Destructive Wave Interference

[waynet]

I was wondering about this one day and perhaps someone here can supply an explanation. Let’s assume you can get two waves on the same frequency and amplitude perfectly aligned with each other but on opposite phase so they cancel each other out, which from what I understand is called destructive wave interference.

Since the energy from the two waves is canceled out, it can't be measured or used, correct? What happens to that energy? Doesn't that violate the law of conservation of energy?

[pallidin]

Now that's a hell of a good question.
Destructive interferance of sound waves seems to imply that one could have an "action" (generation of sound waves) and a cancelled "reaction" (due to the interferance).
I could think of a few novel devices if this is true. Surely it must not be that simple.
Curious... perhaps some experts can jump in on this to explain.

[himanshu121]

Energy is redistributed , there is no loss of energy and thus there is no violation of law of conservation of energy

[KLscilevothma]

Originally posted by himanshu121
Energy is redistributed Could you elaborate that please. Energy is redistributed to where ?

[Moni]

Hmm...let me say it...

Imangine two of your friends are trying to pull something...but not from the same side...you two are just 180 deg to each other...and both of you have same strength...then you can't move the thing...can you say...that you aren't loosing energy [;)]

[waynet]

Originally posted by Moni
Imangine two of your friends are trying to pull something...but not from the same side...you two are just 180 deg to each other...and both of you have same strength...then you can't move the thing...can you say...that you aren't loosing energy [;)]
You aren't loosing energy that way, it is changed to heat energy and also you are probably streching the object you are pulling, so it is redistributed there. What I don't understand is where the energy goes with the wave interference.

[himanshu121]

In interference phenomenon there is constructive as well as destructive interfernce. Where in Constructive interference the energy adds up and in destructive interference it subtracts which makes the energy before and after interfernce same

[waynet]

So you are saying that it is impossible to have destructive interference without constructive interference? Is there proof of this somewhere?

[himanshu121]

No. Basically if it(only destructive) would happen then conservation of energy will fail

[KLscilevothma]

I still don't get it. Consider the following picture (see attatchment), when the two pulses meet, only cancellation will occur. So where will be the energy re-distributed to ?

However I understand that the energy at points of cancellation is redistributed to points of reinforcement if the interference pattern contains both destructive and constructive interferences.

[waynet]

At the point the two will meet they will cancel each other out, but the waves will pop back out and continue on, so no energy is lost. My question was if they were going in the same direction and they lined up perfectly.

I would imagine it would be hard to get two waves to do that, but I think it could be possible with light waves and using things like lasers, mirrors and prisms.

[Hydr0matic]

To create destructive interference basically what you need is an interferometer. Here's a typical example of such a setup:
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/MachZehnder/MachZehnder.html

But I don't get why the author claims all light ends up at detector 1, it sounds very contradictory. I mean, if there is no light heading towards detector 2, then which beams are interfering destructively ? .. Do all beams choose to go to D1 because they know in advance that they will interfere destructively towards D2 ? Some quantum weirdness at work here ?

Even though I think the above example doesn't make sense I've constructed another setup yielding destructive interference at both detectors... See bottom picture:
http://hydr0matic.insector.se/fysik/interferometer.jpg

If these dubble-glassed beamsplitters could be produced I believe one could make all light in an interferometer interfere destructively.

[waynet]

Maybe he meant that all visible light ends up at detector 1. Common sense would make you think that it would only be half the intensity, but I guess sometimes stuff like this seems to defy common sense.

Looking at your diagram, you may have to change that first double glassed mirror to a regular beam splitter, because that other page says that the inner surface doesn't cause a phase change. But wouldn't that cause a problem with it lining up correctly?

[Hydr0matic]

Originally posted by waynet
Maybe he meant that all visible light ends up at detector 1. Common sense would make you think that it would only be half the intensity, but I guess sometimes stuff like this seems to defy common sense. Can any experimental results be found on this ? It must have been tested ...

Looking at your diagram, you may have to change that first double glassed mirror to a regular beam splitter, because that other page says that the inner surface doesn't cause a phase change. But wouldn't that cause a problem with it lining up correctly? Why would I have to change it ? Let's analyse ..

