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

[Unes]

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.

This is not true, you are just making a false claim! All of a sudden you have forgotten about the randomness of QED’s paths! The energized electron, at the mirror surface, emits its photon in vast different directions, with varying possibilities. QED’s paths cover vast area, how does QED conveniently ignore all other paths, and just sticks with the optic path!? Isn’t that convenient!?

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.

In our experiment for detector at Y1 location, the produced path by QED all-path calculation it is an optic path, and it has the shortest path from the source to Y1 location. Although this optic path has different direction and different angle of incident, but QED all-path does not consider this information any way. This optic path from source to Y1 location is different from the other optic path that it is resulted by considering the direction and the angle of incident of the incoming photon. Since you are confused, then you keep making false statements, and false claims!

It seems Donald Trump approach it also has infected our scientists.

 

[Polymath257]

This is not true, you are just making a false claim! All of a sudden you have forgotten about the randomness of QED’s paths! The energized electron, at the mirror surface, emits its photon in vast different directions, with varying possibilities. QED’s paths cover vast area, how does QED conveniently ignore all other paths, and just sticks with the optic path!? Isn’t that convenient!?

Because the other paths cancel out.

In our experiment for detector at Y1 location, the produced path by QED all-path calculation it is an optic path, and it has the shortest path from the source to Y1 location. Although this optic path has different direction and different angle of incident, but QED all-path does not consider this information any way. This optic path from source to Y1 location is different from the other optic path that it is resulted by considering the direction and the angle of incident of the incoming photon. Since you are confused, then you keep making false statements, and false claims![/QUOTE]

The optic path to Y1 is eliminated because of the shielding that directs the photons to a different path.

It seems Donald Trump approach it also has infected our scientists.

OK, we are done. You clearly are not interested in the facts of the situation. It is *you* that is making up alternative 'facts' here.

 

[Unes]

This is not true, you are just making a false claim! All of a sudden you have forgotten about the randomness of QED’s paths! The energized electron, at the mirror surface, emits its photon in vast different directions, with varying possibilities. QED’s paths cover vast area, how does QED conveniently ignore all other paths, and just sticks with the optic path!? Isn’t that convenient!?

Because the other paths cancel out.

Such a vague statement, without offering any reason how the other paths are cancelled out!

In our experiment for detector at Y1 location, the produced path by QED all-path calculation it is an optic path, and it has the shortest path from the source to Y1 location. Although this optic path has different direction and different angle of incident, but QED all-path does not consider this information any way. This optic path from source to Y1 location is different from the other optic path that it is resulted by considering the direction and the angle of incident of the incoming photon. Since you are confused, then you keep making false statements, and false claims!

The optic path to Y1 is eliminated because of the shielding that directs the photons to a different path.

I do not see this mysterious shielding! QED all-path also does NOT observe any shielding either!

 

[Polymath257]

Such a vague statement, without offering any reason how the other paths are cancelled out!

Once again, if it isn't the shotest path, there are paths even shorter than cancel it out.

I do not see this mysterious shielding! QED all-path also does NOT observe any shielding either!

If there is no shielding, why is the photon limited to one direction?

 

[Unes]

Once again, if it isn't the shotest path, there are paths even shorter than cancel it out.

In regards that there are many random QED paths, I cannot make any sense out of your statement. The question is: How all other QED’s paths are ignored, in favor of the optic path?

If there is no shielding, why is the photon limited to one direction?

As I explained in my experiment, the source sends its photon in specific direction toward the mirror. The angle of incident is measured very accurately. Then the reflected photon is emitted by the optic law instruction.

Now, your question is the heart of my puzzle; “why does optic law exist for mirror reflection?” And also some people have indicated of the smart light, that light must be smart to find the shortest path.

Science in order to have an answer for these questions, it has offered the QED all-path method. If QED all-path can be verified successfully, then, all those questions have been answered naturally, and with complete satisfaction. So far QED all-path method has been accepted by all physicists that I have encountered with.

