The exhaust is the compressed air stream. Enjoy!
Thursday, March 31, 2016
Wednesday, March 30, 2016
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 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.
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.
(...\(g^+\), \(2T^{+}\), \(p^+\)...)\(\rightarrow\)(...\(g^+\), \(T^{+}\), \(p^+\)...)\(+T^{+}\)
(...\(T^+\), \(2p^{+}\), \(g^+\)...)\(\rightarrow\)(...\(T^+\), \(p^{+}\), \(g^+\)...)\(+p^{+}\)
(...\(p^{+}\), \(2g^+\), \(T^{+}\)...)\(\rightarrow\)(...\(p^+\), \(g^{+}\), \(T^+\)...)\(+g^{+}\)
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,
(...\(p^+\), \(g^+\), \(T^{+}\), \(T^{-}\), \(g^+\), \(T^+\), \(p^{+}\)...)\(\rightarrow\)(...\(p^+\), \(g^+\), \(2T^{+}\), \(p^+\)...)\(+T^{-}+g^{+}\)
(...\(p^+\), \(g^+\), [\(T^{+}\), \(T^{-}\)], \(p^+\), \(g^+\), \(T^{+}\)...)
and
(...\(p^+\), \(g^+\), \(T^{+}\), \(T^{-}\), \(g^+\), \(T^+\), \(p^{+}\)...)
(...\(p^+\), \(g^+\), [\(T^{+}\), \(T^{-}\)], \(p^+\), \(g^+\), \(T^{+}\)...)\(\rightarrow\)(...\(p^+\), \(g^+\), \(T^{+}\), \(p^+\), \(g^+\), \(T^{+}\)...)\(+T^{-}\)
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,
(...\(p^+\), [\(g^{+}\), \(g^{-}\)], \(T^{+}\), \(p^+\), \(g^+\), ...)\(\rightarrow\)(...\(p^+\), \(g^{+}\), \(T^{+}\), \(p^+\), \(g^+\), ...)\(+g^{-}\)
(...\(p^+\), \(g^{+}\), \(g^{-}\), \(p^{+}\), \(g^+\), \(T^+\),...)
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
(...\(T^{+}\), \(p^{+}\), \(e^{-}\), \(T^{+}\), \(p^+\), \(g^{+}\), \(T^{+}\)...)
(...\(T^+\), [\(p^{+}\), \(e^{-}\)], \(g^{+}\), \(T^{+}\), \(p^+\)...)
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.
\(2p^{+}\) 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.
(\(g^{+}\), \(T^{+}\), \(p^{+}\), \(g^{+}\), \(T^{+}\), \(p^+\))
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.
Tuesday, March 29, 2016
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.
[\(T^{+}\), \(T^{-}\)], \(T^{-}\) 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,
[\(p^{+}\), \(e^{-}\)], \(e^{-}\) 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.
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.
(\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\), \(p^+\))
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?
From the case of \(\beta^{+}\) decay, the colliding photon 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.
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.
Monday, March 28, 2016
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.
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.
\(p^+\)+\(P_{g^{+}}\)\(\rightarrow\)\(T^{-}+e^{+}+g^{+}\)
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.
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.
(\(T^+\), \(p^+\), \(g^+\), \(T^+\), \(p^+\))
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, \(c\).
(\(T^+\), \(p^+\))
(\(p^+\))
(\(g^+\), \(T^+\), \(p^+\))
(\(T^+\), \(p^+\), \(g^+\)) 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.
\((e^{-},\,g^{-},\,g^{+})\)
\((e^{-},\,T^{-},\,T^{+})\)
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.
Sunday, March 27, 2016
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...
(\(T^+\), \(p^+\))
(\(p^+\))
(\(g^+\), \(T^+\), \(p^+\))
(\(T^+\), \(p^+\), \(g^+\)) 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 (\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\))
(\(T^+\), \(p^+\))
(\(p^+\))
(\(g^+\), \(T^+\), \(p^+\))
(\(T^+\), \(p^+\), \(g^+\)) 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.
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.
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,
(\(T^+\), \(p^+\), \(g^+\))~\(g^{-}\)
(\(g^+\), \(T^+\), \(p^+\))~\(e^{-}\)
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
(\(T^+\), \(p^+\), \(g^+\))~\(g^{-}\)~\(e^{-}\)~\(T^{-}\)
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.
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.
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.
Saturday, March 26, 2016
Hydrogen Isotopes
Does cyclic permutation of the layered nucleus accounts for isotopes in nature? In the case of three layer nucleus involving all types of positive particles,
Type I (\(p^+\), \(g^+\), \(T^+\))
Type II (\(g^+\), \(T^+\), \(p^+\))
Type III (\(T^+\), \(p^+\), \(g^+\))
Type I (\(p^+\), \(g^+\), \(T^+\))
Type II (\(g^+\), \(T^+\), \(p^+\))
Type III (\(T^+\), \(p^+\), \(g^+\))
Outer layer orbits have larger radii that produce weaker weak fields. So Type I nucleus has the strongest \(g\) field produced by an inner most \(p^{+}\) particle, followed by Type III then Type II nucleus. If this is the main field that interacts with Earth's gravitational field, giving the nucleus weight then Type I nucleus is the heaviest, followed by Type III then Type II.
