Talk:Einstein tensor

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Doesn't the Einstein tensor have a previous name in differential geometry? Is it the curvature form? What exactly does it represent (is it the intrinsic curvature)? Cesiumfrog (talk) 07:37, 8 April 2010 (UTC)

Einstein tensor vanishes at n=2[edit]

BTW, the trace G is zero at n=2 as implied by the formula. I would add a comment that this happens because the Einstein tensor vanishes in two dimensions. TonyMath (talk) 03:41, 2 March 2012 (UTC)

Simpler formulae in standard coordinates?[edit]

Are there simpler formulae for the Einstein tensor when the coordinates are normal or harmonic? —Quantling (talk | contribs) 23:51, 20 March 2014 (UTC)

I think that normal coordinates are the same thing as what the article is referring to as "the special case of a locally inertial reference frame near a point". However in general relativity, the metric is pseudo-Riemannian rather than Riemannian. So one cannot choose the coordinates to make the metric equal to the Kronecker delta at the point, but merely equal to the Minkowski tensor.
The harmonic coordinate condition does not give as much simplification at a point, but it allows one to simplify everywhere at once. The terms involving
 g^{\gamma\zeta} \, \Gamma^\epsilon_{\epsilon\sigma} \Gamma^\sigma_{\gamma\zeta}
will disappear. JRSpriggs (talk) 04:00, 21 March 2014 (UTC)

Use Standard Divergence Notation?[edit]

In the "Definition" section, is there any reason that the standard notation of divergence is not also used? I'm not entirely sure what the semicolon in the equation means. We could just introduce more equalities, like:

∇G = div G = G^(mu)(nu)_(;nu) = 0

Sorry that I don't know how to format things properly, but the bolded stuff is what I propose we add, while the plain-font stuff is what's already there. Popa910 (talk) 22:33, 8 July 2015 (UTC)

As explained at Ricci calculus#Differentiation, the semi-colon indicates a covariant derivative. JRSpriggs (talk) 05:41, 9 July 2015 (UTC)
In principle, also showing an index-free notation would be reasonable. However, its use for the divergence notation of a non-vector is nonstandard, especially since here there is a contraction too; the notation would not work here. —Quondum
Rethinking, the divergence of a symmetric type-(0,p) tensor with p ≥ 1 makes sense (the non-symmetric case works, but becomes ambiguous). The notation G does not make sense (it looks too much like the gradient); ∇ ⋅ G might be suggested by Dyadics § Dyadic algebra §§ Product of dyadic and vector but notation from dyadics is hardly used; div G might have worked (see the end of Divergence § Generalizations). Because these are all relatively nonstandard, it does not make sense to use these notations unless they are fully defined in the article, which sort of negates the point. —Quondum 14:41, 11 July 2015 (UTC)