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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. Polymath257

    Polymath257 Think & Care
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    Once again, if you have 100 detectors, each has a probability of detecting that photon. Each is ost likely to detect that photon from the direction given by the optic law for that detector. BUT it is not guaranteed! It is quite possible to detect a photon from a direction *other* than that given by the optic law. Because paths that are far from the optic path are canceled out when the addition of amplitudes is done, only paths close to the optical path for that detector will have significant probabilities.

    In the case of 100 detectors, you can find the probability of detection at *any* of those detectors using the all-paths formulation with the paths starting from the source and going to the specific detector you want the probability for. In each case, those paths that are NOT the one given by the optic law will be canceled out by other paths.

    Why do you think the all-paths formulation gets prior information about the optic law? it doesn't. It simply computes the probabilities from *all* the paths and is able through adding up those probabilities, to show that the optic law path is selected because it doesn't get canceled out.

    And, again, if the 'mirror' is actually a diffraction grating, which is designed to block some of those canceling paths, then it isn't just possible to find photons that don't follow the optic law: it is very likely.
     
  2. Unes

    Unes Active Member
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    Thank you Polymath257,

    Let us clarify the task that it will be more precise. Let us assume there are 100 detectors, and we label them Y, Y1, Y2, and so on. From the source at point A with proper direction and proper angle of incident the photon can get reflected from the mirror and reach to each one of these detectors. Now a SINGLE photon is sent to the mirror, and the path from the optic law guides that photon to detector Y40.

    Now, we ask QED all-path to figure out where that photon is going to be detected by its natural method.

    First of all, without a target point QED all-path cannot perform any calculation. Just this point should set off the alarm that there is something wrong with QED all-path method. Second, if we offer any of the detectors as the target point, the result of the summation of the amplitude vectors for any of these detectors is EXACTLY the same. This is because QED all-path ignores the direction and the angle of the incident of the incoming photon. (Because, why bother with the information of the incoming photon, while QED all-path will acquire the information of the target; THE EXACT ANSWER IS PACKAGED AT THE TARGET!) QED all-path method cannot differentiate between any of these detectors. As we see with this approach QED all-path is incapable to find the correct path for the reflection to the detector Y40.

    So, please show us the technique for cancelling out the bogus paths, and end up with the path that leads to detector Y40 naturally, and without cheating.

    Now, let us observe how QED all-path operates, this is where the cheating occurs, QED all-path
    REQYUIRES that the correct path from the optic law to be provided for its calculation, in this scenario, the correct reflection path from the optic law that it leads to the detector Y40, is IMPLANTED into the QED all-path methods. Then by operating on this target, QED all-path claims its method finds the path to the detector Y40 naturally!

    This is like that we get the global coordinates for the location of the titanic wreckage from the group who found the wreckage few years ago, and then we claim the credit for finding the titanic wreckage.
     
  3. Polymath257

    Polymath257 Think & Care
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    OK, so you have done an experiment and determined that Y40 has detected a photon and that it came following the optical path.



    OK, QED does *not* predict the specific detector that will pick up the photon. It predicts the *probability* of each detector to pick up a photon. It does not and cannot make the determination that a specific detector will, in fact, pick up that photon.


    Is that clear?

    This is true whether you use the all-paths formulation or not. The wave equation can also be used and it also determines the probabilities, not the specifics. That is the nature of quantum mechanics.

    [QUOTEFirst of all, without a target point QED all-path cannot perform any calculation. Just this point should set off the alarm that there is something wrong with QED all-path method. [/QUOTE]

    Nope. What QED will do, whether you use the PDE or the all-paths formulation, is determine the probability of detection at each of the detectors. That is true of *any* quantum theory, not just QED.


    So, in the all-paths formulation, you get a probability that a photon will be detected at each of the detectors. For each detector, the probability is determined by adding up the amplitudes for all the paths going from the source to that detector.

    This assumes that all the detectors are the same distance from the source. If they are different distances, the amplitudes will not add up to be the same.



    That will not happen. Remember that what we get out of the calculation is a *probability* of detection at the different detectors. In this case, the probability is the same for each detector.



    Again, no it is not. it determines the *probability* of detection at Y40. And, it also calculates the probability of detection at all the other detectors. It does not and cannot predict the specific detector that will get hit.

    Once again, quantum mechanics often cannot predict the specific outcome of an observation. It only predicts probabilities for the different possible outcomes. In this case, *any* quantum theory will only predict the probability of detection at Y40. In your scenario, that probability is the same as for every other detector.
     
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  4. Unes

    Unes Active Member
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    Thank you Polymath257,

    So, you produced the result that the probability of each of the detectors to receive the photon is compatible.
    I totally agree with these statistical results that you have commented.

