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Thread: New Wisdom Unifies Physics

  1. #1 New Wisdom Unifies Physics 
    Forum Freshman andrewgray's Avatar
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    In 2011, I published this paper in (the peer reviewed) Physics Essays (Volume 24: Pages 139-153, 2011):

    "Theory of Intermittent Electrons".


    Not many physicists could understand the paper because a lot of material is covered, and not enough time is spent on each topic. So I have spent some years making 4 videos to explain the paper!


    These four videos explain the paper, "Theory of Intermittent Electrons".

    ("A video says a 1,000,000 words").



    So here are the videos. I encourage you to watch them and let me hear your feedback!
    Summary: This theory is a way to explain microscopic physics without the use of photons.


    Episode 1: The X-Ray Frequency Limit!

    https://drive.google.com/file/d/1lAd...QDh5DZDZu/view



    Episode 2: The PhotoElectric Effect!

    https://drive.google.com/file/d/1tg6...SRUZMa0_q/view


    Episode 3: The Atom! (Part 1)

    https://drive.google.com/file/d/1NCy...xIP4tr9zS/view



    Episode 3: The Atom! (Part 2)

    https://drive.google.com/file/d/1BWn...mPPUcZ6bq/view


    Finally, I am working on the final unifying video:

    Episode 4: Gravity!

    lI need your feedback as I am doing this video. Let's see how the first 3 Episodes go.


    Andrew Ancel Gray


    Last edited by andrewgray; September 29th, 2020 at 12:43 PM.
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    Link doesn't work!


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    Now they do.

    Andrew Ancel Gray
    Last edited by andrewgray; September 29th, 2020 at 12:40 PM.
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    Nope, not going to watch an hour long video. Why not pick on topic and discuss it, this is a discussion forum.
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    Why not pick a topic and discuss it, this is a discussion forum.
    Bufofrog,

    That is an offer I cannot refuse. Let's talk about the double slit experiment:

    This is where you lower the intensity of the incident light on the double slits until you start getting just dots on the film, and over time it builds up into a diffraction pattern.

    "AH... INDIVIDUAL PHOTONS!", the QM people say. What a myth. What you have is simply a very low intensity light wave striking the film at the film's threshold intensity, and in the maxima, the most sensitive silver bromide crystals start to fire. Let me repeat. This is exactly what you would expect if a low intensity light wave struck the film at the threshold intensity for the film. A good analogy would be to position a digital camera instead of film behind the slits! No one in their right mind would think "individual photons" would be setting off the individual transistor cells of a digital camera. BUT YOU WOULD GET THE SAME DOT PATTERN IN A DIGITAL CAMERA as you lowered the light wave intensity to the threshold of the digital light cell transistors in the camera! No "photons" here! Just a low intensity light wave!

    And finally,

    to see how silly this experiment's "single photon" claim is when using film, we need only examine "minimum blackening" of film. This guy tells us what it is:

    Radiometry and photometry in astronomy

    Minimum blackening for film, he says is 0.004 lux-seconds.

    So expose the film at just 1% of this "minimum blackening". Probably no spots on the film from this exposure.
    So we have a 0.00004 lux-second exposure of the film with NO dots at 1%.

    Do you know how many fictional "light particles" this "no-dots-at-all" exposure represents?
    Well, a lux-second is approximately 0.001 Joules/cm2


    A fictional "red light particle" would have about 2 x 10-19 Joules.
    And if we divide it out we see that (.00004 x .001 J/cm2)/(2 x 10-19 J) = 200,000,000,000/cm2.

    So 200,000,000,000 fictional "red light particles" can strike 1 cm2 on the film... and not leave a single dot!

    So claiming that a single dot on a piece of film (oh my goodness!) represents ONE single fictional "light particle" is beyond silly.
    (Oh my goodness!)

    The same is true if you use phtotomultiplier tubes, or any other macroscopic light detector! Just lower the light wave exposure to the threshold of the detector, and you see each individual photomultiplier tube going off near the maxima in the very same way(!) (as a low intensity light wave strikes them near their threshold exposure! )


    As you will understand from my videos,

    "There is no reason to use fictitious light particles for any explanation in physics."



    Andrew Ancel Gray
    Last edited by andrewgray; October 7th, 2020 at 06:30 AM.
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    So, are you saying a single photon or a single electron cannot interfere with itself?

    I think film and photomultiplier tubes are kinda ancient history and CCD detectors are used. Single electrons can certainly be sent through the double slit apparatus.

