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QED’s “All Path Argument” for Mirror Reflection is false, phony, and deceptive.

Discussion in 'Science and Technology' started by Unes, Jun 16, 2017.

  1. Unes

    Unes Active Member
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    QED offers the all-path argument for the optic law for mirror reflection, in my opening post I have explained many problems with the QED all-path model, as we have discussed REPEATEDLY I showed that every one of those problem is valid, and they should not be ignored. I showed that we can put multiple detectors on the receiving side, and all-path is incapable to differentiate between them, and according to QED all-path any one of those detectors, or all of them, have equal probability to receive that SINGLE reflected photon. Anybody can see the absurdity of this suggestion. I showed if we change the direction, or the angle of incident of the incoming photon, it makes no difference to the QED all-path argument, it will produce EXACTLY the same result, regardless of any of those changes. I showed QED all-path requires the receiving detector has to be exactly at right location in order QED all-path result to be valid, and to be in harmony with the optic path. Isn’t that a convenient requirement!? So, QED all-path does not offer us anything new about how and why the Optic Law works.

    Now, let us examine the mirror reflection from the perspective of the energized electron that emits the reflected photon from the surface of the mirror. If this energized electron is a loose electron and it is not bound to an atom, then ACCORDING TO QED the direction of the emitted photon is totally random, and the direction of the optic path is not privileged, and the optic path is just one direction among many others. And if the energized electron is bound to an atom, then the spectroscopy of the atom produces model of the emission for the emitted photon, and again the direction of the optic path is not privileged. So, from the perspective of the emitting electron the path of the optic law is not recognized at all.

    So, no matter how we examine the mirror reflection, we are facing with a big mystery, and our beloved QED theory has no clue why the optic law works the way it does.

    So, QED all-path argument, by its false but extremely bold claim has distracted us, from examining the optic law for the mirror reflection more closely and with fresh new ideas and with an open mind.

    Polymath257, I am grateful for your contributions, please don’t feel bad that you failed to discredit my challenge to QED all-path argument. If I have proved that Prof. Richard Feynman was wrong, then you should not be embarrassed at all! Instead you should join me to correct this mistake from our beloved science, this is very important.
    The ramification of this correction is profound and it is significant.

    I posted my argument in few Physics forums, the only forum that few members responded to my challenge was the following Physics forum: Physics Forums - The Fusion of Science and Community

    My user name in that forum was “Unes”, after few polite exchanges, when I exposed the falsehood of the argument that one of the members had made, all of a sudden, I WAS BANNED from that forum! I could not log in anymore! The Physicists in that forum did not have the courage to face this earth shuddering mistake. So, the scientific community that I have experienced so far they did not have the scientific spirit at all! Science is supposed to test, to examine, to challenge all its theories all the time, but in practice the scientists are as closed-minded as the religious clergies!

    Every one of those QED all-path problems that I have pointed out is very easy to understand, and they are irrefutable, the physicists at the Physic forum they were sharp enough to recognize that right away!

    Who thought that science could be this flaky!?
     
    #41 Unes, Jul 7, 2017
    Last edited: Jul 7, 2017
  2. Polymath257

    Polymath257 Think & Care
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    And you are wrong in these deductions. I have pointed out REPEATEDLY how you are wrong in these ideas and what the correct way is forward.




    And that is true of each and every electron in the material. The sum of the ewffects gives the optic law.


    Your lack of understanding isn't a problem with QED.




    No, the only conclusion is that you don't understand what is being said in spite of many corrections.

    it sounds like you persisted in proclaiming the wrong position you have taken here and failed to learn anything from the responses, just like here. So they banned you for the sake of reasonable discussion. There is no mistake in the physics. The mistake is in your understanding of what the physics can and does say.
     
  3. Unes

    Unes Active Member
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    Polymath257,
    Instead of repeating the same argument again and again, you need to explain exactly which of the QED all-path problems that I have described you disagree with?
    I think we explored those points in details, and I exposed the falsehood of your responses.

    In the fog of summing up the probability of all the possible paths, QED deliberately implements the correct path into its process, and then it announces that it has reached to the correct result by its methods! This is called cheating at any school.
     
