expect that if you (somehow) managed to get left-handed waves, they'd be left-handed travelling both ways and you'd get the same result: either the electric or the magnetic components can cancel, but not both.

No, it's just the simplest way of representing a circularly polarised wave in terms of linear waves. The easy way of seeing this expression is not unique is that you can always take any linear wave $\bar{E}(\bar{x};\,t)$ and decompose it for example as:
$\bar{E}(\bar{x};\,t) \,=\, \frac{1}{2} \bar{E}(\bar{x};\,t) \,+\, \frac{1}{2} \bar{E}(\bar{x};\,t) \,.$​
Apply this trivial decomposition to the two linear waves in the diagram I borrowed off Wikipedia, and you've expressed the circular wave as the sum of four linear waves. It's also possible to construct less trivial examples where you build a linear wave out of two linear waves in phase but with different polarisations, or linear waves with the same polarisation but different phases.

If the two linear waves are orthogonal (and of the same amplitude) then yes, otherwise you get an ellipse instead of a circle.

8. ### RJBeeryNatural PhilosopherValued Senior Member

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Przyk, my ignorance of the field is showing. I was taking what you were saying to mean that ALL light behaves the same way, with a particular handedness as being intrinsic and uniform. Your answer above suggests otherwise, so after some digging it appears that left- or right-handed circular polarized light is common. I found this excerpt interesting
I'm just reading on dextrorotation and chirality, for example. Point being I can't see a reason why we wouldn't be able to have two intersecting EM waves that cancel each other out perfectly, which again raises a variant on James R's OP question.

9. ### Me-Ki-GalBannedBanned

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I don't know . I can tell you this . When we play 2 notes and there is resonance the sound sustains and other intervals on other strings adjacent will start to vibrate of there own accord . If they don't resonate you will get what I call drop off in sound . A dampening effect

10. ### RJBeeryNatural PhilosopherValued Senior Member

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That's a good analogy; what happens to the energy in a sound wave if it contacts a perfectly destructively interfering "anti" soundwave? It seems to be the same question, and the effects of sound being cancelled has to be a well established field of study.

11. ### quadraphonicsBloodthirsty BarbarianValued Senior Member

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They key to that, as in the string case, is to realize that such can only happen at a single point (or some countable collection thereof), and not over the entire sound field in question. If you think on it for a second, you'll see that the only way to cancel out the entire sound field of one source with another source, is for the second source to be exactly the negative of the first, and originate in exactly the same place. Which is a complicated way of saying "there is no sound source in the first place" and so no energy to account for.

If we take the case of active noise-cancellation headphone, note that the goal is only to cancel out the (ambient) sound field in one specific place - right where it meets the ear. In this case, the energy of the ambient sound field is still present everywhere else, as is the energy of the cancellation source, and the total energy in the sound field is the sum of both sources. It's just been carefully arranged so that there is less energy localized in a particular place (at your ear), and more elsewhere (propagating out the back of your headphones, for example). If someone were to place their ear at a different spot than is intended, they'd hear the sum of the "ambient" and "cancellation" source energies, rather then the difference between them.

12. ### przyksquishyValued Senior Member

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Depends on what we're talking about. There's nothing constraining the polarisation of light, so you can have linear or levorotary or dextrorotary circular polarisation or anything in between. But that's just a statement about what the electric field looks like. Once you specify a certain wave in the electric field, circular or planar or otherwise, the orientation of the magnetic field is fixed by a right hand rule.

If you're familiar with vector products, for a propagating wave Maxwell's equations require that $\bar{B} = \frac{1}{c} \hat{k} \times \bar{E}$, where $\hat{k}$ is a unit vector pointing in the direction of propagation of the EM wave. Maxwell's equations don't allow $\bar{B} = - \frac{1}{c} \hat{k} \times \bar{E}$. If you take an electromagnetic wave and consider a similar one only propagating in the opposite direction, then $\hat{k}$ changes sign, so the magnetic field gains an extra sign change compared to the electric field. If you arrange things such that the electric fields cancel out, then the magnetic fields add, and vice-versa. You can't cancel both simultaneously with two counterpropagating electromagnetic waves, no matter the type of polarisation.