    Edit: Looking at your post again I am a bit confused about the point you are trying to make. It seems like you object to the idea of single photons or electrons going through the double slit but not the results the experiments, showing the wave nature of photons, electrons and other particles?
    Last edited by Bufofrog; September 29th, 2020 at 07:33 AM.
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    Bufofrog,

    The point I am trying to make is:

    There are no light particles!

    Once you understand what is actually going on in the double slit experiment, you will see just how silly it has been for about a 100 years!

    Photomultipliers, sensitive film, CCD cameras, etc., etc. , are all macroscopic detectors and if you look at the threshold intensity for them, probably BILLIONS of "fauxtons" could strike them per cm2, and not leave a detection. Do you see how silly the "single-fauxton-at-a-time" claim is yet?

    So the point again, that I am trying to make is:

    There are no light particles!

    Now, there is low current electron interference in an electron microscope! So you put a pico-ammeter in series with your electron gun in the microscope and drop the current to the equivalent of "one-electron-at-a-time!" Do you then go ahead and make the "one-electron-at-a-time" claim? NO YOU DO NOT! Look Bufofrog, when high voltage charge builds up and starts to discharge, does it discharge in a nice smooth even flow???? OF COURSE NOT. When lightning builds up the the thunder clouds, does it discharge nice and evenly to the ground??? OF COURSE NOT!

    So the electrons that build up on the tip of the electron gun in the microscope obviously act the same way. They build up until there is enough of them to have a high enough voltage to overcome the gap, AND THEN THEY DISCHARGE IN A GROUP LIKE LIGHTNING, even though the average current through the microscope is "one-electron-at-a-time!" The electrons actually go through the microscope in high voltage discharge groups. There are plenty of electrons to interfere, so "one-electron-at-a-time" claims are again actually silly.

    It is actually a good thing I got out of physics and got into electrical engineering for a while so I could combine them and actually have the "gumption" to figure out what is actually going on! So here is how electron interference actually works in an electron microscope. The current in the pico-ammeter is set to "one-electron-at-a-time" amps, but actually they are being emitted in high voltage groups as one would expect with high voltage discharges!



    So the electrons are actually pulsating in a coherent group (they all had to be ON when ejected from gun), and interfering as individual PARTICLES as they travel towards the film (which by the same argument CANNOT DETECT SINGLE ELECTRONS!). If two electrons "come ON" at the same time when they cross, they will be radically repelled and not make it to where they were going (a minima). If the two electrons cross while one is ON and the other OFF, then they will continue on to where they were going (a maxima). It is just that simple. Now...


    No matter waves, no light particles, no "single-anything" interference.

    That is what I am trying to say. I hope you now understand what I am trying to tell you, agree with me or not!


    Andrew Ancel Gray
    Last edited by andrewgray; September 30th, 2020 at 09:47 PM.
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    So the electrons are actually pulsating in a coherent group (they all had to be ON when ejected from gun)

    You didn't answer my question, but this seems to indicate that you believe that single electrons don't interfere with themselves.
    Not many physicists could understand the paper because a lot of material is covered,

    Based on this last post, let me assure you, it is not a matter of physicist not understanding, it is a matter of physicist dismissing your conjectures.
    Additionally, the years you have spent on this leave no doubt in my mind that no amount of discussion can sway you from your beliefs. So I won't waste my time in a fruitless discussion.
    Good luck and have fun with your ideas!
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    Bufofrog,

    It would have been better if you would have used logic to try to explain why you disagree with my theory. Remember, it was you who said:

    this is a discussion forum
    I really do not want to leave you behind, Bufofrog. Rather, I would hope you would want to go along for the ride, even if you disagree.


    And yes, that is correct. Single electrons DO NOT interfere with themselves!

    No "single-anything" interference!
    And just like film cannot detect TRILLIONS of "fauxtons-worth" of energy (see above), film definitely IS NOT SENSITIVE ENOUGH to detect single electrons. Probably millions of electrons have to strike the film in a maxima to get a film dot.

    So bufofrog (in reference to above), what do you say about the 1% minimum blackening for film leaving TRILLIONS of FAUXTONs-WORTH OF ENERGY UNDETECTED?

    Andrew Ancel Gray
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  11. #10  
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    Quote Originally Posted by andrewgray View Post
    to see how silly this experiment's "single photon" claim is when using film, we need only examine "minimum blackening" of film. This guy tells us what it is:

    Radiometry and photometry in astronomy

    Minimum blackening for film, he says is 0.004 lux-seconds.

    So expose the film at just 1% of this "minimum blackening". Probably no spots on the film from this exposure.
    So we have a 0.00004 lux-second exposure of the film with NO dots at 1%.

