Neutron charge emission - where does it go?

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Re: Neutron charge emission - where does it go?

Post by Nevyn on Wed Mar 01, 2017 6:37 pm

I don't think the electron is 1/1821 the size of the proton, I think it is 1/1821 the mass of the proton. I could be wrong but if the electron is only 3 or 4 spin levels below the proton (depending on if you believe in higher level axial spins and if those axial spins increase the size or not), then it is only 2^3 smaller than it.

I can't see the electron in that image above. Is it mingled in with the blue charge?
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Re: Neutron charge emission - where does it go?

Post by LloydK on Wed Mar 01, 2017 7:38 pm

I didn't read everything here. Why do you want a neutron in your Hydrogen atom simulation?

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Re: Neutron charge emission - where does it go?

Post by LongtimeAirman on Wed Mar 01, 2017 7:51 pm

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1. I agree with Lloyd. For a hydrogen model, I’m surprised you’ve included a neutron. Hydrogen is usually described as a proton and an electron.

2. I see the electron in the initial setup, but it is soon lost in the photon motions. I appreciate the blue electron charge stream.

3. In my Proposal for Electricity Animation post, I mentioned diatomic Hydrogen, or Hydrogen gas, which is actually two Hydrogens and two electrons. Miles goes into some depth in Diatomic Hydrogen  http://milesmathis.com/diatom.pdf.
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Re: Neutron charge emission - where does it go?

Post by Jared Magneson on Wed Mar 01, 2017 8:29 pm

I included the neutron (deuterium) specifically because I'm studying the neutron more heavily right now, to understand the dynamics and mechanics. Thus, this thread here. Smile

Also I'd already animated the lone proton (no electron, though). It would be very easy to rework both animations and also tritium, so I'll go ahead and get on that.

Here's his paper on the topic including some good stuff on neutrons, if anyone needs a quick link:

http://milesmathis.com/deut.pdf

LongTimeAirman wrote:Miles goes into some depth in Diatomic Hydrogen http://milesmathis.com/diatom.pdf.

Thanks for the link, I'll reread that one again now. Looks particularly useful here. It's hard to digest 300+ papers fully even after multiple reads, thus our analysis here. You guys are very helpful.

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Re: Neutron charge emission - where does it go?

Post by LongtimeAirman on Wed Mar 01, 2017 8:39 pm

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Nevyn, compare the proton and electron, mass and radii from the Mass, Radius, Circumference, Wavelength, and Energy Table.

re, radius electron (m) 2.36E-18, rp,radius proton 4.34E-15. rp/re=1836
me, mass electron (kg) 9.109383E-31, mp mass proton1.672622E-27. mp/me=1836
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Re: Neutron charge emission - where does it go?

Post by Nevyn on Wed Mar 01, 2017 9:40 pm

If we assume the electron is 2.36E-18m in radius, and I assume that means the top level spin radius, then we can't reach the proton in 3 or even 4 spin levels. We have to go through 11 spin levels to reach 4.83328e-15m.

I just don't see how these numbers equate unless those radii are including the charge field of the electron and proton. The proton would have a larger charge field size (relative to its own top level spin radius), since it is bigger and can collide with more ambient charge photons, so we might be able to reach that size with only 3 or 4 spin levels but that is still a huge increase in size for the relative charge fields.

Let's assume we need 4 spin levels to go from an electron to a proton (since that is what Miles has stated, although I don't agree with it) and also we will assume, illogically, that the axial spin doubles the radius.

2.63E-18m * 2 * 2 * 2 * 2 = 3.776e-17m

Let's look at the difference in radius to the stated proton size:

4.34E-15m - 3.776e-17m = 4.30224e-15m

So a very large portion of that size is unaccounted for and I don't think we can use that much space for the emission field.

Let's look at the ratio of that size to the stated proton size:

3.776e-17m / 4.34E-15m = 8.7e+29

That is a huge number. It is 2.9e+21 * c -> 9.667e+12c^2 -> 32200c^3.

