[time 572] Re: [time 546] Fermions & Bosons & Supersymmetry

Wed, 18 Aug 1999 23:43:22 EDT

In a message dated 8/16/99 8:56:14 AM Eastern Daylight Time,
stephenk1@home.com writes:

> 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! :-)

This should help:
What is the entropy in bits at B = 0.123
of the Lorentz factor?
1/(1 - 0.123) = 1.14025
This real number has zero entropy unless
it can be partitioned in some way. It is
decided to partition the log of 1.14025.
Now, writing out only first factors and terms,
1.14025 = (1 + 0.123) * ...
Taking a logarithm,
log(1.14025) = log(1.123) + ...
A distribution whose normalization is then
to be interpreted to be the probability distribution:
1 = log(1.123) / log(1.14025) + ...
1 = 0.883849 + ...
The Shannon entropy in bits uses log base 2 so,
S = -(0.883849) * log(0.883849) / log(2) + ...
S = 0.157438 + ... = 0.6418 bits
Where 0.6418 was obtained by the computer
routine printed near the end of my paper.
Now since B = GM/R/c^2 and v^2/c^2,
this would be equivalent to the mass of our Sun
compacted inside an R = 12 kilometer sphere
where small masses orbit at 105145 km/sec.

I cannot address what happens at B = 0.88
since it is inside the event horizon at B = 0.7035

> :-) 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"?

The method shown above assigns an entropy to all positions in a
space containing mass and assigns an entropy to a test mass
moving relatively inside that space. That is, B =GM/R/c^2 means
that given R what is S(R)? And, B = v^2/c^2 means given v
what is S(v)? Transformations of position and motion then means
that we need to know the inverse of S to get R and v. The computer
finds S given R or v by successively guessing B until the desired
S is found. Circular orbits are the result of zero transformation
of a single point in the S vs B plane. Elliptical orbits appear to be
circular transformations on the S vs B plane.

> 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... :-)

Although I have not addressed the Quantum Mechanics of fermions and
bosons, I see that the ensemble may be analogous to the degenerate
cosmological object in my paper with B very near unity. As I show,
the probabilities become 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... which
yields the degenerate S = 2 = 1/2 + 2/4 + 3/8 + 4/16 + ... + n/2^n + ...
bits. For 2^N particles in this universe we assign an entropy of N,
for 2^N copies in 2^N worlds as N approaches infinity the "weighting"
of "probability halving" results in S = 2 bits instead of many-bits.
Order out of chaos so-to-speak; and we don't get burned to a crisp.

Remember that I am speculating a consistent analysis of a
possible non-fact, a game.



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