# Dark matter is Negative mass!

Discussion in 'Astronomy, Exobiology, & Cosmology' started by icarus2, May 2, 2018.

1. ### icarus2Registered Senior Member

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Dark Matter is Negative Mass

Negative mass is an object whose existence is required by the law of the conservation of energy. The fundamental properties of negative mass can explain important characteristics of dark matter.
1) additional centripetal effects,
2) explanations derived from fundamental principles about the reason why dark matter does not have electromagnetic interaction,
3) repulsive gravity ensuring almost even distribution and lower interaction of dark matter,
4) gravitational lens effect,
5) accelerating expansion of the universe can be explained with negative mass.

I. Introduction

~~~~~~~

II. Main Characteristics of Negative Mass

1. Positive mass is stable at the low energy state, while the negative mass is stable at the high energy state.
One of physics' fundamental principles, "lower energy state is associated with stability" can be only applied to positive mass. However, both negative mass and negative energy level have been denied, as it has been wrongly applied to negative mass.

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Figure 1. When there is negative mass in potential which has a point of maximum value and a point of minimum value.

$\begin{array}{l} \vec F = - {m_ - }\vec a\\ ({m_ - } > 0) \end{array}$

$\vec a = - \frac{{\vec F}}{{{m_ - }}}$

When negative mass exists within potential with maximal and minimal points, different directions of force and acceleration should be considered for negative mass.

The acceleration of negative mass is opposite to the direction of force. Therefore, the negative mass has harmonic oscillation at the maximum point and it is also stable at the maximum point.

In the case of positive mass, it was stable at the minimum point at which energy is the low. However, in case of negative mass, stable equilibrium is a point of maximum value, not a point of minimum value.

It is stable at a low energy state in the case of positive mass. However, it is stable at a high energy state in the case of negative mass. Due to this, "the problem of transition to minus infinite energy level" does not occur, therefore negative mass(energy) and positive mass(energy) can exist stably in our universe.

2. Negative mass has both negative inertial and gravitational mass.
There are repulsive gravitational effects between negative masses.

One of problems regarding movement of negative mass, which researchers are likely to misunderstand is that if there are negative mass particles around the earth (or galaxy), large positive mass, they may not fly to the universe, but freely fall to the earth (or galaxy).

3. See the following video produced to help understand movement of negative mass.

#Paper : Dark matter is Negative mass
https://www.researchgate.net/publication/324525352

Last edited: May 2, 2018
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3. ### icarus2Registered Senior Member

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III. Dark Matter is Negative Mass

1. Negative mass is an object whose existence is required by the law of the conservation of energy.

~~~~~~~

2. Negative mass does not have charges, so is unlikely to have electromagnetic interaction.
Think about whether negative mass has charges.
It is possible that the model designer assumes two models of negative mass with and without charges. However, let's infer what a state is more proper, based on the existing physical laws.

We have observed pair production due to the collision of photon with atomic nucleus. Such phenomena suggest that the process of pair production exists where the charge Q does not exist. Now, lets' examine whether when negative and positive energy exist, each of them can has charges.

1) Electrostatic self-energy $U_{es}$

Since the charge Q is the set of infinitesimal charge dQ, electric force is operated between infinitesimal charges and therefore, electrostatic self-energy exists due to the presence of charge Q itself.

When spherical symmetry and even distribution of charges are assumed, $U_{es}$, the electrostatic self-energy(or electrostatic binding energy($-U_{es}$)) is as follows:

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figure 3. Since all charge Q is a set of infinitesimal charge dQs and each dQ is electromagnetic source, too, there exists electrostatic potential energy among each of dQs.

${U_{es}} = + \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{R_{es}}}})$

However, electrostatic self-energy has positive values for both positive and negative charges.

${U_{es}} = + \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_ + }^2}}{{{R_{es}}}}) = + \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_ - }^2}}{{{R_{es}}}}) > 0$

2) Positive energy can has charges, since a part of its positive energy or mass is able to be transformed into electrostatic self-energy of charges.