Type I hydrogen will also be most heat conductive with \(T^{+}\) particles at the outermost orbit. Type II hydrogen nucleus behave more like a positive charge. And Type III hydrogen nucleus in spin is a magnetic dipole.
Good night.
Follow The Inner
If there is a fourth layer,
(\(p^{+}\), \(g^{+}\), \(T^{+}\), \(p^{+}\))
or
(\(e^{-}\), \(T^{-}\), \(g^{-}\), \(e^{-}\))
we see that the first and fourth particle will both produce the same weak field and interact. Would these two fields reinforce, cancel or be orthogonal?
In all cases, the spin axes of the inner and outer particles (of the same type), will be ordered. Consecutive particles of the same type taking up orbits around the nucleus will align to generate fields that either reinforce, cancel or be orthogonal to the inner layer fields.
It is likely that the outer/added field cancels the inner field for a minimum energy system, such that on removing the outer particle, there is an increase in the inner field strength or a reversal of the observed field.
'Til next time...
(\(p^{+}\), \(g^{+}\), \(T^{+}\), \(p^{+}\))
or
(\(e^{-}\), \(T^{-}\), \(g^{-}\), \(e^{-}\))
we see that the first and fourth particle will both produce the same weak field and interact. Would these two fields reinforce, cancel or be orthogonal?
In all cases, the spin axes of the inner and outer particles (of the same type), will be ordered. Consecutive particles of the same type taking up orbits around the nucleus will align to generate fields that either reinforce, cancel or be orthogonal to the inner layer fields.
It is likely that the outer/added field cancels the inner field for a minimum energy system, such that on removing the outer particle, there is an increase in the inner field strength or a reversal of the observed field.
'Til next time...
Behold Another Onion...
The following is a template for a three layered nucleus of negative particles,
The fields generated by the negative particles point in the opposite direction given by the right hand rule. The whole orbit of the inner particle (in black) spins along a diameter so that the outer particle (in red) spins along the red orbit. In this way, the innermost particle has three spins, the next particle has two spins and the outermost particle has one spin. The three types of negative particle nuclei are,
Type In (\(e^-\), \(T^-\), \(g^-\))
Type IIn (\(T^-\), \(g^-\), \(e^-\))
Type IIIn (\(g^-\), \(e^-\), \(T^-\))
This is not the same cyclic permutation as positive particle nuclei, two particles have swapped layers. The new order is (\(e^-\), \(T^-\), \(g^-\)).
Could this layering go on to four, five and more layers? What about the positive particles?
It is equally likely that a positive particle be caught in the weak field of the spinning negative particle and be sent into circular motion as the weak field spins with the orbit of the negative particle. In general, the interaction with a mix of positive and negative particles is much more complex. However, the general principle is that the particle is caught in a corresponding weak field of a spinning particle. The orbit of the spinning particle itself is spinning about a diameter, and sends the captured particle into spin. The spinning captured particle generates a weak field which in turn attracts another particle, ad infinitum.
A spinning fourth layer will result in the changes in at least one of the three spins of the innermost particle. This fourth spin can be orthogonal to the two spins in the second layer or be orthogonal to the one spin in the third layer, in these cases, spins in the second and third layers are not affected by the addition of the fourth spinning particle to the nucleus.
Science is like a turban, you just know there is an onion inside.
The fields generated by the negative particles point in the opposite direction given by the right hand rule. The whole orbit of the inner particle (in black) spins along a diameter so that the outer particle (in red) spins along the red orbit. In this way, the innermost particle has three spins, the next particle has two spins and the outermost particle has one spin. The three types of negative particle nuclei are,
Type In (\(e^-\), \(T^-\), \(g^-\))
Type IIn (\(T^-\), \(g^-\), \(e^-\))
Type IIIn (\(g^-\), \(e^-\), \(T^-\))
This is not the same cyclic permutation as positive particle nuclei, two particles have swapped layers. The new order is (\(e^-\), \(T^-\), \(g^-\)).
Could this layering go on to four, five and more layers? What about the positive particles?
It is equally likely that a positive particle be caught in the weak field of the spinning negative particle and be sent into circular motion as the weak field spins with the orbit of the negative particle. In general, the interaction with a mix of positive and negative particles is much more complex. However, the general principle is that the particle is caught in a corresponding weak field of a spinning particle. The orbit of the spinning particle itself is spinning about a diameter, and sends the captured particle into spin. The spinning captured particle generates a weak field which in turn attracts another particle, ad infinitum.
A spinning fourth layer will result in the changes in at least one of the three spins of the innermost particle. This fourth spin can be orthogonal to the two spins in the second layer or be orthogonal to the one spin in the third layer, in these cases, spins in the second and third layers are not affected by the addition of the fourth spinning particle to the nucleus.
Science is like a turban, you just know there is an onion inside.
