Why can we see light?

Discussion in 'Highly Technical' started by bwanaaa, Dec 17, 2012.

  1. bwanaaa

    bwanaaa Senior member

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    At any given instant, each photoreceptor in my eye is receiving a multitude of light waves at many frequencies and all out of phase. In other words at any instant, a photoreceptor is getting troughs and peaks and everything in between. The sum of all these waves is zero.*These light waves are all out of phase since they are randomly generated by multiple different tungsten atoms from all over the filament in the light.

    But Throwing thousands of pebbles into a pond still generates waves on the shore - they don't cancel out.

    Why not?
     
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  3. Eureka

    Eureka Diamond Member

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    Because it's not all out of phase. You're assuming a perfect cancellation and it won't be, not when one source is far more biased than all other source (any light source that is pointed directly at you). Also, different frequencies for different sources, so even if they are out of phase they won't be in the same frequency.
     
  4. CycloWizard

    CycloWizard Lifer

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    There are multiple phototransductive molecules each with a different absoprtion spectrum. The system works as follows: a single photon collides with a photosensitive protein. The energy of the photon induces isomerization and releases a "quantum bump" of energy. A stochastic temporal summation of these bumps occurs. If enough occur in a short enough period of time on a given set of photoreceptors, an action potential will fire. The optics of the eye also necessitate some directionality of incident light as light coming from outside the "acceptance angle" is effectively discarded.

    So, I think the problem with your analysis is that you've assumed the net waveform sums to zero at all times in all directions (i.e. perfect destructive interference). The intensity is directly related to the photon flux (i.e. energy per unit area). Since photons are absorbed at the retina rather than reflected, the net flux is non-zero.
     
  5. serpretetsky

    serpretetsky Senior member

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    you need java for this work
    http://www.falstad.com/ripple/

    i suggest playing with that for a while. Try different presets. Put walls in random locations. Try building mufflers, try building horns. Construct a beach front somehow. I spent a lot of hours on that thing.
     
  6. bwanaaa

    bwanaaa Senior member

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    In thinking about the distribution of phases, I can only assume they are random. If the phases are random, then any given light wave should impinge on a detector at the same time as millions of other light waves of al phases in equal amounts . That's why I would expect them to cancel out - because every light wave hits the detector with its inverse concurrently. For there to be a net non zero sum, one phase would need to predominate. Since I am only talking about staring into a headlight, I am not really concerned about angles of incidence, etc.

    Is there a way to quantify the phases of the individual photons arriving at a detector?
     
  7. bwanaaa

    bwanaaa Senior member

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    Wow! Tnx. As I max out the sources the amplitude at the boundary does diminish which supports my question. The likelihood of wave cancellation increases as more waves of random phase are added. ( not truly random but more sources simulates my point)
     
  8. bwanaaa

    bwanaaa Senior member

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    Maybe the photodetectors in our eyes are phase agnostic?
     
  9. Born2bwire

    Born2bwire Diamond Member

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    It's a bit incorrect to think of the photons having phase in the sense that you are thinking it out (and photodetectors by themselves do not detect phase, but that does not mean that they are not subject to the overall effects of phase cancellation as phase is what causes directivity and focusing of light). Photons are the quantum picture and while there is a phase associated with the quantum dynamical treatment, it is not the same phase as when we talk about the classical wave treatment of light. But assuming randomness does not necessarily lead to perfect cancellation. For example, you can look at the case of white noise. White noise is purely random, has all frequencies and is perfectly uncorrelated. Despite this, white noise does not cancel out and it has a power spectrum of unity. It is because the noise is uncorrelated that we do not get cancellation. If you say that we have a signal and randomly have a signal that has opposite phase to cancel it out, then we can just as well say there will be a third signal to reinforce the first. But without correlation, we cannot have the superposition of two waves that cancel each other out perfectly. To do so requires that the phases be correlated so that they are phase locked. The problem with the ripple tank example that you have been playing around with is that there will be strong correlation between your signals. There is both a consistent amplitude and phase to your sources where random sources will have the amplitudes and phases shift in time.
     
    #8 Born2bwire, Dec 18, 2012
    Last edited: Dec 18, 2012
  10. Paperdoc

    Paperdoc Golden Member

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    OP has focused on one aspect - phase - in making his argument. You forgot amplitude.

    Firstly, we can be certain that the amplitudes of all the light wave streams being generated is not the same at all. But thereafter, each wave path ("beam", if you will) travels different distances and through different media, so that both phase and amplitude of the original wave is altered before reaching the receiver in the eye. Thus, even if the phases of all the arriving waves, being randomly distributed, were to sum to zero, the amplitudes would not.

    As OP hints with the comment on waves in water, the same holds true of all electromagnetic wave mixes. Take a quick look at an oscilloscope display of a sound stream. It does not appear to be several clean sine waves mixed together. It appears to be a very complex non-sinusoidal varying signal. Using mathematics we can "un-mix" that complexity to simulate it as a sum of several pure sine waves of different amplitudes and frequencies, but that is not the real nature of the signal. Born2bewire cites the excellent example of "white noise" which contains, ideally, an infinite number of frequencies and amplitudes, and does not cancel itself to silence. Although sound is not electromagnetic, the same holds true of radio waves. We are constantly bathed in a huge mix of such wave signals from radio, TV, cell phones, etc., but the huge mix does not come to a zero sum - all those receivers actually do get their signals cleanly because they each concentrate their "listening" on a very narrow band of the mixture. In essence they each pick out and use only a very tiny portion of the mathematical model of multiple sine waves.