When I watched Prof. Richard Feynman lectures, I detected immediately the flaws in that argument, I started by asking, since all paths are allowed, then what stops us from having multiple detectors, this question exposes QED all-path flaw. Since then I have dissected QED all-path thoroughly. Thanks to you this thread has helped me a lot to streamline my objection. So, my conclusion is QED all-path method is false, phony, and deceptive! After discrediting QED all-path, then we can delve in investigating thoroughly; “why does optic law exist for mirror reflection?”
or , “why is the photon limited to one direction?”

 

[Polymath257]

In regards that there are many random QED paths, I cannot make any sense out of your statement. The question is: How all other QED’s paths are ignored, in favor of the optic path?

They are not 'ignored'. They cancel out in the calculation.

As I explained in my experiment, the source sends its photon in specific direction toward the mirror. The angle of incident is measured very accurately. Then the reflected photon is emitted by the optic law instruction.

The only way to get the source to emit in a psecific direction is through shielding.

Now, your question is the heart of my puzzle; “why does optic law exist for mirror reflection?”

And also some people have indicated of the smart light, that light must be smart to find the shortest path.

yes, it does.

Science in order to have an answer for these questions, it has offered the QED all-path method. If QED all-path can be verified successfully, then, all those questions have been answered naturally, and with complete satisfaction. So far QED all-path method has been accepted by all physicists that I have encountered with.

When I watched Prof. Richard Feynman lectures, I detected immediately the flaws in that argument, I started by asking, since all paths are allowed, then what stops us from having multiple detectors, this question exposes QED all-path flaw. Since then I have dissected QED all-path thoroughly. Thanks to you this thread has helped me a lot to streamline my objection. So, my conclusion is QED all-path method is false, phony, and deceptive! After discrediting QED all-path, then we can delve in investigating thoroughly; “why does optic law exist for mirror reflection?”
or , “why is the photon limited to one direction?”

To get the source to emit in only one direction requires shielding. Without the shielding, the source would emit in all directions with equal probability and then all detectors at equivalent distances *would* be equivalent.

With shielding, the paths that go into the shielding are absorbed and so do not count in the calculation.

I have explained this to you tiple times. I have a feeling others have explained this to you multiple times. It is clear that you are not interested in the specifics of how the formalism works, but only in attempting to get someone to take you criticism seriously. But clearly you are not interested in learning enough to see why the calculations actually work, only in giving your criticism.

 

[Unes]

To get the source to emit in only one direction requires shielding. Without the shielding, the source would emit in all directions with equal probability and then all detectors at equivalent distances *would* be equivalent.

With shielding, the paths that go into the shielding are absorbed and so do not count in the calculation.

I have explained this to you tiple times. I have a feeling others have explained this to you multiple times. It is clear that you are not interested in the specifics of how the formalism works, but only in attempting to get someone to take you criticism seriously. But clearly you are not interested in learning enough to see why the calculations actually work, only in giving your criticism.

Wow! Such an unkind remark! Now, you had to repeat the shielding subject triple times, in order to force me to comprehend it!? I wonder what triggered this raw emotion!? How could a polite scientific discussion be the cause of such resentment and animosity!? Is this maybe, because the frustration has set in, for being unable to poke a single hole in my solid argument? It takes a highly intelligent person to accept failure with grace and humility. I did not say anything regarding the shielding technique. Why do you think that I would dispute it!? Now, you can read my mind!? You are such an extraordinary talent. But I have a bad news for you, whatever you read, it was just in your mind, because, I have no problem with shielding techniques, shielding techniques are fine.

