# A question about the metric in linearised gravity

In linearised gravity the usual approach is to perturb the metric around some fixed background (often taken to be Minkowksi). My question is, does one literally perturb the metric tensor itself, i.e. $g=\eta +h$ where $h$ is a "small" perturbation tensor, or does one simply choose a coordinate system such that the metric can be expressed in terms of the background + a perturbation, i.e. $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, where $\lvert h_{\mu\nu}\rvert <<1$ ?

The reason I ask is that I'm slightly confused when it comes to carrying out a coordinate transformation. Usually, one considers a transformation of the form $x'^{\mu}=x^{\mu}+\xi^{\mu}$ where $\lvert\xi^{\mu}\rvert <<1$, and this induces a transformation of the metric $$g'_{\mu\nu}=\frac{\partial x^{\alpha}}{\partial x'^{\mu}}\frac{\partial x^{\beta}}{\partial x'^{\nu}}g_{\alpha\beta}=\eta_{\mu\nu}+h_{\mu\nu}-\partial_{\mu}\xi_{\nu}-\partial_{\nu}\xi_{\mu}+\mathcal{O}(h^{2})$$ one then identifies the transformed perturbation $h'_{\mu\nu}$ as $$h'_{\mu\nu}=h_{\mu\nu}-\partial_{\mu}\xi_{\nu}-\partial_{\nu}\xi_{\mu}$$ Is the reason for doing this because we can absorb the transformation of $\eta_{\mu\nu}$ into the transformed perturbation $h'_{\mu\nu}$, ignoring higher order changes to $h_{\mu\nu}$ (since they are infinitesimally small), such that the metric in the new coordinates has the same form as it did in the old coordinates, i.e. $$g'_{\mu\nu}=\eta_{\mu\nu}+h'_{\mu\nu}$$ Or is it simply that the coordinate independent form of the metric is $g=\eta +h$ and furthermore, the coordinate transformation correspond to an infinitesimal diffeomorphism, such that the metric changes by a Lie derivative of $\eta$ and of $h$. The Lie derivative of $h$ is a second-order effect and so can be neglected, thus we can absorb the changes in $\eta$ into $h$ such that the metric retains its component form under the diffeomorphism?

Apologies if this is a stupid question, but I'm quite confused by this point at the moment. Any help would be much appreciated.