    #43 Unes, Jul 7, 2017
    Last edited: Jul 7, 2017
  4. Polymath257

    Polymath257 Think & Care
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    No, you did not. You changed the situation, leading to a different problem. The problem when there are multiple detectors is different than the one where there is only one and they are both different than when there is none.

    You assume that a photon can be emitted in a precisely specified direction. That is false. At best, it is emitted in a *range* of possible directions. That is your first mistake. To get it limited to such a range requires shielding that stops certain paths for the photon being emitted. How that shielding is done will affect the calculation.

    So, describe *in detail* what the situation is that you think is problematic: how the emitter is shielded, where the detector(s) are located, etc. Then describe what you think is the problem with the QED description of that situation.

    When you had several detectors, I was assuming the emitter was set up to emit over a range of directions that was very broad, so that each detector has an equal probability of detection. You seemed to not like that concept, but didn't say what you think is wrong with it.

    On the other hand, if the emitter is shielded, then many of the detectors would have a low probability of detection and those in other positions would have a higher probability.

    I even gave a good analogy of the situation: suppose you look at water waves that have been directed by barriers and then go on to reflect off a barrier on the side. You can describe in some detail what the resulting wave spike will do.

    Perhaps the problem is your concept of a photon. There is no difference between a photon and a probability wave of a certain sort. If that wave goes into a detector, that detector has a probability of detecting the photon.

    If you can, describe the situation that you think is problematic in QED in the analogy of the water wave.
     
  5. Unes

    Unes Active Member
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    I am going to deny you any excuse! I only move the ONE detector few inches away from its correct position. QED all-path method and the wave function method cannot tell me that this is a wrong location! According to QED method, they both produce a valid result. But because of the optic law, we do know very well that no photon will be detected at this wrong location.

    In the fog of summing up the probability of all the possible paths, QED deliberately implements the correct path into its process, and then it announces that it has reached to the correct result by its methods! This is called cheating at any school.
     
    #45 Unes, Jul 7, 2017
    Last edited: Jul 7, 2017
  6. Polymath257

    Polymath257 Think & Care
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    What do you mean by the 'wrong location'? In what way is the emitter prevented from emitting a photon in the 'wrong direction'?

    Be precise.
     
  7. Unes

    Unes Active Member
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    The direction of the optic path is known, the detector has to be in that optic path in order to detect the reflected photon. If the detector is moved away from the optic path, then it cannot detect the reflected photon.
     
  8. Polymath257

    Polymath257 Think & Care
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    The only way for a wave's direction to be known precisely is for it to be an infinite plane wave. In that case, the reflected wave is also an infinite plane wave and ALL points have equal probability of detecting the photon.

    If you want a finite wave, then the direction *cannot* be precisely known. Furthermore, the more concentrated the direction of the wave is, the broader the wave has to be and hence the more surface required for the reflection and the wider the reflected wave. So, again, a movement of the detector by a small amount won't affect the probability of detection much.
     
  9. Unes

    Unes Active Member
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    This sounds like that you have never seen a sunshine beam, or a laser pointer beam! You point a laser pointer beam on the wall, it only covers maybe a quarter inch diameter, if your detector is an inch away from that path, then no detection. The optic path is shown with a thin line, when you work with a single photon, the optic path is extremely narrow.
     
    #49 Unes, Jul 8, 2017
    Last edited: Jul 8, 2017
  10. Polymath257

    Polymath257 Think & Care
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    Even lasers have a range of directions because of the wave nature of light. No, a photon does NOT have a very narrow path. A photon has a wave character that is spread out inevitably.

    So, let's assume that we have a laser that is a fairly narrow spread. Suppose we have a detector that is off the optical path by a significant amount.

    What happens?

    Look at all paths from the emitter of the laser to your 'bad' point.

    The wave function at the emitter is fairly narrow (imagine a pulse of a water wave constrained to a long, narrow pathway before opening up).

    So, we are looking at the propagation of that narrow wave function from the emitter to your point. So, for each point in the opening of the emitter, we use the amplitude there as a probability and then look at all paths from that point to your detector, add up the amplitudes, and then add up over all the points in the opening. Are we good so far?

