Wednesday, July 25, 2012

Disproving Newton Pt2: (b) Hollow Sphere (cont.)

Table of Contents
(A) Creating the Tools We Need 4
Introduction: Simplifying the Problem 4
1. Force between a Barbell and a Test-mass 5
2. Getting the Net Force for the Barbell Using the Cosine 6
3. The Equivalent Point Mass is NOT the Geometric Centre 7
4. The Force for a Ring of Negligible Thickness 8
5. Plotting the Force for the Ring 9
(B) The Force for a Hollow Sphere 10
1. Preliminaries for Calculation 10
2. Archimedes’ Theorem 11
3. Defining the Variables 12
4. Integrated Version for Hollow Sphere 14
5. Appendix: Integrating the Hollow Sphere Formula 15
(C) Force for a Solid Sphere 18
1. Force for a Uniform Disk 18
2. Graphing the Force for a Disk 19
3. Force for a Solid Sphere ( Disk Method ) 20
4. Formula and Graph for Solid Sphere ( Disk Method ) 21
5. Force for a Solid Sphere ( Shell Method ) 21
6. Integral and Graph for Solid Sphere ( Shell Method ) 23
7. Summary of Part 2 so far: 24
(D) Failure of Sphere Theorem at Molecular Level 25
1. A Simple Quantization of Mass 25
2. General Method for Discrete Distributions of Mass 26
3. Discrete Form of Hollow Sphere Equation 28
4. Formula and Graph for Quantized Hollow Sphere 30
5. Sample Hollow Spheres of Discrete Particles 31
6. Summary and Conclusions with FAQs 33

(B) The Force for a Hollow Sphere
1. Preliminaries for Calculation
We can get the force that a hollow sphere exerts upon an external test-mass without any calculus at all. All we need is Archimedes’ Theorem, the Pythagorean Theorem, and our formula for a uniform ring. Let’s see how easy it is:
  1. For convenience we replace the GmA mB part of the gravitational formulas with a constant, K. Newton’s formula would now just be K / d2.
  2. The radius R of the sphere we will set to 1 and so all distances will now be in units of R.
  3. The centre-to-centre distance between test-mass and sphere is D, which we leave as a variable but is fixed for any given example. Our formula will be in terms of D, with D in units of R.
The basic idea is simple: We divide the hollow sphere into rings and use our ring formula to get the force from each ring. Then we add them up. 

However, two problems pop up:
  1. How do we figure out the mass for each ring?
  2. How do we figure out the size and distance for each ring?

2. Archimedes’ Theorem
It turns out the mass problem solves itself, if we know Archimedes’ Theorem! Take a sphere and a cylinder with the same diameter. Let two parallel planes cut through them both. Archimedes showed that the surface area of the sections between the planes is the same! But as long as the spacing is fixed, the area between will be the same anywhere on the cylinder, and so the area stays constant for the sphere as well!

That is, rings of equal height, they will automatically have the same area, and mass too. This may seem spooky, but the explanation is simple enough: As the rings’ radius decreases, the tilt of their surface increases just enough to keep the area constant. If the mass is equally spread over the surface, then the area is proportional to mass.

3. Defining the Variables
If we start in the middle and work out toward the edges taking the rings in pairs, each pair will have the same radius, saving some effort.
Ring Mass: Dividing the sphere into 2N rings of equal width ensures that each ring has the same mass. ( via Archimedes’ theorem ) The mass of the whole sphere, mB = 1. This makes the mass of each ring, mi = mB / 2N = 1 / 2N.
( The i subscript identifies each ring for our loop. ) It is true that the curved ‘sphere-slices’ aren’t really rings of negligible thickness, but we can make them as close as we like by increasing the number of slices, 2N.
Coordinates: However, the forces for near and far ring are different, so we have to do each separately. We place the sphere at the origin (0,0), slice the sphere vertically, and use the x-axis for most distances. If i is our counter, and 2N is the number of rings and R = 1 ( the sphere radius ), then we sweep i from – N to + N, and ( i / N ) is just the x-coordinate of each ring, which goes from -1 to +1.

  Ring Distance: Since the test-mass distance will normally be given as centre-to-centre,5 we define ring distance in those terms.
dnear = D - i / N dfar = D + i / N
Conveniently, our choice of origin and using – and + values for i handles the signs automatically, so the form D - i / N is good for both near and far rings.

Ring Radius: ( via Pythagorean theorem )
When setting up our variables, we might be inclined to define the ring radius, ri in terms of the angle f  directly. For instance, ri = sin f, and f = cos-1 ( i / N ).
Rather than insert sin (cos-1 ( i / N ) ) into our formula, we can get ri more simply and directly: Make a standard right triangle around angle f, by using the sphere radius R = 1 as one side and dropping a vertical to the X-axis where the ring will be. ( see diagram. ) The ring radius is then just the y-coordinate. We solve for y using the Pythagorean theorem. : ( y 2 = h 2 - x 2 …h = hypotenuse = 1 )
ri 2 = R 2 - ( i / N )2 = 1 - ( i / N )2 ( we won’t need ri itself, just ri 2 …)
So we only use the perspective of the angle to determine ri 2 in terms of the radius of the sphere. But after that, we are done with it. For the gravity formula we want the perspective of the angle  . But again we can avoid all the trigonometry and angle-juggling by simply working with the distances directly anyway.
Ring Formula:
We also have a choice of forms for the gravity formula for a ring, (see above) but the following version is convenient, substituting in for h via Pythagorean theorem :

Substituting in d and r2 and simplifying the denominator gives the final formula for the hollow sphere: ( note: exponent on r is already handled )
For any given sphere and test-mass at a fixed position, the only variable is i. In fact, the same formula is valid for both inside and outside the hollow sphere, if we allow the (-+) sign to represent the direction of the force along the axis. Good results can be obtained with N > 1000. While this is not practical by hand, it is trivial to write a small computer program to loop the sum and graph the formula. The graph will approximate Newton’s claim regarding the field for a hollow sphere.
We now have a pretty frightening looking formula. But all this is really saying is that we have to integrate our summation formula to make it exact.
Integrating the formula gives us the following simpler formula:
 (We show how we integrated this formula below.) And what is this
formula? It is just Newton's original Inverse Square formula with a toggle factor (Absolute Value function) that turns off the force when the test-particle is inside the sphere.
Is it exact? Mathematically yes. Except as always, the formula has discontinuities and is meaningless when the denominators are zero. (the border of the sphere surface). Does it accurately reflect the physical reality inside the sphere?

4. Integrated Version for Hollow Sphere
The Integral form is of course much faster, and more accurate, provided we accept the assumptions which went into it. ( These will be discussed later. )
The steps for the integration have been skipped. Those who can do the calculus won’t need any help to check the result for themselves. The final integral performs as expected by standard theory, so no controversy with Newton can be found (or resolved) along those lines.


Plotting the Force for the Hollow Sphere
Note the finite limits approaching the vertical asymptote ( which is present ) at the surface, shown on the graph. These are two clues that the mathematics is not a perfect representation of the physical situation.*


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