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Is anti-gravity possible in string theory?

I have read some articles about scientists making assumptions about the existence of anti-gravity, but is it possible in string theory?

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Related: physics.stackexchange.com/q/11542/2451 –  Qmechanic Aug 18 '12 at 18:35
Some supergravity theories have a graviphoton (spin-1 vector field) which could provide a repulsive gravitational force and therefore lead to antigravity, see here. –  Dilaton Dec 6 '12 at 14:38
What is anti-gravity? It seems that it makes no sense in the same way anti-space or anti-time, or anti-spacetime make no sense. –  MBN Dec 6 '12 at 14:48
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1 Answer

up vote 3 down vote accepted

As it happens we do observe anti-gravity in the form of the dark energy. There are speculative ideas to describe this from string theory, though these are far from widely accepted.

In general relativity the curvature of space time is controlled by the stress-energy tensor. To get anti-gravity, or more precisely gravitational repulsion, you just need an appropriate stress-energy tensor. The problem is that normal matter cannot create a stress-energy tensor of the required form: instead you require exotic matter and we have never observed this in a lab.

However you have probably heard of dark energy, which is causing the expansion of the universe to accelerate. This does indeed create a gravitational repulsion. The trouble is that we have no idea what dark energy is.

You specifically mentioned string theory in your question, and one of the ideas from string theory is the string landscape. One possible scenario from this is that our universe is in a metastable state, and if so this would cause a cosmological constant that behaves like the dark energy we have observed.

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but should not gravity depend on the curvature ¿ if the curvature is negative then we would have anti-gravity –  Jose Javier Garcia Aug 20 '12 at 21:37
Are you asking why dark energy causes a repulsion when the universe is flat? If so, I'd post this as a new question because it's quite an interesting point and a bit too involved to answer in a comment. –  John Rennie Aug 21 '12 at 5:52
In more than two dimensions, 'curvature' is a bit more complicated than just a number. The strong energy condition is equivalent to saying 'gravity is locally attractive' or that the Ricci curvature tensor satisfies $R_{\mu\nu}u^\mu u^\nu \geq 0$ for every future-pointing timelike $u$. –  Stan Liou Dec 6 '12 at 16:08
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