Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 Because the energy of an RF photon is way too small to cause ionization My point is you didn't explain why not. There are? What are they? Of course. I'm referring to any quantization effects at radio frequencies. Link to comment Share on other sites More sharing options...
swansont Posted August 19, 2015 Share Posted August 19, 2015 Of course. I'm referring to any quantization effects at radio frequencies. Just a second ago you were talking about the photoelectric effect. Which is it you want to discuss? Link to comment Share on other sites More sharing options...
Sensei Posted August 19, 2015 Share Posted August 19, 2015 (edited) What device? A PMT in avalache mode? Yes they can use a PMT, but it must not be in avalache mode. What is wrong with avalanche mode? Single photon, no matter if it's RF, visible, or x-ray, has very little energy, in comparison to everyday macroscopic scale.. So there is needed amplification. You could miss annihilation of matter with antimatter if you would use just your own eyes (or your devices). It's 400 millions times less energy than battery with U=5 V, I=25 mA, t=1s. Edited August 19, 2015 by Sensei Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 The reason we should not use avalanche devices to detect emr is because it sets a threshold. The old Polaroid film example shows this where you can see what appears just like photons at levels that are in the millions of photons because there's a threshold. Below that threshold you get nothing. Beyond that threshold you get the avalache. It turns the linear world into a digital type world. This quantization describes the atomic world, but from what I'm seeing this quantization is not a law or requirement. An example of that is the experiment I've outlined in this thread, light at say 2mA shined at an effectively far distance, thus allow the emr itself to lower the intensity in a linear fashion, thus bypassing the atomic quantization effects. Just a second ago you were talking about the photoelectric effect. Which is it you want to discuss? Sorry, I'm just replying your posts on the quantization of light. Regarding to your question on who's detailed how the photoelectric effect is explained classically, I'll have I find the reference. I thought it was well known. You have to consider the material your trying to eject electrons and what the spectrum of light the specific electrons in the electron cloud you're ejecting because if your look at the spectrum you'll see it's extremely narrow spectral line. A typical spectral line for various atoms is 10pm to 15pm in the visible spectrum of day 700nm. Do you have any idea how high of a Q that is??? It's outrageous. It means the spectral line of ejecting an electron from metal of say 300nm is going to be oblivious to 660nm red light or even blue light. btw please excuse my iPhone typos. Link to comment Share on other sites More sharing options...
swansont Posted August 19, 2015 Share Posted August 19, 2015 The reason we should not use avalanche devices to detect emr is because it sets a threshold. The old Polaroid film example shows this where you can see what appears just like photons at levels that are in the millions of photons because there's a threshold. Below that threshold you get nothing. Beyond that threshold you get the avalache. It turns the linear world into a digital type world. This quantization describes the atomic world, but from what I'm seeing this quantization is not a law or requirement. An example of that is the experiment I've outlined in this thread, light at say 2mA shined at an effectively far distance, thus allow the emr itself to lower the intensity in a linear fashion, thus bypassing the atomic quantization effects. What if you lower the threshold to one photon? Sorry, I'm just replying your posts on the quantization of light. Regarding to your question on who's detailed how the photoelectric effect is explained classically, I'll have I find the reference. I thought it was well known. You have to consider the material your trying to eject electrons and what the spectrum of light the specific electrons in the electron cloud you're ejecting because if your look at the spectrum you'll see it's extremely narrow spectral line. A typical spectral line for various atoms is 10pm to 15pm in the visible spectrum of day 700nm. Do you have any idea how high of a Q that is??? It's outrageous. It means the spectral line of ejecting an electron from metal of say 300nm is going to be oblivious to 660nm red light or even blue light. Spectral lines are transitions between states. The PEE is ionization. The spectral width of a transition to the the continuum is basically infinitely wide. The atom will be "oblivious" (regarding the PEE) to red light because it has only ~1 eV of energy, and the ionization energy will be several eV, not because the line is only a few MHz wide. A pulse that is close to one wavelength has a bandwidth that is approximately equal to the frequency. For example, green light that is close to one wavelength per pulse will not be a specific frequency. It will have nearly as much red and blue as it has green. A prime example of this is the Scarcelli QM delayed choice quantum eraser experiment where they placed a narrow bandwidth filter before the double slit, thereby making the experiment flawed. Again, you cannot simultaneously have a short pulse and narrow BW. Narrow pulse ≠ single photon. Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 (edited) @Swanson photoelectric effect is complex. Before the atom becomes ionized the electron has to absorb energy. That's where spectral lines are required to analyze the photoelectric effect. Unfortunately, due to its complexity, the photoelectric effect is one of the worst examples to attempt to prove any possible particle nature of light. Edited August 19, 2015 by Theoretical -1 Link to comment Share on other sites More sharing options...
