Question about photon energy

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blaine
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Question about photon energy

Post by blaine »

Yesterday I did some quick calculations of photon energies. I was curious the scale at which the 1/1 vibrational motion exists - because presumably you cannot have photons emitted in between energies (ie between 1/1 and 1/2 or 1/1 and 2/1) unless it is doppler shifted, so if this scale is somewhere in some reasonable energy range it might be possible to check to see if this is the case.

So far I have read through NBM up to the chapter on cosmic rays. It wasn't entirely clear to me how you convert the energy in the s/t notation to an energy in conventional units. At first I tried the natural unit of energy conversion factor (1.49e-3 ergs) multiplied by s/t. Then I compared this value with the value calculated if I use h * s/(t*t_nat), where h is planck's constant and t_nat is the natural unit of time (1.52e-16 s). I was off by a fairly large factor between the two, so I looked at Nehru's paper on calculating planck's constant - http://reciprocalsystem.org/PDFa/Theore ... Nehru).pdf. Turns out if you use the energy conversion factor for the natural units you need to multiply by s_nat which is the natural unit of space (because planck's constant absorbs this factor due to the fact that current theory views the energy of the photon as a function of its frequency and has no concept of temporal coordinates) and divided by the interregional ratio modified by a secondary mass factor. The reasoning for the interregional ratio is that the motion of the photon is within one unit of space (direction reversals are happening in time).

So if this is the case, why does one need the interregional ratio to modify s/1 type vibrations? If you didn't have this modification, wouldn't there be a very large gap in energies between 1/2 1/1 and 2/1? Also, presumably 1/1 would not need this modification either? I guess by Nehru's reasoning, the motion in time for the s/1 type vibrations are accomplished via direction reversals in space, thus the same factor is needed there too?

Here are the results (using h*s/(t*t_nat) ) for the energies around 1/1:
s/t | Energy [eV]
9/1 2.4477e+02
8/1 2.1757e+02
7/1 1.9038e+02
6/1 1.6318e+02
5/1 1.3598e+02
4/1 1.0879e+02
3/1 8.1590e+01
2/1 5.4393e+01
1/1 2.7197e+01
1/2 1.3598e+01
1/3 9.0655e+00
1/4 6.7992e+00
1/5 5.4393e+00
1/6 4.5328e+00
1/7 3.8852e+00
1/8 3.3996e+00
1/9 3.0218e+00

If this is correct, then we would expect to see gaps around 27.2 eV in terms of possible gamma energies... that doesn't seem right given that this is the UV range.

What is interesting is that 1/2 is 13.6 eV (a consequence of using the Rydberg frequency to set the units) - we know that -13.6 eV is the ground state energy of the electron in the hydrogen atom. In other words, a 13.6 eV gamma, if absorbed by the electron, will ionize the hydrogen atom - turning it into a proton and a free electron. Now in the RS, I *think* this would be as follows (using the single displacement notation):

\gamma (1/2) + M 1.5-1.5-(2) \rightarrow

Here we have a few options; from what I understand from NBM, the gamma will get rotated in a probabilistic fashion. Since the hydrogen atom has rotations in each dimension, I think it will either transfer the magnetic or electric rotation depending on which is orthogonal to the photon's translational motion.

If the electric rotation is transferred, then you have
C 0-0-0 + M 1.5-1.5-(1).
Since M 1.5-1.5-(1) is unstable (I think? need to look into this more; not sure what this particle is), a unit of magnetic rotation will be transferred so you end up with
C .5-.5-0 + M 1-1-(1)
C .5-.5-0 is also unstable because you have a negative rotational base with a positive rotational displacement.

That option seems unlikely because of the instability of the products; I wonder if this modifies the probability of this occurrence, or if all possibilities are allowed to occur and within one unit of time the unstable products will dissociate back to their previous components, with the photon exiting in a random direction thus looking like an elastic scatter. Here I'd like to make a quick aside: assuming one can properly calculate the probabilities of these different possibilities, one should be able to convert them to cross sections given the appropriate geometric factors.

