Why is Curium the Last Semi-Stable Element?

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Why is Curium the Last Semi-Stable Element?

Post by Cr6 on Wed Apr 18, 2018 8:34 pm

Miles has an updated/new paper on Curium.


Why is Curium the Last Semi-Stable Element?
That's Cesium, not Curium
by Miles Mathis

First published April 16, 2016

Yes, I am finally back to work on the Periodic Table, after a long break. I saw that Oganesson, number
118, had been added to recent Tables above Radon as a 7th Noble Gas. My readers will know that can't
be right, since I have shown that Radon isn't a Noble Gas itself. So Oganesson can't possibly be one.
The rules of nuclear construction mostly fail after Barium, with the Lanthanides built on a different 5-
proton block, and the Actinides being pieced together from smaller elements. Which led me by
meandering paths back to Curium, the last semi-stable element. It has a stability four orders of
magnitude greater than the elements beyond it, indicating something important.

http://milesmathis.com/curium.pdf

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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Wed Apr 18, 2018 9:00 pm

I just posted in the Graphene thread about this. Happy to see more nuclear structure papers. This should help me build some of the higher elements but it will require a new type of atomic structure. Probably just what I need to pull me back into AV.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Jared Magneson on Thu Apr 19, 2018 2:33 am

An interesting paper, and perhaps a turning point in his analyses. In this one he offered some rather extreme solutions to building larger elemental structures, the kind of stuff that (it seems like) he could probably keep secret and profit from at some point, but I don't think he's into that scene and a good thing, too. His stuff about Darmstadtium was pretty radical to me.

I've also been reading up a lot about Germanium, which can be used in transistor technology. They recently hit 800GHz transfer speeds through that medium (SiGe, "siggy", or Silicon-Germanium) which is pretty astonishing. Granted, it was supercooled, but even 100GHz could be a huge boost for semiconductors if they can figure out how to implement it. I've often recently thought that the ~5GHz barrier of our own CPUs had more to do with a charge throughput barrier than a technological one, given CMOS tech and its evolution. Perhaps Germanium could break us through that barrier - but then again, it's been worked on since the 1970s by IBM and others. And perhaps it's possible that "TPTB" have such speed at their fingertips already, and us lowly peasants just get the CMOS tech as an afterthought or marketing ploy. Sorry for the tin foil there, but sometimes that stuff occurs to me and of course there's nobody else around who would even consider it. Smile

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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 5:48 pm

I am copying a few posts from the Graphene thread (http://milesmathis.forumotion.com/t123-mathis-on-graphene-any-hints) as they are more relevant here than over there.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 5:49 pm

Jared Magneson wrote:So from a physical standpoint, how does this work? Are molecular atoms merely a short distance further apart than Kontatoms? I think we're hitting new ground here, with that latest paper, but I've often had these questions about Uranium too since he first diagrammed it.

That is the million dollar question at the moment. Just how far apart are these bonds and how do we determine that distance? Let's look at how these bonds alter the proton stacks involved.

Broadly speaking, there are 3 possibilities when joining 2 atoms in a nuclear bond:
1) 1 atom has a hook proton stack and the other has no hook stack (such as a noble element);
2) both atoms have hook stacks;
3) neither atom has a hook stack and the joining stack comes from the field.

Case 1) is fairly straight forward and the single hook stack becomes a shared hook stack with the other atom. That is what Miles has been using so far or at least hasn't gone into much detail at this level. He also may be thinking of case 3).

Case 2) is more complicated and may not even be allowed. It involves moving the existing stacks apart to accommodate each other. Molecular bonds do this, or may, but I'm not sure if it is possible for a nuclear bond. I am going to allow it until convinced otherwise.

Case 3) is probably the more likely scenario and is certainly easier to imagine. We start with noble elements, force them together in a field rich in proton stacks of the size we want to bond them. The carousel levels could also fill up with the same stacks. It may be possible to control this. Let's look a little deeper to see how that works.