    But the problem is; this is not what QED all-path claims! QED all-path claims that its statistical result is heavily centered at the path that the optic law projects, and the standard deviation from that path is very low. Here, I exposed how QED all-path achieves that preferred statistical result; it gets it just by conveniently IMPLANTING Y40 as its target, and that is cheating pure and simple. This deception has distracted us from the realization of a very important phenomenon at display in that presumably very simple optic law for reflection.
     
    #24 Unes, Jun 26, 2017
    Last edited: Jun 26, 2017
  5. Polymath257

    Polymath257 Think & Care
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    OK, I am going to try to explain this one more time.

    Suppose we have a source S and 100 detectors Y1 through Y100. What QED does is give the probability of detection of a photon at each of the detectors. So, if you send out a single photon, it will not be able to tell which detector will fire. It will only give the probabilities. Is this OK so far?

    Now, to figure out the probability that detector Y40 will detect a photon, the all-paths calculation looks at all the paths from the source S to the detector Y40. When you add up all the amplitudes from all of those paths, the dominant part of the probability comes from those paths near the optic path from S to Y40.

    If, instead, you calculate the probability of detecting a photon at Y10, you would add up the amplitudes of all the paths from S to Y10. In *that* calculation, the paths closest to the optic path from S to Y10 will dominate the results.

    If, instead, you want to calculate the probability of detecting a photon at Y67, then you look at all paths from S to Y67 and add up their amplitudes. Again, you find that the probability is dominated by those paths close to the optic path from S to Y67.

    So, it gets a probability of detecting a photon at every detector in this way. For each detector, the probability is dominated by paths close to the corresponding optical path. But in *no* case is the optical path required ahead of time to do the calculations. You only 'implant Y40' to find the probability that Y40 will detect a photon.
     
  6. Unes

    Unes Active Member
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    Thank you Polymath257,

    I DID agree with you on all those points! You did not need to repeat them! From all those points that you repeated, without changing any facts, I only rephrased what you had acknowledged. I assume you agree with the following:

    A photon is sent to the mirror, since QED does not know which detector will receive the photon, then in QED, how the photon finds its way to the detector Y40, instead of any other detector is a mystery.
     
    #26 Unes, Jun 27, 2017
    Last edited: Jun 27, 2017
  7. Polymath257

    Polymath257 Think & Care
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    Yes. That is correct. It is the case in any quantum situation, but the way.
     
  8. Unes

    Unes Active Member
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    Thank you Polymath257,
    Once again without changing any fact, I only rephrased the previous statements, then, I assume you still agree with the following:

    This was the optic law that led the photon to detector Y40, then, in this scenario, how optic law found the path to Y40 is a mystery.
     
    #28 Unes, Jun 28, 2017
    Last edited: Jun 28, 2017
  9. Polymath257

    Polymath257 Think & Care
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    No, that is NOT a restatement. The photon going to Y40 need not have gone via the optic path. it could have gone by one close by.

    Furthermore, the 'mystery' is what 'determines' which of the detectors the photon goes to, not that the most likely path is an optic path.
     
  10. Unes

    Unes Active Member
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    Thank you Polymath257,

    I only replaced the phrase: “the photon finds its way to the detector Y40” with this phrase: “how optic law found the path to Y40”, this equivalence it is spelled out in our original scenario: “Now a SINGLE photon is sent to the mirror, and the path from the optic law guides that photon to detector Y40.

    Now, let us explore another scenario, we send a photon to the surface of the mirror, but there is no detector to detect the reflected photon. By examining the entire surface of the mirror, d
    oes QED recognize any specific direction for the reflected photon?
     
    #30 Unes, Jul 1, 2017
    Last edited: Jul 1, 2017
  11. Polymath257

    Polymath257 Think & Care
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    And that is a FALSE statement. The optic law does NOT guide the photon. The photon path is random, but concentrated around the optic path. Do you see the difference?


    No. In quantum mechanics in general, it is meaningless to talk about paths in the absence of detection.
     
  12. Unes

    Unes Active Member
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    Thank you Polymath257,


    Thanks for pointing out the difference from QED perspective.


    You had made this point earlier in this thread, thanks for confirming it again. So, without implementing the path from the optic law, QED all-path can NOT produce any direction for the reflected photon, or make any claim, while we do know this path exists, with or without any detection. This is exactly the point that I intended to prove.

    Here I need to acknowledge the VERY VALUABLE CONTRIBUTIONS that Polymath257 has made so patiently, they helped me to streamline my argument. I am very grateful.
     
    #32 Unes, Jul 5, 2017
    Last edited: Jul 5, 2017
  13. Polymath257

    Polymath257 Think & Care
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    Without a detector, it is meaningless to say the photon went according to a path. We don't know a path exists until we have a detector.

    But even without the all-path formulation, this is the case.
     