The next option is that a unit of positive magnetic rotational displacement gets tranferred to the photon. The problem with this is that the photon is already vibrating with a positive displacement, so it would immediately decay.


Neither of these scenarios result in ionization of the hydrogen atom. Is it possible that I calculated my energies incorrectly? I notice, however, that the 2/1 photon would properly ionize:

\gamma (2/1) + M 1.5-1.5-(2) → M 0-0-0 + M 1-1-(2) → M 0-0-(1) + M 1-1-(1)

Which gives us the desired result; a proton and an electron. What if the electric displacement gets transferred instead? Here we have the inverse of the 1/2 case, the negative rotation + negative vibration is unstable and will dissociate immediately. So for this case it seems the only option is ionization (assuming the motions line up properly). So why doesn't 2/1 have an energy of 13.6 keV? What am I missing here? Any insights are greatly appreciated!

Perhaps the missing ingredient I need is charge? I haven't reached this part of NBM yet, so I will update once I do.
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bperet
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Re: Question about photon energy

Post by bperet »

blaine wrote: Wed Feb 22, 2017 6:56 pm So if this is the case, why does one need the interregional ratio to modify s/1 type vibrations? If you didn't have this modification, wouldn't there be a very large gap in energies between 1/2 1/1 and 2/1? Also, presumably 1/1 would not need this modification either? I guess by Nehru's reasoning, the motion in time for the s/1 type vibrations are accomplished via direction reversals in space, thus the same factor is needed there too?
The shift occurs because Larson uses the "vibrational unit datum" (what separates motion in space from motion in time) for the photon at 256 units (11.664 μm, 25.702 Thz, infrared), not the linear unit speed boundary at 1 unit (45.563 nm, 6.580 Phz, ultraviolet). The shift occurs because of the way photons interact with detectors, which are based on waves, not the rotational components of a birotation.

I've asked Gopi to comment on this, as he got his Ph.D in solar cell engineering and knows a lot about photon energies--way over my head.
If this is correct, then we would expect to see gaps around 27.2 eV in terms of possible gamma energies... that doesn't seem right given that this is the UV range.
There actually IS a gap... it is the boundary between "soft" and "hard" uV (soft being on the visible light side and hard being on the X-ray side).
Since M 1.5-1.5-(1) is unstable (I think? need to look into this more; not sure what this particle is)
Hydrogen is not a "real" atom, as it does not have two, complete double-rotating systems. It is a compound motion of a proton + charged electron neutrino, so you get:

1-1-(1) proton + ½-½-(1)* charged electron neutrino = 1½-1½-(2)* charged hydrogen atom

The neutrino rotation is actually 1-0-(1), but since magnetism is 2-dimensional, it is distributed across the two, magnetic dimensions, being present in each one for half the time. It is basically oscillating between 1-0-(1) and 0-1-(1), so Larson averages it to ½-½-(1) in his notation system.

The neutrino must carry a charge, because its net rotation is zero and there would be no reason for it to "stick" to the proton--it would just pass right through it. The neutrino charge is what keeps hydrogen together.
The next option is that a unit of positive magnetic rotational displacement gets transferred to the photon. The problem with this is that the photon is already vibrating with a positive displacement, so it would immediately decay.

If you convert a magnetic rotation to linear displacement, each magnetic unit produces 16 displacement units (1 magnetic produces 2 electric; 1 electric produces 8 linear; 2 x 8 = 16). There would be "left overs."
So why doesn't 2/1 have an energy of 13.6 keV? What am I missing here? Any insights are greatly appreciated!
It does have that energy, when you use the linear, unit speed datum (from which rotation is displaced):

Natural Units, Wavelength, Frequency, Duration, Energy (J), Hartrees, Electron Volts
(2), 91.127 nm, 3.290 Phz ,303.966 as, 2.179872e-18, 0.5 ,13.606 eV

