The Poison Comes From?

On the way to triple bonds,

\(C\)~\(N\) forms up to a \(Ne\) nucleus with a single unpaired bond, like a Sodium \(Na\) but without the \(g^{+},T^{+}\) inner orbits.  This bond pair with another \(C\)~\(N\) to form \((CN)_2\) a cyanogen.  The single unpaired bond that pairs up two \(CN\)s however, comes from \(N\) not \(C\).

Have a nice day.

Double Bonds, Triple Bonds

Double bonds?  If bonds form from unpaired orbits then parallel double bonds are impossible.  In the case of \(C\), Carbon we first form a stable \(C_2\),

which is just like the \(Ne\), Neon element.  From this we can obtain a ethene, \(C_2H_4\) by unpairing two paired to obtain four unpaired orbits.  To each of these unpaired orbits, a hydrogen atom is bonded to it.  We can image that the horizontal bond that constitute part of the double bond being stretched when the two \(C\) nuclei cannot be concentric, and the vertical bond not effecting the spacial arrangement of the two bonded Hydrogen atoms on each side.  As far as a single \(C\) nucleus is concerned three bonds (a double bond and two bonds with hydrogen) around it distribute themselves coplanar separated by \(120^o\).

This suggests a stronger double bond than a pair of single bonds filling the adjacent vertices of a tetrahedron being twisted into a parallel double bonds.

But why stop at double,

This also suggests a much stronger and stable triple bonds than three twisted single bonds from the corners of a tetrahedron.  A single paired orbit is unpaired and one hydrogen atom bonds with each unpaired orbit.  The resulting structure is linear.

The new interpretation of Quantum Numbers gives flexibility to bond formation as the number of orbits around a nucleus can be reduced by reducing \(n\).  The remaining orbits are unpaired to provide for the number of radicals around the molecules.

For once not twisted!  And among those paired, double and triple bonds, democracy! All equal.  No need for Hybridization.

Hydrogen Bond A Retake

Hydrogen bond,

since Hydrogen is half the size of oxygen covalent radius, the outer orbit (to the left of the dotted line) is relatively more positive (1.5 times).  This part of the orbit is attracted to any region of less potential and forms a weak bond.

Good night.

Bent And Split To Make Water and Ice

If this is how to bent and split to make water,

how to make ice?  Maybe the paired orbits split and cross,

Crossing the split paired orbit, allows all protons in the configuration on the same side of the paired orbit around the Oxygen nucleus, to synchronize.  This makes all the orbit equivalent.  The presence of two anti-phase protons from the two Hydrogen atom bends all the orbits of Oxygen at the center.

There is another way to form water, where \(n=2\) is a valid solution to the quantum system then,

A single paired orbit can enclose a \(He\)~\(C\) nucleus with a \(n=2\) proton shell.  The outer single paired orbit unpairs into two and bond each with a hydrogen atom.  The \(n=2\) shell and the two bonded orbits extend themselves into the vertices of a tetrahedron.

The hydrogen atom being half as small, occupies only one end of the bonded orbit.  This creates an imbalance of charge between the center of the water  molecule and the two outer bonded orbits.

Hydrogen bonding is the attraction of the relatively positive bonded orbits to the center of the water molecule.  Orbits 1, 2, 3 and 4 as shown above, rearrange themselves into the four vertices of a tetrahedron.  What about the \(n=2\) paired orbits of the Oxygen?  These orbits are closer to the Oxygen nucleus and does not interfere with the arrangement of the two water molecules bonded by hydrogen bonding.  All Oxygen orbits revolves over the nucleus and effect each other.

The last scenario is without the complication of bending orbits, but is admissible only if \(n=2\) shell with its new interpretation forms up first and later interacts with the two Hydrogen bonded orbits of the next layer.

Rotating Two Lines

The concept of a force field over space has been to place a unit test particle in location \(s_{x,y,z}\), the force experienced by such a test particle is the strength of the field at \(s_{x,y,z}\) and the direction of the force is the direction of the field line through \(s_{x,y,z}\).

Do such field lines interact?  Image two singular field lines rotated into alignment,

and then rotated out of alignment.  There is no work done.  However when in alignment the potential energy of the system of two field lines is maximum.

This increase in potential energy presents a energy barrier to field rotation.  When the field lines are aligned.

\(U_{max}=2V_r\)

where \(r\) is the distance from the source of the field line to the meet point, \(s_{x,y,z}\) and, \(V_r\) the potential at \(s_{x,y,z}\) due to one source.  When the field lines are not aligned,

\(U=0\)

since these are singular field lines and do not interact except at \(s_{x,y,z}\).

The energy barrier to field rotation is then \(U_{max}=2V_r\).

Homochirality

Three orthogonal directions in space determines a unique direction, and a positive direction is different from the negative direction.  Given that all three type of particles, \(c\), \(g\) and \(T\) are in orbits and are involved in chemical bonds, correspondingly, \(e^{-},p^{+}\), \(g^{-},g^{+}\) and \(T^{-},T^{+}\), then strictly, there can be one approach between two reactive units such that energy required to work against both static fields (three types) and weak fields (also three types) is minimum.

For example, the approach from the top and from the bottom are different,

as the three orthogonal weak fields present a unique direction for which the approach requires minimum energy, the opposite direction need not be of the same energy.

When the disparity in energy required along different (opposing, if one of the reagent is planar) directions is large, the result is homochirality.  Lactic acid is chiral.

What Is The Origin Of The Energy Barrier To Bond Rotation?

Weak fields \(g\) and \(B\), perpendicular to the plane of the parallel protons, \(p^{+}\) orbits.

The Start Of Something Sweet

This is Methane \(CH_4\), where all of carbon proton orbits are unpaired, pulled to off center, where the nucleus is closer to the perimeter of the orbit.  One Hydrogen atom then share an electron with each Carbon orbital proton and the four shared orbits extend into the vertices of a tetrahedron so that they are the furthest apart.

Protons on the Carbon orbits are at \(90^o\) phase to another, so that only one of the four protons in turns is closest to the nucleus at the center.  The proton in the hydrogen orbit is always at \(180^o\) phase with the Carbon proton it is sharing an electron with.

There is however, one problem with this model.  In this particular case, when the top Carbon orbital proton is at the nucleus, the one other orbit at which it is at \(180^o\) phase is at the furthest end of its orbit from the center (middle bottom loop of the diagram above).  The Hydrogen proton sharing electron with this Carbon orbit is also near the center.  Being \(180^o\) out of phase with the Carbon proton it shares an electron with, brings it in phase with the proton now occupying the center.  Two protons at the center!

The Hydrogen covalent radius (\(31\)pm) is however at half of the Carbon covalent radius (\(66\)pm).  The two protons in phase, one from a Carbon orbit at the nucleus, and the other of a Hydrogen orbit will however, not come closer than the radius of the Carbon orbit.

Thanks for coming, but don't waste your time.  Not again.

Water And Bending The New Order

Orbits mark the paths of orbiting electrons.  It is possible to bend the orbits of charge particles with other charge particles as they interact as particles.