    Do you know how many fictional "light particles" this "no-dots-at-all" exposure represents?
    Well, a lux-second is approximately 0.001 Joules/cm2


    A fictional "red light particle" would have about 2 x 10-19 Joules.
    And if we divide it out we see that (.00004 x .001 J/cm2)/(2 x 10-19 J) = 200,000,000,000/cm2.

    So 200,000,000,000 fictional "red light particles" can strike 1 cm2 on the film... and not leave a single dot!

    So claiming that a single dot on a piece of film (oh my goodness!) represents ONE single fictional "light particle" is beyond silly.
    (Oh my goodness!)

    The same is true if you use phtotomultiplier tubes, or any other macroscopic light detector! Just lower the light wave intensity to the threshold of the detector, and you see each individual photomultiplier tube going off near the maxima in the very same way(!) (as a low intensity light wave strikes them near their threshold intensity! )
    I think you misunderstood the notion of "minimum blackening" given in Radiometry and photometry in astronomy. It is not the minimum exposure required to produce a single dot, but the minimum exposure required to produce a visible uniform blackening of the film. It's about the use of film in astronomy, where one is interested in more than single dots produced by single photons.

    Also, you have no basis to say that the same applies to photomultiplier tubes or any other macroscopic light detector.

    Bear in mind that you not only have to explain why a low light intensity pattern is made up of dots, but why the accumulation of dots over time produces a diffraction pattern.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    KJW,

    I understand what you are saying. But the numbers are really against your logic.

    Not satisfied that 1% of "minimum blackening" is below the threshold of the film???

    Well, then... take one-billionth of minimum blackening!

    So this is way below the threshold for the film! (I hope you are not saying that a diffraction pattern will build up
    at an exposure of one-billionth of minimum blackening! )

    Anyway, at this tiny exposure below the film threshold, if we divide it out we see that

    (.000000000004 x .001 J/cm2)/(2 x 10-19 J) = 20,000/cm2

    20,000 fictional "red light particles" can strike 1 cm2 on the film... and not leave a single dot!


    KJW, the numbers for macroscopic light detectors are just against "one-fauxton-at-a-time" claims.

    Now, you want me to elaborate why

    why the accumulation of dots over time produces a diffraction pattern.
    OK. This is fairly simple. The light intensity of the incident light wave is brought up until the film starts producing dots. The most sensitive silver bromide crystals start "to fire". That is, the intensity is brought up and left on until it gets to the threshold for the film. Now where is the exposure of the dim light wave striking the film the greatest??? Well, in the maxima, of course! The film is still below threshold exposure in the minima! So where do you suppose the dim light wave will start producing dots on the film? Well, in the maxima, of course! So it is clear to me that the dim light wave will start producing dots in the maxima of the diffraction pattern (of the dim light wave striking at threshold exposure only in the maxima)!

    OK. Whew.


    Next, KJW, before I go into other "macroscopic light detectors", let me hear you claim that photomultipliers are a billion times more sensitive than film! Otherwise, the logic is just the same. The numbers are against you, KJW.

    Andrew Ancel Gray
    Last edited by andrewgray; October 7th, 2020 at 06:40 AM.
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  13. #12  
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    Quote Originally Posted by andrewgray View Post
    I understand what you are saying.
    Apparently not... because you persist with what you said before, including the "and not leave a single dot". You say that "minimum blackening" is the threshold intensity below which the film will not respond at all. This is not what Radiometry and photometry in astronomy is saying.


    Quote Originally Posted by andrewgray View Post
    I hope you are not saying that a diffraction pattern will build up at an exposure of one-billionth of minimum blackening!
    I am saying that a diffraction pattern will build up regardless of how low the intensity of the light is. There is no threshold intensity below which the film does not respond at all. If the intensity is extremely low, the diffraction pattern will build up extremely slowly, but it will build up. This is a key point that establishes the existence of photons because if light were simply a wave, then at low intensity, there would not be sufficient energy to activate the film at all.

    This was also the problem faced by physicists trying to explain the photoelectric effect prior to Einstein. At the very low light levels used in the experiments, there isn't enough energy to eject the electrons from the metal if the light were simply a wave.


    Quote Originally Posted by andrewgray View Post
    So it is clear to me that the dim light wave will start producing dots in the maxima of the diffraction pattern
    You didn't explain why the diffraction pattern is built up of dots.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    You didn't explain why the diffraction pattern is built up of dots.
    KJW,

    OK. Let us imagine that a low intensity light wave is incident on a CCD camera. Now, let's bring up the exposure from the light wave until some of the most sensitive CCD transistors in the camera start "to fire". No one in there right mind is going to claim these single firings of CCD camera transistors are due to "single fauxtons"! So the CCD camera starts off with "single dots". So why does the CCD camera start off with "single dots" from the diffraction pattern from the dim light wave???? Why does the CCD camera diffraction pattern "build up with dots", when clearly we are NOT talking about "single-fauxton-at-a-time" exposures on macroscopic CCD transistors???

