7.2.28 Thesis On Cosmos

To Understand The Aether
We Need To Understand The Cosmos!				The Cosmos!
		(1,2)		We Understand
28 The Cosmos Exists!
To Understand The Cosmos
We Need Communication Skills.				We Have Communication Skills.
				We Understand

Introduction

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.

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).

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.

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 R_ii and R_ff are locations and r is a displacement. The vector r gives no information about the location of its beginning R_ii or its end R_ff. 

Defining an arbitrary inertial, orthogonal Cartesian coordinate system starting point of R₀₀ as the origin with three perpendicular directions (₁,₂,₃).

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 ≡ R₀₁₁ + R₀₂₂ + R₀₃₃.

We define displacement r_fi in physical space in terms of its components

Definition 2r_fi ≡ R_ff – R_ff.

≡ R_f1f1 – R_i1i1 + R_f2f2 – R_i2i2 + R_f3f3 – R_i3i3

≡ (R_f1 – R_i1)₁ + (R_f2 – R_i2)2 + (R_f3 – R_i3)₃

≡ r_fi11 + r_fi22 + r_fi33

Notice that displacement subtracts out reference to R_ff and R_ii so r_fi has magnitude and direction but not location.

We define area A₁₂ as the cross product

Definition 3A₁₂₁₂ ≡ r₁₁ x r₂₂.

We define volume V as the dot product of area and extension

Definition 4V₁₂₃ ≡ A₁₂₁₂ • r₃₃,

= r₁₁ x r₂₂ • r₃₃.

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 m_i occupies it's own unique location R_ii in space.

Equation 1R_ii ≡ R_i1i1 + R_i2i2 + R_i3i3.

We define density of a volume of space as the mass divided by the volume of space containing it.

Definition 5_x ≡ m_x/V_x

Postulate 3 (The Mobil Property).

Given two elements of Matter m_a and m_b the displacement between them r_abab. Relative motion (velocity) v_abab between them exists!

Given two elements of Matter in space separating at a constant velocity v_abab and the associated displacement r_abab , we operationally define a clock with time t by

Definition 6t_ab ≡ |r_abab| / |v_abab| ,

where r_ab ≡ |r_abab| and v_ab ≡ / |v_abab|.

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.

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.

With a clock to measure time and a length standard to measure distance we operationally define the velocity vector in a given direction v_a for a period of time in space where there are not any collisions as

Definition 7v_a ≡ r_aa/t_a.

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 v_a) of any element of matter m_a 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 V_m in a volume of space V_s

Definition 8 ≡ V_m / V_s 

By Observation 1 we define a universal constant the saturation of the chaos _c  for the ratio of a volume of matter V_mc and the volume of confining space V_sc.

Equation 3.1_c ≡ V_mc / V_sc 

 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.

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.

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.

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 m_ii+ moving in a given direction _i+ in terms of the density (by Observation 1) _o ≡ m_x/V_x is

Equation 2   m_i+ = (1/2)m_i

=(1/2)₀ V_i+ , ( 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 v_k+k+ in a given direction associated with the mass m_ik+ is the average velocity component of the average speed in any given direction. This ranges from 0 to v_o.  An approximate average of this in any direction is half the average speed.

Equation 3 v_k+k+ (1/2)v_ok+ (where k+ indicates the + direction on the k axis.)

The definition of momentum of a mass element m_i is m_i times its velocity v_ik+ 

Definition 9 p_i+k+ ≡  m_i+v_k+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 p_i+k+ m_i+(1/2)v_ok+ , Equation 3 in Definition 9

 (1/2)₀ V_k (1/2)v_ok+ , Equation 2 in Definition 9

(¼)₀ V_k v_ok+ 

 (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 *v_o=dl/dt (*one dimensional).

Figure 4.1 Billiard Balls.

From Figure 4.1, momentum is transmitted the distance L_t in time t_t.

Observable 4.1 L_t =  the distance momentum is transmitted.
Observable 4.2 n =  the numver of elements in length L_t.
Observable 4.3 d =  the diameter of an element.

The length of matter the momentum traveled through instantaneously was

Definition 4.1L_m ≡ nd. 

