Coordinate-free derivation of the Einstein's field equation from the Hilbert action.

2012-09-10T07:38:37Z

It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional (without matter and cosmological constant, which is irrelevant here):
$$S = \int_M R \mu_g,$$ is given by the Einstein's field equation: $$Ric -\frac{1}{2}R g = 0, $$ where $\mu_g$ is the canonical volume form given by the metric $g$, $Ric$ is the Ricci curvature and $R$ is the Ricci scalar.

The standard derivation of the above statement seems to be a not so hard but not so pleasant direct calculation, either in coordinates or abstract indices, expanding everything in terms of the Christoffel symbol and eventually in terms of $g$ and then calculus.

My questions is: is there a more geometric and coordinate-free way to derive this?

Answer by Thomas Richard for Coordinate-free derivation of the Einstein's field equation from the Hilbert action.

2012-09-10T09:04:36Z

This can be found in Besse "Einstein Manifolds", in chapter 4.

The idea is to use Koszul formula for the Levi-Civitta connection to compute the derivative of the curvature with respect to the metric. Bianchi identities also help.

Coordinate-free derivation of the Einstein's field equation from the Hilbert action. - MathOverflow

Coordinate-free derivation of the Einstein's field equation from the Hilbert action.

Answer by Thomas Richard for Coordinate-free derivation of the Einstein's field equation from the Hilbert action.