New concept: CG Fragments

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17 years 9 months ago #16351 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
I suffered a flash of inspiration on this subject over the weekend. I'll have to keep this short, as my home computer is down and I have only 40 minutes left to compose at the keyboard on this public-library computer.

Suppose there are at least four kinds of ultramundane particles; call them A, B, C & D, and three types of ordinary matter; I go with TV's "MI" for particles with weight, Q1 is one type of charge, the opposite charge is Q2; I don't know, yet, which is positive and which negative.

Now, suppose a pair of A's encounter a Q1, coming together at a very narrow angle. The Q1 acts a catallist, enabling the two a's to fuse into a B by adding very slightly to the B's momentum. This imparts a momentum to the Q1 in the direction opposite to the relative velocity.

Applying the same reasoning by which CG's cause gravity by giving positive momentum to an MI, this negative momentum would cause a pair of Q1's to repell eachother.

Hurrying right along: Q2's don't affect A's, but a pair of B's (or maybe it's an A and a B) coming together at a very narrow angle encounter a Q2 and are fused into D (or C), imparting negative momentum to the Q2. So a pair of Q2's will repell eachother.

Since a Q1 is a source of B's and a Q2 converts B's into D's (or c'C), which have no effect on Q1's, a Q1 will attract a Q2 and vice versa. So like charges repel and opposite charges attract. [There is a problem with this paragraph; first time I've put any of this in writing; running out of time; I'll get back to this tomorrow, hopefully.]

D's (or C's) are Classical Gravitons. When a CG encounters an MI it splits into A's. The MI gains positive momentum from the fission of CG's, which accounts for its positive gravity.

A simple momentum vector diagram would clarify how conservation of momentum and energy dictates the above momentum transfers between the ultramundane particles and the ordinary matter particles. Unfortunately, I lack the time and editing power to show that at this time.

As with comets, there are no 2-body captures among ultramundane particles. The presence of the third body---Q1, Q2 or MI---is the catallist necessary for fusion or fission to occur.

This model is in its infancy; there is much hard work to be done in quantifying it. My math skills were never that good and they've been accumulating rust for decades. I suggest that, initially, anyone interested should attempt to make the model work using quanta of momentum. In reality, we're probably looking at normal distributions which look like quanta because of the large numbers of events per second. I suspect that A's, B's, etc. are actually nano-universes, each one as complex as our own bubble-bath of great walls of galaxies.

I better upload this now, incase I have trouble.

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17 years 9 months ago #16402 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Since a Q1 is a source of B's and a Q2 converts B's into D's (or C’s), which have no effect on Q1's, a Q1 will attract a Q2 and vice versa. So like charges repel and opposite charges attract. [There is a problem with this paragraph; first time I've put any of this in writing; running out of time; I'll get back to this tomorrow, hopefully.]<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
The problem I mentioned yesterday is this: If a Q2 can only fuse two B's into a D, then Q2 would be attracted to Q1, but there would be no equal and opposite attraction of Q1 toward Q2. For Q1 to be attracted to Q2, Q2 must appear as a dark spot against the background of A's. This suggests that Q2's do catalyze fusion of an A & B to form a C.

Does an MI split C's into A-B pairs? B's into A-A pairs? Can it also split a C into three A's. Do D's exist, at all?

If an MI is a source of both A's and B's, then both Q1 and Q2 should be attracted to the MI. The MI should also be attracted to Q2. But, since Q1 has no effect on C's, the MI is neither attracted to nor repelled form Q1. This seems to violate the principle that every action has an equal and opposite reaction; perhaps the reaction is indirect--via an intermediate particle. This difference between Q1 and Q2 should hint at which Q is positive and which is negative. Maybe it relates to electro-strong and electro-week forces---about which I know zilch.

Yesterday, I mentioned a simple momentum-vector diagram. To elaborate: Let's suppose that two A's converge on a Q1 at an angle, Theta, in the inertial coordinates with origin at Q1, each having momentum rho. The vector sum of the A's is 2rho cos(Theta/2). After fusion occurs, the B departs along the line bisecting the continuation of the original paths. To conserve energy and momentum, the B must have momentum equal to 2rho, and the balance of the momentum--- 2rho(cos(Theta/2) - 1) is imparted to Q1. Actually, this does give Q1 a slight energy boost, maybe even enough to explode the odd planet---just to make TV happy.

The diagram for Q2 and MI collisions is only slightly more complex, but presently, my poor brain is balking at tackling even that simple geometry.

I realize that my ABC's are a pretty lame excuse for naming these things. I welcome suggestions. Maybe they should have sexy names like Shiva and Vishnu.

Is this my one chance at fame? Just in case this proto-model is ever developed into a full-blown unified field theory or a theory of everything, I think I should reveal my real identity: I am Philip Stanley Janes, 61 years old, presently residing on the Pacific Coast of Washington State, USA. Just remember I planted this seed of an idea, and that I was inspired by TV and, of course, LeSage.

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