Does Destructive Interference Cancel Energy?

[Chris Frisella]

@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.

 

[Charles Link]

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.

 

[Chris Frisella]

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?

 

[Charles Link]

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.

 

[Chris Frisella]

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! ;)

 

[Charles Link]

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! ;)

1) Good question=the top source in both cases has one extra length through the beamsplitter over the source from the right, regardless of which direction it goes. The (phases of the) sources are assumed to be adjusted for any phase difference because of this. The relative phases can be considered to be measured at the point where the sources are each incident on the reflective (lower) beamsplitter surface.$\\$ 2) Correct, but the answer to question (1) answers this. $\\$ 3) Yes. If you used energy coefficients, (energy coefficients work if there is no coherence, e.g. if the alignment is not precise and/or or the sources are not mutually coherent ), the computation says the energy split is 50-50 for each source=no interference. Meanwhile, Maxwell's equations, which govern the electric fields are completely linear (thereby the system is said to be linear), but since the energy equations (i.e. (energy) intensity $I=n E^2$) are second order in the linear parameter $E$, the system is not required to be linear in energy. In cases where there is no interference, the systems are normally also linear in their energy properties.

 

[IllyaKuryakin]

If you call the ground level = 0 energy level, and you dig a hole in it, you can call the whole -1 energy level and the dirt next to the hole +1 energy level. When you put the dirt back in the hole, it all goes back to 0, but you didn't really destroy the dirt. Just another way to look at it.

 

[Chris Frisella]

1) Good question=the top source in both cases has one extra length through the beamsplitter over the source from the right, regardless of which direction it goes. The (phases of the) sources are assumed to be adjusted for any phase difference because of this. The relative phases can be considered to be measured at the point where the sources are each incident on the reflective (lower) beamsplitter surface.$\\$ 2) Correct, but the answer to question (1) answers this. $\\$ 3) Yes. If you used energy coefficients, (energy coefficients work if there is no coherence, e.g. if the alignment is not precise and/or or the sources are not mutually coherent ), the computation says the energy split is 50-50 for each source=no interference. Meanwhile, Maxwell's equations, which govern the electric fields are completely linear (thereby the system is said to be linear), but since the energy equations (i.e. (energy) intensity $I=n E^2$) are second order in the linear parameter $E$, the system is not required to be linear in energy. In cases where there is no interference, the systems are normally also linear in their energy properties.

Good, good! Thanks.

When you are speaking of linear vs non linear you mean in terms of energy distribution within the system, right? The system as a whole would always be linear in the sense that whatever the source energies times time add up to will equal the total energy of the system (considering a closed system where nothing leaks out). However there could be a nonlinear relationship in where the energy is directed; example the beam splitter with two sources and interference.

 

[Chris Frisella]

If you call the ground level = 0 energy level, and you dig a hole in it, you can call the whole -1 energy level and the dirt next to the hole +1 energy level. When you put the dirt back in the hole, it all goes back to 0, but you didn't really destroy the dirt. Just another way to look at it.

I like it :)

 

[Charles Link]

Good, good! Thanks.

When you are speaking of linear vs non linear you mean in terms of energy distribution within the system, right? The system as a whole would always be linear in the sense that whatever the source energies times time add up to will equal the total energy of the system (considering a closed system where nothing leaks out). However there could be a nonlinear relationship in where the energy is directed; example the beam splitter with two sources and interference.

Yes. Correct. The energy is conserved but the energy distribution is not linear. This is actually quite remarkable because if you have a single source on, one half the energy goes to receiver A and one half to receiver B. Turning on a second source can redirect all of the energy to one of the receivers.

 

[Chris Frisella]

Yes. Correct. The energy is conserved but the energy distribution is not linear. This is actually quite remarkable because if you have a single source on, one half the energy goes to receiver A and one half to receiver B. Turning on a second source can redirect all of the energy to one of the receivers.

Ya it is!

 

[Charles Link]

Ya it is!

One more comment on this one: The amplitude of the sum of the two wave components was easy to compute on this one because the two components were either in phase or 180 degrees out of phase. Otherwise, if the two differ by some arbitrary phase, e.g. $E_{sum}=Acos(\omega t)+Bcos(\omega t +\phi)$ you could add them using trigonometric identities and/or a phasor diagram. For arbitrary phase and amplitude, the energy is found to be distributed between the two receivers.

 

[rude man]

Consider two transmitters generating two superimposed electric fields :
E1 = sin(ωt)
E2 = sin(ωt + π).
These fields cancel everywhere yet there are two waves being generated by two transmitters, each with their own Poynting vectors. The Poynting vectors are both positive: P = E x H. Although they cancel everywhere, the two waves exist as if the other weren't there. That's what superposition, the linear addition of functions, is.

Note that they CAN be individually detected, e.g. by a phase-sensitive field strength meter referenced to one of their phases. Assume the meter is referenced to E1, thent the readings of the meters are
V1 = 1
V2 = -1
ignoring constants. Neither meter reads 0.
Same holds for acoustic, light etc. Two waves canceling each other but each individually generating power.These fields are propagating energy thruout space

 

[Charles Link]

Consider two transmitters generating two superimposed electric fields :
E1 = sin(ωt)
E2 = sin(ωt + π).
These fields cancel everywhere yet there are two waves being generated by two transmitters, each with their own Poynting vectors. The Poynting vectors are both positive: P = E x H. Although they cancel everywhere, the two waves exist as if the other weren't there. That's what superposition, the linear addition of functions, is.