    Because the light exposure was brought up to the threshold for firing of the CCD transistors, and the most sensitive transistors in the maxima of the diffraction pattern start "to fire".

    The most sensitive silver bromide crystals in the film act the same way as the most sensitive CCD transistors in the camera. That is why.

    Andrew Ancel Gray
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  15. #14  
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    There are a number of things wrong with your explanation of why the diffraction pattern is built up of dots. Firstly:

    Quote Originally Posted by andrewgray View Post
    Let us imagine that a low intensity light wave is incident on a CCD camera. Now, let's bring up the exposure from the light wave until some of the most sensitive CCD transistors in the camera start "to fire".
    This is not how such experiments are performed. The experiments are performed by decreasing the intensity of the light until there are no more than one photon at a time on average (coherence prevents exact photon numbers while in flight).


    Quote Originally Posted by andrewgray View Post
    the most sensitive transistors in the maxima of the diffraction pattern start "to fire"
    The problem with this is that dots occur everywhere except where the intensity is precisely zero. Also, although the greatest probability of dots occurring is at the maxima, dots can occur near the minima at any time. It isn't the case that dots necessarily occur at the maxima first. In other words, the pattern of dots is not consistent with the most sensitive locations in the maxima of the diffraction pattern.


    Finally, the biggest problem, the problem that you ignored from my previous post, is that at the low intensities used in the experiments, there isn't enough energy at the dot's location to produce the dot.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    KJW,

    I disagree with you, but I can see we will not resolve this here with the double slit experiment. Since this is a friendly discussion, I can see we will have to move to other evidence against "fauxtons" (you say "photons", I say "fauxtons" just to be clear that you believe in them and I do not). You seem fairly sophisticated in your belief in "fauxtons", so let's try the photoelectric effect. I would be very interested in what you have to say about the evidence here!

    First, did you know that the photoelectric effect is a transverse ejection? That is, did you know that most of the photoelectrons in the photoelectric effect are ejected sideways relative to the incoming light polarization? Take for example, the ejections from copper crystals! If you send in a vertically polarized light wave, you get many, many ejections. But if you send in a horizontally polarized light wave, the ejections drop drastically.
    Like this plot from Pedersoli:



    In this plot at 60o, you can see there are about 8 times as many photoejections by the light wave with its electric vector pointed out of the metal, as by the light wave with its electric vector parallel to the surface of the metal.

    Now if you think of "fauxtons" being "light particles" being absorbed by the electrons in the metal, you would immediately think of the electrons being driven deeper into the metal as they absorb particles, and not being ejected sideways out along the polarization of the incident light wave. And from Milton Chaffee,
    The Angular Distribution of Photoelectrons Ejected by Polarized Ultraviolet Light in Potassium Vapor

    https://journals.aps.org/pr/abstract/10.1103/PhysRev.37.1233

    He also finds that the photoelectrons are mostly ejected sideways along the electric vector of the light wave.

    Now in my new theory and in Episode 2 of my videos (see above), I explain very clearly why this is so according to the Theory of Intermittent Electrons. But KJW, I wanted to hear your explanation! That is,

    1. Why are photoelectrons mostly ejected sideways out along the polarization of the incident light wave?


    2. And explain to me why wouldn't the "light particles" being absorbed just drive the electrons deeper into the metal?


    Andrew Ancel Gray
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  17. #16  
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    Quote Originally Posted by andrewgray View Post
    1. Why are photoelectrons mostly ejected sideways out along the polarization of the incident light wave?
    I didn't know that they were. But since I have no particular reason to doubt this, I'll ask "So what?". It should be noted that light also exists as an electromagnetic wave, and that properties of electromagnetic waves, such as polarisation, also give rise to the behaviour of photons. In other words, you cannot disprove the existence of photons by demonstrating the existence of electromagnetic waves.

    For example, the phenomenon of refraction is easiest to explain in terms of electromagnetic waves, and attempting to explain this directly in terms of photons leads to an erroneous explanation. But this doesn't disprove photons, only that a more complication explanation in terms of photons is required.


    Quote Originally Posted by andrewgray View Post
    2. And explain to me why wouldn't the "light particles" being absorbed just drive the electrons deeper into the metal?
    I don't know why you think that it would. When an atom absorbs a photon, an electron in an orbital transitions to an orbital of higher energy, or if the energy of the photon exceeds the ionisation energy, the electron is ejected from the atom with a kinetic energy equal to the excess energy from the photon.