Now, you like to discuss the shielding technique, right!? I guess the shielding technique would suit fine for a diversionary tactic! I wonder what you studied, and which school taught you that technique for debates. I wonder whether you have the temperament to debate any scientific subject. The first rule in any discussion is to have the discipline to focus on the subject, instead of wondering around and discussing unrelated issues, and in this case shielding techniques for sure it is not related to the optic law for mirror reflection. WE DO OBSERVE THE MIRROR REFLECTION ALL THE TIME, AND THERE IS NO SHIELDING INVOLVED IN THIS PROCESS. And regard to the emission of the photon by the energized electron at the surface of the mirror, I do not see any shielding device that you are insisting we need to have in order the photon to be directed toward a specific direction. But you are saying in the case of the mirror reflection this is done by some imaginary QED functions. And you have not elaborated how these imaginary functions are operating. And how do they influence the direction of the reflected photon, despite of the fact that QED is based on randomness in all natural phenomena. BESIDES, THERE IS NO MENTION OF THESE QED IMAGINERY FUNCTIONS IN QED ALL-PATH METHOD at all.

When we are interested to learn something without any prejudice, then both parties genuinely are supposed to stay on the subject to finalize their differences. Unfortunately this thread is turning toward an endless dialogue without aiming toward any specific conclusion, a form of an entertainment without achieving any specific goal. When we want to examine the optic law for mirror reflection, IT IS NATURAL AND IT IS ESSENTIAL THAT WE SHOULD HAVE A SOURCE THAT IT SENDS ITS PHOTONS IN A SPECIFIC DIRECTION, this is an equipment issue, and we let a technician to deal with it. How the technician builds this kind of equipment, that is NOT our concern, and whether he has to use shielding, or any other techniques that is the job of the technician, and it is NOT the subject of this discussion. We ask the technician to provide us with an equipment that it can send ONE photon at a time, and at the specific direction that we choose. Is this point that hard to comprehend, that we were needed to discuss it in these trivial details!? In Prof Richard Feynman lectures in Auckland University the issue of the source was so trivial, that he did not even bother to give any explanation about it. But, for sure, these details are very effective for the diversionary tactic!

It seems when you were cornered to admit the falsehood of QED all-path method, all of sudden you got interested in shielding issue and you needed to emphasize it triple times! Here, I have provided a dynamic and very important subject which it is irrefutable, and a curious scientist would be excited to take advantage of it, instead you have resorted to diversionary tactics avoiding to focus on the subject!

 

[Unes]

Here, I like to present the refined version of my argument.

QED All-Path is Challenged

QED all-path method for mirror reflection claims by adding all the amplitude vectors for all the probable paths, it finds the Optic Path naturally, and independently. This is absolutely a false claim!

In order QED all-path method claim to be valid, QED all-path is supposed to reach to the Optic Path conclusion by its own method independently, and absolutely without any lead from the Optic Path. In the following short passage I am going to expose the flaw within QED’s deceptive argument.

The photon source at point S sends a single photon to the surface of the mirror. This photon has specific direction and specific angle of incident, and the Optic Path defines the path of the reflected photon. Let us assume the reflected photon gets detected by a detector at the target point T1. Since QED all-path method includes all the incoming paths into its configurations, then it ignores the specificity of the direction of the incoming photon. QED all-path does not identify the receiving detector at the target point T1 by its own natural and random method. If I move the receiving detector one inch from the target point T1, side way, to the target point T2, QED all-path method still produces an optic path for the target point T2. In fact if I move the detector to different target points of T3, T4, T5, . . . , T10,000, each of these target points produces its own VALID optic path, and QED all-path can NOT differentiate between any of these target points. QED all-path is incapable to identify which of these 10,000 target points actually receives the reflected photon. This is where QED all-path cheats, QED conveniently implements the correct detector at the target point T1 into its methods, and because of this cheating, it naturally ends up with the correct result! QED all-path is supposed to figure out the reflected path by its own method independently, and not by cheating.

The ramification of this cheating is huge, the energized electron at the surface of the mirror emits its photon at the specific direction that it is defined by the Optic Path, this emission with specified direction rejects QED’s randomness to its core.

 

[Unes]

To present another proof for the falsehood of the QED All-Path Integral approach I have devised the following argument:

I choose a square mirror with dimensions of 1m by 1m. A source at point A sends one photon to the surface of the mirror, this photon hits the mirror at point B, and it gets reflected, optic law for mirror reflection tells us the direction of the reflected photon, let us assume that the reflected photon gets detected at point C.