The length of space the momentum traveled through at *v_o

Definition 4.2L_f ≡ L_t - L_m , or L_m= L_t -L_f .

The observed time t_t for the momentum to be transmitted distance L_t is

Observable 4.4t_t =observed time to transmit momentum distance L_t . 

Matter traveled the free space distance L_f  in time t_t. By Definition 6 t_t ≡ L_t /v_t , the speed of the free matter is

Equation 4.1 v_f = L_f /t_t.

From Observable 4.1 L_t and Observable 4.4 t_t the speed of momentum v_p is

Equation 4.2 v_p=L_t /t_t

From Equation 4.1 v_f =L_f /t_t and Equation 4.2 v_p=L_t /t_t the ratio of the speed of momentum transmission v_m to the average background free space speed v_f  is

Equation 4.3 v_p /v_f =( L_t /t_t)/(L_f /t_t)
= L_t /L_f .

From Equation 4.3 v_p goes to infinity as L_f 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 ≡ m_i/V_i the density of pure matter is its mass divided by its volume

Equation 4.4 _0m =m_i/V_m.

We have Observation 1 ₀ = m_x/V_s but here we are varying the background volume to explore the effect on speed of momentum transmission in the Chaos

Equation 4.5 _b = m_x/V_b

By taking the ratio of Equation 4.4 _0m=m_x/V_m and Equation 4.6 _b = m_x/V_b

Equation 4.6 _0m/_b = (m_x/V_m)/(m_x/V_b) = V_b / V_m = L_m³/L_t³ 

Given the average background free space speed v_o, 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.

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.

Figure 4.3 Graph as the density approaches 100% the speed approaches infinity!

Consider the improbable event of streams forming in the chaos.

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.

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.

Figure 5.3 A vortex cross section.

Consider the improbable event of a vortex ring (toroid) forming from the vortices.

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.

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 v_ii ≡ Limit  (r_i/t_i)_i
t 0 

= Limit  (r_i1/t)_i1+ (r_i2/t)_i2 + (r_i3/t)_i3
t 0 

= (dr_i1/dt)_i1+ (dr_i2/dt)_i2 + (dr_i3/dt)_i3

= v_i1i1+ v_i2i2 + v_i3i3

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 a_ii ≡ Limit   (v_i/t_i)_i
t 0 

= Limit   (v_i1/t)_i1+ (v_i2/t)_i2 + (v_i3/t)_i3
t 0 

 = (dv_i1/dt)_i1+ (dv_i2/dt)_i2 + (dv_i3/dt)_i3.

= a_i1i1+ a_i2i2 + a_i3i3

We can define the three dimensional force F acting on an element of mass m_i with cross section area A_i in the Chaos by time t rate of change matter r₀ V_i in the impacting volume moving at average background speed v_oi .

Definition 12  F_ii ≡ dp_ii /dt
= dp_1i1 /dt + dp_2i2 /dt + dp_3i3 /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 F_ir+ acting in the r+ direction on an element of mass m_ir+. The mass is in a volume with cross section area A_ir. The average momentum p_ir+r+ of the ith element moving in a given direction (r+) is from Equation 4 p_ir+r+ (1/4)_oV_ir+v_or+. The ambient force F_ir+ in any direction _r+ on an arbitrary area of obstruction A_ir is

Equation 5F_ir+r+ ≡ d(p_ir+r+) /dt ,
d((1/4)₀ V_ir+ v_o)/dt_r+ ,      note (1/4)(₀ A_ir v_o) is not a function of time
= (1/4)(₀ A_ir v_o)dr_i/dt_r+ ,      by Definition 7 dr_i/dt_r+ = v_ir+r+
= (1/4)(₀ A_ir v_o) v_ir+r+ ,        by Equation 3 v_ir+r+= (1/2)v_or+
= (1/4)(₀ A_ir v_o)(1/2)v_or+ ,
= (1/8) ₀ A_ir v²_or+ .

We can define the average pressure on an element of mass m_i in a given direction r_r by dividing the average force F_{i ave} over the area A_r .

