A Guess Model of Black Holes and the Evolution of Universe

(1)

A Guess Model of Black Holes and the

Evolution of Universe

Zhenhua Mei1_{, Shuyu Mei}2

1_{College of Chemistry and Molecular Engineering, Qingdao University of Science and Technology, Qingdao, China} 2_{Medical College, Qingdao University, Qingdao, China}

Email: mzh62@qust.edu.cn

Received June 10, 2012; revised July 16, 2012; accepted July 23,2012

ABSTRACT

Based on the gravitational theory, fundamental data, and comprehensible suppositions, an evolution model of the uni-verse was proposed. The uniuni-verse exists in explosion and constringency mobile equilibrium state. The critical sizes of celestial bodies were calculated in their evolution process.

Keywords: Black Hole; Evolution; Universe; Gravitation; Entropy

1. Introduction

As the theoretical ratiocination (Einstein’s theory of rela-tivity) and the cumulating of the more and more obtained data or phenomena from astronomic observation of the celestial bodies, people began to have a general impres-sion of the universe, such as the clangorous words ex-pressed: big bang, expanding universe, and black hole et al. (Ginsburg 1985). From a point of view that the mi-crocosm decides the mami-crocosm, more information of the universe can be predicted from present knowledge of the elementary particles.

In this paper, some parameters of evolution of the uni-verse were calculated. The main thoughts were the New-ton’s universal gravitation and the structures of matter.

2. Preparation of Research

2.1. Fundamental Data and Suppositions

The gravitational constant, G, is 6.670 × 10–11_{N m}2_·kg−2_. The radius of hydrogen atom, a0, is 5.29 × 10−11 m. The mass density of atomic nucleus, ρn, was determined (Er-dei 1976) to be 2 ×1017_kg·m−3_{. The mass of neutron,}_m

n, is 1.67495 × 10−27_{kg, radius}_r

n, 1.2598 × 10−15 m. The combining energy of nucleus is 8 MeV. The gravitational acceleration on earth surface is 9.81 m·s−2_{. The sun has} the mass of 1.989 × 1030_{kg, density of 1.409 × 10}3 kg·m−3_{, radius of 6.960 × 10}8_m.

Suppositions: the combining energy of nucleus (strong interaction in nucleus) came from gravitation, and the nucleus consisted of gravitons; the combinative form between nucleus was its overlaps one another.

2.2. Formula Derivation A—Gravitational Acceleration of a Particle from Solid Sphere According to Newton’s law, the gravitational force be-tween two distant particles has the form:

2 _{F ma}

F GmM r , and  . Hence, the gravitational acceleration, _{a GM r}_ 2_.

However, to a neighboring large body (has a mass of

M, radius of R), the radius of the body cannot be ne-glected; its accurate result can be obtained by processing a mathematical integral.

As Figure 1 expressed, a solid sphere has a radius of R, a homogeneous density of ρ, the particle has a mass, m, and has a distance, nR, from the center, the gravitational acceleration, a, of the particle was derived as follows:

z

m

nR

h _M

O y

x

[image:1.595.339.517.537.706.2]

R φ r

Figure 1. Sketch map of a particle (m), which with a dis-ance, nR, from the center of the solid sphere.

t

(2)









   









   













2 3

2 2

1

3 2

0 0 1 2 2 2

1 1/ 2

2 2 2

0 0 ₁

1/ 2 2

2 2 2 2 2

0 0

d d d d d

4 d d d d

4 d d

4 d d 1 1

R R n R

n R

R R

n R

R R

G mh n

a a M G x y z G

r _mM _x _y _nR

t t nR z

Gk x y t

x y t

Gk x y x y t

Gk x y x y n R x y n R

 



   

 

 





 

   

  

  

 

 

  

       





  

 





3 2 2

1

2 ₂

d d d

R z

x y z z

 

































2 2 2 2 2

0

2 2 2 2 2

0

2 2

2 2 2 2

4 d ln 1 ln

4 d (ln 1 ln

ln 1 ln 1

R

Gk x y x y n R y x y n

a Gk x R x R n R R x R

x n R x n R

 

      

      

 

_      _

 



/ 2

2 2 2

1 1

R

n R

 

 

 

 _ _ 

 

 

  

(1)

where, “M” is the arbitrary point in the solid sphere, “h” the point corresponding to “M” horizontally. “r” is the distance of the point “M” from original point “o”, ρ′ is ideally homogeneous mass density of a solid sphere, ρ

the average one. k stands for the calibration factor of av-erage mass density.