    But perhaps you could explain why increasing the intensity of the light increases the rate that electrons are emitted but not their energy, and increasing the frequency of the light increases the energy of the individual electrons.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    KJW,

    But you are violating a serious principle of QM dogma from the the QM priest himself, Niels Bohr: His principle of complementarity!

    You can explain light as particles or waves, but not both at the same time!
    Naughty! Naughty! I'm just pointing this inconsistency out. My theory has no such silliness. Your QM theory fails because you will have to invoke both a wave and particle explanation at THE SAME TIME!


    Next, you say:
    I don't know why you think that it would... [drive the electrons deeper into the metal]

    KJW, yes you do (know why). You are smart and just feinting ignorance just for the sake of argument! But for the sake of our readers, let me explain why particles that absorb other particles "shot from above" would tend to be driven downward!

    Take for example some styrofoam balls floating in the space station. Now position yourself above them, take a BB gun and fire BBs into the styrofoam balls from above. Do the styrofoam balls instantly go out upwards, perpendicular to the BB firing lines? Of course not. They are driven downward towards the floor of the space station! Like this:



    KJW, you QM'ers take momentum into account for the Compton effect, but "conveniently ignore it" for the photoelectric effect. I am just pointing out all these inconsistencies. My theory does not have this momentum silliness!

    But perhaps you could explain why increasing the intensity of the light increases the rate that electrons are emitted but not their energy, and increasing the frequency of the light increases the energy of the individual electrons.

    KJW, thank you for this! I highly recommend you watch my Episode 2 video, "The Photoelectric Effect" (but if not, let me summarize it for you).

    1. The intermittent electrons in the copper are pulsating their electrical influence ON and OFF.

    2. If these electrons are pulsating in such a way to be in resonance with just the peaks of a vertical light wave (ON with peaks, OFF with valleys) then they will be radically ejected upwards along the polarization E field of the light wave.

    3. If these electrons are pulsating in such a way to be in resonance with just the peaks of a horizontal light wave (ON with peaks, OFF with valleys) then they will be pushed sideways along the surface of the copper (not ejected much).

    4. Now as the electrons (from 2.) get accelerated upwards, they increase their intermittent pulsation rate in accordance with De Broglie. As the pulsation rate increases, eventually these electrons will come into pulsation resonance with both the peaks and valleys of the incident light wave (what we call a non-acceleration resonance). In other words, the acceleration is over and the electrons gain just a finite amount of energy that we see in the photoelectric effect.

    5. If you increase the intensity of the incident light, then the electrons simply reach the non-acceleration resonance faster, but still end up with the same maximum energy! And with the stronger electric field, more electrons would tend to be driven out!

    6. If you increase the frequency of the incident light, then according to De Broglie, the electron reaches a higher energy before getting into the non-acceleration resonance!

    That's the explanation for the bias towards "sideways" ejection along the polarization, the dependence on intensity and frequency, and the kinetic energy limitation seen in the photoelectric effect!

    Andrew Ancel Gray
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  19. #18  
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    Quote Originally Posted by andrewgray View Post
    Niels Bohr: His principle of complementarity!

    You can explain light as particles or waves, but not both at the same time!
    But light can be particles and waves at different times. Actually, light isn't particles or waves... that's too simplistic an interpretation of complementarity. Light can have some particle properties and some wave properties at the same time. But I prefer to think of it more generally than particle properties and wave properties.


    Quote Originally Posted by andrewgray View Post
    Your QM theory fails because you will have to invoke both a wave and particle explanation at THE SAME TIME!
    One thing I neglected to mention is that a wave can be regarded as a superposition of particles, and a particle can be regarded as a superposition of waves. Thus, when I said that refraction is easiest to understand in terms of waves, but can also be understood in terms of particles, I should have said that refraction can be understood in terms of a superposition of particles corresponding to waves.


    Quote Originally Posted by andrewgray View Post
    Next, you say:

    I don't know why you think that it would... [drive the electrons deeper into the metal]
    KJW, yes you do (know why). You are smart and just feinting ignorance just for the sake of argument!
    You are correct, actually. I did know why you would think that. I just wanted you to actually explain it. But with your picture, you didn't complete the trajectory of styrofoam balls. What happens when they hit the floor? They will be reflected upward. So, although the photoelectrons may be initially directed into the metal, the electrons of the bulk metal will force the photoelectrons to be expelled from the metal.


    Quote Originally Posted by andrewgray View Post
    Compton effect
    You mentioned the Compton effect. The Compton effect is an effect that highlights the inadequacy of classical electrodynamics, requiring the existence of photons to explain it. Perhaps you could provide an explanation of this effect without invoking photons. Compton scattering is often said to be inelastic. But when relativity is taken into account, Compton scattering is actually elastic... there exists a frame of reference in which the scattering is elastic.


