| Topic: |
Science > Physics |
| User: |
"OsherD" |
| Date: |
04 Aug 2005 06:46:32 AM |
| Object: |
Feynman Path Integral From T3/T2 = kT1 |
From Osher Doctorow
COPYRIGHT NOTICE
Feynman Path Integral From T3/T2 = kd(T1)
Copyright By Owner Osher Doctorow Ph.D.
First Published 2005
The exponential kernel (integrand) of the Feynman Path Integral
arguably arises as follows.
With T1, T2, T3 respectively the three PI time dimensions of duration,
causation, and transfer of causation, let's take transfer of causation
(T3) as d(causation) = d(T2), and let's take duration as d(T1) instead
of T1 where now the T1 in d(T1) is the "duration-time integrand" which
is the usual clock-time.
Consider the equation:
1) T3/T2 = kd(T1)
which translates from the last paragraph as:
2) d(T2)/T2 = kd(T1)
Integrating both sides yields:
3) T2 = causation = k1 exp(kT1)
where k1 is exp(constant of integration).
The Feynman Path Integral is then arguably the summation/integral of
causations.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Feynman Path Integral From T3/T2 = kT1 |
04 Aug 2005 06:57:14 AM |
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From Osher Doctorow
The dimensionality of d(T2)/T2 = kd(T1) requires that since the left
hand side is dimensionless (because the dimension of T2 "cancels"),
then the right hand side has to be dimensionless, so that k is a
dimensional constant of dimension 1/time. This is not in accord with
avoidance of dimensional constants in my recent PI postings, but other
than regarding duration time T1 as dimensionless, it seems to be
required here.
Osher Doctorow
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