Wednesday, April 16, 2014

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


1 comment:

  1. DavisBJ responded:

    Good, I thank you for the clarification. And I agree with much of what you say. I agree that you have shown that the net gravitational vector of the tilted ellipse is not aligned with the cone axis. I agree that if the net gravitational vector is not aligned with the cone axis, that must mean it is directed at some angle to the cone axis, and that would result in a force towards the nearest wall. And this is mirrored on the “back side” of the disk, as you say, doubling the error.

    I want to use your own scenario, and refer to the CM analysis you show clear back in the very first post a year ago. Your recent post #114 supplements and clarifies the CM proof that the net force is off the cone axis. Now follow me and see if you think I am being misleading (not that I really need to ask, you have kindly been attentive to my posts).

    As I see it, here are the crucial items you use in the CM case:

    1) The ellipse is defined by the intersection of the tilted wall and the cone from the test particle. (I am willing to say “ellipse” in this case, even though we are in a curved wall situation, because the crux of my point will not rely on the figure not being a true ellipse).

    2) The ellipse CM is at some distance, say r, from the test mass.

    3) The tilt of the wall determines the angle the ellipse makes with the cone axis. Here again for the record, if the concave curvature of the wall was included, a slightly bigger or smaller ellipse would have slightly different tilt angles, even though they have the same CM as the original ellipse we are working with. I don’t think that is a significant issue here. I will say the ellipse is tilted at angle theta, period.

    4) We split the ellipse into near and far halves, using a line across the ellipse that goes through the CM and is perpendicular to the direction to the test particle.

    5) By inspection (and by geometry if needed) we can see that the near half of the ellipse is closer to the test particle than is the far half.

    6) Since the two halves of the ellipse are symmetrical, they have the same mass.

    7) It is a reasonable approximation to consider the mass of each half as though it were concentrated at the CM of just that half. This is essentially what you have called the EPM.

    8) The force from the near half will be greater than the force from the far half, as per the inverse square nature of Newton’s Law of Gravity.

    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.

    12) The above is just a baby-step way of saying the vector sum of the near side and far side forces on the test particle will have a component at right angles to the cone axis, resulting in a net finite force on the test particle to the side of the far side of the ellipse.

    I don’t think anything I have said above is substantially different than what you have explained in several ways already. Again, I am trying to be sure we are in step before going forth from here.

    It might sound like, with the above, I have inherently conceded. I have not. I have concerns, but those concerns depend on us clearly agreeing on the above.

    If you agree with the above so signify. If later you find that there are implications that you had not anticipated in the way I summarized my understanding, I will not consider you to be locked into the way I have expressed the ideas. I am not setting a trap, I am trying to define an agreed to base from which we can proceed.

    ReplyDelete