Does Destructive Interference Cancel Energy?

  • #1
Does it cancel it, or just create an equilibrium of forces upon a point(s)?

Example: If you had a string and you sent a wave down one end, and the opposite wave from the other, once they meet there will be an instant when the string is flat as if there had never been waves introduced. But, if you were to analyze the fibers and molecules of the string at the point of the interference, would you find that they were under more "stress" (or just more energized) than a string without any waves in it, or does the energy actually cancel at that moment leaving the string stress/energy free and just as it was before the waves were presented?
 

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  • #2
The question of stresses on the fibres is better framed in terms of forces which do in fact cancel.
 
  • #3
The question of stresses on the fibres is better framed in terms of forces which do in fact cancel.

Ok, so you're saying that the string wouldn't have and forces/stress/whatever at that moment, and the fibers/molecules would basically be like they were before any waves were present?
 
  • #4
For a perfectly uniform string yes, if no then as long as the fibres have not exceeded their elastic limit then yes.
ETA forces are present, the net force is zero.
 
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  • #5
For a perfectly uniform string yes, if no then as long as the fibres have not exceeded their elastic limit then yes.
ETA forces are present, the net force is zero.


Ok. What are ETA forces?
 
  • #6
Note your OP is poorly constructed and leaves a lot of wiggle room. You did not define the phase, frequency and amp.

Google standing wave to see a better question.

ETA, did I just make a pun, wiggle/wave...get it.
 
  • #7
Note your OP is poorly constructed and leaves a lot of wiggle room. You did not define the phase, frequency and amp.

Google standing wave to see a better question.

ETA, did I just make a pun, wiggle/wave...get it.
I'm not talking about standing waves. I just want to know if when two identical waves destructively interfere with each other perfectly, does the energy/force at that moment vanish, or is it just bound up...?
 
  • #8
Two identical waves can never destructively interfere. Tighten up your question.
 
  • #9
Two identical waves can never destructively interfere. Tighten up your question.

Two identical waves 180 degrees out of phase. That better?
 
  • #10
Then there is no energy propagating from the source.
 
  • #11
Would it help to think of a speaker as source of sound and the waves driving the speaker.
 
  • #12
Would it help to think of a speaker as source of sound and the waves driving the speaker.

I don't get the complication you're adding to this. Simple: There is destructive interference. There is total destructive interference. All I want to know is does the energy at that moment vanish or is it bound up in the medium...?
 
  • #14
Bound up??

Yeah, like if two men are pushing against each other (let's imagine sumo wrestlers) with the same force they don't go anywhere. However, there is a lot of pressure and energy being exerted. They are bound up against each other. The force is still there even though they are not moving. In destructive interference, is it like the sumo wrestlers in a sense, or does force actually cancel out at the moment of destruction?
 
  • #15
As previously stated the net force is zero hence the particle feels no interaction.
 
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  • #17
would you find that they were under more "stress" (or just more energized) than a string without any waves in it, or does the energy actually cancel at that moment leaving the string stress/energy free and just as it was before the waves were presented?
The string is moving. It has kinetic energy.

Energy is conserved. In destructive interference energy is always either moved elsewhere or converted into some other form.
 
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  • #18
Your question makes more sense when electromagnetic waves are considered. In quite a number of interference cases, energy is 100% conserved, and whenever there is a cancellation from destructive interference, there are necessarily brighter spots in the energy pattern where constructive interference is occurring. This is the case for both multi-slit interference as well as Fabry-Perot type interference and Michelson interferometer type interference.
 
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  • #19
The string is moving. It has kinetic energy.

Energy is conserved. In destructive interference energy is always either moved elsewhere or converted into some other form.
At last! :smile:
The OP is actually describing of a standing wave, involving two waves, traveling in opposite directions. He is denying it because he isn't coming from that direction in his description but, nonetheless, that's what he is describing. Very often, a standing wave occurs due to reflection at each end. What's being described here is a standing wave that's sustained by two energy sources, one each end. The displacement at each end is kept at zero and energy is fed into the system, building up with no limit (in an ideal case). In practice, the natural losses will impose a maximum amount of energy stored in the standing wave when the power supplied is the power lost to the surroundings. The standing wave is the same as an interference pattern from two sources (only it's in just one dimension). In nodes, the energy density is low (zero), where there is cancellation and maximum in the antinodes where there is enhancement.
 
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  • #20
Something that tends to be ignored about standing waves. You HAVE to acknowledge that Energy must be leaking out of the resonator or it will build up without limit. It's not a problem with 2D Interference patterns because the Energy always goes somewhere (the screen or free space. All resonators lose a fraction of their energy every cycle. Either 1/10, 1/100, 1/1000000 but that is what limits the peak energy.
 
