# Vanishing action integral for Gravitation Field

For a Gravitation Field Action Integral looks like:

$$\begin{equation}\label{1} S_{gravity} = \frac{c^3}{16\pi G}\int R\sqrt{-g} d^4x. \end{equation}$$

А Least Action Principle says the $$\delta S_{gravity} + \delta S_{matter} = 0$$.

But we know the $$\delta S_{gravity}$$ itself is zero: $$\begin{equation} \delta S_{gravity} = \frac{c^3}{16\pi G} \int G_{\mu\nu} \sqrt{-g} \delta g^{\mu\nu}d^4x = \frac{c^3}{16\pi G} \int \sqrt{-g} dx^4 G_{\mu\nu}^{\,\, ;\nu}\xi^{\mu} = 0. \end{equation}$$ Where $$G_{\mu\nu}$$ --- Einstein tensor, $$\xi^{\mu}$$ --- Killing vector.

Also, for matter

$$\begin{equation}\label{2} \delta S_{matter} = \int T_{\mu\nu}\sqrt{-g} \delta g^{\mu\nu}d^4x = \int \sqrt{-g}T^{\mu\nu}_{\,\, ;\nu}\xi_{\mu}d^4x = 0. \end{equation}$$

So, the equation $$\delta S_{gravity} + \delta S_{matter} = 0$$ looks like summation of zeroth $$0 + 0 =0$$ in contrast, for example, with Maxwell's theory where $$\delta S_{EMF} + \delta S_{matter} = 0$$ each therm are non zero.

So, why so for gravity? I really thought that only the sum of two $$\delta S$$ is zero, but not by themselves.

• Principal of least action says that $\delta S_{gravity} + \delta S_{matter} = 0$ for ALL variations, not just specific types of variations (You have shown that $\delta S_{gravity} = 0$ for only a specific type of variation, namely $\delta g_{\mu\nu} = {\cal L}_\xi g_{\mu\nu}$, which is true, but does not violate the variational principal at all. – Prahar Nov 1 '20 at 14:03

1. It seems that OP is only considering infinitesimal gauge transformations $$\delta g_{\mu\nu}~=~\nabla_{\{\mu}\xi_{\nu\}}.$$
2. The stationary action principle for $$S_{\rm gravity}+S_{\rm matter}$$ holds for arbitrary infinitesimal variations $$\delta g_{\mu\nu}$$ (that satisfy appropriate boundary conditions).
In the latter case, $$\delta S_{\rm gravity}$$ is only zero in vacuum without matter.