Behold An Onion...
The following is a template for a three layered nucleus,
of which there can be three types,
Type I (\(p^+\), \(g^+\), \(T^+\))
Type II (\(g^+\), \(T^+\), \(p^+\))
Type III (\(T^+\), \(p^+\), \(g^+\))
assuming all three types of positive particles are involved. It is possible that only the first or first and second layers of the nucleus exist. In both cases, the nucleus can capture the next awaiting particle and be transmuted to a higher layered nucleus. For example,
(\(p^+\), \(g^+\)) two layered nucleus captures a \(T^{+}\) particle \(\rightarrow\)Type I (\(p^+\), \(g^+\), \(T^+\))
Similarly,
(\(g^+\), \(T^+\)) + \(p^+\) \(\rightarrow\)(\(g^+\), \(T^+\), \(p^+\))
(\(T^+\), \(p^+\)) + \(g^+\) \(\rightarrow\)(\(T^+\), \(p^+\), \(g^+\))
Notice that the three types of three layered nucleus are cyclic permutations of (\(p^+\), \(g^+\), \(T^+\)). The generated weak field of a lower layer particle attracts a specific positive particle only.
Could this layering go on to four, five and more layers? What about the negative particles?
Science is like an onion; makes you cry, especially if someone else peeled it apart first.
of which there can be three types,
Type I (\(p^+\), \(g^+\), \(T^+\))
Type II (\(g^+\), \(T^+\), \(p^+\))
Type III (\(T^+\), \(p^+\), \(g^+\))
assuming all three types of positive particles are involved. It is possible that only the first or first and second layers of the nucleus exist. In both cases, the nucleus can capture the next awaiting particle and be transmuted to a higher layered nucleus. For example,
(\(p^+\), \(g^+\)) two layered nucleus captures a \(T^{+}\) particle \(\rightarrow\)Type I (\(p^+\), \(g^+\), \(T^+\))
Similarly,
(\(g^+\), \(T^+\)) + \(p^+\) \(\rightarrow\)(\(g^+\), \(T^+\), \(p^+\))
(\(T^+\), \(p^+\)) + \(g^+\) \(\rightarrow\)(\(T^+\), \(p^+\), \(g^+\))
Could this layering go on to four, five and more layers? What about the negative particles?
Science is like an onion; makes you cry, especially if someone else peeled it apart first.
Friday, March 25, 2016
Partial Cancellation of Weak Fields
The problem with spinning particles at the nucleus is that the field generated are uni-directional as oppose to emanating radially from point source. This means that such weaker fields can be made to cancel partially or even completely in a solid lattice. In the case of gravitational fields for example, partially cancelling these fields in a lattice make the element lighter (less measured mass under Earth's gravity), or equivalently, more dense (more atoms/nuclei in the lattice) given mass.
Some elements then can weigh less but have greater density.
Some elements feels cold others hot.
Some elements are magnetic others not.
Some elements are conductive of electricity others not.
Many physical properties will depend on the interactions of these three weak fields (electric, gravitational and temperature/magnetic) generated as the result of particle spins.
Good Morning.
Some elements then can weigh less but have greater density.
Some elements feels cold others hot.
Some elements are magnetic others not.
Some elements are conductive of electricity others not.
Many physical properties will depend on the interactions of these three weak fields (electric, gravitational and temperature/magnetic) generated as the result of particle spins.
Good Morning.
Psychosis, What's With The Outside?
The outer \(T^{+}\) particle in the nucleus configuration proposed for a \(H\) nucleus might explain hydrogen good heat conductivity.
Which is suggesting that the outer particle effects the physical properties of the element. Magnetic materials/elements will have \(g^{+}\) particles as the outermost particle producing a \(B\) field in spin. Heavy elements will have \(p^{+}\) particles as outermost particle in their nucleus configuration; spinning \(p^{+}\) particles generates a gravitational field that interacts with Earth's gravitational field producing the effect of heavy mass. Metals have \(T^{+}\) particles as the outermost particle, and are conductors of heat and/or electricity. The outermost \(T^{+}\) particle in spin produces an electric field, \(E\); the negative ends of which forms up as electron flow conductive channels in the lattice of electric conductors. When such channels fail to form, the metal is a poor conductor of electricity. This might explain why metals that are good conductors of heat need not be good conductors of electricity also.
Have a psychotic night! Without psychosis I cannot possibly break with conventional science. However, if all these are true, you are the one instead psychotic, who have lost contact with true reality.
Which is suggesting that the outer particle effects the physical properties of the element. Magnetic materials/elements will have \(g^{+}\) particles as the outermost particle producing a \(B\) field in spin. Heavy elements will have \(p^{+}\) particles as outermost particle in their nucleus configuration; spinning \(p^{+}\) particles generates a gravitational field that interacts with Earth's gravitational field producing the effect of heavy mass. Metals have \(T^{+}\) particles as the outermost particle, and are conductors of heat and/or electricity. The outermost \(T^{+}\) particle in spin produces an electric field, \(E\); the negative ends of which forms up as electron flow conductive channels in the lattice of electric conductors. When such channels fail to form, the metal is a poor conductor of electricity. This might explain why metals that are good conductors of heat need not be good conductors of electricity also.