This is what happens what Hydrogen with half the covalent radius \(r_{co}=31\)pm meets Oxygen \(r_{co}=66\)pm.

One of the paired orbits of Oxygen splits at one side.  Two Hydrogen atoms, pair up at each of the split ends and this creates a total of seven paired loops around the nucleus.  The proton-electron pair on each of the hydrogen orbits are in phase and move with constant separation between them.  The proton-electron pair of the Oxygen split pair orbit is in anti-phase, each are at the opposite ends of the orbit.  An electron is always shared by two orbit between an weak field and a proton.  The Hydrogen orbital particles are oscillating at the same frequency as the Oxygen orbital particles.  Half the time, the Hydrogen proton shares an electron with the weak field of one of the Oxygen protons in the split pair orbit.  The other half of the time, the Hydrogen proton shares the electron with the weak field of the other Hydrogen.

The seven pair loops/orbits around the Oxygen nucleus, technically now water \(H_2O\), are spaced equally,

\(\angle Orbit=\cfrac{360}{7}\)

\(\angle Orbit=51.42^o\)

How could this happen?  The diagrams above shows that the orbit is split at \(120^o\).  This happens when a paired orbit of the Oxygen atom first splits into two unpaired orbits.

For clarity, the horizontal paired orbit has been omitted.  The top of the paired orbit between the unpaired orbits splits to share an electron with the unpaired orbits on each side, as two hydrogen atoms share their electron with each of the unpaired orbit.  The angle between the two hydrogen atom is about,

\(\angle H=2*\angle Orbit=102.84^o\)

This compares well with the measured value of \(104.5^o\).

Does the split paired orbit crosses?  Yes (post "Bent And Split To Make Water and Ice" dated 05 Apr 2016).  The hydrogen are at anti-phase such that only one of them is near the center of the Oxygen atom at any one time.  The hydrogen proton nearest the center pushes the Oxygen proton of the other split orbit, bending its trajectory, so that it pairs up its unpaired orbit.  All four orbits spins synchronously, such that all protons of a orbit pair is furthest apart from each other.  The principle discussed previously without an initial orbit unpair-ing still holds.

Have a nice day.

韬光养晦小天地
矮菇虫舍兢相偎
鹏翔静寂撕天裂
鲲迴幽深百川泻
静心镜怀摄众像
温杯浓茶更动情

《打坐》

More Of Six, Heat And Nobel

Another example of \(n=6\) shell, with six orbit pairs equally spaced over a sphere is Oxygen gas, \(O_2\).  The two \(O\) nuclei with three equally spaced orbit pairs, will get as close as possible to complete an \(n=6\) configuration.

The nuclei may not be concentric as the last unpaired orbit aligns.  And since \(n=6\) has higher energy than \(n=3\), energy is required to form \(O_2\) from two \(O\)s.  Oxidation, where \(O_2\) breaks into two \(O\)s occurs readily as the process gives off energy.  Chlorination on the other hand, where \(Cl_2\) breaks into two \(Cl\)s, involves the separation of a orbit pair, will require energy to proceed.

The process of separating an odd number outer shell into two equal lower shells each with an unpaired orbit is endothermic.  The process of separating an even number outer shell into two equal lower shells both with all paired orbits is exothermic.

And that settles that!  Hello, Nobel Prize.

More Of Seven

Another example of \(n=7\) shell, with seven orbits equally spaced over a sphere is Chlorine gas, \(Cl_2\).  The two \(Cl\) nuclei will get as close as possible to complete the last orbital pair.



They may not be concentric as the last unpaired orbit aligns.

In summary, a new periodic table given a slightly different interpretation of Quantum Numbers.

Have a nice day.

Encompassing New Order Is A Mess

Where Nitrogen \(N\) is like a Hydrogen \(H\) and Oxygen \(O\) like a Helium \(He\).



What's make a \(H_2\) and what makes a \(He\)?

Fluorine \(F\) is like a Lithium \(Li\) and Neon \(Ne\) like a Helium \(He\)



And we have \(K\) and \(Ca\),



Once the core of the nucleus becomes bigger, \(n=5\) is a admissible solution and we have the transition metals starting with Scandium.  The are 10 electrons in \(n=5\) to accommodate 10 transitional elements across the series.

\(n=6\) has twelve elements? Two \(n=6\) makes 24 elements for the Lanthanides and transitional metals

\(n=7\) has fourteen elements and \(n=5\) has ten elements makes 24 for the Actinides and transitional elements.  Although it is still possible that across the same series two \(n=6\) shells are being filled instead.

Although we can accommodate transitional metals and both Lanthanides and Actinides series.  The reasoning that orthogonal solutions need not be all included, we have many more possibilities for new elements not included in the periodic table.  For example, since \(n=3\) is a solution, we can have a single electron around an Oxygen \(O\) core as shown below, in stead of a third electron in a \(n=4\) shell as shown above.

It is still the element Fluorine, \(F\) but behave like Lithium and is more ready to donate the single electron and form \(F^{+}\) ion.  It is still Fluorine?  Fluorine with multiple personalities.

Hey, Hey, Hey, an explanation for varied oxidation states!  These variations in atomic structures but with the same atomic and mass number shall be called diffotoms; for different at the bottom.

Good night, and thanks for isotopes.

Orthogonal Solutions, Possible Solutions

We may not be able to solve for the Schrödinger wave equation for the hydrogen atom but we can make a guess at the number of possible solutions and the possible sign (positive/negative values) of the solutions.  The principal quantum number, \(n\) is labelled consecutively, \(n=4\) suggests that there are also valid solutions, \(n=1\), \(n=2\) and \(n=3\).

This solutions for the quantum system of a hydrogen atom extrapolated are orthogonal, they can exist at the same time, and applies to the system simultaneously.  This allows us to find the maximum possible number of electrons given \(n\).

For solution of    Maximum number of possible orbital electrons

\(n=1\)                                 \(2\)
\(n=2\)                                 \(2+4=6\)
\(n=3\)                                 \(2+4+6=12\)
\(n=4\)                                 \(2+4+6+8=20\)
\(n\)                                      \(n*(n+1)\)

So the maximum number of electrons in orbit given \(n\) is \(n(n+1)\), which is obtained by adding solutions from all lower numbered solutions.

And when an \(n=4\) shell is being filled up.

Other possible solutions to the quantum system (hydrogen atom) show up, and seem to duplicate solutions to lower valued \(n\).

These solutions are not orthogonal, they do not exist at the same time and so does not apply to the quantum system simultaneously.

A energy shell cannot at the same time contain both two electrons and three electrons.  But it can contain either two electrons OR three electrons, since both,

(\(n=4, l=0, m=0, s=\pm1/2\)) and

(\(n=4, l=0, m=0, s=\pm1/2\))

(\(n=4, l=1 m=-1\,\, or\,\, 1, s=\pm1/2\))

are valid solutions.

