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building the proton from magnetic monopoles Chapt13.4091 Review of DTW theory #1069 New Physics #1189 ATOM TOTALITY 5th ed
Posted:
Dec 8, 2012 12:49 AM


A single transverse wave is pictured as this with its E and B:
E  ___ B
While a double transverse wave is pictured this with its E and M's:
E M____M+ ?  ? E+
If we examine that structure carefully we see the z and y axis taken up by the E's and M's and the xaxis as the origin point coming out of the screen towards our eyes.
The M's and E's do destructiveinterference and thus allows all photons to have the same constant speed regardless of either wavelength or frequency.
So Special Relativity is insured by a Double Transverse Wave, not by a Single Transverse Wave.
The problem then, is, how do we add M's to the electron to get it from 0.5 MeV to that of the proton of 938 MeV?
Well, the only true answer must come out of the Maxwell Equations and they provide us with the Faraday law as that of the Double Transverse Wave itself.
In the picture above of the E, E+, M and M+ if we deleted the E+ we have a electron as this:
E M____M+
And instead of 4 vertices we have only the 3 vertices. Now actually that picture is not the Faraday law but the Ampere law where the E is a electric current which produces the perpendicular M and M+ as the dipole magnetic field caused by the E current. So if I altered that around a little bit we get the Faraday law on the Double Transverse Wave:
2E M+__  E+
What I have done is still keep a electron with 3 vertices instead of a 4 possible, and where the overall electric charge is still 1 negative. The E's form a closed loop such as a closed loop of wire in Faraday's law and the M+ acts as the bar magnet in Faraday's law, yet always keeping in mind that this is a wave. So the wave itself is conducting and obeying Faraday's law and Ampere's law.
Now how do I get the neutrino, photon and electron out of this wave structure of 4 vertices and double transverse wave? Well, I get it by adding more and more magnetic monopoles of M and M+ and they tend to clump together as dipoles on each vertice, so that 0.5*10^6 M means half of the M's are + and half , except one more M to make the overall number to be  magnetic charge.
So as a pinwheel going around of the 4 vertices, we add more and more M's to form neutrinos, photons, and electrons.
But the difficulty is forming the proton.
And the answer must come from the Maxwell Equations. And not surprisingly the Maxwell Equations do not speak to protons in either the Faraday or Ampere/Maxwell laws. Not even in the symmetrical Maxwell Equations. About the only place that positive electric charge is an issue is in electrochemistry of positive ions in motion.
So, what I am proposing is very bold solution, and I am wavering on whether it stands to be true or false. It is so bold that I am sort of looking for another solution. But to cut the chase, let me outline the bold solution.
The Double Transverse Wave is 2 dimensional of the y and z axis and the particle moving forward along the xaxis. What if the xaxis itself was not a straight line but a curved arc. When we see trajectories of electrons in bubble chambers they are easily recognized since they are arcs, not lines. So that for the proton, the y and z axis are double tranverse waves but the xaxis is also a arc of a tiny circle or sphere.
So that the electron like the photon or neutrino can have MeV of 0.5, whereas no proton can have 0.5 MeV restmass, but must have 938MeV restmass because it has lined up or built up a collection of 938*10^6 M's that forms a arc of a circle that is a sphere of the proton.
The electron or photon or neutrino, in free space can go off in a straight line or a gradual arc for the electron, but the proton nevers leaves its sphere formed from its 938*10^6 M's along its xaxis that imprisons the proton into that sphere space.
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Archimedes Plutonium http://www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electrondotcloud are galaxies



