MetaFrizzics

You can actually see an atom. First you trap it in an EM field.

Then you excite it and make it release light.

You don't need a super-microscope or special eyeballs.

You just need to think outside of the box.

A snapshot of this would be a TRUE picture of an atom.

Summary

 

Of course Newtonian Gravity, including the Sphere Theorem is well substantiated for large quantities of atoms, and is best and most accurately known at the Solar System range of distances. Also, at these distances, electrostatic forces are cancelled out on average quite quickly. In fact, using lasers and complex computations, NASA scientists claim an accuracy of many significant digits for the Inverse Square Law and the Gravitational Constant as we have calculated it. (I will discuss the merits of those claims in another thread.)

 

While the Sphere Theorem and Newtonian Gravitational theory generally can be challenged on any of the three axioms/assumptions previously mentioned above, only on the basis of discrete localization of mass can it be limited in the range of its fitness for physical applications. To challenge Newtonian gravitation on another basis would require formulation of alternate models.

 

QED.

Quantifying the deviation from a Flat Field

 

We can model the quantization of mass by using the pure (unintegrated) form of the equations we have already derived. This will leave the mass discretely distributed at any level of coarseness we wish:

 

 

Of course once you get over a few hundred charges, the perturbations in the field are individually small.

 

However, since we are really discussing the flatness of field inside a hollow sphere it is interesting to note that even with a few thousand charges there is a significant distortion of the flatness resulting in all particles inside quickly accelerating and sticking to the inside surface. Here I show 2000, and 8000 charges:

 

 

One can see that the force near the inside surface of the sphere is as high as the force a few radii away outside, even with as many as 10,000 or 20,000 atoms or charges.

Stability in unequally Spaced Configurations

Counterintuitively, the charges (if constrained to a sphere) could take odd positions, and find a minimal stable energy state, even though in the conventional way of viewing things, the charges are not 'equally' spaced on the surface, but are varying distances apart. What makes the arrangement stable is symmetry of repulsive forces over the surface of the sphere. Being confined to the surface puts additional constraints upon the motion of mobile charges and allows stable configurations not otherwise possible.

 

 

How can repulsive force allow a non uniform arrangement of repulsive particles over a sphere?

 

You will notice that the arrangements of successive rings of charges away from a given particle on the surface are quantized discrete sets of distances, and that the number of particles in each successive ring is odd-even-even-odd-odd. It is especially stable that the first ring is triangular and creates a strongly stable central saddle-point in the centre of the spherical cap on the surface.

 

 

This means for instance, the particle in the first ring (3 surrounding particles) is held in a stable position because of the triangular arrangement, combined with the physical curvature of the proposed constraining sphere (say an aluminium ball). The spherical cap forms a minimal potential energy well, literally! The saddle point is in the middle of the cap.

 

Similarly, the next larger ring of particles also pushes the particle into the centre of the polar cap. It is a physical minimax solution to the potential formed by other charges on the sphere. By simple extrapolation, (all particles on the sphere are geometrically identical) all the the particles are stable and so the whole lattice is stable.

 

Keep in mind that although this symmetrical pattern of charges is stable, this is *not* the general case with an arbitrary number of charges. It only works with certain numbers, like 4, 12, 20 etc. Other numbers will not spread equidistantly and retain correct symmetry.

One of the most elegant and interesting things about dimensions is that they are unique in their attributes and many a theorem or technique that works in one number of dimensions utterly fails in another.

 

Think about this following idea: Although if you had mutually repulsing charges on a ring or circle, they would spread out so that they would be an even distance apart, this is *not* true for a sphere: The particles would indeed spread apart to find a minimal energy rest state, but it would not be a regular or symmetrical pattern with an arbitrary number of particles.

 

Even with the circle where charges naturally space themselves out equidistantly, the interior is not stable.

In the graphs, magnitude of force is shown only, not direction:

 

 

In other words, you can artificially construct a sphere out of points and space them in a symmetrical pattern, but this is artificial. Real point-masses and real charges would not be so cooperative: they would seek the lowest energy state by a relaxation process of the lattice on the surface of the sphere.

 

Only certain special cases (certain fixed numbers of particles) will be able to space themselves equadistantly on the surface of a sphere.

 

In the general case, you can have symmetry, or real charges, but not both.

Now lets show why the Sphere Theorem is false in the real physical situation:

 

The problem here is that simply applying the result of the integration to the sphere is not an accurate description of the physical case.

 

To get right at the issue, since mass is in reality distributed in clumps, and is not a continuum, the Sphere Theorem is only a valid or useful approximation for very large uniform spheres at a macro-level, involving millions of atoms.

