Chapter 16: The Unification of Forces

8/3/2019 Chapter 16: The Unification of Forces

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CHAPTER 16

THE UNIFICATION OF FORCES

3D-SST makes an attempt to unify strong, superstrong (color), electric

and gravitational forces. A proof is provided below that when electric

charges of interacting elements are assumed to be distributed along

circles, the strong, superstrong (color), and gravitational forces can be

explained by applying the Coulombs law for electric charges.

ELECTRIC FORCES BETWEEN TORYCES

Conventionally, we make two assumptions to calculate electric forces

between elementary particles. Firstly, we assume that the entire electric

charge of a particle is concentrated within a dimensionless point located

at its center. We will call this electric charge the point charge.

Secondly, we assume that electric force Fe between two adjacent point

charges e1 and e2 follows the Coulombs law. As we described in

Chapter 5, this force is proportional to the product of electric charges and

inversely proportional to the distance S between them, as given by the

equation below in which 0 is the electric constant:

Fke e

S

e e

Se

1 2

2

1 2

02

4(16-1)

In application to the toryx, we do not assume that its entire electric

charge is concentrated within a dimensionless point located at its center,

but rather is evenly distributed along its leading string. Here is how this

approach works in application to two particles shown in Figure 16.1.

One particle is made of two concentric toryces, outverted toryx X and

inverted toryx x. Similarly, the other particle is made of two concentric

toryces, outverted toryx Y and inverted toryx y. For each toryx, we

divided its entire charge into ne electric mini charges and distributedthem evenly along the toryx leading string.

We calculated electric force between particles in two steps. First,

we determined horizontal components of electric forces Fh between all

electric mini charges of the rings by using the Coulombs law. We then

found the total electric force between particles by summing up the


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337Chapter 16

horizontal components of electric forces applied to all the electric mini

charges of the charged rings.

Fig. 16.1. Electric force between circularly-distributed electric mini charges of

two adjacent toryces.

It is convenient to express electric forces in relative terms. In that

case, the horizontal components of relative electric forces fh between

electric mini charges located at the points a and b, a and B, A and b, and

A and B, are related to the absolute values of horizontal components of

electric forces Fh(ab), Fh(aB), Fh(Ab), andFh(AB), by the equations:

f ab F ab

r

eh hi

( ) ( )

4 02

02

(16-2)

f aB F aBr

eh h

i( ) ( )4 0

2

0

2

(16-3)

1Xb

s

F

Fh

a bA

B

1x

j k

b1Y

b1y

b

Tor x X

Leading Stringof Toryx X

Leading Stringof Toryx x

Leading Stringof Toryx Y

Leading Stringof Toryx y

Tor x x Tor x Y Tor x


3/17

338 The Unification Of Forces

f Ab F Abr

e

h hi( ) ( )

4 02

0

2

(16-4)

f AB F ABr

eh h

i( ) ( )4 0

2

0

2

(16-5)

From Coulombs law and Fig. 16.1, relative electric forces fh betweenelectric mini charges of the toryces X, x and Y, y are equal to:

f ab

n

s b b

s b b b bh

x y

e

y k x j

y k x j y k x j

( )cos cos

( cos cos ) ( sin sin )/

2

1 1

1 12

1 12

3 2(16-6)

f aB

n

s b b

s b b b bh

x Y

e

Y k x j

Y k x j Y k x j

( )cos cos


2

1 1

1 12

12

3 2(16-7)

f Ab

n

s b b

s b b b bh

X y

e

y k X j

y k X j y k X j

( )cos cos


2

1 1

1 12

1 12

3 2(16-8)

f ABn

s b b

s b b b bh

X Y

e

Y k X j

Y k X j Y k X j

( )cos cos


2

1 1

1 12

12 3 2

(16-9)

In the above equations:

ne = the number of distributed electric mini charges in each ring

X= eX/e0 = relative electric charge of toryxX

Y = eY/e0 = relative electric charge of toryx Y

x = ex/e0 = relative electric charge of toryxx

y= ey/e0 = relative electric charge of toryxy

b1X= r1X/ri = relative leading string radius of toryx X

b1Y = r1Y/ri = relative leading string radius of toryx Y

b1x = r1x/ri = relative leading string radius of toryx x

b1y = r1y/ri = relative leading string radius of toryx y

s = S/ri = relative distance between centers of two particlesj = angular position of points A and a on leading strings of toryces Xand x

k = angular position of points B and b on leading strings toryces Yand y.

