# Black hole doesn't have singularity.

Discussion in 'Astronomy, Exobiology, & Cosmology' started by icarus2, Oct 2, 2016.

1. ### icarus2Registered Senior Member

Messages:
101

[ Black hole doesn't have singularity ]

1. Understanding of the problem : Generation of singularity

Generally, stars are under gravitational contraction by their own gravity and it is known that if this is contracted within a certain radius (like Schwarzschild radius), it makes a black hole which even lights cannot escape from.
Since gravity is generally an attractive force, gravitational contraction continues to exist in the black hole, too. Thus, in the central part of black hole exists an area with infinite density of energy, which point we call singularity.
Such singularity denies application of the existing laws of physics and it is unnatural for a certain substantial object to have infinite density of energy. Besides, such singularity has never been observed as substance but is just a mathematical result of general relativity, which is considered a defect or limit of the theory.

We assume that the solution for this singularity consists in quantum mechanics. Though exact explanation is not available because quantum gravity theory in integration of quantum mechanics and gravity has not been completed yet, we are looking for ways to avoid the problem of singularity by supposing the minimum size in the superstring theory.

This writing will prove that an object of positive energy has the minimum size for its existence and that since this size is in proportion to that of energy, there is no singularity with infinite density.

2. Solution of the problem

2-1. Gravitational potential energy with negative values.

~~~~~~~~~~

2-1-1. Gravitational potential energy is an object that has a certain value at all spatial positions.

~~~~~~~~~~

2-1-2. All energies are a gravitational source.

~~~~~~~~~~

2-1-3. $U(r) = - \frac{{GMm}}{r}$ (except r=0) is considered to provide consequently the right explanation for all points.

Teachers and professors have explained that it is alright to set the randomly reference point for gravitational potential energy because, since the variation of gravitational potential energy has caused kinetic change in the problem under review, there was no problem in dealing with only the variation of gravitational potential energy.

From the equation K + U = const. we obtain such equation as $\Delta K = - \Delta U$, which can explain motion with variation, but this neither means that all observers in the same inertial system may set randomly reference point at random nor confirms that U is an object with an optional value.

Let's consider the following case that the value of gravitational potential energy has been fixed for the distance of 0 to infinity from gravitational source.
$U(r) = - \frac{{GMm}}{r}$
$\Delta U = {U_f} - {U_i} = ( - \frac{{GMm}}{{r + h}}) - ( - \frac{{GMm}}{r}) = \frac{{GMm}}{{r(r + h)}}h \approx \frac{{GMm}}{{{r^2}}}h = mgh$
Even though, as above, gravitational potential energy has the value of energy defined for r= 0 to r= infinity from gravitational source, we can obtain the right result in a problem in which its variation matters.

2-1-4. Effect of mass defect in atomic scale caused by binding energy

~~~~~~~~~~

2-2. Gravitational self-energy or Gravitational binding energy

The concept of gravitational self-energy is the total of gravitational potential energy possessed by a certain object M itself. Since a certain object M itself is a binding state of infinitesimal mass dM, it involves the existence of gravitational potential energy among these dMs and is the value of adding up these.

$M = \sum {dM}$

Fig06. Since all mass M is a set of infinitesimal mass dMs and each dM is gravitational source, too, there exists gravitational potential energy among each of dMs. Generally, gravitational potential energy by infinitesimal mass that consists of an object itself is reflected on the mass of the object itself. Mass of an object measured from its outside corresponds to the value of dividing the total of all energy into $c^2$.

Gravitational self-energy or Gravitational binding energy($-U_{gs}$) in case of uniform density is given by:

${U_{gs}} = - \frac{3}{5}\frac{{G{M^2}}}{R}$
(${U_{gs}}$ : gravitational self-energy)

2-3. For black hole or singularity, never fail to consider gravitational self-energy

In the generality of cases, the value of gravitational self-energy is small enough to be negligible, compared to mass energy $mc^2$.

1)The earth's gravitational self-energy is roughly $4.6 \times {10^{ - 10}}$ times as large as the earth's rest mass energy.
2)The moon's gravitational self-energy is roughly $0.2 \times {10^{ - 10}}$ times as large as the moon's rest mass energy.

Therefore, in usual cases, $|{U_{gs}}| < < M{c^2}$, so generally, there was no need to consider gravitational self-energy.

Meanwhile, looking for the size in which gravitational self-energy becomes equal to rest mass energy by comparing both,

${U_{gs}} = | - \frac{3}{5}\frac{{G{M^2}}}{{{R_{gs}}}}| = M{c^2}$
${R_{gs}} = \frac{3}{5}\frac{{GM}}{{{c^2}}}$

This equation means that if infinitesimal mass is uniformly distributed within the radius $R_{gs}$, gravitational self-energy for such an object equals mass energy in size. So, in case of such an object, mass energy and gravitational self-energy can be completely offset while total energy is zero. Since total energy of such an object is 0, gravity exercised on another object outside is also 0.

