Apr
27
2008

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 10^{22} Hz.

Apr
10
2008

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.

Apr
01
2008

The wave function of a free graviton can be written in the form

E_{0} sin (kx-ωt) or B_{0} 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_{g}^{2}/*e*^{2} at 1.088, generally we have

p_{g}^{2}/*e*^{2} = (Const._{1}) _{S}∫ B^{2}∙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(*e*B/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 E^{2} = p^{2}c^{2} + (mc^{2})^{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_{g}^{2}/*e*^{2} 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.

I apologize for the way pi and alpha transfer from my word processor documents into this blog. Pi can look like an n, and alpha can look like an a, and be misread if not compared to n and a elsewhere.

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.”