Science > Physics > Quantum Gravity 149.4: P(F + A --> m) = 1, P(F + A --> q) = 1 in Newton and Coulomb
| Topic: |
Science > Physics |
| User: |
"OsherD" |
| Date: |
02 Jun 2007 02:47:44 PM |
| Object: |
Quantum Gravity 149.4: P(F + A --> m) = 1, P(F + A --> q) = 1 in Newton and Coulomb |
From Osher Doctorow
We saw recently that the Ideal Gas Law in Probable Influence/Causation
terms is equivalent for "all else constant (k)" to:
1) P(F-->temperature or E) = k + P(A-->V)
where A is area, V is volume (so P(A-->V) = 1 is the Holographic
Principle).
Newton's Law of Universal Gravitation and Coulomb's Law for electric
charges yields similar results for temperature or E replaced by mass m
or charge q, except that this time F and A are combined additively:
2) P(F + A --> m) = 1, P(F + A --> q) = 1
We have:
3) F = Gm1m2/r^2 converts in PI to F = G + m1 + m2 - r^2
and therefore:
4) (G + m1 + m2) - (F + r^2) = 0
Add 1 to both sides:
4) 1 + (G + m1 + m2) - (F + r^2) = 1
Therefore:
5) P((F + r^2) --> m1 + m2) = 1
We can leave the equation in this form or combine m1 + m2 into m. r^2
is dimensionally an area A rather than a volume, which we can regard
as proportional to the surface area of a sphere centered at one of the
mass's c.g., so we have:
6) P((F + A) --> m) = 1
Similarly we can convert Coulomb's Law of electrostatics:
7) F = kq1/q2/r^2
to:
8) P((F + A) --> q) = 1
Osher Doctorow
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| User: "chumley" |
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| Title: Re: Quantum Gravity 149.4: P(F + A --> m) = 1, P(F + A --> q) = 1 in Newton and Coulomb |
03 Jun 2007 12:08:32 AM |
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"OsherD" <mdoctorow@comcast.net> wrote in message
news:1180813664.676202.276260@n15g2000prd.googlegroups.com...
From Osher Doctorow
We saw recently that the Ideal Gas Law in Probable Influence/Causation
terms is equivalent for "all else constant (k)" to:
Kosher, you are filled with Ideal Gas, it is causationing out both ends all
the time.
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