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Quantum Gravity 345.3: Resolution of the Quandary By An Arbitrary Probability = 1 or = k(t) constant at time t assignment

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OsherD

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Nov 25, 2009, 8:27:42 AM11/25/09
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From Osher Doctorow

The "quandary", as I refer to it, is the fact that a more complicated
pattern appears for:

1) P(A1 ' --> A2 ' --> ... --> An ' ) for n > 2 integer

than appears for:

2) P(A1 ' --> A2 ' ).

However, we can obtain an approximation for the pattern of P(A1 ' -->
A2 ' ) with n > 2 in (1) by arbitrarily setting:

3) P{(A1 ' --> A2 ' --> ... --> An ' ) U (An ' --> A1 ' ) } = k(t),
where k(t) is a "constant" in (0, 1] at time t (taken as 1 below for
simplicity, although in general it may be different from 1).

We can then prove that:

4) P(A1 ' --> A2 ' --> ... --> An ' ) = 1 - P(An ' --> A1 ' ) + P(A1 '
<--> A2 ' <--> ... <--> An ' )

which resembles the pattern of (2) if we note that P(An ' --> A1 ' ) =
P(A1 --> An). The pattern of equation (2) is from QG 344.91:

5) P(A1 ' --> A2 ' --> A3 ' ) = 1 - P(A1 --> A2 ' ) + P(A2 ' A3 ')
where P(A2 ' A3 ' ) is the unbounded part of P(A2 ' <--> A3 ' ).

To prove (4) using (3), simply set P(E U F) for any (random) sets E,
F) equal to P(E) + P(F) - P(EF) which is a law of probability, and
then set P(E U F) = 1, and replace E U F by the set unon in (3),
etc.

By the way, this also gives us a hint as to the possible error in
Einstein's selection of c = constant for the speed of light. If we
normalize c to 1, then we can regard c = k(t) in (0, 1] as time-
dependent, which although theoretically possible to be constant = 1
throughout time, is not necessarily so (for example, in the early
Universe, although arguably later too). This becomes even more
plausible if we replace c = k(t) by P{c in a < = k(t) < = b for a, b
in (0, 1]} so that we are dealing with probable speeds or velocities
rather than speeds or velocities.

Osher Doctorow

OsherD

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Nov 25, 2009, 8:35:05 AM11/25/09
to
From Osher Doctorow

The pattern of (2) from 344.91 actually refers to n = 3 rather than n
= 2:

1) P(A1 ' --> A2 ' --> A3 ' )

where in 344.91 I used the letters A, B, C instead of A1, A2, A3
respectively.

Osher Doctorow

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