Fruechte Gravity Theory Blog

Earlier it was postulated that the number of turns in an atomic orbital, referenced to the nuclear spin axis of an atom, – or to the nucleus to nucleus axis of a diatomic molecule, is equal to 2n + 1, n being the principal quantum number.  In some instances this may be a little hard to believe, such as when looking at some of the Leighton sketches of Liboff Figure 10.17 [1].  The hydrogen 4F, m = 0, orbital, for example, is one that looks symmetrical with six lobes when sectioned in reference to a plane that includes the z axis of the atom.
Let us say that the electron starts out in the 4F orbital on the negative z axis, which we will also allow to be the negative z axis of the nuclear spin at the same point in time.  It then arcs past the nucleus on its way to the first turn of the orbital where it reverses intrinsic spin orientation as it gives off one or more gravitons.  As the electron spends more time in the areas that show a higher whiteness density in the Leighton sketch, the original -z axis of the atomic nucleus does not have to be aligned with the -z axis of the sketch and the electron spin axis again until the electron enters the 9^th turn.  The nuclear spin axis would be allowed to change azimuthal angle back and forth in quantum angular steps.  The electron can be at the original -z location more than once in the course of the orbital, but not at the same instant in time as the -z axis of the nucleus until the last turn.
The raising and lowering operators of the total angular momentum of the state | E₀, L, S, J, M>, are known as:
J_± | E₀, L, S, J, M> = ћ sqrt[J(J+1) – M(M±1)] | E₀, L, S, J, M±1>  [2] Chpt. 10, Complement D_x , 3.a.
The operators J_± may exist in part to balance the energy so that a steady-state, steady-flow process can occur during the absorption of gravitons by the nucleus and the subsequent release of energy expended by the nucleus in creating a magnetic field that keeps electrons in orbit.  When the electron reverses its intrinsic spin axis at the apex of a turn, it arcs in the opposite direction because its magnetic dipole is reversed, and the magnetic field vector and strength from the nucleus is the same in that locale as just before the turn took place.
From a measurement standpoint E_n looks stable, however the vector operator M ± 1, with 2l + 1 possible integer values of the magnetic quantum number, m_l ([1] Table 10.4, pg 451) , may help keep the energy the same once the uncertainty principle is taken into account.  The shift through integer values of m_l, with [J_z, J_±] = ±ћJ_± ([1] (9.20), pg 358) helps keep the energy of the atom stable.  For example, the energy of the total orbital angular momentum for the D term is 6ћ²/2I ([1] Fig. 9.8, pg 364) through all eigenvalues of L_z, however it cannot be certain at any given instant in time what the value of L_z is for any orbital with higher total orbital angular momentum, L, than the S orbital.
As far as quantum mechanics goes, there does not appear to be anything currently written that will need to change.  For example, take the commutation relations [J_x, J²] = [J_y, J²] = [J_z, J²] = 0  ([1] (9.18), pg 358).  Every soccer match between J² and J_z will continue to result in a tie score and no one will win, or better yet the game is forfeited before it starts.  That is not to say that nothing will get added.  The wave functions going on inside the electron, and how they interact with the Coulomb potential, can be an additional field of physics.
[1] Liboff, Richard L., Introductory Quantum Mechanics, Fourth Edition, Addison Wesley, 2003
[2] Cohen-Tannoudji, Dui, Laloë, Quantum Mechanics, Hermann, 1977, Paris, France

When a stationary permanent magnet holds a steel object, a paper clip for example, against the force of gravity, where does the energy come from to continually hold the clip?  Since work is force times distance, an undergraduate student may say there is no distance moved and therefore no work is done.  On a quantum level however, things are not motionless, and it does take energy to hold the clip.  The answer I think starts with the entry on this blog called “The Nucleus and Gravitons”.  The energy expended through the individual atomic dipole moments of the permanent magnet to hold a ferromagnetic material object against the pull of gravity can come from gravity itself.  The nuclei of the atomic dipoles of the permanent magnet may be absorbing gamma rays at and very near 7.562 x 10²² Hz.