For a general integral

















2 2

2 2 2

2

0d ln , 1 ln 0d

ln 1 arc tg

R

R R x

x x C C n R x x C x

x C

R

R R C R C

C

    



   



₀



(2)

For integral





































































2 2

2 2 2 2

0

2

2 2

0

d ln 1 1 ln 1 1

1

d ln 1 1

R R

2 2

2 2 2 2

2 2

0 2 2 2 2 2 2

2

2 2

1 1 1 1

1 1

1 d 1

1 1 1 1

ln 1 1 1

R R

x R x n R x R x n R

x

x x R x n R

 _ _{ } _ _  _ _{ } _ 

   

   

 

  _     _

 



0





R

R x n R x n R

n R

x

R x n R x n R

R R R n R R

      

 _ _ 



 

 _  _

        

 

 

 _     _  

 



-



























































2 2

2

2 2

0 2 2 2 0

2

2 2

2

0 2 2 2

1 ln 1

1

d d ln 1 1

1 1

1 2 1

ln 1 2 1 1 1 ln

1 1

1 d

1 1

R R

R

n R x x

n R

x x R x n

R x n R

n

R R n R n R

n

n R

x

R x n R

   

 _ _

 _    _

 

   

  

  

 _ _    __   

 

  _ _

 

   



.

2 ₁ 2

R

n R

R

 _ 

 

 



0

(3)

For general integral







_

₁

_

₁

_

2

_

2

_

2

0 2

1

d , ,

R n R

x A R B

A x B



 

 



 n R

Let f =



_A_ _x2__B



1

 





2 1

1 1 1

R n

   .

0

f 





2





_{1 2}

2  _{x x}2__B  _.





f   A x B



 

3



2









1 1 2 3 2

2 2 2 2 2 2 2

2

f  x x B  A x B   A x B  x B  x x B  .

 









3 2

2 2

1 1

0

1 1 1 1 1

f

R

n n

 

    

.





















































4 3

3 2 ₂ ₂ ₂ 1 ₃ ₂ 2

2 _{3 2} _{5 2}

2 2 3 2

3

1 2 3 2 1 2

2 2 2 2 2

6 2 2 2

3 3

2 .

f x x B A x B A x B x x B x x B

A x B x x B x x B

x B x x B A x B x x B

 

  

 _ _



  

          

      

      





3 2













































































5 4 2

4 2 2

4 _{3 2} _{5 2} 3

2 2 2 4 2 2

7 2

24

6 3 3 2

2

f x x B A x B

A x B x x B x x B A x B

 

 _ _ 



    

        

1 2 2 3

2 2 2 2 2 4 2

2 _{3 2} _{5/ 2} _{5 2}

2 2 2 2 2 2 4 2

3

1 2 3 2 5 2

2 2 2 2 3 2 2

2 4 6 8

3 9 9 15

6 3

x B x x B x x B x x B

A x B x B x x B x x B x x B

x x B A x B x B x x B x

   

 _ _ _



  

      

          



       _

















 























1 2 ₂ ₂ 3 2

4 3

2 2 1 2 3 2

2 2 2 2 2 2 2

3 _{1 2} _{5 2}

2 2 2 3 2 2 3 2 2

3

2 ( ) 2 ( ) 3 .

B x x B

x x B A x B A x B x B x x B

A x B x x B x x B x x B x x B

 

  

 _ _

 

  

_ _ _ _ _ _ _ _ _ 

 

 

          

 























5 3 _{3 2} 2

2 2 2 2

1 6 3

0 .

1 1 1 1 1 1 1 1 1 1

f

R

n n n n

 

 

  

 _ _ _ _ _ _ _ _ _ _ 

 

(4)





_

_

_{ }

 









_

_



_

_































2

2 2 4

0 2 0

2 3

3 2

2 ₂ ₂

5 ₂ ₂ 3 ₂ 3/ 2

2

1 0 0

d 1 d 0

2! 4!

