# A ambiguous variation of Einstein-Hilbert action

1. May 19, 2010

### archipatelin

A ambiguous variation of Einstein--Hilbert action

Variation of EH action is:
$$\delta S_{EH}=\int_{\Omega}{\delta(R\sqrt{-g})dx^4}= \int_{\Omega}{G_{\mu\nu}\delta{g^{\mu\nu}}\sqrt{-g}dx^4}=0,$$​
where
$$G_{\mu\nu}:=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R$$​
is symmetric einstein's tensor.

The action have to be extremal for each volume $$\Omega$$. This implicate
$$G_{\mu\nu}\delta{g^{\mu\nu}}=0.$$​
Becouse variation of metric is arbitrary and $$G_{\mu\nu}$$ is principal independed on $$\delta{g^{\mu\nu}}$$, latest equation is equivalent with
$$G_{\mu\nu}=0.$$​
This are ordinary einstein's vacuum equations.

But variation of metric is symmetric tensor, therefore more general form of vacuum
field equations are
$$S_{\mu\nu}:=G_{(\mu\nu)}+F_{[\mu\nu]}=0,$$​
where $$F_{\mu\nu}$$ is whatever antisymmetric tensor build it from metric and its derivations.

Why we can ignore this tensor (it is proven that there is not exist)?

2. May 19, 2010

### haushofer

Re: A ambiguous variation of Einstein--Hilbert action

Because the righthand side is the energy momentum tensor, which per definition is symmetric (well, for special-relativistic field theories this is not entirely true, which is the cause for the need of something called the Belinfante tensor).

Also, the metric is symmetric, so theoretically you could add ANY antisymmetric tensor to your expression, because contracting it with $\delta g_{\mu\nu}$ gives zero anyway. However, this term has to be divergenceless due to covariant energy-momentum conservation. Which tensor should that be?