# Why don't waves erase out each other when looking onto a wall?

If I stand exactly in front of a colorful wall, I imagine the light waves they emit, and they receive should randomly double or erase out each other.

So as a result, I imagine I should see a weird combination of colors, or a full-black/full-white/very lightly perception of the wall, when all the light waves that the wall receives and emits cancel out each other or double each other.

Why doesn't that actually happen? Any time I look into a wall, I never see the wall "cancel out" of my perception. Same for radio waves. Shouldn't radio waves not work at all? There are so many sources where they could reflect and cancel out or annoy each other...

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This looks a bit superficial on the surface, but it actually goes pretty deep. Deeper than I have time to formulate a good answer for right now. –  dmckee May 26 '11 at 6:31
$\langle E^2 \rangle \neq \langle E \rangle^2$ –  genneth May 26 '11 at 8:50
For the wall, because your eye focuses light, like a camera. For the radio, because the signal goes through a frequency band-pass filter before being turned into sound. –  Peter Shor May 26 '11 at 17:50
@Peter what if two waves of the same frequency annoy each other? The receiver couldn't tell them apart I think? –  Johannes Schaub - litb May 26 '11 at 17:52
@Johannes Schaub: When that happens, you get interference. This is why the government assigns frequencies to radio stations so that any two radio stations with the same frequency are far apart. It's also why cellular telephone companies pay large amounts of money in auctions for bandwidth in the frequency spectrum. –  Peter Shor May 26 '11 at 20:18

1) First let us separate colour perception from frequency. Individual frequencies have a color correspondence but the colour the human eye perceives is another story.

2) White light, such as sunlight, is composed of many frequencies.

When the impinging wave hits a wall it can be a) reflected b) absorbed c) scattered incoherently

In order for the light waves to cancel out each other or double each other the photons have to be, within the uncertainty principle, superimposed in time and space. Sometimes it happens, but the probability is small. That is one of the reasons why a reflected beam can never have the same strength as its original beam. If the frequency is the same the probability will be higher than if the frequencies come from a random palette.

This superposition can be achieved with lasers, where there is control of frequency and the beam is coherent, i.e. the phases are preserved upon reflection. A hologram is an example of superposition of same-frequency light to create a three dimensional shape by peaks and dips.

Edit: From a disappeared question the following comment is worth adding:

You can perceive all colours even if only two frequencies are shining on an object. Also in this decade, Land first discovered a two-color system for projecting the entire spectrum of hues with only two colors of projecting light (he later found more specifically that one could achieve the same effect using very narrow bands of 500 nm and 557 nm light).

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Why answer this question in terms of quantum mechanics (photons, uncertainty principle)? In my view, the question has nothing to do with quantum mechanics (at least for the light) --- its just Maxwell's equations. –  Fabian May 26 '11 at 8:56
So that people do not lose sight that everything is quantum mechanical? An additional reason is that if one is thinking of the umpteen zillion photons making up a beam one can intuit how small the probability of superposition is outside coherence of the beam. –  anna v May 26 '11 at 12:14
@Fabian: it's not just Maxwell's equations at all (although sometimes it's okay to use them). Without photons many scattering properties (like back-scattering off deeper layers and scattering off metals) of materials would be quite different and consequently, the world would look quite different. On top of that, what Anna said ;) –  Marek May 26 '11 at 18:11

You asked two questions. First, the light:

As has been pointed out, typically (there are exceptions) the colorful wall is not emitting anything. Let's take blue, for an example. White light shines on the wall. White light is composed of many different frequencies/wavelengths, which we would perceive, if they were busted out with a prism, as many different colors. The blue wall absorbs all these colors except the blue. So there are not a lot of competing wavelengths coming off the wall; the blue paint allowed the wall to absorb all the competing wavelengths of light, and reflect only blue. Another way of looking at it is that the wall is covered with a "blue-reject" absorbtion filter. We get to see the one color that the wall ISN'T.

