Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is antigravity the source of accelerating expansion(dark energy)?

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.

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 $F = + kr$ shaped as a ${\vec F_\Lambda } =\frac{1}{3}\Lambda m{c^2}r\hat r$ shape, antigravity is $F = + \frac{k}{{{r^2}}}$ 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.

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.

${U_S} = - \frac{3}{5}\frac{{G{M^2}}}{r}$ ( r:radius, M: the mass of the sphere )

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,

${U_M} = - \frac{3}{5}\frac{{G{M^2}}}{r}$

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

${U_{DE}} = {k_h}\frac{{G{M^2}}}{r}$

B. Force generated by positive gravitational potential energy.

$\vec F = - \nabla {U_{DE}} = -\frac{{\partial {U_{DE}}}}{{\partial r}}\hat r = - \mathop {\lim}\limits_{\Delta r \to 0} \frac{{{U_{DE}}(r + \Delta r) - {U_{DE}}(r)}}{{\Delta r}}\hat r$

${U_{DE}(r)} = {k_h}\frac{{G{M^2}}}{r} = {k_h}\frac{{G{{(\frac{{4\pi }}{3}{r^3}{\rho _r})}^2}}}{r} ={k_h}G{(\frac{{4\pi }}{3})^2}{\rho _r}^2{r^5}$

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

When considering the law of conservation of mass-energy,

${\rho _r}{r^3} = {\rho _{r + \Delta r}}{(r + \Delta r)^3}$

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

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

${(r + \Delta r)^5} = {r^5}{(1 + \frac{{\Delta r}}{r})^5} = {r^5}(1 + 5\frac{{\Delta r}}{r} + 10{(\frac{{\Delta r}}{r})^2} + \cdots )$

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

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

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

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

$F \approx + {k_h}G{(\frac{{4\pi}}{3})^2}{\rho _r}^2{r^5}(\frac{1}{r})$

use to $\frac{{4\pi }}{3}{r^3}{\rho _r} = M$

Therefore, the force by antigravity source which uniformly distributes

$\vec F = + (\frac{{4\pi G}}{3}){k_h(t)}M{\rho _r}r\hat r$

As a $\vec F = + k\hat r$ 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, $(\frac{{4\pi G}}{3}){k_h (t)}M{\rho _r}r = \frac{1}{3}\Lambda M{c^2}r$

$\Lambda = \frac{{4\pi G{k_h (t)}{\rho_r}}}{{{c^2}}}$

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

Then, let's figure out constant $k_h$ from the current observation results, and verify whether $\Lambda $ 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.

$\frac{{DarkEnergy}}{{Matter}} \approx \frac{{72.1}}{{4.6}} = 15.67$

${k_h} = 15.67 \times \frac{3}{5} = 9.40$

$\Lambda = \frac{{4\pi G{k_h}{\rho _r}}}{{{c^2}}} =\frac{{4 \times 3.14 \times (6.67 \times {{10}^{ - 10}}{m^3}k{g^{ - 1}}{s^{ -2}}) \times {k_h} \times (4.17 \times {{10}^{ - 28}}kg{m^{ - 3}})}}{{9 \times{{10}^{16}}{m^2}{s^{ - 2}}}}$

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

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

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

$F = (\frac{{4\pi G}}{3}){k_h(t)}M{\rho_r}r = \frac{1}{3}\Lambda(t) M{c^2}r$

Since the previous analysis of dark energy was proportional to radial distance r


1) We can consider a general shape, $U = k{r^n}$ (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.

2) 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.

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

Is antigravity the source of accelerating expansion(dark energy)? What do you think of that?

share|improve this question
    
For more on anti-gravity, see Wikipedia, this Phys.SE post, and this MetaPhys.SE post. –  Qmechanic Feb 25 '13 at 19:29
4  
@user12944: It seems like your calculations are Newtonian. It would be important to have a GR based formulation of any phenomenon for it to be acceptable. For eg: if there's any "antigravity", how does it affect Einstein's equation?, What happens to the stress-energy tensor?, etc. –  Siva May 17 '13 at 23:35
4  
Just some notes on presentation: Why all the $\Delta r$ gibberish? If you're just learning calculus this it is fine to do things like this for yourself until you are competent with derivatives, but please don't show it to the rest of us. Differential calculus was invented for a reason... it makes your argument much clearer. It's trivial for most of us to verify if you've taken an ordinary gradient correctly, but I don't want to read of mess of uneccesary $\Delta r$s. It's not impressive either - physicists are impressed by the simplest possible calculations. –  Michael Brown May 18 '13 at 1:25
2  
Here you don't need to do any calculations at all... the two potentials add together and simply renormalise the effective value of $G$, so you clearly made a mistake somewhere. But I'm recovering from a stomach bug so I'm not going to try and track your mistake. –  Michael Brown May 18 '13 at 1:26
4  
Also, the force formula for dark energy you have seen is really quite a fudge. It's best not to take it too literally. You need relativity to really understand this stuff. If you have a new and better theory than general relativity please go ahead and get it published in a reputable journal... but here is not the place to show your work. –  Michael Brown May 18 '13 at 1:31

1 Answer 1

The facts:
No antigravity has been observed ... and no DE has been observed also.
It was discovered (measured) in 1998 an acceleration in the expansion of space.

