# What does vanishing Ricci tensor signify ?

1. Nov 3, 2013

### Genocide

Are Ricci flat manifolds analogous to flat space-time ? Further for Ricci flat manifolds does the Riemann tensor vanish ?

2. Nov 3, 2013

### WannabeNewton

A Ricci flat space-time is definitely not the same thing as flat space-time and obviously Ricci flat space-times don't necessarily have identically vanishing Riemann tensors. Just take any non-trivial vacuum space-time as an example.

Intuitively, if you consider a shear-free, twist-free, geodesic congruence with tangent field $\xi^a$ then by Raychaudhuri's equation we have that $\frac{d\theta}{ds} = \xi^a \nabla_a \theta = -\frac{1}{3}\theta^2$ for a Ricci flat space-time, where $\theta = \nabla_a \xi^a$ is the expansion of the congruence. What this means is that if the expansion is initially zero (i.e. locally the congruence is initially neither diverging nor converging from a point), and we are in a Ricci flat space-time, then it will remain zero for all "time". In other words if you imagine a small sphere (or swarm) of freely falling test particles with zero twist and shear and initially the sphere of freely falling particles are neither falling towards each other radially nor moving away from each other radially then it will remain so for all proper time along the worldline of any one particle in the sphere, if we are in a Ricci flat space-time. Contrast this with what happens when we are not in a Ricci flat space-time and the space-time solves Einstein's equation with the weak energy condition imposed.

Ricci flat space-times do enjoy some nice properties however. For example it is easy to show that the twist $\omega^a = \epsilon^{abcd}\xi_b \nabla_c \xi_d$ of any vector field $\xi^a$ that satisfies killing's equation $\nabla_{(a}\xi_{b)} = 0$ (i.e. $\xi^a$ is a killing field) satisfies $\nabla_{[a}\omega_{b]} = 0$ i.e. there exists some scalar $\omega$ such that $\omega_a = \nabla_a \omega$. Also, for killing fields we always have that $\nabla_a\nabla_b\xi_c = R_{abc}{}{}^{d}\xi_d$ (also easily shown) hence in vacuum killing fields always satisfy $\nabla^a\nabla_b \xi_a = \nabla^a \nabla_a \xi_b = 0$ and they always satisfy $\nabla_a \xi^a = 0$ from killing's equation, point being that the latter is just the Lorenz gauge condition and the former is just the field equation for the electromagnetic 4-potential in the Lorenz gauge i.e. Maxwell's equations are always satisfied in vacuum by killing fields.

There are tons of other interesting and/or useful properties of Ricci flat space-times of course and you can find a profuse of them (including the ones I mentioned above) in Stephani et al. "Exact Solutions of Einstein's Field Equations" which is, to say the least, an incredibly awesome book.

Last edited: Nov 3, 2013