 

It is not in serious dispute that the bulk of the mass of atoms reside in the nucleus and this is demonstrated both mechanically and vis the gravitational field by the scattering matrix. Most scientists agree that this experiment has already been done to death.

 

(1) If we were to make a sphere out of a thin layer of gold atoms for instance, the actual gravitational potential field would not at all reflect the result of an integration of the continuum model proposed by Newton's Sphere Theorem, and so the integral is inapplicable to this problem. The *REAL* field would look more like a golf-ball, and there would be no flat field inside the sphere. All particles floating inside would accelerate outward toward the inside surface due to imbalance of forces, no matter how carefully the sphere was constructed.

 

(2) One might think, "So what? The Sphere Theorem fails at the molecular level. Big deal." But this is not the case at all. The failure is independant of size entirely. It is not tied to physical size, but to the coarseness of the quanization of the mass distribution. This would also be true for charge distribution as in both classical and relativistic electrostatics.

 

(3) You could also have a 3-meter diameter aluminium sphere which would for all intents and purposes would be a continuum. However, once the static charge on it dropped below a few thousand excess electrons or hole charges, even though these charges would spread out as evenly as possible due to repulsion, the electric field would be as lumpy as gravitational potential of the gold-leaf ball we referred to above.

 

This thought experiment is all that is required for any reasonable person to see that the Sphere Theorem is an approximation similar to the Center of Mass theorem, and it fails miserably in many physical situations. you don't have to be a rocket scientist to see this.

 

The Sphere Theorem is not accurate, practical or useful at distances of a few thousand diameters of a typical nucleus.

 

Here is a picture of a discrete distribution of mass or charges to help visualize the situation: clearly the field will be 'lumpy' near the sphere surface (inner or outer)

 

Part Five: The Failure of the Sphere Theorem in Physical Cases

 

Here is an elegant mathematical proof of the failure of the Sphere Theorem at sizes of about 2000 times atomic radii:

 

The failure of the Sphere Theorem is provable mathematically on many levels, the simplest being known mathematical theorems of topology and tesselation of the plane and spherical surface.

 

For instance, of course you can tessellate the Euclidean plane evenly with equal sized spheres (which would represent equal sized/spaced charges). Recall a ball is surrounded by exactly six equal sized balls, and forms a 'honeycomb' pattern of touching balls on the plane. (straight rows oriented 30/60 degrees apart)

 

 

This is impossible on the surface of a sphere. Lets see why:

 

It would require either varying sizes of balls, or symmetry-breaking spaces which could not accomodate a 'whole' ball. The physical result in either case would be an uneven field extending into the hollow area of the sphere.

 

Lets try varying the ball size. This is not possible in the electrostatic case, because charges have fixed discrete values, but with mass-clumps we can vary the density or spread of a clump to allow different size radii of equal gravitational strength in the plane of the tangent to the sphere's surface.

 

The requirement that adjacent balls touch is just the equvalent of equal spacing since the ball surface represents a surface of equal gravitational (or electrostatic) force around a particle.

 

Now for each concentric circle of balls around a given starting point, we can shrink the ballsize for that ring, to squeeze up the previous inner ball/ring. Balls can remain touching, but now notice that one geometric feature, the hexagonal shape of each ring gradually becomes the shape of the spherical approximation. The rings cannot conform to circles and still maintain contact with adjacent balls. The two constraints, keeping balls touching and conforming to the sphere surface are mutually incompatible.

 

 

You can quickly demonstrate this for yourself by taking a billiard ball and spraying it with a layer of mounting glue. Now dip it in a bowl of small ball-bearings, and try to push around the balls to make a uniform surface.

 

No matter how many balls one tries to cover a sphere with (even to the tens of thousands) there would be symmetry breaking, which is all that is required to create an imbalance of forces inside the sphere.

 

Covering the surface with spheres?

 

Recall that equal sized circles cover a fixed percentage of the spherical surface, so shrinking the circles even to infinitesimal sizes has no effect on the problem of covering the entire surface.

 

In fact, even Newton's simple program of using different sized circles to cover the surface is actually not mathematically possible:

 

There is another theorem that says you can't even fill a flat plane with circles, although the set of spaces left behind is a zero-measure set.

 

On the plane it is called an Apollonian Gasket:

 

Appollonian Gasket (click on this link)

 

Do any of you guys actually understand how male - female interactions really work? After all, we are in a science forum

"Oh Ace, you make me laugh!" (Ace Ventura When Nature Calls)

 

Don't you know the internet has changed everything? Men have 24/7 access the the hardest porn they can stand. And they download it by the ton.