I just don't see how these numbers add up. What am I missing?
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Re: Neutron charge emission - where does it go?

Post by LongtimeAirman on Wed Mar 01, 2017 9:51 pm

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Nevyn, Sorry to distress you, please don't look at it that way. There's a simple explanation or two, consider them puzzles. I have an engagement to attend to, and will reply tomorrow.
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Re: Neutron charge emission - where does it go?

Post by Nevyn on Wed Mar 01, 2017 10:02 pm

That's cool, Airman, I like to work from first principles so I want to see all of these values coming from the BPhoton radius, through stacked spins, to some real spin radii. It can be difficult to know what the radius of an electron or proton actually is. Does a given value include the charge field or not? Of course, if they are mainstream values then I do assume they include the charge field, since they can't remove what they don't know.
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Re: Neutron charge emission - where does it go?

Post by Jared Magneson on Wed Mar 01, 2017 10:04 pm

Nevyn wrote:Let's assume we need 4 spin levels to go from an electron to a proton (since that is what Miles has stated, although I don't agree with it) and also we will assume, illogically, that the axial spin doubles the radius.

Can we chart the spin stackings and come up with a solution for this? Perhaps you already have; I know we've talked about it before. Do you think the proton is eight spins above the electron?

From "Unifying the Electron and Proton":

Mathis wrote:We repeat this same math and logic to create the y and z-spins. The radius of the y-spin is 4, so the term will be [1 + (8 x 16 x 32)/22]. We divide by 4 since we must use only half of both end-over-end spins. Likewise, the z-spin is [1 + (8 x 16 x 32 x 64)/24]. We divide by 2 squared squared because we are now in three dimensions. The x-spin is expressing only 1/4 of its strength relative to z, since it is orthogonal twice. The complete equation or representation then becomes:

[1 + 8], [1 + (8 x 16)/2], [1 + (8 x 16 x 32)/22], [1 + (8 x 16 x 32 x 64)/24]
= [1 + 8], [1 + 26], [1 + 210], [1 + 214] = 9, 65, 1025, 16385

The electron with all spins has an energy of 16,385. The electron with no spin has an energy of 1. The electron with axial spin has an energy of 9. If we divide 16,385 by 9 we get 16,385/9 = 1820.56

We may therefore deduce that the electron at rest is spinning only about its own axis. An electron with all possible stable spins is a proton, anti-proton, or neutron. An electron with no z-spin is a meson.

http://milesmathis.com/elecpro.html

Is this merely a matter of nameology? Wink Are we often naming the electron at rest (which doesn't occur naturally, since everything has a linear velocity), or is that where Mathis got fuzzy on us? His numbers are relative to the electron, of course, which doesn't necessarily help us in this case.

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Re: Neutron charge emission - where does it go?

Post by Nevyn on Wed Mar 01, 2017 10:46 pm

Miles is talking about energy there, not radius or mass, although he is using the radii relationships. We could use E=mc^2 to determine the mass difference:

1820.56 / c^2 = 2.02284e-14

I assumed meters but I'm not sure if that value is in g or kg.

Yes, I did write a spreadsheet containing spin radii which can be found at:

http://www.nevyns-lab.com/mathis/math/spin

As I wrote my post above, and found the 11 spin level difference, I originally stopped at 9 thinking that the next doubling would take me over the stated value. I initially took that as evidence of not using higher level axial spins and the proton would be 3 spin sets larger than the electron. Then I thought I should check and found 11 levels difference which goes slightly over the stated value. This could mean something or I could be on the wrong track. Hell, I may be in the wrong ballpark. I just can't help but think that all of these values should be related and derivable from each other and I am determined to find those relationships.

I applaud your coining of the term nameology (which I am pronouncing name-ee-ology, just for kicks). I do this all the time when I can't think of the word I am looking for. It can be quite comical at times. The word you wanted was nomenclature but that doesn't sound anywhere near as cool.
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Re: Neutron charge emission - where does it go?