    I suspect an important part of the answer to this, though, is that light is not waves only. It consists of bundles of energy we call photons which behave a lot like waves as the travel through space. But they still arrive and impart their energy to the receiver. There are no "negative photons" that would cancel out the energy from another photon to yield a zero sum.
     
  11. bwanaaa

    bwanaaa Senior member

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    tnx b2b for your thoughtful reply. leaving aside the ripple tank, and getting back to light, i concede there might be an inequality in the distribution of phases so they dont all have an 'inverse doppleganger'. But taking the the thought a step further, as we turn up the juice on the filament, it gets brighter. If I am generating more photons, the same cancellation should result in a constant brightness. For the light to get brighter, one would have to postulate an increasing disparity in the inequality of phases. Not something that's intellectually satisfying.

    I am beginning to imagine that detection of a photon is an interaction that precludes photon summation. In other words, an atom is stimulated by a photon, and only one photon at a time. The arrival of two photons to an electron orbital results in exclusion of one photon from actually exciting the orbital. I think interference can only be demonstrated/detected/exist when the detector is larger than the excitor. Since the wavelength of light is so much greater than the size of an atom, maybe this is why we do not get interference of the kind that my brain is stuck on.
     
  12. Born2bwire

    Born2bwire Diamond Member

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    Paperdoc's response, as always, is good to think about on. I think that the OP still maintains some misconceptions.

    First, two photons cannot just cancel each other out. True, you may have heard that the photon is its own antiparticle, but that does not come into these discussions of light and superposition. We still require energy conservation. As such, interference of waves can never destroy energy for a lossless medium. Instead, interference directs the energy to specific regions of space. So in the double slit problem, we see areas that experience an increase in energy density (constructive interference) and areas of an associated decrease (reconstructive). But when we talk about photons having a phase, the photons are not canceling each other out. The phase is associated with the wave function. The wave function is an equation that describes the probability density of the problem. By operating on the wave function with the suitable operator, we can get the statistical probability densities or averages of physical values like energy, momentum and spin. We can also get the probability density of the detection of the particle in a given volume. So the phase comes into play in how the full wave function behaves and it is the wave function that undergoes interference. Note that the particles are not necessarily interfering with each other. The actual physical meaning of the phase of the wave function is, if anything, a philosophical consideration that may or may not be defined by the interpretation of quantum physics. Most physicists are of the Copenhagen school and follow a shut up and calculate mentality when it comes to trying to physically define such qualities (the reason being is that while we can find analogies to classical physics, these properties are not experienced in our macroscopic viewpoint. You can interpret them in multiple ways and still end up with the same predictions). The double slit experiment can be performed using individual particles, one at a time, and still produce the same interference pattern that we would expect if we had a steady stream. That is, the effects of the interference can occur even when there are no other particles currently present to interact with (and with photons being bosons, they do not interact with each other very much).

    Second, the ideas that you are attributing to photon behavior applies to classical theories of light. It just happens that the quantum behavior reflects very strongly the classical theory in many ways and people often mistakenly intermix the ideas. The phase of an electromagnetic wave only has meaning when we talk about a steady and large number of photons being emitted. The problem of why a random source emits an efficient power spectrum should be considered from the view of classical physics as the solution here would reflect the solution in the quantum theory. There is no need to consider photons, answering the question interns of the classical picture will answer the question in the quantum picture.

    So looking at it purely as a classical wave, the only real question to answer is why does a random source allow for omnidirectional emission with a nonzero power spectrum? This again has to do with the fact that it is completely uncorrelated merely by the fact that it is random. Over time, the amplitude of a given frequency will shift as will the phase. This results in what is called temporal and spatial deconstruct. If there was an amount of correlation, or coherence, then there would exist some degree of consistent interference. But without this coherence, the waves that make up the spectrum are forever changing and thus cannot have a consistent behavior of interference. Note that this property of coherence is independent of the power of the signal. It matters not if we increase the power density, the incoherence of the random signal persists and thus is not a factor as you suggest.

    Mathematically, we could state that at a given frequency, the wave will look like:

    A(t)*exp{i\omega*t+\phi(t)}

    where the amplitude A and phase \phi change randomly as time progresses.

    I would suggest learning more about noise like uncoloured (white) and coloured (e.g. pink) noise. Other topics that would be relevant are correlation and coherence. As Paperdoc pointed out, these are general properties of signals and while it may be more often discussed in the context of an electrical or audible wave, the electromagnetic waves follow the exact same physics.
     
  13. bwanaaa

    bwanaaa Senior member

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    Tnx again pdoc and b2b. I thought about my question again in light of white noise. It seems I made a fundamental mistake in counting and hence came up with a bad visualization. It's not the case that a photon falls onto a receptor with an inverse doppleganger only. Rather all phases likely bombard the receptor at the same instant because there are so many emitters. By enumerating the photons in pairs I have miscounted. I instead should ask the question what's hitting the receptor at a given instant. Assuming a random distribution of phases, there are only 2 photons that destructively interfere and 2 that constructively interfere within the distance of 1 wavelength. The rest are phase shifted and always have a positive non zero sum.

    Sorry for my phallacy - I feel like a stupid dick.
     
  14. Sleepingforest

    Sleepingforest Platinum Member

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    Don't feel bad. I learned quite a bit skimming through this thread.:thumbsup:
     
  15. MrDudeMan

    MrDudeMan Lifer

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    I'm not sure why no one commented on this post, but it was a great addition to the conversation. Very interesting.