Have a psychotic night! Without psychosis I cannot possibly break with conventional science. However, if all these are true, you are the one instead psychotic, who have lost contact with true reality.
Being Psychotic
We have seen that a positive gravity particle can acquire a negative gravity particle and then attract an electron. If a hydrogen nucleus also contains a gravity particle or two, it is possible that the nucleus attains an extra electron and becomes \(H^{-}\). But how does the nucleus with its singular positive charge particle contains positive gravity particles?
In this dream, these particles do not interact. An electric field do not interact with a gravitational field.
A spinning \(p^{+}\), however generates a gravitational field. The in-going, negative end of this field will attract a positive gravity particle. This is a weak interaction, weaker than the attraction between particles of opposite sign.
This is not a stable configuration. It is possible to image that \(g^{+}\) going into spin but what would cause the orbit of \(p^{+}\) to spin along its diameter? \(p^{+}\) is then in motion with one other degree of freedom, which is permissible in 3D space; it has two spins.
But why?
When \(g^{+}\) goes into a spin, it generates a magnetic, \(B\) field.
Being psychotic, why stop here? So, the negative end of this \(B\) field attracts a positive temperature particle and the whole configuration goes into a spin giving \(p^{+}\) a third spin, and \(g^{+}\) two spins.
The spinning positive temperature particle generates a electric, \(E\) field. This positive temperature particle has only one spin.
The order of these particles is arbitrary, we could have started with a positive gravity particle and build up from there. In which case, \(g^{+}\) has three spins, \(T^{+}\) has 2 spins and the last \(p^{+}\) being attracted to the negative side of the \(E\) field generated by a spinning \(T^{+}\) particle has only one spin.
This last scenario is not at all twisted nor psychotic. Have a cup of NoClass 3 in 1 coffee + psychotics on me!
In this dream, these particles do not interact. An electric field do not interact with a gravitational field.
A spinning \(p^{+}\), however generates a gravitational field. The in-going, negative end of this field will attract a positive gravity particle. This is a weak interaction, weaker than the attraction between particles of opposite sign.
This is not a stable configuration. It is possible to image that \(g^{+}\) going into spin but what would cause the orbit of \(p^{+}\) to spin along its diameter? \(p^{+}\) is then in motion with one other degree of freedom, which is permissible in 3D space; it has two spins.
But why?
When \(g^{+}\) goes into a spin, it generates a magnetic, \(B\) field.
Being psychotic, why stop here? So, the negative end of this \(B\) field attracts a positive temperature particle and the whole configuration goes into a spin giving \(p^{+}\) a third spin, and \(g^{+}\) two spins.
The spinning positive temperature particle generates a electric, \(E\) field. This positive temperature particle has only one spin.
The order of these particles is arbitrary, we could have started with a positive gravity particle and build up from there. In which case, \(g^{+}\) has three spins, \(T^{+}\) has 2 spins and the last \(p^{+}\) being attracted to the negative side of the \(E\) field generated by a spinning \(T^{+}\) particle has only one spin.
This last scenario is not at all twisted nor psychotic. Have a cup of NoClass 3 in 1 coffee + psychotics on me!
Thursday, March 24, 2016
Crystal Balls
Crystal balls; they are balls and they are heavy...
For light rays that originate inside a crystal ball, those that are totally internally refracted will still be totally internally refracted again at the next point they touch the sphere. Part of the ray escapes as the ray is due to particles in circular motion, the orbits of which swing around the apex of a cone, on the surface of the cone, centered at the apex. (From the post "A Bloom Crosses Over" dated 10 Aug 2015.)
Other rays with incident angle at less than the critical angle for total internal refraction will be refracted repeatedly. All such rays will be lost from inside the crystal ball.
A lot of photons collide inside the sphere among the rays that remain inside the crystal ball. This is just like inside a furnace where a lot of photons collide too. Colliding particles generate particles with high energy \(\psi\) and are sent back in time as their \(\psi\)s subsequently collapse.
If a furnace can sent images from a future date back in time, then a crystal ball can do the same. At lower temperature, crystal balls sent images back in time over a shorter period; a few weeks maybe.
(Just cover the other side of the crystal ball, it is not necessary to rotate the ball.)
How much later in time do you need to sign to your crystal ball to receive a message now? Only by experimentation can you tell, given the size, type of material and temperature of your crystal ball.
And you would illuminate the crystal ball such that total internal refraction occurs for most of the rays from the light source. It is the collisions among the photons trapped by total internal refraction inside the crystal ball that produce the time travel phenomenon.
Keeping the temperature low (and constant) allows for a shorter period to your next appointment in the future with your client, who then tells you what happened, as you relay the narratives to the crystal ball (possibly a note or sign) . Back in time, you read the crystal ball as the same client sits before you across the crystal ball. Back in time, you are on the reading side of the crystal ball. In the future appointment, you send images back in time on the receiving side of the crystal ball.
