The Newtonian vacuum field equation $\nabla^2 \phi = \rho$ where $\phi$ is the gravitational potential and $\rho$ is proportional to mass density also has non-trivial vacuum solutions, for example $\phi = -1/r$ for $r$ outside some spherical surface. The Maxwell equations also have non-trivial solutions. In electrostatics, precisely the same as classical gravitation, in elctrodynamics, also radiative solutions of various kinds.
It is not strange that a field theory has non-trivial vacuum solutions. From a mathematical point of view, if it did not, it would not be possible to solve boundary value problems otherwise. Physically, a (local) field theory is supposed to provide a way for spatially separated matter to interact without spooky action at a distance. If interactions were unable to propagate through a region of vacuum we would have a very boring field theory!
If we want to be a little more specific to general relativity, let us note that this theory actually consists of two field equations. The most famous one is Einstein's, $$
R_{\mu\nu} = 8\pi T_{\mu\nu}
$$
which says that matter is the source for the field $\Gamma^{\mu}{}_{\nu\sigma}$ -- the Christoffel symbols. This equation alone does not contain the fundamental characterization of general relativity. It is just an equation for some field. For this field to actually correspond to the curvature it must also satisfy the Bianchi identity $$
R_{\mu\nu[\sigma\tau;\rho]} = 0.
$$
The Bianchi identity is redundant if the Christoffel symbols are defined the way they are in terms of the metric. This is actually analogous with electrodynamics (and for a very good reason, because ED is also a theory of curvature). The Maxwell equations are $$F^{\mu\nu}{}_{,\nu} = j^\mu $$ $$F_{[\mu\nu,\sigma]} = 0$$
and the first equation is the one that couples the electromagnetic field to matter. The second equation is redundant if $F_{\mu\nu}$ is defined in terms of the vector potential.
Now, the electromagnetic field has 6 components but as you can see only 4 of them really couple to matter directly. The second equation represents the freedom for the electromagnetic field to propagate in vacuum. (In fact if you do Fourier analysis to find radiation solutions to the Maxwell equations the first only tells you that radiation is transverse, and the second is the one that actually determines the radiation.) The components are naturally not independent since matter and radiation interact, but I think that this is a nice way to think about why of the classical Maxwell equations $$\begin{matrix}
\nabla \cdot \mathbf{E} = & \sigma\\
\nabla \cdot \mathbf{B} = & 0 \\
\nabla \times \mathbf{E} = & -\frac{\partial \mathbf B}{\partial t} \\
\nabla \times \mathbf{B} = & \mathbf{j} + \frac{\partial \mathbf E}{\partial t}
\end{matrix}$$
two involve only the fields and two involve matter.
Similarly for Einstein's general relativity, in the Einstein field equation $$R_{\mu\nu} = 8\pi T_{\mu\nu}$$ matter only couples to 10 components out of the 20 components in the Riemann curvature tensor. (The Riemann tensor is the physically observable quantity in general relativity.) The other 10 components are in the Weyl tensor. They are the part of the gravitational field that is present in vacuum, so they must include at least the Newtonian potential. By analogy with electrodynamics they also include gravitational radiation.
In the specific case of the Schwarschild and Kerr metrics, not only are all the components of the Ricci tensor 0, one can in fact arrange for all the components of the Weyl tensor except one to be 0 also. This is sort of analogous to how in electrostatics you can always choose the gauge so that the vector potential $\mathbf A = 0$. Perhaps you can think of this as saying that these metrics do not radiate, so only the part of the gravitational field whose limit is the Newtonian potential exists. (But there are radiating metrics with the same property, so maybe this isn't a good way to think.).
There are other vacuum metrics where fewer of the Weyl tensor's components can be made 0, or some gauge freedom remains. It is common to classify metrics along this scheme, which is called Petrov type. In a really famous paper Newman and Penrose show that the Petrov type of gravitational radiation has a near field - transition zone - radiation zone behavior, where more components of the Weyl tensor become irrelevant the further away from the source you go. (This is analogous with electrodynamics again, since in the radiation zone the EM field is transverse, but in the near field it is not.)
The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).
- No, it's not in the Ricci tensor. Any smooth manifold admits a connection $\nabla$, and with respect to this connection we can consider its curvature tensor $R^a_{\,bcd}$, and from this by taking a contraction we can define the Ricci tensor $R_{ab}= R^s_{\,asb}$. And connections can be very arbitrary, not necessarily even arising from a pseudo-Riemannian metric.
- I don't see any direct links.
- No it's not in the stress energy tensor. For instance, suppose we have an arbitrary pseudo-Riemannian metric $g$ (arbitrary signature $(p,q)$). We can still consider its Einstein tensor $R_{ab}-\frac{1}{2}Rg_{ab}$. By setting this to $0$, we see that $\Bbb{R}^4$ with the standard flat pseudo-Riemannian metric $\eta_{p,q}$ is a solution. However, for this to bear any resemblance to (special) relativity we need a $(1,3)$ signature split.
- See (3).
- Yes (though I'd say it's an explicit assumption).
- You mentioned (5), so you're not looking at it wrong.
Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the ansatz already imposes Lorentz signature.
Best Answer
That article's choice of words could certainly be improved. Basically yes, $T$ captures all the non-gravitational "stuff."
The idea of a "gravitational field" doesn't really fit in to GR. There is stress-energy $T$ everywhere, and there is a metric $g$ everywhere, and that's really all you need to define what exists.
The article is trying to say that even if spacetime is more or less empty of stress-energy, there is a potential for stuff to happen thanks to the metric being nontrivial. Really, this is nothing new -- in Newtonian gravity, two separated bodies have as a system a gravitational potential energy that doesn't really reside anywhere. Trying to localize this potential "energy" in space is more difficult/ill-defined in GR than in Newtonian gravity.
In GR you can have two isolated masses sitting in space initially at rest. The stress-energy tensor is nonzero only in the regions occupied by the masses. Exterior to the masses the homogeneous Einstein equation is the same as that for empty Minkowski space, but the solution depends on the boundary conditions imposed by the masses and is in fact not Minkowski but rather some nonlinear superposition of Schwarzschild solutions. The nontrivial nature of the metric (which one could perhaps misleadingly call a "gravitational field") means the masses will start to move toward one another.
Another example is gravitational waves. These are again solutions to the homogeneous Einstein equation (i.e. the equation in vacuum), but they are nontrivial solutions. The metric is not just Minkowski.
To see this more mathematically, one could turn to the ADM equations of motion. Take a hypersurface of constant timelike coordinate $t$. The induced $3$-metric on this surface has components $\gamma_{ij}$ and "conjugate momentum" $\pi^{ij}$. These can be written in terms of $g$ without reference to $T$. Then there are known equations for $\partial_t \gamma_{ij}$ and $\partial_t \pi^{ij}$. In particular, only specially contrived setups will have $\partial_t \gamma_{ij} = 0$.