Newton's Shell Theorem – Bad Mathematics - Bad Physics

Discussion in 'The Cesspool' started by geistkiesel, May 8, 2009.

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  1. geistkiesel Valued Senior Member

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    Newton's Shell Theorem –Bad mathematics - Bad physics

    Take three mass point objects m1 = m2 = m3 = 1 unit mass, G=1 unit gravitation constant, and using init distances the force of attraction between m1 and m3 separated by 10 unit distance is calculated using the universal law of gravity expression, F = Gm1m2/r^2 (minus sign omitted). F12 = (1)(1)(1)/10^2. Similarly, the force generated between m1 and m3 separated by 12 unit distance F13 = 1/144. When the masses are arranged along a common axis the total force of m2 and m3 on m1 is F123 = 1/100 + 1/144 = (1)(1)(1 + 1)/r^2 . Rearranging the terms, 2/r^2 = 244/(100)(144), or, r^2 = (2)(14400)/244 or R^2 = 118.03. The result r = 10.86 and, 10 < r < 11, where 11 is the location of the m1m2 system center of mass (COM).

    The combined forces' COM is located a distance 11 from m1. However, the center of mass-force (CMF) is located at 10.86, which is off set from the COM in the direction of m1. The total forces of the mass of a this spherical shell is calculated manually (see above) using mirrored image pairs of masses on the shell where one membermass of each pair is in opposite hemispheres, one closest to m1, one farthest from m1. Clearly when all forces are calculated, all CMFs of each calculation are located in the nearest sphere segment to m1, contrary to Newton's Shell theorem that without any physical basis, claimed that m1 may consider the mass of the sphere concentrated at the COM of the sphere.

    The links below are consistent examples of the rote acceptance of a developed shell theorem, referenced as an unchallenged law of physics and cheerfully communicated as rigorously copied scientific gospel, chiseled in stone, as it were, and enjoying immunity from heretical thought or criticism by the innoculated consensus of a solemnly deferential scientific community.

    http://en.wikipedia.org/wiki/Shell_theorem
    http://www.physclips.unsw.edu.au/jw/NewtonShell.pdf
    http://www.absoluteastronomy.com/topics/Shell_theorem

    From inspection of the sphere and m1 externally located at some point r from the sphere center it should be obvious that using the concept of "inverse distance squared" as a starting point, the mass of M in the hemisphere closest to m1 will contribute a greater share of the total force on m1 than the farthest hemisphere, hence the CMF for the entire mass M is located on the m-M axis off set from the COM in the direction of m.

    Where, and how, does the Newton Shell theorem (NST) place the CMF at the sphere COM? It doesn't. The question of locating the CMF is not discussed! The NST model begins with a ring on the sphere of differential mass dM oriented perpendicular to m and centered on r. By summing the force for each dM on each ring then integrating over the surface of the sphere, the total force, F = GmM/r^2 results, which says nothing, absolutely nothing, regarding the location of the CMF. The Wikipedia model referenced above states after dF is integrated, that,

    "The shell really does act as though all the mass is concentrated at the center!"

    Some commentators call this Shell theorem result, "proof" of what is claimed.

    The big problem here is that the developed algorithm made no inclusion for determining the location of the CMF. The intuitive assumption that the CMF is located at the COM of the sphere was arbitrarily (instinctively) made (or so this writer has surmised) as clued from the 'r^2' term in the expression, that clearly is an expression for determining the total gravitational force of M on m, only.

    Another flaw peeking from the expression is seen iwhere calculating the net vector force in the m-M direction as derived by taking the cosine projection of the force onto the "r" [m-M] axis and from this, supposedly, the inference was made that the CMF followed the projection of the force onto the m-M axis – the projection of a force vector is mathematically proper (forces perpendicular to the m-M axis 'cancel, or so we are told), but to include the scalar quantity location in the projection of the CMF is just plain "bad mathematics"; not properly placing the CMF in the nearest hemisphere, by inspection, of the conditions, re m and M, is just plain, "bad physics".