  • #21
At last! :smile:
The OP is actually describing of a standing wave, involving two waves, traveling in opposite directions. He is denying it because he isn't coming from that direction in his description but, nonetheless, that's what he is describing. Very often, a standing wave occurs due to reflection at each end. What's being described here is a standing wave that's sustained by two energy sources, one each end. The displacement at each end is kept at zero and energy is fed into the system, building up with no limit (in an ideal case). In practice, the natural losses will impose a maximum amount of energy stored in the standing wave when the power supplied is the power lost to the surroundings. The standing wave is the same as an interference pattern from two sources (only it's in just one dimension). In nodes, the energy density is low (zero), where there is cancellation and maximum in the antinodes where there is enhancement.

Ok, but I'm really just thinking of a single pulse from each end-- one pulse up, and the other down. When they meet there will be destructive interference. I'm still a bit wondering what's going on in that very instant that they perfectly cancel each other out, leaving just a straight string. It makes more sense now though.
 
  • #22
Ok, but I'm really just thinking of a single pulse from each end-- one pulse up, and the other down. When they meet there will be destructive interference. I'm still a bit wondering what's going on in that very instant that they perfectly cancel each other out, leaving just a straight string. It makes more sense now though.
You had the answer to that question up near the top, I think. There may be no displacement where they cross but the string will still be moving and carrying Kinetic Energy in that particular portion, briefly.
 
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  • #23
once they meet there will be an instant when the string is flat

The string is moving. It has kinetic energy.

Chris, you stated yourself that the string is only flat for an instant. That implies that it never stopped moving. Therefore, the energy does not cancel, and it is expressed through the motion of the string. Analysis of the string's molecules at the moment of destructive interference would reveal that they are indeed under stress due to the fact that they are in motion. The forces are what cancel at that moment.
 
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  • #24
Chris, you stated yourself that the string is only flat for an instant. That implies that it never stopped moving. Therefore, the energy does not cancel, and it is expressed through the motion of the string. Analysis of the string's molecules at the moment of destructive interference would reveal that they are indeed under stress due to the fact that they are in motion. The forces are what cancel at that moment.
Yeah, I get this better now :) Thanks
 
  • #25
There may be no displacement where they cross but the string will still be moving and carrying Kinetic Energy in that particular portion, briefly.

Yes, I get this better now. Thank you
 
  • #26
Chris with respect it may help in future questions to realize physics involves real objects with constraints. Answers are specific to the set up of the system and the constraints.
 
  • #27
Chris with respect it may help in future questions to realize physics involves real objects with constraints. Answers are specific to the set up of the system and the constraints.
Of course.
 
  • #28
Does it cancel it, or just create an equilibrium of forces upon a point(s)?

Example: If you had a string and you sent a wave down one end, and the opposite wave from the other, once they meet there will be an instant when the string is flat as if there had never been waves introduced. But, if you were to analyze the fibers and molecules of the string at the point of the interference, would you find that they were under more "stress" (or just more energized) than a string without any waves in it, or does the energy actually cancel at that moment leaving the string stress/energy free and just as it was before the waves were presented?

Please read this:

https://www.physicsforums.com/threads/energy-and-interference.129649/#post-1069202

Zz.
 
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  • #30
@Chris: It may be a bit late to be introducing this fundamental characteristic of waves, in general but all waves involve the variation in time and space of some form of Potential Energy plus some form of Kinetic energy. Water waves, for instance, involve moving water and water being raised and lowered. Sound waves involve variations in pressure and speed of air flow. EM waves involve Electric fields (Potential) and Magnetic fields (a Kinetic form of energy). The Energy is carried in both forms and can often be 'shared' unequally. Hence your zero displacement and high velocity where your two pulses pass each other. The Energy is still there.
 
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  • #31
@Chris: It may be a bit late to be introducing this fundamental characteristic of waves, in general but all waves involve the variation in time and space of some form of Potential Energy plus some form of Kinetic energy. Water waves, for instance, involve moving water and water being raised and lowered. Sound waves involve variations in pressure and speed of air flow. EM waves involve Electric fields (Potential) and Magnetic fields (a Kinetic form of energy). The Energy is carried in both forms and can often be 'shared' unequally. Hence your zero displacement and high velocity where your two pulses pass each other. The Energy is still there.

That's interesting! Cool.
 
