Disproving Newton: Discussion Continues with DavisBJ

Okay, thank you for reading through my last posts.

I think there may still be an issue of language here,
in regard to your point (12) , where you introduce "right-angle".

I think however I can cut to the chase with a new diagram:




This diagram asks the pertinent question,
namely where will the resultant final vector end up?

Could it be that by 'luck' or by the order of the universe,
that after all is said and done, the final vector turns out to
align at least with the Cone-Axis, since we know it won't align
with the original GC axis as posed by the Center of Mass idea.

In fact, there is strong appeal in this argument, on the same
Euclidean basis that we concluded that the Cone-Axis was fixed
by the angle of the plane from the GC line.

We may also (although it will be a leap for some) "by inspection"
conclude that WHATEVER answer the Gravity-Law may finally produce,
the PROPORTION of the Vectors for the Near and Far halves will be constant.

This intuitive eureka is probably quite right (although it may not hold
for just ANY hypothetical Gravity Law).

If this is so, we naturally hold out hope that the final Vector may indeed
be locked onto the Cone Axis.

Reminder: We have shown before that this is NOT a necessary condition for the Parallel Planes / Opposing Cones Law, since wherever the final Vector goes, there is a complimentary tilt on the other cone that keeps the two vectors on a single line.

However, I strongly suggest you look now TWICE at the diagram above:

The following additional observations can be made:

(1) The lower force component Vector will always be BELOW the GC line.

This is a natural result of ALL the mass for the lower half being below the line.

It also incidentally shows the failure of the Center of Mass idea, since even when the ellipse (or circle) is perpendicular to the GC line, the component vectors will not be on the line, and the force will be proportionately weakened, AND the Equivalent Point Mass (EPM) will actually have to be placed on the GC-line FURTHER AWAY than the plane of the disk/ellipse, from the test-mass. This is nonetheless a natural result, since spreading the mass over a plane has the effect of moving the mass further away from the test-mass, and naturally weakens the gravitational force.

(2) This constraint will require the UPPER Force Vector to be a MINIMUM
magnitude and length (for any expected direction), to pull the final
Vector into line with the moving Cone Axis, as the ellipse is tilted or
elongated.

I say this comfortably, but hoping to avoid doing a calculation,
that "by inspection" this cannot be likely for all angles of the ellipse.

In the diagram above, for instance, the Near-force is 3x the magnitude
of the Far-force Vector, so that the Final vector is aligned with the Cone Axis.
Is this a plausible ratio of magnitudes for the diagram before us?
To change the magnitudes to something more sensible, you must
also change the direction of the component vectors to something less plausible...


The reason we ask this question, namely can we align the forces with the Cone Axis,
is because although this isn't sufficient to save Newton's argument,
(it still has other gaping holes), it would be a minimum necessary requirement,
for the analogy to hold.

And it would be a remarkable result in itself, which would go far beyond
proving the Parallel Planes Theorem.

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

Here's another couple of diagrams to assist in visualization:

Remember that it was Newton who composed the original problem,
and used opposing Cones with aligned Cone-axes.

Irrespective of the shape, or regularity of the cones themselves,
(Newton chose circular cones, but elliptical cones could also be used,
in a similar thought experiment),
the final opposing force Vectors must fulfill TWO conditions:

(1) They must be aligned in opposite directions along a single axis.

(2) They must be equal in magnitude.


Now consider the simpler case of parallel planes cut by Newton's
opposing cones, aligned on a single axis:



In the parallel planes case,
the the opposing forces must still share a single axis,
and be equal in magnitude, but that axis can pass through
the test-particle in any direction.
(any direction that is, along the vertical line of symmetry of the ellipses).


In this case, since the forces for each ellipse are complimentary
in the geometrical sense, and equal force components will also align
on a shared axis opposing each other, it is trivial that all the forces
cancel, and there is no NET force on the particle.

The only constraint imposed here is that the planes be parallel,
which forces the alignment of the component Vectors,
and vertical symmetry down the middle of the array on
the plane through both axes.

Now look at the REAL case that Newton proposed, where
the ellipses could actually tilt in different directions and are
NOT simply intersections of parallel planes;



Here a very significant extra constraint must be imposed,
SINCE THE OPPOSING COMPONENT VECTORS ARE NOT EQUAL
in either Magnitude or Direction:

Now they must be coordinated such that however they were acquired,
the RESULTANT FINAL Force Vectors remain on a single shared line,
and equal in magnitude.

That is, removing the constraint that the component vectors be equal and opposite, imposes another two constraints: (1) that they nonetheless combine to align on a single axis shared by each side. (2) that this line pass through the test particle. There is already an unspoken constraint that ALL the final force lines under discussion fall upon the vertical plane passing through the GC line and the Cone Axis, since this is the plane of bilateral symmetry for the system.

To fulfill Newton's original conditions, we must still have opposing cones,
and they must share an axis of symmetry, even if they are not circular
(i.e., radially symmetrical). We assume circular cones for now,
since we need not arbitrarily change the original problem.

In the diagram, it can be seen clearly now that
the Geometric Center line, must now be BENT, into two separate lines,
in order to maintain the condition of opposing cones.

The opposing Ellipse also slides downward, to keep it aligned with
the opposing cone. (we keep the original left GC line and particle fixed,
as per previous convention).

This makes the new GC lines of no use for balancing opposing forces,
and we must look to other lines that can remain straight while
passing through the test particle.

By simple inspection, the only line with the required symmetry
and potential to function as a balance of forces is the CONE axis.

We must presume then intuitively that the component forces for
each half-ellipse must coordinate such that the final Vector is the Cone axis.

Furthermore, we may further reason by inspection that since the
new location of the Cone Axis relative to the ellipses, pierces them
off-center from the original GC of each ellipse, the NEW division
into "portions" of each ellipse must fall on this piercing-point.

This is where your reasoning presumably comes in.

That is, on the one hand, it is trivial that dividing the ellipses
down the middle vertically gives symmetry and equal forces,
due to symmetry.

But it is NOT trivial to show that the "horizontal" division
of the ellipses above and below the Cone Axis must have equal forces,
which is actually REQUIRED if the new orientation is to balance.

Because the new PLANE aligned with the Cone Axis chops the
ellipses into unequal pieces, it must now be shown that their
distances and distribution of mass perfectly balance, which is
equivalent to saying that the force from each of the FOUR 'portions'
are equal in magnitude and direction.

This new strict condition must hold true for every complimentary pair
of angled ellipses.

Essentially we are describing the condition that the Force from
every such ellipse fall upon the Cone Axis at all times.

----------

We should now be ready to trivially disprove the Cone-Axis idea:

If the forces on each side of the new division of the ellipse balance,
and result in a force along the axis, then the two portions are intechangeable.

That is, the left and right side of the diagram below should be equivalent:





But by inspection, this is unlikely...

At this point, the idea that the smaller amount of mass on the left,
NEARER to the test mass, can balance out the larger mass on the right,
farther away, must be abandoned.

Its not about mere proportions anymore.
We leave this proof to the student.

---------

You should also have a second look at these three points you made:

9) Under the premise (which I do not contest) that the near half EPM is above the cone axis, the direction of its force vector to the test particle will be at a finite angle to the cone axis. 10) Similarly, the far half EPM is below the cone axis, and thus its force vector will be directed at an angle to the cone axis. But since it is below the cone axis, the angle will veer off the cone axis on the opposite side from the near side–to-cone axis angle. 11) Geometry dictates that the magnitude of the far side-to-cone axis angle will be smaller than the near side-to-cone axis angle

(9) I think is where you have have gotten a bit muddled again:

I would not assert that the near-half EPM is above the Cone Axis.
That would be yet to be determined.

What we can assert from the initial argument is that the
near-half EPM is above the GC axis.

Whether it coincides with, lags, or surpasses the moving Cone Axis,
is the open question.

I would suggest from the diagrams that although we have been
exaggerating the directions of the component Vectors for visualization
purposes, in fact they are less extreme than the tilt of the Cone Axis
away from the GC axis.

You rightly note that the lower component vector tilts downward at
less of an angle than the upward tilt of the upper component vector,
but we should be talking at this stage in terms of the GC axis, not the
Cone axis, which is also moving as the ellipse is tilted.

Your (11) will stand, provided we are talking about the angle in relation
to the GC axis, not the relation to the cone axis, which is unknown
from the discussion to that point.

Quote:

Originally Posted by DavisBJ I am going to shift gears a little bit. I contend that in broad scope, we are asking what is the net gravitational force on a randomly placed particle within the shell. To do that, we will have to consider every bit of the shell, and ask ourselves what is the vector total of every bit of the shell, no matter how we chose to chop the shell up into manageable pieces. For now, the “forward” and “backward” elliptical pieces will do, as long as we ultimately account for the whole shell.

I wholly understand your natural inclination to revert to a comprehensive approach, in fact a solution via calculus.

Please remember that the modern calculus solution was never in dispute,
as a mathematical structure.

In fact, we went over the mathematical solution to the problem via calculus,
and did the complete process including every stage of integration.

Finding fault with the calculus itself was not our thrust in any way.

As a physicist however, I objected originally to the calculus solution to
to the problem, not as a mathematical techique asserting something
about mathematical entities, but rather as an application to a physical
situation.

My objections there involve in particular the spatial distribution of
mass, which is "quantized" or rather non-smooth at the scale of
atomic particles; i.e., the use of the continuum to solve this problem
fails directly as a result of the clumped distribution of mass as particles.
This quantization of mass distribution results in near-proximity
imbalances and anomalies in the gravitational field, that prevent
the results of the Sphere Theorem from being relevant and applicable
to a physical situation of that type.

The problem there at distances and sizes within range of molecular
structures is perfectly plain, namely that there in fact is no continuum,
and so no balance of forces is possible in spherical structures like
Carbon micro-spheres and 'Bucky-ball' type structures.

The Sphere Theorem fails however, as we've observed, on several levels,
particularly involving misunderstandings regarding the Center of Mass
techniques and reasoning based on an incomplete comprehension of
those approximations.

Thus, other anomalies and inaccuracies are expected, and naturally,
other measures must be taken to bring the early ideas of Newton
in regard to application of gravitational formulas in line with a
properly self-consistent and coherent gravity theory with applications.

Quote:

Now we look at the tilted ellipse that we have discussed at length. We conclude that no way does that piece result in a force aligned with the cone axis that outlines the ellipse. Fine, it is what it is. We have (in abstract) computed the magnitude of the force that ellipse has on the test particle, along with its direction. Part of job done. Get out a tally sheet out, and record “ellipse 1 results = qq Newtons of force directed in such and such a specific direction. There is no restriction on which way I next turn the cone from the test particle, as long as I faithfully follow our procedure of vectorially adding up the forces due to the “near side” and “far side” halves of the enclosed ellipse. Whatever that answer comes out to be, we will add that to the tally sheet and then move on. So, to make sure we don’t leave bothersome gaps, I am going to rotate the cone so the cone axis just skims the edge of the far half of the ellipse we just finished with. Now, we consider the new ellipse, and its two halves. The new ellipse far half is now even farther away than the far half of the prior ellipse. And the near half … the near half … the near half is mostly the old far half of the prior ellipse. But we have already accounted for the effect that has on the test particle, it’s already a major part of the figure on our force tally sheet. We can compute the force contribution of this overlap of the old and the new ellipse in the net force from the first ellipse, or the second, but we don’t get to double dip. That overlapping segment of the wall contributes one force vector, and where do you want it counted – as part of the first ellipse net force, or as part of the second ellipse net force?

Your idea of overlapping ellipses is interesting.
I'm not sure what advantage it is going to procure
in an analysis of the problem, but please go ahead:

In particular, can you sketch a few diagrams of what you are intending?
You can use an online drawing program and link to it if you want to.

--------


Actually one of the things I'd like to do before abandoning this thread
to the other trolls, is to make what should be an obvious observation:

DavisBJ, whoever he may be, appears to me to be 'legit',
in the sense that whether or not he is a physicist of some branch
or other, he appears at least to be an engineer.