    Caveat emptor – beware of standing on the shoulders of giants who have been dead for 300 years. PU!

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    Last edited: May 8, 2009
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  3. D H Some other guy Valued Senior Member

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    What is with this incessant harping against very simple freshman-level mathematics? Learn the math. It isn't that hard.
     
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  5. granpa Registered Senior Member

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    the result of the integration is:

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    which is exactly the same equation you would get if all the mass were at the center. how can you possibly not understand that?
     
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  7. rpenner Fully Wired Valued Senior Member

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    He's just 2300 years behind in mathematics and 320 years behind in physics.
     
  8. AlphaNumeric Fully ionized Registered Senior Member

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    And three points sum to make a spherical shell how.....?

    The fact you find the mathematics completely beyond you does not mean other people do. Infact, as DH has repeatedly said, this is freshman maths, stuff taught to 1st years. It was when I was a freshman. This integration is easy. Ones involving electromagnetic charge are often much harder, due to less pleasant layout of material.

    You are doing the typical crank thing, you find something hard or impossible so you proclaim everyone else must be just blindly accepting it since if you can't understand it surely noone else can. News flash, there's people better and brighter than you. Of course this is true for pretty much everyone but in your case it's particularly true.

    This is basically you saying "I believe gravity behaves like this but this is in contradiction to the Shell Theorem. Therefore the theorem is wrong".

    As your openning paragraph demonstrates, you don't even know how to do the mathematics or physics relating to this. And while we know Newtonian gravity is not perfect, things like the Shell Theorem and Gauss' law for electromagnetism are very very close to being exactly right. Errors in them only creep in when you need relativity or quantum mechanics. On everyday scales they are good physics. Experiment says so.

    But you aren't interested in doing actual experiments, learning actual physics or being open minded. That's too much effort for a crank.
     
  9. geistkiesel Valued Senior Member

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    The expression is an expression for the gravitational force of a shell of mass M. None of the papers I have seen made any attempt to determine the location of the mass force center. The expression F = GmM/r^2 is a an expression indicating the force of a shall of mass M located a distance r from m - the expression describes the force only of the sphere located a distance r from m - the expression does not state, nor can it be inferred, where the center of mass force is located - do you see the difference?

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  10. geistkiesel Valued Senior Member

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    Read the whole Paper. The three masses in a line demonstrated how to determine the center of mass force for three masses. The paper then describes how to properly determine the force center for pair of differential masses and where an additional ring inserted in the opposite hemisphere. There is no corruption of the integral, just a different way of doing the integration.
    If you have a problem with my math why not point to the exact flaw instead of quoting some disgruntled bystander.
    Do you know any physics? If so let's see it.
    Where did I say that "gravity behaves like this"?
    Where did the opening paragraph indicate I don't know the math and physics of the subject matter? Specifically, the exact words, the exact flaw in reasoning, physics, logic . . where's the beef?
    You don't participate here for discussing claims that an ancient dogmatically held and believed 'theorem' is in error, do you? This entire post is mere witless posturing, the high pitched moan of a panicked wanna be scientist --the sweetest of all sounds.... ​
     
  11. Guest254 Valued Senior Member

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    For the record, this result was shown to me and my classmates in high-school (using an argument similar to Newton's original). I feel people are being too fair by classifying it as University level.

    geistkiesel: I think you have something here. I've never trusted integration (and I've got a degree in maths!), so I'm delighted to see you've found a flaw in it. It's amazing how all the great minds: Newton, Gauss, Euler, Riemann, Dirac, Laplace, Einstein; completely missed this massive flaw in mathematics and Newtonian physics!
     
  12. przyk squishy Valued Senior Member

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    http://www.google.com/search?hl=en&...E265&q=define:"mass force center"&btnG=Search
    http://www.google.com/search?hl=en&...5&q=define:"center of mass force"&btnG=Search

    Guess you're on your own there.
     