Post by Jared Magneson on Thu Mar 02, 2017 12:23 am

Nevyn wrote:I applaud your coining of the term nameology (which I am pronouncing name-ee-ology, just for kicks).

That's exactly how it's pronounced, but I didn't coin it! Willem Dafoe's character did, in some old movie entitled, "Boondock Saints".



http://www.moviequotedb.com/movies/boondock-saints-the/quote_4206.html

But you are correct. Mathis was calcing energies, not radius directly. I think I have been conflating the two a lot lately, in my math, so I'm glad you pointed this out before I made even more mistakes.

So on your chart, in the first page, which numbers would be the electron and proton? I was trying to match the data there with your above post's data but couldn't figure it out. Are you saying the proton is 11 spins from the initial Axial spin photon, or eleven higher than the electron? Just trying to clarify. I apologize if I seem muddy on this topic, even after all our conversation. I don't trust myself too deeply on this topic, it's still all very new to me.

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Re: Neutron charge emission - where does it go?

Post by Nevyn on Thu Mar 02, 2017 12:26 am

Jared Magneson wrote:I'm showing the electron as blue, located at the south pole there like it is in the He4 alpha, but I'm also showing it to scale. Maybe not helpful for these diagrams? Maybe I should make it 1/10th the size instead of 1/1821st? Would simple labels to one side be more helpful?


I just watched the video rather than look at the image and I could see the electron before the charge starting flowing.

My question is about that blue charge. Is that coming from the electron? I don't think it should be. It doesn't have enough charge emission, and not in the right direction, to go up through the proton. But if it is just representing the protons through-charge, then everything is fine but with the electron being blue and that charge being blue, I assume they are linked.
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Re: Neutron charge emission - where does it go?

Post by Nevyn on Thu Mar 02, 2017 12:36 am

Jared Magneson wrote:That's exactly how it's pronounced, but I didn't coin it! Willem Dafoe's character did, in some old movie entitled, "Boondock Saints".

Ahh, good movie and great actor.

Jared Magneson wrote:
So on your chart, in the first page, which numbers would be the electron and proton? I was trying to match the data there with your above post's data but couldn't figure it out. Are you saying the proton is 11 spins from the initial Axial spin photon, or eleven higher than the electron? Just trying to clarify. I apologize if I seem muddy on this topic, even after all our conversation. I don't trust myself too deeply on this topic, it's still all very new to me.

Yeah, that was all a bit confusing.

The 11 spin levels comes from the difference between the radii values Airman wrote. I took the electron radius and kept doubling it until it reached somewhere near the value for the proton and it took 11 levels to get there. Since Miles has explicitly stated that the proton is 4 levels above the electron, I couldn't see how to make all of the numbers work.

So, in short, the 11 levels start from the electron and go up to the proton.
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Re: Neutron charge emission - where does it go?

Post by LongtimeAirman on Thu Mar 02, 2017 8:39 pm

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My Mass, Radius, Circumference, Wavelength, and Energy Table is in excel; I’ve played with the numbers and ran across this problem too. I see it takes roughly 11 doublings of an A radius (electron) to reach the next A radius (proton). Please notice that every number on every row is separated by a Dalton (1820.56), (or NIST 1836), or approximately 11 doublings from the A with their adjacent numbers. The Dalton may be a scaler, but how can 1821 be construed as 4 doublings? I agree, what’s going on?

r = mc√c/2
Cγ = λmγ/E = 8rγ
λ = Cc2
Eγ = 2rγ√c = mγc2 = λ/4c√c

The speed of light, mass, radius, circumference, wavelength and energy are all functions of one another. We can easily calculate a column given any single number. I’d say the table’s definitely built from first principles, though my version may be lacking. For example, I used NIST electron and proton values, rather than calculating the radius of the electron to be 1/c2. I obtained Miles’ IR Photon values so I thought it was good enough for an ex govt worker.

