# Destructive interference - where does the energy go?

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

1. ### TachBannedBanned

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Yes, you need to click the links. Anyways, this is well known in MMX. You never looked in the aperture of an MMX interferometer, did you?

3. ### kurrosRegistered Senior Member

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793
Just checking up on this thread and noticed this, sorry if I get the context wrong, I was lazy and didn't read up on everything in the meantime.

To me it sounds like he is describing an effect due to your source beam not being a perfect parallel plane wave. In a real experiment this is the always case, so only in the centre of the output beams do you get the ideal situation, outside of this the path lengths are different due to the curved wavefronts of the beam. You thus get the circular interference fringe patterns he describes, assuming an expanding beam coming from a laser or something.

Even if you did have a perfect parallel wave though your energy is totally accounted for because it exits through one output beam or the other. You probably already discussed that though.

5. ### TachBannedBanned

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Correct. There is always difference in path length, hence, there are always fringes being observed.

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

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Tach:

Yes, I have, as a matter of fact. Not that it matters.

Look, since I know you don't know what causes the circular patterns you observe, and that you won't be able to produce a link that shows why, I'll do the work for you this time. Here's a document that explains the whole thing in detail, with some nice diagrams.

There's little point in going round and round with you and trying to get an actual physical explanation, because you can never manage to produce one when asked. All you do is say "Go and look for yourself", and sometimes you provide a few unhelpful or irrelevant links. You sound like a bit of a wannabe physicist who has picked up some information from pop-science books, but who has not bothered to look at detail or derivations of anything. So, you're reduced to trying to one-up people by throwing the odd buzz-word around and trying to take people down a notch by introducing complications that you don't understand yourself. In short, you're a bit of a fraud, pretending to have actual knowledge that you don't have, when all you've really got is a bunch of second-hand details that you're unable to explain when pushed.

kurros:

Yes. That's all correct. See the linked document for all the gory detail.

8. ### TachBannedBanned

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I don't think so. If you did, you'd have known that you see interference fringes.

I have already done so. TWICE. You need to learn how to click on links before you go ballistic. It is refreshing to see that you changed your tune and that you are admitting that the output is in the form of fringes.

Last edited: May 17, 2011
9. ### James RJust this guy, you know?Staff Member

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I did. Your links were unhelpful and irrelevant in that they gave no explanation of the fringes. Like I said.

And now you know why. That's if you bothered reading my link.

10. ### TachBannedBanned

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I knew why, this is why I corrected you in first place. Had you admitted that you were wrong, it would have saved you the embarassment of contradicting yourself.

It is very similar with the two links that I gave you earlier, all three links contradict your initial claim. It is refreshing to see that you contradicted your earlier false claims. This is the closest you've come to admitting you are wrong. You are making progress.

Last edited: May 17, 2011
11. ### James RJust this guy, you know?Staff Member

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What I said originally is correct.

If you use collimated light and set things up correctly, then no fringes appear. The only way you can get fringes is by having multiple paths of different lengths for the waves. Otherwise, you just see a uniform intensity.

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

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That is / may be / (but I doubt even that) correct in theory but impossible with real light and real light sources as perfectly collimated light does not exist in reality due to "physical optics" and black body radiation laws.

Even light from a long (say 2m - that is really hard to align to make lase, but I have) laser operating in a single mode, with a tiny exit diameter at the laser will spread and make a spot on the moon of several Km diameter (Perhaps more. - How do you think they bounce light off the corner cube reflectors the astronauts left on the moon? That can be done even from satellites above the air.) ALL "collimated" light has some angular divergence, even if the lens is perfect as to avoid that the source must be a zero diameter point – (and neither of these requirements is ever true). If the source were zero diameter, then the intensity would be infinite. As no source can radiate more intensely than a black body that implies the source temperature is also infinite. SUMMARY: Your “collimated” light does not exist even in realistic theory and not even close in practice for use with large path Mach-Zehnder interferometers.

