Antigravity is the source of dark energy (accelerating expansion)

[icarus2]

Hello!
I am sorry. I apologize for my poor English.

 

[ Antigravity is the source of dark energy (accelerating expansion) ]

From the observation of 1998, we found that our universe has been continuing accelerating expansion, and the unknown cause for this accelerating expansion was named dark energy.

 

For the following three reasons, we have been doubtful if dark energy has antigravity.

1) Gravity is the force which has ruled the macroscopic movement of the universe.

2) Dark energy's scale is similar to the magnitude of gravity generated from ordinary matters. (About 15 times)

3) Dark energy’s effect is repulsive.

 

However, nobody is sure whether dark energy truly originates from antigravity, for other several reasons as follows.

 

1) No antigravity has been observed in laboratories or around the earth, thus far.

2) Contrary to that the force coming from dark energy is [math]F = + kr[/math] shaped as a [math]{\vec F_\Lambda } =\frac{1}{3}\Lambda m{c^2}r\hat r[/math] shape, antigravity is [math]F = + \frac{k}{{{r^2}}}[/math] shaped.

3) As dark energy is an unknown effect itself, there is a possibility that other unknown force different from existing ones exists.

 

Since we still have no idea about the source of dark energy, it's been hard to call dark energy "antigravity", even though it was possible to call it "anti-gravitational effect", in the way that its effect is repulsive.

This paper is going to induce a dark energy term from antigravity, and demonstrate that antigravity is the source of dark energy.

A. Gravitational potential energy, when antigravity exists.

 

We are aware of what gravitational self-energy (gravitational binding energy) as the sum of gravitational potential energy is displayed as follows, when matters show a three-dimensional spherical distribution.

[math]{U_S} = - \frac{3}{5}\frac{{G{M^2}}}{r}[/math]

(r: radius, M: the mass of the sphere)

 

=========

http://http://en.wikipedia.org/wiki/Gravitational_binding_energy

=========
Because we are planning to apply this to cosmology,

 

Assumption: For simple modeling, we will suppose that antigravity source has a uniform distribution on a cosmological scale of a level of cluster of galaxies.

 

When gravitational self-energy by ordinary matters is as below in our universe,

[math]{U_M} = - \frac{3}{5}\frac{{G{M^2}}}{r}[/math]

 

Because we don't know how big gravitational potential energy by antigravity is, let's introduce and indicate a constant [math]{k_h}[/math] which is easy for comparison as below, for a simple comparison.

 

[math]{U_{DE}} = {k_h}\frac{{G{M^2}}}{r}[/math]

B. Force generated by positive gravitational potential energy.

 

[math]\vec F = - \Delta {U_{DE}} = -

\frac{{\partial {U_{DE}}}}{{\partial r}}\hat r = - \mathop {\lim}\limits_{\Delta r \to 0} \frac{{{U_{DE}}(r + \Delta r) - {U_{DE}}®}}{{\Delta r}}\hat r[/math]

 

[math]{U_{DE}®} = {k_h}\frac{{G{M^2}}}{r} = {k_h}\frac{{G{{(\frac{{4\pi }}{3}{r^3}{\rho _r})}^2}}}{r} ={k_h}G{(\frac{{4\pi }}{3})^2}{\rho _r}^2{r^5}[/math]

 

[math]{U_{DE}}(r + \Delta r) = {k_h}G{(\frac{{4\pi }}{3})^2}{\rho _{r + \Delta r}}^2{(r + \Delta r)^5}[/math]

 

When considering the law of conservation of mass-energy,

 

[math]{\rho _r}{r^3} = {\rho _{r + \Delta r}}{(r + \Delta r)^3}[/math]

 

[math]{\rho _{r + \Delta r}} = {\rho_r}{(\frac{r}{{r + \Delta r}})^3} = {\rho _r}(1 - 3\frac{{\Delta r}}{r} +
6{(\frac{{\Delta r}}{r})^2}...)[/math]

 

[math]{\rho _{r + \Delta r}}^2 = {\rho_r}^2(1 - 6\frac{{\Delta r}}{r} + 21{(\frac{{\Delta r}}{r})^2}...)[/math]

 

[math]{(r + \Delta r)^5} = {r^5}{(1 + \frac{{\Delta r}}{r})^5} = {r^5}(1 + 5\frac{{\Delta r}}{r} + 10{(\frac{{\Delta r}}{r})^2} + \cdots )[/math]

 

[math]F = - \mathop {\lim }\limits_{\Delta r \to 0} \frac{{{k_h}G{{(\frac{{4\pi }}{3})}^2}[{\rho _{r + \Delta r}}{}^2{{(r + \Delta r)}^5} - \rho _r^2{r^5}]}}{{\Delta r}}[/math]

[math]F \approx - \mathop {\lim }\limits_{\Delta r \to 0} \frac{{{k_h}G{{(\frac{{4\pi }}{3})}^2}{\rho _r}^2{r^5}[(1 - 6\frac{{\Delta r}}{r} + 21{{(\frac{{\Delta r}}{r})}^2})(1 + 5\frac{{\Delta r}}{r} + 10{{(\frac{{\Delta r}}{r})}^2}) - 1]}}{{\Delta r}}[/math]

 

[math]F \approx - \mathop {\lim }\limits_{\Delta r \to 0} \frac{{{k_h}G{{(\frac{{4\pi }}{3})}^2}{\rho _r}^2{r^5}[(1 + (5 -6)\frac{{\Delta r}}{r} + (10 - 30 + 21){{(\frac{{\Delta r}}{r})}^2}) - 1]}}{{\Delta r}}[/math]

 

Let's mark it to the secondary term, for verification of model!

