Fruechte Gravity Theory Blog

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.

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

I think of the strong nuclear force as being due to the vortices* from an electron setting up standing waves within the protons and neutrons within the nucleus of an atom.  For anyone willing to make a try at the math, the Schrödinger equation may be a good place to start.  These standing waves, it is presumed, set up quite nicely within an alpha particle since an alpha particle is very stable.
As a possible consequence, it may be that all nucleons within an intact nucleus have roughly equal positive charges.  In the case where a neutron is ejected from a nucleus, it would gather all the vortices it needs as it takes off and becomes of neutral charge.

*  Kadin, A. M., “Circular Polarization and Quantum Spin: A Unified Real-Space Picture of Photons and Electrons”, ArXiv Quantum Physics preprint, 2005:
http://arxiv.org/ftp/quant-ph/papers/0508/0508064.pdf