We start with a container of, say, Krypton and Xenon.
Apply an electric field to align the atoms or make use of the Earth's charge field to do so.
Introduce H, He, Be, Li, etc, whichever is the right sized stack for your purposes.
Increase electric field to force atoms together.

That gives us the axial (north/south) structure. We can remove the electric field and they should stay fairly stable in a protected or low charge environment. Now we can create the carousel levels, if we want to.

Introduce the desired proton stacks (H, He, etc).
Align stacks to the north/south axis with an electric field.
Apply electric field in a radial pattern in the plane perpendicular to the north/south axis. From outside to inside, forcing the stacks into the carousel positions. This may break the original structure if the electric field is too strong because those central bonds we created are not well protected. This is why such elements are unstable. It may require very precise fields that are targeted at the carousel levels but that seems too difficult.

I am imagining a series of electric fields applied in sequence. Align Z, push, align XY, push, etc, repeated until the desired yield is achieved. Various atoms are added at certain stages to prime the field with the required entities. That actually seems quite feasible, although I'm not sure about some of those steps. For some strange reason, no-one let's me play with billion dollar machines Crying or Very sad .

But I digress. How do nuclear and molecular bonds differ? If we look at case 2), which is the closest to a molecular bond, then we can ask how far do the hook stacks penetrate each other in each scenario? Nuclear bonds have 100% penetration such that the 2 stacks become 1. Molecular bonds have some separation and I imagine it is close to half way but I don't have anything to back that up, it just seems like a nice number. It doesn't really matter too much what the exact value is, any separation means that we are not dealing with a normal proton stack because it does not have an aligned central path for through-charge to flow.

I think this is important because that through-charge is like a locking pin for the stack. It keeps the protons aligned in order to call it a stack. If the 2 stacks penetrate deeper than 50%, then their proton emissions are going to interrupt each others through-charge and they all fall apart. So the 2 stacks must be less than 50% penetration.

50% of what, you may ask? That percentage is applied to the distance from a proton to a point away from that proton that its emission density drops below some value. It is the distance that we could say: within that distance is a proton and outside of that distance is the ambient field. It can be thought of as the size of a proton, but not the size of the proton's stacked spin radius. It includes the protons charge field.

I do not normally define a protons size as this value since I am usually thinking about stacked spins and so I want to use the stacked spin radius for that size, but in this case, where we are using the proton as a charged entity, we can include some of its charge field. I hope that is clear, for it is a bit confusing. I would call the protons' stacked spin size its Quantum Radius and the size I am using here I would call its Charged Radius.

Moving on...

Does everyone agree that there is at least some penetration in the molecular bond? Is it possible to have no penetration? That feels a bit weird to me as there would be two proton stacks emitting directly at each other and the only thing bonding them together would be the atoms through-charge streams.

It is certainly possible to have molecular bonds that do not penetrate and you can look at the Hydrocarbons that I have modeled in the Chemistry section to see some of them. They are not the same type of bond though, as they have 1 stack emitting into the input of another stack, not 2 stacks emitting at each other.


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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 5:49 pm

Cr6 wrote:
I'd like to see too "Molecular bonding" via Left/Right spins in a sense. What's incoming/outgoing with charge/photons that can be shared via protons and unbalanced hooks? What really are the "basics" to allow joining/alignment into a stable mass (proton-wise/stacked photon spin-wise/alpha-wise-"hook" wise)? Maybe this question is too simple but these are perspectives I'm trying to reconcile in the model..."bonding" references these perspectives.

As far as I can tell, the actual bonds are caused by the charge channels of the atoms, not the emission of the protons that make those atoms. In most cases, where the atoms bond along the north/south axis, it is the through-charge of each atom that aligns allowing a longer path. That path represents a direct route for charge photons but how that creates a bond I'm not really sure.