Now that we have a side business reading crystal balls as gypsies, have a nice day.
For light rays that originate inside a crystal ball, those that are totally internally refracted will still be totally internally refracted again at the next point they touch the sphere. Part of the ray escapes as the ray is due to particles in circular motion, the orbits of which swing around the apex of a cone, on the surface of the cone, centered at the apex. (From the post "A Bloom Crosses Over" dated 10 Aug 2015.)
Other rays with incident angle at less than the critical angle for total internal refraction will be refracted repeatedly. All such rays will be lost from inside the crystal ball.
A lot of photons collide inside the sphere among the rays that remain inside the crystal ball. This is just like inside a furnace where a lot of photons collide too. Colliding particles generate particles with high energy \(\psi\) and are sent back in time as their \(\psi\)s subsequently collapse.
If a furnace can sent images from a future date back in time, then a crystal ball can do the same. At lower temperature, crystal balls sent images back in time over a shorter period; a few weeks maybe.
(Just cover the other side of the crystal ball, it is not necessary to rotate the ball.)
How much later in time do you need to sign to your crystal ball to receive a message now? Only by experimentation can you tell, given the size, type of material and temperature of your crystal ball.
And you would illuminate the crystal ball such that total internal refraction occurs for most of the rays from the light source. It is the collisions among the photons trapped by total internal refraction inside the crystal ball that produce the time travel phenomenon.
Now that we have a side business reading crystal balls as gypsies, have a nice day.
Personal Time Travel Device And Nostradamus Watching TV
From the previous post "When Time Is Slow With A Big Brinjal" dated 20 Mar 2016, it is then possible to generate high value of \(\psi\) by colliding specific particles at specific orientations. As \(\psi\) collapses, a time force is generated that allows time travel.
The direction of travel, forward or backward depends on the second order change in \(\psi\), as suggested in the post "My Dream, Time Travel And Temperature Particles" dated 7 Dec 2014.
Which bring us to the story of Nostradamus. His dwelling was supposed to be now, a glass workshop
with a high temperature furnace. This furnace has many colliding particles generating high \(\psi\) particles that can be sent back in time. Could it be possible, that he obtained his visions of the future by looking into the direction of the furnace of the future? The furnace sends images possibly from a TV, from the future, into his view.
In the same delusion that I can explain everything, I shall next explain crystal balls.
Til next time...
The direction of travel, forward or backward depends on the second order change in \(\psi\), as suggested in the post "My Dream, Time Travel And Temperature Particles" dated 7 Dec 2014.
Which bring us to the story of Nostradamus. His dwelling was supposed to be now, a glass workshop
with a high temperature furnace. This furnace has many colliding particles generating high \(\psi\) particles that can be sent back in time. Could it be possible, that he obtained his visions of the future by looking into the direction of the furnace of the future? The furnace sends images possibly from a TV, from the future, into his view.
In the same delusion that I can explain everything, I shall next explain crystal balls.
Til next time...
Tuesday, March 22, 2016
Sunday, March 20, 2016
When Time Is Slow With A Big Brinjal
If it is possible that as the axes of the wave reshuffle, the oscillatory component of the wave be swapped with a time dimension along which the wave is at light speed. Then,
it is possible to create a particle of high energy (oscillating between a time dimension and a space dimension) but low speed, (\(v\lt c\)) along a time dimension.
If this wave is able to accelerate to light speed along a time dimension (\(t_g\) in the example above) what would the mechanism be?
Does the circular/spherical wave collapse to a lower radius till it gain light speed? (The momentum of the created particle remained unchanged in space.) This, surprisingly, coincides with the post "My Dream, Time Travel And Temperature Particles" dated 7 Dec 2014, where it was suggested that a collapsing \(\psi\) creates a time force.
Such particles will provide a glimpse of the time dimension at low speed \(v\lt c\). If at this time, \(t_c\) and \(t_g\) were to swap again, in the example above, the particle is sent back in time with time speed \(v\lt c\).
it is possible to create a particle of high energy (oscillating between a time dimension and a space dimension) but low speed, (\(v\lt c\)) along a time dimension.
If this wave is able to accelerate to light speed along a time dimension (\(t_g\) in the example above) what would the mechanism be?
Does the circular/spherical wave collapse to a lower radius till it gain light speed? (The momentum of the created particle remained unchanged in space.) This, surprisingly, coincides with the post "My Dream, Time Travel And Temperature Particles" dated 7 Dec 2014, where it was suggested that a collapsing \(\psi\) creates a time force.
Such particles will provide a glimpse of the time dimension at low speed \(v\lt c\). If at this time, \(t_c\) and \(t_g\) were to swap again, in the example above, the particle is sent back in time with time speed \(v\lt c\).
Time Will Tell
If the time dimension component of a wave along which the wave has light speed, can be aligned and made to cancel with another colliding wave, then it is possible by inspecting the particles created after such collisions to determine how the time dimensions \(t_c\), \(t_g\) and \(t_T\) are ordered.
