# Is it possible for a Ricci-flat manifold to have a negative energy measured by distant observers?

Supposing that there is a stationary sphere made of regular matters, the external spacetime geometry can be described by the Schwarzschild metric, which describes a spacetime with vanishing Ricci curvature but nonzero tidal force. However, the Schwarzschild metric does not prohibit the mass term to be negative. If there is a sphere made of the hypothetical negative matters, the external spacetime can still be described by the Schwarzschild metric. In this case the sphere produces an outwardly directed gravity pull, but other than that, it possesses zero Ricci curvature and nonzero tidal force which is nothing unusual.

We know that this won’t happen because inside the sphere the spacetime is no longer Ricci-flat due to the $${T_{\mu\nu}}$$ term. Because negative energy hasn’t been observed, $${T_{\mu\nu}}$$ can never be negative. The problem is that $${T_{\mu\nu}}$$ is not the sole cause of gravity. We can find a solution which is Ricci flat everywhere but has a nonzero ADM mass (geon). For example, when two beams of gravitational waves collide, there is a nonzero invariant mass of the system despite the zero $${T_{\mu\nu}}$$ everywhere. Can we use a similar trick to find a Ricci-flat manifold with a negative energy? If the answer is negative, what causes the asymmetry in the Einstein field equation?

No, because a negative energy star is not in the state of equilibrium. If you solve Einstein field equations for static spherically symmetric constant negative energy density sphere you will see that the pressure gradient in and on the matter sphere is positive. For static case (=equilibrium) it has to be negative. If you like look into corresponding equations (1) - (3), and boundary conditions (7) - (8), for the case of $$\alpha \rightarrow -\alpha$$.
• You mean plain vacuum spacetime? I have always though that without matter (nonzero $T_{\mu \nu}$ somewhere) there is no spacetime (physically) at all.