swansont Posted August 19, 2015 Share Posted August 19, 2015 @Swanson photoelectric effect is complex. Before the atom becomes ionized it has to absorb energy. That's where spectral lines are required to analyze the photoelectric effect. Unfortunately, due to its complexity, the photoelectric effect is one of the worst examples to attempt to prove any possible particle nature of light. No, really, you don't. In the PEE you are by definition ionizing the atom; you measure a current of emitted electrons. There is only a cutoff wavelength (or frequency) beyond which the effect does not happen. Spectral line widths don't enter into the analysis at all. Unless you are doing it wrong, that is. Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 No, really, you don't. In the PEE you are by definition ionizing the atom; you measure a current of emitted electrons. There is only a cutoff wavelength (or frequency) beyond which the effect does not happen. Spectral line widths don't enter into the analysis at all. Unless you are doing it wrong, that is. You're contradicting yourself. You say the it's caused by ionization, which is true of course, but you don't acknowledge the simple fact that spectral lines (obviously absorption spectra lines) is what determines how much energy is absorbed by the matter. And again it's complex, since the spectral lines which pertain to the ejection of electrons causing ionization, as opposed to spectral lines which pertaining to electron which merely heat up the matter without ejection. So you can heat the matter all day till you're blue in the face with light that does not eject electrons which merely heats up the matter and you only end up ionizing the matter to a small degree as common joule meeting. It's a known fact that joule heating also ejects electrons in the vacuum tubes, but by an infinitesimal amount of energy compared to UV light. BTW there is indeed a peak wavelength to the photoelectric effect. Again it depends on how well the light at a given frequency causes ionization. If you've seen the UV spectral lines you would see how many there are. Anyhow I'm surprised you haven't seen the papers on how the photoelectric effect is explained classically. Link to comment Share on other sites More sharing options...
swansont Posted August 19, 2015 Share Posted August 19, 2015 You're contradicting yourself. You say the it's caused by ionization, which is true of course, but you don't acknowledge the simple fact that spectral lines (obviously absorption spectra lines) is what determines how much energy is absorbed by the matter. And again it's complex, since the spectral lines which pertain to the ejection of electrons causing ionization, as opposed to spectral lines which pertaining to electron which merely heat up the matter without ejection. So you can heat the matter all day till you're blue in the face with light that does not eject electrons which merely heats up the matter and you only end up ionizing the matter to a small degree as common joule meeting. It's a known fact that joule heating also ejects electrons in the vacuum tubes, but by an infinitesimal amount of energy compared to UV light. BTW there is indeed a peak wavelength to the photoelectric effect. Again it depends on how well the light at a given frequency causes ionization. If you've seen the UV spectral lines you would see how many there are. The spectral lines are for transitions involving intermediate states, NONE of which are involved in the PEE. It's the ground state and the continuum. That's it. The light that causes the PEE is not resonant with any of the transitions for which you would have spectral lines. It's at higher energy than all of them. If there is a peak wavelength to the photoelectric effect, you should have no trouble providing a citation You have it backwards with regard to joule heating — the amount of energy you need to add to "boil off" electrons is much, much higher than what you need with photons. I don't see where I have contradicted myself. I've only contradicted you. Anyhow I'm surprised you haven't seen the papers on how the photoelectric effect is explained classically. Since the PEE is quantum-mechanical in nature, and QM is well-established, why would you expect this? I haven't read any papers on phlogiston, either. Why would I, other than as an historical investigation? Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 The spectral lines are for transitions involving intermediate states, NONE of which are involved in the PEE. It's the ground state and the continuum. That's it. The light that causes the PEE is not resonant with any of the transitions for which you would have spectral lines. It's at higher energy than all of them.Nonsense. Absorption spectral lines are self explanatory. If the materials doesn't absorb the light for a given frequency then it doesn't absorb the energy. You have it backwards with regard to joule heating the amount of energy you need to add to "boil off" electrons is much, much higher than what you need with photons.That's what I said. Link to comment Share on other sites More sharing options...
John Cuthber Posted August 19, 2015 Share Posted August 19, 2015 Nonsense. Absorption spectral lines are self explanatory. If the materials doesn't absorb the light for a given frequency then it doesn't absorb the energy. Plain wrong. Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 Plain wrong.What are you people smoking over here? Lol. If the material reflects the light then it's not absorbed. Furthermore, as stated, absorption is no guarantee an electron will eject. Enough time wasted on nonsense. Unfortunately it seems apparent nobody here can perform the request an experiment. -1 Link to comment Share on other sites More sharing options...