So in reality, what James states does not exist / is not true. I.e. never will one screen, (assuming it is thousands of wavelengths wide and not a zero diameter point) of Mach-Zender interferometer be black and the other have the full intensity (as if the "collimated" beam were falling on it directly).

Also note that if one is speaking of photo-electric detectors instead of screens, then it makes little sense to speak of fringes, and even the PE detectors are not points. I.e. effectively they are, at best, just very tiny screens. That makes adjusting the interferometer for equal path lengths (to the two detectors) impossible - I.e. if you get full intensity on one point and no light on another point, you have no way to know these points are "looking" at the same order of the pattern - for example one point may be at order n and another at order (n+m +0.5) to get one bright and the other dark but not with equal path lengths. Or in other words, you don't have any way to know your two "point detectors" are symetrically placed for the same path lengths - it is entirely possible that both are seeing bright fringes (or a dark ones). Getting one dark and one bright does NOT mean they are equal distant from the source, etc. In fact as the source is of finite size the very concept of equal distance from it is quite vague.

Thus I side with Tach in this discussion and until you counter the above facts, will count this as your physics error number 5.

Some details of my experience with a "meter scale" Mach-Zehnder interferometer:

(1) I have been able to briefly adjust one so that there was a noticeable curvature of the fringes on the screen and relative wide* fringes. I.e. center of the circular fringes was only 10 or less screen widths off to one side of the screen (but of course there was no light there so even if the screen were much wider, I would not have seen the bright central spot. (I have seen circular fringes with a small plate spaced Fabry-Perot interferometer, for ~100 hours total, as part of my Ph.D, research.) Also with a Mach-Zehnder interferometer, even slight temperature changes immediately destroy that near perfect alignment so that "straight" line fringes immediately appeared. Of course what appears as "straight" bright and dark lines on the screen are really arcs of a circle with center far off the screen.

(2) The two paths need not be the same length to produce the interference pattern. If they are made unequal, one can measure the length of the photons. The ones I measured were about 30cm long. Below the drawing is how to do that:

Give the beam splitter a very slight counter clock wise rotation – that has very little effect on the beam it transmits, but will swing the beam labeled “SB” a little to the left of the top left mirror. So you pull that mirror to the left to intercept the beam SB again, and of course rotate it slightly clockwise to send the beam it reflects back exactly on the original path. Now beam path SB is slightly longer than the path of beam which passes thru the beam splitter.

What you then observe on the screen is that the black fringes are no longer completely without light (and the bright fringes are not as bright, but with the unaided by instruments eye that small decrease in their intensity will be hard to notice.)

What you must know to understand this (and what follows) is that each photon ONLY interferes with itself. (This is easily proven with long exposures of film at the screen with such weak light that most of the time not even one photon exists – I.e. the interference pattern still forms in the exposed film.) I.e. in some way impossible for humans to really understand, each photon goes thru both paths of the interferometer.

When the path difference is small fraction of the photon length, most photons interfere with themselves as if the path lengths were equal,** but some do not. They fall upon one screen chosen by chance – I.e. may fall on a part of the screen that was dark for the equal path interferometer.

One progressively rotates the beam splitter more and adjusts the top left mirror as before to increase the path length difference and observes the interference pattern is “washing out” – become weaker. When the path length difference is greater than the photon length, the interference pattern is gone – equal intensity everywhere on both screens. This is because the photon “sensing” the longer path arrives at the screen too late to interfere with that part of itself which is “sensing” the shorter path. A path difference of ~30cm completely “washed out” the interference pattern I had been observing with more nearly equal path lengths.

Photons from different sources have different lengths. Some are several meters long. It is complex, and related to the QM’s uncertainty principle (and Fourier analysis). I.e. if the frequency of the photon is very well defined then there are many cycles in it (Fourier analysis shows this). When that is true, the momentum of the photon is also precisely defined so where it is at any time is poorly defined – I.e. it appears to be long if you measure, by any means, where it is at. Fast electronic shutter can show this also, but varying the interferometer path difference is cheaper and an easy lab experiment, I have done.

