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.
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.