**Stephen P. King** (*stephenk1@home.com*)

*Mon, 16 Aug 1999 08:59:05 -0400*

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Hi Bill,

My response is at the end.

WDEshleman@aol.com wrote:

*>
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*> In a message dated 8/13/99 11:26:39 PM Eastern Daylight Time,
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*> stephenk1@home.com writes:
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*>
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*> > What I speculate on is the possibility that
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*> > > signals
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*> > > are copied into MWs; ie, matter reveals itself as photons in other
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*> worlds
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*> > and
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*> > > photons in our world reveal matter in other worlds.
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*> >
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*> > Umm, are you familiar with the supersymmetry transformation that
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*> > involves the transformation of bosons (such as photons) into fermions
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*> > (such as electrons, protons, etc.) and vise versa. I have always
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*> > wondered why such a beautiful symmetry is not experimentally obvious.
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*> > Maybe because we are looking too hard for it! :-) In my thinking the
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*> > Universe objects are composed of quantum systems (no "ultimate
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*> > indivisible particles") to for local systems, these quantum systems
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*> > would, if we suppose that the "Super Poincare" symmetry is real, have
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*> > both "matter" and "photon" properties. Now, what if we fail to see the
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*> > multitude of particles that the usual interpretation of supersymmetry
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*> > generates for the same reason that we do not see the other worlds?
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*>
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*> Stephen,
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*> Please, my mind cannot explain as fast as you can understand. :-)
*

Ok, I just see think in pictures. My "resolution" is not too great

sometimes. :-)

*> I prefer to think of the sequence forms of my infinite product as
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*> "Broken Lorentz Symmetry Transformations" which leave invariant
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*> the form:
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*>
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*> (x^(2^n))^2 + (y^(2^n))^2 + (z^(2^n))^2 - (R^(2^n))^2.
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*>
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*> The infinite group of Lorentz factors is therefore:
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*>
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*> 1/(1 - (B^(2^n)))^(1/2^n) where n=0,inf & 0<B<1
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*>
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*> or,
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*>
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*> 1/(1-B), 1/(1-B^2)^(1/2), 1/(1-B^4)^(1/4), 1/(1-B^8)^(1/8),
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*> 1/(1-B^16)^(1/16), ...
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*>
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*> The infinite product for 1/(1-B) contains all of these factors and is easily
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*> transformed into them by moving factors from the right side to the left
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*> side of the identity Theorem VI.
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Umm, Could you give us some explicit examples of calculations using a

value of B, say .123 and .88? I have a local off-line friend that is

very interested in this idea! :-)

*> I presently think of the bosons as being represented by the factors of the
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*> infinite product for 1/(1-B), not the Lorentz factors. Feel free to speculate
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*> if you would, as I am in much need of some creativity. Some sites
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*> concerning super-symmetry would be good too. I'm not dumb,
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*> just ignorant.
*

:-) To tell the truth, I have a bit of trouble understanding completely

the picture represented by your infinite product, but it is fascinating!

I start with your notion "that the Lorentz factor is itself the result

of a product of many small signals reflected from the multitude." Now,

can we think of this as how the point to point translations (or motion

transformations in general) are 'built' up from the infinite number of

contributions of the "multitude" of "other worlds"?

Question is: How do the contributions "add up" so that only a finite

quantity remains? Are the contributions "weighted" such that a finite

number add and the remainder destructively interfere with each other?

Can we think of the worlds as the Fourier transforms of the particle

distribution "patterns" and superpose (Fourier synthesis) them and

transform back to get the result? How do we think of the statistics

(Fermi & Bose) with regard to this operation, if it works? :-) I an just

trying to eliminate possibilities... :-)

Here are some sites on supersymmetry that I found:

http://kaluza.physik.uni-konstanz.de/2MS/vn/vn/node2.html

http://www.suite101.com/article.cfm/advanced_physics/20702

http://supersymmetry.com/

http://203.120.175.2/journals/mpla/1337/0150.html

http://aj.encyclopedia.com/articles/12487.html

http://wwwhephy.oeaw.ac.at/p3w/theory/susy/intro.html

http://www.desy.de/user/projects/Physics/casimir.html (this might be

useful in testing your notion!?)

I believe that Matti understands the mathematics of supersymmetry

better that I do, I am just an ignorant philosopher. :-)

Onward,

Stephen

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