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Why gravity is an attractive force?


One may say that it is because of space time curvature but General Relativity is built on this law: $\displaystyle G \frac{m_1 \times m_2}{r^2}$ (To be more precise, it is derived from it's potentiel form known as Gauss's law for gravity that can be written like this: $\nabla \cdot g = - 4\pi Gp $). So GR can't explain why gravity is attractive. General Relativity only explains how gravity occurs in terms of space time.


Another may stick with the Quantum theory stating that gravitons are responsible for the force of gravity, but why a massless particle with spin 2 will exchange positive momentum?

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Possible duplicate: physics.stackexchange.com/q/11542/2451 –  Qmechanic Sep 29 '13 at 14:46
    
@Qmechanic I think that my question is a bit different from this physics.stackexchange.com/q/11542/2451 –  user29727 Sep 29 '13 at 14:51
    
@Qmechanic Should I edit the question as follows: "Do General Relativity and Quantum Theory explain why gravity is attractive?" –  user29727 Sep 29 '13 at 14:56
    
Which quantum theory of gravity are you referring to? –  Qmechanic Sep 29 '13 at 14:57
    
@Qmechanic I'm not referring to a Quantum theory of gravity but to the standard model of particle physics. –  user29727 Sep 29 '13 at 17:32
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marked as duplicate by Qmechanic Oct 2 '13 at 14:35

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Read A.Zee, Quantum Field Theory In a Nutshell, Princeton, Chapter I.5, p 30 (first edition)

In Quantum Field Theory, "forces" between 2 "charged" particles correspond to an exchange of "virtual gauge bosons". For instance, the repulsive force between 2 electrons, corresponds to an "exchange" of a "virtual photon" (a perturbation of the photon field). Here the gauge boson is the photon, of spin 1.

We may consider a graviton theory as a QFT, in this case, the charge is the momentum/energy of the particle, and the gauge boson is the graviton, of spin 2.

In these theories, you have to write a Lagrangian, and you have to respect a sign coherence about the euclidean action which has to be positive. This constraint gives you the correct sign for the Lorentz-invariant Lagrangian.

Now, you may calculate the interaction energy between 2 "charged" particles. In fact, one use currents instead of charges. So, for instance, for Quantum Eelectrodynamics, the interaction energy is a functional of currents and a gauge boson quantity named propagator.

For a graviton theory, the "current" is the stress-energy tensor, and the propagator is the graviton propagator.

The constraints about the sign of the action I speak above, have a direct consequence on the sign of the propagator. The consequence of that, is that particles with same charge exchanging a particle of odd spin, repeal each other, while particles with same charge exchanging a particle of even spin, attract each other.

So, for a graviton theory, with spin 2, we see that particles with same charge attract each other.

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It's not true to say that gravity is always attractive. The discovery of dark energy provides experimental evidence for gravitational repulsion. There is also indirect evidence for cosmic inflation, which currently holds the record for gravitational repulsion.

However it is true to say that the force between positive masses is always attractive, in contrast to electromagnetism where charges of the same sign repel. This is a consequence of the spin 2 of the gravitational field, as discussed in Trimok's answer. No-one has ever observed negative mass, and there are some pragmatic reasons for doubting it exists, but nothing in GR prevents you from putting in the matter with a negative sign.

Your comments on GR are at best misleading. GR equates curvature (strictly speaking the Einstein tensor) to the stress-energy tensor, and mass (strictly speaking energy density) makes up only one part of this tensor. Pressure also contributes to the stress-energy, and indeed dark energy creates a gravitational repulsion because it behaves as if it had a negative pressure.

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GR can't explain why space time curvature will cause two massive bodies to attract each other. –  user29727 Sep 30 '13 at 17:42
    
@Adobe: the two massive bodies follow geodesics, but because spacetime has been curved the two geodesics converge. That's how GR describes the gravitational force between the two masses. –  John Rennie Sep 30 '13 at 18:47
    
GR is based on a law that tell us that the gravitational force is attractive. So it can't explain why gravity is attractive. The only thing that it can explain is, as you said, how [...] describes the gravitational force between two masses. –  user29727 Sep 30 '13 at 18:49
    
It's not clear why you say "GR is based on a law that tell us that the gravitational force is attractive". GR is a conjectured theory that embodies various plausible principles like general covariance and has been (so far) confirmed by experiment. It is not based on, nor derived from, Newton or Gauss' theories of gravity. Rather the reverse, since the Newton and Gauss laws are low speed/density limits of GR. –  John Rennie Oct 1 '13 at 7:44
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No, that blog post derives the field equations from the Hilbert action. The field equations and the Hilbert action are alternative but equivalent ways of expressing GR. The blog just shows you how to get one given the other. –  John Rennie Oct 1 '13 at 18:28
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Gravity may be repulsive to anti-matter. There are many experiments carried in a variety of groups that try to study the effect of gravity on anti-matter (I saw that in the last SPS and APS conference, in the particle physics seminars). The adventure with gravity has just started, and we still need a lot of experimental data to give an informed decision on why gravity is only attractive.

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Anti-matter is just like matter, i.e, baryonic matter. Why would it be repulsive since the only difference between AM and M is charge? Does it say something about the nature of charge? –  user29727 Sep 29 '13 at 14:44
    
That's very not true. There's a fundamental difference between matter and anti-matter. Have you heard about the concept called CPT symmetry which is proven to hold with Lorentz invariance, and the experimentally proven concept called CP Violation? Matter is equivalent to anti-matter only in a world reversed in space parity and time symmetries. But in general, if you use anti-matter in our world, you could have different dynamics. Read more about CPT Symmetry and CP Violation. –  The Quantum Physicist Sep 29 '13 at 14:56
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If gravity is repulsive to antimatter, you could use that fact to create a perpetual motion machine--create a positron on the ground, let it "fall" upward into the upper atmosphere, have it annhilate with an electron, and send that energy back down to the surface, with the photons blue-shifting. The end result is that you'd have more than the $2mc^{2}$ of rest energy that you started out with. –  Jerry Schirmer Sep 29 '13 at 15:42
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@SamerAfach: testing the hypothesis is a valid useful thing to do, of course. There are strong theoretical reasons to expect a null result to that experiment, as stated above. Additionally, anti-quarks contribute to the parton functions for protons and neutrons, and contribute in different amounts for different nuclei. If anti-quarks were repelled by the Earth's gravitational field, you would find a different "weight" for different nuclei. –  Jerry Schirmer Sep 29 '13 at 16:47
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And snide crap aside, do you have an answer for the above objection? That construction IS a perpetual motion machine, it's not as stupid as one--if antimatter is repelled by ordinary matter, you can create a perpetual motion machine. –  Jerry Schirmer Sep 29 '13 at 16:48
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