1/ 1 1 1

1

3 2!

1 1 1 ₁ ₁ ₁ ₁ ₁

1 1 6 3

5 4! ₁ ₁ ₁ ₁ ₁ ₁ ₁ ₁ ₁

1 1 1

1

6

1 1 1 ₁

R n R R f f

x R n x f x x

A x B

R

R n R R

R

n _n _n

R _n _n _n

R n

n

 

  



      

   

   

  



    _ _ _ _ _

 



5 2 2 1

R n

 



 



  



 

  

 

 



         





  

  







































2

2 2

3 _{3 2} 2

2 2 2

1 1 1 1

1 2 1 _.

40

1 1 1 1 1 1 1 1 1 1

n n

n n n n

  

 _ _ _ _ _

 



 



 



 

 



 _ _ _ _ _ _ _ _ _ _ 

 

 

2 ₍₄₎

Substituted (3) with (4)





































_

_



_

_























2 2

0

2

2 2

3 _{3 2} 2

2 2 2 2

d ln 1 1 ln 1 2 1

1 2 1 ₁ ₁

1 1 ln 1

6

1 1 1 1 1 ₁ ₁ ₁

1 2 1

.

40 ₁ ₁ ₁ ₁ ₁ ₁ ₁ ₁ ₁ ₁

R

x R x n R R R n

n

R n R R n

n n

n n n n

 

 _ _{ } _ _  _ _ _ 

    

    



   _



      



  _    _ _ _



 



 



 

 



 _ _ _ _ _ _ _ _ _ _ 

 

 



2

2 2

1

1 1

n  n

(5)

As the results of (2) and (5) expressed, thus, (1) becomes





























_

_



_

_











































2

3 2

2 1

1 1

1

1 n 1 

2 2

2

2 ₂ ₂

3 2

2 2 2

2

2 2

2 1 2

4 ln 1 2 1 1 1 1 ln

1 1 1

1

6

1 1 1 ₁ ₁ ₁ ₁ ₁

1 1 1

20 40

1 1 1 1 1 1 1 1

1 2 1

ln 1 2 1 1 1 1 ln

1 1

n

a Gk R R n n

n

n _n _n

n n n

n

R n n

n

     

 _    _  

  

 _ _

   

  

    _ _ _ _ _

 

   

  _

 

       



  

 

 _    _   

(5)









_

_



_

_



































_

_





2

2 ₂ ₂

3 _{3 2} 2

2 2 2 2

2 2

1 1 1

1

6

1 1 1 ₁ ₁ ₁ ₁ ₁

1 1 1 1

20 40

1 1 1 1 1 1 1 1 1 1

ln 1 1 1 1 arc tg ln 1 1 1 ar

1

n

n _n _n

n n n n

R n n R n n

n

  

  

    _ _ _ _ _

 

  

 



          _



1 1

c tg .

1

n



          

 -1   _

 

4Gk RF n , then, Let a n

 



 













































































_

_



_

_



2 2

2

2 2 2 2

3 _{3 2} 2

2 2 2

2

2 ₂ ₂

1 2 1 1 2 1 1

ln 1 1 ln 1 1 ln

1 2 1 1 1

1 1 1

1

6

1 1 1 1 1 1 1 1

1 1 1 1

20 40

1 1 1 1 1 1 1 1 1 1

1 1 1

1

6

1 1 1 ₁ ₁ ₁ ₁ ₁

n n

F n n n

n n

n

n n n

n n n n

n

n _n _n

 _ _ _ _ _ _



      

 _ _ _ _ _

  

   

        

 

 

         

  

   _ _ _ _ _

2

2 1

1 1

n

  

 

     





























3 _{3 2} 2

2 2 2

2

1 1 1 1

20 40

1 1 1 1 1 1 1 1 1 1

1 1 ₁ ₁

ln arc tg arc tg .