    In your explanation of the photoelectric effect, you haven't addressed the problem that at the extremely low light levels used in the experiments, there isn't enough energy in the electromagnetic wave at the location of the electron to eject it from the metal.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  20. #19  
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    One thing I neglected to mention is that a wave can be regarded as a superposition of particles, and a particle can be regarded as a superposition of waves...
    KWJ,

    Even if you regard the incident light wave somehow as a superposition of particles, it still will not explain the sideways ejections in potassium vapor. Your "bounce off the floor" analogy implying that the electrons are somehow bounced back upwards "off the floor" gets rejected because of the the sideways along-the-polarization ejections from the potassium vapor (thank goodness we have experimental evidence, eh!!???)! No "floor there" to bounce off of and all directions are the same except for the polarization! The QM theory for the photoelectric ejections is in logical trouble!

    So KJW, how do you explain the sideways-ejection-along-the-polarization" from the potassium vapor???
    And that's not all. Now that we have "our heads together" on the "electrons driven downward" silliness, that's not the only momentum silliness that QM has. Let's take a look at photoelectric momentum considerations:

    So consider a faux light particle, a 10 eV UV fauxton! It has a faux energy of

    Euv = 10 eV = 1.6 x 10-18 Joules. Then we can get its faux momentum
    Puv = E/c = 5.3 x 10-27 kg m/s
    .

    So if we give all that faux energy to the photoelectron that absorbs it ( minus the energy needed to escape the surface ) we get for the photoelectron:

    Ee = 10 eV - 3 eV = 7 eV = 1.12 x 10-18 Joules. So for the photoelectron momentum we get:
    Pe = = 1.42 x 10 -24 kg m/s !!!!

    In other words, the photoelectron has obtained about 268 times the momentum of the faux light particle that it supposedly absorbed!
    |Pe| = 268 |Puv| !!! ???

    DO THE ARITHMETIC YOURSELF. CONVINCE YOURSELF! THIS IS NO JOKE!


    These momentum values are pure QM silliness. One would expect (from a logical light particle theory), that the free conduction photoelectron would gain about the same momentum as the faux light particle that is absorbed. BUT NO! So again we see why QM'ers "conveniently ignore" momentum considerations in the photoelectric effect:

    Don't ignore it any more! The Intermittent Electron Theory explanation is much more logical!

    As far as there "not being enough energy" to eject the particle... Remember, this is an acceleration resonance! It is like a father pushing his daughter on a swing. He need only use his pinky every time she comes by to add a minute amount of energy to her swing and eventually she goes over the top!


    Next, the Compton Effect!

    Andrew Ancel Gray
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  21. #20  
    KJW
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    Quote Originally Posted by andrewgray View Post
    Even if you regard the incident light wave somehow as a superposition of particles, it still will not explain the sideways ejections in potassium vapor. Your "bounce off the floor" analogy implying that the electrons are somehow bounced back upwards "off the floor" gets rejected because of the the sideways along-the-polarization ejections from the potassium vapor (thank goodness we have experimental evidence, eh!!???)! No "floor there" to bounce off of and all directions are the same except for the polarization!
    You asked me about why the electrons aren't pushed further into the metal, and my "bounce off the floor" analogy addressed that question. I already answered the question about the sideways ejection, but let me repeat: you cannot disprove the existence of photons by demonstrating the existence of electromagnetic waves, because any phenomenon that can be explained in terms of electromagnetic waves can also be explained in terms of the corresponding superposition of photons. However, I should also point out that a potassium atom is more than 70000 times as massive as an electron, so even a single potassium atom can push the photoelectron in a different direction to the photon's momentum.


    Quote Originally Posted by andrewgray View Post
    In other words, the photoelectron has obtained about 268 times the momentum of the faux light particle that it supposedly absorbed!
    You almost convinced me that something was wrong, but I realised that the error was this:

    Quote Originally Posted by andrewgray View Post
    One would expect (from a logical light particle theory), that the free conduction photoelectron would gain about the same momentum as the faux light particle that is absorbed.
    Your error appears to me to result from a misunderstanding of collisions between objects of very different mass. Physics requires that both energy and momentum are conserved. If two objects have the same kinetic energy, then the less massive faster object has less momentum. Thus, it is to be expected that the more massive slower electron will have a much greater momentum than a photon of the same energy. Momentum is conserved because the collision also involves a metal atom. But, because the metal atom has a much greater mass than both the electron and the photon, though the metal atom will have a change in momentum comparable to the electron and photon, its gain in kinetic energy will be negligible. Therefore, except for the energy required to remove the electron from the atom, the energy of the electron will be the same as the energy of the photon, and therefore the momentum of the electron will be much greater than the momentum of the photon.


