Chapter 18. Black Holes

| Title Page | Table of Contents | Preface | <<<< | >>>> | Appendixes | References | To Order This Book | WritWord Homepage |

In this chapter, we present Bekenstein and Hawking's equation for the entropy of a Black Hole in terms of both SI and SG units of measure.

Table X in Appendix E pertains to the contents of this chapter.

A. Black-Hole Entropy with Constants

The historical equation uses SI units for the entropy of a Black Hole and is:

Sbh = A k c3 (4 hbar G)-1                               (186

where:

A is the area of the Black Hole,

k is Boltzmann's constant,

c is the velocity of light,

hbar is Planck's constant, h, divided by (2 π), and

G is Newton's gravitational constant.

B. Black-Hole Entropy without Constants

Using the quantum attributes of the masson, the values of these factors are, as follows:

Factor Unitary Attributes of the Masson
A |A|g λg2
k Eg·kg-1
c λg·tg-1
hbar Eg·tg (2 π)-1
G 2 (4 π β)-1 λg3·mg-1·tg-2

Substituting these SG units into the entropy equation, gives:

Sbh = (|A|g λg2) (Eg·kg-1) (λg·tg-1)3 ×

  {4 [Eg·tg (2 π)-1] [2 (4 π β)-1 λg3·mg-1·tg-2]}-1     (187

which, when rearranged to separate factors with positive exponents from whose with negative ones, gives:

Sbh = (|A|g λg2 Eg λg3 2 π 4 π β mg tg2) ×

                          (kg tg3 4 Eg tg 2 λg3)-1     (188

and cancelling factors, gives:

Sbh = |A|g λg2 π π β mg (kg tg2)-1 = |A π2 β|g Eg·kg-1 =

                                 |A π2 β|g sg         (189

where the area now includes the factors, π2 and β. The SG unit of entropy, sg , is comprised of the SG unit of energy, Eg , (Planck energy) per SG unit of temperature, kg , (Planck temperature), which is the same as Boltzmann's constant, k.

C. Acceleration of Gravity in SI Units

The historical equation for the SI value of the magnitude of the acceleration of gravity on the surface of a homogeneous spherical mass is:

gm = G Mm rm-2                                       (190

where:

G = Newton's gravitation constant,

Mm = SI mass of the sphere, and

rm = SI radius of the sphere.

Earth's Gravity

We can calculate the SI value of the acceleration of gravity on the surface of the earth by knowing that the average density of the earth, dEm, is about 5.5 metric tons per cubic meter, which is 5.5 × 103 kg·m-3. By the definition of the meter, we already know that the circumferenceof the earth, cEm, is 4.0 × 107 m and its radius, rEm, is 4.0 × 107 (2 π)-1 m. Therefore, the SI volume of the earth is:

VEm = 4(3)-1 π rEm3 = 4(3)-1 π [(4.0 × 107)(2 π)-1 m]3 =

         (4.0 × 107 m)3 (6 π2)-1 = 1.1 × 1021 m3       (191

The SI mass of the earth, MEm, is its average density times its volume:

MEm= dEm VEm = (5.5 × 103 kg·m-3)(1.1 × 1021 m3) =

                            6.0 × 1024 kg             (192

Using Equation 190, the SI value of the magnitude of the acceleration of gravity at the surface of the earth is:

gEm = G MEm rEm-2 = (6.7 × 10-11 m3·kg-1·s-2) ×

     (6.0 × 1024 kg)[4.0 × 107 (2 π)-1 m]-2 =

                                          9.8 m·s-2   (193

D. Acceleration of Gravity in SG Units

We calculate the SG value of the acceleration equation by modifying Equation 190:

gg = G Mg rg-2 = [2 (β 4 π)-1 λg3·mg-1·tg-2] ×

                      [|M|g mg][|r|g-2 λg-2] =

                2 |M|g (β 4 π |r|g2)-1 λg·tg-2          (194

which we can read as:  The SG value of the magnitude of the acceleration of gravity on the surface of a homogeneous spherical ball of matter is twice its mass divided by 137 times the area of its surface.

E. Escape Velocity in SI Units

The historical equation for the SI value of the magnitude of the velocity of escape from the surface of a spherical mass is:

vm = (2 G Mm rm-1)0.5                                   (195

Earth's Escape Velocity

We can calculate the velocity of escape from the surface of the earth, as follows:

vEm = {2 (6.7 × 10-11 m3·kg-1·s-2)(6.0 × 1024 kg) ×

                          [4.0 × 107 (2 π)-1 m]-1}0.5 =

                               1.1 × 104 m·s-1        (196

which is equal to approximately 25,000 miles per hour.

