 |
|
| Science Forum Index » Physics Forum » Quantum Gravity 341.9: Proof That Probable... |
|
Page 1 of 1 |
|
| Author |
Message |
| OsherD... |
 Posted: Mon Nov 02, 2009 5:30 pm |
|
|
|
Guest
|
From Osher Doctorow
Newton's Law of Universal Gravitation says:
1) F = Gm1m2/r^2 (or Gm1m2/d^2, d = r)
I will use r or d interchangeably.
In one dimension, appropriate along a radius vector, d is given by:
2) d = |x - y|, x and y points along the radius line regarded as
abscissa.
Let x > = y. Then:
3) d = x - y, x > = y
However, Probable Causation/Influence is:
4) 1 + y - x = 1 - (x - y) = 1 - d for x > = y, and we must have x -
y < = 1.
Since PI is 1 - d, and F of (1) is proportional to 1/d^2, let us see
whether PI and F increase together. We have:
5) d(1/d)/d(d) = -1/d^2 < 0
6) d(1 - d)/d(d) = -d(d)/d(d) = -1 < 0
so both 1/d and 1 - d decrease as d increases and vice versa. But the
same thing happens for 1/d^2 in (1) and 1 - d by differentiation.
When d = 1, then 1 - d = 0, and when d = 0, then 1 - d = 1. So by
(4) we only need to worry about d in [0, 1], and PI increases with
Newtonian Gravitational Force F of (1), as does 1/d and 1/d^2, with
m1, m2 constant.
Osher Doctorow |
|
|
| Back to top |
|
|
|
| OsherD... |
| Posted: Mon Nov 02, 2009 5:42 pm |
|
|
|
Guest
|
From Osher Doctorow
Can PI be taken with x and y as spatial coordinates? Yes, since
implicitly spatial coordinates x = x(t), y = y(t). There is no
difficulty in regarding x as the Cause of y and y as the Effect of x
for any two spatial coordinates such that 0 < = y < = x < = 1.
If it had turned out that 0 < = x < = y < = 1, then we would have used
1 + x - y as the relevant expression with y > = x, corresponding to
reversing PI from P ' (A-->B) to P ' (B --> A). If both hold, then x
= y, which satisfies both inequalities.
Osher Doctorow |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Sun Mar 21, 2010 5:17 am
|
|