Comparing $R_{gs}$ with $R_B$, the radius of Schwarzschild black hole,

${R_{gs}} = \frac{3}{5}\frac{{GM}}{{{c^2}}} < {R_B} = \frac{{2GM}}{{{c^2}}}$
${R_{gs}} = 0.3{R_B}$

This means that there exists the point where gravitational self-energy becomes equal to mass energy within the radius of black hole, and that, supposing a uniform distribution, the value exists at the point $0.3R_{B}$, a 30% level of the black hole radius.

Since this value is on a level not negligible against the size of black hole, we should never fail to consider gravitational self-energy'' for case of black hole.

2-4. Black hole doesn't have singularity.

~~~~~~~~~~

==========
#Paper
On Problems and Solutions of General Relativity. (Commemoration of the 100th Anniversary of General Relativity)

Last edited: Oct 2, 2016

3. ### Dr_ToadIt's green!Valued Senior Member

Messages:
1,847
Totally. Awesome kimchi.

5. ### icarus2Registered Senior Member

Messages:
101
2-4. Black hole doesn't have singularity.

From the equation above, even if some particle comes into the radius of black hole, it is not a fact that it contracts itself infinitely to the point R=0. From the point $R_{gs}$, gravity is 0, and when it enters into the area of $R_{gs}$, total energy within $R_{gs}$ region corresponds to negative values enabling antigravity to exist.

This $0.3R_{B}$ region comes to exert repulsive effects of gravity on the particles outside of it, therefore it interrupting the formation of singularity at the near the area R=0.

Fig07. Considering gravitational potential energy for black hole, the area of within $R_{gs}$ has gravitational self-energy of negative value, which is larger than mass energy of positive value. This area(within $R_{gs}$) exercises antigravity on all particles entering this area anew, and accordingly prevents all masses from gathering to r=0.

However, it still can perform the function as black hole because $R_{gs}$ is only 30% of $R_{B}$ with a large difference in volume and, comparing total mass, it still can correspond to a very large quantity of mass. Therefore, it still can perform the function as black hole on the objects outside of $R_{B}$.