The horizontal components of electric forces Fh(ab), Fh(aB), Fh(Ab), and

Fh(AB) between the toryces Xx and Yy are calculated from Equations (16-


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339Chapter 16

2) (16-5). Finally, the total horizontal component of relative electric

force Fh between the toryces X, x and Y, y is calculated from the

equation: Fh = Fh(ab) + Fh(aB) + Fh(Ab) + Fh(AB) (16-10)

CRITICAL DISTANCESBETWEEN CHARGED RINGS

When the distance between two interacting elements is much greater than

the radii of charged rings, the total electric force between the rings is

inversely proportional to the square of the distance, as prescribed by the

Coulombs law given by Equation (16-1). The electric force between the

charged rings, however, deviates explicitly from the Coulombs inverse

square relationship when the distance between the centers of the rings is

comparable the ring radii. This distance is conveniently described inrelative term by a so-called relative critical distance sc = b1A b1B.

Similarly to the other relative dimensions, the relative critical distance scis calculated in respect to the inversion radius of toryx trailing string ri.

Figure 16.2 shows the charged rings A and B at various relative

distances between the ring centers s. The ring A (solid line) is charged

negatively, while the ring B (dotted line) is positively charged. When the

relative distance between the ring centers s is much greater than the

relative critical distance sc, (s >> sc) as shown in Figure 16.2a, calculated

electric forces are practically the same as these calculated from Equation

(16-1) applied to the point electric charges.

The results of calculations, however, change significantly when s

becomes comparable with sc. When s is equal to the relative criticaldistance sc = b1A + b1B (Fig. 16.2b), the attraction electric force is

maximum. When s is equal to the relative critical distance sc = b1A - b1B(Fig. 16.2c), the ring B is inside the ring A. In that case, the electric

forces will try to hold the rings together. This would produce the

maximum repulsion electric force which will decrease as the distance

between the ring centers decreases. Finally, at s = 0 (Fig. 16.2d), the

electric forces will hold both rings in equilibrium at a common central

point.

Depending on the sizes of charged rings and the distances between

their centers, 3D-SST identifies three kinds of electric forces: Coulombs

forces, strong forces, and superstrong forces. For all three kinds of

electric forces we can use a general equation, in which the relative

electric force fh is a function of relative distance s:

f ab

h x (16-11)

where a, b are constants.


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Figure 16.2. Electric forces between two oppositely charged rings.

Coulombs forces - Coulombs forces are inversely proportional to

the square of the distance between centers of the charged rings. Thus,

for Coulombs forces, the exponent of Equation (16-11) x = 2. This

relationship is valid when the relative distance between the ring centers

is much greater than the relative critical distance (s >> sc).

Strong forces There are two conditions for developing strong

forces. Firstly, the relative distances between the ring centers s must be

comparable with the relative critical distances sc. Secondly, the radii of

the charged rings must be comparable with the inversion radius of toryx

trailing string ri, such as for the cases when b1 1, 2, 3. . .. The above

two conditions may exist, for instance, during interaction between

matched toryces of harmonic a-trons, electrons and positrons that formnucleon shells.

Superstrong forces There are two conditions for developing

superstrong forces. Firstly, the relative distances between the ring

centers s must be comparable with the relative critical distances sc.

Secondly, the radii of the charged rings must be much smaller than the

1Ab

1B

1B1Ac

c

b

A

A

A

B

B

B

s = b + b

1B1As = b - b

s

s

s >> s

s

s = 0

s = 0

B

a)

b)

c)

d)


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341Chapter 16

inversion radius of toryx trailing string ri (b1 2, and it

further increases as the relative distance between the ring centers s

becomes closer to the relative critical distances sc.

STRONG FORCES

We will analyze now the strong forces between both toryces and

elementary particles.

Strong forces between toryces - As an example, let us review a

graph presented in Figure 16.3. This graph shows the calculated relative

strong forces fh between two toryces, real positive toryx A0 1 01 4, ,

/ and real

negative toryx A0 1 01 4

, ,

/ , making up a neutral harmonic a-tron a0 1 00

, , as a

function of relative distance between the toryx centers s. In this case, we

can identify two critical distances: sc = 2 (point a) and sc = 1 (point

b).