T = Transmitted beam
R = Reflected beam
C = some constant phase change due to refraction in glass.

TT: 4*C
RR: 4*C + 2*\frac{\lambda}{2}

TR: 4*C
RT: 4*C + 2*\frac{\lambda}{2}

The situation is the same at both ends due to the dubble-glassed beamsplitters. The numbers above will give constructive interference at both ends, but if you add another phase shift of one-half a wavelength due to distance traveled at the RX paths, there will be destructive interference at both ends.

[Thallium]

Forgive me for posting this stupid question: Why can two waves on the same frequency and applitude cancel eachother out?

[waynet]

Can any experimental results be found on this ? It must have been tested ...Probably, but I wouldn't know where to look for it, I'm not a physicist.
but if you add another phase shift of one-half a wavelength due to distance traveled at the RX paths, there will be destructive interference at both ends.That is why I suggested the change. I guess it could be done several ways.

So, you are going to try building such a device?

[waynet]

Originally posted by Thallium
Forgive me for posting this stupid question: Why can two waves on the same frequency and applitude cancel eachother out? I found this page that shows how you can have waves cancel each other out:
http://www.colorado.edu/physics/2000/applets/fourier.html

One time I've experienced this is when a friend of mine didn't wire his car stereo correctly, so that both front car door speakers were wired on the same channel, but on opposite phase. (one speaker wired backwards) He had both car doors open and turned up the stereo really loud. If you stood somewhere on a line directly behind the car, you couldn't hear the stereo, except for the echo bouncing off the hills in the distance.

[Hydr0matic]

Originally posted by waynet
I guess it could be done several ways.Perhaps, I'm not sure. I tried to figure out a setup using single-glassed splitters but I couldn't find one. I think it's essential for the setup that the beams experience equal refraction shifts.

So, you are going to try building such a device?If I sometime in the distant future get access to the resources, yes. I'm not exactly a physicist either you know [;)]

[edwardone]

Hi i m looking for the software/shareware that let me to do destructive sound interference.Where can i find it?
Thanks

[jdavel]

waynet asked: "What happens to that energy?"

What energy?

[Kurdt]

I believe the electric fied vector is zero and the magnetic field vector is zero thus we have nothing to detect the energy by in terms of 'seeing' but in terms of E and B field vetors they are still oscillating with a definiet frequncy and therefore definite energy. Just so happens the vector add to zero when 180 degrees out of phase.

[JerryMac]

Sorry to dig this thread back up but I have been thinking about this same issue a lot. I wanted to question something further.

The pulse example he gave was something I thought about in particular myself. What confused me is once the waves cancel each other out, what attributes of the system allow them to "know" where to continue after time T where they cancel.

In other words if we could freeze a moment in time where they are cancelled, what can we observe to tell us that the two pulses' magnitudes would return and continue on as before? How could we tell the difference between this two pulse system and a system where there were never any pulses at all?

Another way a look at it... even if there was one pulse and I froze time, how could I tell if the pulse was moving in a positive or negative direction along the X axis. If I can't tell, how can the Universe?

Is this all really a version of the Uncertainty Principal?

[DaleSpam]

Are you interested in EM waves or sound waves, and what is the geometry of the pulses? I know more about EM than sound, and you never get any conservation of energy issues in EM.

[JerryMac]

EM are what interests me. I've been trying to grasp things at that level and even further with things like the double slit experiment, especially how a single electron wave interfers with itself. Or maybe even deeper with how fundamental particles interact and move at a quantum level.

[DaleSpam]

With EM waves you have to remember that there is energy in both the E-field and the B-field. So whenever you have destructive interference in one you have constructive interference in the other such that the total energy is conserved.

For the rest of what you describe you really need to learn the details of QM, specifically QED.

[carrotstien]

Actually, DaleSpam...this isn't always true. What you speak of is true only in oppositely moving waves. However, when they move in the same direction, a flipped E-field results in a flipped B-field, so that 180-degree shifted waves traveling in the same direction would cancel each other out.

as such (with theoretically example)...
http://i50.tinypic.com/1zl9opl.jpg

This is hard to do, but can be done..sort of as shown in the picture (in theory)
The lasers are the same frequency - the color is just used to differentiate them.