1 1

n n n n

n

n n

n

     

 

         

 

    

 

  1 1

2

     

(6)

F(n) is a function of n. The, F – n, diagram was showed igure 2.

the earth, the averag density ρis 5.5153 × 103_kg·m−3_{, the radius}_R_{is 6.356078 × 10}6_{m. The} gravi-tational acceleration on earth surface was calculated (n = 1) to be 12.68 m·s−2_{. However, the real value of} gravita-tional acceleration on earth is 9.81 m·s−2_{. It is 0.7736} times than the calculated one. It is because of the earth that does not have an ideally homogeneous density used in above formula derivation process. The ratio can be used as calibration factor, k, of average density of a solid sphere (ρ′ = kρ. For earth, k = 0.7736; and it was

sup-posed to be fit for any other celestial bodies).

2.3. Formula Derivation B—Self-Gravitational Pressure in Center of Solid Sphere

The gravitational pressure in center from solid sphere can be calculated according to Equation (6).

in F

For e mass

1 1

dFa m ad  dV SRa nd 3, (where, V SnR 3)

1 3

dpSRadn

 

1 2

1 2 ₀

4

d d .

3

(6)

The integral, 1

 

0F n nd



, can be calculated diagram-matically. The diagram showed in Figure 3.

The area beneath the curve calculated by computer gave the result of 0.62129. Thus, p = 0.62129 × 4Gkρ2_R2_{/3. Supposing the considered celestial body has} the same value of calibration factor k of earth, then we obtained

p = 0.6408Gρ2_R2_{. (7.1)}

2.4. Formula Derivation C—Gravitational Potential Energy from Solid Sphere

The gravitational potential energy, Eg, from a solid sphere can be calculated.

 

2 _{d .}

n F n n





(8) According to relation (6), t

 

1 d

g

d d 4

g

E 

_

E 

_

ma l mGk R

he integral,



F n n, can be calculated diagramm gram showed in Figure 4.

The area beneath the curve calculated by computer atically. The dia

0 2 4 6 8 10 1.4

1.2 1.0 0.8 0.6 0.4 0.2 0.0

Distance from center/R

Co

gr

av

ita

tio

[image:6.595.312.534.81.244.2]

n

Figure 2. Diagram of function F(n).

ef

t of

at

io

n/m

·s

–2

fi

ci

en

al a

ccel

er

0 2 4 6 8 10 1.4

1.2 1.0 0.8 0.6 0.4 0.2 0.0

Distance from center/R

Co

of

gr

av

itat

io

na

tion

Fi iagram of

1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0

0 5 10 15 20 Distance from center/R

C

oe

ff

ici

en

t o

f

gr

av

itat

io

na

l a

ccel

er

ati

on

/m

·s

–2

Figure 4. Partial diagram of function F(n). The variable n changes from 1 to ∞.

gave the result of 1.7193. Therefore, we have

Eg = 1.7193 × 4GkρmR2 = 5.320GρmR2. (8.1)

3. Target Calculation

3.1. Maximal Radius of Celestial Bodies Atoms have similar energy demand whe

compressed in nucleus forming a new atom (which has lower atomic number). Taking atom zinc as an example, when one electron of Zn is compressed in nucleus, the is

ef

fi

ci

en

t

l ac

ce

le

ra

/m

·s

–2

gure 3. Partial d function F(n). The variable n changes from zero to one.

n one electron is

otopic atom Cu is then formed. The energy demand, ΔEa, can be calculated. According to quantum theory, the energy of an outer electron of a atom has the form E ≈ −13.6Z2_/_n2_{(in ground state). Thus,}

2 2

13.

4

Zn

E   6Z₂ 13.630₂ 765 eV

n     .

2 2

29

13.6 13.6 715 eV

4

cu

Z E

n

      .

1 cu Zn 50 eV

E E E

    .

For an extreme hot electron of atom of hydrogen, ΔEa = ΔE1 + 13.6 = 63.6 eV = 1.02 × 10−

radioactive

is decreasing

adius a, we guessed that the hydrogen at

nucleus. Then, the maximu pressure, pmax,a, for com-pressing a hydrogen atom to neutron could be derived as: ΔEa = FΔr = pSΔr = pSa0.