Figure 16.3. Relative strong force fh between toryces of a neutral a-tron a0 1 00

, , .

When the relative distance between the toryx centers s decreases and

approaches the first critical distance sc = 2, the relative strong

attraction (positive) force fh increases exponentially, reaching its

maximum positive value at sc = 2. As the relative distance between the

Relative distance between toryces, s

0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4

-0.2

-0.16

-0.12

-0.08

-0.04

0

0.04

0.08

0.12

0.16

.2

a

oryx

oryx Y

b = 2.0X= /4

b = 2/3 = +1/4

Relativestrongf

orce,f h


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toryx centers s continues to decrease, the attraction force sharply

decreases and becomes a repulsion (negative) force. Within the distance

range between toryces from the first critical distance sc = 2 to thesecond critical distance sc = 1, the relative strong force fh oscillates up

and down, acting intermittently as either attraction or repulsion force. At

the second critical distance sc = 1, the relative strong force reaches its

last negative value and then exponentially reduces and becomes

negligibly small as the distance between the toryx centers reduces to

zero.

When the relative distances between toryces s > 25, the exponent x

of Equation (16-11) is close to 2. Thus, the strong force reduces to a

conventional Coulombs force. For s < 25, the exponent x increases as

the distance between the toryx centers decreases and approaches the first

relative critical distance sc = 2, as shown in Table 16.1.

Table 16.1. Parameters of Equation (16-11) for relative strong

force fhbetween toryces of a neutral a-tron a0 1 00

, , .

Range of s Parameters of Equation (16-11)

From To a b x

20.000 15.000 0.074314 4.509 10-6 2.06434

15.000 12.640 0.077369 7.461 10-6

2.08316

12.640 10.100 0.084727 1.716 10-5 2.12884

10.100 7.490 0.100549 4.867 10-5

2.22556

7.490 5.000 0.147782 1.815 10-4 2.47843

5.000 3.960 0.350112 7.803 10-4

3.16774

3.960 3.040 4.836712 2.919 10-3 5.547743.040 2.843 31529.51 8.799 10

-314.22765

2.843 2.660 103818480 1.116 10-2 22.26000

Strong forces between elementary particles - As an example, let us

review a graph presented in Figure 16.4. The graph shows the calculated

relative strong forces fh between two neutral harmonic a -trons a0 1 00

, , as a

function of relative distance between centers of a-trons s. In that case,

we can identify three relative critical distances between a-trons: sc = 4.0

(point a), sc = 2 (point b), and sc = 1 (point c). When s > 4.0, fh is a

repulsion (negative) force that increases as s decreases, reaching its

maximum negative value at sc = 4.0. As s continues to decrease, the

repulsion strong force decreases and then becomes attraction (positive)

force that increases exponentially until it reaches its maximum positive

value at sc = 2. Within the distance range between the two relative

critical distances sc = 2 and sc = 1, the relative strong force fhoscillates up and down, acting intermittently as either attraction or

repulsion force. At sc = 1, the force reaches its last maximum negative


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343Chapter 16

value and then sharply decreases. Subsequently, it approaches slowly a

zero value as the distance between the centers of a-trons gets closer to

zero.

Figure 16.4. Relative strong force fh between harmonic neutral a-trons a 0 1 00

, , .

SUPERSTRONG FORCES

We will analyze below the superstrong forces between toryces and

between elementary particles.

Superstrong forces between toryces - As an example, let us review a

graph presented in Figure 16.5. The graph shows the calculated

superstrong forces fh between two imaginary toryces, positive toryx

Z1 1 01

, , and negative toryx Z1 1 0

1, ,

, making up an excited neutral z-tron z0 1 00

, ,

as a function of relative distance between the toryx centers s. There are

two relative critical distances in this case: sc = 0.000026626 (not shown),

and sc = 0.00729745 (see point a). At the distances s > 0.00729745,

relative superstrong attraction (positive) force fh increases exponentially

with the decrease of the distance. At s = 0.00729745, fh reaches its peakvalue. After a sharp initial decrease, fh then gradually reduces to zero,

while oscillating with increasing amplitude.

-8

-6

-4

-2

0

2

4

6

0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 4.4

Relative distance between a-trons, s

a

c

b = 2.0 = - /4b = 2/31 = +1/4x

b = 2.0 = - /4Yb = 2/31 = +1/4

oryces Xand x

oryces Y nd y

R

elativestrongforce,

f h


9/17


Figure 16.5. Relative superstrong force fhbetween toryces of an excited

neutral z-tron z1 1 00

, , .

