Consequence of diffeomorphisms invariance in General Relativity

Question

Let's consider a theory with gravity and matter field(s) $\Phi$. The action of this theory is the following: \begin{equation} S[g,\Phi] = S_g[g]+S_m[g,\Phi] = \frac{1}{16\pi G}\int_Md^4x\sqrt{-g}R+\int_Md^4x\sqrt{-g}\mathcal{L}_m[x,\Phi]. \end{equation} The fact the the Einstein-Hilbert action $S_g[g]$ is invariant under diffeomorphisms implies the conservation of the Einstein tensor $G_{\mu\nu}$: \begin{equation} \delta_\xi S_g[g]=0\Longrightarrow \nabla_\mu G^{\mu\nu}=0. \end{equation} The fact that the matter action $S_m[g,\Phi]$ is invariant under diffeomorphism implies the conservation of the energy-momentum tensor $T^{\mu\nu}$: \begin{equation} \delta_\xi S_m[g,\Phi]=0\Longrightarrow \nabla_\mu T^{\mu\nu}=0. \end{equation} But in the end, isn't it the total action $S[g,\Phi]$ that is supposed to be invariant under diffeomorphisms? So could we imagine some kind of transformations where $S_g[g]$ and $S_m[g,\Phi]$ are not invariant but the sum the two is? So the transformation of the two terms would cancel each other.

Is this precisely what Einstein's equations are saying?

Answer 1

The matter action must be invariant under diffeomorphisms, so we have $$ \delta_{\xi} S_M [g,\phi^{I}] = 0 = \int \frac{\delta S_M}{\delta g^{\mu \nu}}\delta g^{\mu \nu} + \int \frac{\delta S_M}{\delta \Phi^I}\delta \phi^I \ . $$ The last term, the variation w.r.t to the matter fields, is zero if $\phi$ satisfies the matter equations of motion. Then using the definition of $T^{\mu \nu}$ and knowing to use the Lie derivative of the metric, we arrive at $\nabla^{\mu}T_{\mu \nu} = 0$ (I assume you know how to fill in those steps).

The Einstein Hilbert action $S_{EH}$ is diffeomorphism invariant, not just because we want it to be. You can do the calculation in full (i.e. without assuming it). So you'd need to use a different gravitational action in order to be in the situation you're describing.