The hook stacks are located where these channels enter/exit the nucleus. They do provide a source of charge for these channels and they can also deplete some of the charge as their own emission. You can look at the hook stacks as an indication of how well fed the atom already is, before any other atom might bond with it. If the core of the atom contains stacks with 6 protons each, then they can pass a lot of charge through. If the hook stack is only 1 proton, then it is only being fed 1/6th of the charge that it could handle, 2 protons and it has 1/3rd. The core of the atom does not attract that charge, but it will accept it if supplied.

That's why the hooks are important. They essentially set the base charge capability at that point in the nucleus. For charge coming in at that location they act as an intake fan, pushing some of the charge that it receives down into the nucleus. When that charge reaches the other side of the nucleus it encounters another hook stack which will take some of it as its own emission field and allow most to pass through as a coherent stream. That is a charge channel of the atom. Each hook stack acts as both intake and exhaust, depending on which way the charge is coming from, but on the exhaust side it also allows most of it through because the charge is a more focused stream than what arrives from outside of the nucleus which is random ambient field charge.

Why do the nuclei come together to bond? I think this is caused by the ambient field pushing from all directions. When two atoms are near each other they will each shield the other from a certain amount of that ambient field. This causes less force from the direction of the other atom and therefore they will move together. Some may try to use gravity here but I don't find it necessary. It may play a part but the field already contains enough forces to do it.

Once they are close and aligned, the charge channels of each atom will work together. This results is less repulsion since the charge is being channeled, not collided with or redirected. An analogy for this is a hose and a metal plate. Given enough pressure in the hose, the metal plate can be pushed away from the hose. However, if there is a hole in the middle of the plate then some water goes straight through and does not provide any force on the plate. If we make the hole large enough to take all of the water from the hose, then we are only left with outside forces which are pushing the two together.

Cr6 wrote:
Like is "doping" just a form of a charge flow around larger frozen "crystal" structures with certain atoms positioned in edge positions that optimize with the large "crystal" structure. Does a super-cold atoms create a 4th path for bonding via "hardened crystals"...path channeling outside of "blocks" like whiskey poured on blocks of ice? Thinking of Noble gases primarily (Xe) as you mentioned.

I think doping is just adding certain atoms that provide some desired property. They will fit into the molecule in such a way that they enhance it. Maybe providing more charge channeling capability, maybe better conduction or maybe it smooths the internal charge pulses of otherwise incompatible atoms. I'm not sure how crystals fit into all of this yet. They are an interesting topic but I think I need to understand basic bonds before I can have any confidence in creating theory for them. Graphene and other so-called 2D molecules are a good stepping stone into the 3D structure of a crystal. If we can figure out these bonds then I think we will be able to model graphene and nano-tubes, etc.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 5:50 pm

LongtimeAirman wrote:.
Nevyn, thanks for the extensive detailed descriptions, I can now properly appreciate how the Curium paper offers many additional atomic possibilities. That thread has been robbed, but you did refer to this thread from that one.

Given your request for opinions, I like your second idea – there are no hook stack penetrations. A partial penetration of two hook stacks would interfere with the proton separations within each atom’s stack, breaking the integrity of each of those pole hook stacks before a bond can form – no halfway positions allowed(?). Spins need to be taken into consideration, here alphas complicate things, how can a proton penetrate an alpha? Any penetrations would leave at least leave the alphas intact, I don't see how penetrations can work. 1) Molecular bond – no interpenetration of hook stacks, just two stacked hook stacks where each stack adjusts slightly one way or the other, preventing the two from completely merging their hook charge currents. Given enough ambient energy the two atoms will separate at that bond. 2) Nuclear bond – the hook stack bond is indivisible.
.

When I first starting thinking about these bonds and their penetration into each other, I saw each proton moving apart and then the 2 stacks sliding together seamlessly. I think that is very naive now. I think the stacks must remain a whole entity and just move over a bit so that the other stack can sit along side of it and then they can create a complete stack if their centers align.