Figure 16.6. Relative superstrong force fh between two excited neutral

z-trons z1 1 00

, , .

-1.0E+09

-5.0E+08

0.0E+00

5.0E+08

1.0E+09

1.5E+09

2.0E+09

2.5E+09

3.0E+09

3.5E+09

.0E+09

0.006 0.0064 0.0068 0.0072 0.0076 0.008

Relative distance between toryces, s

a

tron toryx Xb = 0.003662038

= 37.035999X

tron toryx Yb = 0.003635412Y

= 0.137.035999Y

Relativesuperstrongforce,f h

-1.6E+09

-1.2E+09

-8.0E+08

-4.0E+08

0.0E+00

4.0E+08

8.0E+08

1.2E+09

1.6E+09

2.0E+09

2.4E+09

0.006 0.0064 0.0068 0.0072 0.0076 0.008

Relative distance between z-trons, s

- ron

tron

b = .003662038X b = .003635412x= 37.035999 = +137.035999

b = .003662038Y b = .003635412y= 37.035999Y = +137.035999

Relativesuperstrongforce,f

h


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345Chapter 16

Superstrong forces between elementary particles - As an example,

let us review a graph presented in Figure 16.6. The graph shows the

calculated relative superstrong forces fh between two excited neutral z-trons as a function of relative distance between the z-trons centers s. In

this case, when s > 0.00735, fh is a repulsion (negative) force that

increases as s decreases, reaching its maximum negative value at s =

0.00735. As the relative distance between z-trons s continues to

decrease, the repulsion force decreases and then becomes attraction

(positive) force that increases sharply until it reaches its maximum value

at s 0.00730. The attraction force then sharply decreases, and then,

after passing a crossover point a, it becomes a repulsion (negative) force.

After reaching its maximum negative value at s 0.00726, the repulsion

force first sharply decreases and then exponentially approaches zero,

while oscillating with increasing amplitude.

INTERACTIONS BETWEEN ELECTRONS

The following example shows the electric forces between two harmonic

electrons e0 1 01

, , calculated based on two methods. According to the first

method, the electrons are presented as the point charges. According to

the second method, the electrons are presented as the charged rings

substituting the harmonic electron toryces.

Figure 16.7. Electric forces between two electrons calculated for the point and

circularly-distributed electric charges.

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

1 1.5 2 2.5 3 3.5 4 4.5 5

elative distance between electrons, s

Distributed charges

Point charges

ritical distance

b = b = 2.0

= = -1.0

n = 360

-0.6

1X Y

X Y

e

Relativeelectricforce,f h


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According to Table 14.5 of Chapter 14, each harmonic electron e0 1 01

, , is

made of one negative real and one negative imaginary harmonic toryx.

In this case, there are two critical distances:sc = 0 and sc = 4.Figure 16.7 illustrates the results of calculations for the range of s

from 1 to 5. As expected, for the point charges the electric repulsion

(negative) force increases inversely proportional to the square of the

distance s between the point charges. For the charged rings, however,

the electric force changes with distance s in a completely different

manner. Within the range ofs from 5 to 4, the rate of change of electric

force with distance s becomes noticeably greater than for the point

charges, and no longer follows the inverse square relationship. As the

distance s decreases and approaches the critical distance sc = 4, the rate

increases exponentially. At s = 4, the electric force reaches a peak value,

and then decreases abruptly and becomes even less than the force

calculated for the point charges at the same distances.

Figure 16.8. Electric forces between two electrons calculated for point and

distributed electric charges (0 < s < 0.4).

Within the range of s from 0 to 0.4, a difference between the results

of calculations based on two methods is also very explicit. As shown inFigure 16.8, within this range of s the electric repulsion force between

the point charges increases smoothly inversely proportional to the square

of s. For the charged rings, however, the electric force no longer

increases with the decrease of s. Instead, it oscillates around its zero

value with increasing amplitude as the distance s decreases. Thus, when

the relative distance s between electrons becomes less than 4, the

0

-250

-200

-150

-100

-50

0

50

100

150

200

250

0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

Relative distance between electrons, s

oint charges

istributed charges

b = b = 2.0

= = - .0

n = 360

1X Y

Y

Relativeelectricforce,

f h


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347Chapter 16

electrons are no longer rejecting one another but get into a state of

oscillating equilibrium. This may explain the reported existence of

electron clusters.