Definition 13     P ≡ F_ir+aver+●A_irr+ / (A_irr+ ● A_irr+ )   
=  F_ir+ / A_ir ,
 
Using Equation 5 F_ir+r+ = (1/8)_oA_ir v²_or+  or F_ir+ = (1/8) _o A_ir v²_o in Definition 13 P = F_ir+/A_ir gives

Equation 6     P = (1/8) _oA_ir v²_o / A_ir
= (1/8) _o v²_o ,

Note that in the primitive (no vortexes) Chaos this is universal and for a vortex it is the external pressure. 

For uniform circular motion we study the diagram below.

Figure 6.1 A point in uniform circular motion.

We have a point moving at a constant speed v_c. 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.

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 m_i 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 F_i=m_ia_i.
=
Using Equation 6.2 a_i=v_i²/r_i in Equation 6.3 F=ma gives

Equation 6.4 F_i =m_iv_i²/r_i.

Using Equation 4.5 m_i = m_i/V_mi in Equation 6.4 F_i =m_iv_i²/r_i gives

Equation 6.5 F_i =_miV_mi v_i²/r_i.

Looking at the segment in the diagram above we have the volume V_i

Equation 6.6 V_i = d_lA_i 

Using Equation 6.6 V_i=d_lA_i in Equation 6.5 F_i =_miV_mi v_i²/r_i gives

Equation 6.7 F_i =_md_lA_i v_i²/r_i.

We have the pressure from the Chaos by Equation 5.2 P_i = (1/8) ₀ v²_o .
Using this in Definition 13 P_i = F_i /A_i gives the external force as

Equation 6. 8 F_ext = PA =(1/8)₀v²_oA

We apply Newton's third law (to every force there is an equal and opposite force)

Equation 6.9 F_r-= -F_r+ . 

Using  Definition 5   Density  _x ≡ m_x/V_x in Equation 6.4 F_c =mv²/r gives

Equation 6.10F_c = _md_lA_iv_i² /r_i

Using Equation 6.8 F_ext =PA =(1/8)_ov²_oA and Equation 6.10 F_c = _md_lA_iv_i² /r_i  in Equation 6.9 Fr-=-Fr+ gives

Equation 6.11 _md_lA_iv_i² /r_i = (1/8)₀v_o²A_i ,  

_md_l v_i² /r_i = (1/8)₀v_o² ,

d_lv_i² /r = (1/8)(₀/_m)v_o² .

If we assume that the driving speed of the Chaos v_o² is the same as the tangential speed of the vortex v_i²
we have d_l/r =(1/8)( _o/_m), or:

Equation 6.12 r =8d_l(_m/_o)

This Equation 6.12 r =8d_l(_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 = 8d_l.
If the pure density _m was twice that of the chaos the ratio of r = 16d_l.
This is a limit. the minor radius r (the minor radius of the vortex must be smaller than 8d_l (the thickness of the vortex wall).  We have:

Equation 6.13 r > 8d_l .

The internal energy is kinetic. The energy of a segment is (1/2)mv², m= 2rd_lm ,

Equation 6.14 K.E._R/z =rd_lmv². 

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.

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 = 2R.

Using Equation 7.1 z = 2R in Equation 6.14 K.E._R/z =rd_lmv²,  the total energy K.E._R would be

Equation 7.2 K.E._R = rd_lmv²(z) ,solving Equation 6.14 for K.E._R ,

= rd_lmv²(2R)

= 2²Rrd_lmv².

Using Equation 6.12 r =8d_l(_m/_o)  in Equation 7.2 K.E = 2²Rrd_lmv² gives

Equation 7.3K.E=2²R {8d_l(_m/_o)} d_lmv²

= 16²Rd²_lv²²_m /_o

This show us the variables while energy is conserved is such large change as the major diameter changes in magnetic fields.
 

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.

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 a_o. The average cross sectional area for the n center rings is

Equation 8.1 A_center = n a_o .

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 F_center from all directions. We use Equation 8.1 A_center = n a_o to multiply the pressure in Equation 6  P_i =  (1/8) _o v²_o 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   F_{o center} = P A_center

=  (1/8) _o v²_o n_center a_o

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   A_top = ka_o .

For a sphere of projection with radius R_s we have the projected spherical surface area of 4R²_s or

Equation 7 A_sphere = 4R².