I acknowledge this from his plain pattern of behaviour:

(1) Although not all engineers/physicists are equal, especially in areas
outside of their expertise and interest, all engineers that I've met,
don't let a matter of science rest until they understand it, and typically
they will try and solve the problem, albeit in their own way.

This is precisely what DavisBJ has done.

And on that note, I have to confess that his first pass at explaining
his own calculations are not at all clear (I didn't expect them to be),
and he has not provided diagrams. This is unfortunate, for it is also
unclear to me (and I am at least very familiar with the problem),
whether he has indeed got a proof, and even whether in his own way,
he has come around to acknowledging that there in fact is some pull
on a particle inside a uniform hollow sphere.

It would really appreciated by me (although that is not essential),
if I could get him to clarify just what he has seen in his own calculations,
because at least others, for instance Stripe and many readers,
will want to know what side of the question he comes down on,
and will want to also see why.

We can put DavisBJ's actions in direct contrast to those of gcthomas,
who simply assumes he understands the questions, makes no analysis
of his own, shows no interest in the problem other than to contradict
the claim without proofs or mathematical evidence, only seeks to
argue a smokescreen about Newton's Principia in Latin, and finally,
simply asserts my discussion is wrong and the 'status quo' as he
thinks he understands it is 'right'. This kind of behaviour is completely
unscientific, unconvincing to the readers he attempts to divert, and
rather blatantly shows he is the laziest 'scientist' around, or simply a fake.


(2) DavisBJactually did struggle with unfamiliar material, making some
mistakes, and asking questions, and clarifying his own understanding of
what the thread is about. He was polite and actually did read the stuff,
spent a lot of effort trying to understand the arguments, and was very
interested and concerned about the conclusions and claims.

Again this is precisely what another scientist or technician would do,
as opposed to the behaviour of others. His actual success in his
endeavour is not so relevant, because of course there are all kinds of
engineers and scientists with specialist training and holes in their skillset.
Once again DavisBJ comes across as convincing, whereas gcthomas
fails and faceplants himself again.


(3) DavisBJ showed the typical confidence of an engineer/physicist,
utterly convinced he would have no real issue solving the problem,
and answering for himself any lingering questions. Also typically, it
naturally turned out to be harder and more complex than he anticipated,
but that hardly phased him and he pressed ahead to his own solution,
applying such tools as he found himself with.

Again, a point awarded for realism and authenticity. Its hard not to
believe that DavisBJ is a scientist of some kind. He is utterly consistent
in his method, activity, and attitudes. Maybe a bit heavier on the
'beer-drinking engineer' side of the scale than the 'tea-totaling physicist'
side, but more than in the ballpark.

Contrast that again with gcthomas, who can't make the effort to
crawl out of his armchair far enough to reach for a pencil or even
an online calculator, and try to get it, yet has ample energy for
troll-like contradicting, mockery, and insults.

I give 3 out of 3 for DavisBJ as being the more authentic scientist/engineer.

The peanut gallery crowd is self-evident too,
but I feel obligated to give recognition where due,
even if DavisBJ isn't on our team.

Newton: Sphere Theorem, Continued Response to DavisBJ

Easy and simple:

(1) "geometry" , i.e., Euclidean properties.


Lets measure the angle of the ellipse to the test particle in the following
manner:

Lets take the angle between the ellipse and the GC line to the particle,
to define the angle of the ellipse to the particle.

To illustrate, when the angle between the ellipse plane and the GC line
is 90 degrees, we will say the ellipse is perpendicular to the particle,
and we can say the ellipse is "head-on" to the particle,
(i.e., not at an angle in ordinary parlance). Call this the 'normal position'.

In this 'normal position', if the cone is circular, so is the ellipse,
and also the GC line and the Cone-axis are identical (superimposed).
This can be called the 'normal', 'start', or 'root' or 'base' position.

In that case, we are letting the circular cone define the shape of the ellipse,
by intersection.

Now, any 'angle' of the ellipse will be measured "from" this base position.



For instance, we could talk of the ellipse being 45 degrees from base.

Suppose then that the plane was tilted in just this way: 45 degrees away
from its base position which is perpendicular or 90 degrees from the
Center of Gravity (CG) line to the particle.

We choose still to use the CG line as the 'standard', because it keeps the thought experiments simple, and makes the calculations and arguments simple too. Before you reject this choice, keep in mind two things here: (a) Using the axis through the center of a circle or ellipse will keep the averaged or effective distance of the object constant when we rotate it, for purposes of the thought experiments to test the Center of Mass theorem/idea. (b) Dividing the circle or ellipse at its center keeps the calculations for each half simple, by making the mass exactly half, and also makes comparisons and logic regarding relative strength of force easy.

If we let the intersection of the plane with a circular cone define
the shape of the ellipse, then we will get a moderate ellipse as in the
diagrams we have been using.

In this case, as we observed above, the center-line of axis for the cone
drifts off from the GC line.

Now we are ready to talk about the "Geometry" part of this thought experiment.

First let us define what we are going to hold constant:
Namely

(a) the shape (circular) and the breadth (edge to axis angle)
of the CONE, and
(b) the distance of the GC of the ellipse to the test particle, and

(c) the orientation of our view of the GC and particle, by fixing
the GC line as a horizontal line on the page, for convenience.

Starting at the base position, with a perpendicular circle,
and the GC-line and Cone-axis congruent, we note that
the Cone-Axis drifts upward as we rotate the PLANE at the GC intersection.

And here is the "Euclidean Geometry" part of the observation:

The ANGLE of the GC-line to the Cone-Axis is entirely determined
by the ANGLE of the ellipse, as measured and defined from the
GC-line, irrespective of scale.

We can now say that the cone-axis angle is absolutely FIXED by
the plane-angle for the ellipse, irrespective of the scale or size.

____________________________

Suppose now instead of a changing shape, defined by the cone
cutting the plane, we choose again for simplicity in discussion,
a fixed moderate ellipse, as we both want to explore.

It is this fixed moderate ellipse that I'd like to consider for a moment,
to clarify another idea:

We rotate now this fixed moderate ellipse on the bisecting axis,
through its GC as we have been doing previously, but keep its GC
fixed in space as we also fix the test-particle.

This is the experiment with which we can test the Center of Mass idea
in the simplest manner possible, because we can keep the MASS of the
Ellipse CONSTANT (=1), and divide the near and far half exactly (1/2),
and we can observe whether the EFFECTIVE Center of Mass moves off
from the Geometric Center (GC) of the ellipse, which it obviously does.

In fact, there are TWO observations to be made here:

(a) The DIRECTION of the Force Vector on the particle moves (upward).

(b) The MAGNITUDE of the Force Vector on the particle changes (increases).

(This note should clarify the imprecise words I used in the previous posts
concerning the Force Vector as well. I certainly don't want to muddy
the waters. My point was that the Force Vector must be examined
as a 'two-part' entity, having both direction and magnitude.
We are interested not just in the fact that the vector changes, but also
HOW it changes, specifically in the first instance, the DIRECTION.
Later, when comparing the forces from opposing disks/ellipses,
the MAGNITUDE will also be of interest, because they are supposed
balance, not just in direction but in magnitude resulting in NET ZERO
FORCE.)

These profound results show that the standard "Center of Mass" idea
as normally calculated (i.e., weighted average of mass and position),
FAILS for nearby particles, because the actual EFFECTIVE 'Center of Mass'
MOVES as the object is rotated on its GC in space.

This result is significant for the Sphere Theorem also, because
Newton used circular disks in his own analogical argument for
the Sphere Theorem.

We have ourselves pointed out elsewhere (check all the gravity threads)
that just because Newton's analogy was wrong, doesn't mean
the Sphere Theorem is wrong. However, because his arguments
fail in the manner we have shown, in fact the Sphere Theorem fails.

We should also point out that not only does the Center of Mass concept fail,
and only remain valid as an approximation when the object is at a great
enough distance so that it effectively acts as a 'point particle',

But there is a further concept that comes out of that failure:

We have substituted the "Center of Mass" with a more flexible and
useful concept, namely the "Equivalent Point-Mass" (EPM), defined as
the position one would place a point-mass of the same mass as
an object with extension in space to give the same Force Vector
on the test particle.

As it happens, the EPM moves around in the space around an object
which is not radially symmetrical like a sphere. It is totally dependent
upon the position of the test-particle.

As a result, we were able to show that Gravity Waves are a natural
result of Newtonian Gravitational theory, and that General Relativity
is not needed in order to have Gravity Waves.

Newton's Gravity Waves

First of all, every concern you express in this last post (#115)
has been addressed already in the one immediately previous, #114.

For instance, you say here in #115:

Quote:

Show us where, in either one, you have shown the cone centerline. I am not asking to see that cone centerline just because it would help me to visualize things. That cone centerline points directly at the point in the ellipse that defines the division between the “near side” and the “far side” portions of the ellipse.

But in the previous post, I deliberately created two new diagrams,
showing the cone and cone centerline for you (and everyone else).
You just didn't bother to look at that post, before posting your new post.

No one is hiding anything.

The reason I didn't bother with the cone centerline is that
it is not relevant to making a simple calculation as to whether
the disk pulls off-center toward the near end.

Your question is a different question,
namely does the off-center pull or line of force track the cone-centerline.

Whether it does or does not, has no bearing on the question of
whether the direction of pull veers off the Center of Mass.

The pull direction certainly DOES veer off the Center of Mass.

For the purposes of proving that, we MUST use the line that
actually passes through the Center of Mass for comparison.

You have raised a new question, one that it turns out still has
no relevance to either the question of the failure of the Center of Mass concept,
or the question of whether or not the forces 'balance'.

Why does your point have no relevance?

Here is why:

Newton proposed that the forces balanced from opposing cone areas,
in his analogy, for two reasons:

(1) The pull from each was in the exact opposite direction.

For this purpose it is not necessary that the two forces
follow the cone center-line. And in fact they don't.

But as it turns out, the two forces DON'T pull along the same line,
and since these forces don't follow the same line,
these forces DON'T follow the cone center-line either.
The centerline for the TWO cones is the same line, by choice
in the original definition and claim.


(2) The pull from each side was exactly equal.

This was a kludge by Newton caused by a misunderstanding by him
of how the process of calculus using infinitesimals worked.
This is not really surprising, since Newton was actually 'inventing'
the Calculus as he worked, and it is very common among pioneers
and inventors that they don't fully understand their own discoveries.

In fact, it took many more mathematicians centuries to sort out what
Newton and Leibnitz had begun to uncover. But it was a long time
before mathematicians had a good grasp on what the Calculus could do,
and how to apply it, and what it meant. Newton did not have the
understanding of his own rough ideas that we have now.

Our discussion revealed already that Newton's idea fails on both counts.

(1) The pulls from each side do NOT pull in exactly opposite directions.

(2) The pulls from each side are NOT exactly equal.

This failure can be shown:

First, because of the 'twist' of angle on opposing sides,
the two disks of Newton do not pull in the same direction.

Secondly, the use of circular disks by Newton is actually
a poor approximation, so that the pull is not the same strength either.

This second cause of inaccuracy comes from two problems:

(a) The concavity of the real sections, making the flatness a mirage.

(b) The impossibility of tiling the surface of a sphere with circles.

In both cases, Newton had misunderstood his own arguments regarding
the Calculus. Reducing the size of the circular disks in his argument
had no effect on the incompleteness of the tiling problem:

When we tile a plane (or even a uniformly curved surface) with packed circles,
there is always the same percentage of area unused and unaccounted for.
Shrinking the circles makes no difference, no matter how small we make
them. The areas NOT covered by the packed circles remain uncovered,
and the relative sizes of the two areas, covered/uncovered doesn't change.



Newton mistakenly thought that they vanished when the circles were made
'vanishingly small' (i.e., infinitesimals).

There are Chaotic solutions to tiling a plane with different sized circles,
but these don't apply simply to spheres, and in any case, Newton had
no knowledge of those and made no use of them:



secondly, Newton imagined also that making the circles smaller made
the curvature of the sphere negligible. This also was a case where
Newton had fooled himself, because he did not understand his own
discovery and application of the Calculus.

