Wednesday, April 16, 2014

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 View Post
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'.