Theoretical

Dude this is absolutely ridiculous. It's obvious what your trying to do here. It's a freaking dipole radio antenna transmitting at a frequency. At some far distance away is another freaking radio dipole antenna that only receives. Go get a freaking antenna software if you actually think there's not going to be oscillating current in it. Good god I'm done here. I won't allow you to waste more of my. It's obvious you people who live in these forums are experts at what you do. Go ahead and make a mockery of this. Don't care anymore. I'm going to swamp YouTube and forums with overwhelming jaw dropping experiments and mathematical evidence of what Quantum Mechanics can't predict. How are you thugs going to stop that?

Wow, it speaks. No it's not zero. Lookup radiation resistance and electrical resistivity of wire, and read up on dipole antennas.

Absolutely not lol. If emr is traveling along x axis and b-field is polarized on z axis then the induced voltage is on the y axis. Current is then induced on y axis as seen in any antenna software. Current is oscillating on y axis, b-field still polarized on z axis, force is applied on x axis. Bingo, moment. Works out exactly to p=h/wavelength Anyway, with those with at least a half brain but not quiet full: You claim the momentum from hf amount of energy per wavelength of light is h/wavelength. Great, classical mechanics can correctly predict whatever you throw at it. One hf. A million hf. Don't matter. Pick a number. The math in the top post uses one hf per wavelength. So the requirement is that the receiving antenna absorbs one hf of energy per wavelength. To do that you need to adjust the transmitting antenna power or move it farther enough away. Doesn't matter. All that matters is the receiving antenna absorbs one hf so as to see if classical mechanics gets the correct momentum of h/wavelength. You can clearly see in my top post where the receiving antenna input is adjusted so that the antenna absorb one hf per wavelength. Haha I have no idea what you want. Is there a neanderthal to human translator in the room? Anyhow enough of this nonsense. I get the feeling I could hand you ToE and you would sneeze and continue eating your burrito. This thread will never make it back to the Quantum Mechanics section. Stop asking me questions! I've given enough to convince a zombie.

God who opened the playpen. I'm out of here.

There's no physicist who has even a half a brain would consider your response valid. If you know anything about radio electromagnetism you will know that electromagnetism is a B-field moving at the speed of light, which induces a voltage.

To know what the symbols mean see the top post. The equation works under all known conditions given known space properties such as permittivity and permeability.

Admin fine if I answer the questions and prove its not speculative then are you going to move this thread back to the Quantum Mechanics section? The current is adjusted by the following equation to that the receiving antenna is receiving one photon worth of energy per wavelength, which is power in watts: ((h*f^2) / (V^2/R)) The h*f^2 is the energy per wavelength of one photon per wavelength. The V^2/R is power, see ohms laws. So the current is therefore equal to V/R, which is equal to sqrt(h*f^2 / R) This is not difficult to understand. In the antenna software one can see the receiving antenna received power, which equals h*f^2. And the antenna software gives the current, and so then you can calculate the force on the current by using the well known classical equation F=I*L*B The equation V=v*B*L is a well known classical equation to calculate the voltage caused by a moving magnetic field.

Swanson I'm not going to converse with someone who accuses you of jerry rigging but then follows there statement by asking a question about the very line you accused me of being a jerry rigger. ... And actually I'll no longer answer anyone's questions here because the last guy, "Strange" asked me a question in a sincere fashion but when I spent thr time to type all of the math equations for him he turned around and spit on me in an emotional outburst. That's it. If you can't figure it out then too bad. You'll have k wait for the video or send me a private message convincing me your sincere and not like all these truth firefighters that hang out on these forums to suppress truth. Who knows what their intent is. Maybe they're government thugs. :/

I wanted to copy and paste this since I already took the time to type it. Unfortunately I won't have time to reply anytime soon. The following shows how classical mechanics on a macro scale proves an hf amount of electromagnetic has h/λ momentum, derived from classical mechanics. Well established antenna software confirms. Hopefully within a few months I'll take the time to document all of the other issues academic community has had in getting classical mechanics to show other effects such as Compton scatter, blackbody radiation, mass inertia, etc. The following uses electromagnetism at radio frequencies to derive the equation. Such a method is 100% evidence since we're dealing with macro scale experiments using well confirmed electromagnetism effects such as the electromotive force on charge from a moving field. There are two antennas, transmitting and receiving. There's absolutely no Jerry rigging required to derive this equation. The resulting momentum is relative to how much electromagnetic radiation is absorbed, which is photon momentum. Enjoy! Photon/light momentum: V=v*B*L Solve for B, substitute v for c: B=V/L/c F=I*L*B I=V/R F=(V/R)*L*B Scale to one photon per wavelength: F=(V/R)*L*B * ((h*f^2) / (V^2/R)) Substitute B for V/L/c F=(V/R)*L*(V/L/c) * ((h*f^2)/(V^2/R)) Reduces to: F=h*f*(f/c) F=h*f/λ p=F*s One photon of energy takes 1/f seconds: p=(h*f/λ) * (1/f) p=h/λ c=speed of light f=frequency h=Planck constant v=velocity p=momentum Receiving antenna: V=voltage caused by B-field B=B-field I=current caused by B-field R=resistance L=length of receiving antenna F=resulting forward force r=distance away from dipole ps, I've built numerous antennas using the well known and established NEC antenna engine. When I set the transmitter so the receiving antenna absorbs one hf of energy per wavelength of time (1/f) then the resulting momentum on the antenna as calculated by simple classical mechanics equations is p=h/λ Don't blame classical mechanics. pss, for those who will cry the above math is nothing but slight of hand trickery, I want to make it very clear to open-minded scientists that make no mistake that electro motive force on the oscillating current in the receiving in Tina according to classical mechanics is precisely equal to the correct amount. When you multiply that force by the amount of time it takes for one wavelength we get the well-known photon momentum equation. Momentum equals force * time. Remember that the force is calculating by 100% classical mechanics. We can then take this one step further in a very clear macro scale classical mechanics mathematics derive Compton scanning. typo on my iPhone: Tina-=antenna Compton scanning=Compton scattering. I use iPhone voice to text dictation. It's not the best lol. psss lol One final gift. Take the electromotive force in wire caused by electromagnetic induction, fill the empty space with further electron charges, scale down to a single particle that has a sphere of the classical electron radius, calculate the electromotive force by using classical mechanics induction equation, and this result is the precise force caused by mass inertia.