The implication of this interpretation is not that there are no sub-shells, but that they are filled up in time and do not exist simultaneously.  The sub-shell evolves as electrons are added,

\(4s\rightarrow4p\rightarrow4d\rightarrow4f\)

When \(n=4\) presents a \(4d\) sub-shell configuration/solution, \(4s\) do not exist!

 Just as \(m=-1\) or \(m=1\) but not both for the same electron!  One electron can only travel in one direction around its orbit.

Just when you think mistakes on \(_2He\), \(_3Li\) and \(_4Be\) was funny.

Loops Not...

This interpretation means that electron orbits are circular and electron clouds should only be spherical.

The orbital loops observed experimentally of \(p\), \(d\) and \(f\) orbits are due to the weak \(E\) fields generated by spinning \(T^{+}\) particles.  These fields have a positive and negative end (two opposing loops to one field) and the number of loop pairs increases with the number of unpaired \(T^{+}\) particles in the nucleus.

The loops appears when the \(T^{+}\) are in unpaired orbits.  In such cases, under experimental conditions to measure electric fields, their orbits space out such that the generated \(E\) fields have the least interaction and do not overlap.

Those distorted elongated spheres are not electron clouds.

Quantum Numbers And Sun Flowers

Quantum Number assignment,

where \(m\), the projection of the angular momentum is perpendicular to \(l=0\) azimuth and so, \(m=0\) always, for \(l=0\).  Spin number \(s\) is associated with individual particle in the nucleus; for electrons, they spin around the positive particles.  \(n\) is assigned to all particles of the same orbital radius.

As such we have three sets of orthogonal Quantum Number,

\(n_g\), \(l_g\), \(m_g\) and \(s_g\) for gravity particles

\(n_T\), \(l_T\), \(m_T\) and \(s_T\) for temperature particles

\(n_c\), \(l_c\), \(m_c\) and \(s_c\) for charge particles

In this interpretation, \(l\) is assigned to equally space orbits around the nucleus.  For \(n_{or}\) number of orbits,

\(0\le l\le n_{or}-1\).

\(m\) is the projection of \(l\) perpendicular to \(l=0\), arbitrarily.

Why then are petals of number 2, 8, 18, 32 preferred on the sun flowers of the main periodic table?

UFO Pizza

My very own UFO pizza.

The exhaust is the compressed air stream.  Enjoy!

Building Them Up...

We have seen previously (post "Alpha Decay" dated 29 Mar 2016), that the weak fields generated by spinning positive particles can interact and cancels each other.  Each of such weak field generated by a particle holds onto another positive particle which in turn generates a weak field.  The nucleus is build up in layers of positive particles.  The nucleus is complete when it acquires negative particles that neutralizes the positive particles.  Does the weak field (positive end) acquires the negative particle or does the spinning positive particle acquires the negative particle?

In the case of Hydrogen molecules, \(H_2\),

the electron is shared by a weak field and a positive particle across two hydrogen atoms.  The two electrons are furthest apart from each other, across the orbit of the positive particle.  All the orbital planes of corresponding positive particles of the two nuclei are in parallel, which allows the two other types of negative particle to be shared in a similar way.

The negative particles are held in place by both a weak field and a positive particle.  The negative particle spins in an orbit around the positive particle slightly displaced by the weak field.

The situation is much more complex as we double the number of positive particles to twelve and set the paired orbits as far apart as possible, orthogonal to each other,

In this way, the positive particle as furthest apart from each other. And as we double the nucleus again, we have the SunFlower atom (that don't exist),

These are the cases where the orbits intersect as more hydrogen nuclei are added.  It is possible that the next tuple (\(g^{+}\), \(T^{+}\), \(p^{+}\)) are added beyond the first \(p^{+}\),

in which case all nuclei from such a series have a Hydrogen-2 nucleus core.  And Neon, \(Ne\) is

and a Carbon, \(C\) nucleus is,

and Oxygen, \(O\),

and Lithium with an unpaired electron in orbit \(Li\) is,

but it is more likely that \(Li\) has equal orbits.

And Beryllium \(Be\),

It's symmetrical shape and paired orbital electrons might account for its relative stability.

And Nitrogen, \(N\) with an unpaired electron in orbit is,

And the list goes on...

Have a Sun Flower day.

Even More Radioactivity

And so,

(...\(g^+\), \(2T^{+}\), \(p^+\)...)\(\rightarrow\)(...\(g^+\), \(T^{+}\), \(p^+\)...)\(+T^{+}\)

(...\(T^+\), \(2p^{+}\), \(g^+\)...)\(\rightarrow\)(...\(T^+\), \(p^{+}\), \(g^+\)...)\(+p^{+}\)

(...\(p^{+}\), \(2g^+\), \(T^{+}\)...)\(\rightarrow\)(...\(p^+\), \(g^{+}\), \(T^+\)...)\(+g^{+}\)

where the merged particles separate into two (posts "Two Quantum Wells, Quantum Tunneling, " dated 19 Jul 2015) and one of the two leaves the nuclei.

Good night.

More Radioactive Decays, You Will Pew...

The existence of (\(T^{+}\),\(T^{-}\)) pair within the nucleus sequence,

(...\(p^+\), \(g^+\), [\(T^{+}\), \(T^{-}\)], \(p^+\), \(g^+\), \(T^{+}\)...)

and

(...\(p^+\), \(g^+\), \(T^{+}\), \(T^{-}\), \(g^+\), \(T^+\), \(p^{+}\)...)

also opens up new possibilities when \(T^{-}\) escapes,

(...\(p^+\), \(g^+\), [\(T^{+}\), \(T^{-}\)], \(p^+\), \(g^+\), \(T^{+}\)...)\(\rightarrow\)(...\(p^+\), \(g^+\), \(T^{+}\), \(p^+\), \(g^+\), \(T^{+}\)...)\(+T^{-}\)

and,

(...\(p^+\), \(g^+\), \(T^{+}\), \(T^{-}\), \(g^+\), \(T^+\), \(p^{+}\)...)\(\rightarrow\)(...\(p^+\), \(g^+\), \(2T^{+}\), \(p^+\)...)\(+T^{-}+g^{+}\)

And in an analogous way, the existence of a (\(g^{+}\),\(g^{-}\)) pair within the nucleus sequence,

(...\(p^+\), [\(g^{+}\), \(g^{-}\)], \(T^{+}\), \(p^+\), \(g^+\), ...)

and

(...\(p^+\), \(g^{+}\), \(g^{-}\), \(p^{+}\), \(g^+\), \(T^+\),...)

suggests,

(...\(p^+\), [\(g^{+}\), \(g^{-}\)], \(T^{+}\), \(p^+\), \(g^+\), ...)\(\rightarrow\)(...\(p^+\), \(g^{+}\), \(T^{+}\), \(p^+\), \(g^+\), ...)\(+g^{-}\)

and

(...\(p^+\), \(g^{+}\), \(g^{-}\), \(p^{+}\), \(g^+\), \(T^+\),...)\(\rightarrow\)(...\(p^+\), \(2g^{+}\), \(T^+\),...)\(+g^{-}+p^{+}\)

In all the above four cases, when the remaining nuclei collapse, there is a release of energy.  Have a nice day.