  13. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    The problem with your math is that you're not adding up all the points of a sphere in doing your calculation. You're just picking two points, "2" and "3" that form a line which passes through "1". You have to add up all the points over the entire shell or else you won't get Newton's result.
     
  14. geistkiesel Valued Senior Member

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    In developing the Shell Theorem there is he step in projecting the force (F) vectors associated with each dM on the shell surface, onto (F(cos(phi) the r axis running from the test mass m to the center of the sphere. Integrating the function produces ,, F = GmM/d^2, where d is the distance between m and the sphere center. The expression is one of force, not location. The expression, in words, says just what you have read and heard many times – The total force associated of the total mass on the sphere, located a distance d from m.
    This says the center of the sphere imposing a force on m is at d, the distance of the sphere from m, and does not say, the mass center of force is located at d.
    Taking the projected force onto the r axis is mathematically permissible as the force term is a vector, however, the corresponding 'center of force' is a scalar and may not be projected, mathematically, or physically.

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  15. geistkiesel Valued Senior Member

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    Guest254, Has anbody ever said to you, "You made my day?" Thank you sir and give my hurrahs to your high school class and that I accept the pat on the back posted by the their teacher. I emphacise a very important guide to life which I mentioned elsewhere -"Caveat Emptor [of] standing on the shoulders of giants that have been dead for 300 years"

    You statement regarding the list of the geniuses missing Newton's error - the story I understand was that the shell theorem was the device Newton used to develop calculus, which he needed for the 'shell problem'. So subjects like projecting a scalar using vector analysis was probably not considered by Newton, or any of the others mentioned they were literally all novices to the math and the physics. physics itself was in a state of early growth (some say that currently the state of growth is still early) so some possiblilities surely escaped N. I refer to those results of modern measurement of the 'speed of the force vector' being no less than 10^10(c), some say it out loud, the force vector is instantaneous. If the latter, which seems the best probability, there is no reason to discuss a force field betyween the sphere and the test mass. As a friend told me when I was explaining my thesis, the "force just is".

    Here (THE LINK) is Tom Ahern ( a rogue of sorts) but easy to understand and as an aside see how diplomatically he handles relativity theory objections, he isn't a big fan, and other physical dogma. And remember, the 'instantaneous' appearance of forces linking two physical objects is intensely studied currently - see Einstein, Podowski and Rosen Experiment, AKA 'EPR experiment' for a discussion that lays out sooe fundamental physics re quantum mechanics without getting put to sleep by the math. The EPR experiment was purelY a 'thought experiment'.
    http://www.metaresearch.org/cosmology/gravity/speed_limit.asp

    Another rule of life, apparently, is, "don't mess with the badddest dog on the block." and one didn't, mess with Isaac (some say a master at Plagerism, but this is to me just a rumor, but what the hell Isaac has been dead a long time and this insult will not disturb his current dream - Ah yes perchance to dream, aye that's the rub;
    Years later [after science] when he was Exchequer of the Treasury he set a record for the number of hangings of coin counterfeiters. I hear the record stands today.

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  16. geistkiesel Valued Senior Member

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    Thanks, I should have stuck to the original language the "location where the total mass is concentrated and considering this as the source of the gravitational force." -- I am paraphrasinge here. Perhaps some got the meaning, I think you did, that I ntended, some didn't and I may have confused a few. That is what the forum is all about isn't it? I mean getting the descriptions properly set and then we can all join in and talk about the same thing, what a wickedly brilliant concept!

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  17. geistkiesel Valued Senior Member

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    No, I am adding all the points - the same points in the standard shell integral, only I am adding the points in a different order for the for the purpose of determining where m sees the force originating. I need two points for this. Even so, the integral needs further modification in order to calculate the location of the weighted average of all the individjual points - let me call the points the 'mass-force-center'

    I assume you checked my math on the two arbitrary points and that all was proper there. Check the thread starter.
    Calculate all the forces on m for each pair of dM'. One dM is in the nearest shell half to m the other in the farthest shell half from m. Then calculate the location m sees where the the force of each pair originates.

