Coordinate-free derivation of the Einstein's field equation from the Hilbert action. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:55:27Z http://mathoverflow.net/feeds/question/106786 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti Coordinate-free derivation of the Einstein's field equation from the Hilbert action. Lizao Li 2012-09-10T07:38:37Z 2012-09-10T09:04:36Z <p>It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional (without matter and cosmological constant, which is irrelevant here):<br> $$S = \int_M R \mu_g,$$ 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.</p> <p>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.</p> <p>My questions is: is there a more geometric and coordinate-free way to derive this? </p> http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti/106792#106792 Answer by Thomas Richard for Coordinate-free derivation of the Einstein's field equation from the Hilbert action. Thomas Richard 2012-09-10T09:04:36Z 2012-09-10T09:04:36Z <p>This can be found in Besse "Einstein Manifolds", in chapter 4.</p> <p>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.</p>