Second Take On \(\beta^{-}\) Decays Again

The existence of electron inside the nucleus allows for an alternate view of \(\beta^{-}\) decays.

(...\(T^+\), [\(p^{+}\), \(e^{-}\)],  \(g^{+}\), \(T^{+}\), \(p^+\)...)

where \(p^{+}\) in the pair [\(p^{+}\), \(e^{-}\)] holds the next particle, \(g^{+}\).  And

(...\(T^{+}\), \(p^{+}\), \(e^{-}\), \(T^{+}\), \(p^+\), \(g^{+}\), \(T^{+}\)...)

where \(e^{-}\) in orbit around the \(p^{+}\) generates a weak \( B\) field holds the next particle, \(T^{+}\).

When \(e^{-}\) is release from it orbit in the first sequence,

(...\(T^+\), [\(p^{+}\), \(e^{-}\)],  \(g^{+}\), \(T^{+}\), \(p^+\)...)\(\rightarrow(...T^+, p^{+}, g^{+}, T^{+}, p^+...)+e^{-}\)

the electron is emitted together with a packet of energy as the positive particles from higher orbits in the nucleus collapse inward.  [\(p^{+}\), \(e^{-}\)] might be seem as a neutron that decayed into a proton and a \(\beta^{-}\) particle with an release of energy.

When \(e^{-}\) is release from it orbit in the second sequence,

(...\(g^{+}\), \(T^{+}\), \(p^{+}\), \(e^{-}\), \(T^{+}\), \(p^+\), \(g^{+}\), \(T^{+}\)...)\(\rightarrow(...T^+, 2p^{+}, g^{+}, T^{+}...)+e^{-}+T^{+}\)

Energy is also released as positive particles from higher orbits in the nucleus collapse inward. Two particles, \(e^{-}\) and, \(T^{+}\) from the next higher layer is also released.  The resulting nucleus is unstable with a \(2p^{+}\) particle,

(...\(T^+\), \(2p^{+}\), \(g^{+}\), \(T^{+}\)...)

\(2p^{+}\) may separate into two \(p^{+}\) particles (posts "Two Quantum Wells, Quantum Tunneling, " dated 19 Jul 2015), one of which is ejected from the nucleus and so results in \(\beta\)-delayed proton emission decay.

Electron Capture And Energy Accounting

Consider the nucleus sequence derived from considering weak field interactions,

(\(g^{+}\), \(T^{+}\), \(p^{+}\), \(g^{+}\), \(T^{+}\), \(p^+\))

when a \(p^{+}\) particle captures an electron \(e{^{-}}\), the weak field from the spinning \(p^{+}\) particle holds on to \(g^{+}\) of the next higher layer still.

(\(T^+\), [\(p^{+}\), \(e^{-}\)],  \(g^{+}\), \(T^{+}\), \(p^+\))

as  \(g^{+}\) is moved to a further orbit, energy is absorbed.  This energy absorbed may be previously been accounted as an electron neutrino.

If \(e^{-}\) in spin around \(p^{+}\) is able to establish a \(B\) weak field that holds the \(T^{+}\) particle from higher layers then it is possible,

(\(g^{+}\), \(T^{+}\), \(p^{+}\), \(e^{-}\), \(T^{+}\), \(p^+\))\(+g^{+}\)

that the \(g^{+}\) particle from the next higher layer is ejected.

It is also possible that the capture of an electron happens via the weak \(E\) field established by a spinning \(T^{+}\) particle as the weak field is a directed field with a positive and negative end,

(\(g^{+}\), [\(T^{+}\),\(e^{-}\)], \(p^{+}\), \(g^{+}\), \(T^{+}\), \(p^+\))

some energy is absorbed as the next particle is pushed to higher orbit radius.

It is possible that, subsequent to the electron capture,

(\(g^{+}\), \(T^{+}\), \(e^{-}\), \(T^{+}\), \(p^+\))\(+p^{+}+g^{+}\),

both \(p^{+}\) and \(g^{+}\) are ejected and that \(e^{-}\) in orbit establishes a \(B\) field that holds onto \(T^{+}\).  A proton and a neutron is ejected with the release of energy, as \(T^{+}\) collapses inwards.

There are more possibilities here than what was observed of electron capture in the laboratory.  In these cases, the electron neutrino and electron anti-neutrino are energies involved in moving the positive particles in the nucleus from lower orbits to higher orbits (energy absorbed) and from higher orbits to lower orbits (energy release), respectively.  This explanation does not require the awkwardness of a particle to carry away negative energy.

Heart Of The Matter

Then we come to the very first issue that was side-stepped.  Why is the very first particle spinning?  The first particle is spinning because there is an opposite particle spinning around it.  It is also possible that, it is the weak field generated by the spinning negative particle that attracts the next particle.

[\(T^{+}\), \(T^{-}\)],  \(T^{-}\) spins around \(T^{+}\).  \(T^{+}\) spins about a smaller radius and acts as the first particle in the nucleus set,

([\(T^{+}\), \(T^{-}\)], \(p^+\), \(g^+\)).

It may also be possible that \(T^{-}\) spinning establishes a weak \(g\) field that attracts a \(g^{+}\) in which case we have,

(\(T^{+}\), \(T^{-}\), \(g^+\), \(T^+\), \(p^{+}\))

[\(p^{+}\), \(e^{-}\)], \(e^{-}\) spins around \(p^{+}\).  \(p^{+}\) spins about a smaller radius and acts as the first particle in the nucleus set,

([\(p^{+}\), \(e^{-}\)], \(g^+\), \(T^+\)).

It may be possible that \(e^{-}\) spinning establishes a weak \(B\) field that attracts a \(T^{+}\) in which case we have,

(\(p^{+}\), \(e^{-}\), \(T^+\), \(p^{+}\), \(g^+\))

and

[\(g^{+}\), \(g^{-}\)] spins around \(g^{+}\).  \(g^{+}\) spins about a smaller radius and acts as the first particle in the nucleus set,

([\(g^{+}\), \(g^{-}\)], \(T^+\), \(p^{+}\)).

It may be possible that \(g^{-}\) spinning establishes a weak \(E\) field that attracts a \(p^{+}\) in which case we have,

(\(g^{+}\), \(g^{-}\), \(p^{+}\), \(g^+\), \(T^+\))

Such variations may account for nuclei phenomenon at the center of the nucleus;   Issues of unexplained attraction and love.

Please note that the interaction within [\(p^{+}\), \(e^{-}\)], is not weak field interaction.  As with any inter-layer negative particle,

(\(T^+\), [\(p^{+}\), \(e^{-}\)],  \(g^{+}\))

the particle \(e^{-}\) puts more distance between \(p^{+}\) and \(g^{+}\).  Or

(\(T^+\), \(p^{+}\), \(e^{-}\),  \(T^{+}\))

where \(e^{+}\) via its weak field captures a \(T^{+}\).

