# Could dark energy be negative gravity?

Main question: Could dark energy (the mysterious accelerating expansion of the universe) be explained by "negative gravity"?

"Spin off" questions:

1. Does antimatter have negative gravity?
2. If antimatter has negative gravity, and is present in the universe with equal amounts of matter and antimatter, would it produce the observed amount of dark energy in the universe?

My theory that I am trying to validate:

Gravity is the flow of space into matter, and it flows out of antimatter. If this is true, antimatter should have "negative gravity", I would expect it to have a repelling effect, while matter has an attracting effect. For example, if antimatter were on earth (somehow without annihilating), it would fall to the planet. However, if there were an antimatter planet, objects would "fall up" off it. In fact, I believe the antimatter planet would break apart because of this quality. Furthermore, I don't believe antimatter planets or stars would form because of this quality (e.g. it would remain anti-hydrogen). I also theorize that it would be in parts of the universe where matter isn't (e.g. in the voids outside the galactic clusters), and would be very diffuse (because it would repel itself). I believe the theorized repelling effect of antimatter could be what we are observing as dark energy (hence why I am asking these questions). I just searched for this online to see if anyone else has this theory, and yes, someone has something very similar here: http://www.universetoday.com/84934/antigravity-could-replace-dark-energy-as-cause-of-universes-expansion/

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You're not "theorizing", you're speculating. You'd be theorizing when you propose a testable mathematical model. –  Florin Andrei Jun 21 '14 at 1:21

Could dark energy (the mysterious accelerating expansion of the universe) be explained by "negative gravity"?

But it already is "negative gravity". In general relativity, the stress-energy tensor $T_{\mu\nu}$ describes the energy, momentum, and stress of matter in spacetime. Through the Einstein field equation, it is connected with Ricci curvature $R_{\mu\nu}$, which consists of half of the twenty independent degrees of freedom of spacetime curvature.

The intuitive geometrical meaning of Ricci curvature is as follows: suppose you have a small ball of test particles, initially all at the same velocity with respect to one another, and allow them to fall freely under gravity. Then the Ricci curvature (contracted with their intitial velocity) measures the initial acceleration of the volume of the ball of test particles divided by the initial volume. Thus, this is a good way to tell if something is locally "attractive" or "repulsive", based on whether a ball of test particles around it starts shrinking or expanding.

Through the Einstein field equation, stress-energy is connected to Ricci curvature. As a result, the volume of the small ball of test particles shrinks proportionally to $\rho+3p$, where $\rho$ is the energy density and $p$ is the average of the principal stresses--for a perfect fluid, the pressure. Naturally, if $\rho+3p < 0$, this is "negative gravity" in the sense that the test ball will start expanding instead. This quantity is also what's seen in the cosmological Friedmann equations. Since the Friedmann-Robertson-Walker family of solutions assume that the universe is homogeneous and isotropic on the large scale, it's sensible that treat the large-scale the same as the above local condition.

Dark energy has $\rho+3p<0$, and in particular (for cosmological constant), $\rho = -p > 0$.

To recap: (1) in general relativity, gravity depends on the stress-energy tensor, which has more than just the mass-energy density, and numerically pressure is more important because there are three spatial dimensions while only one time dimension, (2) we can characterize locally "negative gravity" as $\rho+3p < 0$, or, more generally, as any the violation of the strong energy condition, because this characterizes whether a small ball of test particles is attracted or repulsed by the local stress-energy content.

Does antimatter have negative gravity?

Not in general relativity. Energy is the gravitational charge, and antimatter has positive energy; thus it should gravitate in the same way.

Gravity is the flow of space into matter, and it flows out of antimatter.

If you're thinking of some analogy with electric charge, with electric field lines flowing out of positive charge and into negative charge, antimatter has positive mass, which is required by quantum field theory.

... I just searched for this online to see if anyone else has this theory, and yes, someone has something very similar here: ...

Some particles, such as photons, are their own anti-particle. This has immediate implication for Villata's claim that particles and antiparticles behave oppositely in a (normal) matter-produced gravitational field: it predicts that gravitational lensing cannot happen, since if light is bent one way by a gravitational field, Villata's stance predicts it's also bent in the the reverse direction.

Your stance is different from Villata's, but the corresponding statement for your theory is that light cannot produce a gravitational field. That's a lot harder to specifically test for, but it's clear that it's massively inconsistent with general relativity.

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