  14. Unes

    Unes Active Member
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    The optic law for mirror reflection was realized because it HAS BEEN OBSERVED numerous times, so, no detector is required to claim that the path of the reflection is known.
     
  15. Polymath257

    Polymath257 Think & Care
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    Since the direction of the photon is not known, neither is the path.
     
  16. Unes

    Unes Active Member
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    Both the direction and the angle of incident of the in-comping photon are known ahead of time, and the optic law without using any detector provides the reflection path. But my argument was that QED is using this result from the optic law VERY CONVENIENTLY to make its claim that it finds the reflection path naturally. TO FIND THE REFLECTION PATH NATURALLY, QED SHOULD NOT CHEAT, AND IT SHOULD NOT USE THE PROVIDED PATH FROM THE OPTIC LAW, INSTEAD IT SHOULD DEMONSTRATE THAT IT’S OWN TECHNIQUE CAN FIGURE OUT THE REFLECTION PATH. WHICH ALREADY IT WAS CONCEDED THAT QED WAS INCAPABLE OF DOING THAT!
     
    #36 Unes, Jul 5, 2017
    Last edited: Jul 5, 2017
  17. Polymath257

    Polymath257 Think & Care
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    The exact direction *cannot* be known ahead of time. There is an uncertainty principle for such angles in a similar way to the uncertainty principle for position and momentum. You have a *range* of possible directions provided by shielding. That provides a *range* of possible indexes of reflection.

    If you have no detector, it is meaningless to take about what the photon is doing. Period. Only detection of the photon gives information about that photon.

    Now, if you have a shielded source that limits the direction it can emit photons, and if you have no detector, we can *still* compute the probabilities of detection *if* we were to put a detector at various points. For each such point, there is only a probability of detection. And for each such point, the majority of the probability comes from the optic path and paths close to it.


    QED does NOT use the optic path when doing the calculations. I have no idea why you seem to think it does. The optic path *is* selected as the most likely path for a photon to take to a point where you do the calculation. You get the probability a detector would have if placed at that point for finding a photon.

    You seem to think a photon *must* take the optic path. That is false. And, in fact, we can build diffraction gratings where other paths give significant probabilities. In such, paths that are not close to the optic path no longer get canceled out by the formulation and so detection from those paths is possible (and likely, and desired).

    So, once again, why are you of the opinion that QED *uses* the optic path to do its calculations as opposed to *deriving* the optic path from those calculations?
     
  18. Polymath257

    Polymath257 Think & Care
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    I'd also point out that we never detect a *path* for a photon. What we detect is a photon at some point from a direction.

    Here is an analogy: consider a water wave coming from something bobbing in the water. That wave spreads out from that location in a circular pattern unless it is directed by barriers to a particular direction. But even when there are barriers, the water wave will spread out once it is past those barriers. So it is impossible to force a specific direction from the wave.

    Next, if a water wave comes to a barrier, it will get reflected. if you look at the reflected wave, the direction of travel will obey the optic law for the path back to the barrier and then back to the source. The angle of incidence of the wave will equal the angle of reflection of that wave. And this will be true at every point. The wave still goes to all other points, though.

    The main difference between this and light (other than the dimension) is that for light, the photon is a probability wave. The wave itself still is reflected in the same fashion. If you attempt to limit the direction, the wave will still spread out after the barrier is past. And even if you send out only a single photon, you still get a full probability wave that represents the probability to detect the photon.
     
  19. Unes

    Unes Active Member
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    I wonder how many times we have to examine this VERY SIMPLE LOGIC that we do NOT have to revisit it again and again and again!?

    Assume the optic law does NOT exist. A photon is sent to a mirror, with a SPECIFIC DIRECTION, please use QED’s technique, and tell me the direction of the reflected photon. You already acknowledged that QED without the target point is incapable of doing any calculation. Which statement in this argument you are disagreeing with? Please be precise.
     
    #39 Unes, Jul 5, 2017
    Last edited: Jul 5, 2017
  20. Polymath257

    Polymath257 Think & Care
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    OK, let's assume for the same of argument that you have shielding on your emitter and that it only allows photons to go out over a small range of angles. To limit to a specific angle is impossible because of the wave nature of a photon (first place I have issues).

    So, we calculate the probabilities of detecting a photon at two points: one close to the optic path and the other not close to such a path.

    In the first, when we do the calculation over all possible paths from the emitter, the shielding eliminates many of the paths (the photon is absorbed) and no probability is contributed. After that, *because the optic path is shortest*, the contribution from all paths is not zero and is even fairly high.

    In the second, the paths remaining after absorption by the shielding are all canceled out by other paths close by. So the probability of detection at such a point is very low: close to zero.

    The 'path' is the collection of points with high probability of detection. it is spread out because of the wave nature of things.
     
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