There is another possibility, as opposed to total deflection, for what happens when gravitons encounter a nucleus.  Since the makings of an electron exist inside a neutron, and neutrons and protons have their own spin functions going on inside, it is not outside the realm of possibilities that atomic nuclei, each when part of its own functional atom, absorb gravitons.
What a nucleus in a gravitational field, the earth’s let’s say, would do with all this absorbed energy is not too hard to imagine.  The Coulomb field of an atom is full of electromagnetic wave activity involved with keeping electrons in orbit, and some of that energy may escape.  Therefore we would have gravitational energy replenishing Coulomb energy through both the electrons and the nucleus of an atom.
The synchronization involved as a graviton enters a nucleus would have to be just as smooth as when one enters an electron in a quantum atomic orbital in order for the nucleus to not have been pushed by the absorbed gravitons.  Also, inflow of energy must balance outflow in order for there to be a steady-state, steady-flow process.  In this case it is only those gravitons which pass near the edge of a nucleus that would have to be deflected by its local magnetic field.
If you have read it and remember, my April 2007 paper already has the Coulomb force as the final mediator of the gravitational force.

The wave function of a free graviton can be written in the form
E₀ sin (kx-ωt) or B₀ sin (kx-ωt)
where the wave number k = p/ћ, and the electric and magnetic field components are in phase with each other.
The momentum of a free graviton, which we will now call p_g for purposes seen later, is written as p_g = h/λ_g = 1.671 x 10^-19 kg-m/s.  The fundamental charge, termed e, that of an electron or a proton, is 1.602 x 10^-19 C.  Linear momentum from a free graviton must be transferred into angular momentum as it is absorbed by an electron in a quantum atomic orbital.
In terms of energy, with the numerical value of p_g²/e² at 1.088, generally we have
p_g²/e² = (Const.₁) _S∫ B²∙n dA
The rotational wave function of a graviton as an internal component of an electron may be of the form:
Ψ_rot = (Const.₂) exp [i2πα(eB/p_g)ct],
which would be orthogonal to a corresponding standing wave function.
In Ψ_rot , p_g is the same momentum of the graviton when free and traveling at the speed of light, B is the internal magnetic field of the electron that is perpendicular to a plane that passes through the center of the electron and is perpendicular to its spin axis, and α is the fine-structure constant.  Once released again however, the free graviton likely has no rotational component.  In fact, it must not if a graviton is to be absorbed by an electron in either of two spin orientations.
The imaginary component of the complex wave function of an absorbed graviton will be large enough so that it takes several gravitons to make a full charge.  Additionally, not all the graviton energy will go into charge contribution; some will contribute to the mass of the electron.
Not including spin energy, the equation E² = p²c² + (mc²)² represents the relationship between total energy of a particle, the electron in this case, and its linear momentum and rest mass.  It is fortunate that the numerical value of p_g²/e² is greater than 1 so that we can be more confident that there is an uptick in both mass and linear momentum for an electron in a quantum atomic orbital when a graviton is absorbed.  An interesting concept here is that the electron gains linear momentum in the opposite direction, within a certain conical angle, that the free graviton was traveling.
The rotational energy of an electron was estimated in my April 2007 paper as being 3.06 x 10^-9 J, and elsewhere on this blog the number of gravitons in an electron was roughly estimated as 3.06 x 10^-9 J / 5.011 x 10^-11 J = 61.  Using the reciprocal of the dimensionless fine-structure constant, the average number of gravitons in an electron, in an atomic orbital at the face of the earth, could possibly be 137/2 = 68.5.

 Apr 14 2008, Rev. A: Added a subscript “g” to the representation of the momentum of a free graviton after first mention, and also to the wavelength of a graviton.  Added the paragraph “Not including … was traveling.”

Does an electron swell slightly with mass in the gravitational field of the earth as it scoops up gravitons while traveling in a quantum orbital toward the earth?  Does it then give off multiple gravitons as it makes its next turn?  If so, my guess is that they all fly off with one flip of the electron.  Certainly a slightly more massive electron will need additional momentum balance as it makes a turn, than an electron that is less massive.  I would estimate that an electron in an atomic orbital is never at a point where it cannot accept any more gravitons, because otherwise we could not use the gravitational constant so consistently.
Since the electron is traveling at maybe 0.2c in an atomic orbital, it can scoop up gravitons a lot more often than the nucleus must deflect them by its magnetic field, if that is what the nucleus does.  Since a magnetic field does no work, it is the electric fields and currents inside a nucleus, – those also producing the internal nuclear forces, that would expend the energy to deflect gravitons.  It must then recover that energy through the Coulomb field, which would then recover its energy through the gravitational field.  This is an interesting concept because of the unification principles involved.

For those who come to the knowledge of this theory by this web site, without knowing about the YouTube videos, the YouTube links are at: 

http://www.youtube.com/watch?v=47JNFaltiN4 

http://www.youtube.com/watch?v=wgFBocci8NI&feature=related 

http://www.youtube.com/watch?v=ce1JqoCjjZE&feature=related 

Thank you Dan Fordice, Jeff Fordice, and Fairmont High School for the time spent on this! 

