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

According to Einstein, the space-time is curved and the origin of the curvature is the presence of matter i.e. the presence of the energy-momentum tensor $T_{ab}$ in Einstein's field equations. If our universe were empty (i.e. $T_{ab}=0$ and the cosmological constant $\Lambda$ is setted to be $0$) then I would expect only the flat solution to the vacuum field equations $$R_{ab}=0$$ Surprisingly there are non flat (or non trivial) solutions to the above equations, for example the Schwarzschild solution. This conflicts with the fact that the matter curves the spacetime, so what is the origin of the curvature for these non trivial solutions? I understand that mathematically $R_{ab}=0$ (Ricci-flatness) doesn't imply that the metric is flat, i.e. non trivial solutions are formally admissible, but I don't understand how this is explained physically.

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Ricci flatness doesn't mean no curvature. The Riemann tensor can still describe a curved space even when the Ricci tensor is zero. –  John Rennie Nov 21 '13 at 11:13
Yes I know this, but I don't understand what is the physical origin of this curvature in absence of matter. I've explained this in my question. –  Galoisfan Nov 21 '13 at 11:19
I think that the Schwarzschild solution is analogous to the 1/r potential of a point charge in electrostatics. In both cases, the solutions satisfy empty space versions their respective field equations (i.e., Einstein field eq. and Poisson eq.) except at the origin. –  higgsss Nov 21 '13 at 11:33
Yes, but the point is that Schwarzschild solution implies a curvature of the spacetime, where does this curvature come from? –  Galoisfan Nov 21 '13 at 11:37
@Galoisfan, surely you see that the Schwarzschild solution has a mass parameter and that the spacetime is flat when the that parameter is zero? Does that fact not illuminate the answer? –  Alfred Centauri Nov 21 '13 at 12:09
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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.)

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Robin Ekman: You answer starts out with: "The Newtonian vacuum field equation $\nabla^2 \phi = \rho$ where $\phi$ is the gravitational potential and $\rho$ is proportional to mass density [...]" -- Is this equation as such not more correctly called "the Newtonian field equation"? (p.s. I noticed a typo in the first paragraph, and in the penultimate.) –  user12262 Mar 29 at 7:42
Yes, that's correct, the 'vacuum' should not be there. –  Robin Ekman Mar 29 at 8:08

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.

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Morally speaking, in the Schwarzschild metric , the stress-energy tensor is a distribution... –  Trimok Nov 21 '13 at 11:40
@Trimok: if you say so :-) I'm at the limit of my comfort zone here! –  John Rennie Nov 21 '13 at 11:41
Why is it a distribution? –  MBN Nov 21 '13 at 12:47
@MBN, see Trimok's comment to his answer. –  Alfred Centauri Nov 21 '13 at 13:51
–  John Rennie Nov 21 '13 at 15:48

In addition to the other answers here, I would like to add that your intuition would have been right on target in lower dimensions ($2+1$ and $1+1$). There, $T_{ab}=0$ and $\Lambda=0$ does imply (locally) flat space (although nontrivial topological effects are still possible).

In $3+1$ (and higher) dimensions the geometry is not uniquely defined by Ricci tensor. Here, we can separate from the curvature tensor the fully traceless part: Weyl tensor that can be thought of as describing purely gravitational dynamical degrees of freedom.

The existence of such degrees of freedom is exemplified in the phenomenon of gravitational radiation.

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So, these gravitational waves curve the four dimensional space time independently from the presence of matter in the region that we are considering. As a sort of backgrond curvature.... –  Galoisfan Nov 21 '13 at 15:02
Not independently. The field equations are nonlinear, after all. So the 'matter' and 'space-time geometry' interact. –  user23660 Nov 21 '13 at 15:14
In quantum language this means that a material system can absorb graviton increasing its energy -- we have conversion of purely gravitational entity into matter. Reversely, there is nonzero (although very small) probability that electron-positron pair annihilate into a pair of gravitons: we have conversion of matter into pure gravity. –  user23660 Nov 21 '13 at 15:21

A point is that the stress-energy tensor $T_{ab}$ is not conserved (see Wiki). This is because gravitation has its own stress-energy (pseudo) tensor $t_{ab}$, even if we may always choose a frame such as $t_{ab}=0$ for some particular space-time point. Only the sum of the two, (up to a factor $g$ if we take the Landau–Lifshitz pseudotensor) is conserved : $(-g (T^{ab} + t^{ab}_{LL})),_b=0$

If we look at the complicated expression of $t^{ab}_{LL}$, one sees that it involves quadratic quantities of first derivatives of the metric tensor $g_{ab}$ (and the metric tensor itself and its inverse)

The Riemann curvature $R_{abcd}$, by definition, depends on second derivatives, and quadratic quantities of first derivatives, of the metric tensor (and the metric tensor itself and its inverse).

So, if we consider a "vacuum" solution $R_{ab}=0$, so $T_{ab}=0$, but with $t^{ab}_{LL}$ not being zero and not being constant, for instance gravitational waves, this means that the first derivatives of the metric tensor are not zero, and there is no general reason, why the Riemann curvature $R_{abcd}$, made of first and second derivatives of the metric tensor, should be zero (except trivial cases).

[EDIT]

Rigourously speaking, one has to compute the freedom degrees for the curvature tensor $R_{abcd}$ and the Ricci tensor $R_{ab}$. It turns, that, in a space-time of dimension $\geq 4$, the freedom degrees for $R_{abcd}$ are greater than the freedom degrees for $R_{ab}$, so there is enough "place" for Ricci-flat space-time with non-null curvature.

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So the origin of the space-time curvature is not only due to the matter present in the spacetime... There is no matter, i.e. $T_{ab}=0$, (nearly) outside the surface of a planet, but in this region the space-time is curved because of the presence of the gravitational field of the planet. Is this a rough explaination? –  Galoisfan Nov 21 '13 at 12:18
This is a different problem (see my comment on the answer of John Rennie). In my answer, I consider $T_{ab}=0$ everywhere. In the Schwarzschild metric, $T_{00}$ should be considered as a distribution (a delta-function) –  Trimok Nov 21 '13 at 12:24
Good point, and hence the geon, though I don't think anyone has come up with a stable geon so far. –  John Rennie Nov 21 '13 at 15:50

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.

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