# Destructive interference - where does the energy go?

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

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

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Well these two dual beam laser pencils, pointing at each other do fill the space between with what seems to be no energy anywhere as well as one does. Single beam lasers pointing at each other can not do that even if of exactly the same polarization (E field directions) and frequency.

When "clashing head on" each beam will feed into the other laser's source. This could make one cease to lase as inside each laser (When lasing.) there is stronger E fields than in the external beams and the entering beam of one can be in principle, be a negative feedback, as if mirror reflected the laser's own beam back inside it.

If the stimulated emission is thus killed by the other laser's beam entering, and the non-thermal excitation of the upper state continues, then the intensity of the random direction light produced by that upper to lower state transition will increases (by the energy of the beam being feed back into the laser from the other laser which is still functioning.)

With exactly opposite pointed lasers you really only have one laser system.
How much laser light part "A" produces, if any, will depend upon part "B" (and conversely). The total energy pumping up the non-thermal population upper states and decaying radiatively to the lower states, will ultimately exit in random directions mainly thru the sides of the laser (This because the laser beams are not sending the energy anywhere out side of the integrated Part "A" / Part "B" single laser.

BTW, I would expect that this integrated laser is an oscillator probably with frequency in the audio range. I.e. when one part can not lase because of the negative feed back from the other the non-thermal upper state population will be increasing as the strong E field it has internally (When it is lasing) is not causing stimulated radiative depopulation of the upper state. In the other lasing laser, in contrast, it is as they say "burning a hole" in the Doppler distribution of atoms /molecules which can be stimulated to radiate in phase. I.e. its "lasable" upper state population is decreasing. Soon it will go off and the other laser, now with no negative feed back will begin to lase. - I.e.alternation of which is lasing is very likely. - an (I would guess) audio frequency oscillator.

Last edited by a moderator: May 17, 2011

3. ### RJBeeryNatural PhilosopherValued Senior Member

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Billy T, that was a dizzying explanation but I think I digested some of it.

Turn laser on inside a sealed, perfect black-body oven with dimensions less than the wavelength of the stimulated light. What happens?

5. ### Me-Ki-GalBannedBanned

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I think it is also understood by structural engineers familiar with bridge construction. As a few bridges have collapsed from vibrational resonance in the past, It forced the Engineers to come up with solutions to deal with the stack up of vibration from wind , cars and what have you. I don't know if it gets you any closer to the question at hand , but maybe it is some kind of sink hole kind of thing . I don't know . Sound has made Me very curious why and how resonance works . What excites adjacent objects and causes vibration? I have heard rumors that adjacent contained water will vibrate when another container of water vibrates at a specific rate . I never did it and it does sound a little bit like quackery . Don't know

7. ### Me-Ki-GalBannedBanned

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O.K. you all know I ain't a scientist, but I think it is a gain in rate of decay . The destructive interference is closer to a natural state of rest and what you would call non destructive is a gain achieved by compatibility. I don't understand gain that well in effects for guitars . How the signal is looped or compressed to create sustain ?

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

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quote me back a sentence or two that confused you and I will say more words about what it means - I like to teach and you seem like you want to learn.
Part I made bold certainly rules out laser existing so I have no idea what you are asking. Atoms are much smaller than most radiation, but can interact with it.

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

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Billy T:

In principle, zero diameter sources and perfect lenses are possible. They are not possible in practice. That's the difference between principle and practice.

But all of this is really just buying into Tach's pointless hair-splitting. Fundamentally, it's a waste of time.

A plane wave does not spread as it travels. Collimation takes a spherical wave and makes it a plane wave.

alfa brane:

The mean is defined as:

$\bar{x} = (1/N)\sum_i x_i$

where N is the number of samples. Re-arranging this, we see:

$(\sum_i x_i) / \bar{x} = N$

In other words, the total divided by the mean is necessarily equal to the number of samples.

10. ### MikeORegistered Senior Member

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Please forgive me if what I have here has been already been brought up. Reading the first page was fun and it got me thinking, but I’m out of time and have to shut down soon. I promise to read the other 5 pages tomorrow.

I just can’t wait to suggest this. It’s been many years since my E&M class but I seem to remember that the energy density (ergs per cubic cm?) of a light wave was proportional to the SQUARE of the amplitudes of the E and/or M fields’ intensities or amplitudes.

Funny thing, how the square function was a big deal on another thread this evening.

