What should I do with $\frac{\partial x'^\lambda}{\partial x'^\mu}$ in a Gauge transformation in GR? I am learning about gauge transformations in GR. As I understand, we write the metric (to first order) as $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ where $\eta_{\mu\nu}$ is Minkowski and $h_{\mu\nu}$ is a small perturbation. Upon making a change of coordinates
\begin{equation}
x'^{\mu}\rightarrow x^{\mu} + \xi^\mu
\end{equation}
the metric in the primed coordinates becomes
\begin{equation}
g'_{\mu\nu}=\eta_{\mu\nu}' + h'_{\mu\nu}=\eta_{\mu\nu} +\delta_{\mu\nu} + h_{\mu\nu}.
\end{equation}
Due to the new (primed) coordinates, $\eta_{\mu\nu}$ will not necessarily be Minkowski anymore ($\eta_{\mu\nu}\rightarrow\eta'_{\mu\nu}$), but because the perturbation is small, I can absorb the perturbation of $\eta_{\mu\nu}$ away from Minkowski (which I called $\delta_{\mu\nu}$) into $h'_{\mu\nu}$. Also, because perturbations away from $h_{\mu\nu}$ will be second order in small quantities, we can leave leave h_{\mu\nu} as it is.


*

*Is this reasoning correct? Is there anything that I am missing or a better way to think about this?

*I am trying to work through demonstrating that $h'_{\mu\nu}=h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu$. When I write $g'_{\mu\nu} = g_{\lambda\sigma}\frac{\partial x^\lambda}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}$, and plug in $x^\lambda = x'^\lambda - \xi^\lambda(x')$, I get a lot of terms contain derivatives that look like $\frac{\partial x^\lambda}{\partial x'^\mu}=\frac{\partial x'^\lambda}{\partial x'^\mu}-\frac{\partial \xi^\lambda}{\partial x'^\mu}$. How should I treat/think about the $\frac{\partial x'^\lambda}{\partial x'^\mu}$ terms? 
 A: If you have $x'^{\mu}=x^\mu+\epsilon\xi^\mu$ (where only $\epsilon$ needs to be small - it is better to separate 'smallness" this way imo), then $$ \partial x'^{\mu'}/\partial x^\mu=\delta^{\mu'}_{\mu}+\epsilon\partial_\mu\xi^{\mu'}. $$
You should learn Lie derivatives however, because then you'd reaize that the transformation $x'^\mu=x^\mu+\epsilon\xi^\mu$ is an example of a smooth 1-parameter family of point tranformations and the behaviour of tensor fields under this transformation is given by the Lie derivative.
For your first question, the reasoning is essentially correct, however I have found that (in my opinion) the best way to look at perturbations in GR/differential geometry is by how it is detailed in Stewart's Advanced General Relativity.
Basically, imagine you have a background spacetime $(M^0,g^0)$, and you have a one-parameter family of spacetimes that "deviate only slightly" from this background spacetime, let's denote them by $(M^\epsilon,g^\epsilon)$. Because this is a small deviation, we assume the two manifolds are diffeomorphic and one such diffeo is given by $\phi^\epsilon:M^0\rightarrow M^\epsilon$. It should be understood that $\epsilon$ is variable, and "smallness of the deviation" is measured by the fact that the pullback $(\phi^\epsilon)^*g^\epsilon$ is only different to first order from $g^0$, eg. you can approximate well as $$ g=(\phi^\epsilon)^*g^\epsilon=g^0+\epsilon g^{(1)}. $$
I have called this quantity $g$ because in practice, when dealing with perturbations, you want to work on the unperturbed spacetime $M^0$ via pullbacks, so $g$ will be the working metric.
However, the diffeomorphism $\phi^\epsilon$ is not unique. Let $\psi^\epsilon:M^0\rightarrow M^\epsilon$ also be a diffeo, and let $$ g'=(\psi^\epsilon)^*g^\epsilon=g^0+\epsilon g'^{(1)} $$ be the perturbed metric given by this map.
The difference between $\phi$ and $\psi$ can be quantified on $M^0$ by letting $\chi^\epsilon:M^0\rightarrow M^0$ be a 1-parameter family of diffeos on $M^0$ given by $$ \psi^\epsilon=\phi^\epsilon\circ\chi^\epsilon $$
or explicitly, as $$ \chi^\epsilon=(\phi^\epsilon)^{-1}\circ\psi^\epsilon. $$
Then we can express the perturbed metric $g'$ with the other perturbed metric $g$ by noting that $$ g'=(\psi^\epsilon)^*g^\epsilon=(\phi^\epsilon\circ\chi^\epsilon)^*g^\epsilon=(\chi^\epsilon)^*(\phi^\epsilon)^*g^\epsilon=(\chi^\epsilon)^*g. $$
But 1) we are expanding to first order in $\epsilon$, 2) $\chi^0=\text{Id}$ by continuity, so we have $$ g'=g+\epsilon\frac{d}{d\epsilon}(\chi^\epsilon)^*g|_{\epsilon=0}=g+\epsilon\frac{d}{d\epsilon}(\chi^\epsilon)^*(g^0+\epsilon g^{1})|_{\epsilon=0}=g+\epsilon\mathcal{L}_X(g^0+\epsilon g^{(1)}), $$ where $X$ is the vector field which generates the 1-parameter map $\chi^\epsilon$, so we have $$ g'=g+\epsilon\mathcal{L}_Xg^{0}+O(\epsilon^2)=g^0+\epsilon(g^{(1)}+\mathcal{L}_X g^{0}). $$
So different perturbations differ by Lie derivatives of the unperturbed metric with respect to an arbitrary vector field, and $X$ is indeed arbitrary, since any smooth $\chi^\epsilon$ will describe the difference between two diffeomorphisms of $M^0$ and $M^\epsilon$.
In your case, $g^0=\eta$, $g^{(1)}=h$ and $X=\xi$, so we have $$ h_{\mu\nu}'=h_{\mu\nu}+\mathcal{L}_\xi\eta_{\mu\nu}, $$ where $$ \mathcal{L}_\xi\eta_{\mu\nu}=\xi^\sigma\partial_\sigma\eta_{\mu\nu}+\eta_{\mu\sigma}\partial_\nu\xi^\sigma+\eta_{\sigma\nu}\partial_\mu\xi^\sigma=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu. $$