In fact, since the distribution of mass of the sphere does not change,
regardless of how we chop it up, or how many pieces we chop it into,
the force errors caused by the curvature do not vanish.

Newton's error in thought was that if he made the circles small enough,
the error caused by the tilt of each circle became insignificant. In fact,
Newton was right on this, but missed the big picture. Its not the
individual tilts of each circle that matter, but the total distribution of mass
for the surface, which NEVER CHANGES, and this error never decreases
nor can it vanish.

The failure of Newton to understand the effects of the Calculus on
the problem caused him to err in using his analogy as an argument
for the Hollow Sphere Theorem.

Newton was right however, in saying that if the Hollow Sphere Theorem
held, then one could move on to the Solid Sphere Theorem, by constructing
a solid sphere out of shells.

However, a much larger caveat must be stated against your objections;

You have continued to make a big deal about ellipses, and how I made no mention of them and did not address them in my disproof of Newton's theorem. What you have failed to note is that Newton himself insisted on tiling the surface with circles, which results in the CONES in his argument being NON-circular and irregular, i.e., not radially symmetric. The main reason I used CIRCLES in my disproof, was because NEWTON used CIRCLES in his proof. You can attempt to write your own unique 'proof' of the Sphere Theorem, using ellipses if you want to, and I will be happy to disprove and debunk that too. I'll warn you in advance, that if you don't address the clear criticisms we have made with Newton's original "proof", you are not likely to compose a successful "proof" based on ellipses rather than circles.

Your criticisms of the other post, far too late and irrelevant,
are not worth addressing.

Disproving Newton: Answering Objections (cont. pt 3)

I've decided to give you one more shot at this, with diagrams.

The ideas may be a bit slippery, but they only require a grasp
of Euclidean geometry:

Consider the following NEW diagram, this time with ellipses:



Here we have for argument's sake AN ELLIPSE, rotated on an axis through
its Geometric Center (GC), dividing the equally distributed mass on either side
of the line.

Intuitively we know that the near side, being closer, will exert more
gravitational force than the far side.

So far, so good. This means that the DIRECTION OF FORCE,
must necessarily tilt away from the GC axis, and toward the top (upward)
in the diagram, which is basically a SIDE VIEW.

However, as the ellipse is tilted, the Cone of intersection with a plane
coinciding with the flat ellipse (of uniform density) tilts, distorts, and shrinks.
(The max excursion of the cone will be when the elliptical disk is perpendicular).

So now lets look at the diagram with the Cone of Intersection superimposed:



Immediately we should notice that the CONE axis, has drifted away from the GC Axis.
This corresponds to your previous observation that a larger part of the ellipse
is now BELOW the half-way point of the cone.

Put another way, as the Cone tilts to accommodate the rotating ellipse,
its Center Axis, tilts upward. Remember that the Cone Axis of Symmetry,
being always centered in the cone, will be equal angles from every side.
From the side view, when the cone is projected flat, the Cone Axis will
always BISECT the angle between the upper and lower edge of the cone.

We already know that relative to the GC axis, the direction of force also tilts
away upward from it. The question is, will it follow exactly the Cone Axis?

There are a few observations from Euclidean geometry that will help here.

(1) Regardless of the distance of an ellipse intersecting the cone
(which is now held constant for the thought experiment),
the angle of the Cone Axis to the GC Axis will stay the same.

That is, by the law of Euclidean proportion, (regardless of how big we draw),
the angles and position of each axis stay the same. The only thing
that would move would be the size and position of the ellipse, but NOT
its own angle re: the GC axis and the test particle.

(2) Regardless of the size and distance of the ellipse,
the position of the Cone axis and the PROPORTION of the Ellipse on
either side of the Cone Axis will stay the same, again by the Law of
Euclidean proportions.

As long as the ellipse stays at the same angle, we can simply increase
its size and move it to the left, or decrease its size and move it to the
right. The Cone Axis and cone don't change, and neither does the GC Axis.

(3) The force itself is not dependent upon the geometry.

That is, only its direction is dependent, dependent upon the angle of the Ellipse.

The actual force could be defined by any Law, not just inverse square,
and yet the result should be the same:

(4) The DIRECTION of the force (only) IS dependent on the geometry.

That is, once the actual direction is determined, by whatever law,
the ANGLE of the LINE OF FORCE will not change. It will be strictly
dependent upon the actual law, when we calculate the force for each
side of the ellipse. This is done by dividing the ellipse as usual,
exactly in half for easy calculation, and assuming the force is
strictly dependent upon distance/direction for an equivalent point-particle.

That is, even if the 'Center of Mass' idea strictly fails as a generalization
for shapes that are not radially symmetric in proximity, it still enables us to ballpark:

The approximate Center of Mass (CM) for each half will be located
somewhere near the physical geometric average of the positions of
all the components of each half (i.e., somewhere near the GC of each half).

Making the diagram larger or smaller won't change the angle of
direction for the combined forces for each half. This is a simple
"by inspection" line of reasoning based on the Euclidean Law of proportions
once again.

The final question then is:

(A) Does the real direction of force track the Cone Axis?

Because if it does, we could make a much larger generalization
than that of the parallel planes law. In fact, it would mean that
even the ANGLE of an ellipse didn't matter. it would act as if
it were a point particle at the point where the Cone Axis pierced it.

This would in essence mean that the force was only dependent upon density of mass.

In this case, the 'tilt' of an ellipse wouldn't matter for the purposes
of balancing the forces between two opposing ellipses. Their effective
Center of Mass would always be on the CONE axis, and the force would
always be along that axis, balanced in direction at least, if not strength.


It would follow that the forces would only depend upon the density value.

The problem of course is that since the force is ALWAYS (also) dependent
upon the DENSITY VALUE, the direction of force for a given configuration
is not fixed, even though the CONE axis IS fixed.

This means that the Direction of Force CANNOT track the Cone Axis,
and we already know that it can't track the GC axis, so ...

Both the Center of Mass (approximation) and the argument of Newton
regarding the balancing of opposing 'cone intersections' on the sphere
surface are false, and the Sphere Theorem must fail.

Disproving Newton: Answering Objections (cont.)

Problem is, the issue I am diligently trying to get him to address, is that nowhere has he proven that indeed the net gravitational force due to the “near side” is stronger that the net gravitational force due to the “far side”. The pseudo-proof that led him to that conclusion in his first post was predicated on the cone centerline remaining directed at the center of the “disk”, even after the disk was tilted. In post 108 he concedes that the disk distorts into an ellipse (in the parallel plane case), and that the “far half” (bad choice of words, since it is clearly not half) “grows quite large”. Nary a whisper from him about how to evaluate whether this “growing quite large” of the distant half is enough to compensate for it being farther away. He assumes the answer he has to have to salvage his train wreck. And nowhere, nowhere in his first post is there even any inkling that he realized that tilting the disk would both alter where the cone centerline intersects the ellipse (formerly disk), as well as alter the mass it enclosed. Lets focus right on this paragraph:

There was no need to focus on the fact that the center of the ellipse
does not lay on the center-line of the (circular) cone.

Instead I focused on the much more practical and easy problem
of determining empirically whether or not the near edge of a circle
(or ellipse for it is the same result) does indeed pull "off-center".

For that, we have the following perfectly accurate diagrams,
which will work equally for an ellipse as for a circle.

Try to follow the simple argument below.
I would say it is self-evident "by inspection", but clearly that seems
not to be the case for some:

5. How the Center of Mass Theorem Fails The balance of forces from all parts of the disk is only strictly maintained along the axis of rotation. Only from a spot along this axis does the distance to moving parts stay constant. For test-masses facing the spin, the movement can be roughly represented simply as a pair of point-masses rotating around the GC in the same plane. Although changes in distance seem balanced, the forces acting on the off-center test-mass are not balanced at all, since they follow the inverse-square law. To show this without mathematics, you can simulate the gravitational field using a compass and a ball of unmagnetized iron in the same horizontal plane. If the force between ball and south pole, and between ball and north pole were equal regardless of compass swing (corresponding to tilt above), the needle would sit wherever you placed it, instead of turning toward the ball. Magnetism also decreases as the square of the distance, and gives us a feel for gravity. Gravity is simply too weak between small objects to measure at home, without extreme cleverness. 6. Watching the Direction of Force change We can easily show the change by using the parallelogram rule. The final vector will always point toward the true location for an equivalent point-mass ( EPM ), and this is not usually at the GC. During a tilt, the closer part (a) will always be a greater angle from the test-mass, since both (a) and (b) are always an equal vertical distance from the GC. For the final vector to point along the GC line, the furthest part would have to give the greatest force, to counteract the smaller angle. This is clearly impossible. This is independent of the inverse-square law, and is true for any force which decreases with distance. 7. 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. Opposing disks on a sphere have the wrong orientation when they tilt:

What I would bring your special attention to is the following diagram:



Please note that here we can substitute an ellipse for the 'barbell' idealization.

Here the Ellipse would then be rotated on its axis, preserving its distance from its own center to the test-particle.

This does NOT mean that this line remains in the center of the 'cone' in other diagrams.

In fact, 'by inspection' you can see that the angles above and below the horizontal
indicate that this line between the center of the ellipse and the particle CANNOT
fall on the centerline of the cone.

The original purpose of the diagram is strictly to prove that the "center of mass"
idea (theorem/hypothesis/approximation) does not hold, since force from the near
half (yes half!) mass increases disproportionately to the decrease in the far half.

And it does prove that, in conjunction with the second diagram below:



Here the Vector Addition Law is used (albeit abstractly) to prove graphically
that the DIRECTION of the pull cannot be along the line to the geometrical center.

But in doing so, it ALSO proves graphically that the DIRECTION of PULL
cannot be along the cone centerline (since then the two angles between
the near and far vectors would have to be equal if the central summed
vector fell on the centerline of the circular cone).

That is, by inspection you should be able to see that the ratio of the two angles
between the vectors added and the final vector can be anything besides 1.

Since this ratio is VARIABLE (based on the relative pull of each half),
the final vector can swing all over the place inside the cone,
depending upon the tilt of the ellipse.

Note there are several lines under discussion:

(1) The centerline of the cone.

(2) The line from particle to the center of the ellipse.

(3) The final vector line indicating the summed direction of the pull.

These three only coincide when the disk is a circle, and is perpendicular
to the line of pull (which is the line from the center to the particle).

At other times, the DIRECTION OF FORCE may vary, always toward
the nearest end of the circle/ring/disk/elliptical disk.

Because the direction of force veers off the centerline of the cone,
AND ALSO the line to the center of the ellipse,
and always in a predictable direction (toward the nearest end of object),

It means that for forces to balance, the line of pull from the opposing
circle/ring/disk/elliptical disk must tilt the same number of degrees
in the opposite direction (still passing through the particle point),
exactly the same amount.

The forces can be balanced if the planes are parallel, whether or not
the cones cut circles or disks. It is in part a question of "equal effective
area"/distance = force, but also quite visibly a question of geometrical
symmetry and balance.

The theorem concerning equal force from disks cut by cones
on parallel planes ALREADY accounts for the fact that the disks
will be ellipses when the cones are on an angle to the planes.

This point was never disputed. It was always accepted both by the
originators of the parallel plane theorem and myself, that the cones
would actually cut ellipses, and that the GCs of the ellipses would not
fall on the axis of symmetry of the cones themselves.

It is worthwhile to speak a bit more on this important theorem:

(1) It is a statement about the ANGLE of the planes to the cones,
not particularly the DISTANCE of the ellipses from the particle in the center.

That is, it is a theorem that is rather elegantly, INDEPENDENT of DISTANCE.

The ellipses are so defined by the cut of the cones on the surface, and nothing more.

The particle can be anywhere between the two parallel planes,
and naturally unequal distances from each plane.

___________________________

Your main objection to the observations we have made here seems still to be this:

You are having a problem with the idea that the ellipse in these illustrations
is rotated on a line through its own Geometric center, instead of say,
on a line across the ellipse that falls upon the line of axis of the cone.