17_J As the gravitational force increasing, many

otopes formed with the atomic number . Considering the atom of hydrogen, that has the maximum compression r 0

oms would be the last element to be compressed to m





17

max, 3 ₁₁ 3

0

13 2

3 3 1.02 10

4π _{3.14159 4} _{5.29 10}

1.64 10 Nm

a a

E

p Ea V

a

  

  

   

  

 

[image:6.595.63.284.361.689.2] [image:6.595.337.503.463.537.2]

(7)

where, Δr was replaced with the radius of hydrogen atom

a0.

The pressure comes from the gravitation of celestial body. Combining Equations (7.1) and (9), we have

0.6408Gρ2_R2_{= 1.64 × 10}13_, then,

6 1

10

G



  .

If a solid celestial body has the same average mass density of earth (5.5153), then

max 5.07

R

6

max, ₃ ₁₁

1 5.07 10

5.5153 10 6.670 10

solid

R   _

  

To a hot gas celestial body, which has the same aver-age mass density of sun (1.409), then, we have

8

1.12 10 m.

 

6 max,

8

1 5.07 10

1.409 10 4.41 10 m.

gas

R  

 

 

3 _{6.670 10}_ 11

29

5.06 10 kg . 3.2. Minimum Neutron Star

A model of neutron star was showed in Figure 5. A neu-tron body with the radius of R is in the center of the star. A general matter layer with the thickness of (n − 1)R is around the neutron body.

contains two parts.

Equation (6) can express linearly in a short segment.

F(n) = 1.355571n, (n = 0 → 1). According to Equation (7),

max

m V

The gravitational pressure on the surface of neutron body can be calculated. It

 

2

d kp R F n nd .



1 2



4 3 i

p  p G

 





2 2

2 2 ₁ 2_.

kR

kG n  R

1 1 _{2 1}

1

4 4

d

3 3

1.355571 d 0.9037 n

n

p Gkp F n n G

n n

 



 



 

2 1

4 _{d .}

3

n

nkGR F n n





_

2

2 1 2 ₁

4

d 3

. n

p Gk R F n n

p p p

 



 



layer

1 2

where, ρn is the nuclear mass density, ρthe mass density matter. Substituting the parameters, we have





3 _n2 ₁ _R2

 _

(10)

 

10 2 1

7.589 10 R nF n nd

 

_

1.1418 10

p 

y z

(n - 1)R

[image:7.595.354.506.86.211.2]

x

Figure 5. A model of neutron star. Combining Equations (9) and ( 0), we have 1





 

3 2 2 10 2

1.1418 1.64

n



  



)

7.1) and (13), we have

1 13

10 n 1 R 7.589 10 R



F n nd ₍₁₁

10 . 



According to Equations (6) and (11), we obtained fol-lowing results:

R = 11.2 m, nR = 1121 m

R = 16 m, nR = 32 m

R = 261 m, nR = 261.6 m

R = 15257 m, nR = 15257 m. Combining Equations (

32





2

32

2

11 17

1.99 10 0.6408

1.99 10

0.6408 6.670 10 2 10

n

R

G



 

   

3.3. Minimal

the radius of a neutron star achieved its maximal value, as the mass continuing accumu ing, a graviton body began growing, the radius of neutron star then de-cr

eutron star began to have the ability to draw back a photon; we call it becoming a body of black hole. The maximum radius of black hole can be calculated by imitating the process of what in paragraph 3.2. Just replacing ρ and ρn with the value of

ρn





 

24 2 2

24 2 1

1.865234 10 1

8.682 10 n d .

p n R

R F n n

  

 

_

(12)

According to Equations (12) and (13), following

rela-30

max n 2.98 10 kg.

m  p V  

15257m. 

Mass of Black Hole When

lat

eased correspondingly. When the mass of graviton body grew large enough, the n

(8)

tions were obtained:

R = 0.971 m, nR = 9714 m, M = 7.679 ×1029_kg = 7.679 ×1029_kg

m, M = 4.537 ×10 kg

R = 4200 10 kg

, M = 1.114×1033_kg.

