Would an inverse-cube law obey Newton's shell theorem?

Discussion in 'Physics & Math' started by eram, Mar 15, 2013.

  1. RJBeery Natural Philosopher Valued Senior Member

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    If you're still butt-hurt by the whole experience, I'm sorry. I appreciate you sharing your knowledge, and I think I've demonstrated that I thank people and also admit I'm wrong when I believe it to be true.
     
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  3. Guest254 Valued Senior Member

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    That's about the level of response we've come to expect. The apology is noted.
     
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  5. Guest254 Valued Senior Member

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    Hi Billy T,

    The point I tried to allude to earlier is that the material in post 13 does not constitue a proof. The argument is certainly suggestive, but that's about it. The intersection of the cone you speak of, with the spherical shell, is not a flat disc, no matter how thin a cone you choose to consider. We're not to know that this approximation doesn't affect the total contribution of all the forces in such a way that the theorem fails.

    You should forget the fact you know the theorem is true. When you examine a proof, you need to put yourself in the position of a skeptic.
     
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  7. RJBeery Natural Philosopher Valued Senior Member

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    Guest, I'm curious as to your reaction to this
    which is something I provided Tach over a week ago, immediately dismissed by him as inapplicable.
     
  8. Guest254 Valued Senior Member

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    It is stating that you can write the gravitational field as the gradient of a scalar function and since the field is divergence free away from the source, the corresponding scalar field is harmonic. We know what spherically symmetric functions that are harmonic at infinity look like. Taking the gradient of these functions tells us what the gravitational field will fall off like.

    Short version: reconstruct gravitational field by solving Poisson's equation \(\Delta \phi = \rho\) for the gravitational potential. Outside the support of source, the gravitational field is harmonic.

    This requires some knowledge of the theory of harmonic functions. If we are to assume people know this, then the entire thread is answered in one tiny argument: inside the shell the gravitational potential is harmonic, and by spherical symmetry it must be constant on the shell. The weak maximum principle implies the gravitational potential is constant everywhere inside the sphere, hence the gravitational force is zero. This holds in any dimension, but my guess is people wont understand the argument.
     
  9. Tach Banned Banned

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    Back to "proof" by cherry-picking bits and pieces of wiki, eh? Your time would be better spent learning topology (the difference between 1-ball and 1-sphere, between disc and circle) and calculus (the difference between integrand and integral).
     
  10. Guest254 Valued Senior Member

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    Why on earth would learning what those two words mean require someone to learn topology? Most people know the difference between an apple and a doughnut without knowing the first thing about topology.
     
  11. Tach Banned Banned

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    The point is that he demonstrated that he DOESN'T know the difference, mixing them up repeatedly in this thread.
     
  12. RJBeery Natural Philosopher Valued Senior Member

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    Cling to what you can, Tach. I'm content to use the wrong terminology occasionally as long as I get the big picture. It should scare you a bit that I deduced the truth from my armchair without misusing online math tools to come to the wrong conclusion.
     
  13. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    That "not a flat disk" is true, but the poof does not assume the surfaces of the sphere included inside the "double cone" is a flat surface. It only assumes that the mass contained inside is proportional to r^2 (or R^2) by the postulated "uniform surface density."

    Consider a very tiny sphere with small radius of curvature. For example with diameter circle like this: o
    Also let the point where field is to be evaluated, A, be off center so the R/r = 2 then the area (or mass) on the surface at the more distant point inside the double cone is 4 times greater than the mass closer to A inside the smaller cross section cone. So the proof still holds, despite the small radius of curverature.

    Do you agree? I.e. that this geometric proof is valid regardless of the sphere size.

    In fact the proof would fail if the surface of the "sphere" was made of flat segments like a socker ball. To see this with ease, consider a cube. There is small gravity force inside a cube with uniform density walls. It should be directed towards the center of the surface square closest to point A if the line thru both A and the center of the cube passes thru the center of that surface square too. That is easy to understand if you imagine A is only microns away from that square and the cube is huge. If A is on a "long diagonal" and very near a corner of the huge cube then the force is towards that corner. I´m not 100% sure of these last claims, but do believe them to be true.

    One reason why I believe that is I considered a cube of edge length 2 with all the mass uniflormly concentrated along the 12 edges and A very near (even at) the center of one square surface. Then some of the edge mass is only 1 away from A and the greatest separation of mass on those four edges is only squre root of 2. All the mass of the opposite sides´s four edges is more than 2 away from A and some of it (at corners) is square root of 4+2 = 6 away from A. The net force from the four edges not part of either the closest or farthest way surfaces are allat least root 2 and up to root 6 way from point A and cancel each other out except for components on the line thru A and the two centeres (cube & near square).
     
    Last edited by a moderator: Mar 29, 2013
  14. Tach Banned Banned

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    Being incapable of expressing anything in a formal way makes you incapable of producing any physics, this is why you are relegated at the position of "natural philosopher".