John Cuthber Posted August 19, 2015 Share Posted August 19, 2015 An electron that's been knocked free from a piece of metal is a free electron. in the absence of an electromagnetic field it only has kinetic energy. and it can have pretty much any amount of that which it likes. So there's no defined energy and thus there's no resonant transition. Also re. "If the material reflects the light then it's not absorbed. " Thanks for clarification of the fact that you don't know how mirrors work. Perhaps you should stop worrying about our smoking habits and go and learn some science Actually knowing what you are talking about can be very rewarding. perhaps you might try it some time. Link to comment Share on other sites More sharing options...
Sensei Posted August 19, 2015 Share Posted August 19, 2015 Unfortunately it seems apparent nobody here can perform the request an experiment. You can/should do this experiment by yourself. PMT cost $30-$40 http://www.ebay.com/bhp/photomultiplier Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 (edited) An electron that's been knocked free from a piece of metal is a free electron. in the absence of an electromagnetic field it only has kinetic energy. and it can have pretty much any amount of that which it likes. So there's no defined energy and thus there's no resonant transition. Also re. "If the material reflects the light then it's not absorbed. " Thanks for clarification of the fact that you don't know how mirrors work. Perhaps you should stop worrying about our smoking habits and go and learn some science Actually knowing what you are talking about can be very rewarding. perhaps you might try it some time. The heart of reflection is best understood with macro scale radio frequencies. It is electromagnetic induction where the charge accelerates/decelerates, reflecting the emr energy. It has absolutely nothing to do with the absorption and immediate emission of so called photons. Edited August 19, 2015 by Theoretical Link to comment Share on other sites More sharing options...
Sensei Posted August 19, 2015 Share Posted August 19, 2015 (edited) The heart of reflection is best understood with macro scale radio frequencies. It is electromagnetic induction where the charge accelerates/decelerates, reflecting the emr energy. It has absolutely nothing to do with the absorption and immediate emission of so called photons. Induction in non-metals? How do you explain bouncing off radiowave from walls in your room in that case? They're made of non-metals (or metals bound in compounds such as white paint Zinc Oxide or Titanium Oxide). Edited August 19, 2015 by Sensei Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 Induction in non-metals? How do you explain bouncing off radiowave from walls in your room in that case? They're made of non-metals (or metals bound in compounds such as white paint Zinc Oxide or Titanium Oxide). Very simple to understand, but complex to describe. Basically you need to know the frequency, the VO the radiation resistance of the object, the resistivity and permeability and dielectric of the material. 4nec software is a nice starting pointing, but you'll need to learn how to construct grids with reactive components. I can assure you that I and many others have done such calculations with countless different materials other than copper or other common metals. It works out perfectly classically. To everyone: I came across this note among others. Just do some research. There are lot of modern articles and books that detail exactly how and why the so-called photon is absolutely not required to explain before electric effect. In fact there are people who have gone through a lot of calculations, and they match experiments that do not require the photon. If you do the research, you will see it is a fact that it turns out the so-called photon is not required to explain the photoelectric effect. Anyhow, here's a quote from a physicsforum.com Insights article. https://www.physicsforums.com/insights/sins-physics-didactics/ "Modern understanding of the photoelectric effect. Let us discuss the photoelectric effect in the most simple approximation, but in terms of modern quantum theory. From this modern point of view the photoelectric effect is the induced transition of an electron from a bound state in the metal (or any other bound system, e.g., a single atom or molecule) to a scattering state in the continuous part of the energy spectrum. *****To describe induced transitions, in this case the absorption of a photon by an atom, molecule, or solid, we do not need to quantize the electromagnetic field at all***** but a classical electromagnetic wave will do, which we shall prove now in some detail." Link to comment Share on other sites More sharing options...
Strange Posted August 19, 2015 Share Posted August 19, 2015 Interesting that you choose an article on how to teach quantum theory in an attempt to discredit quantum theory! From that article: One can derive Planck’s black-body-radiation formula (1900) only under the assumption that despite the absorption and stimulated emission of energy quanta ℏω of the electromagnetic field, there is also a spontaneous emission, and from a modern point of view, this can indeed only be explained from the quantization of the electromagnetic field (in addition to the quantization of the electron). Then indeed, for the free quantized electromagnetic field, there is a particle-like interpretation, leading to a consistent picture of the electromagnetic field, interacting with charged particles, Quantum Electrodynamics. (emphasis added) Link to comment Share on other sites More sharing options...