To be exhaustive it's due to a convention that $\bar{B} = + \frac{1}{c} \hat{k} \times \bar{E}$ and not $\bar{B} = - \frac{1}{c} \hat{k} \times \bar{E}$, but it's not something you can tamper with for individual waves. It follows from a convention set for the whole of electrodynamics by the right hand rule in the Lorentz force law. You could make it a left hand rule instead and replace $\bar{B}$ with $-\bar{B}$ everywhere in electrodynamics, but this wouldn't let you cancel out both the electric and magnetic parts of two counterpropagating electromagnetic waves.

13. ### RJBeeryNatural PhilosopherValued Senior Member

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OK my initial research suggests that a classically physical process of "perfectly destructive" wave interference (such as waves in a rope, sound, etc) produces heat in the medium, while an analogous effect of "perfectly destructive" wave interference with light DOES eliminate the wave, but with the effect of creating a higher energy wave elsewhere in the system...
This answers the questions about using reflection to have an EM wave interfere with itself (for the most part), but not the question about two perfectly destructive lasers being directed towards one another. Maybe it isn't possible for some reason?

14. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I came late to thread, but started at the OP. Here is what I found:

Farsight later guessed incorrectly:
His text, now bold, is however 100% correct. But he must have missed Pete’s earlier post as Pete answered that correctly:
but was not thinking when he replied:
To:
As, I see it, the illustration correctly shows sum wave with 1.4142 times amplitude of either wave. (The energy of a wave is proportional to the square of the amplitude.)

I only skimmed the remaining post of thread, but as James R summarized them in post 46 (plus earlier correct posts especially by Pete) they seemed to be OK.

Last edited by a moderator: May 16, 2011
15. ### RJBeeryNatural PhilosopherValued Senior Member

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Przyk, your follow-up explanation confirms my original understanding of your earlier post (before I confused myself), thanks for taking the time. Can you explain how we can cancel out complete paths of laser light from the two-armed interferometer mentioned in my last post without generating the double-magnitude magnetic field that you were talking about? Or am I being presumptuous that the magnetic field does not exist there?

16. ### James RJust this guy, you know?Staff Member

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In the case of a linearly polarised wave, the answer is no. The phase of the entire wave (i.e. both E and B components) does shift on reflection, by 180 degrees if the reflection is at normal incidence. But the E and B components shift together.

In the interferometer, when the E fields cancel out so do the B fields, since the two field components are in phase with one another. In one exit aperture of the interferometer, you end up with zero amplitude; at the other exit you get double the amplitude.

17. ### arfa branecall me arfValued Senior Member

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When you divide a total by its mean, what do you have, statistically?

18. ### James RJust this guy, you know?Staff Member

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Nothing significant.

For example, take the numbers 1, 2, 3, 4, 5. The mean is 3. The total is 15. Divide 15 by 3 and you get 5. What's the significance of 5?

19. ### TachBannedBanned

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No, if you ever ran the experiment you would have known that you get interference fringes (alternating zero amplitude and maximum amplitude) at the same aperture. This is the very basis of this class of experiments.

Last edited: May 17, 2011
20. ### James RJust this guy, you know?Staff Member

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No you don't. If you adjust the lengths of the interferometer arms so that the waves intefere completely destructively at one output, then they'll interfere completely constructively at the other output. Like I said.

By the way, here's a picture of what we're talking about: the Mach-Zehnder interferometer.

21. ### TachBannedBanned

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You said wrong. The interferometer arms have nothing to do with this, you get fringe patterns whether the arms are exactly equal (as in MMX) or unequal (as in KTX). The reason is that there are multiple length light paths no matter what, this is how you get the fringes.

22. ### TachBannedBanned

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Mach-Zander is a 4 arm interferometer, first used by Hammar (who got the idea from Fizeau), RJBerry asked about a 2 arm interferometer as in MMX/KTX.

23. ### James RJust this guy, you know?Staff Member

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Tach:

Where are these multiple-length light paths you speak of? Got a diagram?

Oh, and a Michelson-Morley interferometer is not really different from a Mach-Zehnder. It just has the light paths folded back on themselves. The basic theory is the same.