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

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

  1. Tach Banned Banned

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    I don't buy your "literature", it isn't physics. So, could you please quit waving your arms and try using math?
     
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  3. Tach Banned Banned

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    I very much doubt it since math says that:

    \(F(r) = \frac{Gm}{2 \pi} \int_{0}^{2 \pi} { \frac{(r-Rcos \theta) d \theta}{R^2+r^2-2Rr cos \theta}\)

    and this ain't zero unless \(r=0\).

    If the law of attraction goes with the inverse square power then:


    \(F(r) = \frac{Gm}{2 \pi} \int_{0}^{2 \pi} { \frac{(r-Rcos \theta) d \theta}{(R^2+r^2-2Rr cos \theta)^{3/2}}\)
     
    Last edited: Mar 20, 2013
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  5. RJBeery Natural Philosopher Valued Senior Member

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    Define your variables please. Also, I'll bother with the TEX on one condition: you agree in advance to publicly recognize that they are correct (if they indeed are). Otherwise this is a pointless proposition.
     
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  7. Tach Banned Banned

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    \(R=\)radius of the circle
    \(r=\)distance of the test probe from the center of the circle
    \(m=\)mass of test probe
    \(G=\)universal constant
    \(\theta=\)angle that sweeps the circle, it is the angle made by \(R\) with the x-axis.

    Sure, go ahead. Though, based on past performance I very much doubt that you'll provide any math (or any math that is correct).
     
  8. RJBeery Natural Philosopher Valued Senior Member

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    Tach, I started this last night but I'm not done; hopefully tonight. My analysis is not parallel with yours because I'm to formalize what I wrote (which is a generalization of what Billy T originally wrote). If you need things in the precise form above it would help if you showed your work.

    Also, my logic earlier brought me to the stated conclusion:
    And, from here we find:
     
  9. Tach Banned Banned

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    I showed my work, it already proves that you are wrong, now it is time to show yours. Math only, it is a simple problem to solve, you only need to integrate the elementary force along the circle, any freshman can do it in 15 minutes, don't try to bullshit your way through "armchair philosophy", your standard armwaving will NOT cut it.

    Err, the wiki quote talks about a special case of irrotational fields that happen to have zero divergence, what does all this have to do with the gravitational force being zero? You are trawling non-sequitur quotes from wiki just to demonstrate that you don't know the difference between a field and a force. Like you did not know the difference between 1-sphere and 1-ball earlier in this thread?
     
    Last edited: Mar 21, 2013
  10. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I´ll just note thatTach´s generalization of my real 3D hollow sphere is correct and sort of obvious. (Peplace my 2 which is how the surface scales for the 3D sphere with P scaling of the (P+1) dimention hollow sphere. I did not make that generalization as wanted to answer the question of thread with least confusion but rigor.

    I will add now that implict in my demonstration was the assumption that the axis of the "double cone" with apex at A, the point where gravity field was to be evaluated, passed thru the center of the 3D hollow sphere. I now call this orientation of the cone axis the "zero degree" orinentation. The value of R/r is greatest for this zero orientation and R/r falls smoothly to unity as axis approaches a 90 degree orientation.

    But for any orientation not zero there is a left and right symetrical set of double cones making less attractive force towards their include small disk masses of the wall. The left and right componets of these forces exactly cancel, leaving the net force, if any, ALWAYS DIRECTED as is stated in the original post (#13, as I reacall).
     
  11. eram Sciengineer Valued Senior Member

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    Can you provide a diagram to illustrate this scenario?
     
  12. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    No. I think text is clear, but ask about what you are uncertain about and I will try other words.

    Here are a few: A cone has an axis with definite fixed orientaion in space. It is "generated" by a line passing thru one point on that axis and rotating about that axis with a constant angle (not 0 nor 90 degees) wrt that axis. Thus, all cones are actually what I called a "double cone" but because people without considerable knowlede of geometry, think of cones as like the one you get icecream in, I called my cone a "double cone".

    What I said was implict in my post 13 proof was that the cone axia pased thru the center of the hollow sphere. In post you are asking about I considered other orientation of the cone axis. Noting that the ratio R/r would decrease to unity as the cone axis angle approached 90 degrees.* Also noting that for every angle B there was a equal but opposite angle, -B and pair wise, the componets of their forces of their attraction to the wall points within their cones cancelled out but added to the wall point the cone with B = 0 was attracted to.

    * Don´t be confused. The internal angle of the double cone considered is AWAYS very very small - it is not growing to 90 degrees. But the axis of the "double cone" can go to 90 degrees from the 0 degree orientation assumed tacitly in post 13.
     
  13. RJBeery Natural Philosopher Valued Senior Member

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    This is really all you need:

    Please Register or Log in to view the hidden image!


    You can see that the area of mass in a cone increases with the square of the distance, just as the gravitational force decreases in an inverse fashion. This is why Newton's Sphere works the way it does on the inner shell.

    Apply this logic to other dimensions and the veracity of what I said below becomes obvious.
    Why Tach can't see it, who knows. Why Billy T seems to be agreeing with Tach...who knows. Using Newton's analysis on a circle is quite difficult though. My integral is daunting and I would love to see the solution...especially if a "freshman could do it in 15 minutes". Doing it on a point probe perpendicular to the center of the circle is trivial, but within the circle itself is not (to me). I do have another way to tackle it, though, and I'll try posting it shortly.

