most recent 30 from http://mathoverflow.net 2013-06-19T04:58:34Z http://mathoverflow.net/feeds/question/106847 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106847/geometric-derivation-of-the-einsteins-field-equation-from-the-hilbert-action Lizao Li 2012-09-10T21:33:08Z 2012-09-10T21:33:08Z

<p>It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional is given by the Einstein field equation (for a statement, see <a href="http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti" rel="nofollow">previous question</a>). The standard derivation of this is through <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry#TheKoszulformula" rel="nofollow">Koszul's formulae</a> either in coordinates (for example <a href="http://en.wikipedia.org/wiki/Einstein%25E2%2580%2593Hilbert_action" rel="nofollow">wikipedia</a>), or in abstract index notation (for example, in Wald's General Relativty), or in coordinate-free notation (for example, as pointed out by <a href="http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti" rel="nofollow">Thomas Richard</a> in Besse - Einstein manifold). This approach is mainly algebraic by using the definition in terms of Koszul's formulae and then calculus in various notations. Essentially the derivation is a direct calculation without the need to even mention the manifold. </p> <p>I am wondering if there is a way to derive/interprete the statement refered to at the beginning using an alternative method which is more geometric, ie. using parallel transport or alike. The criterion for "geometric" being (a) a direct reference to the manifold is necessary; or (b) a picture, at least in principle a mental picture, can be drawn to at least carry the main idea of the derivation (of course, pictures of formulae don't count).</p>