 

20 years ago they were worrying about too many men turning gay. It's time to worry about intra-species STDs.

 

In the 80's Women's Liberationists dreamed of making men obselete, and keeping them as pets for studding purposes. Now everyone is obselete.

 

Jesus has less than a generation to come back and patch things up, or the big guy might as well stand aside and let Lucasfilms do the job for him.

 

Once again technology has come in to solve the problem of ensuring the rich get into your daughter's pants before anyone else.

3. Defining the Variables

 

Constants:

 

D = centre to centre Distance from sphere to test-mass.

R = Radius of the sphere = 1.

 

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 is 1. This makes the mass of each ring a simple constant:

 

[math]m_{sphere} = 1[/math]

 

[math] m_{ring} = \frac{m_{sphere}}{2N} = \frac{1}{2N} [/math]

 

Ring Force Formula: Substituting ring mass into our general ring formula we get:

 

[math]F_{ring} = \frac{m_{ring} * d}{(r^2 + d^2)^{3/2}} = \frac{1}{2N}*\frac{d}{(r^2 + d^2)^{3/2}}[/math]

 

Here now only d and r are variables.

 

Setting up Counters and Coordinates

 

( We will use the variable i as a counting variable for our adding 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 to this by increasing the number of slices, 2N. )

 

 

Coordinates: Although ring sizes come in pairs, the force for each ring is 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 in 2N steps.

 

Ring Distance:

Given the centre to centre distance D to the test-mass is just its own x-coordinate, we define ring distance in those terms, that is the difference between the x-coordinate of the ring and the x-coordinate of the test-mass:

 

[math]d = ( D - i / N ) [/math]

 

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 on either side of the origin.

 

Substituting into our formula:

 

[math]F_{ring} = \frac{1}{2N}*\frac{( D - i / N )}{(r^2 + ( D - i / N )^2)^{3/2}}[/math]

 

 

Ring Radius: ( via Pythagorean theorem )

When setting up our variables, we might be inclined to define the ring radius, r in terms of the angle phi directly. For instance,

 

[math]r = \sin{(\phi)}[/math] , and [math]\phi = cos^{-1} ( i / N )[/math].

 

Rather than insert the horrible [math]r = \sin{(cos^{-1} ( i / N ))}[/math] into our formula, we can get r more simply and directly:

 

Make a standard right triangle around angle [math]\phi[/math] (see diagram), by using the sphere radius R = 1 as one side and dropping a vertical to the X-axis where the ring will be. The ring radius is then just the y-coordinate. We solve for y using the Pythagorean theorem. :

 

[math] y^2 = h^2 - x^2 [/math] (…h = hypotenuse = 1 )

 

[math] r^2 = R^2 - ( i / N )^2 = 1 - ( i / N )^2 [/math]

 

( we won’t need r itself, just [math]r^2[/math] …)

 

So we only use the perspective of the angle [math]\phi[/math] to determine [math]r^2[/math] 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 [math]\theta[/math]. But again we can avoid all the trigonometry and angle-juggling by simply working with the distances directly anyway.

 

Substituting into our formula:

 

[math]F_{ring} = \frac{1}{2N}*\frac{( D - i / N )}{(1 - ( i / N )^2 + ( D - i / N )^2)^{3/2}}[/math]

 

Now the denominator will simplify further (squaring right-hand inner term):

 

[math]F_{ring} = \frac{1}{2N}*\frac{( D - i / N )}{(D^2 - 2D( i / N ) + 1)^{3/2}}[/math]

 

The total force, summing up all the rings (using our counter i) is now:

 

[math]F_{sphere} = \frac{1}{2N}*[\sum_{i=-N}^{+N}{[\frac{( D - i / N )}{(D^2 - 2D( i / N ) + 1)^{3/2}}] }][/math]

 

The summation sweeps i from -N to +N in 2n steps. We are able to pull the constant outside the summation as a multiplier. Finally, to make the formula more and more accurate, we want to cut the sphere into finer and finer slices. So the formula becomes exact when we take N to infinity as a limit:

 

[math]F_{sphere} = [\lim_{N \rightarrow \infty}[\frac{1}{2N}*[\sum_{i=-N}^{+N}{[\frac{( D - i / N )}{(D^2 - 2D( i / N ) + 1)^{3/2}}] }]]][/math]

 

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.