Focusing on Unifying the Electron and Proton: http://milesmathis.com/elecpro.html. I’ve read it many times before today. My latest effort hasn’t gotten me any further. Here’s one quote – “Well, as the radius is to the velocity, the circumference will be to the spin”; someone needs to explain what that means. I can follow the energy calculation expressions - finally divide 16,385 by 9, and we get about 1821.

I'll throw in a scientific wag - straight doubling makes no sense, maybe the radii are stretched out by the speed of light similar to mass, wavelength and energy.  
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Re: Neutron charge emission - where does it go?

Post by Nevyn on Thu Mar 02, 2017 10:29 pm

Airman wrote:Here’s one quote – “Well, as the radius is to the velocity, the circumference will be to the spin”; someone needs to explain what that means.

It took me quite a few reads of that section but I think I may have some meaning for it.

Miles is talking about the relationships between these things. We can express the velocity in radii just by dividing the velocity by the radius. That gives us the number of radii per second or whatever our units are.

We can do the same thing with the spin but we can't use the radius, we have to use the circumference. So we divide the orbital velocity by the circumference and that value should be the same as the linear velocity divided by the radius. Note that the linear velocity and the tangential velocity are equal in this instance.

At first, it doesn't seem to make sense to express the velocity in radii, but the reason he is doing that is because the radius is common to both equations.

This is actually what I did when I developed my spin equation a few weeks ago. I took Miles angular velocity equation (which gives the tangential velocity as traveled on the circumference), calculated a value with it and then divided that by the circumference. This gives me the relationship between the tangential velocity and the circumference which can then be converted into an angle by multiplying by 2pi because 2pi is an expression of the circumference in angles.

You have to remember that a circumference is not a circle because a circumference is a circle at a certain radius. For my needs, I only wanted the angle because I was already using the radius in the translation and just needed the rotation to go with it. Once I put the translation and the rotation into the same matrix (the translation and rotation are separate parts of a 4x4 matrix) I had created a matrix that put the particle in the correct location for that spin level. Multiplying all of those spin level matrices together gave me a single matrix that took the BPhoton from the particles location (its center) to the correct position relative to that center.

Another way to look at it is that the linear velocity and the radius contain straight distances. The orbital velocity and the circumference contain curved distances. That is why we can't just divide the orbital (or angular) velocity by the radius, because it isn't curved and you can't divide by something different to what you are dividing.

This is just really saying that the distance traveled by the linear velocity, in some given amount of time, must match the amount of spin about the circumference in that same time. In other threads I have made posts saying that you have to make sure that your relative velocities are all equal rather than care about real time. We can speed up and slow down our animations as long as the relationships between things remain the same. This is what Miles is saying here.

Given an axial spin, the particle must make 1 complete revolution in the same time that the linear velocity has traveled 8 times the radius. If that is the case, then the linear velocity and tangential velocities are equal. It took me quite some time to figure that out on my own. I didn't realise that Miles had written it in not-so-straight-forward terms already.
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Re: Neutron charge emission - where does it go?

Post by Nevyn on Thu Mar 02, 2017 10:39 pm

Airman wrote:I'll throw in a scientific wag - straight doubling makes no sense, maybe the radii are stretched out by the speed of light similar to mass, wavelength and energy.

No, there is no way for the radius to be stretched out like that. The radius is a real thing that does not rely on time but mass, wavelength and energy do, even if the mainstream don't realise it. Because they rely on time, they can be stretched out since you are looking at them with respect to their previous position or value but the radius is relative to the center and that is moving with the velocity too.

Mass and energy are a little different than wavelength but they still rely on time and they both have the velocity wrapped up inside of them. The radius is what it is no matter what the velocity is. It is completely independent of anything else.
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Re: Neutron charge emission - where does it go?