Anyway, the net amplitude of the E & M fields is zero where the two beams are sucessfully alligned to interfere, but the energy densities of each wave do not interfere. The energy densities are positive since the squaring of the field amplitudes removes negative sign.

The field amplitudes are positive and negative at the same points so their net is zero, but the net energy from each wave is positive everwhere, and still present and being conducted towards it eventual absorbtion elsewhere.

The fields cancel but the energy does not.

***

Now I wonder if it matters if the waves are traveling in the same directions or in opposite directions. :shrug:

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

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MikeO:

When waves are added together, what gets added is amplitudes, not intensities directly. The intensity of the resultant wave at any point is proportional to the square of the resultant amplitude at that point, which is NOT the same as being proportional to the sum of the intensities of the original waves.

For example, if at a particular point wave 1 has amplitude A and wave 2 has amplitude -A, then the resultant amplitude when we add the two waves is zero. The resultant intensity is the square of zero, which is zero of course. So if A is amplitude and I is intensity:
$A_{1+2} = A_1 + A_2 = A - A = 0$
$I_{1+2} \ne I_1 + I_2 = A^2 + (-A)^2 = 2A^2$

Rather,

$I_{1+2} = (A_1 + A_2)^2 = (A - A)^2 = 0$

12. ### MikeORegistered Senior Member

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Is the Intensity the same thing as what I was trying to describe as energy density?

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

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What is the temperature of a zero diameter source that is radiating energy? Answer: infinity.

This follows from the fact that no source can radiate greater intensity than a black body at the same temperature. If you have a very small diameter black body at temperature T radiating I and reduce its diameter by factor of four, (1/16 of the surface area) then for it to still radiate I, the temperature must be increased to 2T as energy radiated goes as T^4. Do you agree with this?

You are smart. -You know where I am is going with this, but for others here it is:
I.e. even after cutting diameter by factor of 4 a zillion times the diameter is still not a point, but the Temperature required to still radiate I is a zillion times T. In the limit of a point, the temperature required is infinite.

Ergo: Point sources do not exist, IN PRINCIPLE. This is NOT a limit of technology or just a practical problem.

-----
One mathematical proof that you were in error by saying that “perfect collimation” is possible in principle is sufficient, but I will argue that even the “perfect lens” is also impossible, IN PRINCIPLE because its diameter is finite:

True a plane wave does not spread, but we don't have a plane wave. We have a segment of one and the cross section area of it does grow larger as it travels. One proof of this is diffraction by small circular hole (or any lens). - The diffraction spreading depends only upon the hole's (or lens’) diameter - not on the opaque material the hole has been punched in. Diffraction is spreading.

If the hole has the diameter of a wavelength, then on the back side of the hole energy will be found over a quite a wide angle. This is hard to demonstrate at optical frequencies, but easy to demonstrate at radar or short wave length radio frequencies.

For example if 10cm wavelength radiation is incident upon a meter, or larger, diameter copper plate with a 10 cm diameter hole in the center of it, then in the region on the “back side” of the hole the RF energy is NOT confined to a 10cm diameter beam. In fact it is impossible to make a beam with only a few wavelengths diameter which does not grow progressively and detectably grow larger as it travels.

I will admit that when the diameter of the beam is many wavelengths, then the spreading of it as it travels makes very little percentage increase in its size, and that this almost always the case for application like the Mach-Zender interferometer we were discussing. Thus this argument is not as powerful as the mathematical one first presented.

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14. ### RJBeeryNatural PhilosopherValued Senior Member

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I guess it's either a malformed or nonsensical question. I'll try again: take a laser capable of emanating light, say, in the 1000nm range and encase the tip with a black-body material at a distance than 1000nm from the point of emission; turn on the laser...what happens?

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

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The black body absorbs all the radiation so from the laser's POV, it is not even there - laser behaves as if its radiation was traveling to infinity.

This assumes that the black body remains cold. If it is radiating, some of its radiation will have the wavelength of the laser light but it will not be a coherent feed back into the laser. Thus not likely to make much difference as already inside the laser there is radiation of this frequency which is not coherent. This energy feed back can excite atoms in the lower state of the laser transition, so effective is an increase in the pumping up population in the excited state - I.e. may make a slight increase in the laser output, and that would slightly increase the temperature of the black body, but I am 99.9% sure this is not an instability which has greater than unity loop gain. I.e. both laser output and black body temperature come to some steady conditions of temperature and output. All lasers have energy loss mechanisms. They would be called upon to dissipate the energy returned by the black body in the steady state that this system would reach.