${E_T} = + {E_0} = + {E_0} + ( - {U_{es}}) + ( + {U_{es}}) = ( + {E_0} - {U_{es}}) + \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{R_{es}}}})$

However, negative energy does not have positive energy to produce charges, so it is very unlikely to have charges.

3) Limit of spontaneous process by the principle of negative mass

In positive mass, the transformation for ${E_0}$ to ${E_0} - {U_{es}}$ is the process in which energy is lowered, which corresponds to the fundamental principle that positive mass is stable at the low energy state.

${E_T} = - {E_0} = - {E_0} + ( - {U_{es}}) + ( + {U_{es}}) = - ({E_0} + {U_{es}}) + \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{R_{es}}}})$

Meanwhile, in negative mass, the transformation for ${-E_0}$ to ${-E_0} - {U_{es}}$ is the process in which energy is lowered, which does not correspond to the fundamental principle that the negative mass is stable at the high energy state.

Therefore, negative mass is unlikely to have charges, according to this principle.

4) Electrostatic self-energy has a potential to explain substantial parts of elementary particle mass.

Most parts of elementary particles’ rest mass are likely to be originated from the electrostatic (or electromagnetic) self-energy, due to the presence of charges. Although there are many points to be additionally considered, such a model’s suggestions for the magnitude of quark should be currently examined with electrostatic self-energy equivalent to mass energy.

$E = m{c^2} = \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{R_{es}}}})$

${R_{es}} = \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{m{c^2}}})$

For convenience of calculation, it is defined as follows $Q = ke$, $m{c^2} = \alpha (eV)$

${R_{es}} = \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{m{c^2}}}) = \frac{3}{5}(\frac{1}{{4\pi {\varepsilon _0}}})(\frac{{{k^2}{e^2}}}{{m{c^2}}}) = \frac{{{k^2}}}{\alpha }(8.64 \times {10^{ - 10}})[m]$

If the charge is distributed only in the spherical shell, the radius is reduced by approximately 16.7%.
${U_{es - shell}} = \frac{1}{2}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{R_{es}}}})$

${R_{es - shell}} = \frac{1}{2}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{m{c^2}}}) = \frac{{{k^2}}}{\alpha }(7.20 \times {10^{ - 10}})[m]$

Since mass is $2.4MeV/c^2$ and quantity of electrical charge is (+2/3)e, for up quark,
if k=2/3 and $\alpha = 2.4 \times {10^6}$ are substituted,

${R_{es - U}} = \frac{{{k^2}}}{\alpha }(8.64 \times {10^{ - 10}})[m] = \frac{{{{(\frac{2}{3})}^2}}}{{2.4 \times {{10}^6}}}(8.64 \times {10^{ - 10}})[m] = 1.6 \times {10^{ - 16}}m$

By interpreting this, it can be inferred that the mass of up quark is derived from electrostatic self-energy, if the radius of up quark is approximately ${R_{es - U}} = 1.6 \times {10^{ - 16}}m$ and here, charges are evenly distributed.

If inferences of other quarks can be acquired,
Down quark's mass: $4.8MeV/c^2$, charge: (-1/3)e
${R_{es - D}} = 2.0 \times {10^{ - 17}}m$

Strange quark's mass: $95MeV/c^2$, charge: (-1/3)e
${R_{es - S}} = 1.01 \times {10^{ - 18}}m$

Charm quark's mass: $1.275GeV/c^2$, charge: (2/3)e
${R_{es - C}} = 3.01 \times {10^{ - 19}}m$

Bottom quark's mass: $4.18GeV/c^2$, charge: (-1/3)e
${R_{es - B}} = 2.29 \times {10^{ - 20}}m$

Top quark's mass: $172.44GeV/c^2$, charge: (2/3)e
${R_{es - T}} = 2.22 \times {10^{ - 21}}m$

It is desirable to think that the magnitude of quark from such a simple model is smaller than that of an atomic nucleus.

The fact that the six quark radii (or diameters) vary by roughly $10^1$ orders of magnitude, suggests that there is some principle or rule. We considered only the electrostatic self-energy value. Nonetheless, some rules seem to suggest that the self-energy of the charge may be the most important term.

