Squaring the Einstein-Hilbert action - how much damage can a simple squaring do? I am interested in understanding the consequences of squaring the Einstein Hilbert action:
$$
S=\int \sqrt{\left( \left(R\sqrt{|\det g_{\mu\nu}|} \right)^2 + L_m \right)}d^4x
$$
The field equations resulting from the variational approach leads to
$$
R\sqrt{|\det g_{\mu\nu}|} \left( R_{\mu\nu} -\frac{1}{2}g_{\mu\nu} R \right) = \frac{8 \pi G}{c^4}T_{\mu\nu}
$$
What are the consequences of having the extra term   $R\sqrt{|\det g_{\mu\nu}|}$. Can it simply be absorbed into the stress-energy tensor without having too much of an effect?
 A: To start, your proposed action is not coordinate invariant. The determinant $\sqrt{|g|}$ properly belongs together with $d^4x$, because the combination $\sqrt{|g|}\, d^4x$ is invariant. If you don't do that, you get equations of motion that are only invariant under transformations that preserve four-volume.
So if you do the "correct" change, which is replacing $R$ by $R^2$, you get a special case of $f(R)$ gravity, in which you just replace the Ricci scalar by some arbitrary function on it. If you do this, the field equations end up being
$$ 2R R_{\mu\nu} - \frac12 R^2 g_{\mu\nu} + 2(g_{\mu\nu} \square - \nabla_\mu \nabla_\nu)R = \kappa T_{\mu\nu},$$
which, as you can see, are quite different from the usual equations, and are fourth order in the metric. As an immediate example, we can have vacuum solutions with $R=0$ but $R_{\mu\nu} \neq 0$.
This is not usually done this way, because the result is very different from GR. It's more common to instead regard the quadratic term as a perturbation, and use $f(R) = R + \alpha R^2$, which reduces to GR when $\alpha = 0$.