    Now, take any path from the opening to your detector. It is not a shortest path. There are close-by paths that are both longer and shorter than the path you are considering. So close by paths will have a wide range of amplitudes to add up when computing the probability of detection. But that means those amplitudes add up to be zero (those that are shorter cancel those that are longer). So the probability of detection at your detector is zero.

    On the other hand, if we have a detector on the optical path, then the paths from the emitter to the detector are *all* longer than the path, so we don't get cancellation from shorter paths. This leaves a significant probability of detection.

    The point is that the optical path is *also* the shortest path from the detector to a point. Such shortest paths don't get canceled by even shorter paths, leaving a large probability. Those paths that are not shortest paths do get canceled by shorter paths leading to low probabilities.
     
  11. Unes

    Unes Active Member
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    The last few exchanges sound like we are at the level below the high school!

    Light manifests many phenomena and it has many properties. Here we are ONLY examining the mirror reflection phenomenon. In this phenomenon light moves in straight line, and light wave properties is a minor issue, that it is not even considered. QED has offered the all-path method to explain this phenomenon. So, please focus on the subject, and examine the validity or the falsehood of QED all-path method.

    Light is not a theoretical thing that we can manipulate it with our wild theories, it is a real phenomenon that it has specific properties.
     
    #51 Unes, Jul 8, 2017
    Last edited: Jul 8, 2017
  12. Polymath257

    Polymath257 Think & Care
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    Sorry, but if you are doing QED, you *have* to consider the wave aspect. Light is a wave phenomenon. And QED looks at it as such. The optic path is selected out because it is a 'shortest path' and hence the amplitudes don't get canceled out by paths nearby.

    You are doing geometric optics. That is a *first* approximation for how light moves and works. The next stage up is to consider the wave aspects. The next step after that is to consider polarization. And the next step up after that is to consider quantum effects. QED lives at that level.

    But even the wave level of approximation shows that many of your ideas and assumptions are simply false.

    Yes, light has specific properties: it is a wave phenomenon that shows polarization and quantum effects. Geometric optics is the most trivial aspect of this and is often simply wrong.

    For example, there are many situations where light does NOT simply go in a straight line. There are many situations where the optic path is not the correct one. You are so fixated by the first approximation that you seem to want to ignore the actual behavior of the light.
     
  13. Unes

    Unes Active Member
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    Thank you Polymath257,
    Your last post showed me why you and other physicists have been so defiant in light of such overwhelming and irrefutable evidence in regard to QED all-path cheating. I like to thank you especially for that information.
    I am NOT ignoring the actual behavior of light. I will explain shortly where we differ. You are correct one hundred percent to say that QED has many levels, and it covers many subjects, I do acknowledge that WITHOUT ANY DISPUTE. We both agree, that there are EXPERTS in each one of those specialized subjects. And I, by focusing on the QED all-path cheating problems, I am becoming an expert in this subject, and THIS IS THE SUBJECT OF THIS THREAD. So, being fixated and focused on the subject in science and in business, that is a good thing, and I highly recommend it. But in social life, that would be a disaster and an agony!
    I am sure there are many phenomena that QED explains them based on the wave property of photon. But in QED all-path method there is no mention of photon’s wave property! The amplitude vectors method is all classical physics. WHEN WE ARE TALKING ABOUT THE OPTIC PATH, WE ARE IN THE DOMAIN OF THE CLASSICAL PHYSICS. DON’T YOU THINK TALKING ABOUT HIGHLY REFINED AND VERY LOW MARGINAL EFFECT IS A MISPLACED ARGUMENT!?

    IF QED ALL-PATH LIKES TO IMPOSE WILD QUANTUM REFINED PHENOMENA WHICH THEIR IMPACTS ARE VERY MARGINAL INTO ITS ARGUMENTS, THEN IT BETTER OFFERS ITS OWN SOLUTIONS OF THE OPTIC PATH, AND NOT TO BRING THE SOLUTION FROM THE OPTIC PATH INTO ITS METHOD. AND PROVE IT TO US THAT ITS METHODS DO PRODUCE THE RESULT THAT IT MATCHES WITH THE OPTIC PATH NATURALLY.

    So, mixing the issues will only confuse us and we lose the sight of the “first approximation”.

