Mathematical falsification of Newton's Shell Theorem

Discussion in 'The Cesspool' started by geistkiesel, Sep 7, 2009.

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  1. geistkiesel Valued Senior Member

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    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:
     
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  3. D H Some other guy Valued Senior Member

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    ... and you never, ever, ever learn, do you?
     
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  5. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    No vector summation, no integrals, not even a proper explanation of all the variables and step-by-step deductions. This is a science forum, not clown school.
     
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  7. 1100f Banned Registered Senior Member

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    Asking for proper explanation of all the variables and step-by-step deductions is mainstream dogma. So why do you need them?

    Isn't it better to write expressions like these: K = GmM/(d^2)4R ?

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  8. §outh§tar is feeling caustic Registered Senior Member

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    imo that expression is far too mainstream to be of any use in intellectual discussion :shrug:
     
  9. temur man of no words Registered Senior Member

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    Shell is done. Let's do the flesh now.
     
  10. rpenner Fully Wired Valued Senior Member

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    I'm reaching for the ban button, but I can't find it. Oh, wait. I'm not a moderator here.

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  11. James R Just this guy, you know? Staff Member

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    geistkiesel:

    You seem obsessed with this.

    The wikipedia link provides a VERY clear PROOF of the shell theorem:

    http://en.wikipedia.org/wiki/Shell_theorem

    Please explain where the error is in that proof, exactly.

    ---

    You have not defined your symbols. What is d here? Is R the radius of the shell? Are you copying one of the links you gave, or making this up yourself?

    What is X?

    ---

    Perhaps the best way forward at this point is for you to explain the error in the wikipedia calculation, using the terminology used there - if you can.
     
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