 

We can't teach here how to integrate a summation like this, but I have created a pdf format document to show those who want to see how it is done:

 

How to Integrate the Sphere Equation (.pdf) (click here for link)

 

Here ends the modern 'proof' for the Sphere Theorem. That is, the mathematical integration of the formula: The result of the integration gives us the following simpler formula:

 

[math]F_{sphere} = \frac{1}{D^2}*(\frac{D-1}{|D-1|}+\frac{D+1}{|D+1|})[/math]

 

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).

 

Is it an accurate reflection of the physical reality inside the sphere? Stay tuned.

 

I hope someone will benefit from the Herculean effort this took...

 

This concludes Part (4) of the first post (my outline).

© The Force for a Hollow Sphere

 

1. Preliminaries for Calculation

 

We can get the force that a hollow sphere exerts upon a 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:

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 the forces.

 

 

1) The radius R of the main sphere we will set to 1 and so all distances will now be in units of R. Little r will be used for the radii of the rings we will use.

2) 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 also in units of R. Little d will be used for the distance to each ring.

 

However, two problems pop up:

a) How do we figure out the mass for each ring?

b) 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, not only are the areas equal between the cylinder and sphere, but all the slices are equal to each other as well. If we chop the sphere into pieces of equal width, they will all have the same area.

 

 

That is, with rings of equal height 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.

 

This gives us the excuse to simply slice the sphere into N equally wide slices, knowing the mass of each slice will be 1/N, and they will add up to one.

You are right again. I used the term rather loosely for a high-school audience. The main point is to handle the vectors purely algebraically, rather than take time to cover vectors, when we can get by without such a course. The same tactic is used with trigonometry here: while we can't really avoid introducing it, we can keep the math simple enough to not require calculators with special functions, and allow even high-school drop-outs to follow along without too many problems.

 

I will probably make a few more technical slips, as I try to keep the entire thread informal enough for the widest audience. I hope the serious mathematicians don't have to suffer too much in this process.

 

(Also I'm still learning the version of 'Tex' used here...

 

Thank you again for your comments and interest.

(B) Force between a Barbell and a Test-mass

 

1. Setting up the Barbell

 

Take a barbell made of two spheres connected by a rod of negligible mass, and place it vertically on the y-axis, centred at the origin. This will be system A, with the mass = mA. The mass is divided equally so each end is ½ mA.

 

 

Now we want to move a test mass B along the x-axis. It should be obvious that the total force on the test-mass is just the sum of the pull from each end of the barbell. We can work out the force from each end separately, and then just add them. However, since each end pulls in a slightly different direction, the total force will be the vector sum of the forces from each end. We could use the parallelogram rule each time we move the test-mass, but we really just want one equation for every position (x) of the test-mass.

 

Now d is the basic centre-to-centre distance between the two systems. In the diagram r is the radius of system A, or the distance from the centre of mass to each end. And h is the actual direct distance from the test-mass to each end of the barbell. We need h to use in place of 'd' in Newton's equation. The angle Theta shows the actual direction of pull from each end, relative to the overall direction of pull. This pull happens to be toward the centre of mass of the barbell due to the balance of forces. We need the angle to correct for the direction of pull.

 

2. Net Force for the Barbell Using the Cosine

 

Any vector can be split up into a vertical and horizontal component. Here the vertical components from each end are opposing and equal so they cancel, leaving only the horizontal components. We can ignore the vertical parts.

 

 

The horizontal parts are in the same direction and we know what that is, so we just add them. We have reduced a vector problem to a simple addition. But what are the horizontal parts? The horizontal component is just the whole vector scaled down by the proportion shown in the triangle above.

To scale down the force we multiply it by the simple ratio or fraction formed by the horizontal distance d over the hypotenuse h , = d / h.

Those familiar with trigonometry will recognize this as the cosine = adj / hyp .

 

[math]F_{total} = F_{top} * \cos{\theta} + F_{bottom} * \cos{\theta} .........( eq. 1.3 )[/math]

 

Multiplying by the Cosine function correctly scales down the force from each end to the net horizontal component only. We then just add to get the total force in the X direction. Since angles for each vector are equal, we can simplify again, and rather than use cosine functions we prefer to work directly with distances:

 

[math] F_{total} = ( F_{top} + F_{bottom} ) * \frac{d}{h}............... ( eq. 1.4 )[/math]

 

The forces from each end are actually equal since the angles, distances, and masses are equal. The horizontal components are equal as well.

 

Now we're going to use Newton's Formula for a point-mass for each end of the barbell. We will simplify it after making a few calculations. (Notice we use the direct distance h , not d for the force calculation.)