But, there may be many competing light sources, with random orientations and intensities, illuminating the blue wall. Again, everything is absorbed except blue. The blue light we see is randomly oriented and phased. No problem. The sum of the blue light vectors from any given point will still add up to a resultant wave that is of blue frequency. Yes, if you planned it very, very carefully (e.g. used polarization equipment and lasers with controllable phase offsets) you could engineer some strange effects. Normal (non-laser) light is much more random than that.

Sometimes, ESPECIALLY with AM radios, you will get fade. Sometimes this is just because of blind spots due to terrain and atmospheric bending. Sometimes it is one of the phenomena you described: multipath interference. The radio wave which reaches your antenna has followed two or more separate paths, with at least one of those paths being longer than the other(s) by odd multiples (1, 3, 5, ...) of one-half a wavelength. The waves hitting the antenna add to zero, and you hear nothing.

I hope that helps.

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I'd like to take this in three parts, and I am going to stick to a classical description, though the quantum mechanical description is in some ways easier and just as valid.

1. Superposition is the principle that the amplitudes due to two waves incident on the same point in space at the same time can be naively added together, but the waves do not affect each other.{#}

That last bit is important enough to write again: the waves do not affect each other. They can, in fact pass right through one another.

For the purposes below we can treat every wave as being a plane wave and thus being described by an amplitude $A$, a wave number $\vec{k}$, and a phase $\phi$. The wave number encodes both the direction of propagation (that why it's a vector) and the frequency (in the absolute value) $|k| = \frac{\omega}{c} = \frac{f}{2 \pi c}$.

This description is not complete because (a) the planar nature is only approximate and (b) even allowing for that the phase can be a function of position on the wave front; but we can safely neglect these issues.

So at any given point in space and time we have a big jumble of different waves all added together $$\sum_{\text{directions}} \int d\omega \int d\phi A(\vec{k},\phi) e^{i(\vec{k}\cdot\vec{x} - \omega t + \phi)}$$

2. You eye is a camera: it has a small opening with a focusing element (lens) at the front and a light sensitive projection surface (retina) at the back. I sorts out incoming light waves two ways.

First, the lens focuses light to a particular point on the retina according to it's direction. That makes the sum over direction go away, because each patch on the retina sees only one direction (well, a very small range of directions). If the lens isn't the right shape you lose some of this and the image becomes blurry. Then you need glasses.

Secondly individual patches of the retina are sensitive to light of different frequency ranges{*}. So for each rod or cone, the integral over angular frequencies $\omega$ is reduced to quite moderate limits (human vision extends over roughly one octave in the electromagnetic spectrum).

That still leaves us with the sum over the phases, and I am going to elide this part of the problem by saying that they are roughly constant over the small dimensions of the eye's lens and time-scale of you retinal response; if that doesn't make you happy I will refer you to genneth's comment: $\langle E^2 \rangle \neq \langle E \rangle^2$.

3. Radios live in Fourier space:{+} They typically accept signals from a very wide set of direction, so we don't get the spacial filtering that you get with the eye. However, the electronics that back them select only a very narrow band of frequencies. In essence inspecting the incoming signals in Fourier space. And the signal you are looking for has an additional periodicity (either amplitude in AM or frequency in FM) in the audible range as well, so another round of filtering is done and you generally have a dominate remaining signal. Intentional signals have constant phases, while unintentional signals (i.e. background or noise) have random phases and they tend to cancel out.

{*} The mechanism by which the rods and cones respond to light in inherently quantum mechanical, the means by which those signals are propagated to the brain is inherently biochemical and the post processing done by the brain is huge topic in and of itself; all of which is beyond the cope of this discussion.

{+} Traditional radios. Like the AM/FM job in your grandfather's car (since we're all 21th century people and have spiffy combined GPS-mp3 player widget or something). Ultra-wide band like your wireless base station is another matter.

