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Natural Science Forum / Physics / General Physics / July 2007



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Quantum Gravity 168.0: Cross-Phase Desingularizing/Decoding

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OsherD - 31 Jul 2007 02:17 GMT
>From Osher Doctorow

In Section 167, we saw that eliminating the square root (that is, half
power) in x^(1/2) to obtain x^1 = x optimizes Probable Causation/
Influence (PI) in the Special Relativity beta or gamma factor.   The
function which does this when x > = 0 is:

1) y = f(x) = x^2

This gives us an additional insight into the Riccati Differential
Equation which is the fundamental differential equation of PI:

2) dy/dt = A(t) + B(t)y + C(t)y^2

at least regarding the y^2 term.   Here x = t.

For real variables f(x) which are centered at 0 (any expression like
f(x - k) for k constant can be translated to 0), y = x^(1/2) is one of
several functions which yield singularities at or below 0, the main
ones being:

3) f(x) = x^(1/2)
4) f(x) = 1/x or k/x (k real constant)
5) f(x) = log(x)

Since there arguably really are singularities in physics, mathematical
singularities don't have to be eliminated at least for real variables,
but they do demarcate different phases.   To get the analog of a
particular variable in a different phase, we arguably eliminate the
singularity in the simplest way possible.

Thus, letting --> indicate what something becomes in the PI "phase",
we have respectively for (3)-(5) above:

6) x^(1/2) --> x by y = f(x) = x^2 for x > = 0
7) 1/x --> 1 - x by u = f(y/x) = y - x
8) log(x) --> x by y = f(x) = exp(x), since exp(log(x)) = x if x > = 1

Not only is the function y = x^2 of (6) found in the Riccati
Differential Equation, but the function y = exp(x) is a solution of
the special Riccati Differential Equation (with x = t):

9) dy/dt = B(t)y,  B(t) = k (real constant)

That leaves only f(y/x) = y - x, which is how conditional probability
y/x for x not 0 becomes Probable Causation/Influence 1 + y - x for 0 <
= y < = x < = 1.  Note that in this case only, y and x have to have no
common factors for uniqueness.

Osher Doctorow
Dear Leader - 31 Jul 2007 03:06 GMT
> >From Osher Doctorow
>
> In Section 167, we saw that eliminating the square root (that is, half
> power) in x^(1/2) to obtain x^1 = x

very clever, but is it the positive root or negative root?
 
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