    I ask you to make any two mirror image points, calculate the force on m separately, then, work the other way around and add the two forces (total force on m) and calculate r^2 (r). All points where m 'sees ' the force originating is in the closest spherical half.

    Start with the closest point to m with is mate, the farthest from m, and then move up the sphere. You only need three points. When calculating points nearest the center of the sphere. then half way up the sphere and finally the last two points on the rings now facing each other as close as mathematical zeros will allow. This last calculation places the center of this force very near the , but integral calculus isn't horse shoes. The point m sees for the total the force is in the near half, always.

    Thje easiest way to see it is to recognize that the inverse square of the distance means that for any two equal masses coincidental to the a line through m that the closest of the equal masses will contribute more force on m than does the farthest mass with the result that m does not see the force emanating from the center of mass of the system. For sure m sees the mass comin at it on the line connecting the masses, but m is not justified in assuming the origin of the force is located at the center of mass of the sphere system.

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  18. geistkiesel Valued Senior Member

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    You know of the droll manner, for sure. Can you point to a specific flaw in the reasoning, mathematics, physics or logic? Or are you really just a member of the audience that likes to share 'eye winking' chum-bonding commeradie with those of similar characteristics?

    Whatever, I would appreciate a comment directly focussing on the post and away from what you see as mistaken mental aparitions, you know the drill. You know I am not a demagoue, in fact I'm not even an angel, oh no, I've led a full life, drinking, gambling and once, well, once I talked back to my first wife.

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  19. AlphaNumeric Fully ionized Registered Senior Member

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    Firstly, that's not a paper. Don't insult people who actually write papers. Secondly, you've failed to do any maths.

    I can't see where you've done any maths which is anything coherent and valid and right.

    Infinitely more than you. How's this?

    So if I give you an integral to do which is from a 1st year mathematical methods course which does multidimensional integrals (which I just happen to teach

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    ) you'll be able to do it, is that what you're saying?

    And you can't participate here if you're unwilling to do any science or read anything before proclaiming 300 years of mathematics and physics which you haven't read, can't do, don't understand and which has a lot of experimental validation is wrong.

    So, are you up for doing an integral?
     
  20. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    The problem is you haven't done the actual addition yet, which has to be done as a vector sum in which the forces cancel in all but one direction.

    I did check it and for the most part it's ok (given its simplicity, there's not much that can go wrong, especially since you haven't even tried to integrate it yet). You're wrong about the hemisphere thing though- your scheme does not match points in the closest hemisphere with points in the farthest hemisphere in a 1-1 ratio. Some of the points you match will both lie on the closer hemisphere, some will be tangential and not matched (although the contribution from the tangential points is infinitesimal and doesn't really matter).

    As I mentioned above, your assertion that each of your pairings couples a mass point in the farthest hemisphere to a point in the nearest hemisphere is incorrect. Regardless, this is not the key problem with your argument- I'll get to that below.

    And I ask you to perform a vector sum over all of the forces from all of the infinitesimal mass points in the shell. Even if, for every individual pairing, the effective centre of the force lies in the closest hemisphere, when you do the sum over all forces in each direction, the vertical and horizontal forces cancel each other out and you find the effective centre of the force lies at the centre of the shell (as seen by an outsider, since inside the shell everything cancels and there is zero force). In a sense, this cancellation pushes the "effective centre" further back and away from the closer hemisphere.

    If you still don't like my reasoning, do the integral yourself and we can see whether your assertions hold true in the actual calculation- that's the easiest and really the only way to silence your critics, since their criticisms are based on mathematics that have been proven down to the level of a small set of simple real number axioms.
     
  21. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Thread closed.

    geistkiesel banned three days for reposting the same crap here over and over.
     
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