And suddenly, life is unpredictable with so much love.

Prevailing science would insist on weak interactions within the nucleus such that the charge of the nucleus is the atomic number and is also the number of protons in the nucleus.  The presence of an electron, ie. non weak interaction, reduces the net charge of the nucleus.  The atomic number then is one less and does not reflect the number of protons in the nucleus.

Pulling From The Outside

The weak fields holding the Helium-4, \(^4He\) nucleus in place in a larger nucleus are gravitational.  This means that any acceleration, and high gravity acting through the center of the larger nucleus will aid the ejection of the Helium-4 nucleus.

A centrifuge at high spin can speed alpha decay.

Alpha Decay

This is the Helium-4 \(^4He\) nucleus,

(\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\), \(p^+\))

For large nuclei that are built up from the cyclic permutation set of positive particles based on weak force interactions, it is possible that,

(...\(p^+\), (\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\), \(p^+\)), \(g^+\), \(T^{+}\)...)

a Helium-4 nucleus is embedded within the large nucleus.  When this group of particles are removed,

(...\(p^+\), \(g^+\), \(T^{+}\)...)

the large nucleus collapses to a well ordered nucleus set still.

Why would the series of particles be ejected?  This embedded nucleus is held in place by the weak fields produced by two spinning \(p^{+}\) particles, one at each ends.  If these weak fields are disrupted it is then possible to eject the embedded nucleus.

The simplest scenario by which this can happen without the involvement of outside particle/photon is for the weak fields to cancel each other at least partially.  This can happen as both particles' orbits are also spinning about their diameters.

When their orbital planes are parallel and their spins are in opposite directions, the weak fields they generate cancel partially.  The hold on the embedded helium-4 nucleus weakens on both ends.  Since, the inner particle generate a stronger field, the field at the outer particle (\(p^{+}\) on \(g^{+}\) outside of the \(^4He\) nucleus) can be reduced to zero first.  The remaining force at the inner end (\(p^{+}\) outside of the \(^4He\) nucleus, on \(g^{+}\)) pulls the \(^4He\) nucleus inwards towards the center of the large nucleus.  The \(^4He\) nucleus escape through the center of the larger nucleus.

The above happens over a range of distances from the center of the nucleus.  All nuclei with radii with this range of distances will alpha decay, ie radii of this range and bigger.

This is one way, purely speculative, how alpha decay can happen.  Have a nice day.

Note:  The mechanism of weak fields alignment and cancellation might also explain spontaneous fission where nucleus disintegrates into two or more smaller nuclei and other particles, and cluster decay where nucleus emits a specific type of smaller nucleus that is larger than an alpha particle.

Other Possible Colliding Particles

It is possible that the photon/particles suffer a time axis swap after the collision, in which case to obtain the correct colliding photon, the time axes on the present photons are swapped.

From the case of \(\beta^{+}\) decay, the colliding photon is,

where \(t_g\leftrightarrow t_c\) from a \(P_{g^{+}}\) particle.  This is a \(P_{e^{-}}\), an electro-magnetic wave.  The radioactive decay after the collision is,

where a time axis swap results in the expected \(g^{+}\) particle.

And in the case of \(\beta^{-}\) decay, the colliding photon is,

where \(t_T\leftrightarrow t_c\) from a \(P_{p^{+}}\) particle.  This is a \(P_{T^{-}}\) photon or a magnetic-gravito wave.  The radioactive decay is then,

where a time axis swap produces the \(p^{+}\) particle as expected.

Which are the actual photon pair involved in \(\beta\) decay?

Mechanism For Radioactive Decay

It is really arbitrary how the mechanisms for \(\beta\) decays are cooked up using particle collisions.

Mathematically, the resulting wave after the time axes swap is equivalent to the wave before the swap.  The time axes variables are arbitrary, but we know that, either,

\(t_c=i.t_T\)  ot  \(t_T=i.t_c\)

that the time axes are orthogonal.

How much energy is expended when we multiply a time axis \(t\) by \(i\)?  For the swap process as a whole, nothing, since mathematically the waves are equivalent, no net energy input is required.

Which brings us to the other colliding particle,

On collision, the photon \(P_{g^{+}}\) stops along the space dimension \(x\) but has speed \(v=c\) along \(t_c\).  It then splits into two along the \(t_g\) axis.  Part of it \(g^{+}\) travels along the positive \(t_g\), \(v=c\) and the other part travels along \(t_g\) negative; ie. \(v=-c\).  The momentum of the photon is either split equally between the particle and the antiparticle or, only the anti-particle has \(v=c\) in space when \(g^{+}\) is captured (\(p^{+}\) is originally in orbit).

Why should one particle be sent back in time?  This particle is otherwise an electron, the negative particle of \(p^{+}\).

One particle is sent back in time because \(P_{g^{+}}\) is stationary along \(t_c\) before the collision and \(p^{+}\) is at light speed along \(t_c\).  After the collision, part of \(P_{g^{+}}\) is propelled forward, the other part is repelled backwards along \(t_c\).  Then the time axes swapped, \(t_g\leftrightarrow t_c\).  \(t_g\) now has two velocity vectors positive and negative that sum to zero, and \(t_c\) has a vector component \(v=c\).

On the \(p^{+}\) particle \(t_c\leftrightarrow t_T\) and on the photon, \(P_{g^{+}}\), \(t_c\leftrightarrow t_g\).

In the case of \(\beta^{-}\) decay, there is only one time axis swap,

and the photon stopped along the space dimension after the collision,

In this case both colliding particles have speed \(v=c\) along \(t_c\) when they collided.  The photon \(P_{g^{-}}\) has \(v=0\) on \(t_T\), in the particle \(g^{+}\), \(t_T\) is the oscillatory component of the wave.  In the previous case,  the photon \(P_{g^{+}}\) is stationary along \(t_c\), \(v=0\) and the particle \(p^{+}\) is at light speed, \(v=c\) along \(t_c\).  Momentum along \(t_c\) is split into two, a negative part and a positive part, they sum to zero.

In both cases, the oscillatory components remain intact.  After the collision, except for the photon/particle that slowed, all non oscillatory time axes swapped.  If one of the colliding, non oscillatory time axes has zero velocity, momentum along that time axes splits into a negative and positive part and an anti-particle is created.

In summary...

In both cases, the particles involved in the decays in the nucleus is first identified.  And appropriate photon that provides for the resultant particles after the decay is made to collide with the nucleus particle.  The nucleus particle is transmuted by swapping non oscillatory time axes.  If between the photon and the nucleus particle, any of the non-oscillatory time axes has zero velocity, a anti-particle is produced.  This anti-particle is identified after the time axis swap.

Is this a scheme for general radioactive decay?  Maybe.

Magnetic Monopoles

In the presented schemes for \(\beta\) decays are correct, \(T^{+}\) emitted from the nucleus during \(\beta^{-}\) decay is the electron anti-neutrino and \(T^{-}\) changed from \(p^{+}\) and emitted during \(\beta^{+}\) decay is the electron neutrino.