With activity increasing on the use of my calculations, it may be a good time to make a statement on the copyright of my main calculation.
The formula G = 4hf/3 was derived in November 2005.  The formula appears in an autumn 2006 family coat of arms done in a Fairmont public school art class by my daughter, and which hangs at the end of our hallway at home.  The formula also appeared in a January 8, 2007 article in the Fairmont Sentinel, and on a sheet of paper I held up in a KEYC – Mankato television interview aired on January 18, 2007.
The formula G = 4hf/3 can be used one of two ways.  One can use a conventional, measured value of the gravitational constant for G, and calculate the frequency of a graviton.  Alternatively, one can convert exactly one third the proton mass to equivalent energy as a massless photon through Einstein’s equation E = mc², solve for f, the frequency of the photon, through the formula E = hf, and come up with a value of the gravitational constant through my formula G = 4hf/3.  This value of G will then be as accurate as the measured value of the mass of the proton.
For those just starting to learn physics, “h” is Planck’s constant and is equal to 6.626 x 10^-34 J-s.  It is easiest if you keep all units in the ‘kg-m-s’ system for these calculations.

With the emission of a graviton, an electron in an atomic quantum orbit must either undergo a nutation, or reverse its spin orientation as it contributes to the particle angular momentum.
The shape of an orbital that you see at the top of this blog is similar to what many of us picture, however reality may be something different.  Let us say, for discussion purposes, that an electron reverses its spin orientation when it gives off a graviton.  The electron as a particle with half-integer spin changes sign “when the system of coordinates is completely rotated about an axis” ([1], §54).   Should each orbital then include graviton emitting turns of quantity 2n+1, n=1,2,3, the spin orientation of the electron would reverse for every course that brings it back to its orientation and position relative to starting coordinates within d(q-q’).  The total particle angular momentum will then alternate along any prescribed axis between j = l + ½ and j = l – ½.
Another way of stating it is that for half-integer j, Χ(Ф+2π) = – Χ (Ф).  The base function changes sign under a rotation of 2π ([1], §95).
It is the statistical nature of quantum mechanics that has allowed its angular momentum and energy eigenvalue determinations to be very useful in physics.  The average intrinsic spin angular momentum of the electron has thus been able to be used without respect to the emission and absorption of gravitons.
 

[1] Landau, L. D. and Lifshitz, E. M., Quantum Mechanics, Non-relativistic Theory, Translated from the Russian by J. B. Sykes and J. S. Bell, Addison-Wesley Publishing, 1958

The current electron spin g factor given by NIST is -2.002 319 304 3622 *.  Since the earth is in orbit around the sun, and receives its replacement gravitational energy from the sun, this may be why the g factor is greater than 2 in magnitude.  It makes me wonder if the g factor at Jupiter is closer to -2, and greater in magnitude than the Earth g factor at Venus because it is closer to the sun.
The magnetic moment of the electron is tied to this g factor:
μ_S = g_eμ_B(S/ћ),
where μ_S is the magnetic moment of the electron, with units of Amp–meter², and μ_B represents the Bohr magneton.  The spin angular momentum of the electron, S, is ћ/2.
It may be that when l g l, for electrons in a macroscopic mass, drops below 2, atoms start to collapse.  If so, it is certainly possible that electrons near the center of our galaxy have l g l < 2, with atoms collapsing at a slow rate. * http://physics.nist.gov/cgi-bin/cuu/Value?gem|search_for=g+factor

According to a Wikipedia article on Comet McNaught, NASA and ESA’s Ulysses spacecraft “made an unexpected pass through the tail of the comet”, where its Solar Wind Ion Composition Spectrometer (SWICS) “measured the composition speed of the comet tail and solar wind”.
The article explains that sensor data showed that oxygen ions in the solar wind had picked up electrons to replace some of their missing electrons as the ions passed through the tail of the comet.  It goes on to say that “the tail had slowed the solar wind to half its normal speed.”
Space science professor Michael Combi is quoted as saying:  “This was very surprising to me. Way past the orbit of Mars, the solar wind felt the disturbance of this little comet. It will be a serious challenge for us theoreticians and computer modellers to figure out the physics,” *
Here we learn that as atomic ions traveling radially outward from the sun in the solar wind pick up electrons and incorporate them into atomic orbitals, the ions slow down.  For anyone willing to learn some physics, this is because after the ions pick up electrons, they collect more gravitions.  The ions will slow down due to increased gravitational acceleration back toward the sun.
* http://en.wikipedia.org/wiki/Comet_McNaught