ELECTROMASS DENSITY

As we described in Chapter 5, scientists noticed in the mid-eighteenth

century a close correlation between gravitational and electric forces.

This notion was supported by a seeming similarity between the equations

describing the two forces. According to Newtons law of universal

gravitation, the gravitational force Fg between two adjacent bodies is

proportional to the product of masses m1 and m2 of these bodies and

inversely proportional to the square of the distance Sbetween them:

F m m GS

g 1 2

2(16-12)

But, in spite of the similarity between the above equation and the

equation for the Coulombs electric force defined by Equation (16-1),

physical meaning of these forces is different. The gravitational force is

known as an attraction (positive) force, always pulling the bodies

together. The electric force, however, works both ways. It can be either

attraction (positive) force or repulsion (negative) force. 3D-SST ties the

electric and gravitational forces together by introducing a term called the

universal electromass density u . As follows from Equations (16-1) and

(16-12), when gravitational force Fg is equal to electric attraction force

Fe, the electromass density is equal to:

u

e

m

e

mG

1

1

2

2

02 (16-13)

Since both the gravitational constant G and the electric constant 0 do

not vary, one may conclude from Equation (16-13) that the ratio of

electric charge e to gravitational mass m is constant for all bodies. In

electron and positron the electromass density reaches its maximum value

defined as the elementary electromass density 0 :

0

0

0

e

m (16-14)

From Equations (16-13), (16-14), and Equation (14-18) of Chapter 14,

we find that the ratio of the electromass density u to the electromass

density 0 is equal to the universal polarization factoru:


13/17


um

eGu

0

0

0

0

2 (16-15)

In the past, a common way of explaining attraction gravitational

forces between two bodies A and B was by assigning, for instance, a

positive charge to the body A and a negative charge to the body B. But,

this idea works only for two adjacent bodies. Once you added one more

body, this logic falls apart. If, for instance, the third body has a negative

charge then it will be attracted by the body A and repulsed by the bodyB.

Reversing the sign of electric charge of the third body will not help,

because it will then be attracted by the body B and repulsed by the body

A. 3D-SST finds a solution to this problem by applying the concept of

helicola to gravity.

GRAVITATIONAL FORCE

According to 3D-SST, gravitational field of a body is presented by the

matched charge-polarized toryces that form a so-called g-tron. One of

the toryces is outverted (negative), while the other one is inverted

(positive). Figure 16.9 shows two g-trons Xand Yseparated by a relative

distance s. Relative radii of leading strings of their constituent outverted

toryces are respectively equal to b1X and b1Y. Relative radii of leading

strings of their constituent inverted toryces are respectively equal to b1xand b1y .

To apply the same equations that are used to calculate electric forces

between toryces making up elementary particles, we expressed all thedimensions related to the gtrons in relative terms in respect to the

inversion radius of the toryx trailing string ri:

re

m ci

02

0 0

28

The outverted toryces forming the g-trons have the same structures

as the toryces making up excited electrons, while the inverted torycesforming the g-trons have the same structure as the toryces making up

excited positrons. The properties of both outverted and inverted toryces

of the g-trons are also described by the same equations, except for the

inversion radii of trailing string rj that for the g-trons with gravitationalmasses mX and mY are given by the equations:

rm G

jXX

2

2; r

m G

cjY

Y

22

(16-16)


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349Chapter 16

Figure 16.9. Two adjacent g-trons.

It is worth to point out that the inversion radii riX and riY are two

times smaller than the Schwarzchild radius (see Chapter 10). There is a

principal difference between inversion radii rjX and rjY of the toryces

composing the g-tron and inversion radii ri of the excited toryces thatmake up the elementary particles. Whereas, the inversion radius ri is

constant for all excited toryces making up elementary particles, the

inversion radii rjX and rjY are proportional to the gravitational masses mXand mY of respective g-trons.

According to 3D-SST, each celestial body contains a black hole

which gravitational field is defined by the inverted toryx of the g-tron.