3.4. Calculation of Graviton

Nuclear has the combining energy of about 8 MeV (Wichmann 1971). Deducting the influence of proton, the maximum combining energy of nuclear would be 9.15 MeV, i.e., 1.47 × 10−12_{J. The parameter of the radius of a} graviton, r0, is the first needs to obtain. Depending on above suppositions described in Section 2.1, and the normal gravitational potential energy formula, E = Gm1m2/r, substituting m1, m2 with the mass of neutron, E with the combining energy in nucleus, then we have

R = 9.714 m, nR = 9714 m, M

R = 97.10 m, nR = 9710 m, M = 7.671 ×1029_kg

R = 1154 m, nR = 9279 m, M = 6.722 ×1029_kg

R = 4078 m, nR = 7340 29

R = 10793 m, nR = 11873 m, M = 3.672 ×1030_kg

R = 14090 m, nR = 14908 m, M = 7.825 ×1030_kg 32 4 m, nR = 42273 m, M = 1.971 ×

R = 74934 m, nR = 75084 m

,



 



0 1 2

r Gm m E

Supposing the energy demand for dest ΔEn, is equal to its maximum combining en have

ΔEn = 1.47 ×10−12 J.

Following the derivate process in Section 3.1, but replaced ΔEa with ΔEn, a0 with rn,Equation (9) becomes

2

11 27 12

52

6.670 10 1.67495 10 1.47 10 1.273 10 m.

  



    

 

roy a neutron, ergy, then we





max, 3

12 3 15 3

4π 10

98 10 n

n

E

p En V

r

 



  

 (13)

but re- laced ρand max, n, Equa- io





 

24 2 2 7 2

1 32

1.865 10 1 1.376 10 d

1.99 10

n g

n R  R F n n



   

 



32 2

3 1.47 4 3.14159 1.25 1.76 10 Nm .

 



 

 

Following the derivate process in Section 3.2

ρn with ρn, and ρg, pmax, a with p , p

t n (11) became Equation (14).

(14)







 



32 24 2

1.76 10 1.865 10 1

   

1 2 7

1

0 g



nF n nd . 1.376 1

R n

  (15)

A Smalles ed XTE J1650-

500) has the g (Robert & Rob 2008). Perha ole

theoreti-cal

empo-ra have





 

t Known Black Hole (nam mass, Mb, of 7.558 × 1030 k ps it is not the minimal black h er, it was considered to be true

ly; howev here t

rily. According to the data and Equation (8), we

2 2 2 2

1.7193 4

4 d

2.

g n

g n _n

E Gkp mn R

mGk R F n n

mc

  

 

 









 

17 2 2 2 26

3.4386 10 n R g n R _n n nd 2.180 10



  

_

 





 

26 17

2.180 10 3.4386 10

R



17

2 10 d .

g _n F n n

 

  

_

  

(16)













17 3 3 3

30

2 10 4π 3 4π 3

7.558 10 .

b g n

M    n R     R

 













17 3 3 3

30

2 10 4π 3 4π 3

7.558 10 .

g n

n R   R

    

 





17 3 3 3

30

8.37766 10 4.1888

7.558 10 .

g n

n R   R

  

 





30 17 3

7.558 10 8.3776 10

R   n

1 3 17

4.1888 g 2 10 .

   (17)

Combining Equations (15), (16) and (17), we have Equations (18) and (19).





 







26 17 2

2.180

R 

1 2 17

2 10 n nd

 

  

30 17 3

7.558 10 8.3776 10 n

  

10 3.4386 10 n

18)

1 3 17 4.1888.  2 10

  

g



_n

(

g





 









 



26 17 2

2.180 10 3.4386 10 n

1 2 17 



32 24 2

1.76 10 1.865 10 n 1

1 2

.

7

1.376 10 g n nd



2 10 d

g _n n n



  

n

   

 

_



(19)

Combining Equations (18) and (19) diagrammatically, we obtained

n = 1.0190.

(9)

nR = 14358 m.

ρg = 6.3096 × 1017 kg·m−3. (20)











₅₂ 3



₁₃₈

4π 1.273 10 3 5.452 10 kg.

    (21)

where mg is stands for the mass of a graviton.