If \(T^{+}\) and \(T^{-}\) are particles that produces \(B\) fields, they are then the magnetic monopoles.

But under normal circumstances these particles behave as waves not as particles.  This is the reason why they are not noticeable as opposing particles, like opposing charges.

Furthermore, \(g^{+}\) particles are neutrons.  But what are then, \(g^{-}\) particles?

If all nuclei have \(g^{+}\) particles, then Earth should be a positive gravity particle not a negative particle.
We may have a problem.

Correcta. Correcta.

\(\beta^{+}\) Decay

If this is \(\beta^{+}\) decay,

\(p^+\)+\(P_{g^{+}}\)\(\rightarrow\)\(T^{-}+e^{+}+g^{+}\)

where the collision of photon, \(P_{g^{+}}\) with \(p^{+}\) reverses \(t_c\) and \(t_T\) on \(p^{+}\) changing it into a \(T^{-}\) particle that is detected as the electron neutrino.  The photon is completely slowed in space to give an \(g^{+}\) particle and, to provide more energy to the collision, a \(e^{+}\) particle is produced also.  On the \(e^{+}\) particle, velocity along \(t_g\) is completely reversed.



The problem with \(\beta^{+}\) decay is that it normally occurs with the emission of two \(g^{+}\) particles also.  For example,

(\(T^+\), \(p^+\), \(g^+\), \(T^+\))\(\rightarrow\)(\(2T^{+})+2g^{+}+e^{+}+T^{-}\)

Only when the \(p^{+}\) particle involved are at the innermost end of the nucleus set, eg.

(\(p^+\), \(g^+\), \(T^+\))\(\rightarrow\)(\(2g^{+},\,\,T^{+})+e^{+}+T^{-}\)

does \(\beta^{+}\) decay not emit the two \(g^{+}\) particle and it would seem that the \(p^{+}\) particle has converted into a \(g^{+}\) particle.

What?  Radioactive decays do not involved photons?  Maybe.

Photons Destroy Everything

The scheme for \(\beta^{-}\) decays suggests that all \(g^{+}\) particles in the nucleus are susceptible to radioactive decay when bombarded with photons of high enough energy.  And that all nuclei with \(g^{+}\) particles can decay.  The nucleus collapses with the release of a electron or positron.

The resulting nucleus set up is unstable.

Photons destroy everything except nuclei without \(g^{+}\) particles, ie. hydrogen.  So, the only effective sun block is hydrogen.

Helium Isotopes And General Periodicities

Stable Helium-3, \(^3He\)

(\(T^+\), \(p^+\), \(g^+\), \(T^+\), \(p^+\))

and

(\(p^+\), \(g^+\), \(T^+\), \(p^+\))

where the spin of \(p^{+}\) particles contributes to atomic mass.

Maybe possible Helium-3, \(^3He\)

(\(g^+\), \(T^+\), \(2p^+\))

where the last particle is doubled in numbers.

Stable Helium-4, \(^4He\)

(\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\), \(p^+\))

Unstable Helium-2, \(^2He\)

(\(T^+\), \(2p^+\))

where the \(T^{+}\) particle left behind after decay has it mistaken as \(\beta\) decay.

(\(2p^+\))

this is more likely, and it splits into two \(^1H\).

The problem is Hydrogen has many isotopes.  All of which can be turned into a Helium isotopes by adding to the hydrogen nucleus set (\(p^{+}\)) or (\(g^+\), \(T^{+}\), \(p^+\)) when the hydrogen nucleus set ends in \(p^{+}\) or, by adding \(p^{+}\) when the hydrogen nucleus set ends in \(T^{+}\) and, by adding (\(T^{+}\), \(p^+\)) when the hydrogen nucleus set ends in \(g^{+}\).

The simple idea of building the nucleus up from weak fields holds.  Although this view includes isotopes naturally as the nucleus is built up, there is no simple periodicity.  Periodicity in chemical reactions involving charges occurs when we consider the addition of \(p^{+}\) particles only and  group all nuclei with the same number of \(p^{+}\) particles together.

So there are two other periodicities, when we group nuclei with the same number of \(g^{+}\) or the same number of \(T^{+}\) particles.  These are periodicities of chemical reactions involving \(g^{-}\) and \(g^{+}\) particles, and \(T^{+}\) and \(T^{-}\) particles separately.

Reconsidering \(\beta^{-}\) Decay

Consider all the stable isotopes of hydrogen,

(\(T^+\), \(p^+\))

(\(p^+\))

(\(g^+\), \(T^+\), \(p^+\))

(\(T^+\), \(p^+\), \(g^+\))  and  (\(T^+\), \(p^+\), \(g^+\), \(T^+\))

(\(p^+\), \(g^+\))  and  (\(p^+\), \(g^+\), \(T^+\))

(\(g^+\), \(T^+\), \(p^+\), \(g^+\))  and  (\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\))

If \(\beta^{-}\) decay is still centered around \(g^{+}\) particles then the following scheme maybe possible,

Where a photon, \(P_{p^{+}}\) collides with a \(g^{+}\) particle.  The two time axes, \(t_c\) and \(t_g\) of \(g^{+}\) swapped and transmute it to a \(e^{-}\) particle which is ejected from the nucleus.  The photon slows down and becomes a proton, \(p^{+}\).  This proton merged with the lower or higher layer proton, \(p^{+}\) in the nucleus to give \(2p^{+}\). When the proton merged with  a higher particle, the falling particle will emit a small amount of energy.  For example,

(\(g^+\), \(T^+\), \(p^+\))\(\rightarrow\)(\(2p^{+}\)) + \(e^{-}\) + \(T^{+}\)

in this case, \(T^{+}\) is the electron anti-neutrino and the captured photons merge with a higher layer proton to give two protons.  Furthermore,

(\(T^+\), \(p^+\), \(g^+\))\(\rightarrow\)(\(T^{+}\), \(2p^{+}\)) + \(e^{-}\)

without the emission of an electron anti-neutrino.  And,

(\(T^+\), \(p^+\), \(g^+\), \(T^+\))\(\rightarrow\)(\(T^{+}\), \(2p^{+}\)) + \(e^{-}\) + \(T^{+}\)

with the emission of an electron anti-neutrino and the photons merged down one layer.

The set (\(T^+\), \(p^+\), \(g^+\)) arises from considering positive particles being captured by weak fields due to positive particle spins in the hydrogen nucleus.  It is a repeating series that occurs in the nuclei of other elements.  If this \(\beta^{-}\) decay scheme is true, it will also apply to all nuclei susceptible to such decays.

The \(T^{+}\) particle originates from the nucleus.  It is released as the weak field holding it disappeared when the spinning \(g^{+}\) particle generating the field is transmuted to a \(e^{-}\) particle after colliding with a photon.

