Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

(propose a free ad)

Let $(M,g)$ be a riemannian manifold with riemannian curvature $R$. Then the Riemann-Ricci curvature is:

$$RR(X,Y)=tr(R(X,e_i)R(Y,e_i))$$

A Riemann-Einstein metric is such that:

$$RR(X,Y)=\lambda g(X,Y)$$

with $\lambda$ a scalar.

Can we have spherical solutions of the Riemann-Einstein?

Doesn't the Schwarzschild-deSitter solution qualify? Or am I missing something?

In a vacuum, the gravitational field equations take the form $$RR(X,Y) - \frac{1}{2}tr(RR(e,e))g(X,Y) + \Lambda g(X,Y) = 0$$ By taking the trace reverse they can be written as $$RR(X,Y)=\Lambda g(X,Y)$$ so any vacuum solution with a cosmological constant is a Riemann-Einstein metric. As user AzureSea states the deSitter-Schwartzchild solution is a spherically symmetric example.

user contributions licensed under cc by-sa 3.0 with attribution required