Theoretical Posted August 19, 2015 Author Share Posted August 19, 2015 Interesting that you choose an article on how to teach quantum theory in an attempt to discredit quantum theory! From that article: (emphasis added) No, i'm pointing out that your QM does not require the photon to explain the photoelectric effect. So we can put an end to the photoelectric effect debate. So you're saying an article says QM needs to quantize light to explain blackbody radiation? So are your trying to start another debate lol? I can assure you classical mechanics does a perfect job at explaining blackbody radiation. ps, don't blame classical mechanics. Blame those who cannot use it to explain the universe. Link to comment Share on other sites More sharing options...
John Cuthber Posted August 20, 2015 Share Posted August 20, 2015 The heart of reflection is best understood with macro scale radio frequencies. It is electromagnetic induction where the charge accelerates/decelerates, reflecting the emr energy. It has absolutely nothing to do with the absorption and immediate emission of so called photons. It is best understood by some model that actually gives the right answer. Since you say that your model won't work for a strong absorber, your model doesn't work. Link to comment Share on other sites More sharing options...
ajb Posted August 20, 2015 Share Posted August 20, 2015 (edited) Okay, so I'll settle for ***any*** type of experiment that indicates single photons. Please, anyone??? The existence of photons is not really in doubt. While, photoelectric bunching can be given a semi-classical interpretation, the claim is that antibunching has no semi-classical interpretation and requires that the electromagnetic field be quantised. Experimental results on antibunching are generally understood as providing the best evidence that the electromagnetic field is quantised. The classical reference here is [1]. I am sure one can find more recent experiments that back this up. Reference [1] H. J. Kimble, M. Dagenais, and L. Mandel, Photon Antibunching in Resonance Fluorescence, Phys. Rev. Lett. 39, 691, 1977. Edited August 20, 2015 by ajb Link to comment Share on other sites More sharing options...
Strange Posted August 20, 2015 Share Posted August 20, 2015 Okay, so I'll settle for ***any*** type of experiment that indicates single photons. . I'm fairly sure the Compton effect can't be explained classically. No, i'm pointing out that your QM does not require the photon to explain the photoelectric effect. Could you explain the math in that article for me, I couldn't quite follow it. Link to comment Share on other sites More sharing options...
ajb Posted August 20, 2015 Share Posted August 20, 2015 I'm fairly sure the Compton effect can't be explained classically. True, but there are semi-classical explanations that are okay. Schroedinger published a paper on this in 1927 (I think). So things like Compton scattering and the photoelectric effect can be explained reasonably using semi-classical arguments, that is you do not quantise the electromagnetic field. Lots of things in quantum optics can be explained semi-classically. However, for more modern experiments you do need the full theory of QED and photons. The fact that the electromagnetic field is quantised is not really in any doubt. Link to comment Share on other sites More sharing options...
imatfaal Posted August 20, 2015 Share Posted August 20, 2015 ! Moderator Note Moved to Speculations. This topic has moved from being a request for information and clarification to being an argument against a well accepted and exceedingly well evidenced areas of physics. Please take a moment to read the rules and guidelines of the Speculations Forum Link to comment Share on other sites More sharing options...
swansont Posted August 20, 2015 Share Posted August 20, 2015 https://www.physicsforums.com/insights/sins-physics-didactics/ "Modern understanding of the photoelectric effect. Let us discuss the photoelectric effect in the most simple approximation, but in terms of modern quantum theory. From this modern point of view the photoelectric effect is the induced transition of an electron from a bound state in the metal (or any other bound system, e.g., a single atom or molecule) to a scattering state in the continuous part of the energy spectrum. *****To describe induced transitions, in this case the absorption of a photon by an atom, molecule, or solid, we do not need to quantize the electromagnetic field at all***** but a classical electromagnetic wave will do, which we shall prove now in some detail." You should read through to the end. In addition to what Strange quoted, there is also the bit about spontaneous emission. There are individual results one can try and explain with classical physics instead of QM, but one tends to find that the contortions you have to go through to explain the result cause other results to be unexplainable. The quantized EM field is the best explanation that works under the widest range of application. No, i'm pointing out that your QM does not require the photon to explain the photoelectric effect. So we can put an end to the photoelectric effect debate. So you're saying an article says QM needs to quantize light to explain blackbody radiation? So are your trying to start another debate lol? I can assure you classical mechanics does a perfect job at explaining blackbody radiation. ps, don't blame classical mechanics. Blame those who cannot use it to explain the universe. The article is very clear that QM is required to explain the PEE. It's just saying that the quantization is of the electron states is sufficient. Link to comment Share on other sites More sharing options...
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