Now, I black out a tiny circle from the surface of the mirror, this little circle has a radius of just 1 mm and it is centered at point B. This tiny black out only covers 0.0314% of the mirror surface, and still more than 99.96% of the mirror surface is intact. The source at point A send another photon to point B, but there is no reflection from point B any more, and Optic Law for mirror reflection tells us, that the detector at point C no longer will detect any photon reflection. No matter how we manipulate the rest of the mirror surface to maximize the summation of the amplitude vectors, there will be no reflection from the rest of the mirror surface in any direction, none whatsoever.


This is quite obvious if point B reflects 100% of the incident photon, then naturally there is nothing left for the rest of the mirror to reflect. However, if still some physicists who are advocating for QED insist that there is still some left over for the rest of the mirror to reflect, then I would point out that just the assertion of the left-over drastically affects the value of the amplitude vectors everywhere, and the presented calculation for the summation of the amplitude vectors is valid no more. This analysis concludes that only point B on the mirror surface reflects the incoming photon, and the rest of the mirror surface contributes nothing, in the mirror reflection process. In conclusion we can flatly reject and dismiss the fancy QED All-Path Integral argument as being a bogus argument.


If still there are some physicists who doubt my conclusion, in order to settle all doubts, we can prove the validity of my conclusion with an experiment. The experiment will show that the reflection happens only based on the Optic Law, and there is no reflection from the rest of the mirror.

 

[Polymath257]

To present another proof for the falsehood of the QED All-Path Integral approach I have devised the following argument:

I choose a square mirror with dimensions of 1m by 1m. A source at point A sends one photon to the surface of the mirror, this photon hits the mirror at point B, and it gets reflected, optic law for mirror reflection tells us the direction of the reflected photon, let us assume that the reflected photon gets detected at point C.

Already you have a problem. You cannot localize a photon to one point, so claiming it is reflected from point B is going against the physics. Also, it is not the *same* photon that is detected at C as the one emitted by A. The original photon is absorbed and a new one is emitted.

Now, I black out a tiny circle from the surface of the mirror, this little circle has a radius of just 1 mm and it is centered at point B. This tiny black out only covers 0.0314% of the mirror surface, and still more than 99.96% of the mirror surface is intact. The source at point A send another photon to point B, but there is no reflection from point B any more, and Optic Law for mirror reflection tells us, that the detector at point C no longer will detect any photon reflection. No matter how we manipulate the rest of the mirror surface to maximize the summation of the amplitude vectors, there will be no reflection from the rest of the mirror surface in any direction, none whatsoever.

First of all, this isn't even a quantum effect. It is a wave effect. And the experiment has been done. The part corresponding to the blacked out portion of the mirror *does* show a reflection. Look up 'Fresnel diffraction from an occluded mirror'. If you direct light to the center of a mirror and occlude the center of the mirror, there will still be a bright dot at the center of the image because of diffraction.

This is quite obvious if point B reflects 100% of the incident photon, then naturally there is nothing left for the rest of the mirror to reflect. However, if still some physicists who are advocating for QED insist that there is still some left over for the rest of the mirror to reflect, then I would point out that just the assertion of the left-over drastically affects the value of the amplitude vectors everywhere, and the presented calculation for the summation of the amplitude vectors is valid no more. This analysis concludes that only point B on the mirror surface reflects the incoming photon, and the rest of the mirror surface contributes nothing, in the mirror reflection process. In conclusion we can flatly reject and dismiss the fancy QED All-Path Integral argument as being a bogus argument.

If still there are some physicists who doubt my conclusion, in order to settle all doubts, we can prove the validity of my conclusion with an experiment. The experiment will show that the reflection happens only based on the Optic Law, and there is no reflection from the rest of the mirror.

The experiment was done long ago and the results don't agree with your intuition.

 

[Unes]

Polymath257,
Thank you for pointing out the flaws in my novice assumptions, I really appreciate that. Could you please also explain the assumption that QED All-Path Integral approach makes, that all the reflections from the mirror surface are directed toward the detecting photon-multiplier, these emissions are supposed to be random and to go in all directions, and not conveniently toward our detecting device. To me this false assumption neutralizes a glaring contradiction that exists within the QED theory, and makes the result to get compatible with the Optic Law.