Post by LongtimeAirman on Thu Mar 02, 2017 11:21 pm

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Nevyn wrote. Given an axial spin, the particle must make 1 complete revolution in the same time that the linear velocity has traveled 8 times the radius. If that is the case, then the linear velocity and tangential velocities are equal. It took me quite some time to figure that out on my own. I didn't realise that Miles had written it in not-so-straight-forward terms already.

Thanks Nevyn. I must admit, its hard for me to ask for help at times, and sometimes even harder receiving it. This paragraph is especially helpful. I do seem to have a bit of difficulty with combining linear and curved motion.

If mass is time dependent, it would seem to me that radius would be time dependent too. I'll chew on it.

I've been thinking about the dalton a lot lately. If I get any inspiration I promise I'll share.
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Re: Neutron charge emission - where does it go?

Post by Nevyn on Fri Mar 03, 2017 12:17 am

Don't feel bad about misunderstanding curved motion or distances. The mainstream still hasn't understood it with 100's of years of effort!

The important thing to remember when dealing with linear and curved motions is to keep like things together and un-like things apart. You can't divide a curve by a linear distance and expect a meaningful answer. This is why Miles often works in relative values. You can exchange relationships but you can't exchange values.

I might have thrown you a curve ball regarding mass. Mass is time dependent when discussing charge since we need to know the mass of that charge and that relies on time. However, if there is such a thing as intrinsic mass (which only the BPhoton could have) then this is not time dependent. I think it is the intrinsic mass that you are thinking of where-as I was thinking of mass by charge emission. I was wrong to use that kind of mass because we are not talking about charge emission here (although it is difficult to know if those mass values include the charge field mass and if they do then they are time dependent).

The mainstream separate the linear velocity from mass, only because they can calculate the linear velocity (and had that before the concept of mass) but they don't know what mass is or where it comes from. They can see that there is more to a collision than just the linear velocity and have attached this to mass (yet another hole filled with a word and little more). If you subscribe to my idea of stacked spin velocities being the mass, then mass is velocity, expressed in a collision. What else could it be? Mass only effects velocity, so why not assume it is a velocity (or a few)? Of course, you can't make that connection without stacked spins since they provide a different kind of velocity than the linear version we are all used to.

So to be clear and try to clean up my mess, intrinsic mass, if it exists, is not time dependent and will not stretch by the linear velocity either. But if you are thinking of mass increase in a particle accelerator, then this is time dependent or more importantly, velocity dependent (which includes time inside of it). Of course, Miles has described this kind of mass increase as stacking more spins so that comes right back to my idea of stacked spins being the mass.
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Re: Neutron charge emission - where does it go?

Post by Jared Magneson on Fri Mar 03, 2017 12:47 am

Good stuff, guys. I got this question the other day in an online debate with some of my good Facebook pals. An apologist asked why doubling the radius in the next stacked spin would increase the mass. I didn't have a great answer for him, except to point out that the proton was already considered more massive than the electron in the standard model and that's the data that shows a mass difference right there.

We know the proton is more massive. Is this because it has more velocity (non-linear) in the field, due to those additional spins? It collides with other particles with more strength in the field, relative to the electron, because it's harder for another particle to divert the proton from its path (unstack a spin)?

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Re: Neutron charge emission - where does it go?

Post by Nevyn on Fri Mar 03, 2017 12:59 am

That's close to my theory. In any collision, and mass is only expressed in a collision, all of the velocities have to be taken into consideration so we must include the spin velocities. The mainstream fall back to momentum: p = mv. They don't know what that mass is, but they know they need it. Since v expresses the linear velocity, then m must express the spin velocity. To be more precise, m expresses the sum of spin velocities. Since mass is only expressed in a collision, and collisions are about velocities, then why not express mass as a velocity. Seems pretty straight forward to me but I haven't tried to play with the numbers. I find it difficult to sort through the mainstream mess. This is why I prefer to work from first principles so that I can skate right around their mess.

There are a few posts about my idea somewhere on this site. I can't remember which thread it is in though. It could be either the Atomic Viewer or Spin Sim threads.
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Re: Neutron charge emission - where does it go?