 

[math]F \approx - \mathop {\lim}\limits_{\Delta r \to 0} {k_h}G{(\frac{{4\pi }}{3})^2}{\rho _r}^2{r^5}[ -\frac{1}{r} + (\frac{{\Delta r}}{{{r^2}}})][/math]

 

[math]F \approx + {k_h}G{(\frac{{4\pi}}{3})^2}{\rho _r}^2{r^5}[\frac{1}{r} - (\frac{{\Delta r}}{{{r^2}}})][/math]

use to [math]\frac{{4\pi }}{3}{r^3}{\rho _r} = M[/math]

 

Therefore, the force by antigravity source which uniformly distributes

 

[math]\vec F = + (\frac{{4\pi G}}{3}){k_h}M{\rho _r}r(1 - \frac{{\Delta r}}{r})\hat r[/math]

As a [math]\vec F = + k\hat r[/math] shape, this force is repulsive force, and is proportional to r like dark energy.

 

If we assume that this force is the same as the existing force related to dark energy,

 

[math](\frac{{4\pi G}}{3}){k_h}M{\rho _r}r = \frac{1}{3}\Lambda M{c^2}r[/math]

 

[math]\Lambda = \frac{{4\pi G{k_h}{\rho_r}}}{{{c^2}}}[/math]

 

Here, the total mass, M was used, as the force by gravitational potential energy affects all particles in the three-dimensional sphere.

 

We can see that [math]\Lambda [/math] is also a constant, since [math]G,{k_h},{\rho _r}[/math] and c are all constants. This is an evidence of why cosmological constant related to the present dark energy looks like a constant.

 

Then, let's figure out constant [math]k_h[/math] from the current observation results, and verify whether [math]\Lambda [/math] calculated by us is a right value.

 

Not mass-energy, we are measuring a gravitational effect from the observation of the universe, and supposing the existence of mass-energy corresponding to the gravitational effect.

 

The ratio of magnitude of gravitational effects of the present dark energy and matters can be yielded as below.

 

[math]\frac{{DarkEnergy}}{{Matter}} \approx \frac{{72.1}}{{4.6}} = 15.67 = |\frac{{{U_{ - + }}}}{{{U_{ + + }}}}|[/math]

 

[math]{k_h} = 15.67 \times \frac{3}{5} = 9.40[/math]

 

[math]\Lambda = \frac{{4\pi G{k_h}{\rho_r}}}{{{c^2}}} = 3.64 \times {10^{ - 52}}[\frac{1}{{{m^2}}}][/math]

 

This value is in accord with the dimension of cosmological constant that is being inferred from the existing observation results, and is similar with the prediction, too

 

http://http://en.wikipedia.org/wiki/Cosmological_constant

=========

Thus, the current standard model of cosmology, the Lambda-CDM model, includes the cosmological constant, which is measured to be on the order of 10^−52 m^−2, in metric units.

=========

 

Anyways, we can see that the force generated from gravitational potential energy by antigravity has the same shape as the force by dark energy, and that it is possible to accurately explain its magnitude and repulsive effect.

 

Moreover, we figured out the secondary term to verify whether this model would be right,

 

The force generated from gravitational potential energy by antigravity can be indicated like

[math]F = (\frac{{4\pi G}}{3}){k_h}M{\rho_r}r(1 - \frac{{\Delta r}}{r}) = \frac{1}{3}\Lambda M{c^2}r(1 - \frac{{\Delta r}}{r})[/math]

Since the previous analysis of dark energy was proportional to radial distance r, we can find out whether the model is right or wrong using [math](\frac{{\Delta r}}{r})[/math] term of the above formula.

C. Meanings including proof

1) The essence of dark energy is antigravity.

2) The force by [math]U_{-+}[/math](GPE between negative mass and positive mass, positive gravitational potential energy) is

[math]\vec F = (\frac{{4\pi G}}{3}){k_h}M{\rho _r}r(1 - \frac{{\Delta r}}{r})\hat r = \frac{1}{3}\Lambda M{c^2}r(1 - \frac{{\Delta r}}{r})\hat r[/math]

shaped, and it is needed to conduct an observatory experiment of [math](\frac{{\Delta r}}{r})[/math] term, for verification of this model.

3) The above evidence explains why dark energy looks like a constant.

4) "Negative energy(mass)" and "antimatter with a possibility of generating antigravity" can be candidates of antigravity source.

5) The above evidence only considered forms of matters, but implies that dark energy is a function of time, considering radiation or the secondary term. There is a need of figuring out a relativistic formula including radiation.