    Oh, and Tach, you did not "show your work". You simply posted the conclusion, which you probably just copied from the generalization shown above this section. It doesn't matter though, I'm still having fun sharpening my skills and proving you wrong just adds to it.

    You can't win, Tach. As I'm forced to (re)learn trig and calculus and TEX I shall become more powerful than you can possibly image.
     
  14. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I must have mis-spoke - I. e. said Tach instead of you. I was refering to who ever extended my post 13 to other than 3D. I.e. with:

    "for the ratios of (R/r)^(P-p) to be equal to unity.

    In other words:
    - a 1-ball obeys the inverse law to the power of 0 since the "area of a point" is constant
    - a 2-ball (circle) obeys the inverse law to the power of 1 since the "area of a line" is given in one dimension
    - a 3-ball (sphere) obeys the inverse square law
    - a 4-ball obeys the inverse law to the power of 3
    - etc. "
     
  15. Tach Banned Banned

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    Three days, and you are still struggling with the elementary math, this is really amusing.

    I wrote that section, clown.


    Delusions, delusions.....you are still struggling with a freshman problem.
     
  16. eram Sciengineer Valued Senior Member

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    I get it now.
     
  17. eram Sciengineer Valued Senior Member

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    Are you guys still dealing with a 1-ball?
     
  18. Tach Banned Banned

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    Not 1-ball, 1-sphere (circle), RJ is his struggling with it. Three days and still no math.

     
  19. eram Sciengineer Valued Senior Member

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    Is it the same as the problem I came up with?
     
  20. AlphaNumeric Fully ionized Registered Senior Member

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    Yes.

    Sorry in the delay in replying to your PM.The 1d case, ie a ring of matter, has basically the same mathematical form as the 2d case, except that the dM expression has no sin term, \(dM = \frac{M}{2\pi}d\theta\), which means there is no sin in the numerator of the appropriate integral, making the integrand something you cannot directly integrate.

    I was thinking about whether there is a way for someone to explore the result for themselves if they don't know the maths but can do basic coding in R or Mathematica etc. The force between two objects of mass M and m a distance r apart is \(\frac{GMm}{r^{2}}\) along their joining line. Now suppose you want to see what the gravitational force of a shell or ball is. Well you can approximate a continuous material using a uniformly distributed set of point masses. Suppose you have some way of sampling uniformly from an n-sphere (I will explain the method in a moment). You pick the location you're interested in, say inside the sphere or outside it at some distance, and then compute the pull on an object at that location due to the pull of all the sampled points you have. If you have N points and each has a mass of M/N then you can compare the result with the expression for the pull due to a single point at the centre of the sphere with mass M. If you make N bigger you should tend to the result you get if you did the integral. This is essentially a Monte Carlo method for solving integrals and is common place in many particle physics or fluid mechanics algorithms, which often involve integrals too difficult to do any other way.

    So how do you sample points uniformly from a sphere or a ball? The ball is easy, just sample uniformly D numbers from [-R,R] to make a vector in \(\mathbb{R}^{D}\) and throw away if it does not satisfy \(||\mathbf{x}||<R\), ie outside the ball's surface. The sphere is less obvious, as you cannot use polar coordinates as they do not uniformly cover a sphere (clustering at poles) or uniformly from [-R,R] (clustering at corners) but still easy. Pick D numbers from the Gaussian distribution \(\mathcal{N}(0,1)\) to give you a vector in \(\mathbb{R}^{D}\) and then rescale the vector to have length R. Do this a few times and you'll see it leads to a uniform spread over the sphere. Do this N times to get N points and then repeat the sum of forces procedure. You can check your algorithms by setting D=3 and seeing if you recover the Shell theorem for both inside and outside the sphere.

    I'd program it myself but I've got work to do, plus my programming time is currently spent trying to use a Wii remote to control a Raspberry Pi controlling a set of motors. Damn Python...
     
  21. eram Sciengineer Valued Senior Member

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    1,877
    I wonder how many points one needs to get an accurate value.

    On a side note, I found this: http://www.nowykurier.com/toys/gravity/gravity.html


    Why is it that after creating a proto-disk and plopping an OMFG mass into it, some particles are ejected?
     
    Last edited: Mar 26, 2013
  22. RJBeery Natural Philosopher Valued Senior Member

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    OK that's pretty cool! The answer to your question has to do with initial rotation velocity, which I wish the simulation had a parameter for.
    EDIT: Wait, I was only producing the proto disk, I didn't know you could also introduce arbitrary particles with velocity. Those ejected particles are basically using gravity assist to shoot off like that.
     
  23. Tach Banned Banned

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    Aren't you supposed to be working on your math? It has been 5 days and you are still unable to produce a solution that usually should take a freshman 15 minutes to produce.
    You threatened that you are becoming a mathematical giant, where is the math to back up your claim, RJ? How many more days do you need to solve the simple exercise? Or is it weeks? Months?
     
    Last edited: Mar 26, 2013

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