 

So ignoring direction, the raw force from one end will be as follows . (We will set G and mB to 1 at the end as explained before):

 

[math]F_{half} = \frac{G ( \frac{1}{2}* m_A ) m_B}{h^2} = \frac{mA }{2h^2}...................( eq. 1.5 )[/math]

 

Since each barbell end includes only ½ mA, and we want 2 x this amount (the force from both ends), the two's cancel out.

 

[math] F_{total} = \frac{m_A}{h^2} * \frac{d}{h} = \frac{m_A * d}{h^3}.....................( eq. 1.6 )[/math]

 

This is almost right. But we want a version in terms of d and r without having to deal with angles or h . Now we get rid of h by substituting d and r back in using the Pythagorean Theorem:

 

[math]h = ( r^2 + d^2 )^{1 / 2}...............( Pythagorean Theorem )[/math]

 

 

[math]F = \frac{ m_{barbell} * d }{( r^2 + d^2 )^{3 / 2}}.............. ( eq. 1.7 )[/math]

 

 

This at last is the form of the equation that we want. We will be using this as the core piece to place inside other expressions and combine with other items later. Well, it wasn't too bad. Mostly a bit of algebra and some common sense.

 

( For those whose algebra is a bit rusty, recall that multiplication and division can usually be done in any order, and we can freely move multipliers and divisors in and out of brackets. ab + ac = a ( b + c ) etc. There is nothing special that prevents us from doing so here, since we have been careful in handling the direction of the forces. )

 

3. The Force for a Ring of Negligible Thickness

 

Ok, enough nasty algebra: Lets solve the next step with almost no math at all. We'll replace the shaft of the barbell with a perpendicular ring. The spheres can slide around like beads on the ring. If we keep splitting each mass in half and spread the halves out equal distances around the ring, the force will actually stay the same!

 

 

As long as pairs stay opposite one another the total force does not change, since all the mass remains at the same distance and angle from the test-mass.

 

Sliding a bead around the ring makes a cone of equal angle to the test-mass about the x-axis. Only the perpendicular components of the force change, but these stay balanced as long as we have even numbers and equal spacing, pairing off the perpendicular forces.

 

 

Although the case is now 3-dimensional, the equations stay the same. We can keep dividing the mass until it is equally spread around the ring as finely as we wish.

 

The real value of our equations is now apparent. They also define exactly the force from a perpendicular uniform ring of negligible cross-section. This is the kind of tool we can use to analyze hollow cylinders, spherical shells, and other related shapes. The equation we want for a 3-d ring is the same as the 2-d barbell.

 

[math]\displaystyle{ F = \frac{m_{ring} * d}{( r^2 + d^2 )^{\frac{3}{2}} }..............( eq. 1.8 )[/math]

 

 

Now we are ready to move onto the main task: figuring out the force for a sphere.

2. Simplifying the Problem

 

Part of solving any tough task is simplifying the problem any way we can.

Newton's equation for the force between two masses looks like this:

 

[math] F = \frac{G m_A m_B}{d^2} .......................( eq. 1.1 )[/math]

 

For our purposes we can simplify this equation quite a lot before we even start:

 

(1) 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.

 

(2) We can set the Gravitational Constant (G) 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.

 

(3) 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.

 

(4) We will place 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).

 

(5) 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:

 

[math] F = \frac{1}{d^2} ................................( eq. 1.2 )[/math]

 

This is the essential and active part of the equation, which says the force falls off according to the distance squared. For now we are going to leave in the 'mass' variables however, to make a few steps in the next section clear.

Calculating the Actual Equation for the Force of a Hollow Sphere

 

Now we begin (4) of our first post, showing the modern mathematical structure.

 

If you know a bit of algebra, you can understand the Sphere Theorem Equation. As promised there, we will present the math at a high-school level (no mean feat in itself), so that anyone should be able to follow all the steps. We will avoid Calculus entirely (until the very last step). This will require a bit of special preparation for ordinary readers. Here is the outline for the next section:

 

1. Overview

 

(1) First we will figure out the equation for the force between a perpendicular barbell and a test-mass. The reason we need to do this is because it will give us a piece of the puzzle we need to calculate the force for a hollow sphere.

 

(2) Next we show that indeed the equation for a barbell is exactly the same as the equation for a perpendicular ring! This is a surprising but simple result that requires no difficult calculus.

 

(3) We will plot the way the force changes as we move a test mass closer to the ring, and even through it and out the other side along the x-axis. This will give us a feel for how the force can behave for different shapes.

 

(4) Then we will begin setting up the job of calculating the force for a sphere. This will involve picking a method of attack, and working out some ways to simplify it.

