Getting the Lagrangian from the action in curved spacetime Suppose I have this action:
$$ S = \int \mathrm d^4 x\sqrt{-g}\times \text{something}$$
where $g$ is the determinant of the metric.
Should I take the Lagrangian to be:
$$ \mathcal L = \sqrt{-g} \times \text{something} $$
or:
$$ \mathcal L = \text{something}$$
instead? Yes, this is a stupid question.
 A: It is, in a sense, just semantics but I'd say the natural choice is $\mathcal{L}=\sqrt{-g}\times \text{something}$. If you take this definition, the general form of the equations of motion is the same as when doing QFT in Minkowski, with the appropriate generalizations to account for curvature. Furthermore, I think it is standard practice to define the action by
$$S\equiv \int dt L=\int d^4x \mathcal{L}$$
This form is also preserved when the convention I propose is accepted.
A: The natural choice is actually $\mathcal{L}=\text{something}$, the reason being is that the $\sqrt{-g}$ term is naturally paired with the volume form $d^4x$.
Even before considering curved spacetime, consider non Cartesian coordinates.  For example spherical coordinates
$$ dt\,dx\,dy\,dz = r^2 \sin\theta\ dt\,dr\,d\theta\,d\phi$$
Where does that term $r^2 \sin\theta$ come from?  That is the determinant of the Jacobian matrix of the change of coordinates.  This is the role $\sqrt{-g}$ plays, without which the volume form would not be invariant to coordinate changes.
If you separated it the other way, neither the Lagrangian density nor the volume form would be Lorentz invariant.  So it makes much more sense to keep the $\sqrt{-g}$ with the integration, and keep $\mathcal{L}$ just a Lagrangian density.