6) Even if mass-energy of antigravity source is equal to mass-energy of gravity source, the repulsive gravity effect could be 15 times bigger than the gravitational effect resulting from gravity source. (Refer to paper)

7) We can answer the CCC(Cosmological Constant Coincidence) problem of "Why does dark energy have the similar scale with matters?". It is because it has the same gravitational effect as them.

8) We can consider a general shape, [math]U = k{r^n}[/math] (n is real number) as a cause for dark energy, and thus there is a high possibility that it is no accident that the above evidence is valid.

9) While the existing cosmological constant or vacuum energy is a concept not to conserve energy, gravitational potential energy is conserved.

--- Icarus2

The Change of Gravitational Potential Energy And Dark Energy in the Zero Energy Universe

http://http://vixra.org/abs/1110.0019

 

[icarus2]

I'm sorry. typo.

 

[math]\vec F = -\nabla {U_{DE}} = -\frac{{\partial {U_{DE}}}}{{\partial r}}\hat r[/math]

 

 

[elfmotat]

The Newtonian approximation of gravity with nonzero Cosmological Constant is well-known, and falls directly out of linearized General Relativity:

 

[math]S=\int \left [\frac{1}{8\pi G}|\nabla \phi|^2+ \left (\rho-\frac{c^2\Lambda}{4\pi G} \right ) \phi \right ]d^4x[/math]

 

This action yields the field equations for Newtonian gravity:

 

[math]\nabla^2 \phi +c^2\Lambda =4\pi G\rho[/math]

 

If you assume that all of the mass in the universe is just a single point particle at the origin of our coordinate system, i.e. [math]\rho =M\delta (\mathbf{r})[/math], then by applying Gauss' Law to the field equations with the point particle density (and assuming the equations of motion for a test particle are [math]\frac{d^2 \mathbf{r}}{dt^2}=-\nabla \phi[/math]) you arrive at the following force law between the gravitating mass [math]M[/math] and a test particle of mass [math]m[/math]:

 

[math]\frac{\mathbf{F}}{m}=-\frac{GM}{|\mathbf{r}|^3}\mathbf{r}+\frac{c^2\Lambda}{3}\mathbf{r}[/math]

 

Since our field equations are linear, the superposition principle applies and the force contribution from multiple point particles may be simply added with vector addition. From this, the shell theorem follows, etc., and we see that in a universe filled with complicated arrangements of matter ([math]\rho(\mathbf{r})[/math] can be any complicated function) all matter acts on all other all other matter according to the above force law. So clumps of matter which are close enough together will be attracted to each other, while clumps spread out far enough will actually repel each other.

 

This means that in the context of Newtonian gravitation, the presence of a positive nonzero Cosmological Constant is interpreted as a repulsive term in the force law. This means that, as far as Newton is concerned, gravity is indeed repulsive once you're far enough away from sources of gravitation (masses). So in this context "Dark Energy" is simply the result of a modification of Newtonian gravity. The observation that far-away galaxies are accelerating away from us is all explained by our force law.

 

 

 

In full-fledged General Relativity, the action is:

 

[math]S=\int \left [ \frac{c^4}{16\pi G}(R-2\Lambda )+\mathcal{L}_{matter} \right ]\sqrt{-g}d^4x[/math]

 

This yields the following (nonlinear) field equations:

 

[math]G_{\mu \nu}+\Lambda g_{\mu \nu }=\frac{8\pi G}{c^4}T_{\mu \nu}[/math]

 

From these field equations you can get exact solutions for the universe (using assumptions like isotropy and homogeneity) such as the FLRW metric. In this solution the spacial part of the metric (i.e. how spacial distances between points are determined) is dependent on a function of time [math]a(t)[/math] called the "scale factor." Plugging this solution into the field equations yields a few useful equations (called the Fiedmann equations), one of which is:

 

[math]\frac{\ddot a}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^{2}}\right) + \frac{\Lambda c^{2}}{3}[/math]

 

where [math]\ddot{a}=\frac{d^2}{dt^2}[a(t)][/math]. Here, we can see that ordinary energy density and pressure are acting to "contract" the distances between points (i.e. trying to make [math]\ddot{a}[/math] negative). On the other hand, the term involving the Cosmological constant is working to increase the distances between points.

 

We can also see from the Friedmann equations that the Cosmological Constant term acts as some sort of energy with negative pressure (which is dubbed "Dark Energy"). So in the context of GR, the Cosmological Constant is something which works to expand the universe itself.

 

 

 

The point of this post was to show you that your idea that Dark Energy is actually a form of "antigravity" actually does work, to a certain approximation. But what's "really" going on is that the distances between points in the universe is increasing. GR is a much more accurate model of gravity than Newtonian gravitation, especially when things are moving fast compared to light and when gravitational fields are very strong. But the effects of a nonlinear gravity theory are also apparent within our own solar system, for example with the precession of Mercury. Applying Newtonian gravity on the cosmological scale is bound to give lots of inaccurate predictions. Your idea works to develop an intuition about the expansion of space, but it's bound to fail if you try to use it to make predictions about the behavior of the universe.

Edited February 13, 2013 by elfmotat