 

(5) We will look at Archimedes' Theorem, which will help us understand why we can keep the problem simple, and divide the sphere in a straightforward way.

 

(6) Finally, we will set up the equation and procedure, by building up the pieces needed, using some trigonometry and summation methods.

 

(7) We will look at how we can improve the accuracy of our approximation by increasing the number of slices taken to a limit. This leaves us ready to do a simple integration, which will provide us with an exact equation for the sphere!

 

Some conventions: We will use Capital Letters for main objects, and small letters for parts of objects, distances and masses. We will use subscripts to identify quantities belonging to various objects, to distinguish them. Greek letters will stand for angles. Special cases will be explained as we go.

Thanks for the corrections of detail.

I thought these pictures would be helpful for those just starting out.

I modified some original pics found on the web (quite significantly) for clarity,

and added pics and text of my own. Textbooks are usually so dry that the reader is left with no picture at all (figuratively as well!).

Algebra, Field Theories, Physical Applications and Superposition

 

Algebras

One universal problem with using algebra to formulate physical laws is the mathematical quirk of the 'divide by zero' error and its nonsensical meaning or lack of definition. The moment we define a relationship with a simple equation involving multiplication, there will usually be a way to rearrange it to form a fraction, creating a problem where there was none before, and disallowing certain values. Newton's Inverse Square Law of gravity is no exception:

 

[math]F = \frac{1}{d^2}[/math]

 

Here the equation has no meaning for a distance of zero. This does not imply anything about the physical state of the field: It's just an artifact of the mathematical rules of algebra.

 

Classical Field Theories

Let's reformulate Newton's Inverse Square Law to take into account direction of force (make it a vector equation using the x-coordinate). We do this by placing the point-particle at the origin, and imagine moving our test-particle along the x-axis.

 

[math]F = \frac{1 * (x/|x|)}{x^2}[/math]

 

(Here we use the Absolute Value function to toggle the sign and give direction to the force).

 

 

(1) Notice here the 'divide by zero' problem creates a vertical asymptote, where the equation has no meaning. If we approach the origin from one direction or the other, however, we notice that the force goes off to infinity in opposite directions. They cannot converge. (We may assume either a discontinuity or a failure of the equation at close distances for the physical case.)

 

(2) The second problem with Newton's simple Inverse Square Law is not so easy to deal with: That is the infinities themselves. All simple point-particle/field theories suffer from these 'infinities' blowing up the equation as limits are approached. Interestingly, the extension into space of the particle (making it no longer a true 'point'-mass) eliminates the infinities, because we can invoke Newton's Sphere Theorem in this case! The Inverse Square Law gets substituted by a new equation (for a solid sphere) derived from the original law plus the Sphere Theorem.

 

In the following graph, the infinities are eliminated by chopping them off at the extended particle surface. For a solid sphere, the force decreases linearly below the surface instead of increasing in a runaway fashion:

 

 

Physical Assumptions & Superposition

We can however make reasonable assumptions about the force as the test-particle occupies the origin.

 

(1) The reasonable assumption is the forces are balanced whatever they might be, and the net force is zero (not +/- infinity!). This is the assumption behind the 'axiom' that a particle's own field has no effect on itself, and this allows us to add forces as simple vectors without worrying about or supposing complex interactions between the 'fields' of different particles, like refraction, diffraction, interference etc. (This assumption isn't out of nowhere, but is 'weakly' based upon the known symmetry of the situation.)

 

Some more Assumptions

Even later when Newtonian Gravitation was reformulated as a Classical Field Theory, several simplifications were inevitably retained:

 

(2) Instantaneous Action at a Distance: Like the Classical Electrostatic Field, speed of propagation or application of the force was not taken into consideration, and time effects were ignored.

 

(3) Assumption of Continuous Mass: Little was known of Atomic Theory as well when Newtonian gravity was given its Classical Field formulation, and it is not surprising that using the new methods of calculus, objects with a physical volume in space were treated as if the 'mass' itself was essentially a continuum, not a discrete collection of particles.

 

Now we can see three significant 'axioms' or assumptions that went into the formulation of Classical Newtonian gravity, all of which could be challenged on straightforward scientific grounds reflecting our current state of knowledge.

Just because some particular explanation is flawed does not mean a statement is incorrect.

Quite true. However one quite valid purpose to showing the errors of previous arguments and explanations is to understand them and improve them. This is precisely how Einstein eliminated the aether and reformulated Maxwell's electromagnetism into SRT. This is in fact how science generally advances: by successful challenge and correction of popular ideas.

Why go to all these lengths of proving that different proofs are flawed?