~~~~~~~~~~

2-5. The minimal size of existence

~~~~~~~~~~

2-6. Expansion of general relativity

2-6-1. We can solve the problem of singularity by separating the term($- {M_{gs}} = \frac{{{U_{gs}}}}{{{c^2}}}$) of gravitational self-energy from mass and including it in the solutions of field equation.

M --> $M + (-M_{gs})$

In the Schwarzschild solution,

$d{s^2} = - (1 - \frac{{2GM}}{{{c^2}r}}){c^2}d{t^2} + \frac{1}{{(1 - \frac{{2GM}}{{{c^2}r}})}}d{r^2} + {r^2}d{\theta ^2} + {r^2}{\sin ^2}\theta d{\phi ^2}$

For the sphere with uniform density,

$- {M_{gs}} = - \frac{3}{5}\frac{{G{M^2}}}{{{R}{c^2}}}$

$d{s^2} = - (1 - \frac{{2G(M - {M_{gs}})}}{{{c^2}r}}){c^2}d{t^2} + \frac{1}{{(1 - \frac{{2G(M - {M_{gs}})}}{{{c^2}r}})}}d{r^2} + {r^2}d{\theta ^2} + {r^2}{\sin ^2}\theta d{\phi ^2}$

In general, $M \gg {M_{gs}}$. But we should never fail to consider "gravitational self-energy'' for case of black hole.

1)If $M \gg M_{gs}$, we get the equation (46).
2)If $M \ll {M_{gs}}$,

$d{s^2} \simeq - (1 + \frac{{2G{M_{gs}}}}{{{c^2}r}}){c^2}d{t^2} + \frac{1}{{(1 + \frac{{2G{M_{gs}}}}{{{c^2}r}})}}d{r^2} + {r^2}d{\theta ^2} + {r^2}{\sin ^2}\theta d{\phi ^2}$

i) $r \ge R$, By eq.(47)

$d{s^2} \simeq - (1 + \frac{{6{G^2}{M^2}}}{{5{c^4}Rr}}){c^2}d{t^2} + \frac{1}{{(1 + \frac{{6{G^2}{M^2}}}{{5{c^4}Rr}})}}d{r^2} + {r^2}d{\theta ^2} + {r^2}{\sin ^2}\theta d{\phi ^2}$

ii) $0 \le r < R$
The area of within $R_{gs}$ has gravitational self-energy of negative value, which is larger than mass energy of positive value. This area(within $R_{gs}$) exercises antigravity on all particles entering this area anew, and accordingly prevents all masses from gathering to r=0.

Besides, negative mass has gravitation effect which is repulsive to each other. Therefore, we can assume that $-M_{gs}$ is almost evenly distributed. Therefore ${\rho _{gs}}$ is constant. And we must consider the Shell Theorem.

$- {M_{gs}} = - \frac{{4\pi {r^3}}}{3}{\rho _{gs}}$

$(1 + \frac{{2G{M_{gs}}}}{{{c^2}r}}) = 1 + \frac{{2G(\frac{{4\pi }}{3}{r^3}{\rho _{gs}})}}{{{c^2}r}} = 1 + \frac{{8\pi G{\rho _{gs}}{r^2}}}{{3{c^2}}}$

$d{s^2} \simeq - (1 + \frac{{8\pi G{\rho _{gs}}{r^2}}}{{3{c^2}}}){c^2}d{t^2} + \frac{1}{{(1 + \frac{{8\pi G{\rho _{gs}}{r^2}}}{{3{c^2}}})}}d{r^2} + {r^2}d{\theta ^2} + {r^2}{\sin ^2}\theta d{\phi ^2}$

If $r \to 0$,

$d{s^2} \simeq - {c^2}d{t^2} + d{r^2} + {r^2}d{\theta ^2} + {r^2}{\sin ^2}\theta d{\phi ^2}$

There is no singularity.

For Schwarzschild black hole, the Kretschmann scalar is,

${R^{\mu \nu \rho \sigma }}{R_{\mu \nu \rho \sigma }} = \frac{{48{G^2}{M^2}}}{{{c^4}{r^6}}}$

If $r \to 0$,
${R^{\mu \nu \rho \sigma }}{R_{\mu \nu \rho \sigma }} = \frac{{48{G^2}{{(M - {M_{gs}})}^2}}}{{{c^4}{r^6}}} \approx \frac{{48{G^2}{{( - {M_{gs}})}^2}}}{{{c^4}{r^6}}} = \frac{{{{(16\pi G{\rho _{gs}})}^2}}}{3{c^4}}$

It does not diverge. Therefore,
Black hole doesn't have singularity.

==========
#Paper
On Problems and Solutions of General Relativity. (Commemoration of the 100th Anniversary of General Relativity)
https://www.researchgate.net/publication/301696194

Last edited: Oct 2, 2016

Messages:
21,703
As commonly understood, most cosmologists do not believe that the Singularity exists, other then as defined by the limits of GR.
The singularity as commonly understood, need not have infinite quantities such as density and spacetime curvature, although it could lead to such concepts.

Last edited: Oct 3, 2016
8. ### SimonsCatRegistered Member

Messages:
213
I won't go too deep into the work in the OP... I just don't have the time to read it all today.

I will go to say that, singularities are unphysical, they probably don't exist in nature and nature will find other solutions! And boy they are plenty to look through... for instance, the holographic black hole doesn't even have an interior! All the information on a holographic black hole is stored on its horizon, so there will not be any singular solutions for a holographic black holes. Other types of models, use gravitational quantum effects (like a Planck star) to stop gravitational collapse into singular regions of spacetime.

9. ### icarus2Registered Senior Member

Messages:
101

Waiting for quantum gravity theory to be completed to solve the singularity issue in a black hole is wrong as it was made by our stereotypes.

When there occurred a problem in singular point, one dimensional idea that problems should be solved from $\lambda$, wavelength that has a little bigger than “point” was partially acted. Of course, we should try to establish a quantum gravity theory for other reasons.

We can think of a black hole of big size and approach this problem by reducing the mass of this black hole. In other words, we should form a certain internal structure of usual size and apply the experience that we had applied the limit.

$\mathop {\lim }\limits_{M \to small} {R_{gs}}$

If you are still uncomfortable with $R_{gs}$, think about a black hole with the size 10 billion times bigger than the solar mass. Schwarzschild radius of this black hole is ${R_S} = 3 \times {10^{10}}km$ and $R_{gs}$ of this black hole $1 \times {10^{10}}km$.

${\rho _ {BH}}$ = Average mass density of black hole.

${\rho _ {BH}} = \frac{{3{c^2}}}{{8\pi G}}\frac{1}{{{R^2}}} = \frac{{3{c^6}}}{{32\pi {G^3}}}\frac{1}{{{M^2}}}$

${\rho _ {10Bil - BH}} = 1.81(kg/{m^3}) < < {\rho _ {Earth}} = 5500(kg/{m^3})$

Is it a size that requires quantum mechanics?
Is it a high density state that requires quantum mechanics?
Black hole of this size is Newtonian mechanics' object and therefore, gravitational potential energy must be considered.

Let’s reduce the mass of this black hole gradually and approach three times the solar mass, the smallest size of black hole where stars can be formed!

In case of the smallest black hole with three times the solar mass, ${R_S} = 9km$. $R_{gs}$ of this object is as far as 3km. In other words, even in a black hole with smallest size that is made by the contraction of a star, the distribution of internal energy can't be reduced to at least radius 3km.

==========
# Paper :
1.On Problems and Solutions of General Relativity: Commemoration of the 100th Anniversary of General Relativity
https://www.researchgate.net/publication/301696194

2.Size and Expansion of the Universe in Zero Energy Universe
https://www.researchgate.net/publication/309786718

Last edited: Feb 12, 2017
10. ### KittamaruNever cruel nor cowardly...Staff Member

Messages:
13,666
icarus2 - do you plan to actually reply to any of the other members comments and/or hold conversation, or are you simply using this as a whiteboard? Just curious...

11. ### icarus2Registered Senior Member

Messages:
101
Dear Kittamaru,

I am very sorry! I can’t English well. T.T
So, my conversation is very limited.