And maybe determine whether travelling along the negative direction along any of the time dimensions is possible.
Om...Brinjal...Om...
Big brinjal takes a long time to finish...Om...
And maybe determine whether travelling along the negative direction along any of the time dimensions is possible.
Om...Brinjal...Om...
Big brinjal takes a long time to finish...Om...
Banging Creation
There can be three ways, two particles colliding and interacting as waves, temporarily superimpose and cancel part of their wave components and collapse,
The oscillatory components of the waves in the same direction and opposing phase, or in opposing directions but in phase, will cancel when superimposed and so temporarily collapse the resultant wave. This is type 1.
In type 2 collision, the velocity component along a time dimension of the wave cancels. Assuming that it is possible to cancel the velocity component of the wave along a time dimension. Although the time dimensions are not accessible to us as the three space dimensions, it is still be possible to orientate the wave component triplets, which forms a orthogonal corner, such that the time dimensions of the colliding waves at light speed, are in opposing directions.
In type 3 collision, both oscillatory component and the velocity component of the wave cancel.
Collapsing the wave temporarily allows the components of the wave to be reshuffled and so create a new type of particle after the collision.
Creation, but first destruction, temporarily...
The oscillatory components of the waves in the same direction and opposing phase, or in opposing directions but in phase, will cancel when superimposed and so temporarily collapse the resultant wave. This is type 1.
In type 2 collision, the velocity component along a time dimension of the wave cancels. Assuming that it is possible to cancel the velocity component of the wave along a time dimension. Although the time dimensions are not accessible to us as the three space dimensions, it is still be possible to orientate the wave component triplets, which forms a orthogonal corner, such that the time dimensions of the colliding waves at light speed, are in opposing directions.
In type 3 collision, both oscillatory component and the velocity component of the wave cancel.
Collapsing the wave temporarily allows the components of the wave to be reshuffled and so create a new type of particle after the collision.
Creation, but first destruction, temporarily...
Saturday, March 19, 2016
Big Brinjal And A Cashier Wanting To Know Where
Just when I thought I can be as happy as the Danes, I remember colliding electrons to obtain gravity waves,
Since, space dimensions are orthogonal, these particles collides with energy oscillating in the space dimension perpendicular to the direction of travel. These oscillating energies are not necessarily in opposite direction to each other. If it is necessary that the waves be first destroyed by cancelling energies along \(t_T\), then the oscillating energies must oppose each other when the waves collide. Furthermore, these energies must be in phase given their opposing orientations. The probability of a collision satisfying these two requirements of orientation and phase is,
\(P_c=\cfrac{1}{2\pi}.\cfrac{1}{2\pi}\)
but since the particles can be in the same direction with \(\pi\) phase difference and still collide and temporarily cancel energy along \(t_T\),
\(Pro_c=2P_c=\cfrac{1}{2\pi^2}\)
After the waves' destruction, it reforms into a wave that exists in \(t_g\), and travels along \(t_c\) at light speed. When the particle/wave depart after reforming, they have oscillatory energy along \(t_T\), ie. a positive gravity particle. The waves' destruction during the superposition is only temporary, but is necessary, for the particles to reform. During the process, \(t_c\) and \(t_g\) swap roles.
The new particle is accelerated in the third space dimension perpendicular to the other two. This is because, momentum along the line of collision/travel is destroyed so the resulting wave will not have any velocity component along this direction; energy along \(t_T\) is perpendicular to the line of collision/travel and the new wave is accelerated along a direction perpendicular to the oscillating energy. Given many collisions along the same line of travel, the positive gravity particles emerge perpendicular to the line of collision, at the point of collision (and/or the velocity component at light speed).
Since the collision can result in no swapping of roles between \(t_c\) and \(t_g\), the probability of generating positive gravity particles after the collision, \(P\) is,
\(P=\cfrac{3}{4}Pro_c=\cfrac{3}{8\pi^2}\)
There can be two positive gravity particles, or one electron and one positive gravity particle, or two electrons going in opposite directions after the collision.
This is a description of what might happen during the collision of two electrons head on. It is not an explanation of how gravity particles are created (Why \(t_c\) and \(t_g\) can swap roles?). However it does suggest that, by similar processes, other types of particle can be created by banging other particles of the same type or not. The salient point is that the colliding waves (the particles interact as waves when closer together) be temporarily destroyed by cancelling their oscillatory energy components.
I cannot afford tall burgers, only big brinjal on offer. And if you must know, I cut the eggplant into small pieces, steam them over rice and stuff my mouth with them. That's where I stuff myself with it, ma'am.
Now, I am still a happy person. Om...brinjal...Om...
Note: It is also possible to temporarily destroy the waves by aligning them such that the momentum along \(t_g\) cancels when they super-impose. In this case there is no phase requirement, only an orientation requirement for the temporary destruction of the waves.
\(P_c=Pro_c=\cfrac{1}{2\pi}\)
the rest follows. The probability of generating positive gravity particles after the collision, \(P\), considering both possibilities, is,
\(P=\cfrac{3}{8}\left(\cfrac{1}{\pi^2}+\cfrac{1}{\pi}\right)\)
Om...brinjal...Om...