Lets think this through together carefully.

It is most natural to rotate the disk/ring/barbell/ellipse through
its OWN geometric center, in testing the Center of Mass Idea.

The Center of Mass is always (for any shaped object) defined
as the weighted average of the masses and their position in the object,
because the total force is the vector sum of all the individual forces.

This coincides always with the Geometrical Center (GC) for all
regular and symmetrical objects of constant density.

Thus if the Center of Mass idea holds at all, it should hold for
a rotation around the object's GC.

If this fails, the Center of Mass idea fails.

Remember the whole point of the Center of Mass idea,
is to simplify the problem of force by replacing the object
with an Equivalent Point-Mass (EPM) at an appropriate position,
so that the simple Gravitational formula can be reliably used.

It so happens, that in testing the Center of Mass idea,
it is shown by inspection via Vector diagrams,
that the direction of pull will not stay on the axis of symmetry of
a circular cone from the ellipse.

The rotation of the ellipse on an axis through its own GC
is not relevant or remarkable to the independent fact that
that near end will always change the direction of pull away from the GC.

One can by a successive series of approximations (replacing portions of
the ellipse with point-masses) or by calculus show that
indeed the VECTOR changes direction when the ORIENTATION
of the Ellipse departs the perpendicular position relative to a line
drawn from a test-point to its own GC.

This is entirely independent of any cone one might care to draw or consider.

Any ellipse in any absolute position in Euclidean space can be wrapped in a
cone of absolute radial symmetry such that the point of the cone is at the test particle.

This 'fixes' the position of the cone, and if the test-particle is not on the
axis perpendicular to the plane of the ellipse, then neither will the
axis of symmetry of the cone pass through the GC of the ellipse.

The Axis of symmetry for the cone will ALWAYS pass through the near side
of the ellipse off-line from the GC. But the axis perpendicular to the
Axis of symmetry, piercing it, and cutting across the near end of the disk
is not an axis of interest in testing whether or not the near and far halves
of the ellipse balance.

The axis crossing the GC of the ellipse is chosen instead, because it
neatly divides the mass into equal halves (on a uniform density ellipse),
and so the near/far hypothesis can be tested, independently of other issues.

Disproving Newton: Answering some objections

I'll happily address the oval shape now.

(1) Assume a flat plane which is tilted.

(2) The cone is correctly noted to cut an ellipse rather than a circle, when the plane is tilted.

(3) This makes no difference to the argument that the near end of the (ellipse) has greater change in pull than the far end of the (ellipse).

(4) That is, it makes no difference whether the flat cut-out is a circle tilted or becomes an ellipse.

(5) In either case, due to the Inverse-Square law, the INCREASE in pull from the near side of the object, cannot balance the DECREASE in pull from the far side.

(6) Remember that when the (circle or ellipse) tilts on an axis through its center,
the exact same amount of mass will be on each side of the line of hinging.

(7) Since the masses ARE identical on the near and far halves of the object,
the FORCES CANNOT be identical, due to the non-linear action of the distance law.


No flaw in the argument ensues.

Yes, a circular cone cuts an ellipse on a tilted plane.
No, this cannot save Newton's argument and reasoning (which was actually only
an analogy) in support of the Hollow Sphere Theorem.


note:

I'll try and make this even easier.

We don't need to consider the complex argument of Newton at all.

The Center of Mass hypothesis naturally suggests that a disk or a ring, like a sphere,
acts the same as if all the mass were concentrated at a point at its geometrical center.
Tilting a disk, a ring, or even a flat (uniform density) ellipse on an axis through its center
changes the force it has on a nearby stationary object.

That alone is enough to destroy the Hollow Sphere Theorem Hypothesis.

It is in fact empirically known that rotating a charged or mass ring or disk changes the force on a stationary object outside its perimeter.

Newton's hypothesis (and Gauss's "proof") is literally a shell game.




Remember also that the fundamental disproof for Newton's argument is this:



This will not be affected by the fact that you wish to replace disks with ellipses, or
displace the 'effective center of mass' off-center on an ellipse.
Had the ellipses been twisting in opposite directions, this argument would be clinching,
in that any tilt of the line passing through the centers of both ellipses and through the point inside the sphere,
would be moved in opposing directions, keeping the line straight, and the pulls in direct opposition.

But since in BOTH cases (circles or ellipses) the 'line of sight' through the particle is "bent" toward
the nearest inside of the hollow sphere, and the NET pull is not opposing but toward the near side,
no amount of 'adjusting' the strengths of the pulls from each disk will save the situation.

This really is a 'qualitative' aspect of the problem of balancing forces, namely a NET Direction error,
not a 'quantitative' problem, i.e., 'balancing forces from opposing disks/ellipses/caps.

Look at the diagram again: If the Left side DOES balance, the Right side CANNOT.

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

Lets try again from another angle:

Yes, you are absolutely right that if you merely tilt the intersection of the plane
with the cone, the "far half" of the ellipse grows quite large (ignore a spherical surface here).

Now, where do you want to draw the line of force for this ellipse?

Suppose you don't move the center-line (of force) of the cone at all.
You assume that the pull from each side of the ellipse is balanced,
namely by the accelerating increase in the size of the 'far half' of
the ellipse, and the (slower accelerating area) of the near side.

Now, and this must be your argument,
the force cannot change at all.
The planes can tilt INDEPENDENTLY as long as they cut the central line of force
at the same point.

That is, although an ellipse changes in size and mass as you tilt the plane,
there is no change in the NET force (in either strength or direction!)
as you tilt your plane.

That would be wonderful, but is a far stronger assertion than the
usual Euclidean assertion that the 'force' balances between parallel planes
when double-cones intersect them as in the diagram above/below.

Your argument in essence is that the forces balance in BOTH sides of the picture below,
because the LINES OF FORCE don't change direction or strength, when you change the tilt.




Remember that for the forces to balance, they must stay on a SINGLE line pulling in opposing directions.

If that line is not the center-line of the cone, then where is it? It MUST pass through the cone point(s).

If it goes off-center of the cone-center line, on one side, it MUST go in the OPPOSITE direction in the other cone,
to stay straight.

The So-Called "Proof" of the Shell Theorem

The So-Called "Proof" of the Shell Theorem


Here is the typical "proof" for the Shell Theorem (the Hollow Sphere Theorem), as presented by university math faculties (this one was from saddleback.edu).

It was presented by a reader as "proof" that Newton was right about the force inside a uniform hollow sphere. (post #73) Of course the 'proof' does nothing of the kind.

As another reader (post #69) posted, a "theorem" in mathematical parlance is a mathematical statement considered to be "proven true" only in the sense that the mathematical result follows logically from the starting premises and the application of appropriate algebraic and rules and conventions. Even this idea of 'proven mathematically' isn't always clear, and theorems are constantly modified, limited, and expanded in the field of mathematics.

And a mathematical "proof" is not a in any sense a "physics" proof, which is based upon how well a given piece of mathematics actually applies to a physical situation, and its predictive power, when the mathematical variables and other elements are assigned to physical entities and measurements.

The so-called "proof" given begins by implicitly claiming to model a vague physical situation. None of the items below are properly defined, and none of the physical theory relating the idealized abstract model and its elements and diagram to a physical reality is ever presented or articulated. All of that remains not only unproven, but actually unstated. The premises and assumptions remain unexposed. Its appropriateness and accuracy of the theorem in encompassing a physical situation is left completely unexamined.


Essentially, the "proof" begins by ignoring the main physical problem. The author might have happily begun by explicitly making the admission:

"This 'proof' does not expound any physical theory. Nor does it examine the strength of the connection of its elements to physical realities. For that see a physics textbook. We are here only outlining a mathematical theorem:

Unlike our own diagram, nothing here is properly explained or labelled. None of the mathematical entities are described, and even key mathematical axioms and premises are left unacknowledged. For instance, the application of Sin and Cos functions have no meaning, even in mathematical proof, without the assumption of a Euclidean Spacetime manifold. But this is one of the very things to be proved in a gravity theory, and something that requires explicit discussion. We can't fault Newton for his ignorance of spacetime manifold options, but we can certainly fault a modern proponent of the "Shell Theorem" for failing to qualify his 'proofs' in the modern context.




Remarkably, a form of Newton's classic gravity equation simply appears in the third line here without any qualification or explanation. None of steps of the algebraic manipulation is shown, and none of the theorems or axioms required for these steps is given. No derivation or justification for any of the steps that are actually shown are explained.

We now have an equation, an integral, to which someone has assigned a physical meaning, by identifying the elements in the equation with elements in the original physical problem. While the substitution of variables and elements may be acceptable, the student/reader is left abandoned to the "God-like" presentation of the 'teacher'. The only hope is memorization of the forms acceptable on the exam, when the question is posed.

Now comes the hilarious part of the "proof":



"From the Tables:..." yes, that's right, the four difficult pages of real hardcore integration, and their theoretical underpinnings are skipped entirely, unlike our own appendix (see post #13) on the integration of this difficult integral (previous step). Naturally the 1st year student is not expected to know how to actually integrate a function like this. Like a trained monkey he must look it up in an advanced volume/table of Integrals, and simply jump to the next part of the problem.

This is not a formal 'proof' in fact of the Shell theorem or any other theorem. Its a "memorize" the entry in a table method, and don't ask any questions style of teaching suitable only for dummies.

Now, magically, the Force Vector F just appears. An algebraic equation has been derived (skipping all critical steps and techniques) for the force exerted on the object (not the sphere). No explanation for how the mathematics has been connected to a real physical vector is given. Essentially, all that has happened is that an equation from a book has been assigned a task.

A similar phoney implicit step has been performed for the Force vector in the separate case when the point-object is inside the shell/sphere. No notice of the failure of Newton's equation in the case where the particle is actually at the border of the shell is given. No explanation for what happens when a particle pierces the shell, exits/enters, is attempted.

The student is left with two equations, one of which reduces to a constant (0), but he has no idea of why there is no 'force' inside a 'uniform shell of insignificant thickness with its mass evenly distributed on its surface', or whether it is even true.

This is a classic example of the academic "bluster".

A bunch of algebra and some calculus references are waved in front of the inquirer, and they are just left feeling stupid, because in fact it is impossible to derive any proof of the Shell Theorem from this terse, highly condensed and deficient presentation of a complex operation which invokes literally hundreds of mathematical and physical assumptions.

The student is told to "go read the textbook discussion" on the Shell method of integration, or 'Newtonian Gravity'. He does the only thing possible in such a ridiculous situation: He memorizes the steps approved by the lecturer, gets his grade, and has no clue what is really transpiring in gravitational theory.

This is a classic example of what I would call the worst form of teaching imaginable, in spite of how common it is in institutions where bodies are pushed through the door and large amounts of money are collected.

Is the math incorrect?

 

What is disputed is a claim in a physical theory,
namely that the gravitational field inside any practical hollow sphere is zero,
either as a physical fact or as an approximation.

We are well aware of the mathematical apparatus and its 'value',
(for instance in the context of a developing algebraic Group),
but categorically reject any claim that any calculus operation can
intrinsically in any significant way 'prove' the validity of a physical theory.

Yet people as yourself keep focussing on whether or not the math is 'correct',
when the issue is actually,

"Does this cute mathematical artefact have any physical meaning?"

The question for a physicist will always be,

"Is this equation going to be accurate for my physical experiment,
in the manner in which I'm going to use it?",

not,

"Is this equation adequately defined as a mathematical entity,
to the satisfaction of esoteric mathematical theorists?"

As I already stated in post #65:

Quote:

This misunderstanding in regard to the purpose and meaning of Calculus, and indeed any mathematical result generally, i.e., its 'truth-content' in regard to reality, is one of the most common logical and epistemological errors in engineering, physics, and even amateur mathematics, and runs rampant in 'pop-science'.

Proving Ideas: Wikipedia vs. Reality

These results were important to Newton's analysis of planetary motion; they are not immediately obvious, but they can be proven with calculus. (Alternatively, Gauss's law for gravity offers a much simpler way to prove the same results.)Source: http://en.wikipedia.org/wiki/Shell_theorem The wikipedia entry is false, according to its own obvious premise.
Physical theories are not proven by calculus in any way.
They are proven (if at all) by accurate physical measurement.