If there is a fourth-generation quark, we can predict the mass of the fourth-generation quark by using this reasoning. In the history of science, there is a case of Titius-Bode' law.

The above calculation suggests the possibility that electrostatic self-energy due to the presence of charges can explain some parts of elementary particle mass. In order to make a precise model, the model related with kinetic and other energy, relativistic effects and orbital model(in case of electron) related with charge distribution can be also considered.

However, this seems to deny the logic of the Higgs mechanism or Higgs field, which gives mass to the elementary particles. This model claims that most of the mass of a charged particle is from the charge distribution. Electrostatic self-energy comes from the definition of electric force or electrostatic potential energy.

Perhaps, is the Higgs mechanism a mechanism that only applies to weak interaction particles?

This inference also suggests that negative energy is unlikely have charges, because electrostatic self-energy is positive mass (energy), which provides an evidence about the fact that electromagnetic interaction does not occur.

Hence, negative mass does not have any charges and is very unlikely to have electromagnetic interaction. In addition, even if other fundamental forces are applied to the negative mass, the fundamental charges underlying the forces may be supposed to have positive self-energy.

This corresponds to characteristics required for the dark matter, and also gives some explanations to “why does the dark matter not have electromagnetic interaction?”

Last edited: May 2, 2018

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5. ### icarus2Registered Senior Member

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3. Centripetal force effects that negative mass outside the galaxy creates inside it
If negative mass(energy) and positive mass(energy) were came into being together at the beginning of universe, since positive masses have attractive gravitational effects with each other, so it forms stars and galaxy. However, negative masses have repulsive gravitational effects towards each other, so it cannot form any giant structure and may spread out almost uniformly across the whole area of universe.

Owing to the effect of negative mass and positive mass, negative mass disappears near massive positive mass structures (such as the galaxy, etc.) after meeting positive mass. However, negative mass, which came into existence at the beginning of universe, can still exist in a vacuum state outside of general galaxy.

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Figure 4. The structure of the galaxy surrounded by negative mass that is distributed equally.(fig. a)) The white area is the area where negative mass almost does not exist.(fig. a))

Let's examine the effect of the centripetal force of negative mass that is outside the galaxy on mass m, which is located within the galaxy.

1) If we assume that the empty space is filled with both negative mass and positive mass of the same density then,

Total energy of the white empty space = $0 = ( + mc^2 ) + ( - mc^2 ) = 0$

2) Negative mass is now uniformly distributed over the whole area so the effect of negative mass on mass m becomes 0. (fig. b))

3) The remaining positive mass is distributed over the white area at the density of negative mass. (fig. c)) Gravity that uniformly distributes positive mass works on positive mass m located on radius r is worked upon only by the distribution of mass within radius r. -Shell Theorem.(fig. d))

Therefore, the effect of negative mass that remains outside of the galaxy is compressive to the distribution of positive mass within the radius r in the galaxy. This means that the dark matter, consisting of negative mass outside galaxy, has additional effect of centripetal force on stars within the galaxy.

$\frac{{m{v^2}}}{r} = \frac{{G({M_{matter}} + {M_{dark}})m}}{{{r^2}}}$

$v = \sqrt {\frac{{G({M_{matter}} + {M_{dark}})}}{r}} = \sqrt {\frac{{G({M_{matter}} + \frac{{4\pi {r^3}{\rho _{dark}}}}{3})}}{r}} \approx \sqrt {\frac{{4\pi G{\rho _{dark}}}}{3}} r = \sqrt {\frac{{4\pi G{\rho _ - }}}{3}} r$

This effect suggests that the further from the center of the galaxy, the more mass effect exists and agrees with the current situation where the further from the center of the galaxy, the more dark matter exists.

If dark matter is exist in the galaxy, dark matter will be affected by the distribution of stars and interstellar gas in the galaxy. That is, the density of dark matter around the solar system will be higher than the average of the galaxy. The existing CDM model suggests that there is a local distribution of dark matter around the local gravitational source (such as solar system).

Some models offer a theoretical background on how dark matter is not detected in the solar system by lowering the total amount of dark matter around the solar system by assuming that dark matter is completely evenly distributed in the galaxy.