4.

x parts; and the process goes a circle (a → f →a). a) Growing of y. Preponderant no

kg, radius, 4.41 × 108_km.

b) Formation of neutron star. As the continuously ac-cumulating of matter, the normal molecular celestial body begins to compress accompanied with

engendered in the center. The compressing process looks like a process of phase variation. In the process, it will release large

c) Growing of graviton star. With continuously com-pressing, a graviton body would begin to grow center. The graviton star has a critical mass of 7.558 × 1030_{kg, radius of 14.9 km.}

d) Formation of black hole. When graviton star grows to its critical value, it becomes a mi

the mean time. It can capture photons just near its surface. Afterward, it gradually grows in mass with the accumu-la

e) Explosion of black hole. When the density of black hole increased large enough, the distance between gravi-tons going over a critical small value, the gravitational mechanism destroyed. The gravitation becomes repulsive force. Then, explosion occurred; the entropy increased sharply. Neutrons, protons, electrons, etc. particles an afterward atoms were produced subsequently in the ex plosion process. After explosion, this part of universe exists in expansion state in a unabidin

f) Constringency of partial univers

of explosion, the running matters slowed down, spread in wide space, mixed, and became part of cosmic dusts. Then, partial universe would exist in a contractive state in a long run. In intermediate constringency, the celestial bodies formed.

5. Discussion

When a celestial body has abundant material source in its surroundings, it will grow quickly and has a larger mass than we calculated. If the celestial body exists in process b, it would be a fixed star. The larger mass it has, the more quickly it will be compressed, and more efficiency the energy would release. i.e., the larger mass it owns, the higher brightness it will has. In th

star could burn with no existence of special fuels of nu-clear fusion, but merely existence of normal atoms.

will have a big black hole in its center.

The universe consists of vast cosmic dusts and galax-ies. They go along circle evolutive processes described above. Occasionally a limit big black ho

causing the expansion in a partial space.

long time scale, most space contracted slowly. The uni-verse exists in a mobile equilibrium state of explosion and constringency (Thomas & Hermann 1948). Those observed stars, which leave away at acceleratory velocity (Hubble 1929), might result in other reasons. It is by no means that the whole universe is expanding now. Re-versely, our surrounding universe showed us its contrac-tive views. We cannot imagine that the

Milky Galaxy is formed in an expansion pr

6. Tags

A roughly information about graviton was derived in this pa

cs. Future observation of existence of new smaller bla parently meaningful.

NCES

Press, 1985.

[2] G. Erdei, “Principles of Structure of Matter,” 1971.

[3] E. H. Wichmann, “Quantum Physics,” Berkeley Physics [4] N. Robert and G. Rob, “NASA Scientists Identify Small- ddard Space Flight

[5]

3 17

0

4π 3 6.3096 10

g g

m  r  

is way, a shining

Evolvement of the Universe

The evolutive process of the universe was divided by us

into si _{evident that every galaxy}According to the calculated results in this paper, it is normal celestial bod

rmal celestial bodies will attract substances from its surroundings. It gradually grows as it accumulates the mass. It would have the maximum mass of 5.06 × 1029

le explodes, However, in a

neutron body

quantity of energy (lose of masses), become an extremely hot body and shine. A maximum neutron star has a maximum mass of 2.98 ×1030_{kg, maximum}

radius of 15.3 km. shape of our _ocess.

in the

per. It is an exiting result to the researchers in funda-mental physi

nimum black hole in

ck hole is ap ting of matter. As the mass increasing, far more distant

photons can be hold in. The compressing energy that released absorbed by itself. As an isolated system, the entropy of the black hole decreased spontaneously in its growing process.

REFERE

[1] V. L. Ginsburg, “On Physics and Astrophysics,” Nauka Mir Press,

Course, McGraw Hill, 1971. est Known Black Hole,” NASA’s Go Center, 2008.

d

- http://www.nasa.gov/centers/goddard/news/topstory/2008_{/smallest_blackhole.html} G. Thomas and B. Hermann, “The Steady-State Theory of the Expanding Universe,” Royal Astronomical Society, Vol. 108, 1948, p. 252.

[6] E. Hubble, “A Relation between Distance and Radial Velocity among Extra-Galactic Nebulae,” Proceedings of g period.

e. In the late period