The calculus, which by the way we fully provided in posts #10 and #13,
proves nothing at all, except that the demonstration
that mathematical reasoning follows well-understood rules.

The mistake made in the wikipedia article, and probably also made
in many 1st year Physics texts and popular science articles,
is that the successful completion of a math problem in integration,
proves anything at all for the science of physics, or the operation of
classical gravitational and electrostatic forces.
It doesn't and can't because the problems and their solution are
in entirely different categories of science.

Integration is used to 'prove' mathematical results, for the mathematical 'world',
in a very narrow and abstract mathematical sense.
Calculus is not any kind of tool that can bridge or connect the worlds of mathematics to the world of physics.
Those connections are made by axioms, premises, hypotheses, definitions,
and conventions, in conjunction with higher analytical theories of meaning.

A calculus result only has a tentative meaning and existence
in the mind of a mathematician. Its value is not based on 'truth-content'
in regard to the real world, but is rather based on 'consistency' results
through Group Theory and Theory of Algebra, such as Galois and the Lebesque field.

This misunderstanding in regard to the purpose and meaning of Calculus,
and indeed any mathematical result generally, i.e., its 'truth-content' in
regard to reality, is one of the most common logical and epistemological errors
in engineering, physics, and even amateur mathematics, and runs rampant in 'pop-science'.

Real physicists and mathematicians don't "prove" theorems
in the sense of their applicability to physical problems by using math.
The create rather a tentative 'plausibility' to mathematical tools,
based on their apparent usefulness in solving physical problems,
and accurately predicting results.

Feynman for instance, would never make any truth-claim whatever about QED.
He would only call it
"the most accurate predictor of experimental measurement we have,
good for about 8 digits of accuracy, in proper use."

Nor does Gauss' law prove anything at all, except in the mathematical realm.
Its an analysis and methodology which results from other mathematical theorems.

Experimental Verification of Gravitational Discussion

In our case also, there are physical experiments that can verify
various aspects of the discussion as we have posed.
Gravity itself is too weak a force for direct measurements,
and the scales also make experiments and measurements normally difficult and cost-prohibitive.
However, the same mathematics is used in Classical Electrodynamics (CED),
so electrical experiments can be devised that are easier and cheaper.
As many are aware, CED has been replaced formally by QED,
(Quantum Electrodynamics), an entirely different beast.
But both GR and QED are what is called "overkill" tools
for much ordinary work in both gravity and electricity.
For most engineering jobs, Newtonian Physics (NP) and CED
are the right tool for the right job.
This is scientific pragmatism, not ideology.

The most important points of our discussion however,
are not based upon the known differences between NP/CED and SR/GR/QED.
Our work is not a replacement for any of these theories,
but a refinement of all of them, and all of them are affected.

The most important points of our discussion also however,
only require a few thought experiments, and some reminders
regarding already established scientific facts which have been left unapplied.

This makes the work important enough to look at,
while also gratefully easy to 'prove' or 'disprove'
to the satisfaction of most physicists.
All that is required is:

(1) Checking the math, and its appropriateness to the physical applications.

(2) Carrying out the thought-experiments, and applying known and undisputed physical aspects to the problems.

(3) Proposing and executing physical experiments, specially designed to expose predicted effects and differences from previous 'modeling'.

We can say that (1) has been repeatedly done, (2) can be done by the reader, and (3) is open to scientific investigators and researchers, with no limits.

Here are some specific points:

(1) The 'continuum' model is quite useful and fruitful as a theoretical construct in dealing with fields and spacetime, but quantization of energy etc. is also a daily fact. It is not surprising that at a molecular level, where discrete masses and charges are actually heavily concentrated point-particles distributed across relatively vast distances, that simplified 'continuum' treatments fail.

(2) A Thought experiment such as the firing of a mass or charged particle at a hollow sphere made of a low number of discrete particles (like a molecular carbon-ball) is enough to carry the point home that the inside of a structure made of such crudely distributed masses/charges cannot have a 'flat' gravitational or EM field. The trajectory of the fired particle will be bent.

We don't need to do a real experiment, because thousands of such trajectory experiments with electrons and neutrons have already been done, and exhibit the expected patterns of the thought-experiment, suggesting that Carbon-balls would not act like perfect "continuum"-style hollow spheres. If they did, many other experiments would have had different outcomes.

Example Scattering Pattern for X-ray bombardments:


Figure 6. a, Experimental corrected X-ray scattering pattern obtained from an orthorhombic hen egg-white lysozyme. This picture has been previously published and described by Faure et al. (1994). It was collected using synchrotron radiation with image plates and has had parasitic scattering due to solvent, air and the capillary subtracted. b, Very diffuse X-ray scattering pattern obtained from the intramolecular protein motion from the full simulation, including all protein degrees of freedom.

Disproving Newton Part 3 The Death of the Center of Mass Theorem

Disproving Newton
Part 3
The Death of the Center of Mass Theorem




by Nazaroo


Newton’s Centre of Mass Theorem is shown to be self-contradictory


(With a New Introduction)

Last Updated: June 19, 2005


Synopsis

This paper examines the Centre of Mass Theorem ( CMT ), originated by Newton, as it is currently formulated and presented in physics textbooks around the world today.
Specifically, after examining the details of the theorem, we show:

1) That CMT is incapable of properly handling external forces exerted upon a system, because its method discards crucial information regarding those forces.
2) That CMT cannot be patched, modified or extended simply and easily, because its flaws are deeply entrenched and are systemic to its derivation and formulation.
3) That CMT is self-contradictory and mathematically incoherent. Its theoretical foundations are misguided and confused.
4) No sensible or useful reformulation of CMT is really possible, and its only value is historical. It cannot be retained as an integral part of modern Newtonian mechanics.
5) That the few valid components contained in CMT should be left to stand alone under restricted circumstances, such as the Conservation of Linear Momentum ( CLM ), or else be absorbed into other valid theorems and methods, such as those covering Angular Momentum and Equilibrium of forces.
6) That CMT should be abandoned as a teaching tool, since it misleads people regarding both its meaning and value for Newtonian mechanics.
7) That CMT cannot be used either as a theoretical foundation or even an auxiliary support for the Sphere Theorem ( ST ) in regard to Newtonian gravitation.

In summary, the Centre of Mass Theorem is essentially incorrect, and flawed as a theoretical basis for understanding the physical phenomena it is expected to account for. Teachers and students are better off avoiding this theorem entirely and waiting for a better theoretical overview or presentation to be offered.

Table of Contents



Introduction
1. Quick Review: Newtonian Gravity
2. Gravity made Practical
3. The Background and Purpose of the Centre of Mass Theorem
4. The Centre of Mass Concept ( CM )
5. Special Properties of the Centre of Mass
6. The Centre of Mass and the Geometric Centre
7. The Generic Nature of the Centre of Mass
8. Practical Methods of Calculating the CM
9. Derivation of the Centre of Mass

The Basic Theorem of the Centre of Mass ( CMT )
1. A Typical Formulation
2. The Scope and Claims of CMT
3. Interpretation of CMT
4. Counterexamples contradicting CMT
5. Realistic Restrictions for CMT
6. Re-interpretation of CMT ( ‘no free lunch’ theorem )

Analysis and Criticism of CMT
1. Flaw A: Failure to distinguish between types of systems
2. Flaw B: Non Uniqueness of the CM
3. CMT : Blow by Blow Critique

Demonstration of the Failure of CMT
1. Analyzing Newton’s Balls
2. CMT contradicts NTF
3. CMT contradicts Itself
4. The 'Success' of CMT: A Cosmological Coincidence



Introduction


1. Quick Review: Newtonian Gravity

In brief, Newton viewed gravity as a force of attraction between bodies. Matter is made of basic particles, such as protons and electrons. Each particle attracts every other, and so there is a small force between every pair of particles.


These forces are independent, in the sense that they do not interfere with each other as sound waves or light waves can. That is, unlike waves that vary over time, gravity is a static force that only varies with distance. Gravitational forces pass through one another in a ghostlike fashion, only acting on actual particles.
Gravity apparently reaches infinite distances and penetrates everything. Interposing a third mass between two objects does not diminish or deflect the force. In fact it has no effect at all, except to add two more paired forces. The gravitational force between each pair of objects is dependent upon their own mass and distance alone. The total net force is just the superimposing and adding up of all these forces between pairs.

Every object exerts the same total force 100 light-years away as it does next door (!):
Although gravity weakens as distance squared, this really just reflects a fixed amount of force spreading out over an increasing spherical surface area. It can be represented by a fixed number of lines of force spreading out like spokes of a wheel. This essentially lossless spread could even be called the Conservation of Gravity.

Even more remarkably, gravity seems able to draw upon infinite amounts of energy to apply itself. One object exerts the same force upon one other object, or an infinite number of them, as necessary. Gravity doesn’t run down like a clock, or drain like a battery. The force doesn’t become defocused, diluted, or randomized as it stretches across space. These basic discoveries are mystifying. Nor is there any obvious mechanism or medium for the actual transmission of the force. This led Newton to view gravity as a spooky, instantaneous action at a distance, and led Einstein to interpret it as a geometric property of space-time itself.

2. Gravity made Practical

Although mysterious in terms of ordinary notions of cause and effect, at least gravitational forces are easy to handle in principle. We don’t need to worry about extra interactions or side-effects. For any given pair of particles, Newton’s basic formula could be used to calculate the force. With more particles we just add all the gravitational forces together like we do other forces as vectors to get the total net force on any one particle or object.
Easy in principle however becomes impractical with billions of particles per handful of matter. Useful formulas must apply to large groups of particles at a time. Newton addressed this problem with two claims:

(1) The Sphere Theorem: Uniform spheres act as if they were point-masses, and so the simple formula for point-masses can be used ‘as is’ with celestial bodies. One assumes the mass is concentrated at the geometric centre (GC), and distances are measured centre-to-centre between spheres.

(2) The Centre Of Mass Theorem: Similarly, any object’s centre of mass ( a fixed point of balance inside it1) also moves according to the laws of motion, as if all the mass were concentrated there. External forces can be summed and applied to the centre of mass (CM) to predict its new position etc.2

These two key theorems are the basis for practical calculations and predictions for the motion of large masses under Newtonian gravity.
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1. An object’s centre of mass ( the weighted average of position of all particles ) can be found by measuring weight, volume, composition, density & shape, without descending to the atomic level. 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 is determined by the distribution of mass. If the system is symmetrical in both shape and distribution of mass, its centre of mass coincides with its geometric centre. These two are not the same as the centre of gravity, which is only relevant in a uniform field. Nor are they the same as an equivalent point-mass position, which is what we have to calculate.