It does get tricky if they are not completely penetrating, though, as in molecular bonds. Could the stacks not move at all, and just start mingling their charge fields? If we assume less than 50% penetration then I think this might be feasible. Each stack is emitting directly into the other stack which feeds both the stacks and the charge channel of the atoms.

My question is: Is the hook stack, when looking at it as an input, the only source of charge for the charge channel behind it? I imagine that some stray charge will skip by the stack and make it into the charge channel all on its own, but the hook stack provides a more coherent stream. It is acting like a little laser aimed into the nucleus.


Last edited by Nevyn on Sat Apr 21, 2018 5:59 pm; edited 1 time in total
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 5:50 pm

Jared Magneson wrote:I'm having a hard time visualizing "hook stacks" in this context. I mean, visualizing the various atoms is easy (thanks to your Atomic Viewer and Periodic Table, which I use daily!) but I don't quite grasp the concept of the "hook" here. Going to study it more, but maybe you could draw up something quick and simple, Nevyn? It doesn't have to be fancy, maybe just Photoshop or MSpaint or something, or even (heaven help us) on paper with one of those pencil thingies?

Perhaps starting small, with say the water bond or something might be helpful?

I have been thinking about creating some images to show this stuff. I'll see what I can come up with.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 6:08 pm

I think there is a danger is assuming that the emission of a proton or stack is always a repulsion. It seems very logical at first, but the emission is not a constant force, it is lots and lots of little forces provided by each and every charge photon. Since there are gaps between the charge photons, with varying intervals, it is not a situation that gives 100% repulsion. It allows some squishy-ness, for want of a better word. This can allow the stacks to come quite close to each other, and in fact they would even enhance each other, some-what, since they each provide a source of charge for the other.

Also, I mentioned above how the exhaust process of a stack allows most of the charge channel to go by without interaction. When 2 atoms bond, they essentially make the input process do the same. Most of the charge from the other stack is going to just go straight through and into the nucleus. Effectively, in this scenario, the hook stacks almost become irrelevant. It does need further study though. I think Jared is right, we need diagrams.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Jared Magneson on Sat Apr 21, 2018 6:34 pm

Perhaps for now diagrams could be something as simple as screenshots from Atomic Viewer, with arrows or loops in Photoshop to show how they tie together? Easy work, but I don't quite see how hooks work yet so I'd surely get it wrong. I'm following you logically but just don't know what to make of it. Not saying you're wrong by a long shot, just - extremely curious!

For example, I can easily visualize the alpha's proton emissions "turning" or holding those two neutrons in place, even before my animations. Mathis's diagrams were enough. It doesn't have to be complex to be accurate. Smile

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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 8:01 pm

All images will be positioned side-ways so that the normal north/south axis is actually east/west. This is just to fit the screen.

Here we have case 1)



There is a Barium atom on the left which already has a single hook stack. On the right we have a Krypton atom that has no hook stacks at all because it is a noble element.

The single hook stack of Barium will be shared with Krypton to create Uranium. I placed Krypton just a little bit away so that you can see that the hook stack belongs to Barium. In a Uranium atom it would be so close that you couldn't tell which atom initially had the shared hook stack.

Notice how isolated that middle hook stack is. It has no protection from external charge influences or neutrons or whatever decides to move into that area.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 8:04 pm

Here we have case 2)



I had to use 2 Barium atoms because I found a bug in AV that stops some elements from loading. I'll fix that up shortly.

You can see that each Barium atom provides a hook stack, so these will need to merge together to create a nuclear bond. That would create something that looked like a Helium atom in between the main nuclei, but it wouldn't have any neutrons in it, so it is a little different.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 8:06 pm

Here we have case 3)



On the left we have a Xenon atom and on the right Krypton. Between them is a Helium atom which will become the shared hook stack, if we can still call it a hook stack.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 8:42 pm

To look closer at the bonding, I am going to use Oxygen atoms just because they are smaller and easier to work with.

