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Why does a mass attract all the masses around it? Why should't it repel or just stay calm? Why should it be like that?

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possible duplicate of Why is gravitation force always attractive? – jinawee Aug 8 '14 at 9:03
@jinawee: I don't think that is a duplicate because the OP is not just asking why the force can't be repulsive, but also why it can't be zero. – John Rennie Aug 8 '14 at 9:53
up vote 5 down vote accepted

Gravity may be treated as a quantum field theory. In this kind of theory, interactions are represented by field correlations, more known as "virtual particles", "virtual gravitons" in the case of gravity. The fact that two charges (more precisely, in the case of the gravitation, $2$ positive energy densities) attract each other is due to the sign associated to a quantity named propagator which describes these field correlations. The sign of the propagator depends on the relation between the Lagrangian and an invariant LorentZ quantity, described below, with the constraint that the time derivative of the spatial components of the fields get a positive sign in the Lagrangian.

Take a metric $g_{ij} = Diag (1,-1,-1,-1)$. This metrics is used to raise or lower indices.

With a scalar field (spin 0), the invariant lorentz quantity is $\partial_\mu \phi \partial^\mu \phi = (\partial_0 \phi)^2 - \sum\limits_{i=1}^3(\partial_i \phi)^2$, the Lagrangian is then $L = + \partial_\mu \phi \partial^\mu \phi $

With a spin one field (electromagnetic field), the invariant Lorentz quantity is proportionnal to $(\partial_\mu A_\nu - \partial_\nu A_\mu) (\partial^\mu A^\nu - \partial^\nu A^\mu) = - 2 \sum\limits_{i=1}^3 (\partial_0 A_i)^2 + ....$

The minus sign here comes from $g^{ii}=-1$, for $i=1;2,3$. We must have a positive sign for the time derivatives of the spatial components, so we have to put a minus sign, that is $L $ proportionnal to $- (\partial_\mu A_\nu - \partial_\nu A_\mu) (\partial^\mu A^\nu - \partial^\nu A^\mu)$

With the graviton, which has spin $2$, and is representing gravity interactions, the interesting part of the invariant Lorentz quantity is : $(\partial_\mu h_{\nu \rho} \partial^\mu h^{\nu \rho}) = + \sum\limits_{i,j=1}^3 (\partial_0 h_{ij})^2 + ....$

The $+$ sign comes from : $g^{ii} g^{jj} = (-1)^2= +1$, so the Lagrangian is $L = \partial_\mu h_{\nu \rho} \partial_\mu h^{\nu \rho}$ + other terms

This game with the sign has a direct physical consequence. When you have a $+$ sign, charges of same nature attract, and when you have a $-$ sign, charges of same nature repell. So, positive energy densities ("masses") attract each other by gravitational interaction.

This is the general idea. Well, I made some simplifications... , for a complete discussion, see : Zee (Quantum Field Theory in a nutshell), Chapter 1.5.

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The masses can't repel each other because gravity is mediated by a spin 2 field, and for spin 2 the force between charges of equal signs is attractive. See the question Why is gravitation force always attractive? for an explanation of this.

But it's impossible to say why the force can't be zero. Experiment shows that masses do attract each other, and General Relativity tells us how masses attract each other. But as to why this happens, that's essentially asking why the value of Newton's constant, $G$, isn't zero. There is no answer to this other than the obvious one that if $G$ was zero we wouldn't be here to measure it. We don't have any theory that predicts the value of $G$.

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Isn't it provable that spin 2 particles like the graviton generate attractive forces between object of similar charge? – Whelp Aug 8 '14 at 8:44
@Whelp: for spin 2 bosons the force between charges of equal sign is attractive, but that doesn't mean the force can't be zero, only that it can't be repulsive. – John Rennie Aug 8 '14 at 8:49

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