    You mentioned: “You are so fixated by the first approximation”
    So, if QED made a mistake at its first approximation (this is where the things are much easier to comprehend), then for sure we will never catch that mistake in much more refined and very complex configuration.

    Now, I do exactly know why at QED’s more refined and complex configurations we do not see those easy to understand problems at all, this is because by adopting the optic path at our first approximation, we already have implemented the correct answer into our processes, then by more elaborate methods we are only refining the correct result, and the correct result will become finer and more detailed. So, my objection is about the method that QED has used at its first approximation; how does QED find the optic path?

    Let us examine a scenario. On the detecting side, I put the detector at point Y1 and I preform QED all-path calculation, QED all-path calculation produces a valid result. Then, I move the detector to point Y2 and I preform QED all-path calculation, QED all-path calculation produces a valid result. Then, again I move the detector to point Y3, and so on, until I finish with point Y100, or maybe with Y100,000 location. Every one of these results looks valid, and QED all-path cannot differentiate between them. And QED has no way to know which one is the optic path, unless the optic path is identified by the experiment or by the optic law. If we use the optic law, we know the direction of the incoming photon and its angle of incident, WHICH THEY ALL HAVE BEEN IGNORED IN ALL-PATH METHOD. These information will give us the path of the reflected photon, which in our previous example we assumed to be the path to the detector at Y40. According to optic path detector Y40 will detect the reflected photon.


    Now let us examine an aspect of the wave property for this scenario. You are suggesting that the wave property of photon creates some possibility that it might be detected by detector Y17 location, instead of Y40 location. Let us consider such possibility, but how can you compare such low and marginal possibility with the almost the certainty that exist at Y40 location!? This is so absurd that we give these two possibilities comparable weight! The difference between these possibilities is so vast, that Y17 possibility might be ignored altogether. I do expect much more plausible argument from a physicist.

    QED all-path on its own, it does not know where the optic path is. So, in the process of its first approximation the optic path is implemented into QED processes. And we do know exactly, this is cheating. This cheating will never get exposed by doing more refined configuration. And here we are trying to stop this scientific deception.
     
    #53 Unes, Jul 10, 2017
    Last edited: Jul 10, 2017
  14. Polymath257

    Polymath257 Think & Care
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    You have presented no such thing. All you have presented is vague issues that show you lack of understanding as much as anything else. You have also consistently ignored the explanations as to why you are wrong.




    You are nowhere close to being an expert. There simply is no cheating problem.



    Wrong. The whole goal of the all-paths formulation is to compute the wave function. The rotating vectors are part of that wave nature. Also, the end result is a probability and not a path.




    Given that you are talking about the all-paths formulation, and that *is* the highly refined description that takes into account all those effects, no it isn't a misplaced argument. If you are doing QED, you are looking at the quantum effects, which is way beyond the simply wave effects.




    I already did that. The amplitudes for nearby paths will cancel each other out *except* for those paths that are 'shortest' from the emitter to the detector. That is what gives the optic path.




    There is no mistake except in your understanding.




    And I have answered that several times. The optic path is the path that does not get canceled out because there are no shorter paths.

    When you first presented this, I was under the assumption that your emitter gave out light equally in all directions. Since then, you have said that a laser light is used to provide directionality. Is this still the case?


    If so, then yes, the all-paths formulation *does* distinguish between them. Those detectors that are not on a shortest path will have their probabilities cancel out to a very high degree. So they will be unlikely to detect light. The detectors that *are* on shortest paths will not have their amplitudes canceled out and so they *will* have high probabilities of detecting the light.



    You are ignoring a vast number of aspects of light that need to be considered. For example, how does the opening in the mitter compare to the wavelength of the light used? If it is about the same, then there *will* be a spreading that can affect the calculations (it does so by changing the rotation rate of the amplitude).

    But, given the overall level of your question, let's assume that the opening is large compared to the wavelength of the light so the spreading is minor. Then the all-paths formulation would have the probabilities cancel out for Y17 and not for Y40. It is that simple. The path to Y40 will be close to a shortest path and that to Y17 will not be.

    And this is simply wrong. QED does not need to know where the optic path is. It gives that path probability simply because it is the one that does not get canceled out by shorter paths.

    And, once again, it is easy enough to create a physical situation where paths other than the optic path are followed. This is a property of the light. You ignore this.
     