pss id someone could be so kind as to send me a message containing a link to a good open-minded science forum Thanks!

If admins move anything that questions the standard model to a speculative area, which is what you did, then this forum is not what I'm looking for. I will not participate in discussions at this forum given is closed-minded nature, but I will at some point post a link to my video that will go over all of the math and experents details which clearly show that light is not quantized. Please see the first page of this discussion to see an outline what areas my video and pdf will cover. ps, no more time for anyone here. I'm extremely confident that I just cracked the code on the discovery of exactly what the electric and gravitational forces are. If true, it will be the discovery of all discoveries. If true, then it means the electric and gravitational forces are 4th dimensionallaly polarized radiation. More specifically, the electric field is a certain type of 4th dimensional radiation at the Compton wavelength which causes higher frequency harmonics. Gravity is caused by the non-linearity of space caused by radiation. Matter is circulating condensed light due to the non-linearity of space at intense delta-fields. We'll see if that's all correct.

Good, you seem to be the only one here interested in finding truth. I will go over the math, the derivations. Everything will be detailed in pdfs and videos. Compton scattering is 100% classical, just as is light, photon momentum, blackbody radiation, photoelectric effect, inertia. This should be my last post in this thread unless someone is up to the challenge of doing my experiments. Very briefly, to understand Compton scatter you first need to understand photon/emr momentum. In photon momentum the emr travels through the charge, which causes the charge to oscillate back and forth transversely. The oscillation causes the charge to move through the emr B-field, which *always* causes a net forward force, which is photon momentum. Here's the math derivation: Photon/light momentum: V=v*B*L Solve for B, substitute v for c: B=V/L/c F=I*L*B I=V/R F=(V/R)*L*B Scale to one photon per wavelength: F=(V/R)*L*B * ((h*f^2) / (V^2/R)) Substitute B for V/L/c F=(V/R)*L*(V/L/c) * ((h*f^2)/(V^2/R)) Reduces to: F=h*f*(f/c) F=h*f/λ p=F*s One photon of energy takes 1/f seconds: p=(h*f/λ) * (1/f) p=h/λ c=speed of light f=frequency h=Planck constant v=velocity p=momentum Receiving antenna: V=voltage caused by B-field B=B-field I=current caused by B-field R=resistance L=length of receiving antenna F=resulting forward force r=distance away from dipole This is easily seen and confirmed in well established antenna software. Explaining Compton scattering is more involved so for now I'll only outline it. So now if the charge can move in the direction the light is traveling, the charge will move forward in the direction the light is traveling, which was caused by photon momentum. As the charge moves in the direction with light, the charge continues to oscillate, but since it's moving, there's a Doppler effect which is found from the c - v, where v is the electron's velocity. This Doppler effects causes the electron to radiate lower frequency radiation. The end result is the particle scattering and some light converted to lower frequency electromagnetism radiation. Sorry I won't have time to answer more questions unless it's regarding the experiment.

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.

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."

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.

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.

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. That's what I said.

Nice idea. The atomic world is a beautiful world of light circulating dancing around each other in a beautiful tune in warped space fields.

I would highly recommend you read my previous thread. The atomic world by its very nature produces pulses of light which you could call packets, but this is not a requirement or universal law. I've done numerous experiments at the visible light region which shows intense light shined far away to the point of where quantum mechanics predicts it should be quantized "packets" is not true. Light is classical. Atomic world is quantized in nature, but the very nature of light itself is not quantized. Also I've done extensive so-called "single photon" experiments at radio frequencies below 50 MHz. The results show light is not quantized. In due time all of this and a whole lot more will be documented in PDFs and video.

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.

@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.

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. 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.

Of course. I'm referring to any quantization effects at radio frequencies.

What device? A PMT in avalache mode? Yes they can use a PMT, but it must not be in avalache mode.

Why isn't it showing up at radio frequencies? Why are there valid classical explanations for all photoelectric effects? Aw I was looking for someone who has one capable of detecting "single photons" without an avalanche effect.