2. That is, the centre of mass traces out a parabola or moves in a straight line, even if the object appears to move eccentrically.
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3. The Background and Purpose of the Centre of Mass Theorem

To assess the impact of a flaw in the Centre of Mass Theorem ( CMT ), we need to review its function. Newton’s theoretical foundation rests upon atoms exerting mutual forces of attraction upon one another. That is, his gravity formula actually only applies to particles. This would be hopelessly impractical unless we could economically deal with large groups of particles at once.
The whole purpose of CMT is to enable us to do so. CMT then is a key part of Newton’s complete theory. Without CMT or some other practical technique any gravity theory is just a curiosity. Newton’s spooky notion of invisible, instantaneous action-at-a-distance would probably have been branded as mere magical thinking long ago, if not for the spectacular success bought for it by CMT. 3
That is, people accepted Newton’s theory, because his formulas and methods actually worked, or seemed to. Even those today who reject Newton’s theory presumably don’t object to engineers continuing to use Newton’s techniques ( i.e. CMT ) to build bridges and to calculate rocket trajectories.
It is hard to deny that CMT is still presented as a currently valid theorem, and that most scientists assume the consistency and validity of its derivation and formulation, at least within the sphere of Newtonian mechanics.4 Now and then CMT is called an ‘approximation’, although no quantification is offered. The basic position given on CMT is that it is accurate. 5
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3. Even with the success of Newton’s techniques, many brilliant minds questioned his theoretical foundations. Newton’s success regarding celestial bodies relied upon the Sphere Theorem ( ST ) but this appears to be just a special case of CMT.
4. Angular Momentum in particular has been modified to accommodate Relativity, but this is really only relevant for particle physics. For ordinary objects Newtonian versions of angular momentum are expected to be valid and applicable to gravitational problems also.
5.
For instance see Elements of Newtonian Mechanics Knudsen/Hjorth, Springer, 2000 pg. 195



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The Current State of Affairs as to Gravitational Theories

The only viable alternative is General Relativity ( GR ). But GR is still in its infancy and in its current form is simply impractical. For most purposes Newton’s CMT is all we have. Even those who prefer GR as a theoretical foundation still use CMT for ordinary calculations. This is no surprise, since Einstein designed GR to mimic or reduce to CMT in ordinary situations. For these cases GR then offers no increased accuracy or practical advantage, nor does it make any alternate predictions. Apparently CMT (with a few corrections for relativistic effects when necessary) is all we really need. So Newton’s method ( CMT ) is used for accuracy and convenience, while Einstein’s theory ( GR ) is preferred as a theoretical foundation for other reasons, such as compatibility with Special Relativity.

Other Related Theorems and Methods

To complete the picture presented so far, we must mention the Angular Momentum and Equilibrium theorems. These handle rotation and the application of forces to rigid bodies, and they also handle balance under uniform gravitational fields. Both theorems make use of the Centre of Mass concept, ( CM ) but do not directly rely upon the Centre of Mass Theorem. ( CMT ). One might argue that most engineers rely upon these other theorems more heavily than they rely upon CMT .

Perceived Agenda for Gravitational Theories and Techniques

Naturally, Newton’s theoretical foundation ( NTF ) is expected to fade away because it can’t explain new discoveries. It wasn’t written to explain them, and would need ongoing revision to do so. In hindsight, NTF appears not to be a real explanation at all, but simply the ad hoc concept Newton needed to justify CMT , in the same way he used ‘Absolute Space’ to justify inertial frames and simplify the laws of motion. That is, NTF really depends upon CMT for its existence, not vise versa. Both NTF and Absolute Space were necessary for Newton, but are no longer so for us. If this is so, we can discard NTF and keep CMT as one removes the scaffold once the house is built. If CMT doesn’t require NTF, or they were found incompatible in some unforeseen way, we could even now choose CMT over NTF on more than empirical grounds.
In a happy world, so it would be. Newton needs CMT , and we need CMT, but we don’t need Newton’s NTF. It could be left as a learning tool, or retired to a history shelf, and CMT would be retained in the honourable service of GR.

Probable Outcome of a New Analysis

This however appears not to be the future of the Centre of Mass theorem. The reader should discover as we did that CMT has fatal flaws, if they can find the patience to step carefully through the arguments below. Is this an exaggeration? Surely people will go on using CMT as they always have, in the same manner as they did after GR. Not exactly.
GR was clearly revolutionary, overturning theoretical foundations. Yet, for ordinary mechanics, life went on pretty much undisturbed: In this case, the theoretical modifications are obviously far more modest. Yet something very different has happened qualitatively. The function of CMT has inevitably changed, because it can no longer be trusted. Its results, even its successes, require careful reinterpretation. CMT must now be applied more cautiously, keeping an eye out for much more than just relativistic effects. Many results and theorems will need reformulating in cases where it is clear that errors may become significant.

4. The Centre of Mass Concept ( CM )

The centre of mass ( CM ) is not a vague concept in Newtonian mechanics. It is a unique point related to a system, and is located within the volume of space that encompasses the system. A system in this context is a group of particles, each having a definite mass and position at some instant in time.6 A system need not be a rigid body, and systems can overlap and contain one another. However, the instantaneous position of the CM for a system is always defined in principle, because a reliable method is given for calculating it. The CM is called the ‘weighted average of position’7 of the particles, and each coordinate is defined by a calculation averaging the coordinates and masses of the particles in the system:






This is simply an arithmetic mean of position, weighted by mass.

This procedure always gives an unambiguous location, which is fixed relative to the rest frame of the system it belongs to. This relation can be called an ‘internal position’ to the system.
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7' e.g. University Physics 7th Ed. Sears/Zemansky/Young, Addison Wesley 1987 pg. 197

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5. Special Properties of the Centre of Mass

The CM cannot change without a change in the distribution of mass.

So the CM is always a fixed position for a rigid object, which is defined as a system with a fixed relative distribution of mass. This allows us to relax the simultaneity restriction for rigid objects, and we can track the motion of the CM as easily as other parts of a system without having to constantly recalculate it.
It is trivially true and desirable that a change of coordinate system, or scaling of all the masses and/or distances by some factor (i.e., a change of units) has no effect on the internal position of the CM. For instance, since the CM is only a point, altering the orientation of a system has no effect on the CM.
For the CM to hold its position relative to other systems however, the system must be rotated only on an axis through the CM itself. It is also true that changing the orientation, distance, mass or size of a system relative to a second system, or changing the second system’s orientation to the first, also has no effect on the internal location of the CM. But this latter case is not a mere change of coordinates, scales or units.
We expect the CM to be unaffected by trivial transforms like rotations, scaling and reflections. However, a system can actually be squashed flat, exploded, or stretched symmetrically in any number of dimensions without affecting the CM. One can even add or subtract mass, particles, combinations, whole systems, provided the effect is symmetrical.

Conversely however, even extreme changes in the distribution of mass don’t necessarily affect the position of the CM.

Any number of systems could average out to the same CM. Although dependent upon distribution, the CM is a one-way trapdoor, preserving no information whatsoever as to the actual distribution of mass. Neither its location nor its motion can really tell us anything more than the general point of balance within a system.

6. The Centre of Mass and the Geometric Centre

The CM is closely related to the geometric centre ( GC ), because for systems of point-masses, the GC can be similarly defined as the ‘unweighted’ average of position, which simply means the ‘equally weighted’ average.

When dealing with systems of point-masses, the CM and GC will be the same as long as all the points have equal mass.
Where the mass can be considered continuous because of the quantity of point-masses, all that is really required is uniform density for the CM and GC to coincide.
If the mass is distributed symmetrically around either the CM or GC, then it is so around both, and they are identical. For many symmetrical objects, such as spheres, disks, cylinders, and regular polygons, the CM and GC are again the same, as long as the objects have merely a radially symmetrical gradient.

Because of this, the CM and GC can share both topological and geometric properties. For instance, if the system’s point-masses sit on a straight line, the CM is also on that line. If they lie on a plane, the CM is also on the plane. Like the CM, the GC is also indifferent to orientation, is a fixed internal point in a rigid object, and lacks unique association to a specific point-mass distribution.

7. The Generic Nature of the Centre of Mass

Above, we referred to different systems having the same CM. This is only meaningful with an independent way of defining the position of a system, which the GC provides. The GC gives an alternate means of defining position, and helps to locate the CM, but does little else for us. The CM is the more relevant concept for gravitational forces, and for rotation as well.
Using the CM and the GC alone only allows us to distinguish two kinds of systems topologically, those in which they coincide, and those in which they don’t. For example, take any two non-symmetrical systems, each of which has non-coinciding CM and GC. Superimposing them with their GC at same location, we can rotate and scale one of them until their CM s also coincide, making them equivalent systems according to CMT. Clearly, even when the GC and CM do not coincide there is little to distinguish between systems, applying these concepts alone.

If the current claims found in CMT actually hold in the physical world, we would expect diverse systems to behave in an identical manner, namely the manner predicted by CMT.

8. Practical Methods of Calculating the CM

Obviously we can’t know the instantaneous position and mass of billions of particles, let alone calculate a weighted average from the data. The whole interest in the CM is to develop practical methods to measure and predict motion of large systems, without having to handle individual point-masses.

Ultimately, people rarely use the formal method of calculating the CM , except perhaps for a system with only a handful of particles or celestial bodies. Otherwise, matter is treated as though its mass were continuously distributed, and cases are limited to rigid symmetrical shapes with simple density patterns. However, the CM is invisible, often sits in empty space between masses in a system, and sometimes can only be found through the GC. If they don’t coincide, the orientation of the system is also needed as well as the GC, in order to find the CM. A combination of geometry and calculus is used along with assumptions about the behaviour of the CM found in explanations of CMT stated below.
In a locally uniform gravitational field, such as near the earth with small objects, a simple method for finding the CM is to exploit symmetry (it must be on an axis of symmetry in an object with a known symmetrical distribution of mass). One can also hang an object from some point two times, and in each case extend the vertical line through the object or along its surface, and mark their intersection.

9. Derivation of the Centre of Mass

Modern treatments derive the CM by first applying Newton’s 2nd law (F = mA) to each particle in a system. Position is substituted for A as a derivative (acceleration is the 2nd derivative of position). Mass is allowed to slip inside the derivative, since it is assumed to be a constant for each particle: 8




Next, an important declaration is made:

‘The total force on the system 9 is the vector sum of all the forces on all the particles.’ 10

This is such a remarkable theoretical step that it deserves a comment: The idea is that just as we get the net force on a particle by summing the forces, so we should calculate the final net force upon a system by summing the forces applied to each particle in it, regardless of where in the system they are actually applied.
Thus, the total force of the system is written as the vector sum of the individual forces on each particle. Using the sum of derivatives = derivative of the sum rule, the summation is also slipped inside:





Finally, multiplying by M/M ( M = total mass ) and slipping the 1/M portion inside the derivative as well, puts it all back into the form of Newton’s 2nd law, ( F = MA ) only this time A is clearly seen to be the 2nd derivative of the formula given earlier for the CM, namely the weighted arithmetic mean average:



From Newton’s third law it is noted that all the internal forces between particles within the system come in equal and opposite pairs, and so they cancel leaving zero net force. Only forces from outside the system are unpaired and unbalanced and so represent a leftover net force.

That is, while Ftotal = Fexternal + Finternal , Finternal = 0 and so Ftotal = Fexternal

Thus the final equation Fexternal = Mtotal * AccelerationCM seems to imply that the CM actually obeys Newton’s 2nd law as long as Fexternal is understood to be the vector sum of all the external forces, as we originally defined it in the derivation.


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8. This is a minor fudge. Mass varies slightly as energy is stored or released in chemical bonding or atomic decay.
9. Physics 3rd ed. Wolfson/Pasachoff, Addison Wesley, 1999 pg. 239
10. University Physics 7th ed. Sears/Zemansky/Young, Addison Wesley, 1987 pg. 198 etc.
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The Basic Theorem ( CMT )

We are now ready to state clearly what CMT essentially asserts: Every system of particles having mass acts as a single point-mass concentrated at its CM, and external forces act upon the system as if they were all acting upon the CM.

1. A Typical Formulation

1) Suppose we have a suitable means of measuring the position and time of point-masses in any direction and at any distance.
2)We establish a suitable rest-frame by measuring the motion of a sufficiently isolated point-mass and assuming Newton’s first law of motion.
3)We define the CM of a system as the weighted average of position (as above.)
4)The displacement of the CM is then the weighted average of the displacements.
5)The velocity of the CM is then the weighted average of all the velocities.
6)Momentum is defined for particles as mv, and the vector sum of all the particle momenta in an isolated system is shown to be conserved. m1v1 + m2v2 + … = ( a constant vector )
7)This vector sum is identified as a momentum, that of the system. (MVsystem ) = m1v1 + m2v2 + … and we get a law of Conservation of Momentum .
8)The total mass M is defined as the arithmetic sum of all the masses.
9)The velocity of the system is defined and isolated by dividing both sides of the momentum equation by M.
10)The velocity of the system then turns out to be the velocity of the CM. (!)
11)The total mass M is then assigned to the CM.
12)The momentum of the CM is then the momentum of the system.
13)The law of conservation of Momentum is applied to the CM.
14)The CM appears to obey the first law of motion when the system is isolated, and it has uniform motion, even if parts of the system have complex motion. For instance, if the system is rotating, it rotates around the CM.
15)The system appears to behave as though it were a single point-mass located at the CM, with all its mass concentrated there. This assumption is extended to non-isolated systems, under acceleration from the effects of external forces.
16)Newton’s second law (F = mA) is applied to the system as a whole (as above).
17)Just as the net force upon an individual point-mass is the vector sum of all the forces, the net total force on the system/CM is assumed to be the simple vector sum of all the external forces. (In this case, the forces are those upon the individual particles at different locations however.)
18)The acceleration of the CM is calculated by dividing the total force on the system by the total mass ( by rearranging the equation to A = F / M ).
19)It turns out that with force so defined, the CM is the position found on the right side of the equation following Newton’s 2nd law (on paper), and vise versa.
20)Therefore, the CM acts as a point-mass of the same mass as the system, and all external forces can be simply applied there (as a vector sum). We can either predict the CM motion from known forces, or the forces from the CM motion.

