Quantum Gravity 260.0: The Real Interpretation of F = Gm1m2/r^2
- From: OsherD <mdoctorow@xxxxxxxxx>
- Date: Sun, 25 May 2008 14:37:54 -0700 (PDT)
From Osher Doctorow
A rather surprising fact about variables that are normalized to take
values in [0, 1] is the fact that the equation:
1) u = y/x
attains its maximum 1 for u when y = x, regardless of how big or small
x or y are (except that the equation does not allow x = 0). This is
not the way that most people read such an equation, since they are
used to assuming that the bigger y and/or the small x, the bigger u
is.
The same thing happens for the Probable Causation/Influence version of
(1), namely:
2) u = 1 + y - x
which attains its maximum when y = x regardless of what values x and y
have, although this time x can be 0 (and so can y).
The Newtonian equation:
3) F = Gm1m2/r^2
therefore, when interpreted as an equation between normalized
variables in [0, 1], has the rather surprising implication that:
4) max(F) occurs when Gm1m2 = r^2
and in the PI version, r can even be 0 if Gm1m2 is 0, which would
occur for one or both of the masses m1, m2 = 0..
Let us ignore G momentarily (say, by setting units in which G = 1) and
examine (4):
5) m1m2 = r^2 (G = 1)
It is quite a curious coincidence that the mass dimensions (2) equal
the length or distance dimensions (2). If m1 were equal to m2, we
would get:
6) m1^2 = r^2 (G = 1, m1 = m2)
which would lead us to suspect (to say the least) that:
7) m1 = r
If m1 does not equal m2, then from (5), we get:
8) r = sqrt(m1m2) (a type of mean value of m1 and m2)
Once again, mass seems to be fundamentally related to distance in each
of these equations, namely the distance between the masses with the
"deepest" case in a sense occurring when the masses are both equal.
What in the world does this mean?
Notice that if one of the masses, say m1, is taken at the center of a
sphere and m2 is taken at its surface, then r^3 times a constant is
the volume of the sphere, and r^2 times a constant is the surface area
of the sphere. The curvature of the sphere increases as the size of
the sphere increases. Could the "curvature of space" in GR be nothing
but an artifact of the sphere defined by two masses? And if the
sphere were a solid ball filled with mass of constant magnitude per
unit volume, then the volume would be proportional to the mass.
To get "minimal" gravitation F, Readers can easily prove that r^2 = 1
and m1m2 = 0 is required.
Readers are left to analyze this latter physically as homework.
Osher Doctorow
.
- Follow-Ups:
- Prev by Date: Re: It is time, yet again. I am back.
- Next by Date: Re: It is time, yet again. I am back.
- Previous by thread: Score another for LeSage's Theory of Gravity
- Next by thread: Re: Quantum Gravity 260.0: The Real Interpretation of F = Gm1m2/r^2
- Index(es):
Relevant Pages
|