This model explains the presence of additional gravitational effects in the galaxy and the reason why the dark matter is not found on the earth or in the vicinity of the solar system.

In this model, dark matter (negative mass) exists outside the galaxy, and there is only a gravitational effect in the galaxy}} that is not affected by the local gravitational source (such as solar system) in the galaxy.

We observed only the effect of matter in the Earth. However, it is presumed that the observation of the gravitational effect of dark matter and dark energy at the galactic scale or above, is due to the existence of negative energy and negative mass outside the galactic structure.

4) Computer simulation

6m : 00s ~

If the negative mass is disposed at the outline, the positive mass vibrates. Therefore a kind of centripetal force exists. This corresponds to the “centripetal force” when considering rotation of the galaxy.

5) The model above is a principle explanation. Moreover, axisymmetric distribution due to rotation of the galaxy should be also considered. It is also possible to consider repulsive force affected by negative mass as well as negative mass’s free fall onto the center of the galaxy.

4. Low interaction between dark matters: occurring from repulsive gravitational effects between dark matters

The existing CDM models argue that dark matters are almost evenly distributed as they only have gravitational interaction, except for electromagnetic one, even though dark matters are five times or hundreds times more than other masses. However, is this process true?

There are more certain explanations and rules in the model of negative mass.

Positive masses have attractive gravitational effects with each other, so it forms stars and galaxy. However, negative masses have repulsive gravitational effects towards each other, so it cannot form any giant structure and may spread out almost uniformly across the whole area of universe. Also, negative mass exists outside the galaxy, and there is only a gravitational effect inside the galaxy.

This provides strong explanations to the phenomenon that current dark matters do not strongly interact with each other, but are almost evenly distributed.

#Paper : Dark matter is Negative mass
https://www.researchgate.net/publication/324525352

Last edited: May 2, 2018

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7. ### NotEinsteinValued Senior Member

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Why?

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8. ### icarus2Registered Senior Member

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5. Negative mass can also explain dark energy.

Negative mass engenders antigravity which is consistent with effects of dark energy that allows for accelerating expansion of the universe.

Friedman equation can be induced from 00 component of field equation. But we can also induce this from conservation of energy in classical mechanics, which helps capture the situation definitely.

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Figure 6. The Friedmann equation can derive from field equation or mechanical energy conservation equation.

What we are looking for in cosmology is one attraction term and one repulsion term. If a negative mass exists, we can get it.

If negative mass and positive mass coexist, gravitational potential energy consists of the following three items.

$U_T = \sum\limits_{i < j}^{} {( - \frac{{Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }}) + \sum\limits_{i < j}^{} {( - \frac{{G( - m_{ - i} )( - m_{ - j} )}}{{r_{ - - ij} }})} + \sum\limits_{i,j}^{} {( - \frac{{G( - m_{ - i} )m_{ + j} }}{{r_{ - + ij} }})}}$

$U_T = \sum\limits_{i < j}^{} {( - \frac{{Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }}) + \sum\limits_{i < j}^{} {( - \frac{{Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})} + \sum\limits_{i,j}^{} {( + \frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})} }$

$U_T = U_{ + + } + U_{ - - } + U_{ - + }$

The present cosmological constant can be obtained by adding potential energy ${U_\Lambda } = - \frac{1}{6}\Lambda m{c^2}{r^2}$ to mechanical energy conservation equation.

If we insert "new potential energy term" into mechanical energy conservation equation, we will get a dark energy term and dark matter term. And if $U_{++}$, $U_{--}$, $U_{-+}$ has a ratio(4.9% : 26.8% : 68.3%) between each other, maybe, we will estimate that ratio of energy density such as 4.9% : 26.8% : 68.3% exist.

This model can prove the energy composition(Matter : Dark matter : Dark energy) ratio of the universe and CCC (Cosmological Constant Coincident) Problem.