Post by Jared Magneson on Fri Mar 03, 2017 1:30 am

That makes a tremendous amount of sense to me, and it solves several mysteries (from the standard model) at once. I've often been curious about how our quanta can leap into larger, higher states of energy and mass, even while being the same fundamental "particle". What that is made of, we don't know yet. It's a postulate. But given we have matter existing at all, it's really cool to unwind how it behaves. And this model is far more elegant and "sensical" than the standard ones, with all their fudgings and dodgings and probability clouds.

In other news, here's my latest video on Hydrogen, showing kinda how the electron gets pushed into the charge potential pole of the proton. I'm not happy with the Vimeo compression though, as it's darkened the ambient purple charge and the through-charge to the point of invisibility, nearly. I'm going to render it with larger particles and try again tonight.

I might try overlaying a screenshot of your Spin Sim app at the proton level as a texture map for my proton sphere, too. Might look cooler, or help visually explain that the proton isn't exactly a big yellow sphere. Hope you don't mind; your Spin Sim is still way beyond what I'm doing in Maya.

https://vimeo.com/206370190


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Re: Neutron charge emission - where does it go?

Post by Nevyn on Fri Mar 03, 2017 3:08 am

I don't mind at all. I'll be amazed if it works.

I've been thinking of writing a shader that takes in the location of the particle and calculates the position of the BPhoton based on spin levels. If Maya gives you the power to set custom attributes for that shader, then you can use it.

I have all of the math already, just need to port it to a shader language (not sure if Maya uses GLSL, the OpenGL shader language, or Cs, the NVidia language but I do know that you can add custom shaders).

It will have to make some assumptions but I can produce an easy to use shader for a proton/electron type and another for the neutron/nectron type. Should make it simple for you to put them wherever you want them just by specifying the position and spin level parameters.

I already have charge emission shaders in Atomic Viewer, but that might be a bit hard to set up in Maya as you need a few values per charge photon. That's easy to do in a programming language but maybe not so easy in a UI. Scripting would help there.
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Re: Neutron charge emission - where does it go?

Post by Nevyn on Fri Mar 03, 2017 3:51 am

Here is that link to my spin as mass theory: http://milesmathis.the-talk.net/t65-stacked-spin-breakthrough#600

There is some really good stuff in that thread.
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Re: Neutron charge emission - where does it go?

Post by Jared Magneson on Mon Mar 13, 2017 10:40 pm

Making a bit of progress here. Bear with me: I know the charge profiles for these are all wrong, as they're only emitting spherically and not recycling in this demonstration, and there's no atomic spin involved just yet. But I was able to create repulsions and almost build the alpha using dynamics alone, no animation tricks.

https://vimeo.com/208245448



A lot of refining left to do, including getting the numbers precise and importing our actual stacked spins to replace the spherical particles I'm showing. Guess I'm just excited since this is what I've been trying at for some time, now. The N pole electron for example isn't emitting as much as the S pole, but I wanted to share this and get some feedback anyway.

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Re: Neutron charge emission - where does it go?

Post by LongtimeAirman on Tue Mar 14, 2017 6:06 pm

.
Jared, Here's my response to your request for feedback (with caveat more needs to be done):

Nice overall effect, surprising what a little motion can do. First time I saw it my stomach contracted along with the protons; yeah, yeah, and wobbled with the neutrons.  

Ok, I see the electrons emitting blue charge.

Why do we see so much charge (magenta) originating at the two maximum vertical extents? Are those electrons there? I guess it makes sense if the maximum extents are the electrons associated with adjacent alphas, a vertical nuclear charge channel; I don’t imagine lone alphas encountering such strong currents otherwise.

Any defining equations you care to share, i.e. What are you using for repulsion?

When replacing charged particles with their B_photon spins, I would expect one to stop at the previous A spin. In which an electron sized B_photon moved through stacked spins within the proton (or neutron). Your thoughts?
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