  15. Unes

    Unes Active Member
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    There are many confusions and claims in these statements.

    After this many exchanges we are still in disagreement whether QED all-path method cheats about finding the optic path NATURALLY, or it is not. We need to settle this issue with utmost clarity. We have to find out who is the confused party in this discussion. So, for clarity I am going to spell out the following straight forward experiment with precision.

    The source sends a SINGLE photon to the reflecting mirror. Location of the source and the direction that it sends the photon is fixed. We can measure the angle of incident with accuracy of fraction of one second. This information produces a classical optic path. Since you have been insisting the optic path is NOT required, in advance, then please disregard this information about this classical optic path. Classical optic path is defined by a fine line from the source to the mirror, and the reflected line out of the mirror, and with specificity that I mentioned before. I needed to mention this, because you questioned the accuracy level of the optic path.

    I place a photon detector at Y1 location. Y1 location is ONE inch away from the classical optic path. Like this: __.__

    We do all-path calculation from this source to the detector at Y1 location. As it is explained with QED all-path method, we use all possible paths that start from the source to the detector at Y1 location. The amplitude vectors for the off center paths they cancel out each other, and the amplitude vectors for the paths at the center they add up and give us the amplitude vectors summation. This calculation gives us the probability for the reflected photon detection at Y1 location.

    Now, you are free to choose any property of photon, or any phenomena that are related to this experiment. You can analyze this experiment any way that you like. At the end, you need to answer the following question with clarity and without any confusion: does the detector at Y1 location detect the reflected photon?
     
  16. Polymath257

    Polymath257 Think & Care
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    OK, I shall do that.


    No, this is wrong as explained below.


    OK, first, I am going to assume that the wavelength of the photon is much smaller than one inch (if you use radio waves, the results will be different).

    Since, as you state, Y1 is one inch away from the optic path, no matter what path you take from the source to Y1, there will be other paths close by that are both shorter and longer. Now, remember that the amount of rotation of the amplitude vectors depends on the length of the path concerned. Because of this, the rotating amplitude vectors will point in different directions for the longer and shorter paths (this is where I use the fact that the wavelength is small: for longer wavelengths, the rotation of the amplitude is not as rapid) and so they will cancel.

    So, because Y1 is not on a shortest path, the amplitudes from all the paths will cancel out to a very good approximation, so the probability of detection will be very low.

    Is that clear enough?
     
  17. Unes

    Unes Active Member
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    It sounds that you are totally out of your senses, WE ARE TALKING ABOUT THE OPTIC LAW AND MIRROR REFLECTION, the light wave lengths range between 0.0004 and 0.0007 mm!!!

    Second, in QED all-path we INCLUDED ALL THE PATHS FROM THE SOURCE TO THE DETECTOR, where do you get these other paths!!??

    Third, why probability is so low, we have included all the paths as it is instructed by QED all-path method?
     
    #57 Unes, Jul 12, 2017
    Last edited: Jul 12, 2017
  18. Polymath257

    Polymath257 Think & Care
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    The wavelengths being used were not clearly stated.


    Yes, for each path from the source to the detector, we get an amplitude. We then add up these amplitudes for all those paths. In the case of the detector at Y1 that is not on the optic path, every path has a corresponding one that cancels it. This is not the case for those points on the optic path. And that is what distinguishes the optic path.
     
  19. Unes

    Unes Active Member
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    Till now, you have been saying the optic path is not required! Now, why are you comparing the path to the detector at YI with the optic path!? I DO KNOW THAT OPTIC PATH IS THE SHORTEST PATH, but you have been claiming that you will acquire that information by the QED all-path method independently. So, why are you cheating?
     
    #59 Unes, Jul 12, 2017
    Last edited: Jul 12, 2017
  20. Polymath257

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    I am not cheating. The optic path is selected out *because* it is the shortest path and thereby doesn't have all nearby paths cancel out. All other paths are not shortest so do get canceled out.

    The QED formulation shows that the shortest path is the one that has the high probabilities of detection for the standard setup. That doesn't go into the calculation. It comes out of them.

    And, in those situations where the other paths do not get canceled out (like diffraction gratings), paths other than the shortest path can have high probabilities for detection.
     
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