QED says that the incident photon gets absorbed by an electron and then this energized electron emits the reflected photon, if this assumption is correct then we should not have Optic Law for mirror reflection, because we have learned the direction that the energized electron emits its photon cannot be predicted, and the emission is random in all directions. Clearly something very important is missing in this QED analysis, and our QED All-path Integral approach deceptively hides this glaring contradiction that exists between what the theory suggests and what we observe in nature.

 

[Unes]

Last week I suggested an experiment to prove the falsehood of QED All-Path Integral approach, now I would like to make some changes to that experiment, in order to address the Fresnel diffraction issue.

I choose a square mirror with dimensions of 1m by 1m. A source at point A sends one photon to the surface of the mirror, this photon hits the mirror at point B, and it gets reflected, optic law for mirror reflection gives us the direction of the reflected photon, the reflected photon gets detected at point C.

Now, I blackout a tiny circle from the surface of the mirror, this little circle has a radius of just 1 cm and it is centered at point B. With this blackout area still 99.96% of the mirror surface is intact. Even if we increase the radius of the blackout area to 5cm, still more than 99.2% of the mirror surface is intact. And for this large blackout area, the diffraction may no longer be an issue.

Now I need to answer the following objections:

Already you have a problem. You cannot localize a photon to one point, so claiming it is reflected from point B is going against the physics. Also, it is not the *same* photon that is detected at C as the one emitted by A. The original photon is absorbed and a new one is emitted.

I did not say that the reflected photon is the “same” photon. I interjected that the incident photon is reflected from point B, and that is how it is explained by the Optic Law, apparently point B is a small area, and for sure it is not a mathematical point.

First of all, this isn't even a quantum effect. It is a wave effect. And the experiment has been done. The part corresponding to the blacked out portion of the mirror *does* show a reflection. Look up 'Fresnel diffraction from an occluded mirror'. If you direct light to the center of a mirror and occlude the center of the mirror, there will still be a bright dot at the center of the image because of diffraction.

QED All-Path Integral approach is a quantum effect, and that is the subject of this analysis. We are discussing whether; after we create a blackout circle, with radius of 5cm, centered at point B, do we still get any reflection due to the effect of the QED All-Path Integral approach, that our photon detector at point C can detect it?

According to the point that you made, the reflection would be from Fresnel diffraction from an occluded mirror, and that is not the reflection that it would be compatible with the Optic Law that our detector at point C can detect it. So, no matter how we manipulate the mirror surface in order to maximize the summation of the amplitude vectors, still there would be NO reflection from the mirror surface, that our detector at point C can detect it. So, we can confidently conclude that QED All-Path Integral approach is false.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Here I would like to suggest a different experiment; instead of making a blackout circle centered at point B, we make a hole in the mirror, centered at point B, with the radius of 5cm; large enough that the incident photons pass through the center of the hole, and the effect of diffraction from the edge of the hole would be negligible. Now, I set up a second detecting device at point D on the back side of the mirror to detect these pass through incident photons. Again like the blackout case still 99.2% of the mirror surface is intact. And I assume that each incident photon that the source emits it reaches to the second detector without any incident.

Now, we allow the surface of the mirror to be manipulated in order to maximize the summation of the amplitude vectors for the QED’s “All-Pass Integral approach”. For any configuration that we can imagine for the surface of the mirror, there can NOT be any photon detected by the first detector at point C any more; this is because any photon detected by this detector, that will violate the law of the conservation of energy. So, QED “All-Pass Integral approach” can NOT produce any reflection, and it is a false argument.

 

[Polymath257]

If the size of the hole is larger than the wavelength of the light, the path integral formulation gives the same result as classical optics.

You have to have mirrors or obstructions, or holes with sizes similar to the wavelength of the photons used to get anything other than the classical results. But, when you *do* have such, the results agree with classical optics. But they do agree with the wave theories of light. And the path integral formulation reduces to the wave version when enough photons are used.

