By Rogue Physicist, Mastervalver and Metafrizzics
(C) 1999, Updated Aug 24, 2005
Synopsis: Part 2 In this section we derive some tools and give a mathematical treatment of both hollow and solid spheres. Then follows an analysis, which tries to lay bare the assumptions supporting the bridge between the mathematics and the physical reality. In the following section, we show in detail how the Hollow Sphere Theorem breaks down at sizes smaller than about 10,000 times the diameter of an atom. The results also apply to spheres of any size having low numbers of charges, in the range of < 10,000 electrons or holes. This is obviously a significant finding, for which test by experiment is a reasonable expectation.
(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
Introduction: Simplifying the Problem
Part of solving any tough task is simplifying the problem any way we can.
Newton's equation for the force looks like this, which seems complicated:
For our purposes we can simplify this equation quite alot before we even start:
- We can set the radius of the sphere we are going to measure to be '1' . We will measure all other distances in units of this radius. This is important, because the geometric features we'll find are all relative to sphere size.
- We can set the Gravitational Constant to be '1', by choosing the right units of mass. The Gravitational Constant is not an 'absolute' constant. It really coordinates mass with units of distance, and is set by our choice of units.
- We can set the mass of each body to be '1' as well. The mass of an object is usually constant. Later we can extend our findings to bodies with different masses and arbitrary sizes.
- We will centre the Sphere at the Origin of our coordinate axis, so the geometric centre (and centre of mass) will be: (x,y,z) = (0, 0, 0).
- Our Test-mass will just move along the X-axis. The position of our test-mass will just be the X-coordinate. This will greatly simplify our calculations.
Since G x ma x mb = 1 x 1 x 1 = 1
Newton's equation is now simply the pure Inverse Square Law:
1. Force between a Barbell and a Test-mass
Take a barbell made of two equal uniform spheres, rigidly connected by a rod of negligible mass, system A. A fixed distance away from the barbell, is another uniform sphere, test-mass B, the same mass as the barbell. 1 It should be obvious that the total force on B is just the sum of the pull from each end of the barbell. We can work out the pull from each end separately, and then just add them. But since each end pulls in a slightly different direction the total force is the vector sum of the forces for A1 and for A2. 2
r is the radius of system A, and d is the basic centre-to-centre distance between the two systems. h is the direct distance to each end, and the angle Theta q shows the actual direction of pull from each end relative to the overall pull, which happens to be toward the geometric centre of system A, due to symmetry.
1. The geometric centre (GC) of an object is determined by its shape, i.e., distribution of volume and extension in space. The centre of mass (CM) is determined by the distribution of mass. The CM happens to coincide with the GC in system A , since it is symmetrical in both shape and distribution of mass . These two are not the same as the centre of gravity, which is only relevant in a uniform gravitational field.
Different again is an equivalent point-mass position, which we have to calculate. By equivalent point-mass, (or EPM), we mean a point-mass having the same mass as the system we are replacing. By EPM position, we mean where we must put the EPM to generate exactly the same force as the system. The Sphere Theorem (ST) only claims that the EPM is at the GC for spheres, not barbells or other shapes, so it is natural that we have to calculate it.
2. If we were just solving an example, we could combine the vectors using the parallelogram rule on graph paper, but since we want a general solution for all cases, we want to solve algebraically. This is easily done by using trigonometry.
2. Getting the Net Force for the Barbell Using the Cosine
Any vector can be split up into a vertical and horizontal component. Here the vertical components for each sphere in system A are opposing and equal so they cancel, leaving only the horizontal components. We can ignore the vertical parts.
Multiplying by the Cosine function correctly scales down the force from each sphere to the net horizontal component only. We then just add these components to get the total force in the X direction.
Substituting the definition of cosine (adj / hyp = d / h), and using Newton,3
These are good given ( h and d ) or ( h and q ). To get it for ( d and q ) we pick a multiplier and combine it in:
What is the physical difference between equation (1.2) and (1.3) ? In terms of h, if we spread the barbell ends apart in an orbit around sphere B, keeping the distance h to each ball constant, the force is proportional to the cosine. But if we keep d constant (the distance between systems) and just lengthen the barbell, the spheres spread apart vertically, and the force drops as the cube of the cosine!
3. Note that we use the hypotenuse for the direct distance to each sphere from B, and then get the horizontal component by multiplying by the cosine of the angle from the horizontal.
3. The Equivalent Point Mass is NOT the Geometric CentreThe last result is practical, but if we want a version in terms of d and r without having to deal with angles we can take equation (1.1) and do this:
What does it mean? As we shorten the barbell the angle q shrinks, r vanishes, h approaches d, d in the numerator cancels out and we simply get Newton’s formula. If we lengthen the barbell q approaches 90o , 1 the force falls away to nearly zero, even though the GC and CM hold position and the mass also stays constant.
Clearly, when the force is dropping then the EPM must be moving further away, off of the geometric centre and CM. So generally, the centre of mass is not relevant for calculating the force between the two systems.
The key here is that the EPM is simply the viewpoint from sphere B. It is only meaningful in relation to some fixed external position. If we swung sphere B around to the other side, the EPM would also move to the opposite side: A non-spherical object has a different EPM for every possible position around it, and each object will feel a force based upon its own personal viewpoint.
1 Keep in mind that the angle is not between the two spheres from B , but between the horizontal line ( d ) and each sphere. The angle is always < 90o and the cosine is always positive.
4. The Force for a Ring of Negligible Thickness
Lets replace the shaft of the barbell with a perpendicular ring around the axis. The spheres can be anywhere on the ring, as long as they are opposite one another. The spheres can be replaced by equivalent point-masses. We can even split each point-mass in half, assign each ¼ the total mass, and spread them out equally around the ring. We can do this without changing the total force, since all the mass remains at the same distance and angle from sphere B. Only the vertical components of the force are affected by rearranging, but these stay balanced and cancel. Although the case is now three-dimensional, the equations remain the same. This process can be continued ad infinitum until the mass is equally spread around the ring in increments as fine as we wish.
The real value of our equations is now apparent. They also define exactly the force of a uniform ring of negligible cross-section perpendicular to sphere B. This is the kind of tool we can use to analyze hollow cylinders, spherical shells, and other related shapes.
5. Plotting the Force for the Ring
We can now plot the force for the ring with a radius of one unit, placed at (0,0,0) on our coordinate axis. The test-particle is simply swept along the x-axis, ( d = x ) and we can plot the results by superimposing the graph and the ring in space.
It took a couple of pages, but it was worth it. We have shown that the force weakens as a system’s mass spreads out perpendicularly. We have also shown that the true EPM is not at the GC but further away. And we have an elegant and powerful equation for the force of a ring. No fancy calculus was needed.