geistkiesel
Valued Senior Member
Mathematical falsification of Newton’s Shell Theorem
The shell theorem has been discussed to death in this forum, see,
http://en.wikipedia.org/wiki/Shell_theorem
http://www.freebase.com/view/en/shell_theorem
http://www.answers.com/topic/shell-theorem
But remember these three references all state the mainstream dogma and are subject to mathematical falsification below.
Starting with the integrated shell integral, but before it is evaluated, the force expression is,
For K = GmM/(d^2)4R and k = ( d^2 - R^2) then the force expression is,
F = K{-k/X +X} which will evaluated in two parts, F1 = K{-k/X +X} evaluated from d – R to d, which produces,
F1 = K {k(1/d(d – R) – R)} (1)
and from symmetry considerations,
F2 = K{k(1/d(d + R) – R)}. (2)
You may assure your self that F1 + F2 results in the integrand being 4R and hence cancels from K. Here when making an actual calculation the embedded 4R term must be used. If assuming R = 1, the division by 4[R] acts as a normalization term which reduces the actual share of the total force that each contribute to the total force.
In comparing F1 and F2 for size by asking, is F1 > F2, or ‘is all the mass justifiably concentrated at the shell center, the COM’? If FR1 > F1 there is no scientific justification for maintaining that the shell behaves as stated. K, k, the common d multiplier, and the trailing ‘-R’ in F1 and F2 may be ignored as these terms are common to both F1 and F2 and do bear on the relative sizes of F1 and F2.
Let k1 =1/(d – R) and k2 = 1/(d + R). If k1 > k2, then F1 > F2. As (d – R) and (d + R) are both denominator terms, then because (d – R) < (d + R) then necessarily, k1 > k2; therefore, F1 > F2. QED.
:shrug:
The shell theorem has been discussed to death in this forum, see,
http://en.wikipedia.org/wiki/Shell_theorem
http://www.freebase.com/view/en/shell_theorem
http://www.answers.com/topic/shell-theorem
But remember these three references all state the mainstream dogma and are subject to mathematical falsification below.
Starting with the integrated shell integral, but before it is evaluated, the force expression is,
For K = GmM/(d^2)4R and k = ( d^2 - R^2) then the force expression is,
F = K{-k/X +X} which will evaluated in two parts, F1 = K{-k/X +X} evaluated from d – R to d, which produces,
F1 = K {k(1/d(d – R) – R)} (1)
and from symmetry considerations,
F2 = K{k(1/d(d + R) – R)}. (2)
You may assure your self that F1 + F2 results in the integrand being 4R and hence cancels from K. Here when making an actual calculation the embedded 4R term must be used. If assuming R = 1, the division by 4[R] acts as a normalization term which reduces the actual share of the total force that each contribute to the total force.
In comparing F1 and F2 for size by asking, is F1 > F2, or ‘is all the mass justifiably concentrated at the shell center, the COM’? If FR1 > F1 there is no scientific justification for maintaining that the shell behaves as stated. K, k, the common d multiplier, and the trailing ‘-R’ in F1 and F2 may be ignored as these terms are common to both F1 and F2 and do bear on the relative sizes of F1 and F2.
Let k1 =1/(d – R) and k2 = 1/(d + R). If k1 > k2, then F1 > F2. As (d – R) and (d + R) are both denominator terms, then because (d – R) < (d + R) then necessarily, k1 > k2; therefore, F1 > F2. QED.
:shrug:
