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Aether 7.2.27 |
Created Dec 3-13, 2003 Stavropol, RU.
To Understand The Aether |
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We Need To Understand The Cosmos! |
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The Cosmos! | ||
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We Understand | |
28 The Cosmos Exists! | ||||
To Understand The Cosmos |
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We Need Communication Skills. |
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We Have Communication Skills. | ||
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We Understand |
What is the reality that we have to start with to reason on, to investigate and interpret our results to discover the true science of physical reality and discover the eternal unchanging laws.
When I say reason on I do not mean scientific babble, I mean the logical foundation of physical reality that can guide our investigation and discover mathematical, computerizied models that prdict outcomes in reality.founded on e if we reason on and with correctly we can dis
Space, Aether in Space and Motion of the Aether in Space.
Insights |
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*Aether (From Wikipedia, the free encyclopedia) In Plato's Timaeus (St-55c) Plato described aether as "that which God used in the delineation of the universe."Logos (pronounced /ˈloʊɡɒs/ or /ˈlɒgɒs/; Greek λόγος logos) is an important term. Heraclitus (ca. 535–475 BCE) established the term as meaning the source and fundamental order of the cosmos. John 1:1-5 In the beginning was the Logos, and the Logos was with God, and the Logos was God. The same was in the beginning with God. All things were made by him; and without him was not any thing made that was made. In him was life; and the life was the light of men. And the light shineth in darkness; and the darkness comprehended it not. |
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Logos - the source and fundamental order of the cosmos.
Cosmos - the universe regarded as an orderly, harmonious whole.
Universe - All matter and energy, including the earth, the galaxies, and the contents of intergalactic space, regarded as a whole.
Chaos - The background random matter in transnational equilibrium (an ideal fluid in thermal equilibrium).
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Figure 1.1 Random background motion of elements of matter in space.
The Physical Universe is the Space, Matter and Motion in which and from which our visible and invisible physical Universe is formed. Aether is the one and only formless substance of the Logos from which and in which all material (Matter) bodies are formed and exist. The background Aether is in constant motion with collisions from all directions transferring momentum in all directions uniformly.
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Postulate 1 (The Extensive Property).
The extensive property of the Logos exists and is called Space! Space is continuous in three linear, independent, infinite dimensions.
Note: a vector is an arrow going from a starting point to an end. A vector can be a location (given an arbitrary starting point) or displacement . An example of the use of terms would be the corner stone in a building as the origin.
Locations are vectors from the origin to points. Displacements are vectors from one point to another. Displacements subtract out reference to the origin.
In the Figure below Rii and Rf
f are locations and
r
is a displacement. The vector
r
gives no information about the location of its beginning Ri
i or its end Rf
f.
Defining an arbitrary inertial, orthogonal Cartesian coordinate system starting point of R00 as the origin with three perpendicular directions (
1,
2,
3).
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Figure 2.1 Location r in an orthogonal Cartesian coordinate system.
We define location R in physical space in terms of its components
Definition 1R0
≡ R01
1 + R02
2 + R03
3.
We define displacement rfi
in physical space in terms of its components
Definition 2rfi
≡ Rf
f – Rf
f.
≡ Rf1
f1 – Ri1
i1 + Rf2
f2 – Ri2
i2 + Rf3
f3 – Ri3
i3
≡ (Rf1 – Ri1)
1 + (Rf2 – Ri2)
2 + (Rf3
– Ri3)
3
≡
rfi1
1 +
rfi2
2 +
rfi3
3
Notice that displacement subtracts out reference to Rff and Ri
i so rfi
has magnitude and direction but not location.
We define area A12 as the cross product
Definition 3A12
12 ≡
r1
1 x
r2
2.
We define volume V as the dot product of area and extension
Definition 4V123 ≡ A12
12 •
r3
3,
=
r1
1 x
r2
2 •
r3
3.
Postulate 2 (The Inertial Property).
The inertial property of the Logos called Matter exists! Matter is not created or destroyed. Each element of Matter is infinitely divisible. Each element of Matter occupies its own space uniquely. Basic Matter is a non-viscous, inelastic fluid. Out of the Logos Matter all material things formed and in the unformed chaos the move and have their being. The inertial property of Logos is called mass m.
Logos Matter is not just a medium in which electro-magnetic radiation propagates and is effected (like the imagined Aether or Dark Matter). The radiation and all material objects are forms created of the Logos Matter and swimming in the Logos Matter.
Each point of Logos Matter mi occupies it's own unique location Rii in space.
Equation 1Ri
i ≡ Ri1
i1 + Ri2
i2 + Ri3
i3.
We define density of a volume of space as the mass divided by the volume of space containing it.
Definition 5x ≡ mx/Vx
Postulate 3 (The Mobil Property).
Given two elements of Matter ma and mb the displacement between them rabab. Relative motion (velocity) vab
ab
between them exists!
Given two elements of Matter in space separating at a constant velocity vabab and the associated displacement rab
ab
, we operationally define a clock with time t by
Definition 6tab ≡ |
rab
ab| /
|vab
ab| ,
where
rab
≡ |
rab
ab| and vab ≡ / |vab
ab|.
With a clock defined we can graph the position of an element of Matter as a function of time. Collisions in the Matter change the position of an element of Matter instantaneously so the graph is jagged as represented in one dimension below.
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Figure 3.1 Position as a function of time.
This kind of graph has no well defined slope at the points of collision. In calculus terms r(t) is not a differentiable function of time.
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With a clock to measure time and a length standard to measure distance we operationally define the velocity vector in a given direction va for a period of time in space where there are not any collisions as
Definition 7va
≡
ra
a/
ta.
In the inertial frame of reference the net flow in a given direction is equal to the net flow in the opposite direction because as with the one dimensional physics toy motion flow is possible simultaneously in all directions (meaning + and - directions along each axis).
The magnitude of velocity (speed va) of any element of matter ma in space exists with an average speed v0. The average speed
v0 is uniform through out background matter. From this we conclude there exists an inertial frame of reference R0 where the average speed v0 is zero.
Given the random distribution of background matter, with a large enough volume V, we observe the effects of chaotic motion matter is in
equilibrium giving an average background density0 that is constant.
Observation 1 The average background density of matter c is constant.
The dimensionless saturation is the ration of the volume of mass Vm
in a volume of space Vs
Definition 8 ≡ Vm / Vs
By Observation 1 we define a universal constant the saturation of the chaos c for the ratio of a volume of matter Vmc and the volume of confining space Vsc.
Equation 3.1c ≡ Vmc / Vsc
With space, density, and motion we can define momentum passing through a given area or impinging on a given surface. The definition of momentum as mass m times its velocity v needs some insight when talking about a fluid like the Logos Matter. What mass are we talking about and what velocity are we talking about?
Figure 4.1 Source of basic Insight
Basic matter is not viscous and not compressible:
This means that if we had two pieces of pure basic matter with no internal motion, in a collision the momentum of one could be transmitted through the other without friction. For example see the diagram below showing a collision.
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Figure 3.2 Basic matter is not viscous and not compressible.
In Figure 3.3 we observe the mechanism for keeping a form in the chaos.
Observation 2 In collision momentum is conserved but instantaneously transferred ahead with a space discontinuity in the path of the center of mass of momentum. In the chaos the shape depends on the shapes and mass distributions in the collision. Since the shape is chaotic so is the outcome.
Momentum can move in opposite directions at the same time in the same pure matter. Below is a model of a Physics Toy with hanging steel balls.
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Figure 3.3 Physics Toy with hanging steel balls.
In Figure 3.3 we observe the mechanism for keeping a form in the chaos.
Observation 3 Matter deposited on one side of a stream is removed by collisions on the opposite side!
When we consider that the pure matter is not viscous and not compressible, we see that momentum freely moves in all directions at the same time.
Question: In background matter, what is the mass moving in a given direction?
Considering uniform random motion (similar to a gas) where half the matter has a component moving in any given direction and half the matter has a component
moving in the opposite direction.
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Figure 3.4 Elements of matter moving through random collisions.
Answer: At any given time, half the mass has a motion component in any direction and the other half of the mass has a component of motion in the opposite direction. With the infinitesimal sizes in the Logos Matter, this is on a scale large enough that random fluctuations are averaged out.
The expression of the half of the mass mii+
moving in a given direction
i+ in terms of the density (by Observation 1)
o ≡ mx/Vx is
Equation 2 mi+ = (1/2)mi
=(1/2)
0
Vi+ , ( i+ is direction on the i axis.)
Question: What is the average component in any given direction at any given time of the average speed?
The average speed associated with any momentum is distributed in all directions equally, so we need to find the average component of the average speed in any
given direction (note only one direction along the axis, both ways average to zero).
Answer: The velocity vk+k+ in a given direction associated with
the mass mik+ is the average velocity component of the average speed in any given direction. This ranges from 0 to vo. An
approximate average of this in any direction is half the average speed.
Equation 3 vk+
k+
(1/2)vo
k+ (where k+ indicates the + direction on the k axis.)
The definition of momentum of a mass element mi is mi times its velocity vik+
Definition 9 pi+
k+ ≡ mi+vk+
k+
For a component of momentum passing through a given area in a given direction, we use Equation 2 and Equation
3 in Definition 9 to get:
Equation 4 pi+
k+
mi+(1/2)vo
k+ , Equation 3 in Definition 9
(1/2)
0 Vk (1/2)vo
k+ , Equation 2 in Definition 9
(¼)
0 Vk vo
k+
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(The speed of a longitudinal wave as a function of density)
Matter transfers its momentum in collision. The collisions are of incompressible fluid matter. Consider a set of billiard balls in a row. An element of matter traveling with the average background speed *vo=dl/dt (*one dimensional).
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Figure 4.1 Billiard Balls.
From Figure 4.1, momentum is transmitted the distance Lt in time tt.
Observable 4.1 Lt = the distance momentum is
transmitted.
Observable 4.2 n = the numver of elements in length Lt.
Observable 4.3 d = the diameter of an element.
The length of matter the momentum traveled through instantaneously was
Definition 4.1Lm ≡ nd.
The length of space the momentum traveled through at *vo
Definition 4.2Lf ≡ Lt - Lm , or Lm=
Lt -Lf .
The observed time tt for the momentum to be transmitted distance Lt is
Observable 4.4tt =observed time to transmit momentum distance Lt
.
Matter traveled the free space distance Lf in time tt. By Definition 6 tt ≡ Lt
/vt , the speed of the free matter is
Equation 4.1 vf = Lf /tt.
From Observable 4.1 Lt and Observable 4.4 tt
the speed of momentum vp is
Equation 4.2 vp=Lt /tt
From Equation 4.1 vf =Lf /tt and Equation 4.2 vp=Lt /tt the ratio of the speed of momentum transmission vm to the average background
free space speed vf is
Equation 4.3 vp /vf =( Lt /tt)/(Lf
/tt)
= Lt /Lf .
From Equation 4.3 vp goes to infinity as Lf goes to zero. That is as the density increases the speed of transmission
of momentum increases while the speed of free motion is unchanged.
By Definition 5 i ≡ mi/Vi
the density of pure matter is its mass divided by its volume
Equation 4.4 0m =mi/Vm.
We have Observation 1 0 = mx/Vs
but here we are varying the background volume to explore the effect on speed of momentum transmission in the Chaos
Equation 4.5 b = mx/Vb
By taking the ratio of Equation 4.4 0m=mx/Vm
and Equation 4.6
b = mx/Vb
Equation 4.6 0m/
b = (mx/Vm)/(mx/Vb)
= Vb / Vm = Lm3/Lt3
Given the average background free space speed vo, consider average speed as a function of density. We look at this system
for an arbitrary unit of distance (large enough that the averages are not invalidated). We use 1000 units of total space and very the free space 1000 to 0.01.
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Figure 4.2 The Table uses a 1000 units total distance Lt and varies the free space Lf 1000 to 0.01.
The graph below represents the data table above.
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Figure 4.3 Graph as the density approaches 100% the speed approaches infinity!
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Consider the improbable event of streams forming in the chaos.
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Figure 5.1 A stream forming in the chaos.
Our ideal matter has its own characteristics. To have intuitive insight into behavior of a stream on the chaos we need to have the reality experiences that reflect these characteristics.
A stream is a group organization out of the chaotic matter. A steam travels with a group velocity with respect to the background which averages to zero.
Observation 4 From Observation 3 says that the stream shape will tend to be maintained while Observation 2 says there is a difference in the speed of any element and the external speed of momentum as the momentum advances through transfer. An unverified thought is that the average speed of an element of matter participating in the stream is the same as the ambient average speed. This would mean that there is a direct relationship between that average ambient speed and the speed of any stream.
The next level of organization is a stream forming a vortex. Consider the improbable event of a vortex forming from streams. Below we consider a section of a vortex about the axis of rotation.
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Figure 5.2 A vortex forming in a stream.
What is the difference in effect of ambient chaos on the inner vortex boundary and the outer vortex boundary?
Observation 5 The need here is to discover the mechanics that give identity and multiplicity there is some condition (probably density dependent) that gives a size constraint. The spiral form of solar systems and galaxies may also be an outcome of the same mechanics on a larger scale. Here I am not talking about some abstract statistical formulation but a concrete visible mechanics. What is happening differently on the inside surface and the outside surface interaction with the chaos is a focus worth attention.
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Figure 5.3 A vortex cross section.
Consider the improbable event of a vortex ring (toroid) forming from the vortices.
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Figure 5.4 A toroid cut out.
With vortex rings and multi-ring objects formed from them in the Chaos, we can find objects and observe their position as a function of time. The position as a function of time is smoother than collisions in the Chaos. An example would be throwing a ball. up as shown below.
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Figure 5.5 Position as a function of time.
Using the limit as t
0 in Definition 7 operationally defines the three-dimensional velocity of and
object as the slope of the displacement time graphs.
Definition 10 vi
i ≡ Limit (
ri/
ti)
i
t
0
= Limit (
ri1/
t)
i1+ (
ri2/
t)
i2 + (
ri3/
t)
i3
t
0
= (dri1/dt)
i1+ (dri2/dt)
i2 + (dri3/dt)
i3
= vi1
i1+ vi2
i2 + vi3
i3
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Figure 5.6 Velocity as a function of time.
We operationally define the three-dimensional acceleration of and object as the slope of the velocity time graphs
Definition 11 ai
i ≡ Limit (
vi/
ti)
i
t
0
= Limit (
vi1/
t)
i1+ (
vi2/
t)
i2 + (
vi3/
t)
i3
t
0
= (dvi1/dt)
i1+ (dvi2/dt)
i2 + (dvi3/dt)
i3.
= ai1
i1+ ai2
i2 + ai3
i3
We can define the three dimensional force F acting on an element of mass mi with cross section area Ai in the Chaos by
time t rate of change matter r0 Vi in the impacting volume moving at average background speed voi .
Definition 12 Fi
i ≡ dpi
i
/dt
= dp1
i1
/dt + dp2
i2 /dt + dp3
i3 /dt
Now we need to recognize the infinite divisibility of the Chaos. This means that when two particles of matter collide the momentum of the contacting parts is
not lost like in the crash of to cars but as with the physics toy continues in exchange and jumps ahead into the forward element in both directions. Since Chaos
is ubiquitous (everywhere like a fog) this continuation of momentum and in fact jumping ahead was accounted for in the average speed in any given direction.
We have the net force Fir+ acting in the r+ direction on an element of mass mir+. The mass is in a volume
with cross section area Air. The average momentum pir+r+ of the ith element moving in a given direction (r+) is from Equation 4 pir+
r+
(1/4)
oVir+vo
r+. The ambient force Fir+ in any direction
r+ on an arbitrary area of obstruction Air is
Equation 5Fir+
r+ ≡ d(pir+
r+) /dt ,
d((1/4)
0 Vir+ vo)/dt
r+ , note (1/4)(
0 Air vo) is not a function of time
= (1/4)(
0
Air vo)dri/dt
r+ , by Definition
7 dri/dt
r+ = vir+
r+
= (1/4)(
0
Air vo) vir
+r+ , by Equation 3 vir+
r+= (1/2)vo
r+
= (1/4)(
0
Air vo)(1/2)vo
r+ ,
= (1/8)
0
Air v2o
r+ .
We can define the average pressure on an element of mass mi in a given direction rr by dividing the average force Fi ave over the area Ar
.
Definition 13 P ≡ Fir+ave
r+●Air
r+ / (Air
r+ ● Air
r+ )
= Fir+ / Air ,
Using Equation 5 Fir+r+ = (1/8)
oAir v2o
r+ or Fir+ = (1/8)
o Air v2o in Definition 13 P = Fir+/Air
gives
Equation 6 P = (1/8)
oAir v2o / Air
= (1/8)
o
v2o ,
Note that in the primitive (no vortexes) Chaos this is universal and for a vortex it is the external pressure.
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For uniform circular motion we study the diagram below.
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Figure 6.1 A point in uniform circular motion.
We have a point moving at a constant speed vc. At time t1 the point is at r1. At time t2 the point is at r2. In the change of time is dt, the change in position is dr and the change in velocity is dv.
We have similar triangles that give equal ratios for dr/r and dv/v
Observation 6 dr/r = dv/v.
Solving Definition 10 v=dr/dt for dt gives:
Equation 6.1 dt = dr/v
Rearranging Observation 6 dr/r = dv/v gives vdr/r = dv.
Then dividing both sides by Equation 6.1 dt = dr/v gives (vdr/r)/(dr/v)=dv/dt=a then reducing gives centripetal acceleration
Equation 6.2 a=v2/r.
Consider a vortex section shown below.
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Figure 6.2 A vortex cross section.
The pressure on the surface area exerts the force that is equal to the centripetal (inward) force of rotation. Note that momentum in a stable vortex behaves
like an element of matter mi and momentum is associated with the same element as it moves.
We apply Newton's second law to a segment of the vortex cut out
Equation 6.3 Fi=miai.
=
Using Equation 6.2 ai=vi2/ri in Equation 6.3 F=ma gives
Equation 6.4 Fi =mivi2/ri.
Using Equation 4.5 mi = mi/Vmi
in Equation 6.4 Fi =mivi2/ri gives
Equation 6.5 Fi =
miVmi vi2/ri.
Looking at the segment in the diagram above we have the volume Vi
Equation 6.6 Vi = dlAi
Using Equation 6.6 Vi=dlAi in Equation 6.5 Fi
=miVmi vi2/ri gives
Equation 6.7 Fi =
mdlAi vi2/ri.
We have the pressure from the Chaos by Equation 5.2 Pi = (1/8) 0 v2o .
Using this in Definition 13 Pi = Fi /Ai gives the external force as
Equation 6. 8 Fext = PA =(1/8)
0v2oA
We apply Newton's third law (to every force there is an equal and opposite force)
Equation 6.9 Fr-= -Fr+ .
Using Definition 5 Density x
≡ mx/Vx in Equation 6.4 Fc =mv2/r gives
Equation 6.10Fc =
mdlAivi2 /ri
Using Equation 6.8 Fext =PA =(1/8)ov2oA
and Equation 6.10 Fc =
mdlAivi2
/ri in Equation 6.9 Fr-=-Fr+ gives
Equation 6.11 mdlAivi2 /ri = (1/8)
0vo2Ai ,
mdl
vi2 /ri = (1/8)
0vo2 ,
dlvi2 /r = (1/8)(
0/
m)vo2
.
