Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: building the proton from magnetic monopoles Chapt13.4091 Review of
DTW theory #1069 New Physics #1189 ATOM TOTALITY 5th ed

Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  
plutonium.archimedes@gmail.com

Posts: 9,924
Registered: 3/31/08
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
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply


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 x-axis as the origin point coming out of the
screen towards our eyes.

The M's and E's do destructive-interference 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 x-axis. What if the x-axis
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 x-axis 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 rest-mass, but must have 938MeV
rest-mass 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 x-axis that
imprisons the proton into that sphere space.

Google's New-Newsgroups censors AP posts and halted a proper
archiving of author, but Drexel's Math Forum does not and my posts?in
archive form is seen here:

http://mathforum.org/kb/profile.jspa?userID=499986

Archimedes Plutonium
http://www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.