Roughly calculation:
Average of WMAP and Planck - Matter : Dark Matter : Dark Energy = 4.75% : 25.05% : 70.20%
Average of Pair Creation Model - Matter : Dark Matter : Dark Energy = 4.75% : 25.00% : 70.25%

Refer to my paper
1) https://www.researchgate.net/publication/263468413
2) https://www.researchgate.net/publication/287217009

Last edited: May 2, 2018
9. ### NotEinsteinValued Senior Member

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Is this a blog, or do you intend to respond to my post?

10. ### icarus2Registered Senior Member

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I'm sorry. I do not speak English well. Thus, the answers are very limited.

There is a little explanation in my paper.
1. Negative mass is an object whose existence is required by the law of the conservation of energy.

If the law of energy conservation that “energy will be always conserved” is applied to the initial state of the universe, this gives rise to the question of “where the energy (of matter) of our universe did come from?” The most natural answer to this question is the assumption that the energy itself was not something created and that zero-energy state went through phase transition keeping conservation of energy and generating negative and positive energy.

Therefore, in order to offset the positive energy of matter, negative energy (mass) is needed.

Suppose that the universe started with a particular energy value $E_0$ rather than Zero Energy.
Logically, this universe model allows all real-valued times energy states of this particular energy value $E_0$, thus allowing for the possibility of existence of infinite universe with infinitely many different energy values. It also requires a rational explanation for the fact that a particular $E_0$ value has been selected in the process.

IMO,
The negative energy (mass) is the demand and outcome of the law of conservation of energy.

Last edited: May 3, 2018
11. ### exchemistValued Senior Member

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But have you not read of the hypothesis that the positive energy of matter and radiation may be balanced by the "negative" energy represented by gravitational potential, which rises to zero only at infinite separation?

12. ### Michael 345In Aust : found it :)Valued Senior Member

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So to get negative mass you start by taking away all the mass you have until you have zero mass

And then you take MORE mass away from zero mass until you have negative mass

Ok

First question :- is negative mass the stuff you take OUT of zero mass

or

is negative mass the HOLE you leave behind in zero mass?

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13. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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It sounds like you are making a lot of assumptions with no logical reasons.
I wouldn't say a photon collides with a nucleus, but that is ok. What do you mean charge does not exist? An electron and a positron certainly have charges.
How do you think that energy can have a charge? That in nonsensical. Energy is not a substance that you can put in a cup. It is not a substance that can have a charge.
Dark matter was first theorized because of the motion of galaxies. There appears to be a large amount of dark matter in galaxies! You are claiming the opposite of observations.
This assumption is wildly outside of the accepted physics. In general the smallest change is e ( 1.602176634×10−19 Coulombs). You could say, I suppose that the smallest charge is actually e/3 because that is the charge carried by some types of quarks, but they do not exist in isolation. But claiming that a charge is a set of infinitesimals is wrong.

14. ### exchemistValued Senior Member

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This is interesting - at the risk of going off at a tangent to the thread subject. I heard a physicist the other day on the radio saying that the fact that the charge on quarks is multiples of +/- 1/3, while the charge on the electron is 1, strongly hints at a deeper connection between leptons and quarks than physics has yet uncovered. Why should they be different and yet in exact ratios involving 1/3?

I thought that was a rather insightful remark - it had not occurred to me before. I wonder if anyone here can comment?

15. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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I can't comment on that, but let's hope Nebel doesn't see this because he will probably say it is because the earth is the 3rd planet from the sun or something...

16. ### exchemistValued Senior Member

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You're a mind reader. I damned nearly added a remark to my post saying that I must sound a bit like nebel!

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But in fact this one is curious, given that those ratios are what is needed for exact electrical neutrality, and yet we do not have a model that accounts for why this happens - it just is, so far as we can tell at present.

17. ### NotEinsteinValued Senior Member

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True.

This doesn't follow. Not all energy is expressed as mass. It would be true is all energy was in the form of rest-mass ($E=mc^2$), but it's not. In GR, there are contributions to the stress-energy tensor from negative energy, without the need to introduce negative masses. You really need to justify this jump in logic.

I suggest you use the word multiverse, because universe-in-universe makes little sense.

It doesn't; look up the anthropological argument.

You keep claiming that, but it's up to you to actually demonstrate that it does.