    Quote Originally Posted by andrewgray View Post
    As far as there "not being enough energy" to eject the particle... Remember, this is an acceleration resonance! It is like a father pushing his daughter on a swing. He need only use his pinky every time she comes by to add a minute amount of energy to her swing and eventually she goes over the top!
    No... this won't work. The accumulation of sufficient energy to eject an electron would require a long time, implying a delay between the exposure of the metal to light and the emission of photoelectrons. There is no such delay, photoelectrons are emitted immediately upon exposure to light.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  22. #21  
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    KJW,

    There is no doubt that the electron ends up with 268 times the faux momentum of the faux light particle it supposedly absorbed. But... like you, QM doesn't see how illogical it is, and must bring in some atom-supposedly-bound-to-the-electron, even though it is the free conduction electrons in the metal we are talking about (not bound to any atom), and give this supposed atom -267 kg m/s amount of momentum (i.e., backwards) to balance the momentum out! Why would the atom go backwards if the electron, bound to it, went forward? Seems to me, the electron would just "drag it along with it forward as well". Wow. What a dumb theory (glad its not mine).

    And you did NOT explain the ejection along the polarization angle. Just having "a wave" from a superposition of particles explains nothing. Just like a cork in the ocean, if a wave strikes the cork, it goes up and down, GOING NOWHERE. THE CORK DOES NOT GET TOSSED INTO THE AIR 1000 feet! You have to have some asymmetry towards one direction along the polarization. You have explained nothing. You have not already answered the question.


    So let me put you on the spot!
    Will you please explain why the electrons are ejected sideways from the potassium vapor along the polarization of the incident light wave?


    And... I will have to disagree with you about "not enough energy". At resonance, there will be plenty of energy available to send some electrons on their way immediately in even dim light. Only 1.6 x 10-18 Joules are needed to send an electron out. You are wrong about this KJW. If you were able to use a scope to look at the dim light intensity at any one point, then it would be very noisy with random spikes and valleys bustling on the scope. There would be plenty of energy in one of these spikes to eject a single electron immediately somewhere with a resonance! You have "ideal conditions" too much on the mind, KJW! Build up to 1.6 x 10-18 Joules??? It seems like you would have a hard time keeping your light wave down below that kind of "immediate" energy delivery.

    Whew! Nice thought-provoking, friendly discussion! Let me say thank you KJW for that!

    Andrew Ancel Gray
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    OK. Now on to Compton Scattering!

    You mentioned the Compton effect. The Compton effect is an effect that highlights the inadequacy of classical electrodynamics, requiring the existence of photons to explain it. Perhaps you could provide an explanation of this effect without invoking photons.
    Well of course we can KJW! Thank you for that! On with Compton Scattering!

    Arthur Compton directed x-rays into carbon and noticed that some of the x-rays that re-emerged from the carbon had shifted frequencies. Like seen in the diagram below on the right side. Compton's description of this uses "faux x-ray particles colliding with electrons!" No! Not again!


    X-rays of about .71 Angstroms (4.2 exaHertz) are incident onto the carbon, and x-rays of about .71, .73, and .75 Angstroms ( 4.2, 4.1, and 4.0 exaHertz ) emerge from the carbon. The Theory of Intermittent Electrons explains this in the following way:

    1. The inner electrons of the carbon are intermittently turning their electrical influence ON and OFF very rapidly out of sync with the nucleus (see Episode 3 above for detailed explanation) and at x-ray frequencies. One of the orbits has an electron pulsating at a frequency just slightly greater than the 4.2 exaHertz.
    2. When this electron (in its orbit) comes around and is moving at 0.035c away from the source, the two frequencies match, and there is an acceleration resonance between the incident x-rays and the orbital electron's pulsation. The electron is ON in sync with both the peaks and valleys of the incident x-rays while moving away from the incident x-rays. This causes the orbiting electron to re-radiate the incident x-rays with a Doppler shift since it is receding from the source.
    3. And indeed! If one puts V=0.035c as the receding velocity in a Doppler Shift formula and work it out, the resulting Doppler Shifted frequencies align perfectly with Compton's data at all angles!

    So Compton Scattering is simply a Doppler shifted re-radiation off of a receding resonant inner electron orbital that matches Compton's data perfectly!