Here we have 2 O atoms in a pre-bond position.



In order to create a nuclear bond, the hook stacks in the middle will need to move up/down a bit to accommodate each other.



Unfortunately, my paint program (or my usage of it) isn't handling transparency very well and it overwrites the other part of the image if I move it over to see the bonded molecule. Just imagine them close so that the middle hook stacks have their red protons above/below each other.

For a molecular bond, they would not be so close, but how close can they get? If you look at the protons in those hook stacks, then the radius of the red sphere is the Quantum Radius that I mentioned above. The gray disc is the Charged Radius, or at least from the center to the edge is the radius. We could align these such that the edge of each disc is roughly near the edge of the red sphere of the other proton. That would be 50% penetration. This could occur even if the stacks do not move up/down and the protons were emitting directly at each other because that charge disc is really the charge emission of the protons which is a photon field. Not a solid structure but a fairly continuous force, none-the-less.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 9:02 pm

Let's have a look at the charge channels of an Oxygen atom.

To do that, it will be easier to turn the atom around a bit so that we can see the charge discs of those hooks.



Oops, forgot to rotate it.

The gray discs represent the charge field of the hook stacks, which in this case are just a single proton each. They represent a field moving from the red sphere to the outside in all directions in the equatorial plane of the proton. You can see that most of that is going to be lost to the ambient field but some of it will be emitted towards the internal stack of the atom (the 6 proton stack).



That charge becomes the through-charge of the nucleus. It has a nice clear path through that 6-stack but once beyond it, it encounters the other hook stack. Most of it is going to miss the protons BPhoton and continue on to the ambient field as a coherent stream. Some of it will be used for emission of that proton and some of that emission will be sent back into the nucleus as through-charge again, but in the opposite direction. This allows an atom to store charge as its internal stream gains density. I also think that this is how an atom heats up, or it is closely related.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 9:07 pm

Looking at the through-charge stream we can see that this creates an internal resonance. This resonance has a frequency set by the size of the nucleus. The larger the nucleus, or the longer the path through it, the slower that frequency becomes. This may help explain why certain atoms emit a certain frequency of light or why they react to certain frequencies in an electric field.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Jared Magneson on Sat Apr 21, 2018 9:11 pm

Something like this?




(I inverted the background so it would be a little easier to see, hope you don't mind. Probably just my plasma.)

Or to keep up with you in near-realtime, is this what you meant by 50% penetration?



Last edited by Jared Magneson on Sat Apr 21, 2018 9:16 pm; edited 1 time in total

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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 9:14 pm

Yep, that's it (a 50% penetration).

Does the white background look better for you? It feels washed out to me. I think the black background makes the colors pop out. What does everyone else think about the background colors?
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Re: Why is Curium the Last Semi-Stable Element?

Post by Jared Magneson on Sat Apr 21, 2018 9:18 pm

Let's just stick with your color conventions, especially on YOUR stuff. It's fine either way to be honest, like I said my gamma and curves on this plasma TV are very hot so on a regular LCD or LED it probably looks fine on black. Smile

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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 9:29 pm

Does anyone have any questions at this point? Anything that could use diagrams or a better explanation? Have I covered what was discussed in my earlier posts? Honestly, if I just keep going we are likely to end up so far off the path that we won't know how to get back!
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Re: Why is Curium the Last Semi-Stable Element?

Post by LongtimeAirman on Sat Apr 21, 2018 9:38 pm

.
The only thing interpenetrating at the 50% point are the emission planes between the protons of the two hook stacks. When a nuclear bond is being formed the two stacks may shuffle or slide.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 10:05 pm

Everything I talked about above assumed a normal environment with just the ambient field and no external electric fields so I thought I'd talk a bit about how an electric field effects the situation.