You did it correctly.
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Gopi
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Re: Question about photon energy

Post by Gopi »

There are two aspects of the photon: Real and imaginary. Larson focuses on the real part alone, so you get the clear 1, 1/2, 2/1 demarcations. But those jumps are not seen. Nehru identified that there is a scale change, so the "unit size" becomes smaller, but it still doesn't cover all of it. This is because the projection of the imaginary onto the real provides the continuity. The mistake conventional science makes is it postulates a continuum to begin with, in RS we clarify the real parts, and identify the continuum as the interpolation between two real values. Larson uses "probability", a dubious approach, on this one.

To get a more visual description of what I mean, look at the color wheel:
Image

The line-like structure, on the cyan magenta and yellow, are the "discrete" points of Larson, and they are true visually as well. In reactions, those are the ones that take part. The rest are still there, as interpolations.

You can also imagine a book on a table, which you can move on the table not by sliding, but by flipping it repeatedly on its edge. The book is not in full contact with the table as it is being flipped, but only periodically. The shadow of the book however, moves continuously.
blaine
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Re: Question about photon energy

Post by blaine »

Hi Bruce and Gopi,

Coming back to this because I was fairly confused about the responses and I stopped working on it for a while and I've been coming back to RS concepts recently to try and work out spectroscopy again.

In regards to the energy calculation, you said that a photon with displacement (2) has energy 13.6 eV. But (2) means two negative displacement units, meaning a speed of 3/1. I calculated 3/1 as having three times the energy of 1/1, so 81.6 eV. I calculated a displacement of 1 (speed of 1/2) having an energy of 13.6 eV. I must have done something incorrectly - I was thinking perhaps that the energies are reversed from how I had them listed? In other words, a photon with speed 2/1 would have half the energy of 1/1, and 1/2 would have twice that of 1/1? It doesn't make sense intuitively to me (I would think a larger speed means larger energy) but I could be misunderstanding the nature of the vibrational motion.

Gopi, you mentioned that phase is what allows energies measured between the fractional gaps of 1/1, 2/1, etc. I think this is different from the concept of phase in conventional physics where phase is akin to an angle, and has no effect on the energy of the particle. It does impact the energies imparted but not the overall energy. There are several well measured lines for hydrogen such as the lyman alpha line at 10.2 eV which are between these fractions. How does the photon obtain such an energy in RS2?

Something I've been wondering is what would be the result of defining the base units of time and space in terms of plancks constant and the gravitational constant instead of rydberg constant? In other words, setting the base unit of time and the base unit of space to planck time and planck space (or perhaps using h instead of h bar as argued by xavier borg here: http://blazelabs.com/f-u-const.asp). This would definitely allow for any measured energy of photons. But many other things would have to be re-derived in light of such a radical change. Specifically, the calculations done "within unit space" would no longer be within unit space so that would have to be redone. Anyway, something that I've been thinking about.
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bperet
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Re: Question about photon energy

Post by bperet »

I am actually looking at this right now (my white board in the office is covered with photon stuff), trying to separate out what is actually the photon, from what is the photon interacting with matter. Basically, we cannot do ANYTHING with the photon, directly, so we have to infer its structure by HOW it interacts with matter. Nehru did the initial work on this with his birotation papers (which you should read).

The data is a bit confusing, as it was not what I expected. The photon distribution is NOT linear, and appears to have 8 dimensions to it (an octonian). Bit puzzled by this, at the moment... was expecting 4, a quaternion structure where you have i.j.(-k) = +1 rotations, creating an "internal" birotation since i.j=k, so k.(-k) = birotation.

Appears to be an exponential series... the octonian 28 = 256 (very "computer"), that "breaks down," not "builds up"...