Note that they CAN be individually detected, e.g. by a phase-sensitive field strength meter referenced to one of their phases. Assume the meter is referenced to E1, thent the readings of the meters are
V1 = 1
V2 = -1
ignoring constants. Neither meter reads 0.
Same holds for acoustic, light etc. Two waves canceling each other but each individually generating power.These fields are propagating energy thruout space

This one I disagree with. If they are spatially separated, they do not cancel everywhere. There will be regions of constructive interference, and the total energy/power will be conserved. If the two sources are at the same location in space, nothing is being generated.

 

[rude man]

This one I disagree with. If they are spatially separated, they do not cancel everywhere. There will be regions of constructive interference, and the total energy/power will be conserved. If the two sources are at the same location in space, nothing is being generated.

2 identical transmitters face each other, pi phase difference. At the exact mid-point the resultant is zero for all time yet the signals can independently be detected by phase-sensitive meters.

 

[Charles Link]

2 identical transmitters face each other, pi phase difference. At the exact mid-point the resultant is zero for all time yet the signals can independently be detected by phase-sensitive meters.

They are thereby spatially separated. It will be possible to find locations where the signals from the two sources are 90 degrees out of phase. At these locations the phase sensitive detection will be able to observe a single source and not see the other one. (This is not the case at a location where the signals are 180 degrees out of phase. When they are 180 degrees apart at the same amplitude, nothing will be observed, with or without phase sensitive detection. If they are different amplitudes and 180 degrees apart, the phase detector will not know what the amplitude of the individual sources is . Meanwhile, if the signals are in phase with each other at a given location, phase sensitive detection can not separate them.)

 

[rude man]

They are thereby spatially separated. It will be possible to find locations where the signals from the two sources are 90 degrees out of phase. At these locations the phase sensitive detection will be able to observe a single source and not see the other one. (This is not the case at a location where the signals are 180 degrees out of phase. When they are 180 degrees apart at the same amplitude, nothing will be observed, with or without phase sensitive detection. If they are different amplitudes and 180 degrees apart, the phase detector will not know what the amplitude of the individual sources is . Meanwhile, if the signals are in phase with each other at a given location, phase sensitive detection can not separate them.)

I have to admit, that makes sense.

 

[Chris Frisella]

One more comment on this one: The amplitude of the sum of the two wave components was easy to compute on this one because the two components were either in phase or 180 degrees out of phase. Otherwise, if the two differ by some arbitrary phase, e.g. $E_{sum}=Acos(\omega t)+Bcos(\omega t +\phi)$ you could add them using trigonometric identities and/or a phasor diagram. For arbitrary phase and amplitude, the energy is found to be distributed between the two receivers.

What if you had 2 sources and beam combiner, making only one beam, which was adjusted to create total destructive interference? Since the beams don't overlap anywhere until they pass through the combiner, and there is only one beam made by the combiner, I don't know if you'd see any constructive interference to balance the loss from destructive. Perhaps the combiner element would just heat up?

 

[Charles Link]

What if you had 2 sources and beam combiner, making only one beam, which was adjusted to create total destructive interference? Since the beams don't overlap anywhere until they pass through the combiner, and there is only one beam made by the combiner, I don't know if you'd see any constructive interference to balance the loss from destructive. Perhaps the combiner element would just heat up?

Perhaps the easiest way to create two mutually coherent sources is to first split a source with a beamsplitter and then recombine the beams with a second beamsplitter (using, of course, mirrors to steer the beams, etc.). You never get destructive interference on both receivers after the second beamsplitter. The $\pi$ phase change for the reflection that starts with the incident beam in air (as opposed to the reflection inside the beamsplitter) makes it so that the case of complete destructive interference on both receiver arms never occurs. The power distribution at the receivers is determined by the relative phases of the two sources. There is complete energy conservation for every possible choice of phase. For a zero phase difference, the beams recombine completely at one receiver, and for a $\pi$ phase difference, they completely recombine at the other receiver. $\\$ Incidentally, you may find it of interest that the Michelson interferometer is an apparatus just like the 2-beamsplitter apparatus just mentioned where the same beamsplitter is used twice: first, to split the beam, and then again, to recombine the beams after they each reflect off of one of the mirrors. The presence of the outgoing beam (=beam being split) on the beamsplitter does not affect the reflections that take place for the beams as they are being recombined=the fresnel coefficients work for the individual components, just as if it were an apparatus with two separate beamsplitters.

 

[sophiecentaur]

There is an equivalent to this with radio frequency signals. To get the geometry for cancellation to occur everywhere, it is obvious that the two sources have to be co-sited. Optically, this is not achievable perfectly. But it can the achieved at RF by having two antenna (say dipoles) very close together - or, better still, use the same antenna. So you need no power to be transmitted from this single antenna, yet have two transmitters 'feeding' it. That means all the power from each transmitter needs to be absorbed by the other transmitter. That only requires the transmitters to produce the same amplitude signals in antiphase. There will be zero volts driving the antenna so no power will be radiated. But energy has not been destroyed or eliminated. All the Power will have been dissipated within the output resistances of two transmitter amplifiers. Not the first instance of a transmitter with zero efficiency!