Let mirror 1 be a perfect mirror which is at 45 degrees from the red laser. (again, color is just for the diagram)

Let Mirror B be semi transparent (50:50)

As far as I remember, reflected beams get a 180 degree shift.

So, imagine that, at point B (right before reflection), both beams are in phase.

Vertically at B: blue beam goes through, red beam get reflect + shifted..results, 2 lasers with 180 degree offset with equal power...total destructive interference.

Horizontally at B: Red beam goes through, blue beam gets reflected + 180 degree shift...results, 2 lasers with 180 degree offset with equal power...total destructive interference.



Now, I have heard some arguement a while back that in such a case, with such geometry..the light wouldn't go through, or get reflected - it would actually become completely absorbed in the mirror(s) as heat. In terms of conservation of energy, i can believe that (if we assume that the two 180-degree offset same-direction beams don't have any energy), that the energy must stay in the mirrors. However, in terms of intuition, something is wrong. If the energy stays in the mirrors during the perfect geometrical set up...then moving mirror2 up by have of a wavelength would then let the light go through - and all of a sudden the mirrors aren't horrible by absorbing the energy.

I think, a much easier explanation is that it can theoretically (not counting for the difficulty in creating perfect mirrors or lining up everything perfectly), be that coupled off-set like-directional em-waves can exist...they contain energy, but cannot lose in any classical manner. I like this explanation more...not because it's right or wrong, but because it introduces a bunch of interesting concepts like something containing bound energy - an energy which can't normally leave (reminds u of matter by any chance?)

[DaleSpam]

Actually, DaleSpam...this isn't always true. ...
Let mirror 1 be a perfect mirror which is at 45 degrees from the red laser. (again, color is just for the diagram)

Let Mirror B be semi transparent (50:50)...

in such a case, with such geometry..the light wouldn't go through, or get reflected - it would actually become completely absorbed in the mirror(s) as heat. In terms of conservation of energy, i can believe that (if we assume that the two 180-degree offset same-direction beams don't have any energy), that the energy must stay in the mirrors. You are correct. What I described applies specifically to EM waves in free space. The kind of interference you mention can only happen in the presence of matter so that there is something for the fields to do work on.

[carrotstien]

and by matter u mean the mirrors?

i guess...i mean, theoretically you can use a gravity lens to a similar effect.

But still, does the resultant propagating wave contain energy or not?

[DaleSpam]

But still, does the resultant propagating wave contain energy or not?If you have total destructive interference then there is no propagating wave and all of the energy goes into the mirror. If the destructive interference is partial then you have part of the energy in the resultant wave and the rest in the mirror.

By the way, above you expressed disbelief that moving the mirror a quarter wavelength could change the result so dramatically, but this kind of thing happens all of the time for antenna design and for resonant cavity design. The only difference is that you are dealing with shorter wavelengths.

[carrotstien]

Agreed...I was thinking about the case that my professor told me about. Something that looked like light going from some source through A, and then destructively interfering at point B - and I remember him telling me that in such a case, light would never even leave point A. But I may be mistaken.

What about if you use non-matter means of bending the light? I haven't figured out a way, or a proof, but i'm pretty sure you could use a gravity mirror (such as a black hole) in such a way to get the two beams moving on top of each other - without any em-wave to matter to em-wave events

[DaleSpam]

My GR knowledge is not sufficient to answer. I doubt it is possible in a static spacetime, and in a non-static spacetime energy is notoriously hard to even define, let alone conserve.

[|squeezed>]

Recall the Michelson interferometer:

.....(c)........(d)

...................BS2
-->..\...-->....\.....| (a)
......BS1....... __
....................(b)

(a),(b) : mirrors

For convenience i placed an extra beam splitter (BS1) before the beam enters the interferometer. In this way you can see the overlap from the mirrors' back-reflections at (c). When you have a bright fringe (constructive interference) at (d) you get a dark fringe at (c) (destructive interference) and vice versa. You cannot have a dark fringe in both arms.

Hope that helps.

[carrotstien]

[cool un-allowed theory was here]

[DaleSpam]

First, what you are describing is called a dipole, and it is well understood. The field is non-zero always, but it decays faster than a monopole field.