One of the most valuable reasons to do so, is to open the minds of others to the possibility of an error in the result as well as in the proof itself. This is the correct and only way to proceed if one wants to convince others of a truth one has discovered.

That does NOTHING to prove that the property in question is untrue. The ONLY WAY to disprove the theorem is to calculate the field inside and show that it has a non-zero value somewhere. I have no idea if your intention is to disprove the theorem...

True again. If you refer back to my outline in the first post, you will note that we are about to begin step (4), which will involve a modern mathematical formulation of the Sphere Theorem and an analysis.

Are you going to calculate the field inside a sphere and show us how big it is - in other words are you going to show us the magnitude of the error?...

In a word, yes. As previously stated, we will show that contrary to popular beliefs and claims printed in some textbooks, the Sphere Theorem is an APPROXIMATION just like the Centre of Mass Method. There are cases and ranges in which it does not apply to physical situations, and we will explore those.

 

In doing all of this, we will not be significantly straying in any sense from 'mainstream' scientific method or theory. We will be doing this using standard methods and uncontroversial reasoning processes. We will simply be correcting and explaining some faulty notions about a physical description of gravitational forces.

what does your tirade have to do with us arguing against this guy's desire to seduce a young girl so that he might be able to score a mint condition toy?

How we wish it were: "Talk this guy into taking one of the used models. Sidetrack him with this video and then headfake him with these mail-order bride catalogs."

 

How it really is: 14 yr old girl to friend: "Is he rich? I mean really rich? Yeah okay"

Alternative Ways of Saving Newton's Sphere Theorem

 

All is not lost yet however, because we can relax our requirements: The force does not need to be balanced at the vertex of the cones, because we still have the rest of the sphere to account for. We could still save the Sphere Theorem for Newton if we could show that any net force produced by the polar caps was counter-balanced by forces from the ‘pumpkin’ shaped remainder of the spherical surface:

In the diagram, the polar caps are exerting a net pull to the right on the test-mass. To compensate for this, we must propose an equal and opposite counter-force pulling to the left, from the remainder of the sphere, as in the diagram.

The curves below each picture show the strength and direction of the force from the system as the test-mass is moved back and forth along the horizontal axis. Upward on the graph means a pull to the right, and downward means a pull to the left.

 

We know what the basic forces will look like between the caps. The position will be unstable: pull will increase as the test-mass approaches either end, and will accelerate toward the nearest pole.

 

To counter this, the forces inside the midsection must do the opposite and possess a stable equilibrium point. Again, due to symmetry we assume only horizontal net forces are possible. But now we have moved from Newton’s incredible idea that polar caps balance, and hence are equal to flat disks, to a new idea that can only be described as fantastic:

 

The changes in force on the test-mass from any arbitrary pair of polar caps are exactly matched by the changes in counter-force from the remaining mid-section. This is the hidden demand of the Sphere Theorem!

 

That there could be a finite construction that could satisfy this demand is incredible. That it should be a radially symmetrical sphere seems miraculous and fortunate. It goes without saying that a proof for this solution is hardly trivial.

Disproving Newton's Argument for the Sphere Theorem

 

(1) Not all points in the sphere are equal.

 

Firstly, all points inside the shell are not equal. There are two distinct sets of points inside a sphere. Set 1 contains only one point, the origin, or geometric centre (GC), equidistant from all points on the surface. Set 2 contains all the other points. All the other key geometric features follow from this.

 

Let P be a point anywhere inside a sphere, but not at the centre. Only lines passing through a sphere's centre will pierce the surface perpendicularly. All other lines pierce the surface at some other angle. So a line passing through P must also pass through the centre to be perpendicular at the surface. Only one line passes through both P and the centre, and is perpendicular at the surface.

 

(2) Most of Newton’s Cones Cut Sphere on an Angle

 

Similarly, since a cone-axis is a line, only one cone-axis passing through P is perpendicular at the surface. Only cones formed on this axis will make perpendicular disks or spherical caps. Cones on some other cone-axis will cut tilted ellipses or spherical caps.

 

Newton’s scheme is to cover the surface using cone-pairs. Each pair must have a different axis, so only one pair can be perpendicular at the surface if P is not at the centre. All the other cone-pairs pierce the surface on an angle, and make caps or disks which are tilted relative to P.

 

(3) Tilted Disks Pull Off-Centre

 

A point-mass is only pulled directly toward the centre of a uniform disk when the point-mass lies in the same plane as the disk or when the point-mass lies on the axis of the disk. That is, whenever a disk has any other tilt relative to the point-mass, there is a residual force toward the nearest disk edge, pulling the point-mass off-course from the disk centre. This is because the Centre of Mass theorem fails in close proximity or for significant spreads in distribution of mass, since it is only an approximation.