256/2 = 128
128/2 = 64
64/2 = 32
32/2 = 16
16/2 = 8
8/2 = 4
4/2 = 2
2/2 = 1 (cannot have less than 1 unit, so stops here)

Plot these out on a line, as a repeating series of 256, and curiously enough, you get the EM band structure. (Bands are defined by waves that behave with similar characteristics, such as X-ray, uV, visible, iR, and all the various radio bands.)

The repeating series is defined by s×256n, where n=-8..+8. In the region of n=0, we find visible light in the octave between s=8 (violet) and s=16 (dark red):

7 is where ultraviolet ends.
8=violet, 9=indigo, 10=blue, 11=green, 12=yellow, 13=orange, 14=light red, 15=dark red
16 begins the transition to infrared.

I'll post more as I understand what the data is telling me.
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blaine
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Re: Question about photon energy

Post by blaine »

Thanks for the response, I'll definitely take a closer look at Nehru's papers. I do remember him talking about a birotation and I sort of visualized it as two helices rotating at the same speed but opposite direction... Not sure if that is accurate.

When you are talking about these numbers, are you referring to the energy of the photon? Some s/t or t/s ratio perhaps?

Something to mention is that when I was first looking at the energy of the photon, I thought well the units of energy are t/s, so why not just take the speed of oscillation and invert it (so a photon with vibrational speed 1/2 would have energy 2/1) , then just convert to whatever units... This would basically just reverse the table that I calculated in my first post, and it would make the hydrogen ionization work out fine. But upon more in depth examination of Nehru's calculation of planck's constant I came up with the table in my original post, which personally makes more sense to me since I think of a motion of 2/1 having more energy than a motion of 1/2.

Something that I've been thinking of regarding the RS2 model of the photon is, what if rotation is actually prime, and linear motion is somewhat of an illusion? In other words, what if the universe is cyclic; its birth and death constitute a full circle, and everything within that large circle is just rotations compounded upon rotations. The progression of the reference frame that Dewey Larson described would then just be a "geodesic" along that greater rotation.
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Re: Question about photon energy

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blaine wrote: Wed Jan 10, 2018 4:22 pm When you are talking about these numbers, are you referring to the energy of the photon? Some s/t or t/s ratio perhaps?
The numbers are natural units of space.
blaine wrote: Wed Jan 10, 2018 4:22 pm Something to mention is that when I was first looking at the energy of the photon, I thought well the units of energy are t/s, so why not just take the speed of oscillation and invert it (so a photon with vibrational speed 1/2 would have energy 2/1) , then just convert to whatever units...
The problem with that is when you flip speed to energy, you also change from linear velocity, v1, to orbital velocity, v2. That is why Plank's constant is needed to convert linear to angular velocity in E (angular) = h (rotational operator) f (linear).[/quote]
blaine wrote: Wed Jan 10, 2018 4:22 pm Something that I've been thinking of regarding the RS2 model of the photon is, what if rotation is actually prime, and linear motion is somewhat of an illusion? In other words, what if the universe is cyclic; its birth and death constitute a full circle, and everything within that large circle is just rotations compounded upon rotations. The progression of the reference frame that Dewey Larson described would then just be a "geodesic" along that greater rotation.
You are basically talking about a Riemann sphere here. Looked at it before, but Larson was insistent that the universe did not have a birth or death; it was steady-state. The evidence I have to date confirms that.

The RS2 research differs from Larson's RS in that the photon is NOT a primary motion, but a consequence of shear strain between two or more motions (much like the 19th century aether researchers looked at it). Larson used it because he could not accept angular velocity as primary--he needed a "line" (the SHM of the photon) to spin around to make a particle. RS2 accepts linear (yang) and angular (yin) velocities as primary, so the first "manifestation" is the positron and electron--NOT the photon.

So that goes back to what I am researching now--how to differentiate photon motion from the objects that created the shear, producing it.
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blaine
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Re: Question about photon energy

Post by blaine »

Perhaps the difficulty lies from the fact that the photon is a boson in the Standard Model, which means its a force carrying particle. It carries the electromagnetic force, which follows a U(1) symmetry group. In RS terms, the photon is simply the transfer of a 1 dimensional linear type motion between two existing motions.