On the Earth, we already have a weak electric field pointing up from the surface and up into space. This is enough to generally align atoms into the north/south arrangement and is part of the reason that most bonds are formed along that axis (the other being that the carousel level is difficult to work with since it can spin and it is perpendicular to the main alignment).

We generally have twice as much charge coming up from the Earth as we do down from space. This is enough to flip atoms over such that if one end of the atom has a larger hook stack than the other (is a conductor), then the larger end will point downwards so that it has the most charge input from the ambient field.

This means that the hook stacks do not have a truly random distribution of charge input. The input is weighted to that north/south field. This helps the through-charge streams as they can get more charge to increase their density since they are aligned in the same direction.

If we introduce our own electric field(s), then we can increase these effects. We can turn atoms if we want to. We can even turn selected atoms (or types of atoms) and not others by applying just the right strength field. It may even be possible to rotate atoms to any degree we desire, but I don't know if we have that level of precision at this point in time.

The strongest field is going to align everything to its axis. Weaker fields will be able to move the atoms around in order to increase bond potential as long as they are near perpendicular to the main field.

How does an electric field effect the hook stacks?

First and foremost, it will increase their strength. It will cause the hooks to have a denser emission field but it will not effect the core stacks in the same way because the internal charge streams are flowing through them, for the most part. All stacks in an atom will be strengthened but the hooks will get the most benefit and the stack that is closer to the source of the electric field will get the greatest benefit of all. If you are thinking of an atom with a fire under its butt, then you are on the right track Very Happy .
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 10:08 pm

What do you mean by shuffle or slide?

I see a nuclear bond like this as having the red spheres of the hook stacks aligned such that they can have a through-charge of their own and truly be considered a stack and not 2 separate ones. This will involve the 2 stacks sliding over a bit (up and down in the images of O above) so they they can both fit together.
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Re: Why is Curium the Last Semi-Stable Element?

Post by LongtimeAirman on Sat Apr 21, 2018 10:21 pm

.
I wrote. “The only thing interpenetrating at the 50% point are the emission planes between the protons of the two hook stacks. When a nuclear bond is being formed the two stacks may shuffle or slide”.

Nevyn wrote. What do you mean by shuffle or slide?

Slide means the two hook stacks slide in two different directions despite the two penetrating emission fields. I was trying to accomodate penetrating hook stack emission fields. Nevermind, the more I think of it the less I like the idea of overlapping hook stacks – there’s too big a chance of blowing protons out of the stack when hook stack emissions penetrate each other so closely. I don’t believe a stable molecular bond can exist as 50% overlapping hook stacks between two atoms since those two penetrating stacks would not share charge, their emissions will add in the main charge channel directions, somewhat but mostly bypassing each other. I like the idea that a molecular bond is a  less than straight hook stack charge alignment, and a nuclear bond has a straight stack charge channel.

Nevyn wrote. My question is: Is the hook stack, when looking at it as an input, the only source of charge for the charge channel behind it? I imagine that some stray charge will skip by the stack and make it into the charge channel all on its own, but the hook stack provides a more coherent stream. It is acting like a little laser aimed into the nucleus.

I see the hook stacks as a firewall for the nuclear charge channel. Yes they are a primary source for charge  - not the only source, ambient photons still enter the atom from many directions. I agree that the Hook stacks will largely determine the charge capacity of the main charge channel as well as resonant characteristics.
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Re: Why is Curium the Last Semi-Stable Element?

Post by Nevyn on Sat Apr 21, 2018 10:30 pm

Ok, I think we are on the same track with respect to stacks moving around and penetrating each other. I realised yesterday that I had been ignoring the possibility of the stacks directly aligning (and not moving over) so I thought I'd mention it and see how others felt.

A firewall is a pretty good analogy, actually. Most traffic has to go through the hook stacks to enter the channel. I like it.
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Re: Why is Curium the Last Semi-Stable Element?

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