Yep. It wouldn’t impact entanglement experiments, however, and wouldn’t impact wave physics characteristics, but rather the particle physics characteristics of the experiment.
The two-slit experiment depends upon the assumption that photons (wave or particle) are above the Planck threshold—if they’re beneath it, they wouldn’t have sufficient energy to reliably induce a reaction in the screen behind the plate, meaning a strict wave interpretation could be valid (the intermittent reactions could be the product of sufficient energy build-up in the receiving electrons, rather than photons intermittently striking different parts of the screen). In regard to wave characteristics of photons, statistically this would be nearly identical to particle emissions—we should expect “blips” in a distribution roughly equal to the distribution we should expect from particle emissions. I say nearly identical because I assume some underlying mechanism by which electrons lose energy over time, meaning the least-heavily radiated areas to lose energy at a rate rapid enough to prevent valence shell shifting and hence fewer blips.
...which might be evidence for my theory, actually, since we do indeed see fewer reactions than we might expect in the least-radiated portions of the screen, per that open problem/unexplained phenomenon whose name I can’t recall that Eliezer goes on about a bit in one of the sequences. (The observed reactions are the square of the probability, rather than the probability itself, of a particle hitting a given section of the screen. I’m mangling terminology, I know.) Laziness is now competing with curiosity on whether I go and actually pull out one of my mathematics textbooks. If I were in therapy for crackpottery this would set me back months.
(Note: Having looking up the experiment to try to get the proper name for the screen behind the plate (without any success), it appears I was mistaken in my initial claim—the -original- intended purpose of the experiment, demonstrating wave characteristics of light, remains intact. It’s merely wave-particle duality, a later adaptation of the experiment, which loses evidence. Retracting that comment as invalid.)
I have a question. My meta-question is whether the question makes sense in light of what you said. (I like working in low-information conditions, downside being dumb questions.)
Wouldn’t this still be a testable difference? If electrons can briefly store energy, you could send a steady stream of below-Planck photons. Standard QM predicts no spots on the photoplate, but you predict spots, right?
The question becomes—how do you create a steady stream of below-Planck photons? In the current model, photons are only emitted when electrons shift valence shells—these photons start, at least, as above-Planck.
Rhydberg’s model (assuming I understand where he was going with it correctly) asserts that photons are -also- emitted when the electrons are merely energetic—black-body radiation, essentially. However, if your electrons are energetic, and at least 50% of all photons are being shared by the emitting medium, you’re going to get above-Planck photons anyways. (If you’re emitting enough radiation to create spots in the receiving medium, you’re dealing with energy that is at least occasionally above Planck scales, and this energy is already in the emitting medium.)
An important thing to remember is that the existing model was devised to explain black-body radiation. The Planck scale is really really low, low enough that the bar can be cleared by (AFAIK) any material with an energy level meaningfully above absolute zero. (And maybe even there, I’ve never looked into blackbody radiation of Einstein-Bose condensates.)
So in principle, for a sensitive enough photoplate (it’s currently nonreactive to blackbody radiation), for a dark enough room (so as not to set the photoplate off constantly), yes.
However, that ties us into another problem, which you may have sensed coming—the photoplate would be setting -itself- off constantly.
I assume light is a wave, not is a particle, which gives a little more wiggle-room on the experiment; sections of the plate experience distributed energy build-up, which is released all at once in a cascade reaction when a sufficiently large (still quite small) region of the photoplate has amassed sufficient energy to react with only a small amount (say, a nearby atom reacting) of energy.
Cyclotron radiation wavelengths can be tuned, as they aren’t tied to valence shells.
The number of spots per second from thermal statistics plus harmonics on the cyclotron radiation can be calculated. If the electrons are also absorbing photons classically, you should get extra spots when they happen to add up.
I think you’re going to see Rhydberg-OrphanWilde-interpretation blackbody radiation anyway. When an electron bounces off another, it counts as acceleration and produces cyclotron radiation. It might be different in magnitude, though.
I think photoplates can be tuned too. It should have to be hit by a single particle with more than the activation energy for the light-sensitive reaction. (Neglecting tunneling.) Therefore, it should be possible to pick a compound with a suitably high activation energy.
If you’re emitting enough radiation to create spots in the receiving medium, you’re dealing with energy that is at least occasionally above Planck scales, and this energy is already in the emitting medium.
But it will look statistically different. From what I understand, photons below the necessary energy will just bounce off or get absorbed by some other process. That’s how the photoelectric effect is supposed to work, anyway.
I’m not familiar enough with cyclotron radiation (read: I’m not familiar with it at all; my understanding of cyclotrons is limited to “They’re the things hospitals and labs use to produce small amounts of radioactive isotopes”) to be able to contribute to this discussion, so I’m afraid I’ll have to tap out due to ignorance I currently don’t have time to rectify.
Also, keep in mind that the only was an experiment can fail is if it provides no new information; the only way to render an experiment invalid or less useful is to show that you don’t know as much more as you thought you did.
But I thought you were referring to the modification of the two-slit experiment where electrons were the wave being measured, not photons.
And an experiment can’t fail to provide new information, because you thought it would provide information and then it didn’t, which means it has something to teach you about experiment design. Unless you’re proposing that an experiment that goes exactly as expected is a waste of time?
That said I think what Wilde means by ‘invalid’ is that a strong conclusion that resulted from the experiment is invalid in light of the fact that an entirely different model is consistent with the evidence.
An experiment that fails would be “I was trying to measure the speed of neutrinos, but I measured lab errors instead.”, or “I tried to titrate a solution, but used an excess of phenolphthalein accidentally.”
You are not a crackpot, unless there are major factual errors in your explanation of your theory.