If we assume that the driving speed of the Chaos vo2 is the same as the tangential speed of the vortex vi2
we have dl/r =(1/8)( o/
m), or:
Equation 6.12 r =8dl(
m/
o)
This Equation 6.12 r =8dl(m/
o) would mean that the ratio of the wall radius and the wall thickness would be a
constant. Since vortex rings are the building blocks of magnetic fields, this could explain how they can range so much in major radius. That is that as the
major radius expands conservation of energy is accomplished by the shrinking the minor radius and wall thickness.
The question remains, what limits the decrease in the radius of the vortex ring? We could speculate on the ratio of the densities, say if
the pure density is equal that of the chaos the ratio of r = 8dl.
If the pure density m was twice that of the chaos the ratio of r = 16dl.
This is a limit. the minor radius r (the minor radius of the vortex must be smaller than 8dl (the thickness of the vortex wall). We have:
Equation 6.13 r > 8dl .
The internal energy is kinetic. The energy of a segment is (1/2)mv2, m= 2rdl
m ,
Equation 6.14 K.E.R/z =
rdl
mv2.
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We start with a toroid cut out shown below. We assume this is a closed system in the sense the the mass is constant and the internal energy is constant.
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Figure 7.1 A toroid cut out.
The total length of z (the axis of the toroid segments) is the center circumference 2R
Equation 7.1 z = 2
R.
Using Equation 7.1 z = 2R in Equation
6.14 K.E.R/z =
rdl
mv2, the total energy K.E.R would be
Equation 7.2 K.E.R =
rdl
mv2(z)
,solving Equation 6.14 for K.E.R ,
=
rdl
mv2(2
R)
= 2
2Rrdl
mv2.
Using Equation 6.12 r =8dl(m/
o) in Equation 7.2 K.E = 2
2Rrdl
mv2
gives
Equation 7.3K.E=2
2R {8dl(
m/
o)} dl
mv2
= 16
2Rd2lv2
2m /
o
This show us the variables while energy is conserved is such large change as the major diameter changes in magnetic fields.
8 |
Consider two groups of vortex rings (making up visible matter) as depicted in the diagram below. The center and top spots represents the cross sectional area of a spherical object.
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Figure8.1 Two groups (center and top) of vortex rings (making up visible matter).
The idea here is that the pressure of the chaos is ubiquitous but clusters of vortex rings deflect the incoming momentum causing a shadow effect. This shadow effect causes the shadowed surface to experience decrease in momentum impacting on the shadowed side. This decrease in pressure between the objects causes them to experience a net external force toward each other. Groups of rings (objects) are forced toward each other rather than the false concept of "action at a distance" mediated by some imaginary thing called a field.
Given the average cross section area of a single vortex ring is ao. The average cross sectional area for the n center rings is
Equation 8.1 Acenter = n ao .
We assume that the chaos is fine enough that the cross section area of every ring is seen.
The free vortex rings at the center experience pressure from the momentum of the chaos impacting giving a force Fcenter from all
directions. We use Equation 8.1 Acenter = n ao to multiply the pressure in Equation
6 Pi = (1/8) o v2o get the force
impinging on the surface of the center group from any direction. The ambient force on the center object is
Equation 8.2 Fo center = P Acenter
= (1/8)
o
v2o ncenter ao
What part of that force is missing due to the shielding or shadow effect of the object at the top?
Given the shielding and its shadow the decrease in force in that direction is proportional to the ration of the area of the obstruction divided by the area of the sphere on which it lies with the shadowed object at the center of the sphere and the shadowing object at the top of the sphere.
When object at the top has k vortex rings it casts a shadow of
Equation 8.3 Atop = kao .
For a sphere of projection with radius Rs we have the projected spherical surface area of 4R2s or
Equation 7 Asphere = 4
R2.
Where the center object has n rings of cross-sectional area of ao. The shadow effect causes reduction in pressure from that
direction. The net force pushing the two groups together is proportional to the ratio of the area of the cross section of a single ring ao times the
number k of rings and the area of the sphere of projection radius Rs and projected surface area of 4R2s or
Equation 8.4 Atop / Asphere or kao
/(4
R2 ) .