18. ### Michael 345In Aust : found it :)Valued Senior Member

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Now you sound like Jan

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19. ### icarus2Registered Senior Member

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I know the model.
We can propose two models as an example for zero energy universe model.

Model-1. Model that considers gravitational potential energy as negative energy.

${E_T} = 0 = ( + E) + ( - E) = \sum { + m{c^2}} + \sum { - \frac{{G{m_i}{m_j}}}{{{r_{ij}}}}} = 0$

Model-2. Model that considers negative mass as negative energy.

${E_T} = 0 = ( + E) + ( - E) = \sum { + {m_ + }{c^2}} + \sum { - {m_ - }{c^2}} + \sum U = 0$

Stephen Hawking, Alan Guth, and Alexander Vilenkin are the pioneers who advocated that positive mass energy could be offset by gravitational potential energy only, like in Model-1.

However, Model-1 has several disadvantages.

Problem-1. It is possible for the negative gravitational potential energy to offset the positive mass energy. By the way~

Looking for the size in which total gravitational potential energy becomes equal to total rest mass energy by comparing both,

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

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

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

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

In other words, in order for Model-1 to be established, our universe must exist in a black hole.
For some people, this is a big problem.

Problem-2. The universe is expanding.
The universe is expanding. Mass term $\sum { + m{c^2}}$ is constant, whereas R increases, and so the absolute value of gravitational potential energy $|- \frac{3}{5}\frac{{G{M^2}}}{R}|$ gets smaller and finally energy state of 0 is broken. In general, the kinetic energy is small compared to the mass energy and can be ignored. And, it's a positive energy. Anyway, focusing on the gravitational potential energy~

To look at how equation is valid by unfolding the formula,

$M{c^2} - \frac{3}{5}\frac{{G{M^2}}}{R} = 0$

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

In this model-1, to establish energy conservation law while the universe is expanding, (mass) energy needs to be increased, which increases R of the universe.

In my opinion, this is a fatal problem in Model-1. So, I have a doubt in Model-1.

#Paper
https://www.researchgate.net/public...s_for_the_Model_We_are_living_in_a_black_hole

Last edited: May 4, 2018
20. ### icarus2Registered Senior Member

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In your writings, it seems that I am in medieval times when negative numbers were introduced ~

Simply put, negative mass indicates that negative energy can be locally distributed and has characteristics as particle. Negative mass has both negative inertial and gravitational mass.

21. ### icarus2Registered Senior Member

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I do not agree with you at all.
It is a general statement, logical, and does not contain many assumptions.

If the word collision is inconvenient, write the word interaction ~
There was no charge before the collision. There was no charge before pair production.

We do not know it. However, we know the state before and after pair production.

I do not agree with you at all.
Energy or mass is a physical quantity that was already present before pair production, it also exists after pair production. Whereas charge is a physical quantity that did not exist before pair production. So reasoning is possible. If the phrase is uncomfortable, think of it as an object with energy~

Ok.

Not at all. To date, no one has ever observed dark matter directly. We do not yet know what the source of the (gravitational) effect we observed.

We have observed that the rotation curve of the galaxy is different from that of the solar system, and we are looking for ways to explain it. This may be explained by the presence of additional gravitational sources and may be explained by the presence of additional gravitational effects. There are other ways, of course.

Not at all. Much of mainstream physics is based on a reductionist perspective. In addition, the differential view has been a great success. It is not only a common sense analysis that does not require explanation, but also there are many examples from the major books.

If you search by electrostatic self-energy or gravitational self-energy, you will know. You can also see the problem of handling the uniformly distribution of charge Q.

Not at all.

Last edited: May 4, 2018

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Ok.

23. ### NotEinsteinValued Senior Member

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This is wrong; the sign of the mass doesn't matter in $E=mc^2$. That's because the full equation is: $E^2=(pc)^2+(mc^2)^2$. The famous $E=mc^2$ is only valid for cases where $p=0$ and $m>0$. You should be using: $E=|m|c^2$ in your case.
See: https://en.wikipedia.org/wiki/Energy–momentum_relation

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