Since, space dimensions are orthogonal, these particles collides with energy oscillating in the space dimension perpendicular to the direction of travel. These oscillating energies are not necessarily in opposite direction to each other. If it is necessary that the waves be first destroyed by cancelling energies along \(t_T\), then the oscillating energies must oppose each other when the waves collide. Furthermore, these energies must be in phase given their opposing orientations. The probability of a collision satisfying these two requirements of orientation and phase is,
\(P_c=\cfrac{1}{2\pi}.\cfrac{1}{2\pi}\)
but since the particles can be in the same direction with \(\pi\) phase difference and still collide and temporarily cancel energy along \(t_T\),
\(Pro_c=2P_c=\cfrac{1}{2\pi^2}\)
After the waves' destruction, it reforms into a wave that exists in \(t_g\), and travels along \(t_c\) at light speed. When the particle/wave depart after reforming, they have oscillatory energy along \(t_T\), ie. a positive gravity particle. The waves' destruction during the superposition is only temporary, but is necessary, for the particles to reform. During the process, \(t_c\) and \(t_g\) swap roles.
The new particle is accelerated in the third space dimension perpendicular to the other two. This is because, momentum along the line of collision/travel is destroyed so the resulting wave will not have any velocity component along this direction; energy along \(t_T\) is perpendicular to the line of collision/travel and the new wave is accelerated along a direction perpendicular to the oscillating energy. Given many collisions along the same line of travel, the positive gravity particles emerge perpendicular to the line of collision, at the point of collision (and/or the velocity component at light speed).
Since the collision can result in no swapping of roles between \(t_c\) and \(t_g\), the probability of generating positive gravity particles after the collision, \(P\) is,
\(P=\cfrac{3}{4}Pro_c=\cfrac{3}{8\pi^2}\)
There can be two positive gravity particles, or one electron and one positive gravity particle, or two electrons going in opposite directions after the collision.
This is a description of what might happen during the collision of two electrons head on. It is not an explanation of how gravity particles are created (Why \(t_c\) and \(t_g\) can swap roles?). However it does suggest that, by similar processes, other types of particle can be created by banging other particles of the same type or not. The salient point is that the colliding waves (the particles interact as waves when closer together) be temporarily destroyed by cancelling their oscillatory energy components.
I cannot afford tall burgers, only big brinjal on offer. And if you must know, I cut the eggplant into small pieces, steam them over rice and stuff my mouth with them. That's where I stuff myself with it, ma'am.
Now, I am still a happy person. Om...brinjal...Om...
Note: It is also possible to temporarily destroy the waves by aligning them such that the momentum along \(t_g\) cancels when they super-impose. In this case there is no phase requirement, only an orientation requirement for the temporary destruction of the waves.
\(P_c=Pro_c=\cfrac{1}{2\pi}\)
the rest follows. The probability of generating positive gravity particles after the collision, \(P\), considering both possibilities, is,
\(P=\cfrac{3}{8}\left(\cfrac{1}{\pi^2}+\cfrac{1}{\pi}\right)\)
Om...brinjal...Om...
Wednesday, March 16, 2016
I Shall Explain Everything...
This is why the Danish are happiest,
they stack up.
And this is why we can have stack up electron orbits,
they do not stack up, but
\(q+q=2q\)
they merged.
Which suggests that ionization energy increases with the number of electrons in orbit, as the forces between the electrons, interacting as waves, are attractive.
Have a tall Danish burger. Be happy.
they stack up.
And this is why we can have stack up electron orbits,
\(q+q=2q\)
they merged.
Which suggests that ionization energy increases with the number of electrons in orbit, as the forces between the electrons, interacting as waves, are attractive.
Have a tall Danish burger. Be happy.
Tuesday, March 15, 2016
CP, Check Please!
If there are particles that exhibit a weak electric field when in spin but themselves not a charge, then there is no charge symmetry. Since two other particles \(\small{-g,\,\,+T}\) exhibit a positive electric field when in spin, \(\small{\cfrac{1}{3}}\) or \(\small{3}\) is the new game considering only weak electric interactions.
For that matter, no thermal particle symmetry nor gravitational particle symmetry; both of which are particles with opposites and are able to establish a force field.
For that matter, no thermal particle symmetry nor gravitational particle symmetry; both of which are particles with opposites and are able to establish a force field.
Sunday, March 13, 2016
Free Energy
What is this fragment of of energy previously oscillating along an orthogonal dimension of a wave at light speed? This fragment belongs to the energy of the photon given by Planck's equation,
\(E=h.f\)
It is not a part separate from the total energy of the wave. When the photon is absorbed in its totality, this fragment is absorbed also.
There is no free lunch! Not yet.
\(E=h.f\)
It is not a part separate from the total energy of the wave. When the photon is absorbed in its totality, this fragment is absorbed also.
There is no free lunch! Not yet.
Entanglement And Free Energy
Are the two particles after the split entangled?