 

(4) Tilted Disk Pairs on Sphere Don’t Balance Out

 

Could disks balance their forces in spite of tilt? Yes, but only between two (infinite) uniform parallel planes, where the tilts cancel. Then masses between the planes can indeed experience zero net force. Newton's claim for hollow spheres actually turns out to be true (in theory) for parallel planes

 

 

We can understand intuitively that even though the direction of force is off-center because of tilt, we can exactly counter that with an equal and opposite tilt on the opposing disk, without having to correct the direction change from the tilt. But on the sphere this is impossible: The disks actually double the error. This is why Newton's thought experiment works with parallel planes, but not with spheres.

 

 

(5) Infinitesimal Limits Don't make the Imbalance Vanish

 

Newton himself posed that making the cones very small (infinitesimal) would eliminate any accuracies, (one problem is the fact that spherical caps aren't really flat disks at any size), but Newton didn't foresee the problem of tilt.

 

- - - - - - - - - - - - - - - - -

Tilt indeed vanishes or becomes insignificant if the disks are made very small relative to their distance from the test-particle. However this is just a kind of sleight of hand;

 

 

(a) The problem isn't tilt per se, but the actual distribution of mass, in this case manifested as tilt with larger disks.

 

(b) But just as a curve can be chopped into infinitesimally small straight lines, while it's curvature remains constant, this has no effect on the actual distribution of mass, which is fixed. We can measure things many different ways, but this doesn't move the mass.

 

The information concerning distribution of mass can be contained by the tilt, or translated into other forms, but problem of the imbalance of forces remains, and Newton's argument is a failure.

 

This section shows that Newton's argument is wrong. If the Sphere Theorem is correct after all, it must be correct for reasons other than those that Newton gave.

Waves are waves. What is a probablity wave?

This seems like a fun question to answer:

 

You have a wave function (a function of time or position).

This represents in a 'one step removed' fashion the probability of a particle's position being someplace or in some state. You get the actual probability by squaring the instantaneous amplitude of the wave at a given time or place.

 

This (always positive) number which is less than or equal to 'one' represents the probability of the particle being found at a certain location and/or time. i.e., .5 means a 50% chance of the particle being found there.

 

 

Notice that the Wave Function isn't directly describing the photons travelling through space: Instead the wave equation (properly set up and with the portion of interest isolated) describes the probability of a photon hitting a given surface, defined some fixed distance away from the two slits. So if enough photons hit the surface, a spotted pattern develops which approximates the probability wave (wave function squared) for that surface:

 

Wow everybody's position is so far from any kind of moral or ethical high-ground that it really doesn't matter what Hwang thinks does it?

 

Marriage - biblical tradition: meaningless for atheists. Why do you want it?

 

Biblical Law - you break you buy. marriage for non-virgins *not* equal to marriage. There's another ceremony for premarital sex: its called stoning.

 

Only place to find a virgin in USA is in the under 15 category? Oh well, that's what happens when an entire society abandons religious values for hedonism. See you boys in hell. Shouldn't have sold those condoms and birth control pills over the counter like candy, and pretended they were safe substitutes for common sense and caution.

 

Will civilization be destroyed by atom bombs? No. It will be destroyed by STDs (AIDS is wiping out a continent now), and breast cancer caused by yet another product (the pill) that "doesn't cause cancer".

 

The latest horror movie in 3-d: Return of the cigarette company Ad agencies, working for German drug companies.

The 'multiple realities' idea is an interpretation of Quantum Mechanics by Everett: his 'many worlds' interpretation. It is totally foreign to SRT.

 

The only multiple thing in Minkowski space is the 'movable' plane of simultaneity, which cuts a different slice in spacetime for each observer with a different velocity and location.

Another short answer is that according to current SRT theory, once the first train is moving at .9999C all the trains are. And it would take an almost infinite amount of energy to cause the second (model) train to move at all, relative to the carrier train.

 

Remember that in both SRT and GRT that while intertial frames are 'relative' (moving fixed velocities relative to each other), acceleration is ABSOLUTE.

i was dating a senior when i was a sophomore(major cool points, btw). i was 15 and she was nineteen.

"You lucky, lucky b******!. Someone must think the sun shines out your ***" - Life of Brian (fellow prisoner)

Okay, 15 is too young, and 19 in California means you've sailed every porn studio casting-couch in town (twice)... Let's split the difference, and say 17: old enough to know all the tricks but not confident enough to charge the full price.