Second, this forum is not the appropriate place for personal theories. Please re-read the rules that you agreed to when you signed up.

[hangover]

I was wondering about this one day and perhaps someone here can supply an explanation. Let’s assume you can get two waves on the same frequency and amplitude perfectly aligned with each other but on opposite phase so they cancel each other out, which from what I understand is called destructive wave interference.

Since the energy from the two waves is canceled out, it can't be measured or used, correct? What happens to that energy? Doesn't that violate the law of conservation of energy?

sorry for posting my opinion. To my knowledge(General Physics),"energy" is not cancel out in destructive interference, is it?
In thin film interference(2 wave travelling in the same direction), the energy is redistributed to the wave involving in constructive interference.
In the case of 2 waves travelling in opposite direction, destructive interference occurs; when crest of 1 wave meets the the trough of another wave, the wave seems to be disappeared. However, when this process finishes, wave appears again and the 2 wave keeps on moving. Therefore, according to what i learnt, energy is conserved and, in my opinion, energy is stored in the moment of cancellation.
Sorry, it is only my explanation and may be misleading.

[Claude Bile]

Sorry to dig this thread back up but I have been thinking about this same issue a lot. I wanted to question something further.

The pulse example he gave was something I thought about in particular myself. What confused me is once the waves cancel each other out, what attributes of the system allow them to "know" where to continue after time T where they cancel.

Consider a standing wave on a string; at some time, then the displacement will be zero everywhere. Where has the energy gone? Nowhere, because while the displacement is zero, the velocity is not. You can make a similar argument for EM waves; while the E field may be zero at some time, dE/dt and d^2E/dt^2 are not zero, which is important. (Keep in mind too, that quantities like energy density and Poynting vector are time-averaged, which is why time derivatives do not appear in these expressions).

In other words if we could freeze a moment in time where they are cancelled, what can we observe to tell us that the two pulses' magnitudes would return and continue on as before? How could we tell the difference between this two pulse system and a system where there were never any pulses at all?

Another way a look at it... even if there was one pulse and I froze time, how could I tell if the pulse was moving in a positive or negative direction along the X axis. If I can't tell, how can the Universe?

The momentum of the wave will determine the direction that it propagates.

Is this all really a version of the Uncertainty Principal?

Nope, you don't need to resort to quantum physics!

Claude.

[Claude Bile]

sorry for posting my opinion. To my knowledge(General Physics),"energy" is not cancel out in destructive interference, is it?
In thin film interference(2 wave travelling in the same direction), the energy is redistributed to the wave involving in constructive interference.
In the case of 2 waves travelling in opposite direction, destructive interference occurs; when crest of 1 wave meets the the trough of another wave, the wave seems to be disappeared. However, when this process finishes, wave appears again and the 2 wave keeps on moving. Therefore, according to what i learnt, energy is conserved and, in my opinion, energy is stored in the moment of cancellation.
Sorry, it is only my explanation and may be misleading.

To add to this;

In the instance of thin-films, you can either suppress transmission or reflection but never both. Energy (and momentum) is always conserved.

Claude.

[Darmog]

Palladin would be interested to learn that I have noise cancelling headphones that work rather well to make listening to music much better in car, train, and plane. The background roar and rumble is much reduced. Air pilots use a similar article. I presume the energy is mostly absorbed in the shells and partly dissipated in all directions.
Darmog

[petelewis]

Palladin would be interested to learn that I have noise cancelling headphones that work rather well to make listening to music much better in car, train, and plane. The background roar and rumble is much reduced. Air pilots use a similar article. I presume the energy is mostly absorbed in the shells and partly dissipated in all directions.
Darmog

I'd be interested to hear more about this from someone in the know. I'm considering buying noise-cancelling headphones, but it seems to me that they would actually be sending more sound energy into my ears, even though they would sound quieter. Am I right?

Are there health implications of this?

[Darmog]

from Darmog to Petelewis and other interested in noise cancelling earphones.
I'm sure manufacturers of these phones would be pleased to reassure you that there is no extra energy entering the ears and may offer to send you a set on approval.