2. The Scope and Claims of CMT

The scope and generality of CMT is as follows: It applies to all systems of point-masses, rigid or not, isolated or not. It is not dependent upon ‘uniform fields’ or special cases. This is often explicitly stated in physics texts, for instance:

‘ ( in contrast to the centre of gravity concept, ) …the centre of mass, conversely, is defined independently of any gravitational effect.’ 11

At this point the following claims are typically made:

‘(The CM ) does obey Newton’s law…We have defined the CM so that we can apply the 2nd law to the entire system rather than each … particle.12 …As far as its overall motion is concerned, a complex system acts as though all its mass were concentrated at the CM…we defined the CM so that its motion obey(s)…F = MA. ’ 13
‘When a (system) of particles is acted on by external forces, the CM moves just as though all the mass were concentrated at that point and it were acted on by a resultant force equal to the (vector) sum of the external forces on the system’14
‘The ( CM ) moves as if the entire mass were concentrated in that point, and all external forces act there.’15
‘The ( CM ) of the system moves as if all the mass of the system were concentrated at that point. … the system moves as if the resultant external force were applied to a single particle of (equal) mass … located at the centre of mass.’16

Physical support for CMT is usually offered in the form of examples such as non-uniformly weighted hammers or wrenches. When rotating, they revolve around their CM. The CM can be marked on the object and observed to move in a straight line as the object spins through space or slides across a table.










3. Interpretation of CMT

One can search through the most recent physics texts, yet not find any qualification on the claims for CMT. The word ‘approximation’ does not appear there, although liberally used in other contexts. The one time we have found it used in relation to CMT is as follows:

‘Newton’s second law guides the motion of the CM just as it guides the motion of a single mass point. This is the justification for the often used approximation of an extended body as a mass-point.’ 17

This is hardly a statement that CMT is an ‘approximation’ in any significant or quantifiable sense. If anything, CMT is once again asserted to be accurate. There is no hint of when CMT might prove inaccurate, or by how much. There is no mechanism, no example offered, no acknowledgment of any failure of CMT.

4. Counterexamples contradicting CMT

Ordinary physics students can be forgiven for thinking that here at last is some promise of a general method for dealing with systems and forces that is practical, reliable, and easy to use. They will not stay misled for long however, once they experience problems involving rotation and torque.
Note that if one applies equal and opposite force vectors sideways to each end of a simple rod, it spins (about its CM in fact! ). But if we were to follow CMT as presented, and simply add them vectorally and apply the resultant zero vector to the CM of the rod, we would never predict rotation or discover the rules of torque.





This isn’t a new discovery, obviously. But we would do well to discover the nature of the failure of CMT in cases such as this, so that we can modify its scope and its interpretation to reflect reality. One simple way to do this would be to just include theories of torque and equilibrium within CMT as a basic modification.



This would be unsatisfactory for several reasons. For one thing, it assumes the validity of CMT in other cases not involving torque. But we have already seen that CMT has been poorly defined as to scope and interpretation, and there may be other flaws in CMT.


5. Realistic Restrictions for CMT

CMT only deals with linear (translational) motion.

How did CMT fail in the previous example? Because in fact it has no mechanism at all to deal with rotation and associated forces. It cannot, because by nature a point-mass has no orientation, nor any mechanism to connect to, store, or even measure relative rotation. Rotation can be thought of as balanced acceleration. CMT ‘handles’ balanced acceleration by canceling it out and ignoring it. One way to modify CMT to handle rotation would be to convert angular momentum into a kind of mass vector. In this case however, mass becomes a variable, and some of the mathematical steps in CMT obviously become invalid.

CMT only deals with vectors passing through a CM axis

Rigid bodies can be thought of as machines able to convert linear motion into rotational motion. As such, they are unsuitable for a method using a simple summation of vectors regardless of application point. Arbitrary forces often produce a mixture of translational and rotational motion. One solution would limit external forces under consideration to those applied along an axis passing through the CM. Then at least one point-mass in the system needs to be on this axis to receive the force. We could also handle cases having a secondary mechanism that distributes and delivers force in a parallel direction, such as a flat plunger.
In other special cases, CMT could handle balanced forces acting on parts of a system. For instance, one could simultaneously apply two vectors having appropriate directions and proportionate magnitude relative to a CM axis. This would work like a uniform gravitational field, where forces trivially move the system as if the sum of forces were applied to the CM.

Otherwise, CMT can only handle forces (or fields) acting symmetrically upon the system, around the axis of net direction through the CM.

These restrictions limit forces and combinations of forces to those that produce pure linear (translational) motion without rotation. The careful expansion beyond uniform fields allows for the kind of spatial distortion of a system that doesn’t affect the relative location of the CM on any axis of symmetry.

6. Re-interpretation of CMT ( ‘no free lunch’ theorem )

With a sober assessment of the restraints upon CMT , we can now carefully reinterpret its claims.

‘The ( CM ) moves as if the entire mass were concentrated in that point, and all external forces act there.’18 And,
‘… the system moves as if the resultant external force were applied to a single particle of (equal) mass … located at the centre of mass.’19

These statements are true (and only true) if by ‘resultant external force’ we mean the result after we filter all those external forces through a technique which calculates and discards all torque and rotation effects. This technique would leave us with a net linear force related the system as a whole through the CM.
External forces act on the CM indirectly since they are altered by mechanisms unique to each particular system, and their effects rely upon things like rigidity, interconnection of parts, and distribution of mass. There is no universal or simple method to convert real external forces into linear motion of the system or CM.

‘the CM moves just as though …it were acted on by a resultant force equal to the (vector) sum of the external forces on the system’20

This statement is blatantly false. In fact, if a force is applied to a point on a rigid body so that its directional axis does not pierce the CM, it produces a torque, making a variable amount of its magnitude available for application to the overall linear motion of the system. There is no way of knowing the net contribution of the force to linear (translational) motion without calculating and subtracting the torque.






The example above is even more damning for CMT than the last. The ratio of translational to rotational motion which vector force F gives to the rod depends upon the distance R from the CM. The vector itself gives no indication of linear motion, and a sum of such vectors can’t either.21




Analysis and Criticism of CMT

We will skip general criticisms of Newtonian mechanics regarding relativistic effects. Our concern is whether CMT is consistent with the rest of Newton or even self-consistent. Granted that it appears to give good results in many instances, we want to know why, and if it fails or can fail, where and why it fails.

1. Flaw A: Failure to distinguish between types of systems

When CMT leaves the definition of ‘system’ open, allowing both arbitrary groups of point-masses in space, and also rigid bodies, it clearly fails:
The vector sum approach of dealing with external forces is only valid If we have a system of truly independent point-masses. That is, if we have a purely classical system of elastic particles having no significant gravitational, electromagnetic or other forces connecting them. In this case, each particle is truly isolated and has a definite uniform velocity, which cannot change without a collision and an exchange of energy, preserving the law of conservation of momentum.
External forces then act upon the system by direct application of a force vector to an individual point-mass, say through a collision. In this case, the definition of the velocity of the system as the weighted average of the individual velocities will naturally and correctly reflect the application of a force vector to any particle of the system. This force vector, although only physically affecting one particle, will be correctly assimilated into the average position, velocity, and momentum of the system.
But what systems conform to this classical billiard-ball style model? A bag of billiard balls released into deep space. Each exerts negligible gravitational and electrostatic force upon the others. We can even correctly handle uniform gravitational fields, since each ball receives a force proportional to its mass independent of position, and the force vectors can be simply summed. Adding the individual masses gives a total mass independent of position as well, but that is fine. Thus, if the particles are spread apart far enough so that they act independently, we can use CMT safely, and the CM will reflect the system’s velocity and momentum.
But this is precisely not the type of system we want to apply CMT to! All large masses of interest consist of closely grouped particles which are bound together by strong electromagnetic and gravitational forces. These lumps of actual matter form a class of rigid or ‘semi-rigid’ bodies in which the particles interact strongly through fields, and are capable of converting linear forces into rotation and storing energy through angular momentum.

Thus a group of (non rotating) billiard balls is a legitimate ‘system’ for CMT, but an individual billiard ball is not. Real billiard balls and other objects can rotate.

.

Since a raw vector sum of external forces on the system cannot distinguish between rotational or translational results, it cannot reliably represent the translational motion of any rigid or semi-rigid system of closely interacting particles. The vector sum cannot properly reflect the motion of the system or the CM, and is a meaningless value. To work, any version of CMT would require pre-treatment of external forces via torque handling techniques, and a consideration of the effects of rotation on the inertia of a system.
An easy fix is not possible for the current version of CMT however, because the vector sum is already deeply imbedded in its structure and formulation! The current version of CMT is clearly and fatally flawed. The next question is where exactly, and can it be fixed?

2. Flaw B: Non Uniqueness of the CM

Suppose we try to extend CMT to non-rotating rigid bodies. In this case, we would still need the CM concept in order to apply force without causing rotation. Thus the CM itself is retained in the theory of torque. We would also have to restrict external forces to those essentially applied along an axis of direction through the CM, but this generalizes the concept from a point to an axis.
The body would then move according to the laws of motion. However, now so does every particle in the rigid body, and nothing distinguishes the CM. In fact, part of the interpretation in the Centre of Gravity Theorem ( CGT ) predicts that all the particles in a system have the same velocity and parabolic trajectory, even when separated in the air, excluding any extra motion imparted in separation. The function of the CM is then reduced to an aid in applying force, but has no physical uniqueness in a rigid system , and no other physical meaning.
What is now left in our hands, after all these exceptions, constraints and limits? We had to abandon the summation of external forces, except for ‘systems’ of free non-interacting particles. We had to abandon rotating and rotatable systems to methods devised to handle torque. We had to abandon applying forces except uniformly or on a special axis, according to laws of equilibrium and balance. We had to abandon uniqueness and other overstated claims about the CM. The only substantive item left is a basic conservation of momentum law.

The theorem of Conservation of (linear) Momentum:
‘When the external force is zero, the total momentum P (the vector sum of the individual momenta) remains constant.’22
And, ‘the velocity of the CM is the same before and after (an interaction) in which the total linear momentum is conserved.’23

An understated, (and possibly unintended) sober assessment of these results is found in Knudsen:

‘The two theorems, the conservation of (linear) momentum for a closed (isolated) system, and ( CMT ) have the same physical content.’24

Note particularly the necessary word ‘closed’, i.e., completely isolated with no external forces. What is implied for systems with external forces? Not much.

Where does this leave us?

It leaves us with the perhaps daunting task of explaining the successes of CMT, removing the contradictions in it, and reformulating it in a manner that saves the phenomena, gives us a more solid theoretical foundation. Also desirable is increased accuracy and reliability, some practical methods, and some new and interesting predictions regarding Newtonian Kinematics and Dynamics. CMT itself now lies in shreds before us. The consequences for the Sphere Theorem ( ST ) which is a special case of CMT will be considered later.