 

[Unes]

If the size of the hole is larger than the wavelength of the light, the path integral formulation gives the same result as classical optics.

You have to have mirrors or obstructions, or holes with sizes similar to the wavelength of the photons used to get anything other than the classical results. But, when you *do* have such, the results agree with classical optics. But they do agree with the wave theories of light. And the path integral formulation reduces to the wave version when enough photons are used.

Since in our experiments the radius of the blackout area (centered at point B), and the radius of the hole (centered at point B), are much greater than the wavelength of the incident photon (r >> λ), then, no matter how the surface of the rest of the mirror gets manipulated in order to maximize the summation of the amplitude vectors, there will be no reflection from more than 99.2% surface of the mirror, for either cases of these two experiments. So, we proved that QED All-Path integral approach is a bogus argument and it has no validity.

Even though we proved the falsehood of QED “All-Path Integral approach, but my focal point is: it has deceptively hidden the glaring contradiction that exists within the process of the mirror reflection.

 

[Polymath257]

What you seem to miss is that the optic law (equal incidence and reflection angles) is an *approximation*. And it is an approximation that works only when the wavelength of the light is small compared to all other sizes in the problem (mirror size, patch size, etc).

Once we get to the point where the patch or mirror is about the same size as the wavelength, the optic law is routinely violated. And this is verified by actual observations.

This isn't even a quantum effect. it is a wave effect. Quantum effects have to do with the discreteness of photons, which have both wave and particle aspects.

In your case, where the wavelength is much smaller than other relevant sizes, the path formulation will give results that cancel out and no reflection will be seen.

I would suggest that you read Feynman's book 'QED' for more details.

 

[Unes]

I would suggest that you read Feynman's book 'QED' for more details.

I bought and I read Feynman’s book “QED: The Strange Theory of Light and Matter”; the content of the book is very similar to his lecture at Auckland University on 1979.

In my suggested experiments the size of the Blackout area is a circle with radius of 5 cm, and the size of the hole is a circle with radius of 5 cm, do these sizes still produce diffraction with photons of the visible light? If they do not, then why are you keep discussing diffraction, and wave properties of light? With all due respect, the wave properties of light are NOT the subject of this discussion.

Polymath257, you are so knowledgeable, I desperately want to know the answer to the following question, I will be eternally grateful to you for solving this puzzle for me; QED says that the incident photon gets absorbed by an electron and then this energized electron emits the reflected photon. Now the problem is; since the energized electron emits photon in a random direction, then how can we have the Optic Law for mirror reflection, where the reflection happens is in a specific direction?

 

[Polymath257]

I bought and I read Feynman’s book “QED: The Strange Theory of Light and Matter”; the content of the book is very similar to his lecture at Auckland University on 1979.

In my suggested experiments the size of the Blackout area is a circle with radius of 5 cm, and the size of the hole is a circle with radius of 5 cm, do these sizes still produce diffraction with photons of the visible light? If they do not, then why are you keep discussing diffraction, and wave properties of light? With all due respect, the wave properties of light are NOT the subject of this discussion.

Polymath257, you are so knowledgeable, I desperately want to know the answer to the following question, I will be eternally grateful to you for solving this puzzle for me; QED says that the incident photon gets absorbed by an electron and then this energized electron emits the reflected photon. Now the problem is; since the energized electron emits photon in a random direction, then how can we have the Optic Law for mirror reflection, where the reflection happens is in a specific direction?

The wave nature is essential when doing the path formulation. The phase of the wave is what goes into the final sum over paths. In the case of reflection, the sums from non-optic law paths cancel out due to phase changes in nearby paths.

As in ALL quantum calculations, you need to find the probability of detection. In the case of photons being reflected that probability is low except on the optic path.

 

[Unes]

The wave nature is essential when doing the path formulation. The phase of the wave is what goes into the final sum over paths. In the case of reflection, the sums from non-optic law paths cancel out due to phase changes in nearby paths.