Note: How does a photon slow down?  The time dimension wrap around a space dimension.  When the particle has light speed in space, its time speed is zero.  It is a photon.  When the photon slows down in space, its time speed increases towards light speed.  When its speed in space is zero, it becomes a particle, its speed along \(t_T\) is light speed, \(c\).

Told You It Is Just For Fun

From the post "How Much Further Still Can Gravity Particles Go?" dated 29 Jun 2015, it was proposed that a nucleus can be made up of hydrogen particles (neutral hydrogen nuclei),

\((e^{-},\,g^{-},\,g^{+})\)

\((e^{-},\,T^{-},\,T^{+})\)

and

\((e^{-},\,p^{+})\)

and that the pairs,

\((g^{-},\,g^{+})\) and \((T^{-},\,T^{+})\) are equivalent to \(p^{+}\) and are called proton pair.

In particular \(g^{+}\) acts like a electron anti-neutrino AND a electron neutrino in some radioactive decays.

In the previous posts "Stable, Unstable, All Mental", "Where's Sneezy?", "Order, Order Please!",etc the role of the negative particles are ignored and the focus is on positive particles in the nucleus of hydrogen.  The atomic mass is solely contributed by \(g^{+}\) particles and \(p^{+}\) particles in spin and the nucleus is built up from weak fields due to particle spins.

The latter posts concerns the hydrogen nucleus only, consideration given to the weak fields suggests that the previous posts are too simplistic.  In particular, the addition of a proton, \(p^{+}\) to the nucleus also requires the addition of other particles, (\(g^{+}\), \(T^{+}\)) or (\(T^{+}\)), in view of the weak fields.  Unless, the nucleus set ends with a \(p^{+}\) particle, for example,

(\(g^+\), \(T^+\), \(p^+\))

then on receiving an extra \(p^{+}\), becomes,

(\(g^+\), \(T^+\), \(2p^+\))

where the weak field due to the spinning \(T^{+}\) particle attracts two \(p^{+}\) particles.

Thus \(\beta\) decays have to reconsidered in this new light.

Note:  The post "How Much Further Still Can Gravity Particles Go?" dated 29 Jun 2015 and other posts suggesting that the hydrogen nucleus can be,

\((e^{-},\,g^{-},\,g^{+})\) or

\((e^{-},\,T^{-},\,T^{+})\)

are defunct.

Stable, Unstable, All Mental

Oh no, these are all stable isotopes,

(\(T^+\), \(p^+\))

(\(p^+\))

(\(g^+\), \(T^+\), \(p^+\))

(\(T^+\), \(p^+\), \(g^+\))  and  (\(T^+\), \(p^+\), \(g^+\), \(T^+\))

(\(p^+\), \(g^+\))  and  (\(p^+\), \(g^+\), \(T^+\))

(\(g^+\), \(T^+\), \(p^+\), \(g^+\))  and  (\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\))

the spins of \(p^{+}\) particles do not contribute to the mass of the nuclei, only the presence of \(g^{+}\) particle.  In which case, hydrogen isotopes have zero mass, mass of one \(g^{+}\) and the mass of two \(g^{+}\).  The relative abundance of these stable isotopes give rise to the decimals in hydrogen mass that can not be factored.

The notion of \(p^{+}\) having mass \(g^{+}\) is the result of having to add the tuple (\(g^+\), \(T^+\), \(p^+\)) to any nucleus ending with a \(p^{+}\) particle in the cyclic permutation set.  The tuple must contain a \(g^{+}\) particle.  Such an array of stable isotopes with different positions of \(p^{+}\) in the nucleus may also add to the decimal points in experimental isotope mass measurements, if the weak \(g\) field generated by the spinning \(p^{+}\) particles also contribute to mass.  What about spins?  Spins seem to be associated with \(g^{+}\) particles only.

Unstable nuclei are not any members of the cyclic permutation set.  For example,

(\(3g^+\), \(T^+\), \(p^+\))

(\(T^+\), \(p^+\), \(3g^+\))  and  (\(T^+\), \(p^+\), \(3g^+\), \(T^+\))

(\(p^+\), \(3g^+\))  and  (\(p^+\), \(3g^+\), \(T^+\))

are all Hydrogen-3, \(^3H\) with spin \(\small{\cfrac{1}{2}}^{+}\); the group \(3g^{+}\) spins as one.  What about \(^3H\) with \(2^{-}\) spin?

What is \(g^{+}\) and how can it transmute to a charge?

The time axes, \(t_g\) and \(t_c\) of a \(g^{+}\) particle swapped.  The result is an electron that leaves the nucleus; \(\beta^-\) decay.

How does such a swap occurs?  'Til next time...

Where's Sneezy?

And if we order the Hydrogen nuclei written down so far by their masses,

(\(T^+\), \(p^+\))

(\(p^+\))

(\(g^+\), \(T^+\), \(p^+\))

(\(T^+\), \(p^+\), \(g^+\))  and  (\(T^+\), \(p^+\), \(g^+\), \(T^+\))

(\(p^+\), \(g^+\))  and  (\(p^+\), \(g^+\), \(T^+\))

(\(g^+\), \(T^+\), \(p^+\), \(g^+\))  and  (\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\))

keeping in mind that given equal number of \(g^{+}\) particles, the lower position of \(p^{+}\) in the set increases mass, and assuming that the addition of a \(T^{+}\) particle does not change mass.

Here, we are one short of the seven discovered Hydrogen isotopes.  Where's the last?

Note: Cyclic permutation, the particles are added in specific order.  No swapping of position nor adding other types of particle please.

Order, Order Please!

Are other nuclei made up of basic hydrogen nuclei Type I and III,

Type I (\(p^+\), \(g^+\), \(T^+\))

Type III (\(T^+\), \(p^+\), \(g^+\))??

No, because we have Helium-3 \(^3He\), by adding \(p^{+}\) to the Type I hydrogen nucleus \(^3H\)

(\(p^+\), \(g^+\), \(T^+\))+\(p^{+}\)\(\rightarrow\)(\(p^+\), \(g^+\), \(T^+\), \(p^{+}\))

and Helium-2 \(^2He\), by adding (\(T^+\), \(p^{+}\)) to the Type III hydrogen nucleus \(^2H\)

(\(T^+\), \(p^+\), \(g^+\), \(T^+\), \(p^{+}\))

Remember that the hydrogen nucleus Type I nucleus is the heaviest, followed by Type III then Type II, from which we made the correspondence,

Type I (\(p^+\), \(g^+\), \(T^+\)) is the Hydrogen-3 nucleus

Type III (\(T^+\), \(p^+\), \(g^+\)) is the Hydrogen-2 nucleus

and

Type II (\(g^+\), \(T^+\), \(p^+\)) is the Hydrogen nucleus

and we may have, without adding \(p^{+}\) particles which would increase the atomic number, create other Hydrogen isotopes from the Type II and III nuclei types,

From the Type II (\(g^+\), \(T^+\), \(p^+\)) nucleus,

(\(g^+\), \(T^+\), \(p^+\), \(g^+\))

and

(\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\))

From the Type III (\(T^+\), \(p^+\), \(g^+\)) nucleus

(\(T^+\), \(p^+\), \(g^+\), \(T^+\))

It is not possible to add to a Type I (\(p^+\), \(g^+\), \(T^+\)) nucleus because the next particle to be added is \(p^{+}\) which would increment the atomic number.