There are several other bosons in physics because of the possible transfer of rotational motion; hence the SU(2) and SU(3) symmetries in the Standard Model - there are 3 types of bosons for the weak force which mediates the SU(2) symmetry, and there are 8 types of "gluons" for the strong force which mediate the SU(3) symmetry. These correspond to two dimensional and three dimensional rotational motions respectively. These bosons have mass because they constitute the change in motion for rotational type motions. thus are themselves a rotational motion. Perhaps this is where your 8 variables are coming from - accounting for the possible changes in motion between two complex 3d motions. If this is the case, then you are describing the generic "boson" rather than just the photon. But I could be misinterpreting your meaning.

If this is the case, then understanding the properties of the photon (or arbitrary boson) would the same as enumerating the possible transfers of motion available between any two arbitrary motions. The energy of the boson, being a scalar property, is just translating this motion change into a number that represents the potential for moving an electron (ie translational motion of the basic rotational unit).

Something that I've been thinking alot about is what type of motions are the atomic excitations and the nuclear excitations? Both of these excitations will emit photons when their "excited motion" leaves the entity (ie is transferred to another entity) and the entity goes back to the ground state (as Larson put it, the more stable configuration). The wavelength of the photon from nuclear excitation (MeV range) tends to be around the size of the nucleus, (ie picometers) whereas the wavelength for photons from electronic excitation (keV range) tends to be the size of the element (angstroms). I think this gives a clue as to the type of motions involved.
blaine
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Re: Question about photon energy

Post by blaine »

In regards to SU(3), I found some papers showing the connection between octonions and the SU(3) basis elements (the Gell-Mann matrices) and their connection to particle physics

https://arxiv.org/pdf/1204.0242.pdf
https://arxiv.org/pdf/1107.1559.pdf

Perhaps these could be useful for making a mathematical system in terms of octonions work for describing the various motions of RS2
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bperet
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Re: Question about photon energy

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Unfortunately, when you start talking hypothetical particles (gluons and such) that have no "natural consequence" derivation in the RS, I get a bit lost... I prefer just to take a fresh look at the data to see just what IS a natural consequence, then, like Larson, try to match it up with something observed. (Cannot really do that with a lot of this, because there is little difference between "theoretical" and "fantasy" physics nowadays.)
blaine wrote: Thu Jan 11, 2018 6:35 pm Something that I've been thinking alot about is what type of motions are the atomic excitations and the nuclear excitations? Both of these excitations will emit photons when their "excited motion" leaves the entity (ie is transferred to another entity) and the entity goes back to the ground state (as Larson put it, the more stable configuration). The wavelength of the photon from nuclear excitation (MeV range) tends to be around the size of the nucleus, (ie picometers) whereas the wavelength for photons from electronic excitation (keV range) tends to be the size of the element (angstroms). I think this gives a clue as to the type of motions involved.
I have not dug in to the origin of isomers yet, in the context of the RS. What are your thoughts on it?

Certain things can be eliminated. It cannot be a change in the magnetic or electric rotational speeds, as that would change the atomic number. It could be a property of magnetic ionization (captured neutrinos, unknown to conventional science) or electric ionization (captured, uncharged electrons).

I took a quick look at isomers, which leads directly to the nuclear shell model, where they treat the protons and neutrons in the nucleus much like they treat the electrons outside of it. Electron quantum numbers are created by the interaction of photons and atoms--there are no physical "shells" or "orbitals," but just harmonically resonant interactions that give the appearance of physical shells. The same is probably true for the nuclear shells--they are a product of how they are being measured, not of an actual, physical structure.

Isomers act like something captured in the nucleus, that can be emitted as a birotation in complex, quaternion or octonian form. Since the RS model is nothing more than a compound rotation, there is no shell structure to it to explain the behavior, hence the need for an additional component.
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