# Is there only one vacuum solution of the Einstein equations?

I am thinking about this: A vacuum solution means vanishing Ricci tensor. The Ricci tensor is a contraction of the Riemann, which itself involves contains second derivatives of the metric. Thus they are different Lorentzian metrics with vanishing Ricci tensor. But is there ultimately one metric which has all other vacuum solutions as a special case? Like Kerr has Schwarzschild has Minkowski, by setting the free parameters correctly. Furthermore, how many free parameters can a metric in the vacuum have?

I hope this make sense.

Edit: I guess I am not sure if I am thinking about it in the right way. The most general solution of the vacuum Eq. is a metric with 10 degrees of freedom. By, for example, fixing 9 of them to certain values, we obtain a metric like Schwarzschild, modeling the exterior of a spherical object. Fixing 8 of them in a certain way gives a metric which models the exterior of central, spinning object. Is there now, theoretically, a way to fix some of those freedoms to model every other possible vacuum scenario imaginable? Like binaries etc.

Still not sure whether my question actually makes sense.

• You obviously know there is more than one vacuum solution because you mention three of them (Minkowksi, Schwarzschild and Kerr), so I guess your real question is about the degrees of freedom. Perhaps you could clarify this. For a vacuum solution the Riemann and Weyl tensors are equal, so the number of degrees of freedom in the Riemann tensor will be the ten degrees of freedom associated with the Weyl tensor. Mar 3 '19 at 10:19
• Whenever you have some given vacuum solution, there are others that are like it but with more or different gravitational waves. Jun 29 '21 at 8:20

• @schoeni I am not sure why you are surprised. The wave equation $u_{tt}-u_{xx}=0$ has solution $u(t,x)=f(x+t)+g(x-t)$, where $f$ and $g$ are arbitrary functions. This is already infinitely many parameters, infinitely dimenssional space of solutions.