Vacuum and repulsive gravity

Question

How can one show from General Relativity that gravity is attractive force, and under which conditions it becomes repulsive, also why positive energy vacuum drives repulsive gravity?

Answer 1

The Einstein field equations actually don't say anything at all about the nature of matter. Their structure is that they relate a certain measure of spacetime curvature G to the stress-energy tensor T: $G_{ab}=8\pi T_{ab}$. The stress-energy tensor describes any matter that is present; it's zero in a vacuum. Trivially, you can write down any equations you like describing an arbitrary spacetime that you've made up, and then by calculating G you can find the T that is required in order to allow the existence of that spacetime. However, that T may have properties that are different from those of any known type of matter. T has a very specific structure for certain types of matter, such as electromagnetic radiation or "dust" (meaning a perfect fluid made of particles that have velocities $\ll c$ relative to one another). There are various conjectures, called energy conditions, about what types of stress-energy tensors are physically possible for realistic types of matter. They have names like the weak energy condition (WEC), strong energy condition (SEC), etc. The WEC amounts to a statement that the energy density is never negative in any frame of reference. If it was violated, then you could get repulsive gravity. Basically all energy conditions are known to be violated under some circumstances. Here is a nice discussion of that: http://arxiv.org/abs/gr-qc/0205066

The cosmological constant is sometimes taken as a separate term in the Einstein field equations, but it can also be treated as a type of matter with a certain contribution to the stress-energy tensor. A spacetime with only a cosmological constant, and nothing else in it, violates various energy conditions.

Answer 2

We need a clear operational meaning about what it means for gravity to be "repulsive". If we think of it too naively... say a distant observer in Schwarzschild spacetime viewing a particle orbit, for whom Schwarzschild time $t$ has a clear operational meaning and Schwarzschild radial coordinate $r$ is good enough. Can the orbit have positive $d^2r/dt^2$, i.e., have outward rather than inward acceleration? Yes, absolutely: in fact, it must be the case, because in those coordinates a radially freefalling particle never reaches the horizon.

But that's just perverse. Let's look for a local operational definition. For example, take a small (near each other) collection of comoving test particles, four-velocity $u$, and see what they do. Their geodesic deviation is given by the Riemann curvature tensor, and if we have a small ball of volume $V$, then $$\lim_{V\to 0}\frac{\ddot{V}}{V}\vert_{t=0} = -R^\alpha{}_{\mu\alpha\nu} u^\mu u^\nu,$$ and this contracted Riemann tensor is the Ricci curvature $R_{\mu\nu}$. Thus, the attractive or repulsive behavior of the ball of test particles is given by the Ricci curvature.

Take a gander at the strong energy condition: for every future-pointing timelike vector $u$, $$\left(T_{\mu\nu}-\frac{1}{2}Tg_{\mu\nu}\right)u^\mu u^\nu \geq 0.$$ It's not immediately obvious as to what this actually means, but contracting the Einstein field equation $R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}$ gets you $R - \frac{1}{2}Rg^\mu{}_\mu = 8\pi T$, i.e., $R = -8\pi T$: the Ricci scalar is just the negative trace of the stress-energy tensor, up to a constant. Back-substituting into the field equation, and voila: $$R_{\mu\nu} = 8\pi\left(T_{\mu\nu} - \frac{1}{2}Tg_{\mu\nu}\right),$$ so the SEC just says that $R_{\mu\nu}u^\mu u^\nu \geq 0$.

Conclusion: for this reasonable local notion of what it means for gravity to be attractive/repulsive, gravity is non-repulsive if, and only if, the strong energy condition holds.

In the case of a perfect fluid, $T^{\mu\nu} = (\rho+p)u^\mu u^\nu + pg^{\mu\nu}$, so that $T = T^\mu{}_\mu = -\rho + 3p$. From the frame comoving with the fluid, substitution gives $\rho + 3p\geq 0$. On the other hand, approaching a lightlike $u$ in one of the spatial directions gives $\rho + p\geq 0$, this step being justified by continuity.

Since the Einstein field equation with a cosmological constant just adds a $\Lambda g_{\mu\nu}$ opposite the stress-energy tensor, it is equivalent to a perfect fluid with density $\rho = \Lambda/8\pi$ and pressure $p = -\Lambda/8\pi$. Thus, we need positive vacuum energy density to break the SEC.

Answer 3

In the context of General Relativity, small test particles move on geodesics. A geodesic is a generalization of a straight line (for example, on a curved surface like a football). The geodesics are determined by the metric $g_{\mu \nu}$.

The Einstein equation is $G_{\mu \nu} = 8 \pi T_{\mu \nu}$.

$G_{\mu \nu}$ consists of the metric $g_{\mu \nu}$ and its derivatives.

$T_{\mu \nu}$ is matter, and has specific forms depending on what matter is present, like dust, radiation, or lambda. You can think of $T_{\mu \nu}$ as the input to the problem, like a heavy star sitting at the center of empty space.

Due to the concrete form of $G_{\mu \nu}$ as written down by Einstein, when you solve the equation for dust or radiation, $g_{\mu \nu}$ will be such that test particles will appear to be attracted to the matter described by $T_{\mu \nu}$. For a heavy star sitting at the center of empty space, the geodesics of test particles coming from very far away will bend toward the heavy star. This solution is actually famous for being the first explicit solution of a problem in the framework of General Relativity, and is called the Schwarzschild solution or Schwarzschild metric after Karl Schwarzschild.