Given the mass mo of a single ring (constant mass of a single ring). The mass of the top object with n rings is
Equation 8.5 mcenter = nmo .
The mass of the center object with k rings is
Equation 8.6 mtop= kmo .
So the net force from the shielding imbalance is
Equation 8.7 Fcenter net = Fo center Atop
/ Asphere , shielding from Equation 8.4
= (1/8)
o
v2o Acenter Atop / Asphere , using Equation 8.2
= (1/8)
o
v2o kao nao /(4
R2 ) ,
using Equation 7 ,8.5 , and 8.6
= (1/32
)
o v2o a2o k n / R2 ,
simplifying
= (1/32
)
o v2o a2o (mtop/mo)
(mcenter/mo) / R2 ,
= (1/32
m2o)
o v2o a2o
mtop mcenter / R2 , using Definition 14
= G mtop mcenter / R2 .
We define the Universal Gravitational Constant
Definition 14G ≡ (1/32
m2o)
o v2o a2o then = G (mkmn /R2)
This is then a derivation Newton's general law of gravitation!
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Figure List | Foundation Items |
Figure 1.1 Random background motion of Matter in space. Figure 2.1 Location r in an orthogonal Cartesian coordinate system. Figure 3.1 Position as a function of time. Figure 3.2 Basic Matter is not viscous and not compressible. Figure 3.3 Physics Toy with hanging steel balls Figure 3.4 Elements of Matter moving through random collisions. Figure 4.1 Billiard Balls Figure 4.2 As the density approaches 100% speed approaches infinity! Figure 5.1 A stream forming in the chaos. Figure 5.2 A vortex forming in a stream. Figure 5.3 A vortex cross section. Figure 5.4 A toroid cut out. Figure 5.5 Position as a function of time. Figure 5.6 Velocity as a function of time. Figure 6.1 A point in uniform circular motion. Figure 6.2 A vortex cross section. Figure 7.1 A toroid cut out. Figure 8.1 Two groups of vortex rings (making up visible Matter) |
Postulate 1 Space R0 Definition 1 Location R0 Postulate 2 Matter m Exists! Definition 5 Density Postulate 3 Motion v Definition 6 Time t |
10 |
Postulate 1 Space R0 |
Equation 1 Ri![]() ![]() ![]() ![]() Equation 2 mi+ =(1/2) ![]() Equation 3 vk+ ![]() ![]() Equation 3.1 saturation ![]() Equation 4 pi+ ![]() ![]() ![]() Equation 4.1 velocity in free space vf = Lf /tt. Equation 4.2 velocity of momentum in the chaos vp ![]() ![]() Equation 4.3 speed of momentum / speed in free space vp /vf = Lt /Lf . Equation 4.4 density of pure matter ![]() Equation 4.5 varying the background volume ![]() Equation 4.6 ratio of constant and varied density ![]() ![]() Observation 4 difference in speed of element and speed of momentum Observation 5 What's happening differently inside surface and outside Equation 5 Force Fir+ ![]() ![]() ![]() Equation 6 Pressure Pi = (1/8) ![]() Observation 6 similar triangles dr/r = dv/v. Equation 6.1 dt = dr/v Equation 6.2 centripetal acceleration a=v2/r. Equation 6.3 Newton's second law Fi=miai. Equation 7 surface area of a sphere of radius R Asphere = 4 ![]() Equation 7.1 Major diameter of ring ztotal = 2 ![]() Equation 8.1 Acenter = n ao . Equation 8.2 Fo center = P Acenter= (1/8) ![]() Equation 8.3 Atop = kao . Equation 8.4 Ratio of view obstruction Atop / Asphere or kao /(4 ![]() Equation 8.5 center object mcenter = nmo . Equation 8.6 top object mtop= kmo . Equation 8.7 Force Fcenter net = G mtop mcenter / R2 |