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16. ### RJBeeryNatural PhilosopherValued Senior Member

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Billy T, I asked the question from a emitter/absorber POV. One of the characteristics of a black-body oven is that their physical dimensions must be "many multiples" of the wavelengths of EM radiation they can contain. Combining this with your bowl of jello analogy...where does the energy go?

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

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In 2nd version of your question, which was understandable, you asked about " black-body material" close to the exit of the laser. Now you seem to be thinking of a large box, with tiny hole - the way one approximates a black body in practice. In this case you need to tell the internal wall temperature to know what radiation emerges for the hole into the laser but most of what I said before still applies.

It is true that the oven box realization of a black body needs the hole to be a tiny fraction of the internal wall area. I am not sure what you mean by a black body - there is the concept of perfect absorber and there is the physical approximation (small hole in large box). You seem to have switched to the oven box in this third edition of you question. Is that correct? Anyway that is what I answered in the first paragraph above.

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

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Billy T:

Yes, I agree with you.

Every model situation in physics omits some details in order to concentrate on others. You can progressively add more and more features to any model to make it more and more "real" - more in accordance with what you'd actually observe in a real experiment. But often it's helpful in developing a theory to just leave out features that are considered not to have an important influence on the observed results.

So, when we're dealing with diffraction or interference of light, we're usually more interested in the angles of diffraction, where the interference fringes are etc. than we are about details of the source. We commonly idealise the source to a point source of zero size, or a source that produces plane waves, for example. The reason is simple: the source is not the topic of interest. It is how the light from the source behaves.

So, I agree that if we want to increase the complexity of the model, we can insist that there are no true point sources, because every source must have a temperature etc. etc. But that just muddies the waters as far as looking at the interference effects goes.

Tach's whole aim with his initial smart-arse comment that "you always see fringes in a Michelson interferometer" was to muddy the waters - effectively to wreck a simple explanation by making it more "real" by adding unnecessary complexity to the model.

For practical purposes, many light sources can be modelled as point sources. For example, in the two-slit experiment (Young's experiment), you normally start with a theoretical point source of spherical waves, then allow those waves to spread out sufficiently far from the source so that it is reasonable to approximate the spherical wavefront by a plane wave. And that's where you put your slits. End result: you have a two-slit experiment with plane waves incident on the slits, and you can do all the interference theory under that assumption (i.e. the assumption of constant phase of the wave across each slit). In practical terms, you can quantify exactly how good or bad an approximation this is, just by looking at by how much a spherical wavefront differs from a plane wavefront at a particular source distance and slit separation.

Similarly, when you say that every real source (like a light bulb or laser) is not in fact a point source, we can exactly quantify the extent to which the wavefronts will planar at the slits depending on the distance from the slits to the source. The point is that every source looks like a point source provided you're far enough away from it. You need to decide in advance your acceptable criterion for "far enough".

Even a perfect plane wave diffracts through an aperture. That's the condition under which all of the basic physics is derived - check any textbook. Diffraction by an aperture does not rely in any way on the incident wave having a spherical wavefront.

19. ### TachBannedBanned

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No, it wasn't. I simple pointed out that your claim of total destructive interference at one end of the interferometer arm is false and that it is contradicted by experiment. I have also pointed out that , based on your repeated statements on the subject, you never ran the experiment. If you did, you would have noticed the interference fringes.

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

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JamesIt is true that NEVER is it possible to get one screen dark and the other with full intensity free of fringes. To achieve that you do need a point source and no spreading to various directions of propagation of the light by the diffraction due to the finite aperature lens.

Also that is only possible in theory if the path lengths differ by only a small fraction of a wave length - As I noted earlier that could happen by chance but does so less than 1 in a million times.

Thus Tach's statement that fringes are always seen is a much better approximation of the truth than your claims. So should not be described as: "smart-arse comment". I doubt if anyone has ever observed what you claim, even if they tried hard to produce it.

Another point: Because you will see fringes, and and you want them to be equally strong / intense one very often does not use the collimated beam you speak of. Instead the light source illuminates the back side of a frosted glass plate which is at the focus of the lens - I.e. intentionally avoids anything even approximating a "collimated beam" or a "point source." I HAVE ALWAYS USED A FROSTED GLASS PLATE because then there is no need to accurately position it.

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21. ### temurman of no wordsRegistered Senior Member

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Kinetic energy