It is possible to reduce the existing nuclei types I and III by one particle without changing the atomic number,

From Type I (\(p^+\), \(g^+\), \(T^+\)),

(\(p^+\), \(g^+\))

and

(\(p^+\)), which is mass-less unless the single \(p^{+}\) spins and generates a \(g\) field.

From Type III (\(T^+\), \(p^+\), \(g^+\))

(\(T^+\), \(p^+\))

which is also mass-less unless the single \(p^{+}\) spins and generates a \(g\) field.

Do all these fit well with hydrogen isotopes already discovered?  No!  There are nine different nuclei here but only seven discovered isotopes.  The assumption that the basic nuclei types are made up of all three positive particles may be wrong.  It could be that, the nucleus

(\(T^+\), \(p^+\))

reduced from a Type III nucleus (\(T^+\), \(p^+\), \(g^+\)) is Hydrogen-2, \(^2H\),

(\(p^+\), \(g^+\))

reduced from Type I (\(p^+\), \(g^+\), \(T^+\)) nucleus is Hydrogen-3 \(^3H\), and

(\(p^+\))

reduced from (\(p^+\), \(g^+\)) is Hydrogen-1, \(^1H\).

In which case, there is still two extra isotopes.

Order, order, order in the court of hydrogen please.

Note:  The difference in mass among the isotopes was previously attributed to the strength of the weak \(g\) field, which depended on the position of \(p^{+}\) particles (thus orbital radii) in the ordered set.

And The Word "Creation"

\(T^{+}\)~\(T^{-}\) bondings result in solids, which is broken by the application of heat.  Free \(T^{-}\) particles disrupt \(T^{-}\) particle sharing between \(T^{+}\) particles, and break the bond.  As the \(T^{+}\)~\(T^{-}\) bonding are broken discretely, one at a time with temperature, the solid collapses beyond an abrupt threshold number of \(T^{+}\)~\(T^{-}\) bondings broken.  The pure solid melts at a sharp temperature.  With less \(T^{+}\)~\(T^{-}\) bondings, the lattice flows as a liquid.

With even higher temperature (more \(T^{+}\) particles), \(T^{-}\) particles in spin around a nucleus with more \(T^{+}\) particles, generates a greater \(g\) field.  These stronger \(g\) fields have a anti-gravity effect on the atoms/molecules.  They turn into the gaseous state.  The liquid boils.

The introduction of \(T^{-}\) can also occurs at low temperature. \(T^{+}\)~\(T^{-}\) bondings are also broken at low temperature and results in cracks in the solid.  For the solid to liquidize, the atoms/molecules must have increase buoyancy as the result of greater \(g\) field.  This occurs only with increasing temperature; ie. an abundance of \(T^{+}\) particles.

It is this ability of \(T^{-}\) particles in spin to generate \(g\) fields that allows them to influence nuclear reaction that involves only gravity particles.

We did not detect gravity and temperature particles directly, but we have been living with their effects since Creation.

There is it again, the word "Creation".  And again.

Note:  Increasing pressure increases the density of participating particles at the reaction site.  The secondary effects of increasing uni-directional \(g\) fields due to increased spins on reactions depend on the other factors.  The absorption of a free \(g^{-}\) particles is impeded in the inward direction of the \(g\) fields, but is increased in the outward direction of the field.  Orientation is one such factor.

With increasing \(g\) field, high temperature can eject a \(g^{-}\) particle from the nucleus.

In Plain Sight...

From the Type II nucleus, the element Hydrogen, \(H\),

(\(g^+\), \(T^+\), \(p^+\))~\(e^{-}\)

But what are these?

From the Type I nucleus,

(\(p^+\), \(g^+\), \(T^+\))~\(T^{-}\)

From the Type III nucleus,

(\(T^+\), \(p^+\), \(g^+\))~\(g^{-}\)

May be all particle types must be paired, \(g^{+}\)~\(g^{-}\), \(T^{+}\)~\(T^{-}\) and \(p^{+}\)~\(e^{-}\)

And so, the Hydrogen element is actually,

(\(g^+\), \(T^+\), \(p^+\))~\(e^{-}\)~\(T^{-}\)~\(g^{-}\)

and the other isotopes are,

(\(p^+\), \(g^+\), \(T^+\))~\(T^{-}\)~\(g^{-}\)~\(e^{-}\)

and

(\(T^+\), \(p^+\), \(g^+\))~\(g^{-}\)~\(e^{-}\)~\(T^{-}\)

And where have \(T^{-}\) and \(g^{-}\) been hiding? 

Why Reactions Occurs And Nuclear Reactions

The presence of the three weak fields \(E\), \(g\), \(B\) is the reasons why reactions take place.

And we expand reactions to include two other types of interactions, that between bonded and orbiting gravitational particles and, between bonded and orbiting temperature particles, in addition to charge particles.

At this point however, there is no explanation for the relative abundance of hydrogen isotopes.  One possibility is that, Type I and Type III nucleus interacting through their outermost orbiting particle, \(T^{+}\) and \(g^{+}\) respectively, with the corresponding negative particles, \(T^{-}\) and \(g^{-}\), create other nuclei.  Both temperature and gravitational particles give rise to nuclear reactions.

Good night.

Hydrogen Gas

And behold, after much thought, the Hydrogen element, \(H\),

of the lightest Type II nucleus.  The electron is attracted to the outermost proton, it is not spinning under the influence of the \(E\) field generated by the spinning \(T^{+}\) particle.

And the weak fields around the nucleus is simplified to an triplet corner,

as each of the weak field emerge parallel to the axis of rotation of the previous orbiting particle which is along a diameter of the orbit of the previous particle.  Two consecutive weak fields are orthogonal.

This \(E\) field however can be attracted to the electron around another charge neutral hydrogen nucleus.  The geometry of their union however, depends on the interaction of all three weak fields.

The opposing \(E\) fields keeps the nuclei apart, the two \(g\) fields in parallel doubles its mass under gravity, and the aligned \(B\) fields result in a weak resultant magnetic field around the molecule.

Along the \(E\) fields the nuclei behave as particles, the weak fields are attracted to the electron around the other hydrogen nucleus.  Along the \(g\) fields the nuclei behave as waves and merged in parallel.  Along the \(B\) fields, alignment suggests that the axes of particles' spins are parallel and that the particles spin in the same sense.  This requires minimum energy.

The presence of weak fields around the nucleus that can be rotated, aligned and made to cancel (opposing spins) or add (parallel spins), provides explanations to other characteristics of a molecule such as bond angle, magnetic properties and dipoles.

Have a nice day...

Note: We have not included \(g^{-}\) and \(T^{-}\) particles in this model for Hydrogen.