It does render the two-slit experiment invalid for its intended purpose, even if it doesn’t affect the data, if that helps my case.
The two-slit experiment which acts exactly as wave physics suggests it should, even where it is used to demonstrate entaglement?
Yep. It wouldn’t impact entanglement experiments, however, and wouldn’t impact wave physics characteristics, but rather the particle physics characteristics of the experiment.
The two-slit experiment depends upon the assumption that photons (wave or particle) are above the Planck threshold—if they’re beneath it, they wouldn’t have sufficient energy to reliably induce a reaction in the screen behind the plate, meaning a strict wave interpretation could be valid (the intermittent reactions could be the product of sufficient energy build-up in the receiving electrons, rather than photons intermittently striking different parts of the screen). In regard to wave characteristics of photons, statistically this would be nearly identical to particle emissions—we should expect “blips” in a distribution roughly equal to the distribution we should expect from particle emissions. I say nearly identical because I assume some underlying mechanism by which electrons lose energy over time, meaning the least-heavily radiated areas to lose energy at a rate rapid enough to prevent valence shell shifting and hence fewer blips.
...which might be evidence for my theory, actually, since we do indeed see fewer reactions than we might expect in the least-radiated portions of the screen, per that open problem/unexplained phenomenon whose name I can’t recall that Eliezer goes on about a bit in one of the sequences. (The observed reactions are the square of the probability, rather than the probability itself, of a particle hitting a given section of the screen. I’m mangling terminology, I know.) Laziness is now competing with curiosity on whether I go and actually pull out one of my mathematics textbooks. If I were in therapy for crackpottery this would set me back months.
(Note: Having looking up the experiment to try to get the proper name for the screen behind the plate (without any success), it appears I was mistaken in my initial claim—the -original- intended purpose of the experiment, demonstrating wave characteristics of light, remains intact. It’s merely wave-particle duality, a later adaptation of the experiment, which loses evidence. Retracting that comment as invalid.)
I have a question. My meta-question is whether the question makes sense in light of what you said. (I like working in low-information conditions, downside being dumb questions.)
Wouldn’t this still be a testable difference? If electrons can briefly store energy, you could send a steady stream of below-Planck photons. Standard QM predicts no spots on the photoplate, but you predict spots, right?
The answer to that is a firm “Maybe.”
The question becomes—how do you create a steady stream of below-Planck photons? In the current model, photons are only emitted when electrons shift valence shells—these photons start, at least, as above-Planck.
Rhydberg’s model (assuming I understand where he was going with it correctly) asserts that photons are -also- emitted when the electrons are merely energetic—black-body radiation, essentially. However, if your electrons are energetic, and at least 50% of all photons are being shared by the emitting medium, you’re going to get above-Planck photons anyways. (If you’re emitting enough radiation to create spots in the receiving medium, you’re dealing with energy that is at least occasionally above Planck scales, and this energy is already in the emitting medium.)
An important thing to remember is that the existing model was devised to explain black-body radiation. The Planck scale is really really low, low enough that the bar can be cleared by (AFAIK) any material with an energy level meaningfully above absolute zero. (And maybe even there, I’ve never looked into blackbody radiation of Einstein-Bose condensates.)
So in principle, for a sensitive enough photoplate (it’s currently nonreactive to blackbody radiation), for a dark enough room (so as not to set the photoplate off constantly), yes.
However, that ties us into another problem, which you may have sensed coming—the photoplate would be setting -itself- off constantly.
I assume light is a wave, not is a particle, which gives a little more wiggle-room on the experiment; sections of the plate experience distributed energy build-up, which is released all at once in a cascade reaction when a sufficiently large (still quite small) region of the photoplate has amassed sufficient energy to react with only a small amount (say, a nearby atom reacting) of energy.
Cyclotron radiation wavelengths can be tuned, as they aren’t tied to valence shells.
The number of spots per second from thermal statistics plus harmonics on the cyclotron radiation can be calculated. If the electrons are also absorbing photons classically, you should get extra spots when they happen to add up.
I think you’re going to see Rhydberg-OrphanWilde-interpretation blackbody radiation anyway. When an electron bounces off another, it counts as acceleration and produces cyclotron radiation. It might be different in magnitude, though.
I think photoplates can be tuned too. It should have to be hit by a single particle with more than the activation energy for the light-sensitive reaction. (Neglecting tunneling.) Therefore, it should be possible to pick a compound with a suitably high activation energy.
But it will look statistically different. From what I understand, photons below the necessary energy will just bounce off or get absorbed by some other process. That’s how the photoelectric effect is supposed to work, anyway.
I’m not familiar enough with cyclotron radiation (read: I’m not familiar with it at all; my understanding of cyclotrons is limited to “They’re the things hospitals and labs use to produce small amounts of radioactive isotopes”) to be able to contribute to this discussion, so I’m afraid I’ll have to tap out due to ignorance I currently don’t have time to rectify.
Also, keep in mind that the only was an experiment can fail is if it provides no new information; the only way to render an experiment invalid or less useful is to show that you don’t know as much more as you thought you did.
But I thought you were referring to the modification of the two-slit experiment where electrons were the wave being measured, not photons.
And an experiment can’t fail to provide new information, because you thought it would provide information and then it didn’t, which means it has something to teach you about experiment design. Unless you’re proposing that an experiment that goes exactly as expected is a waste of time?
That said I think what Wilde means by ‘invalid’ is that a strong conclusion that resulted from the experiment is invalid in light of the fact that an entirely different model is consistent with the evidence.
An experiment that fails would be “I was trying to measure the speed of neutrinos, but I measured lab errors instead.”, or “I tried to titrate a solution, but used an excess of phenolphthalein accidentally.”