  • #32
An interesting case can occur with electromagnetic waves where the destructive interference between two waves occurs for an entire region, e.g. on a Michelson interferometer or a beamsplitter with two separate coherent sources. The destructive interference can occur over the entire direction so that no energy goes to that side, and all the energy winds up on the other side. By adjusting the phase of one of the sources (by ## \pi ##), all of the energy can be made to go to the side that originally had zero energy. Here we have complete wave cancellation (from two sources going in the same direction that are ## \pi ## out of phase with each other) without any energy passing into the region at all. Unlike the case of the waves on a string, where energy is necessarily passing through the point, in this case there simply is no energy present because of the destructive interference. The energy occurs in the region where the waves have constructive interference, where the sources are in phase with each other. ## \\ ## (Additional detail: For the case of two coherent sources incident on a beamsplitter, it is an asymmetric problem because one face of the beamsplitter typically has an AR (anti-reflection coating). All of the partial reflection/partial transmission occurs off of only one face. The other face has 100% transmission. The reflection incident from the air side gets an extra ## \pi ## phase change because it is off of a more dense medium. Essentially, the two coherent beams can be combined into one by such a configuration. In one direction off of the beamsplitter, the partially transmitted wave and the other partially reflected wave are made to be ## \pi ## out of phase. In the other direction they will be in phase. For a 50-50 energy split (R=1/2), the fresnel reflection coefficient ## \rho=+/- 1/\sqrt{2} ##.) ## \\ ## One additional item: For a single source, the energy gets split 50-50 between the two directions from the beamsplitter. When the second source is turned on, coherent effects occur and there is no longer a 50-50 energy split for each source. Complete wave cancellation can occur in one direction with 100% of the energy going to the other direction. It is wave cancellation, but could not be called energy cancellation. (There's no such thing as negative energy in a wave.) The problem is a very simple linear one for the wave from each source. The resulting waves from each source are superimposed and then the energy is computed.
 
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  • #33
Charles, That's some revealing data. Thanks...

I get the essence of all that, however I don't quite get some of the finer points you made:

(Additional detail: For the case of two coherent sources incident on a beamsplitter, it is an asymmetric problem because one face of the beamsplitter typically has an AR (anti-reflection coating). All of the partial reflection/partial transmission occurs off of only one face. The other face has 100% transmission. The reflection incident from the air side gets an extra π \pi phase change because it is off of a more dense medium.

Perhaps a diagram would be helpful here so I could see the arrangement of this device. Wouldn't all beams experience the same amount of phase change if they are all interacting with the same medium in the splitter?
 
  • #34
Charles, That's some revealing data. Thanks...

I get the essence of all that, however I don't quite get some of the finer points you made:



Perhaps a diagram would be helpful here so I could see the arrangement of this device. Wouldn't all beams experience the same amount of phase change if they are all interacting with the same medium in the splitter?
The beamsplitter is at a 45 degree angle. (slope=+1) One beam comes from the top of the paper. One beam from the right. Two receivers: One on the left and the second at the bottom. The beams are mutually coherent plane waves. One finer point you might have missed is mentioned in the details of the previous post that one of the reflections gets a ## \pi ## phase shift, which is essentially a minus sign. This can be seen at normal incidence where ## \rho=(n_1-n_2)/(n_1+n_2) ## (n's are indices of refraction). ## \rho ## from one direction is positive and from the other it is negative. ## \rho=E_r/E_i ## is the fresnel reflection coefficient for the electric field amplitude. The negative ## \rho ## (the minus sign ) is the same as a ## \pi ## phase shift. The same thing happens at 45 degree angle of incidence (the minus sign for ## n_2>n_1 ##), that happens at normal incidence,although the expressions for ## \rho ## become more complex (at 45 degrees and at any other angles of incidence). The materials can be chosen to make the energy reflection coefficient ## R=1/2 ## at 45 degrees, so that ## \rho=+1/\sqrt{2} ## or ## \rho=-1/\sqrt{2} ##. It turns out, the composite transmission coefficient ##\tau=E_t/E_i=1/\sqrt{2} ## for both beams (when ## R=1/2 ## ) so that you can readily compute the emerging E fields. The two sources can be assumed to have some arbitrary phase difference ## \phi ## , although the calculation is simplest if they are either assumed to be in phase with each other or ## \pi ## out of phase. In one case all the energy goes to one receiver, and in the other case it all goes to the other receiver. You can assume the top surface has the AR coating (beamsplitter is finite width), so that the reflection of the beam incident from the right will have an extra ## \pi ## phase change. When the two sources start out in phase, this makes them destructively interfere on the path to the bottom receiver, so that all the energy goes to the left where the two sources are in phase. One thing I should add is to compute the intensity (power) ## I=n E^2 ## (i.e. it is proportional to ## n E^2 ##. (In optics, the calculations for intensity are often done in this fashion without the additional constants that are in the Poynting vector for power. And n=1 for air.) You can write the amplitudes for the different wave components and compute the energy at the two receivers. e.g. for the bottom receiver ## E_{Bottom}=(1/\sqrt{2}-1/\sqrt{2})E_o=0 ##(one component is transmitted from the top and the other reflected from the right). For the left receiver ## E_{Left}=(1/\sqrt{2}+1/\sqrt{2})E_0=\sqrt{2}E_0 ## (one component is transmitted from the right and the other reflected from the top) and we see our final intensity at the left receiver ## I_{Left}=(\sqrt{2}E_0)^2=2 E_0^2 ## which is the sum of the intensities of the two initial beams ## I_0=E_o^2 ## for each. (Note for the above energy reflection coefficient ## R=I_r/I_i=E_r^2/E_i^2=\rho^2 ##. Notice also when there are two coherent sources, the energy reflection coefficients ## R ## can not be used to compute the resulting energy distribution (this system is not linear in its energy properties, but is completely linear in regard to the E fields), but the ## \rho ## 's computed from the ## R ## are completely valid for doing the computation.) @Chris Frisella It's quite a lot of detail, but hopefully you can at least follow some of it.
 