The mass (M) is the source of gravity (G). Every force has a source of it, charges, masses,... .
The source of antigravity (AG) is ??????????? (SoAG).
In the absence of apparent source of AG we can assume that antigravity was imprinted in space long time ago and the sources disappeared. It is as the SoAG were converted in AG preserving the energy balance.
In this way, and it is all I can conceive, the AG is a field, and it is the gradient of field that produces a force.
In a expanding space it is expectable that any gradient will have to decay, the same as with the decrease of photon energy, by the reddening of light, as time goes by.
But it is the contrary effect that is needed: to motivate an accelerated expansion the gradient must grow. How can it be growing ? Can we imagine any probable cause? We have the same problem with the space expansion: no probable cause.
To name the source of the effect as DE or AG is not important because none offer a solution. They are just names of the probable causes of the symptoms (the measures we have).

There are sources of gravity so distant of us that the associated gravity field is reaching Earth only now. In physics it is strongly beleived that gravity propagates away of sources at c speed. Following Einstein we have no reason to doubt.
I've no way of convincing myself that the distant stars have the same energetic content now, billions of years past the moment they released the gravitational field, that only now we sense. If here and now the sorrounding energy content is different than here in the past then I will have to assume that the source of this change (the distant stars) have to change also as time go by.
If it is not so then the SoAG can not have disappeared.

Busted by the mesuring process ?
I am saying that stars in the past had more massive atoms than the ones we see around and the mass is beeing converted in gravitational/electrostatic fields as field spread away. I know that it is hard to beleive, but can you think that the universe is still viable if all atoms have the double of the mass thay show now? Of course that they have to be also twice as large. And, if c is constant, the time unit is also twice as large. The light emmited in the past by those larger atoms have longer wavelengths (red) compared to the light emmited in the same process.
We measure the world with atoms, i.e. properties derived from them and this fact make impossible an independent measure in the lab, because we measure ratios and those are kept constant. The measured space expansion is an artifact of the fact that the 'rule' we use is shrinking: for the same physical distance we have now the double of the measure if the atoms shrinked to half.
The law that relates the evolution of the source is naturally exponential (think of radioactivity) and that gives us the illusion of an accelerated expansion.

With this model A self-similar model of the Universe unveils the nature of dark energy (pdf), there are no free lunches at all, no DE, no inflation era, no space expansion, no... new particle, no new force, no new field.
With my modest contribution a new model is presented formally, with all equations you like, with all physical reasonings you deserve, and compared with the official $\Lambda CDM$. I still have expectations that sometime, someone will present argumentation ($*$) and I will be around to discuss the subject.
($*$) as usual downvotes are expected but, in the absence of any argument, I can consider it an implicit acceptance of impotence.

At least I expect that the author of the question will understand why I'm frontally against any kind of AG/DE, and I hope he/she keeps the motivation in the search of the hidden reality.

PS:
GR is a gravitational theory only (mass/field problem). In a spherical symmetry the distance between two points have two solutions: the short one and the other one (ex: I can go directly from London to Casablanca or using the same meridian go from London to North pole, then thru the equator, next South pole and finally Casablanca). Einstein formulated GR thinking that the matter was ageless and, as part of the solution he incorporated the $\Lambda$ to account for the longer path. I've several problems with the appropriation of GR to other purposes that gravitation because: The more we measure the space at large the more 'flat' it is. It his accepted the matter had a beginning and the GR model was consistent with an ageless universe. After 1998 $\Lambda$ was refurbished to explain DE, without explaining a single bit. The last reason is that data agrees with the model of shrinking matter without need of $\Lambda$ (It has only one parameter $H_0$) and BBT has six (?) and by WP "and have made the unexpected discovery that the expansion of the Universe appears to be accelerating." and I read "unexpected discovery" == "wrong model".

share|improve this answer
    
Atoms were bigger and had more mass and as a result, every frequency got shifted down? In all of this, how did the fine structure constant stay constant? –  Brandon Enright May 18 '13 at 3:45
    
@Brandon: pag 10, table I, dimensionally fine structure y is a constant( for a M,L,T equal change) but Planck constant is not a constant. Bigger atoms -> larger electronic orbitals ,for equal $c$ more time is needed for a round trip equals a lower frequency. –  Helder Velez May 18 '13 at 4:24
    
@Brandon ; I meant 'dimensionless' instead of the word 'constant' –  Helder Velez May 18 '13 at 4:35
2  
Electrons do not orbit atoms in the classic sense of something orbiting. Also, hydrogen spectral lines (how the redshift is measured) are about the difference in energy for an electron orbital. Have you worked out the math to show that changing the mass of particles makes bigger atoms or changes emission lines in the way you claim? –  Brandon Enright May 18 '13 at 4:37
2  
"as usual downvotes are expected but, in the absence of any argument, I can consider it an implicit acceptance of impotence." No. This is why: from the FAQ: "Pitches for your own personal theories or work We deal with mainstream physics here. Anything that couldn't be published in a reputable journal is not appropriate on this site." You're just trolling with your snipes. BTW, I'm not one of the down-voters. I like crazy theories. –  Michael Brown May 18 '13 at 5:05

protected by Qmechanic May 18 '13 at 0:07

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.