In page-319 of Visual Differential Geometry, the following is written:

When we speak of a solution to Einstein's equation, we mean a geometry of space time (defined by it's metric) that satisfies the Einstein Vacuum equation:


I am a bit confused in the above. How does one known they have the right solution to the Ricci equation which will give equations agreeing with experiments?

  • $\begingroup$ There are many different vacuum solutions other than Minkowski spacetime, you know... $\endgroup$ May 17, 2022 at 16:40
  • $\begingroup$ I do know other exist, but that's not really my doubt $\endgroup$ May 17, 2022 at 16:41
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    $\begingroup$ There are many different metrics $g$ (aside from the Minkowski metric) which satisfy Ricci=0. Each such metric is a different solution to Einstein's equation. Three such examples are 1) Minkowski space, 2) Schwarzschild black hole 3) Kerr black hole. $\endgroup$
    – Prahar
    May 17, 2022 at 16:44
  • $\begingroup$ @Prahar The black hole examples are misleading, if not wrong; the stress-energy tensor is nonzero at the singularity. Locally a black hole solution is a vacuum solution, but it is not true globally. $\endgroup$ May 17, 2022 at 16:47
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    $\begingroup$ @AlexNelson I disagree. The singularity is not actually part of the manifold so we definitely have a solution on the entire manifold globally. $\endgroup$
    – Prahar
    May 17, 2022 at 16:49

2 Answers 2


Don't we we already have a geometry of space time as soon as we write down the Minkowski metric?

Yes, but that's putting the cart before the horse. Einstein's equations are a system of differential equations for the (a priori unknown) components of the metric tensor, which take the form

$$\mathrm{Ric}-\frac{1}{2}g R \propto T$$

where $\mathrm{Ric}$ is the Ricci tensor, $R$ is the Ricci scalar, $g$ is the metric tensor, and $T$ is the stress-energy tensor. Both $\mathrm{Ric}$ and $R$ are built from the metric tensor and its first and second derivatives, so this constitutes a system of PDE's for the components of $g$.

In vacuum, $T=0$. One can show that this also implies that $R=0$ by taking the trace of both sides with respect to $g$, and so Einstein's equations reduce to $$\mathrm{Ric}=0$$

There are many distinct vacuum solutions - your example of Minkowski space is only one of them. In order to explore the space of possible solutions, we generally impose symmetry constraints which allow us to simplify the equation $\mathrm{Ric}=0$, and then proceed from there. For example, if we require that the metric be static and spherically symmetric about the origin, then (in the corresponding coordinate system) the possible metric tensors form a family

$$g = -\left(1-\frac{r_0}{r}\right)\mathrm dt^2 + \frac{\mathrm dr^2}{1-\frac{r_0}{r}} + r^2\big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)$$

indexed by the unknown constant $r_0$. If $r_0=0$ then we recover Minkowski space in spherical coordinates; otherwise, this family of solutions describes the spacetime outside of a spherically symmetric body of mass $M \equiv \frac{r_0 c^2}{2G}$.

Hmm say for newton's theory we can calculate the forces and predict the future. If we have to wait for future to know the equation doesn't that make it useless?

If I hand you a spring and a block and ask you to compute the period of its oscillation, can you do this without making any measurements? No, of course not. You'll need to measure the spring constant and the mass of the block; once you have established values for the unknown parameters, then you can use Newton's laws to predict the future.

The same is true here. In order to decide which vacuum solution is appropriate, we need to make some measurements to understand what the symmetries and unknown parameters of the metric are. Once we have done so, we can write down the appropriate solution and then predict e.g. how test masses will move through the curved spacetime, or how the curvature of the spacetime itself will evolve with time.

Regarding the back-and-forth in the comments:

The black hole examples are misleading, if not wrong; the stress-energy tensor is nonzero at the singularity. Locally a black hole solution is a vacuum solution, but it is not true globally.

I disagree. The singularity is not actually part of the manifold so we definitely have a solution on the entire manifold globally.

