# Physical meaning of non-trivial solutions of vacuum Einstein's field equations

I'm amazed that there are all of these answers here and no one has pointed out that a spacetime containing nothing but (caustic-free) gravitational radiation will be singularity-free and will be Ricci flat but not Riemann flat, which is a cleaner example than the Schwarzschild metric.

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.)

If you consider first the Schwarzschild metric describing the space outside some sphere on non-zero radius, the Ricci curvature is indeed zero outside the sphere and indeed the stress-energy tensor is also zero there.

The black hole is the limit of taking the radius of the sphere to zero. In that case both the Ricci curvature and stress-energy tensor are zero outside the singularity, but mass can still be present because the Ricci tensor is undefined at the singularity, which is the only place that the stress-energy tensor is non-zero.