Jun 30 2010

## Goce satellite

Hi Dan,

Traveling and the email does not go through. Thanks for the BBC article link: http://news.bbc.co.uk/2/hi/science/nature/8767763.stm

Jun 30 2010

Hi Dan,

Traveling and the email does not go through. Thanks for the BBC article link: http://news.bbc.co.uk/2/hi/science/nature/8767763.stm

Since the earth has a magnetic field and gravitions like to follow magnetic field lines, it makes sense that the gravitational field varies slightly over the surface of the earth. The article says that gravitational acceleration varies from 9.78 m/s^2 (minimum) at the equator to 9.83 m/s^2 (maximum) at the poles. Looking at a typical pattern of magnetic field lines of a dipole magnet, one would expect a greater gravitational acceleration at the poles of the earth compared to the equator.

Jun 23 2010

It happened again; a Greek letter did not come through properly. In “Subatomic Particle Structure” it was the letter nu. I try to be very careful when copying referenced formulas into the blog, and the character was not a mistake, just a software incompatibility. Sorry.

Jun 22 2010

Research on the internal structure of subatomic particles has been ongoing since the early days of particle accelerators and cosmic ray experiments, and there are several points of view that merit study. The view promoted here has not much new, and should sound familiar. The very stable proton, as CERN would agree, is a worthy starting point.

First of all, “there is *no hard core* to the proton. … It could be like jelly, or it could be like a strawberry, with seeds scattered throughout, but *no accumulation* of them at the centre.” When scattering e^{–} + p is elastic, the resulting Bjorken scaling “is interpreted as indicating that the scattering takes place off point like constituents of the proton, called partons.”, where “the structure factors are functions of ω only.” ([1], pgs 16-18).

Since “q is the photon 4-momentum”, for “an individual parton of mass m” we can use the relativistic mass of the graviton, m_{g} = 5.575 x 10^{-28} kg [2], in which case the dimensionless ω = (2Mν)/q^{2}, with q^{2} = 2mν ([1], pg 17), gives ω = m_{p}/m_{g} = 3.

One may think of the internal structure of the proton as a crystal lattice, with nodes, or partons, that are basically zero points of eigenvectors, where internal and pass through wave functions add or subtract momentum. If you are a string theorist, added head to tail you may choose to look at the vector structure as vibrating strings. If you have a Mechanical or Civil engineering degree you may think of the structure as finite elements, – not necessarily tetrahedral however. The nodes are the seeds of Ryder’s strawberry.

The defined boundary of a subatomic particle results from gravitational pressure at its barrier domain. Let us view then the particle as a wave packet “when the wave packet is subject to the influence of a parabolic potential. Physically, this result arises from the fact that the tendency of the wave packet to spread is compensated by the potential, whose effect is to push the wave packet towards the origin from regions where V(x) is large.” ([3], Compl. G_{v}, 3.c., pg 572) Due to gravitational pressure, V(x) would be large at the barrier domain of the particle. With consistent frequency, ω_{g} = 4.75 x 10^{23} rad. sec^{-1} [2], gravitons step through the barrier easily.

Baryons and fermions would be similar in structure, though differing greatly in density. Scattering can happen through barrier elasticity or partial penetration with node to node scattering. Deeper node penetration on a large nucleus can result in alpha decay.

[1] Ryder, Lewis H., Quantum Field Theory, Second Edition, Cambridge University Press, 1996

[2] https://www.fruechtetheory.com/blog/2010/06/15/muonic-states/

[3] Cohen-Tannoudji, Dui, Laloë, Quantum Mechanics, Hermann, 1977, Paris, France

Jun 15 2010

In converting the graviton energy to mass, we can apply 5.011 x 10^{-11} J = m_{g}c^{2}, or m_{g} = 5.575 x 10^{-28} kg, though due to relativistic effects the graviton is never converted completely to mass in the natural world. The muon mass, m_{μ} = 207 m_{e} = 1.86 x 10^{-28} kg ([1], Compl. A_{v}, 4.c., pg 527), is m_{g}/3 and m_{p}/9, where m_{p} is the mass of the proton.

To quote the Frenchmen: “A muon μ^{–} which has been slowed down in matter can be attracted by the Coulomb field of an atomic nucleus and can form a bound state with the nucleus.” ([1], Compl. A_{v}, 4., pg 525). Before going any further then, and at the risk of sounding trite, like the quark and positron, the muon’s real identity may be that of a graviton.

Taking the concept of mass further, a conjugate wave graviton would slow down much more when passing through a proton or another nucleus than when it passes through an electron. This increased compaction helps explain why the proton mass is 1836.5 times the electron mass.

Again relating to the muon, “the spread of the ground state, if the well were perfectly parabolic, would be on the order of: √(ћ/(2m_{μ}ω)) ≈ 4.7 x 10^{-13} cm” ([1], Compl. A_{v}, 4.c., pg 528). This of course is very close to the wavelength of a free graviton: 3.965 x 10^{-13} cm.

The frequency of a muon is also close to that of a graviton:

ω ≈ 1.3 x 10^{22} rad. sec^{-1 }([1], Compl. A_{v}, 4.c., pg 527),

whereas equivalent units for a graviton come out as:

ω_{g} = (7.562 x 10^{22} Hz) (2π rad/cycle) = 4.75 x 10^{23} rad. sec^{-1 }

It cannot be overlooked that the rotational component of a conjugate wave graviton passing through a fundamental particle may be zero when considering gravity alone. Similar to the metal pump tops sold in the ‘50’s and ‘60’s, of which I remember operating one as a boy, the pump action is linear and the top spin rotational. Should there be a rotational action of a pass through graviton, it may be related only to Coulomb field production. This Coulomb field action, produced by charge, is in total balance. Griffiths states it this way: “… plus and minus charges occur in *exactly* equal amounts, to fantastic precision, in bulk matter, so that their effects are almost completely neutralized. Were it not for this, we would be subjected to enormous forces: a potato would explode violently if the cancellation were imperfect by as little as one part in 10^{10}.” ([2], pg xiv)

While it is difficult to find in books or on the web how fast the Coulomb force is transmitted, my guess is that it is nearly instantaneous, transmitting through the gravitational field in wave packets at group velocity, by phase shift and chirality, the combination of which determines positive and negative charge.

[1] Cohen-Tannoudji, Dui, Laloë, Quantum Mechanics, Hermann, 1977, Paris, France

[2] Griffiths, David J., Introduction to Electrodynamics, Third Edition, c. 1999, Prentice-Hall, Inc.