These are two reasonable ways of interpreting the situation - the former is somewhat less rigorous, but that's not meant as an insult. In electrodynamics, we say that a point charge has electric field $\mathbf E \propto \hat r/r^2$. Early on in our studies, however, we observe that the divergence of this vector field appears to vanish everywhere. If we consider the Maxwell equation $\mathrm{div}(\mathbf E) \propto \rho$ and the subsequent $\oint \mathbf E \cdot \mathrm d\mathbf S \propto \int \rho \mathrm dV$ obtained via the divergence theorem, the left-hand side is non-zero but the right-hand side appears to vanish, leading to an apparent paradox.

The resolution to this apparent paradox can take two forms. In the first, we say that $\hat r/r^2$ diverges at the origin, so we cannot trust our naive computation of the divergence of $\mathbf E$ there; we use the divergence theorem to demonstrate that $\mathrm{div}(\mathbf E) \propto \delta(\mathbf r)$, where $\delta$ is the Dirac delta function. This is not actually a genuine function, and requires the extra machinery of distribution theory to understand properly, but at the physicists level of rigor we simply sweep these additional technicalities under the rug.

In the more mathematically rigorous resolution to the paradox, we similarly observe that $\hat r/r^2$ diverges at the origin, but now we conclude that the space on which we seek a solution is not $\mathbb R^3$ but rather $\mathbb R^3-\{0,0,0\}$ - in other words, we remove the origin from the space. In this case we indeed have that $\rho=0$ at every point under consideration, but there is no paradox because the divergence theorem simply does not hold if our space has a hole in it.

The same rough idea applies to black hole solutions. In a rigorous sense, the points at which the curvature becomes singular are not actually part of the manifold and therefore it would be perfectly correct to regard these solutions as vacuum solutions. On the other hand, it is perfectly common at the physicist's level of rigor to say that the mass is all concentrated at the points at which the curvature becomes infinite, so these solutions are vacuum solutions except at those points.

  • $\begingroup$ I was debating whether to write an answer or not, but I think this is an excellent answer. $\endgroup$ May 17, 2022 at 17:35
  • $\begingroup$ QUestion: What are the things you need to measure to fix the solution of the Einstein field equation? $\endgroup$ May 17, 2022 at 21:04
  • $\begingroup$ @Aplateofmomos It depends on what kind of model you'd like to build. If you'd like to model the (vacuum) spacetime outside of a static, spherically symmetric body, then it suffices to measure the body's total mass. If the body is rotating (so it is merely axisymmetric, not fully spherically symmetric) you'd need to know its mass and angular momentum. If it is uniformly charged, you'd need to know its total charge, and so on. $\endgroup$
    – J. Murray
    May 17, 2022 at 21:10
  • $\begingroup$ How would you measure these things in relativity? To my understanding there is no absolute reference frame? So how would one even describe these things... $\endgroup$ May 17, 2022 at 21:12
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    $\begingroup$ @Aplateofmomos I don't really know what you mean. If we model the Earth as a static, spherically symmetric mass distribution, then we can calculate the trajectory of say, a stone that I throw. That trajectory will depend on the total mass of the planet, so a measurement of the former constitutes an indirect measurement of the latter. $\endgroup$
    – J. Murray
    May 17, 2022 at 21:15

We are used to call Newton's law of gravity to the equation:$$\mathbf a = -\frac{GM\mathbf r}{r^3}$$ for a point in the vacuum at a distance $r$ from the center of a spherical body.

Taking $\mathbf a = -\nabla \phi$, where $$\phi = \frac{GM}{r}$$ we get the Lapacian equation: $$\nabla^2 \phi = 0$$ that is the Newtonian equivalent to $Ricci = 0$.

Of course, the equation is also solved for $\phi = 0$, that corresponds to a situation without gravity. If the Laplacian equation is presented as "Newtonian gravity field equation" it will also seem too general and abstract.


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