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3. CMT : Blow by Blow Critique

Referring to the boxed and numbered steps in the formulation:
On 3) : The choice of weighted arithmetic mean could be challenged. It is the ‘first moment’ or centroid. This comes out of applications involving balance in a uniform field or equilibrium under rotation. Because of the physical properties of levers, torque is proportional to m x d. But field forces are proportional to m / d2 or q / d2 and other ways of weighting particle position could easily be proposed.
Take a non-uniform hollow sphere: At one point inside the gravitational field is zero ( ZG ) . This is not the GC or the CM. For obvious reasons, one may prefer this as the fundamental location of the sphere. The theoretical justification for the CM therefore is weak for both rigid and non-rigid systems.
Any supplementary argument drawing on rotation or equilibrium is irrelevant if we restrict CMT to systems only having linear motion. That the CM can actually sit in space between particles should remind us that it is not a real physical entity.

On 5) : Having the velocity etc. of the system dependent on the same definition gives consistency and beauty. But the potential circularity from interdependence throughout is far more important and worrisome. A solid physical basis for every observable, not relying upon the properties of numbers, vectors and algebraic systems, is desirable.

On 6) & 7) : Momentum is a powerful and deep concept, and total momentum of a system is also clearly a useful number, but are they fundamental physical entities? Does adding bound vectors together give us another bound vector, or an unbound abstraction? We don’t need the vector sum of individual momenta to be a ‘real’ vector that can be decomposed into ‘Mass’ and ‘Velocity’, in order to have Conservation of Momentum. We could just as well have Conservation of a Sum of Momenta.

On 8) : The total mass may be the most questionable entity of all, although on the surface it looks unassailable. Yet can we just add the masses together, when they are really spread all over space? This is actually one of the very things that CMT is supposed to demonstrate, not assume.

On 10) - 13): velocity CM = momentumsystem / Mtotal Is this miraculous confirmation of a theory? … or just an illusion predetermined by bad definitions and pulled off by taking abstractions too literally and allowing nonsensical operations?
On 14) & 15): The characterization of the CM seems extreme and overextended.25 But does the behaviour of the CM at rest or in uniform motion really tell us anything about how the system will accelerate under external forces? Should a baboon observe a sleeping lion to determine its temper before poking it?

On 16) - 19): This result is probably the fatal flaw in CMT and seems to have been completely misunderstood. Force applied from outside to systems as simple as a rigid stick have variable effects on linear motion, and the effects are not reflected in vector sums. We know that just adding the forces gives ludicrous results. The fact that combining such a sum with another suspect entity, total mass , results in a ‘solution’ for Newton’s 2nd law ought to make us jump out of our skins. - Not in the direction of making magical claims for the CM, but in the direction of questioning the applicability of the 2nd Law to abstract non-physical entities, and toward abandoning the current formulation of CMT.

Generally, force vectors are a restricted type of vector. They are not ‘free vectors’, such as displacement vectors. They are linked to a location and time, or to a particle, and so they are called ‘bound vectors’. Normally, they can only be added vectorally (i.e., by parallelogram law) if they are bound to the same location and time, or the same particle. The directional sense of a vector often only has physical meaning when assigned to point to or from a location or source of the force.
The resultant vector is also bound to a particle or location in the same way and it also can only act upon the system through actual physical mechanisms.
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25. Theorists seem to have stopped just short of giving it intelligence and a will of its own! Not bad for an abstract point-mass whose location in real space is actually empty.






Demonstrating CMT Failure



1. Analyzing Newton’s Balls

First, let's take a pair of equal point-masses joined by a rod of negligible mass (a barbell), and call it System A. The centre of mass ( CM ) for system A is at the geometrical centre (GC). For clarity we draw our point-masses as a small balls. To get the total force on another test-mass, (System B), aligned on the same x-axis we just add the forces from each end of the barbell using Newton’s formula. We don't even need the Sphere Theorem or vectors with the point-masses are all in a row in a single dimension.







For clarity we normalize the equations by making the masses of each system 1 unit, and choose distance units to make the gravitational constant G = 1 as well. This lets us drop the Gmamb part of Newton's formula. Now the formulas simply become 1/ d2 for each part, the d in this case is the distance between the test-mass (system B) and each each end of the barbell. The new equation for the force between the systems is restated as




In this form we can keep the system-to-system distance constant while spreading our balls. We could normalize d as well, ( i.e., d = 1 ) but we leave it in to show that the formula reduces to the inverse square law. We construct a simple triangle in the diagram to identify Theta in the equation. By inspection the angle Theta in the diagram is always less than 45o because the right triangle is formed using the distance d , the distance between the geometrical centre ( GC ) of system A and the test-mass.

The force depends upon the spread in system A and becomes Newton’s formula when the two point-masses merge as r = 0 and Theta = 0 o.

We use this formula for the force of system A to calculate the position of a singular equivalent point-mass ( EPM ). That is, we wish to replace system A with a single point-mass having exactly the same mass as system A. We need to place it so that it exerts the very same force on B. By inspection we know that this location will be somewhere along the axis of alignment. We only need to find the distance from test-mass B.

Since we want a new distance ( dnew ), giving the same force, and the EPM is just a point-mass, we simply rearrange Newton’s point-mass formula ( NGF ) to solve for distance, and substitute our new equation for force into it. The Gravitational Constant and the masses cancel and drop out, leaving a general formula for the distance from the test-mass, based on Theta alone. Remember, Theta is just a convenient way to express the spread of the balls in system A:



Keeping the centre-to-centre distance d constant, we spread our balls. Naturally, the EPM distance ( dnew ) decreases as the spread and the force increases. The position of the EPM drifts closer to test-mass B , catching up with the inner point-mass by the time it reaches test-mass B.

This is precisely as it should be. Gravity is an inverse-square law. It is non-linear (varying exponentially) by nature. The increase in force as one point-mass approaches test-mass B is not balanced by a complimentary decrease from the retreating point-mass, because the rate of change slows down farther away. This would be true of any non-linear distance law, and is not remarkable.

…Except that the system does not behave ‘as though all its mass was concentrated at the CM’ at all! At least as to magnitude of force, it behaves as though its mass were located at an entirely different position. Rotating system A will change the direction of force as well. This will be generally true of any object with a mass distributed in space in a non-spherically symmetric manner.

It is important to grasp the quality of the variation here: It is not a small error of approximation, or even adjustable with a simple factor. The force can fluctuate almost infinitely in any direction < 90o. It is virtually unpredictable without exact information as to both the distribution of mass within systems and the distances between the systems.


2. CMT contradicts NTF

It is important at this point to carefully explain what actually went wrong here. We calculated the correct location for an equivalent point-mass the hard way, by actually computing and adding the (two) individual forces acting on the system from outside. It is not the location predicted by CMT. That is, CMT contradicts results using Newton’s Gravity Formula for point-masses ( NGF ), which is the core of NTF. But CMT was supposed to be a shortcut for NGF, to save us having to use it.
Is NGF at fault, or CMT? In fact, the fault is clearly with CMT. It simply provides no mechanism to accommodate the vast fluctuations in force possible with even the simplest gravitational objects. The failure of CMT is built in. Creating a weighted average of position just throws away critical information as to distribution of mass right from the start. This self-mutilation fatally cripples CMT. It simply can’t do what is being asked of it, and so CMT invariably produces force vectors which are incorrect in magnitude and/or direction. It is no surprise that CMT is internally inconsistent. As a result, CMT not only contradicts NGF, but also ST , as we will see below. Since ST is just a special case of CMT, it simply contradicts itself!
Although we’d prefer CMT to NGF, we can’t. Abandoning NGF wouldn’t save CMT in any case. There is no way to relax the absolute requirement that force vectors sum in a coherent and self-consistent way. Couldn’t CMT be valid for some other unknown reason? No. CMT itself is an invalid method for calculating external gravitational forces and predicting motion, because it wipes out critical information.

3. CMT contradicts Itself

A simple example shows that CMT and ST contradict one another. Take a solid uniform half-sphere, and calculate its centre of mass ( CM ). For our purposes, it is irrelevant where it actually is, but by inspection it must be somewhere on the axis of symmetry. We can mark it with a dot . The important thing is that the CM is definitely a fixed point inside the object, since it is the weighted average of all the atoms making it up, and its a rigid solid.




Placing two half-spheres together to make a whole sphere puts the CM for each half at a fixed, equal opposing distance from the GC of the whole sphere. By symmetry, the CM for the whole sphere is at the GC itself.



Now add a test-mass. This is a precise parallel to the case above with the balls. The forces upon each half should add up to the force upon the whole sphere. Remember, this is just a static case: there is no motion involved. CMT should enable us to replace the halves with point-masses at their respective CM s , and add the vector forces. If CMT is self-consistent, an EPM for the halves should be at the same location as the CM of the whole sphere. But we already know this is not the case. Each half contributes unequally to the force, because of its position. The resultant force gives an EPM location that is closer to the test-mass than the CM of the sphere. Clearly CMT can’t deliver the goods.




Similarly, One could split up the sphere horizontally: Again treating the halves separately and using the Centre of Mass Theorem diverges from the result predicted by the Sphere Theorem:

According to CMT, any half-sphere, spinning or not, and regardless of orientation, will follow the laws of motion as if its mass were concentrated at that point. This must entail applying vectors to the CM and using actual masses for practical calculation. For such purposes, one is to treat the half-sphere as a point-mass of exactly the same mass located at the CM. That is, if CMT has any value, it must mean that we can use it along with NGF to calculate the external force between objects, and predict the resulting motion. This is precisely how Newton uses it to calculate and predict projectile motion under earth gravity. There, the gravitational field is virtually uniform for small objects near the surface.
This is not a trivial point, nor can one get around it by insisting that the centre of mass concept is only meaningful as a centre of gravity and ‘only applicable in a uniform field’. The gravity field in the isolation of deep space would presumably be the most uniform field of all. There, the mass of the universe (the distant backdrop of stars) is randomly distributed but essentially spread evenly in all directions.
If you can use CMT out in space, and also next to a huge mass like the earth, you must be able to use it pretty well everywhere. But if we can use it almost anywhere, what are the real restrictions as to when and how we can use it? CMT itself offers no help or explanation for such internal contradictions as plainly exist.

4. The 'Success' of CMT: A Cosmological Coincidence

As it turns out, both Newton’s and our good fortune has more to do with circumstantial factors than universal physical principles. The apparent success of CMT is just that: an appearance created by special circumstance. The vast distances of space dwarf the actual size of stars so much that those spherical bodies are essentially point-masses, and ST while trivially true is also untestable as such. On the other hand, gravity is so weak a force that it is virtually immeasurable between even adjacent objects at the normal human scale of experience. Earth gravity originates at an average distance roughly the radius of the earth, again dwarfing the size of ordinary objects.
Thus the mechanism for CMT ‘s apparent success is really just a set of special circumstances and cases. The root of the matter is simply that for most objects considered, the mass is sufficiently concentrated and point-like for any errors in CMT to pass undetected under the umbrella of imprecision of measurement.

CMT and GR


But does this sufficiently explain the success and accuracy of CMT ? In fact, no it does not. There remains another aspect of CMT not yet discussed, and a hidden mechanism to explain it. Basing CMT on NTF alone is a failure, because any simple version of CMT fails a compatibility test with NGF, the heart of NTF, as well as failing the basic self-consistency test.
Looking back at examples given to justify CMT, one notices that rotation is always involved, although not necessarily gravity. This indicates there is real physical content behind the mistaken current formulation. If our only real knowledge of the CM comes through rotation or interaction with a uniform gravitational field, we should look more closely here for a better foundation than CMT can offer for both phenomena in relation to the CM.
But a new formulation of CMT is necessary and desirable, to account for rotation, moments of inertia and angular momentum. That is, paradoxically, while rotation effects are the downfall of CMT as currently formulated, and must be excluded from it, rotation is probably the only justification for any continued use of the CM concept! Basing some new version of CMT on GR or a revised version of Newtonian Mechanics offers promise.


Hidden Penalties and Dangers

Our good practical good fortune has its flip side in our theoretical misfortune and the potential dangers in assuming the validity of the concepts packaged in CMT. It seems reasonable to address these issues.