As in ALL quantum calculations, you need to find the probability of detection. In the case of photons being reflected that probability is low except on the optic path.

As it is shown in the Feynman’s lectures, we do NOT need the Optic Path in order to have reflection; by removing some strips from the surface of the mirror, we can remove the negative fazed amplitude vectors, and that process generates a big value for the summation of the amplitude vectors for the rest of the mirror. So, in my experiments where I have blackout a circle, centered at point B, or cut-out a circle, centered at point B, we have removed the Optic Paths, but according to Prof. Feynman’s lectures we should observe reflections from the stripped out mirror. And the predictions from my suggested experiments are; that we will NOT observe any reflection. As a result my experiments prove beyond any doubt the faulty nature of the QED All-Path Integral approach.

In any case, even with the help of the faulty QED All-path Integral approach, still there is no answer for the glaring contradiction that exists for the mirror reflection, then, it is safe to say that physics yet to admit for this colossal failure.

 

[Polymath257]

As it is shown in the Feynman’s lectures, we do NOT need the Optic Path in order to have reflection; by removing some strips from the surface of the mirror, we can remove the negative fazed amplitude vectors, and that process generates a big value for the summation of the amplitude vectors for the rest of the mirror. So, in my experiments where I have blackout a circle, centered at point B, or cut-out a circle, centered at point B, we have removed the Optic Paths, but according to Prof. Feynman’s lectures we should observe reflections from the stripped out mirror. And the predictions from my suggested experiments are; that we will NOT observe any reflection. As a result my experiments prove beyond any doubt the faulty nature of the QED All-Path Integral approach.

In any case, even with the help of the faulty QED All-path Integral approach, still there is no answer for the glaring contradiction that exists for the mirror reflection, then, it is safe to say that physics yet to admit for this colossal failure.

In order to not have cancellation in the striped out mirror, the size of the stripes need to be about the same as the wavelength of the light involved. That is what is required to eliminate the 'negative amplitude vectors'.

And, when you that condition met, you *do* get 'reflection' at angles that are not the optical path angle. That is what happens in a diffraction grating, after all.

Which means your prediction is wrong (and it is even well-known to be wrong).

 

[Unes]

In order to not have cancellation in the striped out mirror, the size of the stripes need to be about the same as the wavelength of the light involved. That is what is required to eliminate the 'negative amplitude vectors'.

And, when you that condition met, you *do* get 'reflection' at angles that are not the optical path angle. That is what happens in a diffraction grating, after all.

Which means your prediction is wrong (and it is even well-known to be wrong).

In my suggested experiment; the mirror is striped out according to the specification of diffraction grating, and it has a hole in it, with radius of 5 cm, centered at point B. Now, when the incident photon passes through the hole, and it is captured by the second detector, then, there is no energy left to perform a diffraction grating. As matter of fact if in this situation, any diffraction grating can be detected, then we have to throw out the Law of the conservation of energy. For this scenario the equation of energy according to what you say, it will be as follow:

One incident photon (E1) = One captured photon by the second detector (E1) + the energy from the diffraction grating

This equation cannot be right, because on the right side we do have an extra energy (from the diffraction grating). So, by claiming that I am wrong, that does not make it so; your claim should also satisfy the Law of conservation of energy, which it does NOT.

This conclusion simply means; that in All-Path Integral approach those paths are just imaginary paths, and they do NOT carry any energy, and QED All-Path integral approach is false.

This is a strange behavior from the physicists, it looks like they have been bewitched by QED All-Path Integral approach, that they refuse to see these obvious flaws.

As I have said repeatedly in my previous posts, physicists by introducing the faulty QED All-path Integral approach, they have tried to ignore the glaring contradiction that exists for the mirror reflection. One should tell to these physicists, never mind with the idea of QED All-Path Integral approach, please answer the following puzzle:

QED says that the incident photon gets absorbed by an electron and then this energized electron emits the reflected photon, since the energized electron emits photon in a random direction, then how can we have the Optic Law for mirror reflection, where the reflection happens is in a specific direction?