Where the center object has n rings of cross-sectional area of a_o. 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 a_o times the number k of rings and the area of the sphere of projection radius R_s and projected surface area of 4R²_s or

Equation 8.4 A_top / A_sphere or ka_o /(4R² ) .

Given the mass m_o of a single ring (constant mass of a single ring). The mass of the top object with n rings is

Equation 8.5 m_center = nm_o .

The mass of the center object with k rings is

Equation 8.6 m_top= km_o .

So the net force from the shielding imbalance is

Equation 8.7 F_{center net} = F_{o center} A_top / A_sphere ,                    shielding from Equation 8.4

= (1/8) _o v²_o A_center A_top / A_sphere ,  using Equation 8.2

= (1/8) _o v²_o ka_o na_o /(4R² ) ,          using Equation 7 ,8.5 , and 8.6

= (1/32) _o v²_o a²_o k n / R² ,              simplifying

= (1/32) _o v²_o a²_o (m_top/m_o) (m_center/m_o) / R² ,

= (1/32 m²_o) _o v²_o a²_o  m_top m_center  / R² ,  using Definition 14

= G  m_top m_center  / R² .

We define the Universal Gravitational Constant

Definition 14G ≡ (1/32 m²_o) _o v²_o a²_o   then = G (mkmn /R2)

This is then a derivation Newton's general law of gravitation!

Postulate 1 Space R0 Exists! Postulate 2 Matter m Exists in Space! Postulate 3 Motion v of Matter Exists! Observation 1 Background density o is constant. Observation 2 collision momentum with discontinuity Observation 3 deposited one side removed on opposite! Definition 1 Location R0 ≡ R011 + R022 + R033. Definition 2 Displacement rfi  ≡ Rff – Rff. Definition 3 Area A1212 ≡ r11 x r22. Definition 4 Volume V123 ≡ A1212 • r33, Definition 5 Density x ≡ mx/Vx . Definition 6 Time  tab ≡ |rabab| / |vabab| . Definition 7 Velocity va ≡ raa/ta. Definition 8 Saturation o ≡ Vm / Vs  Definition 9 Momentum pi+k+ ≡  mi+vk+k+ Definition 4.1 Lm ≡ nd.  Definition 4.2 Lf ≡ Lt - Lm , or Lm= Lt -Lf . Definition 10 vii ≡ Limit t 0 (ri/ti)i Definition 11 aii ≡ Limit t 0 (vi/ti)i Definition 12  Fii ≡ dp1i1 /dt + dp2i2 /dt + dp3i3 /dt Definition 13 Pressure P ≡ Fir+ / Air Definition 14 G ≡ (1/32m2o)ov2oa2o Observable 4.1 Lt = distance momentum transmitted. Observable 4.2 n =  number of elements in length Lt. Observable 4.3 d =  diameter of an element.

Equation 1 Rii ≡ Ri1i1 + Ri2i2 + Ri3i3. Equation 2 mi+ =(1/2)0 Vi+ , ( i+ is direction on the i axis.)  Equation 3 vk+k+=(1/2)vok+ (where k+ indicates +direction on k axis.) Equation 3.1 saturation c ≡ Vmc / Vsc  Equation 4 pi+k+ = (¼)0 Vk vok+  Equation 4.1 velocity in free space  vf = Lf /tt. Equation 4.2 velocity of momentum in the chaos  vp=Lt /tt Equation 4.3  speed of momentum / speed in free space   vp /vf = Lt /Lf . Equation 4.4 density of pure matter 0m =mi/Vm. Equation 4.5  varying the background volume  b = mx/Vb Equation 4.6 ratio of constant and varied density  0m/b = Lm3/Lt3  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+r+ ≡ (1/8) 0 Air v2or+ . Equation 6 Pressure Pi = (1/8) 0 v2o , 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 = 4R2. Equation 7.1 Major diameter of ring ztotal = 2R. Equation 8.1  Acenter = n ao . Equation 8.2   Fo center = P Acenter=  (1/8) o v2o ncenter ao Equation 8.3  Atop = kao . Equation 8.4 Ratio of view obstruction Atop / Asphere or kao /(4R2 ) . Equation 8.5  center object mcenter = nmo . Equation 8.6 top object  mtop= kmo . Equation 8.7 Force Fcenter net = G  mtop mcenter  / R2