Can Oscillating Charges Emit a Hydrogen-Like Spectrum Through Prism Dispersion?

[T.Roc]

hydro,

are you aware that each spectral line (color) is created by a drop in "orbit" from an electron previously energised by a harmonic frequency (photon)?

i don't think the doppler effect (red/blue shift) is related to the spectrum. doppler is moving toward, then away from, the observer. in emission / absorption, the direction is from one to the other.

TRoc

 

[Hydr0matic]

are you aware that each spectral line (color) is created by a drop in "orbit" from an electron previously energised by a harmonic frequency (photon)?

Omg, you must really think I'm an idiot. Yes, I'm very familiar with the atomic model... Thomson's, Bohr's, Schrödinger's. I've read a lot of QM, although mostly on a conceptual level. But I've just taken a full semester of university math last spring, so any math you'd like to use in explaning would not go over my head.

i don't think the doppler effect (red/blue shift) is related to the spectrum. doppler is moving toward, then away from, the observer. in emission / absorption, the direction is from one to the other.

I know what I'm trying to expain isn't what is usually considered doppler. Doppler is considered a macro effect, and now I'm applying it on oscillators. I know it seems crazy, but I don't know how else to explain it.
Think about it, when exactly does doppler apply ? Does it matter on what distances the relative motion takes place ? Does it matter if the relative motion is over distances of 1 lightyear or a few nano meters ? .. As I interpret it, ANY motion (of the source) towards or away from the observer should be considered doppler. Even if this relative motion is a few nanometers back and forth during the time span of an oscillator's period. It shouldn't matter. The oscillator is moving at such speeds there's bound to be major wavelength shifts as a result of any z-movement relative the observer.

 

[reilly]

Hydr0matic -- I'm an ex-professor of physics, so I answer questions with questions. You'll find your answers if you will spend an hour or two reading about radiation in a physics book or two. I will, once again, suggest a clue: waves can be sinusoidal in time, and, independently sinusoidal or not in space as well.
Regards,
Reilly Atkinson

 

[Hydr0matic]

Hydr0matic -- I'm an ex-professor of physics, so I answer questions with questions. You'll find your answers if you will spend an hour or two reading about radiation in a physics book or two.

I doubt it. Any answers I'll find there should also be available on the internet. However, I have spent countless hours and long nights searching for it, unsuccessfully.
All I can find is descriptions of the radiation emitted perpendicular to the oscillator.

I will, once again, suggest a clue: waves can be sinusoidal in time, and, independently sinusoidal or not in space as well.

When you say "or not", are you referring to "independently" och "sinusoidal" ?
I already know that EM waves can be non-sinusoidal. The question is whether or not the non-perpendicular waves emitted by an oscillating charge are sinusoidal. I know they're not, I just want it confirmed so we can move on with this discussion.

 

[Hydr0matic]

the link for the animation is exactly what i was talking about. the waves are sinusoidal, it is only the perspective of each, according to their angle viewed, that changes. if you could rotate the image one "degree" clockwise, you would see that they are the same form when they are perpendicular to the direction viewed. notice the parallel lines (0' and 180') don't look like waves at all, just as the donut wouldn't look round at those angles.

T.Roc, I understand what you're talking about. And the animation is first and formost meant to be viewed like you describe - all waves are sinusoidal ones emitted perpendicular to the oscillator, just viewed from different angles. However, this animation could also serve a different purpose as a cross-section of the radiation emitted at all angles. In this view, the lines 0° and 180° illustrate the fact that no radiation is emitted along the oscillating axle.

[ Since the cross-section view isn't the original purpose of this animation, the speed of light is too slow, almost making the crusts "tip over" the troughs. ]

 

[T.Roc]

hyrdo,

ok - photons(n energy) are created when an electron fall back to its' previous orbit after being bumped to a higher orbit by a photon of the same energy (n) - we are on the same page, and i didn't think you are an idiot.

the other lines in the spectrum (photons) are EACH created by subsequent changes in the possible orbits for the nucleus. every line you see is created by a different photon (frequency).

your idea seems to be that one oscillation is making all these photons? not so. if you want to look at this with wave forms, that is fine. the straight line between the emitter and receiver (90') is all that is required because for every angle above 90', there is a symmetrical angle below 90', and they cancel each other out.

even if this were not the case, the harmonics created by such a small range of frequency change is not enough to produce the wide range seen in the hydrogen spectrum (or others). ie. - you might get a "red" to "red-orange" change, but not orange to violet, and definitely not IR through visible and into UV.

TRoc

 

[Hydr0matic]

your idea seems to be that one oscillation is making all these photons?

Well that's the beauty of it .. By explaining discrete spectra as a result of atomic motion, you don't actually need to infer any energy levels at all. ... Of all the crazy statements you've ever heard, this takes the cake huh ? ... "We don't need atomic energy levels" .. wow, I even amaze my self.

Ironically, this isn't that far from our current model, is it? We started out with planetary orbits, then energy levels were introduced with fixed orbits. Now, there's a wavefunction for each electron describing the probability of finding it in a certain position.
Is it just me or is this another way of saying "- the electrons are somewhere within the atom" .. Were it not for the fact that each electron is associated with a certain energy level...

not so. if you want to look at this with wave forms, that is fine. the straight line between the emitter and receiver (90') is all that is required because for every angle above 90', there is a symmetrical angle below 90', and they cancel each other out.

Yes, I realize that. Of course, you need more than one oscillating atom to get a visible spectrum. Like, perhaps, a cloud of atoms oscillating in random directions.

even if this were not the case, the harmonics created by such a small range of frequency change

What makes you say it's small? The atoms are oscillating at very high speeds. And the amount of shifting is obviously very depent on the oscillator's speed.