Destructive Wave Interference

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  • #26
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)...
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?)
 
  • #27
Dale
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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.
 
  • #28
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?
 
  • #29
Dale
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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.
 
  • #30
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
 
  • #31
Dale
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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.
 
  • #32
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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.
 
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  • #33
[cool un-allowed theory was here]
 
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  • #34
Dale
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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.
 
  • #35
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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.
 
  • #36
Claude Bile
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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.
 
  • #37
Claude Bile
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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.
 
  • #38
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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
 
  • #39
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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?
 
  • #40
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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.
 

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