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"Radios live in Fourier space..." I LOVE that, and intend to steal it from you! :-) –  Vintage May 26 '11 at 18:33

I will like to point out that the walls do not emit any light but instead they reflect what shines on them. You may also be interested in reading more about diffuse reflection which roughly and non mathematically attempts at explaining some of the physics at work here. If you are looking for a more mathematical treatment, this buffalo lecture on interference may help you --esp. the coherence bit.

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This is on Color contancy: en.wikipedia.org/wiki/Color_constancy –  Georg May 26 '11 at 14:06
+1 Very interesting. –  Monster Truck May 26 '11 at 14:08
The mechanism used to reflect light is absorption-radiation. I can see the walls as emitters, in that sense. –  Derek E Oct 19 '13 at 18:00

Some good quick answers here. I just saw this years later and felt like taking a crack. :P

There are so many sources where they could reflect and cancel out or annoy each other..

Yes, and that's exactly what they do. They step on each other's toes and, in general, obey the principle of superposition.

Shouldn't radio waves not work at all?

You know they do. (It's a useful guide, sometimes.) So, the question is, why do they still superimpose in a "coherent" way? The following will use a radio as an example.

Ideally, an antenna is going to get hit harder with the more direct path from the source, than from indirect paths that bounced around the neighboring geometry. Depending on the frequencies you are tuning to and the time-of-flight delay for these indirect paths, this might not be a problem. Radio waves from other sources could interfere, as well. In the end, you get a stronger signal, superimposed with maybe somewhat-off and not as strong signals, superimposed with some weaker random noise. (again, ideally)

At least, this is what things look like to the antenna. The antenna, being a transducer, changes the problem for us at this point. From the antenna on, we start dealing with current being driven through circuits, which has it's own noise contributions, but also it's own mechanisms for signal processing, that produce what you hear out of the radio.

Basically, all that complicated superposition wave stuff that you thought was happening actually is happening. It's just that it can be dealt with.

[color perception part]

Why doesn't that actually happen? Any time I look into a wall, I never see the wall "cancel out" of my perception.

Your eye has a lensing structure at the front, a water-like fluid in the middle, and a bio-transducer mechanism at the back. (I like to think of light in terms of rays/photon-paths when thinking about lenses.)

The lensing structure at the front acts as a a kind of filter. All the background stuff is targeted to miss the back of your eye, so only the photons following a particular path will hit on "that particular part of the back of your eye". (Note: maybe some degenerate paths here.) This corresponds to receiving some bunch of photons from a perceived ray in the world.

There are different kinds of transducers in the back of your eye and they have different spectral responses. The game here is then statistics on the kinds of photons received and the functioning of your transducers at that location. (See, for example, color blindness.) Regardless, you are biologically wired (from the brain to the eye) to figure out some light/color perception from this bunch of photons at that spot at the back of your eye. (And you have two eyes, see: binocular vision.)

So, again, all that complicated stuff is happening. In this case, it is you that has evolved to deal with it.

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The waves do interfere with each other as you expect, but the interference pattern alternates between constructive interference and destructive interference in such a rapid fashion that your eye can't perceive it.

If you want to be able to see the interference, take a laser pointer and use that to illuminate the wall. You should see a pattern called "speckle". (It's easiest to see if the beam spot is big; use a lens to do this if you have one. Moving your head back and forth transversely also helps to make the speckle obvious.)

The speckle pattern comes from the constructive and destructive interference of the light reflecting from different parts of the wall. The reason why it looks static (if you can hold your laser pointer still enough) is because the laser has a very narrow spread of frequencies (and, to get more technical, also has transverse phase-coherence). If you were to change the wavelength of the laser, the pattern would change. If you were to change the wavelength back and forth very rapidly, the pattern would change very rapidly, your eye would average it out, and you wouldn't be able to see the speckle anymore.

This is what's going on with broadband light - it contains a whole bunch of different frequencies, so the interference "averages out". That's why (unless you're using a laser) you never see it.

As for radio waves: yeah, they interfere too. That's why your microwave has a turntable in it.

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