F. Escape Velocity in SG Units

The equation for the SG value of the magnitude of the velocity of escape from the surface of a spherical mass is:

vg = (2 G Mm rm-1)0.5 = [|M|g (π β |r|g)-1]0.5 λg·tg-1     (197

G. Schwarzschild Radius in SI Units

By definition, the Schwarzschild radius, r0, is the radius of a homogeneous, spherical mass, M, that is compressed to the point that it becomes a Black Hole from which light cannot escape because the velocity of escape from the surface of the spherical mass is greater than the velocity of light, c. The historical SI equation for the speed-of-light escape velocity is derived from Equation 195, where (vm = c):

c = (2 G Mm r0m-1)0.5                                    (198

Solving for the SI value of the Schwarzschild radius gives:

r0m = 2 G Mm c-2                                       (199

H. Schwarzschild Radius in SG Units

The SG equation for the speed-of-light escape velocity is based upon Equation 197, where (vg = c):

c = 1 λg·tg-1 = [|M|g (π β |r0|g)-1]0.5 λg·tg-1           (200

Solving for the SG value of the Schwarzschild radius gives:

r0g = |M|g (π β)-1 λg                                  (201

In essence, if a homogeneous, spherical entity's mass-to-radius ratio in SG units is greater than (π β), the entity is a Black Hole.

I. Density of a Black Hole

The equation for the density of a homogeneous sphere of radius, r, is mass, M, per volume, V:

d = M V-1 = M [4 π r3(3)-1]-1 = 3 M (4 π r3)-1         (202

The SG equation for the density of a Black Hole with a Schwarzschild radius is:

dbh = 3 Mg {4 π [r0g]3}-1 =

     3 |Mg| mg {4 π [|M|g (π β)-1 λg]3}-1 =

           3 π2 β3 (4 |M|g2)-1 mg·λg-3               (203

where the density is inversely proportional to the square of the mass. Therefore, we realize that the less massive the Black Hole, the more dense it must be.

J. Earth as a Black Hole

A Black Hole as massive as the earth possesses an SI Schwarzschild radius of:

rEm = 2 G MEm c-2 =

     2 (6.7 × 10-11 m3kg-1s-2) (6.0 × 1024 kg) ×

                          (3.0 × 108 ms-1)-2 =

                                 8.9 × 10-3 m         (204

Can you imagine--the mass of the earth contained within a ball of bubble gum?

K. Black Hole as Dense as Water

The equation for the density of a Black Hole that is as dense as water is:

dbh = 1.0 × 103 kg·m-3 = 3 π2 β3 (4 |M|g2)-1 mg·λg-3      (205

The SI unit of density, kg·m-3, converts to the SG unit, mg·λg-3, by using the conversion factors in Equations 80 and 82:

kg·m-3 = [(1.9 × 10-9)-1 mg] [(1.2 × 10-33)-1 λg]-3 =

                               9.1 × 10-91 mg·λg-3      (206

Substituting, gives:

dbh = (1.0 × 103)(9.1 × 10-91) mg·λg-3 =

                    3 π2 β3 (4 |M|g2)-1 mg·λg-3          (207

Removing the units of measure to simplify, gives:

dbhg = (9.1 × 10-88) = 3 π2 β3 (4 |M|g2)-1               (208

Solving for the SG mass of that enormous Black Hole gives:

|M|g = [3 π2 β3 (4)-1 (9.1 × 10-88)-1]0.5 = 1.4 × 1047

or

Mbh = 1.4 × 1047 mg                                    (209

A Black Hole with a density of water would contain an SI mass of:

Mbh = (1.4 × 1047 mg)(1.9 × 10-9 kgmg-1) =

                                  2.7 × 1038 kg       (210

which is over a hundred-million times more massive than our solar system. Its circumference would be:

cbh = 4 π G Mbh c-2 = 4 π (6.7 × 10-11 m3·kg-1·s-2) ×

                    (2.7 × 1038 kg) (3.0 × 108 m·s-1)-2 =

                   2.5 × 1012 m                      (211

which is almost three times the length of earth's orbit about the sun.

| Title Page | Table of Contents | Preface | <<<< | >>>> | Appendixes | References | To Order This Book | WritWord Homepage |