Only if the photon that drives the particles apart by \(x=\pi\) does not also impart energy onto the orthogonal dimension along which energy are oscillating, by which the particle would be entangled. As such, in order that the split particles be entangled, this photon must firstly, exists in the same time dimension as \(2q\) and secondly has no energy in the oscillating dimension. There can only be one possibility, a photon of the opposite particle. For example, in the case of a big electron,
If the momentum of such a photon is expended in driving the split particles apart, what then happened to the energy in the orthogonal oscillatory dimension? In the above example, where do the energy along \(t_g\) go, after the collision? （This photon, when \(x,\,\, v=c\) is replaced with \(t_T,\,\,v=c\) becomes a proton.)
If energy along \(t_g\) is still oscillating, then it is still a photon but at lower speed. Only part of the energy of the photon has been transfer to the big particle. The photon reduced speed after the collision, and is accelerated to light speed again. Photons are self-propelling. If this energy along \(t_g\) is not oscillating, it is no longer a wave. Can such dissipative, free energy exist?
It is very likely that only the former case applies and that such free energy fragments does not exist, only waves/particles.
Only if the photon that drives the particles apart by \(x=\pi\) does not also impart energy onto the orthogonal dimension along which energy are oscillating, by which the particle would be entangled. As such, in order that the split particles be entangled, this photon must firstly, exists in the same time dimension as \(2q\) and secondly has no energy in the oscillating dimension. There can only be one possibility, a photon of the opposite particle. For example, in the case of a big electron,
If the momentum of such a photon is expended in driving the split particles apart, what then happened to the energy in the orthogonal oscillatory dimension? In the above example, where do the energy along \(t_g\) go, after the collision? （This photon, when \(x,\,\, v=c\) is replaced with \(t_T,\,\,v=c\) becomes a proton.)
If energy along \(t_g\) is still oscillating, then it is still a photon but at lower speed. Only part of the energy of the photon has been transfer to the big particle. The photon reduced speed after the collision, and is accelerated to light speed again. Photons are self-propelling. If this energy along \(t_g\) is not oscillating, it is no longer a wave. Can such dissipative, free energy exist?
It is very likely that only the former case applies and that such free energy fragments does not exist, only waves/particles.
Saturday, March 12, 2016
Friday, March 11, 2016
Negavitve Space?
Intuitively, because,
\(\psi=-\cfrac{\partial\psi}{\partial\,x}\) --- (*)
and that \(\psi\) exist as discrete frequencies in the frequency domain (ie as Dirac Delta functions). With the Inverse Fourier Transform of such frequency \(\psi\), back to the space domain which is an integration of all frequencies with a subsequent substitution,
\(c.t=x\)
where \(c\) is light speed, spreads \(\psi\) over all space. This might suggest that \(\psi\) when non zero, would require an infinite amount of energy. When we integrate \(\psi\) over all space however, the differentiation of \(\psi\) in expression (*) cancels with the integral of \(\psi\) over all space and we are left with an finite expression in \(\psi\) which is also \(F_{\rho}\), as
\(F_{\rho}=\psi\)
\(F_{\rho}\) that spreads from \(+\infty\) to \(-\infty\) and is finite at each value of \(x\). In this way, the field around a particle extends to infinity in space but does not require an infinite amount of energy; \(\psi\) in the frequency domain exists as discrete frequencies.
Quantum mechanics do not blow up to infinity as the result of expression (*) and Fourier transform from the frequency domain. This however, also implies that negative space must exist! But, what is negative space?!
Note: Negative frequencies need not exist as \(\psi\) is zero elsewhere but the positive discrete frequency value. Since we do not count cycles in the negative, negative frequencies implies negative time.
One big fat zero is still finite.
\(\psi=-\cfrac{\partial\psi}{\partial\,x}\) --- (*)
and that \(\psi\) exist as discrete frequencies in the frequency domain (ie as Dirac Delta functions). With the Inverse Fourier Transform of such frequency \(\psi\), back to the space domain which is an integration of all frequencies with a subsequent substitution,
\(c.t=x\)
where \(c\) is light speed, spreads \(\psi\) over all space. This might suggest that \(\psi\) when non zero, would require an infinite amount of energy. When we integrate \(\psi\) over all space however, the differentiation of \(\psi\) in expression (*) cancels with the integral of \(\psi\) over all space and we are left with an finite expression in \(\psi\) which is also \(F_{\rho}\), as
\(F_{\rho}=\psi\)
\(F_{\rho}\) that spreads from \(+\infty\) to \(-\infty\) and is finite at each value of \(x\). In this way, the field around a particle extends to infinity in space but does not require an infinite amount of energy; \(\psi\) in the frequency domain exists as discrete frequencies.
Quantum mechanics do not blow up to infinity as the result of expression (*) and Fourier transform from the frequency domain. This however, also implies that negative space must exist! But, what is negative space?!
Note: Negative frequencies need not exist as \(\psi\) is zero elsewhere but the positive discrete frequency value. Since we do not count cycles in the negative, negative frequencies implies negative time.
One big fat zero is still finite.