* All interference fringes are arcs of a circle (or the full circle) and each has the same total area. Thus the smallest diameter arcs are wider than than the large diameter arcs.

** Actually on the length scale that is important (the wavelength) less than 1 in million Mach-Zender interferometer set ups is ever with exactly, by chance, with the same path lengths.

Last edited by a moderator: May 17, 2011
13. ### RJBeeryNatural PhilosopherValued Senior Member

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Billy T and Tach, I don't know your histories with James R but I think it's unfair to call his post an "error".
He qualified his statement correctly; you pointing out that collimated light is not possible is like me pointing out that travelling at .9c is not possible in a discussion of the Twins Paradox. It's a technicality that occurs due to our technology and practical physics, but it isn't strictly physically forbidden. The benefit of a thought experiment is that we can idealize the conditions.

I do have a question though, James, which is this: you agree that a perfectly destructive laser is possible with an interferometer, but this apparently contradicts what Przyk said earlier (regarding the RHR nature of light, and how if the E is cancelled, the M is doubled, etc). What am I missing?

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

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No. what James claims (perfect collimation) is impossible in principle - it requires a zero diameter source at the focus of a perfect lens. The "perfect lens" is just a technology limit, but the zero diameter source is a limit IN PRINCIPLE, because as I explained that implies the source temperature is infinite. See part of my post which explains this based on the black body radiation laws.

It is also false /impossible as "physical optics" is the reality - a light beam spreads as it travels (EVEN IN PERFECT VACUUM)*. For both these reasons "perfect collimation" is impossible IN PRINCIPLE.

* Hygen's POV about how light propagates explains why this is so.

PS I have great admiration for James - I make at least 5 errors in physic for everyone of his. My only and very minor dislike of James is due to him giving me a zero points warning for a poll post I started. He said it was for "improper language" but I NEVER use that. The real reason fro my warning was James did not like what the poll was about, and perhaps thought it could even cause some legal problems for Sciforums. (Poll asked if the US should destroy the factories in China that US investors, like Warren Buffett,** had recently made as they gave China more modern factories than the US had. I.e. US could never compete economical with China so long as its factories were more modern / efficient and salaries were much lower there.)

** He used the tax reduction saving GWB gave to the rich to become owner of 10+% of BYD motors new factory and has now recovered, in stock price rise not counting his dividends, more than three times his original investment. I was doubly angry about GWB's tax reductions because of how they were adding to the debt and making the US less competitive, not stimulating the US economy as claimed but making jobs in Asia. That anger made me ask in the poll post if the US should at least undo the damage to US's competitive position GWB's tax relief for the already wealthy had done.

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15. ### RJBeeryNatural PhilosopherValued Senior Member

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Billy, I will clearly defer to you on this, but I still feel that it's a bit of a sidetrack to the discussion at hand. Can't we discuss the possibility of "perfectly destructive interference" of lasers GIVEN that they are perfectly collimated, etc?

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

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Yes, but there is really very little of interest to discuss. It (destructive interference) is like is like making a hole in a bowl of jello - yes it happens but the energy removed at that location appears elsewhere, just as the jello will rise up some where else.

The only thing difficult in both cases is to compute exactly where the energy, or jello, will appear.

17. ### RJBeeryNatural PhilosopherValued Senior Member

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Yes, I don't question that you're right on this and I cannot speak for anyone else but I'm learning quite a bit here. What I want to know is, if such a theoretical "antiwave" exists that can remove a laser from existing on an arm of an interferometer, what is preventing that "antiwave" from existing on its own, allowing us to annihilate a laser without having an avenue for the energy (i.e. the 2nd interferometer arm) to express itself? There is something that I'm missing because Przyk said such an "antiwave" is not possible but it appears to be possible under certain circumstances; I just want to know what those circumstances are (and please don't simply respond with "it's only possible IF the energy has another way of expressing itself!

