Re: E = wLF Derived By Modified Quantum Logic
From: OsherD (mdoctorow_at_comcast.net)
Date: 01/24/05
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Date: 23 Jan 2005 19:38:13 -0800
Notice that:
1) p(x,z) = 1 + z - x < = p(x,y) + p(y,z)
since we haveL
2) 1 + z - x < = (1 + y - x) + (1 + z - y) = 2 + z - x
Similarly:
3) p2(x, y) = 1 + (y1 - x1)/2 + (y2 - x2)/2 < =
p2(x,z) + p2(z,y) = (1 + (z1 - x1)/2) + (1 + (z2-x2)/2)
+ (1 + (y1 - z1)/2) + (1 + (y2 - z2)/2) = 4 + (y1-x1)/2 + (y2 - x2)/2
and it is easy to prove similarly the triangle inequality for pn(x,y) =
pn((x1,...,xn), (y1,...,yn)).
So pn(x,y) for n = 1, 2, 3, ... obeys the triangle in-
equality and nonnegativity and is in [0, 1] and increases from 0 to 1
as the Euclidean type distance d(x,y) decreases from 1 to 0.
We could, of course, as in much of functional analysis, concentrate
only on the unit ball or here the unit n-cube in a functional space, or
concentrate on Rare Events which would correspond to P(A) < = .05 and
P(B) < = .05 in the quantity P(A-->B) (or one or both of these last
two) which would select xi and yi in a subset of [0, 1] for i = 1, 2,
..., n. If we want to expand this to the non-Rare scenario without
trying to map [0, infinity) to
[0, 1), we could choose a very large distance (larger than the farthest
observed distance at a particular time in history) D and normalize
distances by D to put them in [0, 1] for Euclidean-type distances. Or
we could agree to use the Euclidean type unit n-cube as a model of a
[0, infinity) X [0, infinity) X ... X [0, infinity) space.
Proximity pn(x, y) involves differences of coordinates yi - xi and so
can be regarded as dimensionally length L (1 would be a unit-valued
quantity with unwritten length dimension in pn(x,y)), although P(A-->B)
is not expected to have all three aspects of dimensionless, probability
dimension P, and length L simultaneously, so we expect a dimensional
constant k1 such that [P(A-->B)], the dimension of P(A-->B) = pn(x,y),
is:
4) [P(A-->B)] = P = k1L = k2(dimensionless probability)
where k2 is another dimensional constant. Since ww* = P, taking [w] =
P^(1/2) would give us (using L for a length/distance variable
corresponding to dimension L if confusion is unlikely):
5) w = P^(1/2) = (k1L)^(1/2)
Continuing in this way in the expression:
6) Ew^(-1)L^(-1)F^(-1)
we could reduce w and L to variable functions of L and therefore for a
constant dimensionless ratio of type (6) express E and F in terms of
only variable L. This puts us in range of our desired results, but
I'll leave that for another time.
Osher Doctorow
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