    Now this is very interesting. Compton's theory itself only predicts longer wavelengths (slower frequencies) with certain wavelength changes. THIS NEW THEORY DOES NOT. If the electron has an acceleration resonance with the incident x-rays while GOING TOWARDS the source, the emerging x-rays can have a HIGHER FREQUENCY! And indeed, we have shown an example of this on the upper left side of the diagram with Copper (we have mirrored the diagram to keep longer wavelengths on the right). And Copper also has a big spike in the middle that is also different than the Compton prediction!

    And there would be no reason that other substances should have the Compton shifted wavelengths! Compton's theory does not allow for that! And indeed: !!!!



    We again see how Compton's QM theory using "fauxton collisions with electrons" is just plain silly, just like the photoelectric effect!
    The explanation from the Theory of Intermittent Electrons is much more logical!


    Andrew Ancel Gray
    Last edited by andrewgray; October 17th, 2020 at 02:47 PM.
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  24. #23  
    KJW
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    Quote Originally Posted by andrewgray View Post
    There is no doubt that the electron ends up with 268 times the faux momentum of the faux light particle it supposedly absorbed. But... like you, QM doesn't see how illogical it is
    It is not illogical. I explained why the photoelectron's momentum is much greater than the original photon's momentum. You are wrong for thinking that they should be equal.


    Quote Originally Posted by andrewgray View Post
    and must bring in some atom-supposedly-bound-to-the-electron, even though it is the free conduction electrons in the metal we are talking about (not bound to any atom)
    Conduction electrons are still bound to the metal as a whole.


    Quote Originally Posted by andrewgray View Post
    and give this supposed atom -267 kg m/s amount of momentum (i.e., backwards) to balance the momentum out! Why would the atom go backwards if the electron, bound to it, went forward? Seems to me, the electron would just "drag it along with it forward as well"
    Because the absorbed photon has supplied the energy necessary to eject the electron from the metal.


    Quote Originally Posted by andrewgray View Post
    Will you please explain why the electrons are ejected sideways from the potassium vapor along the polarization of the incident light wave?
    The electrons would be ejected along the polarisation of the incident light because the electric field of the light imparts a force on the ejected electron.


    Quote Originally Posted by andrewgray View Post
    If you were able to use a scope to look at the dim light intensity at any one point, then it would be very noisy with random spikes and valleys bustling on the scope.
    Why would the intensity be very noisy with random spikes? Surely we are talking about coherent laser light with a smooth sinusoidal waveform.


    Quote Originally Posted by andrewgray View Post
    It seems like you would have a hard time keeping your light wave down below that kind of "immediate" energy delivery.
    Bear in mind that we are talking about single-photon-at-a-time light levels. If it takes the energy of one photon to eject one electron, then consider how much energy is at that electron if the energy of the photon were spread out over the entire surface of the metal.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  25. #24  
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    Quote Originally Posted by andrewgray View Post
    And... I will have to disagree with you about "not enough energy". At resonance, there will be plenty of energy available to send some electrons on their way immediately in even dim light. Only 1.6 x 10-18 Joules are needed to send an electron out.
    Resonance does not automagically create energy where there was none. KJW already pointed out one important consequence of that fact: If a resonance condition were the correct explanation, then there would be an intensity-dependent delay between the application of light, and the time electrons are ejected. Such a delay is not observed, so resonance cannot be at play. I therefore disagree with your disagreement.

    Another problem with invoking resonance as an explanation is the observed experimental fact that the wavelength of the incident light only has to be sufficiently short. If no (shorter) wavelength can ever be off resonance, then there is no meaning to "resonance". There are, in short, no experimentally observed effects that are signatures of resonant phenomena of any kind.

    Quote Originally Posted by andrewgray
    If you were able to use a scope to look at the dim light intensity at any one point, then it would be very noisy with random spikes and valleys bustling on the scope.
    Here you make the fundamental error of conflating the noisy presentation by an instrument, with noise in the parameter being measured. There is a phenomenon known as thermal noise that is responsible for the noise you are describing on the display of an oscilloscope. Noise is introduced downstream of the signal produced by the phenomenon you are instrumenting. Even if the signal itself were noise-free, a scope would present a noisy display. Your argument thus fails. You would do well to study the basics of noise (which, incidentally, is closely related to the work that forced a very reluctant Planck to adopt the quantum hypothesis).

    You would also profit from a study of, e.g., the noise properties of masers and lasers. You are apparently unaware of the existence of devices colloquially known as atomic clocks. The photons that they emit are remarkably free of noise. But if you were to look at the raw signals with an oscilloscope, they could appear noisy, depending on how you arrange the instrumentation. But, again, that noise is coming from exogenous sources. It is not in the emission of the cesium atoms.
    Last edited by tk421; October 18th, 2020 at 03:21 AM.
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