)

18. ### przyksquishyValued Senior Member

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Those are different scenarios. I was talking about two counter-propagating waves: you can't send two EM field pulses toward one another in opposite directions in such a way that both the electric and magnetic fields cancel out at the same time. But if you can have two EM waves propagating in the same direction that cancel out if they're a half cycle out of phase. The EM waves that cancel out in an interferometer are propagating in the same direction. Then, in that case, the way energy is conserved is that if you have destructive interference in one place, such as one output of a beam-splitter, you get constructive interference somewhere else, as your quote in post #70 explains (and as I explained in post #41 and Tach hinted at even earlier).

19. ### RJBeeryNatural PhilosopherValued Senior Member

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Ah yes, it's the direction of travel, got it. You've clarified the difference between the two scenarios but I'm wondering if anyone knows "why" the magnetic field, in relation to the established electric field, is always right-handed? It sounds like that is the crux of the question.

Also, doesn't this violate a rule of symmetry by Nature having a preference? It seems we could simply measure the handedness of an EM wave to determine whether or not we're looking at a true laboratory experiment or the reflection of one..?

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

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Well yes you can have perfectly cancelling wave and anti-wave, if you want to think of nothing as the sum of two equal strength exactly co-propagating waves, one the wave and the other its anti-wave.

In fact, I just looked at my yellow lead pencil and realized it was really a "dual wave laser." I.e. simultaneously, from its tip are emerging your wave and its anti-wave. I'll sell you this amazing pencil for only \$5.

Be careful not to modify, even slightly, the point as then the anti-wave may not come from exactly the same point or may have a slightly different direction of propagation, then you will get some regions in space in front of the pencil with non-zero strength waves, if it still emits these two waves. That could be dangerous as the is no known limit on the strength of the separated wave and anti-wave which add to nothing when co-propagating.

Thus to buy my "dual wave laser" pencil, I will require that you sign a release protecting me from any damage claims it may do when slightly the tip is modified.

Or in terms of my prior jello bowl remarks yes you can make a void in the jello (spot with no jello) without any jello appearing elsewhere, provided that there is no jello in the bowl.

Last edited by a moderator: May 17, 2011
21. ### przyksquishyValued Senior Member

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As opposed to left handed? Ultimately because of a right hand rule in the way the magnetic field is defined. The operational definition of the magnetic field is that it is the field $\bar{B}$ such that
$\bar{F} = q \bar{v} \times \bar{B}$​
where $\bar{F}$, $q$, and $\bar{v}$ are respectively the force on and charge and velocity of a test particle. You could define the magnetic field such that $\bar{F} = - q \bar{v} \times \bar{B}$ instead, and then the magnetic field would flip sign everywhere in electromagnetism and electromagnetic waves would become left-handed with that convention.

No, because the magnetic field isn't actually a vector quantity. It's a type of geometrical object called an axial vector (or pseudo-vector), which means that it behaves like an ordinary vector under rotations, but gains an extra sign flip under improper rotations (such as a mirror image or spatial inversion). This property follows from the definition I cited above. If you define the magnetic field such that $\bar{F} = q \bar{v} \times \bar{B}$ and invert the spatial axes of a system, then $\bar{F} \rightarrow -\bar{F}$ and $\bar{v} \rightarrow - \bar{v}$, but $\bar{B}$ has to keep the same sign in order for the equation to hold.

22. ### arfa branecall me arfValued Senior Member

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About that question of dividing a total by its mean:
Isn't 5 equal to the number of samples? If you divide the total energy of a system by its mean value, don't you have a similar kind of measurement (up to some constant factor), i.e. of the number of statistical samples?

23. ### RJBeeryNatural PhilosopherValued Senior Member

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Billy T: Przyk already clarified this. The issue is related to the direction of wave propagation. If you had TWO pencils, pointing at each other, emitting a laser and an anti-laser respectively, then we could talk.