    You made up a conjecture. Turned out that it was true for dimensions 1 thru 3 but not beyond 3. You couldn't even decide one way or another for a simple case of a circle since you could not even calculate the integrand describing the element of force, dF.
     
  15. RJBeery Natural Philosopher Valued Senior Member

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    You're being willfully ignorant on this. Since you appear to respect Guest's opinion:
    Ask yourself if it is remotely possible that I made up the conjecture based on more than just a wild-ass guess. (hint: there was more than just bourbon involved

    Please Register or Log in to view the hidden image!

    )
     
  16. Guest254 Valued Senior Member

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    Billy, your proof does not hold regardless of the size of sphere. The Newton-esque argument you are trying to reproduce requires that the angle of the cones be "infinitesimal"' so that the resulting triangles are similar and the resulting areas of intersection can be assumed flat. This is, of course, nonsensical. The reason this type of argument gets the right answer is because whilst all the intermediate steps are false, the ratio at the end is equal to that which you would obtain if you had included all the (complicated) error terms, then took the limit as the angle of the cone tends to zero. But then you must add together more and more of the cones that you were considering.... And you end up doing an integral!

    Hope that makes sense!
     
  17. Guest254 Valued Senior Member

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    Come on, you really want credit for saying "hey, it works for (2,3) maybe it works for (n-1,n)"? Ok, go you! God only knows how pleased with yourself you'd be if you'd been able to prove the result yourself! I find it strange you now blindly accept it as true though, given that you can't judge the veracity of my argument.
     
  18. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Yes it does, but there is nothing I see wrong with having the number of cones go to infinity at the beginning of the argument instead of at the end in the integration process. I don´need to actually do any integration, after showing that pair-wise in the limit of vanishing small cones, the force adds to zero by pairwise cancellation.
     
  19. Tach Banned Banned

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    Nope, knowing that you can't provide a proof if your life depended on it , I also know that this is all you did : a wild ass guess.
     
  20. Guest254 Valued Senior Member

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    But Billy, then all you have is that the attraction from two points on the sphere cancel, but the gravitational attraction due to these points would be zero anyway, because they have zero mass (a point on a sphere has what's called an area of "measure zero"). The limiting processes need to be performed in tandem.

    Just take usual 1 dimensional integrals. These are the limit of the area more and more rectangles of smaller and smaller area. If you try to do the limit of the area of the rectangles first, you'd be left with a big sum of zeros, and we both know that not all integrals evaluate to zero!
     
  21. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I´m not sure what you mean by "in tandem" but will try to explain why fact that the mass goes to zero as cones vanish in size does not defeat the proof:

    Consider to quanties, X & Y, that both have limit of 0 as some parameter, which I call "t" goes to 0. If you can show that the ratio X/Y approache say 50 as t appoaches 0 then you can not claim as I did to have proved much of interest; however, I proved that as t approaches 0 that the ratio X/Y approaches unity - i.e. that the forces from the two included masses, although both are vanishing, is doing so in such a way as to be always equal and opposite or "pair wise cancelling" out the net force.

    I hope my point is clear and persuasive to you.
     
  22. RJBeery Natural Philosopher Valued Senior Member

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    No, I "play" with the problem in my mind. I did not verify that it works with (2,3), I came to my conclusion here after digesting Billy T's original assessment. I can "see" it. The field theory interpretation of the inverse-square law was found later, as was the trivial mathematical solution to the 1-ball (and eventually the 2-ball).

    It was Google, I believe, that asked the following question to applicants: If dump trucks came by Mount Fuji every second and were filled with dirt, how long would it take to move the entire mountain? You could certainly estimate the mountain's mass, and the capacity of the trucks, and come up with a time T such that the mountain was moved...but the point of the question was to find those people who could estimate the answer based on nothing but the "feel" of the question, even if you didn't know the mass of Mount Fuji. This means they quickly wanted an answer within a couple of orders of magnitude.

    Without mathematical estimations, how would you answer the question?

    Anyway, if you want another example of intuition (and another chance for me to be proven wrong, which is always a possibility) is this:
    I disagree with this. I believe that the phenomenon inside of a Newton's Sphere does not depend on the massive object's shape (so long as the shell of the massive object is continuous and can be "seen" by the point probe). If the distance between the point probe and the shell is increased through deformation (with an alteration of mass such that it remains constant by area as appropriate), the mass of the shell is proportionally increased and then I believe that the net force inside the massive object remains zero.

    Does anyone care to chime in on this conjecture before it gets analyzed mathematically?
     
  23. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I will: Consider the field point A to be inside a very long cylinder with closed flat ends of same mass density as the walls, but sort of near one end (say R/r = 5 for the distance ratio to the two ends from point A so end forces on point A are not dominate, almost negligible as they are so far away*. - I.e. A is still far from the mass center of the cylinder at point B. I claim that a test particle placed at A will be attracted towards point B. I don´t want to try to prove this, but it seems obvious to me.

    * Say 50 and 10 miles away and cylinder diameter is one inch. To see things clearly with ease it usually helps to go to extremes.
    .
     

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