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  • #35
The beamsplitter is at a 45 degree angle. (slope=+1) One beam comes from the top of the paper. One beam from the right. Two receivers: One on the left and the second at the bottom. The beams are mutually coherent plane waves. One finer point you might have missed is mentioned in the details of the previous post that one of the reflections gets a ## \pi ## phase shift, which is essentially a minus sign. This can be seen at normal incidence where ## \rho=(n_1-n_2)/(n_1+n_2) ## (n's are indices of refraction). ## \rho ## from one direction is positive and from the other it is negative. ## \rho=E_r/E_i ## is the fresnel reflection coefficient for the electric field amplitude. The negative ## \rho ## (the minus sign ) is the same as a ## \pi ## phase shift. The same thing happens at 45 degree angle of incidence (the minus sign for ## n_2>n_1 ##), that happens at normal incidence,although the expressions for ## \rho ## become more complex (at 45 degrees and at any other angles of incidence). The materials can be chosen to make the energy reflection coefficient ## R=1/2 ## at 45 degrees, so that ## \rho=+1/\sqrt{2} ## or ## \rho=-1/\sqrt{2} ##. It turns out, the composite transmission coefficient ##\tau=E_t/E_i=1/\sqrt{2} ## for both beams (when ## R=1/2 ## ) so that you can readily compute the emerging E fields. The two sources can be assumed to have some arbitrary phase difference ## \phi ## , although the calculation is simplest if they are either assumed to be in phase with each other or ## \pi ## out of phase. In one case all the energy goes to one receiver, and in the other case it all goes to the other receiver. You can assume the top surface has the AR coating (beamsplitter is finite width), so that the reflection of the beam incident from the right will have an extra ## \pi ## phase change. When the two sources start out in phase, this makes them destructively interfere on the path to the bottom receiver, so that all the energy goes to the left where the two sources are in phase. One thing I should add is to compute the intensity (power) ## I=n E^2 ## (i.e. it is proportional to ## n E^2 ##. (In optics, the calculations for intensity are often done in this fashion without the additional constants that are in the Poynting vector for power. And n=1 for air.) You can write the amplitudes for the different wave components and compute the energy at the two receivers. e.g. for the bottom receiver ## E_{Bottom}=(1/\sqrt{2}-1/\sqrt{2})E_o=0 ##(one component is transmitted from the top and the other reflected from the right). For the left receiver ## E_{Left}=(1/\sqrt{2}+1/\sqrt{2})E_0=\sqrt{2}E_0 ## (one component is transmitted from the right and the other reflected from the top) and we see our final intensity at the left receiver ## I_{Left}=(\sqrt{2}E_0)^2=2 E_0^2 ## which is the sum of the intensities of the two initial beams ## I_0=E_o^2 ## for each. (Note for the above energy reflection coefficient ## R=I_r/I_i=E_r^2/E_i^2=\rho^2 ##. Notice also when there are two coherent sources, the energy reflection coefficients ## R ## can not be used to compute the resulting energy distribution (this system is not linear in its energy properties, but is completely linear in regard to the E fields), but the ## \rho ## 's computed from the ## R ## are completely valid for doing the computation.) @Chris Frisella It's quite a lot of detail, but hopefully you can at least follow some of it.

Haha, yeah, that is a lot of data! Thank you for your help and energies. I follow it ok. Few fast (hopefully) questions:

1) So one reason for the phase offset between the two sources is one of the reflected beams has to travel through the medium twice, thus giving it greater phase change?

2) The actual amount of phase change is dependent on the width and material properties of the splitter, correct?

3) You say the reflection coefficient R can't be used to compute with when you have two sources. This is because the two sources interfere thus canceling out one of the reflected paths, correct?

Simple answers please! ;)
 
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