Thus, the outer radii of the black holes of the celestial bodies with the

masses mX and mY are given by the equation (16-16). For instance, the

outer radii of the black holes of the Sun, the Jupiter, and the Earth are

equal to:

Celestialbody

Masskg

Outer radius ofblack hole, m

Sun 1.9891 1030

738.558

Jupiter 1.8997 1027 0.70537

Earth 5.9760 1024

0.00222

1Xb

G-Tron X

Inverted(positive)

Toryx

Outverted(negative)

Toryx

G-Tron Y

1Yb

1xb

s

1yb


15/17


The procedure for calculation of forces between the toryces forming

two adjacent g-trons is similar to the procedure for calculation of forces

between the toryces making up two adjacent elementary particles. Here,we also assumed that the electric charges of g-trons are evenly

distributed along the leading strings of their toryces. Figure 16.10 shows

the distribution of electric charges in the toryces of two g-trons X and Y

separated by a relative distance s.

For the two g-trons to represent properly the gravitational properties

of two adjacent bodies, three principal conditions must be met. Firstly,

relative spiral radii of leading strings b1X and b1Y of the outverted toryces

must be equal to the relative distance between interacting g-trons s:

b b sX Y1 1

(16-17)

Figure 16.10. Two adjacent g-trons with distributed electric mini charges.

Secondly, relative spiral radii of leading strings b1x and b1y of inverted

toryces must be equal to:

bs

s

r

rx

jx

i

12 1

; bs

s

r

ry

jy

i

12 1

(16-18)

1Xb

1xb

s

1yb

X

G-TronX

G-TronY

e

xe ye

Ye

1Yb


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351Chapter 16

The third condition stipulates that the total electric charges distributed

along leading strings of both outverted and inverted toryces of the g-

trons X and Y with gravitational masses, mX and mY, are respectivelyequal to:

e em

kx X

u X

d

; e em

ky Y

u Y

d

(16-19)

In equation (16-19), kd is the electric charge density factor that depends

on the selected number of electric mini charges ne distributed along each

toryx leading string. We established that when ne > 20 000, the electric

charge density factorkd is approximately equal to:

k nd e

3 105

exp( )

Thus, as ne , kd 3.Notably, based on Equation (16-19), the total electric charge of each

g-tron X and Y is equal to zero. After establishing the radii of leading

strings of the toryces and the electric charges distributed along these

strings, we are able to calculate electric forces between the g-trons by

employing Equations (16-2) through (16-10) presented in the beginning

of this Chapter. We found that when using the above procedure, the

calculated gravitational forces between the g-trons are the same as the

gravitational forces calculated from Newtons Equation (16-1).

REAL, INVERTED & IMAGINARY STARS

According to 3D-SST, the stars are divided into three types: real,

inverted and imaginary. The type of the star depends on the ratio of its

outer radius rs to the inversion radii of the toryx trailing string of its g-

tron, or to the outer radius of its black hole rj:

br

rs

s

j

Orbital velocity Vb of a celestial body orbiting a star along a circular path

with relative orbital radius bb is defined by the equation:

V b

bcb

b

b

2 1 (16-20)

To see more clearly a difference between various types of stars, let us

imagine an inversion sphere with the radius equal to the outer radius of

the star black hole rj:


17/17


Real star(bs > 1) In a real star, the inversion sphere is below its

surface, like, for instance, in the Sun. Since relative orbital radius bb of a

celestial body moving around the star is greater than relative radius of thestar surface bs, bb is also greater than 1. In that case, according to

Equation (16-20), orbital velocity Vb of a celestial body is subluminal(Vb < c).

Inverted star(0.5 < bs < 1) This is a much denser star than the Sun.

Its inversion sphere expands outside of its surface. Therefore, relative

orbital radius bb of a celestial body moving around the inverted star can

become equal to 1. In that case, according to Equation (16-20), orbital

velocity Vb of the celestial body will be equal to velocity of light (Vb =c). When bb < 1, the celestial body will be turned inside out, and its

orbital velocity Vb will decrease as bs approaches 0.5.Imaginary star(bs < 0.5) This star has extremely high density. Its

inversion sphere expands two or more times beyond its surface.

Therefore, relative orbital radius bb of a celestial body orbiting the

imaginary star can be less than 0.5. In that case, according to Equation

(16-20), orbital velocity Vb of the celestial body will be expressed by the

imaginary number. It means that this body will absorb energy from the

imaginary star.

YOUROWN IDEAS&QUESTIONS

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Chapter 16: The Unification of Forces